diff --git "a/validation.jsonl" "b/validation.jsonl" new file mode 100644--- /dev/null +++ "b/validation.jsonl" @@ -0,0 +1,3102 @@ +{"problem":"a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many r ways can the test be completed if every question is unanswered ?","rationale":"\"5 choices for each of the 4 questions , thus total r of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c .\"","correct":"c","options":{"a":"24 ","b":"120 ","c":"625 ","d":"720","e":"1024"},"options_float":{"a":24.0,"b":120.0,"c":625.0,"d":720.0,"e":1024.0},"annotated_formula":"power(5, 4)","linear_formula":"power(n1,n0)|","chain":"5 ** 4<\/gadget>\n625<\/output>\n625<\/result>","index":0} +{"problem":"a 3 - digit positive integer is chosen at random . what is the probability that the product of its digits is even ?","rationale":"p ( three odd digits ) = 5 \/ 9 * 5 \/ 10 * 5 \/ 10 = 5 \/ 36 ( first digit can not be 0 , so we have a total of 9 digits , 4 even and 5 odd ) . therefore , p ( at least one even digit ) = 1 - 5 \/ 36 = 31 \/ 36 . answer b .","correct":"b","options":{"a":"1 \/ 2 ","b":"31 \/ 36 ","c":"49 \/ 54 ","d":"7 \/ 8","e":"11 \/ 12"},"options_float":{"a":0.5,"b":0.8611111111,"c":0.9074074074,"d":0.875,"e":0.9166666667},"annotated_formula":"subtract(const_1, power(divide(const_1, const_2), 3))","linear_formula":"divide(const_1,const_2)|power(#0,n0)|subtract(const_1,#1)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 3<\/gadget>\n1\/8 = around 0.125<\/output>\n1 - (1\/8)<\/gadget>\n7\/8 = around 0.875<\/output>\n7\/8 = around 0.875<\/result>","index":1} +{"problem":"the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 72 , then 1 \/ m + 1 \/ n is equal to","rationale":"answer we have , m x n = 6 x 210 = 1260 ∴ 1 \/ m + 1 \/ n = ( m + n ) \/ mn = 72 \/ 1260 = 4 \/ 70 = 2 \/ 35 correct option : d","correct":"d","options":{"a":"1 \/ 35 ","b":"3 \/ 35 ","c":"5 \/ 37 ","d":"2 \/ 35","e":"none"},"options_float":{"a":0.0285714286,"b":0.0857142857,"c":0.1351351351,"d":0.0571428571,"e":null},"annotated_formula":"divide(72, multiply(6, 210))","linear_formula":"multiply(n0,n1)|divide(n2,#0)","chain":"6 * 210<\/gadget>\n1_260<\/output>\n72 \/ 1_260<\/gadget>\n2\/35 = around 0.057143<\/output>\n2\/35 = around 0.057143<\/result>","index":3} +{"problem":"in a kilometer race , a beats b by 48 meters or 12 seconds . what time does a take to complete the race ?","rationale":"\"time taken by b run 1000 meters = ( 1000 * 12 ) \/ 48 = 250 sec . time taken by a = 250 - 12 = 238 sec . answer : a\"","correct":"a","options":{"a":"238 sec ","b":"190 sec ","c":"667 sec ","d":"167 sec","e":"176 sec"},"options_float":{"a":238.0,"b":190.0,"c":667.0,"d":167.0,"e":176.0},"annotated_formula":"subtract(divide(multiply(const_1, const_1000), divide(48, 12)), 12)","linear_formula":"divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|","chain":"1 * 1_000<\/gadget>\n1_000<\/output>\n48 \/ 12<\/gadget>\n4<\/output>\n1_000 \/ 4<\/gadget>\n250<\/output>\n250 - 12<\/gadget>\n238<\/output>\n238<\/result>","index":4} +{"problem":"in a school of 650 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?","rationale":"44 + 28 + 10 = 82 % 100 – 82 = 18 % 650 * 18 \/ 100 = 117 answer : e","correct":"e","options":{"a":"173 ","b":"163 ","c":"153 ","d":"143","e":"117"},"options_float":{"a":173.0,"b":163.0,"c":153.0,"d":143.0,"e":117.0},"annotated_formula":"divide(multiply(650, subtract(const_100, add(add(44, 28), 10))), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)","chain":"44 + 28<\/gadget>\n72<\/output>\n72 + 10<\/gadget>\n82<\/output>\n100 - 82<\/gadget>\n18<\/output>\n650 * 18<\/gadget>\n11_700<\/output>\n11_700 \/ 100<\/gadget>\n117<\/output>\n117<\/result>","index":5} +{"problem":"a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , while a and c together can do it 2 hours . how long will b alone take to do it ?","rationale":"\"solution a ’ s 1 hour ’ s work = 1 \/ 4 ; ( b + c ) ' s 1 hour ’ s work = 1 \/ 3 ; ( a + c ) ' s 1 hour ’ s work = 1 \/ 2 ( a + b + c ) ’ s 1 hour ’ s work = ( 1 \/ 4 + 1 \/ 3 ) = 7 \/ 12 b ’ s 1 hour ’ s work = ( 7 \/ 12 - 1 \/ 2 ) = 1 \/ 12 ∴ b alone will take 12 hours to do the work . answer c\"","correct":"c","options":{"a":"8 hours ","b":"10 hours ","c":"12 hours ","d":"24 hours","e":"none of these"},"options_float":{"a":8.0,"b":10.0,"c":12.0,"d":24.0,"e":null},"annotated_formula":"divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 4))))","linear_formula":"divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) - (1\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/3) - (1\/4)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":6} +{"problem":"in a group of 160 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ?","rationale":"\"number of people whose age is less than 20 is given by 160 - 90 = 70 probability p that a person selected at random from the group is less than 20 is gieven by 70 \/ 160 = 0.4375 ~ 0.44 correct answer a\"","correct":"a","options":{"a":"0.44 ","b":"0.55 ","c":"0.65 ","d":"0.75","e":"0.85"},"options_float":{"a":0.44,"b":0.55,"c":0.65,"d":0.75,"e":0.85},"annotated_formula":"divide(subtract(160, 90), 160)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|","chain":"160 - 90<\/gadget>\n70<\/output>\n70 \/ 160<\/gadget>\n7\/16 = around 0.4375<\/output>\n7\/16 = around 0.4375<\/result>","index":7} +{"problem":"an art gallery has only paintings and sculptures . currently , 1 \/ 3 of the pieces of art are displayed , and 1 \/ 6 of the pieces on display are sculptures . if 1 \/ 3 of the pieces not on display are paintings , and 800 sculptures are not on display , how many pieces of art does the gallery have ?","rationale":"\"too many words and redundant info there . ( i ) 1 \/ 3 of the pieces of art are displayed , hence 2 \/ 3 of the pieces of art are not displayed . ( ii ) 1 \/ 6 of the pieces on display are sculptures , hence 5 \/ 6 of the pieces on display are paintings . ( iii ) 1 \/ 3 of the pieces not on display are paintings , hence 2 \/ 3 of the pieces not on display are sculptures . 800 sculptures are not on display , so according to ( iii ) 2 \/ 3 * { not on display } = 800 - - > { not on display } = 1200 . according to ( i ) 2 \/ 3 * { total } = 1200 - - > { total } = 1800 . answer : b .\"","correct":"b","options":{"a":"360 ","b":"1800 ","c":"540 ","d":"640","e":"720"},"options_float":{"a":360.0,"b":1800.0,"c":540.0,"d":640.0,"e":720.0},"annotated_formula":"divide(divide(800, subtract(1, divide(1, 3))), subtract(1, divide(1, 3)))","linear_formula":"divide(n0,n1)|subtract(n0,#0)|divide(n6,#1)|divide(#2,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n800 \/ (2\/3)<\/gadget>\n1_200<\/output>\n1_200 \/ (2\/3)<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":8} +{"problem":"if x and y are integers and | x - y | = 10 , what is the minimum possible value of xy ?","rationale":"\"sometimes the answer choices to a given question provide a big ' hint ' as to how you can go about solving it . this prompt can also be solved without any complex math ideas - you just need to do a bit of ' brute force ' math and you ' ll have the answer relatively quickly . we ' re told that x and y are integers and | x - y | = 10 . we ' re asked for the minimum possible value of ( x ) ( y ) . since all of the answer choices are negative , this tells us that one of the two variables must be negative ( and the other must be positive ) , so we should restrict our work to those options . if . . . x = 9 , y = - 1 , then xy = - 9 x = 8 , y = - 2 , then xy = - 16 x = 7 , y = - 3 , then xy = - 21 x = 6 , y = - 4 , then xy = - 24 x = 5 , y = - 5 , then xy = - 25 x = 4 , y = - 6 , then xy = - 24 x = 3 , y = - 7 , then xy = - 21 from this , we can conclude the xy will start to get bigger as x continues to decrease down to 1 , so there ' s no need to do any additional work . final answer : d\"","correct":"d","options":{"a":"- 9 ","b":"- 18 ","c":"- 24 ","d":"- 25","e":"- 48"},"options_float":{"a":-9.0,"b":-18.0,"c":-24.0,"d":-25.0,"e":-48.0},"annotated_formula":"multiply(negate(add(const_2, const_3)), subtract(10, add(const_2, const_3)))","linear_formula":"add(const_2,const_3)|negate(#0)|subtract(n0,#0)|multiply(#1,#2)|","chain":"2 + 3<\/gadget>\n5<\/output>\n-5<\/gadget>\n-5<\/output>\n10 - 5<\/gadget>\n5<\/output>\n(-5) * 5<\/gadget>\n-25<\/output>\n-25<\/result>","index":9} +{"problem":"two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 25 hours and 16 hours respectively . the ratio of their speeds is ?","rationale":"\"let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 16 : â ˆ š 25 = 4 : 5 . answer : d\"","correct":"d","options":{"a":"4 : 6 ","b":"4 : 3 ","c":"4 : 9 ","d":"4 : 5","e":"4 : 2"},"options_float":{"a":0.6666666667,"b":1.3333333333,"c":0.4444444444,"d":0.8,"e":2.0},"annotated_formula":"divide(sqrt(16), sqrt(25))","linear_formula":"sqrt(n1)|sqrt(n0)|divide(#0,#1)|","chain":"16 ** (1\/2)<\/gadget>\n4<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":10} +{"problem":"at what rate of compound interest per annum will a sum of rs . 1200 become rs . 1348.32 in 2 years ?","rationale":"explanation : let the rate be r % p . a . then , 1200 x ( 1 + r \/ 100 ) 2 = 1348.32 = ( 1 + r \/ 100 ) 2 = 134832 \/ 120000 = 11236 \/ 10000 ( 1 + r \/ 100 ) 2 = ( 106 \/ 100 ) 2 = > 1 + r \/ 100 = 106 \/ 100 r = 6 % answer is a","correct":"a","options":{"a":"6 % ","b":"6.5 % ","c":"7 % ","d":"7.5 %","e":"8 %"},"options_float":{"a":6.0,"b":6.5,"c":7.0,"d":7.5,"e":8.0},"annotated_formula":"subtract(sqrt(divide(multiply(1348.32, const_100), divide(1200, const_100))), sqrt(divide(multiply(1200, const_100), divide(1200, const_100))))","linear_formula":"divide(n0,const_100)|multiply(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|divide(#2,#0)|sqrt(#3)|sqrt(#4)|subtract(#5,#6)","chain":"1_348.32 * 100<\/gadget>\n134_832<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n134_832 \/ 12<\/gadget>\n11_236<\/output>\n11_236 ** (1\/2)<\/gadget>\n106<\/output>\n1_200 * 100<\/gadget>\n120_000<\/output>\n120_000 \/ 12<\/gadget>\n10_000<\/output>\n10_000 ** (1\/2)<\/gadget>\n100<\/output>\n106 - 100<\/gadget>\n6<\/output>\n6<\/result>","index":12} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 6.2 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?","rationale":"\"10 overs - run rate = 6.2 runs scored in first 10 overs = 62 remaining overs 40 total runs to be scored = 282 62 runs already scored 282 - 62 = 220 220 runs to be scored in 40 overs let required runrate be x 40 * x = 220 x = 220 \/ 40 x = 5.5 the required runrate is 5.5 answer : b\"","correct":"b","options":{"a":"6.25 ","b":"5.5 ","c":"6.75 ","d":"7","e":"8"},"options_float":{"a":6.25,"b":5.5,"c":6.75,"d":7.0,"e":8.0},"annotated_formula":"divide(subtract(282, multiply(10, 6.2)), 40)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 6.2<\/gadget>\n62<\/output>\n282 - 62<\/gadget>\n220<\/output>\n220 \/ 40<\/gadget>\n11\/2 = around 5.5<\/output>\n11\/2 = around 5.5<\/result>","index":13} +{"problem":"during a car trip , maria stopped to rest after she traveled 1 \/ 2 of the total distance to her destination . she stopped again after she traveled 1 \/ 4 of the distance remaining between her first stop and her destination , and then she drove the remaining 135 miles to her detination . what was the total distance , in miles from maria ' s starting point to her destination ?","rationale":"easy 360 is the answer . 3 \/ 4 ( x \/ 2 ) = 135 x = 135 * 8 \/ 3 = 360 . answer c","correct":"c","options":{"a":"280 ","b":"320 ","c":"360 ","d":"420","e":"480"},"options_float":{"a":280.0,"b":320.0,"c":360.0,"d":420.0,"e":480.0},"annotated_formula":"divide(135, subtract(1, add(divide(1, 2), multiply(divide(1, 2), divide(1, 4)))))","linear_formula":"divide(n0,n1)|divide(n0,n3)|multiply(#0,#1)|add(#0,#2)|subtract(n0,#3)|divide(n4,#4)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) * (1\/4)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/2) + (1\/8)<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n135 \/ (3\/8)<\/gadget>\n360<\/output>\n360<\/result>","index":15} +{"problem":"a and b go around a circular track of length 600 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ?","rationale":"\"time taken to meet for the first time at the starting point = lcm { length of the track \/ speed of a , length of the track \/ speed of b } = lcm { 600 \/ ( 36 * 5 \/ 18 ) , 600 \/ ( 54 * 5 \/ 18 ) } = lcm ( 60 , 40 ) = 120 sec . answer : a\"","correct":"a","options":{"a":"120 ","b":"775 ","c":"555 ","d":"288","e":"278"},"options_float":{"a":120.0,"b":775.0,"c":555.0,"d":288.0,"e":278.0},"annotated_formula":"divide(600, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778)))","linear_formula":"multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n54 * (5\/18)<\/gadget>\n15<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n15 - 10<\/gadget>\n5<\/output>\n600 \/ 5<\/gadget>\n120<\/output>\n120<\/result>","index":19} +{"problem":"the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 132 . what is the number of frogs in the pond ?","rationale":"\"solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 132 . so , total number of ducks and frogs in the pond , = 2 * 132 = 262 . therefore , number of frogs , = ( 262 * 39 ) \/ 76 = 134 . answer : option b\"","correct":"b","options":{"a":"148 ","b":"134 ","c":"156 ","d":"144","e":"none"},"options_float":{"a":148.0,"b":134.0,"c":156.0,"d":144.0,"e":null},"annotated_formula":"multiply(39, divide(132, divide(add(add(37, 39), const_2), const_2)))","linear_formula":"add(n0,n1)|add(#0,const_2)|divide(#1,const_2)|divide(n2,#2)|multiply(n1,#3)|","chain":"37 + 39<\/gadget>\n76<\/output>\n76 + 2<\/gadget>\n78<\/output>\n78 \/ 2<\/gadget>\n39<\/output>\n132 \/ 39<\/gadget>\n44\/13 = around 3.384615<\/output>\n39 * (44\/13)<\/gadget>\n132<\/output>\n132<\/result>","index":21} +{"problem":"a certain car can travel 32 kilometers on a liter of fuel . if the fuel tank ’ s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers )","rationale":"\"fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 \/ 1.6 mile speed ( km \/ hour ) = d \/ t = 32 ( km * ) x 3.9 x 3.8 \/ 5.7 replace ( km * ) to miles ; multiply by 1 \/ 1.6 mile speed ( miles \/ hour ) = 32 x 3.9 x 3.8 \/ 5.7 x 1.6 = 78 miles \/ hour ans : a ps : i felt the factors were easy to cancel out , so did n ' t require much rounding off = 32 x 3.9 x 3.8 \/ 5.7 x 1.6 = 52 a\"","correct":"a","options":{"a":"52 ","b":"65 ","c":"78 ","d":"91","e":"104"},"options_float":{"a":52.0,"b":65.0,"c":78.0,"d":91.0,"e":104.0},"annotated_formula":"divide(multiply(divide(multiply(32, 3.8), 1.6), 3.9), 5.7)","linear_formula":"multiply(n0,n4)|divide(#0,n6)|multiply(n1,#1)|divide(#2,n2)|","chain":"32 * 3.8<\/gadget>\n121.6<\/output>\n121.6 \/ 1.6<\/gadget>\n76<\/output>\n76 * 3.9<\/gadget>\n296.4<\/output>\n296.4 \/ 5.7<\/gadget>\n52<\/output>\n52<\/result>","index":23} +{"problem":"the ratio of the number of females to males at a party was 1 : 2 but when 7 females and 7 males left , the ratio became 1 : 3 . how many people were at the party originally ?","rationale":"the total number of people are x females + 2 x males . 3 * ( x - 7 ) = 2 x - 7 x = 14 there were 3 x = 42 people at the party originally . the answer is d .","correct":"d","options":{"a":"28 ","b":"32 ","c":"36 ","d":"42","e":"50"},"options_float":{"a":28.0,"b":32.0,"c":36.0,"d":42.0,"e":50.0},"annotated_formula":"add(divide(subtract(multiply(multiply(7, 2), 3), multiply(7, 2)), const_2), subtract(multiply(multiply(7, 2), 3), multiply(7, 2)))","linear_formula":"multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,const_2)|add(#3,#2)","chain":"7 * 2<\/gadget>\n14<\/output>\n14 * 3<\/gadget>\n42<\/output>\n42 - 14<\/gadget>\n28<\/output>\n28 \/ 2<\/gadget>\n14<\/output>\n14 + 28<\/gadget>\n42<\/output>\n42<\/result>","index":25} +{"problem":"the compound ratio of 5 : 6 , 3 : 2 and 4 : 5 is ?","rationale":"\"answer : option c 5 \/ 6 : 3 \/ 2 : 4 \/ 5 = 1 : 1\"","correct":"c","options":{"a":"3 : 5 ","b":"2 : 7 ","c":"1 : 1 ","d":"1 : 4","e":"none"},"options_float":{"a":0.6,"b":0.2857142857,"c":1.0,"d":0.25,"e":null},"annotated_formula":"divide(divide(multiply(5, 3), multiply(6, 2)), divide(multiply(3, 4), multiply(2, 5)))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)|","chain":"5 * 3<\/gadget>\n15<\/output>\n6 * 2<\/gadget>\n12<\/output>\n15 \/ 12<\/gadget>\n5\/4 = around 1.25<\/output>\n3 * 4<\/gadget>\n12<\/output>\n2 * 5<\/gadget>\n10<\/output>\n12 \/ 10<\/gadget>\n6\/5 = around 1.2<\/output>\n(5\/4) \/ (6\/5)<\/gadget>\n25\/24 = around 1.041667<\/output>\n25\/24 = around 1.041667<\/result>","index":26} +{"problem":"a woman complete a journey in 20 hours . she travels first half of the journey at the rate of 21 km \/ hr and second half at the rate of 24 km \/ hr . find the total journey in km .","rationale":"\"0.5 x \/ 21 + 0.5 x \/ 24 = 20 - - > x \/ 21 + x \/ 24 = 40 - - > x = 448 km . d\"","correct":"d","options":{"a":"334 km . ","b":"216 km . ","c":"314 km . ","d":"448 km .","e":"544 km ."},"options_float":{"a":334.0,"b":216.0,"c":314.0,"d":448.0,"e":544.0},"annotated_formula":"multiply(multiply(divide(multiply(20, 24), add(24, 21)), 21), const_2)","linear_formula":"add(n1,n2)|multiply(n0,n2)|divide(#1,#0)|multiply(n1,#2)|multiply(#3,const_2)|","chain":"20 * 24<\/gadget>\n480<\/output>\n24 + 21<\/gadget>\n45<\/output>\n480 \/ 45<\/gadget>\n32\/3 = around 10.666667<\/output>\n(32\/3) * 21<\/gadget>\n224<\/output>\n224 * 2<\/gadget>\n448<\/output>\n448<\/result>","index":27} +{"problem":"find the compound interest on $ 1200 for 6 years at 20 % p . a . if ci is component yearly ?","rationale":"\"a = p ( 1 + r \/ 100 ) ^ t = 1200 ( 1 + 20 \/ 100 ) ^ 6 = $ 3583 ci = $ 2383 answer is c\"","correct":"c","options":{"a":"$ 120 ","b":"$ 150 ","c":"$ 2383 ","d":"$ 250","e":"$ 300"},"options_float":{"a":120.0,"b":150.0,"c":2383.0,"d":250.0,"e":300.0},"annotated_formula":"subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 6)), 1200)","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 6<\/gadget>\n46_656\/15_625 = around 2.985984<\/output>\n1_200 * (46_656\/15_625)<\/gadget>\n2_239_488\/625 = around 3_583.1808<\/output>\n(2_239_488\/625) - 1_200<\/gadget>\n1_489_488\/625 = around 2_383.1808<\/output>\n1_489_488\/625 = around 2_383.1808<\/result>","index":28} +{"problem":"you need to unlock a secret code using following clues , can you ? here you have the clues : clue - 1 : 0 7 9 ( one of the numbers is correct and is placed in its correct position ) clue - 2 : 0 3 2 ( nothing is correct ) clue - 3 : 1 0 8 ( two numbers are correct but not placed at its correct position . ) clue - 4 : 9 2 6 ( one number is correct but not placed at its correct position . ) clue - 5 : 6 7 8 ( one number is correct but not placed at its correct position . )","rationale":"0 7 9 : 9 is placed correctly * 0 3 2 : none of the numbers is in the code * 1 0 8 : 1 and 8 are correct number of code but placed at wrong position * 9 2 6 : number 9 is there but placed at wrong position * 6 7 8 : number 8 is there but placed at wrong position correct answer is a ) 819","correct":"a","options":{"a":"819 ","b":"918 ","c":"198 ","d":"189","e":"891"},"options_float":{"a":819.0,"b":918.0,"c":198.0,"d":189.0,"e":891.0},"annotated_formula":"add(add(multiply(8, const_100), multiply(1, const_10)), 9)","linear_formula":"multiply(n11,const_100)|multiply(n0,const_10)|add(#0,#1)|add(n3,#2)","chain":"8 * 100<\/gadget>\n800<\/output>\n1 * 10<\/gadget>\n10<\/output>\n800 + 10<\/gadget>\n810<\/output>\n810 + 9<\/gadget>\n819<\/output>\n819<\/result>","index":29} +{"problem":"set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 7 , while all other numbers remain the same , what is the new average of set s ?","rationale":"\"old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 7 then total increased to 62 + 7 = 69 new avg - 69 \/ 10 = 6.9 . hence answer is e .\"","correct":"e","options":{"a":"6.6 ","b":"6.7 ","c":"6.8 ","d":"6.85","e":"6.9"},"options_float":{"a":6.6,"b":6.7,"c":6.8,"d":6.85,"e":6.9},"annotated_formula":"divide(add(multiply(10, 6.2), 7), 10)","linear_formula":"multiply(n0,n1)|add(n2,#0)|divide(#1,n0)|","chain":"10 * 6.2<\/gadget>\n62<\/output>\n62 + 7<\/gadget>\n69<\/output>\n69 \/ 10<\/gadget>\n69\/10 = around 6.9<\/output>\n69\/10 = around 6.9<\/result>","index":31} +{"problem":"if p ( a ) = 1 \/ 5 and p ( b ) = 2 \/ 5 , find p ( a n b ) if a and b are independent events .","rationale":"\"p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 1 \/ 5 . 2 \/ 5 p ( a n b ) = 2 \/ 25 . d\"","correct":"d","options":{"a":"7 \/ 25 ","b":"3 \/ 25 ","c":"8 \/ 25 ","d":"2 \/ 25","e":"3 \/ 17"},"options_float":{"a":0.28,"b":0.12,"c":0.32,"d":0.08,"e":0.1764705882},"annotated_formula":"multiply(divide(1, 5), divide(2, 5))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/5) * (2\/5)<\/gadget>\n2\/25 = around 0.08<\/output>\n2\/25 = around 0.08<\/result>","index":32} +{"problem":"a train is 320 meter long is running at a speed of 45 km \/ hour . in what time will it pass a bridge of 140 meter length ?","rationale":"\"speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 320 + 140 = 460 meter time = distance \/ speed = 460 * ( 2 \/ 25 ) = 36.8 seconds answer : e\"","correct":"e","options":{"a":"11 seconds ","b":"38 seconds ","c":"40 seconds ","d":"88 seconds","e":"36.8 seconds"},"options_float":{"a":11.0,"b":38.0,"c":40.0,"d":88.0,"e":36.8},"annotated_formula":"divide(add(320, 140), divide(multiply(45, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"320 + 140<\/gadget>\n460<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n460 \/ (25\/2)<\/gadget>\n184\/5 = around 36.8<\/output>\n184\/5 = around 36.8<\/result>","index":33} +{"problem":"by the how much is two - fifth of 300 greater than 3 - fifths of 125 ?","rationale":"reqd no . = 2 ⠁ „ 5 ã — 300 - 3 ⠁ „ 5 ã — 125 = 120 - 75 = 45 answer d","correct":"d","options":{"a":"15 ","b":"3 ","c":"5 ","d":"45","e":"none of these"},"options_float":{"a":15.0,"b":3.0,"c":5.0,"d":45.0,"e":null},"annotated_formula":"subtract(multiply(divide(const_2, add(const_2, const_3)), 300), multiply(divide(3, add(const_2, const_3)), 125))","linear_formula":"add(const_2,const_3)|divide(const_2,#0)|divide(n1,#0)|multiply(n0,#1)|multiply(n2,#2)|subtract(#3,#4)","chain":"2 + 3<\/gadget>\n5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 300<\/gadget>\n120<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 125<\/gadget>\n75<\/output>\n120 - 75<\/gadget>\n45<\/output>\n45<\/result>","index":35} +{"problem":"excluding stoppages , the speed of a train is 60 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ?","rationale":"\"explanation : t = 20 \/ 60 * 60 = 20 answer : option e\"","correct":"e","options":{"a":"12 ","b":"14 ","c":"16 ","d":"18","e":"20"},"options_float":{"a":12.0,"b":14.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"subtract(const_60, multiply(const_60, divide(40, 60)))","linear_formula":"divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|","chain":"40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n60 * (2\/3)<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20<\/result>","index":37} +{"problem":"an error 10 % in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ?","rationale":"\"percentage error in calculated area = ( 10 + 10 + ( 10 ã — 10 ) \/ 100 ) % = 21 % answer : d\"","correct":"d","options":{"a":"6.64 % ","b":"16.64 % ","c":"15.64 % ","d":"21 %","e":"10.64 %"},"options_float":{"a":6.64,"b":16.64,"c":15.64,"d":21.0,"e":10.64},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 10)), square_area(const_100)), const_100), square_area(const_100))","linear_formula":"add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|","chain":"100 + 10<\/gadget>\n110<\/output>\n110 ** 2<\/gadget>\n12_100<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n12_100 - 10_000<\/gadget>\n2_100<\/output>\n2_100 * 100<\/gadget>\n210_000<\/output>\n210_000 \/ 10_000<\/gadget>\n21<\/output>\n21<\/result>","index":38} +{"problem":"if 2 a = 3 b and ab ≠ 0 , what is the ratio of a \/ 3 to b \/ 2 ?","rationale":"\"a nice fast approach is the first find a pair of numbers that satisfy the given equation : 2 a = 3 b here ' s one pair : a = 3 and b = 2 what is the ratio of a \/ 3 to b \/ 3 ? in other words , what is the value of ( a \/ 3 ) \/ ( b \/ 2 ) ? plug in values to get : ( a \/ 3 ) \/ ( b \/ 2 ) = ( 3 \/ 3 ) \/ ( 2 \/ 2 ) = 1 \/ 1 = 1 c\"","correct":"c","options":{"a":"27 \/ 8 ","b":"8 \/ 27 ","c":"1 ","d":"9 \/ 4","e":"81 \/ 64"},"options_float":{"a":3.375,"b":0.2962962963,"c":1.0,"d":2.25,"e":1.265625},"annotated_formula":"divide(multiply(2, 3), multiply(3, 2))","linear_formula":"multiply(n0,n1)|divide(#0,#0)|","chain":"2 * 3<\/gadget>\n6<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 \/ 6<\/gadget>\n1<\/output>\n1<\/result>","index":42} +{"problem":"each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 3 and the 4 can not be adjacent digits in the integer , how many 5 - digit integers are possible ?","rationale":"total number of 5 digit numbers that can be formed = 5 ! = 120 . total number of 5 digit numbers in which 3 and 4 are adjacent = 48 . this can be calculated by taking 3 and 4 as a pair and considering positions where they can appear . number of 5 digit numbers in which 3 and 4 are not adjacent = total number of 5 digit numbers - total number of 5 digit numbers in which 3 and 4 are adjacent . ans = 120 - 48 = 72 c","correct":"c","options":{"a":"48 ","b":"66 ","c":"72 ","d":"78","e":"90"},"options_float":{"a":48.0,"b":66.0,"c":72.0,"d":78.0,"e":90.0},"annotated_formula":"subtract(multiply(multiply(multiply(5, 4), 3), const_2), multiply(multiply(multiply(4, 3), const_2), const_2))","linear_formula":"multiply(n1,n4)|multiply(n3,n4)|multiply(n3,#0)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|subtract(#4,#5)","chain":"5 * 4<\/gadget>\n20<\/output>\n20 * 3<\/gadget>\n60<\/output>\n60 * 2<\/gadget>\n120<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\n120 - 48<\/gadget>\n72<\/output>\n72<\/result>","index":43} +{"problem":"in a 200 m race , a beats b by 20 m and c by 38 m . in a race of 600 m b will beat c by","rationale":"\"it means , when a covers 200 m , b covers only ( 200 - 20 ) = 180 m and c covers only ( 200 - 38 ) = 162 m = > when c covers 162 m , b covers 180 m = > when b covers 600 m , c covers 162 \/ 180 × 600 = 540 m hence , b beats c by 600 - 540 = 60 metre answer : e\"","correct":"e","options":{"a":"20 m ","b":"30 m ","c":"25 m ","d":"15 m","e":"60 m"},"options_float":{"a":20.0,"b":30.0,"c":25.0,"d":15.0,"e":60.0},"annotated_formula":"subtract(600, multiply(inverse(multiply(divide(200, subtract(200, 38)), divide(subtract(200, 20), 200))), 600))","linear_formula":"subtract(n0,n2)|subtract(n0,n1)|divide(n0,#0)|divide(#1,n0)|multiply(#2,#3)|inverse(#4)|multiply(n3,#5)|subtract(n3,#6)|","chain":"200 - 38<\/gadget>\n162<\/output>\n200 \/ 162<\/gadget>\n100\/81 = around 1.234568<\/output>\n200 - 20<\/gadget>\n180<\/output>\n180 \/ 200<\/gadget>\n9\/10 = around 0.9<\/output>\n(100\/81) * (9\/10)<\/gadget>\n10\/9 = around 1.111111<\/output>\n1 \/ (10\/9)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 600<\/gadget>\n540<\/output>\n600 - 540<\/gadget>\n60<\/output>\n60<\/result>","index":44} +{"problem":"in what time will a train 120 m long cross an electric pole , it its speed be 121 km \/ hr ?","rationale":"\"speed = 121 * 5 \/ 18 = 33.6 m \/ sec time taken = 120 \/ 33.6 = 3.6 sec . answer : d\"","correct":"d","options":{"a":"2.5 ","b":"2.7 ","c":"2.9 ","d":"3.6","e":"2.1"},"options_float":{"a":2.5,"b":2.7,"c":2.9,"d":3.6,"e":2.1},"annotated_formula":"divide(120, multiply(121, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n121 * (5\/18)<\/gadget>\n605\/18 = around 33.611111<\/output>\n120 \/ (605\/18)<\/gadget>\n432\/121 = around 3.570248<\/output>\n432\/121 = around 3.570248<\/result>","index":45} +{"problem":"at a recent small town election for mayor a total of 963 votes were cast for the 4 candidates , the winner exceding his opponents by 53 , 79 and 105 votes , respectively . how many votes were cast for each candidate ?","rationale":"c 195 the number of votes the winning candidate received was 963 + 53 + 79 + 105 \/ 4 = 300 . the second received 300 – 53 = 247 , the third received 300 – 105 = 195 .","correct":"c","options":{"a":"134 ","b":"178 ","c":"195 ","d":"166","e":"194"},"options_float":{"a":134.0,"b":178.0,"c":195.0,"d":166.0,"e":194.0},"annotated_formula":"subtract(divide(add(add(963, 53), add(79, 105)), const_4), 105)","linear_formula":"add(n0,n2)|add(n3,n4)|add(#0,#1)|divide(#2,const_4)|subtract(#3,n4)","chain":"963 + 53<\/gadget>\n1_016<\/output>\n79 + 105<\/gadget>\n184<\/output>\n1_016 + 184<\/gadget>\n1_200<\/output>\n1_200 \/ 4<\/gadget>\n300<\/output>\n300 - 105<\/gadget>\n195<\/output>\n195<\/result>","index":46} +{"problem":"in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 5 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?","rationale":"\"total fish = x percentage of second catch = ( 5 \/ 50 ) * 100 = 10 % so , x * 10 % = 50 x = 500 ans b .\"","correct":"b","options":{"a":"400 ","b":"500 ","c":"1,250 ","d":"2,500","e":"10,000"},"options_float":{"a":400.0,"b":500.0,"c":1250.0,"d":2500.0,"e":10000.0},"annotated_formula":"divide(50, divide(5, 50))","linear_formula":"divide(n2,n1)|divide(n0,#0)|","chain":"5 \/ 50<\/gadget>\n1\/10 = around 0.1<\/output>\n50 \/ (1\/10)<\/gadget>\n500<\/output>\n500<\/result>","index":48} +{"problem":"together , 15 type a machines and 7 type b machines can complete a certain job in 4 hours . together 8 type b machines and 15 type c machines can complete the same job in 11 hours . how many q hours would it take one type a machine , one type b machine , and one type c machine working together to complete the job ( assuming constant rates for each machine ) ?","rationale":"\"say the rates of machines a , b and c are a , b , and c , respectively . together 15 type a machines and 7 type b machines can complete a certain job in 4 hours - - > 15 a + 7 b = 1 \/ 4 ; together 8 type b machines and 15 type c machines can complete the same job in 11 hours - - > 8 b + 15 c = 1 \/ 11 . sum the above : 15 a + 15 b + 15 c = 1 \/ 4 + 1 \/ 11 = 15 \/ 44 - - > reduce by 15 : a + b + c = 1 \/ 44 - - > so , the combined rate of the three machines is 1 \/ 44 job \/ hour - - > time is reciprocal of the rate , thus machines a , b and c can do the job q in 44 hours . answer : c .\"","correct":"c","options":{"a":"22 hours ","b":"30 hours ","c":"44 hours ","d":"60 hours","e":"it can not be determined from the information above ."},"options_float":{"a":22.0,"b":30.0,"c":44.0,"d":60.0,"e":null},"annotated_formula":"divide(const_1, divide(add(divide(const_1, 4), divide(const_1, 11)), 15))","linear_formula":"divide(const_1,n2)|divide(const_1,n5)|add(#0,#1)|divide(#2,n0)|divide(const_1,#3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 11<\/gadget>\n1\/11 = around 0.090909<\/output>\n(1\/4) + (1\/11)<\/gadget>\n15\/44 = around 0.340909<\/output>\n(15\/44) \/ 15<\/gadget>\n1\/44 = around 0.022727<\/output>\n1 \/ (1\/44)<\/gadget>\n44<\/output>\n44<\/result>","index":50} +{"problem":"rectangular floors x and y have equal area . if floor x is 10 feet by 18 feet and floor y is 9 feet wide , what is the length of floor y , in feet ?","rationale":"the area of a rectangle is : area = length x width we are given that floor x is 10 feet by 18 feet and that floor y is 9 feet wide . so we can say : length of x = 10 width of x = 18 width of y = 9 length of y = n we also can say : area of floor x = area of floor y ( length of x ) ( width of x ) = ( length of y ) ( width of y ) ( 10 ) ( 18 ) = 9 n ( 10 ) ( 2 ) = n 20 = n answer b .","correct":"b","options":{"a":"13 1 \/ 2 ","b":"20 ","c":"18 3 \/ 4 ","d":"21","e":"24"},"options_float":{"a":13.0,"b":20.0,"c":18.0,"d":21.0,"e":24.0},"annotated_formula":"divide(rectangle_area(10, 18), 9)","linear_formula":"rectangle_area(n0,n1)|divide(#0,n2)","chain":"10 * 18<\/gadget>\n180<\/output>\n180 \/ 9<\/gadget>\n20<\/output>\n20<\/result>","index":51} +{"problem":"a distributor sells a product through an online store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 18 per item . what is the price that the buyer observers online if the distributor wants to maintain a 20 % profit on the cost of the item ?","rationale":"\"let x be the price that buyers see online . the distributor wants to receive 1.2 ( original price ) which should be 80 % of x . 1.2 ( 18 ) = 0.8 x x = 1.2 ( 18 ) \/ 0.8 = 1.5 ( 18 ) = $ 27 the answer is c .\"","correct":"c","options":{"a":"$ 25 ","b":"$ 26 ","c":"$ 27 ","d":"$ 28","e":"$ 29"},"options_float":{"a":25.0,"b":26.0,"c":27.0,"d":28.0,"e":29.0},"annotated_formula":"divide(add(multiply(divide(20, const_100), 18), 18), divide(subtract(const_100, 20), const_100))","linear_formula":"divide(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(n1,#0)|add(n1,#3)|divide(#4,#2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 18<\/gadget>\n18\/5 = around 3.6<\/output>\n(18\/5) + 18<\/gadget>\n108\/5 = around 21.6<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(108\/5) \/ (4\/5)<\/gadget>\n27<\/output>\n27<\/result>","index":53} +{"problem":"the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 47 kg , then the weight of b is :","rationale":"\"explanation : let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 … . ( i ) a + b = ( 40 x 2 ) = 80 … . ( ii ) b + c = ( 47 x 2 ) = 94 … . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 174 … . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 39 . b ’ s weight = 39 kg . answer : a\"","correct":"a","options":{"a":"39 kg ","b":"32 kg ","c":"33 kg ","d":"34 kg","e":"none of these"},"options_float":{"a":39.0,"b":32.0,"c":33.0,"d":34.0,"e":null},"annotated_formula":"subtract(add(multiply(40, const_2), multiply(47, const_2)), multiply(45, const_3))","linear_formula":"multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|","chain":"40 * 2<\/gadget>\n80<\/output>\n47 * 2<\/gadget>\n94<\/output>\n80 + 94<\/gadget>\n174<\/output>\n45 * 3<\/gadget>\n135<\/output>\n174 - 135<\/gadget>\n39<\/output>\n39<\/result>","index":54} +{"problem":"6 women can do 75 units of work in 8 days by working 5 hrs \/ day . in how many days can 4 women do 30 units of work by working 8 hrs \/ day ?","rationale":"6 x 8 x 5 \/ 75 = 4 x 8 xa \/ 30 solve this for a then we get a = 3 days answer : a","correct":"a","options":{"a":"3 days ","b":"4 days ","c":"5 days ","d":"6 days","e":"7 days"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(divide(multiply(multiply(6, 8), 5), 75), divide(multiply(4, 8), 30))","linear_formula":"multiply(n0,n2)|multiply(n2,n4)|divide(#1,n5)|multiply(n3,#0)|divide(#3,n1)|divide(#4,#2)","chain":"6 * 8<\/gadget>\n48<\/output>\n48 * 5<\/gadget>\n240<\/output>\n240 \/ 75<\/gadget>\n16\/5 = around 3.2<\/output>\n4 * 8<\/gadget>\n32<\/output>\n32 \/ 30<\/gadget>\n16\/15 = around 1.066667<\/output>\n(16\/5) \/ (16\/15)<\/gadget>\n3<\/output>\n3<\/result>","index":55} +{"problem":"the speed of a boat in still water is 20 km \/ hr and the rate of current is 5 km \/ hr . the distance travelled downstream in 18 minutes is :","rationale":"\"explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 18 \/ 60 ) ) km = 7.5 km . answer : c\"","correct":"c","options":{"a":"9.5 ","b":"5.5 ","c":"7.5 ","d":"6.5","e":"5.1"},"options_float":{"a":9.5,"b":5.5,"c":7.5,"d":6.5,"e":5.1},"annotated_formula":"multiply(divide(18, const_60), add(20, 5))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"18 \/ 60<\/gadget>\n3\/10 = around 0.3<\/output>\n20 + 5<\/gadget>\n25<\/output>\n(3\/10) * 25<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":56} +{"problem":"michelle deposited a certain sum of money in a savings account on july 1 st , 2007 . she earns an 14 % interest compounded semiannually . the sum of money in the account on december 31 st , 2009 is approximately what percent of the initial deposit ?","rationale":"\"since michelle earns 14 % interest compounded semiannually , then she earns 7 % interest every 6 months . now , the simple interest earned in 5 periods ( 30 months = 5 * 6 months ) would be 7 % * 5 = 35 % . but , since the interest iscompoundedevery 6 months , then there would be interest earned on interest ( very small amount ) thus the actual interest earned would be a little bit more than 35 % , only answer choice d fits . answer : d\"","correct":"d","options":{"a":"117 % ","b":"120 % ","c":"121 % ","d":"138 %","e":"145 %"},"options_float":{"a":117.0,"b":120.0,"c":121.0,"d":138.0,"e":145.0},"annotated_formula":"multiply(power(add(1, divide(divide(14, const_100), const_2)), add(const_2, const_3)), const_100)","linear_formula":"add(const_2,const_3)|divide(n2,const_100)|divide(#1,const_2)|add(#2,n0)|power(#3,#0)|multiply(#4,const_100)|","chain":"14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n(7\/50) \/ 2<\/gadget>\n7\/100 = around 0.07<\/output>\n1 + (7\/100)<\/gadget>\n107\/100 = around 1.07<\/output>\n2 + 3<\/gadget>\n5<\/output>\n(107\/100) ** 5<\/gadget>\n14_025_517_307\/10_000_000_000 = around 1.402552<\/output>\n(14_025_517_307\/10_000_000_000) * 100<\/gadget>\n14_025_517_307\/100_000_000 = around 140.255173<\/output>\n14_025_517_307\/100_000_000 = around 140.255173<\/result>","index":57} +{"problem":"the average of the marks of 30 students in a class is 45 . if the marks of each student are doubled , find the new average ?","rationale":"\"sum of the marks for the 30 students = 30 * 45 = 1350 . the marks of each student are doubled , the sum also will be doubled . the new sum = 1350 * 2 = 2700 . so , the new average = 2700 \/ 30 = 90 . answer : d\"","correct":"d","options":{"a":"85 ","b":"77 ","c":"33 ","d":"90","e":"75"},"options_float":{"a":85.0,"b":77.0,"c":33.0,"d":90.0,"e":75.0},"annotated_formula":"multiply(45, const_2)","linear_formula":"multiply(n1,const_2)|","chain":"45 * 2<\/gadget>\n90<\/output>\n90<\/result>","index":58} +{"problem":"mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 25 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ?","rationale":"\"lets assume that candidate got 25 % votes and total votes is 100 . candidate won = 25 remaining = 75 to get 50 % , candidate requires 25 votes from 100 which is 25 % and 25 votes from 75 . 25 \/ 75 = 33.33 % which is approx 33 % . hence the answer is e\"","correct":"e","options":{"a":"10 % ","b":"12 % ","c":"15 % ","d":"17 %","e":"33 %"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":17.0,"e":33.0},"annotated_formula":"multiply(divide(subtract(divide(50, const_100), divide(25, const_100)), subtract(const_1, divide(25, const_100))), const_100)","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) - (1\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n(1\/4) \/ (3\/4)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":59} +{"problem":"a certain company expects quarterly earnings of $ 0.80 per share of stock , half of which will be distributed as dividends to shareholders while the rest will be used for research and development . if earnings are greater than expected , shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings . if quarterly earnings are $ 1.10 per share , what will be the dividend paid to a person who owns 100 shares of the company ' s stock ?","rationale":"\"eps actual > eps expected . each gets and additional . 12 per share . thus . 52 * 100 - - > $ 52 answer is c\"","correct":"c","options":{"a":"$ 92 ","b":"$ 96 ","c":"$ 52 ","d":"$ 120","e":"$ 240"},"options_float":{"a":92.0,"b":96.0,"c":52.0,"d":120.0,"e":240.0},"annotated_formula":"multiply(add(divide(0.80, const_2), multiply(multiply(subtract(1.10, 0.80), const_10), 0.04)), 100)","linear_formula":"divide(n0,const_2)|subtract(n3,n0)|multiply(#1,const_10)|multiply(n1,#2)|add(#0,#3)|multiply(n4,#4)|","chain":"0.8 \/ 2<\/gadget>\n0.4<\/output>\n1.1 - 0.8<\/gadget>\n0.3<\/output>\n0.3 * 10<\/gadget>\n3<\/output>\n3 * 0.04<\/gadget>\n0.12<\/output>\n0.4 + 0.12<\/gadget>\n0.52<\/output>\n0.52 * 100<\/gadget>\n52<\/output>\n52<\/result>","index":60} +{"problem":"the price of precious stone is directly proportional to square of its weight . what will be the loss incurred if a stone weighing 28 gm and costing 28,000 breaks into two pieces whose weights are in the ratio 15 : 13 ?","rationale":"x = 28000 \/ 28 ^ 2 = 35.71 cost of 15 g = 35.71 * 15 ^ 2 = 8035.714 cost of 13 g = 35.71 * 13 ^ 2 = 6035.714 total cost = 14071.428 loss = 28000 - 14071.428 = 13928.5 = 14000 answer : e","correct":"e","options":{"a":"no loss ","b":"10,000 ","c":"28,000 ","d":"9,500","e":"14,000"},"options_float":{"a":null,"b":10000.0,"c":28000.0,"d":9500.0,"e":14000.0},"annotated_formula":"subtract(multiply(28, multiply(const_10, const_100)), add(multiply(divide(multiply(28, multiply(const_10, const_100)), multiply(28, 28)), multiply(13, 13)), multiply(divide(multiply(28, multiply(const_10, const_100)), multiply(28, 28)), multiply(15, 15))))","linear_formula":"multiply(const_10,const_100)|multiply(n0,n0)|multiply(n3,n3)|multiply(n2,n2)|multiply(n0,#0)|divide(#4,#1)|multiply(#5,#2)|multiply(#5,#3)|add(#6,#7)|subtract(#4,#8)","chain":"10 * 100<\/gadget>\n1_000<\/output>\n28 * 1_000<\/gadget>\n28_000<\/output>\n28 * 28<\/gadget>\n784<\/output>\n28_000 \/ 784<\/gadget>\n250\/7 = around 35.714286<\/output>\n13 * 13<\/gadget>\n169<\/output>\n(250\/7) * 169<\/gadget>\n42_250\/7 = around 6_035.714286<\/output>\n15 * 15<\/gadget>\n225<\/output>\n(250\/7) * 225<\/gadget>\n56_250\/7 = around 8_035.714286<\/output>\n(42_250\/7) + (56_250\/7)<\/gadget>\n98_500\/7 = around 14_071.428571<\/output>\n28_000 - (98_500\/7)<\/gadget>\n97_500\/7 = around 13_928.571429<\/output>\n97_500\/7 = around 13_928.571429<\/result>","index":61} +{"problem":"roy is now 6 years older than julia and half of that amount older than kelly . if in 4 years , roy will be twice as old as julia , then in 4 years what would be roy ’ s age multiplied by kelly ’ s age ?","rationale":"r = j + 6 = k + 3 r + 4 = 2 ( j + 4 ) ( j + 6 ) + 4 = 2 j + 8 j = 2 r = 8 k = 5 in 4 years ( r + 4 ) ( k + 4 ) = 12 * 9 = 108 the answer is e .","correct":"e","options":{"a":"72 ","b":"84 ","c":"90 ","d":"96","e":"108"},"options_float":{"a":72.0,"b":84.0,"c":90.0,"d":96.0,"e":108.0},"annotated_formula":"multiply(add(add(6, subtract(6, 4)), 4), add(subtract(add(6, subtract(6, 4)), divide(6, const_2)), 4))","linear_formula":"divide(n0,const_2)|subtract(n0,n1)|add(n0,#1)|add(n1,#2)|subtract(#2,#0)|add(n1,#4)|multiply(#3,#5)","chain":"6 - 4<\/gadget>\n2<\/output>\n6 + 2<\/gadget>\n8<\/output>\n8 + 4<\/gadget>\n12<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n8 - 3<\/gadget>\n5<\/output>\n5 + 4<\/gadget>\n9<\/output>\n12 * 9<\/gadget>\n108<\/output>\n108<\/result>","index":62} +{"problem":"the mean of 50 observations was 36 . it was found later an observation 48 was wrongly taken as 23 . the corrected new means is :","rationale":"\"solution correct sum = ( 36 × 50 + 48 - 23 ) = 1525 . ∴ correct mean = 1825 \/ 50 = 36.5 answer c\"","correct":"c","options":{"a":"35.2 ","b":"36.1 ","c":"36.5 ","d":"39.1","e":"none of these"},"options_float":{"a":35.2,"b":36.1,"c":36.5,"d":39.1,"e":null},"annotated_formula":"divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)","linear_formula":"multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|","chain":"36 * 50<\/gadget>\n1_800<\/output>\n50 - 2<\/gadget>\n48<\/output>\n48 - 23<\/gadget>\n25<\/output>\n1_800 + 25<\/gadget>\n1_825<\/output>\n1_825 \/ 50<\/gadget>\n73\/2 = around 36.5<\/output>\n73\/2 = around 36.5<\/result>","index":67} +{"problem":"there are 750 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ?","rationale":"\"let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 750 . hence , we have m + f = 750 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f \/ 2 + m \/ 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 750 \/ 3 = 250 solving the three equations yields the solution f = 250 , m = 500 , and d = 250 . the number of female democratic participants equals half the female participants equals 250 \/ 2 = 125 . answer : c\"","correct":"c","options":{"a":"75 ","b":"100 ","c":"125 ","d":"175","e":"225"},"options_float":{"a":75.0,"b":100.0,"c":125.0,"d":175.0,"e":225.0},"annotated_formula":"divide(subtract(multiply(divide(750, const_3), const_4), 750), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)|","chain":"750 \/ 3<\/gadget>\n250<\/output>\n250 * 4<\/gadget>\n1_000<\/output>\n1_000 - 750<\/gadget>\n250<\/output>\n250 \/ 2<\/gadget>\n125<\/output>\n125<\/result>","index":68} +{"problem":"in what ratio must water be mixed with milk to gain 16 2 \/ 3 % on selling the mixture at cost price ?","rationale":"required ratio is cheaper quantity : dearer quantity = ( d - m ) : ( m - c ) answer with explanation : step i ) let c . p of 1 litre of mild be rs . 1 and s . p of 1 litre of mild be rs . 1 gain = 50 \/ 3 per cent  c . p of 1 litre of mixture =              1 3 50 100 100 x = re . 100 x 3 \/ 350 x 1 = 6 \/ 7 ( mean price ) step ( ii ) by the rule of allegation i . c . p of 1 liter of water = 0 ii . c . p of 1 liter of milk = 1 iii . mean price ( p ) = 6 \/ 7 iv . d – m = 1 – 6 \/ 7 = 1 \/ 7 v . m – c = 6 \/ 7 - 0 = 6 \/ 7  ratio of water and milk = 1 \/ 7 : 6 \/ 7 = 1 : 6 answer : d","correct":"d","options":{"a":"1 : 9 ","b":"1 : 1 ","c":"1 : 2 ","d":"1 : 6","e":"1 : 5"},"options_float":{"a":0.1111111111,"b":1.0,"c":0.5,"d":0.1666666667,"e":0.2},"annotated_formula":"divide(add(16, divide(2, 3)), const_100)","linear_formula":"divide(n1,n2)|add(n0,#0)|divide(#1,const_100)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n16 + (2\/3)<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) \/ 100<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":70} +{"problem":"9 friends decided to divide the hotel bill evenly . if the bill was $ 124.15 dollars , how much money did they pay if 1 cent is the smallest unit ?","rationale":"this is equivalent to finding the first number that is divisible by 9 that occurs after 12415 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 12415 = 13 and 9 is divisible by 9 . hence , we need to remove 4 to this number for it to be divisible by 9 correct option : a","correct":"a","options":{"a":"$ 124.11 ","b":"$ 124.12 ","c":"$ 124.13 ","d":"$ 124.14","e":"$ 124.15"},"options_float":{"a":124.11,"b":124.12,"c":124.13,"d":124.14,"e":124.15},"annotated_formula":"subtract(124.15, divide(const_4, const_100))","linear_formula":"divide(const_4,const_100)|subtract(n1,#0)","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n124.15 - (1\/25)<\/gadget>\n124.11<\/output>\n124.11<\/result>","index":71} +{"problem":"on sunday , bill ran 4 more miles than he ran on saturday . julia did not run on saturday , but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of 36 miles on saturday and sunday , how many miles did bill run on sunday ?","rationale":"\"let bill run x on saturday , so he will run x + 4 on sunday . . julia will run 2 * ( x + 4 ) on sunday . . totai = x + x + 4 + 2 x + 8 = 36 . . 4 x + 12 = 36 . . x = 6 . . ans = x + 4 = 6 + 4 = 10 answer e\"","correct":"e","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"10"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":10.0},"annotated_formula":"add(divide(subtract(36, add(4, multiply(const_2, 4))), 4), 4)","linear_formula":"multiply(n0,const_2)|add(n0,#0)|subtract(n1,#1)|divide(#2,n0)|add(n0,#3)|","chain":"2 * 4<\/gadget>\n8<\/output>\n4 + 8<\/gadget>\n12<\/output>\n36 - 12<\/gadget>\n24<\/output>\n24 \/ 4<\/gadget>\n6<\/output>\n6 + 4<\/gadget>\n10<\/output>\n10<\/result>","index":72} +{"problem":"the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 2000 resolutions ?","rationale":"\"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 2000 resolutions . = 2000 * 2 * 22 \/ 7 * 22.4 = 281600 cm = 2816 m answer : c\"","correct":"c","options":{"a":"1187 m ","b":"1704 m ","c":"2816 m ","d":"2827 m","e":"2897 m"},"options_float":{"a":1187.0,"b":1704.0,"c":2816.0,"d":2827.0,"e":2897.0},"annotated_formula":"divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 2000), const_100)","linear_formula":"add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)|","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21 + 1<\/gadget>\n22<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n(22\/7) * 22.4<\/gadget>\n70.4<\/output>\n70.4 * 2<\/gadget>\n140.8<\/output>\n140.8 * 2_000<\/gadget>\n281_600<\/output>\n281_600 \/ 100<\/gadget>\n2_816<\/output>\n2_816<\/result>","index":74} +{"problem":"earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 300 envelopes","rationale":"\"earl takes 1 min . for 36 envelopes . ellen takes 3 \/ 2 mins for the same . so ellen can stuff ( ( 36 ) \/ ( 3 \/ 2 ) ) in 1 min . i . e . , 24 envlpes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 300 envelopes they will take 300 \/ 60 mins . i . e . , 5 mins . answer : b\"","correct":"b","options":{"a":"6 minutes ","b":"5 minutes ","c":"7 minutes ","d":"3 minutes","e":"4 minutes"},"options_float":{"a":6.0,"b":5.0,"c":7.0,"d":3.0,"e":4.0},"annotated_formula":"divide(300, add(36, divide(36, divide(const_3, const_2))))","linear_formula":"divide(const_3,const_2)|divide(n0,#0)|add(n0,#1)|divide(n1,#2)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n36 \/ (3\/2)<\/gadget>\n24<\/output>\n36 + 24<\/gadget>\n60<\/output>\n300 \/ 60<\/gadget>\n5<\/output>\n5<\/result>","index":76} +{"problem":"the sum of first 45 natural numbers is ?","rationale":"\"let sn = ( 1 + 2 + 3 + . . . + 45 ) . this is an a . p . in which a = 1 , d = 1 , n = 45 . sn = n [ 2 a + ( n - 1 ) d ] = 45 x [ 2 x 1 + ( 45 - 1 ) x 1 ] = 45 x 46 = ( 45 x 23 ) 2 2 2 = 45 x ( 20 + 3 ) = 45 x 20 + 45 x 3 = 900 + 135 = 1035 . c )\"","correct":"c","options":{"a":"1030 ","b":"1034 ","c":"1035 ","d":"1037","e":"1040"},"options_float":{"a":1030.0,"b":1034.0,"c":1035.0,"d":1037.0,"e":1040.0},"annotated_formula":"divide(multiply(45, add(45, const_1)), const_2)","linear_formula":"add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|","chain":"45 + 1<\/gadget>\n46<\/output>\n45 * 46<\/gadget>\n2_070<\/output>\n2_070 \/ 2<\/gadget>\n1_035<\/output>\n1_035<\/result>","index":77} +{"problem":"vehicle x is 22 miles ahead of vehicle y , which is traveling the same direction along the same route as vehicle x . if vehicle x is traveling at an average speed of 36 miles per hour and vehicle y is traveling at an average speed of 45 miles per hour , how long will it take vehicle y to overtake and drive 23 miles ahead of vehicle x ?","rationale":"relative speed = 45 - 36 = 9 miles per hour dist required = 22 + 23 = 45 miles time taken to overtake = 45 \/ 9 = 5 hours . a is the answer .","correct":"a","options":{"a":"5 hours ","b":"7 hours 20 minutes ","c":"4 hours ","d":"3 hours 45 minutes","e":"2 hours 30 minutes"},"options_float":{"a":5.0,"b":7.0,"c":4.0,"d":3.0,"e":2.0},"annotated_formula":"divide(add(22, 23), subtract(45, 36))","linear_formula":"add(n0,n3)|subtract(n2,n1)|divide(#0,#1)","chain":"22 + 23<\/gadget>\n45<\/output>\n45 - 36<\/gadget>\n9<\/output>\n45 \/ 9<\/gadget>\n5<\/output>\n5<\/result>","index":78} +{"problem":"in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 5 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ?","rationale":"\"sugar : flour = 5 : 5 = 10 : 10 ; flour : soda = 10 : 1 = 10 : 1 ; thus we have that sugar : flour : soda = 10 x : 10 x : 1 x . also given that 10 x \/ ( 1 x + 60 ) = 8 \/ 1 - - > x = 240 - - > sugar = 10 x = 2400 answer : e .\"","correct":"e","options":{"a":"600 ","b":"1200 ","c":"1500 ","d":"1600","e":"2400"},"options_float":{"a":600.0,"b":1200.0,"c":1500.0,"d":1600.0,"e":2400.0},"annotated_formula":"multiply(divide(5, 5), multiply(10, divide(multiply(8, 60), subtract(10, 8))))","linear_formula":"divide(n0,n1)|multiply(n4,n5)|subtract(n2,n5)|divide(#1,#2)|multiply(n2,#3)|multiply(#0,#4)|","chain":"5 \/ 5<\/gadget>\n1<\/output>\n8 * 60<\/gadget>\n480<\/output>\n10 - 8<\/gadget>\n2<\/output>\n480 \/ 2<\/gadget>\n240<\/output>\n10 * 240<\/gadget>\n2_400<\/output>\n1 * 2_400<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":79} +{"problem":"the average weight of 8 person ' s increases by 3 kg when a new person comes in place of one of them weighing 70 kg . what might be the weight of the new person ?","rationale":"\"total weight increased = ( 8 x 3 ) kg = 24 kg . weight of new person = ( 70 + 24 ) kg = 94 kg . answer : a\"","correct":"a","options":{"a":"94 kg ","b":"85 kg ","c":"90 kg ","d":"100 kg","e":"110 kg"},"options_float":{"a":94.0,"b":85.0,"c":90.0,"d":100.0,"e":110.0},"annotated_formula":"add(multiply(8, 3), 70)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"8 * 3<\/gadget>\n24<\/output>\n24 + 70<\/gadget>\n94<\/output>\n94<\/result>","index":80} +{"problem":"the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 10 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ?","rationale":"\"area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 10 = rs . 9060 answer : a\"","correct":"a","options":{"a":"s . 9060 ","b":"s . 4520 ","c":"s . 4527 ","d":"s . 4530","e":"s . 4521"},"options_float":{"a":9060.0,"b":4520.0,"c":4527.0,"d":4530.0,"e":4521.0},"annotated_formula":"multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 10)","linear_formula":"add(n0,n1)|multiply(n2,const_2)|multiply(n4,n5)|multiply(n5,n6)|multiply(#0,#1)|multiply(n5,#3)|subtract(#4,#2)|subtract(#6,#5)|multiply(n3,#7)|","chain":"2 * 12<\/gadget>\n24<\/output>\n15 + 25<\/gadget>\n40<\/output>\n24 * 40<\/gadget>\n960<\/output>\n6 * 3<\/gadget>\n18<\/output>\n960 - 18<\/gadget>\n942<\/output>\n4 * 3<\/gadget>\n12<\/output>\n3 * 12<\/gadget>\n36<\/output>\n942 - 36<\/gadget>\n906<\/output>\n906 * 10<\/gadget>\n9_060<\/output>\n9_060<\/result>","index":82} +{"problem":"a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 7 days . b can do the work alone in ?","rationale":"\"suppose a , b and c take x , x \/ 2 and x \/ 3 respectively to finish the work . then , ( 1 \/ x + 2 \/ x + 3 \/ x ) = 1 \/ 7 6 \/ x = 1 \/ 7 = > x = 42 so , b takes 21 hours to finish the work . answer : d\"","correct":"d","options":{"a":"19 ","b":"12 ","c":"11 ","d":"42","e":"114"},"options_float":{"a":19.0,"b":12.0,"c":11.0,"d":42.0,"e":114.0},"annotated_formula":"multiply(add(add(const_1, const_2), const_3), 7)","linear_formula":"add(const_1,const_2)|add(#0,const_3)|multiply(n0,#1)|","chain":"1 + 2<\/gadget>\n3<\/output>\n3 + 3<\/gadget>\n6<\/output>\n6 * 7<\/gadget>\n42<\/output>\n42<\/result>","index":83} +{"problem":"a man can swim in still water at 4.5 km \/ h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?","rationale":"\"m = 4.5 s = x ds = 4.5 + x us = 4.5 + x 4.5 + x = ( 4.5 - x ) 2 4.5 + x = 9 - 2 x 3 x = 4.5 x = 1.5 answer : d\"","correct":"d","options":{"a":"1.0 ","b":"1.9 ","c":"1.6 ","d":"1.5","e":"1.1"},"options_float":{"a":1.0,"b":1.9,"c":1.6,"d":1.5,"e":1.1},"annotated_formula":"divide(4.5, const_3)","linear_formula":"divide(n0,const_3)|","chain":"4.5 \/ 3<\/gadget>\n1.5<\/output>\n1.5<\/result>","index":86} +{"problem":"ifaequals the sum of the even integers from 2 to 120 , inclusive , andbequals the sum of the odd integers from 1 to 119 , inclusive , what is the value of a - b ?","rationale":"\"this is a solution from beatthegmat : even numbers : ( 120 - 2 ) \/ 2 + 1 = 60 even integers . ( 120 + 2 ) \/ 2 = 61 is the average of the even set . sum = avg * ( # of elements ) = 61 * 60 = 3660 = a odd numbers : ( 119 - 1 ) \/ 2 + 1 = 60 odd integers . ( 119 + 1 ) \/ 2 = 60 is the average of the odd set . sum = avg * ( # of elements ) = 60 * 60 = 3600 = b a - b = 3660 - 3600 = 60 . answer : a\"","correct":"a","options":{"a":"60 ","b":"10 ","c":"19 ","d":"20","e":"21"},"options_float":{"a":60.0,"b":10.0,"c":19.0,"d":20.0,"e":21.0},"annotated_formula":"subtract(multiply(divide(120, 2), add(divide(120, 2), 1)), multiply(divide(add(119, 1), 2), add(divide(subtract(119, 1), 2), 1)))","linear_formula":"add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)|","chain":"120 \/ 2<\/gadget>\n60<\/output>\n60 + 1<\/gadget>\n61<\/output>\n60 * 61<\/gadget>\n3_660<\/output>\n119 + 1<\/gadget>\n120<\/output>\n119 - 1<\/gadget>\n118<\/output>\n118 \/ 2<\/gadget>\n59<\/output>\n59 + 1<\/gadget>\n60<\/output>\n60 * 60<\/gadget>\n3_600<\/output>\n3_660 - 3_600<\/gadget>\n60<\/output>\n60<\/result>","index":90} +{"problem":"at what price must an article costing rs . 47.50 be marked in order that after deducting 7 % from the list price . it may be sold at a profit of 25 % on the cost price ?","rationale":"\"cp = 47.50 sp = 47.50 * ( 125 \/ 100 ) = 59.375 mp * ( 93 \/ 100 ) = 59.375 mp = 63.84 answer : a\"","correct":"a","options":{"a":"63.84 ","b":"62.6 ","c":"62.1 ","d":"62.7","e":"62.2"},"options_float":{"a":63.84,"b":62.6,"c":62.1,"d":62.7,"e":62.2},"annotated_formula":"divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 7))","linear_formula":"multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)|","chain":"47.5 * 25<\/gadget>\n1_187.5<\/output>\n1_187.5 \/ 100<\/gadget>\n11.875<\/output>\n47.5 + 11.875<\/gadget>\n59.375<\/output>\n59.375 * 100<\/gadget>\n5_937.5<\/output>\n100 - 7<\/gadget>\n93<\/output>\n5_937.5 \/ 93<\/gadget>\n63.844086<\/output>\n63.844086<\/result>","index":92} +{"problem":"in may mrs lee ' s earnings were 50 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ?","rationale":"lets say the family income is 100 in may , lee earned 50 family income is 50 in june , lee earned 20 % more than may , so it is ( 50 + 20 * 50 \/ 100 = 60 ) family income is same 50 in june lee ' s income percent is 60 * 100 \/ 110 ~ 55 ans is a","correct":"a","options":{"a":"55 ","b":"56 ","c":"34 ","d":"78","e":"23"},"options_float":{"a":55.0,"b":56.0,"c":34.0,"d":78.0,"e":23.0},"annotated_formula":"multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100)","linear_formula":"add(n1,const_100)|add(#0,const_100)|divide(#0,#1)|multiply(#2,const_100)","chain":"100 + 20<\/gadget>\n120<\/output>\n120 + 100<\/gadget>\n220<\/output>\n120 \/ 220<\/gadget>\n6\/11 = around 0.545455<\/output>\n(6\/11) * 100<\/gadget>\n600\/11 = around 54.545455<\/output>\n600\/11 = around 54.545455<\/result>","index":93} +{"problem":"a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 80 percent of books that were loaned out are returned and there are 69 books in the special collection at that time , how many books of the special collection were loaned out during that month ?","rationale":"\"there are 6 books less ( 75 - 69 ) which represents 20 % of the loaned books ( 100 - 80 ) so total loaned out books = 30 answer b\"","correct":"b","options":{"a":"20 ","b":"30 ","c":"35 ","d":"40","e":"55"},"options_float":{"a":20.0,"b":30.0,"c":35.0,"d":40.0,"e":55.0},"annotated_formula":"divide(subtract(75, 69), subtract(const_1, divide(80, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"75 - 69<\/gadget>\n6<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n6 \/ (1\/5)<\/gadget>\n30<\/output>\n30<\/result>","index":95} +{"problem":"if an object travels 300 feet in 6 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet )","rationale":"\"1 mile = 5280 feet = > 1 feet = 1 \/ 5280 miles if the object travels 300 feet in 6 sec then it travels 300 \/ 6 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 50 feet in 1 hour = 180000 feet in 1 hr = 180000 \/ 5280 miles in 1 hour = 18000 \/ 528 miles \/ hr ~ 34 miles \/ hr answer - e\"","correct":"e","options":{"a":"30 ","b":"31 ","c":"32 ","d":"33","e":"34"},"options_float":{"a":30.0,"b":31.0,"c":32.0,"d":33.0,"e":34.0},"annotated_formula":"divide(divide(300, 5280), multiply(6, divide(1, const_3600)))","linear_formula":"divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)|","chain":"300 \/ 5_280<\/gadget>\n5\/88 = around 0.056818<\/output>\n1 \/ 3_600<\/gadget>\n1\/3_600 = around 0.000278<\/output>\n6 * (1\/3_600)<\/gadget>\n1\/600 = around 0.001667<\/output>\n(5\/88) \/ (1\/600)<\/gadget>\n375\/11 = around 34.090909<\/output>\n375\/11 = around 34.090909<\/result>","index":96} +{"problem":"in a sports club with 80 members , 48 play badminton and 46 play tennis and 7 do not play either . how many members play both badminton and tennis ?","rationale":"48 + 46 = 94 but where as total number is 80 - 7 = 73 therefore answer is 94 - 73 = 21 hence answer is b","correct":"b","options":{"a":"23 ","b":"21 ","c":"20 ","d":"26","e":"27"},"options_float":{"a":23.0,"b":21.0,"c":20.0,"d":26.0,"e":27.0},"annotated_formula":"subtract(add(add(48, 46), 7), 80)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)","chain":"48 + 46<\/gadget>\n94<\/output>\n94 + 7<\/gadget>\n101<\/output>\n101 - 80<\/gadget>\n21<\/output>\n21<\/result>","index":98} +{"problem":"in a fuel station the service costs $ 2.10 per vehicle and every liter of fuel costs $ 0.60 . assuming that you fill up 3 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ?","rationale":"\"the service cost of 3 vans and 2 trucks is 5 * 2.10 $ 10.50 the fuel in 3 vans is 3 * 65 = 195 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 481 liters the total fuel cost is 481 * 0.6 = $ 288.60 the total cost is $ 288.60 + $ 10.50 = $ 299.10 the answer is e .\"","correct":"e","options":{"a":"$ 287.50 ","b":"$ 290.40 ","c":"$ 293.30 ","d":"$ 296.20","e":"$ 299.10"},"options_float":{"a":287.5,"b":290.4,"c":293.3,"d":296.2,"e":299.1},"annotated_formula":"add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.60), multiply(multiply(3, 65), 0.60)), multiply(2.10, add(3, 2)))","linear_formula":"add(n2,n3)|multiply(n4,n5)|multiply(n2,n4)|divide(#1,const_100)|multiply(n1,#2)|multiply(n0,#0)|add(n4,#3)|multiply(n3,#6)|multiply(n1,#7)|add(#8,#4)|add(#9,#5)|","chain":"120 * 65<\/gadget>\n7_800<\/output>\n7_800 \/ 100<\/gadget>\n78<\/output>\n78 + 65<\/gadget>\n143<\/output>\n143 * 2<\/gadget>\n286<\/output>\n286 * 0.6<\/gadget>\n171.6<\/output>\n3 * 65<\/gadget>\n195<\/output>\n195 * 0.6<\/gadget>\n117<\/output>\n171.6 + 117<\/gadget>\n288.6<\/output>\n3 + 2<\/gadget>\n5<\/output>\n2.1 * 5<\/gadget>\n10.5<\/output>\n288.6 + 10.5<\/gadget>\n299.1<\/output>\n299.1<\/result>","index":99} +{"problem":"half a number plus 7 is 17 . what is the number ?","rationale":"\"let x be the number . always replace ` ` is ' ' with an equal sign ( 1 \/ 2 ) x + 7 = 17 ( 1 \/ 2 ) x = 17 - 7 ( 1 \/ 2 ) x = 20 x = 20 correct answer is e\"","correct":"e","options":{"a":"8 ","b":"9 ","c":"10 ","d":"11","e":"20"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":20.0},"annotated_formula":"multiply(subtract(17, 7), const_2)","linear_formula":"subtract(n1,n0)|multiply(#0,const_2)|","chain":"17 - 7<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":103} +{"problem":"if the ratio of the sum of the first 6 terms of a g . p . to the sum of the first 3 terms of the g . p . is 343 , what is the common ratio of the g . p ?","rationale":"\"343 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) \/ ( a 1 + a 2 + a 3 ) factorize the same terms 343 = 1 + ( a 4 + a 5 + a 6 ) \/ ( a 1 + a 2 + a 3 ) write every term with respect to r a 1 = a 1 a 2 = a 1 * r ^ 1 a 3 = a 1 * r ^ 2 . . . . . . . . . 343 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) \/ ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 342 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) \/ ( ( 1 + r ^ 1 + r ^ 2 ) ) 342 = r ^ 3 r = 7 d\"","correct":"d","options":{"a":"3 ","b":"1 \/ 3 ","c":"2 ","d":"7","e":"1 \/ 7"},"options_float":{"a":3.0,"b":0.3333333333,"c":2.0,"d":7.0,"e":0.1428571429},"annotated_formula":"power(subtract(343, const_1), divide(const_1, 3))","linear_formula":"divide(const_1,n1)|subtract(n2,const_1)|power(#1,#0)|","chain":"343 - 1<\/gadget>\n342<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n342 ** (1\/3)<\/gadget>\n342**(1\/3) = around 6.993191<\/output>\n342**(1\/3) = around 6.993191<\/result>","index":105} +{"problem":"bert and rebecca were looking at the price of a condominium . the price of the condominium was 50 % more than bert had in savings , and separately , the same price was also 25 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings .","rationale":"\"suppose bert had 100 so price becomes 150 , this 150 = 1.25 times r ' s saving . . so r ' s saving becomes 120 so required ratio is 100 : 120 = 5 : 6 answer : e\"","correct":"e","options":{"a":"1 : 6 ","b":"1 : 3 ","c":"1 : 2 ","d":"3 : 4","e":"5 : 6"},"options_float":{"a":0.1666666667,"b":0.3333333333,"c":0.5,"d":0.75,"e":0.8333333333},"annotated_formula":"divide(divide(const_100, add(const_100, 50)), divide(const_100, add(const_100, 25)))","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(const_100,#0)|divide(const_100,#1)|divide(#2,#3)|","chain":"100 + 50<\/gadget>\n150<\/output>\n100 \/ 150<\/gadget>\n2\/3 = around 0.666667<\/output>\n100 + 25<\/gadget>\n125<\/output>\n100 \/ 125<\/gadget>\n4\/5 = around 0.8<\/output>\n(2\/3) \/ (4\/5)<\/gadget>\n5\/6 = around 0.833333<\/output>\n5\/6 = around 0.833333<\/result>","index":106} +{"problem":"a shopkeeper sells 500 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ?","rationale":"sp per metre = 18000 \/ 500 = rs . 36 loss per metre = rs . 5 cp per metre = 36 + 5 = rs . 41 . answer : e","correct":"e","options":{"a":"12 ","b":"27 ","c":"29 ","d":"50","e":"41"},"options_float":{"a":12.0,"b":27.0,"c":29.0,"d":50.0,"e":41.0},"annotated_formula":"add(divide(18000, 500), 5)","linear_formula":"divide(n1,n0)|add(n2,#0)|","chain":"18_000 \/ 500<\/gadget>\n36<\/output>\n36 + 5<\/gadget>\n41<\/output>\n41<\/result>","index":107} +{"problem":"the difference of two numbers is 1565 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ?","rationale":"\"let the smaller number be x . then larger number = ( x + 1565 ) . x + 1565 = 6 x + 15 5 x = 1550 x = 310 smaller number = 310 . e )\"","correct":"e","options":{"a":"270 ","b":"280 ","c":"290 ","d":"300","e":"310"},"options_float":{"a":270.0,"b":280.0,"c":290.0,"d":300.0,"e":310.0},"annotated_formula":"divide(add(1565, 15), subtract(6, const_1))","linear_formula":"add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|","chain":"1_565 + 15<\/gadget>\n1_580<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_580 \/ 5<\/gadget>\n316<\/output>\n316<\/result>","index":108} +{"problem":"how many even number in the range between 15 to 100 inclusive are not divisible by 3","rationale":"\"we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 16,18 . . . 100 using ap formula , we can say 100 = 16 + ( n - 1 ) * 2 or n = 43 . for 6 , 18 , . . . 96 using ap formula , we can say 96 = 18 + ( n - 1 ) * 6 or n = 14 . hence , only divisible by 2 but not 3 = 43 - 14 = 29 . hence , answer a\"","correct":"a","options":{"a":"29 ","b":"30 ","c":"31 ","d":"33","e":"46"},"options_float":{"a":29.0,"b":30.0,"c":31.0,"d":33.0,"e":46.0},"annotated_formula":"subtract(divide(subtract(subtract(100, 15), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2))","linear_formula":"multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)|","chain":"100 - 15<\/gadget>\n85<\/output>\n85 - 2<\/gadget>\n83<\/output>\n83 \/ 2<\/gadget>\n83\/2 = around 41.5<\/output>\n100 - 2<\/gadget>\n98<\/output>\n3 * 4<\/gadget>\n12<\/output>\n98 - 12<\/gadget>\n86<\/output>\n86 - 3<\/gadget>\n83<\/output>\n83 - 3<\/gadget>\n80<\/output>\n80 \/ 3<\/gadget>\n80\/3 = around 26.666667<\/output>\n(80\/3) \/ 2<\/gadget>\n40\/3 = around 13.333333<\/output>\n(83\/2) - (40\/3)<\/gadget>\n169\/6 = around 28.166667<\/output>\n169\/6 = around 28.166667<\/result>","index":109} +{"problem":"the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 7 per metre approximately","rationale":"\"explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 \/ 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 \/ 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 7 ) = rs . 10400 . answer : option b\"","correct":"b","options":{"a":"4457 ","b":"10400 ","c":"4235 ","d":"4547","e":"4675"},"options_float":{"a":4457.0,"b":10400.0,"c":4235.0,"d":4547.0,"e":4675.0},"annotated_formula":"multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 7)","linear_formula":"divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)|","chain":"17.56 \/ pi<\/gadget>\n17.56\/pi = around 5.589522<\/output>\n(17.56\/pi) ** (1\/2)<\/gadget>\n4.19046536795139\/sqrt(pi) = around 2.364217<\/output>\n(4.19046536795139\/sqrt(pi)) * 100<\/gadget>\n419.046536795139\/sqrt(pi) = around 236.421691<\/output>\n2 * pi * (419.046536795139\/sqrt(pi))<\/gadget>\n838.093073590278*sqrt(pi) = around 1_485.481296<\/output>\n(838.093073590278*sqrt(pi)) * 7<\/gadget>\n5866.65151513195*sqrt(pi) = around 10_398.36907<\/output>\n5866.65151513195*sqrt(pi) = around 10_398.36907<\/result>","index":110} +{"problem":"if 5400 mn = k ^ 4 , where m , n , and k are positive integers , what is the least possible value of 2 m + n ?","rationale":"k ^ 4 = 5400 ∗ mn k ^ 4 = 3 ^ 3.2 ^ 3.5 ^ 2 . mn in order to make rhs a perfect power of 4 , we require it to be multiplied by 3 , 2 & 5 ^ 2 mn = 3 . 2.5 ^ 2 mn = 150 = 10 * 15 ( least possible ) answer = 2 * 10 + 15 = 35 answer : e","correct":"e","options":{"a":"11 ","b":"18 ","c":"20 ","d":"25","e":"35"},"options_float":{"a":11.0,"b":18.0,"c":20.0,"d":25.0,"e":35.0},"annotated_formula":"multiply(4, power(const_3, const_2))","linear_formula":"power(const_3,const_2)|multiply(n1,#0)","chain":"3 ** 2<\/gadget>\n9<\/output>\n4 * 9<\/gadget>\n36<\/output>\n36<\/result>","index":112} +{"problem":"tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 7 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ?","rationale":"\"( 1 mph + 7 mph ) \/ 2 = 4 mph correct option is : d\"","correct":"d","options":{"a":"8 mph ","b":"5.25 mph ","c":"3.5 mph ","d":"4 mph","e":"0.5 mph"},"options_float":{"a":8.0,"b":5.25,"c":3.5,"d":4.0,"e":0.5},"annotated_formula":"divide(add(1, 7), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"1 + 7<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":113} +{"problem":"a train of 25 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.0 km long ?","rationale":"\"d = 25 * 60 + 1000 = 2500 m t = 2500 \/ 60 * 18 \/ 5 = 150 sec = 2.5 mins answer : a\"","correct":"a","options":{"a":"2.5 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":2.5,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"add(divide(multiply(add(25, const_1), 60), const_1000), 1.0)","linear_formula":"add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)|","chain":"25 + 1<\/gadget>\n26<\/output>\n26 * 60<\/gadget>\n1_560<\/output>\n1_560 \/ 1_000<\/gadget>\n39\/25 = around 1.56<\/output>\n(39\/25) + 1<\/gadget>\n64\/25 = around 2.56<\/output>\n64\/25 = around 2.56<\/result>","index":114} +{"problem":"find the principle on a certain sum of money at 9 % per annum for 2 2 \/ 5 years if the amount being rs . 1120 ?","rationale":"1120 = p [ 1 + ( 9 * 12 \/ 5 ) \/ 100 ] p = 921.05 answer : e","correct":"e","options":{"a":"1000 ","b":"912.88 ","c":"927.78 ","d":"926.82","e":"921.05"},"options_float":{"a":1000.0,"b":912.88,"c":927.78,"d":926.82,"e":921.05},"annotated_formula":"divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 9), const_100), const_1))","linear_formula":"multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)","chain":"2 * 5<\/gadget>\n10<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 \/ 5<\/gadget>\n12\/5 = around 2.4<\/output>\n(12\/5) * 9<\/gadget>\n108\/5 = around 21.6<\/output>\n(108\/5) \/ 100<\/gadget>\n27\/125 = around 0.216<\/output>\n(27\/125) + 1<\/gadget>\n152\/125 = around 1.216<\/output>\n1_120 \/ (152\/125)<\/gadget>\n17_500\/19 = around 921.052632<\/output>\n17_500\/19 = around 921.052632<\/result>","index":116} +{"problem":"a man whose bowling average is 12.4 , takes 5 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ?","rationale":"\"12.4 x + 26 \/ ( 5 + x ) = 12 12.4 x + 26 = 60 + 12 x x = 34 \/ 0.4 = 85 answer : d\"","correct":"d","options":{"a":"82 ","b":"83 ","c":"84 ","d":"85","e":"86"},"options_float":{"a":82.0,"b":83.0,"c":84.0,"d":85.0,"e":86.0},"annotated_formula":"divide(subtract(multiply(floor(12.4), 5), 26), subtract(12.4, floor(12.4)))","linear_formula":"floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2)|","chain":"floor(12.4)<\/gadget>\n12<\/output>\n12 * 5<\/gadget>\n60<\/output>\n60 - 26<\/gadget>\n34<\/output>\n12.4 - 12<\/gadget>\n0.4<\/output>\n34 \/ 0.4<\/gadget>\n85<\/output>\n85<\/result>","index":119} +{"problem":"anand finishes a work in 7 days , bittu finishes the same job in 8 days and chandu in 6 days . they take turns to finish the work . anand on the first day , bittu on the second and chandu on the third day and then anand again and so on . on which day will the work get over ?","rationale":"they completed 146 \/ 168 work in 6 days working one by one on eacg day . on 7 th day , 22 \/ 168 work is left which will be completed by anand on 7 th day . answer : d","correct":"d","options":{"a":"3 rd ","b":"6 th ","c":"9 th ","d":"7 th","e":"8 th"},"options_float":{"a":3.0,"b":6.0,"c":9.0,"d":7.0,"e":8.0},"annotated_formula":"add(floor(multiply(inverse(add(add(divide(const_1, 7), divide(const_1, 8)), divide(const_1, 6))), const_3)), const_1)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)|multiply(#5,const_3)|floor(#6)|add(#7,const_1)","chain":"1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/7) + (1\/8)<\/gadget>\n15\/56 = around 0.267857<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(15\/56) + (1\/6)<\/gadget>\n73\/168 = around 0.434524<\/output>\n1 \/ (73\/168)<\/gadget>\n168\/73 = around 2.30137<\/output>\n(168\/73) * 3<\/gadget>\n504\/73 = around 6.90411<\/output>\nfloor(504\/73)<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7<\/result>","index":120} +{"problem":"if the product of two numbers is 45276 and their h . c . f . is 22 , find their l . c . m .","rationale":"explanation : hcf * lcm = 45276 , because we know product of two numbers = product of hcf and lcm lcm = 45276 \/ 22 = 2058 option d","correct":"d","options":{"a":"2574 ","b":"2500 ","c":"2100 ","d":"2058","e":"none of these"},"options_float":{"a":2574.0,"b":2500.0,"c":2100.0,"d":2058.0,"e":null},"annotated_formula":"divide(45276, 22)","linear_formula":"divide(n0,n1)","chain":"45_276 \/ 22<\/gadget>\n2_058<\/output>\n2_058<\/result>","index":121} +{"problem":"okeydokey paid 5 apples at the apple investment booth for an unopened box of earthworms . artichokey paid 7 apples towards the same lot . the booth pays off earthworms proportional to the apples invested . the box turned out to contain 60 earthworms . how many of them should okeydokey receive ?","rationale":"i 1 = 5 i 2 = 7 i 1 share 5 parts & i 2 share 7 parts total 12 parts - - - - - > 60 - - - - > 1 part - - - - - - - > 5 i 1 share = 5 parts - - - - - > 25 a","correct":"a","options":{"a":"25 ","b":"27 ","c":"28 ","d":"29","e":"30"},"options_float":{"a":25.0,"b":27.0,"c":28.0,"d":29.0,"e":30.0},"annotated_formula":"divide(multiply(60, 5), add(5, 7))","linear_formula":"add(n0,n1)|multiply(n0,n2)|divide(#1,#0)","chain":"60 * 5<\/gadget>\n300<\/output>\n5 + 7<\/gadget>\n12<\/output>\n300 \/ 12<\/gadget>\n25<\/output>\n25<\/result>","index":123} +{"problem":"a glass was filled with 10 ounces of water , and 0.01 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ?","rationale":"\"in 20 days 20 * 0.01 = 0.2 ounces of water evaporated , which is 0.2 \/ 10 ∗ 100 = 2 of the original amount of water . answer : d .\"","correct":"d","options":{"a":"0.002 % ","b":"0.02 % ","c":"0.2 % ","d":"2 %","e":"20 %"},"options_float":{"a":0.002,"b":0.02,"c":0.2,"d":2.0,"e":20.0},"annotated_formula":"multiply(divide(multiply(0.01, 20), 10), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"0.01 * 20<\/gadget>\n0.2<\/output>\n0.2 \/ 10<\/gadget>\n0.02<\/output>\n0.02 * 100<\/gadget>\n2<\/output>\n2<\/result>","index":124} +{"problem":"a 64 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ?","rationale":"amount of salt = 6.4 assume x gallons of water are added . 6.4 \/ 64 + x = 8 \/ 100 640 = 8 x + 512 8 x = 128 x = 16 correct option : e","correct":"e","options":{"a":"8 ","b":"12 ","c":"13 ","d":"14","e":"16"},"options_float":{"a":8.0,"b":12.0,"c":13.0,"d":14.0,"e":16.0},"annotated_formula":"divide(multiply(64, subtract(10, 8)), 8)","linear_formula":"subtract(n1,n2)|multiply(n0,#0)|divide(#1,n2)","chain":"10 - 8<\/gadget>\n2<\/output>\n64 * 2<\/gadget>\n128<\/output>\n128 \/ 8<\/gadget>\n16<\/output>\n16<\/result>","index":126} +{"problem":"there are 200 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ?","rationale":"\"there are 198 managers and 2 others . the 2 others would compose 2 % of the total number of people if there were 100 people in the room . thus 100 managers must leave . the answer is b .\"","correct":"b","options":{"a":"150 ","b":"100 ","c":"99 ","d":"50","e":"25"},"options_float":{"a":150.0,"b":100.0,"c":99.0,"d":50.0,"e":25.0},"annotated_formula":"divide(subtract(multiply(200, divide(99, const_100)), multiply(200, divide(98, const_100))), subtract(const_1, divide(98, const_100)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)|","chain":"99 \/ 100<\/gadget>\n99\/100 = around 0.99<\/output>\n200 * (99\/100)<\/gadget>\n198<\/output>\n98 \/ 100<\/gadget>\n49\/50 = around 0.98<\/output>\n200 * (49\/50)<\/gadget>\n196<\/output>\n198 - 196<\/gadget>\n2<\/output>\n1 - (49\/50)<\/gadget>\n1\/50 = around 0.02<\/output>\n2 \/ (1\/50)<\/gadget>\n100<\/output>\n100<\/result>","index":131} +{"problem":"if a and b are the roots of the equation x 2 - 9 x + 9 = 0 , then the value of a 2 + b 2 is :","rationale":"\"sol . ( b ) the sum of roots = a + b = 9 product of roots = ab = 9 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 81 - 18 = 63 answer a\"","correct":"a","options":{"a":"63 ","b":"24 ","c":"17 ","d":"6","e":"5"},"options_float":{"a":63.0,"b":24.0,"c":17.0,"d":6.0,"e":5.0},"annotated_formula":"add(power(divide(subtract(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2), power(divide(add(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2))","linear_formula":"multiply(n1,const_4)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|add(n1,#3)|subtract(n1,#3)|divide(#5,n0)|divide(#4,n0)|power(#6,n0)|power(#7,n0)|add(#8,#9)|","chain":"9 ** 2<\/gadget>\n81<\/output>\n4 * 9<\/gadget>\n36<\/output>\n81 - 36<\/gadget>\n45<\/output>\n45 ** (1\/2)<\/gadget>\n3*sqrt(5) = around 6.708204<\/output>\n9 - (3*sqrt(5))<\/gadget>\n9 - 3*sqrt(5) = around 2.291796<\/output>\n(9 - 3*sqrt(5)) \/ 2<\/gadget>\n9\/2 - 3*sqrt(5)\/2 = around 1.145898<\/output>\n(9\/2 - 3*sqrt(5)\/2) ** 2<\/gadget>\n(9\/2 - 3*sqrt(5)\/2)**2 = around 1.313082<\/output>\n9 + (3*sqrt(5))<\/gadget>\n3*sqrt(5) + 9 = around 15.708204<\/output>\n(3*sqrt(5) + 9) \/ 2<\/gadget>\n3*sqrt(5)\/2 + 9\/2 = around 7.854102<\/output>\n(3*sqrt(5)\/2 + 9\/2) ** 2<\/gadget>\n(3*sqrt(5)\/2 + 9\/2)**2 = around 61.686918<\/output>\n((9\/2 - 3*sqrt(5)\/2)**2) + ((3*sqrt(5)\/2 + 9\/2)**2)<\/gadget>\n(9\/2 - 3*sqrt(5)\/2)**2 + (3*sqrt(5)\/2 + 9\/2)**2 = around 63<\/output>\n(9\/2 - 3*sqrt(5)\/2)**2 + (3*sqrt(5)\/2 + 9\/2)**2 = around 63<\/result>","index":133} +{"problem":"a light has a rating of 80 watts , it is replaced with a new light that has 25 % higher wattage . how many watts does the new light have ?","rationale":"\"final number = initial number + 25 % ( original number ) = 80 + 25 % ( 80 ) = 80 + 20 = 100 . answer e\"","correct":"e","options":{"a":"105 ","b":"95 ","c":"80 ","d":"60","e":"100"},"options_float":{"a":105.0,"b":95.0,"c":80.0,"d":60.0,"e":100.0},"annotated_formula":"multiply(80, add(const_1, divide(25, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n80 * (5\/4)<\/gadget>\n100<\/output>\n100<\/result>","index":134} +{"problem":"in the class of 40 students , 30 speak hindi and 20 speak english . what is the lowest possible number of students who speak both the languages ?","rationale":"let the student who speaks hindi - x let the student who speaks english - y as ( xuy ) - ( xny ) = total 30 + 20 - ( xny ) = 40 = 10 a )","correct":"a","options":{"a":"10 ","b":"15 ","c":"20 ","d":"25","e":"30"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"subtract(add(30, 20), 40)","linear_formula":"add(n1,n2)|subtract(#0,n0)","chain":"30 + 20<\/gadget>\n50<\/output>\n50 - 40<\/gadget>\n10<\/output>\n10<\/result>","index":135} +{"problem":"if k is an integer and 0.00010101 x 10 ^ k is greater than 100 , what is the least possible value of k ?","rationale":"0.00010101 * 10 ^ k > 100 we need to move the decimal point to the right 6 places to get 101.01 this is equivalent to multiplying by 10 ^ 6 . the answer is d .","correct":"d","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(log(divide(100, 0.00010101)), log(const_10))","linear_formula":"divide(n2,n0)|log(const_10)|log(#0)|divide(#2,#1)","chain":"100 \/ 0.000101<\/gadget>\n990_099.009901<\/output>\nlog(990_099.009901)<\/gadget>\n13.80556<\/output>\nlog(10)<\/gadget>\nlog(10) = around 2.302585<\/output>\n13.80556 \/ log(10)<\/gadget>\n13.80556\/log(10) = around 5.995679<\/output>\n13.80556\/log(10) = around 5.995679<\/result>","index":137} +{"problem":"how many numbers from 10 to 50 are exactly divisible by 3 .","rationale":"\"12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45,48 . 13 numbers . 10 \/ 3 = 3 and 50 \/ 3 = 16 = = > 16 - 3 = 13 . therefore 13 digits . a )\"","correct":"a","options":{"a":"13 ","b":"14 ","c":"16 ","d":"17","e":"19"},"options_float":{"a":13.0,"b":14.0,"c":16.0,"d":17.0,"e":19.0},"annotated_formula":"add(divide(subtract(multiply(floor(divide(50, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1)","linear_formula":"divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)|","chain":"50 \/ 3<\/gadget>\n50\/3 = around 16.666667<\/output>\nfloor(50\/3)<\/gadget>\n16<\/output>\n16 * 3<\/gadget>\n48<\/output>\n10 \/ 3<\/gadget>\n10\/3 = around 3.333333<\/output>\nfloor(10\/3)<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n48 - 12<\/gadget>\n36<\/output>\n36 \/ 3<\/gadget>\n12<\/output>\n12 + 1<\/gadget>\n13<\/output>\n13<\/result>","index":138} +{"problem":"the ratio of laxmi ' s age to the age of her mother is 3 : 11 . the difference of their ages is 24 years . the ratio of their ages after 3 years will be ?","rationale":"suppose age of laxmi is x years . the difference of their ages \\ inline \\ therefore ratio of their ages after 3 years = = = 12 : 36 = 1 : 3 answer : b","correct":"b","options":{"a":"1 : 9 ","b":"1 : 3 ","c":"1 : 2 ","d":"1 : 1","e":"3 : 3"},"options_float":{"a":0.1111111111,"b":0.3333333333,"c":0.5,"d":1.0,"e":1.0},"annotated_formula":"divide(add(multiply(3, divide(24, subtract(11, 3))), 3), add(multiply(11, divide(24, subtract(11, 3))), 3))","linear_formula":"subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(n0,#2)|add(n0,#3)|divide(#4,#5)","chain":"11 - 3<\/gadget>\n8<\/output>\n24 \/ 8<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 + 3<\/gadget>\n12<\/output>\n11 * 3<\/gadget>\n33<\/output>\n33 + 3<\/gadget>\n36<\/output>\n12 \/ 36<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":139} +{"problem":"solution y is 30 percent liquid x and 70 percent water . if 2 kilograms of water evaporate from 6 kilograms of solution y and 2 kilograms of solution y are added to the remaining 4 kilograms of liquid , what percent of this new solution is liquid x ?","rationale":"\"in 8 kilograms of solution y there are 0.3 * 6 = 1.8 kilograms of solution x ; after 2 kilograms of water are replaced by 2 kilograms of solution y , to the existing 1.8 kilograms of solution x , 0.3 * 2 = 0.6 kilograms of solution x are added , so in the new solution of 6 kilograms there are 1.8 + 0.6 = 2.4 kilograms of solution x , which is 2.4 \/ 6 * 100 = 40 % of this new solution . answer : d .\"","correct":"d","options":{"a":"33 % ","b":"35 1 \/ 2 % ","c":"37 % ","d":"40 %","e":"42 %"},"options_float":{"a":33.0,"b":35.0,"c":37.0,"d":40.0,"e":42.0},"annotated_formula":"multiply(const_100, divide(add(multiply(6, divide(30, const_100)), multiply(2, divide(30, const_100))), add(add(multiply(6, divide(30, const_100)), multiply(2, divide(30, const_100))), add(subtract(multiply(6, divide(70, const_100)), 2), multiply(2, divide(70, const_100))))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(n3,#0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n3,#1)|add(#2,#3)|subtract(#5,n2)|add(#4,#7)|add(#6,#8)|divide(#6,#9)|multiply(#10,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n6 * (3\/10)<\/gadget>\n9\/5 = around 1.8<\/output>\n2 * (3\/10)<\/gadget>\n3\/5 = around 0.6<\/output>\n(9\/5) + (3\/5)<\/gadget>\n12\/5 = around 2.4<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n6 * (7\/10)<\/gadget>\n21\/5 = around 4.2<\/output>\n(21\/5) - 2<\/gadget>\n11\/5 = around 2.2<\/output>\n2 * (7\/10)<\/gadget>\n7\/5 = around 1.4<\/output>\n(11\/5) + (7\/5)<\/gadget>\n18\/5 = around 3.6<\/output>\n(12\/5) + (18\/5)<\/gadget>\n6<\/output>\n(12\/5) \/ 6<\/gadget>\n2\/5 = around 0.4<\/output>\n100 * (2\/5)<\/gadget>\n40<\/output>\n40<\/result>","index":142} +{"problem":"10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 3312 , how much was it in the beginning ?","rationale":"\"x * ( 90 \/ 100 ) * ( 80 \/ 100 ) = 3312 x = 4600 answer : e\"","correct":"e","options":{"a":"3800 ","b":"4200 ","c":"4400 ","d":"4500","e":"4600"},"options_float":{"a":3800.0,"b":4200.0,"c":4400.0,"d":4500.0,"e":4600.0},"annotated_formula":"floor(divide(3312, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))))","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(9\/10) * (4\/5)<\/gadget>\n18\/25 = around 0.72<\/output>\n3_312 \/ (18\/25)<\/gadget>\n4_600<\/output>\nfloor(4_600)<\/gadget>\n4_600<\/output>\n4_600<\/result>","index":143} +{"problem":"a is twice as good a workman as b and they took 11 days together to do the work b alone can do it in .","rationale":"\"wc = 2 : 1 2 x + x = 1 \/ 11 x = 1 \/ 33 = > 33 days answer : e\"","correct":"e","options":{"a":"17 days ","b":"12 days ","c":"29 days ","d":"25 days","e":"33 days"},"options_float":{"a":17.0,"b":12.0,"c":29.0,"d":25.0,"e":33.0},"annotated_formula":"multiply(divide(multiply(11, add(const_2, const_1)), const_2), const_2)","linear_formula":"add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)|","chain":"2 + 1<\/gadget>\n3<\/output>\n11 * 3<\/gadget>\n33<\/output>\n33 \/ 2<\/gadget>\n33\/2 = around 16.5<\/output>\n(33\/2) * 2<\/gadget>\n33<\/output>\n33<\/result>","index":144} +{"problem":"if the price of 357 apples is rs . 1517.25 , what will be the approximate price of 49 dozens of such apples ?","rationale":"explanation : price of 357 apples = rs . 1517.25 price of 1 apple = rs . 1517.25 \/ 357 price of 49 dozens apples = rs ( 49 × 12 × 1517.25 \/ 357 ) ≈ rs . 2500 answer : option a","correct":"a","options":{"a":"rs . 2500 ","b":"rs . 2300 ","c":"rs . 2200 ","d":"rs . 1400","e":"rs . 1600"},"options_float":{"a":2500.0,"b":2300.0,"c":2200.0,"d":1400.0,"e":1600.0},"annotated_formula":"multiply(multiply(49, const_12), divide(1517.25, 357))","linear_formula":"divide(n1,n0)|multiply(n2,const_12)|multiply(#0,#1)","chain":"49 * 12<\/gadget>\n588<\/output>\n1_517.25 \/ 357<\/gadget>\n4.25<\/output>\n588 * 4.25<\/gadget>\n2_499<\/output>\n2_499<\/result>","index":147} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 3.8 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?","rationale":"\"10 overs - run rate = 3.8 runs scored in first 10 overs = 38 remaining overs 40 total runs to be scored = 282 38 runs already scored 282 - 38 = 244 244 runs to be scored in 40 overs let required runrate be x 40 * x = 244 x = 244 \/ 40 x = 6.1 the required runrate is 6.1 answer : d\"","correct":"d","options":{"a":"6.25 ","b":"6.5 ","c":"6.75 ","d":"6.1","e":"8"},"options_float":{"a":6.25,"b":6.5,"c":6.75,"d":6.1,"e":8.0},"annotated_formula":"divide(subtract(282, multiply(10, 3.8)), 40)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 3.8<\/gadget>\n38<\/output>\n282 - 38<\/gadget>\n244<\/output>\n244 \/ 40<\/gadget>\n61\/10 = around 6.1<\/output>\n61\/10 = around 6.1<\/result>","index":148} +{"problem":"a person buys an article at rs . 480 . at what price should he sell the article so as to make a profit of 24 % ?","rationale":"\"cost price = rs . 480 profit = 24 % of 480 = rs . 115 selling price = cost price + profit = 480 + 115 = 595 answer : a\"","correct":"a","options":{"a":"595 ","b":"882 ","c":"772 ","d":"662","e":"521"},"options_float":{"a":595.0,"b":882.0,"c":772.0,"d":662.0,"e":521.0},"annotated_formula":"add(480, multiply(480, divide(24, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n480 * (6\/25)<\/gadget>\n576\/5 = around 115.2<\/output>\n480 + (576\/5)<\/gadget>\n2_976\/5 = around 595.2<\/output>\n2_976\/5 = around 595.2<\/result>","index":149} +{"problem":"what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 110 ?","rationale":"surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) \/ 2 = ( second element + last but one element ) \/ 2 = ( third element + third last element ) \/ 2 etc . etc . so mean = median = 110 \/ 2 = 55 answer is c","correct":"c","options":{"a":"10 ","b":"25 ","c":"55 ","d":"75","e":"100"},"options_float":{"a":10.0,"b":25.0,"c":55.0,"d":75.0,"e":100.0},"annotated_formula":"divide(110, const_2)","linear_formula":"divide(n0,const_2)","chain":"110 \/ 2<\/gadget>\n55<\/output>\n55<\/result>","index":150} +{"problem":"roses can be purchased individually for $ 6.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ?","rationale":"\"buy as many $ 50 deals as possible . we can by 650 \/ 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 \/ 6.3 = ~ 4 roses for that amount . total = 312 + 4 = 316 . answer : c .\"","correct":"c","options":{"a":"156 ","b":"162 ","c":"316 ","d":"324","e":"325"},"options_float":{"a":156.0,"b":162.0,"c":316.0,"d":324.0,"e":325.0},"annotated_formula":"add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 6.30)))","linear_formula":"divide(n3,n2)|multiply(const_12,const_2)|floor(#0)|multiply(n2,#2)|multiply(#2,#1)|subtract(n3,#3)|divide(#5,n0)|floor(#6)|add(#7,#4)|","chain":"680 \/ 50<\/gadget>\n68\/5 = around 13.6<\/output>\nfloor(68\/5)<\/gadget>\n13<\/output>\n12 * 2<\/gadget>\n24<\/output>\n13 * 24<\/gadget>\n312<\/output>\n13 * 50<\/gadget>\n650<\/output>\n680 - 650<\/gadget>\n30<\/output>\n30 \/ 6.3<\/gadget>\n4.761905<\/output>\nfloor(4.761905)<\/gadget>\n4<\/output>\n312 + 4<\/gadget>\n316<\/output>\n316<\/result>","index":151} +{"problem":"the sub - duplicate ratio of 25 : 49 is","rationale":"\"root ( 25 ) : root ( 49 ) = 5 : 7 answer : b\"","correct":"b","options":{"a":"4 : 3 ","b":"5 : 7 ","c":"6 : 4 ","d":"1 : 4","e":"2 : 3"},"options_float":{"a":1.3333333333,"b":0.7142857143,"c":1.5,"d":0.25,"e":0.6666666667},"annotated_formula":"divide(sqrt(25), sqrt(49))","linear_formula":"sqrt(n0)|sqrt(n1)|divide(#0,#1)|","chain":"25 ** (1\/2)<\/gadget>\n5<\/output>\n49 ** (1\/2)<\/gadget>\n7<\/output>\n5 \/ 7<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":152} +{"problem":"a man walking at a rate of 10 km \/ hr crosses a bridge in 24 minutes . the length of the bridge is ?","rationale":"\"speed = 10 * 5 \/ 18 = 50 \/ 18 m \/ sec distance covered in 24 minutes = 50 \/ 18 * 24 * 60 = 4000 m answer is d\"","correct":"d","options":{"a":"1521 ","b":"1492 ","c":"1667 ","d":"4000","e":"1112"},"options_float":{"a":1521.0,"b":1492.0,"c":1667.0,"d":4000.0,"e":1112.0},"annotated_formula":"multiply(divide(multiply(10, const_1000), const_60), 24)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 60<\/gadget>\n500\/3 = around 166.666667<\/output>\n(500\/3) * 24<\/gadget>\n4_000<\/output>\n4_000<\/result>","index":153} +{"problem":"find the ratio in which rice at rs . 7.10 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg ?","rationale":"\"solution required ratio = 60 : 80 = 3 : 4 answer c\"","correct":"c","options":{"a":"1 : 3 ","b":"2 : 3 ","c":"3 : 4 ","d":"4 : 5","e":"none of these"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":0.75,"d":0.8,"e":null},"annotated_formula":"divide(subtract(6.30, 5.70), subtract(7.10, 6.30))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"6.3 - 5.7<\/gadget>\n0.6<\/output>\n7.1 - 6.3<\/gadget>\n0.8<\/output>\n0.6 \/ 0.8<\/gadget>\n0.75<\/output>\n0.75<\/result>","index":154} +{"problem":"a class consists of 15 biology students and 10 chemistry students . if you pick two students at the same time , what ' s the probability that one is maths and one is chemistry ?","rationale":"p ( 1 st b , 2 nd c ) = 15 \/ 25 * 10 \/ 24 = 150 \/ 600 = 1 \/ 4 ; p ( 1 st c , 2 nd b ) = 10 \/ 25 * 15 \/ 24 = 150 \/ 600 = 1 \/ 4 . p = 1 \/ 4 + 1 \/ 4 = 1 \/ 2 . answer : c .","correct":"c","options":{"a":"2 \/ 7 ","b":"5 \/ 7 ","c":"1 \/ 2 ","d":"3 \/ 7","e":"1 \/ 2"},"options_float":{"a":0.2857142857,"b":0.7142857143,"c":0.5,"d":0.4285714286,"e":0.5},"annotated_formula":"add(multiply(divide(15, add(15, 10)), divide(10, subtract(add(15, 10), const_1))), multiply(divide(15, subtract(add(15, 10), const_1)), divide(10, add(15, 10))))","linear_formula":"add(n0,n1)|divide(n0,#0)|divide(n1,#0)|subtract(#0,const_1)|divide(n1,#3)|divide(n0,#3)|multiply(#1,#4)|multiply(#5,#2)|add(#6,#7)","chain":"15 + 10<\/gadget>\n25<\/output>\n15 \/ 25<\/gadget>\n3\/5 = around 0.6<\/output>\n25 - 1<\/gadget>\n24<\/output>\n10 \/ 24<\/gadget>\n5\/12 = around 0.416667<\/output>\n(3\/5) * (5\/12)<\/gadget>\n1\/4 = around 0.25<\/output>\n15 \/ 24<\/gadget>\n5\/8 = around 0.625<\/output>\n10 \/ 25<\/gadget>\n2\/5 = around 0.4<\/output>\n(5\/8) * (2\/5)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":156} +{"problem":"an aeroplane covers a certain distance of 420 kmph in 6 hours . to cover the same distance in 4 2 \/ 3 hours , it must travel at a speed of","rationale":"\"speed of aeroplane = 420 kmph distance travelled in 6 hours = 420 * 6 = 2520 km speed of aeroplane to acver 2520 km in 14 \/ 3 = 2520 * 3 \/ 14 = 540 km answer b .\"","correct":"b","options":{"a":"440 ","b":"540 ","c":"640 ","d":"740","e":"250"},"options_float":{"a":440.0,"b":540.0,"c":640.0,"d":740.0,"e":250.0},"annotated_formula":"divide(multiply(420, 6), divide(add(multiply(3, 4), 2), 3))","linear_formula":"multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3)|","chain":"420 * 6<\/gadget>\n2_520<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 + 2<\/gadget>\n14<\/output>\n14 \/ 3<\/gadget>\n14\/3 = around 4.666667<\/output>\n2_520 \/ (14\/3)<\/gadget>\n540<\/output>\n540<\/result>","index":157} +{"problem":"the average mark of the students of a class in a particular exam is 80 . if 8 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam .","rationale":"\"let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 8 ) students = 90 ( x - 8 ) 80 x - ( 8 * 20 ) = 90 ( x - 8 ) 560 = 10 x = > x = 56 answer : e\"","correct":"e","options":{"a":"15 ","b":"25 ","c":"35 ","d":"45","e":"56"},"options_float":{"a":15.0,"b":25.0,"c":35.0,"d":45.0,"e":56.0},"annotated_formula":"divide(subtract(multiply(90, 8), multiply(8, 20)), subtract(90, 80))","linear_formula":"multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|","chain":"90 * 8<\/gadget>\n720<\/output>\n8 * 20<\/gadget>\n160<\/output>\n720 - 160<\/gadget>\n560<\/output>\n90 - 80<\/gadget>\n10<\/output>\n560 \/ 10<\/gadget>\n56<\/output>\n56<\/result>","index":159} +{"problem":"what is the smallest integer k for which 64 ^ k > 4 ^ 17 ?","rationale":"\"64 ^ k > 4 ^ 17 4 ^ ( 3 k ) > 4 ^ 17 3 k > 17 k = 6 the answer is c .\"","correct":"c","options":{"a":"4 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"add(floor(divide(17, const_3)), const_1)","linear_formula":"divide(n2,const_3)|floor(#0)|add(#1,const_1)|","chain":"17 \/ 3<\/gadget>\n17\/3 = around 5.666667<\/output>\nfloor(17\/3)<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6<\/result>","index":160} +{"problem":"one side of a rectangular field is 14 m and one of its diagonal is 17 m . find the area of the field .","rationale":"\"solution other side = √ ( 17 ) 2 - ( 14 ) 2 = √ 289 - 196 = √ 93 = 9.6 m . ∴ area = ( 14 x 9.6 ) m 2 = 134.4 m 2 . answer a\"","correct":"a","options":{"a":"134.4 ","b":"120 ","c":"150 ","d":"180","e":"none"},"options_float":{"a":134.4,"b":120.0,"c":150.0,"d":180.0,"e":null},"annotated_formula":"rectangle_area(14, sqrt(subtract(power(17, const_2), power(14, const_2))))","linear_formula":"power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)|","chain":"17 ** 2<\/gadget>\n289<\/output>\n14 ** 2<\/gadget>\n196<\/output>\n289 - 196<\/gadget>\n93<\/output>\n93 ** (1\/2)<\/gadget>\nsqrt(93) = around 9.643651<\/output>\n14 * (sqrt(93))<\/gadget>\n14*sqrt(93) = around 135.011111<\/output>\n14*sqrt(93) = around 135.011111<\/result>","index":165} +{"problem":"a hiker walking at a constant rate of 4 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 24 kilometers per hour . the cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate . how many minutes must the cyclist wait until the hiker catches up ?","rationale":"\"in 5 minutes , the cyclist travels a distance of ( 5 \/ 60 ) * 24 = 2 km . the time it takes the hiker to complete this distance is 2 \/ 4 hours = 30 minutes the cyclist needs to wait 30 - 5 = 25 minutes the answer is d .\"","correct":"d","options":{"a":"10 ","b":"15 ","c":"20 ","d":"25","e":"30"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"multiply(divide(subtract(multiply(24, divide(5, const_60)), multiply(4, divide(5, const_60))), 4), const_60)","linear_formula":"divide(n2,const_60)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|divide(#3,n0)|multiply(#4,const_60)|","chain":"5 \/ 60<\/gadget>\n1\/12 = around 0.083333<\/output>\n24 * (1\/12)<\/gadget>\n2<\/output>\n4 * (1\/12)<\/gadget>\n1\/3 = around 0.333333<\/output>\n2 - (1\/3)<\/gadget>\n5\/3 = around 1.666667<\/output>\n(5\/3) \/ 4<\/gadget>\n5\/12 = around 0.416667<\/output>\n(5\/12) * 60<\/gadget>\n25<\/output>\n25<\/result>","index":166} +{"problem":"the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.1 kg . the weight of the new student is","rationale":"\"exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.1 weight of the new student = ( 30 * 27.1 – 29 * 28 ) = 813 - 812 = 1 answer : b\"","correct":"b","options":{"a":"22 kg ","b":"1 kg ","c":"22.4 kg ","d":"21 kg","e":"none of these"},"options_float":{"a":22.0,"b":1.0,"c":22.4,"d":21.0,"e":null},"annotated_formula":"subtract(multiply(add(29, const_1), 27.1), multiply(29, 28))","linear_formula":"add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|","chain":"29 + 1<\/gadget>\n30<\/output>\n30 * 27.1<\/gadget>\n813<\/output>\n29 * 28<\/gadget>\n812<\/output>\n813 - 812<\/gadget>\n1<\/output>\n1<\/result>","index":167} +{"problem":"what will be in unit ' s place digit of 3 to the power 34 .","rationale":"as the unit place digit of 3 to the power of 1 is 3 , to the power of 2 is 9 . to the power of 3 is 7 , to the power of 4 is 1 and it ' ll be repeated again in the same order . . . . so the unit place digit of 3 to the power of 34 is 9 . . answer : e","correct":"e","options":{"a":"1 ","b":"6 ","c":"3 ","d":"0","e":"9"},"options_float":{"a":1.0,"b":6.0,"c":3.0,"d":0.0,"e":9.0},"annotated_formula":"power(3, const_2)","linear_formula":"power(n0,const_2)","chain":"3 ** 2<\/gadget>\n9<\/output>\n9<\/result>","index":168} +{"problem":"a group of 8 friends want to play doubles tennis . how many different ways can the group be divided into 4 teams of 2 people ?","rationale":"c 28 ∗ c 26 ∗ c 24 ∗ c 22 \/ 4 ! = 105 , we are dividing by 4 ! ( factorial of the # of teams ) as the order of the teams does not matter . if 8 people are - 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , then ( 12 ) ( 34 ) ( 56 ) ( 78 ) would be the same 4 teams as ( 56 ) ( 78 ) ( 12 ) ( 34 ) , as we do n ' t have team # 1 , team # 2 , . . . you can think about this in another way . for the first person we can pick a pair in 7 ways ; for the second one in 5 ways ( as two are already chosen ) ; for the third one in 3 ways ( as 4 people are already chosen ) ; for the fourth one there is only one left . so we have 7 * 5 * 3 * 1 = 105 answer : e","correct":"e","options":{"a":"420 ","b":"2520 ","c":"168 ","d":"90","e":"105"},"options_float":{"a":420.0,"b":2520.0,"c":168.0,"d":90.0,"e":105.0},"annotated_formula":"subtract(subtract(multiply(multiply(multiply(8, 4), 2), 2), const_10), const_10)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|multiply(n2,#1)|subtract(#2,const_10)|subtract(#3,const_10)","chain":"8 * 4<\/gadget>\n32<\/output>\n32 * 2<\/gadget>\n64<\/output>\n64 * 2<\/gadget>\n128<\/output>\n128 - 10<\/gadget>\n118<\/output>\n118 - 10<\/gadget>\n108<\/output>\n108<\/result>","index":169} +{"problem":"a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 10 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ?","rationale":"\"wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 10 employee paid = . 9 * 240 = 216 $ answer b\"","correct":"b","options":{"a":"$ 198 ","b":"$ 216 ","c":"$ 220 ","d":"$ 230","e":"$ 240"},"options_float":{"a":198.0,"b":216.0,"c":220.0,"d":230.0,"e":240.0},"annotated_formula":"subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 10))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4)|","chain":"200 \/ 100<\/gadget>\n2<\/output>\n2 * 20<\/gadget>\n40<\/output>\n200 + 40<\/gadget>\n240<\/output>\n240 \/ 100<\/gadget>\n12\/5 = around 2.4<\/output>\n(12\/5) * 10<\/gadget>\n24<\/output>\n240 - 24<\/gadget>\n216<\/output>\n216<\/result>","index":170} +{"problem":"an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 1,455 , what was the total number of pairs of contact lenses that she sold ?","rationale":"( x + 5 ) * 150 + x * 85 = 1455 = > x = 3 total lens = 3 + ( 3 + 5 ) = 11 answer a","correct":"a","options":{"a":"11 ","b":"13 ","c":"15 ","d":"17","e":"19"},"options_float":{"a":11.0,"b":13.0,"c":15.0,"d":17.0,"e":19.0},"annotated_formula":"add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5)","linear_formula":"add(n0,n1)|multiply(const_100,const_4)|multiply(n2,const_10)|multiply(n0,n2)|add(#1,#2)|add(n2,#4)|add(#5,const_1000)|subtract(#6,#3)|divide(#7,#0)|multiply(#8,const_2)|add(n2,#9)","chain":"4 * 100<\/gadget>\n400<\/output>\n5 * 10<\/gadget>\n50<\/output>\n400 + 50<\/gadget>\n450<\/output>\n450 + 5<\/gadget>\n455<\/output>\n455 + 1_000<\/gadget>\n1_455<\/output>\n150 * 5<\/gadget>\n750<\/output>\n1_455 - 750<\/gadget>\n705<\/output>\n150 + 85<\/gadget>\n235<\/output>\n705 \/ 235<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 + 5<\/gadget>\n11<\/output>\n11<\/result>","index":171} +{"problem":"it costs $ 5 for the first 1 \/ 3 hour to use the laundry machine at the laundromat . after the first ¼ hour it costs $ 12 per hour . if a certain customer uses the laundry machine for 2 hours and 30 minutes , how much will it cost him ?","rationale":"2 hrs 30 min = 150 min first 20 min - - - - - - > $ 5 time left is 130 min . . . now , 60 min costs $ 12 1 min costs $ 12 \/ 60 150 min costs $ 12 \/ 60 * 130 = > $ 26.00 so , total cost will be $ 26 + $ 5 = > $ 31 hence answer will be c","correct":"c","options":{"a":"$ 27 . ","b":"$ 29 . ","c":"$ 31 . ","d":"$ 33 .","e":"$ 35 ."},"options_float":{"a":27.0,"b":29.0,"c":31.0,"d":33.0,"e":35.0},"annotated_formula":"add(add(multiply(12, 2), 5), multiply(divide(12, const_60), divide(30, 3)))","linear_formula":"divide(n3,const_60)|divide(n5,n2)|multiply(n3,n4)|add(n0,#2)|multiply(#0,#1)|add(#3,#4)","chain":"12 * 2<\/gadget>\n24<\/output>\n24 + 5<\/gadget>\n29<\/output>\n12 \/ 60<\/gadget>\n1\/5 = around 0.2<\/output>\n30 \/ 3<\/gadget>\n10<\/output>\n(1\/5) * 10<\/gadget>\n2<\/output>\n29 + 2<\/gadget>\n31<\/output>\n31<\/result>","index":172} +{"problem":"a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 1881 , what is the value of n ?","rationale":"\"factorise 1881 . . 3 * 11 * 57 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 57 = 114 . . ans : c\"","correct":"c","options":{"a":"22 ","b":"38 ","c":"114 ","d":"122","e":"672"},"options_float":{"a":22.0,"b":38.0,"c":114.0,"d":122.0,"e":672.0},"annotated_formula":"multiply(2, divide(divide(1881, add(const_10, const_1)), const_3))","linear_formula":"add(const_1,const_10)|divide(n4,#0)|divide(#1,const_3)|multiply(n0,#2)|","chain":"10 + 1<\/gadget>\n11<\/output>\n1_881 \/ 11<\/gadget>\n171<\/output>\n171 \/ 3<\/gadget>\n57<\/output>\n2 * 57<\/gadget>\n114<\/output>\n114<\/result>","index":173} +{"problem":"if 8 workers can build 8 cars in 8 days , then how many days would it take 4 workers to build 4 cars ?","rationale":"8 workers can build 1 car per day on average . 1 worker can build 1 \/ 8 of a car per day . 4 workers can build 1 \/ 2 car per day . the time required to build 4 cars is 4 \/ ( 1 \/ 2 ) = 8 days the answer is c .","correct":"c","options":{"a":"2 ","b":"4 ","c":"8 ","d":"12","e":"16"},"options_float":{"a":2.0,"b":4.0,"c":8.0,"d":12.0,"e":16.0},"annotated_formula":"multiply(divide(multiply(8, 8), 8), divide(4, 4))","linear_formula":"divide(n3,n3)|multiply(n0,n0)|divide(#1,n0)|multiply(#2,#0)","chain":"8 * 8<\/gadget>\n64<\/output>\n64 \/ 8<\/gadget>\n8<\/output>\n4 \/ 4<\/gadget>\n1<\/output>\n8 * 1<\/gadget>\n8<\/output>\n8<\/result>","index":175} +{"problem":"in 10 years , a will be twice as old as b was 10 years ago . if a is now 9 years older than b , the present age of b is","rationale":"\"solution let b ' s present age = x years . then , a ' s present age ‹ = › ( x + 9 ) years . ∴ ( x + 9 ) + 10 = 2 ( x - 10 ) ‹ = › x + 19 = 2 x - 20 ‹ = › x = 39 . answer d\"","correct":"d","options":{"a":"9 years ","b":"19 years ","c":"29 years ","d":"39 years","e":"none"},"options_float":{"a":9.0,"b":19.0,"c":29.0,"d":39.0,"e":null},"annotated_formula":"add(add(multiply(const_2, 10), 10), 9)","linear_formula":"multiply(n0,const_2)|add(n0,#0)|add(n2,#1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 10<\/gadget>\n30<\/output>\n30 + 9<\/gadget>\n39<\/output>\n39<\/result>","index":176} +{"problem":"if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 8 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ?","rationale":"\"50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 \/ 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 8 * 2 = 16 hours ; in 1 hour 15 journeymen can finish 1 \/ 16 of the job . therefore , in 1 hour 10 apprentices and 15 journeymen can finish 1 \/ 20 + 1 \/ 16 = 9 \/ 80 of the job . answer : c\"","correct":"c","options":{"a":"1 \/ 9 ","b":"29 \/ 180 ","c":"9 \/ 80 ","d":"1 \/ 5","e":"39 \/ 121"},"options_float":{"a":0.1111111111,"b":0.1611111111,"c":0.1125,"d":0.2,"e":0.3223140496},"annotated_formula":"add(divide(const_1, divide(multiply(50, 4), 10)), divide(const_1, divide(multiply(30, 8), 15)))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|divide(#0,n4)|divide(#1,n5)|divide(const_1,#2)|divide(const_1,#3)|add(#4,#5)|","chain":"50 * 4<\/gadget>\n200<\/output>\n200 \/ 10<\/gadget>\n20<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n30 * 8<\/gadget>\n240<\/output>\n240 \/ 15<\/gadget>\n16<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/20) + (1\/16)<\/gadget>\n9\/80 = around 0.1125<\/output>\n9\/80 = around 0.1125<\/result>","index":177} +{"problem":"two ships are sailing in the sea on the two sides of a lighthouse . the angle of elevation of the top of the lighthouse is observed from the ships are 30 º and 45 º respectively . if the lighthouse is 100 m high , the distance between the two ships is :","rationale":"\"xplanation : let ab be the lighthouse and c and d be the positions of the ships . then , ab = 100 m , < acb = 30 º and < adb = 45 º . ab \/ ac = tan 30 answer : c\"","correct":"c","options":{"a":"173 m ","b":"200 m ","c":"273 m ","d":"300 m","e":"none of these"},"options_float":{"a":173.0,"b":200.0,"c":273.0,"d":300.0,"e":null},"annotated_formula":"multiply(add(sqrt(const_3), const_1), 100)","linear_formula":"sqrt(const_3)|add(#0,const_1)|multiply(n2,#1)|","chain":"3 ** (1\/2)<\/gadget>\nsqrt(3) = around 1.732051<\/output>\n(sqrt(3)) + 1<\/gadget>\n1 + sqrt(3) = around 2.732051<\/output>\n(1 + sqrt(3)) * 100<\/gadget>\n100 + 100*sqrt(3) = around 273.205081<\/output>\n100 + 100*sqrt(3) = around 273.205081<\/result>","index":178} +{"problem":"in a class , 5 people like baseball and football , 2 people only like baseball , 3 people only like football , and 6 people like neither . how many people are in the class ?","rationale":"16 . to get the answer , you can simply add the different types of students , as there are no overlaps in the categories : 5 + 2 + 3 + 6 = 16 . as such , the correct answer is c : 16 .","correct":"c","options":{"a":"10 ","b":"14 ","c":"16 ","d":"20","e":"22"},"options_float":{"a":10.0,"b":14.0,"c":16.0,"d":20.0,"e":22.0},"annotated_formula":"add(add(add(5, 2), 3), 6)","linear_formula":"add(n0,n1)|add(n2,#0)|add(n3,#1)","chain":"5 + 2<\/gadget>\n7<\/output>\n7 + 3<\/gadget>\n10<\/output>\n10 + 6<\/gadget>\n16<\/output>\n16<\/result>","index":179} +{"problem":"how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8.5 m x 6 m x 22.5 cm ?","rationale":"\"number of bricks = volume of wall \/ volume of bricks = 850 x 600 x 22.5 \/ 25 x 11.25 x 6 = = 6800 answer : e\"","correct":"e","options":{"a":"6400 ","b":"6410 ","c":"6440 ","d":"6500","e":"6800"},"options_float":{"a":6400.0,"b":6410.0,"c":6440.0,"d":6500.0,"e":6800.0},"annotated_formula":"divide(multiply(multiply(multiply(8.5, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6))","linear_formula":"multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|","chain":"8.5 * 100<\/gadget>\n850<\/output>\n6 * 100<\/gadget>\n600<\/output>\n850 * 600<\/gadget>\n510_000<\/output>\n510_000 * 22.5<\/gadget>\n11_475_000<\/output>\n25 * 11.25<\/gadget>\n281.25<\/output>\n281.25 * 6<\/gadget>\n1_687.5<\/output>\n11_475_000 \/ 1_687.5<\/gadget>\n6_800<\/output>\n6_800<\/result>","index":180} +{"problem":"given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 4 ?","rationale":"p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit of p + 4 is 6 + 4 = 10 . answer : c .","correct":"c","options":{"a":"7 ","b":"8 ","c":"10 ","d":"11","e":"12"},"options_float":{"a":7.0,"b":8.0,"c":10.0,"d":11.0,"e":12.0},"annotated_formula":"add(add(3, 3), 4)","linear_formula":"add(n0,n0)|add(n3,#0)","chain":"3 + 3<\/gadget>\n6<\/output>\n6 + 4<\/gadget>\n10<\/output>\n10<\/result>","index":181} +{"problem":"the profit obtained by selling an article for rs . 56 is the same as the loss obtained by selling it for rs . 42 . what is the cost price of the article ?","rationale":"\"s . p 1 - c . p = c . p – s . p 2 56 - c . p = c . p - 42 2 c . p = 56 + 42 ; c . p = 98 \/ 2 = 49 answer : c\"","correct":"c","options":{"a":"rs . 40 ","b":"rs . 50 ","c":"rs . 49 ","d":"rs . 59","e":"none of these"},"options_float":{"a":40.0,"b":50.0,"c":49.0,"d":59.0,"e":null},"annotated_formula":"divide(add(56, 42), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"56 + 42<\/gadget>\n98<\/output>\n98 \/ 2<\/gadget>\n49<\/output>\n49<\/result>","index":183} +{"problem":"an 80 - liter solution of cool - drink is made from 10 % jasmine water . if 5 liters of jasmine and 15 liters of water were added to the solution , what percent of the solution is jasmine ?","rationale":"the percent of jasmine in the resulting solution is : ( amount of jasmine ) \/ ( total volume ) ( 0.1 ( 80 ) + 5 ) \/ 100 = 13 \/ 100 = 13 % the answer is b .","correct":"b","options":{"a":"11 % ","b":"13 % ","c":"15 % ","d":"17 %","e":"19 %"},"options_float":{"a":11.0,"b":13.0,"c":15.0,"d":17.0,"e":19.0},"annotated_formula":"multiply(divide(add(multiply(80, divide(10, const_100)), 5), const_100), const_100)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n2,#1)|divide(#2,const_100)|multiply(#3,const_100)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n80 * (1\/10)<\/gadget>\n8<\/output>\n8 + 5<\/gadget>\n13<\/output>\n13 \/ 100<\/gadget>\n13\/100 = around 0.13<\/output>\n(13\/100) * 100<\/gadget>\n13<\/output>\n13<\/result>","index":185} +{"problem":"an error 3 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is","rationale":"\"100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 103 x 103 ) cm 2 . ( a 2 - a 1 ) = [ ( 103 ) 2 - ( 100 ) 2 ] = ( 103 + 100 ) x ( 103 - 100 ) = 609 cm 2 . percentage error = 6.09 e\"","correct":"e","options":{"a":"6.04 % ","b":"6.14 % ","c":"6.23 % ","d":"6.26 %","e":"6.09 %"},"options_float":{"a":6.04,"b":6.14,"c":6.23,"d":6.26,"e":6.09},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 3)), square_area(const_100)), const_100), square_area(const_100))","linear_formula":"add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|","chain":"100 + 3<\/gadget>\n103<\/output>\n103 ** 2<\/gadget>\n10_609<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n10_609 - 10_000<\/gadget>\n609<\/output>\n609 * 100<\/gadget>\n60_900<\/output>\n60_900 \/ 10_000<\/gadget>\n609\/100 = around 6.09<\/output>\n609\/100 = around 6.09<\/result>","index":186} +{"problem":"share rs . 6600 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ?","rationale":"\"amount received by sanjay . 4 \/ 12 x 6600 = 2200 = ( related ratio \/ sum of ratio ) x total amount so , the amount received by sanjay is 2200 . a\"","correct":"a","options":{"a":"2200 ","b":"980 ","c":"1200 ","d":"1240","e":"1400"},"options_float":{"a":2200.0,"b":980.0,"c":1200.0,"d":1240.0,"e":1400.0},"annotated_formula":"subtract(divide(6600, 2), divide(6600, 6))","linear_formula":"divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)|","chain":"6_600 \/ 2<\/gadget>\n3_300<\/output>\n6_600 \/ 6<\/gadget>\n1_100<\/output>\n3_300 - 1_100<\/gadget>\n2_200<\/output>\n2_200<\/result>","index":187} +{"problem":"the cash realised on selling a 14 % stock is rs . 105.25 , brokerage being 1 \/ 4 % is ?","rationale":"\"cash realised = rs . ( 105.25 - 0.25 ) = rs . 106 . answer : d\"","correct":"d","options":{"a":"366 ","b":"106 ","c":"102 ","d":"105","e":"122"},"options_float":{"a":366.0,"b":106.0,"c":102.0,"d":105.0,"e":122.0},"annotated_formula":"subtract(105.25, divide(1, 4))","linear_formula":"divide(n2,n3)|subtract(n1,#0)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n105.25 - (1\/4)<\/gadget>\n105<\/output>\n105<\/result>","index":188} +{"problem":"there are 6 boxes numbered 1 , 2 , … , . 6 . each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is","rationale":"list down possibilities : from only 1 box all the way to all 6 if only one of the boxes has a green ball , it can be any of the 6 boxes . so , we have 6 possibilities . if two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes . 12 , 23 , 34 , 45 , 56 . if 3 of the boxes have green balls , there are 4 possibilities : 123 , 234 , 345 , 456 . if 4 boxes have green balls , there are 3 possibilities : 1234 , 2345 , 3456 . if 5 boxes have green balls , there are 2 possibilities : 12345 , 23456 . if all 6 boxes have green balls , there is just 1 possibility . total number of possibilities = 6 + 5 + 4 + 3 + 2 + 1 = 21 . ans : b","correct":"b","options":{"a":"22 ","b":"21 ","c":"45 ","d":"27","e":"25"},"options_float":{"a":22.0,"b":21.0,"c":45.0,"d":27.0,"e":25.0},"annotated_formula":"divide(multiply(6, add(6, 1)), 2)","linear_formula":"add(n0,n1)|multiply(n0,#0)|divide(#1,n2)","chain":"6 + 1<\/gadget>\n7<\/output>\n6 * 7<\/gadget>\n42<\/output>\n42 \/ 2<\/gadget>\n21<\/output>\n21<\/result>","index":189} +{"problem":"a , b , c and d enter into partnership . a subscribes 1 \/ 3 of the capital b 1 \/ 4 , c 1 \/ 5 and d the rest . how much share did a get in a profit of rs . 2490 ?","rationale":"\"let the total amount in the partnership be ' x ' . then a ' s share = x \/ 3 b ' s share = x \/ 4 c ' s share = x \/ 5 d ' s share = x - ( x \/ 3 + x \/ 4 + x \/ 5 ) = 13 x \/ 60 a : b : c : d = x \/ 3 : x \/ 4 : x \/ 5 : 13 x \/ 60 = 20 : 15 : 12 : 13 a ' s share in the profit of rs . 2490 = 20 ( 2490 \/ 60 ) = rs . 830 . answer : b\"","correct":"b","options":{"a":"rs . 832 ","b":"rs . 830 ","c":"rs . 822 ","d":"rs . 812","e":"rs . 810"},"options_float":{"a":832.0,"b":830.0,"c":822.0,"d":812.0,"e":810.0},"annotated_formula":"multiply(2490, divide(1, 3))","linear_formula":"divide(n0,n1)|multiply(n6,#0)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n2_490 * (1\/3)<\/gadget>\n830<\/output>\n830<\/result>","index":191} +{"problem":"a 270 metres long bullet train running at the speed of 120 kmph crosses another bullet train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other bullet train ?","rationale":"b 230 relative speed = ( 120 + 80 ) km \/ hr = 200 x 5 \/ 18 = 500 \/ 9 m \/ sec then , ( x + 270 ) \/ 9 = 500 \/ 9 - - > x + 270 = 500 - - > x = 230 .","correct":"b","options":{"a":"226 ","b":"230 ","c":"430 ","d":"310","e":"260"},"options_float":{"a":226.0,"b":230.0,"c":430.0,"d":310.0,"e":260.0},"annotated_formula":"multiply(subtract(multiply(add(120, 80), const_0_2778), divide(270, 9)), 9)","linear_formula":"add(n1,n2)|divide(n0,n3)|multiply(#0,const_0_2778)|subtract(#2,#1)|multiply(n3,#3)","chain":"120 + 80<\/gadget>\n200<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n200 * (5\/18)<\/gadget>\n500\/9 = around 55.555556<\/output>\n270 \/ 9<\/gadget>\n30<\/output>\n(500\/9) - 30<\/gadget>\n230\/9 = around 25.555556<\/output>\n(230\/9) * 9<\/gadget>\n230<\/output>\n230<\/result>","index":195} +{"problem":"winson traveled the entire 60 miles trip . if he did the first 10 miles of at a constant rate 20 miles per hour and the remaining trip of at a constant rate 48 miles per hour , what is the his average speed , in miles per hour ?","rationale":"average speed = sum of distance \/ sum of time . if he travelled the first 10 miles at 20 miles \/ hr , it would take 0.5 hr . for the remaining trip , if he went at 48 miles \/ 1 hr , it would take 1 hour . then , the average speed is 60 miles \/ ( 0.5 + 1 ) hrs = 40 miles \/ 1 hr . therefore , the answer is e .","correct":"e","options":{"a":"20 mph ","b":"24 mph ","c":"30 mph ","d":"32 mph","e":"40 mph"},"options_float":{"a":20.0,"b":24.0,"c":30.0,"d":32.0,"e":40.0},"annotated_formula":"divide(60, add(divide(subtract(60, 10), 48), divide(10, 20)))","linear_formula":"divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#2,#0)|divide(n0,#3)","chain":"60 - 10<\/gadget>\n50<\/output>\n50 \/ 48<\/gadget>\n25\/24 = around 1.041667<\/output>\n10 \/ 20<\/gadget>\n1\/2 = around 0.5<\/output>\n(25\/24) + (1\/2)<\/gadget>\n37\/24 = around 1.541667<\/output>\n60 \/ (37\/24)<\/gadget>\n1_440\/37 = around 38.918919<\/output>\n1_440\/37 = around 38.918919<\/result>","index":196} +{"problem":"a pet store holds cats and dogs . if the difference between the number of cats and the number of dogs is 7 . what could be the ratio of cats to dogs in the pet store ?","rationale":"\"say theratioof cats to dogs is a \/ b . then thenumberof cats would be ax and thenumberof dogs bx , for some positive integer x . we are told that ax - bx = 7 - - > x ( a - b ) = 7 . since 7 is a prime number it could be broken into the product of two positive multiples only in one way : x ( a - b ) = 1 * 7 . the above implies that either x = 1 and a - b = 7 or x = 7 and ( a - b ) = 1 . therefore the correct answer should have the difference between numerator and denominator equal to 1 or 7 . for the original question only option which fits is e , 4 : 5 . cats = 7 * 4 = 28 and dogs = 7 * 5 = 35 . answer : c .\"","correct":"c","options":{"a":"1 : 3 ","b":"1 : 4 ","c":"28 : 35 ","d":"2 : 5","e":"4 : 6"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.8,"d":0.4,"e":0.6666666667},"annotated_formula":"divide(multiply(7, const_4), add(multiply(7, const_4), 7))","linear_formula":"multiply(n0,const_4)|add(n0,#0)|divide(#0,#1)|","chain":"7 * 4<\/gadget>\n28<\/output>\n28 + 7<\/gadget>\n35<\/output>\n28 \/ 35<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":197} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 18 seconds . find the length of the train ?","rationale":"\"speed = 60 * ( 5 \/ 18 ) m \/ sec = 50 \/ 3 m \/ sec length of train ( distance ) = speed * time ( 50 \/ 3 ) * 18 = 300 meter answer : d\"","correct":"d","options":{"a":"150 meter ","b":"170 meter ","c":"156 meter ","d":"300 meter","e":"154 meter"},"options_float":{"a":150.0,"b":170.0,"c":156.0,"d":300.0,"e":154.0},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 18)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"60 * 1_000<\/gadget>\n60_000<\/output>\n60_000 \/ 3_600<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 18<\/gadget>\n300<\/output>\n300<\/result>","index":198} +{"problem":"the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 13 and 14 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 23 * 13 ) and ( 23 * 14 ) . larger number = ( 23 * 14 ) = 322 . answer : c\"","correct":"c","options":{"a":"338 ","b":"278 ","c":"322 ","d":"231","e":"121"},"options_float":{"a":338.0,"b":278.0,"c":322.0,"d":231.0,"e":121.0},"annotated_formula":"multiply(23, 14)","linear_formula":"multiply(n0,n2)|","chain":"23 * 14<\/gadget>\n322<\/output>\n322<\/result>","index":199} +{"problem":"the value of ( . 02 ) 2 + ( 0.52 ) 2 + ( 0.035 ) 2 \/ ( 0.002 ) 2 + ( 0.052 ) 2 + ( 0.0035 ) 2","rationale":"( . 02 ) 2 + ( 0.52 ) 2 + ( 0.035 ) 2 \/ ( 0.002 ) 2 + ( 0.052 ) 2 + ( 0.0035 ) 2 = a 2 + b 2 + c 2 \/ ( a \/ 10 ) 2 + ( b \/ 10 ) 2 + ( c \/ 10 ) 2 where a = . 02 , b = . 52 , c = . 035 = 100 ( a 2 + b 2 + c 2 ) \/ a 2 + b 2 + c 2 = 100 answer : a","correct":"a","options":{"a":"100 ","b":"1000 ","c":". 001 ","d":". 0001","e":". 01"},"options_float":{"a":100.0,"b":1000.0,"c":1.0,"d":1.0,"e":1.0},"annotated_formula":"divide(add(add(power(divide(2, const_100), 2), power(0.52, 2)), power(0.035, 2)), add(add(power(0.002, 2), power(0.052, 2)), power(0.0035, 2)))","linear_formula":"divide(n0,const_100)|power(n2,n0)|power(n4,n0)|power(n6,n0)|power(n8,n0)|power(n10,n0)|add(#3,#4)|power(#0,n0)|add(#7,#1)|add(#6,#5)|add(#8,#2)|divide(#10,#9)","chain":"2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) ** 2<\/gadget>\n1\/2_500 = around 0.0004<\/output>\n0.52 ** 2<\/gadget>\n0.2704<\/output>\n(1\/2_500) + 0.2704<\/gadget>\n0.2708<\/output>\n0.035 ** 2<\/gadget>\n0.001225<\/output>\n0.2708 + 0.001225<\/gadget>\n0.272025<\/output>\n0.002 ** 2<\/gadget>\n0.000004<\/output>\n0.052 ** 2<\/gadget>\n0.002704<\/output>\n0.000004 + 0.002704<\/gadget>\n0.002708<\/output>\n0.0035 ** 2<\/gadget>\n0.000012<\/output>\n0.002708 + 0.000012<\/gadget>\n0.00272<\/output>\n0.272025 \/ 0.00272<\/gadget>\n100.009191<\/output>\n100.009191<\/result>","index":202} +{"problem":"how much time does a train 160 metres long running at 72 km \/ hr take to pass a pole ?","rationale":"\"explanation : 72 km \/ hr = 72 * 5 \/ 18 = 20 m \/ s speed = distance \/ time ; v = d \/ t 20 = 160 \/ t t = 8 s answer : a\"","correct":"a","options":{"a":"8 s ","b":"2.5 s ","c":"7.5 s ","d":"7.6 s","e":"7.4 s"},"options_float":{"a":8.0,"b":2.5,"c":7.5,"d":7.6,"e":7.4},"annotated_formula":"multiply(divide(divide(160, const_1000), 72), const_3600)","linear_formula":"divide(n0,const_1000)|divide(#0,n1)|multiply(#1,const_3600)|","chain":"160 \/ 1_000<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) \/ 72<\/gadget>\n1\/450 = around 0.002222<\/output>\n(1\/450) * 3_600<\/gadget>\n8<\/output>\n8<\/result>","index":203} +{"problem":"find the smallest 4 - digit number which is a multiple of 112 .","rationale":"the smallest four digit number is 1000 . if 1000 is divided by 112 , the remainder is 104 . 112 - 104 = 8 , if 8 is added to 1000 , it will become the smallest four digit number and a multiple of 112 . answer : b","correct":"b","options":{"a":"896 ","b":"1008 ","c":"1120 ","d":"1024","e":"none of these"},"options_float":{"a":896.0,"b":1008.0,"c":1120.0,"d":1024.0,"e":null},"annotated_formula":"multiply(subtract(const_10, const_1), 112)","linear_formula":"subtract(const_10,const_1)|multiply(n1,#0)","chain":"10 - 1<\/gadget>\n9<\/output>\n9 * 112<\/gadget>\n1_008<\/output>\n1_008<\/result>","index":208} +{"problem":"the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 15 th student is","rationale":"\"explanation : age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = ( 225 - 214 ) = 11 years . answer : a ) 11\"","correct":"a","options":{"a":"11 ","b":"15 ","c":"6 ","d":"8","e":"76"},"options_float":{"a":11.0,"b":15.0,"c":6.0,"d":8.0,"e":76.0},"annotated_formula":"subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n5 * 14<\/gadget>\n70<\/output>\n9 * 16<\/gadget>\n144<\/output>\n70 + 144<\/gadget>\n214<\/output>\n225 - 214<\/gadget>\n11<\/output>\n11<\/result>","index":210} +{"problem":"the sale price of a trolley bag including the sale tax is rs . 560 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is :","rationale":"\"explanation : 112 % of s . p . = 560 s . p . = rs . ( 560 x 100 \/ 112 ) = rs . 500 . c . p . = rs ( 100 \/ 125 x 500 ) = rs 400 answer : c\"","correct":"c","options":{"a":"rs 1000 ","b":"rs 900 ","c":"rs 400 ","d":"rs 600","e":"none of these"},"options_float":{"a":1000.0,"b":900.0,"c":400.0,"d":600.0,"e":null},"annotated_formula":"divide(subtract(560, multiply(560, divide(12, const_100))), add(divide(25, const_100), const_1))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|add(#1,const_1)|multiply(n0,#0)|subtract(n0,#3)|divide(#4,#2)|","chain":"12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n560 * (3\/25)<\/gadget>\n336\/5 = around 67.2<\/output>\n560 - (336\/5)<\/gadget>\n2_464\/5 = around 492.8<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(2_464\/5) \/ (5\/4)<\/gadget>\n9_856\/25 = around 394.24<\/output>\n9_856\/25 = around 394.24<\/result>","index":211} +{"problem":"stacy and heather are 40 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour faster than heather ' s constant rate of 5 miles \/ hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ?","rationale":"\"original distance between s and h = 40 miles . speed of s = 5 + 1 = 6 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 24 \/ 60 = t + 2 \/ 5 hours . now , the total distances traveled by s and h = 40 miles - - - > 6 * ( t + 2 \/ 5 ) + 5 * t = 40 - - - > t = 188 \/ 55 hours . thus h has traveled for 188 \/ 55 hours giving you a total distance for h = 5 * 188 \/ 55 = 17 miles . c is thus the correct answer . p . s . : based on the wording of the question , you should calculatehow far from theoriginal destination has heather walkedwhen the two meet . ' original destination ' for h does not make any sense . original destination for h was situated at a distance of 20 miles .\"","correct":"c","options":{"a":"4 miles ","b":"8 miles ","c":"17 miles ","d":"6 miles","e":"12 miles"},"options_float":{"a":4.0,"b":8.0,"c":17.0,"d":6.0,"e":12.0},"annotated_formula":"multiply(divide(subtract(40, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5)","linear_formula":"add(n1,n2)|add(n2,#0)|divide(#0,const_60)|multiply(n3,#2)|subtract(n0,#3)|divide(#4,#1)|multiply(n2,#5)|","chain":"1 + 5<\/gadget>\n6<\/output>\n6 \/ 60<\/gadget>\n1\/10 = around 0.1<\/output>\n24 * (1\/10)<\/gadget>\n12\/5 = around 2.4<\/output>\n40 - (12\/5)<\/gadget>\n188\/5 = around 37.6<\/output>\n5 + 6<\/gadget>\n11<\/output>\n(188\/5) \/ 11<\/gadget>\n188\/55 = around 3.418182<\/output>\n(188\/55) * 5<\/gadget>\n188\/11 = around 17.090909<\/output>\n188\/11 = around 17.090909<\/result>","index":212} +{"problem":"a motorcyclist goes from bombay to pune , a distance of 128 kms at an average of 32 kmph speed . another man starts from bombay by car 2 ½ hours after the first , and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ?","rationale":"\"t = 128 \/ 32 = 4 h t = 4 - 3 = 1 time ratio = 4 : 1 = 4 : 1 speed ratio = 1 : 4 answer : c\"","correct":"c","options":{"a":"1 : 3 ","b":"1 : 5 ","c":"1 : 4 ","d":"1 : 1","e":"1 : 8"},"options_float":{"a":0.3333333333,"b":0.2,"c":0.25,"d":1.0,"e":0.125},"annotated_formula":"divide(divide(128, divide(128, 32)), divide(128, subtract(divide(128, 32), const_3)))","linear_formula":"divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3)|","chain":"128 \/ 32<\/gadget>\n4<\/output>\n128 \/ 4<\/gadget>\n32<\/output>\n4 - 3<\/gadget>\n1<\/output>\n128 \/ 1<\/gadget>\n128<\/output>\n32 \/ 128<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":214} +{"problem":"a man whose speed is 5.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ?","rationale":"\"explanation : speed of man = 5.5 kmph speed of stream = 1.5 kmph speed in downstream = 7 kmph speed in upstream = 4 kmph average speed = ( 2 x 7 x 4 ) \/ 11 = 5 kmph . answer : b\"","correct":"b","options":{"a":"1 kmph ","b":"5 kmph ","c":"6 kmph ","d":"4 kmph","e":"7 kmph"},"options_float":{"a":1.0,"b":5.0,"c":6.0,"d":4.0,"e":7.0},"annotated_formula":"divide(add(5.5, subtract(5.5, 1.5)), const_2)","linear_formula":"subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)|","chain":"5.5 - 1.5<\/gadget>\n4<\/output>\n5.5 + 4<\/gadget>\n9.5<\/output>\n9.5 \/ 2<\/gadget>\n4.75<\/output>\n4.75<\/result>","index":215} +{"problem":"when positive integer x is divided by 11 , the quotient is y and the remainder is 4 . when 2 x is divided by 5 , the quotient is 3 y and the remainder is 4 . what is the value of 4 y – x ?","rationale":"\"( 1 ) x = 11 y + 4 ( 2 ) 2 x = 15 y + 4 let ' s subtract equation ( 1 ) from equation ( 2 ) . 4 y = x 4 y - x = 0 the answer is c .\"","correct":"c","options":{"a":"- 2 ","b":"- 1 ","c":"0 ","d":"1","e":"2"},"options_float":{"a":-2.0,"b":-1.0,"c":0.0,"d":1.0,"e":2.0},"annotated_formula":"subtract(multiply(4, divide(subtract(multiply(2, 4), 4), subtract(multiply(5, 3), multiply(2, 11)))), add(multiply(11, divide(subtract(multiply(2, 4), 4), subtract(multiply(5, 3), multiply(2, 11)))), 4))","linear_formula":"multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n2)|subtract(#0,n5)|subtract(#1,#2)|divide(#3,#4)|multiply(n6,#5)|multiply(n0,#5)|add(n1,#7)|subtract(#6,#8)|","chain":"2 * 4<\/gadget>\n8<\/output>\n8 - 4<\/gadget>\n4<\/output>\n5 * 3<\/gadget>\n15<\/output>\n2 * 11<\/gadget>\n22<\/output>\n15 - 22<\/gadget>\n-7<\/output>\n4 \/ (-7)<\/gadget>\n-4\/7 = around -0.571429<\/output>\n4 * (-4\/7)<\/gadget>\n-16\/7 = around -2.285714<\/output>\n11 * (-4\/7)<\/gadget>\n-44\/7 = around -6.285714<\/output>\n(-44\/7) + 4<\/gadget>\n-16\/7 = around -2.285714<\/output>\n(-16\/7) - (-16\/7)<\/gadget>\n0<\/output>\n0<\/result>","index":216} +{"problem":"martin bought 10 concert tickets , some at the full price of $ 2.00 per ticket , and some at a discounted price of $ 1.60 per ticket . if he spent a total of $ 18.40 , how many discounted tickets did he buy ?","rationale":"let x be the number of tickets he bought at $ 2 per ticket . then 2 x + ( 10 - x ) 1.6 = 18.4 0.4 x = 2.4 = > x = 6 discounted tickets = 10 - x = 4 ans : b","correct":"b","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"subtract(multiply(10, 2), multiply(1.6, 10))","linear_formula":"multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)","chain":"10 * 2<\/gadget>\n20<\/output>\n1.6 * 10<\/gadget>\n16<\/output>\n20 - 16<\/gadget>\n4<\/output>\n4<\/result>","index":217} +{"problem":"the area of a square field is 625 km 2 . how long will it take for a horse to run around at the speed of 25 km \/ h ?","rationale":"explanation area of field = 625 km 2 . then , each side of field = √ 625 = 25 km distance covered by the horse = perimeter of square field = 25 × 4 = 100 km ∴ time taken by horse = distances \/ peed = 100 \/ 25 = 4 h answer d","correct":"d","options":{"a":"12 h ","b":"10 h ","c":"8 h ","d":"4 h","e":"none of these"},"options_float":{"a":12.0,"b":10.0,"c":8.0,"d":4.0,"e":null},"annotated_formula":"divide(multiply(const_4, sqrt(625)), 25)","linear_formula":"sqrt(n0)|multiply(#0,const_4)|divide(#1,n2)|","chain":"625 ** (1\/2)<\/gadget>\n25<\/output>\n4 * 25<\/gadget>\n100<\/output>\n100 \/ 25<\/gadget>\n4<\/output>\n4<\/result>","index":218} +{"problem":"how is 3 % expressed as a decimal fraction ?","rationale":"\"3 \/ 100 = 0.03 answer : b\"","correct":"b","options":{"a":"0.3 ","b":"0.03 ","c":"0.003 ","d":"0.0003","e":"3"},"options_float":{"a":0.3,"b":0.03,"c":0.003,"d":0.0003,"e":3.0},"annotated_formula":"divide(3, const_100)","linear_formula":"divide(n0,const_100)|","chain":"3 \/ 100<\/gadget>\n3\/100 = around 0.03<\/output>\n3\/100 = around 0.03<\/result>","index":219} +{"problem":"the cross section of a canal is trapezium in shape . the canal is 12 m wide at the top and 8 m wide at the bottom . if the area of the cross section is 840 sq . m , the depth of the canal is :","rationale":"½ ( 12 + 8 ) d = 840 or d = 84 m answer : b","correct":"b","options":{"a":"42 m ","b":"84 m ","c":"63 m ","d":"8.75 m","e":"9.75 m"},"options_float":{"a":42.0,"b":84.0,"c":63.0,"d":8.75,"e":9.75},"annotated_formula":"divide(840, divide(add(12, 8), const_2))","linear_formula":"add(n0,n1)|divide(#0,const_2)|divide(n2,#1)","chain":"12 + 8<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n840 \/ 10<\/gadget>\n84<\/output>\n84<\/result>","index":220} +{"problem":"if x \/ 4 years ago roger was 21 years old and x \/ 4 years from now he will be 4 x years old , how old will he be 5 x years from now ?","rationale":"assume the current age = a a - x \/ 4 = 28 ( i ) a + x \/ 4 = 4 x or a = 15 x \/ 4 ( ii ) putting the value of a from ( ii ) in ( i ) 15 x \/ 4 - x \/ 4 = 28 or 14 x \/ 4 = 28 therefore x = 8 and a = 30 5 x years from now , age will be 30 + 5 * 8 = 70 option e","correct":"e","options":{"a":"32 ","b":"35 ","c":"30 ","d":"40","e":"70"},"options_float":{"a":32.0,"b":35.0,"c":30.0,"d":40.0,"e":70.0},"annotated_formula":"subtract(multiply(multiply(4, 5), 4), divide(subtract(21, const_1), const_2))","linear_formula":"multiply(n0,n4)|subtract(n1,const_1)|divide(#1,const_2)|multiply(n0,#0)|subtract(#3,#2)","chain":"4 * 5<\/gadget>\n20<\/output>\n20 * 4<\/gadget>\n80<\/output>\n21 - 1<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n80 - 10<\/gadget>\n70<\/output>\n70<\/result>","index":221} +{"problem":"a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 21 per kg , then his gain percent is","rationale":"\"explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 21 ) = rs . 168 . profit = 168 - 150 = 18 so , profit % = ( 18 \/ 150 ) * 100 = 12 % option a\"","correct":"a","options":{"a":"12 % ","b":"13 % ","c":"14 % ","d":"15 %","e":"16 %"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"divide(multiply(subtract(multiply(21, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20)))","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3)|","chain":"5 + 3<\/gadget>\n8<\/output>\n21 * 8<\/gadget>\n168<\/output>\n5 * 18<\/gadget>\n90<\/output>\n3 * 20<\/gadget>\n60<\/output>\n90 + 60<\/gadget>\n150<\/output>\n168 - 150<\/gadget>\n18<\/output>\n18 * 100<\/gadget>\n1_800<\/output>\n1_800 \/ 150<\/gadget>\n12<\/output>\n12<\/result>","index":222} +{"problem":"if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 54 , then * is equal to :","rationale":"\"45 - [ 28 - { 37 - ( 15 - * ) } ] = 54 = > 45 - [ 28 - { 37 - 15 + * } ] = 54 45 - [ 28 - 37 + 15 - * ] = 54 = > 45 [ 43 - 37 - * ] = 54 45 - [ 6 - * ] = 54 = > 45 - 6 + * = 54 39 + * = 54 = > * = 54 - 39 = 19 answer : c\"","correct":"c","options":{"a":"- 29 ","b":"- 19 ","c":"15 ","d":"29","e":"39"},"options_float":{"a":-29.0,"b":-19.0,"c":15.0,"d":29.0,"e":39.0},"annotated_formula":"subtract(54, subtract(45, add(subtract(28, 37), 15)))","linear_formula":"subtract(n1,n2)|add(n3,#0)|subtract(n0,#1)|subtract(n4,#2)|","chain":"28 - 37<\/gadget>\n-9<\/output>\n(-9) + 15<\/gadget>\n6<\/output>\n45 - 6<\/gadget>\n39<\/output>\n54 - 39<\/gadget>\n15<\/output>\n15<\/result>","index":224} +{"problem":"if 2 tic equals 3 tacs and 2 tacs equal 7 tocs , what is the ratio of one tic to one toc ?","rationale":"2 tic = 3 * tac and 2 * tac = 7 * toc ; 4 * tic = 6 * tac and 6 * tac = 21 * toc - - > 4 * tic = 21 * toc - - > tic \/ toc = 21 \/ 4 . answer : c .","correct":"c","options":{"a":"15 \/ 2 ","b":"6 \/ 5 ","c":"21 \/ 4 ","d":"3 \/ 10","e":"1 \/ 15"},"options_float":{"a":7.5,"b":1.2,"c":5.25,"d":0.3,"e":0.0666666667},"annotated_formula":"add(divide(7, 2), divide(3, 2))","linear_formula":"divide(n3,n0)|divide(n1,n0)|add(#0,#1)","chain":"7 \/ 2<\/gadget>\n7\/2 = around 3.5<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(7\/2) + (3\/2)<\/gadget>\n5<\/output>\n5<\/result>","index":225} +{"problem":"how many prime numbers are between 29 \/ 3 and 87 \/ 6 ?","rationale":"\"29 \/ 3 = 58 \/ 6 = 10 - 87 \/ 6 = 15 - prime numbers between 10 and 15 are 11 and 13 - sign signifies that the number is marginally less . answer b\"","correct":"b","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"floor(const_2)","linear_formula":"floor(const_2)|","chain":"floor(2)<\/gadget>\n2<\/output>\n2<\/result>","index":226} +{"problem":"find the greatest number which leaves the same remainder when it divides 25 , 57 and 105 .","rationale":"105 - 57 = 48 57 - 25 = 32 105 - 25 = 80 the h . c . f of 32 , 48 and 80 is 16 . answer : d","correct":"d","options":{"a":"18 ","b":"8 ","c":"12 ","d":"16","e":"none of these"},"options_float":{"a":18.0,"b":8.0,"c":12.0,"d":16.0,"e":null},"annotated_formula":"divide(subtract(57, 25), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,const_2)","chain":"57 - 25<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16<\/result>","index":227} +{"problem":"find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e\"","correct":"e","options":{"a":"1235 ","b":"1456 ","c":"1567 ","d":"1678","e":"1635"},"options_float":{"a":1235.0,"b":1456.0,"c":1567.0,"d":1678.0,"e":1635.0},"annotated_formula":"add(multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6), 15)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)|","chain":"1_365 - 15<\/gadget>\n1_350<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_350 \/ 5<\/gadget>\n270<\/output>\n270 * 6<\/gadget>\n1_620<\/output>\n1_620 + 15<\/gadget>\n1_635<\/output>\n1_635<\/result>","index":228} +{"problem":"a furniture store owner decided to drop the price of her recliners by 20 % to spur business . by the end of the week she had sold 50 % more recliners . what is the percentage increase of the gross ?","rationale":"\"say a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 % , she will earn $ 80 per recliner and she sells 50 % more ie . , 150 recliners hence her sales yields 150 * 80 = $ 12000 increase in sales = 12000 - 10000 = $ 2000 so % increase = 2000 * 100 \/ 10000 = 20 % c is the answer\"","correct":"c","options":{"a":"10 % ","b":"15 % ","c":"20 % ","d":"25 %","e":"50 %"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":50.0},"annotated_formula":"subtract(multiply(multiply(subtract(const_1, divide(20, const_100)), add(const_1, divide(50, const_100))), const_100), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n(4\/5) * (3\/2)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 100<\/gadget>\n120<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20<\/result>","index":229} +{"problem":"108 ã — 108 + 92 ã — 92 = ?","rationale":"explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1082 + 922 = ( 100 + 8 ) 2 + ( 100 â ˆ ’ 8 ) 2 = 2 ( 1002 + 82 ) = 2 ( 10000 + 64 ) = 20128 . answer : option c","correct":"c","options":{"a":"20120 ","b":"11500 ","c":"20128 ","d":"20660","e":"20228"},"options_float":{"a":20120.0,"b":11500.0,"c":20128.0,"d":20660.0,"e":20228.0},"annotated_formula":"add(subtract(power(108, const_2), 108), subtract(power(92, const_2), 92))","linear_formula":"power(n0,const_2)|power(n2,const_2)|subtract(#0,n0)|subtract(#1,n2)|add(#2,#3)","chain":"108 ** 2<\/gadget>\n11_664<\/output>\n11_664 - 108<\/gadget>\n11_556<\/output>\n92 ** 2<\/gadget>\n8_464<\/output>\n8_464 - 92<\/gadget>\n8_372<\/output>\n11_556 + 8_372<\/gadget>\n19_928<\/output>\n19_928<\/result>","index":230} +{"problem":"1 - 1 - 1 - 1 - 1 - 1 do any operations between the numbers to make the sum as 37 ?","rationale":"1 + 1 + 1 = 3 3 factorial = 6 6 to the power 1 + 1 = 6 ^ 2 = 36 36 + 1 = 37 answer : d","correct":"d","options":{"a":"34 ","b":"35 ","c":"36 ","d":"37","e":"38"},"options_float":{"a":34.0,"b":35.0,"c":36.0,"d":37.0,"e":38.0},"annotated_formula":"add(multiply(add(1, 1), power(add(1, 1), power(add(1, 1), add(1, 1)))), add(power(add(1, 1), add(1, 1)), 1))","linear_formula":"add(n0,n0)|power(#0,#0)|add(n0,#1)|power(#0,#1)|multiply(#0,#3)|add(#2,#4)","chain":"1 + 1<\/gadget>\n2<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n2 * 16<\/gadget>\n32<\/output>\n4 + 1<\/gadget>\n5<\/output>\n32 + 5<\/gadget>\n37<\/output>\n37<\/result>","index":232} +{"problem":"two trains of equal length are running on parallel lines in the same directions at 46 km \/ hr . and 36 km \/ hr . the faster trains pass the slower train in 18 seconds . the length of each train is :","rationale":"\"explanation : the relative speed of train is 46 - 36 = 10 km \/ hr = ( 10 x 5 ) \/ 18 = 25 \/ 9 m \/ s 10 × 518 = 259 m \/ s in 18 secs the total distance traveled is 18 x 25 \/ 9 = 50 m . therefore the length of each train is = 50 \/ 2 = 25 m . answer a\"","correct":"a","options":{"a":"25 m ","b":"50 m ","c":"72 m ","d":"80 m","e":"none of these"},"options_float":{"a":25.0,"b":50.0,"c":72.0,"d":80.0,"e":null},"annotated_formula":"divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 18), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|","chain":"46 - 36<\/gadget>\n10<\/output>\n10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 3_600<\/gadget>\n25\/9 = around 2.777778<\/output>\n(25\/9) * 18<\/gadget>\n50<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n25<\/result>","index":234} +{"problem":"275 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees","rationale":"\"26 trees have 25 gaps between them , required distance ( 275 \/ 25 ) = 11 a\"","correct":"a","options":{"a":"11 ","b":"12 ","c":"14 ","d":"16","e":"17"},"options_float":{"a":11.0,"b":12.0,"c":14.0,"d":16.0,"e":17.0},"annotated_formula":"divide(275, add(subtract(26, 2), const_1))","linear_formula":"subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)|","chain":"26 - 2<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n275 \/ 25<\/gadget>\n11<\/output>\n11<\/result>","index":238} +{"problem":"a group of boy scouts and girls scouts is going on a rafting trip . 60 % of the scouts arrived with signed permission slips . if 45 % of the scouts were boy scouts and 50 % of the boy scouts arrived with signed permission slips , then what percentage of the girl scouts arrived with signed permission slips ? round to the nearest percent .","rationale":"\"45 % were boy scouts so 55 % ( 100 - 45 = 55 ) were girl scouts . # of boy scouts with permission slips signed + # of girl scouts with permission slips signed = total # with permission slip signed ( 50 % of 45 % of the total going ) + ( ? % of 55 % of the total going ) = 60 % of the total going we can let the ` ` total going ' ' = 1,000 arbitrarily since we only care about the percent ( of girl scouts w \/ permission signed ) . 50 * . 45 * 1,000 + x * . 55 * 1,000 = . 60 * 1,000 225 + x * 550 = 600 550 x = 375 x = 375 \/ 550 x = . 6818181818 which converts to a percent of 68.18181818 % which rounds to 68 % b\"","correct":"b","options":{"a":"64 ","b":"68 ","c":"72 ","d":"76","e":"80"},"options_float":{"a":64.0,"b":68.0,"c":72.0,"d":76.0,"e":80.0},"annotated_formula":"multiply(divide(subtract(60, divide(multiply(45, 50), const_100)), subtract(const_100, 45)), const_100)","linear_formula":"multiply(n1,n2)|subtract(const_100,n1)|divide(#0,const_100)|subtract(n0,#2)|divide(#3,#1)|multiply(#4,const_100)|","chain":"45 * 50<\/gadget>\n2_250<\/output>\n2_250 \/ 100<\/gadget>\n45\/2 = around 22.5<\/output>\n60 - (45\/2)<\/gadget>\n75\/2 = around 37.5<\/output>\n100 - 45<\/gadget>\n55<\/output>\n(75\/2) \/ 55<\/gadget>\n15\/22 = around 0.681818<\/output>\n(15\/22) * 100<\/gadget>\n750\/11 = around 68.181818<\/output>\n750\/11 = around 68.181818<\/result>","index":239} +{"problem":"the length of the bridge , which a train 155 metres long and travelling at 45 km \/ hr can cross in 30 seconds , is ?","rationale":"\"speed = [ 45 x 5 \/ 18 ] m \/ sec = [ 25 \/ 2 ] m \/ sec time = 30 sec let the length of bridge be x metres . then , ( 155 + x ) \/ 30 = 25 \/ 2 = > 2 ( 155 + x ) = 750 = > x = 220 m . answer : c\"","correct":"c","options":{"a":"76 m ","b":"178 m ","c":"220 m ","d":"187 m","e":"176 m"},"options_float":{"a":76.0,"b":178.0,"c":220.0,"d":187.0,"e":176.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 155)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 155<\/gadget>\n220<\/output>\n220<\/result>","index":240} +{"problem":"the shopkeeper sale , all of the prices of the items sold were different . if the price of a radio sold at the shopkeeper sale was both the 9 th highest price and the 14 th lowest price among the prices of the items sold , how many items were sold at the shopkeeper sale ?","rationale":"8 + 13 + 1 = 22 answer : b","correct":"b","options":{"a":"23 ","b":"22 ","c":"32 ","d":"26","e":"17"},"options_float":{"a":23.0,"b":22.0,"c":32.0,"d":26.0,"e":17.0},"annotated_formula":"add(9, 14)","linear_formula":"add(n0,n1)","chain":"9 + 14<\/gadget>\n23<\/output>\n23<\/result>","index":241} +{"problem":"find the value of 2 x [ ( 3.6 x 0.48 x 2.50 ) \/ ( 0.12 x 0.09 x 0.5 ) ]","rationale":"\"answer 2 x [ ( 3.6 x 0.48 x 2.50 ) \/ ( 0.12 x 0.09 x 0.5 ) ] = 2 x [ ( 36 x 48 x 250 ) \/ ( 12 x 9 x 5 ) ] = 2 x 4 x 4 x 50 = 1600 correct option : d\"","correct":"d","options":{"a":"800 ","b":"500 ","c":"900 ","d":"1600","e":"none"},"options_float":{"a":800.0,"b":500.0,"c":900.0,"d":1600.0,"e":null},"annotated_formula":"multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 2)","linear_formula":"multiply(n1,n2)|multiply(n4,n5)|multiply(n3,#0)|multiply(n6,#1)|divide(#2,#3)|multiply(n0,#4)|","chain":"3.6 * 0.48<\/gadget>\n1.728<\/output>\n1.728 * 2.5<\/gadget>\n4.32<\/output>\n0.12 * 0.09<\/gadget>\n0.0108<\/output>\n0.0108 * 0.5<\/gadget>\n0.0054<\/output>\n4.32 \/ 0.0054<\/gadget>\n800<\/output>\n800 * 2<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":242} +{"problem":"20 is subtracted from 80 % of a number , the result is 60 . find the number ?","rationale":"\"( 80 \/ 100 ) * x – 20 = 60 8 x = 800 x = 100 answer : c\"","correct":"c","options":{"a":"150 ","b":"997 ","c":"100 ","d":"288","e":"271"},"options_float":{"a":150.0,"b":997.0,"c":100.0,"d":288.0,"e":271.0},"annotated_formula":"divide(add(20, 60), divide(80, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"20 + 60<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n80 \/ (4\/5)<\/gadget>\n100<\/output>\n100<\/result>","index":243} +{"problem":"a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 40 paise per sq . metre","rationale":"area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 40 = 34000 p in rs = 34000 \/ 100 = rs 340 answer : a","correct":"a","options":{"a":"s 340 ","b":"s 780 ","c":"s 880 ","d":"s 480","e":"s 980"},"options_float":{"a":340.0,"b":780.0,"c":880.0,"d":480.0,"e":980.0},"annotated_formula":"divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 40), const_100)","linear_formula":"multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)","chain":"110 * 65<\/gadget>\n7_150<\/output>\n2.5 * 2<\/gadget>\n5<\/output>\n110 - 5<\/gadget>\n105<\/output>\n65 - 5<\/gadget>\n60<\/output>\n105 * 60<\/gadget>\n6_300<\/output>\n7_150 - 6_300<\/gadget>\n850<\/output>\n850 * 40<\/gadget>\n34_000<\/output>\n34_000 \/ 100<\/gadget>\n340<\/output>\n340<\/result>","index":244} +{"problem":"there are 280 doctors and nurses at a hospital . if the ratio of doctors to nurses is 5 to 9 , how many nurses are at the hospital ?","rationale":"the number of nurses at the hospital is ( 9 \/ 14 ) * 280 = 180 . the answer is d .","correct":"d","options":{"a":"165 ","b":"170 ","c":"175 ","d":"180","e":"185"},"options_float":{"a":165.0,"b":170.0,"c":175.0,"d":180.0,"e":185.0},"annotated_formula":"multiply(divide(280, add(5, 9)), 9)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)","chain":"5 + 9<\/gadget>\n14<\/output>\n280 \/ 14<\/gadget>\n20<\/output>\n20 * 9<\/gadget>\n180<\/output>\n180<\/result>","index":245} +{"problem":"a man sells a car to his friend at 10 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was :","rationale":"\"explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 \/ 120 × 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 10 % . now s . p = rs . 45000 and loss = 10 % c . p . rs . [ 10090 × 45000 ] = rs . 50000 . correct option : c\"","correct":"c","options":{"a":"rs . 25000 ","b":"rs . 37500 ","c":"rs . 50000 ","d":"rs . 60000","e":"none of these"},"options_float":{"a":25000.0,"b":37500.0,"c":50000.0,"d":60000.0,"e":null},"annotated_formula":"divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 10))","linear_formula":"add(n2,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|divide(#1,#0)|multiply(#3,const_100)|divide(#4,#2)|","chain":"54_000 * 100<\/gadget>\n5_400_000<\/output>\n100 + 20<\/gadget>\n120<\/output>\n5_400_000 \/ 120<\/gadget>\n45_000<\/output>\n45_000 * 100<\/gadget>\n4_500_000<\/output>\n100 - 10<\/gadget>\n90<\/output>\n4_500_000 \/ 90<\/gadget>\n50_000<\/output>\n50_000<\/result>","index":246} +{"problem":"jim ’ s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.35 for each 2 \/ 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ?","rationale":"\"let the fixed charge of jim ’ s taxi service = 2.25 $ and charge per 2 \/ 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 \/ . 4 ) * . 35 = 2.25 + 9 * . 35 = 5.4 $ answer e\"","correct":"e","options":{"a":"$ 3.15 ","b":"$ 4.45 ","c":"$ 4.80 ","d":"$ 5.05","e":"$ 5.40"},"options_float":{"a":3.15,"b":4.45,"c":4.8,"d":5.05,"e":5.4},"annotated_formula":"add(2.25, multiply(0.35, divide(3.6, divide(2, 5))))","linear_formula":"divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n3.6 \/ (2\/5)<\/gadget>\n9<\/output>\n0.35 * 9<\/gadget>\n3.15<\/output>\n2.25 + 3.15<\/gadget>\n5.4<\/output>\n5.4<\/result>","index":248} +{"problem":"fred and sam are standing 55 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 6 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ?","rationale":"\"relative distance = 55 miles relative speed = 6 + 5 = 11 miles per hour time taken = 55 \/ 11 = 5 hours distance travelled by sam = 5 * 5 = 25 miles = c\"","correct":"c","options":{"a":"5 ","b":"9 ","c":"25 ","d":"30","e":"45"},"options_float":{"a":5.0,"b":9.0,"c":25.0,"d":30.0,"e":45.0},"annotated_formula":"multiply(5, divide(55, add(6, 5)))","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"6 + 5<\/gadget>\n11<\/output>\n55 \/ 11<\/gadget>\n5<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25<\/result>","index":253} +{"problem":"a prize of $ 1200 is to be distributed among 40 winners , each of whom must be awarded at least $ 40 . if 2 \/ 5 of the prize will be distributed to 3 \/ 5 of the winners , what is the greatest possible individual award ?","rationale":"\"total value of the prize = $ 1200 number of people = 40 2 \/ 5 of 1200 ( = $ 480 ) should be distributed among 3 \/ 5 of 40 ( = 24 people ) with each getting $ 40 each . remaining money = 1200 - 480 = $ 720 . now in order to ' maximize ' 1 prize , we need to minimise the others and we have been given that each should get $ 40 . thus , minimising the remaining 15 people ( = 40 - 24 - 1 . ' - 1 ' to exclude 1 that needs to be maximised ) = 15 * 40 = 600 . thus the maximum award can be = 720 - 600 = $ 120 , hence b is the correct answer .\"","correct":"b","options":{"a":"$ 100 ","b":"$ 120 ","c":"$ 280 ","d":"$ 300","e":"$ 360"},"options_float":{"a":100.0,"b":120.0,"c":280.0,"d":300.0,"e":360.0},"annotated_formula":"subtract(subtract(1200, divide(multiply(1200, 2), 5)), multiply(subtract(subtract(40, divide(multiply(40, 3), 5)), const_1), 40))","linear_formula":"multiply(n0,n3)|multiply(n1,n5)|divide(#0,n4)|divide(#1,n4)|subtract(n0,#2)|subtract(n1,#3)|subtract(#5,const_1)|multiply(n1,#6)|subtract(#4,#7)|","chain":"1_200 * 2<\/gadget>\n2_400<\/output>\n2_400 \/ 5<\/gadget>\n480<\/output>\n1_200 - 480<\/gadget>\n720<\/output>\n40 * 3<\/gadget>\n120<\/output>\n120 \/ 5<\/gadget>\n24<\/output>\n40 - 24<\/gadget>\n16<\/output>\n16 - 1<\/gadget>\n15<\/output>\n15 * 40<\/gadget>\n600<\/output>\n720 - 600<\/gadget>\n120<\/output>\n120<\/result>","index":257} +{"problem":"the profits of qrs company rose 20 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ?","rationale":"\"assume 100 in march , then 120 in april as 20 % increase , then 96 in may as 20 % decrease from april , and then 144 in june which is 150 % of 96 . so overall increase is from 100 to 144 is 44 % answer c\"","correct":"c","options":{"a":"15 % ","b":"32 % ","c":"44 % ","d":"62 %","e":"80 %"},"options_float":{"a":15.0,"b":32.0,"c":44.0,"d":62.0,"e":80.0},"annotated_formula":"multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100)))), const_1))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(6\/5) * (4\/5)<\/gadget>\n24\/25 = around 0.96<\/output>\n(3\/2) * (24\/25)<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n100 * (11\/25)<\/gadget>\n44<\/output>\n44<\/result>","index":258} +{"problem":"if the speed of a man is 63 km per hour , then what is the distance traveled by him in 25 seconds ?","rationale":"\"the distance traveled in 25 sec = 63 * ( 5 \/ 18 ) * 25 = 437.5 m answer : d\"","correct":"d","options":{"a":"275 m ","b":"360 m ","c":"375 m ","d":"437.5 m","e":"440 m"},"options_float":{"a":275.0,"b":360.0,"c":375.0,"d":437.5,"e":440.0},"annotated_formula":"multiply(multiply(63, const_0_2778), 25)","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n63 * (5\/18)<\/gadget>\n35\/2 = around 17.5<\/output>\n(35\/2) * 25<\/gadget>\n875\/2 = around 437.5<\/output>\n875\/2 = around 437.5<\/result>","index":259} +{"problem":"if a sum of money doubles itself in 10 years at simple interest , the ratepercent per annum is","rationale":"\"explanation : let sum = x then simple interest = x rate = ( 100 * x ) \/ ( x * 10 ) = 10 option a\"","correct":"a","options":{"a":"10 ","b":"12.5 ","c":"13 ","d":"13.5","e":"14"},"options_float":{"a":10.0,"b":12.5,"c":13.0,"d":13.5,"e":14.0},"annotated_formula":"divide(divide(const_2, divide(10, const_100)), const_2)","linear_formula":"divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n2 \/ (1\/10)<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":261} +{"problem":"a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 7 miles per day and it will take them two days , how many miles long is the route down the mountain ?","rationale":"\"on the way down , the rate is 1.5 * 7 = 10.5 miles per day . the distance of the route down the mountain is 2 * 10.5 = 21 miles . the answer is d .\"","correct":"d","options":{"a":"18 ","b":"19 ","c":"20 ","d":"21","e":"22"},"options_float":{"a":18.0,"b":19.0,"c":20.0,"d":21.0,"e":22.0},"annotated_formula":"multiply(multiply(7, const_2), divide(const_3, const_2))","linear_formula":"divide(const_3,const_2)|multiply(n0,const_2)|multiply(#0,#1)|","chain":"7 * 2<\/gadget>\n14<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n14 * (3\/2)<\/gadget>\n21<\/output>\n21<\/result>","index":262} +{"problem":"2.08 can be expressed in terms of percentage as","rationale":"\"explanation : while calculation in terms of percentage we need to multiply by 100 , so 2.08 * 100 = 208 answer : option c\"","correct":"c","options":{"a":"2.08 % ","b":"20.8 % ","c":"208 % ","d":"0.208 %","e":"none of these"},"options_float":{"a":2.08,"b":20.8,"c":208.0,"d":0.208,"e":null},"annotated_formula":"multiply(2.08, const_100)","linear_formula":"multiply(n0,const_100)|","chain":"2.08 * 100<\/gadget>\n208<\/output>\n208<\/result>","index":264} +{"problem":"the population of a town is 8000 . it decreases annually at the rate of 30 % p . a . what will be its population after 2 years ?","rationale":"\"formula : ( after = 100 denominator ago = 100 numerator ) 8000 ã — 70 \/ 100 ã — 70 \/ 100 = 3920 answer : a\"","correct":"a","options":{"a":"3920 ","b":"5120 ","c":"5200 ","d":"5400","e":"5500"},"options_float":{"a":3920.0,"b":5120.0,"c":5200.0,"d":5400.0,"e":5500.0},"annotated_formula":"subtract(subtract(8000, multiply(8000, divide(30, const_100))), multiply(subtract(8000, multiply(8000, divide(30, const_100))), divide(30, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n8_000 * (3\/10)<\/gadget>\n2_400<\/output>\n8_000 - 2_400<\/gadget>\n5_600<\/output>\n5_600 * (3\/10)<\/gadget>\n1_680<\/output>\n5_600 - 1_680<\/gadget>\n3_920<\/output>\n3_920<\/result>","index":265} +{"problem":"a hall is 15 meters long and 10 meters wide . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , what is the volume of the hall ( in cubic meters ) ?","rationale":"\"2 hl + 2 hw = 2 lw h = lw \/ ( l + w ) volume = lwh = ( lw ) ^ 2 \/ ( l + w ) = 900 the answer is d .\"","correct":"d","options":{"a":"600 ","b":"700 ","c":"800 ","d":"900","e":"1000"},"options_float":{"a":600.0,"b":700.0,"c":800.0,"d":900.0,"e":1000.0},"annotated_formula":"volume_rectangular_prism(15, 10, divide(multiply(rectangle_area(15, 10), const_2), rectangle_perimeter(15, 10)))","linear_formula":"rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|multiply(#0,const_2)|divide(#2,#1)|volume_rectangular_prism(n0,n1,#3)|","chain":"15 * 10<\/gadget>\n150<\/output>\n150 * 2<\/gadget>\n300<\/output>\n2 * (15 + 10)<\/gadget>\n50<\/output>\n300 \/ 50<\/gadget>\n6<\/output>\n15 * 10 * 6<\/gadget>\n900<\/output>\n900<\/result>","index":266} +{"problem":"a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points e that an individual player might have scored ?","rationale":"\"general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points e for 12 th player is 100 - 77 = 23 . answer : e .\"","correct":"e","options":{"a":"7 ","b":"13 ","c":"16 ","d":"21","e":"23"},"options_float":{"a":7.0,"b":13.0,"c":16.0,"d":21.0,"e":23.0},"annotated_formula":"add(subtract(100, multiply(12, 7)), 7)","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|","chain":"12 * 7<\/gadget>\n84<\/output>\n100 - 84<\/gadget>\n16<\/output>\n16 + 7<\/gadget>\n23<\/output>\n23<\/result>","index":267} +{"problem":"a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1990 on her 32 nd birthday first be eligible to retire under this provision ?","rationale":"she must gain at least 70 points , now she has 31 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1990 + 19 = 2009 . answer : b .","correct":"b","options":{"a":"2003 ","b":"2009 ","c":"2005 ","d":"2006","e":"2007"},"options_float":{"a":2003.0,"b":2009.0,"c":2005.0,"d":2006.0,"e":2007.0},"annotated_formula":"add(1990, divide(subtract(70, 32), const_2))","linear_formula":"subtract(n0,n3)|divide(#0,const_2)|add(n2,#1)","chain":"70 - 32<\/gadget>\n38<\/output>\n38 \/ 2<\/gadget>\n19<\/output>\n1_990 + 19<\/gadget>\n2_009<\/output>\n2_009<\/result>","index":268} +{"problem":"the average age of 17 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ?","rationale":"\"age of the 15 th student = 17 * 15 - ( 14 * 5 + 16 * 9 ) = 255 - 214 = 41 years answer is b\"","correct":"b","options":{"a":"39 ","b":"41 ","c":"45 ","d":"42","e":"50"},"options_float":{"a":39.0,"b":41.0,"c":45.0,"d":42.0,"e":50.0},"annotated_formula":"subtract(subtract(multiply(17, 15), multiply(5, 14)), multiply(9, 16))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|subtract(#0,#1)|subtract(#3,#2)|","chain":"17 * 15<\/gadget>\n255<\/output>\n5 * 14<\/gadget>\n70<\/output>\n255 - 70<\/gadget>\n185<\/output>\n9 * 16<\/gadget>\n144<\/output>\n185 - 144<\/gadget>\n41<\/output>\n41<\/result>","index":269} +{"problem":"ajay can ride 50 km in 1 hour . in how many hours he can ride 1500 km ?","rationale":"\"1 hour he ride 50 km he ride 1500 km in = 1500 \/ 50 * 1 = 30 hours answer is a\"","correct":"a","options":{"a":"30 hrs ","b":"15 hrs ","c":"20 hrs ","d":"25 hrs","e":"18 hrs"},"options_float":{"a":30.0,"b":15.0,"c":20.0,"d":25.0,"e":18.0},"annotated_formula":"divide(1500, 50)","linear_formula":"divide(n2,n0)|","chain":"1_500 \/ 50<\/gadget>\n30<\/output>\n30<\/result>","index":270} +{"problem":"a square mirror has exactly half the area of the rectangular wall on which it is hung . if each side of the mirror is 54 inches and the width of the wall is 68 inches , what is the length of the wall , in inches ?","rationale":"since the mirror is 42 inches in all sides , it must be a square . area of a square is a = a ^ 2 ; 54 ^ 2 = 2916 . area of rectangle is double of that 2 * 2916 = 5832 . now a = lw and we need find w so a \/ l = w ; 5832 \/ 68 = 85.7 answer ! answer is d","correct":"d","options":{"a":"25.7 ","b":"72.7 ","c":"70.7 ","d":"85.7","e":"56.7"},"options_float":{"a":25.7,"b":72.7,"c":70.7,"d":85.7,"e":56.7},"annotated_formula":"divide(multiply(const_2, square_area(54)), 68)","linear_formula":"square_area(n0)|multiply(#0,const_2)|divide(#1,n1)","chain":"54 ** 2<\/gadget>\n2_916<\/output>\n2 * 2_916<\/gadget>\n5_832<\/output>\n5_832 \/ 68<\/gadget>\n1_458\/17 = around 85.764706<\/output>\n1_458\/17 = around 85.764706<\/result>","index":271} +{"problem":"the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee ’ s report , what is the probability that the two members selected will both be english teachers ?","rationale":"\"probability of first member an english teacher = 3 \/ 9 probability of second member an english teacher = 2 \/ 8 probability of both being english teacher = 3 \/ 9 x 2 \/ 8 = 1 \/ 12 ( d ) answer : d\"","correct":"d","options":{"a":"2 \/ 3 ","b":"1 \/ 3 ","c":"2 \/ 9 ","d":"1 \/ 12","e":"1 \/ 24"},"options_float":{"a":0.6666666667,"b":0.3333333333,"c":0.2222222222,"d":0.0833333333,"e":0.0416666667},"annotated_formula":"multiply(divide(3, add(add(3, 4), 2)), divide(2, subtract(add(add(3, 4), 2), const_1)))","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n0,#1)|subtract(#1,const_1)|divide(n2,#3)|multiply(#2,#4)|","chain":"3 + 4<\/gadget>\n7<\/output>\n7 + 2<\/gadget>\n9<\/output>\n3 \/ 9<\/gadget>\n1\/3 = around 0.333333<\/output>\n9 - 1<\/gadget>\n8<\/output>\n2 \/ 8<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/3) * (1\/4)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1\/12 = around 0.083333<\/result>","index":275} +{"problem":"a woman swims downstream 45 km and upstream 15 km taking 3 hours each time , what is the speed of the woman in still water ?","rationale":"45 - - - 3 ds = 15 ? - - - - 1 15 - - - - 3 us = 5 ? - - - - 1 m = ? m = ( 15 + 5 ) \/ 2 = 10 answer : a","correct":"a","options":{"a":"10 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"divide(add(divide(45, 3), divide(15, 3)), const_2)","linear_formula":"divide(n0,n2)|divide(n1,n2)|add(#0,#1)|divide(#2,const_2)","chain":"45 \/ 3<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n15 + 5<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":276} +{"problem":"in order to obtain an income of rs . 1000 from 50 % stock at rs . 500 , one must make an investment of","rationale":"\"explanation : market value = rs . 500 . required income = rs . 1000 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 50 ( ie , 50 % of the face value 100 ) , investment = rs . 500 to obtain rs . 1000 , investment = 500 \/ 50 ã — 1000 = 10000 answer : option d\"","correct":"d","options":{"a":"rs . 17000 ","b":"rs . 16050 ","c":"rs . 15400 ","d":"rs . 10000","e":"rs . 10500"},"options_float":{"a":17000.0,"b":16050.0,"c":15400.0,"d":10000.0,"e":10500.0},"annotated_formula":"multiply(divide(500, 50), 1000)","linear_formula":"divide(n2,n1)|multiply(n0,#0)|","chain":"500 \/ 50<\/gadget>\n10<\/output>\n10 * 1_000<\/gadget>\n10_000<\/output>\n10_000<\/result>","index":277} +{"problem":"jack and jill are marathon runners . jack can finish a marathon ( 40 km ) in 4.5 hours and jill can run a marathon in 4.0 hours . what is the ratio of their average running speed ? ( jack : jill )","rationale":"\"average speed of jack = distance \/ time = 40 \/ ( 9 \/ 2 ) = 80 \/ 9 average speed of jill = 40 \/ ( 4.0 ) = 10 ratio of average speed of jack to jill = ( 80 \/ 9 ) \/ 10 = 8 \/ 9 answer c\"","correct":"c","options":{"a":"14 \/ 15 ","b":"15 \/ 14 ","c":"8 \/ 9 ","d":"5 \/ 4","e":"can not be determined"},"options_float":{"a":0.9333333333,"b":1.0714285714,"c":0.8888888889,"d":1.25,"e":null},"annotated_formula":"divide(divide(40, 4.5), divide(40, 4.0))","linear_formula":"divide(n0,n1)|divide(n0,n2)|divide(#0,#1)|","chain":"40 \/ 4.5<\/gadget>\n8.888889<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n8.888889 \/ 10<\/gadget>\n0.888889<\/output>\n0.888889<\/result>","index":278} +{"problem":"5 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 3553 , how much was it in the beginning ?","rationale":"\"x * ( 95 \/ 100 ) * ( 80 \/ 100 ) = 3553 x = 4675 answer : d\"","correct":"d","options":{"a":"2988 ","b":"2776 ","c":"4400 ","d":"4675","e":"881"},"options_float":{"a":2988.0,"b":2776.0,"c":4400.0,"d":4675.0,"e":881.0},"annotated_formula":"floor(divide(3553, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 20), const_100))))","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(19\/20) * (4\/5)<\/gadget>\n19\/25 = around 0.76<\/output>\n3_553 \/ (19\/25)<\/gadget>\n4_675<\/output>\nfloor(4_675)<\/gadget>\n4_675<\/output>\n4_675<\/result>","index":281} +{"problem":"what is the sum of the odd integers from 35 to 55 , inclusive ?","rationale":"\"the mean is 45 . sum = mean ( # of elements ) there are 11 odd numbers between 35 - 55 inclusive . 11 * 45 = 495 a\"","correct":"a","options":{"a":"495 ","b":"550 ","c":"555 ","d":"600","e":"605"},"options_float":{"a":495.0,"b":550.0,"c":555.0,"d":600.0,"e":605.0},"annotated_formula":"divide(multiply(35, 55), const_4)","linear_formula":"multiply(n0,n1)|divide(#0,const_4)|","chain":"35 * 55<\/gadget>\n1_925<\/output>\n1_925 \/ 4<\/gadget>\n1_925\/4 = around 481.25<\/output>\n1_925\/4 = around 481.25<\/result>","index":282} +{"problem":"every monday , marina eats one croissant and every tuesday she eats two croissants . on each subsequent day of the week , she eats a number of croissants equal to the sum of the croissants eaten on the two previous days , with the exception that if she eats more than 4 croissants on any particular day , the next day she will eat only one croissant . at the end of the week ( which runs from monday through sunday ) , the cycle resets , and marina goes back to eating one croissant on monday , two on tuesday , and so forth . if a particular month begins on a saturday , how many croissants will marina eat on the 30 th of that month ?","rationale":"she eat as follow mon - 1 tue - 2 wed - 3 thr - 5 fri - 1 ( since she had more than 4 previous day ) sat - 6 sunday - 1 ( since she had more than 4 previous day ) so 30 th day of month she will have 1 . answer is a","correct":"a","options":{"a":"1 ","b":"2 ","c":"3 ","d":"5","e":"6"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"subtract(subtract(30, multiply(const_4, add(const_3, const_4))), const_1)","linear_formula":"add(const_3,const_4)|multiply(#0,const_4)|subtract(n1,#1)|subtract(#2,const_1)","chain":"3 + 4<\/gadget>\n7<\/output>\n4 * 7<\/gadget>\n28<\/output>\n30 - 28<\/gadget>\n2<\/output>\n2 - 1<\/gadget>\n1<\/output>\n1<\/result>","index":283} +{"problem":"if the sides of a cube are in the ratio 5 : 3 . what is the ratio of their diagonals ?","rationale":"explanation : diagonal of a cube = a √ 3 where a is side a 1 : a 2 = 5 : 3 d 1 : d 2 = 5 : 3 where √ 3 cancelled both side answer : e","correct":"e","options":{"a":"5 : 9 ","b":"9 : 3 ","c":"5 : 4 ","d":"3 : 5","e":"5 : 3"},"options_float":{"a":0.5555555556,"b":3.0,"c":1.25,"d":0.6,"e":1.6666666667},"annotated_formula":"divide(5, 3)","linear_formula":"divide(n0,n1)","chain":"5 \/ 3<\/gadget>\n5\/3 = around 1.666667<\/output>\n5\/3 = around 1.666667<\/result>","index":285} +{"problem":"the average weight of 26 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class .","rationale":"\"explanation : average weight of 26 boys = 50.25 total weight of 26 boys = 50.25 × 26 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 26 ) + ( 45.15 × 8 ) total boys = 26 + 8 = 34 average weight of all the boys = ( 50.25 × 26 ) + ( 45.15 × 8 ) \/ 34 = 49.05 answer : option a\"","correct":"a","options":{"a":"49.05 ","b":"42.25 ","c":"50 ","d":"51.25","e":"52.25"},"options_float":{"a":49.05,"b":42.25,"c":50.0,"d":51.25,"e":52.25},"annotated_formula":"divide(add(multiply(26, 50.25), multiply(8, 45.15)), add(26, 8))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"26 * 50.25<\/gadget>\n1_306.5<\/output>\n8 * 45.15<\/gadget>\n361.2<\/output>\n1_306.5 + 361.2<\/gadget>\n1_667.7<\/output>\n26 + 8<\/gadget>\n34<\/output>\n1_667.7 \/ 34<\/gadget>\n49.05<\/output>\n49.05<\/result>","index":286} +{"problem":"a one - foot stick is marked in 1 \/ 4 and 1 \/ 5 portion . how many total markings will there be , including the end points ?","rationale":"lcm of 20 = 20 1 \/ 4 marking are ( table of 5 ) 0 . . . . . . 5 . . . . . . 10 . . . . . 15 . . . . . . 20 ( total = 5 ) 1 \/ 5 marking are ( table of 4 ) 0 . . . . . . . 4 . . . . . . 8 . . . . . . 12 . . . . . . . . 16 . . . . . . 20 ( total = 6 ) overlapping markings are 0 . . . . . . . . 20 ( total = 2 ) total markings = 5 + 6 - 2 = 9 answer = d","correct":"d","options":{"a":"18 ","b":"16 ","c":"13 ","d":"9","e":"11"},"options_float":{"a":18.0,"b":16.0,"c":13.0,"d":9.0,"e":11.0},"annotated_formula":"add(4, 5)","linear_formula":"add(n1,n3)","chain":"4 + 5<\/gadget>\n9<\/output>\n9<\/result>","index":287} +{"problem":"find out the c . i on rs . 5000 at 4 % p . a . compound half - yearly for 1 1 \/ 2 years ?","rationale":"\"explanation : a = 5000 ( 51 \/ 50 ) 3 = 5306.04 5000 - - - - - - - - - - - 306.04 answer : c\"","correct":"c","options":{"a":"272.22 ","b":"128.29 ","c":"306.04 ","d":"198.39","e":"212.22"},"options_float":{"a":272.22,"b":128.29,"c":306.04,"d":198.39,"e":212.22},"annotated_formula":"subtract(multiply(5000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 5000)","linear_formula":"divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(26\/25) ** (3\/2)<\/gadget>\n26*sqrt(26)\/125 = around 1.060596<\/output>\n5_000 * (26*sqrt(26)\/125)<\/gadget>\n1040*sqrt(26) = around 5_302.980294<\/output>\n(1040*sqrt(26)) - 5_000<\/gadget>\n-5000 + 1040*sqrt(26) = around 302.980294<\/output>\n-5000 + 1040*sqrt(26) = around 302.980294<\/result>","index":288} +{"problem":"the average score of a cricketer in 2 matches is 30 and in other 3 matches is 40 . then find the average score in all the 5 matches ?","rationale":"average in 5 matches = ( 2 * 30 + 3 * 40 ) \/ 2 + 3 = 60 + 120 \/ 5 = 180 \/ 5 = 36 answer is e","correct":"e","options":{"a":"25 ","b":"27 ","c":"30 ","d":"35","e":"36"},"options_float":{"a":25.0,"b":27.0,"c":30.0,"d":35.0,"e":36.0},"annotated_formula":"divide(add(multiply(2, 30), multiply(3, 40)), 5)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)","chain":"2 * 30<\/gadget>\n60<\/output>\n3 * 40<\/gadget>\n120<\/output>\n60 + 120<\/gadget>\n180<\/output>\n180 \/ 5<\/gadget>\n36<\/output>\n36<\/result>","index":289} +{"problem":"in what ratio must tea at rs . 62 per kg be mixed with tea at rs . 72 per kg so that the mixture must be worth rs . 64.50 per kg ?","rationale":"mean price = ( 72 - 64.5 = 7.5 ) ( 64.5 - 62 = 2.5 ) req ratio = 7.5 : 2.5 = 3 : 1 answer a","correct":"a","options":{"a":"3 : 1 ","b":"2 : 1 ","c":"5 : 2 ","d":"3 : 2","e":"3 : 4"},"options_float":{"a":3.0,"b":2.0,"c":2.5,"d":1.5,"e":0.75},"annotated_formula":"divide(subtract(72, 64.5), subtract(64.5, 62))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)","chain":"72 - 64.5<\/gadget>\n7.5<\/output>\n64.5 - 62<\/gadget>\n2.5<\/output>\n7.5 \/ 2.5<\/gadget>\n3<\/output>\n3<\/result>","index":291} +{"problem":"a solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm . the height of the cone is :","rationale":"sol . let the height of the cone be h cm . then , 1 \/ 3 ∏ * 6 * 6 * h = 4 \/ 3 ∏ * 3 * 3 * 3 ⇒ h = [ 36 * 3 \/ 36 ] = 3 cm . answer d","correct":"d","options":{"a":"8 cm ","b":"6 cm ","c":"4 cm ","d":"3 cm","e":"none"},"options_float":{"a":8.0,"b":6.0,"c":4.0,"d":3.0,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(const_4, const_3), const_3), const_3), multiply(6, 6))","linear_formula":"multiply(const_3,const_4)|multiply(n0,n0)|multiply(#0,const_3)|multiply(#2,const_3)|divide(#3,#1)","chain":"4 * 3<\/gadget>\n12<\/output>\n12 * 3<\/gadget>\n36<\/output>\n36 * 3<\/gadget>\n108<\/output>\n6 * 6<\/gadget>\n36<\/output>\n108 \/ 36<\/gadget>\n3<\/output>\n3<\/result>","index":292} +{"problem":"if x is the median of the set { 8 \/ 2 , 11 \/ 3 , 27 \/ 9 , 21 \/ 5 , x } , x could be","rationale":"the median is the middle number once all the numbers are arranged in increasing \/ decreasing order . we see that 11 \/ 3 = 3 . something , 28 \/ 9 = 3 . something 21 \/ 5 = 4 . something 9 \/ 2 = 4 . something so x should greater than the smallest two numbers and smaller than the greatest two numbers . we can see that x = 4 is possible . ( first look at the simplest option or the middle option since options are usually arranged in increasing \/ decreasing order ) answer ( a )","correct":"a","options":{"a":"4 ","b":"17 \/ 5 ","c":"16 \/ 5 ","d":"30 \/ 7","e":"31 \/ 7"},"options_float":{"a":4.0,"b":3.4,"c":3.2,"d":4.2857142857,"e":4.4285714286},"annotated_formula":"add(add(divide(add(add(add(divide(8, 2), divide(11, 3)), divide(27, 9)), divide(21, 5)), const_4), divide(const_10, const_100)), divide(const_10, const_100))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|divide(n6,n7)|divide(const_10,const_100)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,const_4)|add(#8,#4)|add(#9,#4)","chain":"8 \/ 2<\/gadget>\n4<\/output>\n11 \/ 3<\/gadget>\n11\/3 = around 3.666667<\/output>\n4 + (11\/3)<\/gadget>\n23\/3 = around 7.666667<\/output>\n27 \/ 9<\/gadget>\n3<\/output>\n(23\/3) + 3<\/gadget>\n32\/3 = around 10.666667<\/output>\n21 \/ 5<\/gadget>\n21\/5 = around 4.2<\/output>\n(32\/3) + (21\/5)<\/gadget>\n223\/15 = around 14.866667<\/output>\n(223\/15) \/ 4<\/gadget>\n223\/60 = around 3.716667<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(223\/60) + (1\/10)<\/gadget>\n229\/60 = around 3.816667<\/output>\n(229\/60) + (1\/10)<\/gadget>\n47\/12 = around 3.916667<\/output>\n47\/12 = around 3.916667<\/result>","index":294} +{"problem":"how many seconds will a train 100 meters long take to cross a bridge 140 meters long if the speed of the train is 36 kmph ?","rationale":"\"explanation : d = 100 + 140 = 240 s = 36 * 5 \/ 18 = 10 mps t = 240 \/ 10 = 24 sec answer : option b\"","correct":"b","options":{"a":"26 ","b":"24 ","c":"25 ","d":"82","e":"27"},"options_float":{"a":26.0,"b":24.0,"c":25.0,"d":82.0,"e":27.0},"annotated_formula":"divide(add(140, 100), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"140 + 100<\/gadget>\n240<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n240 \/ 10<\/gadget>\n24<\/output>\n24<\/result>","index":295} +{"problem":"while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 6 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ?","rationale":"\"both computers together process files at a rate of 240 \/ 12 + 240 \/ 6 = 20 + 40 = 60 files per hour . the time required to process 240 files is 240 \/ 60 = 4 hours the answer is c .\"","correct":"c","options":{"a":"3 ","b":"3.5 ","c":"4 ","d":"4.5","e":"5"},"options_float":{"a":3.0,"b":3.5,"c":4.0,"d":4.5,"e":5.0},"annotated_formula":"divide(240, add(divide(240, 12), divide(240, 6)))","linear_formula":"divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(n0,#2)|","chain":"240 \/ 12<\/gadget>\n20<\/output>\n240 \/ 6<\/gadget>\n40<\/output>\n20 + 40<\/gadget>\n60<\/output>\n240 \/ 60<\/gadget>\n4<\/output>\n4<\/result>","index":296} +{"problem":"in town x , 64 percent of the population are employed , and 46 percent of the population are employed males . what percent of the employed people in town x are females ?","rationale":"we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 46 are employed males , hence 18 % are employed females . ( employed females ) \/ ( total employed people ) = 18 \/ 64 = 28 % answer : c .","correct":"c","options":{"a":"16 % ","b":"25 % ","c":"28 % ","d":"40 %","e":"52 %"},"options_float":{"a":16.0,"b":25.0,"c":28.0,"d":40.0,"e":52.0},"annotated_formula":"multiply(divide(subtract(64, 46), 64), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)","chain":"64 - 46<\/gadget>\n18<\/output>\n18 \/ 64<\/gadget>\n9\/32 = around 0.28125<\/output>\n(9\/32) * 100<\/gadget>\n225\/8 = around 28.125<\/output>\n225\/8 = around 28.125<\/result>","index":298} +{"problem":"the lcm of two numbers is 2310 and hcf is 55 . if one of the numbers is 210 . then what is the other number ?","rationale":"\"first number * second number = lcm * hcf other number = 2310 * 55 \/ 210 = 11 * 55 = 605 answer : d\"","correct":"d","options":{"a":"715 ","b":"825 ","c":"330 ","d":"605","e":"465"},"options_float":{"a":715.0,"b":825.0,"c":330.0,"d":605.0,"e":465.0},"annotated_formula":"divide(multiply(2310, 55), 210)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"2_310 * 55<\/gadget>\n127_050<\/output>\n127_050 \/ 210<\/gadget>\n605<\/output>\n605<\/result>","index":299} +{"problem":"a and b invests rs . 2500 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?","rationale":"\"( 2.5 * 6 + 5 * 6 ) : ( 4 * 12 ) 45 : 48 = > 15 : 16 answer : b\"","correct":"b","options":{"a":"11 : 5 ","b":"15 : 16 ","c":"15 : 2 ","d":"9 : 9","e":"11 : 1"},"options_float":{"a":2.2,"b":0.9375,"c":7.5,"d":1.0,"e":11.0},"annotated_formula":"divide(add(multiply(2500, 6), multiply(multiply(2500, const_2), 6)), multiply(4000, add(6, 6)))","linear_formula":"add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)|","chain":"2_500 * 6<\/gadget>\n15_000<\/output>\n2_500 * 2<\/gadget>\n5_000<\/output>\n5_000 * 6<\/gadget>\n30_000<\/output>\n15_000 + 30_000<\/gadget>\n45_000<\/output>\n6 + 6<\/gadget>\n12<\/output>\n4_000 * 12<\/gadget>\n48_000<\/output>\n45_000 \/ 48_000<\/gadget>\n15\/16 = around 0.9375<\/output>\n15\/16 = around 0.9375<\/result>","index":301} +{"problem":"a car traveled from san diego to san francisco at an average speed of 54 miles per hour . if the journey back took twice as long , what was the average speed of the trip ?","rationale":"let the time taken be = x one way distance = 54 x total distance traveled = 2 * 54 x = 108 x total time taken = x + 2 x = 3 x average speed = 108 x \/ 3 x = 36 answer : c","correct":"c","options":{"a":"24 . ","b":"32 . ","c":"36 . ","d":"42 .","e":"44 ."},"options_float":{"a":24.0,"b":32.0,"c":36.0,"d":42.0,"e":44.0},"annotated_formula":"inverse(add(inverse(54), divide(inverse(54), const_2)))","linear_formula":"inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)","chain":"1 \/ 54<\/gadget>\n1\/54 = around 0.018519<\/output>\n(1\/54) \/ 2<\/gadget>\n1\/108 = around 0.009259<\/output>\n(1\/54) + (1\/108)<\/gadget>\n1\/36 = around 0.027778<\/output>\n1 \/ (1\/36)<\/gadget>\n36<\/output>\n36<\/result>","index":303} +{"problem":"a runner runs the 40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot , and continues to run at half her previous speed . if the second half takes her 8 hours longer than the first half , how many hours did it take the runner to run the second half ?","rationale":"\"the runner runs the first 20 miles at speed v and the second 20 miles at speed v \/ 2 . the time t 2 to run the second half must be twice the time t 1 to run the first half . t 2 = 2 * t 1 = t 1 + 8 t 1 = 8 and so t 2 = 16 . the answer is c .\"","correct":"c","options":{"a":"10 ","b":"12 ","c":"16 ","d":"20","e":"24"},"options_float":{"a":10.0,"b":12.0,"c":16.0,"d":20.0,"e":24.0},"annotated_formula":"divide(40, divide(divide(40, const_2), 8))","linear_formula":"divide(n0,const_2)|divide(#0,n1)|divide(n0,#1)|","chain":"40 \/ 2<\/gadget>\n20<\/output>\n20 \/ 8<\/gadget>\n5\/2 = around 2.5<\/output>\n40 \/ (5\/2)<\/gadget>\n16<\/output>\n16<\/result>","index":304} +{"problem":"a pump will fill a tank in 3 hours . because of a leak in the tank , it took 3 hours 30 min to fill the tank . in what time the leak can drain out all the water of the tank and will make tank empty ?","rationale":"work done for 1 hr without leak = 1 \/ 3 work done with leak = 31 \/ 2 = = > 7 \/ 2 work and leak in 1 hr = 2 \/ 7 work leak in 1 hr = 1 \/ 3 - 2 \/ 7 = 1 \/ 21 so 21 hrs answer b","correct":"b","options":{"a":"20 hrs ","b":"21 hrs ","c":"22 hrs ","d":"23 hrs","e":"24 hrs"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":23.0,"e":24.0},"annotated_formula":"divide(multiply(divide(add(multiply(3, const_60), 30), 30), multiply(3, const_60)), const_60)","linear_formula":"multiply(n0,const_60)|add(n2,#0)|divide(#1,n2)|multiply(#2,#0)|divide(#3,const_60)","chain":"3 * 60<\/gadget>\n180<\/output>\n180 + 30<\/gadget>\n210<\/output>\n210 \/ 30<\/gadget>\n7<\/output>\n7 * 180<\/gadget>\n1_260<\/output>\n1_260 \/ 60<\/gadget>\n21<\/output>\n21<\/result>","index":305} +{"problem":"the area of a square garden is q square feet and the perimeter is p feet . if q = p + 21 , what is the perimeter of the garden in feet ?","rationale":"let x be the length of one side of the square garden . x ^ 2 = 4 x + 21 x ^ 2 - 4 x - 21 = 0 ( x - 7 ) ( x + 3 ) = 0 x = 7 , - 3 p = 4 ( 7 ) = 28 the answer is b .","correct":"b","options":{"a":"24 ","b":"28 ","c":"32 ","d":"36","e":"40"},"options_float":{"a":24.0,"b":28.0,"c":32.0,"d":36.0,"e":40.0},"annotated_formula":"multiply(divide(21, subtract(const_4, const_1)), const_4)","linear_formula":"subtract(const_4,const_1)|divide(n0,#0)|multiply(#1,const_4)","chain":"4 - 1<\/gadget>\n3<\/output>\n21 \/ 3<\/gadget>\n7<\/output>\n7 * 4<\/gadget>\n28<\/output>\n28<\/result>","index":306} +{"problem":"how many multiples of 3 are there between 10 and 100 , exclusive ?","rationale":"\"3 * 4 = 12 3 * 33 = 99 total multiples = ( 33 - 4 ) + 1 = 30 exclude 10 and 100 = 30 answer is a\"","correct":"a","options":{"a":"30 ","b":"29 ","c":"28 ","d":"31","e":"33"},"options_float":{"a":30.0,"b":29.0,"c":28.0,"d":31.0,"e":33.0},"annotated_formula":"add(divide(subtract(100, 10), 3), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n30 + 1<\/gadget>\n31<\/output>\n31<\/result>","index":309} +{"problem":"an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 28 gallons of ethanol , how many gallons of fuel a were added ?","rationale":"\"say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 28 gallons then 0.12 a + 0.16 ( 200 - a ) = 28 - - > a = 100 . answer : c .\"","correct":"c","options":{"a":"160 ","b":"150 ","c":"100 ","d":"80","e":"50"},"options_float":{"a":160.0,"b":150.0,"c":100.0,"d":80.0,"e":50.0},"annotated_formula":"divide(subtract(multiply(200, divide(16, const_100)), 28), subtract(divide(16, const_100), divide(12, const_100)))","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)|","chain":"16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n200 * (4\/25)<\/gadget>\n32<\/output>\n32 - 28<\/gadget>\n4<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n(4\/25) - (3\/25)<\/gadget>\n1\/25 = around 0.04<\/output>\n4 \/ (1\/25)<\/gadget>\n100<\/output>\n100<\/result>","index":310} +{"problem":"maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 14 kilometers , maxwell ' s walking speed is 4 km \/ h , and brad ' s running speed is 6 km \/ h . what is the total time it takes maxwell before he meets up with brad ?","rationale":"total distance = 14 kms maxwell speed = 4 kms \/ hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance \/ relative speed total distance after brad started = 10 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms \/ hr time taken to meet brad after brad started = 10 \/ 10 = 1 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 1 = 4 + 4 = 8 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell \/ maxwell ' s speed = 8 \/ 4 = 2 hrs . . . answer a","correct":"a","options":{"a":"2 ","b":"4 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(add(14, 6), add(4, 6))","linear_formula":"add(n0,n2)|add(n1,n2)|divide(#0,#1)","chain":"14 + 6<\/gadget>\n20<\/output>\n4 + 6<\/gadget>\n10<\/output>\n20 \/ 10<\/gadget>\n2<\/output>\n2<\/result>","index":311} +{"problem":"lloyd normally works 7.5 hours per day and earns $ 3.50 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 1.5 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ?","rationale":"\"daily working hour * regular rate + overtime * increased rate 7.5 * 3.5 + 3 * 3.5 * 1.5 = 42 answer b\"","correct":"b","options":{"a":"$ 33.75 ","b":"$ 42.00 ","c":"$ 51.75 ","d":"$ 54.00","e":"$ 70.00"},"options_float":{"a":33.75,"b":42.0,"c":51.75,"d":54.0,"e":70.0},"annotated_formula":"add(multiply(7.5, 3.50), multiply(multiply(subtract(10.5, 7.5), 1.5), 3.50))","linear_formula":"multiply(n0,n1)|subtract(n4,n0)|multiply(n3,#1)|multiply(n1,#2)|add(#0,#3)|","chain":"7.5 * 3.5<\/gadget>\n26.25<\/output>\n10.5 - 7.5<\/gadget>\n3<\/output>\n3 * 1.5<\/gadget>\n4.5<\/output>\n4.5 * 3.5<\/gadget>\n15.75<\/output>\n26.25 + 15.75<\/gadget>\n42<\/output>\n42<\/result>","index":314} +{"problem":"tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , how many e kilometers per hour did tammy walk on the second day ?","rationale":"ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its simple algebra for s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 but for some reason im getting 3.5 and not 4.5 . 8 s + 6 s + 3 = 52 14 s = 49 s = 3.5","correct":"c","options":{"a":"3 ","b":"3.5 ","c":"4 ","d":"4.5","e":"6"},"options_float":{"a":3.0,"b":3.5,"c":4.0,"d":4.5,"e":6.0},"annotated_formula":"add(divide(subtract(52, divide(subtract(divide(add(14, 2), const_2), 2), const_2)), add(subtract(divide(add(14, 2), const_2), 2), divide(add(14, 2), const_2))), divide(const_1, const_2))","linear_formula":"add(n0,n1)|divide(const_1,const_2)|divide(#0,const_2)|subtract(#2,n1)|add(#2,#3)|divide(#3,const_2)|subtract(n2,#5)|divide(#6,#4)|add(#7,#1)","chain":"14 + 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8 - 2<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n52 - 3<\/gadget>\n49<\/output>\n6 + 8<\/gadget>\n14<\/output>\n49 \/ 14<\/gadget>\n7\/2 = around 3.5<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(7\/2) + (1\/2)<\/gadget>\n4<\/output>\n4<\/result>","index":315} +{"problem":"the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 462 the other is","rationale":"\"the other number = l . c . m * h . c . f \/ given number = 2310 * 30 \/ 462 = 150 answer is d .\"","correct":"d","options":{"a":"330 ","b":"300 ","c":"270 ","d":"150","e":"350"},"options_float":{"a":330.0,"b":300.0,"c":270.0,"d":150.0,"e":350.0},"annotated_formula":"divide(multiply(30, 2310), 462)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"30 * 2_310<\/gadget>\n69_300<\/output>\n69_300 \/ 462<\/gadget>\n150<\/output>\n150<\/result>","index":316} +{"problem":"a person incurs a loss of 5 % be selling a watch for rs . 1140 . at what price should the watch be sold to earn 5 % profit .","rationale":"explanation : let the new s . p . be x , then . ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) = > ( 951 \/ 140 = 105 \/ x ) = > x = 1260 answer is c","correct":"c","options":{"a":"rs . 1200 ","b":"rs . 1230 ","c":"rs . 1260 ","d":"rs . 1290","e":"rs . 1270"},"options_float":{"a":1200.0,"b":1230.0,"c":1260.0,"d":1290.0,"e":1270.0},"annotated_formula":"multiply(divide(1140, divide(subtract(const_100, 5), const_100)), divide(add(const_100, 5), const_100))","linear_formula":"add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|divide(n1,#2)|multiply(#4,#3)","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n1_140 \/ (19\/20)<\/gadget>\n1_200<\/output>\n100 + 5<\/gadget>\n105<\/output>\n105 \/ 100<\/gadget>\n21\/20 = around 1.05<\/output>\n1_200 * (21\/20)<\/gadget>\n1_260<\/output>\n1_260<\/result>","index":318} +{"problem":"patanjali walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day , but at the same speed as on the second day . how many miles in total did she walk ?","rationale":"patanjali walked 18 miles on the first day , walking 3 miles per hour i . e . total time of walk on day - 1 = 18 \/ 3 = 6 hours second day time of walk = 6 - 1 = 5 hours and speed = 3 + 1 = 4 miles per hour i . e . distance walked on second day = 5 * 4 = 20 miles third day time of walk = 6 hours and speed = 4 miles per hour i . e . distance walked on second day = 6 * 4 = 24 miles total distance travelled on three days = 18 + 20 + 24 = 62 answer : option e","correct":"e","options":{"a":"24 ","b":"44 ","c":"58 ","d":"60","e":"62"},"options_float":{"a":24.0,"b":44.0,"c":58.0,"d":60.0,"e":62.0},"annotated_formula":"add(add(18, multiply(add(const_2, const_3), add(3, const_1))), multiply(add(add(const_2, const_3), const_1), add(3, const_1)))","linear_formula":"add(const_2,const_3)|add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)|add(n1,#3)|multiply(#2,#1)|add(#4,#5)","chain":"2 + 3<\/gadget>\n5<\/output>\n3 + 1<\/gadget>\n4<\/output>\n5 * 4<\/gadget>\n20<\/output>\n18 + 20<\/gadget>\n38<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6 * 4<\/gadget>\n24<\/output>\n38 + 24<\/gadget>\n62<\/output>\n62<\/result>","index":319} +{"problem":"the current in a river is 10 mph . a boat can travel 50 mph in still water . how far up the river can the boat travel if the round trip is to take 5 hours ?","rationale":"upstream speed = 50 - 10 = 40 mph downstream speed = 50 + 10 = 60 mph d \/ 40 + d \/ 60 = 5 hours solving for d we get d = 120 answer : a","correct":"a","options":{"a":"120 ","b":"100 ","c":"96 ","d":"85","e":"150"},"options_float":{"a":120.0,"b":100.0,"c":96.0,"d":85.0,"e":150.0},"annotated_formula":"divide(multiply(multiply(subtract(50, 10), add(50, 10)), 5), add(add(50, 10), subtract(50, 10)))","linear_formula":"add(n0,n1)|subtract(n1,n0)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)","chain":"50 - 10<\/gadget>\n40<\/output>\n50 + 10<\/gadget>\n60<\/output>\n40 * 60<\/gadget>\n2_400<\/output>\n2_400 * 5<\/gadget>\n12_000<\/output>\n60 + 40<\/gadget>\n100<\/output>\n12_000 \/ 100<\/gadget>\n120<\/output>\n120<\/result>","index":320} +{"problem":"60 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is :","rationale":"\"explanation : let the two parts be ( 60 - x ) and x . then , 10 ( 60 - x ) + 22 x = 780 = > 12 x = 180 = > x = 15 . bigger part = ( 60 - x ) = 45 . answer : b ) 45\"","correct":"b","options":{"a":"33 ","b":"45 ","c":"26 ","d":"28","e":"19"},"options_float":{"a":33.0,"b":45.0,"c":26.0,"d":28.0,"e":19.0},"annotated_formula":"subtract(60, divide(subtract(780, multiply(60, 10)), subtract(22, 10)))","linear_formula":"multiply(n0,n1)|subtract(n2,n1)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3)|","chain":"60 * 10<\/gadget>\n600<\/output>\n780 - 600<\/gadget>\n180<\/output>\n22 - 10<\/gadget>\n12<\/output>\n180 \/ 12<\/gadget>\n15<\/output>\n60 - 15<\/gadget>\n45<\/output>\n45<\/result>","index":321} +{"problem":"a library branch originally contained 18360 volumes , 30 % of which were fiction novels . 1 \/ 3 of the volumes were transferred to another location and 1 \/ 5 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ?","rationale":"fiction novels = 5,508 transferred to another location = 6,120 transferred fiction novels = 1,101 non transferred fiction novels = 4,407 percent of the remaining collection was fiction novels = 4,407 \/ ( 18360 - 6120 ) * 100 = > 36.004 . . . % hence answer will be ( d )","correct":"d","options":{"a":"2.5 % ","b":"17.67 % ","c":"28.3 % ","d":"36 %","e":"73.6 %"},"options_float":{"a":2.5,"b":17.67,"c":28.3,"d":36.0,"e":73.6},"annotated_formula":"multiply(divide(multiply(divide(30, const_100), subtract(const_1, divide(1, 5))), subtract(const_1, divide(const_1, const_3))), const_100)","linear_formula":"divide(n1,const_100)|divide(n2,n5)|divide(const_1,const_3)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#0,#3)|divide(#5,#4)|multiply(#6,const_100)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(3\/10) * (4\/5)<\/gadget>\n6\/25 = around 0.24<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(6\/25) \/ (2\/3)<\/gadget>\n9\/25 = around 0.36<\/output>\n(9\/25) * 100<\/gadget>\n36<\/output>\n36<\/result>","index":323} +{"problem":"there is one 4 - digit whole number n , such that the last 4 digits of n 2 are in fact the original number n","rationale":"c 9376 looking at the last digit , the last digit must be either 0 , 1 , 5 or 6 . then looking at the last two digits , the last two digits must be either 0001 , 25 or 76 . then looking at the last three digits , the last three digits must be either 00001 , 625 or 376 . then looking at the last four digits , the last four digits must be either 0000001 , 0625 or 9376 . out of those , only 9376 is a 4 digit number","correct":"c","options":{"a":"8476 ","b":"7376 ","c":"9376 ","d":"7378","e":"9466"},"options_float":{"a":8476.0,"b":7376.0,"c":9376.0,"d":7378.0,"e":9466.0},"annotated_formula":"add(multiply(subtract(const_12, const_3), const_1000), add(multiply(const_3, const_100), multiply(subtract(multiply(const_12, const_4), const_10), 2)))","linear_formula":"multiply(const_100,const_3)|multiply(const_12,const_4)|subtract(const_12,const_3)|multiply(#2,const_1000)|subtract(#1,const_10)|multiply(n2,#4)|add(#0,#5)|add(#6,#3)","chain":"12 - 3<\/gadget>\n9<\/output>\n9 * 1_000<\/gadget>\n9_000<\/output>\n3 * 100<\/gadget>\n300<\/output>\n12 * 4<\/gadget>\n48<\/output>\n48 - 10<\/gadget>\n38<\/output>\n38 * 2<\/gadget>\n76<\/output>\n300 + 76<\/gadget>\n376<\/output>\n9_000 + 376<\/gadget>\n9_376<\/output>\n9_376<\/result>","index":324} +{"problem":"when two dice are tossed , what is the probability that the total score is a prime number ?","rationale":"total number of outcomes possible = 36 probability of total score is a prime number = 15 p ( e ) = 15 \/ 36 = 5 \/ 12 answer : c","correct":"c","options":{"a":"1 \/ 6 ","b":"3 ","c":"5 \/ 12 ","d":"7 \/ 12","e":"9 \/ 12"},"options_float":{"a":0.1666666667,"b":3.0,"c":0.4166666667,"d":0.5833333333,"e":0.75},"annotated_formula":"divide(add(multiply(const_6, const_2), const_3), multiply(const_6, const_6))","linear_formula":"multiply(const_2,const_6)|multiply(const_6,const_6)|add(#0,const_3)|divide(#2,#1)","chain":"6 * 2<\/gadget>\n12<\/output>\n12 + 3<\/gadget>\n15<\/output>\n6 * 6<\/gadget>\n36<\/output>\n15 \/ 36<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":325} +{"problem":"running at the same constant rate , 6 identical machines can produce a total of 420 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ?","rationale":"\"let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 420 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 420 x = ( 10 x 4 x 420 ) \/ ( 6 ) x = 2800 . answer : c\"","correct":"c","options":{"a":"648 ","b":"1800 ","c":"2800 ","d":"10800","e":"10900"},"options_float":{"a":648.0,"b":1800.0,"c":2800.0,"d":10800.0,"e":10900.0},"annotated_formula":"multiply(multiply(divide(420, 6), 4), 10)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"420 \/ 6<\/gadget>\n70<\/output>\n70 * 4<\/gadget>\n280<\/output>\n280 * 10<\/gadget>\n2_800<\/output>\n2_800<\/result>","index":326} +{"problem":"there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 5 days more ?","rationale":"\"760 - - - - 22 760 - - - - 20 x - - - - - 5 x * 5 = 760 * 20 x = 3040 760 - - - - - - - 2280 answer : a\"","correct":"a","options":{"a":"2280 ","b":"2240 ","c":"2299 ","d":"2288","e":"2266"},"options_float":{"a":2280.0,"b":2240.0,"c":2299.0,"d":2288.0,"e":2266.0},"annotated_formula":"subtract(divide(multiply(760, subtract(22, const_2)), 5), 760)","linear_formula":"subtract(n1,const_2)|multiply(n0,#0)|divide(#1,n2)|subtract(#2,n0)|","chain":"22 - 2<\/gadget>\n20<\/output>\n760 * 20<\/gadget>\n15_200<\/output>\n15_200 \/ 5<\/gadget>\n3_040<\/output>\n3_040 - 760<\/gadget>\n2_280<\/output>\n2_280<\/result>","index":327} +{"problem":"the ratio of the volumes of two cubes is 216 : 343 . what is the ratio of their total surface areas ?","rationale":"\"ratio of the sides = ³ √ 216 : ³ √ 343 = 6 : 7 ratio of surface areas = 62 : 72 = 31 : 36 answer : b\"","correct":"b","options":{"a":"81 : 12 ","b":"31 : 36 ","c":"81 : 22 ","d":"8 : 19","e":"1 : 11"},"options_float":{"a":6.75,"b":0.8611111111,"c":3.6818181818,"d":0.4210526316,"e":0.0909090909},"annotated_formula":"power(divide(216, 343), divide(const_1, const_3))","linear_formula":"divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)|","chain":"216 \/ 343<\/gadget>\n216\/343 = around 0.629738<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(216\/343) ** (1\/3)<\/gadget>\n6\/7 = around 0.857143<\/output>\n6\/7 = around 0.857143<\/result>","index":328} +{"problem":"the length and breadth of a rectangle is increased by 10 % and 30 % respectively . what is the increase in the area ?","rationale":"\"100 * 100 = 10000 110 * 130 = 14300 - - - - - - - - - - 4300 10000 - - - - - - - 4300 100 - - - - - - - 43.00 answer a\"","correct":"a","options":{"a":"43.0 ","b":"44.0 ","c":"43.5 ","d":"45.0","e":"42.0"},"options_float":{"a":43.0,"b":44.0,"c":43.5,"d":45.0,"e":42.0},"annotated_formula":"divide(multiply(subtract(rectangle_area(add(10, const_100), add(30, const_100)), rectangle_area(const_100, const_100)), const_100), rectangle_area(const_100, const_100))","linear_formula":"add(n0,const_100)|add(n1,const_100)|rectangle_area(const_100,const_100)|rectangle_area(#0,#1)|subtract(#3,#2)|multiply(#4,const_100)|divide(#5,#2)|","chain":"10 + 100<\/gadget>\n110<\/output>\n30 + 100<\/gadget>\n130<\/output>\n110 * 130<\/gadget>\n14_300<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n14_300 - 10_000<\/gadget>\n4_300<\/output>\n4_300 * 100<\/gadget>\n430_000<\/output>\n430_000 \/ 10_000<\/gadget>\n43<\/output>\n43<\/result>","index":329} +{"problem":"if k is the greatest positive integer such that 3 ^ k is a divisor of 15 ! then k =","rationale":"for solving this type of tasks we can use such approach 15 \/ 3 = 5 15 \/ 3 ^ 2 = 1 and remainder and 3 ^ 3 is more than 15 so we stop on second exponent 5 + 1 = 6 so maximum value of k = 6 and answer is d","correct":"d","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"add(const_1, divide(15, 3))","linear_formula":"divide(n1,n0)|add(#0,const_1)","chain":"15 \/ 3<\/gadget>\n5<\/output>\n1 + 5<\/gadget>\n6<\/output>\n6<\/result>","index":332} +{"problem":"how many of the integers between 15 and 55 are even ?","rationale":"\"number start between 15 to 55 is 40 numbers half of them is even . . which is 20 answer : b\"","correct":"b","options":{"a":"21 ","b":"20 ","c":"11 ","d":"10","e":"9"},"options_float":{"a":21.0,"b":20.0,"c":11.0,"d":10.0,"e":9.0},"annotated_formula":"divide(subtract(55, 15), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,const_2)|","chain":"55 - 15<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":333} +{"problem":"if ryan and smith car ratio is 5 : 8 . ryan bought 6 new car and smith bought 2 new car . now , smith has 18 car . what is their car ratio now ?","rationale":"current no . of smith car = 8 x + 2 = 18,8 x = 16 , x = 2 . current car ratio = 10 + 6 \/ 18 = 16 \/ 18 = 8 : 9 . answer c","correct":"c","options":{"a":"5 : 4 ","b":"3 : 2 ","c":"8 : 9 ","d":"6 : 5","e":"7 : 5"},"options_float":{"a":1.25,"b":1.5,"c":0.8888888889,"d":1.2,"e":1.4},"annotated_formula":"divide(add(multiply(divide(5, 8), subtract(18, 2)), 6), 18)","linear_formula":"divide(n0,n1)|subtract(n4,n3)|multiply(#0,#1)|add(n2,#2)|divide(#3,n4)","chain":"5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n18 - 2<\/gadget>\n16<\/output>\n(5\/8) * 16<\/gadget>\n10<\/output>\n10 + 6<\/gadget>\n16<\/output>\n16 \/ 18<\/gadget>\n8\/9 = around 0.888889<\/output>\n8\/9 = around 0.888889<\/result>","index":336} +{"problem":"the grade point average of one third of the classroom is 45 ; the grade point average of the rest is 60 . what is the grade point average of the whole class ?","rationale":"let n = total students in class total points for 1 \/ 3 class = 45 n \/ 3 = 15 n total points for 2 \/ 3 class = 60 * 2 n \/ 3 = 40 n total points for whole class = 15 n + 40 n = 55 n 63 n total class points \/ n total students = 55 grade point average for total class answer : b","correct":"b","options":{"a":"54 ","b":"55 ","c":"60 ","d":"65","e":"70"},"options_float":{"a":54.0,"b":55.0,"c":60.0,"d":65.0,"e":70.0},"annotated_formula":"add(multiply(divide(60, const_3), const_2), divide(45, const_3))","linear_formula":"divide(n0,const_3)|divide(n1,const_3)|multiply(#1,const_2)|add(#0,#2)","chain":"60 \/ 3<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n45 \/ 3<\/gadget>\n15<\/output>\n40 + 15<\/gadget>\n55<\/output>\n55<\/result>","index":339} +{"problem":"63 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ?","rationale":"\"according to the chain rule , m 1 x t 1 = m 2 x t 2 therefore , 63 x 12 x 5 = 30 x 6 x x x = 21 hence , the number of days = 21 . answer : a\"","correct":"a","options":{"a":"21 ","b":"23 ","c":"24 ","d":"25","e":"26"},"options_float":{"a":21.0,"b":23.0,"c":24.0,"d":25.0,"e":26.0},"annotated_formula":"divide(multiply(multiply(63, 12), 5), multiply(30, 6))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|","chain":"63 * 12<\/gadget>\n756<\/output>\n756 * 5<\/gadget>\n3_780<\/output>\n30 * 6<\/gadget>\n180<\/output>\n3_780 \/ 180<\/gadget>\n21<\/output>\n21<\/result>","index":340} +{"problem":"marcella has 24 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ?","rationale":"\"marcella has 24 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 24 pairs of shoes minus the 5 ' destroyed ' pairs yields 19 pairs that still fulfill the requirements . answer : c\"","correct":"c","options":{"a":"21 ","b":"20 ","c":"19 ","d":"16","e":"15"},"options_float":{"a":21.0,"b":20.0,"c":19.0,"d":16.0,"e":15.0},"annotated_formula":"subtract(24, add(floor(divide(9, const_2)), const_1))","linear_formula":"divide(n1,const_2)|floor(#0)|add(#1,const_1)|subtract(n0,#2)|","chain":"9 \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\nfloor(9\/2)<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n24 - 5<\/gadget>\n19<\/output>\n19<\/result>","index":341} +{"problem":"two trains are approaching each other at 60 mph each . the length of each train is 1 \/ 6 miles . when the two locomotives ( engines ) meet , how many seconds will it take for them to pass each other completely ?","rationale":"total distance to cover = 2 * ( 1 \/ 6 ) = 2 \/ 6 miles relative speed = 60 + 60 = 120 mph time taken = ( 2 \/ 6 ) \/ 120 * 3600 = 10 sec . answer : d","correct":"d","options":{"a":"8 sec ","b":"11 sec ","c":"12 sec ","d":"10 sec","e":"15 sec"},"options_float":{"a":8.0,"b":11.0,"c":12.0,"d":10.0,"e":15.0},"annotated_formula":"multiply(60, multiply(60, divide(multiply(divide(1, 6), const_2), multiply(60, const_2))))","linear_formula":"divide(n1,n2)|multiply(n0,const_2)|multiply(#0,const_2)|divide(#2,#1)|multiply(n0,#3)|multiply(n0,#4)","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 2<\/gadget>\n1\/3 = around 0.333333<\/output>\n60 * 2<\/gadget>\n120<\/output>\n(1\/3) \/ 120<\/gadget>\n1\/360 = around 0.002778<\/output>\n60 * (1\/360)<\/gadget>\n1\/6 = around 0.166667<\/output>\n60 * (1\/6)<\/gadget>\n10<\/output>\n10<\/result>","index":343} +{"problem":"a courtyard is 25 meter long and 15 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ?","rationale":"\"number of bricks = courtyard area \/ 1 brick area = ( 2500 × 1500 \/ 20 × 10 ) = 18750 answer : c\"","correct":"c","options":{"a":"22877 ","b":"27778 ","c":"18750 ","d":"27999","e":"17799"},"options_float":{"a":22877.0,"b":27778.0,"c":18750.0,"d":27999.0,"e":17799.0},"annotated_formula":"divide(multiply(multiply(25, const_100), multiply(15, const_100)), multiply(20, 10))","linear_formula":"multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)|","chain":"25 * 100<\/gadget>\n2_500<\/output>\n15 * 100<\/gadget>\n1_500<\/output>\n2_500 * 1_500<\/gadget>\n3_750_000<\/output>\n20 * 10<\/gadget>\n200<\/output>\n3_750_000 \/ 200<\/gadget>\n18_750<\/output>\n18_750<\/result>","index":344} +{"problem":"a certain elevator has a safe weight limit of 1950 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 190 pounds and the average weight of the others being 200 pounds ?","rationale":"\"lets assume there are 2 x people . half of them have average weight of 190 and other half has 200 . maximum weight is = 1950 so 190 * x + 200 * x = 1950 = > 390 x = 1950 = > x is approximately equal to 5 . so total people is 2 * 5 = 10 answer d .\"","correct":"d","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"multiply(divide(multiply(const_10, 190), add(190, 200)), const_2)","linear_formula":"add(n1,n2)|multiply(n1,const_10)|divide(#1,#0)|multiply(#2,const_2)|","chain":"10 * 190<\/gadget>\n1_900<\/output>\n190 + 200<\/gadget>\n390<\/output>\n1_900 \/ 390<\/gadget>\n190\/39 = around 4.871795<\/output>\n(190\/39) * 2<\/gadget>\n380\/39 = around 9.74359<\/output>\n380\/39 = around 9.74359<\/result>","index":348} +{"problem":"a and b invests rs . 8000 and rs . 12000 in a business . after 4 months , a withdraws half of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ?","rationale":"a : b ( 8000 * 4 ) + ( 4000 * 8 ) : ( 12000 * 6 ) + ( 9000 * 6 ) 64000 : 120000 8 : 15 answer : c","correct":"c","options":{"a":"32 : 99 ","b":"82 : 31 ","c":"8 : 15 ","d":"34 : 89","e":"35 : 21"},"options_float":{"a":0.3232323232,"b":2.6451612903,"c":0.5333333333,"d":0.3820224719,"e":1.6666666667},"annotated_formula":"divide(add(multiply(8000, 4), multiply(divide(12000, const_3), multiply(const_2, const_4))), add(multiply(12000, multiply(const_2, const_3)), multiply(subtract(12000, divide(12000, const_3)), multiply(const_2, const_3))))","linear_formula":"divide(n1,const_3)|multiply(n0,n2)|multiply(const_2,const_4)|multiply(const_2,const_3)|multiply(#0,#2)|multiply(n1,#3)|subtract(n1,#0)|add(#1,#4)|multiply(#3,#6)|add(#5,#8)|divide(#7,#9)","chain":"8_000 * 4<\/gadget>\n32_000<\/output>\n12_000 \/ 3<\/gadget>\n4_000<\/output>\n2 * 4<\/gadget>\n8<\/output>\n4_000 * 8<\/gadget>\n32_000<\/output>\n32_000 + 32_000<\/gadget>\n64_000<\/output>\n2 * 3<\/gadget>\n6<\/output>\n12_000 * 6<\/gadget>\n72_000<\/output>\n12_000 - 4_000<\/gadget>\n8_000<\/output>\n8_000 * 6<\/gadget>\n48_000<\/output>\n72_000 + 48_000<\/gadget>\n120_000<\/output>\n64_000 \/ 120_000<\/gadget>\n8\/15 = around 0.533333<\/output>\n8\/15 = around 0.533333<\/result>","index":349} +{"problem":"the height of the wall is 6 times its width and length of the wall is 7 times its height . if the volume of the wall be 86436 cu . m . its width is","rationale":"\"explanation : let width = x then , height = 6 x and length = 42 x 42 x ã — 6 x ã — x = 86436 x = 7 answer : d\"","correct":"d","options":{"a":"4 m ","b":"5 m ","c":"6 m ","d":"7 m","e":"8 m"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"power(divide(86436, multiply(multiply(6, 7), 6)), divide(const_1, const_3))","linear_formula":"divide(const_1,const_3)|multiply(n0,n1)|multiply(n0,#1)|divide(n2,#2)|power(#3,#0)|","chain":"6 * 7<\/gadget>\n42<\/output>\n42 * 6<\/gadget>\n252<\/output>\n86_436 \/ 252<\/gadget>\n343<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n343 ** (1\/3)<\/gadget>\n7<\/output>\n7<\/result>","index":350} +{"problem":"money invested at x % , compounded annually , triples in value in approximately every 112 \/ x years . if $ 2000 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ?","rationale":"\"x = 8 % 112 \/ x years = 112 \/ 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 2000 triples to $ 6000 , in 28 years , it will again triple to $ 6000 * 3 = $ 18,000 answer d\"","correct":"d","options":{"a":"$ 3,750 ","b":"$ 5,600 ","c":"$ 8,100 ","d":"$ 18,000","e":"$ 22,500"},"options_float":{"a":3750.0,"b":5600.0,"c":8100.0,"d":18000.0,"e":22500.0},"annotated_formula":"multiply(2000, power(const_3, divide(28, divide(112, 8))))","linear_formula":"divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2)|","chain":"112 \/ 8<\/gadget>\n14<\/output>\n28 \/ 14<\/gadget>\n2<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n2_000 * 9<\/gadget>\n18_000<\/output>\n18_000<\/result>","index":351} +{"problem":"a man buys rs . 44 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is :","rationale":"\"dividend on rs . 44 = rs . 9 \/ 100 x 44 = rs . 3.96 . rs . 12 is an income on rs . 100 . rs . 3.96 is an income on rs . 100 \/ 12 x 3.96 = rs . 33 . answer : option e\"","correct":"e","options":{"a":"s . 12 ","b":"s . 15 ","c":"s . 18 ","d":"s . 21","e":"s . 33"},"options_float":{"a":12.0,"b":15.0,"c":18.0,"d":21.0,"e":33.0},"annotated_formula":"multiply(divide(const_100, 12), multiply(divide(9, const_100), 44))","linear_formula":"divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2)|","chain":"100 \/ 12<\/gadget>\n25\/3 = around 8.333333<\/output>\n9 \/ 100<\/gadget>\n9\/100 = around 0.09<\/output>\n(9\/100) * 44<\/gadget>\n99\/25 = around 3.96<\/output>\n(25\/3) * (99\/25)<\/gadget>\n33<\/output>\n33<\/result>","index":352} +{"problem":"the duplicate ratio of 4 : 5 is ?","rationale":"4 ^ 2 : 5 ^ 2 = 16 : 25 answer : e","correct":"e","options":{"a":"1 : 2 ","b":"1 : 4 ","c":"1 : 8 ","d":"1 : 18","e":"16 : 25"},"options_float":{"a":0.5,"b":0.25,"c":0.125,"d":0.0555555556,"e":0.64},"annotated_formula":"divide(power(4, const_2), power(5, const_2))","linear_formula":"power(n0,const_2)|power(n1,const_2)|divide(#0,#1)","chain":"4 ** 2<\/gadget>\n16<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n16 \/ 25<\/gadget>\n16\/25 = around 0.64<\/output>\n16\/25 = around 0.64<\/result>","index":354} +{"problem":"how many different ways can 2 students be seated in a row of 5 desks , so that there is always at least one empty desk between the students ?","rationale":"\"6 ways to seat the students : with two empty seats between 1 empty w \/ one student on the left most 1 empty . . . . right most two students can be interchanged 6 x 2 = 12 e\"","correct":"e","options":{"a":"2 ","b":"3 ","c":"4 ","d":"6","e":"12"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":6.0,"e":12.0},"annotated_formula":"permutation(subtract(5, const_1), 2)","linear_formula":"subtract(n1,const_1)|permutation(#0,n0)|","chain":"5 - 1<\/gadget>\n4<\/output>\nfactorial(4) \/ factorial(4 - 2)<\/gadget>\n12<\/output>\n12<\/result>","index":355} +{"problem":"he average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 70 kg . what might be the weight of the new person ?","rationale":"\"explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 70 + 20 ) kg = 90 kg . answer : a\"","correct":"a","options":{"a":"90 kg ","b":"55 kg ","c":"45 kg ","d":"85 kg","e":"25 kg"},"options_float":{"a":90.0,"b":55.0,"c":45.0,"d":85.0,"e":25.0},"annotated_formula":"add(multiply(2.5, 8), 70)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"2.5 * 8<\/gadget>\n20<\/output>\n20 + 70<\/gadget>\n90<\/output>\n90<\/result>","index":356} +{"problem":"if the function q is defined by the formula q = 5 w \/ ( 4 h ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , h is doubled , and z is tripled ?","rationale":"\"we just need to find the factor thats all , w - > quadrupled - > 4 w h - > doubled - > 2 h z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and h * z in denominator . hence , additional factor being introduced = 4 \/ 2 * 9 = 4 \/ 18 = 2 \/ 9 = b\"","correct":"b","options":{"a":"1 \/ 9 ","b":"2 \/ 9 ","c":"4 \/ 9 ","d":"3 \/ 9","e":"2 \/ 27"},"options_float":{"a":0.1111111111,"b":0.2222222222,"c":0.4444444444,"d":0.3333333333,"e":0.0740740741},"annotated_formula":"divide(4, multiply(2, power(const_3, 2)))","linear_formula":"power(const_3,n2)|multiply(n2,#0)|divide(n1,#1)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n2 * 9<\/gadget>\n18<\/output>\n4 \/ 18<\/gadget>\n2\/9 = around 0.222222<\/output>\n2\/9 = around 0.222222<\/result>","index":357} +{"problem":"the radius of a wheel is 20.4 cm . what is the distance covered by the wheel in making 400 resolutions ?","rationale":"\"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 400 resolutions . = 400 * 2 * 22 \/ 7 * 20.4 = 51244.8 cm = 512.448 m answer : a\"","correct":"a","options":{"a":"512.448 m ","b":"704 m ","c":"510.448 m ","d":"706 m","e":"204 m"},"options_float":{"a":512.448,"b":704.0,"c":510.448,"d":706.0,"e":204.0},"annotated_formula":"divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 20.4), const_2), 400), const_100)","linear_formula":"add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)|","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21 + 1<\/gadget>\n22<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n(22\/7) * 20.4<\/gadget>\n64.114286<\/output>\n64.114286 * 2<\/gadget>\n128.228572<\/output>\n128.228572 * 400<\/gadget>\n51_291.4288<\/output>\n51_291.4288 \/ 100<\/gadget>\n512.914288<\/output>\n512.914288<\/result>","index":359} +{"problem":"a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 20 % . what was its profit on the items sold in february ?","rationale":"\"assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.8 * 150 x = 120 x hence total profit = 20 % option a\"","correct":"a","options":{"a":"20 % ","b":"30 % ","c":"35 % ","d":"37.5 %","e":"40 %"},"options_float":{"a":20.0,"b":30.0,"c":35.0,"d":37.5,"e":40.0},"annotated_formula":"subtract(multiply(divide(subtract(const_100, 20), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100)","linear_formula":"add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n100 + 20<\/gadget>\n120<\/output>\n100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n120 * (5\/4)<\/gadget>\n150<\/output>\n(4\/5) * 150<\/gadget>\n120<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20<\/result>","index":361} +{"problem":"if the cost price of 50 articles is equal to the selling price of 20 articles , then the gain or loss percent is ?","rationale":"\"percentage of profit = 30 \/ 20 * 100 = 150 % answer : e\"","correct":"e","options":{"a":"16 ","b":"127 ","c":"12 ","d":"18","e":"150"},"options_float":{"a":16.0,"b":127.0,"c":12.0,"d":18.0,"e":150.0},"annotated_formula":"multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 20), 50)), divide(multiply(const_100, 20), 50)))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 * 20<\/gadget>\n2_000<\/output>\n2_000 \/ 50<\/gadget>\n40<\/output>\n100 - 40<\/gadget>\n60<\/output>\n60 \/ 40<\/gadget>\n3\/2 = around 1.5<\/output>\n100 * (3\/2)<\/gadget>\n150<\/output>\n150<\/result>","index":362} +{"problem":"what is rate of interest if principal . amount be 400 , simple interest 160 and time 2 year .","rationale":"s . i = ( p * r * t ) \/ 100 160 = 800 r \/ 100 r = 160 \/ 8 = 20 % answer c","correct":"c","options":{"a":"10 ","b":"12.5 ","c":"20 ","d":"12","e":"14.5"},"options_float":{"a":10.0,"b":12.5,"c":20.0,"d":12.0,"e":14.5},"annotated_formula":"multiply(divide(160, multiply(400, 2)), const_100)","linear_formula":"multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)","chain":"400 * 2<\/gadget>\n800<\/output>\n160 \/ 800<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":363} +{"problem":"the radius of a sphere is increased by 50 % . the increase in surface area of the sphere is :","rationale":"\"explanation : take radius 100 . then surface area is 4 × π π × 100 × 100 . after increase radius by 50 % the radius becomes 100 + 50 % of 100 = 150 then new surface area is 4 × π π × 150 × 150 then put the values into formula of percentage = 4 π 1502 − 4 π 10024 π 1002 × 1004 π 1502 − 4 π 10024 π 1002 × 100 = 125 % answer : b\"","correct":"b","options":{"a":"168 ","b":"125 ","c":"186 ","d":"197","e":"191"},"options_float":{"a":168.0,"b":125.0,"c":186.0,"d":197.0,"e":191.0},"annotated_formula":"multiply(divide(subtract(surface_sphere(add(const_1, divide(50, const_100))), surface_sphere(const_1)), surface_sphere(const_1)), const_100)","linear_formula":"divide(n0,const_100)|surface_sphere(const_1)|add(#0,const_1)|surface_sphere(#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n4 * pi * ((3\/2) ** 2)<\/gadget>\n9*pi = around 28.274334<\/output>\n4 * pi * (1 ** 2)<\/gadget>\n4*pi = around 12.566371<\/output>\n(9*pi) - (4*pi)<\/gadget>\n5*pi = around 15.707963<\/output>\n(5*pi) \/ (4*pi)<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 100<\/gadget>\n125<\/output>\n125<\/result>","index":365} +{"problem":"a man can row 4 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?","rationale":"\"m = 4 s = 1.2 ds = 4.8 us = 2.8 x \/ 4.8 + x \/ 2.8 = 1 x = 1.77 d = 1.77 * 2 = 3.54 answer : c\"","correct":"c","options":{"a":"5.98 ","b":"3.98 ","c":"3.54 ","d":"3.87","e":"3.82"},"options_float":{"a":5.98,"b":3.98,"c":3.54,"d":3.87,"e":3.82},"annotated_formula":"multiply(divide(multiply(add(4, 1.2), subtract(4, 1.2)), add(add(4, 1.2), subtract(4, 1.2))), const_2)","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|","chain":"4 + 1.2<\/gadget>\n5.2<\/output>\n4 - 1.2<\/gadget>\n2.8<\/output>\n5.2 * 2.8<\/gadget>\n14.56<\/output>\n5.2 + 2.8<\/gadget>\n8<\/output>\n14.56 \/ 8<\/gadget>\n1.82<\/output>\n1.82 * 2<\/gadget>\n3.64<\/output>\n3.64<\/result>","index":366} +{"problem":"in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive difference w between n and m ?","rationale":"\"we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result w will be 8 - 5 = 3 and answer b\"","correct":"b","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"subtract(5, 2)","linear_formula":"subtract(n1,n0)|","chain":"5 - 2<\/gadget>\n3<\/output>\n3<\/result>","index":367} +{"problem":"h is a positive integer and 225 and 216 are both divisors of h . if h = ( 2 ^ a ) * ( 3 ^ b ) * ( 5 ^ c ) , where a , b and c are positive integers , what is the least possible value of a + b + c ?","rationale":"lets make factorization of 225 and 216 . . 225 = 5 x 5 x 3 x 3 x 3 216 = 2 x 2 x 2 x 3 x 3 x 3 h would have to have 3 two ' s , 225 has 3 threes and and so does 216 but they can be the same three threes so we count them only once . . . 225 has 2 fives . . . so we had them together and we get 3 + 3 + 2 = 8 ( b ) ( answer ) . . .","correct":"b","options":{"a":"4 ","b":"8 ","c":"6 ","d":"7","e":"5"},"options_float":{"a":4.0,"b":8.0,"c":6.0,"d":7.0,"e":5.0},"annotated_formula":"add(add(3, 3), 2)","linear_formula":"add(n3,n3)|add(n2,#0)","chain":"3 + 3<\/gadget>\n6<\/output>\n6 + 2<\/gadget>\n8<\/output>\n8<\/result>","index":368} +{"problem":"if p and q are both odd prime numbers andp < q , then how many different positive integer factors does 2 pqhave ?","rationale":"p and q are both odd prime numbers - it means either p or q is not 2 and since prime numbers have only two factors - 1 and the number itself p and q each will have ( 1 + 1 ) = 2 factors hence 2 pq will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer","correct":"d","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"12"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":12.0},"annotated_formula":"power(2, const_3)","linear_formula":"power(n0,const_3)","chain":"2 ** 3<\/gadget>\n8<\/output>\n8<\/result>","index":369} +{"problem":"? x 12 = 173 x 240","rationale":"\"let y x 12 = 173 x 240 then y = ( 173 x 240 ) \/ 12 = 3460 answer : a\"","correct":"a","options":{"a":"3460 ","b":"685 ","c":"865 ","d":"495","e":"534"},"options_float":{"a":3460.0,"b":685.0,"c":865.0,"d":495.0,"e":534.0},"annotated_formula":"divide(multiply(173, 240), 12)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"173 * 240<\/gadget>\n41_520<\/output>\n41_520 \/ 12<\/gadget>\n3_460<\/output>\n3_460<\/result>","index":371} +{"problem":"a man goes to shopkeeper and says that if u give me as much as i have , i will spend rs . 10 \/ - on ur shop . shopkeeper agrees and man moves to another shopkeeper ans asked same . at 3 rd shop after all the transaction , he was left with no money . find the initial amount with the man","rationale":"man has rs . x he leaves - 1 st shop wid rs . ( 2 x - 10 ) 2 nd shop wid rs . [ 2 ( 2 x - 10 ) - 10 ] 3 rd shop wid rs . { 2 [ 2 ( 2 x - 10 ) - 10 ] - 10 } = 0 ( since he has no money left ) so solving the eqn x = 8.75 answer : d","correct":"d","options":{"a":"5.75 ","b":"6.75 ","c":"7.75 ","d":"8.75","e":"9.75"},"options_float":{"a":5.75,"b":6.75,"c":7.75,"d":8.75,"e":9.75},"annotated_formula":"divide(add(divide(add(divide(10, const_2), 10), const_2), 10), const_2)","linear_formula":"divide(n0,const_2)|add(n0,#0)|divide(#1,const_2)|add(n0,#2)|divide(#3,const_2)","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5 + 10<\/gadget>\n15<\/output>\n15 \/ 2<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) + 10<\/gadget>\n35\/2 = around 17.5<\/output>\n(35\/2) \/ 2<\/gadget>\n35\/4 = around 8.75<\/output>\n35\/4 = around 8.75<\/result>","index":372} +{"problem":"a starts a business with $ 3500 . after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . b ' s contribution in the capital is","rationale":"ratio in which profit is divided = 2 : 3 assume that b ' s contribution to the capital = b 3500 × 12 : b × 7 = 2 : 3 ⇒ 3500 × 12 × 3 = 2 × b × 7 ⇒ b = 3500 × 12 × 3 \/ 2 × 7 = 500 × 6 × 3 = 9000 answer : b","correct":"b","options":{"a":"7000 ","b":"9000 ","c":"8000 ","d":"9900","e":"7900"},"options_float":{"a":7000.0,"b":9000.0,"c":8000.0,"d":9900.0,"e":7900.0},"annotated_formula":"divide(divide(multiply(multiply(3500, const_12), 3), 2), subtract(const_12, 5))","linear_formula":"multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|divide(#2,n2)|divide(#3,#1)","chain":"3_500 * 12<\/gadget>\n42_000<\/output>\n42_000 * 3<\/gadget>\n126_000<\/output>\n126_000 \/ 2<\/gadget>\n63_000<\/output>\n12 - 5<\/gadget>\n7<\/output>\n63_000 \/ 7<\/gadget>\n9_000<\/output>\n9_000<\/result>","index":374} +{"problem":"two trains of length 180 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?","rationale":"\"relative speed = ( 42 + 30 ) * 5 \/ 18 = 4 * 5 = 20 mps . distance covered in passing each other = 180 + 280 = 400 m . the time required = d \/ s = 460 \/ 20 = 23 sec . answer : b\"","correct":"b","options":{"a":"22 sec ","b":"23 sec ","c":"76 sec ","d":"20 sec","e":"66 sec"},"options_float":{"a":22.0,"b":23.0,"c":76.0,"d":20.0,"e":66.0},"annotated_formula":"divide(add(180, 280), multiply(add(42, 30), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"180 + 280<\/gadget>\n460<\/output>\n42 + 30<\/gadget>\n72<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n460 \/ 20<\/gadget>\n23<\/output>\n23<\/result>","index":375} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 7 sec . what is the length of the train ?","rationale":"\"speed = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec length of the train = speed * time = 50 \/ 3 * 7 = 116.66 answer : e\"","correct":"e","options":{"a":"112.21 ","b":"27.21 ","c":"117.66 ","d":"277","e":"116.66"},"options_float":{"a":112.21,"b":27.21,"c":117.66,"d":277.0,"e":116.66},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 7)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"60 * 1_000<\/gadget>\n60_000<\/output>\n60_000 \/ 3_600<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 7<\/gadget>\n350\/3 = around 116.666667<\/output>\n350\/3 = around 116.666667<\/result>","index":376} +{"problem":"sand is poured into a box so that the box is being filled at the rate of 4 cubic feet per hour . if the empty rectangular box is 7 feet long , 6 feet wide , and 2 feet deep , approximately how many hours does it take to fill the box ?","rationale":"the volume the box is : length * width * depth = 7 * 6 * 2 = 84 cubic feet . 84 cubic feet \/ 4 cubic feet per hour = 21 hours . it will take 21 hours to fill the box . the answer is b .","correct":"b","options":{"a":"20 ","b":"21 ","c":"22 ","d":"23","e":"24"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":23.0,"e":24.0},"annotated_formula":"divide(multiply(multiply(7, 6), 2), 4)","linear_formula":"multiply(n1,n2)|multiply(n3,#0)|divide(#1,n0)","chain":"7 * 6<\/gadget>\n42<\/output>\n42 * 2<\/gadget>\n84<\/output>\n84 \/ 4<\/gadget>\n21<\/output>\n21<\/result>","index":377} +{"problem":"two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 40 km and the other at the rate of 50 kmph . in what time will they be completely clear of each other from the moment they meet ?","rationale":"\"t = ( 111 + 165 ) \/ ( 40 + 50 ) * 18 \/ 5 t = 11.04 answer : a\"","correct":"a","options":{"a":"11.04 ","b":"10.0 ","c":"6.85 ","d":"5.85","e":"6.15"},"options_float":{"a":11.04,"b":10.0,"c":6.85,"d":5.85,"e":6.15},"annotated_formula":"divide(add(111, 165), multiply(add(40, 50), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"111 + 165<\/gadget>\n276<\/output>\n40 + 50<\/gadget>\n90<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n90 * (5\/18)<\/gadget>\n25<\/output>\n276 \/ 25<\/gadget>\n276\/25 = around 11.04<\/output>\n276\/25 = around 11.04<\/result>","index":380} +{"problem":"a and b walk around a circular track . they start at 8 a . m from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 9 a . m","rationale":"explanation : relative speed = ( 2 + 3 ) = 5 rounds per hour so , they cross each other 5 times in an hour hence , they cross 5 times before 9 a . m answer : option a","correct":"a","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"add(2, 3)","linear_formula":"add(n1,n2)","chain":"2 + 3<\/gadget>\n5<\/output>\n5<\/result>","index":382} +{"problem":"10 men and 15 women together can complete a work in 6 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ?","rationale":"1 man ' s 1 day work = 1 \/ 100 ( 10 men + 15 women ) ' s 1 day work = 1 \/ 6 15 women ' s 1 day work = ( 1 \/ 6 - 10 \/ 100 ) = 1 \/ 15 1 woman ' s 1 day work = 1 \/ 225 1 woman alone can complete the work in 225 days . answer : e","correct":"e","options":{"a":"127 days ","b":"667 days ","c":"177 days ","d":"187 days","e":"225 days"},"options_float":{"a":127.0,"b":667.0,"c":177.0,"d":187.0,"e":225.0},"annotated_formula":"multiply(divide(multiply(const_1, 100), subtract(multiply(const_1, 100), multiply(10, 6))), multiply(15, 6))","linear_formula":"multiply(n3,const_1)|multiply(n0,n2)|multiply(n1,n2)|subtract(#0,#1)|divide(#0,#3)|multiply(#4,#2)","chain":"1 * 100<\/gadget>\n100<\/output>\n10 * 6<\/gadget>\n60<\/output>\n100 - 60<\/gadget>\n40<\/output>\n100 \/ 40<\/gadget>\n5\/2 = around 2.5<\/output>\n15 * 6<\/gadget>\n90<\/output>\n(5\/2) * 90<\/gadget>\n225<\/output>\n225<\/result>","index":384} +{"problem":"in a basket of apples , 12 % of them are rotten and 66 are in good condition . find the total number of apples in the basket .","rationale":"let the total number of apples in the basket be m 12 % of the apples are rotten , and apples in good condition are 66 therefore , according to the question , 88 % of m = 66 88 \/ 100 × m = 66 m = ( 66 × 100 ) \/ 88 m = 3 × 25 m = 75 therefore , total number of apples in the basket is 75 . answer : b","correct":"b","options":{"a":"33 ","b":"75 ","c":"23 ","d":"20","e":"287"},"options_float":{"a":33.0,"b":75.0,"c":23.0,"d":20.0,"e":287.0},"annotated_formula":"divide(66, subtract(const_1, divide(12, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)","chain":"12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n1 - (3\/25)<\/gadget>\n22\/25 = around 0.88<\/output>\n66 \/ (22\/25)<\/gadget>\n75<\/output>\n75<\/result>","index":385} +{"problem":"a piece of cloth cost rs 35 . if the length of the piece would have been 4 m longer and each meter cost re 1 less , the cost would have remained unchanged . how long is the piece ?","rationale":"if length of piece is x and cost per mtr is c , then x * c = 35 or c = 35 \/ x ( x + 4 ) * ( c - 1 ) = 35 or x * c + 4 c - x - 4 = 35 or 35 + 4 c - x - 4 = 35 4 c - x = 4 4 * ( 35 \/ x ) - x = 4 x = 10 mtrs answer : b","correct":"b","options":{"a":"5 mtrs ","b":"10 mtrs ","c":"12 mtrs ","d":"15 mtrs","e":"18 mtrs"},"options_float":{"a":5.0,"b":10.0,"c":12.0,"d":15.0,"e":18.0},"annotated_formula":"divide(35, divide(add(4, sqrt(add(multiply(multiply(4, 4), 35), multiply(4, 4)))), multiply(const_2, 4)))","linear_formula":"multiply(n1,n1)|multiply(n1,const_2)|multiply(n0,#0)|add(#2,#0)|sqrt(#3)|add(n1,#4)|divide(#5,#1)|divide(n0,#6)","chain":"4 * 4<\/gadget>\n16<\/output>\n16 * 35<\/gadget>\n560<\/output>\n560 + 16<\/gadget>\n576<\/output>\n576 ** (1\/2)<\/gadget>\n24<\/output>\n4 + 24<\/gadget>\n28<\/output>\n2 * 4<\/gadget>\n8<\/output>\n28 \/ 8<\/gadget>\n7\/2 = around 3.5<\/output>\n35 \/ (7\/2)<\/gadget>\n10<\/output>\n10<\/result>","index":387} +{"problem":"what is the average of 4 tenths and 5 thousandths ?","rationale":"four tenths = 0.4 five thousandths = 0.005 the average is ( 0.4 + 0.005 ) \/ 2 = 0.2025 answer : c","correct":"c","options":{"a":"0.2022 ","b":"0.2012 ","c":"0.2025 ","d":"0.2021","e":"0.2029"},"options_float":{"a":0.2022,"b":0.2012,"c":0.2025,"d":0.2021,"e":0.2029},"annotated_formula":"divide(add(divide(4, const_10), divide(5, multiply(const_10, const_100))), const_2)","linear_formula":"divide(n0,const_10)|multiply(const_10,const_100)|divide(n1,#1)|add(#0,#2)|divide(#3,const_2)","chain":"4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n5 \/ 1_000<\/gadget>\n1\/200 = around 0.005<\/output>\n(2\/5) + (1\/200)<\/gadget>\n81\/200 = around 0.405<\/output>\n(81\/200) \/ 2<\/gadget>\n81\/400 = around 0.2025<\/output>\n81\/400 = around 0.2025<\/result>","index":388} +{"problem":"the average weight of a , b and c is 43 kg . if the average weight of a and b be 48 kg and that of b and c be 42 kg , then the weight of b is :","rationale":"\"explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 43 x 3 ) = 129 â € ¦ . ( i ) a + b = ( 48 x 2 ) = 96 â € ¦ . ( ii ) b + c = ( 42 x 2 ) = 84 â € ¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 180 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 51 . b â € ™ s weight = 51 kg . answer e\"","correct":"e","options":{"a":"55 kg ","b":"53 kg ","c":"49 kg ","d":"52 kg","e":"51 kg"},"options_float":{"a":55.0,"b":53.0,"c":49.0,"d":52.0,"e":51.0},"annotated_formula":"subtract(add(multiply(48, const_2), multiply(42, const_2)), multiply(43, const_3))","linear_formula":"multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|","chain":"48 * 2<\/gadget>\n96<\/output>\n42 * 2<\/gadget>\n84<\/output>\n96 + 84<\/gadget>\n180<\/output>\n43 * 3<\/gadget>\n129<\/output>\n180 - 129<\/gadget>\n51<\/output>\n51<\/result>","index":389} +{"problem":"pipe p can fill a tank in 3 hours , pipe q in 9 hours and pipe r in 18 hours . if all the pipes are open , in how many hours will the tank be filled ?","rationale":"\"explanation : part filled by ( p + q + r ) in 1 hour = ( 1 \/ 3 + 1 \/ 9 + 1 \/ 18 ) = 1 \/ 2 all the three pipes together will fill the tank = 2 \/ 1 = 2 hours answer a\"","correct":"a","options":{"a":"2 hours ","b":"2.4 hours ","c":"3 hours ","d":"3.5 hours","e":"4.5 hours"},"options_float":{"a":2.0,"b":2.4,"c":3.0,"d":3.5,"e":4.5},"annotated_formula":"inverse(add(divide(const_1, 18), add(divide(const_1, 3), divide(const_1, 9))))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)|","chain":"1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/3) + (1\/9)<\/gadget>\n4\/9 = around 0.444444<\/output>\n(1\/18) + (4\/9)<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ (1\/2)<\/gadget>\n2<\/output>\n2<\/result>","index":390} +{"problem":"if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 25 minutes ?","rationale":"\"since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 25 minutes the population increases by 25 x 4 = 100 people . answer . b .\"","correct":"b","options":{"a":"80 ","b":"100 ","c":"150 ","d":"240","e":"300"},"options_float":{"a":80.0,"b":100.0,"c":150.0,"d":240.0,"e":300.0},"annotated_formula":"multiply(divide(const_60, 15), 25)","linear_formula":"divide(const_60,n0)|multiply(n1,#0)|","chain":"60 \/ 15<\/gadget>\n4<\/output>\n4 * 25<\/gadget>\n100<\/output>\n100<\/result>","index":391} +{"problem":"a monkey ascends a greased pole 10 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ?","rationale":"the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 10 meters . i mean this will continue for first 8 * 2 = 16 mins . he would have reached 8 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 16 + 1 = 17 mins . so , asnwer will be c","correct":"c","options":{"a":"10 th minute ","b":"21 st minute ","c":"17 th minute ","d":"22 nd minute","e":"13 th minute"},"options_float":{"a":10.0,"b":21.0,"c":17.0,"d":22.0,"e":13.0},"annotated_formula":"add(multiply(multiply(const_4, const_2), 2), 1)","linear_formula":"multiply(const_2,const_4)|multiply(n1,#0)|add(n2,#1)","chain":"4 * 2<\/gadget>\n8<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16 + 1<\/gadget>\n17<\/output>\n17<\/result>","index":392} +{"problem":"the number 35 is equal to the sum of the cubes of two integers . what is the product of those integers ?","rationale":"\"2 ^ 3 + 3 ^ 3 = 35 therefore product is 2 * 3 = 6 a\"","correct":"a","options":{"a":"6 ","b":"15 ","c":"21 ","d":"27","e":"39"},"options_float":{"a":6.0,"b":15.0,"c":21.0,"d":27.0,"e":39.0},"annotated_formula":"multiply(floor(power(divide(35, const_2), divide(const_1, const_3))), power(subtract(35, power(floor(power(divide(35, const_2), divide(const_1, const_3))), const_3)), divide(const_1, const_3)))","linear_formula":"divide(n0,const_2)|divide(const_1,const_3)|power(#0,#1)|floor(#2)|power(#3,const_3)|subtract(n0,#4)|power(#5,#1)|multiply(#3,#6)|","chain":"35 \/ 2<\/gadget>\n35\/2 = around 17.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(35\/2) ** (1\/3)<\/gadget>\n2**(2\/3)*35**(1\/3)\/2 = around 2.596247<\/output>\nfloor(2**(2\/3)*35**(1\/3)\/2)<\/gadget>\n2<\/output>\n2 ** 3<\/gadget>\n8<\/output>\n35 - 8<\/gadget>\n27<\/output>\n27 ** (1\/3)<\/gadget>\n3<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6<\/result>","index":393} +{"problem":"fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 22 percent , what is the discount rate on pony jeans ?","rationale":"\"let x be the percentage decrease in fox jeans pony jeans = 22 - x ( 15 * x \/ 100 ) * 3 + ( 18 ( 22 - x ) \/ 100 ) * 2 = 9 . . . . . . . . . . . given 45 x \/ 100 + 792 - 36 x \/ 100 = 9 9 x + 792 \/ 100 = 9 9 x + 792 = 900 9 x = 108 x = 12 = fox jeans hence pony jeans = 22 - 12 = 10 answer : b\"","correct":"b","options":{"a":"9 % ","b":"10 % ","c":"11 % ","d":"12 %","e":"13 %"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":12.0,"e":13.0},"annotated_formula":"multiply(subtract(divide(22, const_100), divide(subtract(9, multiply(divide(22, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100)","linear_formula":"divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)|","chain":"22 \/ 100<\/gadget>\n11\/50 = around 0.22<\/output>\n18 * 2<\/gadget>\n36<\/output>\n(11\/50) * 36<\/gadget>\n198\/25 = around 7.92<\/output>\n9 - (198\/25)<\/gadget>\n27\/25 = around 1.08<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 - 36<\/gadget>\n9<\/output>\n(27\/25) \/ 9<\/gadget>\n3\/25 = around 0.12<\/output>\n(11\/50) - (3\/25)<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":394} +{"problem":"if i walk at 4 km \/ h , i miss the bus by 10 minutes . if i walk at 5 km \/ h , i reach 5 minutes before the arrival of the bus . how far i walk to reach the bus stand ?","rationale":"\"d = product of speed difference of time \/ difference of speed d = 4 × 5 \/ 60 [ 10 − ( − 5 ) \/ 5 − 4 ] [ here , – ve sign indicates before the schedule time ] ⇒ d = 5 km answer a\"","correct":"a","options":{"a":"5 km ","b":"4.5 km ","c":"5 1 ⁄ 4 km \/ h ","d":"can not be determined","e":"none of these"},"options_float":{"a":5.0,"b":4.5,"c":5.0,"d":null,"e":null},"annotated_formula":"multiply(divide(multiply(4, 5), subtract(5, 4)), divide(add(10, 5), const_60))","linear_formula":"add(n1,n3)|multiply(n0,n2)|subtract(n2,n0)|divide(#1,#2)|divide(#0,const_60)|multiply(#3,#4)|","chain":"4 * 5<\/gadget>\n20<\/output>\n5 - 4<\/gadget>\n1<\/output>\n20 \/ 1<\/gadget>\n20<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15 \/ 60<\/gadget>\n1\/4 = around 0.25<\/output>\n20 * (1\/4)<\/gadget>\n5<\/output>\n5<\/result>","index":396} +{"problem":"a jogger running at 9 kmph along side a railway track is 260 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ?","rationale":"\"speed of train relative to jogger = ( 45 – 9 ) km \/ h = 36 km \/ h = ( 36 × 5 ⁄ 18 ) m \/ sec = 10 m \/ sec distance to be covered = ( 260 + 120 ) m = 380 m . ∴ time taken = ( 380 ⁄ 10 ) sec = 38 sec . answer b\"","correct":"b","options":{"a":"3.6 sec ","b":"38 sec ","c":"36 sec ","d":"72 sec","e":"none of these"},"options_float":{"a":3.6,"b":38.0,"c":36.0,"d":72.0,"e":null},"annotated_formula":"multiply(multiply(divide(divide(add(260, 120), const_1000), subtract(45, 9)), const_60), const_60)","linear_formula":"add(n1,n2)|subtract(n3,n0)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_60)|multiply(#4,const_60)|","chain":"260 + 120<\/gadget>\n380<\/output>\n380 \/ 1_000<\/gadget>\n19\/50 = around 0.38<\/output>\n45 - 9<\/gadget>\n36<\/output>\n(19\/50) \/ 36<\/gadget>\n19\/1_800 = around 0.010556<\/output>\n(19\/1_800) * 60<\/gadget>\n19\/30 = around 0.633333<\/output>\n(19\/30) * 60<\/gadget>\n38<\/output>\n38<\/result>","index":397} +{"problem":"mr . evans will states that each of his children will receive an equal share of his estate and that his grandchildren will split a portion of the estate that is equal to the share received by each of his children . if mr . evans has 2 children and 6 grandchildren , then approximately what percentage of mr . evans estate will each grandchild receive ?","rationale":"33.33 \/ 6 = 5.5 answer : c","correct":"c","options":{"a":"20 % ","b":"17 % ","c":"5.5 % ","d":"3.3 %","e":"2.8 %"},"options_float":{"a":20.0,"b":17.0,"c":5.5,"d":3.3,"e":2.8},"annotated_formula":"multiply(divide(divide(divide(const_100, const_3), 6), const_100), const_100)","linear_formula":"divide(const_100,const_3)|divide(#0,n1)|divide(#1,const_100)|multiply(#2,const_100)","chain":"100 \/ 3<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) \/ 6<\/gadget>\n50\/9 = around 5.555556<\/output>\n(50\/9) \/ 100<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/18) * 100<\/gadget>\n50\/9 = around 5.555556<\/output>\n50\/9 = around 5.555556<\/result>","index":398} +{"problem":"find the quotient when total no . of divisors of 16 ! is divided by right most non - zero digit of 15 !","rationale":"first find the factors of 16 ! and then divide with non zero digit of 15 ! factors of 16 ! prime no below 16 are 2 3 5 7 11 13 now highest power of 2 in 16 ! is 15 3 is 65 is 37 is 211 , is 113 is 1 so no of factors is 16 * 7 * 4 * 3 * 2 * 2 = 5376 now last digit in 15 ! it can be written as 2 ^ 11 * 3 ^ 6 * 5 ^ 3 * 7 ^ 2 * 11 ^ 1 * 13 * 1 5 ^ 3 * 2 ^ 3 gives 0 so remaining is 2 ^ 3 = 8 therefore 5376 \/ 8 = 672 answer : a","correct":"a","options":{"a":"672 ","b":"682 ","c":"692 ","d":"745","e":"845"},"options_float":{"a":672.0,"b":682.0,"c":692.0,"d":745.0,"e":845.0},"annotated_formula":"add(subtract(multiply(multiply(16, 15), const_3), const_100), multiply(const_10, const_4))","linear_formula":"multiply(const_10,const_4)|multiply(n0,n1)|multiply(#1,const_3)|subtract(#2,const_100)|add(#0,#3)","chain":"16 * 15<\/gadget>\n240<\/output>\n240 * 3<\/gadget>\n720<\/output>\n720 - 100<\/gadget>\n620<\/output>\n10 * 4<\/gadget>\n40<\/output>\n620 + 40<\/gadget>\n660<\/output>\n660<\/result>","index":399} +{"problem":"if the wheel is 14 cm then the number of revolutions to cover a distance of 1408 cm is ?","rationale":"\"2 * 22 \/ 7 * 14 * x = 1408 = > x = 16 answer : a\"","correct":"a","options":{"a":"16 ","b":"10 ","c":"14 ","d":"12","e":"11"},"options_float":{"a":16.0,"b":10.0,"c":14.0,"d":12.0,"e":11.0},"annotated_formula":"divide(1408, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14))","linear_formula":"multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)|","chain":"3 * 100<\/gadget>\n300<\/output>\n1 * 10<\/gadget>\n10<\/output>\n300 + 10<\/gadget>\n310<\/output>\n310 + 4<\/gadget>\n314<\/output>\n314 \/ 100<\/gadget>\n157\/50 = around 3.14<\/output>\n2 * (157\/50)<\/gadget>\n157\/25 = around 6.28<\/output>\n(157\/25) * 14<\/gadget>\n2_198\/25 = around 87.92<\/output>\n1_408 \/ (2_198\/25)<\/gadget>\n17_600\/1_099 = around 16.014559<\/output>\n17_600\/1_099 = around 16.014559<\/result>","index":400} +{"problem":"calculate the area of a triangle , if the sides of are 65 cm , 60 cm and 25 cm , what is its area ?","rationale":"\"the triangle with sides 65 cm , 60 cm and 25 cm is right angled , where the hypotenuse is 65 cm . area of the triangle = 1 \/ 2 * 60 * 25 = 750 cm 2 answer : b\"","correct":"b","options":{"a":"850 cm 2 ","b":"750 cm 2 ","c":"450 cm 2 ","d":"550 cm 2","e":"650 cm 2"},"options_float":{"a":850.0,"b":750.0,"c":450.0,"d":550.0,"e":650.0},"annotated_formula":"multiply(divide(60, const_2), 25)","linear_formula":"divide(n1,const_2)|multiply(n2,#0)|","chain":"60 \/ 2<\/gadget>\n30<\/output>\n30 * 25<\/gadget>\n750<\/output>\n750<\/result>","index":401} +{"problem":"the admission charge in a water park is $ 1 by adult and $ 0.75 by child accompanying an adult . how much was there of child with an adult who paid in quite $ 3.25 to have access to the water park ?","rationale":"a = # of adults price for on adult = $ 1.00 e = # of kids price for on kid ( accompanying an adult ) = $ 0.75 t = total ( $ ) $ 3.25 t = price for an adult * a + price for on kid * e 3.25 = $ 1 * a + $ 0.75 * e 3.25 = a + $ 0.75 * e because the kids have to be accompanying by an adult , a = 1 $ 3.25 = 1 + $ 0.75 * e $ 3.25 - 1 = 1 - 1 + $ 0.75 * e $ 2.25 = $ 0.75 * e $ 2.25 \/ $ 0.75 = ( $ 0.75 * e ) \/ $ 0.75 3 = e = number of kids accompanying an adult . correct option : c ) 3 kids","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(add(subtract(subtract(multiply(multiply(3.25, 1), const_2), const_0_25), const_0_25), 1), const_4)","linear_formula":"multiply(n0,n2)|multiply(#0,const_2)|subtract(#1,const_0_25)|subtract(#2,const_0_25)|add(n0,#3)|subtract(#4,const_4)","chain":"3.25 * 1<\/gadget>\n3.25<\/output>\n3.25 * 2<\/gadget>\n6.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n6.5 - (1\/4)<\/gadget>\n6.25<\/output>\n6.25 - (1\/4)<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7 - 4<\/gadget>\n3<\/output>\n3<\/result>","index":403} +{"problem":"the area of a square is 4225 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square .","rationale":"\"let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4225 a = 65 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 41 : 130 answer : e\"","correct":"e","options":{"a":"5 : 28 ","b":"5 : 19 ","c":"5 : 12 ","d":"5 : 13","e":"41 : 130"},"options_float":{"a":0.1785714286,"b":0.2631578947,"c":0.4166666667,"d":0.3846153846,"e":0.3153846154},"annotated_formula":"divide(subtract(sqrt(4225), 24), multiply(sqrt(4225), const_2))","linear_formula":"sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)|","chain":"4_225 ** (1\/2)<\/gadget>\n65<\/output>\n65 - 24<\/gadget>\n41<\/output>\n65 * 2<\/gadget>\n130<\/output>\n41 \/ 130<\/gadget>\n41\/130 = around 0.315385<\/output>\n41\/130 = around 0.315385<\/result>","index":407} +{"problem":"jisha walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day , but at the same speed as on the second day . how many miles in total did she walk ?","rationale":"jisha walked 18 miles on the first day , walking 3 miles per hour i . e . total time of walk on day - 1 = 18 \/ 3 = 6 hours second day time of walk = 6 - 1 = 5 hours and speed = 3 + 1 = 4 miles per hour i . e . distance walked on second day = 5 * 4 = 20 miles third day time of walk = 6 hours and speed = 4 miles per hour i . e . distance walked on second day = 6 * 4 = 24 miles total distance travelled on three days = 18 + 20 + 24 = 62 answer : option e","correct":"e","options":{"a":"24 ","b":"44 ","c":"58 ","d":"60","e":"62"},"options_float":{"a":24.0,"b":44.0,"c":58.0,"d":60.0,"e":62.0},"annotated_formula":"add(multiply(speed(18, 3), add(const_1, 3)), add(18, multiply(subtract(speed(18, 3), const_1), add(const_1, 3))))","linear_formula":"add(n0,const_1)|speed(n1,n0)|multiply(#0,#1)|subtract(#1,const_1)|multiply(#0,#3)|add(n1,#4)|add(#5,#2)","chain":"18 \/ 3<\/gadget>\n6<\/output>\n1 + 3<\/gadget>\n4<\/output>\n6 * 4<\/gadget>\n24<\/output>\n6 - 1<\/gadget>\n5<\/output>\n5 * 4<\/gadget>\n20<\/output>\n18 + 20<\/gadget>\n38<\/output>\n24 + 38<\/gadget>\n62<\/output>\n62<\/result>","index":408} +{"problem":"the spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls . the diameters of two of these are 1 1 \/ 2 cm and 2 cm respectively . the diameter of third ball is ?","rationale":"\"4 \/ 3 π * 3 * 3 * 3 = 4 \/ 3 π [ ( 3 \/ 2 ) 3 + 23 + r 3 ] r = 1.25 d = 2.5 answer : b\"","correct":"b","options":{"a":"2.8 ","b":"2.5 ","c":"2.2 ","d":"2.0","e":"2.1"},"options_float":{"a":2.8,"b":2.5,"c":2.2,"d":2.0,"e":2.1},"annotated_formula":"multiply(power(subtract(subtract(power(divide(3, 1), 3), power(divide(add(1, divide(1, 1)), 1), 3)), 1), inverse(3)), 1)","linear_formula":"divide(n0,n3)|divide(n2,n3)|inverse(n0)|add(n2,#1)|power(#0,n0)|divide(#3,n3)|power(#5,n0)|subtract(#4,#6)|subtract(#7,n2)|power(#8,#2)|multiply(#9,n3)|","chain":"3 \/ 1<\/gadget>\n3<\/output>\n3 ** 3<\/gadget>\n27<\/output>\n1 \/ 1<\/gadget>\n1<\/output>\n1 + 1<\/gadget>\n2<\/output>\n2 \/ 1<\/gadget>\n2<\/output>\n2 ** 3<\/gadget>\n8<\/output>\n27 - 8<\/gadget>\n19<\/output>\n19 - 1<\/gadget>\n18<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n18 ** (1\/3)<\/gadget>\n18**(1\/3) = around 2.620741<\/output>\n(18**(1\/3)) * 1<\/gadget>\n18**(1\/3) = around 2.620741<\/output>\n18**(1\/3) = around 2.620741<\/result>","index":409} +{"problem":"ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 2 percent and the mobile phone at a profit of 10 percent . overall he make a .","rationale":"\"let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 2 \/ 100 ) = 15000 - 300 m = 8000 ( 1 + 10 \/ 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 300 + 800 = rs . 500 as this is positive , an overall profit of rs . 500 was made . answer : c\"","correct":"c","options":{"a":"228 ","b":"288 ","c":"500 ","d":"200","e":"881"},"options_float":{"a":228.0,"b":288.0,"c":500.0,"d":200.0,"e":881.0},"annotated_formula":"subtract(add(multiply(15000, subtract(const_1, divide(2, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0)|","chain":"2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n1 - (1\/50)<\/gadget>\n49\/50 = around 0.98<\/output>\n15_000 * (49\/50)<\/gadget>\n14_700<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n8_000 * (11\/10)<\/gadget>\n8_800<\/output>\n14_700 + 8_800<\/gadget>\n23_500<\/output>\n15_000 + 8_000<\/gadget>\n23_000<\/output>\n23_500 - 23_000<\/gadget>\n500<\/output>\n500<\/result>","index":411} +{"problem":"on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 28 inch from the base of the mountain . how many km is he from the base of the mountain ?","rationale":"\"explanation : since 312 inch = 136 km so 1 inch = 136 \/ 312 km so 28 inch = ( 136 ã — 28 ) \/ 312 = 12.2 km answer : b\"","correct":"b","options":{"a":"14.83 ","b":"12.2 ","c":"14.8 ","d":"14.82","e":"14.12"},"options_float":{"a":14.83,"b":12.2,"c":14.8,"d":14.82,"e":14.12},"annotated_formula":"divide(multiply(28, 136), 312)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"28 * 136<\/gadget>\n3_808<\/output>\n3_808 \/ 312<\/gadget>\n476\/39 = around 12.205128<\/output>\n476\/39 = around 12.205128<\/result>","index":412} +{"problem":"positive integer y is 50 percent of 25 percent of positive integer x , and y percent of x equals 100 . what is the value of y ?","rationale":"y = 50 % of 25 % 0 f x = x \/ 8 and y \/ 100 of x = 100 y \/ 100 * 8 y = 100 y = 125 answer - b","correct":"b","options":{"a":"50 ","b":"125 ","c":"200 ","d":"1,000","e":"2,000"},"options_float":{"a":50.0,"b":125.0,"c":200.0,"d":1000.0,"e":2000.0},"annotated_formula":"multiply(multiply(divide(50, 100), divide(25, const_100)), const_1000)","linear_formula":"divide(n0,n2)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_1000)","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) * (1\/4)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 1_000<\/gadget>\n125<\/output>\n125<\/result>","index":413} +{"problem":"in simple interest what sum amounts of rs . 1120 \/ - in 4 years and rs . 1200 \/ - in 5 years ?","rationale":"for 1 year interest = 1200 - 1120 = 80 for 4 year 320 then sum = 1120 - 320 = 800 answer : c","correct":"c","options":{"a":"rs . 500 ","b":"rs . 600 ","c":"rs . 800 ","d":"rs . 900","e":"rs . 700"},"options_float":{"a":500.0,"b":600.0,"c":800.0,"d":900.0,"e":700.0},"annotated_formula":"subtract(1120, multiply(4, subtract(1200, 1120)))","linear_formula":"subtract(n2,n0)|multiply(n1,#0)|subtract(n0,#1)","chain":"1_200 - 1_120<\/gadget>\n80<\/output>\n4 * 80<\/gadget>\n320<\/output>\n1_120 - 320<\/gadget>\n800<\/output>\n800<\/result>","index":414} +{"problem":"the current of a stream runs at the rate of 4 kmph . a boat goes 6 km and back to the starting point in 2 hours , then find the speed of the boat in still water ?","rationale":"s = 4 m = x ds = x + 4 us = x - 4 6 \/ ( x + 4 ) + 6 \/ ( x - 4 ) = 2 x = 8 answer : c","correct":"c","options":{"a":"5 ","b":"6 ","c":"8 ","d":"4","e":"1"},"options_float":{"a":5.0,"b":6.0,"c":8.0,"d":4.0,"e":1.0},"annotated_formula":"divide(power(4, 2), 2)","linear_formula":"power(n0,n2)|divide(#0,n2)","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":416} +{"problem":"cathy and jim begin running at the same time and they start at the same place on a straight path . cathy runs at 10 miles per hour , and jim runs at 6 miles per hour . after 18 minutes , cathy stops to stretch . if it takes cathy 27 minutes to stretch and jim continues to run during this time , how many minutes will it take cathy to catch up to jim ?","rationale":"in 18 minutes , cathy runs 3 miles . in 45 minutes , jim runs 4.5 miles . cathy can catch jim at a rate of 4 miles per hour . since jim is ahead by 1.5 miles , it will take cathy 1.5 \/ 4 hours = 22.5 minutes to catch jim . the answer is c .","correct":"c","options":{"a":"20.5 ","b":"21.5 ","c":"22.5 ","d":"23.5","e":"24.5"},"options_float":{"a":20.5,"b":21.5,"c":22.5,"d":23.5,"e":24.5},"annotated_formula":"divide(subtract(multiply(6, divide(add(18, 27), const_60)), multiply(10, divide(18, const_60))), divide(subtract(10, 6), const_60))","linear_formula":"add(n2,n3)|divide(n2,const_60)|subtract(n0,n1)|divide(#0,const_60)|divide(#2,const_60)|multiply(n0,#1)|multiply(n1,#3)|subtract(#6,#5)|divide(#7,#4)","chain":"18 + 27<\/gadget>\n45<\/output>\n45 \/ 60<\/gadget>\n3\/4 = around 0.75<\/output>\n6 * (3\/4)<\/gadget>\n9\/2 = around 4.5<\/output>\n18 \/ 60<\/gadget>\n3\/10 = around 0.3<\/output>\n10 * (3\/10)<\/gadget>\n3<\/output>\n(9\/2) - 3<\/gadget>\n3\/2 = around 1.5<\/output>\n10 - 6<\/gadget>\n4<\/output>\n4 \/ 60<\/gadget>\n1\/15 = around 0.066667<\/output>\n(3\/2) \/ (1\/15)<\/gadget>\n45\/2 = around 22.5<\/output>\n45\/2 = around 22.5<\/result>","index":417} +{"problem":"the speed of a car is 65 km in the first hour and 45 km in the second hour . what is the average speed of the car ?","rationale":"\"s = ( 65 + 45 ) \/ 2 = 55 kmph answer : b\"","correct":"b","options":{"a":"72 kmph ","b":"55 kmph ","c":"50 kmph ","d":"80 kmph","e":"82 kmph"},"options_float":{"a":72.0,"b":55.0,"c":50.0,"d":80.0,"e":82.0},"annotated_formula":"divide(add(65, 45), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"65 + 45<\/gadget>\n110<\/output>\n110 \/ 2<\/gadget>\n55<\/output>\n55<\/result>","index":418} +{"problem":"set x consists of all two - digit primes and set y consists of all positive multiples of 4 less than 100 . if the two sets are combined into one , what will be the range of the new set ?","rationale":"\"set x = { 11 , 13 , 17 , . . . . . . . . . . . . . , 83 , 89 , 97 } set y = { 4 , 8 , 12 , . . . . . . . . . . . . . . . , 88 , 92 , 96 } combining two sets , say set z set z = { 4,8 , 11 , 12,13 , 17 , . . . . . . . . . . . . . . . . . . . , 83 , 88 , 89 , 92 , 96 , 97 } range = max value - min value range ( z ) = 97 - 4 = 93 oa b is the answer .\"","correct":"b","options":{"a":"84 ","b":"93 ","c":"90 ","d":"92","e":"95"},"options_float":{"a":84.0,"b":93.0,"c":90.0,"d":92.0,"e":95.0},"annotated_formula":"subtract(subtract(100, const_3), 4)","linear_formula":"subtract(n1,const_3)|subtract(#0,n0)|","chain":"100 - 3<\/gadget>\n97<\/output>\n97 - 4<\/gadget>\n93<\/output>\n93<\/result>","index":419} +{"problem":"car z travels 50 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ?","rationale":"\"the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 50 miles \/ hour , the car will run 40 miles \/ gallon and at 60 miles \/ hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 40 miles \/ gallon at a constant speed of 60 miles \/ gallon . with 10 gallons of fuel , the car can therefore travel 40 miles \/ gallon * 10 gallons = 400 miles . answer c .\"","correct":"c","options":{"a":"320 ","b":"375.2 ","c":"400 ","d":"408.3","e":"440"},"options_float":{"a":320.0,"b":375.2,"c":400.0,"d":408.3,"e":440.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), 50), 10)","linear_formula":"divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 50<\/gadget>\n40<\/output>\n40 * 10<\/gadget>\n400<\/output>\n400<\/result>","index":420} +{"problem":"30 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ?","rationale":"\"i came up with ( e ) 75 think of 100 people total : from the first fact , 30 of these are women with fair hair . from the second fact , these 20 women make up 40 % of the total fair haired population . we can then make a ratio of 60 : 40 fair haired men to fair haired women . this means that ( 60 \/ 40 ) * 30 equals the number of fair haired men , which is 45 men with fair hair . add this 45 to the 30 women and get 75 fair haired men and women out of 100 total men and women . 75 % e\"","correct":"e","options":{"a":"25 ","b":"30 ","c":"50 ","d":"55","e":"75"},"options_float":{"a":25.0,"b":30.0,"c":50.0,"d":55.0,"e":75.0},"annotated_formula":"multiply(divide(30, 40), const_100)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|","chain":"30 \/ 40<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n75<\/result>","index":421} +{"problem":"the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 255 , then the mean profit for the last 15 days would be","rationale":"\"average would be : 350 = ( 255 + x ) \/ 2 on solving , x = 445 . answer : b\"","correct":"b","options":{"a":"rs . 200 ","b":"rs . 445 ","c":"rs . 275 ","d":"rs . 425","e":"none of these"},"options_float":{"a":200.0,"b":445.0,"c":275.0,"d":425.0,"e":null},"annotated_formula":"divide(subtract(multiply(30, 350), multiply(15, 255)), 15)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n3)|","chain":"30 * 350<\/gadget>\n10_500<\/output>\n15 * 255<\/gadget>\n3_825<\/output>\n10_500 - 3_825<\/gadget>\n6_675<\/output>\n6_675 \/ 15<\/gadget>\n445<\/output>\n445<\/result>","index":422} +{"problem":"a rectangle the length of which is 8 inches and the width of which is 6 inches is made up 48 1 inch by 1 inch squares . through how many of the squares does a diagonal of the rectangle pass ?","rationale":"1 ) if the two sides are co - prime , sum of sides - 1 . . 2 ) and if not co - prime , sum of sides - hcf of sides . . the difference is related to the diagonal passing through a vertex of squares at some place . . here 6 and 8 are not co - primes , so ans = 6 + 8 − hcf ( 6,8 ) = 14 − 2 = 12 answer : d","correct":"d","options":{"a":"6 ","b":"8 ","c":"10 ","d":"12","e":"16"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":12.0,"e":16.0},"annotated_formula":"subtract(add(8, 6), const_2)","linear_formula":"add(n0,n1)|subtract(#0,const_2)","chain":"8 + 6<\/gadget>\n14<\/output>\n14 - 2<\/gadget>\n12<\/output>\n12<\/result>","index":423} +{"problem":"if x + y = 3 and x 2 y 3 + y 2 x 3 = 27 , what is the value of xy ?","rationale":"\"xy = 3 as x + y = 3 x 2 y 3 + y 2 x 3 = 27 x 2 y 2 ( y + x ) = 27 substituting x + y x 2 y 2 = 9 xy = 3 answer : c\"","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"sqrt(divide(27, 3))","linear_formula":"divide(n5,n0)|sqrt(#0)|","chain":"27 \/ 3<\/gadget>\n9<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":424} +{"problem":"a 600 meter long train crosses a signal post in 40 seconds . how long will it take to cross a 7.2 kilometer long bridge , at the same speed ?","rationale":"\"s = 600 \/ 40 = 15 mps s = 7200 \/ 15 = 480 sec = 8 min . answer : c\"","correct":"c","options":{"a":"4 min ","b":"2 min ","c":"8 min ","d":"9 min","e":"3 min"},"options_float":{"a":4.0,"b":2.0,"c":8.0,"d":9.0,"e":3.0},"annotated_formula":"divide(divide(multiply(7.2, const_1000), speed(600, 40)), const_60)","linear_formula":"multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)|","chain":"7.2 * 1_000<\/gadget>\n7_200<\/output>\n600 \/ 40<\/gadget>\n15<\/output>\n7_200 \/ 15<\/gadget>\n480<\/output>\n480 \/ 60<\/gadget>\n8<\/output>\n8<\/result>","index":428} +{"problem":"a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 100 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?","rationale":"\"many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 100 kilometers per hour is 1 \/ 100 hours = 3,600 \/ 100 seconds = 36 seconds ; time to cover this distance at regular speed is 36 + 2 = 38 seconds = 38 \/ 3,600 hours = 4 \/ 379 hours ; so , we get that to cover 1 kilometer 1 \/ ( 4 \/ 379 ) hours is needed - - > regular speed 94.7 kilometers per hour ( rate is a reciprocal of time or rate = distance \/ time ) . answer : e .\"","correct":"e","options":{"a":"71.5 ","b":"72 ","c":"72.5 ","d":"73","e":"94.7"},"options_float":{"a":71.5,"b":72.0,"c":72.5,"d":73.0,"e":94.7},"annotated_formula":"divide(1, divide(add(multiply(const_3600, divide(1, 100)), 2), const_3600))","linear_formula":"divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)|","chain":"1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n3_600 * (1\/100)<\/gadget>\n36<\/output>\n36 + 2<\/gadget>\n38<\/output>\n38 \/ 3_600<\/gadget>\n19\/1_800 = around 0.010556<\/output>\n1 \/ (19\/1_800)<\/gadget>\n1_800\/19 = around 94.736842<\/output>\n1_800\/19 = around 94.736842<\/result>","index":430} +{"problem":"a can do a piece of work in 5 hours ; b and c together can do it in 3 hours , while a and c together can do it in 2 hours . how long will b alone take to do it ?","rationale":"\"a ' s 1 hour ' s work = 1 \/ 5 ; ( b + c ) ' s 1 hour ' s work = 1 \/ 3 ; ( a + c ) ' s 1 hour ' s work = 1 \/ 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 \/ 5 + 1 \/ 3 ) = 8 \/ 15 . b ' s 1 hour ' s work = ( 8 \/ 15 - 1 \/ 2 ) = 1 \/ 30 . therefore a alone will take 30 hours to do the work . a\"","correct":"a","options":{"a":"30 ","b":"20 ","c":"10 ","d":"6","e":"5"},"options_float":{"a":30.0,"b":20.0,"c":10.0,"d":6.0,"e":5.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 5))))","linear_formula":"divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/2) - (1\/5)<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/3) - (3\/10)<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ (1\/30)<\/gadget>\n30<\/output>\n30<\/result>","index":433} +{"problem":"lally ' s internet provider offers unlimited traffic which costs $ 0.5 per day charged off at 12 a . m . the service is discontinued when client ' s debt exceeds $ 5 . if today in the morning lally makes a payment of $ 7 , for how many days , including today , will she be connected to the internet without any additional payments , given her initial balance was $ 0 ?","rationale":"day 0 - payment of $ 7 is made in the morning ( after 12 am midnight ) day 1 onwards ( after 12 am midnight ) internet service will begin and will run for 7 * 2 = 14 days i . e . at the end of day 15 balance will be zero . now lally can still use the internet till her debt reaches $ 5 which means total of 5 * 2 = 10 days . now the question states for how many days , including today , will she be connected to the internet without any additional payments i . e . we will have to include day 0 also in our final answer . therefore total no . of days = 1 + 14 + 10 = 25 = c","correct":"c","options":{"a":"15 ","b":"24 ","c":"25 ","d":"26","e":"13"},"options_float":{"a":15.0,"b":24.0,"c":25.0,"d":26.0,"e":13.0},"annotated_formula":"subtract(multiply(12, 5), multiply(multiply(7, 0.5), const_10))","linear_formula":"multiply(n1,n2)|multiply(n0,n3)|multiply(#1,const_10)|subtract(#0,#2)","chain":"12 * 5<\/gadget>\n60<\/output>\n7 * 0.5<\/gadget>\n3.5<\/output>\n3.5 * 10<\/gadget>\n35<\/output>\n60 - 35<\/gadget>\n25<\/output>\n25<\/result>","index":435} +{"problem":"a retailer bought a machine at a wholesale price of $ 108 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ?","rationale":"\"my solution : wholesale price = 108 retail price , be = x he provides 10 % discount on retail price = x - 10 x \/ 100 this retail price = 20 % profit on wholesale price x - 10 x \/ 100 = 108 + 1 \/ 5 ( 108 ) x = 144 ; answer : e\"","correct":"e","options":{"a":"81 ","b":"100 ","c":"120 ","d":"135","e":"144"},"options_float":{"a":81.0,"b":100.0,"c":120.0,"d":135.0,"e":144.0},"annotated_formula":"divide(multiply(add(108, divide(multiply(108, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10))","linear_formula":"multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)|","chain":"108 * 20<\/gadget>\n2_160<\/output>\n2_160 \/ 100<\/gadget>\n108\/5 = around 21.6<\/output>\n108 + (108\/5)<\/gadget>\n648\/5 = around 129.6<\/output>\n(648\/5) * 100<\/gadget>\n12_960<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n12_960 \/ 90<\/gadget>\n144<\/output>\n144<\/result>","index":436} +{"problem":"in one day , juan sends keith 8 times as many messages as he sends laurence , and laurence sends missy 4.5 times as many messages as he receives from juan . if missy received 18 messages from laurence , how many messages does keith receive from juan ?","rationale":"juan - - > laurence = x messages ; juan - - > keith = 8 x messages ( juan sends keith eight times as many messages as he sends laurence ) ; laurence - - > missy = 4.5 x = 18 ( laurence sends missy 4.5 times as many messages as he receives from juan ) . 4.5 x = 18 - - > x = 4 - - > 8 x = 32 . answer : c .","correct":"c","options":{"a":"13 ","b":"26 ","c":"32 ","d":"48","e":"56"},"options_float":{"a":13.0,"b":26.0,"c":32.0,"d":48.0,"e":56.0},"annotated_formula":"multiply(divide(18, 4.5), 8)","linear_formula":"divide(n2,n1)|multiply(n0,#0)","chain":"18 \/ 4.5<\/gadget>\n4<\/output>\n4 * 8<\/gadget>\n32<\/output>\n32<\/result>","index":438} +{"problem":"edward earns $ 7.00 per hour for the first 40 hours he works per week , and twice this rate for overtime . if michael earned $ 210 last week , how many hours did he work ?","rationale":"$ 7 * 40 + $ 12 * x = $ 210 - - > x = 5 hours . total working hours = 40 + 5 = 45 answer : c .","correct":"c","options":{"a":"43 ","b":"44 ","c":"45 ","d":"46","e":"47"},"options_float":{"a":43.0,"b":44.0,"c":45.0,"d":46.0,"e":47.0},"annotated_formula":"add(divide(subtract(multiply(7, 40), 210), multiply(7, const_2)), 40)","linear_formula":"multiply(n0,n1)|multiply(n0,const_2)|subtract(#0,n2)|divide(#2,#1)|add(n1,#3)","chain":"7 * 40<\/gadget>\n280<\/output>\n280 - 210<\/gadget>\n70<\/output>\n7 * 2<\/gadget>\n14<\/output>\n70 \/ 14<\/gadget>\n5<\/output>\n5 + 40<\/gadget>\n45<\/output>\n45<\/result>","index":440} +{"problem":"i bought two books ; for rs . 470 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ?","rationale":"\"x * ( 85 \/ 100 ) = ( 470 - x ) 119 \/ 100 x = 274.17 answer : a\"","correct":"a","options":{"a":"274.17 ","b":"280.17 ","c":"299.17 ","d":"266.17","e":"279.17"},"options_float":{"a":274.17,"b":280.17,"c":299.17,"d":266.17,"e":279.17},"annotated_formula":"divide(multiply(470, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19)))","linear_formula":"add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|","chain":"100 + 19<\/gadget>\n119<\/output>\n470 * 119<\/gadget>\n55_930<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 + 119<\/gadget>\n204<\/output>\n55_930 \/ 204<\/gadget>\n1_645\/6 = around 274.166667<\/output>\n1_645\/6 = around 274.166667<\/result>","index":441} +{"problem":"18 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ?","rationale":"\"1 women ' s 1 day work = 1 \/ 126 1 child ' s 1 day work = 1 \/ 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 \/ 126 + 10 \/ 140 ) = 1 \/ 9 5 women and 10 children will complete the work in 9 days . answer : d\"","correct":"d","options":{"a":"8 days ","b":"6 days ","c":"7 days ","d":"9 days","e":"1 days"},"options_float":{"a":8.0,"b":6.0,"c":7.0,"d":9.0,"e":1.0},"annotated_formula":"inverse(add(divide(5, multiply(18, 7)), divide(18, multiply(18, 14))))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)|","chain":"18 * 7<\/gadget>\n126<\/output>\n5 \/ 126<\/gadget>\n5\/126 = around 0.039683<\/output>\n18 * 14<\/gadget>\n252<\/output>\n18 \/ 252<\/gadget>\n1\/14 = around 0.071429<\/output>\n(5\/126) + (1\/14)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9<\/result>","index":442} +{"problem":"when positive integer x is divided by positive integer y , the remainder is 9 . if x \/ y = 96.45 , what is the value of y ?","rationale":"\"guys , one more simple funda . 5 \/ 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 \/ 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 \/ 5 = 6.4 now . 4 x 5 = 2 is the remainder given x \/ y = 96.45 and remainder is 9 so . 45 x y = 9 hence y = 20 ans d\"","correct":"d","options":{"a":"96 ","b":"75 ","c":"48 ","d":"20","e":"12"},"options_float":{"a":96.0,"b":75.0,"c":48.0,"d":20.0,"e":12.0},"annotated_formula":"divide(9, subtract(96.45, floor(96.45)))","linear_formula":"floor(n1)|subtract(n1,#0)|divide(n0,#1)|","chain":"floor(96.45)<\/gadget>\n96<\/output>\n96.45 - 96<\/gadget>\n0.45<\/output>\n9 \/ 0.45<\/gadget>\n20<\/output>\n20<\/result>","index":443} +{"problem":"an employee ’ s annual salary was increased $ 5000 . if her new annual salary now equals $ 25000 what was the percent increase ?","rationale":"new annual salary = $ 25000 salary increase = $ 5000 . original salary = $ 25000 - $ 5000 . = $ 20000 % increase = ( $ 15000 \/ $ 75000 ) * 100 = 80 % hence b .","correct":"b","options":{"a":"15 % ","b":"80 % ","c":"20 % ","d":"22 %","e":"24 %"},"options_float":{"a":15.0,"b":80.0,"c":20.0,"d":22.0,"e":24.0},"annotated_formula":"multiply(divide(subtract(25000, 5000), 25000), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)","chain":"25_000 - 5_000<\/gadget>\n20_000<\/output>\n20_000 \/ 25_000<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":444} +{"problem":"of the 200 employees in a certain company , 30 percent will be relocated to city x and the remaining 70 percent will be relocated to city y . however , 40 percent of the employees prefer city y and 60 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ?","rationale":"120 prefer x ( group 1 ) ; 80 prefer y ( group 2 ) . city y needs 140 people : letall 80 who prefer y ( entire group 2 ) be relocated there , the rest 60 will be those who prefer x from group 1 ; city x needs 60 people : 120 - 60 = 60 from group 1 will be relocated to x , which they prefer . so , the highest possible number of employees who will be relocated to the city they prefer is 80 + 60 = 140 . answer : c .","correct":"c","options":{"a":"115 ","b":"120 ","c":"140 ","d":"152","e":"165"},"options_float":{"a":115.0,"b":120.0,"c":140.0,"d":152.0,"e":165.0},"annotated_formula":"add(multiply(40, const_2), 60)","linear_formula":"multiply(n3,const_2)|add(n4,#0)","chain":"40 * 2<\/gadget>\n80<\/output>\n80 + 60<\/gadget>\n140<\/output>\n140<\/result>","index":446} +{"problem":"the arithmetic mean and standard deviation of a certain normal distribution are 14.5 and 1.7 , respectively . what value is exactly 2 standard deviations less than the mean ?","rationale":"\"mean = 14.5 two standard deviations is 1.7 + 1.7 = 3.4 there could be two calues for this . mean + two standard deviations = 17.9 mean - two standard deviations = 11.1 answer choice has 11.1 and so c is the answer .\"","correct":"c","options":{"a":"10.5 ","b":"11 ","c":"11.1 ","d":"12","e":"12.5"},"options_float":{"a":10.5,"b":11.0,"c":11.1,"d":12.0,"e":12.5},"annotated_formula":"subtract(14.5, multiply(2, 1.7))","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|","chain":"2 * 1.7<\/gadget>\n3.4<\/output>\n14.5 - 3.4<\/gadget>\n11.1<\/output>\n11.1<\/result>","index":447} +{"problem":"what is the x - intercept of a line that passes through the point ( 3 , 4 ) and has a slope of 2 ?","rationale":"first , plot the point ( 3 , 4 ) on a graph . a slope of 2 means “ up 2 , over 1 , ” so plot this point on a graph , too . here ’ s the resulting graph : y over 1 up 2 ( 1 , 0 ) ( 2 , 2 ) ( 3 , 4 ) ( 4 , 6 ) x the line passes through the x - axis at 1 , making the correct answer d ) 1","correct":"d","options":{"a":"- 2 ","b":"- 1 ","c":"0 ","d":"1","e":"2"},"options_float":{"a":-2.0,"b":-1.0,"c":0.0,"d":1.0,"e":2.0},"annotated_formula":"divide(subtract(multiply(2, 3), 4), 2)","linear_formula":"multiply(n0,n2)|subtract(#0,n1)|divide(#1,n2)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 - 4<\/gadget>\n2<\/output>\n2 \/ 2<\/gadget>\n1<\/output>\n1<\/result>","index":450} +{"problem":"raja spends 35 % of his monthly income on household items , 20 % of his monthly income on buying cloths , 5 % of his monthly income on medicines and saves the remaining amount which is rs . 15000 . find his monthly income .","rationale":"savings 40 % - 15000 expenditure 60 % - 22500 total - 37500 answer : c","correct":"c","options":{"a":"rs . 40000 ","b":"rs . 36000 ","c":"rs . 37500 ","d":"rs . 45000","e":"none of these"},"options_float":{"a":40000.0,"b":36000.0,"c":37500.0,"d":45000.0,"e":null},"annotated_formula":"divide(15000, subtract(const_1, add(add(divide(35, const_100), divide(20, const_100)), divide(5, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|add(#0,#1)|add(#3,#2)|subtract(const_1,#4)|divide(n3,#5)","chain":"35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(7\/20) + (1\/5)<\/gadget>\n11\/20 = around 0.55<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(11\/20) + (1\/20)<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n15_000 \/ (2\/5)<\/gadget>\n37_500<\/output>\n37_500<\/result>","index":451} +{"problem":"what least number should be subtracted from 13600 such that the remainder is divisible by 87 ?","rationale":"\"13600 ÷ 87 = 156 , remainder = 28 hence 28 is the least number which can be subtracted from 13600 such that the remainder is divisible by 87 answer is c\"","correct":"c","options":{"a":"27 ","b":"29 ","c":"28 ","d":"30","e":"31"},"options_float":{"a":27.0,"b":29.0,"c":28.0,"d":30.0,"e":31.0},"annotated_formula":"reminder(13600, 87)","linear_formula":"reminder(n0,n1)|","chain":"13_600 % 87<\/gadget>\n28<\/output>\n28<\/result>","index":452} +{"problem":"list k consists of 12 consecutive integers . if - 5 is the least integer in list k , what is the range of the positive integers in list k ?","rationale":"\"answer = a = 5 if least = - 5 , then largest = 6 range = 6 - 1 = 5\"","correct":"a","options":{"a":"5 ","b":"6 ","c":"7 ","d":"11","e":"12"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":11.0,"e":12.0},"annotated_formula":"subtract(subtract(subtract(12, 5), const_1), const_1)","linear_formula":"subtract(n0,n1)|subtract(#0,const_1)|subtract(#1,const_1)|","chain":"12 - 5<\/gadget>\n7<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6 - 1<\/gadget>\n5<\/output>\n5<\/result>","index":455} +{"problem":"a dishonest dealer professes to sell goods at the cost price but uses a weight of 920 grams per kg , what is his percent ?","rationale":"\"920 - - - 80 100 - - - ? = > 8.7 % answer : e\"","correct":"e","options":{"a":"22 % ","b":"25 % ","c":"77 % ","d":"99 %","e":"8.7 %"},"options_float":{"a":22.0,"b":25.0,"c":77.0,"d":99.0,"e":8.7},"annotated_formula":"subtract(multiply(divide(const_100, 920), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)","linear_formula":"add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|","chain":"100 \/ 920<\/gadget>\n5\/46 = around 0.108696<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n(5\/46) * 1_000<\/gadget>\n2_500\/23 = around 108.695652<\/output>\n(2_500\/23) - 100<\/gadget>\n200\/23 = around 8.695652<\/output>\n200\/23 = around 8.695652<\/result>","index":456} +{"problem":"a case contains c cartons . each carton contains b boxes , and each box contains 600 paper clips . how many paper clips are contained in 2 cases ?","rationale":"2 cases * c cartons \/ case * b boxes \/ carton * 600 clips \/ box = 1200 bc paper clips the answer is c .","correct":"c","options":{"a":"600 bc ","b":"600 b \/ c ","c":"1200 bc ","d":"1200 b \/ c","e":"1200 \/ bc"},"options_float":{"a":600.0,"b":600.0,"c":1200.0,"d":1200.0,"e":1200.0},"annotated_formula":"multiply(2, 600)","linear_formula":"multiply(n0,n1)","chain":"2 * 600<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":458} +{"problem":"an army ’ s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 10000 people who applied for the army and 350 were finally selected , what was the value of n ?","rationale":"\"fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 16000 = 8000 step ( 4 ) 50 % of 8000 = 4000 step ( 5 ) 50 % of 4000 = 2000 step ( 6 ) 50 % of 2000 = 1000 step ( 7 ) 50 % of 1000 = 500 last step of accepting 70 % of 500 = 350 ( our target ) total 8 steps required . ans d\"","correct":"d","options":{"a":"4 ","b":"5 ","c":"6 ","d":"9","e":"8"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":9.0,"e":8.0},"annotated_formula":"add(add(const_2, add(const_1, const_4)), const_2)","linear_formula":"add(const_1,const_4)|add(#0,const_2)|add(#1,const_2)|","chain":"1 + 4<\/gadget>\n5<\/output>\n2 + 5<\/gadget>\n7<\/output>\n7 + 2<\/gadget>\n9<\/output>\n9<\/result>","index":459} +{"problem":"a ladder 30 feet long is leaning against a wall that is perpendicular to level ground . the bottom of the ladder is 10 feet from the base of the wall . if the top of the ladder slips down 5 feet , how many feet will the bottom of the ladder slip ?","rationale":"30 ^ 2 - 10 ^ 2 = 800 it means that the height is equal to 28.28 . since the top of the ladder slips down 5 feet , then the height of the wall = 28.28 - 5 = 23.28 the bottom = sqrt ( 30 ^ 2 - 23.28 ^ 2 ) = sqrt ( 900 - 541.95 ) = 18.92 ans is c","correct":"c","options":{"a":"14 ","b":"15 ","c":"18.92 ","d":"19","e":"25"},"options_float":{"a":14.0,"b":15.0,"c":18.92,"d":19.0,"e":25.0},"annotated_formula":"sqrt(subtract(power(30, const_2), power(subtract(sqrt(subtract(power(30, const_2), power(10, const_2))), 5), const_2)))","linear_formula":"power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|subtract(#3,n2)|power(#4,const_2)|subtract(#0,#5)|sqrt(#6)","chain":"30 ** 2<\/gadget>\n900<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n900 - 100<\/gadget>\n800<\/output>\n800 ** (1\/2)<\/gadget>\n20*sqrt(2) = around 28.284271<\/output>\n(20*sqrt(2)) - 5<\/gadget>\n-5 + 20*sqrt(2) = around 23.284271<\/output>\n(-5 + 20*sqrt(2)) ** 2<\/gadget>\n(-5 + 20*sqrt(2))**2 = around 542.157288<\/output>\n900 - ((-5 + 20*sqrt(2))**2)<\/gadget>\n900 - (-5 + 20*sqrt(2))**2 = around 357.842712<\/output>\n(900 - (-5 + 20*sqrt(2))**2) ** (1\/2)<\/gadget>\nsqrt(900 - (-5 + 20*sqrt(2))**2) = around 18.916731<\/output>\nsqrt(900 - (-5 + 20*sqrt(2))**2) = around 18.916731<\/result>","index":461} +{"problem":"if 3 workers collect 48 kg of cotton in 4 days , how many kg of cotton will 9 workers collect in 2 days ?","rationale":"( 3 * 4 ) \/ 48 = ( 9 * 2 ) \/ x x = 72 kg answer : c","correct":"c","options":{"a":"88 kg ","b":"65 kg ","c":"72 kg ","d":"44 kg","e":"55 kg"},"options_float":{"a":88.0,"b":65.0,"c":72.0,"d":44.0,"e":55.0},"annotated_formula":"multiply(multiply(divide(divide(48, 3), 4), 9), 2)","linear_formula":"divide(n1,n0)|divide(#0,n2)|multiply(n3,#1)|multiply(n4,#2)","chain":"48 \/ 3<\/gadget>\n16<\/output>\n16 \/ 4<\/gadget>\n4<\/output>\n4 * 9<\/gadget>\n36<\/output>\n36 * 2<\/gadget>\n72<\/output>\n72<\/result>","index":462} +{"problem":"evaluate : | 8 - 8 ( 3 - 12 ) | - | 5 - 11 | = ?","rationale":"\"according to order of operations , inner brackets first . hence | 8 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 8 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 8 + 72 | - | 5 - 11 | = | 80 | - | - 6 | = 80 - 6 = 74 correct answer b ) 74\"","correct":"b","options":{"a":"40 ","b":"74 ","c":"60 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":74.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"subtract(subtract(8, multiply(8, subtract(3, 12))), negate(subtract(5, 11)))","linear_formula":"subtract(n2,n3)|subtract(n4,n5)|multiply(n1,#0)|negate(#1)|subtract(n0,#2)|subtract(#4,#3)|","chain":"3 - 12<\/gadget>\n-9<\/output>\n8 * (-9)<\/gadget>\n-72<\/output>\n8 - (-72)<\/gadget>\n80<\/output>\n5 - 11<\/gadget>\n-6<\/output>\n-(-6)<\/gadget>\n6<\/output>\n80 - 6<\/gadget>\n74<\/output>\n74<\/result>","index":467} +{"problem":"heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 25 litres of tomato juice ?","rationale":"answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 25 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 3.125 liters . answer : b","correct":"b","options":{"a":"2.8 liters . ","b":"3.125 liters . ","c":"8.5 liters . ","d":"2.6 liters .","e":"2.1 liters ."},"options_float":{"a":2.8,"b":3.125,"c":8.5,"d":2.6,"e":2.1},"annotated_formula":"divide(multiply(divide(subtract(const_100, 90), const_100), 25), divide(subtract(const_100, 20), const_100))","linear_formula":"subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)","chain":"100 - 90<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 25<\/gadget>\n5\/2 = around 2.5<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(5\/2) \/ (4\/5)<\/gadget>\n25\/8 = around 3.125<\/output>\n25\/8 = around 3.125<\/result>","index":468} +{"problem":"triathlete dan runs along a 4 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour , what is his average rate for the entire trip in miles per minute ?","rationale":"\"dan travels 8 miles round trip . running part : ( 4 \/ 10 * 60 = 24 minutes ) swimming part : ( 4 \/ 6 * 60 = 40 minutes ) 8 miles in ( 24 + 40 ) minutes 8 \/ 64 = 1 \/ 8 mile per minute answer : 1 \/ 8 mile per minute\"","correct":"a","options":{"a":"1 \/ 8 ","b":"2 \/ 15 ","c":"3 \/ 15 ","d":"1 \/ 4","e":"3 \/ 8"},"options_float":{"a":0.125,"b":0.1333333333,"c":0.2,"d":0.25,"e":0.375},"annotated_formula":"divide(add(4, 4), add(multiply(divide(4, 6), const_60), multiply(divide(4, 10), const_60)))","linear_formula":"add(n0,n0)|divide(n0,n2)|divide(n0,n1)|multiply(#1,const_60)|multiply(#2,const_60)|add(#3,#4)|divide(#0,#5)|","chain":"4 + 4<\/gadget>\n8<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 60<\/gadget>\n40<\/output>\n4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 60<\/gadget>\n24<\/output>\n40 + 24<\/gadget>\n64<\/output>\n8 \/ 64<\/gadget>\n1\/8 = around 0.125<\/output>\n1\/8 = around 0.125<\/result>","index":469} +{"problem":"at a small company , 58 percent of the employees are women , and 60 percent of the employees are married . if 2 \/ 3 of the men are single , what fraction of the women are married ?","rationale":"lets take total employees are 100 . given that , total women = 58 and total married = 60 . total men = 100 - 58 = 42 and single men = 2 \/ 3 * 42 = 28 . married men = total men - single men = 42 - 28 = 14 . married women = total married - married men = 60 - 14 = 46 . fraction of women are married = married women \/ total women = 46 \/ 58 = 23 \/ 29 . ans c","correct":"c","options":{"a":"5 \/ 16 ","b":"1 \/ 3 ","c":"23 \/ 29 ","d":"7 \/ 10","e":"5 \/ 7"},"options_float":{"a":0.3125,"b":0.3333333333,"c":0.7931034483,"d":0.7,"e":0.7142857143},"annotated_formula":"divide(subtract(divide(60, const_100), multiply(subtract(const_1, divide(58, const_100)), subtract(const_1, divide(2, 3)))), divide(58, const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n2,n3)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|subtract(#0,#5)|divide(#6,#1)","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n58 \/ 100<\/gadget>\n29\/50 = around 0.58<\/output>\n1 - (29\/50)<\/gadget>\n21\/50 = around 0.42<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(21\/50) * (1\/3)<\/gadget>\n7\/50 = around 0.14<\/output>\n(3\/5) - (7\/50)<\/gadget>\n23\/50 = around 0.46<\/output>\n(23\/50) \/ (29\/50)<\/gadget>\n23\/29 = around 0.793103<\/output>\n23\/29 = around 0.793103<\/result>","index":470} +{"problem":"if 10 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ?","rationale":"\"explanation : let the required no of hectares be x . then men - - - hectares - - - days 10 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 \/ 10 * 30 \/ 24 * 80 x = 360 answer : c\"","correct":"c","options":{"a":"127 ","b":"237 ","c":"360 ","d":"450","e":"281"},"options_float":{"a":127.0,"b":237.0,"c":360.0,"d":450.0,"e":281.0},"annotated_formula":"divide(multiply(36, 30), divide(multiply(10, 24), 80))","linear_formula":"multiply(n3,n4)|multiply(n0,n2)|divide(#1,n1)|divide(#0,#2)|","chain":"36 * 30<\/gadget>\n1_080<\/output>\n10 * 24<\/gadget>\n240<\/output>\n240 \/ 80<\/gadget>\n3<\/output>\n1_080 \/ 3<\/gadget>\n360<\/output>\n360<\/result>","index":472} +{"problem":"a man can row a distance of 5 km in 60 min with the help of the tide . the direction of the tide reverses with the same speed . now he travels a further 20 km in 10 hours . how much time he would have saved if the direction of tide has not changed ?","rationale":"\"explanation : he covered 5 km in 1 hour , so he might cover 20 km in 4 hours . but he took 10 hours . he would have saved 10 – 4 = 6 hours . answer : d\"","correct":"d","options":{"a":"2 ","b":"8 ","c":"1 ","d":"6","e":"5"},"options_float":{"a":2.0,"b":8.0,"c":1.0,"d":6.0,"e":5.0},"annotated_formula":"subtract(10, divide(20, 5))","linear_formula":"divide(n2,n0)|subtract(n3,#0)|","chain":"20 \/ 5<\/gadget>\n4<\/output>\n10 - 4<\/gadget>\n6<\/output>\n6<\/result>","index":473} +{"problem":"a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 15 % . their gross income from sales for one month increases by what percent ?","rationale":"\"let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.15 x = 1.035 * ( p * x ) , an increase of 3.5 % . the answer is c .\"","correct":"c","options":{"a":"1.5 % ","b":"2.5 % ","c":"3.5 % ","d":"4.5 %","e":"5.5 %"},"options_float":{"a":1.5,"b":2.5,"c":3.5,"d":4.5,"e":5.5},"annotated_formula":"subtract(subtract(15, 10), divide(15, 10))","linear_formula":"divide(n1,n0)|subtract(n1,n0)|subtract(#1,#0)|","chain":"15 - 10<\/gadget>\n5<\/output>\n15 \/ 10<\/gadget>\n3\/2 = around 1.5<\/output>\n5 - (3\/2)<\/gadget>\n7\/2 = around 3.5<\/output>\n7\/2 = around 3.5<\/result>","index":475} +{"problem":"if a man lost 8 % by selling oranges at the rate of 25 a rupee at how many a rupee must he sell them to gain 25 % ?","rationale":"\"92 % - - - - 25 125 % - - - - ? 92 \/ 125 * 25 = 18.4 answer : c\"","correct":"c","options":{"a":"20 ","b":"15 ","c":"18.4 ","d":"20","e":"12.8"},"options_float":{"a":20.0,"b":15.0,"c":18.4,"d":20.0,"e":12.8},"annotated_formula":"divide(multiply(subtract(const_100, 8), 25), add(const_100, 25))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|","chain":"100 - 8<\/gadget>\n92<\/output>\n92 * 25<\/gadget>\n2_300<\/output>\n100 + 25<\/gadget>\n125<\/output>\n2_300 \/ 125<\/gadget>\n92\/5 = around 18.4<\/output>\n92\/5 = around 18.4<\/result>","index":476} +{"problem":"express a speed of 90 kmph in meters per second ?","rationale":"\"90 * 5 \/ 18 = 25 mps answer : e\"","correct":"e","options":{"a":"10 mps ","b":"76 mps ","c":"26 mps ","d":"97 mps","e":"25 mps"},"options_float":{"a":10.0,"b":76.0,"c":26.0,"d":97.0,"e":25.0},"annotated_formula":"multiply(const_0_2778, 90)","linear_formula":"multiply(n0,const_0_2778)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 90<\/gadget>\n25<\/output>\n25<\/result>","index":480} +{"problem":"together , 15 type a machines and 7 type b machines can complete a certain job in 4 hours . together 8 type b machines and 15 type c machines can complete the same job in 11 hours . how many w hours would it take one type a machine , one type b machine , and one type c machine working together to complete the job ( assuming constant rates for each machine ) ?","rationale":"\"say the rates of machines a , b and c are a , b , and c , respectively . together 15 type a machines and 7 type b machines can complete a certain job in 4 hours - - > 15 a + 7 b = 1 \/ 4 ; together 8 type b machines and 15 type c machines can complete the same job in 11 hours - - > 8 b + 15 c = 1 \/ 11 . sum the above : 15 a + 15 b + 15 c = 1 \/ 4 + 1 \/ 11 = 15 \/ 44 - - > reduce by 15 : a + b + c = 1 \/ 44 - - > so , the combined rate of the three machines is 1 \/ 44 job \/ hour - - > time is reciprocal of the rate , thus machines a , b and c can do the job w in 44 hours . answer : c .\"","correct":"c","options":{"a":"22 hours ","b":"30 hours ","c":"44 hours ","d":"60 hours","e":"it can not be determined from the information above ."},"options_float":{"a":22.0,"b":30.0,"c":44.0,"d":60.0,"e":null},"annotated_formula":"divide(const_1, divide(add(divide(const_1, 4), divide(const_1, 11)), 15))","linear_formula":"divide(const_1,n2)|divide(const_1,n5)|add(#0,#1)|divide(#2,n0)|divide(const_1,#3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 11<\/gadget>\n1\/11 = around 0.090909<\/output>\n(1\/4) + (1\/11)<\/gadget>\n15\/44 = around 0.340909<\/output>\n(15\/44) \/ 15<\/gadget>\n1\/44 = around 0.022727<\/output>\n1 \/ (1\/44)<\/gadget>\n44<\/output>\n44<\/result>","index":481} +{"problem":"a man traveled from the village to the post - office at the rate of 25 kmph and walked back at the rate of 4 kmph . if the whole journey took 5 hours 48 minutes , find the distance of the post - office from the village ?","rationale":"explanation : average speed = , here a = 25 b = 4 = 2 x 25 x 4 \/ ( 25 + 4 ) = 200 \/ 29 km \/ hr . distance covered in 5 hours 48 minutes = speed x time = ( 200 \/ 29 ) x ( 29 \/ 5 ) distance covered in 5 hours 48 minutes = 40 kms . distance of the post office from the village = ( 40 \/ 2 ) = 20 km . answer : c","correct":"c","options":{"a":"70 m ","b":"80 km ","c":"20 km ","d":"10 km","e":"73 km"},"options_float":{"a":70.0,"b":80.0,"c":20.0,"d":10.0,"e":73.0},"annotated_formula":"divide(multiply(divide(multiply(multiply(const_2, 25), 4), add(25, 4)), divide(add(25, 4), 5)), const_2)","linear_formula":"add(n0,n1)|multiply(n0,const_2)|divide(#0,n2)|multiply(n1,#1)|divide(#3,#0)|multiply(#4,#2)|divide(#5,const_2)","chain":"2 * 25<\/gadget>\n50<\/output>\n50 * 4<\/gadget>\n200<\/output>\n25 + 4<\/gadget>\n29<\/output>\n200 \/ 29<\/gadget>\n200\/29 = around 6.896552<\/output>\n29 \/ 5<\/gadget>\n29\/5 = around 5.8<\/output>\n(200\/29) * (29\/5)<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":483} +{"problem":"a bag contains 21 toys numbered 1 to 21 . a toy is drawn and then another toy is drawn without replacement . find the probability that both toys will show even numbers .","rationale":"since there are 10 even numbers in the given range , the probability that the first toy shows an even number = 10 \/ 21 since the toy is not replaced , there are now 9 even numbered toys and total 20 toys left . hence , probability that the second toy shows an even number = 9 \/ 20 . required probability = 10 \/ 21 x 9 \/ 20 = 9 \/ 42 . answer : b","correct":"b","options":{"a":"5 \/ 21 ","b":"9 \/ 42 ","c":"6 \/ 22 ","d":"4 \/ 21","e":"3 \/ 21"},"options_float":{"a":0.2380952381,"b":0.2142857143,"c":0.2727272727,"d":0.1904761905,"e":0.1428571429},"annotated_formula":"multiply(divide(subtract(floor(divide(21, const_2)), const_1), subtract(21, const_1)), divide(floor(divide(21, const_2)), 21))","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|floor(#0)|divide(#2,n0)|subtract(#2,const_1)|divide(#4,#1)|multiply(#5,#3)","chain":"21 \/ 2<\/gadget>\n21\/2 = around 10.5<\/output>\nfloor(21\/2)<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n21 - 1<\/gadget>\n20<\/output>\n9 \/ 20<\/gadget>\n9\/20 = around 0.45<\/output>\n10 \/ 21<\/gadget>\n10\/21 = around 0.47619<\/output>\n(9\/20) * (10\/21)<\/gadget>\n3\/14 = around 0.214286<\/output>\n3\/14 = around 0.214286<\/result>","index":484} +{"problem":"trapezoid jklm in the x - y plane has coordinates j = ( – 2 , – 4 ) , k = ( – 2 , 1 ) , l = ( 6 , 7 ) , and m = ( 6 , – 4 ) . what is its perimeter ?","rationale":"\"this one may look challenging initially , but all kinds of simplifying tricks . first of all , j & k have the same x - coordinate , so that ’ s a vertical segment , and the length is jk = 1 – ( – 4 ) = 5 . l & m also have the same x - coordinate , so that ’ s also a vertical segment , and the length is lm = 7 – ( – 4 ) = 11 . j & m have the same y - coordinate , so that ’ s also a horizontal segment , and the length is jm = 6 – ( – 2 ) = 8 . right there , we have the lengths of three of the four sides . the last side to find is kl . from k = ( – 2 , 1 ) , we have to go over 8 and up 6 to get to l = ( 6 , 7 ) , so 6 and 8 the legs of the right triangle , and kl is the hypotenuse . we recognize another common pythagorean triplet — ( 6 , 8 , 10 ) , which is a multiple of the ( 3 , 4 , 5 ) triplet . thus , without further calculations , we immediately know kl = 10 . perimeter = jk + kl + lm + jm = 5 + 10 + 11 + 8 = 34 answer = a\"","correct":"a","options":{"a":"34 ","b":"36 ","c":"38 ","d":"40","e":"42"},"options_float":{"a":34.0,"b":36.0,"c":38.0,"d":40.0,"e":42.0},"annotated_formula":"add(add(add(const_10, add(1, 4)), add(2, 6)), add(7, 4))","linear_formula":"add(n1,n3)|add(n0,n4)|add(n1,n5)|add(#0,const_10)|add(#3,#1)|add(#4,#2)|","chain":"1 + 4<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n2 + 6<\/gadget>\n8<\/output>\n15 + 8<\/gadget>\n23<\/output>\n7 + 4<\/gadget>\n11<\/output>\n23 + 11<\/gadget>\n34<\/output>\n34<\/result>","index":485} +{"problem":"a tradesman by means of his false balance defrauds to the extent of 20 % ? in buying goods as well as by selling the goods . what percent does he gain on his outlay ? a . 20 % b . 45 %","rationale":"g % = 20 + 20 + ( 20 * 20 ) \/ 100 = 44 % answer : c","correct":"c","options":{"a":"77 ","b":"66 ","c":"44 ","d":"99","e":"11"},"options_float":{"a":77.0,"b":66.0,"c":44.0,"d":99.0,"e":11.0},"annotated_formula":"add(add(20, 20), divide(multiply(20, 20), const_100))","linear_formula":"add(n0,n0)|multiply(n0,n0)|divide(#1,const_100)|add(#0,#2)","chain":"20 + 20<\/gadget>\n40<\/output>\n20 * 20<\/gadget>\n400<\/output>\n400 \/ 100<\/gadget>\n4<\/output>\n40 + 4<\/gadget>\n44<\/output>\n44<\/result>","index":486} +{"problem":"a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability e that all 3 jellybeans removed from the bag are blue ?","rationale":"\"method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes \/ total outcomes = 10 c 3 \/ 20 c 3 probability e = 120 \/ 1140 = 2 \/ 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 \/ 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 \/ 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 \/ 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 \/ 20 ) * ( 9 \/ 19 ) * ( 8 \/ 18 ) = 2 \/ 19 answer : option b\"","correct":"b","options":{"a":"9 \/ 100 ","b":"2 \/ 19 ","c":"1 \/ 8 ","d":"3 \/ 20","e":"3 \/ 10"},"options_float":{"a":0.09,"b":0.1052631579,"c":0.125,"d":0.15,"e":0.3},"annotated_formula":"divide(choose(10, 3), choose(add(10, 10), 3))","linear_formula":"add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2)|","chain":"binomial(10, 3)<\/gadget>\n120<\/output>\n10 + 10<\/gadget>\n20<\/output>\nbinomial(20, 3)<\/gadget>\n1_140<\/output>\n120 \/ 1_140<\/gadget>\n2\/19 = around 0.105263<\/output>\n2\/19 = around 0.105263<\/result>","index":488} +{"problem":"a train 50 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ?","rationale":"\"let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 \/ 18 m \/ sec 50 \/ [ ( x + 5 ) * 5 \/ 18 ] = 6 30 ( x + 5 ) = 900 x = 25 kmph answer is b\"","correct":"b","options":{"a":"15 kmph ","b":"25 kmph ","c":"35 kmph ","d":"45 kmph","e":"55 kmph"},"options_float":{"a":15.0,"b":25.0,"c":35.0,"d":45.0,"e":55.0},"annotated_formula":"subtract(divide(divide(50, 6), const_0_2778), 5)","linear_formula":"divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2)|","chain":"50 \/ 6<\/gadget>\n25\/3 = around 8.333333<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(25\/3) \/ (5\/18)<\/gadget>\n30<\/output>\n30 - 5<\/gadget>\n25<\/output>\n25<\/result>","index":489} +{"problem":"on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with 5 times the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.80 per glass on the first day , what was the price per glass on the second day ?","rationale":"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 5 units of water was used to make 6 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 6 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 6 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 6 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.8 = 6 * x - - > x = $ 0.27 . answer : b .","correct":"b","options":{"a":"$ 015 ","b":"$ 0.27 ","c":"$ 0.30 ","d":"$ 0.40","e":"$ 0.45"},"options_float":{"a":15.0,"b":0.27,"c":0.3,"d":0.4,"e":0.45},"annotated_formula":"divide(0.8, const_3)","linear_formula":"divide(n1,const_3)","chain":"0.8 \/ 3<\/gadget>\n0.266667<\/output>\n0.266667<\/result>","index":492} +{"problem":"the average of 11 10 prime numbers which are odd is ?","rationale":"explanation : sum of first 11 prime no . which are odd = 195 average = 195 \/ 11 = 17.72 answer : a","correct":"a","options":{"a":"17.72 ","b":"18.52 ","c":"19.63 ","d":"16.53","e":"15.63"},"options_float":{"a":17.72,"b":18.52,"c":19.63,"d":16.53,"e":15.63},"annotated_formula":"add(divide(divide(11, const_0_33), const_2), const_1)","linear_formula":"divide(n0,const_0_33)|divide(#0,const_2)|add(#1,const_1)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n11 \/ (1\/3)<\/gadget>\n33<\/output>\n33 \/ 2<\/gadget>\n33\/2 = around 16.5<\/output>\n(33\/2) + 1<\/gadget>\n35\/2 = around 17.5<\/output>\n35\/2 = around 17.5<\/result>","index":494} +{"problem":"the first doughnut is priced at $ 1 and then if you purchase additional doughnuts as dozens then the price is $ 6 \/ dozen . what is the total number of doughnuts purchased if you paid $ 25 ?","rationale":"\"$ 25 = 4 * $ 6 + $ 1 the number of doughnuts is 4 * 12 + 1 = 49 the answer is b .\"","correct":"b","options":{"a":"45 ","b":"49 ","c":"53 ","d":"57","e":"61"},"options_float":{"a":45.0,"b":49.0,"c":53.0,"d":57.0,"e":61.0},"annotated_formula":"multiply(divide(25, 6), const_12)","linear_formula":"divide(n2,n1)|multiply(#0,const_12)|","chain":"25 \/ 6<\/gadget>\n25\/6 = around 4.166667<\/output>\n(25\/6) * 12<\/gadget>\n50<\/output>\n50<\/result>","index":495} +{"problem":"during the first two weeks of january , the total rainfall in springdale was 25 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall during the second week of january ?","rationale":"\"total rainfall in 2 weeks = 25 inches . assume the rainfall in second week = 1 . x rainfall in first week = x total rainfall = 2.5 x = 25 inches x = 10 and 1.5 x = 15 rainfall during second week = 9 inches option e\"","correct":"e","options":{"a":"5 inches ","b":"6 inches ","c":"9 inches ","d":"10 inches","e":"15 inches"},"options_float":{"a":5.0,"b":6.0,"c":9.0,"d":10.0,"e":15.0},"annotated_formula":"multiply(25, divide(1.5, add(const_1, 1.5)))","linear_formula":"add(n1,const_1)|divide(n1,#0)|multiply(n0,#1)|","chain":"1 + 1.5<\/gadget>\n2.5<\/output>\n1.5 \/ 2.5<\/gadget>\n0.6<\/output>\n25 * 0.6<\/gadget>\n15<\/output>\n15<\/result>","index":496} +{"problem":"a train crosses a bridge of length 200 m in 10 seconds and a lamp post on the bridge in 5 seconds . what is the length of the train in metres ?","rationale":"\"let length of train = l case - 1 : distance = 200 + l ( while crossing the bridge ) time = 10 seconds i . e . speed = distance \/ time = ( 200 + l ) \/ 10 case - 2 : distance = l ( while passing the lamp post ) time = 5 seconds i . e . speed = distance \/ time = ( l ) \/ 5 but since speed has to be same in both cases so ( 200 + l ) \/ 10 = ( l ) \/ 5 i . e . 2 l = l + 200 i . e . l = 200 answer : option c\"","correct":"c","options":{"a":"375 m ","b":"750 m ","c":"200 m ","d":"800 m","e":"300 m"},"options_float":{"a":375.0,"b":750.0,"c":200.0,"d":800.0,"e":300.0},"annotated_formula":"multiply(divide(200, subtract(10, 5)), 5)","linear_formula":"subtract(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"10 - 5<\/gadget>\n5<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n40 * 5<\/gadget>\n200<\/output>\n200<\/result>","index":497} +{"problem":"the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 385 the other is","rationale":"the other number = l . c . m * h . c . f \/ given number = 2310 * 30 \/ 385 = 180 answer is a .","correct":"a","options":{"a":"180 ","b":"300 ","c":"270 ","d":"250","e":"350"},"options_float":{"a":180.0,"b":300.0,"c":270.0,"d":250.0,"e":350.0},"annotated_formula":"divide(multiply(30, 2310), 385)","linear_formula":"multiply(n0,n1)|divide(#0,n2)","chain":"30 * 2_310<\/gadget>\n69_300<\/output>\n69_300 \/ 385<\/gadget>\n180<\/output>\n180<\/result>","index":499} +{"problem":"mary and mike enter into a partnership by investing $ 550 and $ 450 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if mary received $ 1000 more than mike did , what was the profit made by their business in that year ?","rationale":"\"explanatory answer let the profit made during the year be $ 3 x therefore , $ x would have been shared equally and the remaining $ 2 x would have been shared in the ratio 5.5 : 4.5 i . e . , 55 % of 2 x would go to mary and 45 % of 2 x would go to mike . hence , mary would get ( 55 - 45 ) % of 2 x more than mike or 10 % of 2 x = $ 1000 i . e . , ( 60 \/ 100 ) * 2 x = 1000 or 2 x = 10000 . hence , the profit made by the company during the year $ 3 x = $ 15000 . answer : c\"","correct":"c","options":{"a":"$ 8000 ","b":"$ 10000 ","c":"$ 15000 ","d":"$ 12000","e":"$ 12500"},"options_float":{"a":8000.0,"b":10000.0,"c":15000.0,"d":12000.0,"e":12500.0},"annotated_formula":"divide(1000, multiply(subtract(divide(550, const_1000), divide(450, const_1000)), subtract(const_1, divide(const_1, const_3))))","linear_formula":"divide(n0,const_1000)|divide(n1,const_1000)|divide(const_1,const_3)|subtract(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)|divide(n2,#5)|","chain":"550 \/ 1_000<\/gadget>\n11\/20 = around 0.55<\/output>\n450 \/ 1_000<\/gadget>\n9\/20 = around 0.45<\/output>\n(11\/20) - (9\/20)<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/10) * (2\/3)<\/gadget>\n1\/15 = around 0.066667<\/output>\n1_000 \/ (1\/15)<\/gadget>\n15_000<\/output>\n15_000<\/result>","index":501} +{"problem":"a goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 24 seconds . what is the length of the goods train ?","rationale":"\"explanation : speed = [ 72 x ( 5 \/ 18 ) ] m \/ sec = 20 m \/ sec . time = 24 sec . let the length of the train be x metres . then , [ ( x + 250 ) \/ 24 ] = 20 = > x + 250 = 480 = > x = 230 . answer : a\"","correct":"a","options":{"a":"230 m ","b":"240 m ","c":"260 m ","d":"270 m","e":"none of these"},"options_float":{"a":230.0,"b":240.0,"c":260.0,"d":270.0,"e":null},"annotated_formula":"multiply(subtract(24, divide(250, multiply(const_0_2778, 72))), multiply(const_0_2778, 72))","linear_formula":"multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 72<\/gadget>\n20<\/output>\n250 \/ 20<\/gadget>\n25\/2 = around 12.5<\/output>\n24 - (25\/2)<\/gadget>\n23\/2 = around 11.5<\/output>\n(23\/2) * 20<\/gadget>\n230<\/output>\n230<\/result>","index":502} +{"problem":"20 men can dig a tunnel in 30 days . if after 10 days , 20 more men are deployed to dig the tunnel , how many more days it will take to dig the whole tunnel ?","rationale":"a remaining work can be completed by 20 men in 30 - 10 = 20 days . if 20 men can do x work in 20 days , and let us say 40 men take n days to do the same work . now , more men , less number of days ( inversely proportion ) 20 : 40 : : n : 20 n = 20 x 20 \/ 40 = 10 days .","correct":"a","options":{"a":"10 ","b":"15 ","c":"20 ","d":"25","e":"30"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"divide(subtract(const_1, multiply(divide(const_1, 30), 10)), multiply(divide(const_1, 30), const_2))","linear_formula":"divide(const_1,n1)|multiply(n2,#0)|multiply(#0,const_2)|subtract(const_1,#1)|divide(#3,#2)","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/30) * 10<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/30) * 2<\/gadget>\n1\/15 = around 0.066667<\/output>\n(2\/3) \/ (1\/15)<\/gadget>\n10<\/output>\n10<\/result>","index":503} +{"problem":"if the product of 10 integers is negative , at most how many of the integers can be negative ?","rationale":"\"the product of 10 integers is negative thus an odd number of integers need to be negative to have a negative product we are asked at most how many are required . so , the highest odd integer before 6 , i . e . 9 correct option : d\"","correct":"d","options":{"a":"2 ","b":"3 ","c":"4 ","d":"9","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":9.0,"e":6.0},"annotated_formula":"subtract(10, const_1)","linear_formula":"subtract(n0,const_1)|","chain":"10 - 1<\/gadget>\n9<\/output>\n9<\/result>","index":504} +{"problem":"the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 72 . find the principal ?","rationale":"\"p = 72 ( 100 \/ 5 ) ^ 2 = > p = 1800 answer : e\"","correct":"e","options":{"a":"2277 ","b":"2667 ","c":"3600 ","d":"2766","e":"1800"},"options_float":{"a":2277.0,"b":2667.0,"c":3600.0,"d":2766.0,"e":1800.0},"annotated_formula":"divide(72, subtract(power(add(divide(20, const_100), const_1), 2), add(multiply(divide(20, const_100), 2), const_1)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#0)|add(#2,const_1)|power(#1,n0)|subtract(#4,#3)|divide(n2,#5)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 2<\/gadget>\n36\/25 = around 1.44<\/output>\n(1\/5) * 2<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) + 1<\/gadget>\n7\/5 = around 1.4<\/output>\n(36\/25) - (7\/5)<\/gadget>\n1\/25 = around 0.04<\/output>\n72 \/ (1\/25)<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":506} +{"problem":"in a division , a student took 63 as divisor instead of 36 . his answer was 24 . the correct answer is -","rationale":"\"x \/ 63 = 24 . x = 24 * 63 . so correct answer would be , ( 24 * 63 ) \/ 36 = 42 . answer : a\"","correct":"a","options":{"a":"42 ","b":"32 ","c":"48 ","d":"28","e":"38"},"options_float":{"a":42.0,"b":32.0,"c":48.0,"d":28.0,"e":38.0},"annotated_formula":"divide(multiply(63, 24), 36)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"63 * 24<\/gadget>\n1_512<\/output>\n1_512 \/ 36<\/gadget>\n42<\/output>\n42<\/result>","index":507} +{"problem":"the sum of the numbers is 550 . if the first number be twice the second and third number be one - third of the first , then the second number is :","rationale":"\"let the second number be x . then , first number = 2 x and third number = 2 x \/ 3 . 2 x + x + 2 x \/ 3 = 550 11 x \/ 3 = 550 x = 150 answer : b\"","correct":"b","options":{"a":"48 ","b":"150 ","c":"72 ","d":"84","e":"27"},"options_float":{"a":48.0,"b":150.0,"c":72.0,"d":84.0,"e":27.0},"annotated_formula":"divide(multiply(550, const_3), add(const_10, const_1))","linear_formula":"add(const_1,const_10)|multiply(n0,const_3)|divide(#1,#0)|","chain":"550 * 3<\/gadget>\n1_650<\/output>\n10 + 1<\/gadget>\n11<\/output>\n1_650 \/ 11<\/gadget>\n150<\/output>\n150<\/result>","index":508} +{"problem":"if ( 1 – 1.25 ) n = 4 , then n =","rationale":"\"( 1 – 1.25 ) n = 4 simplify to get : - 0.25 n = 4 rewrite as ( - 1 \/ 4 ) n = 4 multiply both sides by - 4 to get : n = - 16 answer : b\"","correct":"b","options":{"a":"− 400 ","b":"− 16 ","c":"− 4 ","d":"4","e":"400"},"options_float":{"a":-400.0,"b":-16.0,"c":-4.0,"d":4.0,"e":400.0},"annotated_formula":"negate(multiply(4, const_4))","linear_formula":"multiply(n2,const_4)|negate(#0)|","chain":"4 * 4<\/gadget>\n16<\/output>\n-16<\/gadget>\n-16<\/output>\n-16<\/result>","index":510} +{"problem":"a team p of 20 engineers can complete work or task in 32 days . another team q of 16 engineers can complete same task in 30 days . then the ratio of working capacity of 1 member of p to the 1 member of q is a . 3 : 2","rationale":"explanation : let the capacity of an engineer in p = x units , and in q = y units . working capacity of p = x × 32 × 20 working capacity of q = y × 16 × 30 as the total work is same , we equate the above equations . ⇒ x × 32 × 20 = y × 16 × 30 ⇒ xy = 16 × 3032 × 20 = 34 answer : b","correct":"b","options":{"a":"22 ","b":"34 ","c":"77 ","d":"29","e":"21"},"options_float":{"a":22.0,"b":34.0,"c":77.0,"d":29.0,"e":21.0},"annotated_formula":"add(multiply(3, const_10), const_4)","linear_formula":"multiply(n6,const_10)|add(#0,const_4)","chain":"3 * 10<\/gadget>\n30<\/output>\n30 + 4<\/gadget>\n34<\/output>\n34<\/result>","index":514} +{"problem":"a certain kind of necklace is made from 9 green , 6 white , and 3 orange beads . what is the maximum number of these necklaces that can be made from 45 beads of each color ?","rationale":"atleast 9 green breads are needed to make a necklace , so irrespective of the combinations of white and orange beads , the number of necklaces that can be formed will be 45 \/ 9 = 5 answer : - c","correct":"c","options":{"a":"7 ","b":"8 ","c":"5 ","d":"6","e":"2"},"options_float":{"a":7.0,"b":8.0,"c":5.0,"d":6.0,"e":2.0},"annotated_formula":"divide(45, 9)","linear_formula":"divide(n3,n0)","chain":"45 \/ 9<\/gadget>\n5<\/output>\n5<\/result>","index":515} +{"problem":"the average of 45 results is 25 and the average of other 25 results is 45 . what is the average of all the results ?","rationale":"\"answer sum of 70 result = sum of 45 result + sum of 25 result . = 45 x 25 + 25 x 45 = 2250 \/ 70 correct option : b\"","correct":"b","options":{"a":"24 ","b":"32 ","c":"48 ","d":"50","e":"none"},"options_float":{"a":24.0,"b":32.0,"c":48.0,"d":50.0,"e":null},"annotated_formula":"divide(add(multiply(45, 25), multiply(25, 45)), add(45, 25))","linear_formula":"add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)|","chain":"45 * 25<\/gadget>\n1_125<\/output>\n25 * 45<\/gadget>\n1_125<\/output>\n1_125 + 1_125<\/gadget>\n2_250<\/output>\n45 + 25<\/gadget>\n70<\/output>\n2_250 \/ 70<\/gadget>\n225\/7 = around 32.142857<\/output>\n225\/7 = around 32.142857<\/result>","index":517} +{"problem":"10 friends meet for movie 20 for picnic and 5 for games 4 for movie & picnic 2 for movie & games , 0 for picnic & games , 2 for all , how many are students are there in all ?","rationale":"n ( a u b u c ) = n ( a ) + n ( b ) + n ( c ) - n ( a intersection b ) - n ( a intersection c ) - n ( c intersection b ) + n ( a intersection b - intersection c ) hence , total no . of students = 10 + 20 + 5 - 4 - 0 - 2 + 2 = 31 . answer : b","correct":"b","options":{"a":"30 ","b":"31 ","c":"32 ","d":"33","e":"34"},"options_float":{"a":30.0,"b":31.0,"c":32.0,"d":33.0,"e":34.0},"annotated_formula":"add(subtract(add(subtract(add(10, 20), 4), 5), 2), 2)","linear_formula":"add(n0,n1)|subtract(#0,n3)|add(n2,#1)|subtract(#2,n4)|add(n4,#3)","chain":"10 + 20<\/gadget>\n30<\/output>\n30 - 4<\/gadget>\n26<\/output>\n26 + 5<\/gadget>\n31<\/output>\n31 - 2<\/gadget>\n29<\/output>\n29 + 2<\/gadget>\n31<\/output>\n31<\/result>","index":519} +{"problem":"a can give b 100 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"explanation : a runs 1000 m while b runs 900 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) \/ 900 = 8000 \/ 9 = 888.88 m . b can give c = 1000 - 888.88 = 111.12 m . answer : a\"","correct":"a","options":{"a":"111.12 m ","b":"221.12 m ","c":"271.12 m ","d":"611.12 m","e":"191.12 m"},"options_float":{"a":111.12,"b":221.12,"c":271.12,"d":611.12,"e":191.12},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 100)))","linear_formula":"multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 200<\/gadget>\n800<\/output>\n1_000 * 800<\/gadget>\n800_000<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n800_000 \/ 900<\/gadget>\n8_000\/9 = around 888.888889<\/output>\n1_000 - (8_000\/9)<\/gadget>\n1_000\/9 = around 111.111111<\/output>\n1_000\/9 = around 111.111111<\/result>","index":523} +{"problem":"alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 961 . find the cost of each ice - cream cup ?","rationale":"\"let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 961 96 + 225 + 490 + 6 x = 961 6 x = 150 = > x = 25 . answer : a\"","correct":"a","options":{"a":"rs . 25 ","b":"rs . 22.50 ","c":"rs . 20 ","d":"rs . 17.50","e":"none of these"},"options_float":{"a":25.0,"b":22.5,"c":20.0,"d":17.5,"e":null},"annotated_formula":"divide(subtract(subtract(subtract(961, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6)","linear_formula":"multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)|","chain":"16 * 6<\/gadget>\n96<\/output>\n961 - 96<\/gadget>\n865<\/output>\n5 * 45<\/gadget>\n225<\/output>\n865 - 225<\/gadget>\n640<\/output>\n7 * 70<\/gadget>\n490<\/output>\n640 - 490<\/gadget>\n150<\/output>\n150 \/ 6<\/gadget>\n25<\/output>\n25<\/result>","index":525} +{"problem":"a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 20 - sided polygon have ?","rationale":"there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] \/ 2 , n = number of sides of the regular polygon . here , n = 20 . plugging it in , we get 170 diagonals ! answer ( b ) .","correct":"b","options":{"a":"875 ","b":"170 ","c":"1425 ","d":"2025","e":"2500"},"options_float":{"a":875.0,"b":170.0,"c":1425.0,"d":2025.0,"e":2500.0},"annotated_formula":"divide(multiply(subtract(20, const_3), 20), const_2)","linear_formula":"subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)","chain":"20 - 3<\/gadget>\n17<\/output>\n17 * 20<\/gadget>\n340<\/output>\n340 \/ 2<\/gadget>\n170<\/output>\n170<\/result>","index":526} +{"problem":"a number is said to be prime saturated if the product of all the different positive prime factors of z is less than the square root of z . what is the greatest two digit prime saturated integer ?","rationale":"\"clearly d a number is said to be prime saturated if the product of all the different positive prime factors of z is less than the square root of z . 96 has more number of smaller prime factor thats the clue ! ! = d\"","correct":"d","options":{"a":"99 ","b":"98 ","c":"97 ","d":"96","e":"95"},"options_float":{"a":99.0,"b":98.0,"c":97.0,"d":96.0,"e":95.0},"annotated_formula":"add(multiply(const_2, const_3), subtract(const_100, const_10))","linear_formula":"multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)|","chain":"2 * 3<\/gadget>\n6<\/output>\n100 - 10<\/gadget>\n90<\/output>\n6 + 90<\/gadget>\n96<\/output>\n96<\/result>","index":528} +{"problem":"find large number from below question the difference of two numbers is 1311 . on dividing the larger number by the smaller , we get 11 as quotient and the 11 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1311 ) . x + 1311 = 11 x + 11 10 x = 1300 x = 130 large number = 130 + 1311 = 1441 answer : c\"","correct":"c","options":{"a":"1209 ","b":"1324 ","c":"1441 ","d":"1300","e":"1635"},"options_float":{"a":1209.0,"b":1324.0,"c":1441.0,"d":1300.0,"e":1635.0},"annotated_formula":"multiply(divide(subtract(1311, 11), subtract(11, const_1)), 11)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_311 - 11<\/gadget>\n1_300<\/output>\n11 - 1<\/gadget>\n10<\/output>\n1_300 \/ 10<\/gadget>\n130<\/output>\n130 * 11<\/gadget>\n1_430<\/output>\n1_430<\/result>","index":530} +{"problem":"1 blue dice and 2 red dice are thrown . . . . what is the probability that the no . appeared on blue dice is always greater than both the no . appeared on the 2 red dice . .","rationale":"if the no . on blue dice is 2 then 2 red dice can take ( 1,1 ) value = 1 ^ 2 if the no . on blue dice is 3 then 2 red dice can take ( 1,1 ) ( 1,2 ) ( 2,1 ) ( 2,2 ) = 2 ^ 2 if the no . on blue dice is 4 then 2 red dice can take ( 1,1 ) ( 1,2 ) ( 2,1 ) ( 2,2 ) ( 1,3 ) ( 2,3 ) ( 3,2 ) ( 3,1 ) ( 3,3 ) = 3 ^ 2 similarly , if the no . on blue dice is 5 then 2 red dice can take = 4 ^ 2 and if the no . on blue dice is 6 then 2 red dice can take = 5 ^ 2 therefor , probability = total no . of favourable outcome \/ total no . of possible outcome = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 \/ 216 = 55 \/ 216 answer : a","correct":"a","options":{"a":"55 \/ 216 ","b":"56 \/ 216 ","c":"57 \/ 216 ","d":"54 \/ 216","e":"53 \/ 216"},"options_float":{"a":0.2546296296,"b":0.2592592593,"c":0.2638888889,"d":0.25,"e":0.2453703704},"annotated_formula":"divide(add(add(add(add(power(1, const_2), power(2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_3, const_2), const_2)), add(power(multiply(multiply(const_3, const_2), const_2), const_2), add(multiply(multiply(multiply(const_3, const_2), const_2), add(power(1, const_2), power(2, const_2))), multiply(multiply(const_3, const_2), const_2))))","linear_formula":"add(const_2,const_3)|multiply(const_2,const_3)|power(n0,const_2)|power(n1,const_2)|power(const_3,const_2)|power(const_4,const_2)|add(#2,#3)|multiply(#1,const_2)|power(#0,const_2)|add(#6,#4)|multiply(#6,#7)|power(#7,const_2)|add(#9,#5)|add(#10,#7)|add(#12,#8)|add(#13,#11)|divide(#14,#15)","chain":"1 ** 2<\/gadget>\n1<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n5 + 9<\/gadget>\n14<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n14 + 16<\/gadget>\n30<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n30 + 25<\/gadget>\n55<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n12 * 5<\/gadget>\n60<\/output>\n60 + 12<\/gadget>\n72<\/output>\n144 + 72<\/gadget>\n216<\/output>\n55 \/ 216<\/gadget>\n55\/216 = around 0.25463<\/output>\n55\/216 = around 0.25463<\/result>","index":531} +{"problem":"excluding stoppages , the speed of the bus is 54 kmph and including stoppages , it is 36 kmph . for how many minutes does the bus stop per hr","rationale":"due to stoppages , it covers 18 km less . time taken to cover 18 km is [ 18 \/ 54 * 60 ] min = 20 min answer is e .","correct":"e","options":{"a":"12 min ","b":"11 min ","c":"13 min ","d":"19 min","e":"20 min"},"options_float":{"a":12.0,"b":11.0,"c":13.0,"d":19.0,"e":20.0},"annotated_formula":"multiply(subtract(const_1, divide(36, 54)), const_60)","linear_formula":"divide(n1,n0)|subtract(const_1,#0)|multiply(#1,const_60)","chain":"36 \/ 54<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 60<\/gadget>\n20<\/output>\n20<\/result>","index":533} +{"problem":"a 150 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ?","rationale":"\"speed = 150 \/ 18 = 25 \/ 3 m \/ sec . let the length of the platform be x meters . then , ( x + 150 ) \/ 39 = 25 \/ 3 = > x = 325 m . l = 325 - 150 = 175 m answer : option b\"","correct":"b","options":{"a":"300 ","b":"175 ","c":"360 ","d":"370","e":"380"},"options_float":{"a":300.0,"b":175.0,"c":360.0,"d":370.0,"e":380.0},"annotated_formula":"subtract(multiply(speed(150, 18), 39), 150)","linear_formula":"speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|","chain":"150 \/ 18<\/gadget>\n25\/3 = around 8.333333<\/output>\n(25\/3) * 39<\/gadget>\n325<\/output>\n325 - 150<\/gadget>\n175<\/output>\n175<\/result>","index":535} +{"problem":"a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 35 , how many coins of each type are there ?","rationale":"\"let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 35 ⇒ 1.75 x = 35 ⇒ x = 20 coins answer a\"","correct":"a","options":{"a":"20 coins ","b":"30 coins ","c":"28 coins ","d":"25 coins","e":"none of these"},"options_float":{"a":20.0,"b":30.0,"c":28.0,"d":25.0,"e":null},"annotated_formula":"divide(35, add(add(inverse(const_4), inverse(const_2)), const_1))","linear_formula":"inverse(const_4)|inverse(const_2)|add(#0,#1)|add(#2,const_1)|divide(n2,#3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/4) + (1\/2)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) + 1<\/gadget>\n7\/4 = around 1.75<\/output>\n35 \/ (7\/4)<\/gadget>\n20<\/output>\n20<\/result>","index":536} +{"problem":"a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 588.25 . how many boxes of plain cookies did she sell ?","rationale":"\"let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1588.25 = > x = 786 first get the difference of the revenue to the number of boxes . 1588.25 - 1585 = 3.25 now you know that the difference between plain cookies boxes and chocolate cookies boxes is small . we now divide the revenue by two . 1588.25 \/ 2 = ~ 794 because the difference between the possible answers is big , this leaves us with choice e .\"","correct":"e","options":{"a":"0 ","b":"233 ","c":"500 ","d":"695","e":"786"},"options_float":{"a":0.0,"b":233.0,"c":500.0,"d":695.0,"e":786.0},"annotated_formula":"divide(add(const_1000, 588.25), const_2)","linear_formula":"add(n4,const_1000)|divide(#0,const_2)|","chain":"1_000 + 588.25<\/gadget>\n1_588.25<\/output>\n1_588.25 \/ 2<\/gadget>\n794.125<\/output>\n794.125<\/result>","index":537} +{"problem":"how many positive even integers less than 100 contain digits 5 or 7 ?","rationale":"\"two digit numbers : 5 at tens place : 50 , 52,54 , 56,58 7 at tens place : 70 , 72,74 , 76,78 if 5 and 7 is at units place , the number cant be even total : 5 + 5 = 10 answer a\"","correct":"a","options":{"a":"10 ","b":"17 ","c":"18 ","d":"19","e":"20"},"options_float":{"a":10.0,"b":17.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"multiply(5, const_2)","linear_formula":"multiply(n1,const_2)|","chain":"5 * 2<\/gadget>\n10<\/output>\n10<\/result>","index":538} +{"problem":"a person can swim in still water at 4 km \/ h . if the speed of water 2 km \/ h , how many hours will the man take to swim back against the current for 10 km ?","rationale":"\"m = 4 s = 2 us = 4 - 2 = 2 d = 10 t = 10 \/ 2 = 5 answer : b\"","correct":"b","options":{"a":"3 ","b":"5 ","c":"8 ","d":"9","e":"6"},"options_float":{"a":3.0,"b":5.0,"c":8.0,"d":9.0,"e":6.0},"annotated_formula":"divide(10, subtract(4, 2))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"4 - 2<\/gadget>\n2<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":539} +{"problem":"the price of a certain painting increased by 25 % during the first year and decreased by 15 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ?","rationale":"\"easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 25 = 125 price at the end of year 2 = 125 - 125 * 0.15 = 125 * 0.85 = 106.25 hence required answer = ( 106.25 \/ 100 ) * 100 % = 106.25 % answer is b .\"","correct":"b","options":{"a":"109 % ","b":"106.25 % ","c":"120 % ","d":"115 %","e":"110 %"},"options_float":{"a":109.0,"b":106.25,"c":120.0,"d":115.0,"e":110.0},"annotated_formula":"add(subtract(subtract(25, 15), divide(multiply(15, 25), const_100)), const_100)","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|add(#3,const_100)|","chain":"25 - 15<\/gadget>\n10<\/output>\n15 * 25<\/gadget>\n375<\/output>\n375 \/ 100<\/gadget>\n15\/4 = around 3.75<\/output>\n10 - (15\/4)<\/gadget>\n25\/4 = around 6.25<\/output>\n(25\/4) + 100<\/gadget>\n425\/4 = around 106.25<\/output>\n425\/4 = around 106.25<\/result>","index":540} +{"problem":"rs . 1010 \/ - distributed among a , b and c such that on decreasing their shares by rs . 25 , rs . 10 and rs . 15 respectively , the balance money would be divided among them in the ratio 3 : 2 : 5 . then , c ’ s share is : ?","rationale":"a : b : c = 3 : 2 : 5 total parts = 10 c ' s share is = 5 parts 10 - - - - - > rs . 960 \/ - 3 - - - - - > rs . 480 \/ - a ' s total = 480 + 15 = rs . 495 \/ - b )","correct":"b","options":{"a":"rs . 485 ","b":"rs . 495 \/ - ","c":"rs . 575 ","d":"rs . 685","e":"rs . 592"},"options_float":{"a":485.0,"b":495.0,"c":575.0,"d":685.0,"e":592.0},"annotated_formula":"add(multiply(5, divide(subtract(1010, add(add(25, 10), 15)), add(add(3, 2), 5))), 15)","linear_formula":"add(n1,n2)|add(n4,n5)|add(n3,#0)|add(n6,#1)|subtract(n0,#2)|divide(#4,#3)|multiply(n6,#5)|add(n3,#6)","chain":"25 + 10<\/gadget>\n35<\/output>\n35 + 15<\/gadget>\n50<\/output>\n1_010 - 50<\/gadget>\n960<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 + 5<\/gadget>\n10<\/output>\n960 \/ 10<\/gadget>\n96<\/output>\n5 * 96<\/gadget>\n480<\/output>\n480 + 15<\/gadget>\n495<\/output>\n495<\/result>","index":542} +{"problem":"a is twice as good a work man as b and together they finish a piece of work in 36 days . the number of days taken by b alone to finish the work is :","rationale":"solution ( a ’ s 1 day ’ s work ) : ( b ’ s 1 day ’ s work ) = 2 : 1 . ( a + b ) ' s 1 day ’ s work = 1 \/ 36 divide 1 \/ 36 in the ratio 2 : 1 . ∴ b ’ s 1 day ’ s work = ( 1 \/ 36 x 1 \/ 3 ) = 1 \/ 108 hence , b alone can finish the work in 108 days . answer b","correct":"b","options":{"a":"100 days . ","b":"108 days . ","c":"98 days . ","d":"47 days .","e":"103 days ."},"options_float":{"a":100.0,"b":108.0,"c":98.0,"d":47.0,"e":103.0},"annotated_formula":"inverse(divide(inverse(36), add(const_2, const_1)))","linear_formula":"add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)","chain":"1 \/ 36<\/gadget>\n1\/36 = around 0.027778<\/output>\n2 + 1<\/gadget>\n3<\/output>\n(1\/36) \/ 3<\/gadget>\n1\/108 = around 0.009259<\/output>\n1 \/ (1\/108)<\/gadget>\n108<\/output>\n108<\/result>","index":543} +{"problem":"5 % of ( 25 % of $ 1600 ) is ?","rationale":"5 % of ( 25 % of $ 1600 ) = 5 \/ 100 ( 25 \/ 100 * 1600 ) = $ 20 answer is a","correct":"a","options":{"a":"$ 20 ","b":"$ 30 ","c":"$ 45 ","d":"$ 15","e":"$ 60"},"options_float":{"a":20.0,"b":30.0,"c":45.0,"d":15.0,"e":60.0},"annotated_formula":"divide(multiply(divide(multiply(1600, 25), const_100), 5), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)","chain":"1_600 * 25<\/gadget>\n40_000<\/output>\n40_000 \/ 100<\/gadget>\n400<\/output>\n400 * 5<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n20<\/result>","index":544} +{"problem":"how many diagonals does a 60 - sided convex polygon have ?","rationale":"\"a 60 - sided convex polygon has 60 vertices . if we examine a single vertex , we can see that we can connect it with 57 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 60 vertices can be connected with 57 vertices to create a diagonal then the total number of diagonals would be ( 60 ) ( 57 ) = 3420 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3420 by 2 to get 1710 . the answer is c .\"","correct":"c","options":{"a":"960 ","b":"1280 ","c":"1710 ","d":"2560","e":"3420"},"options_float":{"a":960.0,"b":1280.0,"c":1710.0,"d":2560.0,"e":3420.0},"annotated_formula":"divide(factorial(60), multiply(factorial(subtract(60, const_2)), factorial(const_2)))","linear_formula":"factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)|","chain":"factorial(60)<\/gadget>\n8_320_987_112_741_390_144_276_341_183_223_364_380_754_172_606_361_245_952_449_277_696_409_600_000_000_000_000<\/output>\n60 - 2<\/gadget>\n58<\/output>\nfactorial(58)<\/gadget>\n2_350_561_331_282_878_571_829_474_910_515_074_683_828_862_318_181_142_924_420_699_914_240_000_000_000_000<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n2_350_561_331_282_878_571_829_474_910_515_074_683_828_862_318_181_142_924_420_699_914_240_000_000_000_000 * 2<\/gadget>\n4_701_122_662_565_757_143_658_949_821_030_149_367_657_724_636_362_285_848_841_399_828_480_000_000_000_000<\/output>\n8_320_987_112_741_390_144_276_341_183_223_364_380_754_172_606_361_245_952_449_277_696_409_600_000_000_000_000 \/ 4_701_122_662_565_757_143_658_949_821_030_149_367_657_724_636_362_285_848_841_399_828_480_000_000_000_000<\/gadget>\n1_770<\/output>\n1_770<\/result>","index":546} +{"problem":"rs . 1210 were divided among 3 person p , q , r so that p : q = 5 : 4 and q : r = 9 : 10 . then r gets the amount ?","rationale":"p : q = 5 : 4 , q : r = 9 : 10 = ( 9 x 4 \/ 9 ) : ( 10 x 4 \/ 9 ) = 4 : 40 \/ 9 . so , p : q : r = 5 : 4 : 40 \/ 9 = 45 : 36 : 40 sum of ratio terms is = ( 45 + 36 + 40 ) = 121 . r share of amount is rs ( 1210 x 40 \/ 121 ) = rs . 400 . b","correct":"b","options":{"a":"rs . 300 ","b":"rs . 400 ","c":"rs . 480 ","d":"rs . 500","e":"rs . 520"},"options_float":{"a":300.0,"b":400.0,"c":480.0,"d":500.0,"e":520.0},"annotated_formula":"multiply(divide(1210, add(add(divide(5, 4), divide(10, 9)), const_1)), divide(10, 9))","linear_formula":"divide(n2,n3)|divide(n5,n4)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)|multiply(#4,#1)","chain":"5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n10 \/ 9<\/gadget>\n10\/9 = around 1.111111<\/output>\n(5\/4) + (10\/9)<\/gadget>\n85\/36 = around 2.361111<\/output>\n(85\/36) + 1<\/gadget>\n121\/36 = around 3.361111<\/output>\n1_210 \/ (121\/36)<\/gadget>\n360<\/output>\n360 * (10\/9)<\/gadget>\n400<\/output>\n400<\/result>","index":547} +{"problem":"a fruit drink is made of orange , watermelon , and grape juice , where 35 percent of the drink is orange juice and 35 percent is watermelon juice . if the drink is made with 45 ounces of grape juice , how many ounces is the drink total ?","rationale":"\"let the total number of ounces in the drink be x . % of orange = 35 % % of watermelon = 35 % % of grape = 100 % - 70 % = 30 % 0.30 x = 45 x = 150 therefore there are a total of 150 ounces in the drink . the answer is c .\"","correct":"c","options":{"a":"120 ","b":"140 ","c":"150 ","d":"180","e":"200"},"options_float":{"a":120.0,"b":140.0,"c":150.0,"d":180.0,"e":200.0},"annotated_formula":"add(add(45, divide(multiply(45, 35), subtract(const_100, add(35, 35)))), divide(multiply(45, 35), subtract(const_100, add(35, 35))))","linear_formula":"add(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|subtract(const_100,#0)|divide(#1,#3)|divide(#2,#3)|add(n2,#4)|add(#6,#5)|","chain":"45 * 35<\/gadget>\n1_575<\/output>\n35 + 35<\/gadget>\n70<\/output>\n100 - 70<\/gadget>\n30<\/output>\n1_575 \/ 30<\/gadget>\n105\/2 = around 52.5<\/output>\n45 + (105\/2)<\/gadget>\n195\/2 = around 97.5<\/output>\n(195\/2) + (105\/2)<\/gadget>\n150<\/output>\n150<\/result>","index":548} +{"problem":"for all real numbers v , an operation is defined by the equation v * = v - v \/ 3 . if ( v * ) * = 24 , then v =","rationale":"\"( v * ) * = ( v - v \/ 3 ) - ( v - v \/ 3 ) \/ 3 24 = 2 v \/ 3 - 2 v \/ 9 = 4 v \/ 9 v = 54 the answer is d .\"","correct":"d","options":{"a":"45 ","b":"48 ","c":"51 ","d":"54","e":"57"},"options_float":{"a":45.0,"b":48.0,"c":51.0,"d":54.0,"e":57.0},"annotated_formula":"divide(divide(24, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3)))","linear_formula":"divide(const_1,n0)|subtract(const_1,#0)|divide(n1,#1)|divide(#2,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n24 \/ (2\/3)<\/gadget>\n36<\/output>\n36 \/ (2\/3)<\/gadget>\n54<\/output>\n54<\/result>","index":549} +{"problem":"the least perfect square , which is divisible by each of 21,36 and 66 is","rationale":"l . c . m . of 21 , 36 , 66 = 2772 . now , 2772 = 2 x 2 x 3 x 3 x 7 x 11 to make it a perfect square , it must be multiplied by 7 x 11 . so , required number = 22 x 32 x 72 x 112 = 213444 answer a","correct":"a","options":{"a":"213444 ","b":"214344 ","c":"214434 ","d":"213443","e":"none of these"},"options_float":{"a":213444.0,"b":214344.0,"c":214434.0,"d":213443.0,"e":null},"annotated_formula":"multiply(multiply(multiply(power(add(const_3, const_4), const_2), power(divide(divide(66, const_3), const_2), const_2)), power(const_2, const_2)), power(const_3, const_2))","linear_formula":"add(const_3,const_4)|divide(n1,const_3)|power(const_2,const_2)|power(const_3,const_2)|divide(#1,const_2)|power(#0,const_2)|power(#4,const_2)|multiply(#5,#6)|multiply(#7,#2)|multiply(#8,#3)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 ** 2<\/gadget>\n49<\/output>\n66 \/ 3<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n11 ** 2<\/gadget>\n121<\/output>\n49 * 121<\/gadget>\n5_929<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n5_929 * 4<\/gadget>\n23_716<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n23_716 * 9<\/gadget>\n213_444<\/output>\n213_444<\/result>","index":550} +{"problem":"if w \/ x = 2 \/ 3 and w \/ y = 6 \/ 15 , then ( x + y ) \/ y =","rationale":"w \/ x = 2 \/ 3 = > x = 3 \/ 2 w and w \/ y = 6 \/ 15 = > y = 15 \/ 6 w ( x + y ) \/ y = ( 3 \/ 2 w + 15 \/ 6 w ) \/ 15 \/ 6 w = 4 w \/ ( 15 \/ 6 w ) = 8 \/ 5 correct option : d","correct":"d","options":{"a":"4 \/ 5 ","b":"6 \/ 5 ","c":"7 \/ 5 ","d":"8 \/ 5","e":"9 \/ 5"},"options_float":{"a":0.8,"b":1.2,"c":1.4,"d":1.6,"e":1.8},"annotated_formula":"add(divide(divide(6, 2), divide(15, 3)), const_1)","linear_formula":"divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)","chain":"6 \/ 2<\/gadget>\n3<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) + 1<\/gadget>\n8\/5 = around 1.6<\/output>\n8\/5 = around 1.6<\/result>","index":555} +{"problem":"a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 500,000 ?","rationale":"the first digit and last digit are the same so the 2 possibilities are 6 or 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 2 * 10 * 10 = 200 the answer is b .","correct":"b","options":{"a":"160 ","b":"200 ","c":"240 ","d":"300","e":"480"},"options_float":{"a":160.0,"b":200.0,"c":240.0,"d":300.0,"e":480.0},"annotated_formula":"multiply(multiply(subtract(6, const_4), const_10), const_10)","linear_formula":"subtract(n2,const_4)|multiply(#0,const_10)|multiply(#1,const_10)","chain":"6 - 4<\/gadget>\n2<\/output>\n2 * 10<\/gadget>\n20<\/output>\n20 * 10<\/gadget>\n200<\/output>\n200<\/result>","index":556} +{"problem":"a trader sold an article on a certain price with 40 % profit . if he sold double of previous selling price then find its profit %","rationale":"\"let cost price = 100 % selling price = 140 % new s . p . = 280 % p % = 280 - 100 = 180 % answer is b\"","correct":"b","options":{"a":"140 % ","b":"180 % ","c":"120 % ","d":"100 %","e":"150 %"},"options_float":{"a":140.0,"b":180.0,"c":120.0,"d":100.0,"e":150.0},"annotated_formula":"subtract(multiply(const_2, add(const_100, 40)), const_100)","linear_formula":"add(n0,const_100)|multiply(#0,const_2)|subtract(#1,const_100)|","chain":"100 + 40<\/gadget>\n140<\/output>\n2 * 140<\/gadget>\n280<\/output>\n280 - 100<\/gadget>\n180<\/output>\n180<\/result>","index":557} +{"problem":"the average price of an decoration piece increases over the years . if from 1990 to 1996 , the price of the decoration piece increased by 13 % and from 1996 to 2001 it increased by 20 % , what is the price of the decoration piece in 2001 if the price in 1990 was $ 11500 ?","rationale":"whenever you have a question in which a value is increased by x % and then increased by y % then the overall effective increase = ( x + y + xy \/ 100 ) % the effective increase = 13 + 20 + ( 13 × 20 \/ 100 ) = 35.6 % this is a little more 33 % or 1 \/ 3 the original value . the quickest way to solve this question is to use approximation . 11500 + 1 \/ 3 ( 12000 ) = 15500 therefore option a is the answer . i selected 12000 because it ' s value is a little more 11500 and it ' s easier to divide by 3 ( ie to calculate 1 \/ 3 or 33 % ) .","correct":"a","options":{"a":"$ 15594 . ","b":"$ 15322 . ","c":"$ 14786 . ","d":"$ 14543 .","e":"$ 12988 ."},"options_float":{"a":15594.0,"b":15322.0,"c":14786.0,"d":14543.0,"e":12988.0},"annotated_formula":"multiply(add(const_1, divide(20, const_100)), multiply(add(const_1, divide(13, const_100)), 11500))","linear_formula":"divide(n5,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n8,#3)|multiply(#2,#4)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n13 \/ 100<\/gadget>\n13\/100 = around 0.13<\/output>\n1 + (13\/100)<\/gadget>\n113\/100 = around 1.13<\/output>\n(113\/100) * 11_500<\/gadget>\n12_995<\/output>\n(6\/5) * 12_995<\/gadget>\n15_594<\/output>\n15_594<\/result>","index":558} +{"problem":"a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many h cubes are painted on only 2 faces ?","rationale":"\"1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all on the edge of the cube , in themiddleof the edge . there are 4 in front , 4 in back and 4 more on thestripthat runs around the left \/ top \/ right \/ bottom of the cube . h = 4 + 4 + 4 = 12 . answer a\"","correct":"a","options":{"a":"12 ","b":"8 ","c":"6 ","d":"10","e":"16"},"options_float":{"a":12.0,"b":8.0,"c":6.0,"d":10.0,"e":16.0},"annotated_formula":"multiply(const_4, power(27, divide(const_1, const_3)))","linear_formula":"divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n27 ** (1\/3)<\/gadget>\n3<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12<\/result>","index":560} +{"problem":"the length of a rectangle is increased by 15 % and its breadth is decreased by 20 % . what is the effect on its area ?","rationale":"\"100 * 100 = 10000 115 * 80 = 9200 answer : e\"","correct":"e","options":{"a":"1288 ","b":"1299 ","c":"1000 ","d":"10000","e":"9200"},"options_float":{"a":1288.0,"b":1299.0,"c":1000.0,"d":10000.0,"e":9200.0},"annotated_formula":"multiply(add(15, const_100), subtract(const_100, 20))","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|","chain":"15 + 100<\/gadget>\n115<\/output>\n100 - 20<\/gadget>\n80<\/output>\n115 * 80<\/gadget>\n9_200<\/output>\n9_200<\/result>","index":561} +{"problem":"how many seconds will a train 150 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?","rationale":"\"d d = 150 + 150 = 300 s = 36 * 5 \/ 18 = 10 mps t = 300 \/ 10 = 30 sec\"","correct":"d","options":{"a":"28 sec ","b":"23 sec ","c":"29 sec ","d":"30 sec","e":"36 sec"},"options_float":{"a":28.0,"b":23.0,"c":29.0,"d":30.0,"e":36.0},"annotated_formula":"divide(add(150, 150), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"150 + 150<\/gadget>\n300<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n300 \/ 10<\/gadget>\n30<\/output>\n30<\/result>","index":563} +{"problem":"the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 8.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 \/ 2 ) - x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ?","rationale":"\"i tried using time \/ rate approach : - initial price difference = 8.80 - 3.20 = 5.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled so , 5.6 = t ( 5.4 x ) - > t = ( 5.6 ) \/ 5.4 x final price = 3.20 + 5 x * t - > 3.20 + 5 * 5.6 \/ 5.4 = 8.4 answer a .\"","correct":"a","options":{"a":"$ 8.40 ","b":"$ 5.10 ","c":"$ 5.30 ","d":"$ 5.50","e":"$ 5.60"},"options_float":{"a":8.4,"b":5.1,"c":5.3,"d":5.5,"e":5.6},"annotated_formula":"add(3.20, multiply(divide(subtract(8.80, 3.20), add(5, subtract(sqrt(2), 1))), 5))","linear_formula":"sqrt(n3)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5)|","chain":"8.8 - 3.2<\/gadget>\n5.6<\/output>\n2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n(sqrt(2)) - 1<\/gadget>\n-1 + sqrt(2) = around 0.414214<\/output>\n5 + (-1 + sqrt(2))<\/gadget>\nsqrt(2) + 4 = around 5.414214<\/output>\n5.6 \/ (sqrt(2) + 4)<\/gadget>\n5.6\/(sqrt(2) + 4) = around 1.034315<\/output>\n(5.6\/(sqrt(2) + 4)) * 5<\/gadget>\n28.0\/(sqrt(2) + 4) = around 5.171573<\/output>\n3.2 + (28.0\/(sqrt(2) + 4))<\/gadget>\n3.2 + 28.0\/(sqrt(2) + 4) = around 8.371573<\/output>\n3.2 + 28.0\/(sqrt(2) + 4) = around 8.371573<\/result>","index":564} +{"problem":"find the area of the quadrilateral of one of its diagonals is 28 cm and its off sets 9 cm and 6 cm ?","rationale":"\"1 \/ 2 * 28 ( 9 + 6 ) = 210 cm 2 answer : a\"","correct":"a","options":{"a":"210 cm 2 ","b":"150 cm 2 ","c":"168 cm 2 ","d":"198 cm 2","e":"987 cm 2"},"options_float":{"a":210.0,"b":150.0,"c":168.0,"d":198.0,"e":987.0},"annotated_formula":"multiply(multiply(divide(const_1, const_2), add(6, 9)), 28)","linear_formula":"add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n6 + 9<\/gadget>\n15<\/output>\n(1\/2) * 15<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) * 28<\/gadget>\n210<\/output>\n210<\/result>","index":565} +{"problem":"when the price of a product was decreased by 20 % , the number sold increased by 60 % . what was the effect on the total revenue ?","rationale":"let the price of the product be $ 100 and let original sale be 100 pieces then , total revenue = 100 * 100 = $ 10000 new revenue = 80 * 160 = $ 12800 increase in revenue = 2800 * 100 \/ 10000 = 28 % answer is a","correct":"a","options":{"a":"28 % ","b":"32 % ","c":"35 % ","d":"45 %","e":"52 %"},"options_float":{"a":28.0,"b":32.0,"c":35.0,"d":45.0,"e":52.0},"annotated_formula":"multiply(subtract(multiply(add(const_1, divide(60, const_100)), divide(subtract(const_100, 20), const_100)), const_1), const_100)","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|add(#0,const_1)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(8\/5) * (4\/5)<\/gadget>\n32\/25 = around 1.28<\/output>\n(32\/25) - 1<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) * 100<\/gadget>\n28<\/output>\n28<\/result>","index":567} +{"problem":"in a rare coin collection , all coins are either pure gold or pure silver , and there is initially one gold coin for every 3 silver coins . with the addition of 15 more gold coins to the collection , the ratio of gold coins to silver coins is 1 to 2 . based on this information , how many total coins are there now in this collection ( after the acquisition ) ?","rationale":"initially the number of gold coins and silver coins is 2 k and 6 k . after adding gold coins , the number of coins is 3 k and 6 k . thus k = 15 and the total number of coins is 9 k = 135 . the answer is b .","correct":"b","options":{"a":"124 ","b":"135 ","c":"146 ","d":"157","e":"168"},"options_float":{"a":124.0,"b":135.0,"c":146.0,"d":157.0,"e":168.0},"annotated_formula":"add(add(multiply(multiply(2, 15), 3), multiply(2, 15)), 15)","linear_formula":"multiply(n1,n3)|multiply(n0,#0)|add(#1,#0)|add(n1,#2)","chain":"2 * 15<\/gadget>\n30<\/output>\n30 * 3<\/gadget>\n90<\/output>\n90 + 30<\/gadget>\n120<\/output>\n120 + 15<\/gadget>\n135<\/output>\n135<\/result>","index":568} +{"problem":"a student has to obtain 35 % of the total marks to pass . he got 100 marks and failed by 40 marks . the maximum marks are ?","rationale":"\"let the maximum marks be x then , 35 % of x = 100 + 40 35 x \/ 100 = 140 35 x = 140 * 100 = 14000 x = 400 answer is b\"","correct":"b","options":{"a":"280 ","b":"400 ","c":"420 ","d":"480","e":"520"},"options_float":{"a":280.0,"b":400.0,"c":420.0,"d":480.0,"e":520.0},"annotated_formula":"divide(add(100, 40), divide(35, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"100 + 40<\/gadget>\n140<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n140 \/ (7\/20)<\/gadget>\n400<\/output>\n400<\/result>","index":571} +{"problem":"if a : b = 3 : 4 , b : c = 7 : 9 , c : d = 5 : 7 , find a : d ?","rationale":"\"a \/ d = ( 3 \/ 4 ) * ( 7 \/ 9 ) * ( 5 \/ 7 ) = > 5 \/ 12 answer : a\"","correct":"a","options":{"a":"5 \/ 12 ","b":"5 \/ 15 ","c":"5 \/ 10 ","d":"5 \/ 11","e":"5 \/ 13"},"options_float":{"a":0.4166666667,"b":0.3333333333,"c":0.5,"d":0.4545454545,"e":0.3846153846},"annotated_formula":"divide(multiply(multiply(3, 7), 5), multiply(multiply(4, 9), 7))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|","chain":"3 * 7<\/gadget>\n21<\/output>\n21 * 5<\/gadget>\n105<\/output>\n4 * 9<\/gadget>\n36<\/output>\n36 * 7<\/gadget>\n252<\/output>\n105 \/ 252<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":572} +{"problem":"the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ?","rationale":"\"let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 ⇒ 40 y + 384 = 36 y + 432 ⇒ 4 y = 48 ∴ y = 12 b\"","correct":"b","options":{"a":"8 ","b":"12 ","c":"15 ","d":"17","e":"18"},"options_float":{"a":8.0,"b":12.0,"c":15.0,"d":17.0,"e":18.0},"annotated_formula":"divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 32)), 4)","linear_formula":"multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)|","chain":"40 - 4<\/gadget>\n36<\/output>\n12 * 36<\/gadget>\n432<\/output>\n12 * 32<\/gadget>\n384<\/output>\n432 - 384<\/gadget>\n48<\/output>\n48 \/ 4<\/gadget>\n12<\/output>\n12<\/result>","index":573} +{"problem":"a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 12 days before the completion of the work . in how many days was the work completed ?","rationale":"\"one day work of a , b and c = 1 \/ 24 + 1 \/ 30 + 1 \/ 40 = 1 \/ 10 work done by a and b together in the last 12 days = 12 * ( 1 \/ 24 + 1 \/ 30 ) = 9 \/ 10 remaining work = 1 \/ 10 the number of days required for this initial work = 1 days . the total number of days required = 12 + 1 = 13 days . answer : a\"","correct":"a","options":{"a":"13 days ","b":"16 days ","c":"18 days ","d":"11 days","e":"38 days"},"options_float":{"a":13.0,"b":16.0,"c":18.0,"d":11.0,"e":38.0},"annotated_formula":"add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 12)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 12))), add(divide(const_1, 24), divide(const_1, 30)))), 12)","linear_formula":"add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,n3)|add(#1,#2)|multiply(#0,#3)|divide(const_1,#5)|multiply(n3,#4)|add(#4,#6)|subtract(const_1,#7)|divide(#9,#8)|add(n3,#10)|","chain":"1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/24) + (1\/30)<\/gadget>\n3\/40 = around 0.075<\/output>\n(3\/40) * 12<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n2 + 3<\/gadget>\n5<\/output>\n2 * 12<\/gadget>\n24<\/output>\n5 * 24<\/gadget>\n120<\/output>\n1 \/ 120<\/gadget>\n1\/120 = around 0.008333<\/output>\n(1\/120) + (3\/40)<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/10) \/ (1\/12)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) + 12<\/gadget>\n66\/5 = around 13.2<\/output>\n66\/5 = around 13.2<\/result>","index":574} +{"problem":"the total marks obtained by a student in mathematics and physics is 30 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .","rationale":"\"let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 30 and c - p = 20 m + c \/ 2 = [ ( m + p ) + ( c - p ) ] \/ 2 = ( 30 + 20 ) \/ 2 = 25 . answer : c\"","correct":"c","options":{"a":"40 ","b":"30 ","c":"25 ","d":"data inadequate","e":"none of these ."},"options_float":{"a":40.0,"b":30.0,"c":25.0,"d":null,"e":null},"annotated_formula":"divide(add(30, 20), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"30 + 20<\/gadget>\n50<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n25<\/result>","index":577} +{"problem":"a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1540 in all from both of them as interest . the rate of interest per annum is ?","rationale":"\"let the rate be r % p . a . then , ( 5000 * r * 2 ) \/ 100 + ( 3000 * r * 4 ) \/ 100 = 1540 100 r + 120 r = 1540 r = 7 % answer : a\"","correct":"a","options":{"a":"7 % ","b":"12 % ","c":"74 % ","d":"10 %","e":"45 %"},"options_float":{"a":7.0,"b":12.0,"c":74.0,"d":10.0,"e":45.0},"annotated_formula":"multiply(divide(1540, add(multiply(5000, 2), multiply(3000, 4))), const_100)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)|","chain":"5_000 * 2<\/gadget>\n10_000<\/output>\n3_000 * 4<\/gadget>\n12_000<\/output>\n10_000 + 12_000<\/gadget>\n22_000<\/output>\n1_540 \/ 22_000<\/gadget>\n7\/100 = around 0.07<\/output>\n(7\/100) * 100<\/gadget>\n7<\/output>\n7<\/result>","index":578} +{"problem":"solve for x and check : 25 x = 675","rationale":"\"solution : dividing each side by 25 , we obtain ( 25 x \/ 675 ) = ( 675 \/ 25 ) therefore : x = 27 check : 25 x = 675 ( 25 * 27 ) = 675 675 = 675 answer : a\"","correct":"a","options":{"a":"675 ","b":"725 ","c":"895 ","d":"1025","e":"none of these"},"options_float":{"a":675.0,"b":725.0,"c":895.0,"d":1025.0,"e":null},"annotated_formula":"multiply(25, divide(675, 25))","linear_formula":"divide(n1,n0)|multiply(n0,#0)|","chain":"675 \/ 25<\/gadget>\n27<\/output>\n25 * 27<\/gadget>\n675<\/output>\n675<\/result>","index":579} +{"problem":"given that x and y are events such that z ( x ) = 0.02 z ( y ) = 0.10 z ( xny ) = 0.10 z ( x | y ) = find z ( x | y ) and z ( x | y )","rationale":"here , x and y are events z ( x | y ) = z ( xny ) \/ z ( y ) = 0.10 \/ 0.10 = 1 q ( g | h ) = z ( xny ) \/ z ( x ) = 0.4 \/ 0.02 = 1 \/ 5 answer is a","correct":"a","options":{"a":"1 \/ 5 ","b":"1 \/ 6 ","c":"1 \/ 6 ","d":"1 \/ 4","e":"1 \/ 2"},"options_float":{"a":0.2,"b":0.1666666667,"c":0.1666666667,"d":0.25,"e":0.5},"annotated_formula":"inverse(divide(0.1, 0.02))","linear_formula":"divide(n1,n0)|inverse(#0)","chain":"0.1 \/ 0.02<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":580} +{"problem":"a jar contains a mixture of ab in the ratio 4 : 1 . when 40 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially .","rationale":"40 litres of mixture that is replaced will contain 32 litres of a and 8 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 32 ] \/ [ k - 8 + 40 ] = 2 \/ 3 = > 12 k - 96 = 2 k - 16 + 80 = > 10 k = 160 solve for k which is k = 16 so initial volume of liquid a = 4 k = 64 litres answer : d","correct":"d","options":{"a":"12 ","b":"15 ","c":"16 ","d":"64","e":"25"},"options_float":{"a":12.0,"b":15.0,"c":16.0,"d":64.0,"e":25.0},"annotated_formula":"multiply(divide(multiply(add(multiply(3, const_2), multiply(2, const_2)), divide(40, subtract(multiply(3, const_2), 1))), add(4, 1)), 4)","linear_formula":"add(n0,n1)|multiply(n4,const_2)|multiply(n3,const_2)|add(#1,#2)|subtract(#1,n1)|divide(n2,#4)|multiply(#3,#5)|divide(#6,#0)|multiply(n0,#7)","chain":"3 * 2<\/gadget>\n6<\/output>\n2 * 2<\/gadget>\n4<\/output>\n6 + 4<\/gadget>\n10<\/output>\n6 - 1<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n10 * 8<\/gadget>\n80<\/output>\n4 + 1<\/gadget>\n5<\/output>\n80 \/ 5<\/gadget>\n16<\/output>\n16 * 4<\/gadget>\n64<\/output>\n64<\/result>","index":581} +{"problem":"the length of a rectangle is 3 times its width . if the width of the rectangle is 6 inches , what is the rectangle ' s area , in square inches ?","rationale":"\"if the width is 6 in and the length is 3 times the width , then the length is 3 * 6 = 18 in the area is given by 6 * 18 = 108 square inches correct answer b\"","correct":"b","options":{"a":"30 square inches ","b":"108 square inches ","c":"68 square inches ","d":"89 square inches","e":"92 square inches"},"options_float":{"a":30.0,"b":108.0,"c":68.0,"d":89.0,"e":92.0},"annotated_formula":"rectangle_area(6, multiply(3, 6))","linear_formula":"multiply(n0,n1)|rectangle_area(n1,#0)|","chain":"3 * 6<\/gadget>\n18<\/output>\n6 * 18<\/gadget>\n108<\/output>\n108<\/result>","index":582} +{"problem":"a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 2500 . the difference between the profit shares of a and c is ?","rationale":"\"explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 2500 = > 5 parts out of 15 parts is rs . 2500 now , required difference is 6 - 4 = 2 parts required difference = 2 \/ 5 ( 2500 ) = rs . 1000 answer : d\"","correct":"d","options":{"a":"621 ","b":"276 ","c":"236 ","d":"1000","e":"211"},"options_float":{"a":621.0,"b":276.0,"c":236.0,"d":1000.0,"e":211.0},"annotated_formula":"multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 2500)","linear_formula":"divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2)|","chain":"12_000 \/ 10_000<\/gadget>\n6\/5 = around 1.2<\/output>\n8_000 \/ 10_000<\/gadget>\n4\/5 = around 0.8<\/output>\n(6\/5) - (4\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 2_500<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":583} +{"problem":"a truck drives 80 miles on local roads at 30 mph , and 50 miles on the highway at 60 mph , the truck also take a hour lunch break . what is the average speed of the entire trip ?","rationale":"so the answer is plainly a . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 80 miles , and v = 30 mph so we can obtain time : 30 * t = 80 or t = 80 \/ 30 = 2.6 hours the needed time to cover 80 miles in the same way we should divide 50 to 60 to find the needed time to cover 50 miles , so t = 0.8 hours so the total time for covering total distance would be 2.6 + 0.8 + 1 ( lunch time ) = 4.4 hours and total distance is 80 + 50 = 130 miles final stage : average speed is total distance divide to total time : 130 \/ 4.4 = 29.5 miles per hour rounding up to 30 . . . .","correct":"a","options":{"a":"30 mph ","b":"29 mph ","c":"28 mph ","d":"31 mph","e":"23 mph"},"options_float":{"a":30.0,"b":29.0,"c":28.0,"d":31.0,"e":23.0},"annotated_formula":"divide(add(80, 60), add(add(divide(50, 60), divide(80, 30)), const_1))","linear_formula":"add(n0,n3)|divide(n2,n3)|divide(n0,n1)|add(#1,#2)|add(#3,const_1)|divide(#0,#4)","chain":"80 + 60<\/gadget>\n140<\/output>\n50 \/ 60<\/gadget>\n5\/6 = around 0.833333<\/output>\n80 \/ 30<\/gadget>\n8\/3 = around 2.666667<\/output>\n(5\/6) + (8\/3)<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) + 1<\/gadget>\n9\/2 = around 4.5<\/output>\n140 \/ (9\/2)<\/gadget>\n280\/9 = around 31.111111<\/output>\n280\/9 = around 31.111111<\/result>","index":584} +{"problem":"carrey rented a car for rs . 20 plus rs . 0.25 per kilometer driven . samuel rented a car for rs . 24 plus rs . 0.16 per kilometer driven . if each drove d km . and each was charged exactly the same amount for the rental , then d equals ?","rationale":"explanation : 20 + 0.25 × d = 24 + 0.16 d solving we get d = 44.4 answer : a","correct":"a","options":{"a":"44.4 ","b":"44.8 ","c":"1 ","d":"19","e":"44.11"},"options_float":{"a":44.4,"b":44.8,"c":1.0,"d":19.0,"e":44.11},"annotated_formula":"divide(subtract(24, 20), subtract(0.25, 0.16))","linear_formula":"subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)","chain":"24 - 20<\/gadget>\n4<\/output>\n0.25 - 0.16<\/gadget>\n0.09<\/output>\n4 \/ 0.09<\/gadget>\n44.444444<\/output>\n44.444444<\/result>","index":585} +{"problem":"the first doughnut is priced at $ 1 and then if you purchase additional doughnuts as dozens then the price is $ 6 \/ dozen . what is the total number of doughnuts purchased if you paid $ 24 ?","rationale":"$ 24 = 4 * $ 6 the number of doughnuts is 4 * 12 = 48 the answer is c .","correct":"c","options":{"a":"44 ","b":"46 ","c":"48 ","d":"50","e":"52"},"options_float":{"a":44.0,"b":46.0,"c":48.0,"d":50.0,"e":52.0},"annotated_formula":"multiply(divide(24, 6), const_12)","linear_formula":"divide(n2,n1)|multiply(#0,const_12)","chain":"24 \/ 6<\/gadget>\n4<\/output>\n4 * 12<\/gadget>\n48<\/output>\n48<\/result>","index":586} +{"problem":"the speed of the boat in still water in 12 kmph . it can travel downstream through 48 kms in 3 hrs . in what time would it cover the same distance upstream ?","rationale":"\"still water = 12 km \/ hr downstream = 48 \/ 3 = 16 km \/ hr upstream = > > still water = ( u + v \/ 2 ) = > > 12 = u + 16 \/ 2 = 8 km \/ hr so time taken in upstream = 48 \/ 8 = 6 hrs answer : e\"","correct":"e","options":{"a":"8 hours ","b":"6 hours ","c":"4 hours ","d":"5 hours","e":"6 hours"},"options_float":{"a":8.0,"b":6.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"divide(48, subtract(12, subtract(divide(48, 3), 12)))","linear_formula":"divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)|","chain":"48 \/ 3<\/gadget>\n16<\/output>\n16 - 12<\/gadget>\n4<\/output>\n12 - 4<\/gadget>\n8<\/output>\n48 \/ 8<\/gadget>\n6<\/output>\n6<\/result>","index":587} +{"problem":"on a sum of money , the simple interest for 2 years is rs . 326 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is","rationale":"\"explanation : the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r ã — si ) \/ ( 2 ã — 100 ) difference between the compound interest and simple interest = 340 - 326 = 14 ( r ã — si ) \/ ( 2 ã — 100 ) = 14 ( r ã — 326 ) \/ ( 2 ã — 100 ) = 14 r = 8.58 % answer : option c\"","correct":"c","options":{"a":"15 % ","b":"14.25 % ","c":"8.58 % ","d":"10.5 %","e":"11.5 %"},"options_float":{"a":15.0,"b":14.25,"c":8.58,"d":10.5,"e":11.5},"annotated_formula":"divide(multiply(const_100, subtract(subtract(340, divide(326, 2)), divide(326, 2))), divide(326, 2))","linear_formula":"divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)|","chain":"326 \/ 2<\/gadget>\n163<\/output>\n340 - 163<\/gadget>\n177<\/output>\n177 - 163<\/gadget>\n14<\/output>\n100 * 14<\/gadget>\n1_400<\/output>\n1_400 \/ 163<\/gadget>\n1_400\/163 = around 8.588957<\/output>\n1_400\/163 = around 8.588957<\/result>","index":588} +{"problem":"in a certain supermarket , a triangular display of cans is arranged in 9 rows , numbered 1 through 9 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 120 cans in the entire display , how many cans are in the seventh row ?","rationale":"\"let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 24 ) = 9 x + 3 ( 1 + 2 + . . . + 8 ) = 9 x + 3 ( 8 ) ( 9 ) \/ 2 = 9 x + 108 since the total is less than 120 , x must equal 1 . the number of cans in the 7 th row is 1 + 3 ( 6 ) = 19 the answer is d .\"","correct":"d","options":{"a":"13 ","b":"15 ","c":"17 ","d":"19","e":"21"},"options_float":{"a":13.0,"b":15.0,"c":17.0,"d":19.0,"e":21.0},"annotated_formula":"add(multiply(add(3, 3), 3), floor(divide(subtract(120, multiply(divide(multiply(3, subtract(9, 1)), const_2), 9)), 9)))","linear_formula":"add(n3,n3)|subtract(n0,n1)|multiply(n3,#1)|multiply(n3,#0)|divide(#2,const_2)|multiply(n0,#4)|subtract(n4,#5)|divide(#6,n0)|floor(#7)|add(#8,#3)|","chain":"3 + 3<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n9 - 1<\/gadget>\n8<\/output>\n3 * 8<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12 * 9<\/gadget>\n108<\/output>\n120 - 108<\/gadget>\n12<\/output>\n12 \/ 9<\/gadget>\n4\/3 = around 1.333333<\/output>\nfloor(4\/3)<\/gadget>\n1<\/output>\n18 + 1<\/gadget>\n19<\/output>\n19<\/result>","index":589} +{"problem":"the length of a rectangle is 3 times its width . if the width of the rectangle is 4 inches , what is the rectangle ' s area , in square inches ?","rationale":"\"if the width is 4 in and the length is 3 times the width , then the length is 3 * 4 = 12 in the area is given by 4 * 12 = 48 square inches correct answer d\"","correct":"d","options":{"a":"30 square inches ","b":"75 square inches ","c":"68 square inches ","d":"48 square inches","e":"92 square inches"},"options_float":{"a":30.0,"b":75.0,"c":68.0,"d":48.0,"e":92.0},"annotated_formula":"rectangle_area(4, multiply(3, 4))","linear_formula":"multiply(n0,n1)|rectangle_area(n1,#0)|","chain":"3 * 4<\/gadget>\n12<\/output>\n4 * 12<\/gadget>\n48<\/output>\n48<\/result>","index":590} +{"problem":"the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.7 years . the ration of the number of boys to the number of girls in the class is ?","rationale":"\"let the ratio be k : 1 . then , k * 16.4 + 1 * 15.7 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.7 ) = k = 0.1 \/ 0.6 = 1 \/ 6 required ratio = 1 \/ 6 : 1 = 1 : 6 . answer : a\"","correct":"a","options":{"a":"1 : 6 ","b":"2 : 3 ","c":"2 : 5 ","d":"2 : 1","e":"2 : 4"},"options_float":{"a":0.1666666667,"b":0.6666666667,"c":0.4,"d":2.0,"e":0.5},"annotated_formula":"divide(subtract(15.8, 15.7), subtract(16.4, 15.8))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|","chain":"15.8 - 15.7<\/gadget>\n0.1<\/output>\n16.4 - 15.8<\/gadget>\n0.6<\/output>\n0.1 \/ 0.6<\/gadget>\n0.166667<\/output>\n0.166667<\/result>","index":591} +{"problem":"a man walking at a rate of 5 km \/ hr crosses a bridge in 15 minutes . the length of the bridge is ?","rationale":"\"speed = 5 * 5 \/ 18 = 25 \/ 18 m \/ sec distance covered in 15 minutes = 25 \/ 18 * 15 * 60 = 1250 m answer is a\"","correct":"a","options":{"a":"1250 m ","b":"1310 m ","c":"1500 m ","d":"1000 m","e":"1195 m"},"options_float":{"a":1250.0,"b":1310.0,"c":1500.0,"d":1000.0,"e":1195.0},"annotated_formula":"multiply(divide(multiply(5, const_1000), const_60), 15)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|","chain":"5 * 1_000<\/gadget>\n5_000<\/output>\n5_000 \/ 60<\/gadget>\n250\/3 = around 83.333333<\/output>\n(250\/3) * 15<\/gadget>\n1_250<\/output>\n1_250<\/result>","index":592} +{"problem":"the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 11 . if p invested the money for 5 months , find for how much time did q invest the money ?","rationale":"\"7 * 5 : 5 * x = 7 : 11 x = 11 answer : d\"","correct":"d","options":{"a":"19 ","b":"17 ","c":"13 ","d":"11","e":"12"},"options_float":{"a":19.0,"b":17.0,"c":13.0,"d":11.0,"e":12.0},"annotated_formula":"multiply(multiply(divide(7, 5), divide(11, 7)), 5)","linear_formula":"divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)|","chain":"7 \/ 5<\/gadget>\n7\/5 = around 1.4<\/output>\n11 \/ 7<\/gadget>\n11\/7 = around 1.571429<\/output>\n(7\/5) * (11\/7)<\/gadget>\n11\/5 = around 2.2<\/output>\n(11\/5) * 5<\/gadget>\n11<\/output>\n11<\/result>","index":594} +{"problem":"a fair price shopkeeper takes 10 % profit on his goods . he lost 30 % goods during theft . his loss percent is :","rationale":"\"explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 70 . s . p . of each item = rs . 1.10 total sale = 1.10 * 70 = rs . 77 hence , loss % = 23 \/ 100 * 100 = 23 % answer : a\"","correct":"a","options":{"a":"23 % ","b":"42 % ","c":"32 % ","d":"12 %","e":"22 %"},"options_float":{"a":23.0,"b":42.0,"c":32.0,"d":12.0,"e":22.0},"annotated_formula":"subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 30), const_100)))","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)|","chain":"100 + 10<\/gadget>\n110<\/output>\n110 * 30<\/gadget>\n3_300<\/output>\n3_300 \/ 100<\/gadget>\n33<\/output>\n110 - 33<\/gadget>\n77<\/output>\n100 - 77<\/gadget>\n23<\/output>\n23<\/result>","index":595} +{"problem":"how many seconds will a 500 m long train take to cross a man walking with a speed of 3 km \/ hr in the direction of the moving train if the speed of the train is 63 km \/ hr ?","rationale":"\"speed of train relative to man = 63 - 3 = 60 km \/ hr . = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec . time taken to pass the man = 500 * 3 \/ 50 = 30 sec . answer : b\"","correct":"b","options":{"a":"22 ","b":"30 ","c":"88 ","d":"26","e":"12"},"options_float":{"a":22.0,"b":30.0,"c":88.0,"d":26.0,"e":12.0},"annotated_formula":"divide(500, multiply(subtract(63, 3), const_0_2778))","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"63 - 3<\/gadget>\n60<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n500 \/ (50\/3)<\/gadget>\n30<\/output>\n30<\/result>","index":597} +{"problem":"the average age of 10 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is","rationale":"\"total age increased = ( 10 * 2 ) years = 20 years . sum of ages of two new men = ( 21 + 23 + 20 ) years = 64 years average age of two new men = ( 64 \/ 2 ) years = 32 years . answer : e\"","correct":"e","options":{"a":"22 ","b":"30 ","c":"99 ","d":"38","e":"32"},"options_float":{"a":22.0,"b":30.0,"c":99.0,"d":38.0,"e":32.0},"annotated_formula":"add(divide(add(21, 23), const_2), multiply(const_1, 10))","linear_formula":"add(n1,n2)|multiply(n0,const_1)|divide(#0,const_2)|add(#2,#1)|","chain":"21 + 23<\/gadget>\n44<\/output>\n44 \/ 2<\/gadget>\n22<\/output>\n1 * 10<\/gadget>\n10<\/output>\n22 + 10<\/gadget>\n32<\/output>\n32<\/result>","index":600} +{"problem":"if 64 ( 8 ^ x ) = 1 then x =","rationale":"\"8 ^ x = 1 \/ 64 8 ^ x = 1 \/ 8 ^ 2 8 ^ x = 8 ^ - 2 x = - 2 e\"","correct":"e","options":{"a":"2 ","b":"– 1 ","c":"0 ","d":"1","e":"- 2"},"options_float":{"a":2.0,"b":1.0,"c":0.0,"d":1.0,"e":-2.0},"annotated_formula":"divide(log(divide(1, 64)), log(8))","linear_formula":"divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|","chain":"1 \/ 64<\/gadget>\n1\/64 = around 0.015625<\/output>\nlog(1\/64)<\/gadget>\n-log(64) = around -4.158883<\/output>\nlog(8)<\/gadget>\nlog(8) = around 2.079442<\/output>\n(-log(64)) \/ log(8)<\/gadget>\n-log(64)\/log(8) = around -2<\/output>\n-log(64)\/log(8) = around -2<\/result>","index":601} +{"problem":"the base and height of a parallelogram are 12 m and 6 m respectively . then its area is = = = = = = ?","rationale":"a = b * h = > a = 12 * 6 = 72 sq . m . answer : ( e )","correct":"e","options":{"a":"27 sq . m ","b":"126 sq . m ","c":"84 sq . m ","d":"18 sq . m","e":"72 sq . m"},"options_float":{"a":27.0,"b":126.0,"c":84.0,"d":18.0,"e":72.0},"annotated_formula":"rectangle_area(12, 6)","linear_formula":"rectangle_area(n0,n1)","chain":"12 * 6<\/gadget>\n72<\/output>\n72<\/result>","index":604} +{"problem":"pipe a can fill a tank in 6 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ?","rationale":"\"let the pipe b be closed after x minutes . 30 \/ 6 - x \/ 24 = 1 = > x \/ 24 = 30 \/ 6 - 1 = 4 = > x = 4 * 24 = 96 . answer : c\"","correct":"c","options":{"a":"18 ","b":"27 ","c":"96 ","d":"27","e":"21"},"options_float":{"a":18.0,"b":27.0,"c":96.0,"d":27.0,"e":21.0},"annotated_formula":"multiply(subtract(divide(30, 6), const_1), 24)","linear_formula":"divide(n2,n0)|subtract(#0,const_1)|multiply(n1,#1)|","chain":"30 \/ 6<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n4 * 24<\/gadget>\n96<\/output>\n96<\/result>","index":607} +{"problem":"the number of diagonals of a polygon of n sides is given by the formula q = n ( n - 3 ) \/ 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ?","rationale":"\"q = n ( n - 3 ) q = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b\"","correct":"b","options":{"a":"3 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"add(3, 2)","linear_formula":"add(n0,n1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5<\/result>","index":610} +{"problem":"a fair price shopkeeper takes 10 % profit on his goods . he lost 70 % goods during theft . his loss percent is :","rationale":"\"explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 30 . s . p . of each item = rs . 1.10 total sale = 1.10 * 30 = rs . 33 hence , loss % = 67 \/ 100 * 100 = 67 % answer : d\"","correct":"d","options":{"a":"72 % ","b":"42 % ","c":"32 % ","d":"67 %","e":"22 %"},"options_float":{"a":72.0,"b":42.0,"c":32.0,"d":67.0,"e":22.0},"annotated_formula":"subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 70), const_100)))","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)|","chain":"100 + 10<\/gadget>\n110<\/output>\n110 * 70<\/gadget>\n7_700<\/output>\n7_700 \/ 100<\/gadget>\n77<\/output>\n110 - 77<\/gadget>\n33<\/output>\n100 - 33<\/gadget>\n67<\/output>\n67<\/result>","index":611} +{"problem":"the age of somu is one - third his father ' s . 8 years back he was one - fifth of his father ' s age . what is his persent age ?","rationale":"\"explanation : let somu ' s age be x and that of his father be 3 x . so , x - 8 = 3 x - 8 \/ 5 = x = 16 answer : option d\"","correct":"d","options":{"a":"11 ","b":"13 ","c":"14 ","d":"16","e":"10"},"options_float":{"a":11.0,"b":13.0,"c":14.0,"d":16.0,"e":10.0},"annotated_formula":"divide(subtract(multiply(add(const_4, const_1), 8), 8), subtract(add(const_4, const_1), const_3))","linear_formula":"add(const_1,const_4)|multiply(n0,#0)|subtract(#0,const_3)|subtract(#1,n0)|divide(#3,#2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 8<\/gadget>\n40<\/output>\n40 - 8<\/gadget>\n32<\/output>\n5 - 3<\/gadget>\n2<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16<\/result>","index":612} +{"problem":"13 buckets of water fill a tank when the capacity of each bucket is 42 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 6 litres ?","rationale":"\"capacity of the tank = ( 13 ã — 42 ) litre number of buckets required of capacity of each bucket is 17 litre = 13 ã — 42 \/ 6 = 13 ã — 7 = 91 answer is a\"","correct":"a","options":{"a":"91 ","b":"80 ","c":"96 ","d":"98","e":"90"},"options_float":{"a":91.0,"b":80.0,"c":96.0,"d":98.0,"e":90.0},"annotated_formula":"divide(multiply(42, 13), 6)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"42 * 13<\/gadget>\n546<\/output>\n546 \/ 6<\/gadget>\n91<\/output>\n91<\/result>","index":613} +{"problem":"two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 22 seconds . the ratio of their speeds is ?","rationale":"\"let the speeds of the two trains be x m \/ sec and y m \/ sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) \/ ( x + y ) = 22 = = > 27 x + 17 y = 22 x + 22 y = = > 5 x = 5 y = = > x \/ y = 1 \/ 1 . answer : c\"","correct":"c","options":{"a":"3 \/ 1 ","b":"3 \/ 1 ","c":"1 \/ 1 ","d":"3 \/ 5","e":"5 \/ 2"},"options_float":{"a":3.0,"b":3.0,"c":1.0,"d":0.6,"e":2.5},"annotated_formula":"divide(subtract(27, 22), subtract(22, 17))","linear_formula":"subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)|","chain":"27 - 22<\/gadget>\n5<\/output>\n22 - 17<\/gadget>\n5<\/output>\n5 \/ 5<\/gadget>\n1<\/output>\n1<\/result>","index":615} +{"problem":"exactly 3 \/ 7 of the people in the room are under the age of 21 , and exactly 5 \/ 9 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ?","rationale":"\"the total number of the people in the room must be a multiple of both 7 and 9 ( in order 3 \/ 7 and 5 \/ 9 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 9 , which is 63 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 91 people in the room . therefore there are 3 \/ 7 * 63 = 27 people in the room under the age of 21 . answer : a .\"","correct":"a","options":{"a":"27 ","b":"35 ","c":"39 ","d":"60","e":"65"},"options_float":{"a":27.0,"b":35.0,"c":39.0,"d":60.0,"e":65.0},"annotated_formula":"divide(multiply(multiply(7, 9), 3), 7)","linear_formula":"multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)|","chain":"7 * 9<\/gadget>\n63<\/output>\n63 * 3<\/gadget>\n189<\/output>\n189 \/ 7<\/gadget>\n27<\/output>\n27<\/result>","index":616} +{"problem":"the product of two successive numbers is 7832 . which is the smaller of the two numbers ?","rationale":"\"c 88 from the given alternatives , 88 × 89 = 7832 ∴ smaller number = 88\"","correct":"c","options":{"a":"78 ","b":"68 ","c":"88 ","d":"97","e":"37"},"options_float":{"a":78.0,"b":68.0,"c":88.0,"d":97.0,"e":37.0},"annotated_formula":"sqrt(7832)","linear_formula":"sqrt(n0)|","chain":"7_832 ** (1\/2)<\/gadget>\n2*sqrt(1958) = around 88.498588<\/output>\n2*sqrt(1958) = around 88.498588<\/result>","index":618} +{"problem":"the value of 3.6 x 0.48 x 2.50 \/ 0.12 x 0.09 x 0.5 is","rationale":"\"solution therefore = 3.6 x 0.48 x 2.50 \/ 0.12 x 0.09 x 0.5 = 36 x 48 x 250 \/ 12 x 9 x 5 = 800 answer b\"","correct":"b","options":{"a":"80 ","b":"800 ","c":"8000 ","d":"80000","e":"none of these"},"options_float":{"a":80.0,"b":800.0,"c":8000.0,"d":80000.0,"e":null},"annotated_formula":"divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3)|","chain":"3.6 * 0.48<\/gadget>\n1.728<\/output>\n1.728 * 2.5<\/gadget>\n4.32<\/output>\n0.12 * 0.09<\/gadget>\n0.0108<\/output>\n0.0108 * 0.5<\/gadget>\n0.0054<\/output>\n4.32 \/ 0.0054<\/gadget>\n800<\/output>\n800<\/result>","index":619} +{"problem":"two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 30 hr . if all pipes are opened simultaneously , then the cistern will be filled in","rationale":"\"solution : work done by all the tanks working together in 1 hour . 1 \/ 10 + 1 \/ 12 − 1 \/ 30 = 3 \/ 20 hence , tank will be filled in 20 \/ 3 = 6.67 hour option ( a )\"","correct":"a","options":{"a":"6.67 hr ","b":"8 hr ","c":"8.5 hr ","d":"10 hr","e":"none of these"},"options_float":{"a":6.67,"b":8.0,"c":8.5,"d":10.0,"e":null},"annotated_formula":"inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 30)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)|","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/10) + (1\/12)<\/gadget>\n11\/60 = around 0.183333<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(11\/60) - (1\/30)<\/gadget>\n3\/20 = around 0.15<\/output>\n1 \/ (3\/20)<\/gadget>\n20\/3 = around 6.666667<\/output>\n20\/3 = around 6.666667<\/result>","index":620} +{"problem":"if 1 \/ 8 of a pencil is black 1 \/ 2 of the remaining is white and the remaining 7 \/ 2 is blue find the total length of the pencil ?","rationale":"( 1 \/ 8 ) p + ( 1 \/ 2 ) p + ( 7 \/ 2 ) = p on solving the above for p we get 9.333 now checking for the answer : 9.333 \/ 8 = 1.166 cm 9.333 \/ 2 = 4.665 cm now 1.166 + 4.665 + ( 7 \/ 2 ) = 9.33 cm answer : d","correct":"d","options":{"a":"7.33 cm ","b":"6.33 cm ","c":"5.33 cm ","d":"9.33 cm","e":"8.33 cm"},"options_float":{"a":7.33,"b":6.33,"c":5.33,"d":9.33,"e":8.33},"annotated_formula":"add(divide(multiply(divide(7, 2), add(divide(1, 8), divide(1, 2))), subtract(const_1, add(divide(1, 8), divide(1, 2)))), divide(7, 2))","linear_formula":"divide(n0,n1)|divide(n0,n3)|divide(n4,n3)|add(#0,#1)|multiply(#3,#2)|subtract(const_1,#3)|divide(#4,#5)|add(#6,#2)","chain":"7 \/ 2<\/gadget>\n7\/2 = around 3.5<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/8) + (1\/2)<\/gadget>\n5\/8 = around 0.625<\/output>\n(7\/2) * (5\/8)<\/gadget>\n35\/16 = around 2.1875<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n(35\/16) \/ (3\/8)<\/gadget>\n35\/6 = around 5.833333<\/output>\n(35\/6) + (7\/2)<\/gadget>\n28\/3 = around 9.333333<\/output>\n28\/3 = around 9.333333<\/result>","index":621} +{"problem":"the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 42 degrees . find the temperature on friday ?","rationale":"\"m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 42 tu + w + th = 192 - 42 = 150 f = 184 – 150 = 34 . answer : a\"","correct":"a","options":{"a":"34 ","b":"55 ","c":"66 ","d":"81","e":"21"},"options_float":{"a":34.0,"b":55.0,"c":66.0,"d":81.0,"e":21.0},"annotated_formula":"subtract(42, subtract(multiply(48, const_4), multiply(46, const_4)))","linear_formula":"multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)|","chain":"48 * 4<\/gadget>\n192<\/output>\n46 * 4<\/gadget>\n184<\/output>\n192 - 184<\/gadget>\n8<\/output>\n42 - 8<\/gadget>\n34<\/output>\n34<\/result>","index":622} +{"problem":"a certain taxi company charges $ 2.10 for the first 1 \/ 5 of a mile plus $ 0.40 for each additional 1 \/ 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ?","rationale":"\"a certain taxi company charges $ 2.10 for the first 1 \/ 5 of a mile plus $ 0.40 for each additional 1 \/ 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? a . 15.60 b . 16.00 c . 17.50 d . 18.70 e . 19.10 1 \/ 5 miles = 0.2 miles . the cost of 8 miles long ride would be $ 2.10 for the first 0.2 miles plus ( 8 - 0.2 ) \/ 0.2 * 0.4 = $ 2.1 + $ 15.6 = $ 17.7 . answer : c .\"","correct":"c","options":{"a":"15.6 ","b":"16.0 ","c":"17.7 ","d":"18.7","e":"19.1"},"options_float":{"a":15.6,"b":16.0,"c":17.7,"d":18.7,"e":19.1},"annotated_formula":"add(2.10, multiply(subtract(divide(8, divide(1, 5)), 1), 0.40))","linear_formula":"divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)|","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n8 \/ (1\/5)<\/gadget>\n40<\/output>\n40 - 1<\/gadget>\n39<\/output>\n39 * 0.4<\/gadget>\n15.6<\/output>\n2.1 + 15.6<\/gadget>\n17.7<\/output>\n17.7<\/result>","index":623} +{"problem":"two cyclist start from the same places in opposite directions . one is going towards north at 30 kmph and the other is going towards south 40 kmph . what time will they take to be 50 km apart ?","rationale":"\"to be ( 30 + 40 ) km apart , they take 1 hour to be 50 km apart , they take 1 \/ 70 * 50 = 0.714 hrs answer is b\"","correct":"b","options":{"a":"1.32 hr ","b":"0.714 hrs ","c":"3.231 hrs ","d":"0.515 hrs","e":"6.214 hrs"},"options_float":{"a":1.32,"b":0.714,"c":3.231,"d":0.515,"e":6.214},"annotated_formula":"divide(50, add(30, 40))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"30 + 40<\/gadget>\n70<\/output>\n50 \/ 70<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":624} +{"problem":"in x - y plane , there is a right triangle abc ( ∠ b = 90 o ) . if the length of ac is 25 and the slope of line segment ac is 4 \/ 3 , what is the length of ab ?","rationale":"\"slope = change in vertical direction \/ change in horizontal direction = 4 \/ 3 change in vertical direction = 4 x = ab change in horizontal direction = 3 x = bc ab ^ 2 + bc ^ 2 = 25 ^ 2 16 x ^ 2 + 9 x ^ 2 = 625 25 x ^ 2 = 625 x ^ 2 = 25 x = 5 therefore ab = 5 * 4 = 20 answer : d\"","correct":"d","options":{"a":"12 ","b":"18 ","c":"24 ","d":"20","e":"40"},"options_float":{"a":12.0,"b":18.0,"c":24.0,"d":20.0,"e":40.0},"annotated_formula":"multiply(4, divide(25, sqrt(add(power(4, const_2), power(3, const_2)))))","linear_formula":"power(n2,const_2)|power(n3,const_2)|add(#0,#1)|sqrt(#2)|divide(n1,#3)|multiply(n2,#4)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n16 + 9<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n25 \/ 5<\/gadget>\n5<\/output>\n4 * 5<\/gadget>\n20<\/output>\n20<\/result>","index":625} +{"problem":"the average of 5 consecutive odd numbers a , b , c , d and e is 13 . what percent of b is c ?","rationale":"explanation : in such a case the middle number ( c ) is the average ∴ c = 13 and b = 11 required percentage = 13 \/ 11 x 100 = 118.2 answer : option d","correct":"d","options":{"a":"98.2 ","b":"102.8 ","c":"108.2 ","d":"118.2","e":"122.8"},"options_float":{"a":98.2,"b":102.8,"c":108.2,"d":118.2,"e":122.8},"annotated_formula":"multiply(divide(add(add(divide(multiply(5, 13), 5), const_2), const_2), add(divide(multiply(5, 13), 5), const_2)), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,n0)|add(#1,const_2)|add(#2,const_2)|divide(#3,#2)|multiply(#4,const_100)","chain":"5 * 13<\/gadget>\n65<\/output>\n65 \/ 5<\/gadget>\n13<\/output>\n13 + 2<\/gadget>\n15<\/output>\n15 + 2<\/gadget>\n17<\/output>\n17 \/ 15<\/gadget>\n17\/15 = around 1.133333<\/output>\n(17\/15) * 100<\/gadget>\n340\/3 = around 113.333333<\/output>\n340\/3 = around 113.333333<\/result>","index":626} +{"problem":"a palindrome is a number that reads the same forward and backward , such as 343 . how many odd , 8 - digit numbers are palindromes ?","rationale":"\"first recognize you only need to consider the first four digits ( because the second four are just the first four flipped ) there are 9000 possibilities for the first four digits of a 8 digit number , 1000 - 9999 inclusive . everything starting with a 1 , 3,5 , 7,9 will be odd , which is 5 \/ 9 ths of the combinations . 5 \/ 9 * 9000 = 5000 answer : d\"","correct":"d","options":{"a":"4000 ","b":"4500 ","c":"5055 ","d":"5000","e":"2500"},"options_float":{"a":4000.0,"b":4500.0,"c":5055.0,"d":5000.0,"e":2500.0},"annotated_formula":"divide(power(const_10, divide(8, const_2)), const_2)","linear_formula":"divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\n10 ** 4<\/gadget>\n10_000<\/output>\n10_000 \/ 2<\/gadget>\n5_000<\/output>\n5_000<\/result>","index":627} +{"problem":"a person borrows rs . 7000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year .","rationale":"\"gain in 2 years = [ ( 7000 * 6 * 2 ) \/ 100 ] - [ ( 7000 * 4 * 2 ) \/ 100 ] 840 - 560 = 280 gain in 1 year = ( 280 \/ 2 ) = 140 rs answer : c\"","correct":"c","options":{"a":"100 rs ","b":"150 rs ","c":"140 rs ","d":"180 rs","e":"200 rs"},"options_float":{"a":100.0,"b":150.0,"c":140.0,"d":180.0,"e":200.0},"annotated_formula":"divide(subtract(divide(multiply(multiply(7000, 6), 2), const_100), divide(multiply(multiply(7000, 4), 2), const_100)), 2)","linear_formula":"multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)|","chain":"7_000 * 6<\/gadget>\n42_000<\/output>\n42_000 * 2<\/gadget>\n84_000<\/output>\n84_000 \/ 100<\/gadget>\n840<\/output>\n7_000 * 4<\/gadget>\n28_000<\/output>\n28_000 * 2<\/gadget>\n56_000<\/output>\n56_000 \/ 100<\/gadget>\n560<\/output>\n840 - 560<\/gadget>\n280<\/output>\n280 \/ 2<\/gadget>\n140<\/output>\n140<\/result>","index":628} +{"problem":"what is the difference between the local value of 1 and 2 in the number 2124 ?","rationale":"explanation : 100 â € “ 20 = 80 answer : d","correct":"d","options":{"a":"70 ","b":"75 ","c":"20 ","d":"80","e":"60"},"options_float":{"a":70.0,"b":75.0,"c":20.0,"d":80.0,"e":60.0},"annotated_formula":"subtract(multiply(1, const_100), multiply(2, const_10))","linear_formula":"multiply(n0,const_100)|multiply(n1,const_10)|subtract(#0,#1)","chain":"1 * 100<\/gadget>\n100<\/output>\n2 * 10<\/gadget>\n20<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80<\/result>","index":631} +{"problem":"in 10 years , a will be twice as old as b was 10 years ago . if a is now 7 years older than b the present age of b is","rationale":"\"let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 7 = > 2 b - b - 7 = 30 b = 37 so the present age of b is 37 years answer : a\"","correct":"a","options":{"a":"37 ","b":"38 ","c":"39 ","d":"40","e":"41"},"options_float":{"a":37.0,"b":38.0,"c":39.0,"d":40.0,"e":41.0},"annotated_formula":"add(add(multiply(const_2, 10), 10), 7)","linear_formula":"multiply(n0,const_2)|add(n0,#0)|add(n2,#1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 10<\/gadget>\n30<\/output>\n30 + 7<\/gadget>\n37<\/output>\n37<\/result>","index":632} +{"problem":"you buy a piece of land with an area of â ˆ š 1600 , how long is one side of the land plot ?","rationale":"\"try filling the numbers into the answer y x y = find the closest to 900 . answer b\"","correct":"b","options":{"a":"28 ","b":"40 ","c":"30 ","d":"31","e":"32"},"options_float":{"a":28.0,"b":40.0,"c":30.0,"d":31.0,"e":32.0},"annotated_formula":"sqrt(1600)","linear_formula":"sqrt(n0)|","chain":"1_600 ** (1\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":635} +{"problem":"a can do a work in 15 days b in 14 days and c in 16 days . if they work on it together then in how many days required to complete the work ?","rationale":"person ( a ) ( b ) ( c ) ( a + b + c ) time - ( 15 ) ( 14 ) ( 16 ) rate - ( 224 ) ( 240 ) ( 210 ) ( 674 ) work - ( 3360 ) ( 3360 ) ( 3360 ) ( 3360 ) therefore a + b + c requires ( 3360 \/ 674 ) days to complete entire work = 3360 \/ 674 answer is c","correct":"c","options":{"a":"3360 \/ 210 ","b":"3360 \/ 21 ","c":"3360 \/ 674 ","d":"3360 \/ 240","e":"3360 \/ 224"},"options_float":{"a":16.0,"b":160.0,"c":4.9851632047,"d":14.0,"e":15.0},"annotated_formula":"divide(multiply(multiply(15, 14), 16), multiply(add(add(15, 14), 16), 15))","linear_formula":"add(n0,n1)|multiply(n0,n1)|add(n2,#0)|multiply(n2,#1)|multiply(n0,#2)|divide(#3,#4)","chain":"15 * 14<\/gadget>\n210<\/output>\n210 * 16<\/gadget>\n3_360<\/output>\n15 + 14<\/gadget>\n29<\/output>\n29 + 16<\/gadget>\n45<\/output>\n45 * 15<\/gadget>\n675<\/output>\n3_360 \/ 675<\/gadget>\n224\/45 = around 4.977778<\/output>\n224\/45 = around 4.977778<\/result>","index":636} +{"problem":"tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 9 meters , and the interior of tank b has a height of 9 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?","rationale":"\"the radius of tank a is 9 \/ ( 2 * pi ) . the capacity of tank a is 10 * pi * 81 \/ ( 4 * pi ^ 2 ) = 405 \/ ( 2 * pi ) the radius of tank b is 10 \/ ( 2 * pi ) . the capacity of tank b is 9 * pi * 100 \/ ( 4 * pi ^ 2 ) = 450 \/ ( 2 * pi ) tank a \/ tank b = 405 \/ 450 = 9 \/ 10 = 90 % the answer is b .\"","correct":"b","options":{"a":"75 % ","b":"90 % ","c":"100 % ","d":"120 %","e":"125 %"},"options_float":{"a":75.0,"b":90.0,"c":100.0,"d":120.0,"e":125.0},"annotated_formula":"multiply(multiply(power(divide(9, 10), const_2), divide(10, 9)), const_100)","linear_formula":"divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)|","chain":"9 \/ 10<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) ** 2<\/gadget>\n81\/100 = around 0.81<\/output>\n10 \/ 9<\/gadget>\n10\/9 = around 1.111111<\/output>\n(81\/100) * (10\/9)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 100<\/gadget>\n90<\/output>\n90<\/result>","index":637} +{"problem":"a cistern can be filled by a tap in 6 hours while it can be emptied by another tap in 12 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ?","rationale":"\"net part filled in 1 hour = ( 1 \/ 6 - 1 \/ 12 ) = 1 \/ 12 the cistern will be filled in 12 \/ 1 hrs i . e . , 12 hrs . answer : c\"","correct":"c","options":{"a":"5.2 hrs ","b":"2.9 hrs ","c":"12 hrs ","d":"7.2 hrs","e":"1.4 hrs"},"options_float":{"a":5.2,"b":2.9,"c":12.0,"d":7.2,"e":1.4},"annotated_formula":"divide(const_1, subtract(divide(const_1, 6), divide(const_1, 12)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/6) - (1\/12)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":639} +{"problem":"seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 40 percent ryegrass , what percent of the weight of this mixture is x ?","rationale":"\"- - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 40 % 0.4 x + ( m - x ) 0.25 = 0.4 m 0.15 x = 0.15 m x = 1 m x = 100 % of m a\"","correct":"a","options":{"a":"100 % ","b":"33.33 % ","c":"40 % ","d":"50 %","e":"66.66 %"},"options_float":{"a":100.0,"b":33.33,"c":40.0,"d":50.0,"e":66.66},"annotated_formula":"divide(subtract(40, 25), subtract(divide(40, const_100), divide(25, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n2)|subtract(#0,#1)|divide(#2,#3)|","chain":"40 - 25<\/gadget>\n15<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(2\/5) - (1\/4)<\/gadget>\n3\/20 = around 0.15<\/output>\n15 \/ (3\/20)<\/gadget>\n100<\/output>\n100<\/result>","index":640} +{"problem":"the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if y 12 = 110 , what is the value of y ?","rationale":"ab = ( a - 1 ) ( b - 1 ) y 12 = ( y - 1 ) ( 12 - 1 ) = 110 - - > y - 1 = 10 - - > y = 11 answer : a","correct":"a","options":{"a":"11 ","b":"12 ","c":"13 ","d":"15","e":"18"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":15.0,"e":18.0},"annotated_formula":"add(divide(110, subtract(12, 1)), 1)","linear_formula":"subtract(n2,n0)|divide(n3,#0)|add(n0,#1)","chain":"12 - 1<\/gadget>\n11<\/output>\n110 \/ 11<\/gadget>\n10<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11<\/result>","index":642} +{"problem":"each month a retailer sells 100 identical items . on each item he makes a profit of $ 60 that constitutes 10 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ?","rationale":"\"for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 10 % of the cost : sell price = cost + $ 60 sell price = $ 600 + $ 60 thus , the sell price is $ 660 for each item . selling all 100 items gives the retailer . . . 100 ( $ 60 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 660 ) = $ 33 discount $ 627 = $ 600 + $ 27 now , the retailer makes a profit of just $ 27 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 27 ( x ) = $ 2,000 x = 2,000 \/ 27 x = 222.222222 items you ' ll notice that this is not among the answer choices . . . . 221 and 223 are . selling 221 items would get us 9 ( 221 ) = $ 1989 which is not enough money . to get back to at least $ 2,000 , we need to sell 223 items . final answer : d\"","correct":"d","options":{"a":"191 ","b":"213 ","c":"221 ","d":"223","e":"226"},"options_float":{"a":191.0,"b":213.0,"c":221.0,"d":223.0,"e":226.0},"annotated_formula":"divide(multiply(100, 60), subtract(60, divide(multiply(add(divide(multiply(100, 60), 10), 60), 5), 100)))","linear_formula":"multiply(n0,n1)|divide(#0,n2)|add(n1,#1)|multiply(n3,#2)|divide(#3,n0)|subtract(n1,#4)|divide(#0,#5)|","chain":"100 * 60<\/gadget>\n6_000<\/output>\n6_000 \/ 10<\/gadget>\n600<\/output>\n600 + 60<\/gadget>\n660<\/output>\n660 * 5<\/gadget>\n3_300<\/output>\n3_300 \/ 100<\/gadget>\n33<\/output>\n60 - 33<\/gadget>\n27<\/output>\n6_000 \/ 27<\/gadget>\n2_000\/9 = around 222.222222<\/output>\n2_000\/9 = around 222.222222<\/result>","index":643} +{"problem":"a and b go around a circular track of length 200 m on a cycle at speeds of 36 kmph and 72 kmph . after how much time will they meet for the first time at the starting point ?","rationale":"\"time taken to meet for the first time at the starting point = lcm { length of the track \/ speed of a , length of the track \/ speed of b } = lcm { 200 \/ ( 36 * 5 \/ 18 ) , 200 \/ ( 72 * 5 \/ 18 ) } = lcm ( 20 , 10 ) = 20 sec . answer : d\"","correct":"d","options":{"a":"120 sec ","b":"198 sec ","c":"178 sec ","d":"20 sec","e":"276 sec"},"options_float":{"a":120.0,"b":198.0,"c":178.0,"d":20.0,"e":276.0},"annotated_formula":"divide(200, subtract(multiply(72, const_0_2778), multiply(36, const_0_2778)))","linear_formula":"multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n20 - 10<\/gadget>\n10<\/output>\n200 \/ 10<\/gadget>\n20<\/output>\n20<\/result>","index":644} +{"problem":"if x and y are both odd prime numbers and x < y , how many distinct positive r integer factors does 2 xy have ?","rationale":"since 2 xy prime r factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer .","correct":"d","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"12"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":12.0},"annotated_formula":"multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1))","linear_formula":"add(const_1,const_1)|multiply(n0,#0)|multiply(#0,#1)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":645} +{"problem":"a man can row 8 kmph in still water . when the river is running at 3 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?","rationale":"\"m = 8 s = 3 ds = 11 us = 5 x \/ 11 + x \/ 5 = 1 x = 3.4 d = 3.4 * 2 = 6.8 answer : e\"","correct":"e","options":{"a":"5.75 ","b":"5.7 ","c":"5.76 ","d":"5.74","e":"6.8"},"options_float":{"a":5.75,"b":5.7,"c":5.76,"d":5.74,"e":6.8},"annotated_formula":"multiply(divide(multiply(add(8, 3), subtract(8, 3)), add(add(8, 3), subtract(8, 3))), const_2)","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|","chain":"8 + 3<\/gadget>\n11<\/output>\n8 - 3<\/gadget>\n5<\/output>\n11 * 5<\/gadget>\n55<\/output>\n11 + 5<\/gadget>\n16<\/output>\n55 \/ 16<\/gadget>\n55\/16 = around 3.4375<\/output>\n(55\/16) * 2<\/gadget>\n55\/8 = around 6.875<\/output>\n55\/8 = around 6.875<\/result>","index":646} +{"problem":"the length of a train and that of a platform are equal . if with a speed of 36 k \/ hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ?","rationale":"\"speed = [ 36 * 5 \/ 18 ] m \/ sec = 10 m \/ sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x \/ 60 = 10 ã ¨ x = 10 * 60 \/ 2 = 300 answer : c\"","correct":"c","options":{"a":"752 ","b":"799 ","c":"300 ","d":"750","e":"712"},"options_float":{"a":752.0,"b":799.0,"c":300.0,"d":750.0,"e":712.0},"annotated_formula":"divide(divide(multiply(36, const_1000), divide(const_60, const_1)), const_2)","linear_formula":"divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)|","chain":"36 * 1_000<\/gadget>\n36_000<\/output>\n60 \/ 1<\/gadget>\n60<\/output>\n36_000 \/ 60<\/gadget>\n600<\/output>\n600 \/ 2<\/gadget>\n300<\/output>\n300<\/result>","index":647} +{"problem":"the overall age of x and y is 19 year greater than the overall age of y and z . z is how many decades younger that x ?","rationale":"\"d 19 ( x + y ) â € “ ( y + z ) = 19 x â € “ z = 19\"","correct":"d","options":{"a":"11 ","b":"15 ","c":"12 ","d":"19","e":"20"},"options_float":{"a":11.0,"b":15.0,"c":12.0,"d":19.0,"e":20.0},"annotated_formula":"divide(19, const_1)","linear_formula":"divide(n0,const_1)|","chain":"19 \/ 1<\/gadget>\n19<\/output>\n19<\/result>","index":648} +{"problem":"for all real numbers v , an operation is defined by the equation v * = v - v \/ 3 . if ( v * ) * = 16 , then v =","rationale":"\"( v * ) * = ( v - v \/ 3 ) - ( v - v \/ 3 ) \/ 3 16 = 2 v \/ 3 - 2 v \/ 9 = 4 v \/ 9 v = 36 the answer is c .\"","correct":"c","options":{"a":"30 ","b":"33 ","c":"36 ","d":"39","e":"42"},"options_float":{"a":30.0,"b":33.0,"c":36.0,"d":39.0,"e":42.0},"annotated_formula":"divide(divide(16, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3)))","linear_formula":"divide(const_1,n0)|subtract(const_1,#0)|divide(n1,#1)|divide(#2,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n16 \/ (2\/3)<\/gadget>\n24<\/output>\n24 \/ (2\/3)<\/gadget>\n36<\/output>\n36<\/result>","index":649} +{"problem":"a group of men decided to do a work in 55 days , but 15 of them became absent . if the rest of the group did the work in 60 days , find the original number of men ?","rationale":"\"original number of men = 15 * 60 \/ ( 60 - 55 ) = 180 answer is a\"","correct":"a","options":{"a":"180 ","b":"150 ","c":"140 ","d":"100","e":"25"},"options_float":{"a":180.0,"b":150.0,"c":140.0,"d":100.0,"e":25.0},"annotated_formula":"divide(multiply(15, 60), subtract(60, 55))","linear_formula":"multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"15 * 60<\/gadget>\n900<\/output>\n60 - 55<\/gadget>\n5<\/output>\n900 \/ 5<\/gadget>\n180<\/output>\n180<\/result>","index":651} +{"problem":"a train 250 m long , running with a speed of 50 km \/ hr will pass a tree in ?","rationale":"\"speed = 50 * 5 \/ 18 = 125 \/ 9 m \/ sec time taken = 250 * 9 \/ 125 = 18 sec answer : a\"","correct":"a","options":{"a":"18 sec ","b":"20 sec ","c":"25 sec ","d":"30 sec","e":"1 min"},"options_float":{"a":18.0,"b":20.0,"c":25.0,"d":30.0,"e":1.0},"annotated_formula":"multiply(divide(250, multiply(50, const_1000)), const_3600)","linear_formula":"multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|","chain":"50 * 1_000<\/gadget>\n50_000<\/output>\n250 \/ 50_000<\/gadget>\n1\/200 = around 0.005<\/output>\n(1\/200) * 3_600<\/gadget>\n18<\/output>\n18<\/result>","index":654} +{"problem":"country x imposes a two - tiered tax on imported cars : the first tier imposes a tax of 25 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 15 % . if ron imported a $ 30000 imported car and ended up paying $ 5500 in taxes , what is the first tier ' s price level ?","rationale":"let t be the tier price , p be total price = 30000 per the given conditions : 0.25 t + 0.15 ( p - t ) = 5500 0.25 t + 0.15 * 30000 - 0.15 t = 5500 0.10 t + 4500 = 5500 0.10 t = 5500 - 4500 = 1000 t = 1000 \/ 0.10 = 10000 answer a","correct":"a","options":{"a":"$ 10000 ","b":"$ 11000 ","c":"$ 12000 ","d":"$ 10500","e":"$ 11500"},"options_float":{"a":10000.0,"b":11000.0,"c":12000.0,"d":10500.0,"e":11500.0},"annotated_formula":"divide(subtract(5500, multiply(divide(15, const_100), 30000)), subtract(divide(25, const_100), divide(15, const_100)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n3,#2)|divide(#4,#3)","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 30_000<\/gadget>\n4_500<\/output>\n5_500 - 4_500<\/gadget>\n1_000<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) - (3\/20)<\/gadget>\n1\/10 = around 0.1<\/output>\n1_000 \/ (1\/10)<\/gadget>\n10_000<\/output>\n10_000<\/result>","index":656} +{"problem":"in the faculty of reverse - engineering , 230 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ?","rationale":"\"answer is d : 649 solution : total number of students studying both are 423 + 230 - 134 = 519 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 519 , so 100 % is approx . 649 .\"","correct":"d","options":{"a":"515 . ","b":"545 . ","c":"618 . ","d":"649 .","e":"666 ."},"options_float":{"a":515.0,"b":545.0,"c":618.0,"d":649.0,"e":666.0},"annotated_formula":"add(230, 423)","linear_formula":"add(n0,n1)|","chain":"230 + 423<\/gadget>\n653<\/output>\n653<\/result>","index":657} +{"problem":"the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 48 . what is a possible value for the mean of the distribution ?","rationale":"\"the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 48 : { mean } - 3 * { sd } > 48 ; { mean } - 6 > 48 ; { mean } > 54 . answer : d .\"","correct":"d","options":{"a":"46 ","b":"47 ","c":"48 ","d":"55","e":"50"},"options_float":{"a":46.0,"b":47.0,"c":48.0,"d":55.0,"e":50.0},"annotated_formula":"add(add(48, multiply(3, 2)), const_1)","linear_formula":"multiply(n0,n1)|add(n2,#0)|add(#1,const_1)|","chain":"3 * 2<\/gadget>\n6<\/output>\n48 + 6<\/gadget>\n54<\/output>\n54 + 1<\/gadget>\n55<\/output>\n55<\/result>","index":658} +{"problem":"a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 9 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company charges the same amount under plan a and under plan b ?","rationale":"\"let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 9 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 9 ) = 0.08 t - - > 60 + 6 ( t - 9 ) = 8 t - - > t = 3 . answer : a .\"","correct":"a","options":{"a":"3 ","b":"9 ","c":"15 ","d":"21","e":"30"},"options_float":{"a":3.0,"b":9.0,"c":15.0,"d":21.0,"e":30.0},"annotated_formula":"divide(subtract(0.60, multiply(0.06, 9)), subtract(0.08, 0.06))","linear_formula":"multiply(n1,n2)|subtract(n3,n2)|subtract(n0,#0)|divide(#2,#1)|","chain":"0.06 * 9<\/gadget>\n0.54<\/output>\n0.6 - 0.54<\/gadget>\n0.06<\/output>\n0.08 - 0.06<\/gadget>\n0.02<\/output>\n0.06 \/ 0.02<\/gadget>\n3<\/output>\n3<\/result>","index":659} +{"problem":"zachary is helping his younger brother , sterling , learn his multiplication tables . for every question that sterling answers correctly , zachary gives him 3 pieces of candy . for every question that sterling answers incorrectly , zachary takes away two pieces of candy . after 9 questions , if sterling had answered 2 more questions correctly , he would have earned 31 pieces of candy . how many of the 9 questions did zachary answer correctly ?","rationale":"\"i got two equations : 3 x - 2 y = 25 x + y = 9 3 x - 2 ( 9 - x ) = 25 3 x - 18 + 2 x = 25 5 x = 43 x = 8.6 or between 8 and 9 . ( ans c )\"","correct":"c","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"divide(add(subtract(multiply(2, subtract(9, 2)), multiply(3, 2)), 31), add(3, 2))","linear_formula":"add(n0,n2)|multiply(n0,n2)|subtract(n1,n2)|multiply(n2,#2)|subtract(#3,#1)|add(n3,#4)|divide(#5,#0)|","chain":"9 - 2<\/gadget>\n7<\/output>\n2 * 7<\/gadget>\n14<\/output>\n3 * 2<\/gadget>\n6<\/output>\n14 - 6<\/gadget>\n8<\/output>\n8 + 31<\/gadget>\n39<\/output>\n3 + 2<\/gadget>\n5<\/output>\n39 \/ 5<\/gadget>\n39\/5 = around 7.8<\/output>\n39\/5 = around 7.8<\/result>","index":660} +{"problem":"if the average of 5 positive integers is 60 and the difference between the largest and the smallest of these 5 numbers is 10 , what is the maximum value possible for the largest of these 5 integers ?","rationale":"\"sum of 5 integer ( a , b , c , d , e ) = 5 * 60 = 300 e - a = 10 i . e . e = a + 10 for e to be maximum remaining 4 must be as small as possible since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers i . e . a + a + a + a + ( a + 10 ) = 300 i . e . 5 a = 290 i . e . a = 58 i . e . largest e = 58 + 10 = 68 answer : option d\"","correct":"d","options":{"a":"50 ","b":"52 ","c":"49 ","d":"68","e":"74"},"options_float":{"a":50.0,"b":52.0,"c":49.0,"d":68.0,"e":74.0},"annotated_formula":"add(divide(subtract(multiply(60, 5), 10), 5), 10)","linear_formula":"multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)|","chain":"60 * 5<\/gadget>\n300<\/output>\n300 - 10<\/gadget>\n290<\/output>\n290 \/ 5<\/gadget>\n58<\/output>\n58 + 10<\/gadget>\n68<\/output>\n68<\/result>","index":661} +{"problem":"in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 160 of the workers of that company are men , how many of the workers are women ?","rationale":"\"set up equation : x = total number of workers 160 = 0,4 * 2 \/ 3 * x + 0,8 * 1 \/ 3 * x 160 = 16 \/ 30 x x = 300 300 - 160 = 140 answer e\"","correct":"e","options":{"a":"80 ","b":"95 ","c":"105 ","d":"120","e":"140"},"options_float":{"a":80.0,"b":95.0,"c":105.0,"d":120.0,"e":140.0},"annotated_formula":"multiply(divide(160, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))))","linear_formula":"divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/3) * (1\/5)<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/3) - (1\/15)<\/gadget>\n4\/15 = around 0.266667<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/3) * (2\/5)<\/gadget>\n4\/15 = around 0.266667<\/output>\n(4\/15) + (4\/15)<\/gadget>\n8\/15 = around 0.533333<\/output>\n160 \/ (8\/15)<\/gadget>\n300<\/output>\n(2\/3) - (4\/15)<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/15) + (2\/5)<\/gadget>\n7\/15 = around 0.466667<\/output>\n300 * (7\/15)<\/gadget>\n140<\/output>\n140<\/result>","index":662} +{"problem":"sum of 53 odd numbers is ?","rationale":"\"sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 53 odd numbers = 53 ^ 2 = 2809 answer : b\"","correct":"b","options":{"a":"2709 ","b":"2809 ","c":"2909 ","d":"3009","e":"3109"},"options_float":{"a":2709.0,"b":2809.0,"c":2909.0,"d":3009.0,"e":3109.0},"annotated_formula":"multiply(multiply(53, const_2), divide(53, const_2))","linear_formula":"divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1)|","chain":"53 * 2<\/gadget>\n106<\/output>\n53 \/ 2<\/gadget>\n53\/2 = around 26.5<\/output>\n106 * (53\/2)<\/gadget>\n2_809<\/output>\n2_809<\/result>","index":664} +{"problem":"find the value of 5 + 2 â € ¢ ( 8 â € “ 3 )","rationale":"5 + 2 â € ¢ ( 8 â € “ 3 ) = 5 + 2 ( 5 ) = 5 + 2 * 5 = 5 + 10 = 15 correct answer e","correct":"e","options":{"a":"25 ","b":"13 ","c":"17 ","d":"24","e":"15"},"options_float":{"a":25.0,"b":13.0,"c":17.0,"d":24.0,"e":15.0},"annotated_formula":"add(multiply(subtract(8, 3), 2), 5)","linear_formula":"subtract(n2,n3)|multiply(n1,#0)|add(n0,#1)","chain":"8 - 3<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15<\/result>","index":665} +{"problem":"what least number should be added to 1057 , so that the sum is completely divisible by 23","rationale":"\"explanation : ( 1057 \/ 23 ) gives remainder 22 22 + 1 = 23 , so we need to add 1 answer : option b\"","correct":"b","options":{"a":"4 ","b":"1 ","c":"2 ","d":"3","e":"5"},"options_float":{"a":4.0,"b":1.0,"c":2.0,"d":3.0,"e":5.0},"annotated_formula":"subtract(23, reminder(1057, 23))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|","chain":"1_057 % 23<\/gadget>\n22<\/output>\n23 - 22<\/gadget>\n1<\/output>\n1<\/result>","index":667} +{"problem":". in digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be ?","rationale":"20 * 10 * 5 = 1000 answer : b","correct":"b","options":{"a":"3323 ","b":"1000 ","c":"2877 ","d":"2992","e":"7111"},"options_float":{"a":3323.0,"b":1000.0,"c":2877.0,"d":2992.0,"e":7111.0},"annotated_formula":"multiply(multiply(20, 10), 5)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)","chain":"20 * 10<\/gadget>\n200<\/output>\n200 * 5<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":669} +{"problem":"when 1 \/ 20 % of 8,000 is subtracted from 1 \/ 10 of 8,000 , the difference is","rationale":"1 \/ 20 % of 8000 = 4 1 \/ 10 of 8000 = 800 800 - 2 = 798 ans : d","correct":"d","options":{"a":"50 ","b":"200 ","c":"380 ","d":"798","e":"400"},"options_float":{"a":50.0,"b":200.0,"c":380.0,"d":798.0,"e":400.0},"annotated_formula":"multiply(inverse(10), multiply(multiply(const_100, const_10), add(const_4, const_4)))","linear_formula":"add(const_4,const_4)|inverse(n4)|multiply(const_10,const_100)|multiply(#0,#2)|multiply(#1,#3)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n4 + 4<\/gadget>\n8<\/output>\n1_000 * 8<\/gadget>\n8_000<\/output>\n(1\/10) * 8_000<\/gadget>\n800<\/output>\n800<\/result>","index":670} +{"problem":"35 % of the employees of a company are men . 60 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ?","rationale":"no of employees = 100 ( say ) men = 35 women = 65 men speaking french = 0.6 * 35 = 21 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 21 = 19 and women not speaking french = 65 - 19 = 46 % of women speaking french = 46 \/ 65 * 100 = 70.77 % answer e","correct":"e","options":{"a":"4 % ","b":"10 % ","c":"96 % ","d":"90.12 %","e":"70.77 %"},"options_float":{"a":4.0,"b":10.0,"c":96.0,"d":90.12,"e":70.77},"annotated_formula":"multiply(divide(subtract(divide(subtract(const_100, 35), const_100), subtract(divide(40, const_100), multiply(divide(35, const_100), divide(60, const_100)))), divide(subtract(const_100, 35), const_100)), const_100)","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)","chain":"100 - 35<\/gadget>\n65<\/output>\n65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(7\/20) * (3\/5)<\/gadget>\n21\/100 = around 0.21<\/output>\n(2\/5) - (21\/100)<\/gadget>\n19\/100 = around 0.19<\/output>\n(13\/20) - (19\/100)<\/gadget>\n23\/50 = around 0.46<\/output>\n(23\/50) \/ (13\/20)<\/gadget>\n46\/65 = around 0.707692<\/output>\n(46\/65) * 100<\/gadget>\n920\/13 = around 70.769231<\/output>\n920\/13 = around 70.769231<\/result>","index":671} +{"problem":"a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 45 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ?","rationale":"\"a - b = 120 miles b - c = 60 miles avg speed = 45 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) \/ total time 45 = 180 \/ 4 t t = 60 b - c = 60 mph answer e\"","correct":"e","options":{"a":"40 ","b":"45 ","c":"50 ","d":"55","e":"60"},"options_float":{"a":40.0,"b":45.0,"c":50.0,"d":55.0,"e":60.0},"annotated_formula":"multiply(divide(divide(add(divide(120, const_2), 120), 45), const_4), divide(120, const_2))","linear_formula":"divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|divide(#2,const_4)|multiply(#3,#0)|","chain":"120 \/ 2<\/gadget>\n60<\/output>\n60 + 120<\/gadget>\n180<\/output>\n180 \/ 45<\/gadget>\n4<\/output>\n4 \/ 4<\/gadget>\n1<\/output>\n1 * 60<\/gadget>\n60<\/output>\n60<\/result>","index":673} +{"problem":"dick and jane each saved $ 5,000 in 1989 . in 1990 dick saved 10 percent more than in 1989 , and together he and jane saved a total of $ 10,000 . approximately what percent less did jane save in 1990 than in 1989 ?","rationale":"\"1990 dick saved = $ 5500 jane saved = $ 4500 ( jane saved $ 500 less than she did the prior year ) jane saved approximately $ 500 \/ 5000 $ ( 10 % ) less in 1990 answer : a\"","correct":"a","options":{"a":"10 % ","b":"15 % ","c":"20 % ","d":"25 %","e":"30 %"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"multiply(subtract(const_1, subtract(const_2, add(divide(10, const_100), const_1))), const_100)","linear_formula":"divide(n3,const_100)|add(#0,const_1)|subtract(const_2,#1)|subtract(const_1,#2)|multiply(#3,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n2 - (11\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":675} +{"problem":"robert ate 7 chocolates , nickel ate 5 chocolates . how many more chocolates did robert ate than nickel ?","rationale":"7 - 5 = 2 . answer is e","correct":"e","options":{"a":"4 ","b":"7 ","c":"9 ","d":"5","e":"2"},"options_float":{"a":4.0,"b":7.0,"c":9.0,"d":5.0,"e":2.0},"annotated_formula":"subtract(7, 5)","linear_formula":"subtract(n0,n1)|","chain":"7 - 5<\/gadget>\n2<\/output>\n2<\/result>","index":676} +{"problem":"a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 600 feet . what is its area , in square feet ?","rationale":"you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for sides , and knowing c = 600 , you can deduce the following a = 360 b = 480 360 x 480 = 172,800 a is the answer .","correct":"a","options":{"a":"172,800 ","b":"19,600 ","c":"20,000 ","d":"20,400","e":"20,800"},"options_float":{"a":172800.0,"b":19600.0,"c":20000.0,"d":20400.0,"e":20800.0},"annotated_formula":"multiply(multiply(divide(600, add(const_2, const_3)), const_3), multiply(divide(600, add(const_2, const_3)), const_4))","linear_formula":"add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)","chain":"2 + 3<\/gadget>\n5<\/output>\n600 \/ 5<\/gadget>\n120<\/output>\n120 * 3<\/gadget>\n360<\/output>\n120 * 4<\/gadget>\n480<\/output>\n360 * 480<\/gadget>\n172_800<\/output>\n172_800<\/result>","index":677} +{"problem":"on dividing 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor","rationale":"\"d = ( d - r ) \/ q = ( 109 - 1 ) \/ 9 = 108 \/ 9 = 12 a )\"","correct":"a","options":{"a":"12 ","b":"14 ","c":"18 ","d":"22","e":"24"},"options_float":{"a":12.0,"b":14.0,"c":18.0,"d":22.0,"e":24.0},"annotated_formula":"floor(divide(109, 9))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"109 \/ 9<\/gadget>\n109\/9 = around 12.111111<\/output>\nfloor(109\/9)<\/gadget>\n12<\/output>\n12<\/result>","index":678} +{"problem":"find compound interest on $ 10000 at 15 % per annum for 2 years 4 months , compounded annually .","rationale":"\"time = 2 years 4 months = 2 ( 4 \/ 12 ) years = 2 ( 1 \/ 3 ) years . amount = $ [ 10000 x ( 1 + ­ ( 15 \/ 100 ) ) 2 x ( 1 + ( ( 1 \/ 3 ) * 15 ) \/ 100 ) ] = $ [ 10000 * ( 23 \/ 20 ) * ( 23 \/ 20 ) * ( 21 \/ 20 ) ] = $ 13886.25 . : . c . i . = rs . ( 13886.25 - 10000 ) = $ 3886.25 answer a .\"","correct":"a","options":{"a":"3886.25 ","b":"3286.25 ","c":"3186.25 ","d":"3386.25","e":"3586.25"},"options_float":{"a":3886.25,"b":3286.25,"c":3186.25,"d":3386.25,"e":3586.25},"annotated_formula":"add(add(multiply(10000, divide(15, const_100)), multiply(add(10000, multiply(10000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(10000, multiply(10000, divide(15, const_100))), multiply(add(10000, multiply(10000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const_3)))","linear_formula":"divide(n1,const_100)|divide(#0,const_3)|multiply(n0,#0)|add(n0,#2)|multiply(#3,#0)|add(#2,#4)|add(#3,#4)|multiply(#6,#1)|add(#5,#7)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n10_000 * (3\/20)<\/gadget>\n1_500<\/output>\n10_000 + 1_500<\/gadget>\n11_500<\/output>\n11_500 * (3\/20)<\/gadget>\n1_725<\/output>\n1_500 + 1_725<\/gadget>\n3_225<\/output>\n11_500 + 1_725<\/gadget>\n13_225<\/output>\n(3\/20) \/ 3<\/gadget>\n1\/20 = around 0.05<\/output>\n13_225 * (1\/20)<\/gadget>\n2_645\/4 = around 661.25<\/output>\n3_225 + (2_645\/4)<\/gadget>\n15_545\/4 = around 3_886.25<\/output>\n15_545\/4 = around 3_886.25<\/result>","index":679} +{"problem":"having received his weekly allowance , john spent 3 \/ 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.28 at the candy store . what is john ’ s weekly allowance ?","rationale":"\"total allowance = x amount spent at the arcade = 3 \/ 5 x amount remaining = 2 \/ 5 x amount spent at the toy store = 2 \/ 5 * 1 \/ 3 x = 2 \/ 15 x amount remaining = 2 \/ 5 x - 2 \/ 15 x = 4 \/ 15 x now , 4 \/ 15 x = $ 1.28 therefore , x = $ 4.80 . answer e\"","correct":"e","options":{"a":"$ 2.40 ","b":"$ 3.00 ","c":"$ 3.20 ","d":"$ 3.60","e":"$ 4.80"},"options_float":{"a":2.4,"b":3.0,"c":3.2,"d":3.6,"e":4.8},"annotated_formula":"divide(1.28, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5))))))","linear_formula":"divide(n0,n1)|divide(const_1,n0)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(const_1,#4)|divide(n2,#5)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/3) * (2\/5)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(3\/5) + (2\/15)<\/gadget>\n11\/15 = around 0.733333<\/output>\n1 - (11\/15)<\/gadget>\n4\/15 = around 0.266667<\/output>\n1.28 \/ (4\/15)<\/gadget>\n4.8<\/output>\n4.8<\/result>","index":680} +{"problem":"sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 60 ps per meter is ?","rationale":"\"3 x * 2 x = 3750 = > x = 25 2 ( 75 + 50 ) = 250 m 250 * 0.60 = rs . 150 answer : c\"","correct":"c","options":{"a":"287 ","b":"369 ","c":"150 ","d":"279","e":"361"},"options_float":{"a":287.0,"b":369.0,"c":150.0,"d":279.0,"e":361.0},"annotated_formula":"divide(multiply(60, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)|","chain":"3_750 * 2<\/gadget>\n7_500<\/output>\n7_500 \/ 3<\/gadget>\n2_500<\/output>\n2_500 ** (1\/2)<\/gadget>\n50<\/output>\n3_750 \/ 50<\/gadget>\n75<\/output>\n2 * (50 + 75)<\/gadget>\n250<\/output>\n60 * 250<\/gadget>\n15_000<\/output>\n15_000 \/ 100<\/gadget>\n150<\/output>\n150<\/result>","index":682} +{"problem":"if a person walks at 25 km \/ hr instead of 10 km \/ hr , he would have walked 20 km more . the actual distance traveled by him is :","rationale":"\"let the actual distance travelled be x km . x \/ 10 = ( x + 20 ) \/ 25 25 x = 10 x + 200 15 x = 200 x = 13.3 km . answer : c\"","correct":"c","options":{"a":"50 km ","b":"56 km ","c":"13.3 km ","d":"70 km","e":"80 km"},"options_float":{"a":50.0,"b":56.0,"c":13.3,"d":70.0,"e":80.0},"annotated_formula":"multiply(10, divide(20, subtract(25, 10)))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|","chain":"25 - 10<\/gadget>\n15<\/output>\n20 \/ 15<\/gadget>\n4\/3 = around 1.333333<\/output>\n10 * (4\/3)<\/gadget>\n40\/3 = around 13.333333<\/output>\n40\/3 = around 13.333333<\/result>","index":685} +{"problem":"a big container is 30 % full with water . if 45 liters of water is added , the container becomes 3 \/ 4 full . what is the capacity of the big container ?","rationale":"\"a big container is 30 % full with water and after 45 liters of water is added , the container becomes 75 % full . hence these 45 liters account for 45 % of the container , which means that the capacity of it is 45 \/ 0.45 = 100 liters . or : if the capacity of the container is x liters then : 0.3 x + 45 = 0.75 x - - > x = 100 liters . answer : e .\"","correct":"e","options":{"a":"36 liters ","b":"40 liters ","c":"45 liters ","d":"54 liters","e":"100 liters"},"options_float":{"a":36.0,"b":40.0,"c":45.0,"d":54.0,"e":100.0},"annotated_formula":"divide(45, subtract(divide(3, 4), divide(30, const_100)))","linear_formula":"divide(n2,n3)|divide(n0,const_100)|subtract(#0,#1)|divide(n1,#2)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/4) - (3\/10)<\/gadget>\n9\/20 = around 0.45<\/output>\n45 \/ (9\/20)<\/gadget>\n100<\/output>\n100<\/result>","index":686} +{"problem":"find the smallest number that should be multiplied with 54000 to make it a perfect cube .","rationale":"54000 = 2 ^ 1 * 3 ^ 3 * 10 ^ 3 to make it ps * 2 ^ 2 or 4 answer : a","correct":"a","options":{"a":"4 ","b":"5 ","c":"6 ","d":"3","e":"7"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":3.0,"e":7.0},"annotated_formula":"multiply(divide(divide(divide(divide(54000, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(54000, const_1000), const_3), const_3), const_3))","linear_formula":"divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)","chain":"54_000 \/ 1_000<\/gadget>\n54<\/output>\n54 \/ 3<\/gadget>\n18<\/output>\n18 \/ 3<\/gadget>\n6<\/output>\n6 \/ 3<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4<\/result>","index":688} +{"problem":"a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 500000 ?","rationale":"the first digit and last digit are the same so the 2 possibilities are 6 or 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 2 * 10 * 10 = 200 the answer is b .","correct":"b","options":{"a":"160 ","b":"200 ","c":"240 ","d":"300","e":"480"},"options_float":{"a":160.0,"b":200.0,"c":240.0,"d":300.0,"e":480.0},"annotated_formula":"multiply(divide(500000, multiply(const_1000, const_10)), const_4)","linear_formula":"multiply(const_10,const_1000)|divide(n3,#0)|multiply(#1,const_4)","chain":"1_000 * 10<\/gadget>\n10_000<\/output>\n500_000 \/ 10_000<\/gadget>\n50<\/output>\n50 * 4<\/gadget>\n200<\/output>\n200<\/result>","index":689} +{"problem":"the average of 7 numbers is 25 . if each number be multiplied by 5 . find the average of new set of numbers ?","rationale":"\"explanation : average of new numbers = 25 * 5 = 125 answer : option d\"","correct":"d","options":{"a":"110 ","b":"122 ","c":"120 ","d":"125","e":"145"},"options_float":{"a":110.0,"b":122.0,"c":120.0,"d":125.0,"e":145.0},"annotated_formula":"multiply(25, 5)","linear_formula":"multiply(n1,n2)|","chain":"25 * 5<\/gadget>\n125<\/output>\n125<\/result>","index":690} +{"problem":"a 90 cm long wire is to be cut into two pieces so that one piece will be 2 \/ 5 th of the other , how many centimeters will the shorter piece be ?","rationale":"\"1 : 2 \/ 5 = 5 : 2 2 \/ 7 * 90 = 25 answer : a\"","correct":"a","options":{"a":"25 ","b":"20 ","c":"47 ","d":"36","e":"30"},"options_float":{"a":25.0,"b":20.0,"c":47.0,"d":36.0,"e":30.0},"annotated_formula":"subtract(90, divide(90, add(divide(2, 5), const_1)))","linear_formula":"divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) + 1<\/gadget>\n7\/5 = around 1.4<\/output>\n90 \/ (7\/5)<\/gadget>\n450\/7 = around 64.285714<\/output>\n90 - (450\/7)<\/gadget>\n180\/7 = around 25.714286<\/output>\n180\/7 = around 25.714286<\/result>","index":691} +{"problem":"a courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is :","rationale":"\"explanation : number of bricks = courtyard area \/ 1 brick area = ( 2500 × 1600 \/ 20 × 10 ) = 20000 option c\"","correct":"c","options":{"a":"16000 ","b":"18000 ","c":"20000 ","d":"22000","e":"none of these"},"options_float":{"a":16000.0,"b":18000.0,"c":20000.0,"d":22000.0,"e":null},"annotated_formula":"divide(multiply(multiply(25, const_100), multiply(16, const_100)), multiply(20, 10))","linear_formula":"multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)|","chain":"25 * 100<\/gadget>\n2_500<\/output>\n16 * 100<\/gadget>\n1_600<\/output>\n2_500 * 1_600<\/gadget>\n4_000_000<\/output>\n20 * 10<\/gadget>\n200<\/output>\n4_000_000 \/ 200<\/gadget>\n20_000<\/output>\n20_000<\/result>","index":694} +{"problem":"find the principle on a certain sum of money at 5 % per annum for 2 2 \/ 5 years if the amount being rs . 1008 ?","rationale":"\"1008 = p [ 1 + ( 5 * 12 \/ 5 ) \/ 100 ] p = 900 answer : a\"","correct":"a","options":{"a":"900 ","b":"2217 ","c":"2889 ","d":"2777","e":"2991"},"options_float":{"a":900.0,"b":2217.0,"c":2889.0,"d":2777.0,"e":2991.0},"annotated_formula":"divide(1008, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1))","linear_formula":"multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|","chain":"2 * 5<\/gadget>\n10<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 \/ 5<\/gadget>\n12\/5 = around 2.4<\/output>\n(12\/5) * 5<\/gadget>\n12<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) + 1<\/gadget>\n28\/25 = around 1.12<\/output>\n1_008 \/ (28\/25)<\/gadget>\n900<\/output>\n900<\/result>","index":695} +{"problem":"if a sum of money doubles itself in 20 years at simple interest , the ratepercent per annum is","rationale":"\"explanation : let sum = x then simple interest = x rate = ( 100 * x ) \/ ( x * 20 ) = 5 option d\"","correct":"d","options":{"a":"12 ","b":"12.5 ","c":"13 ","d":"5","e":"14"},"options_float":{"a":12.0,"b":12.5,"c":13.0,"d":5.0,"e":14.0},"annotated_formula":"divide(divide(const_2, divide(20, const_100)), const_2)","linear_formula":"divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n2 \/ (1\/5)<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":696} +{"problem":"at joes steakhouse the hourly wage for a chef is 20 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 7.50 per hour , how much less than a manager does a chef earn each hour ?","rationale":"\"manager wages per hour = $ 7.50 dishwasher wages per hour = half of manager ' s wages . = 1 \/ 2 ( $ 7.50 ) = = > $ 3.75 chef wages per hour = 20 % greater than dishwasher wages - - > 20 % of $ 3.75 = ( 20 * ( $ 3.75 ) ) \/ 100 - - > ( $ 75 ) \/ 100 - - > $ 0.75 therefore , chef wages per hour = $ 3.75 + $ 0.75 = = > $ 4.5 difference of wages between manager and chef = $ 7.50 - $ 4.5 = = > $ 3 answer : b\"","correct":"b","options":{"a":"$ 1.40 ","b":"$ 3 ","c":"$ 3.40 ","d":"$ 4.40","e":"$ 5.40"},"options_float":{"a":1.4,"b":3.0,"c":3.4,"d":4.4,"e":5.4},"annotated_formula":"multiply(subtract(const_1, multiply(divide(add(const_100, 20), const_100), divide(const_1, const_2))), 7.50)","linear_formula":"add(n0,const_100)|divide(const_1,const_2)|divide(#0,const_100)|multiply(#2,#1)|subtract(const_1,#3)|multiply(n1,#4)|","chain":"100 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(6\/5) * (1\/2)<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 7.5<\/gadget>\n3<\/output>\n3<\/result>","index":698} +{"problem":"a person incurs a loss of 5 % be selling a watch for rs . 1140 . at what price should the watch be sold to earn 5 % profit","rationale":"\"explanation : let the new s . p . be x , then . ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) = > ( 95 \/ 1140 = 105 \/ x ) = > x = 1260 option c\"","correct":"c","options":{"a":"rs . 1200 ","b":"rs . 1230 ","c":"rs . 1260 ","d":"rs . 1290","e":"none of these"},"options_float":{"a":1200.0,"b":1230.0,"c":1260.0,"d":1290.0,"e":null},"annotated_formula":"multiply(divide(1140, divide(subtract(const_100, 5), const_100)), divide(add(const_100, 5), const_100))","linear_formula":"add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|divide(n1,#2)|multiply(#4,#3)|","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n1_140 \/ (19\/20)<\/gadget>\n1_200<\/output>\n100 + 5<\/gadget>\n105<\/output>\n105 \/ 100<\/gadget>\n21\/20 = around 1.05<\/output>\n1_200 * (21\/20)<\/gadget>\n1_260<\/output>\n1_260<\/result>","index":699} +{"problem":"a train is 400 meter long is running at a speed of 60 km \/ hour . in what time will it pass a signal post ?","rationale":"speed = 60 km \/ hr = 60 * ( 5 \/ 18 ) m \/ sec = 150 \/ 9 m \/ sec total distance = 400 meter time = distance \/ speed = 400 * ( 9 \/ 150 ) = 24 seconds answer : d","correct":"d","options":{"a":"28 seconds ","b":"30 seconds ","c":"40 seconds ","d":"24 seconds","e":"18 seconds"},"options_float":{"a":28.0,"b":30.0,"c":40.0,"d":24.0,"e":18.0},"annotated_formula":"divide(400, multiply(60, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n400 \/ (50\/3)<\/gadget>\n24<\/output>\n24<\/result>","index":700} +{"problem":"how long does a train 105 m long running at the speed of 68 km \/ hr takes to cross a bridge 90 m length ?","rationale":"\"speed = 68 * 5 \/ 18 = 19 m \/ sec total distance covered = 105 + 90 = 205 m . required time = 205 \/ 19 = 10.7 sec . answer : b\"","correct":"b","options":{"a":"13.9 sec ","b":"10.7 sec ","c":"17.9 sec ","d":"61.9 sec","e":"47.98 sec"},"options_float":{"a":13.9,"b":10.7,"c":17.9,"d":61.9,"e":47.98},"annotated_formula":"divide(add(105, 90), multiply(68, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"105 + 90<\/gadget>\n195<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n68 * (5\/18)<\/gadget>\n170\/9 = around 18.888889<\/output>\n195 \/ (170\/9)<\/gadget>\n351\/34 = around 10.323529<\/output>\n351\/34 = around 10.323529<\/result>","index":703} +{"problem":"there are 6 people in the elevator . their average weight is 152 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person .","rationale":"\"solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 152 ) = 1057 - 912 = 145 . answer a\"","correct":"a","options":{"a":"145 ","b":"168 ","c":"189 ","d":"190","e":"200"},"options_float":{"a":145.0,"b":168.0,"c":189.0,"d":190.0,"e":200.0},"annotated_formula":"subtract(multiply(151, 7), multiply(6, 152))","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|","chain":"151 * 7<\/gadget>\n1_057<\/output>\n6 * 152<\/gadget>\n912<\/output>\n1_057 - 912<\/gadget>\n145<\/output>\n145<\/result>","index":704} +{"problem":"ms . lopez deposits $ 150 in an account that pays 20 % interest , compounded semiannually . how much money will there be in the account at the end of one year ?","rationale":"\"using formula - a = p ( 1 + r \/ n ) ^ nt given p = 150 n = 2 t = 1 r = 0.2 substituting value in formula a = 150 ( 1 + 0.2 \/ 2 ) ^ 2 a = 181.15 $ = c\"","correct":"c","options":{"a":"$ 118.00 ","b":"$ 120.00 ","c":"$ 181.15 ","d":"$ 122.00","e":"$ 140.00"},"options_float":{"a":118.0,"b":120.0,"c":181.15,"d":122.0,"e":140.0},"annotated_formula":"multiply(150, power(add(const_1, divide(divide(20, const_100), const_2)), const_2))","linear_formula":"divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ 2<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n150 * (121\/100)<\/gadget>\n363\/2 = around 181.5<\/output>\n363\/2 = around 181.5<\/result>","index":705} +{"problem":"how much is 70 % of 40 is greater than 4 \/ 5 of 25 ?","rationale":"\"( 70 \/ 100 ) * 40 – ( 4 \/ 5 ) * 25 = 8 answer : e\"","correct":"e","options":{"a":"22 ","b":"67 ","c":"88 ","d":"12","e":"8"},"options_float":{"a":22.0,"b":67.0,"c":88.0,"d":12.0,"e":8.0},"annotated_formula":"subtract(multiply(40, divide(70, const_100)), multiply(divide(4, 5), 25))","linear_formula":"divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)|","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n40 * (7\/10)<\/gadget>\n28<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 25<\/gadget>\n20<\/output>\n28 - 20<\/gadget>\n8<\/output>\n8<\/result>","index":707} +{"problem":"one man adds 3 liters of water to 12 liters of milk and another 4 liters of water to 10 liters of milk . what is the ratio of the strengths of milk in 2 mixtures ?","rationale":"\"strength of milk in the mixture = quantity of milk \/ total quantity of mixture strength of milk in the 1 st mixture = 12 \/ 15 strength of the milk 2 nd mixture = 10 \/ 14 the ratio of their strengths = 12 \/ 15 : 10 \/ 14 = 28 : 25 answer is c\"","correct":"c","options":{"a":"11 : 21 ","b":"15 : 33 ","c":"28 : 25 ","d":"3 : 2","e":"31 : 33"},"options_float":{"a":0.5238095238,"b":0.4545454545,"c":1.12,"d":1.5,"e":0.9393939394},"annotated_formula":"divide(divide(12, add(3, 12)), divide(10, add(4, 10)))","linear_formula":"add(n0,n1)|add(n2,n3)|divide(n1,#0)|divide(n3,#1)|divide(#2,#3)|","chain":"3 + 12<\/gadget>\n15<\/output>\n12 \/ 15<\/gadget>\n4\/5 = around 0.8<\/output>\n4 + 10<\/gadget>\n14<\/output>\n10 \/ 14<\/gadget>\n5\/7 = around 0.714286<\/output>\n(4\/5) \/ (5\/7)<\/gadget>\n28\/25 = around 1.12<\/output>\n28\/25 = around 1.12<\/result>","index":708} +{"problem":"27 is a perfect cube . when x is added to the prime factor of 27 , the result is a prime number . what is x ?","rationale":"27 is 3 * 3 * 3 then a 3 + 1 = 4 b 3 + 3 = 6 c 3 + 5 = 8 d 3 + 8 = 11 e 3 + 11 = 14 here d is the only addition that gives a prime number .","correct":"d","options":{"a":"1 ","b":"3 ","c":"5 ","d":"8","e":"11"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":8.0,"e":11.0},"annotated_formula":"divide(subtract(27, const_3), const_3)","linear_formula":"subtract(n0,const_3)|divide(#0,const_3)","chain":"27 - 3<\/gadget>\n24<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n8<\/result>","index":709} +{"problem":"a part - time employee ’ s hourly wage was increased by 10 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ?","rationale":"\"let ' s say the employee used to make $ 1 \/ hour and worked 100 hours \/ week so , the total weekly income was $ 100 \/ week after the 10 % wage increase , the employee makes $ 1.10 \/ hour we want the employee ' s income to remain at $ 100 \/ week . so , we want ( $ 1.10 \/ hour ) ( new # of hours ) = $ 100 divide both sides by 1.10 to get : new # of hours = 100 \/ 1.10 ≈ 90 hours so , the number of hours decreases from 100 hours to ( approximately ) 90 hours . this represents a 10 % decrease ( approximately ) . answer choice a is the closest answer .\"","correct":"a","options":{"a":"9 % ","b":"15 % ","c":"25 % ","d":"50 %","e":"100 %"},"options_float":{"a":9.0,"b":15.0,"c":25.0,"d":50.0,"e":100.0},"annotated_formula":"multiply(divide(divide(10, const_100), divide(add(10, const_100), const_100)), const_100)","linear_formula":"add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n10 + 100<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n(1\/10) \/ (11\/10)<\/gadget>\n1\/11 = around 0.090909<\/output>\n(1\/11) * 100<\/gadget>\n100\/11 = around 9.090909<\/output>\n100\/11 = around 9.090909<\/result>","index":710} +{"problem":"the price of stock decreased by 5 % last year and rose by 10 % this year . what is the net percentage change in the price of the stock ?","rationale":"\"( 100 % - 5 % ) * ( 100 % + 10 % ) = 0.95 * 1.10 = 1.045 = 104.5 % . the net percentage change in the price of the stock is ( + ) 4.5 % the answer is e\"","correct":"e","options":{"a":"0.2 % ","b":"0.8 % ","c":"1.2 % ","d":"1.6 %","e":"4.5 %"},"options_float":{"a":0.2,"b":0.8,"c":1.2,"d":1.6,"e":4.5},"annotated_formula":"multiply(subtract(multiply(add(divide(10, const_100), const_1), subtract(const_1, divide(5, const_100))), const_1), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (1\/20)<\/gadget>\n19\/20 = around 0.95<\/output>\n(11\/10) * (19\/20)<\/gadget>\n209\/200 = around 1.045<\/output>\n(209\/200) - 1<\/gadget>\n9\/200 = around 0.045<\/output>\n(9\/200) * 100<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":711} +{"problem":"the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds than it was moving after 5 seconds ?","rationale":"\"given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 5 ) = 5 ^ 2 + 2 * 5 = 35 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 35 = 28 km \/ hr option b\"","correct":"b","options":{"a":"4 ","b":"28 ","c":"15 ","d":"48","e":"63"},"options_float":{"a":4.0,"b":28.0,"c":15.0,"d":48.0,"e":63.0},"annotated_formula":"subtract(add(power(7, 2), multiply(7, 2)), add(power(5, 2), multiply(5, 2)))","linear_formula":"multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,n0)|add(#0,#2)|add(#1,#3)|subtract(#4,#5)|","chain":"7 ** 2<\/gadget>\n49<\/output>\n7 * 2<\/gadget>\n14<\/output>\n49 + 14<\/gadget>\n63<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n5 * 2<\/gadget>\n10<\/output>\n25 + 10<\/gadget>\n35<\/output>\n63 - 35<\/gadget>\n28<\/output>\n28<\/result>","index":712} +{"problem":"a soccer team played 280 games and won 65 percent of them . how many games did it win ?","rationale":"\"65 % of 280 = x 0.65 * 280 = x 182 = x answer : a\"","correct":"a","options":{"a":"182 ","b":"100 ","c":"104 ","d":"150","e":"124"},"options_float":{"a":182.0,"b":100.0,"c":104.0,"d":150.0,"e":124.0},"annotated_formula":"divide(multiply(65, 280), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|","chain":"65 * 280<\/gadget>\n18_200<\/output>\n18_200 \/ 100<\/gadget>\n182<\/output>\n182<\/result>","index":713} +{"problem":"the difference between two numbers is 2395 . when the larger number is divided by the smaller one , the quotient is 6 and the remainder is 15 . the smaller number is","rationale":"let the smaller number be x . then , larger number = ( 2395 + x ) therefore 2395 + x = ( 6 x + 15 ) ‹ = › 5 x = 2380 ‹ = › x = 476 . answer : c","correct":"c","options":{"a":"120 ","b":"239 ","c":"476 ","d":"523","e":"526"},"options_float":{"a":120.0,"b":239.0,"c":476.0,"d":523.0,"e":526.0},"annotated_formula":"divide(subtract(2395, 15), subtract(6, const_1))","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)","chain":"2_395 - 15<\/gadget>\n2_380<\/output>\n6 - 1<\/gadget>\n5<\/output>\n2_380 \/ 5<\/gadget>\n476<\/output>\n476<\/result>","index":714} +{"problem":"right triangle xyzs the base of the prism in the figure above . if xy = xz = â ˆ š 9 and the height of the prism is 6 , what is the volume of the prism ?","rationale":"volume of prism = area of base * height = 1 \/ 2 * ( square root of 9 ) * ( square root of 9 ) * 6 = 27 answer : b","correct":"b","options":{"a":"3 ","b":"27 ","c":"13 ","d":"12","e":"15"},"options_float":{"a":3.0,"b":27.0,"c":13.0,"d":12.0,"e":15.0},"annotated_formula":"multiply(triangle_area(sqrt(9), sqrt(9)), 6)","linear_formula":"sqrt(n0)|triangle_area(#0,#0)|multiply(n1,#1)","chain":"9 ** (1\/2)<\/gadget>\n3<\/output>\n(3 * 3) \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\n(9\/2) * 6<\/gadget>\n27<\/output>\n27<\/result>","index":715} +{"problem":"on dividing 22 by a number , the quotient is 7 and the remainder is 1 . find the divisor .","rationale":"\"d = ( d - r ) \/ q = ( 22 - 1 ) \/ 7 = 21 \/ 7 = 3 c\"","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"6","e":"7"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":6.0,"e":7.0},"annotated_formula":"floor(divide(22, 7))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\nfloor(22\/7)<\/gadget>\n3<\/output>\n3<\/result>","index":716} +{"problem":"the smallest number which when diminished by 5 , is divisible by 12 , 16 , 18 , 21 and 28 is","rationale":"\"required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 5 = 1008 + 5 = 1013 answer : b\"","correct":"b","options":{"a":"1008 ","b":"1013 ","c":"1022 ","d":"1032","e":"1043"},"options_float":{"a":1008.0,"b":1013.0,"c":1022.0,"d":1032.0,"e":1043.0},"annotated_formula":"add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 5)","linear_formula":"add(const_3,const_4)|power(const_3,const_2.0)|power(const_2.0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,n0)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n9 * 16<\/gadget>\n144<\/output>\n3 + 4<\/gadget>\n7<\/output>\n144 * 7<\/gadget>\n1_008<\/output>\n1_008 + 5<\/gadget>\n1_013<\/output>\n1_013<\/result>","index":717} +{"problem":"the ratio between x and y is 7 \/ 5 ; x is multiplied by y and y is multiplied by x , what is the ratio between the new values of x and y ?","rationale":"ratio = 7 k \/ 5 k = 7 \/ 5 , 14 \/ 10 , etc . x is multiplied by y and y is multiplied by x - - > ( 7 k * 9 k ) \/ ( 9 k * 7 k ) = 1 answer : c","correct":"c","options":{"a":"7 \/ 5 ","b":"5 \/ 7 ","c":"1 ","d":"49 \/ 25","e":"it can not be determined"},"options_float":{"a":1.4,"b":0.7142857143,"c":1.0,"d":1.96,"e":null},"annotated_formula":"divide(multiply(7, 5), multiply(5, 7))","linear_formula":"multiply(n0,n1)|divide(#0,#0)","chain":"7 * 5<\/gadget>\n35<\/output>\n5 * 7<\/gadget>\n35<\/output>\n35 \/ 35<\/gadget>\n1<\/output>\n1<\/result>","index":718} +{"problem":"when x is multiplied by 3 , the result is 14 more than the result of subtracting x from 26 . what is the value of x ?","rationale":"the equation that can be formed is : 3 x - 14 = 26 - x or , 4 x = 40 or , x = 10 . a answer .","correct":"a","options":{"a":"10 ","b":"- 2 ","c":"11 ","d":"13","e":"22"},"options_float":{"a":10.0,"b":-2.0,"c":11.0,"d":13.0,"e":22.0},"annotated_formula":"divide(add(26, 14), add(const_3, const_1))","linear_formula":"add(n1,n2)|add(const_1,const_3)|divide(#0,#1)","chain":"26 + 14<\/gadget>\n40<\/output>\n3 + 1<\/gadget>\n4<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n10<\/result>","index":719} +{"problem":"raman , lakshmi and muthu enter into partnership . raman invests some money at the beginning , lakshmi invests double the amount after 6 months , and muthu invests thrice the amount after 8 months . if the annual gain be rs . 36000 . lakshmi ' s share is ?","rationale":"x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 \/ 3 * 36000 = 12000 answer : a","correct":"a","options":{"a":"12000 ","b":"14000 ","c":"8000 ","d":"5000","e":"3000"},"options_float":{"a":12000.0,"b":14000.0,"c":8000.0,"d":5000.0,"e":3000.0},"annotated_formula":"multiply(divide(const_2, 6), 36000)","linear_formula":"divide(const_2,n0)|multiply(n2,#0)","chain":"2 \/ 6<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 36_000<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":720} +{"problem":"in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 10 per square m ?","rationale":"length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 10 * 950 = rs . 9500 a","correct":"a","options":{"a":"s . 9500 ","b":"s . 50000 ","c":"s . 57000 ","d":"s . 60000","e":"s . 62000"},"options_float":{"a":9500.0,"b":50000.0,"c":57000.0,"d":60000.0,"e":62000.0},"annotated_formula":"add(add(multiply(multiply(multiply(20, 15), const_2), 10), multiply(multiply(multiply(20, 5), const_2), 10)), multiply(multiply(multiply(15, 5), const_2), 10))","linear_formula":"multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#8)","chain":"20 * 15<\/gadget>\n300<\/output>\n300 * 2<\/gadget>\n600<\/output>\n600 * 10<\/gadget>\n6_000<\/output>\n20 * 5<\/gadget>\n100<\/output>\n100 * 2<\/gadget>\n200<\/output>\n200 * 10<\/gadget>\n2_000<\/output>\n6_000 + 2_000<\/gadget>\n8_000<\/output>\n15 * 5<\/gadget>\n75<\/output>\n75 * 2<\/gadget>\n150<\/output>\n150 * 10<\/gadget>\n1_500<\/output>\n8_000 + 1_500<\/gadget>\n9_500<\/output>\n9_500<\/result>","index":721} +{"problem":"31 of the scientists that attended a certain workshop were wolf prize laureates , and 18 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 50 of the scientists attended that workshop , how many of them were nobel prize laureates ?","rationale":"\"lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 13 . . so total who won n = 11 + 18 = 29 . . d\"","correct":"d","options":{"a":"11 ","b":"18 ","c":"24 ","d":"29","e":"d ) 36"},"options_float":{"a":11.0,"b":18.0,"c":24.0,"d":29.0,"e":36.0},"annotated_formula":"add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 18)","linear_formula":"subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)|","chain":"50 - 31<\/gadget>\n19<\/output>\n19 - 3<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n3 + 8<\/gadget>\n11<\/output>\n11 + 18<\/gadget>\n29<\/output>\n29<\/result>","index":723} +{"problem":"if 4 men can colour 48 m long cloth in 2 days , then 2 men can colour 36 m long cloth in","rationale":"\"the length of cloth painted by one man in one day = 48 \/ 4 × 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 \/ 6 × 2 = 3 day . c\"","correct":"c","options":{"a":"1 day ","b":"2 days ","c":"3 days ","d":"4 days","e":"5 days"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"divide(36, multiply(divide(48, multiply(4, 2)), 2))","linear_formula":"multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)|","chain":"4 * 2<\/gadget>\n8<\/output>\n48 \/ 8<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n36 \/ 12<\/gadget>\n3<\/output>\n3<\/result>","index":724} +{"problem":"a library has an average of 540 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is :","rationale":"\"since the month begins with sunday , to there will be five sundays in the month average required = ( 540 x 5 + 240 x 25 ) \/ 30 ) = 290 answer : option c\"","correct":"c","options":{"a":"300 ","b":"276 ","c":"290 ","d":"285","e":"none"},"options_float":{"a":300.0,"b":276.0,"c":290.0,"d":285.0,"e":null},"annotated_formula":"divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 540), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30)","linear_formula":"add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)|","chain":"3 + 4<\/gadget>\n7<\/output>\n30 \/ 7<\/gadget>\n30\/7 = around 4.285714<\/output>\nfloor(30\/7)<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 540<\/gadget>\n2_700<\/output>\n30 - 5<\/gadget>\n25<\/output>\n25 * 240<\/gadget>\n6_000<\/output>\n2_700 + 6_000<\/gadget>\n8_700<\/output>\n8_700 \/ 30<\/gadget>\n290<\/output>\n290<\/result>","index":726} +{"problem":"when sold at a 70 % discount , a sweater nets the merchant a 40 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ?","rationale":"\"we should be careful about what are we measuring % on \/ what is the base . . let the marked up price = 100 . . selling price = 100 - 70 % of 100 = 30 . . profit = 40 % . . therefore the wholesale purchase cost = x . . . . 1.4 x = 30 or x = 21.42 . . . marked price was 100 so . . . so answer is 78.58 % . . e\"","correct":"e","options":{"a":"20 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"78.58 %"},"options_float":{"a":20.0,"b":40.0,"c":50.0,"d":60.0,"e":78.58},"annotated_formula":"subtract(const_100, divide(subtract(const_100, 70), add(const_1, divide(40, const_100))))","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|add(#0,const_1)|divide(#1,#2)|subtract(const_100,#3)|","chain":"100 - 70<\/gadget>\n30<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n30 \/ (7\/5)<\/gadget>\n150\/7 = around 21.428571<\/output>\n100 - (150\/7)<\/gadget>\n550\/7 = around 78.571429<\/output>\n550\/7 = around 78.571429<\/result>","index":728} +{"problem":"on a certain farm the ratio of horses to cows is 4 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 13 : 7 . after the transaction , how many more horses than cows would the farm own ?","rationale":"\"originally , there were 4 k horses and k cows . 7 ( 4 k - 15 ) = 13 ( k + 15 ) 28 k - 13 k = 195 + 105 15 k = 300 k = 20 the difference between horses and cows is ( 4 k - 15 ) - ( k + 15 ) = 3 k - 30 = 30 the answer is a .\"","correct":"a","options":{"a":"30 ","b":"60 ","c":"75 ","d":"90","e":"105"},"options_float":{"a":30.0,"b":60.0,"c":75.0,"d":90.0,"e":105.0},"annotated_formula":"subtract(subtract(multiply(4, divide(add(multiply(15, 13), multiply(15, 7)), subtract(multiply(4, 7), 13))), 15), add(divide(add(multiply(15, 13), multiply(15, 7)), subtract(multiply(4, 7), 13)), 15))","linear_formula":"multiply(n2,n4)|multiply(n2,n5)|multiply(n0,n5)|add(#0,#1)|subtract(#2,n4)|divide(#3,#4)|add(n2,#5)|multiply(n0,#5)|subtract(#7,n2)|subtract(#8,#6)|","chain":"15 * 13<\/gadget>\n195<\/output>\n15 * 7<\/gadget>\n105<\/output>\n195 + 105<\/gadget>\n300<\/output>\n4 * 7<\/gadget>\n28<\/output>\n28 - 13<\/gadget>\n15<\/output>\n300 \/ 15<\/gadget>\n20<\/output>\n4 * 20<\/gadget>\n80<\/output>\n80 - 15<\/gadget>\n65<\/output>\n20 + 15<\/gadget>\n35<\/output>\n65 - 35<\/gadget>\n30<\/output>\n30<\/result>","index":729} +{"problem":"the total age of a and b is 12 years more than the total age of b and c . c is how many year younger than a","rationale":"\"explanation : given that a + b = 12 + b + c = > a – c = 12 + b – b = 12 = > c is younger than a by 12 years answer : option b\"","correct":"b","options":{"a":"11 ","b":"12 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"multiply(12, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"12 * 1<\/gadget>\n12<\/output>\n12<\/result>","index":730} +{"problem":"if the diameter of circle r is 40 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ?","rationale":"\"let diameter of circle r , dr = 40 and diameter of circle s , ds = 100 radius of circle r , rr = 20 radius of circle s , rs = 50 area of circle r \/ area of circle s = ( pi * rr ^ 2 ) \/ ( pi * rs ^ 2 ) = ( 20 \/ 50 ) ^ 2 = ( 4 \/ 10 ) ^ 2 = 16 % answer : a\"","correct":"a","options":{"a":"16 % ","b":"18 % ","c":"20 % ","d":"22 %","e":"24 %"},"options_float":{"a":16.0,"b":18.0,"c":20.0,"d":22.0,"e":24.0},"annotated_formula":"divide(multiply(circle_area(40), const_100), circle_area(const_100))","linear_formula":"circle_area(n0)|circle_area(const_100)|multiply(#0,const_100)|divide(#2,#1)|","chain":"pi * (40 ** 2)<\/gadget>\n1600*pi = around 5_026.548246<\/output>\n(1600*pi) * 100<\/gadget>\n160000*pi = around 502_654.824574<\/output>\npi * (100 ** 2)<\/gadget>\n10000*pi = around 31_415.926536<\/output>\n(160000*pi) \/ (10000*pi)<\/gadget>\n16<\/output>\n16<\/result>","index":731} +{"problem":"if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer v that must divide n is ?","rationale":"\"q : if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer v that must divide n is : a 6 , b 12 , c 24 , d 36 , e 48 n ^ 2 is divisible by 72 , but it must also be greater than 72 . if n is an integer , then n ^ 2 must be a perfect square . the factorization of 72 is ( 8 ) ( 9 ) , so if it is multiplied by 2 , it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144 , a perfect square . so n ^ 2 must be at least 144 or a multiple of 144 , which means that n must be 12 or a multiple of 12 . b\"","correct":"b","options":{"a":"6 ","b":"12 ","c":"24 ","d":"36","e":"48"},"options_float":{"a":6.0,"b":12.0,"c":24.0,"d":36.0,"e":48.0},"annotated_formula":"multiply(sqrt(divide(72, 2)), 2)","linear_formula":"divide(n1,n0)|sqrt(#0)|multiply(n0,#1)|","chain":"72 \/ 2<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12<\/result>","index":732} +{"problem":"mary can do a piece of work in 26 days . rosy is 30 % more efficient than mary . the number of days taken by rosy to do the same piece of work is ?","rationale":"\"ratio of times taken by mary and rosy = 130 : 100 = 13 : 10 suppose rosy takes x days to do the work . 13 : 10 : : 26 : x = > x = 20 days . hence , rosy takes 20 days to complete the work . answer : b\"","correct":"b","options":{"a":"22 ","b":"20 ","c":"24 ","d":"26","e":"30"},"options_float":{"a":22.0,"b":20.0,"c":24.0,"d":26.0,"e":30.0},"annotated_formula":"divide(26, add(const_1, divide(30, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n26 \/ (13\/10)<\/gadget>\n20<\/output>\n20<\/result>","index":733} +{"problem":"a student has to obtain 33 % of the total marks to pass . he got 175 marks and failed by 56 marks . the maximum marks are ?","rationale":"\"let the maximum marks be x then , 33 % of x = 175 + 56 33 x \/ 100 = 231 x = 700 answer is e\"","correct":"e","options":{"a":"450 ","b":"300 ","c":"500 ","d":"610","e":"700"},"options_float":{"a":450.0,"b":300.0,"c":500.0,"d":610.0,"e":700.0},"annotated_formula":"divide(add(175, 56), divide(33, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"175 + 56<\/gadget>\n231<\/output>\n33 \/ 100<\/gadget>\n33\/100 = around 0.33<\/output>\n231 \/ (33\/100)<\/gadget>\n700<\/output>\n700<\/result>","index":736} +{"problem":"if s and t are both odd prime numbers ands < t , then how many different positive integer factors does 2 sthave ?","rationale":"s and t are both odd prime numbers - it means either s or t is not 2 and since prime numbers have only two factors - 1 and the number itself s and t each will have ( 1 + 1 ) = 2 factors hence 2 st will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer","correct":"d","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"12"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":12.0},"annotated_formula":"multiply(add(2, 2), 2)","linear_formula":"add(n0,n0)|multiply(n0,#0)","chain":"2 + 2<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":737} +{"problem":"the maximum number of student amoung them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :","rationale":"\"solution required number of student = h . c . f of 1001 and 910 = 91 . answer a\"","correct":"a","options":{"a":"91 ","b":"910 ","c":"1001 ","d":"1911","e":"none of these"},"options_float":{"a":91.0,"b":910.0,"c":1001.0,"d":1911.0,"e":null},"annotated_formula":"gcd(1001, 910)","linear_formula":"gcd(n0,n1)|","chain":"gcd(1_001, 910)<\/gadget>\n91<\/output>\n91<\/result>","index":738} +{"problem":"a class of 20 students took a science test . 10 students had an average ( arithmetic mean ) score of 80 . the other students had an average score of 60 . what is the average score of the whole class ?","rationale":"hear , 10 students had average score of 80 , so , 80 x 10 = 800 and 10 students had average score of 60 , so , 60 x 10 = 600 total no of student are 20 ( 800 + 600 ) \/ 20 = 70 correct ans is a","correct":"a","options":{"a":"70 ","b":"60 ","c":"55 ","d":"65","e":"85"},"options_float":{"a":70.0,"b":60.0,"c":55.0,"d":65.0,"e":85.0},"annotated_formula":"divide(add(multiply(10, 80), multiply(subtract(20, 10), 60)), 20)","linear_formula":"multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|add(#0,#2)|divide(#3,n0)","chain":"10 * 80<\/gadget>\n800<\/output>\n20 - 10<\/gadget>\n10<\/output>\n10 * 60<\/gadget>\n600<\/output>\n800 + 600<\/gadget>\n1_400<\/output>\n1_400 \/ 20<\/gadget>\n70<\/output>\n70<\/result>","index":739} +{"problem":"if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 900 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 922.50 . what is the value of x ?","rationale":"simple way to solve this question is to use options . from si , we know that x * y = 45,000 . now , put the value of x = 9000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 9000 + 5 % of 9000 = 9450 . 2 nd year , amount = 9450 + 5 % of 9450 = 9922.50 . we can see after 2 years interest = 9922.50 - 9000 = 922.50 . hence , it satisfies the question . hence a is the correct answer","correct":"a","options":{"a":"9000 ","b":"6000 ","c":"5000 ","d":"4000","e":"3000"},"options_float":{"a":9000.0,"b":6000.0,"c":5000.0,"d":4000.0,"e":3000.0},"annotated_formula":"divide(power(divide(900, const_2), const_2), subtract(922.5, 900))","linear_formula":"divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)","chain":"900 \/ 2<\/gadget>\n450<\/output>\n450 ** 2<\/gadget>\n202_500<\/output>\n922.5 - 900<\/gadget>\n22.5<\/output>\n202_500 \/ 22.5<\/gadget>\n9_000<\/output>\n9_000<\/result>","index":740} +{"problem":"teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 153 per kg , the price of the third variety per kg will be ?","rationale":"\"explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) \/ 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \\ \/ mean price ( rs . 153 ) \/ \\ x − 153 22.50 = > x − ( 153 \/ 22.50 ) = 1 . = > x − 153 = 22.50 . = > x = 175.50 rs . answer : c\"","correct":"c","options":{"a":"rs . 147.50 ","b":"rs . 785.50 ","c":"rs . 175.50 ","d":"rs . 258.50","e":"none of these"},"options_float":{"a":147.5,"b":785.5,"c":175.5,"d":258.5,"e":null},"annotated_formula":"divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 135)), 2)","linear_formula":"add(n2,n2)|add(n0,n1)|add(n4,#0)|multiply(n5,#2)|subtract(#3,#1)|divide(#4,n4)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 + 2<\/gadget>\n4<\/output>\n153 * 4<\/gadget>\n612<\/output>\n126 + 135<\/gadget>\n261<\/output>\n612 - 261<\/gadget>\n351<\/output>\n351 \/ 2<\/gadget>\n351\/2 = around 175.5<\/output>\n351\/2 = around 175.5<\/result>","index":741} +{"problem":"in what time will a railway train 50 m long moving at the rate of 36 kmph pass a telegraph post on its way ?","rationale":"\"t = 50 \/ 36 * 18 \/ 5 = 5 sec answer : a\"","correct":"a","options":{"a":"5 sec ","b":"4 sec ","c":"3 sec ","d":"6 sec","e":"1 sec"},"options_float":{"a":5.0,"b":4.0,"c":3.0,"d":6.0,"e":1.0},"annotated_formula":"divide(50, multiply(36, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n50 \/ 10<\/gadget>\n5<\/output>\n5<\/result>","index":743} +{"problem":"the marks obtained by vijay and amith are in the ratio 8 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ?","rationale":"8 : 5 3 : 2 - - - - - - - 24 : 15 : 10 24 : 10 12 : 5 answer : a","correct":"a","options":{"a":"12 : 5 ","b":"12 : 1 ","c":"6 : 5 ","d":"12 : 2","e":"6 : 3"},"options_float":{"a":2.4,"b":12.0,"c":1.2,"d":6.0,"e":2.0},"annotated_formula":"divide(multiply(8, 3), multiply(5, 2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)","chain":"8 * 3<\/gadget>\n24<\/output>\n5 * 2<\/gadget>\n10<\/output>\n24 \/ 10<\/gadget>\n12\/5 = around 2.4<\/output>\n12\/5 = around 2.4<\/result>","index":744} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 22 overs to reach the target of 282 runs ?","rationale":"\"explanation : runs scored in the first 10 overs = 10 × 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 22 run rate needed = 250 \/ 22 = 11.36 answer : option b\"","correct":"b","options":{"a":"6.25 ","b":"11.36 ","c":"7.4 ","d":"5","e":"6"},"options_float":{"a":6.25,"b":11.36,"c":7.4,"d":5.0,"e":6.0},"annotated_formula":"divide(subtract(282, multiply(10, 3.2)), 22)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 3.2<\/gadget>\n32<\/output>\n282 - 32<\/gadget>\n250<\/output>\n250 \/ 22<\/gadget>\n125\/11 = around 11.363636<\/output>\n125\/11 = around 11.363636<\/result>","index":745} +{"problem":"x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 9 will be equal to :","rationale":"\"explanation : solution : given x = k \/ y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 \/ y ^ 2 = > x = 9 \/ 9 ^ 2 = 1 \/ 9 answer : c\"","correct":"c","options":{"a":"3 ","b":"6 ","c":"1 \/ 9 ","d":"1 \/ 3","e":"9"},"options_float":{"a":3.0,"b":6.0,"c":0.1111111111,"d":0.3333333333,"e":9.0},"annotated_formula":"divide(multiply(1, power(3, const_2)), power(9, const_2))","linear_formula":"power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n1 * 9<\/gadget>\n9<\/output>\n9 ** 2<\/gadget>\n81<\/output>\n9 \/ 81<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":746} +{"problem":"a train running at the speed of 90 km \/ hr crosses a pole in 5 seconds . find the length of the train .","rationale":"\"explanation : speed = 90 * ( 5 \/ 18 ) m \/ sec = 25 m \/ sec length of train ( distance ) = speed * time = 25 * 5 = 125 meter option d\"","correct":"d","options":{"a":"150 meter ","b":"145 meter ","c":"140 meter ","d":"125 meter","e":"none of these"},"options_float":{"a":150.0,"b":145.0,"c":140.0,"d":125.0,"e":null},"annotated_formula":"multiply(divide(multiply(90, const_1000), const_3600), 5)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"90 * 1_000<\/gadget>\n90_000<\/output>\n90_000 \/ 3_600<\/gadget>\n25<\/output>\n25 * 5<\/gadget>\n125<\/output>\n125<\/result>","index":747} +{"problem":"find the mean proportional between 64 & 81 ?","rationale":"\"formula = √ a × b a = 64 and b = 81 √ 64 × 81 = 8 × 9 = 72 d\"","correct":"d","options":{"a":"59 ","b":"61 ","c":"63 ","d":"72","e":"67"},"options_float":{"a":59.0,"b":61.0,"c":63.0,"d":72.0,"e":67.0},"annotated_formula":"divide(add(64, 81), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"64 + 81<\/gadget>\n145<\/output>\n145 \/ 2<\/gadget>\n145\/2 = around 72.5<\/output>\n145\/2 = around 72.5<\/result>","index":748} +{"problem":"if ( a + b ) = 12 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ?","rationale":"\"given a + b = 12 = > a = 12 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 12 - b + 3 - c = > 15 - ( b + c ) = > 15 - 9 = 6 . option d . . .\"","correct":"d","options":{"a":"16 . ","b":"8 . ","c":"7 . ","d":"6 .","e":"- 2 ."},"options_float":{"a":16.0,"b":8.0,"c":7.0,"d":6.0,"e":-2.0},"annotated_formula":"subtract(add(12, 3), 9)","linear_formula":"add(n0,n2)|subtract(#0,n1)|","chain":"12 + 3<\/gadget>\n15<\/output>\n15 - 9<\/gadget>\n6<\/output>\n6<\/result>","index":749} +{"problem":"what is the perimeter of an isosceles right angled triangle . the area of the isosceles right angled triangle is 50 .","rationale":"an isosceles right angled triangle has both the base and height same . the area of an isosceles triangle = 1 \/ 2 * base * height since height = base 1 \/ 2 * height * height = 1 \/ 2 * height ^ 2 = 50 height ^ 2 = 100 height = 10 = base since we know the base and height . we will calculate the hypotenuse using pythagoras base ^ 2 + height ^ 2 = hypotenuse ^ 2 10 ^ 2 + 10 ^ 2 = hypotenuse ^ 2 hypotenuse ^ 2 = 200 hypotenuse = 14.14 perimeter of the triangle = 10 + 10 + 14.14 = 34.14 ans : e","correct":"e","options":{"a":"25 ","b":"36 ","c":"25 ","d":"30","e":"34.14"},"options_float":{"a":25.0,"b":36.0,"c":25.0,"d":30.0,"e":34.14},"annotated_formula":"triangle_perimeter(multiply(sqrt(multiply(50, const_2)), sqrt(const_2)), sqrt(multiply(50, const_2)), sqrt(multiply(50, const_2)))","linear_formula":"multiply(n0,const_2)|sqrt(const_2)|sqrt(#0)|multiply(#2,#1)|triangle_perimeter(#3,#2,#2)","chain":"50 * 2<\/gadget>\n100<\/output>\n100 ** (1\/2)<\/gadget>\n10<\/output>\n2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n10 * (sqrt(2))<\/gadget>\n10*sqrt(2) = around 14.142136<\/output>\n(10*sqrt(2)) + 10 + 10<\/gadget>\n10*sqrt(2) + 20 = around 34.142136<\/output>\n10*sqrt(2) + 20 = around 34.142136<\/result>","index":752} +{"problem":"here 10 programmers , type 10 lines within 10 minutes then 60 lines can type within 60 minutes . how many programmers are needed ?","rationale":"10 10 programmer 10 line 10 mins . 10 programmer 20 line 20 mins . . . . . 10 programmer 60 line 60 mins . answer : d","correct":"d","options":{"a":"16 ","b":"10 ","c":"6 ","d":"60","e":"50"},"options_float":{"a":16.0,"b":10.0,"c":6.0,"d":60.0,"e":50.0},"annotated_formula":"multiply(divide(60, 10), 10)","linear_formula":"divide(n3,n0)|multiply(n0,#0)","chain":"60 \/ 10<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n60<\/result>","index":753} +{"problem":"in a store , the total price for 25 shirts is $ 400 and the total price for 75 sweaters is $ 1500 . by how much does the average ( arithmetic mean ) price of a sweater exceed that of a shirt in this store ?","rationale":"the average price of a shirt is : $ 400 \/ 25 = $ 16 . the average price of a sweater is : $ 1500 \/ 75 = $ 20 . the difference in price is : $ 20 - $ 16 = $ 4 . the answer is d .","correct":"d","options":{"a":"$ 1 ","b":"$ 2 ","c":"$ 3 ","d":"$ 4","e":"$ 5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(divide(1500, 75), divide(400, 25))","linear_formula":"divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)","chain":"1_500 \/ 75<\/gadget>\n20<\/output>\n400 \/ 25<\/gadget>\n16<\/output>\n20 - 16<\/gadget>\n4<\/output>\n4<\/result>","index":755} +{"problem":"jamshid can paint a fence in 50 percent less time than taimour can when each works alone . when they work together , they can paint the fence in 6 hours . how long would it take taimour to paint the fence alone ?","rationale":"\"i believe the answer is c . please see below for explanation . if jamshid can paint a dence in 50 percent less time then taimour we can infer the following rate j = 2 t if working together they can do the job in 8 hours we can infer 1 = 2 t + t * 6 = > 1 \/ 18 working alone taimour can do the job in 1 = 1 \/ 18 * hours = > 18 answer c\"","correct":"c","options":{"a":"6 hours ","b":"8 hours ","c":"18 hours ","d":"24 hours","e":"32 hours"},"options_float":{"a":6.0,"b":8.0,"c":18.0,"d":24.0,"e":32.0},"annotated_formula":"multiply(6, const_3)","linear_formula":"multiply(n1,const_3)|","chain":"6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":756} +{"problem":"if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 800 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 820 . what is the value of x ?","rationale":"\"simple way to solve this question is to use options . from si , we know that x * y = 40,000 . now , put the value of x = 8000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 8000 + 5 % of 8000 = 8400 . 2 nd year , amount = 8400 + 5 % of 8400 = 8820 . we can see after 2 years interest = 8820 - 8000 = 820 . hence , it satisfies the question . hence a is the correct answer\"","correct":"a","options":{"a":"8000 ","b":"6000 ","c":"5000 ","d":"4000","e":"3000"},"options_float":{"a":8000.0,"b":6000.0,"c":5000.0,"d":4000.0,"e":3000.0},"annotated_formula":"divide(power(divide(800, const_2), const_2), subtract(820, 800))","linear_formula":"divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)|","chain":"800 \/ 2<\/gadget>\n400<\/output>\n400 ** 2<\/gadget>\n160_000<\/output>\n820 - 800<\/gadget>\n20<\/output>\n160_000 \/ 20<\/gadget>\n8_000<\/output>\n8_000<\/result>","index":758} +{"problem":"i had 30 donuts when i left home this morning but only 9 when i came back home . what percentage of the donuts are missing ?","rationale":"30 - 9 = 21 21 \/ 30 or 7 \/ 10 7 \/ 10 = 70 \/ 100 = 70 % answer is a","correct":"a","options":{"a":"70 ","b":"65 ","c":"55 ","d":"75","e":"60"},"options_float":{"a":70.0,"b":65.0,"c":55.0,"d":75.0,"e":60.0},"annotated_formula":"add(multiply(30, const_2), const_10)","linear_formula":"multiply(n0,const_2)|add(#0,const_10)","chain":"30 * 2<\/gadget>\n60<\/output>\n60 + 10<\/gadget>\n70<\/output>\n70<\/result>","index":759} +{"problem":"a semicircle has a radius of 10 . what is the approximate perimeter of the semicircle ?","rationale":"\"the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r \/ 2 + 2 r = pi * r + 2 r the perimeter is pi * 10 + 2 * 10 which is about 51 . the answer is e .\"","correct":"e","options":{"a":"27 ","b":"33 ","c":"39 ","d":"45","e":"51"},"options_float":{"a":27.0,"b":33.0,"c":39.0,"d":45.0,"e":51.0},"annotated_formula":"add(divide(circumface(10), const_2), multiply(const_2, 10))","linear_formula":"circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)|","chain":"2 * pi * 10<\/gadget>\n20*pi = around 62.831853<\/output>\n(20*pi) \/ 2<\/gadget>\n10*pi = around 31.415927<\/output>\n2 * 10<\/gadget>\n20<\/output>\n(10*pi) + 20<\/gadget>\n20 + 10*pi = around 51.415927<\/output>\n20 + 10*pi = around 51.415927<\/result>","index":760} +{"problem":"find the average of the first 19 natural numbers ?","rationale":"average of the first ' n ' natural numbers = ( n + 1 ) \/ 2 we get ( 19 + 1 ) \/ 2 = 10 answer : c","correct":"c","options":{"a":"8 ","b":"9 ","c":"10 ","d":"11","e":"12"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":12.0},"annotated_formula":"divide(multiply(divide(add(19, const_1), const_2), 19), 19)","linear_formula":"add(n0,const_1)|divide(#0,const_2)|multiply(n0,#1)|divide(#2,n0)","chain":"19 + 1<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10 * 19<\/gadget>\n190<\/output>\n190 \/ 19<\/gadget>\n10<\/output>\n10<\/result>","index":761} +{"problem":"if the remainder is 11 when positive integer n is divided by 18 , what is the remainder when n is divided by 9 ?","rationale":"\"assume x is quotient here , n = 18 x + 11 - - - - - - - - - - ( 1 ) and n = 9 x + ? we can also write equation ( 1 ) as : n = ( 18 x + 9 ) + 2 . ie 9 ( 2 x + 1 ) + 2 ie the first term is perfectly divisible by 6 . so , the remainder left is 2 . so , answer ( b ) is right choice .\"","correct":"b","options":{"a":"0 ","b":"2 ","c":"1 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":2.0,"c":1.0,"d":3.0,"e":4.0},"annotated_formula":"divide(18, 9)","linear_formula":"divide(n1,n2)|","chain":"18 \/ 9<\/gadget>\n2<\/output>\n2<\/result>","index":762} +{"problem":"if n = 16 × 10 ^ ( - p ) and − 4 < p < 4 , how many different integer values of p will make n a perfect square ?","rationale":"i think the answer should be c . thoose are p - values that satisfy given restriction : - 2 , 0 , 2 ( note , a fraction can be also a perfect square ) - - > 16 * 100 , 16 * 1 , 16 \/ 100","correct":"c","options":{"a":"0 ","b":"2 ","c":"3 ","d":"5","e":"7"},"options_float":{"a":0.0,"b":2.0,"c":3.0,"d":5.0,"e":7.0},"annotated_formula":"add(divide(4, const_2), const_1)","linear_formula":"divide(n2,const_2)|add(#0,const_1)","chain":"4 \/ 2<\/gadget>\n2<\/output>\n2 + 1<\/gadget>\n3<\/output>\n3<\/result>","index":764} +{"problem":"in 10 years , a will be twice as old as b was 10 years ago . if a is now 9 years older than b the present age of b is","rationale":"\"let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 9 = > 2 b - b - 9 = 30 b = 39 so the present age of b is 39 years answer : c\"","correct":"c","options":{"a":"37 ","b":"38 ","c":"39 ","d":"40","e":"41"},"options_float":{"a":37.0,"b":38.0,"c":39.0,"d":40.0,"e":41.0},"annotated_formula":"add(add(multiply(const_2, 10), 10), 9)","linear_formula":"multiply(n0,const_2)|add(n0,#0)|add(n2,#1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 10<\/gadget>\n30<\/output>\n30 + 9<\/gadget>\n39<\/output>\n39<\/result>","index":765} +{"problem":"on the first of the year , james invested x dollars at proudstar bank in an account that yields 2.5 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ?","rationale":"per quarter , interest = 2.5 % so for a year , interest = 10 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 10 % answer = 10.5 % = answer : e","correct":"e","options":{"a":"22.04 % ","b":"16.12 % ","c":"8 % ","d":"18.25 %","e":"10.5 %"},"options_float":{"a":22.04,"b":16.12,"c":8.0,"d":18.25,"e":10.5},"annotated_formula":"multiply(2.5, const_4)","linear_formula":"multiply(n0,const_4)|","chain":"2.5 * 4<\/gadget>\n10<\/output>\n10<\/result>","index":767} +{"problem":"two tains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?","rationale":"\"sol . speed of the first train = [ 120 \/ 10 ] m \/ sec = 12 m \/ sec . speed of the second train = [ 120 \/ 15 ] m \/ sec = 8 m \/ sec . relative speed = ( 12 + 8 ) = m \/ sec = 20 m \/ sec . ∴ required time = ( 120 + 120 ) \/ 20 secc = 12 sec . answer a\"","correct":"a","options":{"a":"12 ","b":"14 ","c":"16 ","d":"20","e":"18"},"options_float":{"a":12.0,"b":14.0,"c":16.0,"d":20.0,"e":18.0},"annotated_formula":"divide(add(120, 120), add(divide(120, 15), divide(120, 10)))","linear_formula":"add(n2,n2)|divide(n2,n1)|divide(n2,n0)|add(#1,#2)|divide(#0,#3)|","chain":"120 + 120<\/gadget>\n240<\/output>\n120 \/ 15<\/gadget>\n8<\/output>\n120 \/ 10<\/gadget>\n12<\/output>\n8 + 12<\/gadget>\n20<\/output>\n240 \/ 20<\/gadget>\n12<\/output>\n12<\/result>","index":769} +{"problem":"in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 82 per kg ?","rationale":"\"( 96 - 82 ) \/ ( 82 - 68 ) = 14 \/ 14 = 1 \/ 1 answer : d\"","correct":"d","options":{"a":"5 : 8 ","b":"4 : 7 ","c":"3 : 7 ","d":"1 : 1","e":"9 : 8"},"options_float":{"a":0.625,"b":0.5714285714,"c":0.4285714286,"d":1.0,"e":1.125},"annotated_formula":"divide(divide(subtract(96, 82), subtract(96, 68)), subtract(const_1, divide(subtract(96, 82), subtract(96, 68))))","linear_formula":"subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|","chain":"96 - 82<\/gadget>\n14<\/output>\n96 - 68<\/gadget>\n28<\/output>\n14 \/ 28<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) \/ (1\/2)<\/gadget>\n1<\/output>\n1<\/result>","index":772} +{"problem":"an article with cost price of 240 is sold at 20 % profit . what is the selling price ?","rationale":"\"sp = 1.20 * 240 = 288 answer : a\"","correct":"a","options":{"a":"288 ","b":"200 ","c":"204 ","d":"207","e":"210"},"options_float":{"a":288.0,"b":200.0,"c":204.0,"d":207.0,"e":210.0},"annotated_formula":"add(240, multiply(240, divide(20, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n240 * (1\/5)<\/gadget>\n48<\/output>\n240 + 48<\/gadget>\n288<\/output>\n288<\/result>","index":774} +{"problem":"a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability r that the first ball picked up is odd numbered ?","rationale":"\"answer - d selecting the balls either even or odd is having probability 50 \/ 100 = 1 \/ 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability r is 2 \/ 3 . d\"","correct":"d","options":{"a":"0 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"2 \/ 3","e":"1"},"options_float":{"a":0.0,"b":0.3333333333,"c":0.5,"d":0.6666666667,"e":1.0},"annotated_formula":"divide(const_2, 3)","linear_formula":"divide(const_2,n3)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":776} +{"problem":"a store reported total sales of $ 460 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ?","rationale":"\"new value – old value ) \/ old value x 100 we are given : february sales this year = 460 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 460 million and the old value = 320 million . let ’ s plug them into our percent change formula . ( new value – old value ) \/ old value x 100 [ ( 460 – 320 ) \/ 320 ] x 100 ≈ 44 % . the answer is c .\"","correct":"c","options":{"a":"2 % ","b":"17 % ","c":"44 % ","d":"65 %","e":"83 %"},"options_float":{"a":2.0,"b":17.0,"c":44.0,"d":65.0,"e":83.0},"annotated_formula":"multiply(divide(subtract(460, 320), 320), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|","chain":"460 - 320<\/gadget>\n140<\/output>\n140 \/ 320<\/gadget>\n7\/16 = around 0.4375<\/output>\n(7\/16) * 100<\/gadget>\n175\/4 = around 43.75<\/output>\n175\/4 = around 43.75<\/result>","index":777} +{"problem":"water boils at 212 ° f or 100 ° c and ice melts at 32 ° f or 0 ° c . if the temperature of a pot of water is 55 ° c , what is the temperature of the pot of water in ° f ?","rationale":"\"let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) \/ ( 212 - 32 ) = ( c - 0 ) \/ ( 100 - 0 ) f = 9 c \/ 5 + 32 f = 9 ( 55 ) \/ 5 + 32 = 131 ° f the answer is c .\"","correct":"c","options":{"a":"116 ° f ","b":"125 ° f ","c":"131 ° f ","d":"138 ° f","e":"145 ° f"},"options_float":{"a":116.0,"b":125.0,"c":131.0,"d":138.0,"e":145.0},"annotated_formula":"add(multiply(divide(subtract(212, 32), 100), 55), 32)","linear_formula":"subtract(n0,n2)|divide(#0,n1)|multiply(n4,#1)|add(n2,#2)|","chain":"212 - 32<\/gadget>\n180<\/output>\n180 \/ 100<\/gadget>\n9\/5 = around 1.8<\/output>\n(9\/5) * 55<\/gadget>\n99<\/output>\n99 + 32<\/gadget>\n131<\/output>\n131<\/result>","index":778} +{"problem":"what distance will be covered by a bus moving at 126 kmph in 10 seconds ?","rationale":"\"126 kmph = 126 * 5 \/ 18 = 35 mps d = speed * time = 35 * 10 = 350 m . answer : c\"","correct":"c","options":{"a":"287 ","b":"600 ","c":"350 ","d":"276","e":"207"},"options_float":{"a":287.0,"b":600.0,"c":350.0,"d":276.0,"e":207.0},"annotated_formula":"multiply(multiply(126, const_0_2778), 10)","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n126 * (5\/18)<\/gadget>\n35<\/output>\n35 * 10<\/gadget>\n350<\/output>\n350<\/result>","index":779} +{"problem":"the dimensions of a room are 10 m x 7 m x 5 m . there are 2 doors and 3 windows in the room . the dimensions of the doors are 1 m x 3 m . one window is of size 2 m x 1.5 m and the other 2 windows are of size 1 m x 1.5 m . the cost of painting the walls at rs . 3 per sq m is","rationale":"explanation : area of 4 walls = 2 ( l + b ) h = 2 ( 10 + 7 ) x 5 = 170 sq m area of 2 doors and 3 windows = 2 ( 1 x 3 ) + ( 2 x 1.5 ) + 2 ( 1 x 1.5 ) = 12 sq m area to be planted = 170 - 12 = 158 sq m cost of painting = rs . 158 x 3 = rs . 474 answer : d","correct":"d","options":{"a":"2488 ","b":"378 ","c":"739 ","d":"474","e":"291"},"options_float":{"a":2488.0,"b":378.0,"c":739.0,"d":474.0,"e":291.0},"annotated_formula":"multiply(subtract(multiply(multiply(2, add(10, 7)), 5), add(add(multiply(2, 3), multiply(2, 1.5)), multiply(2, 1.5))), 3)","linear_formula":"add(n0,n1)|multiply(n3,n4)|multiply(n3,n8)|add(#1,#2)|multiply(n3,#0)|add(#3,#2)|multiply(n2,#4)|subtract(#6,#5)|multiply(n4,#7)","chain":"10 + 7<\/gadget>\n17<\/output>\n2 * 17<\/gadget>\n34<\/output>\n34 * 5<\/gadget>\n170<\/output>\n2 * 3<\/gadget>\n6<\/output>\n2 * 1.5<\/gadget>\n3<\/output>\n6 + 3<\/gadget>\n9<\/output>\n9 + 3<\/gadget>\n12<\/output>\n170 - 12<\/gadget>\n158<\/output>\n158 * 3<\/gadget>\n474<\/output>\n474<\/result>","index":780} +{"problem":"o ( x ) represents the least odd integer greater than x , whereas o ( x ) represents the greatest odd integer less than x . likewise , e ( x ) represents the least even integer greater than x , whereas e ( x ) represents the greatest even integer less than x . according to these definitions , the value of o ( 5.3 ) + e ( – 6.7 ) + o ( – 7.3 ) + e ( 6.6 ) is :","rationale":"o ( 5.3 ) + e ( – 6.7 ) + o ( – 7.3 ) + e ( 6.6 ) = 7 + ( - 6 ) + ( - 9 ) + 6 = - 2 the answer is b .","correct":"b","options":{"a":"- 4 ","b":"- 2 ","c":"0 ","d":"2","e":"4"},"options_float":{"a":-4.0,"b":-2.0,"c":0.0,"d":2.0,"e":4.0},"annotated_formula":"add(add(add(multiply(5.3, const_1), multiply(negate(6.7), const_1)), negate(7.3)), multiply(6.6, const_1))","linear_formula":"multiply(n0,const_1)|multiply(n3,const_1)|negate(n1)|negate(n2)|multiply(#2,const_1)|add(#0,#4)|add(#5,#3)|add(#6,#1)","chain":"5.3 * 1<\/gadget>\n5.3<\/output>\n-6.7<\/gadget>\n-6.7<\/output>\n(-6.7) * 1<\/gadget>\n-6.7<\/output>\n5.3 + (-6.7)<\/gadget>\n-1.4<\/output>\n-7.3<\/gadget>\n-7.3<\/output>\n(-1.4) + (-7.3)<\/gadget>\n-8.7<\/output>\n6.6 * 1<\/gadget>\n6.6<\/output>\n(-8.7) + 6.6<\/gadget>\n-2.1<\/output>\n-2.1<\/result>","index":781} +{"problem":"a quadratic function f ( x ) attains a maximum of 3 at x = 1 . the value of the function at x = 0 is 1 . what is the value of f ( x ) at x = 10 ?","rationale":"explanation : let , the quadratic equation be ax 2 + bx + c . so , f ( x ) = ax 2 + bx + c at x = 0 , the value of function is 1 . x = 0 , f ( x ) = 1 ax 2 + bx + c = a * 0 + b * 0 + c = c c = 1 . at x = 1 , f ( x ) = 3 x = 1 , f ( x ) = 3 a * 1 + b * 1 + c = 3 since c = 1 , a + b = 2 . also , we know that f ( x ) is maximum when x = 1 . if f ( x ) is maximum , ( dx \/ dt ) ( f ( x ) ) = 0 differentiating f ( x ) , we have d \/ dt ( ax 2 + bx + c ) = 2 ax + b at x = 1 , 2 ax + b = 0 . 2 a + b = 0 . b = - 2 a . substituting we have a + b = 2 , or a + - 2 a = 2 . a = - 2 . so , b = 4 . so the equation is - 2 x 2 + 4 x + 1 . at x = 10 , the value is - 2 * 100 + 4 * 10 + 1 i . e - 159 . answer : b","correct":"b","options":{"a":"- 119 ","b":"- 159 ","c":"- 110 ","d":"- 180","e":"- 105"},"options_float":{"a":-119.0,"b":-159.0,"c":-110.0,"d":-180.0,"e":-105.0},"annotated_formula":"add(add(multiply(power(10, const_2), negate(subtract(3, 1))), multiply(subtract(subtract(3, 1), negate(subtract(3, 1))), 10)), 1)","linear_formula":"power(n4,const_2)|subtract(n0,n1)|negate(#1)|multiply(#2,#0)|subtract(#1,#2)|multiply(n4,#4)|add(#3,#5)|add(n1,#6)","chain":"10 ** 2<\/gadget>\n100<\/output>\n3 - 1<\/gadget>\n2<\/output>\n-2<\/gadget>\n-2<\/output>\n100 * (-2)<\/gadget>\n-200<\/output>\n2 - (-2)<\/gadget>\n4<\/output>\n4 * 10<\/gadget>\n40<\/output>\n(-200) + 40<\/gadget>\n-160<\/output>\n(-160) + 1<\/gadget>\n-159<\/output>\n-159<\/result>","index":782} +{"problem":"in a certain company , a third of the workers do not have a retirement plan . 40 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ?","rationale":"\"set up equation : x = total number of workers 120 = 0,4 * 2 \/ 3 * x + 0,6 * 1 \/ 3 * x 120 = 14 \/ 30 x x = 257 257 - 120 = 137 answer e\"","correct":"e","options":{"a":"80 ","b":"95 ","c":"105 ","d":"120","e":"137"},"options_float":{"a":80.0,"b":95.0,"c":105.0,"d":120.0,"e":137.0},"annotated_formula":"multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(40, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(40, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))))","linear_formula":"divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/3) * (2\/5)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(1\/3) - (2\/15)<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * (2\/5)<\/gadget>\n4\/15 = around 0.266667<\/output>\n(1\/5) + (4\/15)<\/gadget>\n7\/15 = around 0.466667<\/output>\n120 \/ (7\/15)<\/gadget>\n1_800\/7 = around 257.142857<\/output>\n(2\/3) - (4\/15)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/15) + (2\/5)<\/gadget>\n8\/15 = around 0.533333<\/output>\n(1_800\/7) * (8\/15)<\/gadget>\n960\/7 = around 137.142857<\/output>\n960\/7 = around 137.142857<\/result>","index":783} +{"problem":"the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 48.0 billion from $ 72.0 billion , a year ago . by what percent did the revenue fall ?","rationale":"\"$ 72 - $ 48 = 24 $ ( 24 \/ 72 ) * 100 = 33.33 % answer : e\"","correct":"e","options":{"a":"20.8 ","b":"30.4 ","c":"31.8 ","d":"32.5","e":"33.3"},"options_float":{"a":20.8,"b":30.4,"c":31.8,"d":32.5,"e":33.3},"annotated_formula":"multiply(divide(subtract(72.0, 48.0), 72.0), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)|","chain":"72 - 48<\/gadget>\n24<\/output>\n24 \/ 72<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":784} +{"problem":"a trader bought a car at 10 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ?","rationale":"\"original price = 100 cp = 90 s = 90 * ( 180 \/ 100 ) = 162 100 - 162 = 62 % answer : b\"","correct":"b","options":{"a":"17 % ","b":"62 % ","c":"12 % ","d":"19 %","e":"22 %"},"options_float":{"a":17.0,"b":62.0,"c":12.0,"d":19.0,"e":22.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 10), add(const_100, 80)), const_100), const_100), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"100 - 10<\/gadget>\n90<\/output>\n100 + 80<\/gadget>\n180<\/output>\n90 * 180<\/gadget>\n16_200<\/output>\n16_200 \/ 100<\/gadget>\n162<\/output>\n162 \/ 100<\/gadget>\n81\/50 = around 1.62<\/output>\n(81\/50) - 1<\/gadget>\n31\/50 = around 0.62<\/output>\n(31\/50) * 100<\/gadget>\n62<\/output>\n62<\/result>","index":785} +{"problem":"a man buys a cycle for rs . 2800 and sells it at a loss of 25 % . what is the selling price of the cycle ?","rationale":"\"s . p . = 75 % of rs . 2800 = rs . 75 x 2800 \/ 100 = rs . 1190 answer : option c\"","correct":"c","options":{"a":"s . 2090 ","b":"s . 2160 ","c":"s . 2100 ","d":"s . 2202","e":"s . 2092"},"options_float":{"a":2090.0,"b":2160.0,"c":2100.0,"d":2202.0,"e":2092.0},"annotated_formula":"divide(multiply(subtract(const_100, 25), 2800), const_100)","linear_formula":"subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|","chain":"100 - 25<\/gadget>\n75<\/output>\n75 * 2_800<\/gadget>\n210_000<\/output>\n210_000 \/ 100<\/gadget>\n2_100<\/output>\n2_100<\/result>","index":787} +{"problem":"the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 25 percent of a , and m equals b decreased by 40 percent of b , what is the value of m \/ x ?","rationale":"a \/ b = 4 \/ 5 m \/ x = ( 3 \/ 5 ) * 5 \/ ( 5 \/ 4 ) * 4 = 3 \/ 5 the answer is e .","correct":"e","options":{"a":"2 \/ 5 ","b":"3 \/ 4 ","c":"4 \/ 5 ","d":"5 \/ 4","e":"3 \/ 5"},"options_float":{"a":0.4,"b":0.75,"c":0.8,"d":1.25,"e":0.6},"annotated_formula":"multiply(divide(subtract(const_100, 40), add(const_100, 25)), divide(5, 4))","linear_formula":"add(n2,const_100)|divide(n1,n0)|subtract(const_100,n3)|divide(#2,#0)|multiply(#3,#1)","chain":"100 - 40<\/gadget>\n60<\/output>\n100 + 25<\/gadget>\n125<\/output>\n60 \/ 125<\/gadget>\n12\/25 = around 0.48<\/output>\n5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n(12\/25) * (5\/4)<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":789} +{"problem":"the area of a triangle is with base 10 m and height 5 m ?","rationale":"\"1 \/ 2 * 10 * 5 = 25 m 2 answer : b\"","correct":"b","options":{"a":"11 ","b":"25 ","c":"787 ","d":"122","e":"12"},"options_float":{"a":11.0,"b":25.0,"c":787.0,"d":122.0,"e":12.0},"annotated_formula":"triangle_area(10, 5)","linear_formula":"triangle_area(n0,n1)|","chain":"(10 * 5) \/ 2<\/gadget>\n25<\/output>\n25<\/result>","index":791} +{"problem":"how many diagonals does a polygon with 13 sides have , if one of its vertices does not connect to any diagonal ?","rationale":"\"if i calculate it using the formulae , # diagonals = n ( n - 3 ) \/ 2 each vertex sends of n - 3 diagonals n = 13 - 1 then 12 * ( 12 - 3 ) \/ 2 = 54 correct option : d\"","correct":"d","options":{"a":"65 ","b":"62 ","c":"58 ","d":"54","e":"60"},"options_float":{"a":65.0,"b":62.0,"c":58.0,"d":54.0,"e":60.0},"annotated_formula":"divide(multiply(subtract(13, const_1), subtract(subtract(13, const_1), const_3)), const_2)","linear_formula":"subtract(n0,const_1)|subtract(#0,const_3)|multiply(#0,#1)|divide(#2,const_2)|","chain":"13 - 1<\/gadget>\n12<\/output>\n12 - 3<\/gadget>\n9<\/output>\n12 * 9<\/gadget>\n108<\/output>\n108 \/ 2<\/gadget>\n54<\/output>\n54<\/result>","index":793} +{"problem":"a car averages 30 miles per hour for the first 5 hours of a trip and averages 42 miles per hour for each additional hour of travel time . if the average speed for the entire trip is 38 miles per hour , how many hours long is the trip ?","rationale":"\"let t be the total time of the trip . 30 * 5 + 42 ( t - 5 ) = 38 t 4 t = 210 - 150 t = 15 the answer is c .\"","correct":"c","options":{"a":"13 ","b":"14 ","c":"15 ","d":"16","e":"17"},"options_float":{"a":13.0,"b":14.0,"c":15.0,"d":16.0,"e":17.0},"annotated_formula":"add(divide(subtract(multiply(38, 5), multiply(30, 5)), subtract(42, 38)), 5)","linear_formula":"multiply(n1,n3)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)|","chain":"38 * 5<\/gadget>\n190<\/output>\n30 * 5<\/gadget>\n150<\/output>\n190 - 150<\/gadget>\n40<\/output>\n42 - 38<\/gadget>\n4<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15<\/result>","index":794} +{"problem":"in a certain country 1 \/ 5 of 8 = 4 . assuming the same proportion , what would be the value of 1 \/ 4 of 16 ?","rationale":"c 10","correct":"c","options":{"a":"13 ","b":"12 ","c":"10 ","d":"16","e":"14"},"options_float":{"a":13.0,"b":12.0,"c":10.0,"d":16.0,"e":14.0},"annotated_formula":"multiply(divide(5, 8), 16)","linear_formula":"divide(n1,n2)|multiply(n6,#0)","chain":"5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n(5\/8) * 16<\/gadget>\n10<\/output>\n10<\/result>","index":795} +{"problem":"in what ratio must a grocer mix two varieties of pulses costing rs . 15 and rs . 20 per kg respectively so as to get a mixture worth rs . 16.50 per kg ?","rationale":"solution required ratio = 3.50 : 1.50 = 35 : 15 = 7 : 3 . answer c","correct":"c","options":{"a":"3 : 7 ","b":"5 : 7 ","c":"7 : 3 ","d":"7 : 5","e":"none"},"options_float":{"a":0.4285714286,"b":0.7142857143,"c":2.3333333333,"d":1.4,"e":null},"annotated_formula":"divide(subtract(20, 16.5), subtract(16.5, 15))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)","chain":"20 - 16.5<\/gadget>\n3.5<\/output>\n16.5 - 15<\/gadget>\n1.5<\/output>\n3.5 \/ 1.5<\/gadget>\n2.333333<\/output>\n2.333333<\/result>","index":796} +{"problem":"a can do a piece of work in 4 days . b can do it in 10 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ?","rationale":"\"c = 1 \/ 2 - 1 \/ 4 - 1 \/ 10 = 3 \/ 20 = > 20 \/ 3 days answer : b\"","correct":"b","options":{"a":"22 days ","b":"20 \/ 3 days ","c":"67 days ","d":"17 \/ 6 days","e":"18 days"},"options_float":{"a":22.0,"b":6.6666666667,"c":67.0,"d":2.8333333333,"e":18.0},"annotated_formula":"divide(multiply(4, 10), divide(subtract(multiply(4, 10), multiply(add(divide(multiply(4, 10), 4), divide(multiply(4, 10), 10)), 2)), 2))","linear_formula":"multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)|","chain":"4 * 10<\/gadget>\n40<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n10 + 4<\/gadget>\n14<\/output>\n14 * 2<\/gadget>\n28<\/output>\n40 - 28<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n40 \/ 6<\/gadget>\n20\/3 = around 6.666667<\/output>\n20\/3 = around 6.666667<\/result>","index":797} +{"problem":"a and b invests rs . 1500 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?","rationale":"\"( 1.5 * 6 + 3 * 6 ) : ( 4 * 12 ) 27 : 48 = > 9 : 16 answer : e\"","correct":"e","options":{"a":"9 : 5 ","b":"9 : 8 ","c":"9 : 2 ","d":"9 : 9","e":"9 : 16"},"options_float":{"a":1.8,"b":1.125,"c":4.5,"d":1.0,"e":0.5625},"annotated_formula":"divide(add(multiply(1500, 6), multiply(multiply(1500, const_2), 6)), multiply(4000, add(6, 6)))","linear_formula":"add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)|","chain":"1_500 * 6<\/gadget>\n9_000<\/output>\n1_500 * 2<\/gadget>\n3_000<\/output>\n3_000 * 6<\/gadget>\n18_000<\/output>\n9_000 + 18_000<\/gadget>\n27_000<\/output>\n6 + 6<\/gadget>\n12<\/output>\n4_000 * 12<\/gadget>\n48_000<\/output>\n27_000 \/ 48_000<\/gadget>\n9\/16 = around 0.5625<\/output>\n9\/16 = around 0.5625<\/result>","index":798} +{"problem":"the length of a rectangular garden is 2 feet longer than 3 times its width . if the perimeter of the garden is 100 feet , find the length of the garden .","rationale":"let l and w be the length and width of the garden . the statement ` ` the length of a rectangular garden is 2 feet longer than 3 times its width ' ' may be formulated by l = 2 + 3 w the formula for the perimeter is given by p = 2 l + 2 w substitute p and l in the above equation by 100 and 2 + 3 w respectively to obtain 100 = 2 ( 2 + 3 w ) + 2 w solve for w and l w = 12 and l = 2 + 3 w = 38 . check that the perimeter of the rectangular garden is 100 p = 2 l + 2 w = 76 + 24 = 100 correct answer b","correct":"b","options":{"a":"42 ","b":"38 ","c":"28 ","d":"57","e":"49"},"options_float":{"a":42.0,"b":38.0,"c":28.0,"d":57.0,"e":49.0},"annotated_formula":"add(multiply(divide(subtract(100, const_4), multiply(2, const_4)), 3), 2)","linear_formula":"multiply(n0,const_4)|subtract(n2,const_4)|divide(#1,#0)|multiply(n1,#2)|add(n0,#3)","chain":"100 - 4<\/gadget>\n96<\/output>\n2 * 4<\/gadget>\n8<\/output>\n96 \/ 8<\/gadget>\n12<\/output>\n12 * 3<\/gadget>\n36<\/output>\n36 + 2<\/gadget>\n38<\/output>\n38<\/result>","index":799} +{"problem":"calculate 24 x 99","rationale":"\"= 24 - 1 = 23 = ( decrement each digit of the number obtained from 9 ) here we got 23 . now = 9 - 2 = 7 and 9 - 3 = 6 . so we have 76 just write these numbers together . that is we have 23 and 76 . hence answer is 2376 answer is e .\"","correct":"e","options":{"a":"2673 ","b":"2763 ","c":"2637 ","d":"2736","e":"2376"},"options_float":{"a":2673.0,"b":2763.0,"c":2637.0,"d":2736.0,"e":2376.0},"annotated_formula":"multiply(24, 99)","linear_formula":"multiply(n0,n1)|","chain":"24 * 99<\/gadget>\n2_376<\/output>\n2_376<\/result>","index":800} +{"problem":"a reduction of 40 % in the price of oil enables a house wife to obtain 8 kgs more for rs . 2400 , what is the reduced price for kg ?","rationale":"\"explanation : 2400 * ( 40 \/ 100 ) = 960 - - - - 8 ? - - - - 1 = > rs . 120 answer : a\"","correct":"a","options":{"a":"rs . 120 ","b":"rs . 125 ","c":"rs . 130 ","d":"rs . 135","e":"rs . 140"},"options_float":{"a":120.0,"b":125.0,"c":130.0,"d":135.0,"e":140.0},"annotated_formula":"divide(divide(multiply(2400, 40), const_100), 8)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|","chain":"2_400 * 40<\/gadget>\n96_000<\/output>\n96_000 \/ 100<\/gadget>\n960<\/output>\n960 \/ 8<\/gadget>\n120<\/output>\n120<\/result>","index":802} +{"problem":"a trail mix company keeps costs down by employing the peanuts : cashews : almonds ratio of 10 : 4 : 1 in each bag of up to 75 total nuts . what is the maximum percentage by which the company could decrease its number of peanuts per bag and still have peanuts constitute more than half the total amount of nuts ?","rationale":"peanuts = 50 cashews = 20 almonds = 5 we want to remove as many peanuts as possible while still having peanuts represent more than half of the mixture . . . the number of cashews and almonds will stay the same though , so we have 20 + 5 = 25 of those non - peanuts in total . if we had 25 peanuts and 25 non - peanuts , then that would be 50 % exactly . we want more than 50 % though , so we need to add in 1 more peanut . this gives us . . . peanuts = 26 cashews = 20 almonds = 5 the question asked for the decrease in the number of peanuts as a percentage . we started with 50 peanuts and removed 24 = 24 \/ 50 = 48 % ; answer : b","correct":"b","options":{"a":"40 % ","b":"48 % ","c":"49 % ","d":"50 %","e":"58 %"},"options_float":{"a":40.0,"b":48.0,"c":49.0,"d":50.0,"e":58.0},"annotated_formula":"multiply(divide(75, add(add(10, 4), 1)), 10)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n0,#2)","chain":"10 + 4<\/gadget>\n14<\/output>\n14 + 1<\/gadget>\n15<\/output>\n75 \/ 15<\/gadget>\n5<\/output>\n5 * 10<\/gadget>\n50<\/output>\n50<\/result>","index":803} +{"problem":"we bought a total of 90 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 390 . how many math books did we buy ?","rationale":"\"m + h = 90 h = 90 - m 4 m + 5 h = 390 4 m + 5 * ( 90 - m ) = 390 m = 60 the answer is b .\"","correct":"b","options":{"a":"57 ","b":"60 ","c":"63 ","d":"66","e":"69"},"options_float":{"a":57.0,"b":60.0,"c":63.0,"d":66.0,"e":69.0},"annotated_formula":"subtract(90, subtract(390, multiply(90, 4)))","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|subtract(n0,#1)|","chain":"90 * 4<\/gadget>\n360<\/output>\n390 - 360<\/gadget>\n30<\/output>\n90 - 30<\/gadget>\n60<\/output>\n60<\/result>","index":805} +{"problem":"a number when divided by 133 leaves 16 as a remainder . what will be the remainder if the number is divided by 50 ?","rationale":"\"133 + 16 = 149 dividing 149 evenly into whole numbers is 2 and leaves 49 remaining . the correct answer is c .\"","correct":"c","options":{"a":"51 ","b":"56 ","c":"49 ","d":"17","e":"7"},"options_float":{"a":51.0,"b":56.0,"c":49.0,"d":17.0,"e":7.0},"annotated_formula":"reminder(add(133, 16), 50)","linear_formula":"add(n0,n1)|reminder(#0,n2)|","chain":"133 + 16<\/gadget>\n149<\/output>\n149 % 50<\/gadget>\n49<\/output>\n49<\/result>","index":807} +{"problem":"the length of a rectangular floor is more than its breadth by 200 % . if rs . 240 is required to paint the floor at the rate of rs . 3 per sq m , then what would be the length of the floor ?","rationale":"\"let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 240 \/ 3 = 80 sq m l b = 80 i . e . , l * l \/ 3 = 240 l 2 = 240 = > l = 15.49 answer : d\"","correct":"d","options":{"a":"65 ","b":"44 ","c":"18 ","d":"15.49","e":"14"},"options_float":{"a":65.0,"b":44.0,"c":18.0,"d":15.49,"e":14.0},"annotated_formula":"multiply(sqrt(divide(divide(240, 3), const_3)), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)|","chain":"240 \/ 3<\/gadget>\n80<\/output>\n80 \/ 3<\/gadget>\n80\/3 = around 26.666667<\/output>\n(80\/3) ** (1\/2)<\/gadget>\n4*sqrt(15)\/3 = around 5.163978<\/output>\n(4*sqrt(15)\/3) * 3<\/gadget>\n4*sqrt(15) = around 15.491933<\/output>\n4*sqrt(15) = around 15.491933<\/result>","index":808} +{"problem":"if the cost price of 50 articles is equal to the selling price of 40 articles , then the gain or loss percent is ?","rationale":"\"c 25 % given that , cost price of 50 article is equal to selling price of 40 articles . let cost price of one article = rs . 1 selling price of 40 articles = rs . 50 but cost price of 40 articles = rs . 40 therefore , the trader made profit . \\ percentage of profit = 10 \/ 40 * 100 = 25 %\"","correct":"c","options":{"a":"33 % ","b":"29 % ","c":"25 % ","d":"55 %","e":"39 %"},"options_float":{"a":33.0,"b":29.0,"c":25.0,"d":55.0,"e":39.0},"annotated_formula":"multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 40), 50)), divide(multiply(const_100, 40), 50)))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 * 40<\/gadget>\n4_000<\/output>\n4_000 \/ 50<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n20 \/ 80<\/gadget>\n1\/4 = around 0.25<\/output>\n100 * (1\/4)<\/gadget>\n25<\/output>\n25<\/result>","index":813} +{"problem":"if 1 \/ ( x + 1 ) + 1 \/ ( x - 1 ) = 1 \/ ( x - 1 ) , what is the value of x ?","rationale":"if we solve the question , we get x = 1 . option : c","correct":"c","options":{"a":"2 ","b":"- 2 ","c":"1 ","d":"- 1","e":"5"},"options_float":{"a":2.0,"b":-2.0,"c":1.0,"d":-1.0,"e":5.0},"annotated_formula":"subtract(add(1, 1), 1)","linear_formula":"add(n0,n0)|subtract(#0,n0)","chain":"1 + 1<\/gadget>\n2<\/output>\n2 - 1<\/gadget>\n1<\/output>\n1<\/result>","index":817} +{"problem":"evaluate 100 ^ 2 - 99 ^ 2 + 98 ^ 2 - 97 ^ 2 + 96 ^ 2 - 95 ^ 2 + . . . + 2 ^ 2 - 1 ^ 2 = ?","rationale":"b 5050 you must have noticed that 50 pairs of n ^ 2 – ( n - 1 ) ^ 2 exist . n ^ 2 – ( n - 1 ) ^ 2 = n + ( n – 1 ) in such case , we may as well write 100 ^ 2 - 99 ^ 2 + 98 ^ 2 - 97 ^ 2 + 96 ^ 2 - 95 ^ 2 + . . . + 2 ^ 2 - 1 ^ 2 as 100 + 99 + 98 + . . . + 2 + 1 = ( 100 x 101 ) \/ 2 = 5050","correct":"b","options":{"a":"3043 ","b":"5050 ","c":"5176 ","d":"9356","e":"4890"},"options_float":{"a":3043.0,"b":5050.0,"c":5176.0,"d":9356.0,"e":4890.0},"annotated_formula":"divide(multiply(100, add(100, const_1)), const_2)","linear_formula":"add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)","chain":"100 + 1<\/gadget>\n101<\/output>\n100 * 101<\/gadget>\n10_100<\/output>\n10_100 \/ 2<\/gadget>\n5_050<\/output>\n5_050<\/result>","index":818} +{"problem":"the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years r in which the population would be doubled assuming there is no emigration or immigration is","rationale":"\"suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 \/ 100 ) ) ^ n n = 35 answer is d .\"","correct":"d","options":{"a":"20 ","b":"r = 25 ","c":"r = 30 ","d":"r = 35","e":"40"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100))","linear_formula":"multiply(const_10,const_3)|subtract(n1,n2)|divide(#1,n0)|subtract(const_100,#0)|multiply(#2,const_100)|divide(#3,#4)|","chain":"10 * 3<\/gadget>\n30<\/output>\n100 - 30<\/gadget>\n70<\/output>\n39.4 - 19.4<\/gadget>\n20<\/output>\n20 \/ 1_000<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 100<\/gadget>\n2<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n35<\/result>","index":820} +{"problem":"there are 15 balls in a jar . you take out 3 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 \/ 3 . how many blue balls were there in the beginning ?","rationale":"\"7 = 3 blue balls + 12 \/ 3 answer : d\"","correct":"d","options":{"a":"12 . ","b":"9 . ","c":"8 . ","d":"7 .","e":"6 ."},"options_float":{"a":12.0,"b":9.0,"c":8.0,"d":7.0,"e":6.0},"annotated_formula":"add(divide(subtract(15, 3), 3), 3)","linear_formula":"subtract(n0,n1)|divide(#0,n3)|add(n1,#1)|","chain":"15 - 3<\/gadget>\n12<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n4 + 3<\/gadget>\n7<\/output>\n7<\/result>","index":821} +{"problem":"the average ( arithmetic mean ) of 8 numbers is 43.1 . if the sum of half of these numbers is 158.4 , what is the average of the other half ?","rationale":"arithmetic mean = sum \/ total numbers sum = 43.1 * 8 = 344.8 sum of half of these numbers is 158.4 . so , 4 numbers sum is 158.4 . rest 4 numbers sum = 344.8 - 158.4 = 186.4 arithmetic mean of the 4 nos = 186.4 \/ 4 = 46.6 hence , c is the answer .","correct":"c","options":{"a":"12.8 ","b":"24.2 ","c":"46.6 ","d":"72.1","e":"96.8"},"options_float":{"a":12.8,"b":24.2,"c":46.6,"d":72.1,"e":96.8},"annotated_formula":"divide(subtract(multiply(8, 43.1), 158.4), divide(8, const_2))","linear_formula":"divide(n0,const_2)|multiply(n0,n1)|subtract(#1,n2)|divide(#2,#0)","chain":"8 * 43.1<\/gadget>\n344.8<\/output>\n344.8 - 158.4<\/gadget>\n186.4<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n186.4 \/ 4<\/gadget>\n46.6<\/output>\n46.6<\/result>","index":822} +{"problem":"the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 45 kms . what was the total distance traveled in 12 hours ?","rationale":"explanation : total distance travelled in 12 hours = ( 45 + 47 + 49 + . . . . . upto 12 terms ) this is an a . p with first term , a = 45 , number of terms , n = 12 , d = 2 . required distance = 12 \/ 2 [ 2 x 45 + { 12 - 1 ) x 2 ] = 6 ( 112 ) = 672 kms . answer : c","correct":"c","options":{"a":"252 kms ","b":"152 kms ","c":"672 kms ","d":"752 kms","e":"152 kms"},"options_float":{"a":252.0,"b":152.0,"c":672.0,"d":752.0,"e":152.0},"annotated_formula":"multiply(add(multiply(2, 45), multiply(subtract(12, const_1), 2)), divide(12, 2))","linear_formula":"divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)","chain":"2 * 45<\/gadget>\n90<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n90 + 22<\/gadget>\n112<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n112 * 6<\/gadget>\n672<\/output>\n672<\/result>","index":826} +{"problem":"n denotes the number obtained when n is rounded to the nearest tenth . for example 4.31 = 4.3 0.089 - 1.135 =","rationale":"here let the first quantity be a = > a rounded to nearest tenths = > 0.1 similarly let the second quantity be b = > b rounded to nearest tenths = > 1.1 = > a - b = 0.1 - 1.1 = - 1 hence a - b = - 1 answer : d","correct":"d","options":{"a":"1.05 ","b":"1.04 ","c":"- 1.05 ","d":"- 1.0","e":"- 0.1"},"options_float":{"a":1.05,"b":1.04,"c":-1.05,"d":-1.0,"e":-0.1},"annotated_formula":"subtract(0.089, 1.135)","linear_formula":"subtract(n2,n3)","chain":"0.089 - 1.135<\/gadget>\n-1.046<\/output>\n-1.046<\/result>","index":827} +{"problem":"initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many w women are currently in the room ?","rationale":"\"the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d\"","correct":"d","options":{"a":"12 ","b":"14 ","c":"15 ","d":"24","e":"36"},"options_float":{"a":12.0,"b":14.0,"c":15.0,"d":24.0,"e":36.0},"annotated_formula":"multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3))","linear_formula":"subtract(n4,n2)|multiply(n1,#0)|divide(#1,n0)|subtract(#2,n3)|multiply(n2,#3)|","chain":"14 - 2<\/gadget>\n12<\/output>\n5 * 12<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15 - 3<\/gadget>\n12<\/output>\n2 * 12<\/gadget>\n24<\/output>\n24<\/result>","index":829} +{"problem":"a certain protective additive increases from 30 days to 45 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ?","rationale":"general formula for percent increase or decrease , ( percent change ) : percent = change \/ original ∗ 100 so , the time between maintenance checks increased by 45 − 30 \/ 30 ∗ 100 = 50 answer : c .","correct":"c","options":{"a":"25 % ","b":"33 1 \/ 3 % ","c":"50 % ","d":"66 2 \/ 3 %","e":"75 %"},"options_float":{"a":25.0,"b":33.0,"c":50.0,"d":66.0,"e":75.0},"annotated_formula":"multiply(divide(subtract(45, 30), 30), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)","chain":"45 - 30<\/gadget>\n15<\/output>\n15 \/ 30<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":831} +{"problem":"a bus trip of 660 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ?","rationale":"\"the time is the distance \/ speed . the time difference is 1 hour . 660 \/ v - 660 \/ ( v + 5 ) = 1 660 ( v + 5 ) - 660 v = ( v ) ( v + 5 ) 3300 = ( v ) ( v + 5 ) 55 * 60 = ( v ) ( v + 5 ) v = 55 mph the answer is d .\"","correct":"d","options":{"a":"40 ","b":"45 ","c":"50 ","d":"55","e":"60"},"options_float":{"a":40.0,"b":45.0,"c":50.0,"d":55.0,"e":60.0},"annotated_formula":"divide(subtract(sqrt(add(multiply(multiply(660, 5), const_4), power(5, const_2))), 5), const_2)","linear_formula":"multiply(n0,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)|","chain":"660 * 5<\/gadget>\n3_300<\/output>\n3_300 * 4<\/gadget>\n13_200<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n13_200 + 25<\/gadget>\n13_225<\/output>\n13_225 ** (1\/2)<\/gadget>\n115<\/output>\n115 - 5<\/gadget>\n110<\/output>\n110 \/ 2<\/gadget>\n55<\/output>\n55<\/result>","index":833} +{"problem":"in an examination , 300 students appeared . out of these students ; 30 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed .","rationale":"the number of students with first division = 30 % of 300 = 30 \/ 100 × 300 = 9000 \/ 100 = 90 and , the number of students with second division = 54 % of 300 = 54 \/ 100 × 300 = 16200 \/ 100 = 162 therefore , the number of students who just passed = 300 – ( 90 + 162 ) = 48 answer : e","correct":"e","options":{"a":"33 ","b":"37 ","c":"54 ","d":"99","e":"48"},"options_float":{"a":33.0,"b":37.0,"c":54.0,"d":99.0,"e":48.0},"annotated_formula":"divide(multiply(300, subtract(const_100, add(30, 54))), const_100)","linear_formula":"add(n1,n2)|subtract(const_100,#0)|multiply(n0,#1)|divide(#2,const_100)","chain":"30 + 54<\/gadget>\n84<\/output>\n100 - 84<\/gadget>\n16<\/output>\n300 * 16<\/gadget>\n4_800<\/output>\n4_800 \/ 100<\/gadget>\n48<\/output>\n48<\/result>","index":835} +{"problem":"5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 45 only . men â € ™ s wages are ?","rationale":"\"5 m = xw = 8 b 5 m + xw + 8 b - - - - - 45 rs . 5 m + 5 m + 5 m - - - - - 45 rs . 15 m - - - - - - 45 rs . = > 1 m = 3 rs . answer : e\"","correct":"e","options":{"a":"6 rs ","b":"7 rs ","c":"8 rs ","d":"4 rs","e":"3 rs"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":4.0,"e":3.0},"annotated_formula":"divide(45, multiply(const_3, 5))","linear_formula":"multiply(n0,const_3)|divide(n2,#0)|","chain":"3 * 5<\/gadget>\n15<\/output>\n45 \/ 15<\/gadget>\n3<\/output>\n3<\/result>","index":837} +{"problem":"two dice are rolled . what is the probability that the sum on the top face of both the dice will be greater than 9 ?","rationale":"total number of outcomes possible when a die is rolled = 6 ( ∵ any one face out of the 6 faces ) hence , total number of outcomes possible two dice are rolled , n ( s ) = 6 × 6 = 36 e = getting a sum greater than 9 when the two dice are rolled = { ( 4 , 6 ) , { 5 , 5 } , { 5 , 6 } , { 6 , 4 } , { 6 , 5 } , ( 6 , 6 ) } hence , n ( e ) = 6 p ( e ) = n ( e ) \/ n ( s ) = 6 \/ 36 = 1 \/ 6","correct":"e","options":{"a":"3 \/ 8 ","b":"2 \/ 3 ","c":"1 \/ 2 ","d":"1 \/ 3","e":"1 \/ 6"},"options_float":{"a":0.375,"b":0.6666666667,"c":0.5,"d":0.3333333333,"e":0.1666666667},"annotated_formula":"multiply(multiply(divide(const_1, multiply(const_2, const_3)), divide(const_1, multiply(const_2, const_3))), multiply(const_2, const_3))","linear_formula":"multiply(const_2,const_3)|divide(const_1,#0)|multiply(#1,#1)|multiply(#2,#0)","chain":"2 * 3<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * (1\/6)<\/gadget>\n1\/36 = around 0.027778<\/output>\n(1\/36) * 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":839} +{"problem":"if x is 12 percent greater than 80 , then x =","rationale":"\"12 % of 80 = ( 80 * 0.12 ) = 9.6 12 % greater than 80 = 80 + 9.6 = 89.6 answer is clearly a .\"","correct":"a","options":{"a":"89.6 ","b":"91.0 ","c":"88.0 ","d":"70.9","e":"71.2"},"options_float":{"a":89.6,"b":91.0,"c":88.0,"d":70.9,"e":71.2},"annotated_formula":"add(80, multiply(divide(12, const_100), 80))","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|","chain":"12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) * 80<\/gadget>\n48\/5 = around 9.6<\/output>\n80 + (48\/5)<\/gadget>\n448\/5 = around 89.6<\/output>\n448\/5 = around 89.6<\/result>","index":841} +{"problem":"a and b complete a work in 6 days . a alone can do it in 8 days . if both together can do the work in how many days ?","rationale":"\"1 \/ 6 + 1 \/ 8 = 7 \/ 24 24 \/ 7 = 3.43 days answer : b\"","correct":"b","options":{"a":"3.75 days ","b":"3.43 days ","c":"2.99 days ","d":"2.98 days","e":"2.44 days"},"options_float":{"a":3.75,"b":3.43,"c":2.99,"d":2.98,"e":2.44},"annotated_formula":"inverse(add(inverse(6), inverse(8)))","linear_formula":"inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/6) + (1\/8)<\/gadget>\n7\/24 = around 0.291667<\/output>\n1 \/ (7\/24)<\/gadget>\n24\/7 = around 3.428571<\/output>\n24\/7 = around 3.428571<\/result>","index":842} +{"problem":"130 kg of an alloy a is mixed with 160 kg of alloy b . if alloy a has lead and tin in the ratio 2 : 3 and alloy b has tin and copper in the ratio 3 : 4 , then the amount of tin in the new alloy is ?","rationale":"\"quantity of tin in 130 kg of a = 130 * 3 \/ 5 = 78 kg quantity of tin in 160 kg of b = 160 * 3 \/ 7 = 68.5 kg quantity of tin in the new alloy = 78 + 68.5 = 146.5 kg answer is c\"","correct":"c","options":{"a":"140.6 kg ","b":"145.3 kg ","c":"146.5 kg ","d":"110.8 kg","e":"114 kg"},"options_float":{"a":140.6,"b":145.3,"c":146.5,"d":110.8,"e":114.0},"annotated_formula":"add(multiply(divide(130, add(2, 3)), 3), multiply(divide(160, add(3, 4)), 3))","linear_formula":"add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)|","chain":"2 + 3<\/gadget>\n5<\/output>\n130 \/ 5<\/gadget>\n26<\/output>\n26 * 3<\/gadget>\n78<\/output>\n3 + 4<\/gadget>\n7<\/output>\n160 \/ 7<\/gadget>\n160\/7 = around 22.857143<\/output>\n(160\/7) * 3<\/gadget>\n480\/7 = around 68.571429<\/output>\n78 + (480\/7)<\/gadget>\n1_026\/7 = around 146.571429<\/output>\n1_026\/7 = around 146.571429<\/result>","index":844} +{"problem":"a can do a piece of work in 10 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 5 days and b leaves after 3 days from the beginning . how long will work lost ?","rationale":"\"5 \/ 10 + 3 \/ 15 + x \/ 20 = 1 x = 18 \/ 3 = 6 answer : d\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"add(divide(subtract(const_1, add(multiply(subtract(3, 5), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(10), inverse(15))), 5))), inverse(15)), const_4.0)","linear_formula":"inverse(n1)|inverse(n2)|inverse(n0)|subtract(n4,n3)|add(#0,#1)|add(#2,#0)|add(#5,#1)|multiply(#4,#3)|multiply(n3,#6)|add(#7,#8)|subtract(const_1,#9)|divide(#10,#0)|add(const_4.0,#11)|","chain":"3 - 5<\/gadget>\n-2<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/15) + (1\/20)<\/gadget>\n7\/60 = around 0.116667<\/output>\n(-2) * (7\/60)<\/gadget>\n-7\/30 = around -0.233333<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + (1\/15)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/20) + (1\/6)<\/gadget>\n13\/60 = around 0.216667<\/output>\n(13\/60) * 5<\/gadget>\n13\/12 = around 1.083333<\/output>\n(-7\/30) + (13\/12)<\/gadget>\n17\/20 = around 0.85<\/output>\n1 - (17\/20)<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) \/ (1\/15)<\/gadget>\n9\/4 = around 2.25<\/output>\n(9\/4) + 4<\/gadget>\n25\/4 = around 6.25<\/output>\n25\/4 = around 6.25<\/result>","index":847} +{"problem":"at the end of year x , automobile installment credit accounted for 35 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 40 billion of credit , or 1 \/ 3 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at that time ?","rationale":"\"system of equations a = ( 35 \/ 100 ) c ( 1 \/ 3 ) a = 40 - - > a = 120 substitution 120 = ( 35 \/ 100 ) c c = ( 100 \/ 35 ) 120 answer - 120 \/ 35 * 100 the correct answer is e . - 342.86\"","correct":"e","options":{"a":"120 ","b":"300 ","c":"401.36 ","d":"350.15","e":"342.86"},"options_float":{"a":120.0,"b":300.0,"c":401.36,"d":350.15,"e":342.86},"annotated_formula":"divide(multiply(40, 3), divide(35, const_100))","linear_formula":"divide(n0,const_100)|multiply(n1,n3)|divide(#1,#0)|","chain":"40 * 3<\/gadget>\n120<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n120 \/ (7\/20)<\/gadget>\n2_400\/7 = around 342.857143<\/output>\n2_400\/7 = around 342.857143<\/result>","index":848} +{"problem":"the length of the bridge , which a train 130 metres long and travelling at 54 km \/ hr can cross in 30 seconds , is :","rationale":"\"speed = [ 54 x 5 \/ 18 ] m \/ sec = [ 15 \/ 1 ] m \/ sec time = 30 sec let the length of bridge be x metres . then , ( 130 + x ) \/ 30 = 15 = > ( 130 + x ) = 450 = > x = 450 - 130 = 320 m answer : c\"","correct":"c","options":{"a":"200 m ","b":"225 m ","c":"320 m ","d":"250 m","e":"240 m"},"options_float":{"a":200.0,"b":225.0,"c":320.0,"d":250.0,"e":240.0},"annotated_formula":"subtract(multiply(divide(multiply(54, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n54 * 1_000<\/gadget>\n54_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n54_000 \/ 3_600<\/gadget>\n15<\/output>\n15 * 30<\/gadget>\n450<\/output>\n450 - 130<\/gadget>\n320<\/output>\n320<\/result>","index":850} +{"problem":"the total age of a and b is 16 years more than the total age of b and c . c is how many year younger than","rationale":"\"given that a + b = 16 + b + c = > a â € “ c = 16 + b â € “ b = 16 = > c is younger than a by 16 years answer : d\"","correct":"d","options":{"a":"14 years ","b":"12 years ","c":"56 years ","d":"16 years","e":"55 years"},"options_float":{"a":14.0,"b":12.0,"c":56.0,"d":16.0,"e":55.0},"annotated_formula":"multiply(16, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"16 * 1<\/gadget>\n16<\/output>\n16<\/result>","index":855} +{"problem":"the area of sector of a circle whose radius is 12 metro and whose angle at the center is 30 ° is ?","rationale":"30 \/ 360 * 22 \/ 7 * 12 * 12 = 37 m 2 answer : c","correct":"c","options":{"a":"35 m 2 ","b":"36 m 2 ","c":"37 m 2 ","d":"38 m 2","e":"40 m 2"},"options_float":{"a":35.0,"b":36.0,"c":37.0,"d":38.0,"e":40.0},"annotated_formula":"divide(multiply(multiply(30, const_pi), power(12, const_2)), const_360)","linear_formula":"multiply(n1,const_pi)|power(n0,const_2)|multiply(#0,#1)|divide(#2,const_360)","chain":"30 * pi<\/gadget>\n30*pi = around 94.24778<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n(30*pi) * 144<\/gadget>\n4320*pi = around 13_571.680264<\/output>\n(4320*pi) \/ 360<\/gadget>\n12*pi = around 37.699112<\/output>\n12*pi = around 37.699112<\/result>","index":856} +{"problem":"find large number from below question the difference of two numbers is 1360 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1360 ) . x + 1360 = 6 x + 15 5 x = 1345 x = 269 large number = 269 + 1360 = 1629 e\"","correct":"e","options":{"a":"1254 ","b":"1376 ","c":"1456 ","d":"1555","e":"1629"},"options_float":{"a":1254.0,"b":1376.0,"c":1456.0,"d":1555.0,"e":1629.0},"annotated_formula":"multiply(divide(subtract(1360, 15), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_360 - 15<\/gadget>\n1_345<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_345 \/ 5<\/gadget>\n269<\/output>\n269 * 6<\/gadget>\n1_614<\/output>\n1_614<\/result>","index":857} +{"problem":"solve the equation 2 – x = 2 b – 2 ax , where a and b are real parameters . find for which values of a the equation has for solution a natural number , if b = 7","rationale":"solution : we write the given equation like this ( 2 a - 1 ) x = 2 ( b - 1 ) the following cases are possible : if 2 a - 1 ≠ 0 , i . e . a ≠ 1 \/ 2 , the equation has single solution x = 2 ( b - 1 ) \/ ( 2 a - 1 ) if a = 1 \/ 2 and b = 1 , the equation gets the kind 0 . x = 0 and every x is solution if a = 1 \/ 2 and b ≠ 1 , we get 0 . x = 2 ( b - 1 ) , where 2 ( b - 1 ) ≠ 0 in this case the equation has no solution . if b = 7 and a ≠ 1 \/ 2 the single solution is x = 2 ( 7 - 1 ) \/ ( 2 a - 1 ) = 12 \/ ( 2 a - 1 ) if a is whole number , then 2 a - 1 is also whole number and the solution x = 12 \/ ( 2 a - 1 ) is natural number when 2 a - 1 is positive divisor of 12 . to have a as whole number , the divisor of 12 mu be odd . but the only whole positive odd numbers divisible by 12 are 1 and 3 . therefore 2 a - 1 = 3 < = > a = 2 or 2 a - 1 = 1 < = > a = 1 a = 2 or 2 a - 1 = 1 < = > a = 1 answer b","correct":"b","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"divide(add(const_1, const_1), 2)","linear_formula":"add(const_1,const_1)|divide(#0,n0)","chain":"1 + 1<\/gadget>\n2<\/output>\n2 \/ 2<\/gadget>\n1<\/output>\n1<\/result>","index":858} +{"problem":"in a caravan , in addition to 50 hens there are 45 goats and 8 camels with some keepers . if the total number of feet be 224 more than the number of heads , find the number of keepers .","rationale":"let the number of keepers be x then , total number of heads = ( 50 + 45 + 8 + x ) = ( 103 + x ) . total number of feet = ( 45 + 8 ) x 4 + ( 50 + x ) x 2 = ( 312 + 2 x ) . ( 312 + 2 x ) - ( 103 + x ) = 224 = = > x = 15 . hence , number of keepers = 15 . answer c .","correct":"c","options":{"a":"25 ","b":"20 ","c":"15 ","d":"13","e":"14"},"options_float":{"a":25.0,"b":20.0,"c":15.0,"d":13.0,"e":14.0},"annotated_formula":"subtract(224, subtract(add(add(multiply(50, const_2), multiply(45, const_4)), multiply(8, const_4)), add(add(50, 45), 8)))","linear_formula":"add(n0,n1)|multiply(n0,const_2)|multiply(n1,const_4)|multiply(n2,const_4)|add(#1,#2)|add(n2,#0)|add(#4,#3)|subtract(#6,#5)|subtract(n3,#7)","chain":"50 * 2<\/gadget>\n100<\/output>\n45 * 4<\/gadget>\n180<\/output>\n100 + 180<\/gadget>\n280<\/output>\n8 * 4<\/gadget>\n32<\/output>\n280 + 32<\/gadget>\n312<\/output>\n50 + 45<\/gadget>\n95<\/output>\n95 + 8<\/gadget>\n103<\/output>\n312 - 103<\/gadget>\n209<\/output>\n224 - 209<\/gadget>\n15<\/output>\n15<\/result>","index":859} +{"problem":"a motorcyclist goes from bombay to pune , a distance of 160 kms at an average of 32 kmph speed . another man starts from bombay by car 2 ½ hours after the first , and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ?","rationale":"\"t = 160 \/ 32 = 5 h t = 5 - 3 = 2 time ratio = 5 : 2 = 5 : 2 speed ratio = 2 : 5 answer : b\"","correct":"b","options":{"a":"1 : 2 ","b":"2 : 5 ","c":"1 : 4 ","d":"1 : 1","e":"1 : 6"},"options_float":{"a":0.5,"b":0.4,"c":0.25,"d":1.0,"e":0.1666666667},"annotated_formula":"divide(divide(160, divide(160, 32)), divide(160, subtract(divide(160, 32), const_3)))","linear_formula":"divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3)|","chain":"160 \/ 32<\/gadget>\n5<\/output>\n160 \/ 5<\/gadget>\n32<\/output>\n5 - 3<\/gadget>\n2<\/output>\n160 \/ 2<\/gadget>\n80<\/output>\n32 \/ 80<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":860} +{"problem":"what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 40 % ?","rationale":"\"the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 40 % increase a surface = 6 * ( ( 1.4 a ) ^ 2 ) = 6 * 1.96 * a ^ 2 the increase in the surface area = ( 6 * 1.96 * a ^ 2 - 6 * a ^ 2 ) \/ 6 * a ^ 2 = ( 6 * a ^ 2 ( 1.96 - 1 ) ) \/ ( 6 * a ^ 2 ) = 1.96 - 1 = 0.96 = 96 % answer : d\"","correct":"d","options":{"a":"95 % ","b":"90 % ","c":"85 % ","d":"96 %","e":"25 %"},"options_float":{"a":95.0,"b":90.0,"c":85.0,"d":96.0,"e":25.0},"annotated_formula":"multiply(subtract(power(add(const_1, divide(40, const_100)), const_2), const_1), const_100)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n(7\/5) ** 2<\/gadget>\n49\/25 = around 1.96<\/output>\n(49\/25) - 1<\/gadget>\n24\/25 = around 0.96<\/output>\n(24\/25) * 100<\/gadget>\n96<\/output>\n96<\/result>","index":861} +{"problem":"how many shots of 1 cm radius can be prepared from a sphere of 5 cm radius ?","rationale":"4 \/ 3 ï € * 5 * 5 * 5 = 4 \/ 3 ï € * 1 * 1 * 1 * x x = 125 answer : d","correct":"d","options":{"a":"33 ","b":"88 ","c":"27 ","d":"125","e":"199"},"options_float":{"a":33.0,"b":88.0,"c":27.0,"d":125.0,"e":199.0},"annotated_formula":"multiply(power(5, const_2), 5)","linear_formula":"power(n1,const_2)|multiply(n1,#0)","chain":"5 ** 2<\/gadget>\n25<\/output>\n25 * 5<\/gadget>\n125<\/output>\n125<\/result>","index":862} +{"problem":"in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 \/ 2 , and in every year men are reduces 10 % . what is population of after 2 year .","rationale":"next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 10 % so total man = 150 * . 9 = 135 women will be = 175 - 135 = 40 so population after two years = [ 135 ^ 2 + 40 ^ 2 ] ^ . 5 = 140.80 = 141 so population after two year = 141 answer : b","correct":"b","options":{"a":"140 ","b":"141 ","c":"142 ","d":"143","e":"144"},"options_float":{"a":140.0,"b":141.0,"c":142.0,"d":143.0,"e":144.0},"annotated_formula":"power(add(power(multiply(subtract(1, divide(1, 10)), 150), const_2), power(subtract(power(add(power(150, const_2), power(90, const_2)), divide(1, 2)), multiply(subtract(1, divide(1, 10)), 150)), const_2)), divide(1, 2))","linear_formula":"divide(n4,n6)|divide(n4,n2)|power(n0,const_2)|power(n1,const_2)|add(#2,#3)|subtract(n4,#0)|multiply(n0,#5)|power(#4,#1)|power(#6,const_2)|subtract(#7,#6)|power(#9,const_2)|add(#8,#10)|power(#11,#1)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 150<\/gadget>\n135<\/output>\n135 ** 2<\/gadget>\n18_225<\/output>\n150 ** 2<\/gadget>\n22_500<\/output>\n90 ** 2<\/gadget>\n8_100<\/output>\n22_500 + 8_100<\/gadget>\n30_600<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n30_600 ** (1\/2)<\/gadget>\n30*sqrt(34) = around 174.928557<\/output>\n(30*sqrt(34)) - 135<\/gadget>\n-135 + 30*sqrt(34) = around 39.928557<\/output>\n(-135 + 30*sqrt(34)) ** 2<\/gadget>\n(-135 + 30*sqrt(34))**2 = around 1_594.289652<\/output>\n18_225 + ((-135 + 30*sqrt(34))**2)<\/gadget>\n(-135 + 30*sqrt(34))**2 + 18225 = around 19_819.289652<\/output>\n((-135 + 30*sqrt(34))**2 + 18225) ** (1\/2)<\/gadget>\nsqrt((-135 + 30*sqrt(34))**2 + 18225) = around 140.780999<\/output>\nsqrt((-135 + 30*sqrt(34))**2 + 18225) = around 140.780999<\/result>","index":863} +{"problem":"a train 124 m long running at 68 kmph crosses a platform in 19 sec . what is the length of the platform ?","rationale":"\"d = 68 * 5 \/ 18 = 19 = 358 – 124 = 234 answer : b\"","correct":"b","options":{"a":"338 ","b":"234 ","c":"240 ","d":"881","e":"271"},"options_float":{"a":338.0,"b":234.0,"c":240.0,"d":881.0,"e":271.0},"annotated_formula":"subtract(multiply(19, multiply(68, const_0_2778)), 124)","linear_formula":"multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n68 * (5\/18)<\/gadget>\n170\/9 = around 18.888889<\/output>\n19 * (170\/9)<\/gadget>\n3_230\/9 = around 358.888889<\/output>\n(3_230\/9) - 124<\/gadget>\n2_114\/9 = around 234.888889<\/output>\n2_114\/9 = around 234.888889<\/result>","index":864} +{"problem":"the hcf of two numbers is 42 and the other two factors of their lcm are 11 and 12 . what is the largest number .","rationale":"\"explanation : hcf of the two numbers = 42 hcf will be always a factor of lcm 42 is factor of lcm other two factors are 11 & 12 then the numbers are ( 42 11 ) and ( 42 x 12 ) = 462 and 504 answer : option d\"","correct":"d","options":{"a":"462 ","b":"450 ","c":"480 ","d":"504","e":"555"},"options_float":{"a":462.0,"b":450.0,"c":480.0,"d":504.0,"e":555.0},"annotated_formula":"multiply(42, 12)","linear_formula":"multiply(n0,n2)|","chain":"42 * 12<\/gadget>\n504<\/output>\n504<\/result>","index":865} +{"problem":"sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 468 per week . how much does she earn in dollars per hour ?","rationale":"\"explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 468 . hourly wage = 468 \/ 36 = 13 answer : e\"","correct":"e","options":{"a":"2 ","b":"8 ","c":"9 ","d":"11","e":"13"},"options_float":{"a":2.0,"b":8.0,"c":9.0,"d":11.0,"e":13.0},"annotated_formula":"divide(468, add(multiply(8, const_3), multiply(6, const_2)))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)|","chain":"8 * 3<\/gadget>\n24<\/output>\n6 * 2<\/gadget>\n12<\/output>\n24 + 12<\/gadget>\n36<\/output>\n468 \/ 36<\/gadget>\n13<\/output>\n13<\/result>","index":867} +{"problem":"a trader bought a car at 20 % discount on its original price . he sold it at a 50 % increase on the price he bought it . what percent of profit did he make on the original price ?","rationale":"\"original price = 100 cp = 80 s = 80 * ( 150 \/ 100 ) = 120 100 - 120 = 20 % answer : c\"","correct":"c","options":{"a":"17 % ","b":"72 % ","c":"20 % ","d":"82 %","e":"16 %"},"options_float":{"a":17.0,"b":72.0,"c":20.0,"d":82.0,"e":16.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 50)), const_100), const_100), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n100 + 50<\/gadget>\n150<\/output>\n80 * 150<\/gadget>\n12_000<\/output>\n12_000 \/ 100<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) - 1<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":868} +{"problem":"17 balls are numbered 1 to 17 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ?","rationale":"p ( 1 st ball is even ) = 8 \/ 17 p ( 2 nd ball is also even ) = 7 \/ 16 p ( both balls are even ) = 8 \/ 17 * 7 \/ 16 = 7 \/ 34 the answer is d .","correct":"d","options":{"a":"1 \/ 7 ","b":"3 \/ 14 ","c":"5 \/ 23 ","d":"7 \/ 34","e":"9 \/ 41"},"options_float":{"a":0.1428571429,"b":0.2142857143,"c":0.2173913043,"d":0.2058823529,"e":0.2195121951},"annotated_formula":"multiply(divide(add(const_4, const_4), 17), divide(subtract(add(const_4, const_4), 1), subtract(17, 1)))","linear_formula":"add(const_4,const_4)|subtract(n0,n1)|divide(#0,n0)|subtract(#0,n1)|divide(#3,#1)|multiply(#2,#4)","chain":"4 + 4<\/gadget>\n8<\/output>\n8 \/ 17<\/gadget>\n8\/17 = around 0.470588<\/output>\n8 - 1<\/gadget>\n7<\/output>\n17 - 1<\/gadget>\n16<\/output>\n7 \/ 16<\/gadget>\n7\/16 = around 0.4375<\/output>\n(8\/17) * (7\/16)<\/gadget>\n7\/34 = around 0.205882<\/output>\n7\/34 = around 0.205882<\/result>","index":873} +{"problem":"aarti can do a piece of work in 9 days . in how many days will she complete three time of work of same type ?","rationale":"\"we have the important relation , more work , more time ( days ) a piece of work can be done in 9 days . three times of work of same type can be done in 9 x 3 = 27 days answer d\"","correct":"d","options":{"a":"6 days ","b":"18 days ","c":"21 days ","d":"27 days","e":"13 days"},"options_float":{"a":6.0,"b":18.0,"c":21.0,"d":27.0,"e":13.0},"annotated_formula":"multiply(const_3, 9)","linear_formula":"multiply(n0,const_3)|","chain":"3 * 9<\/gadget>\n27<\/output>\n27<\/result>","index":874} +{"problem":"what is the value of n if the sum of the consecutive odd intergers r from 1 to n equals 169 ?","rationale":"\"before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n \/ 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is .\"","correct":"b","options":{"a":"47 ","b":"25 ","c":"37 ","d":"33","e":"29"},"options_float":{"a":47.0,"b":25.0,"c":37.0,"d":33.0,"e":29.0},"annotated_formula":"add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1)","linear_formula":"multiply(n0,const_2)|sqrt(n1)|multiply(#1,const_2)|subtract(#2,#0)|add(n0,#3)|","chain":"169 ** (1\/2)<\/gadget>\n13<\/output>\n13 * 2<\/gadget>\n26<\/output>\n2 * 1<\/gadget>\n2<\/output>\n26 - 2<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n25<\/result>","index":875} +{"problem":"there are 7 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .","rationale":"\"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 7 and for second = 6 so total methods = 7 * 6 = 42 answer : e\"","correct":"e","options":{"a":"40 ","b":"49 ","c":"41 ","d":"56","e":"42"},"options_float":{"a":40.0,"b":49.0,"c":41.0,"d":56.0,"e":42.0},"annotated_formula":"add(choose(7, 2), choose(7, 2))","linear_formula":"choose(n0,n1)|add(#0,#0)|","chain":"binomial(7, 2)<\/gadget>\n21<\/output>\n21 + 21<\/gadget>\n42<\/output>\n42<\/result>","index":877} +{"problem":"when a stone is dropped from a building 200 m high , its speed is proportional to the time elapsed after dropping . the distance traveled is proportional to the square of the time elapsed . after 1 second the speed of the train was 10 m \/ sec and it was 190 m above the ground . when its speed is 25 m \/ sec , what would be its distance from the ground ?","rationale":"v = x * t and s = y * t ^ 2 10 = x * 1 , hence x = 10 units ( 200 - 190 ) = y * 1 * 1 y = 10 when speed = 25 m \/ sec , time elapsed = 25 \/ 10 = 2.5 sec distance travelled by stone = 10 * ( 2.5 ) ^ 2 = 62.5 mts so distance of stone above ground = 200 - 62.5 = 137.5 mtrs answer : b","correct":"b","options":{"a":"140 m ","b":"137.5 m ","c":"125.75 m ","d":"142.5 m","e":"152.5 m"},"options_float":{"a":140.0,"b":137.5,"c":125.75,"d":142.5,"e":152.5},"annotated_formula":"subtract(200, multiply(power(divide(25, subtract(200, 190)), const_2), 10))","linear_formula":"subtract(n0,n3)|divide(n4,#0)|power(#1,const_2)|multiply(n2,#2)|subtract(n0,#3)","chain":"200 - 190<\/gadget>\n10<\/output>\n25 \/ 10<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) ** 2<\/gadget>\n25\/4 = around 6.25<\/output>\n(25\/4) * 10<\/gadget>\n125\/2 = around 62.5<\/output>\n200 - (125\/2)<\/gadget>\n275\/2 = around 137.5<\/output>\n275\/2 = around 137.5<\/result>","index":879} +{"problem":"in a bag of small balls 1 \/ 4 are green , 1 \/ 8 are blue , 1 \/ 12 are yellow and the remaining 26 white . how many balls are blue ?","rationale":"let us first find the fraction of green , blue and yellow balls 1 \/ 4 + 1 \/ 8 + 1 \/ 12 = 6 \/ 24 + 3 \/ 24 + 2 \/ 24 , common denominator = 11 \/ 24 , add numerators the fraction of white balls is given by 24 \/ 24 - 11 \/ 24 = 13 \/ 24 so the fraction 13 \/ 24 corresponds to 26 balls . if x is the total number of balls then ( 13 \/ 24 ) of x = 26 balls or ( 13 \/ 24 ) ? x = 26 x = 26 ? ( 24 \/ 13 ) = 48 , total number of balls the fraction of blue balls is 1 \/ 8 of x . the number of blue balls is given by ( 1 \/ 8 ) of 48 = 6 balls . correct answer a","correct":"a","options":{"a":"6 ","b":"5 ","c":"4 ","d":"3","e":"8"},"options_float":{"a":6.0,"b":5.0,"c":4.0,"d":3.0,"e":8.0},"annotated_formula":"multiply(divide(26, subtract(const_1, add(add(divide(1, 4), divide(1, 8)), divide(1, 12)))), divide(1, 8))","linear_formula":"divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(const_1,#4)|divide(n6,#5)|multiply(#6,#1)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/4) + (1\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(3\/8) + (1\/12)<\/gadget>\n11\/24 = around 0.458333<\/output>\n1 - (11\/24)<\/gadget>\n13\/24 = around 0.541667<\/output>\n26 \/ (13\/24)<\/gadget>\n48<\/output>\n48 * (1\/8)<\/gadget>\n6<\/output>\n6<\/result>","index":883} +{"problem":"a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 250 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ?","rationale":"\"the distance between the nest and the ditch is 250 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 250 = 7500 . d = st 7500 \/ 1.5 = t , i think we can take 7500 meters as 7.5 km , then only we get t = 5 . ( 1000 meters = 1 km ) d )\"","correct":"d","options":{"a":"1 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":1.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"divide(divide(multiply(250, multiply(15, const_2)), const_1000), divide(15, const_10))","linear_formula":"divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)|","chain":"15 * 2<\/gadget>\n30<\/output>\n250 * 30<\/gadget>\n7_500<\/output>\n7_500 \/ 1_000<\/gadget>\n15\/2 = around 7.5<\/output>\n15 \/ 10<\/gadget>\n3\/2 = around 1.5<\/output>\n(15\/2) \/ (3\/2)<\/gadget>\n5<\/output>\n5<\/result>","index":884} +{"problem":"solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 300 milliliters of solution x to create a solution that is 15 percent alcohol by volume ?","rationale":"\"15 % is 5 % - points higher than 10 % but 15 % - points lower than 30 % . thus there should be 3 parts of solution x for 1 part of solution y . we should add 100 ml of solution y . the answer is b .\"","correct":"b","options":{"a":"50 ","b":"100 ","c":"150 ","d":"200","e":"250"},"options_float":{"a":50.0,"b":100.0,"c":150.0,"d":200.0,"e":250.0},"annotated_formula":"multiply(divide(subtract(15, 10), subtract(30, 15)), 300)","linear_formula":"subtract(n3,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n2,#2)|","chain":"15 - 10<\/gadget>\n5<\/output>\n30 - 15<\/gadget>\n15<\/output>\n5 \/ 15<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 300<\/gadget>\n100<\/output>\n100<\/result>","index":885} +{"problem":"on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 40 percent . what was the total rent the motel actually charged for that night ?","rationale":"\"let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 40 % . 4 x = 200 = > x = 500 answer a\"","correct":"a","options":{"a":"$ 500 ","b":"$ 800 ","c":"$ 1,000 ","d":"$ 1,600","e":"$ 2,400"},"options_float":{"a":500.0,"b":800.0,"c":1000.0,"d":1600.0,"e":2400.0},"annotated_formula":"divide(multiply(10, subtract(60, 40)), divide(40, const_100))","linear_formula":"divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0)|","chain":"60 - 40<\/gadget>\n20<\/output>\n10 * 20<\/gadget>\n200<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n200 \/ (2\/5)<\/gadget>\n500<\/output>\n500<\/result>","index":886} +{"problem":"the length of the rectangular field is double its width . inside the field there is square shaped pond 9 m long . if the area of the pond is 1 \/ 8 of the area of the field . what is the length of the field ?","rationale":"\"a \/ 8 = 9 * 9 = > a = 9 * 9 * 8 x * 2 x = 9 * 9 * 8 x = 18 = > 2 x = 36 answer : b\"","correct":"b","options":{"a":"54 ","b":"36 ","c":"75 ","d":"28","e":"11"},"options_float":{"a":54.0,"b":36.0,"c":75.0,"d":28.0,"e":11.0},"annotated_formula":"sqrt(divide(multiply(square_area(9), 8), inverse(const_2)))","linear_formula":"inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)|","chain":"9 ** 2<\/gadget>\n81<\/output>\n81 * 8<\/gadget>\n648<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n648 \/ (1\/2)<\/gadget>\n1_296<\/output>\n1_296 ** (1\/2)<\/gadget>\n36<\/output>\n36<\/result>","index":887} +{"problem":"how much time will a train of length 380 m moving at a speed of 72 kmph take to cross another train of length 540 m , moving at 36 kmph in the same direction ?","rationale":"\"the distance to be covered = sum of their lengths = 380 + 540 = 920 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 \/ 18 = 10 mps . time required = d \/ s = 920 \/ 10 = 92 sec . answer : d\"","correct":"d","options":{"a":"50 ","b":"99 ","c":"88 ","d":"92","e":"12"},"options_float":{"a":50.0,"b":99.0,"c":88.0,"d":92.0,"e":12.0},"annotated_formula":"divide(add(380, 540), multiply(subtract(72, 36), const_0_2778))","linear_formula":"add(n0,n2)|subtract(n1,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"380 + 540<\/gadget>\n920<\/output>\n72 - 36<\/gadget>\n36<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n920 \/ 10<\/gadget>\n92<\/output>\n92<\/result>","index":888} +{"problem":"let the number which when multiplied by 16 is increased by 375 .","rationale":"\"solution let the number be x . then , 16 x - x = 375 ‹ = › 15 x = 375 x ‹ = › 25 . answer d\"","correct":"d","options":{"a":"14 ","b":"20 ","c":"26 ","d":"25","e":"30"},"options_float":{"a":14.0,"b":20.0,"c":26.0,"d":25.0,"e":30.0},"annotated_formula":"divide(375, subtract(16, const_1))","linear_formula":"subtract(n0,const_1)|divide(n1,#0)|","chain":"16 - 1<\/gadget>\n15<\/output>\n375 \/ 15<\/gadget>\n25<\/output>\n25<\/result>","index":895} +{"problem":"jill has 14 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ?","rationale":"let the number of each size of jar = wthen 1 \/ 4 w + 1 \/ 2 w + w = 14 1 3 \/ 4 w = 14 w = 8 the total number of jars = 3 w = 24 answer : e","correct":"e","options":{"a":"3 ","b":"6 ","c":"9 ","d":"12","e":"24"},"options_float":{"a":3.0,"b":6.0,"c":9.0,"d":12.0,"e":24.0},"annotated_formula":"multiply(divide(14, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3)","linear_formula":"divide(const_1,const_2)|add(#0,const_0_25)|add(#1,const_1)|divide(n0,#2)|multiply(#3,const_3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/4) + (1\/2)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 + (3\/4)<\/gadget>\n7\/4 = around 1.75<\/output>\n14 \/ (7\/4)<\/gadget>\n8<\/output>\n8 * 3<\/gadget>\n24<\/output>\n24<\/result>","index":897} +{"problem":"a cricket player whose bowling average was 21.5 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.5 . what will be the number of wickets taken by him till the last match ?","rationale":"\"average = total runs \/ total wickets total runs after last match = 21.5 w + 52 total wickets after last match = w + 5 ( 21.5 w + 52 ) \/ ( w + 5 ) = 21.5 - 0.5 = 21 w = 106 so total wickets aftr last match = w + 5 = 111 answer : c\"","correct":"c","options":{"a":"64 ","b":"72 ","c":"111 ","d":"721","e":"108"},"options_float":{"a":64.0,"b":72.0,"c":111.0,"d":721.0,"e":108.0},"annotated_formula":"divide(subtract(multiply(subtract(21.5, 0.5), 5), 52), 0.5)","linear_formula":"subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)|","chain":"21.5 - 0.5<\/gadget>\n21<\/output>\n21 * 5<\/gadget>\n105<\/output>\n105 - 52<\/gadget>\n53<\/output>\n53 \/ 0.5<\/gadget>\n106<\/output>\n106<\/result>","index":898} +{"problem":"8 men , working 8 hours a day can complete a work in 18 days . how many hours a day must 12 men work to complete the same work in 12 days ?","rationale":"\"the number of hours required to complete the work is 8 * 8 * 18 = 1152 12 × 12 × ( x ) = 1152 x = 8 the answer is c .\"","correct":"c","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"divide(multiply(multiply(8, 8), 18), multiply(12, 12))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|","chain":"8 * 8<\/gadget>\n64<\/output>\n64 * 18<\/gadget>\n1_152<\/output>\n12 * 12<\/gadget>\n144<\/output>\n1_152 \/ 144<\/gadget>\n8<\/output>\n8<\/result>","index":899} +{"problem":"if 3 men or 4 women can do a piece of work in 16 days , in how many days can 12 men and 8 women do the same piece of work ?","rationale":"\"3 men can finish a work in 16 days , so in 1 day 3 men finish 1 \/ 16 th of work and in 1 day 1 men can finish , 1 \/ 48 th of work . similarly , in 1 day 1 women can finish 1 \/ 64 th of work . 12 men in 1 day can finish , 12 \/ 48 = 1 \/ 4 , 8 women in 1 day can finish = 8 \/ 64 = 1 \/ 8 1 \/ 4 + 1 \/ 8 = 3 \/ 8 . hence number of days to complete is 8 \/ 3 days . answer : a\"","correct":"a","options":{"a":"8 \/ 3 ","b":"6 ","c":"5 ","d":"4","e":"3"},"options_float":{"a":2.6666666667,"b":6.0,"c":5.0,"d":4.0,"e":3.0},"annotated_formula":"inverse(add(divide(12, multiply(3, 16)), divide(8, multiply(4, 16))))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)|","chain":"3 * 16<\/gadget>\n48<\/output>\n12 \/ 48<\/gadget>\n1\/4 = around 0.25<\/output>\n4 * 16<\/gadget>\n64<\/output>\n8 \/ 64<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/4) + (1\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n1 \/ (3\/8)<\/gadget>\n8\/3 = around 2.666667<\/output>\n8\/3 = around 2.666667<\/result>","index":901} +{"problem":"john has $ 1,600 at the beginning of his trip , after spending money , he still has exactly $ 600 less than he spent on the trip . how much money does john still have ?","rationale":"suppose total money spent = x not spend ( money he still has ) = x - 600 x + x - 600 = 1600 x = 1100 money not spend = 1100 - 600 = 500 answer : e","correct":"e","options":{"a":"$ 200 ","b":"$ 400 ","c":"$ 600 ","d":"$ 800","e":"$ 500"},"options_float":{"a":200.0,"b":400.0,"c":600.0,"d":800.0,"e":500.0},"annotated_formula":"subtract(add(multiply(const_100, const_10), 600), divide(add(600, add(multiply(const_100, const_10), 600)), const_2))","linear_formula":"multiply(const_10,const_100)|add(n1,#0)|add(n1,#1)|divide(#2,const_2)|subtract(#1,#3)","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 + 600<\/gadget>\n1_600<\/output>\n600 + 1_600<\/gadget>\n2_200<\/output>\n2_200 \/ 2<\/gadget>\n1_100<\/output>\n1_600 - 1_100<\/gadget>\n500<\/output>\n500<\/result>","index":903} +{"problem":"a , b and c started a business by investing rs . 800 \/ - , rs . 1000 \/ - and rs . 1200 \/ - respectively for two years . total profit is rs . 1000 \/ - . find the c ' s share ?","rationale":"solution : a : b : c = ( 800 * 2 ) : ( 1000 * 2 ) : ( 12000 * 2 ) = 4 : 5 : 6 . so c ' s share = rs . ( 1000 * 6 \/ 15 ) = rs . 400 . answer : option c","correct":"c","options":{"a":"rs . 600 ","b":"rs . 500 ","c":"rs . 400 ","d":"rs . 800","e":"rs . 900"},"options_float":{"a":600.0,"b":500.0,"c":400.0,"d":800.0,"e":900.0},"annotated_formula":"multiply(1200, divide(1000, add(add(800, 1000), 1200)))","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n1,#1)|multiply(n2,#2)","chain":"800 + 1_000<\/gadget>\n1_800<\/output>\n1_800 + 1_200<\/gadget>\n3_000<\/output>\n1_000 \/ 3_000<\/gadget>\n1\/3 = around 0.333333<\/output>\n1_200 * (1\/3)<\/gadget>\n400<\/output>\n400<\/result>","index":904} +{"problem":"the speed of a car is 85 km in the first hour and 45 km in the second hour . what is the average speed of the car ?","rationale":"s = ( 85 + 45 ) \/ 2 = 65 kmph answer : b","correct":"b","options":{"a":"72 kmph ","b":"65 kmph ","c":"30 kmph ","d":"80 kmph","e":"82 kmph"},"options_float":{"a":72.0,"b":65.0,"c":30.0,"d":80.0,"e":82.0},"annotated_formula":"divide(add(85, 45), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"85 + 45<\/gadget>\n130<\/output>\n130 \/ 2<\/gadget>\n65<\/output>\n65<\/result>","index":905} +{"problem":"the positive value of x that satisfies the equation ( 3 + 2 x ) ^ 5 = ( 1 + 3 x ) ^ 4 is between bunuel , can you please explain this one ?","rationale":"trial and error would probably be the easiest way to solve this problem . when x is large enough positive number , then because of the exponents ( 5 > 4 ) , lhs will be more than rhs ( as you increase the positive value of x the distance between the values of lhs and rhs will increase ) . try x = 1 - - > lhs = 3 ^ 5 = 81 * 3 = 243 and rhs = 4 ^ 4 = 64 * 4 = 256 , so ( 1 + 2 x ) ^ 5 < ( 1 + 3 x ) ^ 4 . as you can see lhs is still slightly less than than rhs . so , the value of x for which ( 1 + 2 x ) ^ 5 = ( 1 + 3 x ) ^ 4 is slightly more than 1 . answer : d .","correct":"d","options":{"a":"0 and 0.5 ","b":"0.5 and 1 ","c":"1 and 1.5 ","d":"1.5 and 2","e":"2 and 2.5"},"options_float":{"a":0.0,"b":0.5,"c":1.0,"d":1.5,"e":2.0},"annotated_formula":"add(1, divide(1, 2))","linear_formula":"divide(n3,n1)|add(n3,#0)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":907} +{"problem":"the h . c . f . of two numbers is 20 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is :","rationale":"\"the numbers are ( 20 x 11 ) and ( 20 x 15 ) . larger number = ( 20 x 15 ) = 300 . answer : b\"","correct":"b","options":{"a":"276 ","b":"300 ","c":"299 ","d":"322","e":"345"},"options_float":{"a":276.0,"b":300.0,"c":299.0,"d":322.0,"e":345.0},"annotated_formula":"multiply(20, 15)","linear_formula":"multiply(n0,n2)|","chain":"20 * 15<\/gadget>\n300<\/output>\n300<\/result>","index":908} +{"problem":"carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 10 inches tall . how tall are the sculpture andbase together ?","rationale":"\"we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 10 = 40 44 \/ 12 = 3.66 feet answer : b\"","correct":"b","options":{"a":"3.1 feet ","b":"3.66 feet ","c":"3.3 feet ","d":"3.4 feet","e":"3.5 feet"},"options_float":{"a":3.1,"b":3.66,"c":3.3,"d":3.4,"e":3.5},"annotated_formula":"divide(add(add(multiply(add(2, 10), 2), 10), 10), add(2, 10))","linear_formula":"add(n0,n1)|multiply(n0,#0)|add(#1,n1)|add(n2,#2)|divide(#3,#0)|","chain":"2 + 10<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 + 10<\/gadget>\n34<\/output>\n34 + 10<\/gadget>\n44<\/output>\n44 \/ 12<\/gadget>\n11\/3 = around 3.666667<\/output>\n11\/3 = around 3.666667<\/result>","index":910} +{"problem":"points a , b , and , c have xy - coordinates ( 2,0 ) , ( 8,12 ) , and ( 14,0 ) , respectively . points x , y , and z have xy - coordinates ( 6,0 ) , ( 8,4 ) , and ( 10,0 ) , respectively . what fraction w of the area of triangle abc is the area of triangle xyz ?","rationale":"\"if you notice , both triangles abc and xyz have a side on x axis . we can take these sides as bases for each triangle , therefore area of abc is 1 \/ 2 * 12 * 12 ( height of abc is the y coordinate of the third point ( 8,12 ) ) similarly area of xyz is 1 \/ 2 * 4 * 4 dividing area of xyz with that of abc gives w = 1 \/ 9 . a\"","correct":"a","options":{"a":"1 \/ 9 ","b":"1 \/ 8 ","c":"1 \/ 6 ","d":"1 \/ 5","e":"1 \/ 3"},"options_float":{"a":0.1111111111,"b":0.125,"c":0.1666666667,"d":0.2,"e":0.3333333333},"annotated_formula":"divide(divide(power(const_4, const_2), const_2), divide(power(add(const_10, const_2), const_2), const_2))","linear_formula":"add(const_10,const_2)|power(const_4,const_2)|divide(#1,const_2)|power(#0,const_2)|divide(#3,const_2)|divide(#2,#4)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n144 \/ 2<\/gadget>\n72<\/output>\n8 \/ 72<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":911} +{"problem":"the cash difference between the selling prices of an article at a profit of 4 % and 2 % is rs . 3 . the ratio of the two selling prices is ?","rationale":"\"let c . p . of the article be rs . x . then , required ratio = 104 % of x \/ 102 % of x = 104 \/ 102 = 52 \/ 51 = 52 : 51 answer : a\"","correct":"a","options":{"a":"52 : 51 ","b":"52 : 53 ","c":"52 : 50 ","d":"22 : 56","e":"52 : 51"},"options_float":{"a":1.0196078431,"b":0.9811320755,"c":1.04,"d":0.3928571429,"e":1.0196078431},"annotated_formula":"divide(add(const_100, 4), add(const_100, 2))","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|","chain":"100 + 4<\/gadget>\n104<\/output>\n100 + 2<\/gadget>\n102<\/output>\n104 \/ 102<\/gadget>\n52\/51 = around 1.019608<\/output>\n52\/51 = around 1.019608<\/result>","index":912} +{"problem":"two whole numbers whose sum is 100 can not be in the ratio","rationale":"d ) 2 : 6","correct":"d","options":{"a":"2 : 23 ","b":"5 : 45 ","c":"10 : 10 ","d":"2 : 6","e":"2 : 3"},"options_float":{"a":0.0869565217,"b":0.1111111111,"c":1.0,"d":0.3333333333,"e":0.6666666667},"annotated_formula":"divide(divide(subtract(divide(100, const_2), const_10), const_2), add(divide(100, const_2), const_10))","linear_formula":"divide(n0,const_2)|add(#0,const_10)|subtract(#0,const_10)|divide(#2,const_2)|divide(#3,#1)","chain":"100 \/ 2<\/gadget>\n50<\/output>\n50 - 10<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n50 + 10<\/gadget>\n60<\/output>\n20 \/ 60<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":913} +{"problem":"in a certain warehouse , 30 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ?","rationale":"\"if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 30 % of the packages weigh less than 75 pounds so , 30 % oft = number of packages that weigh less than 75 pounds 30 % of 240 = 72 , so 72 packages weigh less than 75 pounds of those 72 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 72 - 48 = 24 = c\"","correct":"c","options":{"a":"8 ","b":"64 ","c":"24 ","d":"102","e":"144"},"options_float":{"a":8.0,"b":64.0,"c":24.0,"d":102.0,"e":144.0},"annotated_formula":"subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 30), const_100), 48)","linear_formula":"subtract(const_100,n4)|divide(n2,#0)|multiply(#1,const_100)|multiply(n0,#2)|divide(#3,const_100)|subtract(#4,n2)|","chain":"100 - 80<\/gadget>\n20<\/output>\n48 \/ 20<\/gadget>\n12\/5 = around 2.4<\/output>\n(12\/5) * 100<\/gadget>\n240<\/output>\n240 * 30<\/gadget>\n7_200<\/output>\n7_200 \/ 100<\/gadget>\n72<\/output>\n72 - 48<\/gadget>\n24<\/output>\n24<\/result>","index":916} +{"problem":"a student took 5 papers in an examination , where the full marks were the same for each paper . his marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10 . in all papers together , the candidate obtained 60 % of the total marks then , the number of papers in which he got more than 50 % marks is","rationale":"solution : let the marks obtained in five subjects be 6 x , 7 x , 8 x , 9 x and 10 x . total marks obtained = 40 x max . marks of the five subjects = 40 x \/ 0.6 [ 40 x is 60 % of total marks ] max . marks in each subject = 40 x \/ 0.6 * 5 = 13.33 x hence , % of each subject = 6 x * 100 \/ 13.33 = 45.01 % or , 7 x * 100 \/ 13.33 = 52.51 in same way other percentage are 60.01 % , 67.52 % , 75.01 % . hence , number of subjects in which he gets more than 50 % marks = 4 . answer : option c","correct":"c","options":{"a":"1 ","b":"3 ","c":"4 ","d":"5","e":"7"},"options_float":{"a":1.0,"b":3.0,"c":4.0,"d":5.0,"e":7.0},"annotated_formula":"subtract(5, const_1)","linear_formula":"subtract(n0,const_1)","chain":"5 - 1<\/gadget>\n4<\/output>\n4<\/result>","index":917} +{"problem":"subtracting 5 % of a from a is equivalent to multiplying a by how much ?","rationale":"\"answer let a - 5 % of a = ab . ⇒ ( 95 x a ) \/ 100 = ab ∴ b = 0.95 correct option : a\"","correct":"a","options":{"a":"0.95 ","b":"9.4 ","c":"0.094 ","d":"94","e":"none"},"options_float":{"a":0.95,"b":9.4,"c":0.094,"d":94.0,"e":null},"annotated_formula":"divide(subtract(const_100, 5), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n19\/20 = around 0.95<\/result>","index":918} +{"problem":"a bag consists of 20 marbles , of which 5 are blue , 7 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?","rationale":"\"bag consists of 20 marbles , of which 5 are blue , 7 are red remainder are white . so , white = 20 - 5 - 7 = 8 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 7 \/ 20 + 8 \/ 20 = 15 \/ 20 = 3 \/ 4 hence , answer will be a .\"","correct":"a","options":{"a":"3 \/ 4 ","b":"2 \/ 4 ","c":"1 \/ 4 ","d":"1 \/ 8","e":"1 \/ 16"},"options_float":{"a":0.75,"b":0.5,"c":0.25,"d":0.125,"e":0.0625},"annotated_formula":"divide(add(subtract(20, add(5, 7)), 7), 20)","linear_formula":"add(n1,n2)|subtract(n0,#0)|add(n2,#1)|divide(#2,n0)|","chain":"5 + 7<\/gadget>\n12<\/output>\n20 - 12<\/gadget>\n8<\/output>\n8 + 7<\/gadget>\n15<\/output>\n15 \/ 20<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":919} +{"problem":"a trader bought a car at 20 % discount on its original price . he sold it at a 60 % increase on the price he bought it . what percent of profit did he make on the original price ?","rationale":"\"original price = 100 cp = 80 s = 80 * ( 160 \/ 100 ) = 128 100 - 128 = 28 % answer : a\"","correct":"a","options":{"a":"28 % ","b":"72 % ","c":"12 % ","d":"82 %","e":"16 %"},"options_float":{"a":28.0,"b":72.0,"c":12.0,"d":82.0,"e":16.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 60)), const_100), const_100), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n100 + 60<\/gadget>\n160<\/output>\n80 * 160<\/gadget>\n12_800<\/output>\n12_800 \/ 100<\/gadget>\n128<\/output>\n128 \/ 100<\/gadget>\n32\/25 = around 1.28<\/output>\n(32\/25) - 1<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) * 100<\/gadget>\n28<\/output>\n28<\/result>","index":920} +{"problem":"how many seconds will a 550 m long train take to cross a man walking with a speed of 3 km \/ hr in the direction of the moving train if the speed of the train is 63 km \/ hr ?","rationale":"\"speed of train relative to man = 63 - 3 = 60 km \/ hr . = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec . time taken to pass the man = 550 * 3 \/ 50 = 33 sec . answer : c\"","correct":"c","options":{"a":"12 sec ","b":"30 sec ","c":"33 sec ","d":"16 sec","e":"18 sec"},"options_float":{"a":12.0,"b":30.0,"c":33.0,"d":16.0,"e":18.0},"annotated_formula":"divide(550, multiply(subtract(63, 3), const_0_2778))","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"63 - 3<\/gadget>\n60<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n550 \/ (50\/3)<\/gadget>\n33<\/output>\n33<\/result>","index":923} +{"problem":"jansi went out for shopping . she had in her handbag approximately rs . 15 \/ - in one rupee notes and 20 p . coins . when she returned she had as many one rupee notes as she originally had and as many 20 p . coins as she originally had one rupee notes . she actually came back with about one - third of what she had started out with . how much did she spend and exactly how much did she have with her when she started out ?","rationale":"a 9.60 let us assume that originally jansi had x rupees and y 20 paise coins . going shopping she had ( 100 x + 20 y ) paise . she returned with only ( 100 y + 20 x ) paise . this last sum , as we know , is one - third of the original and therefore 3 ( 100 y + 20 x ) = 100 x + 20 y simplifying we have x = 7 y if y is 1 thenx is 7 . assuming this so jansi had 7.20 rupees when she set out for shopping . this is wrong because jansi actually had about 15 rupees . let us see now what we get if y = 2 . then x = 14 . the original sum was 14.40 rupees which accords with the condition of the problem . if we assume that y = 3 then the sum will be too big — 21.60 rupees . therefore the only suitable answer is 14.40 rupees . after shopping jansi had 2 one rupee notes and 14 twenty paise coins . this is actually l \/ 3 rd of the original sum 1,440 : 3 = 480 . jansi ' s purchases , therefore , cost 14.40 — 4.80 = rs 9.60","correct":"a","options":{"a":"9.6 ","b":"10.6 ","c":"11.0 ","d":"9.4","e":"8.1"},"options_float":{"a":9.6,"b":10.6,"c":11.0,"d":9.4,"e":8.1},"annotated_formula":"subtract(add(multiply(floor(divide(15, divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20))))), divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20)))), divide(multiply(floor(divide(15, divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20))))), 20), const_100)), multiply(add(multiply(floor(divide(15, divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20))))), divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20)))), divide(multiply(floor(divide(15, divide(subtract(multiply(const_3, const_100), 20), subtract(const_100, multiply(const_3, 20))))), 20), const_100)), divide(const_1, const_3)))","linear_formula":"divide(const_1,const_3)|multiply(const_100,const_3)|multiply(n1,const_3)|subtract(#1,n1)|subtract(const_100,#2)|divide(#3,#4)|divide(n0,#5)|floor(#6)|multiply(n1,#7)|multiply(#5,#7)|divide(#8,const_100)|add(#10,#9)|multiply(#11,#0)|subtract(#11,#12)","chain":"3 * 100<\/gadget>\n300<\/output>\n300 - 20<\/gadget>\n280<\/output>\n3 * 20<\/gadget>\n60<\/output>\n100 - 60<\/gadget>\n40<\/output>\n280 \/ 40<\/gadget>\n7<\/output>\n15 \/ 7<\/gadget>\n15\/7 = around 2.142857<\/output>\nfloor(15\/7)<\/gadget>\n2<\/output>\n2 * 7<\/gadget>\n14<\/output>\n2 * 20<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n14 + (2\/5)<\/gadget>\n72\/5 = around 14.4<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(72\/5) * (1\/3)<\/gadget>\n24\/5 = around 4.8<\/output>\n(72\/5) - (24\/5)<\/gadget>\n48\/5 = around 9.6<\/output>\n48\/5 = around 9.6<\/result>","index":924} +{"problem":"at a certain organisation , the number of male members went up by 15 % in the year 2001 from year 2000 , and the number of females members went down by 6 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ?","rationale":"\"men increase by 15 % = = > 1.15 m = males in 2001 women decrease by 6 % = = > 0.94 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.15 m + 0.94 f = 1.012 * ( m + f ) 1.15 m + 0.94 f = 1.012 m + 1.012 f 1.15 m - 1.012 m = 1.012 f - 0.94 f 0.138 m = 0.072 f m \/ f = ( 0.072 ) \/ ( 0.138 ) = 72 \/ 138 = 1 \/ 2 answer = ( a )\"","correct":"a","options":{"a":"1 : 2 ","b":"1 : 3 ","c":"2 : 3 ","d":"3 : 2","e":"2 : 1"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.6666666667,"d":1.5,"e":2.0},"annotated_formula":"divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(6, const_100)), const_1000)), subtract(multiply(add(const_1, divide(15, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000)))","linear_formula":"divide(n4,const_100)|divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#2,const_1)|subtract(const_1,#1)|multiply(#3,const_1000)|multiply(#5,const_1000)|multiply(#4,const_1000)|subtract(#6,#7)|subtract(#8,#6)|divide(#9,#10)|","chain":"1.2 \/ 100<\/gadget>\n0.012<\/output>\n1 + 0.012<\/gadget>\n1.012<\/output>\n1.012 * 1_000<\/gadget>\n1_012<\/output>\n6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n1 - (3\/50)<\/gadget>\n47\/50 = around 0.94<\/output>\n(47\/50) * 1_000<\/gadget>\n940<\/output>\n1_012 - 940<\/gadget>\n72<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n(23\/20) * 1_000<\/gadget>\n1_150<\/output>\n1_150 - 1_012<\/gadget>\n138<\/output>\n72 \/ 138<\/gadget>\n12\/23 = around 0.521739<\/output>\n12\/23 = around 0.521739<\/result>","index":925} +{"problem":"in a party every person shakes hands with every other person . if there were a total of 120 handshakes in the party then what is the number of persons present in the party ?","rationale":"\"explanation : let the number of persons be n ∴ total handshakes = nc 2 = 120 n ( n - 1 ) \/ 2 = 120 ∴ n = 16 answer : option b\"","correct":"b","options":{"a":"15 ","b":"16 ","c":"17 ","d":"18","e":"19"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":19.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(120, const_2), const_4), const_1)), const_1), const_2)","linear_formula":"multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|","chain":"120 * 2<\/gadget>\n240<\/output>\n240 * 4<\/gadget>\n960<\/output>\n960 + 1<\/gadget>\n961<\/output>\n961 ** (1\/2)<\/gadget>\n31<\/output>\n31 + 1<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16<\/result>","index":926} +{"problem":"a square mirror has exactly half the area of the rectangular wall on which it is hung . if each side of the mirror is 21 inches and the width of the wall is 28 inches , what is the length of the wall , in inches ?","rationale":"\"since the mirror is 42 inches in all sides , it must be a square . area of a square is a = a ^ 2 ; 21 ^ 2 = 441 . area of rectangle is double of that 2 * 441 = 882 . now a = lw and we need find w so a \/ l = w ; 882 \/ 28 = 31.5 answer ! answer is b\"","correct":"b","options":{"a":"21.5 ","b":"31.5 ","c":"41.5 ","d":"61.5","e":"71.5"},"options_float":{"a":21.5,"b":31.5,"c":41.5,"d":61.5,"e":71.5},"annotated_formula":"divide(multiply(const_2, square_area(21)), 28)","linear_formula":"square_area(n0)|multiply(#0,const_2)|divide(#1,n1)|","chain":"21 ** 2<\/gadget>\n441<\/output>\n2 * 441<\/gadget>\n882<\/output>\n882 \/ 28<\/gadget>\n63\/2 = around 31.5<\/output>\n63\/2 = around 31.5<\/result>","index":927} +{"problem":"when positive integer x is divided by positive integer y , the remainder is 6 . if x \/ y = 96.15 , what is the value of y ?","rationale":"\"by the definition of a remainder , the remainder here is equal to 6 \/ y . the remainder in decimal form is given as . 15 therefore , 6 \/ y = . 15 solve for y and get 40 . c\"","correct":"c","options":{"a":"96 ","b":"75 ","c":"40 ","d":"25","e":"12"},"options_float":{"a":96.0,"b":75.0,"c":40.0,"d":25.0,"e":12.0},"annotated_formula":"divide(6, subtract(96.15, floor(96.15)))","linear_formula":"floor(n1)|subtract(n1,#0)|divide(n0,#1)|","chain":"floor(96.15)<\/gadget>\n96<\/output>\n96.15 - 96<\/gadget>\n0.15<\/output>\n6 \/ 0.15<\/gadget>\n40<\/output>\n40<\/result>","index":928} +{"problem":"if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 276 games will be played during the tournament . find the number of participants .","rationale":"\"let p be the number of participants . pc 2 = 276 ( p ) ( p - 1 ) = 552 = 24 * 23 p = 24 the answer is c .\"","correct":"c","options":{"a":"20 ","b":"22 ","c":"24 ","d":"26","e":"28"},"options_float":{"a":20.0,"b":22.0,"c":24.0,"d":26.0,"e":28.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(276, const_2), const_4), const_1)), const_1), const_2)","linear_formula":"multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|","chain":"276 * 2<\/gadget>\n552<\/output>\n552 * 4<\/gadget>\n2_208<\/output>\n2_208 + 1<\/gadget>\n2_209<\/output>\n2_209 ** (1\/2)<\/gadget>\n47<\/output>\n47 + 1<\/gadget>\n48<\/output>\n48 \/ 2<\/gadget>\n24<\/output>\n24<\/result>","index":932} +{"problem":"3 friends had dinner at a restaurant . when the bill was received , akshitha paid 3 \/ 4 as much as veena paid and veena paid 1 \/ 2 as much as lasya paid . what fraction of the bill did veena pay ?","rationale":"let veena paid x , so akshitha paid 3 x \/ 4 , and lasya paid 2 x , so total bill paid is given by x + ( 3 x \/ 4 ) + 2 x = 1 , we get i . e . x = 4 \/ 15 answer : a","correct":"a","options":{"a":"4 \/ 15 ","b":"3 \/ 12 ","c":"3 \/ 11 ","d":"3 \/ 10","e":"3 \/ 14"},"options_float":{"a":0.2666666667,"b":0.25,"c":0.2727272727,"d":0.3,"e":0.2142857143},"annotated_formula":"divide(4, add(add(3, 4), multiply(4, 2)))","linear_formula":"add(n0,n2)|multiply(n2,n4)|add(#0,#1)|divide(n2,#2)","chain":"3 + 4<\/gadget>\n7<\/output>\n4 * 2<\/gadget>\n8<\/output>\n7 + 8<\/gadget>\n15<\/output>\n4 \/ 15<\/gadget>\n4\/15 = around 0.266667<\/output>\n4\/15 = around 0.266667<\/result>","index":934} +{"problem":"a number x is 5 times another number y . the percentage that y is less than x is","rationale":"\"say y = 1 and x = 5 . then y = 1 is less than x = 5 by ( 5 - 1 ) \/ 5 * 100 = 4 \/ 5 * 100 = 80 % . answer : c .\"","correct":"c","options":{"a":"12.5 % ","b":"87.5 % ","c":"80 % ","d":"11 %","e":"1 %"},"options_float":{"a":12.5,"b":87.5,"c":80.0,"d":11.0,"e":1.0},"annotated_formula":"multiply(divide(subtract(5, const_1), 5), const_100)","linear_formula":"subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"5 - 1<\/gadget>\n4<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":935} +{"problem":"marketing executives for a certain chewing gum company projected a 40 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 30 % . what percent of the projected revenue was the actual revenue ?","rationale":"last year revenue = 100 ( assume ) ; this year revenue = 70 ; projected revenue = 140 . actual \/ projected * 100 = 70 \/ 140 * 100 = 50 % . answer : a .","correct":"a","options":{"a":"50 % ","b":"58 % ","c":"62.5 % ","d":"64 %","e":"75 %"},"options_float":{"a":50.0,"b":58.0,"c":62.5,"d":64.0,"e":75.0},"annotated_formula":"multiply(divide(subtract(const_100, 30), add(40, const_100)), const_100)","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)|multiply(#2,const_100)","chain":"100 - 30<\/gadget>\n70<\/output>\n40 + 100<\/gadget>\n140<\/output>\n70 \/ 140<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":936} +{"problem":"how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 9 m x 6 m x 22.5 cm ?","rationale":"\"number of bricks = volume of wall \/ volume of bricks = 900 x 600 x 22.5 \/ 25 x 11.25 x 6 = = 7200 answer : d\"","correct":"d","options":{"a":"6400 ","b":"6410 ","c":"6440 ","d":"7200","e":"6800"},"options_float":{"a":6400.0,"b":6410.0,"c":6440.0,"d":7200.0,"e":6800.0},"annotated_formula":"divide(multiply(multiply(multiply(9, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6))","linear_formula":"multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|","chain":"9 * 100<\/gadget>\n900<\/output>\n6 * 100<\/gadget>\n600<\/output>\n900 * 600<\/gadget>\n540_000<\/output>\n540_000 * 22.5<\/gadget>\n12_150_000<\/output>\n25 * 11.25<\/gadget>\n281.25<\/output>\n281.25 * 6<\/gadget>\n1_687.5<\/output>\n12_150_000 \/ 1_687.5<\/gadget>\n7_200<\/output>\n7_200<\/result>","index":937} +{"problem":"at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 65 . if the restaurant serves 120 customers today , raising the average to 90 customers per day , what is the value of x ?","rationale":"withoutusing the formula , we can see that today the restaurant served 30 customers above the average . the total amount above the average must equal total amount below the average . this additional 30 customers must offset the “ deficit ” below the average of 90 created on the x days the restaurant served only 60 customers per day . 30 \/ 30 = 1 day . choice ( a ) . withthe formula , we can set up the following : 90 = ( 60 x + 120 ) \/ ( x + 1 ) 90 x + 90 = 60 x + 120 30 x = 30 x = 1 answer choice ( a )","correct":"a","options":{"a":"1 ","b":"5 ","c":"9 ","d":"15","e":"30"},"options_float":{"a":1.0,"b":5.0,"c":9.0,"d":15.0,"e":30.0},"annotated_formula":"subtract(divide(subtract(120, 90), subtract(90, 65)), divide(subtract(120, const_100), const_100))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)","chain":"120 - 90<\/gadget>\n30<\/output>\n90 - 65<\/gadget>\n25<\/output>\n30 \/ 25<\/gadget>\n6\/5 = around 1.2<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(6\/5) - (1\/5)<\/gadget>\n1<\/output>\n1<\/result>","index":938} +{"problem":"stacy has a 63 page history paper due in 7 days . how many pages per day would she have to write to finish on time ?","rationale":"\"63 \/ 7 = 9 a . 9\"","correct":"a","options":{"a":"9 ","b":"8 ","c":"10 ","d":"8.5","e":"6"},"options_float":{"a":9.0,"b":8.0,"c":10.0,"d":8.5,"e":6.0},"annotated_formula":"divide(63, 7)","linear_formula":"divide(n0,n1)|","chain":"63 \/ 7<\/gadget>\n9<\/output>\n9<\/result>","index":939} +{"problem":"the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 148 , what is the difference between the ages of patrick and monica ?","rationale":"\"ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 148 x = 3 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 3 = 48 answer b\"","correct":"b","options":{"a":"27 ","b":"48 ","c":"45 ","d":"72","e":"18"},"options_float":{"a":27.0,"b":48.0,"c":45.0,"d":72.0,"e":18.0},"annotated_formula":"subtract(multiply(multiply(5, 5), divide(148, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(148, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))))","linear_formula":"multiply(n0,n0)|multiply(n1,n0)|multiply(n1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,#2)|multiply(#5,#0)|subtract(#6,#7)|","chain":"5 * 5<\/gadget>\n25<\/output>\n3 * 3<\/gadget>\n9<\/output>\n3 * 5<\/gadget>\n15<\/output>\n9 + 15<\/gadget>\n24<\/output>\n24 + 25<\/gadget>\n49<\/output>\n148 \/ 49<\/gadget>\n148\/49 = around 3.020408<\/output>\n25 * (148\/49)<\/gadget>\n3_700\/49 = around 75.510204<\/output>\n9 * (148\/49)<\/gadget>\n1_332\/49 = around 27.183673<\/output>\n(3_700\/49) - (1_332\/49)<\/gadget>\n2_368\/49 = around 48.326531<\/output>\n2_368\/49 = around 48.326531<\/result>","index":940} +{"problem":"you collect baseball cards . suppose you start out with 15 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the end ?","rationale":"solution start with 15 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 15 + 1 which is 8 , so you ' re left with 15 - 8 = 7 . peter takes 1 baseball card from you : 7 - 1 = 6 baseball cards . paul triples the number of baseball cards you have : 6 × 3 = 18 baseball cards . so you have 18 at the end . correct answer : b","correct":"b","options":{"a":"17 ","b":"18 ","c":"19 ","d":"20","e":"21"},"options_float":{"a":17.0,"b":18.0,"c":19.0,"d":20.0,"e":21.0},"annotated_formula":"multiply(subtract(subtract(subtract(15, const_4), const_4), 1), const_3)","linear_formula":"subtract(n0,const_4)|subtract(#0,const_4)|subtract(#1,n1)|multiply(#2,const_3)","chain":"15 - 4<\/gadget>\n11<\/output>\n11 - 4<\/gadget>\n7<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":941} +{"problem":"xavier , yvonne , and zelda each try independently to solve a problem . if their individual probabilities for success are 1 \/ 4 , 2 \/ 3 and 5 \/ 8 , respectively , what is the probability that xavier and yvonne , but not zelda , will solve the problem ?","rationale":"p ( xavier will solve ) = 1 \/ 4 p ( yvonne will solve ) = 2 \/ 3 p ( zelda will not solve ) = 1 - 5 \/ 8 = 3 \/ 8 . now , we need to multiply all this ps to find an answer : p = ( 1 \/ 4 ) * ( 2 \/ 3 ) * ( 3 \/ 8 ) = 1 \/ 16 . ans . a .","correct":"a","options":{"a":"1 \/ 16 ","b":"7 \/ 8 ","c":"9 \/ 64 ","d":"5 \/ 64","e":"3 \/ 64"},"options_float":{"a":0.0625,"b":0.875,"c":0.140625,"d":0.078125,"e":0.046875},"annotated_formula":"multiply(multiply(divide(1, 4), divide(2, 3)), subtract(const_1, divide(5, 8)))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|multiply(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/4) * (2\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n(1\/6) * (3\/8)<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":944} +{"problem":"in a certain lottery , the probability that a number between 1 and 15 , inclusive , is drawn is 1 \/ 3 . if the probability that a number 1 or larger is drawn is 2 \/ 3 , what is the probability that a number less than or equal to 15 is drawn ?","rationale":"you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 1 or larger set 2 : number 15 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or less or both ) 2 \/ 3 + p ( set 2 ) - 1 \/ 3 = 1 p ( set 2 ) = 2 \/ 3 answer a","correct":"a","options":{"a":"2 \/ 3 ","b":"1 \/ 3 ","c":"1 \/ 4 ","d":"1 \/ 6","e":"2 \/ 5"},"options_float":{"a":0.6666666667,"b":0.3333333333,"c":0.25,"d":0.1666666667,"e":0.4},"annotated_formula":"subtract(add(1, divide(1, 3)), divide(2, 3))","linear_formula":"divide(n0,n3)|divide(n5,n3)|add(n0,#0)|subtract(#2,#1)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(4\/3) - (2\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":945} +{"problem":"if x is less than y by 25 % then y exceed x by :","rationale":"\"using formula ( x \/ ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 25 % ) = > 25 ( 100 - 25 ) * 100 = 33.33 % answer : a\"","correct":"a","options":{"a":"33.33 % ","b":"25 % ","c":"75 % ","d":"66.66 %","e":"none of these"},"options_float":{"a":33.33,"b":25.0,"c":75.0,"d":66.66,"e":null},"annotated_formula":"multiply(subtract(divide(const_100, subtract(const_100, 25)), const_1), const_100)","linear_formula":"subtract(const_100,n0)|divide(const_100,#0)|subtract(#1,const_1)|multiply(#2,const_100)|","chain":"100 - 25<\/gadget>\n75<\/output>\n100 \/ 75<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) - 1<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":948} +{"problem":"a pharmaceutical company received $ 8 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ?","rationale":"\"solution : this is a percent decrease problem . we will use the formula : percent change = ( new – old ) \/ old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 20 million royalties \/ sales = 8 \/ 20 next 108 million royalties \/ sales = 9 \/ 108 = 1 \/ 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties \/ sales = ( 8 \/ 20 ) x 60 = 24 next 108 million royalties \/ sales = 9 \/ 108 = ( 1 \/ 12 ) x 60 = 5 we can plug 24 and 5 into our percent change formula : ( new – old ) \/ old x 100 [ ( 5 – 24 ) \/ 24 ] x 100 - 475 \/ 6 x 100 at this point we can stop and consider the answer choices . since we know that 475 \/ 6 is just a bit less than ½ , we know that - 475 \/ 6 x 100 is about a 79 % decrease . answer e .\"","correct":"e","options":{"a":"8 % ","b":"15 % ","c":"45 % ","d":"52 %","e":"79 %"},"options_float":{"a":8.0,"b":15.0,"c":45.0,"d":52.0,"e":79.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(8, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(8, 20), const_100)), const_100)","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,const_100)|multiply(#1,const_100)|subtract(#2,#3)|divide(#4,#2)|multiply(#5,const_100)|","chain":"8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n9 \/ 108<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 100<\/gadget>\n25\/3 = around 8.333333<\/output>\n40 - (25\/3)<\/gadget>\n95\/3 = around 31.666667<\/output>\n(95\/3) \/ 40<\/gadget>\n19\/24 = around 0.791667<\/output>\n(19\/24) * 100<\/gadget>\n475\/6 = around 79.166667<\/output>\n475\/6 = around 79.166667<\/result>","index":949} +{"problem":"a square with sides of 15 units each . what would be the area of the square . consider a square as a combination of two right angled triangles .","rationale":"a square can be considered as a set of two right angled triangles joined from the hypotenuse . since the sides are 10 units each . the height and base of both the triangles is 10 each . area of a right angled triangle is 1 \/ 2 * base * height area of both the triangles is 2 * 1 \/ 2 * base * height . since both triangles are similar . = base * height = 15 * 15 = 225 ans : c","correct":"c","options":{"a":"100 ","b":"256 ","c":"225 ","d":"81","e":"144"},"options_float":{"a":100.0,"b":256.0,"c":225.0,"d":81.0,"e":144.0},"annotated_formula":"multiply(triangle_area(15, 15), const_2)","linear_formula":"triangle_area(n0,n0)|multiply(#0,const_2)","chain":"(15 * 15) \/ 2<\/gadget>\n225\/2 = around 112.5<\/output>\n(225\/2) * 2<\/gadget>\n225<\/output>\n225<\/result>","index":950} +{"problem":"how many digits are required to number a book containing 240 pages ?","rationale":"9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 240 - ( 90 + 9 ) = 141 pages will require 141 * 3 = 423 digits . the total number of digits is 9 + 180 + 423 = 612 . the answer is b .","correct":"b","options":{"a":"756 ","b":"612 ","c":"492 ","d":"372","e":"250"},"options_float":{"a":756.0,"b":612.0,"c":492.0,"d":372.0,"e":250.0},"annotated_formula":"add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(240, const_100), const_1), const_3))","linear_formula":"subtract(const_10,const_1)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#4,#0)|add(#6,#5)","chain":"10 - 1<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n90 * 2<\/gadget>\n180<\/output>\n9 + 180<\/gadget>\n189<\/output>\n240 - 100<\/gadget>\n140<\/output>\n140 + 1<\/gadget>\n141<\/output>\n141 * 3<\/gadget>\n423<\/output>\n189 + 423<\/gadget>\n612<\/output>\n612<\/result>","index":952} +{"problem":"out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many e applicants majored in political science and had a gpa higher than 3.0 ?","rationale":"\"total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) e = 20 - 15 = 5 answer : a\"","correct":"a","options":{"a":"5 ","b":"10 ","c":"15 ","d":"25","e":"35"},"options_float":{"a":5.0,"b":10.0,"c":15.0,"d":25.0,"e":35.0},"annotated_formula":"subtract(20, subtract(40, add(10, 15)))","linear_formula":"add(n1,n4)|subtract(n0,#0)|subtract(n2,#1)|","chain":"10 + 15<\/gadget>\n25<\/output>\n40 - 25<\/gadget>\n15<\/output>\n20 - 15<\/gadget>\n5<\/output>\n5<\/result>","index":953} +{"problem":"george does 3 \/ 5 th of a piece of work in 9 days . he then calls in paul and they finish the work in 4 days . how long would paul take to do the work by himself ?","rationale":"\"9 day 3 \/ 5 1 day 1 \/ 15 remaing 1 - 3 \/ 5 = 2 \/ 5 both can do 2 \/ 5 work in 4 day one day work of both = 1 \/ 10 therfor paul would take 1 \/ 10 - 1 \/ 15 = 1 \/ 30 or 30 days answer : c\"","correct":"c","options":{"a":"10 ","b":"20 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"inverse(divide(subtract(subtract(const_1, divide(3, 5)), multiply(4, divide(const_1, multiply(5, 3)))), 4))","linear_formula":"divide(n0,n1)|multiply(n1,n0)|divide(const_1,#1)|subtract(const_1,#0)|multiply(n3,#2)|subtract(#3,#4)|divide(#5,n3)|inverse(#6)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n5 * 3<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n4 * (1\/15)<\/gadget>\n4\/15 = around 0.266667<\/output>\n(2\/5) - (4\/15)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(2\/15) \/ 4<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ (1\/30)<\/gadget>\n30<\/output>\n30<\/result>","index":954} +{"problem":"in august , a cricket team that played 120 matches won 24 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ?","rationale":"let the no of matches played more = x so , ( 120 + x ) * 52 \/ 100 = 28.8 + x by solving we get x = 70 answer : d","correct":"d","options":{"a":"40 ","b":"52 ","c":"68 ","d":"70","e":"98"},"options_float":{"a":40.0,"b":52.0,"c":68.0,"d":70.0,"e":98.0},"annotated_formula":"divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(24, const_100), 120)), subtract(const_1, divide(52, const_100)))","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)","chain":"52 \/ 100<\/gadget>\n13\/25 = around 0.52<\/output>\n(13\/25) * 120<\/gadget>\n312\/5 = around 62.4<\/output>\n24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n(6\/25) * 120<\/gadget>\n144\/5 = around 28.8<\/output>\n(312\/5) - (144\/5)<\/gadget>\n168\/5 = around 33.6<\/output>\n1 - (13\/25)<\/gadget>\n12\/25 = around 0.48<\/output>\n(168\/5) \/ (12\/25)<\/gadget>\n70<\/output>\n70<\/result>","index":955} +{"problem":"in a survey of 340 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ?","rationale":"- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 \/ 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 286 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 340 we have to find not part time and not uninsured . in other words not part time and insured = x \/ 340 = ( 286 - 104 + 13 ) \/ 340 = 39 \/ 68 answer is c .","correct":"c","options":{"a":"7 \/ 12 ","b":"8 \/ 41 ","c":"39 \/ 68 ","d":"1 \/ 8","e":"41 \/ 91"},"options_float":{"a":0.5833333333,"b":0.1951219512,"c":0.5735294118,"d":0.125,"e":0.4505494505},"annotated_formula":"divide(subtract(340, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 340)","linear_formula":"multiply(n1,n3)|multiply(const_1,const_100)|divide(#0,#1)|subtract(n2,#2)|add(n1,#3)|subtract(n0,#4)|divide(#5,n0)","chain":"12.5 * 104<\/gadget>\n1_300<\/output>\n1 * 100<\/gadget>\n100<\/output>\n1_300 \/ 100<\/gadget>\n13<\/output>\n54 - 13<\/gadget>\n41<\/output>\n41 + 104<\/gadget>\n145<\/output>\n340 - 145<\/gadget>\n195<\/output>\n195 \/ 340<\/gadget>\n39\/68 = around 0.573529<\/output>\n39\/68 = around 0.573529<\/result>","index":956} +{"problem":"a ( 3 , w ^ 3 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 - 1 . what is the value of w ?","rationale":"y = x ^ 2 - 1 w ^ 3 = 3 ^ 2 - 1 w ^ 3 = 8 w = 2 answer a","correct":"a","options":{"a":"2 . ","b":"4 . ","c":"5 . ","d":"6 .","e":"9 ."},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":9.0},"annotated_formula":"power(subtract(power(3, 2), 1), divide(1, 3))","linear_formula":"divide(n3,n0)|power(n0,n2)|subtract(#1,n3)|power(#2,#0)","chain":"3 ** 2<\/gadget>\n9<\/output>\n9 - 1<\/gadget>\n8<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n8 ** (1\/3)<\/gadget>\n2<\/output>\n2<\/result>","index":957} +{"problem":"in x game of billiards , x can give y 20 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 120 ?","rationale":"x scores 60 while y score 40 and z scores 30 . the number of points that z scores when y scores 120 = ( 120 * 30 ) \/ 40 = 90 . in x game of 100 points , y gives ( 120 - 90 ) = 30 points to c . a","correct":"a","options":{"a":"30 ","b":"20 ","c":"25 ","d":"40","e":"50"},"options_float":{"a":30.0,"b":20.0,"c":25.0,"d":40.0,"e":50.0},"annotated_formula":"subtract(multiply(subtract(60, 20), const_3), multiply(subtract(60, 30), const_3))","linear_formula":"subtract(n1,n0)|subtract(n1,n2)|multiply(#0,const_3)|multiply(#1,const_3)|subtract(#2,#3)","chain":"60 - 20<\/gadget>\n40<\/output>\n40 * 3<\/gadget>\n120<\/output>\n60 - 30<\/gadget>\n30<\/output>\n30 * 3<\/gadget>\n90<\/output>\n120 - 90<\/gadget>\n30<\/output>\n30<\/result>","index":958} +{"problem":"a house wife saved $ 2.50 in buying an item on sale . if she spent $ 25 for the item , approximately how much percent she saved in the transaction ?","rationale":"actual price = 25 + 2.50 = $ 27.50 saving = 2.50 \/ 27.50 * 100 = 100 \/ 11 = 9 % approximately answer is b","correct":"b","options":{"a":"8 % ","b":"9 % ","c":"10 % ","d":"11 %","e":"12 %"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":12.0},"annotated_formula":"multiply(divide(2.5, add(25, 2.5)), const_100)","linear_formula":"add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)","chain":"25 + 2.5<\/gadget>\n27.5<\/output>\n2.5 \/ 27.5<\/gadget>\n0.090909<\/output>\n0.090909 * 100<\/gadget>\n9.0909<\/output>\n9.0909<\/result>","index":959} +{"problem":"after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6700 . find the actual price of the good .","rationale":"let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6700 1 = 6700 \/ 68.4 100 = ( 6700 * 100 ) \/ 68.4 = rs . 9795 . answer : option d","correct":"d","options":{"a":"s . 6000 ","b":"s . 9000 ","c":"s . 10800 ","d":"s . 9795","e":"s . 9980"},"options_float":{"a":6000.0,"b":9000.0,"c":10800.0,"d":9795.0,"e":9980.0},"annotated_formula":"divide(multiply(6700, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100)))","linear_formula":"multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)","chain":"6_700 * 100<\/gadget>\n670_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 * 10<\/gadget>\n800<\/output>\n800 \/ 100<\/gadget>\n8<\/output>\n80 - 8<\/gadget>\n72<\/output>\n72 * 5<\/gadget>\n360<\/output>\n360 \/ 100<\/gadget>\n18\/5 = around 3.6<\/output>\n72 - (18\/5)<\/gadget>\n342\/5 = around 68.4<\/output>\n670_000 \/ (342\/5)<\/gadget>\n1_675_000\/171 = around 9_795.321637<\/output>\n1_675_000\/171 = around 9_795.321637<\/result>","index":960} +{"problem":"in a factory , an average of 60 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 58 sets \/ day . the average production per day for day last 5 days is ?","rationale":"production during these 5 days = total production in a month - production in first 25 days . = 30 x 58 - 25 x 60 = 240 ∴ average for last 5 days = 240 \/ 5 = 48 c","correct":"c","options":{"a":"20 ","b":"36 ","c":"48 ","d":"50","e":"59"},"options_float":{"a":20.0,"b":36.0,"c":48.0,"d":50.0,"e":59.0},"annotated_formula":"divide(subtract(multiply(add(25, 5), 58), multiply(25, 60)), 5)","linear_formula":"add(n1,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2)","chain":"25 + 5<\/gadget>\n30<\/output>\n30 * 58<\/gadget>\n1_740<\/output>\n25 * 60<\/gadget>\n1_500<\/output>\n1_740 - 1_500<\/gadget>\n240<\/output>\n240 \/ 5<\/gadget>\n48<\/output>\n48<\/result>","index":962} +{"problem":"the ratio between the present ages of a and b is 5 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ?","rationale":"\"let the present ages of a and b be 5 x and 3 x years respectively . then , ( 5 x - 4 ) \/ ( 3 x + 4 ) = 1 \/ 1 2 x = 8 = > x = 4 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 24 : 8 = 3 : 1 answer : c\"","correct":"c","options":{"a":"3 : 9 ","b":"3 : 6 ","c":"3 : 1 ","d":"3 : 4","e":"3 : 2"},"options_float":{"a":0.3333333333,"b":0.5,"c":3.0,"d":0.75,"e":1.5},"annotated_formula":"divide(add(multiply(5, divide(add(5, 5), subtract(5, 3))), 5), subtract(multiply(3, divide(add(5, 5), subtract(5, 3))), 5))","linear_formula":"add(n0,n0)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n0,#3)|subtract(#4,n0)|divide(#5,#6)|","chain":"5 + 5<\/gadget>\n10<\/output>\n5 - 3<\/gadget>\n2<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25 + 5<\/gadget>\n30<\/output>\n3 * 5<\/gadget>\n15<\/output>\n15 - 5<\/gadget>\n10<\/output>\n30 \/ 10<\/gadget>\n3<\/output>\n3<\/result>","index":964} +{"problem":"in n is a positive integer less than 200 , and 14 n \/ 60 is an integer , then n has how many different positive prime factors s ?","rationale":"\"i like to put the numbers in prime factors so it is easier and faster to visualize . 14 * n \/ 60 if we write the factors of 14 - - > 2 , 7 , and the factors of 60 - - > 2 , 2 , 3 , 5 , we have ( 2 * 7 * n ) \/ ( 2 ^ 2 * 3 * 5 ) simplifying 7 * n \/ ( 2 * 3 * 5 ) the only way the equation above has an integer value is if n has at least the factors 2 , 3 and 5 , so we can simplify again and we have the number 7 . the number could be 2 * 3 * 5 , or 2 * 3 * 5 * 2 , or 2 * 3 * 5 * . . . . . however to be less than 200 we can not add any prime number . 2 * 3 * 5 = 120 if we added the next prime factor 7 , we would have s = 2 * 3 * 5 * 7 = 840 thus , answer b\"","correct":"b","options":{"a":"2 ","b":"3 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":2.0,"b":3.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2)","linear_formula":"add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)|","chain":"1 + 4<\/gadget>\n5<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n5 \/ 5<\/gadget>\n1<\/output>\n1 + 2<\/gadget>\n3<\/output>\n3<\/result>","index":966} +{"problem":"an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 40 cm , find the sum of the perimeters of all the triangles .","rationale":"we have 40 for first triangle , when we join mid - points of first triangle we get the second equilateral traingle then the length of second one is 20 and continues . so we have 40 , 20,10 , . . . we have ratio = 1 \/ 2 , and it is gp type . sum of infinite triangle is a \/ 1 - r = 40 \/ 1 - ( 1 \/ 2 ) = 80 equilateral triangle perimeter is 3 a = 3 * 80 = 240 . so option c .","correct":"c","options":{"a":"180 cm ","b":"220 cm ","c":"240 cm ","d":"270 cm","e":"300 cm"},"options_float":{"a":180.0,"b":220.0,"c":240.0,"d":270.0,"e":300.0},"annotated_formula":"add(triangle_perimeter(40, 40, 40), triangle_perimeter(40, 40, 40))","linear_formula":"triangle_perimeter(n5,n5,n5)|add(#0,#0)","chain":"40 + 40 + 40<\/gadget>\n120<\/output>\n120 + 120<\/gadget>\n240<\/output>\n240<\/result>","index":967} +{"problem":"two pipes a and b can fill a tank in 10 hours and 20 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ?","rationale":"\"part filled by a in 1 hour = 1 \/ 10 part filled by b in 1 hour = 1 \/ 20 part filled by ( a + b ) in 1 hour = 1 \/ 10 + 1 \/ 20 = 3 \/ 20 both the pipes together fill the tank in 20 \/ 3 hours answer is b\"","correct":"b","options":{"a":"10 \/ 3 ","b":"20 \/ 3 ","c":"15 \/ 7 ","d":"11 \/ 17","e":"1 \/ 4"},"options_float":{"a":3.3333333333,"b":6.6666666667,"c":2.1428571429,"d":0.6470588235,"e":0.25},"annotated_formula":"divide(const_1, add(divide(const_1, 10), divide(const_1, 20)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/10) + (1\/20)<\/gadget>\n3\/20 = around 0.15<\/output>\n1 \/ (3\/20)<\/gadget>\n20\/3 = around 6.666667<\/output>\n20\/3 = around 6.666667<\/result>","index":968} +{"problem":"the number obtained by interchanging the digits of a two - digit number is less than the original number by 36 . the sum of the digits of the number is 8 . what is the original number ?","rationale":"explanation : let the number be 10 x + y . ∴ x + y = 8 . . . . . ( i ) 10 x + y - ( 10 y + x ) = 36 ∴ 9 ( x - y ) = 3 ∴ x - y = 4 . . . . . . . ( ii ) from equation ( i ) and ( ii ) : x = 6 , y = 2 ∴ number = 62 answer : option b","correct":"b","options":{"a":"71 ","b":"62 ","c":"53 ","d":"35","e":"45"},"options_float":{"a":71.0,"b":62.0,"c":53.0,"d":35.0,"e":45.0},"annotated_formula":"add(multiply(subtract(8, divide(36, add(const_10, 8))), const_10), divide(36, add(const_10, 8)))","linear_formula":"add(n1,const_10)|divide(n0,#0)|subtract(n1,#1)|multiply(#2,const_10)|add(#1,#3)","chain":"10 + 8<\/gadget>\n18<\/output>\n36 \/ 18<\/gadget>\n2<\/output>\n8 - 2<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n60 + 2<\/gadget>\n62<\/output>\n62<\/result>","index":969} +{"problem":"in one alloy there is 10 % chromium while in another alloy it is 6 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy .","rationale":"\"the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.10 * 15 + 0.06 * 35 = 3.6 kg , so the percentage is 3.6 \/ 50 * 100 = 7.2 % . answer : a .\"","correct":"a","options":{"a":"7.2 % ","b":"9 % ","c":"9.2 % ","d":"8.6 %","e":"8.4 %"},"options_float":{"a":7.2,"b":9.0,"c":9.2,"d":8.6,"e":8.4},"annotated_formula":"multiply(divide(add(divide(multiply(10, 15), const_100), divide(multiply(6, 35), const_100)), add(15, 35)), const_100)","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"10 * 15<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n6 * 35<\/gadget>\n210<\/output>\n210 \/ 100<\/gadget>\n21\/10 = around 2.1<\/output>\n(3\/2) + (21\/10)<\/gadget>\n18\/5 = around 3.6<\/output>\n15 + 35<\/gadget>\n50<\/output>\n(18\/5) \/ 50<\/gadget>\n9\/125 = around 0.072<\/output>\n(9\/125) * 100<\/gadget>\n36\/5 = around 7.2<\/output>\n36\/5 = around 7.2<\/result>","index":970} +{"problem":"the food in a camp lasts for 10 men for 20 days . if 30 more men join , how many days will the food last ?","rationale":"\"one man can consume the same food in 10 * 20 = 200 days . 30 more men join , the total number of men = 40 the number of days the food will last = 200 \/ 40 = 5 days . answer : c\"","correct":"c","options":{"a":"8 days ","b":"4 days ","c":"5 days ","d":"10 days","e":"17 days"},"options_float":{"a":8.0,"b":4.0,"c":5.0,"d":10.0,"e":17.0},"annotated_formula":"divide(multiply(10, 20), add(10, 30))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"10 * 20<\/gadget>\n200<\/output>\n10 + 30<\/gadget>\n40<\/output>\n200 \/ 40<\/gadget>\n5<\/output>\n5<\/result>","index":971} +{"problem":"surface area of a sphere is 2464 cm 2 . if its radius is doubled , then the surface area of the new sphere will be","rationale":"sol . let the original radius be r . then , original surface area = 4 π r 2 = 2464 cm 2 ( given ) . new radius = 2 r . ∴ new surface area = 4 π ( 2 r ) 2 = 4 x 4 π r 2 = ( 4 x 2464 ) cm 2 = 9856 cm 2 answer c","correct":"c","options":{"a":"4920 cm 2 ","b":"4727 cm 2 ","c":"9856 cm 2 ","d":"19712 cm 2","e":"none"},"options_float":{"a":4920.0,"b":4727.0,"c":9856.0,"d":19712.0,"e":null},"annotated_formula":"surface_sphere(multiply(sqrt(divide(2464, multiply(const_4, const_pi))), const_2))","linear_formula":"multiply(const_4,const_pi)|divide(n0,#0)|sqrt(#1)|multiply(#2,const_2)|surface_sphere(#3)","chain":"4 * pi<\/gadget>\n4*pi = around 12.566371<\/output>\n2_464 \/ (4*pi)<\/gadget>\n616\/pi = around 196.07889<\/output>\n(616\/pi) ** (1\/2)<\/gadget>\n2*sqrt(154)\/sqrt(pi) = around 14.002817<\/output>\n(2*sqrt(154)\/sqrt(pi)) * 2<\/gadget>\n4*sqrt(154)\/sqrt(pi) = around 28.005634<\/output>\n4 * pi * ((4*sqrt(154)\/sqrt(pi)) ** 2)<\/gadget>\n9_856<\/output>\n9_856<\/result>","index":972} +{"problem":"how long does a train 100 m long travelling at 60 kmph takes to cross a bridge of 80 m in length ?","rationale":"\"d 16.8 sec d = 100 + 80 = 180 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 180 * 3 \/ 50 = 10.8 sec\"","correct":"d","options":{"a":"15.8 sec ","b":"14.9 sec ","c":"12.4 sec ","d":"10.8 sec","e":"11.8 sec"},"options_float":{"a":15.8,"b":14.9,"c":12.4,"d":10.8,"e":11.8},"annotated_formula":"divide(add(100, 80), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"100 + 80<\/gadget>\n180<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n180 \/ (50\/3)<\/gadget>\n54\/5 = around 10.8<\/output>\n54\/5 = around 10.8<\/result>","index":973} +{"problem":"- 74 x 29 + 265 = ?","rationale":"\"given exp . = - 74 x ( 30 - 1 ) + 265 = - ( 74 x 30 ) + 74 + 265 = - 2220 + 339 = - 1881 answer is b\"","correct":"b","options":{"a":"2436 ","b":"- 1881 ","c":"- 2801 ","d":"- 2071","e":"none of them"},"options_float":{"a":2436.0,"b":-1881.0,"c":-2801.0,"d":-2071.0,"e":null},"annotated_formula":"multiply(subtract(const_1, const_2), subtract(multiply(74, 29), 265))","linear_formula":"multiply(n0,n1)|subtract(const_1,const_2)|subtract(#0,n2)|multiply(#1,#2)|","chain":"1 - 2<\/gadget>\n-1<\/output>\n74 * 29<\/gadget>\n2_146<\/output>\n2_146 - 265<\/gadget>\n1_881<\/output>\n(-1) * 1_881<\/gadget>\n-1_881<\/output>\n-1_881<\/result>","index":974} +{"problem":"the population of 6 villages is 803 , 9001100 , 1023945 and 980 . what is the population of the seventh village if the average population of the 7 villages is 1000 ?","rationale":"population of seventh village = 7 * 1000 - ( 803 + 900 + 1100 + 1023 + 945 + 980 ) = 1249 answer is e","correct":"e","options":{"a":"1134 ","b":"1238 ","c":"1348 ","d":"1500","e":"1249"},"options_float":{"a":1134.0,"b":1238.0,"c":1348.0,"d":1500.0,"e":1249.0},"annotated_formula":"subtract(multiply(1000, 7), add(add(const_10, const_10), subtract(subtract(multiply(980, 6), multiply(7, 7)), const_100)))","linear_formula":"add(const_10,const_10)|multiply(n5,n6)|multiply(n0,n4)|multiply(n5,n5)|subtract(#2,#3)|subtract(#4,const_100)|add(#0,#5)|subtract(#1,#6)","chain":"1_000 * 7<\/gadget>\n7_000<\/output>\n10 + 10<\/gadget>\n20<\/output>\n980 * 6<\/gadget>\n5_880<\/output>\n7 * 7<\/gadget>\n49<\/output>\n5_880 - 49<\/gadget>\n5_831<\/output>\n5_831 - 100<\/gadget>\n5_731<\/output>\n20 + 5_731<\/gadget>\n5_751<\/output>\n7_000 - 5_751<\/gadget>\n1_249<\/output>\n1_249<\/result>","index":975} +{"problem":"if 625 ^ ( - x ) + 25 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 14 , what is the value of x ?","rationale":"\"we ' re told that 625 ^ ( - x ) + 25 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 15 . we ' re asked for the value of x . since each of the calculated terms must be positive ( regardless of what the exponent is ) , we can use thebasesto our advantage . . . . . with answer a , we ' d have 625 ^ 4 , which is much bigger than 15 ( and we ' d be adding to that big number ) . eliminate a . with answer e , we ' d have 625 ^ ( - 4 ) , which would create a tiny fraction ( and we ' d add some other fractions to it , so the total would be much too small ) . eliminate e . with answer d , we ' d have 625 ^ ( - 1 \/ 4 ) , which will also be a fraction ( just not as tiny as the one in answer e ) , but the total would still be too small . eliminate d . with answer c , anything to the ' 0 power ' is 1 , so we ' d have 1 + 1 + 1 = 3 . this is not 15 . eliminate c . d\"","correct":"d","options":{"a":"- 4 ","b":"- 1 \/ 4 ","c":"0 ","d":"1 \/ 4","e":"4"},"options_float":{"a":-4.0,"b":-0.25,"c":0.0,"d":0.25,"e":4.0},"annotated_formula":"divide(const_1, 4)","linear_formula":"divide(const_1,n4)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":976} +{"problem":"the radius of the wheel of a bus is 70 cms and the speed of the bus is 66 km \/ h , then the r . p . m . ( revolutions per minutes ) of the wheel is","rationale":"\"radius of the wheel of bus = 70 cm . then , circumference of wheel = 2 π r = 140 π = 440 cm distance covered by bus in 1 minute = 66 ⁄ 60 × 1000 × 100 cms distance covered by one revolution of wheel = circumference of wheel = 440 cm ∴ revolutions per minute = 6600000 \/ 60 × 440 = 250 answer b\"","correct":"b","options":{"a":"200 ","b":"250 ","c":"300 ","d":"330","e":"none of these"},"options_float":{"a":200.0,"b":250.0,"c":300.0,"d":330.0,"e":null},"annotated_formula":"divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 70), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2)))))","linear_formula":"multiply(n1,const_1000)|multiply(const_1,const_60)|multiply(const_3,const_4)|multiply(const_2,const_4)|multiply(n0,const_2)|add(#2,const_2)|add(#3,const_2)|multiply(#0,const_100)|divide(#7,#1)|power(#6,const_2)|divide(#5,#9)|add(#10,const_3)|multiply(#11,#4)|divide(#8,#12)|","chain":"1_000 * 66<\/gadget>\n66_000<\/output>\n100 * 66_000<\/gadget>\n6_600_000<\/output>\n60 * 1<\/gadget>\n60<\/output>\n6_600_000 \/ 60<\/gadget>\n110_000<\/output>\n2 * 70<\/gadget>\n140<\/output>\n3 * 4<\/gadget>\n12<\/output>\n2 + 12<\/gadget>\n14<\/output>\n4 * 2<\/gadget>\n8<\/output>\n2 + 8<\/gadget>\n10<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n3 + (7\/50)<\/gadget>\n157\/50 = around 3.14<\/output>\n140 * (157\/50)<\/gadget>\n2_198\/5 = around 439.6<\/output>\n110_000 \/ (2_198\/5)<\/gadget>\n275_000\/1_099 = around 250.22748<\/output>\n275_000\/1_099 = around 250.22748<\/result>","index":977} +{"problem":"harry started a 4 - mile hike with a full 6 - cup canteen of water and finished the hike in 2 hours with 1 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 1 cups of water during the last mile , how many cups did he drink per mile during the first 3 miles of the hike ?","rationale":"no of cups leaked during the trip = 2 cups . no of cups harry drank = 3 cups . no of cups harry drank during the first 3 miles = 2 . drink \/ mile = 2 \/ 3 answer : b","correct":"b","options":{"a":"1 \/ 3 ","b":"2 \/ 3 ","c":"3 \/ 4 ","d":"5 \/ 4","e":"4 \/ 3"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":0.75,"d":1.25,"e":1.3333333333},"annotated_formula":"divide(subtract(subtract(subtract(6, 2), 1), 1), 3)","linear_formula":"subtract(n1,n2)|subtract(#0,n3)|subtract(#1,n3)|divide(#2,n6)","chain":"6 - 2<\/gadget>\n4<\/output>\n4 - 1<\/gadget>\n3<\/output>\n3 - 1<\/gadget>\n2<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":978} +{"problem":"two taps a and b can fill a cistern in 12 minutes and 18 minutes respectively . they are turned on at the same time . if the tap a is turned off after 4 minutes , how long will tap b take to fill the rest of the cistern ?","rationale":"in one min , ( a + b ) fill the cistern = 1 ⁄ 12 + 1 ⁄ 18 = 5 ⁄ 36 th in 4 min , ( a + b ) fill the cistern = 5 ⁄ 36 × 4 = 5 ⁄ 9 th rest part = 1 - 5 ⁄ 9 = 4 ⁄ 9 th ∵ 1 ⁄ 18 th part is filled by b in one min . ∴ 4 ⁄ 9 th part is filled by b in 18 × 4 ⁄ 9 = 8 min answer a","correct":"a","options":{"a":"8 min . ","b":"9 min . ","c":"10 min . ","d":"7 min .","e":"none of these"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":7.0,"e":null},"annotated_formula":"divide(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 18)), 4)), divide(const_1, 18))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|divide(#4,#1)","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/12) + (1\/18)<\/gadget>\n5\/36 = around 0.138889<\/output>\n(5\/36) * 4<\/gadget>\n5\/9 = around 0.555556<\/output>\n1 - (5\/9)<\/gadget>\n4\/9 = around 0.444444<\/output>\n(4\/9) \/ (1\/18)<\/gadget>\n8<\/output>\n8<\/result>","index":980} +{"problem":"jack and christina are standing 240 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 3 feet per second and christina walks in a straight line toward jack at a constant speed of 3 feet per second . lindy runs at a constant speed of 10 feet per second from christina to jack , back to christina , back to jack , and so forth . what is the total distance , in feet , that lindy has traveled when the three meet at one place ?","rationale":"\"the relative speed of jack and christina is 3 + 3 = 6 feet per second . the distance between them is 210 feet , hence they will meet in ( time ) = ( distance ) \/ ( relative speed ) = 240 \/ 6 = 40 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 10 * 40 = 400 feet . answer : d .\"","correct":"d","options":{"a":"340 ","b":"360 ","c":"380 ","d":"400","e":"420"},"options_float":{"a":340.0,"b":360.0,"c":380.0,"d":400.0,"e":420.0},"annotated_formula":"multiply(divide(240, add(3, 3)), 10)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n3,#1)|","chain":"3 + 3<\/gadget>\n6<\/output>\n240 \/ 6<\/gadget>\n40<\/output>\n40 * 10<\/gadget>\n400<\/output>\n400<\/result>","index":982} +{"problem":"a solution of 70 litres contains oil and water in the ratio 9 : x . if 6 litres of water is added to the solution , the ratio becomes 6 : 4 , find the value of x ?","rationale":"exp . total new quantity = original sol + water = 70 + 6 = 76 new ratio = 6 : 4 , new quantity of oil = 6 \/ 10 * 76 = 45.6 lit , new quantity of water = 4 \/ 10 * 76 = 30.4 lit water present initially = ( 30.4 - 6 ) = 24.4 lit ratio = 45.6 \/ 24.4 = 9 \/ x there for x = 4.81 approx 5 answer : e","correct":"e","options":{"a":"6 ","b":"8.5 ","c":"9 ","d":"6.5","e":"5"},"options_float":{"a":6.0,"b":8.5,"c":9.0,"d":6.5,"e":5.0},"annotated_formula":"divide(9, divide(subtract(70, subtract(multiply(4, divide(add(70, 6), add(6, 4))), 6)), subtract(multiply(4, divide(add(70, 6), add(6, 4))), 6)))","linear_formula":"add(n0,n2)|add(n2,n4)|divide(#0,#1)|multiply(n4,#2)|subtract(#3,n2)|subtract(n0,#4)|divide(#5,#4)|divide(n1,#6)","chain":"70 + 6<\/gadget>\n76<\/output>\n6 + 4<\/gadget>\n10<\/output>\n76 \/ 10<\/gadget>\n38\/5 = around 7.6<\/output>\n4 * (38\/5)<\/gadget>\n152\/5 = around 30.4<\/output>\n(152\/5) - 6<\/gadget>\n122\/5 = around 24.4<\/output>\n70 - (122\/5)<\/gadget>\n228\/5 = around 45.6<\/output>\n(228\/5) \/ (122\/5)<\/gadget>\n114\/61 = around 1.868852<\/output>\n9 \/ (114\/61)<\/gadget>\n183\/38 = around 4.815789<\/output>\n183\/38 = around 4.815789<\/result>","index":984} +{"problem":"if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number d of integers that can be negative ?","rationale":"\"minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning 5 out of 6 also possible as you will see d = 5 or 1 will give you same equation . . ans d\"","correct":"d","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"add(divide(const_10, const_2), 0)","linear_formula":"divide(const_10,const_2)|add(n0,#0)|","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5 + 0<\/gadget>\n5<\/output>\n5<\/result>","index":986} +{"problem":"the length of the rectangular field is double its width . inside the field there is square shaped pond 6 m long . if the area of the pond is 1 \/ 8 of the area of the field . what is the length of the field ?","rationale":"\"a \/ 8 = 6 * 6 = > a = 6 * 6 * 8 x * 2 x = 6 * 6 * 8 x = 12 = > 2 x = 24 answer : a\"","correct":"a","options":{"a":"24 ","b":"32 ","c":"75 ","d":"28","e":"11"},"options_float":{"a":24.0,"b":32.0,"c":75.0,"d":28.0,"e":11.0},"annotated_formula":"sqrt(divide(multiply(square_area(6), 8), inverse(const_2)))","linear_formula":"inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)|","chain":"6 ** 2<\/gadget>\n36<\/output>\n36 * 8<\/gadget>\n288<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n288 \/ (1\/2)<\/gadget>\n576<\/output>\n576 ** (1\/2)<\/gadget>\n24<\/output>\n24<\/result>","index":988} +{"problem":"the area of a triangle is with base 2 m and height 3 m ?","rationale":"\"1 \/ 2 * 2 * 3 = 3 m 2 answer : d\"","correct":"d","options":{"a":"11 ","b":"10 ","c":"787 ","d":"3","e":"12"},"options_float":{"a":11.0,"b":10.0,"c":787.0,"d":3.0,"e":12.0},"annotated_formula":"triangle_area(2, 3)","linear_formula":"triangle_area(n0,n1)|","chain":"(2 * 3) \/ 2<\/gadget>\n3<\/output>\n3<\/result>","index":989} +{"problem":"if the price of a certain computer increased 30 percent from c dollars to 351 dollars , then 2 c =","rationale":"\"before price increase price = c after 30 % price increase price = c + ( 30 \/ 100 ) * c = 1.3 c = 351 ( given ) i . e . c = 351 \/ 1.3 = $ 270 i . e . 2 c = 2 * 270 = 540 answer : option a\"","correct":"a","options":{"a":"540 ","b":"570 ","c":"619 ","d":"649","e":"700"},"options_float":{"a":540.0,"b":570.0,"c":619.0,"d":649.0,"e":700.0},"annotated_formula":"multiply(divide(351, divide(add(const_100, 30), const_100)), 2)","linear_formula":"add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)|","chain":"100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n351 \/ (13\/10)<\/gadget>\n270<\/output>\n270 * 2<\/gadget>\n540<\/output>\n540<\/result>","index":990} +{"problem":"the purchase price of an article is $ 48 . in order to include 5 % of cost for overhead and to provide $ 12 of net profit , the markup should be","rationale":"\"cost price of article = 48 $ % of overhead cost = 5 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 \/ 48 ) * 100 = 25 % total markup should be = 25 + 5 = 30 % answer c\"","correct":"c","options":{"a":"15 % ","b":"25 % ","c":"30 % ","d":"40 %","e":"45 %"},"options_float":{"a":15.0,"b":25.0,"c":30.0,"d":40.0,"e":45.0},"annotated_formula":"add(multiply(divide(12, 48), const_100), 5)","linear_formula":"divide(n2,n0)|multiply(#0,const_100)|add(n1,#1)|","chain":"12 \/ 48<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25 + 5<\/gadget>\n30<\/output>\n30<\/result>","index":991} +{"problem":"a store reduced the price of all items in the store by 12 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ?","rationale":"\"consider price of the all items as $ 100 after a initial reduction of 12 % price becomes = 0.88 * 100 = $ 88 after the final reduction of 10 % price becomes = 0.9 * 88 = $ 79.2 price of all items on second day is 79.2 % of price on first day correct answer option b\"","correct":"b","options":{"a":"80.0 ","b":"79.2 ","c":"81.0 ","d":"81.1","e":"81.9"},"options_float":{"a":80.0,"b":79.2,"c":81.0,"d":81.1,"e":81.9},"annotated_formula":"multiply(multiply(divide(subtract(const_100, 12), const_100), divide(subtract(const_100, 10), const_100)), const_100)","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)|","chain":"100 - 12<\/gadget>\n88<\/output>\n88 \/ 100<\/gadget>\n22\/25 = around 0.88<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n(22\/25) * (9\/10)<\/gadget>\n99\/125 = around 0.792<\/output>\n(99\/125) * 100<\/gadget>\n396\/5 = around 79.2<\/output>\n396\/5 = around 79.2<\/result>","index":992} +{"problem":"darcy lives 1.5 miles from work . she can walk to work at a constant rate of 3 miles per hour , or she can ride the train to work at a constant rate of 20 miles per hour . if she rides the train , there is an additional x minutes spent walking to the nearest train station , waiting for the train , and walking from the final train station to her work . if it takes darcy a total of 10 more minutes to commute to work by walking than it takes her to commute to work by riding the train , what is the value of x ?","rationale":"\"the time it takes darcy to walk to work is ( 1.5 \/ 3 ) * 60 = 30 minutes the time it takes darcy to take the train is ( 1.5 \/ 20 ) * 60 + x = 4.5 + x minutes it takes 15 minutes longer to walk , so 30 = 4.5 + x + 10 x = 15.5 minutes answer : c\"","correct":"c","options":{"a":"10.5 ","b":"15 ","c":"15.5 ","d":"30","e":"60"},"options_float":{"a":10.5,"b":15.0,"c":15.5,"d":30.0,"e":60.0},"annotated_formula":"subtract(subtract(divide(const_60, const_2), 10), divide(const_60, divide(20, 1.5)))","linear_formula":"divide(const_60,const_2)|divide(n2,n0)|divide(const_60,#1)|subtract(#0,n3)|subtract(#3,#2)|","chain":"60 \/ 2<\/gadget>\n30<\/output>\n30 - 10<\/gadget>\n20<\/output>\n20 \/ 1.5<\/gadget>\n13.333333<\/output>\n60 \/ 13.333333<\/gadget>\n4.5<\/output>\n20 - 4.5<\/gadget>\n15.5<\/output>\n15.5<\/result>","index":993} +{"problem":"when the smallest of 3 consecutive odd integers is added to 4 times the largest , it produces a result 732 more than 4 times the middle integer . find the numbers ?","rationale":"x + 4 ( x + 4 ) = 732 + 4 ( x + 2 ) solve for x and find all three numbers x + 4 x + 16 = 732 + 4 x + 8 x = 724 x + 2 = 726 x + 4 = 728 check : the smallest is added to four times the largest 724 + 4 * 728 = 3636 four times the middle 4 * 726 = 2904 3636 is more than 2904 by 3636 - 2904 = 732 a","correct":"a","options":{"a":"732 ","b":"678 ","c":"698 ","d":"710","e":"729"},"options_float":{"a":732.0,"b":678.0,"c":698.0,"d":710.0,"e":729.0},"annotated_formula":"divide(subtract(add(732, multiply(4, const_2)), multiply(4, const_4)), subtract(add(const_1, 4), 4))","linear_formula":"add(n1,const_1)|multiply(n1,const_2)|multiply(n1,const_4)|add(n2,#1)|subtract(#0,n1)|subtract(#3,#2)|divide(#5,#4)","chain":"4 * 2<\/gadget>\n8<\/output>\n732 + 8<\/gadget>\n740<\/output>\n4 * 4<\/gadget>\n16<\/output>\n740 - 16<\/gadget>\n724<\/output>\n1 + 4<\/gadget>\n5<\/output>\n5 - 4<\/gadget>\n1<\/output>\n724 \/ 1<\/gadget>\n724<\/output>\n724<\/result>","index":994} +{"problem":"in a particular city there are 100 homes numbered from 1 , 23 , . . , 100 . the city was build by a builder from chennai . there was 4 to 5 shops in the town which was build by a builder from mumbai . the 2 nd builder can build in 1 ½ times as compared to time by 1 st builder . the 2 nd builder builds in 15 days , then how many 2 ' s are used by the builder from chennai in numbering the 100 homes ?","rationale":"in the last line it should be given the builder from chennai in numbering the 100 homes ? then ans is 20 answer : d","correct":"d","options":{"a":"17 ","b":"18 ","c":"19 ","d":"20","e":"21"},"options_float":{"a":17.0,"b":18.0,"c":19.0,"d":20.0,"e":21.0},"annotated_formula":"multiply(subtract(15, 5), 2)","linear_formula":"subtract(n10,n5)|multiply(n6,#0)","chain":"15 - 5<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":995} +{"problem":"p and q can complete a work in 40 days and 24 days respectively . p alone started the work and q joined him after 8 days till the completion of the work . how long did the work last ?","rationale":"\"explanation : work done by p in 1 day = 1 \/ 40 work done by q in 1 day = 1 \/ 24 work done by p in 8 days = 8 ã — ( 1 \/ 40 ) = 1 \/ 5 remaining work = 1 â € “ 1 \/ 5 = 4 \/ 5 work done by p and q in 1 day = 1 \/ 40 + 1 \/ 24 = 1 \/ 15 number of days p and q take to complete the remaining work = ( 4 \/ 5 ) \/ ( 1 \/ 15 ) = 12 total days = 8 + 12 = 20 answer : option d\"","correct":"d","options":{"a":"5 days ","b":"10 days ","c":"14 days ","d":"20 days","e":"26 days"},"options_float":{"a":5.0,"b":10.0,"c":14.0,"d":20.0,"e":26.0},"annotated_formula":"add(8, divide(subtract(const_1, divide(8, 40)), add(inverse(40), inverse(24))))","linear_formula":"divide(n2,n0)|inverse(n0)|inverse(n1)|add(#1,#2)|subtract(const_1,#0)|divide(#4,#3)|add(n2,#5)|","chain":"8 \/ 40<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/40) + (1\/24)<\/gadget>\n1\/15 = around 0.066667<\/output>\n(4\/5) \/ (1\/15)<\/gadget>\n12<\/output>\n8 + 12<\/gadget>\n20<\/output>\n20<\/result>","index":996} +{"problem":"joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 \/ 4 of all the paint . during the second week , he uses 1 \/ 2 of the remaining paint . how many gallons of paint has joe used ?","rationale":"\"total paint initially = 360 gallons paint used in the first week = ( 1 \/ 4 ) * 360 = 90 gallons . remaning paint = 270 gallons paint used in the second week = ( 1 \/ 2 ) * 270 = 135 gallons total paint used = 225 gallons . option b\"","correct":"b","options":{"a":"18 ","b":"225 ","c":"175 ","d":"216","e":"250"},"options_float":{"a":18.0,"b":225.0,"c":175.0,"d":216.0,"e":250.0},"annotated_formula":"add(multiply(divide(360, 4), 1), divide(subtract(360, multiply(divide(360, 4), 1)), 2))","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)|","chain":"360 \/ 4<\/gadget>\n90<\/output>\n90 * 1<\/gadget>\n90<\/output>\n360 - 90<\/gadget>\n270<\/output>\n270 \/ 2<\/gadget>\n135<\/output>\n90 + 135<\/gadget>\n225<\/output>\n225<\/result>","index":997} +{"problem":"a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 5 per unit . further , the manufacturer has costs of $ 16500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ?","rationale":"by the question , the equation would be 150 p - 85 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 195 . answer e for me .","correct":"e","options":{"a":"$ 28 ","b":"$ 82 ","c":"$ 110 ","d":"$ 138","e":"$ 195"},"options_float":{"a":28.0,"b":82.0,"c":110.0,"d":138.0,"e":195.0},"annotated_formula":"divide(add(multiply(add(80, 5), 150), 16500), 150)","linear_formula":"add(n0,n1)|multiply(n3,#0)|add(n2,#1)|divide(#2,n3)","chain":"80 + 5<\/gadget>\n85<\/output>\n85 * 150<\/gadget>\n12_750<\/output>\n12_750 + 16_500<\/gadget>\n29_250<\/output>\n29_250 \/ 150<\/gadget>\n195<\/output>\n195<\/result>","index":998} +{"problem":"the average runs scored by a batsman in 30 matches is 50 . in the next 15 matches the batsman scored an average of 26 runs . find his average in all the 45 matches ?","rationale":"total score of the batsman in 30 matches = 1500 . total score of the batsman in the next 15 matches = 390 . total score of the batsman in the 30 matches = 1890 . average score of the batsman = 1890 \/ 45 = 42 . answer : b","correct":"b","options":{"a":"31 ","b":"42 ","c":"29 ","d":"97","e":"27"},"options_float":{"a":31.0,"b":42.0,"c":29.0,"d":97.0,"e":27.0},"annotated_formula":"divide(add(multiply(50, 30), multiply(26, 15)), add(30, 15))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"50 * 30<\/gadget>\n1_500<\/output>\n26 * 15<\/gadget>\n390<\/output>\n1_500 + 390<\/gadget>\n1_890<\/output>\n30 + 15<\/gadget>\n45<\/output>\n1_890 \/ 45<\/gadget>\n42<\/output>\n42<\/result>","index":1000} +{"problem":"the owner of a furniture shop charges his customer 60 % more than the cost price . if a customer paid rs . 2000 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 1250 ( 100 \/ 160 ) = rs . 1250 . answer : c\"","correct":"c","options":{"a":"2789 ","b":"2772 ","c":"1250 ","d":"6725","e":"2099"},"options_float":{"a":2789.0,"b":2772.0,"c":1250.0,"d":6725.0,"e":2099.0},"annotated_formula":"divide(2000, add(const_1, divide(60, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n2_000 \/ (8\/5)<\/gadget>\n1_250<\/output>\n1_250<\/result>","index":1001} +{"problem":"a shipment of 240 smartphones contains 84 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ?","rationale":"\"probability of chosing one defective phone from a lot of 240 which ontains 84 defective phones is = ( 84 \/ 240 ) probability of chosing one defective phone from a lot of 239 ( we already picked one ) which ontains 83 ( we already picked one ) defective phones is = ( 83 \/ 239 ) combined probability of series of events = product of the probabilities = ( 84 \/ 240 ) * ( 83 \/ 239 ) 84 \/ 240 is close to ( 7 \/ 20 ) and ( 83 \/ 239 ) = ( 25 \/ 72 ) so answer is ( 7 \/ 20 ) * ( 25 \/ 72 ) ( 9 \/ 74 ) so , answer will be b\"","correct":"b","options":{"a":"1 \/ 250 ","b":"9 \/ 74 ","c":"1 \/ 11 ","d":"1 \/ 9","e":"1 \/ 3"},"options_float":{"a":0.004,"b":0.1216216216,"c":0.0909090909,"d":0.1111111111,"e":0.3333333333},"annotated_formula":"divide(multiply(84, subtract(84, const_1)), multiply(240, subtract(240, const_1)))","linear_formula":"subtract(n1,const_1)|subtract(n0,const_1)|multiply(n1,#0)|multiply(n0,#1)|divide(#2,#3)|","chain":"84 - 1<\/gadget>\n83<\/output>\n84 * 83<\/gadget>\n6_972<\/output>\n240 - 1<\/gadget>\n239<\/output>\n240 * 239<\/gadget>\n57_360<\/output>\n6_972 \/ 57_360<\/gadget>\n581\/4_780 = around 0.121548<\/output>\n581\/4_780 = around 0.121548<\/result>","index":1002} +{"problem":"set a { 3 , 3,3 , 4,5 , 5,5 } has a standard deviation of 1 . what will the standard deviation be if every number in the set is multiplied by 8 ?","rationale":"\"points to remember - 1 . if oneadd \/ subtractthe same amont from every term in a set , sd does n ' t change . 2 . if onemultiply \/ divideevery term by the same number in a set , sd changes by same number . hence the answer to the above question is d\"","correct":"d","options":{"a":"1 ","b":"2 ","c":"4 ","d":"8","e":"16"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":8.0,"e":16.0},"annotated_formula":"multiply(8, 1)","linear_formula":"multiply(n4,n5)|","chain":"8 * 1<\/gadget>\n8<\/output>\n8<\/result>","index":1003} +{"problem":"the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 600 per sq . metre .","rationale":"\"a area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 600 hence , total cost = 5.5 × 3.75 × 600 = $ 12375 a\"","correct":"a","options":{"a":"$ 12375 ","b":"$ 13575 ","c":"$ 16575 ","d":"$ 14575","e":"$ 18575"},"options_float":{"a":12375.0,"b":13575.0,"c":16575.0,"d":14575.0,"e":18575.0},"annotated_formula":"multiply(600, multiply(5.5, 3.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 3.75<\/gadget>\n20.625<\/output>\n600 * 20.625<\/gadget>\n12_375<\/output>\n12_375<\/result>","index":1004} +{"problem":"what is the sum of the multiples of 4 between 63 and 151 inclusive ?","rationale":"\"the fastest way in an ap is to find the average and multiply with total integers . . between 63 and 151 , the smallest multiple of 4 is 64 and largest = 148 . . average = ( 64 + 148 ) \/ 2 = 106 . . total numbers = ( 148 - 64 ) \/ 4 + 1 = 21 + 1 = 22 . . sum = 22 * 106 = 2332 ans e\"","correct":"e","options":{"a":"2240 ","b":"2250 ","c":"2310 ","d":"2330","e":"2332"},"options_float":{"a":2240.0,"b":2250.0,"c":2310.0,"d":2330.0,"e":2332.0},"annotated_formula":"multiply(divide(add(subtract(151, const_3), add(63, const_2)), const_2), add(divide(subtract(subtract(151, const_3), add(63, const_2)), 4), const_1))","linear_formula":"add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)|","chain":"151 - 3<\/gadget>\n148<\/output>\n63 + 2<\/gadget>\n65<\/output>\n148 + 65<\/gadget>\n213<\/output>\n213 \/ 2<\/gadget>\n213\/2 = around 106.5<\/output>\n148 - 65<\/gadget>\n83<\/output>\n83 \/ 4<\/gadget>\n83\/4 = around 20.75<\/output>\n(83\/4) + 1<\/gadget>\n87\/4 = around 21.75<\/output>\n(213\/2) * (87\/4)<\/gadget>\n18_531\/8 = around 2_316.375<\/output>\n18_531\/8 = around 2_316.375<\/result>","index":1007} +{"problem":"if the ratio of apples to bananas is 4 to 3 and the ratio of bananas to cucumbers is 1 to 3 , what is the ratio of apples to cucumbers ?","rationale":"\"the ratio of bananas to cucumbers is 1 to 3 which equals 3 to 9 . the ratio of apples to bananas to cucumbers is 4 to 3 to 9 . the answer is a .\"","correct":"a","options":{"a":"4 : 9 ","b":"4 : 3 ","c":"3 : 7 ","d":"2 : 5","e":"1 : 6"},"options_float":{"a":0.4444444444,"b":1.3333333333,"c":0.4285714286,"d":0.4,"e":0.1666666667},"annotated_formula":"divide(divide(4, 3), 3)","linear_formula":"divide(n0,n1)|divide(#0,n3)|","chain":"4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) \/ 3<\/gadget>\n4\/9 = around 0.444444<\/output>\n4\/9 = around 0.444444<\/result>","index":1009} +{"problem":"a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 46.24 , the number of the member is the group is :","rationale":"\"explanation : money collected = ( 46.24 x 100 ) paise = 4624 paise . ∴ number of members = √ ( 4624 ) = 68 . answer : b\"","correct":"b","options":{"a":"57 ","b":"68 ","c":"77 ","d":"87","e":"97"},"options_float":{"a":57.0,"b":68.0,"c":77.0,"d":87.0,"e":97.0},"annotated_formula":"sqrt(multiply(46.24, const_100))","linear_formula":"multiply(n0,const_100)|sqrt(#0)|","chain":"46.24 * 100<\/gadget>\n4_624<\/output>\n4_624 ** (1\/2)<\/gadget>\n68<\/output>\n68<\/result>","index":1011} +{"problem":"arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 15 from the product of their ages a year from today . what is their age today ?","rationale":"\"ad = ( a + 1 ) ( d + 1 ) - 15 0 = a + d - 14 a + d = 14 a = d ( as they are twin brothers ) a = d = 7 d is the answer\"","correct":"d","options":{"a":"2 . ","b":"4 . ","c":"5 . ","d":"7 .","e":"9 ."},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"divide(subtract(15, const_1), const_2)","linear_formula":"subtract(n0,const_1)|divide(#0,const_2)|","chain":"15 - 1<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":1012} +{"problem":"for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 99 true ?","rationale":"\"answer c ( a + b ) ( a - b ) = 99 3 cases for ( a + b ) , ( a - b ) 99 , 1 33 , 3 11 , 9 answer b\"","correct":"b","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"divide(log(99), log(add(const_4, const_1)))","linear_formula":"add(const_1,const_4)|log(n2)|log(#0)|divide(#1,#2)|","chain":"log(99)<\/gadget>\nlog(99) = around 4.59512<\/output>\n4 + 1<\/gadget>\n5<\/output>\nlog(5)<\/gadget>\nlog(5) = around 1.609438<\/output>\nlog(99) \/ log(5)<\/gadget>\nlog(99)\/log(5) = around 2.855108<\/output>\nlog(99)\/log(5) = around 2.855108<\/result>","index":1013} +{"problem":"when x is multiplied by 2 , the result is 19 more than the result of subtracting x from 26 . what is the value of x ?","rationale":"\"the equation that can be formed is : 2 x - 19 = 26 - x or , 3 x = 45 or , x = 15 . c answer .\"","correct":"c","options":{"a":"- 4 ","b":"- 2 ","c":"15 ","d":"13","e":"22"},"options_float":{"a":-4.0,"b":-2.0,"c":15.0,"d":13.0,"e":22.0},"annotated_formula":"divide(add(26, 19), add(2, const_1))","linear_formula":"add(n1,n2)|add(const_1,n0)|divide(#0,#1)|","chain":"26 + 19<\/gadget>\n45<\/output>\n2 + 1<\/gadget>\n3<\/output>\n45 \/ 3<\/gadget>\n15<\/output>\n15<\/result>","index":1015} +{"problem":"a can give b 60 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"a runs 1000 m while b runs 940 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) \/ 940 = 851.06 m . b can give c = 1000 - 851.06 = 148.94 m . answer : a\"","correct":"a","options":{"a":"148.94 ","b":"148.13 ","c":"148.22 ","d":"111.0","e":"111.12"},"options_float":{"a":148.94,"b":148.13,"c":148.22,"d":111.0,"e":111.12},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 60)))","linear_formula":"multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 200<\/gadget>\n800<\/output>\n1_000 * 800<\/gadget>\n800_000<\/output>\n1_000 - 60<\/gadget>\n940<\/output>\n800_000 \/ 940<\/gadget>\n40_000\/47 = around 851.06383<\/output>\n1_000 - (40_000\/47)<\/gadget>\n7_000\/47 = around 148.93617<\/output>\n7_000\/47 = around 148.93617<\/result>","index":1017} +{"problem":"the average of marks obtained by 120 boys was 38 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ?","rationale":"\"let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 38 24 x = 4560 - 1800 = > x = 2760 \/ 24 x = 115 . hence , the number of boys passed = 115 . answer : c\"","correct":"c","options":{"a":"100 ","b":"110 ","c":"115 ","d":"125","e":"140"},"options_float":{"a":100.0,"b":110.0,"c":115.0,"d":125.0,"e":140.0},"annotated_formula":"divide(subtract(multiply(38, 120), multiply(120, 15)), subtract(39, 15))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|","chain":"38 * 120<\/gadget>\n4_560<\/output>\n120 * 15<\/gadget>\n1_800<\/output>\n4_560 - 1_800<\/gadget>\n2_760<\/output>\n39 - 15<\/gadget>\n24<\/output>\n2_760 \/ 24<\/gadget>\n115<\/output>\n115<\/result>","index":1022} +{"problem":"if x , y , and z are positive integers , and 2 x = 5 y = 8 z , then the least possible value of x + y + z is","rationale":"\"take lcm of 2,5 and 8 = 40 now 2 x = 40 = > x = 20 5 y = 40 = > y = 8 8 z = 40 = > z = 5 20 + 8 + 5 = 33 . option c .\"","correct":"c","options":{"a":"15 ","b":"28 ","c":"33 ","d":"42","e":"60"},"options_float":{"a":15.0,"b":28.0,"c":33.0,"d":42.0,"e":60.0},"annotated_formula":"add(add(divide(divide(multiply(multiply(2, 5), 8), const_2), 2), divide(divide(multiply(multiply(2, 5), 8), const_2), 5)), divide(divide(multiply(multiply(2, 5), 8), const_2), 8))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|divide(#2,n0)|divide(#2,n1)|divide(#2,n2)|add(#3,#4)|add(#6,#5)|","chain":"2 * 5<\/gadget>\n10<\/output>\n10 * 8<\/gadget>\n80<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n20 + 8<\/gadget>\n28<\/output>\n40 \/ 8<\/gadget>\n5<\/output>\n28 + 5<\/gadget>\n33<\/output>\n33<\/result>","index":1023} +{"problem":"john and beth each drove from smallville to crown city by different routes . john drove an an average speed of 40 miles per hour and completed the trip in 30 minutes . beth ' s route was 5 miles longer , and it took her 20 minutes more than john to complete the trip . how many miles per hour was beth ' s average speed on this trip ?","rationale":"speed = distance \/ time distance for john = speed * time = 40 * 1 \/ 2 = 20 miles distance for beth = distance for john + 5 = 20 + 5 = 25 miles time taken by beth = time taken by john + ( 1 \/ 3 hour ) = 1 \/ 2 + 1 \/ 3 = 5 \/ 6 speed of beth = d \/ t = 25 \/ ( 5 \/ 6 ) = 25 * 6 \/ 5 = 30 miles per hour hence , b is the answer .","correct":"b","options":{"a":"24 ","b":"30 ","c":"48 ","d":"54","e":"75"},"options_float":{"a":24.0,"b":30.0,"c":48.0,"d":54.0,"e":75.0},"annotated_formula":"divide(add(multiply(40, divide(30, const_60)), 5), divide(add(30, 20), const_60))","linear_formula":"add(n1,n3)|divide(n1,const_60)|divide(#0,const_60)|multiply(n0,#1)|add(n2,#3)|divide(#4,#2)","chain":"30 \/ 60<\/gadget>\n1\/2 = around 0.5<\/output>\n40 * (1\/2)<\/gadget>\n20<\/output>\n20 + 5<\/gadget>\n25<\/output>\n30 + 20<\/gadget>\n50<\/output>\n50 \/ 60<\/gadget>\n5\/6 = around 0.833333<\/output>\n25 \/ (5\/6)<\/gadget>\n30<\/output>\n30<\/result>","index":1024} +{"problem":"5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 180 only . men â € ™ s wages are ?","rationale":"\"5 m = xw = 8 b 5 m + xw + 8 b - - - - - 180 rs . 5 m + 5 m + 5 m - - - - - 180 rs . 15 m - - - - - - 180 rs . = > 1 m = 12 rs . answer : e\"","correct":"e","options":{"a":"6 rs ","b":"2 rs ","c":"4 rs ","d":"9 rs","e":"12 rs"},"options_float":{"a":6.0,"b":2.0,"c":4.0,"d":9.0,"e":12.0},"annotated_formula":"divide(180, multiply(const_3, 5))","linear_formula":"multiply(n0,const_3)|divide(n2,#0)|","chain":"3 * 5<\/gadget>\n15<\/output>\n180 \/ 15<\/gadget>\n12<\/output>\n12<\/result>","index":1027} +{"problem":"a train is 250 meter long is running at a speed of 36 km \/ hour . in what time will it pass a station of 200 meter length ?","rationale":"\"speed = 36 km \/ hr = 36 * ( 5 \/ 18 ) m \/ sec = 10 m \/ sec total distance = 250 + 200 = 450 meter time = distance \/ speed = 450 * ( 10 ) = 45 seconds answer : c\"","correct":"c","options":{"a":"28 seconds ","b":"27 seconds ","c":"45 seconds ","d":"26 seconds","e":"18 seconds"},"options_float":{"a":28.0,"b":27.0,"c":45.0,"d":26.0,"e":18.0},"annotated_formula":"divide(add(250, 200), divide(multiply(36, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"250 + 200<\/gadget>\n450<\/output>\n36 * 1_000<\/gadget>\n36_000<\/output>\n36_000 \/ 3_600<\/gadget>\n10<\/output>\n450 \/ 10<\/gadget>\n45<\/output>\n45<\/result>","index":1028} +{"problem":"a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 72.25 , the number of the member is the group is :","rationale":"\"explanation : money collected = ( 72.25 x 100 ) paise = 7225 paise . ∴ number of members = √ ( 7225 ) = 85 . answer : d\"","correct":"d","options":{"a":"57 ","b":"67 ","c":"77 ","d":"85","e":"97"},"options_float":{"a":57.0,"b":67.0,"c":77.0,"d":85.0,"e":97.0},"annotated_formula":"sqrt(multiply(72.25, const_100))","linear_formula":"multiply(n0,const_100)|sqrt(#0)|","chain":"72.25 * 100<\/gadget>\n7_225<\/output>\n7_225 ** (1\/2)<\/gadget>\n85<\/output>\n85<\/result>","index":1029} +{"problem":"triangle atriangle b are similar triangles with areas 2023 units square and 1792 units square respectively . the ratio of there corresponding height would be","rationale":"\"let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 \/ y ^ 2 therefore , ( x ^ 2 \/ y ^ 2 ) = 2023 \/ 1792 ( x ^ 2 \/ y ^ 2 ) = ( 17 * 17 * 7 ) \/ ( 16 * 16 * 7 ) ( x ^ 2 \/ y ^ 2 ) = 17 ^ 2 \/ 16 ^ 2 x \/ y = 17 \/ 16 ans = b you need to take a hint from given options to find factors of 2013 and 2527 .\"","correct":"b","options":{"a":"9 : 10 ","b":"17 : 16 ","c":"23 : 27 ","d":"13 : 17","e":"15 : 23"},"options_float":{"a":0.9,"b":1.0625,"c":0.8518518519,"d":0.7647058824,"e":0.652173913},"annotated_formula":"sqrt(divide(2023, 1792))","linear_formula":"divide(n0,n1)|sqrt(#0)|","chain":"2_023 \/ 1_792<\/gadget>\n289\/256 = around 1.128906<\/output>\n(289\/256) ** (1\/2)<\/gadget>\n17\/16 = around 1.0625<\/output>\n17\/16 = around 1.0625<\/result>","index":1030} +{"problem":"the length of a rectangle is reduced by 20 % . by what % would the width have to be increased to maintainthe original area ?","rationale":"\"sol . required change = ( 20 * 100 ) \/ ( 100 - 20 ) = 25 % c\"","correct":"c","options":{"a":"14 % ","b":"15 % ","c":"25 % ","d":"30 %","e":"35 %"},"options_float":{"a":14.0,"b":15.0,"c":25.0,"d":30.0,"e":35.0},"annotated_formula":"multiply(divide(subtract(const_1, divide(subtract(const_100, 20), const_100)), divide(subtract(const_100, 20), const_100)), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ (4\/5)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":1033} +{"problem":"find the area of a parallelogram with base 10 cm and height 20 cm .","rationale":"\"area of a parallelogram = base * height = 10 * 20 = 200 cm 2 answer : option c\"","correct":"c","options":{"a":"200 ","b":"384 ","c":"200 ","d":"244","e":"242"},"options_float":{"a":200.0,"b":384.0,"c":200.0,"d":244.0,"e":242.0},"annotated_formula":"multiply(10, 20)","linear_formula":"multiply(n0,n1)|","chain":"10 * 20<\/gadget>\n200<\/output>\n200<\/result>","index":1034} +{"problem":"a train covers a distance in 50 min , if it runs at a speed of 48 kmph on an average . the speed at which the train must run to reduce the time of journey to 40 min will be .","rationale":"time = 50 \/ 60 hr = 5 \/ 6 hr speed = 48 mph distance = s * t = 48 * 5 \/ 6 = 40 km time = 40 \/ 60 hr = 2 \/ 3 hr new speed = 40 * 3 \/ 2 kmph = 60 kmph answer is a .","correct":"a","options":{"a":"60 kmph ","b":"63 kmph ","c":"57 kmph ","d":"70 kmph","e":"61 kmph"},"options_float":{"a":60.0,"b":63.0,"c":57.0,"d":70.0,"e":61.0},"annotated_formula":"divide(multiply(48, divide(50, const_60)), divide(40, const_60))","linear_formula":"divide(n0,const_60)|divide(n2,const_60)|multiply(n1,#0)|divide(#2,#1)","chain":"50 \/ 60<\/gadget>\n5\/6 = around 0.833333<\/output>\n48 * (5\/6)<\/gadget>\n40<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n40 \/ (2\/3)<\/gadget>\n60<\/output>\n60<\/result>","index":1035} +{"problem":"if y > 0 , ( 2 y ) \/ 10 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"can be reduced to y \/ 5 + 3 y \/ 10 = 2 y \/ 5 = 50 % b\"","correct":"b","options":{"a":"40 % ","b":"50 % ","c":"60 % ","d":"70 %","e":"80 %"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(const_100, add(divide(2, 10), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"2 \/ 10<\/gadget>\n1\/5 = around 0.2<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/5) + (3\/10)<\/gadget>\n1\/2 = around 0.5<\/output>\n100 * (1\/2)<\/gadget>\n50<\/output>\n50<\/result>","index":1036} +{"problem":"200 pupil total , out of 125 like pizza , 115 like burger , then how many like both ?","rationale":"n ( a u b ) = n ( a ) + n ( b ) - n ( a intersection b ) 200 = 125 + 115 - both both = 240 - 200 = 40 . answer : a","correct":"a","options":{"a":"40 ","b":"50 ","c":"60 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"subtract(add(125, 115), 200)","linear_formula":"add(n1,n2)|subtract(#0,n0)","chain":"125 + 115<\/gadget>\n240<\/output>\n240 - 200<\/gadget>\n40<\/output>\n40<\/result>","index":1037} +{"problem":"the area of a rectangular plot is 15 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ?","rationale":"\"l × b = 15 × b ∴ l = 15 m and l – b = 10 ∴ b = 15 – 10 = 5 m answer b\"","correct":"b","options":{"a":"10 metres ","b":"5 metres ","c":"7.5 metres ","d":"data inadequate","e":"none of these"},"options_float":{"a":10.0,"b":5.0,"c":7.5,"d":null,"e":null},"annotated_formula":"subtract(15, 10)","linear_formula":"subtract(n0,n1)|","chain":"15 - 10<\/gadget>\n5<\/output>\n5<\/result>","index":1038} +{"problem":"the compound interest earned by sunil on a certain amount at the end of two years at the rate of 10 % p . a . was rs . 420 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned .","rationale":"\"let the sum be rs . p p { [ 1 + 10 \/ 100 ] 2 - 1 } = 420 p ( 10 \/ 100 ) ( 2 + 10 \/ 100 ) = 420 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 420 \/ ( 0.1 ) ( 2.1 ) = 2000 . amount = 2000 + 420 = rs . 2420 answer : c\"","correct":"c","options":{"a":"rs . 2530 ","b":"rs . 2430 ","c":"rs . 2420 ","d":"rs . 2410","e":"rs . 2440"},"options_float":{"a":2530.0,"b":2430.0,"c":2420.0,"d":2410.0,"e":2440.0},"annotated_formula":"add(divide(420, subtract(power(add(const_1, divide(10, const_100)), const_2), const_1)), 420)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|divide(n1,#3)|add(n1,#4)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n(121\/100) - 1<\/gadget>\n21\/100 = around 0.21<\/output>\n420 \/ (21\/100)<\/gadget>\n2_000<\/output>\n2_000 + 420<\/gadget>\n2_420<\/output>\n2_420<\/result>","index":1040} +{"problem":"sandy walked 20 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 25 meters . what distance is she from the starting point and in which direction ?","rationale":"the net distance is 20 + 25 = 45 meters to the east . the answer is e .","correct":"e","options":{"a":"35 m east ","b":"35 m north ","c":"30 m west ","d":"45 m west","e":"45 m east"},"options_float":{"a":35.0,"b":35.0,"c":30.0,"d":45.0,"e":45.0},"annotated_formula":"add(20, 25)","linear_formula":"add(n0,n3)","chain":"20 + 25<\/gadget>\n45<\/output>\n45<\/result>","index":1041} +{"problem":"the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 4 type b widgets , and the weight of 3 type b widgets is equal to the weight of 7 type c widgets . what is the ratio of the total weight of 1 type a widget and 1 type b widget , to the total weight of 1 type b widget and 1 type c widget ?","rationale":"\"3 b = 7 c and so b = 7 c \/ 3 7 a = 4 b and so a = 4 b \/ 7 = 4 c \/ 3 a + b = 4 c \/ 3 + 7 c \/ 3 = 11 c \/ 3 b + c = 7 c \/ 3 + c = 10 c \/ 3 the ratio of a + b : b + c = 11 : 10 the answer is d .\"","correct":"d","options":{"a":"5 : 4 ","b":"7 : 6 ","c":"9 : 8 ","d":"11 : 10","e":"13 : 12"},"options_float":{"a":1.25,"b":1.1666666667,"c":1.125,"d":1.1,"e":1.0833333333},"annotated_formula":"divide(add(1, divide(7, 4)), add(divide(7, 4), divide(3, 4)))","linear_formula":"divide(n0,n1)|divide(n2,n1)|add(n4,#0)|add(#0,#1)|divide(#2,#3)|","chain":"7 \/ 4<\/gadget>\n7\/4 = around 1.75<\/output>\n1 + (7\/4)<\/gadget>\n11\/4 = around 2.75<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(7\/4) + (3\/4)<\/gadget>\n5\/2 = around 2.5<\/output>\n(11\/4) \/ (5\/2)<\/gadget>\n11\/10 = around 1.1<\/output>\n11\/10 = around 1.1<\/result>","index":1042} +{"problem":"a man can swim in still water at 3 km \/ h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?","rationale":"\"m = 3 s = x ds = 3 + x us = 3 + x 3 + x = ( 3 - x ) 2 3 + x = 6 - 2 x 3 x = 3 x = 1 answer : e\"","correct":"e","options":{"a":"1.7 ","b":"1.4 ","c":"1.1 ","d":"1.5","e":"1"},"options_float":{"a":1.7,"b":1.4,"c":1.1,"d":1.5,"e":1.0},"annotated_formula":"divide(3, const_3)","linear_formula":"divide(n0,const_3)|","chain":"3 \/ 3<\/gadget>\n1<\/output>\n1<\/result>","index":1043} +{"problem":"a man gains 30 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be .","rationale":"\"explanation : let the c . p . = x , then s . p . = ( 130 \/ 100 ) x = 13 x \/ 10 new s . p . = 2 ( 13 x \/ 10 ) = 26 x \/ 10 profit = 26 x \/ 10 - x = 26 x \/ 10 profit % = ( profit \/ c . p . ) * 100 = > ( 26 x \/ 10 ) * ( 1 \/ x ) * 100 = 260 % option d\"","correct":"d","options":{"a":"130 % ","b":"140 % ","c":"150 % ","d":"260 %","e":"170 %"},"options_float":{"a":130.0,"b":140.0,"c":150.0,"d":260.0,"e":170.0},"annotated_formula":"add(multiply(subtract(multiply(add(const_1, divide(30, const_100)), const_2), const_1), const_100), const_100)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|multiply(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|add(#4,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n(13\/10) * 2<\/gadget>\n13\/5 = around 2.6<\/output>\n(13\/5) - 1<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) * 100<\/gadget>\n160<\/output>\n160 + 100<\/gadget>\n260<\/output>\n260<\/result>","index":1044} +{"problem":"2 persons can finish a job in 8 days . first person alone can finish the work in 24 days . how many days does the second person take to finish the job ?","rationale":"2 persons can do work in 1 day = 1 \/ 8 first person can do work in 1 day = 1 \/ 24 second person can do work in 1 day = 1 \/ 8 - 1 \/ 24 = 1 \/ 12 so second person can finish work in 12 days answer : b","correct":"b","options":{"a":"10 days ","b":"12 days ","c":"14 days ","d":"16 days","e":"18 days"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"inverse(subtract(divide(const_1, 8), divide(const_1, 24)))","linear_formula":"divide(const_1,n1)|divide(const_1,n2)|subtract(#0,#1)|inverse(#2)","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/8) - (1\/24)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":1046} +{"problem":"a cistern is filled by pipe a in 12 hours and the full cistern can be leaked out by an exhaust pipe b in 18 hours . if both the pipes are opened , in what time the cistern is full ?","rationale":"\"time taken to full the cistern = ( 1 \/ 12 - 1 \/ 18 ) hrs = 1 \/ 36 = 36 hrs answer : c\"","correct":"c","options":{"a":"50 hrs ","b":"60 hrs ","c":"36 hrs ","d":"80 hrs","e":"90 hrs"},"options_float":{"a":50.0,"b":60.0,"c":36.0,"d":80.0,"e":90.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 12), divide(const_1, 18)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/12) - (1\/18)<\/gadget>\n1\/36 = around 0.027778<\/output>\n1 \/ (1\/36)<\/gadget>\n36<\/output>\n36<\/result>","index":1047} +{"problem":"a number x is multiplied by 3 , and this product is then divided by 7 . if the positive square root of the result of these two operations equals x , what is the value of x if x ≠ 0 ?","rationale":"\"sqrt ( 3 x \/ 7 ) to be perfect square x has to 3 \/ 7 ans : b\"","correct":"b","options":{"a":"25 \/ 9 ","b":"3 \/ 7 ","c":"5 \/ 3 ","d":"3 \/ 5","e":"9 \/ 25"},"options_float":{"a":2.7777777778,"b":0.4285714286,"c":1.6666666667,"d":0.6,"e":0.36},"annotated_formula":"divide(3, 7)","linear_formula":"divide(n0,n1)|","chain":"3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":1050} +{"problem":"the list price of an article is rs . 70 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?","rationale":"\"70 * ( 90 \/ 100 ) * ( ( 100 - x ) \/ 100 ) = 56.16 x = 10.85 % answer : a\"","correct":"a","options":{"a":"10.85 % ","b":"14.85 % ","c":"15.85 % ","d":"16.85 %","e":"17.85 %"},"options_float":{"a":10.85,"b":14.85,"c":15.85,"d":16.85,"e":17.85},"annotated_formula":"multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 56.16), subtract(70, multiply(70, divide(10, const_100)))), const_100)","linear_formula":"divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n70 * (1\/10)<\/gadget>\n7<\/output>\n70 - 7<\/gadget>\n63<\/output>\n63 - 56.16<\/gadget>\n6.84<\/output>\n6.84 \/ 63<\/gadget>\n0.108571<\/output>\n0.108571 * 100<\/gadget>\n10.8571<\/output>\n10.8571<\/result>","index":1051} +{"problem":"a man swims downstream 42 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ?","rationale":"\"42 - - - 3 ds = 14 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 14 + 6 ) \/ 2 = 10 answer : e\"","correct":"e","options":{"a":"2 ","b":"8 ","c":"9 ","d":"6","e":"10"},"options_float":{"a":2.0,"b":8.0,"c":9.0,"d":6.0,"e":10.0},"annotated_formula":"divide(add(divide(18, 3), divide(42, 3)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"18 \/ 3<\/gadget>\n6<\/output>\n42 \/ 3<\/gadget>\n14<\/output>\n6 + 14<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":1052} +{"problem":"a starts business with rs . 3500 and after 8 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b â € ™ s contribution in the capital ?","rationale":"\"explanation : a invested rs . 3500 for 12 months . let b joined with investment x . and he invested for 12 - 8 = 4 months . so there profit ratio = ( 3500 ã — 12 ) : ( 4 x ) = 2 : 3 â ‡ ’ x = 15750 answer : d\"","correct":"d","options":{"a":"24887 ","b":"20778 ","c":"23788 ","d":"15750","e":"2811"},"options_float":{"a":24887.0,"b":20778.0,"c":23788.0,"d":15750.0,"e":2811.0},"annotated_formula":"divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 8), 2))","linear_formula":"multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)|","chain":"3_500 * 12<\/gadget>\n42_000<\/output>\n42_000 * 3<\/gadget>\n126_000<\/output>\n12 - 8<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n126_000 \/ 8<\/gadget>\n15_750<\/output>\n15_750<\/result>","index":1053} +{"problem":"joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 914800 , how many months will it take joan to pay back her mortgage ?","rationale":"joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 914800 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 9148 so we have 1 , 3 , 9 , 27 . . . . . 914800 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) \/ 1 - r using this and plugging in the information we get . . . 9148 = 1 - 3 ^ n \/ 1 - 3 ; 1 - 3 ^ n \/ - 2 cross multiplying we get 9148 x - 2 = 1 - 3 ^ n - 19682 = 1 - 3 ^ n - 19682 = - 3 ^ n 19683 = 3 ^ n ( negatives cancel out ) 19863 can also be re written as 3 ^ 9 therefore ; 3 ^ 9 = 3 ^ n thus n = 9 ( d )","correct":"d","options":{"a":"6 ","b":"8 ","c":"10 ","d":"9","e":"13"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":9.0,"e":13.0},"annotated_formula":"divide(log(add(divide(multiply(914800, const_2), 100), const_1)), log(const_3))","linear_formula":"log(const_3)|multiply(n1,const_2)|divide(#1,n0)|add(#2,const_1)|log(#3)|divide(#4,#0)","chain":"914_800 * 2<\/gadget>\n1_829_600<\/output>\n1_829_600 \/ 100<\/gadget>\n18_296<\/output>\n18_296 + 1<\/gadget>\n18_297<\/output>\nlog(18_297)<\/gadget>\nlog(18297) = around 9.814492<\/output>\nlog(3)<\/gadget>\nlog(3) = around 1.098612<\/output>\nlog(18297) \/ log(3)<\/gadget>\nlog(18297)\/log(3) = around 8.933536<\/output>\nlog(18297)\/log(3) = around 8.933536<\/result>","index":1054} +{"problem":"the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 7 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ?","rationale":"\"area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 7 = rs . 6342 answer : b\"","correct":"b","options":{"a":"s . 4538 ","b":"s . 6342 ","c":"s . 4518 ","d":"s . 4530","e":"s . 4517"},"options_float":{"a":4538.0,"b":6342.0,"c":4518.0,"d":4530.0,"e":4517.0},"annotated_formula":"multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 7)","linear_formula":"add(n0,n1)|multiply(n2,const_2)|multiply(n4,n5)|multiply(n5,n6)|multiply(#0,#1)|multiply(n5,#3)|subtract(#4,#2)|subtract(#6,#5)|multiply(n3,#7)|","chain":"2 * 12<\/gadget>\n24<\/output>\n15 + 25<\/gadget>\n40<\/output>\n24 * 40<\/gadget>\n960<\/output>\n6 * 3<\/gadget>\n18<\/output>\n960 - 18<\/gadget>\n942<\/output>\n4 * 3<\/gadget>\n12<\/output>\n3 * 12<\/gadget>\n36<\/output>\n942 - 36<\/gadget>\n906<\/output>\n906 * 7<\/gadget>\n6_342<\/output>\n6_342<\/result>","index":1056} +{"problem":"the grade point average of the entire class is 80 . if the average of one fourth of the class is 92 , what is the average of the rest of the class ?","rationale":"let x be the number of students in the class . let p be the average of the rest of the class . 80 x = ( 1 \/ 4 ) 92 x + ( 3 \/ 4 ) ( p ) x 320 = 92 + 3 p 3 p = 228 p = 76 . the answer is d .","correct":"d","options":{"a":"73 ","b":"74 ","c":"75 ","d":"76","e":"77"},"options_float":{"a":73.0,"b":74.0,"c":75.0,"d":76.0,"e":77.0},"annotated_formula":"divide(subtract(multiply(80, const_4), 92), subtract(const_4, const_1))","linear_formula":"multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)","chain":"80 * 4<\/gadget>\n320<\/output>\n320 - 92<\/gadget>\n228<\/output>\n4 - 1<\/gadget>\n3<\/output>\n228 \/ 3<\/gadget>\n76<\/output>\n76<\/result>","index":1059} +{"problem":"the compound interest earned on a sum for the second and the third years are $ 1200 and $ 1260 respectively . what is the rate of interest ?","rationale":"\"1260 - 1200 = 60 is the rate of interest on $ 1200 for one year . the rate of interest = ( 100 * 60 ) \/ ( 1200 ) = 5 % the answer is a .\"","correct":"a","options":{"a":"5 % ","b":"8 % ","c":"11 % ","d":"14 %","e":"18 %"},"options_float":{"a":5.0,"b":8.0,"c":11.0,"d":14.0,"e":18.0},"annotated_formula":"divide(multiply(subtract(1260, 1200), const_100), 1200)","linear_formula":"subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)|","chain":"1_260 - 1_200<\/gadget>\n60<\/output>\n60 * 100<\/gadget>\n6_000<\/output>\n6_000 \/ 1_200<\/gadget>\n5<\/output>\n5<\/result>","index":1060} +{"problem":"a fair 2 sided coin is flipped 7 times . what is the probability that tails will be the result at least twice , but not more than 7 times ?","rationale":"\"at least twice , but not more than 7 timesmeans exactly 2 times , 3 times , 4 times , 5 times , 6 times , 7 times the probability of getting exactly k results out of n flips is nck \/ 2 ^ n 7 c 2 \/ 2 ^ 7 + 7 c 3 \/ 2 ^ 7 + 7 c 4 \/ 2 ^ 7 + 7 c 5 \/ 2 ^ 7 + 7 c 6 \/ 2 ^ 7 = 119 \/ 128 option : a\"","correct":"a","options":{"a":"119 \/ 128 ","b":"120 \/ 113 ","c":"114 \/ 210 ","d":"121 \/ 141","e":"104 \/ 117"},"options_float":{"a":0.9296875,"b":1.0619469027,"c":0.5428571429,"d":0.8581560284,"e":0.8888888889},"annotated_formula":"subtract(const_1, add(multiply(inverse(power(2, 7)), 7), add(inverse(power(2, 7)), inverse(power(2, 7)))))","linear_formula":"power(n0,n1)|inverse(#0)|add(#1,#1)|multiply(n1,#1)|add(#2,#3)|subtract(const_1,#4)|","chain":"2 ** 7<\/gadget>\n128<\/output>\n1 \/ 128<\/gadget>\n1\/128 = around 0.007812<\/output>\n(1\/128) * 7<\/gadget>\n7\/128 = around 0.054688<\/output>\n(1\/128) + (1\/128)<\/gadget>\n1\/64 = around 0.015625<\/output>\n(7\/128) + (1\/64)<\/gadget>\n9\/128 = around 0.070312<\/output>\n1 - (9\/128)<\/gadget>\n119\/128 = around 0.929688<\/output>\n119\/128 = around 0.929688<\/result>","index":1061} +{"problem":"in february wilson ’ s earnings were 40 percent of his family ’ s total income . in march wilson earned 25 percent less than in february . if the rest of his family ’ s income was the same in both months , then , in march , wilson ’ s earnings were approximately what percent q of his family ’ s total income ?","rationale":"lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb = 60 x thus wilson ' s earning as % of total family income in march q = 30 x \/ ( 30 + 60 ) x = 30 x \/ 90 x = 33.33 % thus answer is e","correct":"e","options":{"a":"15 % ","b":"17 % ","c":"24 % ","d":"30 %","e":"33 %"},"options_float":{"a":15.0,"b":17.0,"c":24.0,"d":30.0,"e":33.0},"annotated_formula":"multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(#1,#0)|subtract(#0,#2)|subtract(const_1,#2)|divide(#3,#4)|multiply(#5,const_100)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * (2\/5)<\/gadget>\n1\/10 = around 0.1<\/output>\n(2\/5) - (1\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(3\/10) \/ (9\/10)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":1062} +{"problem":"the s . i . on a certain sum of money for 2 years at 4 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ?","rationale":"\"explanation : c . i . = [ 4000 * ( 1 + 10 \/ 100 ) 2 - 4000 ] = ( 4000 * 11 \/ 10 * 11 \/ 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) \/ ( 3 * 8 ) = rs . 5250 answer : a\"","correct":"a","options":{"a":"5250 ","b":"1267 ","c":"5750 ","d":"2267","e":"5262"},"options_float":{"a":5250.0,"b":1267.0,"c":5750.0,"d":2267.0,"e":5262.0},"annotated_formula":"divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(2, divide(4, const_100)))","linear_formula":"divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n4_000 * (121\/100)<\/gadget>\n4_840<\/output>\n4_840 - 4_000<\/gadget>\n840<\/output>\n840 \/ 2<\/gadget>\n420<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n2 * (1\/25)<\/gadget>\n2\/25 = around 0.08<\/output>\n420 \/ (2\/25)<\/gadget>\n5_250<\/output>\n5_250<\/result>","index":1063} +{"problem":"a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes 1 square feet of area in her garden . this year , she has increased her output by 181 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ?","rationale":"let the side for growing cabbages this year be x ft . thus the area is x ^ 2 . let the side for growing cabbages last year be y ft . thus , the area was y ^ 2 . the area would have increased by 181 sq ft as each cabbage takes 1 sq ft space . x ^ 2 - y ^ 2 = 181 ( x + y ) ( x - y ) = 181 181 is a prime number and thus it will be ( 91 + 90 ) * ( 91 - 90 ) . thus x = 91 and y = 90 x ^ 2 = 91 ^ 2 = 8281 the answer is c .","correct":"c","options":{"a":"6564 ","b":"7372 ","c":"8281 ","d":"9331","e":"can not be determined"},"options_float":{"a":6564.0,"b":7372.0,"c":8281.0,"d":9331.0,"e":null},"annotated_formula":"power(add(divide(181, const_2), add(const_0_25, const_0_25)), const_2)","linear_formula":"add(const_0_25,const_0_25)|divide(n1,const_2)|add(#0,#1)|power(#2,const_2)","chain":"181 \/ 2<\/gadget>\n181\/2 = around 90.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n(181\/2) + (1\/2)<\/gadget>\n91<\/output>\n91 ** 2<\/gadget>\n8_281<\/output>\n8_281<\/result>","index":1064} +{"problem":"a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 9.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ?","rationale":"\"the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.095 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 95 f - - > s = f - - > f \/ s = 1 \/ 12 - - > f \/ ( s + f ) = 1 \/ ( 1 + 12 ) = 1 \/ 13 = 0.08 . answer : b .\"","correct":"b","options":{"a":"16.5 % ","b":"8 % ","c":"35 % ","d":"55 %","e":"65 %"},"options_float":{"a":16.5,"b":8.0,"c":35.0,"d":55.0,"e":65.0},"annotated_formula":"multiply(divide(subtract(3.5, 3), subtract(9.5, 3)), const_100)","linear_formula":"subtract(n1,n2)|subtract(n3,n2)|divide(#0,#1)|multiply(#2,const_100)|","chain":"3.5 - 3<\/gadget>\n0.5<\/output>\n9.5 - 3<\/gadget>\n6.5<\/output>\n0.5 \/ 6.5<\/gadget>\n0.076923<\/output>\n0.076923 * 100<\/gadget>\n7.6923<\/output>\n7.6923<\/result>","index":1065} +{"problem":"( p \/ q - q \/ p ) = 21 \/ 10 . then find 4 p \/ q + 4 q \/ p ?","rationale":"let p \/ q = a , then ( a - 1 \/ a ) = 21 \/ 10 ⇒ a 2 − 1 = a . 2110 ⇒ a 2 − 1 = a . 2110 ⇒ 10 a 2 − 21 a − 10 ⇒ 10 a 2 − 21 a − 10 = 0 roots of the equation = − b ± b 2 − 4 ac − − − − − − − √ 2 a − b ± b 2 − 4 ac 2 a a = 21 ± 441 + 400 − − − − − − − − √ 2021 ± 441 + 40020 a = 21 ± 292021 ± 2920 = 5 \/ 2 or - 2 \/ 5 for a = 5 \/ 2 , 4 p \/ q + 4 q \/ p = 58 \/ 5 for a = - 2 \/ 5 , 4 p \/ q + 4 q \/ p = - 58 \/ 5 answer : b","correct":"b","options":{"a":"58 \/ 9 ","b":"58 \/ 5 ","c":"58 \/ 7 ","d":"58 \/ 2","e":"58 \/ 1"},"options_float":{"a":6.4444444444,"b":11.6,"c":8.2857142857,"d":29.0,"e":58.0},"annotated_formula":"divide(add(add(add(add(21, 10), multiply(4, 4)), const_1), const_10), add(4, const_1))","linear_formula":"add(n0,n1)|add(n2,const_1)|multiply(n2,n2)|add(#0,#2)|add(#3,const_1)|add(#4,const_10)|divide(#5,#1)","chain":"21 + 10<\/gadget>\n31<\/output>\n4 * 4<\/gadget>\n16<\/output>\n31 + 16<\/gadget>\n47<\/output>\n47 + 1<\/gadget>\n48<\/output>\n48 + 10<\/gadget>\n58<\/output>\n4 + 1<\/gadget>\n5<\/output>\n58 \/ 5<\/gadget>\n58\/5 = around 11.6<\/output>\n58\/5 = around 11.6<\/result>","index":1067} +{"problem":"a jar was full with honey . a person used to draw out 20 % of the honey from the jar and replaced it with sugar solution . he has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar , the rest part of the jar was filled with the sugar solution . the initial amount of honey in the jar was filled with the sugar solution . the initial amount of honey in the jar was :","rationale":"answer : a ) 1.25 kg","correct":"a","options":{"a":"1.25 kg ","b":"1.29 kg ","c":"1.85 kg ","d":"3.25 kg","e":"1.23 kg"},"options_float":{"a":1.25,"b":1.29,"c":1.85,"d":3.25,"e":1.23},"annotated_formula":"divide(divide(512, power(divide(4, divide(20, 4)), 4)), const_1000)","linear_formula":"divide(n0,n1)|divide(n1,#0)|power(#1,n1)|divide(n2,#2)|divide(#3,const_1000)","chain":"20 \/ 4<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) ** 4<\/gadget>\n256\/625 = around 0.4096<\/output>\n512 \/ (256\/625)<\/gadget>\n1_250<\/output>\n1_250 \/ 1_000<\/gadget>\n5\/4 = around 1.25<\/output>\n5\/4 = around 1.25<\/result>","index":1068} +{"problem":"a pupil ' s marks were wrongly entered as 85 instead of 45 . due to that the average marks for the class got increased by half . the number of pupils in the class is :","rationale":"\"let there be x pupils in the class . total increase in marks = ( x * 1 \/ 2 ) = x \/ 2 . x \/ 2 = ( 85 - 45 ) = > x \/ 2 = 40 = > x = 80 . answer : b\"","correct":"b","options":{"a":"30 ","b":"80 ","c":"20 ","d":"25","e":"26"},"options_float":{"a":30.0,"b":80.0,"c":20.0,"d":25.0,"e":26.0},"annotated_formula":"multiply(subtract(85, 45), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|","chain":"85 - 45<\/gadget>\n40<\/output>\n40 * 2<\/gadget>\n80<\/output>\n80<\/result>","index":1069} +{"problem":"the h . c . f . of two numbers is 20 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 20 x 13 ) and ( 20 x 14 ) . larger number = ( 20 x 14 ) = 280 . answer : option b\"","correct":"b","options":{"a":"276 ","b":"280 ","c":"322 ","d":"345","e":"354"},"options_float":{"a":276.0,"b":280.0,"c":322.0,"d":345.0,"e":354.0},"annotated_formula":"multiply(20, 14)","linear_formula":"multiply(n0,n2)|","chain":"20 * 14<\/gadget>\n280<\/output>\n280<\/result>","index":1072} +{"problem":"two cubes of their volumes in the ratio 64 : 125 . the ratio of their surface area is :","rationale":"\"the ratio of their surface area is 64 : 125 4 : 5 answer is e .\"","correct":"e","options":{"a":"1.5 : 5 ","b":"2 : 5 ","c":"3 : 5 ","d":"1 : 5","e":"4 : 5"},"options_float":{"a":0.3,"b":0.4,"c":0.6,"d":0.2,"e":0.8},"annotated_formula":"divide(power(64, const_0_33), power(125, const_0_33))","linear_formula":"power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n64 ** (1\/3)<\/gadget>\n4<\/output>\n125 ** (1\/3)<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":1073} +{"problem":"if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 22 , what is the value of r - p ?","rationale":"\"arithmetic mean expression for p and q : ( p + q ) \/ 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) \/ 2 = 20 ; q + r = 44 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 24 hence , the correct answer is d\"","correct":"d","options":{"a":"20 ","b":"10 ","c":"30 ","d":"24","e":"5"},"options_float":{"a":20.0,"b":10.0,"c":30.0,"d":24.0,"e":5.0},"annotated_formula":"subtract(multiply(22, const_2), multiply(10, const_2))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|","chain":"22 * 2<\/gadget>\n44<\/output>\n10 * 2<\/gadget>\n20<\/output>\n44 - 20<\/gadget>\n24<\/output>\n24<\/result>","index":1076} +{"problem":"a man is 28 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ?","rationale":"\"let the son ' s present age be x years . then , man ' s present age = ( x + 28 ) years . ( x + 28 ) + 2 = 2 ( x + 2 ) x + 30 = 2 x + 4 = > x = 26 . answer : d\"","correct":"d","options":{"a":"16 ","b":"19 ","c":"15 ","d":"26","e":"10"},"options_float":{"a":16.0,"b":19.0,"c":15.0,"d":26.0,"e":10.0},"annotated_formula":"divide(subtract(28, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))","linear_formula":"multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4 - 2<\/gadget>\n2<\/output>\n28 - 2<\/gadget>\n26<\/output>\n2 - 1<\/gadget>\n1<\/output>\n26 \/ 1<\/gadget>\n26<\/output>\n26<\/result>","index":1077} +{"problem":"in an examination , 30 % of total students failed in hindi , 42 % failed in english and 28 % in both . the percentage of these who passed in both the subjects is :","rationale":"\"explanation : formula n ( a â ˆ ª b ) = n ( a ) + n ( b ) â ˆ ’ n ( a â ˆ © b ) fail in hindi or english = 30 + 42 â € “ 28 = 44 therefore students who passed = 100 â € “ 44 = 56 . answer : c\"","correct":"c","options":{"a":"23 ","b":"37 ","c":"56 ","d":"40","e":"81"},"options_float":{"a":23.0,"b":37.0,"c":56.0,"d":40.0,"e":81.0},"annotated_formula":"subtract(const_100, subtract(add(30, 42), 28))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|","chain":"30 + 42<\/gadget>\n72<\/output>\n72 - 28<\/gadget>\n44<\/output>\n100 - 44<\/gadget>\n56<\/output>\n56<\/result>","index":1078} +{"problem":"a person purchases 90 clocks and sells 40 clocks at a gain of 10 % and 50 clocks at a gain of 20 % . if he sold all of them at a uniform profit of 15 % , then he would have got rs . 40 less . the cost price of each clock is : ?","rationale":"\"let c . p . of clock be rs . x . then , c . p . of 90 clocks = rs . 90 x . [ ( 110 % of 40 x ) + ( 120 % of 50 x ) ] - ( 115 % of 90 x ) = 40 44 x + 60 x - 103.5 x = 40 0.5 x = 40 = > x = 80 answer : c\"","correct":"c","options":{"a":"40 ","b":"60 ","c":"80 ","d":"67","e":"30"},"options_float":{"a":40.0,"b":60.0,"c":80.0,"d":67.0,"e":30.0},"annotated_formula":"divide(40, subtract(add(multiply(40, add(const_1, divide(10, const_100))), multiply(50, add(const_1, divide(20, const_100)))), multiply(90, add(const_1, divide(15, const_100)))))","linear_formula":"divide(n2,const_100)|divide(n4,const_100)|divide(n5,const_100)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|multiply(n1,#3)|multiply(n3,#4)|multiply(n0,#5)|add(#6,#7)|subtract(#9,#8)|divide(n1,#10)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n40 * (11\/10)<\/gadget>\n44<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n50 * (6\/5)<\/gadget>\n60<\/output>\n44 + 60<\/gadget>\n104<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n90 * (23\/20)<\/gadget>\n207\/2 = around 103.5<\/output>\n104 - (207\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n40 \/ (1\/2)<\/gadget>\n80<\/output>\n80<\/result>","index":1079} +{"problem":"two trains travelling in the same direction at 50 and 40 kmph completely pass off another in 1 1 \/ 2 minute . if the length of the first train is 125 m , what is the length of the second train ?","rationale":"rs = 50 â € “ 40 = 10 * 5 \/ 18 = 25 \/ 9 mps t = 90 sec d = 25 \/ 9 * 90 = 250 m 125 - - - - - - - - 125 m answer : c","correct":"c","options":{"a":"300 m ","b":"287 m ","c":"125 m ","d":"250 m","e":"167 m"},"options_float":{"a":300.0,"b":287.0,"c":125.0,"d":250.0,"e":167.0},"annotated_formula":"subtract(multiply(add(multiply(1, const_60), divide(multiply(1, const_60), 2)), multiply(subtract(50, 40), const_0_2778)), 125)","linear_formula":"multiply(n2,const_60)|subtract(n0,n1)|divide(#0,n4)|multiply(#1,const_0_2778)|add(#2,#0)|multiply(#4,#3)|subtract(#5,n5)","chain":"1 * 60<\/gadget>\n60<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n60 + 30<\/gadget>\n90<\/output>\n50 - 40<\/gadget>\n10<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n10 * (5\/18)<\/gadget>\n25\/9 = around 2.777778<\/output>\n90 * (25\/9)<\/gadget>\n250<\/output>\n250 - 125<\/gadget>\n125<\/output>\n125<\/result>","index":1080} +{"problem":"a man can row upstream at 25 kmph and downstream at 31 kmph , and then find the speed of the man in still water ?","rationale":"\"us = 25 ds = 31 m = ( 31 + 25 ) \/ 2 = 28 answer : b\"","correct":"b","options":{"a":"22 ","b":"28 ","c":"30 ","d":"27","e":"18"},"options_float":{"a":22.0,"b":28.0,"c":30.0,"d":27.0,"e":18.0},"annotated_formula":"divide(add(25, 31), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"25 + 31<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":1081} +{"problem":"a cricket player whose bowling average was 23.50 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.50 . what will be the number of wickets taken by him till the last match ?","rationale":"\"average = total runs \/ total wickets total runs after last match = 23.5 w + 52 total wickets after last match = w + 5 ( 23.5 w + 52 ) \/ ( w + 5 ) = 23.5 - 0.5 = 23 w = 126 so total wickets aftr last match = w + 5 = 131 answer : c\"","correct":"c","options":{"a":"64 ","b":"72 ","c":"131 ","d":"456","e":"108"},"options_float":{"a":64.0,"b":72.0,"c":131.0,"d":456.0,"e":108.0},"annotated_formula":"divide(subtract(multiply(subtract(23.50, 0.50), 5), 52), 0.50)","linear_formula":"subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)|","chain":"23.5 - 0.5<\/gadget>\n23<\/output>\n23 * 5<\/gadget>\n115<\/output>\n115 - 52<\/gadget>\n63<\/output>\n63 \/ 0.5<\/gadget>\n126<\/output>\n126<\/result>","index":1083} +{"problem":"on dividing 52 by a number , the quotient is 16 and the remainder is 4 . find the divisor ?","rationale":"\"d = ( d - r ) \/ q = ( 52 - 4 ) \/ 16 = 48 \/ 16 = 3 b )\"","correct":"b","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"floor(divide(52, 16))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"52 \/ 16<\/gadget>\n13\/4 = around 3.25<\/output>\nfloor(13\/4)<\/gadget>\n3<\/output>\n3<\/result>","index":1084} +{"problem":"jim â € ™ s taxi service charges an initial fee of $ 2.5 at the beginning of a trip and an additional charge of $ 0.35 for each 2 \/ 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ?","rationale":"\"let the fixed charge of jim â € ™ s taxi service = 2.5 $ and charge per 2 \/ 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.5 + ( 3.6 \/ . 4 ) * . 35 = 2.5 + 9 * . 35 = 5.65 $ answer d\"","correct":"d","options":{"a":"$ 3.15 ","b":"$ 4.45 ","c":"$ 4.80 ","d":"$ 5.65","e":"$ 5.40"},"options_float":{"a":3.15,"b":4.45,"c":4.8,"d":5.65,"e":5.4},"annotated_formula":"add(multiply(divide(3.6, divide(2, 5)), 0.35), 2.5)","linear_formula":"divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n3.6 \/ (2\/5)<\/gadget>\n9<\/output>\n9 * 0.35<\/gadget>\n3.15<\/output>\n3.15 + 2.5<\/gadget>\n5.65<\/output>\n5.65<\/result>","index":1086} +{"problem":"if the dimensions of a certain rectangular box are 5 inches by 2 inches by 3 inches , then the face of greatest area is of what area ?","rationale":"definitely ( d ) 5 * 3 = 15 sq inches the question asks to find the greatest area of the faces of a cuboid the possible area of the faces are - length * breadth , breadth * height & height * length the possible areas are - 6 , 10 & 15 thus 15 is the greatest possible area . . . . answer : d","correct":"d","options":{"a":"225 square inches ","b":"45 square inches ","c":"25 square inches ","d":"15 square inches","e":"10 square inches"},"options_float":{"a":225.0,"b":45.0,"c":25.0,"d":15.0,"e":10.0},"annotated_formula":"rectangle_area(5, 3)","linear_formula":"rectangle_area(n0,n2)","chain":"5 * 3<\/gadget>\n15<\/output>\n15<\/result>","index":1087} +{"problem":"at a certain university , 66 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ?","rationale":"\"answer is 75 % total women = 66 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 24 % but question wants to know the percent of men that are tenured 24 % \/ 40 % = 60 % c\"","correct":"c","options":{"a":"25 ","b":"37.5 ","c":"60 ","d":"62.5","e":"75"},"options_float":{"a":25.0,"b":37.5,"c":60.0,"d":62.5,"e":75.0},"annotated_formula":"add(subtract(const_100, 66), subtract(90, 66))","linear_formula":"subtract(const_100,n0)|subtract(n2,n0)|add(#0,#1)|","chain":"100 - 66<\/gadget>\n34<\/output>\n90 - 66<\/gadget>\n24<\/output>\n34 + 24<\/gadget>\n58<\/output>\n58<\/result>","index":1088} +{"problem":"a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 50 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ?","rationale":"a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.5 = $ 14 - - > x = $ 28 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 28 + 21 = $ 49 . answer : b .","correct":"b","options":{"a":"$ 37.80 ","b":"$ 49 ","c":"$ 39.20 ","d":"$ 39.50","e":"$ 40.60"},"options_float":{"a":37.8,"b":49.0,"c":39.2,"d":39.5,"e":40.6},"annotated_formula":"add(multiply(35, divide(40, const_100)), 35)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n35 * (2\/5)<\/gadget>\n14<\/output>\n14 + 35<\/gadget>\n49<\/output>\n49<\/result>","index":1089} +{"problem":"excluding stoppages , the speed of a bus is 58 kmph and including stoppages , it is 40 kmph . for how many minutes does the bus stop per hour ?","rationale":"\"due to stoppages , it covers 18 km less . time taken to cover 18 km = ( ( 18 \/ 58 ) ã — 60 ) = 19 min . option ( a ) is correct\"","correct":"a","options":{"a":"19 ","b":"19.13 ","c":"10.5 ","d":"11.35","e":"none of these"},"options_float":{"a":19.0,"b":19.13,"c":10.5,"d":11.35,"e":null},"annotated_formula":"multiply(const_60, divide(subtract(58, 40), 58))","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)|","chain":"58 - 40<\/gadget>\n18<\/output>\n18 \/ 58<\/gadget>\n9\/29 = around 0.310345<\/output>\n60 * (9\/29)<\/gadget>\n540\/29 = around 18.62069<\/output>\n540\/29 = around 18.62069<\/result>","index":1090} +{"problem":"the average salary of the employees in a office is rs . 120 \/ month . the avg salary of officers is rs . 430 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office .","rationale":"\"let no . of non - officers be x 15 * 430 + x * 110 = ( x + 15 ) 120 x = 465 b\"","correct":"b","options":{"a":"400 ","b":"465 ","c":"430 ","d":"450","e":"510"},"options_float":{"a":400.0,"b":465.0,"c":430.0,"d":450.0,"e":510.0},"annotated_formula":"divide(subtract(multiply(15, 430), multiply(15, 120)), subtract(120, 110))","linear_formula":"multiply(n1,n3)|multiply(n0,n3)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)|","chain":"15 * 430<\/gadget>\n6_450<\/output>\n15 * 120<\/gadget>\n1_800<\/output>\n6_450 - 1_800<\/gadget>\n4_650<\/output>\n120 - 110<\/gadget>\n10<\/output>\n4_650 \/ 10<\/gadget>\n465<\/output>\n465<\/result>","index":1091} +{"problem":"the salary of a person was reduced by 14 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ?","rationale":"let the original salary be $ 100 new salary = $ 86 increase on 86 = 14 increase on 100 = 14 \/ 86 * 100 = 16.3 % answer is b","correct":"b","options":{"a":"10 % ","b":"16.3 % ","c":"13.7 % ","d":"15.1 %","e":"12.3 %"},"options_float":{"a":10.0,"b":16.3,"c":13.7,"d":15.1,"e":12.3},"annotated_formula":"multiply(divide(multiply(const_100, divide(14, const_100)), subtract(const_100, multiply(const_100, divide(14, const_100)))), const_100)","linear_formula":"divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)","chain":"14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n100 * (7\/50)<\/gadget>\n14<\/output>\n100 - 14<\/gadget>\n86<\/output>\n14 \/ 86<\/gadget>\n7\/43 = around 0.162791<\/output>\n(7\/43) * 100<\/gadget>\n700\/43 = around 16.27907<\/output>\n700\/43 = around 16.27907<\/result>","index":1093} +{"problem":"a bullet train covers a distance in 50 min , if it runs at a speed of 48 kmph on an average . the speed at which the bullet train must run to reduce the time of journey to 40 min will be .","rationale":"c 60 kmph time = 50 \/ 60 hr = 5 \/ 6 hr speed = 48 mph distance = s * t = 48 * 5 \/ 6 = 40 km time = 40 \/ 60 hr = 2 \/ 3 hr new speed = 40 * 3 \/ 2 kmph = 60 kmph","correct":"c","options":{"a":"80 kmph ","b":"40 kmph ","c":"60 kmph ","d":"90 kmph","e":"30 kmph"},"options_float":{"a":80.0,"b":40.0,"c":60.0,"d":90.0,"e":30.0},"annotated_formula":"divide(multiply(50, 48), 40)","linear_formula":"multiply(n0,n1)|divide(#0,n2)","chain":"50 * 48<\/gadget>\n2_400<\/output>\n2_400 \/ 40<\/gadget>\n60<\/output>\n60<\/result>","index":1095} +{"problem":"the area of a triangle is with base 12 m and height 5 m ?","rationale":"\"1 \/ 2 * 12 * 5 = 30 m 2 answer : b\"","correct":"b","options":{"a":"88 m 2 ","b":"30 m 2 ","c":"66 m 2 ","d":"77 m 2","e":"31 m 2"},"options_float":{"a":88.0,"b":30.0,"c":66.0,"d":77.0,"e":31.0},"annotated_formula":"triangle_area(12, 5)","linear_formula":"triangle_area(n0,n1)|","chain":"(12 * 5) \/ 2<\/gadget>\n30<\/output>\n30<\/result>","index":1096} +{"problem":"what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular decagon ( 10 - sided polygon ) ?","rationale":"from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 \/ 7 . the answer is c .","correct":"c","options":{"a":"3 \/ 10 ","b":"2 \/ 5 ","c":"2 \/ 7 ","d":"10 \/ 21","e":"5 \/ 11"},"options_float":{"a":0.3,"b":0.4,"c":0.2857142857,"d":0.4761904762,"e":0.4545454545},"annotated_formula":"divide(10, divide(multiply(10, subtract(10, const_3)), const_2))","linear_formula":"subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)|divide(n0,#2)","chain":"10 - 3<\/gadget>\n7<\/output>\n10 * 7<\/gadget>\n70<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n10 \/ 35<\/gadget>\n2\/7 = around 0.285714<\/output>\n2\/7 = around 0.285714<\/result>","index":1097} +{"problem":"sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 360 per week . how much does she earn in dollars per hour ?","rationale":"\"let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 360 per week 36 x = 360 x = 10 correct option : b\"","correct":"b","options":{"a":"11 ","b":"10 ","c":"9 ","d":"8","e":"7"},"options_float":{"a":11.0,"b":10.0,"c":9.0,"d":8.0,"e":7.0},"annotated_formula":"divide(360, add(multiply(8, const_3), multiply(6, const_2)))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)|","chain":"8 * 3<\/gadget>\n24<\/output>\n6 * 2<\/gadget>\n12<\/output>\n24 + 12<\/gadget>\n36<\/output>\n360 \/ 36<\/gadget>\n10<\/output>\n10<\/result>","index":1099} +{"problem":"dacid obtained 90 , 92 , 85 , 87 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ?","rationale":"\"average = ( 90 + 92 + 85 + 87 + 85 ) \/ 5 = 439 \/ 5 = 87.8 . answer : c\"","correct":"c","options":{"a":"67 ","b":"26 ","c":"88 ","d":"90","e":"75"},"options_float":{"a":67.0,"b":26.0,"c":88.0,"d":90.0,"e":75.0},"annotated_formula":"divide(add(add(add(add(90, 92), 85), 87), 85), divide(const_10, const_2))","linear_formula":"add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|","chain":"90 + 92<\/gadget>\n182<\/output>\n182 + 85<\/gadget>\n267<\/output>\n267 + 87<\/gadget>\n354<\/output>\n354 + 85<\/gadget>\n439<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n439 \/ 5<\/gadget>\n439\/5 = around 87.8<\/output>\n439\/5 = around 87.8<\/result>","index":1102} +{"problem":"in a group of 6 boys and 4 girls , four children are to be selected . in how many different ways can they be selected ?","rationale":"\"out of 10 students 4 children should to be selected . so the combination will be 10 c 4 . 10 c 4 = 10 ! \/ 6 ! * 4 ! = 10 * 9 * 8 * 7 * 6 ! \/ 6 ! * ( 4 * 3 * 2 * 1 ) 10 c 4 = 210 there are 210 differect possible ways to select 4 children out of 10 students .\"","correct":"c","options":{"a":"159 ","b":"194 ","c":"210 ","d":"209","e":"166"},"options_float":{"a":159.0,"b":194.0,"c":210.0,"d":209.0,"e":166.0},"annotated_formula":"divide(factorial(add(6, 4)), multiply(factorial(4), factorial(6)))","linear_formula":"add(n0,n1)|factorial(n1)|factorial(n0)|factorial(#0)|multiply(#1,#2)|divide(#3,#4)|","chain":"6 + 4<\/gadget>\n10<\/output>\nfactorial(10)<\/gadget>\n3_628_800<\/output>\nfactorial(4)<\/gadget>\n24<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\n24 * 720<\/gadget>\n17_280<\/output>\n3_628_800 \/ 17_280<\/gadget>\n210<\/output>\n210<\/result>","index":1103} +{"problem":"in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 360 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ?","rationale":"the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 \/ 15 of the population . ( 4 \/ 15 ) * 360 = 96 the answer is a .","correct":"a","options":{"a":"96 ","b":"108 ","c":"120 ","d":"132","e":"140"},"options_float":{"a":96.0,"b":108.0,"c":120.0,"d":132.0,"e":140.0},"annotated_formula":"multiply(divide(360, add(add(multiply(3, 2), 4), 5)), 4)","linear_formula":"multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)","chain":"3 * 2<\/gadget>\n6<\/output>\n6 + 4<\/gadget>\n10<\/output>\n10 + 5<\/gadget>\n15<\/output>\n360 \/ 15<\/gadget>\n24<\/output>\n24 * 4<\/gadget>\n96<\/output>\n96<\/result>","index":1104} +{"problem":"the sum of ages of 5 children born at the intervals of 3 years each is 55 years . what is the age of the youngest child ?","rationale":"\"let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 55 5 x = 25 x = 5 . age of the youngest child = x = 5 years . a )\"","correct":"a","options":{"a":"5 years ","b":"4 years ","c":"6 years ","d":"7 years","e":"8 years"},"options_float":{"a":5.0,"b":4.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"subtract(subtract(divide(55, 5), 3), 3)","linear_formula":"divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|","chain":"55 \/ 5<\/gadget>\n11<\/output>\n11 - 3<\/gadget>\n8<\/output>\n8 - 3<\/gadget>\n5<\/output>\n5<\/result>","index":1105} +{"problem":"a man invested rs . 14,400 in rs . 100 shares of a company at 25 % premium . if the company declares 5 % dividend at the end of the year , then how much does he get ?","rationale":"\"solution number of shares = ( 14400 \/ 125 ) = 115.2 . face value = rs . ( 100 x 115.2 ) = rs . 11520 . annual income = rs . ( 5 \/ 100 x 11520 ) = rs . 576 . answer a\"","correct":"a","options":{"a":"rs . 576 ","b":"rs . 600 ","c":"rs . 650 ","d":"rs . 720","e":"none"},"options_float":{"a":576.0,"b":600.0,"c":650.0,"d":720.0,"e":null},"annotated_formula":"multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(25, 100))))), divide(5, 100))","linear_formula":"divide(n3,n1)|divide(n2,n1)|multiply(const_10,const_1000)|multiply(const_1000,const_4)|multiply(n1,const_4)|add(#2,#3)|multiply(n1,#1)|add(#5,#4)|add(n1,#6)|divide(#7,#8)|multiply(n1,#9)|multiply(#0,#10)|","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n4 * 1_000<\/gadget>\n4_000<\/output>\n10_000 + 4_000<\/gadget>\n14_000<\/output>\n4 * 100<\/gadget>\n400<\/output>\n14_000 + 400<\/gadget>\n14_400<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n100 * (1\/4)<\/gadget>\n25<\/output>\n100 + 25<\/gadget>\n125<\/output>\n14_400 \/ 125<\/gadget>\n576\/5 = around 115.2<\/output>\n100 * (576\/5)<\/gadget>\n11_520<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n11_520 * (1\/20)<\/gadget>\n576<\/output>\n576<\/result>","index":1106} +{"problem":"a ladder 22 feet long is leaning against a wall that is perpendicular to level ground . the bottom of the ladder is 7 feet from the base of the wall . if the top of the ladder slips down 3 feet , how many feet will the bottom of the ladder slip ?","rationale":"\"22 ^ 2 - 7 ^ 2 = 435 it means that the height is equal to 20.85 . since the top of the ladder slips down 4 feet , then the height of the wall = 20.85 - 3 = 17.85 the bottom = sqrt ( 22 ^ 2 - 17.85 ^ 2 ) = sqrt ( 484 - 318.62 ) = 12.86 ans is a\"","correct":"a","options":{"a":"12.86 ","b":"15 ","c":"18 ","d":"19","e":"25"},"options_float":{"a":12.86,"b":15.0,"c":18.0,"d":19.0,"e":25.0},"annotated_formula":"sqrt(subtract(power(22, const_2), power(subtract(sqrt(subtract(power(22, const_2), power(7, const_2))), 3), const_2)))","linear_formula":"power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|subtract(#3,n2)|power(#4,const_2)|subtract(#0,#5)|sqrt(#6)|","chain":"22 ** 2<\/gadget>\n484<\/output>\n7 ** 2<\/gadget>\n49<\/output>\n484 - 49<\/gadget>\n435<\/output>\n435 ** (1\/2)<\/gadget>\nsqrt(435) = around 20.856654<\/output>\n(sqrt(435)) - 3<\/gadget>\n-3 + sqrt(435) = around 17.856654<\/output>\n(-3 + sqrt(435)) ** 2<\/gadget>\n(-3 + sqrt(435))**2 = around 318.860078<\/output>\n484 - ((-3 + sqrt(435))**2)<\/gadget>\n484 - (-3 + sqrt(435))**2 = around 165.139922<\/output>\n(484 - (-3 + sqrt(435))**2) ** (1\/2)<\/gadget>\nsqrt(484 - (-3 + sqrt(435))**2) = around 12.850678<\/output>\nsqrt(484 - (-3 + sqrt(435))**2) = around 12.850678<\/result>","index":1107} +{"problem":"water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of hydrogen are there in 117 grams of water ?","rationale":"( 2 \/ 18 ) * 117 = 13 grams the answer is d .","correct":"d","options":{"a":"7 ","b":"9 ","c":"11 ","d":"13","e":"15"},"options_float":{"a":7.0,"b":9.0,"c":11.0,"d":13.0,"e":15.0},"annotated_formula":"multiply(2, divide(117, add(2, 16)))","linear_formula":"add(n0,n1)|divide(n2,#0)|multiply(n0,#1)","chain":"2 + 16<\/gadget>\n18<\/output>\n117 \/ 18<\/gadget>\n13\/2 = around 6.5<\/output>\n2 * (13\/2)<\/gadget>\n13<\/output>\n13<\/result>","index":1108} +{"problem":"if a tire rotates at 400 revolutions per minute when the car is traveling 24 km \/ h , what is the circumference of the tire ?","rationale":"\"400 rev \/ minute = 400 * 60 rev \/ 60 minutes = 24,000 rev \/ hour 24,000 * c = 24,000 m : c is the circumference c = 1 meters correct answer a\"","correct":"a","options":{"a":"1 meters ","b":"6 meters ","c":"5 meters ","d":"3 meters","e":"7 meters"},"options_float":{"a":1.0,"b":6.0,"c":5.0,"d":3.0,"e":7.0},"annotated_formula":"multiply(divide(24, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000)","linear_formula":"multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n60 * 400<\/gadget>\n24_000<\/output>\n24 \/ 24_000<\/gadget>\n1\/1_000 = around 0.001<\/output>\n(1\/1_000) * 1_000<\/gadget>\n1<\/output>\n1<\/result>","index":1109} +{"problem":"a can do a work in 15 days and b in 20 days . if they work on it together for 7 days , then the fraction of the work that is left is","rationale":"\"person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( - ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 \/ 35 ) days to complete entire work for 1 st 4 days they work 35 * 7 = 245 remaining work is 300 - 245 = 55 remaining fraction of work is = 55 \/ 300 = 11 \/ 60 answer e\"","correct":"e","options":{"a":"8 \/ 17 ","b":"7 \/ 15 ","c":"3 \/ 15 ","d":"8 \/ 15","e":"11 \/ 60"},"options_float":{"a":0.4705882353,"b":0.4666666667,"c":0.2,"d":0.5333333333,"e":0.1833333333},"annotated_formula":"subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 7))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/15) + (1\/20)<\/gadget>\n7\/60 = around 0.116667<\/output>\n(7\/60) * 7<\/gadget>\n49\/60 = around 0.816667<\/output>\n1 - (49\/60)<\/gadget>\n11\/60 = around 0.183333<\/output>\n11\/60 = around 0.183333<\/result>","index":1110} +{"problem":"workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.75 lacs . how many workers were they ?","rationale":"n * 50 = ( 375000 - 300000 ) = 75000 n = 75000 \/ 50 = 1500 d","correct":"d","options":{"a":"220 ","b":"230 ","c":"500 ","d":"1500","e":"590"},"options_float":{"a":220.0,"b":230.0,"c":500.0,"d":1500.0,"e":590.0},"annotated_formula":"divide(multiply(multiply(subtract(3.75, 3), const_1000), const_100), 50)","linear_formula":"subtract(n2,n0)|multiply(#0,const_1000)|multiply(#1,const_100)|divide(#2,n1)","chain":"3.75 - 3<\/gadget>\n0.75<\/output>\n0.75 * 1_000<\/gadget>\n750<\/output>\n750 * 100<\/gadget>\n75_000<\/output>\n75_000 \/ 50<\/gadget>\n1_500<\/output>\n1_500<\/result>","index":1111} +{"problem":"40 % of a number is added to 60 , the result is the same number . find the number ?","rationale":"\"( 40 \/ 100 ) * x + 60 = x 4 x + 600 = 10 x 6 x = 600 x = 100 answer : d\"","correct":"d","options":{"a":"300 ","b":"288 ","c":"270 ","d":"100","e":"281"},"options_float":{"a":300.0,"b":288.0,"c":270.0,"d":100.0,"e":281.0},"annotated_formula":"divide(60, divide(60, const_100))","linear_formula":"divide(n1,const_100)|divide(n1,#0)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n60 \/ (3\/5)<\/gadget>\n100<\/output>\n100<\/result>","index":1112} +{"problem":"what is the smallest integer that is multiple of 5 , 711","rationale":"it is the lcm of 5 , 7 and 11 which is 385 . the answer is c .","correct":"c","options":{"a":"70 ","b":"35 ","c":"385 ","d":"280","e":"140"},"options_float":{"a":70.0,"b":35.0,"c":385.0,"d":280.0,"e":140.0},"annotated_formula":"multiply(multiply(subtract(const_12, const_1), add(const_3, const_4)), 5)","linear_formula":"add(const_3,const_4)|subtract(const_12,const_1)|multiply(#0,#1)|multiply(n0,#2)","chain":"12 - 1<\/gadget>\n11<\/output>\n3 + 4<\/gadget>\n7<\/output>\n11 * 7<\/gadget>\n77<\/output>\n77 * 5<\/gadget>\n385<\/output>\n385<\/result>","index":1114} +{"problem":"the average waight of a , b , c is 45 kg . the avgwgt of a & b be 40 kg & that of b , c be 43 kg . find the wgt of b .","rationale":"sol . let a , b , c represent their individual wgts . then , a + b + c = ( 45 * 3 ) kg = 135 kg a + b = ( 40 * 2 ) kg = 80 kg & b + c = ( 43 * 2 ) kg = 86 kg b = ( a + b ) + ( b + c ) - ( a + b + c ) = ( 80 + 86 - 135 ) kg = 31 kg . answer d","correct":"d","options":{"a":"23 kg ","b":"43 kg ","c":"21 kg ","d":"31 kg","e":"43 kg"},"options_float":{"a":23.0,"b":43.0,"c":21.0,"d":31.0,"e":43.0},"annotated_formula":"subtract(multiply(40, const_2), subtract(multiply(45, const_3), multiply(43, const_2)))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_3)|multiply(n2,const_2)|subtract(#1,#2)|subtract(#0,#3)","chain":"40 * 2<\/gadget>\n80<\/output>\n45 * 3<\/gadget>\n135<\/output>\n43 * 2<\/gadget>\n86<\/output>\n135 - 86<\/gadget>\n49<\/output>\n80 - 49<\/gadget>\n31<\/output>\n31<\/result>","index":1116} +{"problem":"a father is now 3 times as old as his son . 8 years back , he was 4 times as old as his son . the age of the son ( in years ) is","rationale":"if father ' s age is f and son ' s age is s , then f = 3 s and f - 8 = 4 * ( s - 8 ) 3 s - 5 = 4 s - 32 s = 24 f = 72 years answer : b","correct":"b","options":{"a":"12 ","b":"24 ","c":"18 ","d":"27","e":"25"},"options_float":{"a":12.0,"b":24.0,"c":18.0,"d":27.0,"e":25.0},"annotated_formula":"subtract(multiply(8, 4), 8)","linear_formula":"multiply(n1,n2)|subtract(#0,n1)","chain":"8 * 4<\/gadget>\n32<\/output>\n32 - 8<\/gadget>\n24<\/output>\n24<\/result>","index":1117} +{"problem":"in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 40 per square m ?","rationale":"\"length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 40 * 950 = rs . 38000 d\"","correct":"d","options":{"a":"s . 43000 ","b":"s . 50000 ","c":"s . 57000 ","d":"s . 38000","e":"s . 62000"},"options_float":{"a":43000.0,"b":50000.0,"c":57000.0,"d":38000.0,"e":62000.0},"annotated_formula":"add(add(multiply(multiply(multiply(20, 15), const_2), 40), multiply(multiply(multiply(20, 5), const_2), 40)), multiply(multiply(multiply(15, 5), const_2), 40))","linear_formula":"multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#8)|","chain":"20 * 15<\/gadget>\n300<\/output>\n300 * 2<\/gadget>\n600<\/output>\n600 * 40<\/gadget>\n24_000<\/output>\n20 * 5<\/gadget>\n100<\/output>\n100 * 2<\/gadget>\n200<\/output>\n200 * 40<\/gadget>\n8_000<\/output>\n24_000 + 8_000<\/gadget>\n32_000<\/output>\n15 * 5<\/gadget>\n75<\/output>\n75 * 2<\/gadget>\n150<\/output>\n150 * 40<\/gadget>\n6_000<\/output>\n32_000 + 6_000<\/gadget>\n38_000<\/output>\n38_000<\/result>","index":1119} +{"problem":"at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses e at the restaurant ? ( assume that all boxes are filled to capacity . )","rationale":"\"most test takers would recognize thesystemof equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e\"","correct":"e","options":{"a":"96 ","b":"240 ","c":"e = 256 ","d":"e = 384","e":"e = 480"},"options_float":{"a":96.0,"b":240.0,"c":256.0,"d":384.0,"e":480.0},"annotated_formula":"multiply(multiply(16, const_2), 15)","linear_formula":"multiply(n1,const_2)|multiply(n2,#0)|","chain":"16 * 2<\/gadget>\n32<\/output>\n32 * 15<\/gadget>\n480<\/output>\n480<\/result>","index":1120} +{"problem":"a sum of money becomes 7 \/ 6 of itself in 7 years at a certain rate of simple interest . the rate per annum is ?","rationale":"\"let sum = x . then , amount = 7 x \/ 6 s . i . = 7 x \/ 6 - x = x \/ 6 ; time = 7 years . rate = ( 100 * x ) \/ ( x * 6 * 7 ) = 50 \/ 21 % . answer : d\"","correct":"d","options":{"a":"50 \/ 13 ","b":"50 \/ 17 ","c":"50 \/ 19 ","d":"50 \/ 21","e":"50 \/ 23"},"options_float":{"a":3.8461538462,"b":2.9411764706,"c":2.6315789474,"d":2.380952381,"e":2.1739130435},"annotated_formula":"multiply(divide(subtract(divide(7, 6), const_1), 7), const_100)","linear_formula":"divide(n0,n1)|subtract(#0,const_1)|divide(#1,n2)|multiply(#2,const_100)|","chain":"7 \/ 6<\/gadget>\n7\/6 = around 1.166667<\/output>\n(7\/6) - 1<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) \/ 7<\/gadget>\n1\/42 = around 0.02381<\/output>\n(1\/42) * 100<\/gadget>\n50\/21 = around 2.380952<\/output>\n50\/21 = around 2.380952<\/result>","index":1123} +{"problem":"a ship 77 km from the shore , springs a leak which admits to 9 \/ 4 tonnes of water in 11 \/ 2 minutes . 92 tonnes of water would sink it . but the pumps can throw out 12 tonnes of water per hour . find the average rate of sailing so that the ship may just reach the shore as it begins to sink .","rationale":"leak admits 9 \/ 4 tonnes water in 11 \/ 2 min = 11 \/ 2 * 4 \/ 9 = 22 \/ 9 leak admits 9 \/ 22 tonne of water in 5 min in 1 min in thrws 1 \/ 5 tonne of water = 9 \/ 22 - 1 \/ 5 = 23 \/ 110 92 \/ 23 \/ 11 min = 440 min 22 \/ 3 min = 7 \/ 22 \/ 22 \/ 3 = > 10.5 km \/ hr answer a","correct":"a","options":{"a":"10.5 ","b":"11 ","c":"12 ","d":"13","e":"15"},"options_float":{"a":10.5,"b":11.0,"c":12.0,"d":13.0,"e":15.0},"annotated_formula":"divide(77, divide(92, subtract(divide(divide(9, 4), divide(divide(11, 2), const_60)), 12)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|divide(#1,const_60)|divide(#0,#2)|subtract(#3,n6)|divide(n5,#4)|divide(n0,#5)","chain":"9 \/ 4<\/gadget>\n9\/4 = around 2.25<\/output>\n11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\n(11\/2) \/ 60<\/gadget>\n11\/120 = around 0.091667<\/output>\n(9\/4) \/ (11\/120)<\/gadget>\n270\/11 = around 24.545455<\/output>\n(270\/11) - 12<\/gadget>\n138\/11 = around 12.545455<\/output>\n92 \/ (138\/11)<\/gadget>\n22\/3 = around 7.333333<\/output>\n77 \/ (22\/3)<\/gadget>\n21\/2 = around 10.5<\/output>\n21\/2 = around 10.5<\/result>","index":1124} +{"problem":"a library has an average of 600 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is :","rationale":"\"since the month begins with sunday , to there will be five sundays in the month average required = ( 600 x 5 + 240 x 25 ) \/ 30 ) = 300 answer : option c\"","correct":"c","options":{"a":"250 ","b":"260 ","c":"300 ","d":"285","e":"none"},"options_float":{"a":250.0,"b":260.0,"c":300.0,"d":285.0,"e":null},"annotated_formula":"divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 600), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30)","linear_formula":"add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)|","chain":"3 + 4<\/gadget>\n7<\/output>\n30 \/ 7<\/gadget>\n30\/7 = around 4.285714<\/output>\nfloor(30\/7)<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 600<\/gadget>\n3_000<\/output>\n30 - 5<\/gadget>\n25<\/output>\n25 * 240<\/gadget>\n6_000<\/output>\n3_000 + 6_000<\/gadget>\n9_000<\/output>\n9_000 \/ 30<\/gadget>\n300<\/output>\n300<\/result>","index":1125} +{"problem":"the lcm of two numbers is 2310 and hcf is 30 . if one of the numbers is 210 . then what is the other number ?","rationale":"\"first number * second number = lcm * hcf other number = 2310 * 30 \/ 210 = 11 * 30 = 330 answer : c\"","correct":"c","options":{"a":"715 ","b":"825 ","c":"330 ","d":"582","e":"465"},"options_float":{"a":715.0,"b":825.0,"c":330.0,"d":582.0,"e":465.0},"annotated_formula":"divide(multiply(2310, 30), 210)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"2_310 * 30<\/gadget>\n69_300<\/output>\n69_300 \/ 210<\/gadget>\n330<\/output>\n330<\/result>","index":1126} +{"problem":"a small college reduced its faculty by approximately 15 percent to 195 professors . what was the original number of faculty members ?","rationale":"\"if x is the original number of faculty members , then after 15 % reduction in faculty members number is . 85 x but we are given . 85 x = 195 x = 229 so the original number of faculty members is 229 correct answer - d\"","correct":"d","options":{"a":"182 ","b":"208 ","c":"224 ","d":"229","e":"302"},"options_float":{"a":182.0,"b":208.0,"c":224.0,"d":229.0,"e":302.0},"annotated_formula":"divide(195, divide(subtract(const_100, 15), const_100))","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)|","chain":"100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n195 \/ (17\/20)<\/gadget>\n3_900\/17 = around 229.411765<\/output>\n3_900\/17 = around 229.411765<\/result>","index":1127} +{"problem":"if the sides of a rectangle are increased by 35 % , what is the percentage increase in the area ?","rationale":"if sides are a and b , after increase sides would be 1.35 a and 1.35 b . percentage increase in area = ( 1.35 a * 1.35 b - ab ) * 100 \/ ab = 82.25 answer : c","correct":"c","options":{"a":"38.25 % ","b":"40.25 % ","c":"82.25 % ","d":"50 %","e":"65 %"},"options_float":{"a":38.25,"b":40.25,"c":82.25,"d":50.0,"e":65.0},"annotated_formula":"multiply(subtract(power(add(divide(35, const_100), const_1), const_2), const_1), const_100)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)","chain":"35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n(7\/20) + 1<\/gadget>\n27\/20 = around 1.35<\/output>\n(27\/20) ** 2<\/gadget>\n729\/400 = around 1.8225<\/output>\n(729\/400) - 1<\/gadget>\n329\/400 = around 0.8225<\/output>\n(329\/400) * 100<\/gadget>\n329\/4 = around 82.25<\/output>\n329\/4 = around 82.25<\/result>","index":1129} +{"problem":"the length of the bridge , which a train 120 m long and traveling at 45 km \/ hr can cross in 30 sec is ?","rationale":"\"speed = 45 * 5 \/ 18 = 25 \/ 2 m \/ sec . time = 30 sec let the length of bridge be x meters . then , ( 120 + x ) \/ 30 = 25 \/ 2 x = 255 m . answer : option e\"","correct":"e","options":{"a":"235 ","b":"240 ","c":"245 ","d":"250","e":"255"},"options_float":{"a":235.0,"b":240.0,"c":245.0,"d":250.0,"e":255.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 120)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 120<\/gadget>\n255<\/output>\n255<\/result>","index":1130} +{"problem":"if a person walks at 20 km \/ hr instead of 10 km \/ hr , he would have walked 40 km more . the actual distance traveled by him is ?","rationale":"\"let the actual distance traveled be x km . then , x \/ 10 = ( x + 40 ) \/ 20 10 x - 400 = > x = 40 km . answer : a\"","correct":"a","options":{"a":"40 km ","b":"76 km ","c":"18 km ","d":"16 km","e":"97 km"},"options_float":{"a":40.0,"b":76.0,"c":18.0,"d":16.0,"e":97.0},"annotated_formula":"multiply(10, divide(40, subtract(20, 10)))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|","chain":"20 - 10<\/gadget>\n10<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n10 * 4<\/gadget>\n40<\/output>\n40<\/result>","index":1132} +{"problem":"there are 840 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ?","rationale":"\"let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 840 . hence , we have m + f = 840 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f \/ 2 + m \/ 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 840 \/ 3 = 280 solving the three equations yields the solution f = 280 , m = 560 , and d = 280 . the number of female democratic participants equals half the female participants equals 280 \/ 2 = 140 . answer : d\"","correct":"d","options":{"a":"75 ","b":"100 ","c":"125 ","d":"140","e":"225"},"options_float":{"a":75.0,"b":100.0,"c":125.0,"d":140.0,"e":225.0},"annotated_formula":"divide(subtract(multiply(divide(840, const_3), const_4), 840), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)|","chain":"840 \/ 3<\/gadget>\n280<\/output>\n280 * 4<\/gadget>\n1_120<\/output>\n1_120 - 840<\/gadget>\n280<\/output>\n280 \/ 2<\/gadget>\n140<\/output>\n140<\/result>","index":1133} +{"problem":"the percentage increase in the area of a rectangle , if each of its sides is increased by 20 % is ?","rationale":"\"let original length = x metres and original breadth = y metres . original area = ( xy ) m 2 . new length = 120 x m = 6 x m . 100 5 new breadth = 120 y m = 6 y m . 100 5 new area = 6 x x 6 y m 2 = 36 xy m 2 . 5 5 25 the difference between the original area = xy and new - area 36 \/ 25 xy is = ( 36 \/ 25 ) xy - xy = xy ( 36 \/ 25 - 1 ) = xy ( 11 \/ 25 ) or ( 11 \/ 25 ) xy increase % = 11 xy x 1 x 100 % = 44 % . a )\"","correct":"a","options":{"a":"44 % ","b":"45 % ","c":"50 % ","d":"55 %","e":"60 %"},"options_float":{"a":44.0,"b":45.0,"c":50.0,"d":55.0,"e":60.0},"annotated_formula":"multiply(subtract(multiply(divide(add(const_100, 20), const_100), divide(add(const_100, 20), const_100)), const_1), const_100)","linear_formula":"add(n0,const_100)|divide(#0,const_100)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|","chain":"100 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * (6\/5)<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 100<\/gadget>\n44<\/output>\n44<\/result>","index":1134} +{"problem":"the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 40 degrees . find the temperature on friday ?","rationale":"\"m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 40 tu + w + th = 192 - 40 = 152 f = 184 – 152 = 32 answer : b\"","correct":"b","options":{"a":"65 degrees ","b":"32 degrees ","c":"37 degrees ","d":"34 degrees","e":"74 degrees"},"options_float":{"a":65.0,"b":32.0,"c":37.0,"d":34.0,"e":74.0},"annotated_formula":"subtract(40, subtract(multiply(48, const_4), multiply(46, const_4)))","linear_formula":"multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)|","chain":"48 * 4<\/gadget>\n192<\/output>\n46 * 4<\/gadget>\n184<\/output>\n192 - 184<\/gadget>\n8<\/output>\n40 - 8<\/gadget>\n32<\/output>\n32<\/result>","index":1135} +{"problem":"difference between the length & breadth of a rectangle is 30 m . if its perimeter is 300 m , then its area is ? ? we have : ( l - b ) = 30 and 2 ( l + b ) = 300 or ( l + b ) = 150 ?","rationale":"solving the two equations , we get : l = 90 and b = 60 . the ratio of length and breath is 90 : 60 = 3 : 2 e","correct":"e","options":{"a":"3 : 5 ","b":"2 : 3 ","c":"3 : 6 ","d":"3 : 4","e":"3 : 2"},"options_float":{"a":0.6,"b":0.6666666667,"c":0.5,"d":0.75,"e":1.5},"annotated_formula":"divide(const_3, 2)","linear_formula":"divide(const_3,n3)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":1136} +{"problem":"what is the maximum number e of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ?","rationale":"27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can have are 4 * 3 * 2 = 24 l * w * h is 8 * 12 * 9 , then max . cube we can have are 2 * 4 * 3 = 24 l * w * h is 8 * 9 * 12 , then max . cube we can have are 2 * 3 * 4 = 24 in all cases we get e = 24 cubes . ans . c","correct":"c","options":{"a":"36 ","b":"32 ","c":"24 ","d":"21","e":"15"},"options_float":{"a":36.0,"b":32.0,"c":24.0,"d":21.0,"e":15.0},"annotated_formula":"multiply(multiply(divide(12, power(27, inverse(const_3))), divide(9, power(27, inverse(const_3)))), floor(divide(8, power(27, inverse(const_3)))))","linear_formula":"inverse(const_3)|power(n0,#0)|divide(n1,#1)|divide(n3,#1)|divide(n2,#1)|floor(#2)|multiply(#3,#4)|multiply(#5,#6)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n27 ** (1\/3)<\/gadget>\n3<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n9 \/ 3<\/gadget>\n3<\/output>\n4 * 3<\/gadget>\n12<\/output>\n8 \/ 3<\/gadget>\n8\/3 = around 2.666667<\/output>\nfloor(8\/3)<\/gadget>\n2<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24<\/result>","index":1137} +{"problem":"find k if 64 \/ k = 4 .","rationale":"since 64 \/ k = 4 and 64 \/ 16 = 4 , then k = 16 correct answer a","correct":"a","options":{"a":"16 ","b":"17 ","c":"18 ","d":"14","e":"13"},"options_float":{"a":16.0,"b":17.0,"c":18.0,"d":14.0,"e":13.0},"annotated_formula":"divide(64, 4)","linear_formula":"divide(n0,n1)","chain":"64 \/ 4<\/gadget>\n16<\/output>\n16<\/result>","index":1140} +{"problem":"if 1 + 3 = 10 ; 3 + 5 = 52 ; 5 + 7 = 174 then 7 + 11 = ?","rationale":"\"1 ^ 3 + 3 ^ 2 = 1 + 9 = 10 3 ^ 3 + 5 ^ 2 = 27 + 25 = 52 5 ^ 3 + 7 ^ 2 = 125 + 49 = 174 and 7 ^ 3 + 11 ^ 2 = 343 + 121 = 464 answer : e\"","correct":"e","options":{"a":"354 ","b":"225 ","c":"654 ","d":"741","e":"464"},"options_float":{"a":354.0,"b":225.0,"c":654.0,"d":741.0,"e":464.0},"annotated_formula":"add(power(7, 3), power(11, const_2))","linear_formula":"power(n9,n3)|power(n10,const_2)|add(#0,#1)|","chain":"7 ** 3<\/gadget>\n343<\/output>\n11 ** 2<\/gadget>\n121<\/output>\n343 + 121<\/gadget>\n464<\/output>\n464<\/result>","index":1141} +{"problem":"10 play kabadi , 30 play kho kho only , 5 play both gmaes . then how many in total ?","rationale":"10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 30 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 30 + 5 = 35 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 35 - 5 = 40 answer : d","correct":"d","options":{"a":"30 ","b":"35 ","c":"38 ","d":"40","e":"45"},"options_float":{"a":30.0,"b":35.0,"c":38.0,"d":40.0,"e":45.0},"annotated_formula":"subtract(add(10, add(30, 5)), 5)","linear_formula":"add(n1,n2)|add(n0,#0)|subtract(#1,n2)","chain":"30 + 5<\/gadget>\n35<\/output>\n10 + 35<\/gadget>\n45<\/output>\n45 - 5<\/gadget>\n40<\/output>\n40<\/result>","index":1142} +{"problem":"14 men make 1400 toy in 5 days 1 day after work 14 more worker join now how may days work can finish","rationale":"14 men make toy in 5 days after 1 day 14 more worker join m 1 \/ m 2 = d 2 \/ d 1 14 \/ 28 = d 2 \/ 4 = 2 day answer : b","correct":"b","options":{"a":"1 day ","b":"2 days ","c":"3 days ","d":"4 days","e":"5 days"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"divide(multiply(multiply(5, 14), subtract(1400, divide(1400, 5))), multiply(multiply(const_2, 14), 1400))","linear_formula":"divide(n1,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n1,#2)|subtract(n1,#0)|multiply(#1,#4)|divide(#5,#3)","chain":"5 * 14<\/gadget>\n70<\/output>\n1_400 \/ 5<\/gadget>\n280<\/output>\n1_400 - 280<\/gadget>\n1_120<\/output>\n70 * 1_120<\/gadget>\n78_400<\/output>\n2 * 14<\/gadget>\n28<\/output>\n28 * 1_400<\/gadget>\n39_200<\/output>\n78_400 \/ 39_200<\/gadget>\n2<\/output>\n2<\/result>","index":1143} +{"problem":"what ratio must a shopkeepermix peas and soybean of rs . 16 and rs . 25 \/ kg , as to obtain a mixture of rs . 19.50 ?","rationale":"\"correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 5.50 : 3.50 = 11 : 7\"","correct":"c","options":{"a":"10 : 7 ","b":"9 : 8 ","c":"11 : 7 ","d":"13 : 11","e":"14 : 8"},"options_float":{"a":1.4285714286,"b":1.125,"c":1.5714285714,"d":1.1818181818,"e":1.75},"annotated_formula":"divide(subtract(25, 19.50), subtract(19.50, 16))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"25 - 19.5<\/gadget>\n5.5<\/output>\n19.5 - 16<\/gadget>\n3.5<\/output>\n5.5 \/ 3.5<\/gadget>\n1.571429<\/output>\n1.571429<\/result>","index":1144} +{"problem":"a sum of rs 468.75 was lent out at simple interest and at the end of 1 year and 8 months , the total amount of rs 500 is received . find the rate of interest ?","rationale":"here p = 468.75 which is the principle . amount = rs 500 amount = si + p 500 = [ 468.75 * ( 20 \/ 12 ) * r ] \/ 100 + 468.75 31.25 = 156.25 * 5 r \/ 100 r = 4 % answer : b","correct":"b","options":{"a":"3 % ","b":"4 % ","c":"5 % ","d":"6 %","e":"7 %"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(divide(multiply(subtract(500, 468.75), const_100), add(1, divide(8, const_12))), 468.75)","linear_formula":"divide(n2,const_12)|subtract(n3,n0)|add(n1,#0)|multiply(#1,const_100)|divide(#3,#2)|divide(#4,n0)","chain":"500 - 468.75<\/gadget>\n31.25<\/output>\n31.25 * 100<\/gadget>\n3_125<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 + (2\/3)<\/gadget>\n5\/3 = around 1.666667<\/output>\n3_125 \/ (5\/3)<\/gadget>\n1_875<\/output>\n1_875 \/ 468.75<\/gadget>\n4<\/output>\n4<\/result>","index":1145} +{"problem":"a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 110 . what was the number he chose ?","rationale":"\"let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 110 x = 124 answer : d\"","correct":"d","options":{"a":"123 ","b":"267 ","c":"277 ","d":"124","e":"120"},"options_float":{"a":123.0,"b":267.0,"c":277.0,"d":124.0,"e":120.0},"annotated_formula":"divide(add(110, 138), 2)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"110 + 138<\/gadget>\n248<\/output>\n248 \/ 2<\/gadget>\n124<\/output>\n124<\/result>","index":1147} +{"problem":"the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 14 cm each . the radius of the sphere is","rationale":"\"solution 4 î r 2 = 2 î 7 x 14 â ‡ ’ r 2 = ( 7 x 14 \/ 2 ) â ‡ ’ 49 â ‡ ’ r = 7 cm . answer d\"","correct":"d","options":{"a":"3 cm ","b":"4 cm ","c":"6 cm ","d":"7 cm","e":"none"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":7.0,"e":null},"annotated_formula":"sqrt(divide(multiply(multiply(const_pi, multiply(14, divide(14, const_2))), const_2), multiply(const_pi, const_4)))","linear_formula":"divide(n0,const_2)|multiply(const_4,const_pi)|multiply(n0,#0)|multiply(#2,const_pi)|multiply(#3,const_2)|divide(#4,#1)|sqrt(#5)|","chain":"14 \/ 2<\/gadget>\n7<\/output>\n14 * 7<\/gadget>\n98<\/output>\npi * 98<\/gadget>\n98*pi = around 307.87608<\/output>\n(98*pi) * 2<\/gadget>\n196*pi = around 615.75216<\/output>\npi * 4<\/gadget>\n4*pi = around 12.566371<\/output>\n(196*pi) \/ (4*pi)<\/gadget>\n49<\/output>\n49 ** (1\/2)<\/gadget>\n7<\/output>\n7<\/result>","index":1150} +{"problem":"if 8 men or 12 women can do a piece of work in 30 days , in how many days can the same work be done by 6 men and 11 women ?","rationale":"\"8 men = 12 women ( i . e 2 men = 3 women ) 12 women 1 day work = 1 \/ 30 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 30 = 1 \/ 360 so , 20 women work = 20 \/ 360 = 1 \/ 18 ans : 18 days answer : e\"","correct":"e","options":{"a":"10 days ","b":"11 days ","c":"13 days ","d":"15 days","e":"18 days"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":15.0,"e":18.0},"annotated_formula":"inverse(add(divide(6, multiply(8, 30)), divide(11, multiply(12, 30))))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)|","chain":"8 * 30<\/gadget>\n240<\/output>\n6 \/ 240<\/gadget>\n1\/40 = around 0.025<\/output>\n12 * 30<\/gadget>\n360<\/output>\n11 \/ 360<\/gadget>\n11\/360 = around 0.030556<\/output>\n(1\/40) + (11\/360)<\/gadget>\n1\/18 = around 0.055556<\/output>\n1 \/ (1\/18)<\/gadget>\n18<\/output>\n18<\/result>","index":1151} +{"problem":"set s consists of integers { 8 , 11 , 12 , 14 , 15 } . if integer n is included in the set , the average ( arithmetic mean ) of set s will increase by 25 % . what is the value of integer n ?","rationale":"the average of the numbers in set s is 12 . if we increase the mean by 25 % , the new mean is 15 . thus , on average , 5 numbers increase by 5 . therefore n = 15 + 25 = 40 the answer is d .","correct":"d","options":{"a":"28 ","b":"32 ","c":"36 ","d":"40","e":"44"},"options_float":{"a":28.0,"b":32.0,"c":36.0,"d":40.0,"e":44.0},"annotated_formula":"add(add(12, 14), 14)","linear_formula":"add(n2,n3)|add(n3,#0)","chain":"12 + 14<\/gadget>\n26<\/output>\n26 + 14<\/gadget>\n40<\/output>\n40<\/result>","index":1153} +{"problem":"the average marks obtained by 50 candidates in a certain examination is 40 . find the total marks .","rationale":"following the above formula , we have the total marks = 50 * 40 = 2000 answer is b","correct":"b","options":{"a":"1800 ","b":"2000 ","c":"1640 ","d":"2200","e":"2400"},"options_float":{"a":1800.0,"b":2000.0,"c":1640.0,"d":2200.0,"e":2400.0},"annotated_formula":"multiply(50, 40)","linear_formula":"multiply(n0,n1)","chain":"50 * 40<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":1154} +{"problem":"the speed of a car is 95 km in the first hour and 60 km in the second hour . what is the average speed of the car ?","rationale":"\"s = ( 95 + 60 ) \/ 2 = 77.5 kmph answer : c\"","correct":"c","options":{"a":"76 kmph ","b":"75 kmph ","c":"77.5 kmph ","d":"56 kmph","e":"86 kmph"},"options_float":{"a":76.0,"b":75.0,"c":77.5,"d":56.0,"e":86.0},"annotated_formula":"divide(add(95, 60), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"95 + 60<\/gadget>\n155<\/output>\n155 \/ 2<\/gadget>\n155\/2 = around 77.5<\/output>\n155\/2 = around 77.5<\/result>","index":1156} +{"problem":"a hiker walking at a constant rate of 4 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 30 kilometers per hour . the cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate . how many minutes must the cyclist wait until the hiker catches up ?","rationale":"\"in 5 minutes , the cyclist travels a distance of ( 5 \/ 60 ) * 30 = 2.5 km . the time it takes the hiker to complete this distance is 2.5 \/ 4 hours = 37.5 minutes the cyclist needs to wait 37.5 - 5 = 32.5 minutes the answer is e .\"","correct":"e","options":{"a":"7.5 ","b":"15 ","c":"17.5 ","d":"25","e":"32.5"},"options_float":{"a":7.5,"b":15.0,"c":17.5,"d":25.0,"e":32.5},"annotated_formula":"multiply(divide(subtract(multiply(30, divide(5, const_60)), multiply(4, divide(5, const_60))), 4), const_60)","linear_formula":"divide(n2,const_60)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|divide(#3,n0)|multiply(#4,const_60)|","chain":"5 \/ 60<\/gadget>\n1\/12 = around 0.083333<\/output>\n30 * (1\/12)<\/gadget>\n5\/2 = around 2.5<\/output>\n4 * (1\/12)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(5\/2) - (1\/3)<\/gadget>\n13\/6 = around 2.166667<\/output>\n(13\/6) \/ 4<\/gadget>\n13\/24 = around 0.541667<\/output>\n(13\/24) * 60<\/gadget>\n65\/2 = around 32.5<\/output>\n65\/2 = around 32.5<\/result>","index":1159} +{"problem":"3 ltr of water is added with 9 ltr of a solution containing 57 % of alcohol in the water . the % of alcohol in the new mixture is ?","rationale":"\"we have a 9 litre solution containing 57 % of alcohol in the water . = > quantity of alcohol in the solution = 9 ã — 57 \/ 100 now 3 litre of water is added to the solution . = > total quantity of the new solution = 9 + 3 = 12 percentage of alcohol in the new solution = 9 ã — 57 \/ 100 12 ã — 100 = 9 ã — 5710012 ã — 100 = 9 ã — 4.75 \/ 100 = 42.75 % b\"","correct":"b","options":{"a":"41.5 % ","b":"42.75 % ","c":"43 % ","d":"45 %","e":"45.75 %"},"options_float":{"a":41.5,"b":42.75,"c":43.0,"d":45.0,"e":45.75},"annotated_formula":"multiply(const_100, divide(multiply(divide(57, const_100), 9), add(3, 9)))","linear_formula":"add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|divide(#2,#0)|multiply(#3,const_100)|","chain":"57 \/ 100<\/gadget>\n57\/100 = around 0.57<\/output>\n(57\/100) * 9<\/gadget>\n513\/100 = around 5.13<\/output>\n3 + 9<\/gadget>\n12<\/output>\n(513\/100) \/ 12<\/gadget>\n171\/400 = around 0.4275<\/output>\n100 * (171\/400)<\/gadget>\n171\/4 = around 42.75<\/output>\n171\/4 = around 42.75<\/result>","index":1160} +{"problem":"a grocer purchased a quantity of bananas at 3 pounds for $ 0.50 and sold the entire quantity at 4 pounds for $ 1.00 . how many pounds did the grocer purchase if the profit from selling the bananas was $ 11.00 ?","rationale":"\"cost price of 1 pound of bananas = 0.5 \/ 3 = 1 \/ 6 selling price of 1 pound of bananas = 1 \/ 4 profit per pound = ( 1 \/ 4 - 1 \/ 6 ) = ( 1 \/ 12 ) total profit is given as 11 ( 1 \/ 12 ) * x = 11 x = 132 answer : d\"","correct":"d","options":{"a":"40 ","b":"60 ","c":"90 ","d":"132","e":"240"},"options_float":{"a":40.0,"b":60.0,"c":90.0,"d":132.0,"e":240.0},"annotated_formula":"divide(11.00, subtract(divide(1.00, 4), divide(0.50, 3)))","linear_formula":"divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|divide(n4,#2)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n0.5 \/ 3<\/gadget>\n0.166667<\/output>\n(1\/4) - 0.166667<\/gadget>\n0.083333<\/output>\n11 \/ 0.083333<\/gadget>\n132.000528<\/output>\n132.000528<\/result>","index":1163} +{"problem":"the length and breadth of a rectangle is increased by 5 % and 15 % respectively . what is the increase in the area ?","rationale":"\"100 * 100 = 10000 105 * 115 = 12075 - - - - - - - - - - - 2075 10000 - - - - - - 2075 100 - - - - - ? = > 20.75 answer d\"","correct":"d","options":{"a":"21.25 ","b":"22.75 ","c":"19.25 ","d":"20.75","e":"22.25"},"options_float":{"a":21.25,"b":22.75,"c":19.25,"d":20.75,"e":22.25},"annotated_formula":"divide(multiply(subtract(rectangle_area(add(5, const_100), add(15, const_100)), rectangle_area(const_100, const_100)), const_100), rectangle_area(const_100, const_100))","linear_formula":"add(n0,const_100)|add(n1,const_100)|rectangle_area(const_100,const_100)|rectangle_area(#0,#1)|subtract(#3,#2)|multiply(#4,const_100)|divide(#5,#2)|","chain":"5 + 100<\/gadget>\n105<\/output>\n15 + 100<\/gadget>\n115<\/output>\n105 * 115<\/gadget>\n12_075<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n12_075 - 10_000<\/gadget>\n2_075<\/output>\n2_075 * 100<\/gadget>\n207_500<\/output>\n207_500 \/ 10_000<\/gadget>\n83\/4 = around 20.75<\/output>\n83\/4 = around 20.75<\/result>","index":1165} +{"problem":"a train running at the speed of 40 km \/ hr crosses a pole in 9 seconds . what is the length of the train ?","rationale":"\"speed = 40 x 5 \/ 18 m \/ sec = 100 \/ 9 m \/ sec . length of the train = ( speed x time ) . length of the train = 100 \/ 9 x 9 m = 100 m . answer : e\"","correct":"e","options":{"a":"120 metres ","b":"180 metres ","c":"324 metres ","d":"150 metres","e":"100 meters"},"options_float":{"a":120.0,"b":180.0,"c":324.0,"d":150.0,"e":100.0},"annotated_formula":"multiply(divide(multiply(40, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"40 * 1_000<\/gadget>\n40_000<\/output>\n40_000 \/ 3_600<\/gadget>\n100\/9 = around 11.111111<\/output>\n(100\/9) * 9<\/gadget>\n100<\/output>\n100<\/result>","index":1166} +{"problem":"a chemical supply company has 60 liters of a 35 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ?","rationale":"\"60 liters of a 35 % hno 3 solution means hno 3 = 21 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 21 + x = 50 % of ( 60 + x ) or x = 18 hence , a\"","correct":"a","options":{"a":"18 ","b":"15 ","c":"20 ","d":"24","e":"30"},"options_float":{"a":18.0,"b":15.0,"c":20.0,"d":24.0,"e":30.0},"annotated_formula":"multiply(subtract(divide(60, const_2), divide(multiply(60, 35), const_100)), const_2)","linear_formula":"divide(n0,const_2)|multiply(n0,n1)|divide(#1,const_100)|subtract(#0,#2)|multiply(#3,const_2)|","chain":"60 \/ 2<\/gadget>\n30<\/output>\n60 * 35<\/gadget>\n2_100<\/output>\n2_100 \/ 100<\/gadget>\n21<\/output>\n30 - 21<\/gadget>\n9<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18<\/result>","index":1167} +{"problem":"the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 140 runs and his average excluding these two innings is 58 runs , find his highest score .","rationale":"\"explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 â € “ 2552 = 208 runs . let the highest score be x , hence the lowest score = x â € “ 140 x + ( x - 140 ) = 208 2 x = 348 x = 174 runs answer : e\"","correct":"e","options":{"a":"179 ","b":"367 ","c":"269 ","d":"177","e":"174"},"options_float":{"a":179.0,"b":367.0,"c":269.0,"d":177.0,"e":174.0},"annotated_formula":"divide(add(140, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2)","linear_formula":"multiply(n0,n1)|subtract(n1,const_2)|multiply(n3,#1)|subtract(#0,#2)|add(n2,#3)|divide(#4,const_2)|","chain":"60 * 46<\/gadget>\n2_760<\/output>\n46 - 2<\/gadget>\n44<\/output>\n58 * 44<\/gadget>\n2_552<\/output>\n2_760 - 2_552<\/gadget>\n208<\/output>\n140 + 208<\/gadget>\n348<\/output>\n348 \/ 2<\/gadget>\n174<\/output>\n174<\/result>","index":1169} +{"problem":"if m ^ 2 - 8 m + 1 = 0 , then m ^ 3 + m ^ ( - 3 ) = ?","rationale":"here from equation 1 we will get m ^ 2 - 8 m + 1 = 0 m ^ 2 = 8 m - 1 m = ( 8 m - 1 ) \/ m . . . . . . . . . ( 1 ) m = 8 - 1 \/ m so m + 1 \/ m = 8 now in equation 2 m ^ 3 + 1 \/ m ^ 3 ( m + 1 \/ m ) ( m ^ 2 - m * 1 \/ m + 1 \/ m ^ 2 ) ( m + 1 \/ m ) ( m ^ 2 - 1 + 1 \/ m ^ 2 ) ( m + 1 \/ m ) ( m ^ 2 + 1 \/ m ^ 2 - 1 ) ( m + 1 \/ m ) { ( m + 1 \/ m ) ^ 2 - 2 m * 1 \/ m - 1 } { since a ^ 2 + b ^ 2 = ( a + b ) ^ 2 - 2 ab } ( m + 1 \/ m ) { ( m + 1 \/ m ) ^ 2 - 2 - 1 } now putting the value of m from equation 1 8 ( 8 ^ 2 - 2 - 1 ) 8 ( 64 - 3 ) = 488 answer : b","correct":"b","options":{"a":"487 ","b":"488 ","c":"489 ","d":"457","e":"495"},"options_float":{"a":487.0,"b":488.0,"c":489.0,"d":457.0,"e":495.0},"annotated_formula":"power(divide(add(add(const_4, const_4), sqrt(add(add(const_10, const_10), multiply(const_10, const_4)))), 2), 3)","linear_formula":"add(const_4,const_4)|add(const_10,const_10)|multiply(const_10,const_4)|add(#1,#2)|sqrt(#3)|add(#0,#4)|divide(#5,n0)|power(#6,n4)","chain":"4 + 4<\/gadget>\n8<\/output>\n10 + 10<\/gadget>\n20<\/output>\n10 * 4<\/gadget>\n40<\/output>\n20 + 40<\/gadget>\n60<\/output>\n60 ** (1\/2)<\/gadget>\n2*sqrt(15) = around 7.745967<\/output>\n8 + (2*sqrt(15))<\/gadget>\n2*sqrt(15) + 8 = around 15.745967<\/output>\n(2*sqrt(15) + 8) \/ 2<\/gadget>\nsqrt(15) + 4 = around 7.872983<\/output>\n(sqrt(15) + 4) ** 3<\/gadget>\n(sqrt(15) + 4)**3 = around 487.997951<\/output>\n(sqrt(15) + 4)**3 = around 487.997951<\/result>","index":1170} +{"problem":"if 3 a – 3 b – 2 c = 32 and √ 3 a - √ ( 2 b + 2 c ) = 4 , what is the value of a + b + c ?","rationale":"when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a – 2 b – 2 c = 32 can be written as √ 3 a ^ 2 - √ ( 2 b + 2 c ) ^ 2 = 32 { √ 3 a - √ ( 2 b + 2 c ) } { √ 3 a + √ ( 2 b + 2 c ) } = 32 . . or 4 * √ 3 a + √ ( 2 b + 2 c ) = 32 . . or √ 3 a + √ ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) √ 3 a - √ ( 2 b + 2 c ) = 4 . . 2 ) √ 3 a + √ ( 2 b + 2 c ) = 8 . . a ) add the two eq . . √ 3 a + √ ( 2 b + 2 c ) + √ 3 a - √ ( 2 b + 2 c ) = 12 . . 2 √ 3 a = 12 . . or √ 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . √ 3 a + √ ( 2 b + 2 c ) - √ 3 a + √ ( 2 b + 2 c ) = 4 . . 2 √ ( 2 b + 2 c ) = 4 . . √ ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . c","correct":"c","options":{"a":"3 ","b":"9 ","c":"10 ","d":"12","e":"14"},"options_float":{"a":3.0,"b":9.0,"c":10.0,"d":12.0,"e":14.0},"annotated_formula":"subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(32, 4), 4), const_2), const_2), const_1), 3), const_1), 2), 3), const_4)","linear_formula":"subtract(n3,n7)|subtract(#0,n7)|subtract(#1,const_2)|subtract(#2,const_2)|subtract(#3,const_1)|subtract(#4,n0)|subtract(#5,const_1)|add(n2,#6)|subtract(#7,n0)|subtract(#8,const_4)","chain":"32 - 4<\/gadget>\n28<\/output>\n28 - 4<\/gadget>\n24<\/output>\n24 - 2<\/gadget>\n22<\/output>\n22 - 2<\/gadget>\n20<\/output>\n20 - 1<\/gadget>\n19<\/output>\n19 - 3<\/gadget>\n16<\/output>\n16 - 1<\/gadget>\n15<\/output>\n15 + 2<\/gadget>\n17<\/output>\n17 - 3<\/gadget>\n14<\/output>\n14 - 4<\/gadget>\n10<\/output>\n10<\/result>","index":1171} +{"problem":"in a group of 100 cars , 47 cars do not have air conditioning . if at least 55 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ?","rationale":"\"lets assume ac = 53 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 55 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 55 . . we have to assign atleast 8 tocars with ac and racing stripes . hence ac = 53 - 8 = 45 . the answer is a\"","correct":"a","options":{"a":"45 ","b":"47 ","c":"49 ","d":"51","e":"53"},"options_float":{"a":45.0,"b":47.0,"c":49.0,"d":51.0,"e":53.0},"annotated_formula":"subtract(100, 55)","linear_formula":"subtract(n0,n2)|","chain":"100 - 55<\/gadget>\n45<\/output>\n45<\/result>","index":1173} +{"problem":"at a certain conference , 78 % of the attendees registered at least two weeks in advance and paid their conference fee in full . if 10 % of the attendees who paid their conference fee in full did not register at least two weeks in advance , what percent of conference attendees registered at least two weeks in advance ?","rationale":"i do not think that venn will be helpful here . a table will make more sense . but here is my approach . 78 % regestered at least 2 weeks and paid full fee . 10 % paid full fee and did not registered at least 2 weeks in advance . then 90 % paid full fee and registered at least 2 weeks before . 90 % * x = 78 % where x is the number of people who registerd 2 weeks in advance and paid full fee . ( answer d )","correct":"d","options":{"a":"18.0 % ","b":"62.0 % ","c":"79.2 % ","d":"86.67 %","e":"82.0 %"},"options_float":{"a":18.0,"b":62.0,"c":79.2,"d":86.67,"e":82.0},"annotated_formula":"divide(multiply(78, const_100), subtract(const_100, 10))","linear_formula":"multiply(n0,const_100)|subtract(const_100,n1)|divide(#0,#1)","chain":"78 * 100<\/gadget>\n7_800<\/output>\n100 - 10<\/gadget>\n90<\/output>\n7_800 \/ 90<\/gadget>\n260\/3 = around 86.666667<\/output>\n260\/3 = around 86.666667<\/result>","index":1174} +{"problem":"if the volume of two cubes are in the ratio 8 : 1 , the ratio of their edges is :","rationale":"explanation : let the edges be a and b of two cubes , then a 3 \/ b 3 = 8 \/ 1 = > ( a \/ b ) 3 = ( 2 \/ 1 ) 3 a \/ b = 2 \/ 1 = > a : b = 2 : 1 option a","correct":"a","options":{"a":"2 : 1 ","b":"3 : 2 ","c":"3 : 5 ","d":"3 : 7","e":"none of these"},"options_float":{"a":2.0,"b":1.5,"c":0.6,"d":0.4285714286,"e":null},"annotated_formula":"cube_edge_by_volume(8)","linear_formula":"cube_edge_by_volume(n0)","chain":"8 ** (1\/3)<\/gadget>\n2<\/output>\n2<\/result>","index":1176} +{"problem":"what is the dividend . divisor 15 , the quotient is 8 and the remainder is 5","rationale":"\"d = d * q + r d = 15 * 8 + 5 d = 120 + 5 d = 125 answer : d\"","correct":"d","options":{"a":"145 ","b":"148 ","c":"150 ","d":"125","e":"158"},"options_float":{"a":145.0,"b":148.0,"c":150.0,"d":125.0,"e":158.0},"annotated_formula":"add(multiply(15, 8), 5)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"15 * 8<\/gadget>\n120<\/output>\n120 + 5<\/gadget>\n125<\/output>\n125<\/result>","index":1177} +{"problem":"x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 20 days . how long will y take to complete the work ?","rationale":"\"work done by x in 8 days = 8 * 1 \/ 40 = 1 \/ 5 remaining work = 1 - 1 \/ 5 = 4 \/ 5 4 \/ 5 work is done by y in 20 days whole work will be done by y in 20 * 5 \/ 4 = 25 days answer is c\"","correct":"c","options":{"a":"10 ","b":"12 ","c":"25 ","d":"18","e":"20"},"options_float":{"a":10.0,"b":12.0,"c":25.0,"d":18.0,"e":20.0},"annotated_formula":"multiply(20, inverse(subtract(const_1, divide(8, 40))))","linear_formula":"divide(n1,n0)|subtract(const_1,#0)|inverse(#1)|multiply(n2,#2)|","chain":"8 \/ 40<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ (4\/5)<\/gadget>\n5\/4 = around 1.25<\/output>\n20 * (5\/4)<\/gadget>\n25<\/output>\n25<\/result>","index":1178} +{"problem":"by selling an article at rs . 500 , a profit of 25 % is made . find its cost price ?","rationale":"\"sp = 500 cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 500 * [ 100 \/ ( 100 + 25 ) ] = 500 * [ 100 \/ 125 ] = rs . 400 answer : a\"","correct":"a","options":{"a":"400 ","b":"267 ","c":"287 ","d":"480","e":"811"},"options_float":{"a":400.0,"b":267.0,"c":287.0,"d":480.0,"e":811.0},"annotated_formula":"divide(multiply(500, const_100), add(const_100, 25))","linear_formula":"add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|","chain":"500 * 100<\/gadget>\n50_000<\/output>\n100 + 25<\/gadget>\n125<\/output>\n50_000 \/ 125<\/gadget>\n400<\/output>\n400<\/result>","index":1180} +{"problem":"in a class of 32 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ?","rationale":"\"total = english + german - both + neither - - > 40 = english + 22 - 12 + 0 - - > english = 22 - - > only english = english - both = 22 - 12 = 10 . answer : b .\"","correct":"b","options":{"a":"30 ","b":"10 ","c":"18 ","d":"28","e":"32"},"options_float":{"a":30.0,"b":10.0,"c":18.0,"d":28.0,"e":32.0},"annotated_formula":"subtract(subtract(add(32, 12), 22), 12)","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)|","chain":"32 + 12<\/gadget>\n44<\/output>\n44 - 22<\/gadget>\n22<\/output>\n22 - 12<\/gadget>\n10<\/output>\n10<\/result>","index":1181} +{"problem":"the ratio of radius of a circle and the side of a square is 2 : 5 . find the ratio of their areas :","rationale":"\"radius \/ side = 2 \/ 5 â ‡ ’ area of circle \/ area of square = 4 \/ 25 answer : b\"","correct":"b","options":{"a":"2 : 1 ","b":"4 : 25 ","c":"8 : 77 ","d":"4 : 121","e":"none"},"options_float":{"a":2.0,"b":0.16,"c":0.1038961039,"d":0.0330578512,"e":null},"annotated_formula":"power(divide(2, 5), 2)","linear_formula":"divide(n0,n1)|power(#0,n0)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) ** 2<\/gadget>\n4\/25 = around 0.16<\/output>\n4\/25 = around 0.16<\/result>","index":1182} +{"problem":"a lion chases a pony 7 hours after the pony runs . lion takes 8 hours to reach the pony . if the average speed of the lion is 49 kmph , what s the average speed of the pony ?","rationale":"pony take 15 hours and lion take 8 hours . . . then distance chased by them is 49 * 8 . so speed of pony is ( 49 * 8 ) \/ 15 = 29.4 kmph . answer is e","correct":"e","options":{"a":"28.3 kmph ","b":"29.8 kmph ","c":"28.4 kmph ","d":"29.5 kmph","e":"29.4 kmph"},"options_float":{"a":28.3,"b":29.8,"c":28.4,"d":29.5,"e":29.4},"annotated_formula":"divide(multiply(49, 8), multiply(7, const_2))","linear_formula":"multiply(n1,n2)|multiply(n0,const_2)|divide(#0,#1)","chain":"49 * 8<\/gadget>\n392<\/output>\n7 * 2<\/gadget>\n14<\/output>\n392 \/ 14<\/gadget>\n28<\/output>\n28<\/result>","index":1183} +{"problem":"on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 360 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ?","rationale":"\"explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x \/ ( x - 360 ) = 2 + 2 ( extra ) = > 2 x - 720 = x = > x = 720 . answer : e\"","correct":"e","options":{"a":"237 ","b":"287 ","c":"197 ","d":"287","e":"720"},"options_float":{"a":237.0,"b":287.0,"c":197.0,"d":287.0,"e":720.0},"annotated_formula":"multiply(360, const_2)","linear_formula":"multiply(n0,const_2)|","chain":"360 * 2<\/gadget>\n720<\/output>\n720<\/result>","index":1184} +{"problem":"the cash realised on selling a 14 % stock is rs . 106.25 , brokerage being 1 \/ 4 % is ?","rationale":"\"cash realised = rs . ( 106.25 - 0.25 ) = rs . 106 . answer : b\"","correct":"b","options":{"a":"366 ","b":"106 ","c":"102 ","d":"192","e":"122"},"options_float":{"a":366.0,"b":106.0,"c":102.0,"d":192.0,"e":122.0},"annotated_formula":"subtract(106.25, divide(1, 4))","linear_formula":"divide(n2,n3)|subtract(n1,#0)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n106.25 - (1\/4)<\/gadget>\n106<\/output>\n106<\/result>","index":1185} +{"problem":"a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 900 , what is the reduced price for kg ?","rationale":"\"a 900 * ( 25 \/ 100 ) = 225 - - - - 5 ? - - - - 1 = > rs . 45\"","correct":"a","options":{"a":"45 ","b":"60 ","c":"70 ","d":"80","e":"20"},"options_float":{"a":45.0,"b":60.0,"c":70.0,"d":80.0,"e":20.0},"annotated_formula":"divide(divide(multiply(900, 25), const_100), 5)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|","chain":"900 * 25<\/gadget>\n22_500<\/output>\n22_500 \/ 100<\/gadget>\n225<\/output>\n225 \/ 5<\/gadget>\n45<\/output>\n45<\/result>","index":1186} +{"problem":"rectangular tile each of size 80 cm by 55 cm must be laid horizontally on a rectangular floor of size 210 cm by 330 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is :","rationale":"\"area of tile = 80 * 55 = 4400 area of floor = 210 * 330 = 72600 no of tiles = 72600 \/ 4400 = 16.5 so , the no of tile = 16 answer : b\"","correct":"b","options":{"a":"26 ","b":"16 ","c":"18 ","d":"19","e":"20"},"options_float":{"a":26.0,"b":16.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"divide(multiply(210, 330), multiply(80, 55))","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)|","chain":"210 * 330<\/gadget>\n69_300<\/output>\n80 * 55<\/gadget>\n4_400<\/output>\n69_300 \/ 4_400<\/gadget>\n63\/4 = around 15.75<\/output>\n63\/4 = around 15.75<\/result>","index":1187} +{"problem":"sowdharya had 90 currency notes in all , some of which are of rs 95 denomination and the remaining of rs 45 denomination . the total amount of all these currency notes was rs . 5000 . how much amount ( in rs ) did she have in the denomination of rs 45 ?","rationale":"let the number of 45 - rupee notes = x then , the number of 95 - rupee notes = ( 90 – x ) 45 x + 95 ( 90 – x ) = 5000 : x = 71 answer : c","correct":"c","options":{"a":"75 ","b":"72 ","c":"71 ","d":"76","e":"73"},"options_float":{"a":75.0,"b":72.0,"c":71.0,"d":76.0,"e":73.0},"annotated_formula":"subtract(90, divide(subtract(5000, multiply(90, 45)), subtract(95, 45)))","linear_formula":"multiply(n0,n2)|subtract(n1,n2)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3)","chain":"90 * 45<\/gadget>\n4_050<\/output>\n5_000 - 4_050<\/gadget>\n950<\/output>\n95 - 45<\/gadget>\n50<\/output>\n950 \/ 50<\/gadget>\n19<\/output>\n90 - 19<\/gadget>\n71<\/output>\n71<\/result>","index":1188} +{"problem":"if an investor puts $ 700 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ?","rationale":"\"1.05 * 1.05 * 700 = $ 771.75 the answer is b .\"","correct":"b","options":{"a":"$ 769.75 ","b":"$ 771.75 ","c":"$ 773.75 ","d":"$ 775.75","e":"$ 777.75"},"options_float":{"a":769.75,"b":771.75,"c":773.75,"d":775.75,"e":777.75},"annotated_formula":"multiply(700, power(add(const_1, divide(divide(10, const_2), const_100)), const_2))","linear_formula":"divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)|","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 + (1\/20)<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n700 * (441\/400)<\/gadget>\n3_087\/4 = around 771.75<\/output>\n3_087\/4 = around 771.75<\/result>","index":1189} +{"problem":"a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ?","rationale":"\"[ 1000 + ( x - 4000 ) * ( 2.5 \/ 100 ) ] - x * ( 5 \/ 100 ) = 600 x = 12000 answer : b\"","correct":"b","options":{"a":"12002 ","b":"12000 ","c":"12028 ","d":"12020","e":"12013"},"options_float":{"a":12002.0,"b":12000.0,"c":12028.0,"d":12020.0,"e":12013.0},"annotated_formula":"divide(600, divide(5, const_100))","linear_formula":"divide(n0,const_100)|divide(n4,#0)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n600 \/ (1\/20)<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":1190} +{"problem":"to deliver an order on time , a company has to make 25 parts a day . after making 25 parts per day for 3 days , the company started to produce 5 more parts per day , and by the last day of work 100 more parts than planned were produced . find how many parts the company made and how many days this took .","rationale":"let x be the number of days the company worked . then 25 x is the number of parts they planned to make . at the new production rate they made : 3 ⋅ 25 + ( x − 3 ) ⋅ 30 = 75 + 30 ( x − 3 ) therefore : 25 x = 75 + 30 ( x − 3 ) − 100 25 x = 75 + 30 x − 90 − 100 190 − 75 = 30 x − 25 115 = 5 x x = 23 so the company worked 23 days and they made 23 ⋅ 25 + 100 = 675 pieces . so answer is e .","correct":"e","options":{"a":"600 ","b":"500 ","c":"575 ","d":"655","e":"675"},"options_float":{"a":600.0,"b":500.0,"c":575.0,"d":655.0,"e":675.0},"annotated_formula":"add(multiply(divide(subtract(add(multiply(add(25, 5), 3), 100), multiply(25, 3)), 5), 25), 100)","linear_formula":"add(n0,n3)|multiply(n0,n2)|multiply(n2,#0)|add(n4,#2)|subtract(#3,#1)|divide(#4,n3)|multiply(n0,#5)|add(n4,#6)","chain":"25 + 5<\/gadget>\n30<\/output>\n30 * 3<\/gadget>\n90<\/output>\n90 + 100<\/gadget>\n190<\/output>\n25 * 3<\/gadget>\n75<\/output>\n190 - 75<\/gadget>\n115<\/output>\n115 \/ 5<\/gadget>\n23<\/output>\n23 * 25<\/gadget>\n575<\/output>\n575 + 100<\/gadget>\n675<\/output>\n675<\/result>","index":1191} +{"problem":"a hollow iron pipe is 21 cm long and its external diameter is 8 cm . if the thickness of the pipe is 1 cm and iron weights 8 g \/ cm cube , then find the weight of the pipe .","rationale":"explanation : in this type of question , we need to subtract external radius and internal radius to get the answer using the volume formula as the pipe is hollow . oh ! line become a bit complicated , sorry for that , lets solve it . external radius = 4 cm internal radius = 3 cm [ because thickness of pipe is 1 cm ] volume of iron = π r 2 h = 22 \/ 7 ∗ [ 42 − 32 ] ∗ 21 cm 3 = 22 \/ 7 ∗ 1 ∗ 21 cm 3 = 462 cm 3 weight of iron = 462 * 8 = 3696 gm = 3.696 kg option a","correct":"a","options":{"a":"3.696 kg ","b":"3.686 kg ","c":"2.696 kg ","d":"2.686 kg","e":"none of these"},"options_float":{"a":3.696,"b":3.686,"c":2.696,"d":2.686,"e":null},"annotated_formula":"divide(multiply(subtract(volume_cylinder(divide(8, const_2), 21), volume_cylinder(subtract(divide(8, const_2), const_1), 21)), 8), const_1000)","linear_formula":"divide(n1,const_2)|subtract(#0,const_1)|volume_cylinder(#0,n0)|volume_cylinder(#1,n0)|subtract(#2,#3)|multiply(n1,#4)|divide(#5,const_1000)","chain":"8 \/ 2<\/gadget>\n4<\/output>\npi * (4 ** 2) * 21<\/gadget>\n336*pi = around 1_055.575132<\/output>\n4 - 1<\/gadget>\n3<\/output>\npi * (3 ** 2) * 21<\/gadget>\n189*pi = around 593.761012<\/output>\n(336*pi) - (189*pi)<\/gadget>\n147*pi = around 461.81412<\/output>\n(147*pi) * 8<\/gadget>\n1176*pi = around 3_694.512961<\/output>\n(1176*pi) \/ 1_000<\/gadget>\n147*pi\/125 = around 3.694513<\/output>\n147*pi\/125 = around 3.694513<\/result>","index":1194} +{"problem":"village x has a population of 72000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ?","rationale":"\"let the population of two villages be equal after p years then , 72000 - 1200 p = 42000 + 800 p 2000 p = 30000 p = 15 answer is a .\"","correct":"a","options":{"a":"15 ","b":"19 ","c":"11 ","d":"18","e":"13"},"options_float":{"a":15.0,"b":19.0,"c":11.0,"d":18.0,"e":13.0},"annotated_formula":"divide(subtract(72000, 42000), add(800, 1200))","linear_formula":"add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|","chain":"72_000 - 42_000<\/gadget>\n30_000<\/output>\n800 + 1_200<\/gadget>\n2_000<\/output>\n30_000 \/ 2_000<\/gadget>\n15<\/output>\n15<\/result>","index":1195} +{"problem":"a person purchases 90 clocks and sells 40 clocks at a gain of 10 % and 50 clocks at a gain of 20 % . if he sold all of them at a uniform profit of 15 % , then he would have got rs . 40 less . the cost price of each clock is :","rationale":"let c . p . of clock be rs . x . then , c . p . of 90 clocks = rs . 90 x . [ ( 110 % of 40 x ) + ( 120 % of 50 x ) ] - ( 115 % of 90 x ) = 40 44 x + 60 x - 103.5 x = 40 0.5 x = 40 = > x = 80 answer : option c","correct":"c","options":{"a":"50 ","b":"90 ","c":"80 ","d":"28","e":"93"},"options_float":{"a":50.0,"b":90.0,"c":80.0,"d":28.0,"e":93.0},"annotated_formula":"divide(40, subtract(add(multiply(40, add(const_1, divide(10, const_100))), multiply(50, add(const_1, divide(20, const_100)))), multiply(90, add(const_1, divide(15, const_100)))))","linear_formula":"divide(n2,const_100)|divide(n4,const_100)|divide(n5,const_100)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|multiply(n1,#3)|multiply(n3,#4)|multiply(n0,#5)|add(#6,#7)|subtract(#9,#8)|divide(n1,#10)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n40 * (11\/10)<\/gadget>\n44<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n50 * (6\/5)<\/gadget>\n60<\/output>\n44 + 60<\/gadget>\n104<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n90 * (23\/20)<\/gadget>\n207\/2 = around 103.5<\/output>\n104 - (207\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n40 \/ (1\/2)<\/gadget>\n80<\/output>\n80<\/result>","index":1197} +{"problem":"find the remainder of the division ( 3 ^ 33 ) \/ 7 .","rationale":"find the pattern of the remainders after each power : ( 3 ^ 1 ) \/ 7 remainder 3 ( 3 ^ 2 ) \/ 7 remainder 2 ( 3 ^ 3 ) \/ 7 remainder 6 ( 3 ^ 4 ) \/ 7 remainder 4 ( 3 ^ 5 ) \/ 7 remainder 5 ( 3 ^ 6 ) \/ 7 remainder 1 - - > this is where the cycle ends ( 3 ^ 7 ) \/ 7 remainder 3 - - > this is where the cycle begins again ( 3 ^ 8 ) \/ 7 remainder 2 continuing the pattern to ( 3 ^ 33 ) \/ 7 gives us a remainder of 6 final answer : e ) 6","correct":"e","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"reminder(power(3, 33), 7)","linear_formula":"power(n0,n1)|reminder(#0,n2)","chain":"3 ** 33<\/gadget>\n5_559_060_566_555_523<\/output>\n5_559_060_566_555_523 % 7<\/gadget>\n6<\/output>\n6<\/result>","index":1199} +{"problem":"an inspector rejects 0.05 % of the meters as defective . how many will he examine to reject 4 ?","rationale":"\"let the number of meters to be examined be x then , 0.05 % of x = 4 ( 5 \/ 100 ) * ( ( 1 \/ 100 ) * x = 4 x = 8000 answer is b\"","correct":"b","options":{"a":"4000 ","b":"8000 ","c":"5000 ","d":"4500","e":"6000"},"options_float":{"a":4000.0,"b":8000.0,"c":5000.0,"d":4500.0,"e":6000.0},"annotated_formula":"divide(multiply(4, const_100), 0.05)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"4 * 100<\/gadget>\n400<\/output>\n400 \/ 0.05<\/gadget>\n8_000<\/output>\n8_000<\/result>","index":1200} +{"problem":"a train passes a platform in 34 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km \/ hr , the length of the platform is","rationale":"\"explanation : speed of the train = 54 km \/ hr = ( 54 × 10 ) \/ 34 m \/ s = 15 m \/ s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) \/ 15 = > ( 300 + l ) \/ 15 = 34 = > 300 + l = 15 × 34 = 510 = > l = 510 - 300 = 210 meter answer : option c\"","correct":"c","options":{"a":"280 meter ","b":"240 meter ","c":"210 meter ","d":"200 meter","e":"none of these"},"options_float":{"a":280.0,"b":240.0,"c":210.0,"d":200.0,"e":null},"annotated_formula":"multiply(multiply(const_0_2778, 54), subtract(34, 20))","linear_formula":"multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 54<\/gadget>\n15<\/output>\n34 - 20<\/gadget>\n14<\/output>\n15 * 14<\/gadget>\n210<\/output>\n210<\/result>","index":1201} +{"problem":"the average age of 15 students of a class is 15 years . out of these , the average age of 3 students is 14 years and that of the other 11 students is 16 years . the age of the 15 th student is","rationale":"\"solution age of the 15 th student = [ 15 x 15 - ( 14 x 3 + 16 x 11 ) ] = ( 225 - 218 ) = 7 years . answer d\"","correct":"d","options":{"a":"9 years ","b":"11 years ","c":"14 years ","d":"7 years","e":"25 years"},"options_float":{"a":9.0,"b":11.0,"c":14.0,"d":7.0,"e":25.0},"annotated_formula":"subtract(multiply(15, 15), add(multiply(3, 14), multiply(11, 16)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n3 * 14<\/gadget>\n42<\/output>\n11 * 16<\/gadget>\n176<\/output>\n42 + 176<\/gadget>\n218<\/output>\n225 - 218<\/gadget>\n7<\/output>\n7<\/result>","index":1202} +{"problem":"a number x is 6 times another number y . the percentage that y is less than x is","rationale":"say y = 1 and x = 6 . then y = 1 is less than x = 6 by ( 6 - 1 ) \/ 6 * 100 = 5 \/ 6 * 100 = 83.3 % . answer : b .","correct":"b","options":{"a":"12.5 % ","b":"83.3 % ","c":"80 % ","d":"11 %","e":"1 %"},"options_float":{"a":12.5,"b":83.3,"c":80.0,"d":11.0,"e":1.0},"annotated_formula":"multiply(divide(subtract(6, const_1), 6), const_100)","linear_formula":"subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)","chain":"6 - 1<\/gadget>\n5<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(5\/6) * 100<\/gadget>\n250\/3 = around 83.333333<\/output>\n250\/3 = around 83.333333<\/result>","index":1203} +{"problem":"liam is pulled over for speeding just as he is arriving at work . he explains to the police officer that he could not afford to be late today , and has arrived at work only four minutes before he is to start . the officer explains that if liam had driven 5 mph slower for his whole commute , he would have arrived at work exactly on time . if liam ' s commute is 10 miles long , how fast was he actually driving ? ( assume that liam drove at a constant speed for the duration of his commute . )","rationale":"\"let t be the number of hours he would need to reach office on time . when he is driving with over speed , he reached office 4 min earlier ! so the equation for this is s ( t - 4 \/ 60 ) = 30 where s is the speed and 30 is the distance . if he decreases his speed by 5 mph then he would have reached his office on time : ( s - 5 ) t = 30 if you solve above equations , you will arrive at t = 2 \/ 3 hr and s = 30 mph therefore answer is d\"","correct":"d","options":{"a":"50 mph ","b":"45 mph ","c":"48 mph ","d":"30 mph","e":"60 mph"},"options_float":{"a":50.0,"b":45.0,"c":48.0,"d":30.0,"e":60.0},"annotated_formula":"add(10, multiply(5, const_4))","linear_formula":"multiply(n0,const_4)|add(n1,#0)|","chain":"5 * 4<\/gadget>\n20<\/output>\n10 + 20<\/gadget>\n30<\/output>\n30<\/result>","index":1204} +{"problem":"what is the compound interest paid on a sum of rs . 3000 for the period of 2 years at 10 % per annum .","rationale":"\"solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 \/ 100 = 11 total % of interest = 10 + 11 = 21 total interest = 21 % 3000 = 3000 * ( 21 \/ 100 ) = 630 answer a\"","correct":"a","options":{"a":"630 ","b":"620 ","c":"610 ","d":"600","e":"none of these"},"options_float":{"a":630.0,"b":620.0,"c":610.0,"d":600.0,"e":null},"annotated_formula":"subtract(multiply(3000, power(add(const_1, divide(10, const_100)), 2)), 3000)","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n3_000 * (121\/100)<\/gadget>\n3_630<\/output>\n3_630 - 3_000<\/gadget>\n630<\/output>\n630<\/result>","index":1205} +{"problem":"a boy perform 5 times a reaction and his reading are 2 , 2.1 , 2 , 2.2 but unfortunately 5 th reading clearly not visible because a drop of chemical fall onto it . but luckily he have mean of all reading which is 2 . please help him in finding 5 th reading .","rationale":"solution let 5 th reading be x ( 2 + 2.1 + 2 + 2.2 + x ) \/ 5 = 2 8.3 + x = 10 x = 1.7 answer b","correct":"b","options":{"a":"2 ","b":"1.7 ","c":"1.9 ","d":"1.8","e":"2.1"},"options_float":{"a":2.0,"b":1.7,"c":1.9,"d":1.8,"e":2.1},"annotated_formula":"subtract(subtract(subtract(subtract(multiply(2, 5), 2), 2.1), 2), 2.2)","linear_formula":"multiply(n0,n1)|subtract(#0,n1)|subtract(#1,n2)|subtract(#2,n1)|subtract(#3,n4)","chain":"2 * 5<\/gadget>\n10<\/output>\n10 - 2<\/gadget>\n8<\/output>\n8 - 2.1<\/gadget>\n5.9<\/output>\n5.9 - 2<\/gadget>\n3.9<\/output>\n3.9 - 2.2<\/gadget>\n1.7<\/output>\n1.7<\/result>","index":1208} +{"problem":"thirty percent of the members of a swim club have passed the lifesaving test . among the members who have not passed the test , 12 have taken the preparatory course and 30 have not taken the course . how many members are there in the swim club ?","rationale":"30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 30 + 12 = 42 members have not passed the test , thus 0.7 * total = 42 - - > total = 60 . answer : a .","correct":"a","options":{"a":"60 ","b":"80 ","c":"100 ","d":"120","e":"140"},"options_float":{"a":60.0,"b":80.0,"c":100.0,"d":120.0,"e":140.0},"annotated_formula":"divide(add(12, 30), divide(subtract(const_100, 30), const_100))","linear_formula":"add(n0,n1)|subtract(const_100,n1)|divide(#1,const_100)|divide(#0,#2)","chain":"12 + 30<\/gadget>\n42<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n42 \/ (7\/10)<\/gadget>\n60<\/output>\n60<\/result>","index":1209} +{"problem":"a certain rectangular window is ( 1 \/ 3 ) times as long as it is wide . if its perimeter is 28 feet , what are its dimensions in terms of length by width ?","rationale":"2 x + 2 y = 28 x + y = 14 x + ( 1 \/ 3 ) x = 14 4 x = 14 * 3 x = 10.5 answer c","correct":"c","options":{"a":"12 by 2 ","b":"11 by 3 ","c":"10.5 by 3.5 ","d":"10 by 4","e":"9 by 3"},"options_float":{"a":12.0,"b":11.0,"c":10.5,"d":10.0,"e":9.0},"annotated_formula":"add(const_10, 1)","linear_formula":"add(n0,const_10)","chain":"10 + 1<\/gadget>\n11<\/output>\n11<\/result>","index":1210} +{"problem":"there are 8 teams in a certain league and each team plays each of the other teams exactly once . if each game is played by 2 teams , what is the total number of games played ?","rationale":"\"number of ways 2 teams can be selected out of 8 teams : c 8 \/ 2 = 8 ∗ 7 \/ 2 = 28 ans : ' ' c ' '\"","correct":"c","options":{"a":"15 ","b":"16 ","c":"28 ","d":"56","e":"64"},"options_float":{"a":15.0,"b":16.0,"c":28.0,"d":56.0,"e":64.0},"annotated_formula":"divide(multiply(8, subtract(8, const_1)), const_2)","linear_formula":"subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|","chain":"8 - 1<\/gadget>\n7<\/output>\n8 * 7<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":1211} +{"problem":"3 candidates in an election and received 1136 , 7636 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ?","rationale":"\"total number of votes polled = ( 1136 + 7636 + 11628 ) = 20400 so , required percentage = 11628 \/ 20400 * 100 = 57 % c\"","correct":"c","options":{"a":"40 % ","b":"55 % ","c":"57 % ","d":"60 %","e":"62 %"},"options_float":{"a":40.0,"b":55.0,"c":57.0,"d":60.0,"e":62.0},"annotated_formula":"multiply(divide(11628, add(add(1136, 7636), 11628)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)|","chain":"1_136 + 7_636<\/gadget>\n8_772<\/output>\n8_772 + 11_628<\/gadget>\n20_400<\/output>\n11_628 \/ 20_400<\/gadget>\n57\/100 = around 0.57<\/output>\n(57\/100) * 100<\/gadget>\n57<\/output>\n57<\/result>","index":1212} +{"problem":"at what rate percent on simple interest will rs . 1750 amount to rs . 2000 in 3 years ?","rationale":"\"explanation : 250 = ( 1750 x 3 xr ) \/ 100 r = 4.76 % answer : option c\"","correct":"c","options":{"a":"4 % ","b":"3 6 \/ 7 % ","c":"4.76 % ","d":"5 %","e":"6 %"},"options_float":{"a":4.0,"b":3.0,"c":4.76,"d":5.0,"e":6.0},"annotated_formula":"multiply(divide(divide(subtract(2000, 1750), 1750), 3), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|","chain":"2_000 - 1_750<\/gadget>\n250<\/output>\n250 \/ 1_750<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/7) \/ 3<\/gadget>\n1\/21 = around 0.047619<\/output>\n(1\/21) * 100<\/gadget>\n100\/21 = around 4.761905<\/output>\n100\/21 = around 4.761905<\/result>","index":1214} +{"problem":"the perimeter of a square is equal to the perimeter of a rectangle of length 19 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places )","rationale":"\"let the side of the square be a cm . perimeter of the rectangle = 2 ( 19 + 14 ) = 66 cm perimeter of the square = 66 cm i . e . 4 a = 66 a = 16.5 diameter of the semicircle = 16.5 cm circumference of the semicircle = 1 \/ 2 ( ∏ ) ( 16.5 ) = 1 \/ 2 ( 22 \/ 7 ) ( 16.5 ) = 25.93 cm to two decimal places answer : e\"","correct":"e","options":{"a":"22.78 ","b":"23.54 ","c":"23.5 ","d":"24.55","e":"25.93"},"options_float":{"a":22.78,"b":23.54,"c":23.5,"d":24.55,"e":25.93},"annotated_formula":"floor(divide(circumface(divide(divide(multiply(const_2, add(14, 19)), const_4), const_2)), const_2))","linear_formula":"add(n0,n1)|multiply(#0,const_2)|divide(#1,const_4)|divide(#2,const_2)|circumface(#3)|divide(#4,const_2)|floor(#5)|","chain":"14 + 19<\/gadget>\n33<\/output>\n2 * 33<\/gadget>\n66<\/output>\n66 \/ 4<\/gadget>\n33\/2 = around 16.5<\/output>\n(33\/2) \/ 2<\/gadget>\n33\/4 = around 8.25<\/output>\n2 * pi * (33\/4)<\/gadget>\n33*pi\/2 = around 51.836279<\/output>\n(33*pi\/2) \/ 2<\/gadget>\n33*pi\/4 = around 25.918139<\/output>\nfloor(33*pi\/4)<\/gadget>\n25<\/output>\n25<\/result>","index":1216} +{"problem":"john and steve are speed walkers in a race . john is 15 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m \/ s , while steve maintains a blistering 3.7 m \/ s speed . if john finishes the race 2 meters ahead of steve , how long was john ’ s final push ?","rationale":"\"let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.7 t + 15 + 2 - - - > 0.5 t = 17 - - - > t = 34 seconds . d is the correct answer .\"","correct":"d","options":{"a":"13 seconds ","b":"17 seconds ","c":"26 seconds ","d":"34 seconds","e":"51 seconds"},"options_float":{"a":13.0,"b":17.0,"c":26.0,"d":34.0,"e":51.0},"annotated_formula":"divide(add(divide(multiply(3.7, add(15, 2)), subtract(4.2, 3.7)), add(15, 2)), 4.2)","linear_formula":"add(n0,n3)|subtract(n1,n2)|multiply(n2,#0)|divide(#2,#1)|add(#0,#3)|divide(#4,n1)|","chain":"15 + 2<\/gadget>\n17<\/output>\n3.7 * 17<\/gadget>\n62.9<\/output>\n4.2 - 3.7<\/gadget>\n0.5<\/output>\n62.9 \/ 0.5<\/gadget>\n125.8<\/output>\n125.8 + 17<\/gadget>\n142.8<\/output>\n142.8 \/ 4.2<\/gadget>\n34<\/output>\n34<\/result>","index":1217} +{"problem":"a ratio between two numbers is 3 : 4 and their l . c . m . is 84 . the first number is","rationale":"\"sol . let the required numbers be 3 x and 4 x . then , their l . c . m . is 12 x . ∴ 12 x = 84 ⇔ x = 7 . hence , the first number is 21 . answer c\"","correct":"c","options":{"a":"60 ","b":"45 ","c":"21 ","d":"15","e":"none"},"options_float":{"a":60.0,"b":45.0,"c":21.0,"d":15.0,"e":null},"annotated_formula":"multiply(divide(84, multiply(3, 4)), 3)","linear_formula":"multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|","chain":"3 * 4<\/gadget>\n12<\/output>\n84 \/ 12<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21<\/result>","index":1218} +{"problem":"the principal that amounts to rs . 4813 in 3 years at 6 1 \/ 4 % per annum c . i . compounded annually , is ?","rationale":"\"principal = [ 4913 \/ ( 1 + 25 \/ ( 4 * 100 ) ) 3 ] = 4813 * 16 \/ 17 * 16 \/ 17 * 16 \/ 17 = rs . 4036 . answer : b\"","correct":"b","options":{"a":"s . 3096 ","b":"s . 4036 ","c":"s . 4085 ","d":"s . 4096","e":"s . 5096"},"options_float":{"a":3096.0,"b":4036.0,"c":4085.0,"d":4096.0,"e":5096.0},"annotated_formula":"divide(4813, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3))","linear_formula":"divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n6 + (1\/4)<\/gadget>\n25\/4 = around 6.25<\/output>\n(25\/4) \/ 100<\/gadget>\n1\/16 = around 0.0625<\/output>\n1 + (1\/16)<\/gadget>\n17\/16 = around 1.0625<\/output>\n(17\/16) ** 3<\/gadget>\n4_913\/4_096 = around 1.199463<\/output>\n4_813 \/ (4_913\/4_096)<\/gadget>\n19_714_048\/4_913 = around 4_012.629351<\/output>\n19_714_048\/4_913 = around 4_012.629351<\/result>","index":1219} +{"problem":"the area of a rectangular plot is 18 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ?","rationale":"\"l × b = 18 × b ∴ l = 18 m and l – b = 10 ∴ b = 18 – 10 = 8 m answer b\"","correct":"b","options":{"a":"10 metres ","b":"8 metres ","c":"7.5 metres ","d":"data inadequate","e":"none of these"},"options_float":{"a":10.0,"b":8.0,"c":7.5,"d":null,"e":null},"annotated_formula":"subtract(18, 10)","linear_formula":"subtract(n0,n1)|","chain":"18 - 10<\/gadget>\n8<\/output>\n8<\/result>","index":1220} +{"problem":"in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the competition and finished first , second , third , or fourth each time . the product of all the points he received was 38500 . how many times did he participate in the competition ?","rationale":"38500 = 2 * 2 * 5 * 5 * 5 * 7 * 11 john participated 7 times . the answer is b .","correct":"b","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"floor(sqrt(divide(38500, multiply(multiply(multiply(11, 7), 5), 2))))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|multiply(n3,#1)|divide(n4,#2)|sqrt(#3)|floor(#4)","chain":"11 * 7<\/gadget>\n77<\/output>\n77 * 5<\/gadget>\n385<\/output>\n385 * 2<\/gadget>\n770<\/output>\n38_500 \/ 770<\/gadget>\n50<\/output>\n50 ** (1\/2)<\/gadget>\n5*sqrt(2) = around 7.071068<\/output>\nfloor(5*sqrt(2))<\/gadget>\n7<\/output>\n7<\/result>","index":1221} +{"problem":"the first flight out of phoenix airport had a late departure . if the next 3 flights departed on - time , how many subsequent flights need to depart from phoenix on - time , for the airport ' s on - time departure rate to be higher than 40 % ?","rationale":"we need on - time departure rate to be higher than 4 \/ 10 , so it should be at least 5 \/ 11 , which means that 5 out of 11 flights must depart on time . since for now 3 out of 4 flights departed on time then 5 - 3 = 2 subsequent flights need to depart on - time . answer : d","correct":"d","options":{"a":"4 ","b":"6 ","c":"5 ","d":"2","e":"8"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":2.0,"e":8.0},"annotated_formula":"add(floor(multiply(divide(40, const_100), 3)), const_1)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|floor(#1)|add(#2,const_1)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 3<\/gadget>\n6\/5 = around 1.2<\/output>\nfloor(6\/5)<\/gadget>\n1<\/output>\n1 + 1<\/gadget>\n2<\/output>\n2<\/result>","index":1222} +{"problem":"two years ago , ram put $ 500 into a savings account . at the end of the first year , his account had accrued $ 100 in interest bringing his total balance to $ 600 . the next year , his account balance increased by 10 % . at the end of the two years , by what percent has ram ' s account balance increased from his initial deposit of $ 500 ?","rationale":"investment 500 dollars 1 st year total gained = 100 total amount end of first year = 600 second year account increased by 10 % = 600 * 0.1 = 60 therefore total amount by second year end = 660 so total percentage increase in money = ( 660 - 500 ) * 100 \/ 500 = 32 % correct answer d = 32 %","correct":"d","options":{"a":"19 % ","b":"28 % ","c":"30 % ","d":"32 %","e":"25 %"},"options_float":{"a":19.0,"b":28.0,"c":30.0,"d":32.0,"e":25.0},"annotated_formula":"multiply(divide(add(multiply(divide(10, const_100), 600), 100), 500), const_100)","linear_formula":"divide(n3,const_100)|multiply(n2,#0)|add(n1,#1)|divide(#2,n0)|multiply(#3,const_100)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 600<\/gadget>\n60<\/output>\n60 + 100<\/gadget>\n160<\/output>\n160 \/ 500<\/gadget>\n8\/25 = around 0.32<\/output>\n(8\/25) * 100<\/gadget>\n32<\/output>\n32<\/result>","index":1225} +{"problem":"if the sum of two numbers is 45 and the h . c . f and l . c . m of these numbers are 3 and 100 respectively , then the sum of the reciprocal of the numbers is equal to :","rationale":"let the numbers be a and b . then , a + b = 45 and ab = 3 * 100 = 300 . required sum = 1 \/ a + 1 \/ b = ( a + b ) \/ ab = 103 \/ 300 answer : a","correct":"a","options":{"a":"103 \/ 300 ","b":"103 \/ 200 ","c":"112 \/ 300 ","d":"154 \/ 140","e":"100 \/ 300"},"options_float":{"a":0.3433333333,"b":0.515,"c":0.3733333333,"d":1.1,"e":0.3333333333},"annotated_formula":"divide(add(3, 100), multiply(3, 100))","linear_formula":"add(n1,n2)|multiply(n1,n2)|divide(#0,#1)","chain":"3 + 100<\/gadget>\n103<\/output>\n3 * 100<\/gadget>\n300<\/output>\n103 \/ 300<\/gadget>\n103\/300 = around 0.343333<\/output>\n103\/300 = around 0.343333<\/result>","index":1227} +{"problem":"mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 45 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ?","rationale":"\"let me try a simpler one . lets assume that candidate got 45 % votes and total votes is 100 . candidate won = 45 remaining = 55 to get 50 % , candidate requires 5 votes from 100 which is 5 % and 5 votes from 55 . 5 \/ 55 = 1 \/ 11 = . 09 = 9 % which is approx 9 % . hence the answer is a .\"","correct":"a","options":{"a":"9 % ","b":"12 % ","c":"15 % ","d":"17 %","e":"20 %"},"options_float":{"a":9.0,"b":12.0,"c":15.0,"d":17.0,"e":20.0},"annotated_formula":"multiply(divide(subtract(divide(50, const_100), divide(45, const_100)), subtract(const_1, divide(45, const_100))), const_100)","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n(1\/2) - (9\/20)<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (9\/20)<\/gadget>\n11\/20 = around 0.55<\/output>\n(1\/20) \/ (11\/20)<\/gadget>\n1\/11 = around 0.090909<\/output>\n(1\/11) * 100<\/gadget>\n100\/11 = around 9.090909<\/output>\n100\/11 = around 9.090909<\/result>","index":1229} +{"problem":"pipe a can fill a tank in 6 hours . due to a leak at the bottom , it takes 12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ?","rationale":"\"let the leak can empty the full tank in x hours 1 \/ 6 - 1 \/ x = 1 \/ 12 = > 1 \/ x = 1 \/ 6 - 1 \/ 12 = ( 2 - 1 ) \/ 12 = 1 \/ 12 = > x = 12 . answer : e\"","correct":"e","options":{"a":"13 ","b":"17 ","c":"18 ","d":"19","e":"12"},"options_float":{"a":13.0,"b":17.0,"c":18.0,"d":19.0,"e":12.0},"annotated_formula":"divide(multiply(12, 6), subtract(12, 6))","linear_formula":"multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|","chain":"12 * 6<\/gadget>\n72<\/output>\n12 - 6<\/gadget>\n6<\/output>\n72 \/ 6<\/gadget>\n12<\/output>\n12<\/result>","index":1230} +{"problem":"w and x start a business with rs . 5000 and rs . 20000 respectively . hoe should they share their profits at the end of one year ?","rationale":"they should share the profits in the ratio of their investments . the ratio of the investments made by w and x = 5000 : 20000 = > 1 : 4 . answer : c","correct":"c","options":{"a":"4 : 1 ","b":"1 : 3 ","c":"1 : 4 ","d":"1 : 5","e":"2 : 1"},"options_float":{"a":4.0,"b":0.3333333333,"c":0.25,"d":0.2,"e":2.0},"annotated_formula":"divide(multiply(5000, const_12), multiply(20000, const_12))","linear_formula":"multiply(n0,const_12)|multiply(n1,const_12)|divide(#0,#1)","chain":"5_000 * 12<\/gadget>\n60_000<\/output>\n20_000 * 12<\/gadget>\n240_000<\/output>\n60_000 \/ 240_000<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":1231} +{"problem":"in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 80 questions and secures 130 marks , the no of questions he attempts correctly is :","rationale":"\"let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 80 – x ) = 130 = > 5 x = 210 = > x = 42 answer : e\"","correct":"e","options":{"a":"35 ","b":"38 ","c":"90 ","d":"88","e":"42"},"options_float":{"a":35.0,"b":38.0,"c":90.0,"d":88.0,"e":42.0},"annotated_formula":"divide(add(130, 80), add(4, 1))","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|","chain":"130 + 80<\/gadget>\n210<\/output>\n4 + 1<\/gadget>\n5<\/output>\n210 \/ 5<\/gadget>\n42<\/output>\n42<\/result>","index":1232} +{"problem":"what is the least number should be added to 1056 , so the sum of the number is completely divisible by 29 ?","rationale":"( 1056 \/ 29 ) gives remainder 12 17 + 12 = 29 , so we need to add 17 a","correct":"a","options":{"a":"17 ","b":"27 ","c":"37 ","d":"47","e":"57"},"options_float":{"a":17.0,"b":27.0,"c":37.0,"d":47.0,"e":57.0},"annotated_formula":"subtract(multiply(add(floor(divide(1056, 29)), const_1), 29), 1056)","linear_formula":"divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)","chain":"1_056 \/ 29<\/gadget>\n1_056\/29 = around 36.413793<\/output>\nfloor(1_056\/29)<\/gadget>\n36<\/output>\n36 + 1<\/gadget>\n37<\/output>\n37 * 29<\/gadget>\n1_073<\/output>\n1_073 - 1_056<\/gadget>\n17<\/output>\n17<\/result>","index":1233} +{"problem":"an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 1455 , what was the total number of pairs of contact lenses that she sold ?","rationale":"( x + 5 ) * 150 + x * 85 = 1455 = > x = 3 total lens = 3 + ( 3 + 5 ) = 11 answer a","correct":"a","options":{"a":"11 ","b":"13 ","c":"15 ","d":"17","e":"19"},"options_float":{"a":11.0,"b":13.0,"c":15.0,"d":17.0,"e":19.0},"annotated_formula":"add(add(divide(subtract(1455, multiply(150, 5)), add(150, 85)), 5), divide(subtract(1455, multiply(150, 5)), add(150, 85)))","linear_formula":"add(n0,n1)|multiply(n0,n2)|subtract(n3,#1)|divide(#2,#0)|add(n2,#3)|add(#4,#3)","chain":"150 * 5<\/gadget>\n750<\/output>\n1_455 - 750<\/gadget>\n705<\/output>\n150 + 85<\/gadget>\n235<\/output>\n705 \/ 235<\/gadget>\n3<\/output>\n3 + 5<\/gadget>\n8<\/output>\n8 + 3<\/gadget>\n11<\/output>\n11<\/result>","index":1234} +{"problem":"if the price of an article went up by 30 % , then by what percent should it be brought down to bring it back to its original price ?","rationale":"\"let the price of the article be rs . 100 . 30 % of 100 = 30 . new price = 100 + 30 = rs . 130 required percentage = ( 130 - 100 ) \/ 130 * 100 = 30 \/ 130 * 100 = 23 1 \/ 13 % answer : b\"","correct":"b","options":{"a":"25 % ","b":"23 1 \/ 13 % ","c":"16 7 \/ 3 % ","d":"16 6 \/ 3 %","e":"16 2 \/ 3 %"},"options_float":{"a":25.0,"b":23.0,"c":16.0,"d":16.0,"e":16.0},"annotated_formula":"multiply(divide(30, add(const_100, 30)), const_100)","linear_formula":"add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)|","chain":"100 + 30<\/gadget>\n130<\/output>\n30 \/ 130<\/gadget>\n3\/13 = around 0.230769<\/output>\n(3\/13) * 100<\/gadget>\n300\/13 = around 23.076923<\/output>\n300\/13 = around 23.076923<\/result>","index":1235} +{"problem":"if n is the product of all the integers from 5 to 20 , inclusive , what is the greatest integer k for which 2 ^ k is a factor of n ?","rationale":"20 \/ 2 = 10 10 \/ 2 = 5 ans : 10 + 5 = 15 answer : e","correct":"e","options":{"a":"11 ","b":"12 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"add(add(add(add(add(add(add(const_1, const_3), const_1), 2), const_1), const_4), const_1), 2)","linear_formula":"add(const_1,const_3)|add(#0,const_1)|add(n2,#1)|add(#2,const_1)|add(#3,const_4)|add(#4,const_1)|add(n2,#5)","chain":"1 + 3<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 + 2<\/gadget>\n7<\/output>\n7 + 1<\/gadget>\n8<\/output>\n8 + 4<\/gadget>\n12<\/output>\n12 + 1<\/gadget>\n13<\/output>\n13 + 2<\/gadget>\n15<\/output>\n15<\/result>","index":1236} +{"problem":"the perimeter of an isosceles right triangle is 14 + 14 sq rt 2 . what is the length of the hypotenuse of the triangle ?","rationale":"\"side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 14 + 14 sqrt 2 or , a . ( 2 + sqrt 2 ) = 14 ( 1 + sqrt 2 ) , a = 14 * ( 1 + sqrt 2 ) \/ 2 + sqrt 2 = 14 * 2.414 \/ 3.414 = then hypotenuse = 9.899 e\"","correct":"e","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9.899"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.899},"annotated_formula":"divide(multiply(14, sqrt(2)), 2)","linear_formula":"sqrt(n2)|multiply(n0,#0)|divide(#1,n2)|","chain":"2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n14 * (sqrt(2))<\/gadget>\n14*sqrt(2) = around 19.79899<\/output>\n(14*sqrt(2)) \/ 2<\/gadget>\n7*sqrt(2) = around 9.899495<\/output>\n7*sqrt(2) = around 9.899495<\/result>","index":1237} +{"problem":"country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 12 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 9 % . if ron imported a $ 18000 imported car and ended up paying $ 1950 in taxes , what is the first tier ' s price level ?","rationale":"let t be the tier price , p be total price = 18000 per the given conditions : 0.12 t + 0.09 ( p - t ) = 1950 0.12 t + 0.09 * 18000 - 0.09 t = 1950 0.03 t + 1620 = 1950 0.03 t = 1950 - 1620 = 330 t = 330 \/ 0.03 = 11000 answer b","correct":"b","options":{"a":"$ 11500 ","b":"$ 11000 ","c":"$ 12000 ","d":"$ 12100","e":"$ 12500"},"options_float":{"a":11500.0,"b":11000.0,"c":12000.0,"d":12100.0,"e":12500.0},"annotated_formula":"divide(subtract(1950, multiply(divide(9, const_100), 18000)), subtract(divide(12, const_100), divide(9, const_100)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n3,#2)|divide(#4,#3)","chain":"9 \/ 100<\/gadget>\n9\/100 = around 0.09<\/output>\n(9\/100) * 18_000<\/gadget>\n1_620<\/output>\n1_950 - 1_620<\/gadget>\n330<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) - (9\/100)<\/gadget>\n3\/100 = around 0.03<\/output>\n330 \/ (3\/100)<\/gadget>\n11_000<\/output>\n11_000<\/result>","index":1240} +{"problem":"a 90 cm long wire is to be cut into two pieces so that one piece will be 2 \/ 7 th of the other , how many centimeters will the shorter piece be ?","rationale":"\"explanation : 1 : 2 \/ 7 = 7 : 2 2 \/ 9 * 90 = 20 answer : option b\"","correct":"b","options":{"a":"73 ","b":"20 ","c":"83 ","d":"21","e":"52"},"options_float":{"a":73.0,"b":20.0,"c":83.0,"d":21.0,"e":52.0},"annotated_formula":"subtract(90, divide(90, add(divide(2, 7), const_1)))","linear_formula":"divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|","chain":"2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n(2\/7) + 1<\/gadget>\n9\/7 = around 1.285714<\/output>\n90 \/ (9\/7)<\/gadget>\n70<\/output>\n90 - 70<\/gadget>\n20<\/output>\n20<\/result>","index":1241} +{"problem":"the sum of 3 integers is 33 . the largest integer is 3 times the middle integer , and the smallest integer is 23 less than the largest integer . what is the product of the 3 integers ?","rationale":"let x be the middle integer . then the largest is 3 x and the smallest is 3 x - 23 . 3 x - 23 + x + 3 x = 33 x = 8 the three integers are 1 , 8 , and 24 . the product is 1 * 8 * 24 = 192 the answer is a .","correct":"a","options":{"a":"192 ","b":"203 ","c":"214 ","d":"225","e":"236"},"options_float":{"a":192.0,"b":203.0,"c":214.0,"d":225.0,"e":236.0},"annotated_formula":"multiply(multiply(divide(multiply(add(33, 23), 3), add(const_3, const_4)), divide(divide(multiply(add(33, 23), 3), add(const_3, const_4)), 3)), subtract(divide(multiply(add(33, 23), 3), add(const_3, const_4)), 23))","linear_formula":"add(n1,n3)|add(const_3,const_4)|multiply(n0,#0)|divide(#2,#1)|divide(#3,n0)|subtract(#3,n3)|multiply(#3,#4)|multiply(#6,#5)","chain":"33 + 23<\/gadget>\n56<\/output>\n56 * 3<\/gadget>\n168<\/output>\n3 + 4<\/gadget>\n7<\/output>\n168 \/ 7<\/gadget>\n24<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n24 * 8<\/gadget>\n192<\/output>\n24 - 23<\/gadget>\n1<\/output>\n192 * 1<\/gadget>\n192<\/output>\n192<\/result>","index":1242} +{"problem":"joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 70 . he was allowed to drop his lowest score . after doing so , his average test score improved to 75 . what is the lowest test score that was dropped ?","rationale":"\"the arithmetic mean of 4 equally weighted tests was 70 . so what we can assume is that we have 4 test scores , each 70 . he dropped his lowest score and the avg went to 75 . this means that the lowest score was not 70 and other three scores had given the lowest score 5 each to make it up to 70 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 70 . so the lowest score = 70 - 15 = 55 answer ( c )\"","correct":"c","options":{"a":"20 ","b":"25 ","c":"55 ","d":"65","e":"80"},"options_float":{"a":20.0,"b":25.0,"c":55.0,"d":65.0,"e":80.0},"annotated_formula":"subtract(multiply(70, 4), multiply(75, const_3))","linear_formula":"multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)|","chain":"70 * 4<\/gadget>\n280<\/output>\n75 * 3<\/gadget>\n225<\/output>\n280 - 225<\/gadget>\n55<\/output>\n55<\/result>","index":1244} +{"problem":"an error 2 % in excess is made while measuring the sideofa square . the % of error in the calculated area of the square is ?","rationale":"\"100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 102 ) 2 - ( 100 ) 2 ] = ( 102 + 100 ) x ( 102 - 100 ) = 404 cm 2 . percentage error = 404 x 100 % = 4.04 % 100 x 100 b\"","correct":"b","options":{"a":"4 % ","b":"4.04 % ","c":"4.15 % ","d":"5 %","e":"5.24 %"},"options_float":{"a":4.0,"b":4.04,"c":4.15,"d":5.0,"e":5.24},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100), square_area(const_100))","linear_formula":"add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|","chain":"100 + 2<\/gadget>\n102<\/output>\n102 ** 2<\/gadget>\n10_404<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n10_404 - 10_000<\/gadget>\n404<\/output>\n404 * 100<\/gadget>\n40_400<\/output>\n40_400 \/ 10_000<\/gadget>\n101\/25 = around 4.04<\/output>\n101\/25 = around 4.04<\/result>","index":1245} +{"problem":"what is the normal price of an article sold at $ 144 after two successive discounts of 10 % and 20 % ?","rationale":"\"0.8 * 0.9 * cost price = $ 144 cost price = $ 200 the answer is c .\"","correct":"c","options":{"a":"$ 184 ","b":"$ 192 ","c":"$ 200 ","d":"$ 208","e":"$ 216"},"options_float":{"a":184.0,"b":192.0,"c":200.0,"d":208.0,"e":216.0},"annotated_formula":"divide(144, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100)))","linear_formula":"subtract(const_100,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n0,#4)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(9\/10) * (4\/5)<\/gadget>\n18\/25 = around 0.72<\/output>\n144 \/ (18\/25)<\/gadget>\n200<\/output>\n200<\/result>","index":1246} +{"problem":"the length of a room is 6.5 m and width is 2.75 m . what is the cost of paying the floor by slabs at the rate of rs . 600 per sq . metre .","rationale":"\"area = 6.5 * 2.75 sq . metre . cost for sq . metre . = rs . 600 hence , total cost = 6.5 * 2.75 * 600 = rs . 10725 answer : d\"","correct":"d","options":{"a":"rs . 18725 ","b":"rs . 16725 ","c":"rs . 30725 ","d":"rs . 10725","e":"rs . 20725"},"options_float":{"a":18725.0,"b":16725.0,"c":30725.0,"d":10725.0,"e":20725.0},"annotated_formula":"multiply(600, multiply(6.5, 2.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"6.5 * 2.75<\/gadget>\n17.875<\/output>\n600 * 17.875<\/gadget>\n10_725<\/output>\n10_725<\/result>","index":1247} +{"problem":"the number of diagonals of a polygon of n sides is given by the formula v = n ( n - 3 ) \/ 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ?","rationale":"\"v = n ( n - 3 ) v = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b\"","correct":"b","options":{"a":"3 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"add(3, 2)","linear_formula":"add(n0,n1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5<\/result>","index":1248} +{"problem":"if 15 % of 30 % of 50 % of a number is 108 , then what is the number ?","rationale":"\"let the number be a given , 15 \/ 100 * 30 \/ 100 * 50 \/ 100 * a = 108 = > 3 \/ 20 * 3 \/ 10 * 1 \/ 2 * a = 108 = > a = 10 * 20 * 10 * 2 = 4800 . answer : d\"","correct":"d","options":{"a":"4000 ","b":"3050 ","c":"4400 ","d":"4800","e":"none of these"},"options_float":{"a":4000.0,"b":3050.0,"c":4400.0,"d":4800.0,"e":null},"annotated_formula":"divide(108, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|divide(n1,const_100)|multiply(#1,#2)|multiply(#0,#3)|divide(n3,#4)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/2) * (3\/10)<\/gadget>\n3\/20 = around 0.15<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * (3\/20)<\/gadget>\n9\/400 = around 0.0225<\/output>\n108 \/ (9\/400)<\/gadget>\n4_800<\/output>\n4_800<\/result>","index":1252} +{"problem":"andy solves problems 75 to 125 inclusive in a math exercise . how many problems does he solve ?","rationale":"\"125 - 75 + 1 = 51 ' c ' is the answer\"","correct":"c","options":{"a":"53 ","b":"52 ","c":"51 ","d":"50","e":"49"},"options_float":{"a":53.0,"b":52.0,"c":51.0,"d":50.0,"e":49.0},"annotated_formula":"add(subtract(125, 75), const_1)","linear_formula":"subtract(n1,n0)|add(#0,const_1)|","chain":"125 - 75<\/gadget>\n50<\/output>\n50 + 1<\/gadget>\n51<\/output>\n51<\/result>","index":1253} +{"problem":"6 ) a marketing firm determined that , of 260 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ?","rationale":"\"180 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 180 = > 4 x = 120 = > x = 30 answer - c\"","correct":"c","options":{"a":"15 ","b":"20 ","c":"30 ","d":"40","e":"45"},"options_float":{"a":15.0,"b":20.0,"c":30.0,"d":40.0,"e":45.0},"annotated_formula":"divide(subtract(subtract(260, 80), 60), const_4)","linear_formula":"subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4)|","chain":"260 - 80<\/gadget>\n180<\/output>\n180 - 60<\/gadget>\n120<\/output>\n120 \/ 4<\/gadget>\n30<\/output>\n30<\/result>","index":1256} +{"problem":"26 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ?","rationale":"\"capacity of the tank = 26 ã — 13.5 = 351 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 351 ⠁ „ 9 = 39 answer b\"","correct":"b","options":{"a":"30 ","b":"39 ","c":"60 ","d":"data inadequate","e":"none of these"},"options_float":{"a":30.0,"b":39.0,"c":60.0,"d":null,"e":null},"annotated_formula":"divide(multiply(13.5, 26), 9)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"13.5 * 26<\/gadget>\n351<\/output>\n351 \/ 9<\/gadget>\n39<\/output>\n39<\/result>","index":1257} +{"problem":"a train is running at a speed of 40 km \/ hr and it crosses a post in 22.5 seconds . what is the length of the train ?","rationale":"\"speed of the train , v = 40 km \/ hr = 40000 \/ 3600 m \/ s = 400 \/ 36 m \/ s time taken to cross , t = 22.5 s distance covered , d = vt = ( 400 \/ 36 ) ã — 22.5 = 250 m distance covered is equal to the length of the train = 250 m correct answer is 250 metres e\"","correct":"e","options":{"a":"190 metres ","b":"160 metres ","c":"200 metres ","d":"120 metres","e":"250 metres"},"options_float":{"a":190.0,"b":160.0,"c":200.0,"d":120.0,"e":250.0},"annotated_formula":"multiply(22.5, multiply(40, const_0_2778))","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n40 * (5\/18)<\/gadget>\n100\/9 = around 11.111111<\/output>\n22.5 * (100\/9)<\/gadget>\n250<\/output>\n250<\/result>","index":1258} +{"problem":"a 70 kg metal bar made of alloy of tin and silver lost 7 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ?","rationale":"\"you can simply use this formula to avoid confusion : w 1 \/ w 2 = ( a 2 - aavg ) \/ ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 7 kg when 70 kg alloy is immersed . this is a loss of ( 7 \/ 70 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 \/ 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 \/ 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin \/ weight of silver = ( silver ' s loss - avg loss ) \/ ( avg loss - tin ' s loss ) x \/ y = ( 7.5 - 10 ) \/ ( 10 - 13.75 ) = 2 \/ 3 e\"","correct":"e","options":{"a":"1 \/ 4 ","b":"2 \/ 5 ","c":"1 \/ 2 ","d":"3 \/ 5","e":"2 \/ 3"},"options_float":{"a":0.25,"b":0.4,"c":0.5,"d":0.6,"e":0.6666666667},"annotated_formula":"divide(divide(subtract(7, multiply(divide(0.375, 5), 70)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(70, divide(subtract(7, multiply(divide(0.375, 5), 70)), subtract(divide(1.375, 10), divide(0.375, 5)))))","linear_formula":"divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)|","chain":"0.375 \/ 5<\/gadget>\n0.075<\/output>\n0.075 * 70<\/gadget>\n5.25<\/output>\n7 - 5.25<\/gadget>\n1.75<\/output>\n1.375 \/ 10<\/gadget>\n0.1375<\/output>\n0.1375 - 0.075<\/gadget>\n0.0625<\/output>\n1.75 \/ 0.0625<\/gadget>\n28<\/output>\n70 - 28<\/gadget>\n42<\/output>\n28 \/ 42<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":1262} +{"problem":"what is the rate percent when the simple interest on rs . 720 amount to rs . 180 in 4 years ?","rationale":"\"180 = ( 720 * 4 * r ) \/ 100 r = 6.25 % answer : a\"","correct":"a","options":{"a":"6.25 ","b":"7.25 ","c":"8.25 ","d":"9.25","e":"10.25"},"options_float":{"a":6.25,"b":7.25,"c":8.25,"d":9.25,"e":10.25},"annotated_formula":"divide(multiply(const_100, 180), multiply(720, 4))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 180<\/gadget>\n18_000<\/output>\n720 * 4<\/gadget>\n2_880<\/output>\n18_000 \/ 2_880<\/gadget>\n25\/4 = around 6.25<\/output>\n25\/4 = around 6.25<\/result>","index":1263} +{"problem":"when 242 is divided by a certain divisor the remainder obtained is 15 . when 698 is divided by the same divisor the remainder obtained is 27 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 5 . what is the value of the divisor ?","rationale":"let that divisor be x since remainder is 15 or 27 it means divisor is greater than 27 . now 242 - 15 = 227 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 27 = 671 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 227 + 671 ) + 15 + 27 = x ( k + l ) + 42 when we divide this number by x then remainder will be equal to remainder of ( 42 divided by x ) = 5 hence x = 42 - 5 = 37 hence e","correct":"e","options":{"a":"11 ","b":"17 ","c":"13 ","d":"23","e":"37"},"options_float":{"a":11.0,"b":17.0,"c":13.0,"d":23.0,"e":37.0},"annotated_formula":"subtract(add(15, 27), 5)","linear_formula":"add(n1,n3)|subtract(#0,n6)","chain":"15 + 27<\/gadget>\n42<\/output>\n42 - 5<\/gadget>\n37<\/output>\n37<\/result>","index":1264} +{"problem":"gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 15 times as heavy as water ?","rationale":"\"let us say , water weighs 1 kg \/ 1 liter . then gold weighs 19 kgs and copper weighs 9 kgs . assume x kgs of gold and y kgs of copper required to get the required alloy . we use weighted average formula = ax = n 1 . a 1 + n 2 . a 2 n 1 + n 2 ax = n 1 . a 1 + n 2 . a 2 n 1 + n 2 therefore , 15 = x . 19 + y . 9 x + y 15 = x . 19 + y . 9 x + y 15 x + 15 y = 19 x + 9 y 4 x = 6 y ⇒ x \/ y = 3 : 2 so gold and copper should be mixed in the ratio 3 : 2 . answer : a\"","correct":"a","options":{"a":"3 : 2 ","b":"3 : 8 ","c":"5 : 7 ","d":"3 : 7","e":"3 : 6"},"options_float":{"a":1.5,"b":0.375,"c":0.7142857143,"d":0.4285714286,"e":0.5},"annotated_formula":"divide(subtract(15, 9), subtract(19, 15))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"15 - 9<\/gadget>\n6<\/output>\n19 - 15<\/gadget>\n4<\/output>\n6 \/ 4<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":1266} +{"problem":"find the cost of fencing around a circular field of diameter 32 m at the rate of rs . 2 a meter ?","rationale":"\"2 * 22 \/ 7 * 16 = 100.5 100.5 * 2 = rs . 201 answer : a\"","correct":"a","options":{"a":"201 ","b":"132 ","c":"772 ","d":"592","e":"261"},"options_float":{"a":201.0,"b":132.0,"c":772.0,"d":592.0,"e":261.0},"annotated_formula":"multiply(circumface(divide(32, const_2)), 2)","linear_formula":"divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|","chain":"32 \/ 2<\/gadget>\n16<\/output>\n2 * pi * 16<\/gadget>\n32*pi = around 100.530965<\/output>\n(32*pi) * 2<\/gadget>\n64*pi = around 201.06193<\/output>\n64*pi = around 201.06193<\/result>","index":1267} +{"problem":"a bookseller has two display windows . she plans to display 3 new fiction books in the left window , and 3 new non - fiction books in the right window . assuming she can put the 4 fiction books in any order , and separately , the 3 non - fiction books in any order , how many total configurations will there be for the two display windows ?","rationale":"the left window will have permutations of the 3 fiction books , so the number of possibilities for that window is permutations = 3 ! = ( 3 ) ( 2 ) ( 1 ) = 6 the right window will have permutations of the 3 non - fiction books , so the number of possibilities for that window is permutations = 3 ! = ( 3 ) ( 2 ) ( 1 ) = 6 any of the 24 displays of the left window could be combined with any of the 6 displays of the right window , so the total number of configurations is 6 * 6 = 36 answer : d .","correct":"d","options":{"a":"24 ","b":"72 ","c":"144 ","d":"36","e":"420"},"options_float":{"a":24.0,"b":72.0,"c":144.0,"d":36.0,"e":420.0},"annotated_formula":"multiply(multiply(4, 3), 3)","linear_formula":"multiply(n0,n2)|multiply(n0,#0)","chain":"4 * 3<\/gadget>\n12<\/output>\n12 * 3<\/gadget>\n36<\/output>\n36<\/result>","index":1269} +{"problem":"if julie rows 32 km upstream and 72 km down steam taking 4 hours each , then the speed of the stream","rationale":"speed upstream = 32 \/ 4 = 8 kmph speed down stream = 72 \/ 8 = 9 kmph speed of stream = ½ ( 9 - 8 ) = 0.5 kmph answer : b","correct":"b","options":{"a":"1.5 kmph ","b":"0.5 kmph ","c":"0.4 kmph ","d":"1.6 kmph","e":"2.5 kmph"},"options_float":{"a":1.5,"b":0.5,"c":0.4,"d":1.6,"e":2.5},"annotated_formula":"subtract(divide(divide(subtract(divide(72, 4), divide(32, 4)), const_2), const_2), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|subtract(#0,#1)|divide(#2,const_2)|divide(#3,const_2)|subtract(#4,const_2)","chain":"72 \/ 4<\/gadget>\n18<\/output>\n32 \/ 4<\/gadget>\n8<\/output>\n18 - 8<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 \/ 2<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) - 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":1272} +{"problem":"sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 \/ 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 \/ 7 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card , what portion of her limit on the platinum card will remain unspent ?","rationale":"\"let s assume the platinum card spending limit = x gold card spending limit will be = x \/ 2 balance on gold card is = x \/ 2 * 1 \/ 3 = x \/ 6 platinum card unspent limit is = x - 1 \/ 7 x = 6 \/ 7 x so if gold card balance is transferred then the rest unspent will be 6 \/ 7 x - x \/ 6 = 29 \/ 42 x so the ans is b\"","correct":"b","options":{"a":"11 \/ 30 ","b":"29 \/ 42 ","c":"17 \/ 30 ","d":"19 \/ 30","e":"11 \/ 15"},"options_float":{"a":0.3666666667,"b":0.6904761905,"c":0.5666666667,"d":0.6333333333,"e":0.7333333333},"annotated_formula":"subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(7)))","linear_formula":"inverse(n3)|inverse(n1)|inverse(const_2)|multiply(#1,#2)|add(#0,#3)|subtract(n0,#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/3) * (1\/2)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/6) + (1\/7)<\/gadget>\n13\/42 = around 0.309524<\/output>\n1 - (13\/42)<\/gadget>\n29\/42 = around 0.690476<\/output>\n29\/42 = around 0.690476<\/result>","index":1275} +{"problem":"of all the homes on gotham street , 1 \/ 3 are termite - ridden , and 4 \/ 7 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ?","rationale":"\"the fraction of homes which are termite - infested but not collapsing is 3 \/ 7 * 1 \/ 3 = 1 \/ 7 the answer is e .\"","correct":"e","options":{"a":"1 \/ 3 ","b":"1 \/ 4 ","c":"1 \/ 5 ","d":"1 \/ 6","e":"1 \/ 7"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.2,"d":0.1666666667,"e":0.1428571429},"annotated_formula":"divide(add(1, const_2.0), multiply(3, 7))","linear_formula":"add(n0,const_2.0)|multiply(n1,n3)|divide(#0,#1)|","chain":"1 + 2<\/gadget>\n3<\/output>\n3 * 7<\/gadget>\n21<\/output>\n3 \/ 21<\/gadget>\n1\/7 = around 0.142857<\/output>\n1\/7 = around 0.142857<\/result>","index":1276} +{"problem":"what is the minimum value of | x - 4 | + | x + 5 | + | x - 5 | ?","rationale":"\"a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 5 | > = 0 - - > minimum occurs at x = - 5 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 5 - - > result = 9 + 0 + 10 = 19 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 9 + 1 = 10 x = 5 - - > result = 1 + 10 + 0 = 11 so minimum value of the expression occurs at x = 4 and the resultant value = 10 answer : e\"","correct":"e","options":{"a":"- 3 ","b":"3 ","c":"5 ","d":"7","e":"10"},"options_float":{"a":-3.0,"b":3.0,"c":5.0,"d":7.0,"e":10.0},"annotated_formula":"add(5, 5)","linear_formula":"add(n1,n2)|","chain":"5 + 5<\/gadget>\n10<\/output>\n10<\/result>","index":1280} +{"problem":"sravan travelled for 15 hours . he covered the first half of the distance at 45 kmph and remaining half of the distance at 30 kmph . find the distance travelled by sravan ?","rationale":"let the distance travelled be x km . total time = ( x \/ 2 ) \/ 45 + ( x \/ 2 ) \/ 30 = 15 = > x \/ 90 + x \/ 60 = 15 = > ( 2 x + 3 x ) \/ 180 = 15 = > x = 540 km answer : b","correct":"b","options":{"a":"520 ","b":"540 ","c":"550 ","d":"560","e":"440"},"options_float":{"a":520.0,"b":540.0,"c":550.0,"d":560.0,"e":440.0},"annotated_formula":"divide(multiply(45, multiply(30, 30)), add(45, 30))","linear_formula":"add(n1,n2)|multiply(n2,n2)|multiply(n1,#1)|divide(#2,#0)","chain":"30 * 30<\/gadget>\n900<\/output>\n45 * 900<\/gadget>\n40_500<\/output>\n45 + 30<\/gadget>\n75<\/output>\n40_500 \/ 75<\/gadget>\n540<\/output>\n540<\/result>","index":1281} +{"problem":"50 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ?","rationale":"\"explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 \/ 600 * 50 = 66.67 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 \/ 800 * 66.67 = 100 litres . answer c\"","correct":"c","options":{"a":"80 litres ","b":"90 litres ","c":"100 litres ","d":"170 litres","e":"none of these"},"options_float":{"a":80.0,"b":90.0,"c":100.0,"d":170.0,"e":null},"annotated_formula":"multiply(1200, multiply(800, divide(50, multiply(800, 600))))","linear_formula":"multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n4,#2)|","chain":"800 * 600<\/gadget>\n480_000<\/output>\n50 \/ 480_000<\/gadget>\n1\/9_600 = around 0.000104<\/output>\n800 * (1\/9_600)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1_200 * (1\/12)<\/gadget>\n100<\/output>\n100<\/result>","index":1282} +{"problem":"a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 \/ 4 - inch increase in the length of the shoe . if the largest size of this shoe is 10 % longer than the smallest size , how long , in inches , is the shoe in size 15 ?","rationale":"\"let x be the length of the size 8 shoe . then 0.1 x = 9 \/ 4 x = 90 \/ 4 inches the size 15 shoe has a length of 90 \/ 4 + 7 \/ 4 = 97 \/ 4 = 24.25 inches the answer is d .\"","correct":"d","options":{"a":"18.25 ","b":"20.75 ","c":"22.25 ","d":"24.25","e":"26.75"},"options_float":{"a":18.25,"b":20.75,"c":22.25,"d":24.25,"e":26.75},"annotated_formula":"add(divide(multiply(divide(1, 4), subtract(17, 8)), divide(10, const_100)), multiply(subtract(15, 8), divide(1, 4)))","linear_formula":"divide(n2,n3)|divide(n4,const_100)|subtract(n1,n0)|subtract(n5,n0)|multiply(#0,#2)|multiply(#0,#3)|divide(#4,#1)|add(#6,#5)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n17 - 8<\/gadget>\n9<\/output>\n(1\/4) * 9<\/gadget>\n9\/4 = around 2.25<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(9\/4) \/ (1\/10)<\/gadget>\n45\/2 = around 22.5<\/output>\n15 - 8<\/gadget>\n7<\/output>\n7 * (1\/4)<\/gadget>\n7\/4 = around 1.75<\/output>\n(45\/2) + (7\/4)<\/gadget>\n97\/4 = around 24.25<\/output>\n97\/4 = around 24.25<\/result>","index":1283} +{"problem":"excluding stoppages , the average speed of a bus is 75 km \/ hr and including stoppages , the average speed of the bus is 40 km \/ hr . for how many minutes does the bus stop per hour ?","rationale":"\"in 1 hr , the bus covers 75 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 75 - 40 ) km i . e 35 km at 75 km \/ hr . stoppage time = 35 \/ 75 hrs = 28 min answer : e\"","correct":"e","options":{"a":"15 min ","b":"18 min ","c":"16 min ","d":"20 min","e":"28 min"},"options_float":{"a":15.0,"b":18.0,"c":16.0,"d":20.0,"e":28.0},"annotated_formula":"subtract(multiply(const_1, const_60), multiply(divide(40, 75), const_60))","linear_formula":"divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)|","chain":"1 * 60<\/gadget>\n60<\/output>\n40 \/ 75<\/gadget>\n8\/15 = around 0.533333<\/output>\n(8\/15) * 60<\/gadget>\n32<\/output>\n60 - 32<\/gadget>\n28<\/output>\n28<\/result>","index":1285} +{"problem":"find the area of a parallelogram with base 26 cm and height 16 cm ?","rationale":"\"area of a parallelogram = base * height = 26 * 16 = 416 cm 2 answer : d\"","correct":"d","options":{"a":"297 cm 2 ","b":"384 cm 2 ","c":"672 cm 2 ","d":"416 cm 2","e":"286 cm 2"},"options_float":{"a":297.0,"b":384.0,"c":672.0,"d":416.0,"e":286.0},"annotated_formula":"multiply(26, 16)","linear_formula":"multiply(n0,n1)|","chain":"26 * 16<\/gadget>\n416<\/output>\n416<\/result>","index":1286} +{"problem":"by selling an article at rs . 600 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ?","rationale":"\"sp = 600 profit = 25 % cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 600 * [ 100 \/ 125 ] = 480 loss = 25 % = 25 % of 480 = rs . 120 sp = cp - loss = 480 - 120 = rs . 360 answer : a\"","correct":"a","options":{"a":"s . 360 ","b":"s . 480 ","c":"s . 500 ","d":"s . 450","e":"s . 550"},"options_float":{"a":360.0,"b":480.0,"c":500.0,"d":450.0,"e":550.0},"annotated_formula":"subtract(divide(multiply(600, const_100), add(25, const_100)), divide(multiply(divide(multiply(600, const_100), add(25, const_100)), 25), const_100))","linear_formula":"add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|","chain":"600 * 100<\/gadget>\n60_000<\/output>\n25 + 100<\/gadget>\n125<\/output>\n60_000 \/ 125<\/gadget>\n480<\/output>\n480 * 25<\/gadget>\n12_000<\/output>\n12_000 \/ 100<\/gadget>\n120<\/output>\n480 - 120<\/gadget>\n360<\/output>\n360<\/result>","index":1287} +{"problem":"the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 600 per sq . metre .","rationale":"\"solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 600 × 20.625 ) = 12375 . answer a\"","correct":"a","options":{"a":"rs . 12375 ","b":"rs . 15550 ","c":"rs . 15600 ","d":"rs . 16500","e":"none of these"},"options_float":{"a":12375.0,"b":15550.0,"c":15600.0,"d":16500.0,"e":null},"annotated_formula":"multiply(600, multiply(5.5, 3.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 3.75<\/gadget>\n20.625<\/output>\n600 * 20.625<\/gadget>\n12_375<\/output>\n12_375<\/result>","index":1289} +{"problem":"if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 14 ?","rationale":"@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 14 . then 4 n < 14 and n < 3.5 . the largest possible integer is n = 3 . the answer is c .","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"floor(divide(14, subtract(5, const_1)))","linear_formula":"subtract(n0,const_1)|divide(n1,#0)|floor(#1)","chain":"5 - 1<\/gadget>\n4<\/output>\n14 \/ 4<\/gadget>\n7\/2 = around 3.5<\/output>\nfloor(7\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":1291} +{"problem":"if 9 workers can build 9 cars in 9 days , then how many days would it take 7 workers to build 7 cars ?","rationale":"\"9 workers can build 1 car per day on average . 1 worker can build 1 \/ 9 of a car per day . 7 workers can build 7 \/ 9 car per day . the time required to build 7 cars is 7 \/ ( 7 \/ 9 ) = 9 days the answer is d .\"","correct":"d","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"multiply(divide(multiply(9, 9), 9), divide(7, 7))","linear_formula":"divide(n3,n3)|multiply(n0,n0)|divide(#1,n0)|multiply(#2,#0)|","chain":"9 * 9<\/gadget>\n81<\/output>\n81 \/ 9<\/gadget>\n9<\/output>\n7 \/ 7<\/gadget>\n1<\/output>\n9 * 1<\/gadget>\n9<\/output>\n9<\/result>","index":1293} +{"problem":"the probability of two events a and b are 0.25 and 0.40 respectively . the probability that both a and b occur is 0.20 . the probability that neither a nor b occur is _________","rationale":"\"we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 40 - . 20 = . 45 but the probability of neither a nor b = 1 - . 45 = 0.55 answer : c\"","correct":"c","options":{"a":"0.45 ","b":"0.4 ","c":"0.55 ","d":"0.05","e":"0.6"},"options_float":{"a":0.45,"b":0.4,"c":0.55,"d":0.05,"e":0.6},"annotated_formula":"subtract(const_1, subtract(add(0.25, 0.40), 0.20))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)|","chain":"0.25 + 0.4<\/gadget>\n0.65<\/output>\n0.65 - 0.2<\/gadget>\n0.45<\/output>\n1 - 0.45<\/gadget>\n0.55<\/output>\n0.55<\/result>","index":1294} +{"problem":"the length of a rectangular floor is more than its breadth by 200 % . if rs . 300 is required to paint the floor at the rate of rs . 5 per sq m , then what would be the length of the floor ?","rationale":"\"let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 300 \/ 5 = 60 sq m l b = 60 i . e . , l * l \/ 3 = 60 l 2 = 180 = > l = 13.42 answer : b\"","correct":"b","options":{"a":"65 ","b":"13.42 ","c":"18 ","d":"16","e":"14"},"options_float":{"a":65.0,"b":13.42,"c":18.0,"d":16.0,"e":14.0},"annotated_formula":"multiply(sqrt(divide(divide(300, 5), const_3)), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)|","chain":"300 \/ 5<\/gadget>\n60<\/output>\n60 \/ 3<\/gadget>\n20<\/output>\n20 ** (1\/2)<\/gadget>\n2*sqrt(5) = around 4.472136<\/output>\n(2*sqrt(5)) * 3<\/gadget>\n6*sqrt(5) = around 13.416408<\/output>\n6*sqrt(5) = around 13.416408<\/result>","index":1296} +{"problem":"the s . i . on a certain sum of money for 5 years at 10 % per annum is half the c . i . on rs . 5000 for 2 years at 12 % per annum . the sum placed on s . i . is ?","rationale":"c . i . = [ 5000 * ( 1 + 12 \/ 100 ) 2 - 5000 ] = ( 5000 * 28 \/ 25 * 28 \/ 25 - 5000 ) = rs . 1272 . sum = ( 636 * 100 ) \/ ( 5 * 10 ) = rs . 1272 answer : e","correct":"e","options":{"a":"1585 ","b":"1454 ","c":"1656 ","d":"2458","e":"1272"},"options_float":{"a":1585.0,"b":1454.0,"c":1656.0,"d":2458.0,"e":1272.0},"annotated_formula":"divide(divide(subtract(multiply(5000, power(add(const_1, divide(12, const_100)), 2)), 5000), 2), multiply(5, divide(10, const_100)))","linear_formula":"divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)|","chain":"12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n1 + (3\/25)<\/gadget>\n28\/25 = around 1.12<\/output>\n(28\/25) ** 2<\/gadget>\n784\/625 = around 1.2544<\/output>\n5_000 * (784\/625)<\/gadget>\n6_272<\/output>\n6_272 - 5_000<\/gadget>\n1_272<\/output>\n1_272 \/ 2<\/gadget>\n636<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n5 * (1\/10)<\/gadget>\n1\/2 = around 0.5<\/output>\n636 \/ (1\/2)<\/gadget>\n1_272<\/output>\n1_272<\/result>","index":1297} +{"problem":"dropbox charges $ 3.10 for first 1 \/ 5 of a month for 500 gb storage , plus $ 0.40 for each additional 1 \/ 5 of a month . what would dropbox charge for a data storage of 500 gb for 8 months ?","rationale":"first 1 \/ 5 month charge = $ 3.10 rest of the months = 8 - ( 1 \/ 5 ) = 39 \/ 5 charge for the rest of the months = 39 * 0.4 = 15.6 total charge = 3.10 + 15.6 = 18.7 answer is d .","correct":"d","options":{"a":"$ 15.60 ","b":"$ 16.00 ","c":"$ 17.80 ","d":"$ 18.70","e":"$ 19.10"},"options_float":{"a":15.6,"b":16.0,"c":17.8,"d":18.7,"e":19.1},"annotated_formula":"add(subtract(divide(multiply(500, 0.4), 8), 8), 1)","linear_formula":"multiply(n3,n4)|divide(#0,n8)|subtract(#1,n8)|add(n1,#2)","chain":"500 * 0.4<\/gadget>\n200<\/output>\n200 \/ 8<\/gadget>\n25<\/output>\n25 - 8<\/gadget>\n17<\/output>\n17 + 1<\/gadget>\n18<\/output>\n18<\/result>","index":1299} +{"problem":"find the simple interest on rs . 72,000 at 16 2 \/ 3 % per year for 9 months .","rationale":"\"p = rs . 72000 , r = 50 \/ 3 % p . a and t = 9 \/ 12 years = 3 \/ 4 years . simple interest = ( p * r * t ) \/ 100 = rs . ( 72,000 * ( 50 \/ 3 ) * ( 3 \/ 4 ) * ( 1 \/ 100 ) ) = rs . 9000 answer is c .\"","correct":"c","options":{"a":"7500 ","b":"6500 ","c":"9000 ","d":"9500","e":"none of them"},"options_float":{"a":7500.0,"b":6500.0,"c":9000.0,"d":9500.0,"e":null},"annotated_formula":"multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100))","linear_formula":"add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600 * 100<\/gadget>\n60_000<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 100<\/gadget>\n900<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n900 * 10<\/gadget>\n9_000<\/output>\n60_000 + 9_000<\/gadget>\n69_000<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48 + 2<\/gadget>\n50<\/output>\n50 \/ 3<\/gadget>\n50\/3 = around 16.666667<\/output>\n69_000 * (50\/3)<\/gadget>\n1_150_000<\/output>\n2 * 6<\/gadget>\n12<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n1_150_000 * (3\/4)<\/gadget>\n862_500<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n862_500 * (1\/100)<\/gadget>\n8_625<\/output>\n8_625<\/result>","index":1300} +{"problem":"if it would take one machine 20 minutes to fill a large production order and another machine 30 minutes to fill the same order , how many minutes would it take both machines working together , at their respective rates , to fill the order ?","rationale":"this question can be solved using the work formula : work = ( a ) ( b ) \/ ( a + b ) where a and b are the individual completion rates of the respective entities ( for the same job ) . we ' re told that one machine takes 10 minutes to complete a job and another machine takes 12 minutes to complete the same job . plugging in those values gets us . . . ( 20 ) ( 30 ) \/ ( 20 + 30 ) = 600 \/ 50 = 12 minutes answer is b","correct":"b","options":{"a":"10 min ","b":"12 min ","c":"15 min ","d":"20 min","e":"18 min"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":20.0,"e":18.0},"annotated_formula":"inverse(add(divide(const_1, 20), divide(const_1, 30)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)","chain":"1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/20) + (1\/30)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":1301} +{"problem":"how many digits are required to number a book containing 220 pages ?","rationale":"\"9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 220 - ( 90 + 9 ) = 121 pages will require 121 * 3 = 363 digits . the total number of digits is 9 + 180 + 363 = 552 . the answer is c .\"","correct":"c","options":{"a":"674 ","b":"604 ","c":"552 ","d":"372","e":"250"},"options_float":{"a":674.0,"b":604.0,"c":552.0,"d":372.0,"e":250.0},"annotated_formula":"add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(220, const_100), const_1), const_3))","linear_formula":"subtract(const_10,const_1)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#4,#0)|add(#6,#5)|","chain":"10 - 1<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n90 * 2<\/gadget>\n180<\/output>\n9 + 180<\/gadget>\n189<\/output>\n220 - 100<\/gadget>\n120<\/output>\n120 + 1<\/gadget>\n121<\/output>\n121 * 3<\/gadget>\n363<\/output>\n189 + 363<\/gadget>\n552<\/output>\n552<\/result>","index":1302} +{"problem":"a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent w of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ?","rationale":"\"cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 \/ 12 = 3 \/ 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is w = . 25 \/ 1 * 100 = 25 % d is the answer .\"","correct":"d","options":{"a":"9 % ","b":"11 % ","c":"15 % ","d":"25 %","e":"90 %"},"options_float":{"a":9.0,"b":11.0,"c":15.0,"d":25.0,"e":90.0},"annotated_formula":"subtract(const_100, multiply(divide(9, 12), const_100))","linear_formula":"divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)|","chain":"9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n100 - 75<\/gadget>\n25<\/output>\n25<\/result>","index":1303} +{"problem":"the ratio between the number of sheep and the number of horses at the stewart farm is 5 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ?","rationale":"\"let the number of sheeps and horses be 4 x and 7 x . now total number of horses = total consumption of horse food \/ consumption per horse = 12880 \/ 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 5 x = 5 * 8 = 40 . hence c .\"","correct":"c","options":{"a":"18 ","b":"28 ","c":"40 ","d":"56","e":"60"},"options_float":{"a":18.0,"b":28.0,"c":40.0,"d":56.0,"e":60.0},"annotated_formula":"multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 5)","linear_formula":"multiply(const_2,const_4)|multiply(const_1000,const_12)|multiply(#0,const_10)|multiply(#0,const_100)|add(#2,#3)|add(#4,#1)|divide(#5,n2)|divide(#6,n1)|multiply(n0,#7)|","chain":"4 * 2<\/gadget>\n8<\/output>\n8 * 10<\/gadget>\n80<\/output>\n8 * 100<\/gadget>\n800<\/output>\n80 + 800<\/gadget>\n880<\/output>\n12 * 1_000<\/gadget>\n12_000<\/output>\n880 + 12_000<\/gadget>\n12_880<\/output>\n12_880 \/ 230<\/gadget>\n56<\/output>\n56 \/ 7<\/gadget>\n8<\/output>\n8 * 5<\/gadget>\n40<\/output>\n40<\/result>","index":1304} +{"problem":"calculate the amount that an investor needs to be invest to earn $ 1005 in interest in 12 months if the investor plans to invest x dollars in a savings account that pays interest at an annual rate of 11 % compounded semi - annually ?","rationale":"\"the approach is substitution , our interest requirement is $ 1005 after 12 months , 2 compounding period . calculate the compound interest on each option and find out the one that yields 460 in 12 months 8900 yielded $ 1005 using the formula a = p ( 1 + r \/ n ) nt hence answer is c\"","correct":"c","options":{"a":"8000 ","b":"8100 ","c":"8900 ","d":"8600","e":"8400"},"options_float":{"a":8000.0,"b":8100.0,"c":8900.0,"d":8600.0,"e":8400.0},"annotated_formula":"divide(multiply(const_100, 1005), 11)","linear_formula":"multiply(n0,const_100)|divide(#0,n2)|","chain":"100 * 1_005<\/gadget>\n100_500<\/output>\n100_500 \/ 11<\/gadget>\n100_500\/11 = around 9_136.363636<\/output>\n100_500\/11 = around 9_136.363636<\/result>","index":1306} +{"problem":"a reduction of 42 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ?","rationale":"\"explanation : 40 * ( 42 \/ 100 ) = 16.8 - - - 64 ? - - - 12 = > rs . 3.15 answer : d\"","correct":"d","options":{"a":"2.15 ","b":"8.15 ","c":"7.15 ","d":"3.15","e":"1.15"},"options_float":{"a":2.15,"b":8.15,"c":7.15,"d":3.15,"e":1.15},"annotated_formula":"multiply(const_12, divide(multiply(42, divide(42, const_100)), 64))","linear_formula":"divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)|","chain":"42 \/ 100<\/gadget>\n21\/50 = around 0.42<\/output>\n42 * (21\/50)<\/gadget>\n441\/25 = around 17.64<\/output>\n(441\/25) \/ 64<\/gadget>\n441\/1_600 = around 0.275625<\/output>\n12 * (441\/1_600)<\/gadget>\n1_323\/400 = around 3.3075<\/output>\n1_323\/400 = around 3.3075<\/result>","index":1307} +{"problem":"a train crosses a bridge of length 1500 m in 70 seconds and a lamp post on the bridge in 20 seconds . what is the length of the train in metres ?","rationale":"\"let length of train = l case - 1 : distance = 1500 + l ( while crossing the bridge ) time = 70 seconds i . e . speed = distance \/ time = ( 1500 + l ) \/ 70 case - 2 : distance = l ( while passing the lamp post ) time = 20 seconds i . e . speed = distance \/ time = ( l ) \/ 20 but since speed has to be same in both cases so ( 1500 + l ) \/ 70 = ( l ) \/ 20 i . e . 1500 + l = 3.5 l i . e . 2.5 l = 1500 i . e . l = 600 answer : option d\"","correct":"d","options":{"a":"375 m ","b":"750 m ","c":"250 m ","d":"600 m","e":"300 m"},"options_float":{"a":375.0,"b":750.0,"c":250.0,"d":600.0,"e":300.0},"annotated_formula":"multiply(divide(1500, subtract(70, 20)), 20)","linear_formula":"subtract(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"70 - 20<\/gadget>\n50<\/output>\n1_500 \/ 50<\/gadget>\n30<\/output>\n30 * 20<\/gadget>\n600<\/output>\n600<\/result>","index":1308} +{"problem":"what will be the cost of building a fence around a square plot with area equal to 289 square feet , if the price per foot of building the fence is rs . 58 ?","rationale":"\"let the side of the square plot be a foot a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 foot cost of building the fence = 68 * 58 = rs . 3944 . answer : a\"","correct":"a","options":{"a":"3944 ","b":"3932 ","c":"3929 ","d":"3926","e":"3902"},"options_float":{"a":3944.0,"b":3932.0,"c":3929.0,"d":3926.0,"e":3902.0},"annotated_formula":"multiply(square_perimeter(sqrt(289)), 58)","linear_formula":"sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|","chain":"289 ** (1\/2)<\/gadget>\n17<\/output>\n4 * 17<\/gadget>\n68<\/output>\n68 * 58<\/gadget>\n3_944<\/output>\n3_944<\/result>","index":1309} +{"problem":"points a , b , and , c have xy - coordinates ( 2,0 ) , ( 8,12 ) , and ( 14,0 ) , respectively . points x , y , and z have xy - coordinates ( 6,0 ) , ( 8,4 ) , and ( 10,0 ) , respectively . what fraction d of the area of triangle abc is the area of triangle xyz ?","rationale":"\"if you notice , both triangles abc and xyz have a side on x axis . we can take these sides as bases for each triangle , therefore area of abc is 1 \/ 2 * 12 * 12 ( height of abc is the y coordinate of the third point ( 8,12 ) ) similarly area of xyz is 1 \/ 2 * 4 * 4 dividing area of xyz with that of abc gives d = 1 \/ 9 . a\"","correct":"a","options":{"a":"1 \/ 9 ","b":"1 \/ 8 ","c":"1 \/ 6 ","d":"1 \/ 5","e":"1 \/ 3"},"options_float":{"a":0.1111111111,"b":0.125,"c":0.1666666667,"d":0.2,"e":0.3333333333},"annotated_formula":"divide(divide(power(const_4, const_2), const_2), divide(power(add(const_10, const_2), const_2), const_2))","linear_formula":"add(const_10,const_2)|power(const_4,const_2)|divide(#1,const_2)|power(#0,const_2)|divide(#3,const_2)|divide(#2,#4)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n144 \/ 2<\/gadget>\n72<\/output>\n8 \/ 72<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":1311} +{"problem":"the dimensions of a field are 20 m by 10 m . a pit 8 m long , 5 m wide and 2 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ?","rationale":"the volume of the earth removed is 8 * 5 * 2 = 80 m ^ 3 . the remaining area of the field is 20 * 10 - 8 * 5 = 160 m ^ 2 . 80 m ^ 3 of the earth evenly spread over the area of 160 m ^ 2 will rise the height by ( height ) = ( volume ) \/ ( area ) = 80 \/ 160 = 0.5 m . answer : c","correct":"c","options":{"a":"1 m ","b":"2 m ","c":"0.5 m ","d":"0.75 m","e":"1.5 m"},"options_float":{"a":1.0,"b":2.0,"c":0.5,"d":0.75,"e":1.5},"annotated_formula":"divide(multiply(10, 8), subtract(rectangle_area(20, 10), rectangle_area(5, 8)))","linear_formula":"multiply(n1,n2)|rectangle_area(n0,n1)|rectangle_area(n2,n3)|subtract(#1,#2)|divide(#0,#3)","chain":"10 * 8<\/gadget>\n80<\/output>\n20 * 10<\/gadget>\n200<\/output>\n5 * 8<\/gadget>\n40<\/output>\n200 - 40<\/gadget>\n160<\/output>\n80 \/ 160<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":1312} +{"problem":"the grade point average of the entire class is 90 . if the average of one third of the class is 96 , what is the average of the rest ?","rationale":"\"let us take number of students on the class as ' x ' . avg grade point = sum of all grade points \/ x sum of all grade points = avg grade point * x = 90 x average of one third of the class = 96 = sum of one third grade points of the class \/ ( x \/ 3 ) sum of one third grade points of the class = 96 * ( x \/ 3 ) = 32 x sum of rest two third grade points of the class = sum of all grade points - sum of one third grade points of the class sum of rest two third grade points of the class = 90 x - 32 x = 58 x avg of two third of the class = sum of rest two third grade points of the class \/ ( x * 2 \/ 3 ) = 58 x \/ ( 2 x \/ 3 ) = 29 * 3 = 87 . hence , answer is d\"","correct":"d","options":{"a":"92 ","b":"89 ","c":"88 ","d":"87","e":"86"},"options_float":{"a":92.0,"b":89.0,"c":88.0,"d":87.0,"e":86.0},"annotated_formula":"divide(subtract(multiply(90, const_4), 96), subtract(const_4, const_1))","linear_formula":"multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)|","chain":"90 * 4<\/gadget>\n360<\/output>\n360 - 96<\/gadget>\n264<\/output>\n4 - 1<\/gadget>\n3<\/output>\n264 \/ 3<\/gadget>\n88<\/output>\n88<\/result>","index":1313} +{"problem":"last year department store x had a sales total for december that was 6 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ?","rationale":"\"let avg for 11 mos . = 10 therefore , dec = 60 year total = 11 * 10 + 60 = 170 answer = 60 \/ 170 = 6 \/ 17 = a\"","correct":"a","options":{"a":"6 \/ 17 ","b":"4 \/ 15 ","c":"1 \/ 3 ","d":"4 \/ 11","e":"4 \/ 5"},"options_float":{"a":0.3529411765,"b":0.2666666667,"c":0.3333333333,"d":0.3636363636,"e":0.8},"annotated_formula":"divide(6, add(subtract(const_12, const_1), 6))","linear_formula":"subtract(const_12,const_1)|add(n0,#0)|divide(n0,#1)|","chain":"12 - 1<\/gadget>\n11<\/output>\n11 + 6<\/gadget>\n17<\/output>\n6 \/ 17<\/gadget>\n6\/17 = around 0.352941<\/output>\n6\/17 = around 0.352941<\/result>","index":1314} +{"problem":"points a , b , and , c have xy - coordinates ( 2,0 ) , ( 8,12 ) , and ( 14,0 ) , respectively . points x , y , and z have xy - coordinates ( 6,0 ) , ( 8,4 ) , and ( 10,0 ) , respectively . what fraction c of the area of triangle abc is the area of triangle xyz ?","rationale":"if you notice , both triangles abc and xyz have a side on x axis . we can take these sides as bases for each triangle , therefore area of abc is 1 \/ 2 * 12 * 12 ( height of abc is the y coordinate of the third point ( 8,12 ) ) similarly area of xyz is 1 \/ 2 * 4 * 4 dividing area of xyz with that of abc gives c = 1 \/ 9 . a","correct":"a","options":{"a":"1 \/ 9 ","b":"1 \/ 8 ","c":"1 \/ 6 ","d":"1 \/ 5","e":"1 \/ 3"},"options_float":{"a":0.1111111111,"b":0.125,"c":0.1666666667,"d":0.2,"e":0.3333333333},"annotated_formula":"divide(divide(power(const_4, const_2), const_2), divide(power(add(const_10, const_2), const_2), const_2))","linear_formula":"add(const_10,const_2)|power(const_4,const_2)|divide(#1,const_2)|power(#0,const_2)|divide(#3,const_2)|divide(#2,#4)","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n144 \/ 2<\/gadget>\n72<\/output>\n8 \/ 72<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":1315} +{"problem":"if x is the median of the set { 9 \/ 2 , 11 \/ 3 , x , 28 \/ 9 , 21 \/ 5 } , x could be","rationale":"\"the median is the middle number once all the numbers are arranged in increasing \/ decreasing order . we see that 11 \/ 3 = 3 . something , 28 \/ 9 = 3 . something 21 \/ 5 = 4 . something 9 \/ 2 = 4 . something so x should greater than the smallest two numbers and smaller than the greatest two numbers . we can see that x = 4 is possible . ( first look at the simplest option or the middle option since options are usually arranged in increasing \/ decreasing order ) answer ( e )\"","correct":"e","options":{"a":"16 \/ 5 ","b":"17 \/ 5 ","c":"4 \/ 7 ","d":"30 \/ 7","e":"4"},"options_float":{"a":3.2,"b":3.4,"c":0.5714285714,"d":4.2857142857,"e":4.0},"annotated_formula":"add(add(divide(add(add(add(divide(9, 2), divide(11, 3)), divide(28, 9)), divide(21, 5)), const_4), divide(const_10, const_100)), divide(const_10, const_100))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|divide(n6,n7)|divide(const_10,const_100)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,const_4)|add(#8,#4)|add(#9,#4)|","chain":"9 \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\n11 \/ 3<\/gadget>\n11\/3 = around 3.666667<\/output>\n(9\/2) + (11\/3)<\/gadget>\n49\/6 = around 8.166667<\/output>\n28 \/ 9<\/gadget>\n28\/9 = around 3.111111<\/output>\n(49\/6) + (28\/9)<\/gadget>\n203\/18 = around 11.277778<\/output>\n21 \/ 5<\/gadget>\n21\/5 = around 4.2<\/output>\n(203\/18) + (21\/5)<\/gadget>\n1_393\/90 = around 15.477778<\/output>\n(1_393\/90) \/ 4<\/gadget>\n1_393\/360 = around 3.869444<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1_393\/360) + (1\/10)<\/gadget>\n1_429\/360 = around 3.969444<\/output>\n(1_429\/360) + (1\/10)<\/gadget>\n293\/72 = around 4.069444<\/output>\n293\/72 = around 4.069444<\/result>","index":1316} +{"problem":"a can do a piece of work in 4 days . b can do it in 7 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ?","rationale":"\"c = 1 \/ 2 - 1 \/ 4 - 1 \/ 7 = 3 \/ 28 = > 28 \/ 3 days answer : b\"","correct":"b","options":{"a":"22 days ","b":"28 \/ 3 days ","c":"67 days ","d":"17 \/ 5 days","e":"18 days"},"options_float":{"a":22.0,"b":9.3333333333,"c":67.0,"d":3.4,"e":18.0},"annotated_formula":"divide(multiply(4, 7), divide(subtract(multiply(4, 7), multiply(add(divide(multiply(4, 7), 4), divide(multiply(4, 7), 7)), 2)), 2))","linear_formula":"multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)|","chain":"4 * 7<\/gadget>\n28<\/output>\n28 \/ 4<\/gadget>\n7<\/output>\n28 \/ 7<\/gadget>\n4<\/output>\n7 + 4<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n28 - 22<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n28 \/ 3<\/gadget>\n28\/3 = around 9.333333<\/output>\n28\/3 = around 9.333333<\/result>","index":1317} +{"problem":"what least no . must be subtracted from 101054 so that remaining no . is divisible by 10 ?","rationale":"\"explanation : on dividing 101054 by 10 we get the remainder 4 , so 4 should be subtracted option b\"","correct":"b","options":{"a":"6 ","b":"4 ","c":"8 ","d":"5","e":"3"},"options_float":{"a":6.0,"b":4.0,"c":8.0,"d":5.0,"e":3.0},"annotated_formula":"subtract(101054, multiply(floor(divide(101054, 10)), 10))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"101_054 \/ 10<\/gadget>\n50_527\/5 = around 10_105.4<\/output>\nfloor(50_527\/5)<\/gadget>\n10_105<\/output>\n10_105 * 10<\/gadget>\n101_050<\/output>\n101_054 - 101_050<\/gadget>\n4<\/output>\n4<\/result>","index":1321} +{"problem":"at what rate percent on simple interest will rs . 750 amount to rs . 900 in 8 years ?","rationale":"150 = ( 750 * 8 * r ) \/ 100 r = 2.5 % answer : b","correct":"b","options":{"a":"6 % ","b":"2.5 % ","c":"4 % ","d":"5 %","e":"3.5 %"},"options_float":{"a":6.0,"b":2.5,"c":4.0,"d":5.0,"e":3.5},"annotated_formula":"multiply(divide(divide(subtract(900, 750), 750), 8), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)","chain":"900 - 750<\/gadget>\n150<\/output>\n150 \/ 750<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ 8<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/40) * 100<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":1322} +{"problem":"machine – x produces 50 % of the total output and machine - y produces 75 % of the total output . an average of 9 units out of a 1000 goods manufactured by machine - x and two unit of 500 units produced by machine - x prove to be defective . what is the probability that a unit chosen at random from the total daily output of the factory is defective ?","rationale":"suppose machine x produce 1000 units total defective = 9 machine y then produces = 1500 units total defective = 6 probability = 15 \/ 2500 = 0.006 answer : d","correct":"d","options":{"a":"0.008 ","b":"0.1 ","c":"0.12 ","d":"0.006","e":"0.007"},"options_float":{"a":0.008,"b":0.1,"c":0.12,"d":0.006,"e":0.007},"annotated_formula":"divide(add(9, divide(divide(9, divide(75, const_100)), const_2)), add(add(1000, multiply(1000, divide(50, const_100))), 1000))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n2,#0)|multiply(n3,#1)|add(n3,#3)|divide(#2,const_2)|add(n2,#5)|add(n3,#4)|divide(#6,#7)","chain":"75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n9 \/ (3\/4)<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n9 + 6<\/gadget>\n15<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1_000 * (1\/2)<\/gadget>\n500<\/output>\n1_000 + 500<\/gadget>\n1_500<\/output>\n1_500 + 1_000<\/gadget>\n2_500<\/output>\n15 \/ 2_500<\/gadget>\n3\/500 = around 0.006<\/output>\n3\/500 = around 0.006<\/result>","index":1323} +{"problem":"find the cost of fencing around a circular field of diameter 22 m at the rate of rs . 2.50 a meter ?","rationale":"\"2 * 22 \/ 7 * 11 = 69 69 * 2 1 \/ 2 = rs . 172 answer : b\"","correct":"b","options":{"a":"188 ","b":"172 ","c":"278 ","d":"279","e":"222"},"options_float":{"a":188.0,"b":172.0,"c":278.0,"d":279.0,"e":222.0},"annotated_formula":"multiply(circumface(divide(22, const_2)), 2.50)","linear_formula":"divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|","chain":"22 \/ 2<\/gadget>\n11<\/output>\n2 * pi * 11<\/gadget>\n22*pi = around 69.115038<\/output>\n(22*pi) * 2.5<\/gadget>\n55.0*pi = around 172.787596<\/output>\n55.0*pi = around 172.787596<\/result>","index":1325} +{"problem":"two cycles are standing side - by - side . one cycle , which is 2.5 feets tall , casts a shadow that is 5 feets long . the other cycle is 2 feets tall . compute , in feets , the length of the shadow that the shorter cycle casts .","rationale":"the ratio of shadow to height is constant , so if x is the length of the shadow , then 2.5 \/ 5 = 2 \/ x and x = 4 . correct answer b","correct":"b","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"10"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":10.0},"annotated_formula":"divide(multiply(5, 2), 2.5)","linear_formula":"multiply(n1,n2)|divide(#0,n0)","chain":"5 * 2<\/gadget>\n10<\/output>\n10 \/ 2.5<\/gadget>\n4<\/output>\n4<\/result>","index":1326} +{"problem":"stacy has a 12 page history paper due in 12 days . how many pages per day would she have to write to finish on time ?","rationale":"\"12 \/ 12 = 1 answer : d\"","correct":"d","options":{"a":"9 ","b":"8 ","c":"10 ","d":"1","e":"6"},"options_float":{"a":9.0,"b":8.0,"c":10.0,"d":1.0,"e":6.0},"annotated_formula":"divide(12, 12)","linear_formula":"divide(n0,n1)|","chain":"12 \/ 12<\/gadget>\n1<\/output>\n1<\/result>","index":1327} +{"problem":"on a two - dimensional coordinate plane , the line a = x ^ 2 - x ^ 3 touches the x - axis in how many places ?","rationale":"\"apparently it ' s a = x ^ 2 - x ^ 3 instead of a = x ^ 2 - a ^ 3 . in this case : the x - intercept is the value ( s ) of x for a = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c .\"","correct":"c","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"subtract(max(3, 2), const_1)","linear_formula":"max(n0,n1)|subtract(#0,const_1)|","chain":"max(3, 2)<\/gadget>\n3<\/output>\n3 - 1<\/gadget>\n2<\/output>\n2<\/result>","index":1329} +{"problem":"if the wages of 6 men for 15 days be rs . 2100 , then find the wages of for 12 days .","rationale":". let the required wages be rs . x . more men , more wages ( direct proportion ) less days , less wages ( direct proportion ) men 6 : 9 : : 2100 : x days 15 : 12 therefore ( 6 x 15 x x ) = ( 9 x 12 x 2100 )  x = ( 9 x 12 x 2100 ) \/ ( 6 x 15 ) = 2520 hence the required wages are rs . 2520 . answer is c","correct":"c","options":{"a":"rs . 2420 ","b":"rs . 2500 ","c":"rs . 2520 ","d":"rs . 2020","e":"rs . 2650"},"options_float":{"a":2420.0,"b":2500.0,"c":2520.0,"d":2020.0,"e":2650.0},"annotated_formula":"multiply(multiply(const_3, const_3), multiply(12, divide(divide(2100, 15), 6)))","linear_formula":"divide(n2,n1)|multiply(const_3,const_3)|divide(#0,n0)|multiply(n3,#2)|multiply(#1,#3)","chain":"3 * 3<\/gadget>\n9<\/output>\n2_100 \/ 15<\/gadget>\n140<\/output>\n140 \/ 6<\/gadget>\n70\/3 = around 23.333333<\/output>\n12 * (70\/3)<\/gadget>\n280<\/output>\n9 * 280<\/gadget>\n2_520<\/output>\n2_520<\/result>","index":1330} +{"problem":"a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 40 percent of the drive from the center , what percent of the round - trip has the technician completed ?","rationale":"\"the complete round trip consists of driving to the service center and then back home again . so , once the technician drives to the service center he \/ she has already competed 50 % of the entire trip . since the technician completes a portion of the trip back home , the correct answer must be greater than 50 % so 0.5 + 0.5 * 0.4 = 0.70 answer : b\"","correct":"b","options":{"a":"5 % ","b":"70 % ","c":"25 % ","d":"40 %","e":"55 %"},"options_float":{"a":5.0,"b":70.0,"c":25.0,"d":40.0,"e":55.0},"annotated_formula":"add(divide(const_100, const_2), divide(multiply(40, divide(const_100, const_2)), const_100))","linear_formula":"divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)|","chain":"100 \/ 2<\/gadget>\n50<\/output>\n40 * 50<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n50 + 20<\/gadget>\n70<\/output>\n70<\/result>","index":1331} +{"problem":"what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 35 ?","rationale":"\"number divisible by 11 : - 1000 \/ 11 = 90 numbers divisible by 35 : - 1000 \/ 35 = 28 numbers divisible by both 11 and 35 = 2 total numbers divisible by both 11 and 28 = 90 + 28 - 2 = 116 ( because we counted 2 in both 90 and 28 ) total numbers not divisible by 11 or 35 = 1000 - 116 = 884 answer : a\"","correct":"a","options":{"a":"884 ","b":"890 ","c":"892 ","d":"910","e":"945"},"options_float":{"a":884.0,"b":890.0,"c":892.0,"d":910.0,"e":945.0},"annotated_formula":"subtract(1000, subtract(add(divide(1000, 11), divide(1000, 35)), divide(1000, multiply(11, 35))))","linear_formula":"divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)|","chain":"1_000 \/ 11<\/gadget>\n1_000\/11 = around 90.909091<\/output>\n1_000 \/ 35<\/gadget>\n200\/7 = around 28.571429<\/output>\n(1_000\/11) + (200\/7)<\/gadget>\n9_200\/77 = around 119.480519<\/output>\n11 * 35<\/gadget>\n385<\/output>\n1_000 \/ 385<\/gadget>\n200\/77 = around 2.597403<\/output>\n(9_200\/77) - (200\/77)<\/gadget>\n9_000\/77 = around 116.883117<\/output>\n1_000 - (9_000\/77)<\/gadget>\n68_000\/77 = around 883.116883<\/output>\n68_000\/77 = around 883.116883<\/result>","index":1332} +{"problem":"3 positive integers a , b , and c are such that their average is 20 and a ≤ b ≤ c . if the median is ( a + 13 ) , what is the least possible value of c ?","rationale":"solution : we know that the average value is 20 therefore ( a + b + c ) \/ 3 = 20 so a + b + c = 60 and b = a + 13 therefore a + ( a + 13 ) + c = 60 the least value of c is when c = b so take c = a + 13 hence , a + ( a + 13 ) + ( a + 13 ) = 60 i . e . , a = 34 \/ 3 a non integer therefore c is not equal to b so take c = b + 1 therefore c = a + 14 a + ( a + 13 ) + ( a + 14 ) = 60 i . e . , a = 33 \/ 3 = 11 hence , a = 11 , b = 24 , and c = 25 answer : e","correct":"e","options":{"a":"21 ","b":"22 ","c":"23 ","d":"24","e":"25"},"options_float":{"a":21.0,"b":22.0,"c":23.0,"d":24.0,"e":25.0},"annotated_formula":"add(divide(subtract(multiply(20, 3), add(13, add(13, const_1))), 3), add(13, const_1))","linear_formula":"add(n2,const_1)|multiply(n0,n1)|add(n2,#0)|subtract(#1,#2)|divide(#3,n0)|add(#0,#4)","chain":"20 * 3<\/gadget>\n60<\/output>\n13 + 1<\/gadget>\n14<\/output>\n13 + 14<\/gadget>\n27<\/output>\n60 - 27<\/gadget>\n33<\/output>\n33 \/ 3<\/gadget>\n11<\/output>\n11 + 14<\/gadget>\n25<\/output>\n25<\/result>","index":1334} +{"problem":"if a certain coin is flipped , the probability that the coin will land heads is 1 \/ 2 . if the coin is flipped 6 times , what is the probability that it will land heads up on the first 4 flips but not on the last 2 flips ?","rationale":"\"p ( hhhhtt ) = 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 = 1 \/ 64 the answer is d .\"","correct":"d","options":{"a":"1 \/ 8 ","b":"1 \/ 16 ","c":"1 \/ 32 ","d":"1 \/ 64","e":"1 \/ 128"},"options_float":{"a":0.125,"b":0.0625,"c":0.03125,"d":0.015625,"e":0.0078125},"annotated_formula":"power(divide(1, 2), 6)","linear_formula":"divide(n0,n1)|power(#0,n2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 6<\/gadget>\n1\/64 = around 0.015625<\/output>\n1\/64 = around 0.015625<\/result>","index":1335} +{"problem":"a man goes downstream at 10 kmph , and upstream 8 kmph . the speed of the stream is","rationale":"\"speed of the stream = 1 \/ 2 ( 10 - 8 ) kmph = 1 kmph . correct option : c\"","correct":"c","options":{"a":"0 kmph ","b":"4 kmph ","c":"1 kmph ","d":"2.5 kmph","e":"26 kmph"},"options_float":{"a":0.0,"b":4.0,"c":1.0,"d":2.5,"e":26.0},"annotated_formula":"divide(subtract(10, 8), const_2)","linear_formula":"subtract(n0,n1)|divide(#0,const_2)|","chain":"10 - 8<\/gadget>\n2<\/output>\n2 \/ 2<\/gadget>\n1<\/output>\n1<\/result>","index":1338} +{"problem":"if shares of two persons in profits are rs . 600 and rs . 1000 then ratio of their capitals is","rationale":"\"profit = time * capital so 600 : 1000 = 3 : 5 answer : a\"","correct":"a","options":{"a":"3 : 5 ","b":"2 : 3 ","c":"4 : 3 ","d":"1 : 3","e":"1 : 2"},"options_float":{"a":0.6,"b":0.6666666667,"c":1.3333333333,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(600, 1000)","linear_formula":"divide(n0,n1)|","chain":"600 \/ 1_000<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":1339} +{"problem":"if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 22 , then n =","rationale":"2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 22 = > 4 x 2 ^ 2 n = 4 ^ 22 = 2 ^ 44 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 44 = > 2 ^ ( 2 n + 2 ) = 2 ^ 44 = > 2 n + 2 = 44 = > n = 21 so . answer will be e .","correct":"e","options":{"a":"3 ","b":"6 ","c":"12 ","d":"23","e":"21"},"options_float":{"a":3.0,"b":6.0,"c":12.0,"d":23.0,"e":21.0},"annotated_formula":"divide(subtract(multiply(22, 2), 2), 2)","linear_formula":"multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)","chain":"22 * 2<\/gadget>\n44<\/output>\n44 - 2<\/gadget>\n42<\/output>\n42 \/ 2<\/gadget>\n21<\/output>\n21<\/result>","index":1341} +{"problem":"a is the average ( arithmetic mean ) of the first 7 positive multiples of 5 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 – b ^ 2 is zero , what is the value of n ?","rationale":"\"if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 20 , which also equals b . b is the second positive multiple of n , thus n = 20 \/ 2 = 10 . the answer is c .\"","correct":"c","options":{"a":"5 ","b":"6 ","c":"10 ","d":"15","e":"20"},"options_float":{"a":5.0,"b":6.0,"c":10.0,"d":15.0,"e":20.0},"annotated_formula":"divide(multiply(5, const_4), 2)","linear_formula":"multiply(n1,const_4)|divide(#0,n3)|","chain":"5 * 4<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":1342} +{"problem":"length of a rectangular plot is 10 mtr more than its breadth . if the cost of fencin g the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ?","rationale":"\"let breadth = x metres . then , length = ( x + 10 ) metres . perimeter = 5300 \/ 26.5 = 200 m . 2 [ ( x + 10 ) + x ] = 200 2 x + 10 = 100 2 x = 90 x = 45 . hence , length = x + 10 = 55 m b\"","correct":"b","options":{"a":"46 m ","b":"55 m ","c":"58 m ","d":"78 m","e":"80 m"},"options_float":{"a":46.0,"b":55.0,"c":58.0,"d":78.0,"e":80.0},"annotated_formula":"divide(add(divide(5300, 26.50), multiply(const_2, 10)), const_4)","linear_formula":"divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)|","chain":"5_300 \/ 26.5<\/gadget>\n200<\/output>\n2 * 10<\/gadget>\n20<\/output>\n200 + 20<\/gadget>\n220<\/output>\n220 \/ 4<\/gadget>\n55<\/output>\n55<\/result>","index":1343} +{"problem":"2 men and 4 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other )","rationale":"the list should be wmwmw . hence , from women 4 ! and men 2 ! , we get ( 4 ! ) ( 2 ! ) = 48 . therefore , the correct answer is e .","correct":"e","options":{"a":"12 ","b":"15 ","c":"18 ","d":"21","e":"48"},"options_float":{"a":12.0,"b":15.0,"c":18.0,"d":21.0,"e":48.0},"annotated_formula":"multiply(factorial(2), factorial(4))","linear_formula":"factorial(n0)|factorial(n1)|multiply(#0,#1)|","chain":"factorial(2)<\/gadget>\n2<\/output>\nfactorial(4)<\/gadget>\n24<\/output>\n2 * 24<\/gadget>\n48<\/output>\n48<\/result>","index":1344} +{"problem":"each day a man meets his wife at the train station after work , and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived 40 minutes earlier than usual . how much time did the man spend walking ?","rationale":"\"as they arrived 40 minutes earlier than usual , they saved 40 minutes on round trip from home to station ( home - station - home ) - - > 20 minutes in each direction ( home - station ) - - > wife meets husband 20 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time , so he must have spent waking the rest of the time before their meeting , which is hour - 20 minutes = 40 minutes . answer : c\"","correct":"c","options":{"a":"45 minutes ","b":"50 minutes ","c":"40 minutes ","d":"55 minutes","e":"35 minutes"},"options_float":{"a":45.0,"b":50.0,"c":40.0,"d":55.0,"e":35.0},"annotated_formula":"subtract(const_60, divide(40, const_2))","linear_formula":"divide(n0,const_2)|subtract(const_60,#0)|","chain":"40 \/ 2<\/gadget>\n20<\/output>\n60 - 20<\/gadget>\n40<\/output>\n40<\/result>","index":1346} +{"problem":"students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 184 cms . what was the ratio a of men to women ?","rationale":"( a ) a = 184 x 5 + 170 x 2 = 1260 . a","correct":"a","options":{"a":"5 : 2 ","b":"5 : 1 ","c":"4 : 3 ","d":"4 : 1","e":"3 : 1"},"options_float":{"a":2.5,"b":5.0,"c":1.3333333333,"d":4.0,"e":3.0},"annotated_formula":"divide(subtract(180, 170), subtract(184, 180))","linear_formula":"subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)|","chain":"180 - 170<\/gadget>\n10<\/output>\n184 - 180<\/gadget>\n4<\/output>\n10 \/ 4<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":1347} +{"problem":"if ( 8000 ) ( 6000 ) = ( 480 ) ( 10 ^ y ) , what is the value of y ?","rationale":"( 8000 ) ( 6000 ) = ( 480 ) ( 10 ^ y ) = > ( 8000 ) ( 6000 ) \/ 480 = 10 ^ y = > 100000 = 100 ^ y = > 10 ^ 5 = 10 ^ y since , base is same so powers will be same too . so , y = 5 answer will be a","correct":"a","options":{"a":"5 ","b":"4 ","c":"3 ","d":"2","e":"1"},"options_float":{"a":5.0,"b":4.0,"c":3.0,"d":2.0,"e":1.0},"annotated_formula":"divide(log(divide(multiply(8000, 6000), 480)), log(const_10))","linear_formula":"log(const_10)|multiply(n0,n1)|divide(#1,n2)|log(#2)|divide(#3,#0)","chain":"8_000 * 6_000<\/gadget>\n48_000_000<\/output>\n48_000_000 \/ 480<\/gadget>\n100_000<\/output>\nlog(100_000)<\/gadget>\nlog(100000) = around 11.512925<\/output>\nlog(10)<\/gadget>\nlog(10) = around 2.302585<\/output>\nlog(100000) \/ log(10)<\/gadget>\nlog(100000)\/log(10) = around 5<\/output>\nlog(100000)\/log(10) = around 5<\/result>","index":1348} +{"problem":"if ( 5 + k ) ( 5 - k ) = ( 5 ^ 2 ) - ( 2 ^ 3 ) , then what is the value of k ?","rationale":"obviously , it is in the formula a ^ 2 - b ^ 2 = ( a + b ) ( a - b ) so k is 2 option d","correct":"d","options":{"a":"1 ","b":"3 ","c":"4 ","d":"2","e":"6"},"options_float":{"a":1.0,"b":3.0,"c":4.0,"d":2.0,"e":6.0},"annotated_formula":"subtract(multiply(5, 3), add(3, const_10))","linear_formula":"add(n5,const_10)|multiply(n0,n5)|subtract(#1,#0)","chain":"5 * 3<\/gadget>\n15<\/output>\n3 + 10<\/gadget>\n13<\/output>\n15 - 13<\/gadget>\n2<\/output>\n2<\/result>","index":1349} +{"problem":"a train consists of 48 boxcars pulled by 2 engines . the boxcars are 15 meters long and the two engines are each 20 meters long . it crosses over a highway 24 meters wide at a speed of 14 m \/ s . how long does it take for the entire train to cross ?","rationale":"a 56 sec d = 48 * 15 m + 2 * 20 m + 24 m = 784 m t = 784 m \/ 14 m \/ s = 56 s","correct":"a","options":{"a":"56 sec ","b":"63 sec ","c":"72 sec ","d":"81 sec","e":"88 sec"},"options_float":{"a":56.0,"b":63.0,"c":72.0,"d":81.0,"e":88.0},"annotated_formula":"divide(add(multiply(48, 15), multiply(20, 2)), 14)","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,n5)","chain":"48 * 15<\/gadget>\n720<\/output>\n20 * 2<\/gadget>\n40<\/output>\n720 + 40<\/gadget>\n760<\/output>\n760 \/ 14<\/gadget>\n380\/7 = around 54.285714<\/output>\n380\/7 = around 54.285714<\/result>","index":1350} +{"problem":"a store reported total sales of $ 416 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ?","rationale":"\"new value – old value ) \/ old value x 100 we are given : february sales this year = 416 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 416 million and the old value = 320 million . let ’ s plug them into our percent change formula . ( new value – old value ) \/ old value x 100 [ ( 416 – 320 ) \/ 320 ] x 100 = 30 % the answer is c .\"","correct":"c","options":{"a":"2 % ","b":"17 % ","c":"30 % ","d":"65 %","e":"83 %"},"options_float":{"a":2.0,"b":17.0,"c":30.0,"d":65.0,"e":83.0},"annotated_formula":"multiply(divide(subtract(416, 320), 320), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|","chain":"416 - 320<\/gadget>\n96<\/output>\n96 \/ 320<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":1352} +{"problem":"the ratio of the radius of two circles is 2 : 5 , and then the ratio of their areas is ?","rationale":"\"r 1 : r 2 = 2 : 5 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 4 : 25 answer : a\"","correct":"a","options":{"a":"4 : 25 ","b":"2 : 9 ","c":"1 : 9 ","d":"3 : 7","e":"3 : 4"},"options_float":{"a":0.16,"b":0.2222222222,"c":0.1111111111,"d":0.4285714286,"e":0.75},"annotated_formula":"divide(circle_area(2), circle_area(5))","linear_formula":"circle_area(n0)|circle_area(n1)|divide(#0,#1)|","chain":"pi * (2 ** 2)<\/gadget>\n4*pi = around 12.566371<\/output>\npi * (5 ** 2)<\/gadget>\n25*pi = around 78.539816<\/output>\n(4*pi) \/ (25*pi)<\/gadget>\n4\/25 = around 0.16<\/output>\n4\/25 = around 0.16<\/result>","index":1353} +{"problem":"of the families in city x in 2000 , 30 percent owned a personal computer . the number of families in city x owning a computer in 2004 was 20 percent greater than it was in 1994 , and the total number of families in city x was 10 percent greater in 2004 than it was in 2000 . what percent of the families in city x owned a personal computer in 1998 ?","rationale":"\"say a 100 families existed in 2000 then the number of families owning a computer in 2000 - 30 number of families owning computer in 2000 = 30 * 120 \/ 100 = 36 number of families in 2004 = 110 the percentage = 36 \/ 110 * 100 = 32.72 % . option : e\"","correct":"e","options":{"a":"55.42 % ","b":"52.25 % ","c":"56.23 % ","d":"30.70 %","e":"32.72 %"},"options_float":{"a":55.42,"b":52.25,"c":56.23,"d":30.7,"e":32.72},"annotated_formula":"multiply(const_100, divide(divide(multiply(add(20, const_100), 30), const_100), add(const_100, 10)))","linear_formula":"add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)|","chain":"20 + 100<\/gadget>\n120<\/output>\n120 * 30<\/gadget>\n3_600<\/output>\n3_600 \/ 100<\/gadget>\n36<\/output>\n100 + 10<\/gadget>\n110<\/output>\n36 \/ 110<\/gadget>\n18\/55 = around 0.327273<\/output>\n100 * (18\/55)<\/gadget>\n360\/11 = around 32.727273<\/output>\n360\/11 = around 32.727273<\/result>","index":1355} +{"problem":"raman mixed 48 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ?","rationale":"\"explanation : cp per kg of mixture = [ 48 ( 150 ) + 36 ( 125 ) ] \/ ( 48 + 36 ) = rs . 139.28 sp = cp [ ( 100 + profit % ) \/ 100 ] = 139.28 * [ ( 100 + 40 ) \/ 100 ] = rs . 195 answer : c\"","correct":"c","options":{"a":"129 ","b":"287 ","c":"195 ","d":"188","e":"112"},"options_float":{"a":129.0,"b":287.0,"c":195.0,"d":188.0,"e":112.0},"annotated_formula":"add(divide(add(multiply(48, 150), multiply(36, 125)), add(36, 48)), multiply(divide(add(multiply(48, 150), multiply(36, 125)), add(36, 48)), divide(40, const_100)))","linear_formula":"add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6)|","chain":"48 * 150<\/gadget>\n7_200<\/output>\n36 * 125<\/gadget>\n4_500<\/output>\n7_200 + 4_500<\/gadget>\n11_700<\/output>\n36 + 48<\/gadget>\n84<\/output>\n11_700 \/ 84<\/gadget>\n975\/7 = around 139.285714<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(975\/7) * (2\/5)<\/gadget>\n390\/7 = around 55.714286<\/output>\n(975\/7) + (390\/7)<\/gadget>\n195<\/output>\n195<\/result>","index":1356} +{"problem":"a certain high school has 500 students . of these students , 30 are taking music , 10 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ?","rationale":"\"we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 30 y = 10 z = 10 so , we have 30 students taking music , 10 taking art and 10 taking both music and art . 20 student taking just music 0 student taking just art 10 student taking both music and art total = 30 students we ' re asked for the total number of students who are taking neither course . that is 500 - 30 = 470 . e\"","correct":"e","options":{"a":"430 ","b":"440 ","c":"450 ","d":"460","e":"470"},"options_float":{"a":430.0,"b":440.0,"c":450.0,"d":460.0,"e":470.0},"annotated_formula":"subtract(500, subtract(add(30, 10), 10))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|","chain":"30 + 10<\/gadget>\n40<\/output>\n40 - 10<\/gadget>\n30<\/output>\n500 - 30<\/gadget>\n470<\/output>\n470<\/result>","index":1359} +{"problem":"in what time will a train 100 metres long cross an electric pole , if its speed be 144 km \/ hr ?","rationale":"solution speed = ( 144 x 5 \/ 18 ) m \/ sec = 40 m \/ sec time taken = ( 100 \/ 40 ) sec = 2.5 sec . answer a","correct":"a","options":{"a":"2.5 sec ","b":"4.25 sec ","c":"5 sec ","d":"12.5 sec","e":"none"},"options_float":{"a":2.5,"b":4.25,"c":5.0,"d":12.5,"e":null},"annotated_formula":"divide(100, multiply(144, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n144 * (5\/18)<\/gadget>\n40<\/output>\n100 \/ 40<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":1360} +{"problem":"a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 70 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ?","rationale":"\"assume the revenue in 2000 to be 100 . then in 2003 it would be 140 and and in 2005 170 , so from 2003 to 2005 it increased by ( 170 - 140 ) \/ 140 = 30 \/ 140 = ~ 21 % . answer : e .\"","correct":"e","options":{"a":"50 % ","b":"40 % ","c":"35 % ","d":"32 %","e":"21 %"},"options_float":{"a":50.0,"b":40.0,"c":35.0,"d":32.0,"e":21.0},"annotated_formula":"multiply(divide(subtract(add(const_1, divide(70, const_100)), add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100)","linear_formula":"divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)|","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n1 + (7\/10)<\/gadget>\n17\/10 = around 1.7<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n(17\/10) - (7\/5)<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) \/ (7\/5)<\/gadget>\n3\/14 = around 0.214286<\/output>\n(3\/14) * 100<\/gadget>\n150\/7 = around 21.428571<\/output>\n150\/7 = around 21.428571<\/result>","index":1363} +{"problem":"a start walking from a place at a uniform speed of 4 kmph in a particular direction . after half an hour , b starts from the same place and walks in the same direction as a at a uniform speed and overtakes a after 1 hour 48 minutes . find the speed of b .","rationale":"\"distance covered by a in 30 min = 1 km b covers extra 1 km in 1 hour 48 minutes ( 9 \/ 5 hr ) i . e . relative speed of b over a = 1 \/ ( 9 \/ 5 ) = 5 \/ 9 so the speed of b = speed of a + 5 \/ 9 = 4 + 5 \/ 9 = 4.55 answer b\"","correct":"b","options":{"a":"4.7 kmph ","b":"4.6 kmph ","c":"4 kmph ","d":"7 kmph","e":"5.3 kmph"},"options_float":{"a":4.7,"b":4.6,"c":4.0,"d":7.0,"e":5.3},"annotated_formula":"add(divide(1, divide(add(const_60, 48), const_60)), 4)","linear_formula":"add(n2,const_60)|divide(#0,const_60)|divide(n1,#1)|add(n0,#2)|","chain":"60 + 48<\/gadget>\n108<\/output>\n108 \/ 60<\/gadget>\n9\/5 = around 1.8<\/output>\n1 \/ (9\/5)<\/gadget>\n5\/9 = around 0.555556<\/output>\n(5\/9) + 4<\/gadget>\n41\/9 = around 4.555556<\/output>\n41\/9 = around 4.555556<\/result>","index":1364} +{"problem":"what is the smallest integer k for which 64 ^ k > 4 ^ 20 ?","rationale":"\"64 ^ k > 4 ^ 20 4 ^ ( 3 k ) > 4 ^ 20 3 k > 20 k = 7 the answer is d .\"","correct":"d","options":{"a":"4 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"add(floor(divide(20, const_3)), const_1)","linear_formula":"divide(n2,const_3)|floor(#0)|add(#1,const_1)|","chain":"20 \/ 3<\/gadget>\n20\/3 = around 6.666667<\/output>\nfloor(20\/3)<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7<\/result>","index":1369} +{"problem":"a cistern 8 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is :","rationale":"\"area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 8 x 1.25 ) + 8 x 4 ] m 2 = 62 m 2 . answer : option a\"","correct":"a","options":{"a":"62 m 2 ","b":"50 m 2 ","c":"53.5 m 2 ","d":"55 m 2","e":"57 m 2"},"options_float":{"a":62.0,"b":50.0,"c":53.5,"d":55.0,"e":57.0},"annotated_formula":"add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 8))), multiply(4, 8))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 4<\/gadget>\n5<\/output>\n(5\/4) * 8<\/gadget>\n10<\/output>\n5 + 10<\/gadget>\n15<\/output>\n2 * 15<\/gadget>\n30<\/output>\n4 * 8<\/gadget>\n32<\/output>\n30 + 32<\/gadget>\n62<\/output>\n62<\/result>","index":1371} +{"problem":"if 2994 ã · 14.5 = 179 , then 29.94 ã · 1.45 = ?","rationale":"29.94 \/ 1.45 = 299.4 \/ 14.5 = ( 2994 \/ 14.5 ) x 1 \/ 10 ) [ here , substitute 179 in the place of 2994 \/ 14.5 ] = 179 \/ 10 = 17.9 answer is e .","correct":"e","options":{"a":"17.1 ","b":"17.3 ","c":"17.5 ","d":"17.7","e":"17.9"},"options_float":{"a":17.1,"b":17.3,"c":17.5,"d":17.7,"e":17.9},"annotated_formula":"divide(179, divide(14.5, 1.45))","linear_formula":"divide(n1,n4)|divide(n2,#0)","chain":"14.5 \/ 1.45<\/gadget>\n10<\/output>\n179 \/ 10<\/gadget>\n179\/10 = around 17.9<\/output>\n179\/10 = around 17.9<\/result>","index":1372} +{"problem":"( √ 27 + √ 243 ) \/ √ 75 = ?","rationale":"\"( √ 27 + √ 243 ) \/ √ 75 = ( 3 √ 3 + 9 √ 3 ) \/ 5 √ 3 = 12 √ 3 \/ 5 √ 3 = 12 \/ 5 . hence , the correct answer is e\"","correct":"e","options":{"a":"2 √ 2 ","b":"2 √ 3 ","c":"3 √ 2 ","d":"3 √ 3","e":"12 \/ 5"},"options_float":{"a":2.0,"b":2.0,"c":3.0,"d":3.0,"e":2.4},"annotated_formula":"divide(add(sqrt(27), sqrt(243)), sqrt(75))","linear_formula":"sqrt(n0)|sqrt(n1)|sqrt(n2)|add(#0,#1)|divide(#3,#2)|","chain":"27 ** (1\/2)<\/gadget>\n3*sqrt(3) = around 5.196152<\/output>\n243 ** (1\/2)<\/gadget>\n9*sqrt(3) = around 15.588457<\/output>\n(3*sqrt(3)) + (9*sqrt(3))<\/gadget>\n12*sqrt(3) = around 20.78461<\/output>\n75 ** (1\/2)<\/gadget>\n5*sqrt(3) = around 8.660254<\/output>\n(12*sqrt(3)) \/ (5*sqrt(3))<\/gadget>\n12\/5 = around 2.4<\/output>\n12\/5 = around 2.4<\/result>","index":1373} +{"problem":"a set s = { x , - 8 , - 7 , - 6 , 6 , 8 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ?","rationale":"median of the set = ( - 6 + 6 ) \/ 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 23 - 21 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer c","correct":"c","options":{"a":"1 ","b":"0 ","c":"2 ","d":"- 1","e":"can not be determined"},"options_float":{"a":1.0,"b":0.0,"c":2.0,"d":-1.0,"e":null},"annotated_formula":"subtract(subtract(subtract(add(add(9, 8), 6), 6), 7), 8)","linear_formula":"add(n0,n5)|add(n2,#0)|subtract(#1,n2)|subtract(#2,n1)|subtract(#3,n0)","chain":"9 + 8<\/gadget>\n17<\/output>\n17 + 6<\/gadget>\n23<\/output>\n23 - 6<\/gadget>\n17<\/output>\n17 - 7<\/gadget>\n10<\/output>\n10 - 8<\/gadget>\n2<\/output>\n2<\/result>","index":1374} +{"problem":"in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 8 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?","rationale":"\"total fish = x percentage of second catch = ( 8 \/ 50 ) * 100 = 16 % so , x * 16 % = 50 x = 312 ans . e\"","correct":"e","options":{"a":"400 ","b":"625 ","c":"1,250 ","d":"2,500","e":"312"},"options_float":{"a":400.0,"b":625.0,"c":1250.0,"d":2500.0,"e":312.0},"annotated_formula":"divide(50, divide(8, 50))","linear_formula":"divide(n2,n1)|divide(n0,#0)|","chain":"8 \/ 50<\/gadget>\n4\/25 = around 0.16<\/output>\n50 \/ (4\/25)<\/gadget>\n625\/2 = around 312.5<\/output>\n625\/2 = around 312.5<\/result>","index":1376} +{"problem":"what is the sum of the digits of integer k , if k = ( 10 ^ 45 - 46 )","rationale":"\"there are 46 digits in 10 ^ 45 when we subtract 46 from it , there will be 45 digits left . 10 ^ 45 can be written as 9999999 . . . . ( 45 times ) + 1 so , 10 ^ 45 - 46 = 9999999 . . . . ( 45 times ) + 1 - 46 = 9999999 . . . . ( 45 times ) - 45 consider the last 2 digits , 99 - 45 = 54 the last 2 digits will be 54 . and our number would be 99999 . . . . . . 99954 with 2 less 9 s after subtraction . number of 9 s left are 43 and the last two digits are 54 the sum of the digits will be ( 43 * 9 ) + 5 + 4 = 396 answer : - d\"","correct":"d","options":{"a":"356 ","b":"366 ","c":"363 ","d":"396","e":"408"},"options_float":{"a":356.0,"b":366.0,"c":363.0,"d":396.0,"e":408.0},"annotated_formula":"add(multiply(subtract(10, const_1), subtract(45, const_2)), 10)","linear_formula":"subtract(n0,const_1)|subtract(n1,const_2)|multiply(#0,#1)|add(#2,n0)|","chain":"10 - 1<\/gadget>\n9<\/output>\n45 - 2<\/gadget>\n43<\/output>\n9 * 43<\/gadget>\n387<\/output>\n387 + 10<\/gadget>\n397<\/output>\n397<\/result>","index":1377} +{"problem":"if 5 a = 3125 , then the value of 5 ( a - 3 ) is :","rationale":"solution 5 a = 3125 ⇔ 5 a = 5 a ⇔ a = 5 . ∴ 5 ( a - 3 ) = 5 ( 5 - 3 ) = 52 = 25 . answer a","correct":"a","options":{"a":"25 ","b":"125 ","c":"625 ","d":"1625","e":"none of these"},"options_float":{"a":25.0,"b":125.0,"c":625.0,"d":1625.0,"e":null},"annotated_formula":"power(5, subtract(5, 3))","linear_formula":"subtract(n0,n3)|power(n0,#0)","chain":"5 - 3<\/gadget>\n2<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n25<\/result>","index":1378} +{"problem":"in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 130 marks , the no of questions he attempts correctly is :","rationale":"\"let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 60 – x ) = 130 = > 5 x = 190 = > x = 38 answer : b\"","correct":"b","options":{"a":"33 ","b":"38 ","c":"99 ","d":"27","e":"71"},"options_float":{"a":33.0,"b":38.0,"c":99.0,"d":27.0,"e":71.0},"annotated_formula":"divide(add(130, 60), add(4, 1))","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|","chain":"130 + 60<\/gadget>\n190<\/output>\n4 + 1<\/gadget>\n5<\/output>\n190 \/ 5<\/gadget>\n38<\/output>\n38<\/result>","index":1379} +{"problem":"a is half good a work man as b and together they finish a job in 20 days . in how many days working alone b finish the job ?","rationale":"\"wc = 1 : 2 2 x + x = 1 \/ 20 = > x = 1 \/ 60 2 x = 1 \/ 60 = > 30 days answer : b\"","correct":"b","options":{"a":"98 days ","b":"30 days ","c":"17 days ","d":"18 days","e":"19 days"},"options_float":{"a":98.0,"b":30.0,"c":17.0,"d":18.0,"e":19.0},"annotated_formula":"multiply(20, divide(const_3, const_2))","linear_formula":"divide(const_3,const_2)|multiply(n0,#0)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n20 * (3\/2)<\/gadget>\n30<\/output>\n30<\/result>","index":1383} +{"problem":"what is the value of n if the sum of the consecutive odd intergers y from 1 to n equals 169 ?","rationale":"\"before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n \/ 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is .\"","correct":"b","options":{"a":"47 ","b":"25 ","c":"37 ","d":"33","e":"29"},"options_float":{"a":47.0,"b":25.0,"c":37.0,"d":33.0,"e":29.0},"annotated_formula":"add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1)","linear_formula":"multiply(n0,const_2)|sqrt(n1)|multiply(#1,const_2)|subtract(#2,#0)|add(n0,#3)|","chain":"169 ** (1\/2)<\/gadget>\n13<\/output>\n13 * 2<\/gadget>\n26<\/output>\n2 * 1<\/gadget>\n2<\/output>\n26 - 2<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n25<\/result>","index":1384} +{"problem":"if log 4 x + log 4 ( 1 \/ 6 ) = 1 \/ 2 then the value of x is ?","rationale":"log 4 x + log 4 ( 1 \/ 6 ) = 1 \/ 2 = > log 4 [ x x 1 \/ 6 ] = 1 \/ 2 = > log 4 [ x \/ 6 ] = 1 \/ 2 = > 41 \/ 2 = x \/ 6 = > 2 = x \/ 6 therefore , x = 12 answer : d","correct":"d","options":{"a":"18 ","b":"24 ","c":"16 ","d":"12","e":"10"},"options_float":{"a":18.0,"b":24.0,"c":16.0,"d":12.0,"e":10.0},"annotated_formula":"multiply(power(4, divide(1, 2)), 6)","linear_formula":"divide(n2,n5)|power(n0,#0)|multiply(n3,#1)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n4 ** (1\/2)<\/gadget>\n2<\/output>\n2 * 6<\/gadget>\n12<\/output>\n12<\/result>","index":1385} +{"problem":"what quantity of water should taken out to concentrate 27 liters of 40 % acidic liquid to 60 % acidic liquid ?","rationale":"\"required answer is = 27 ( 60 - 40 ) \/ 60 = 9 liters answer is d\"","correct":"d","options":{"a":"5 liters ","b":"10 liters ","c":"15 liters ","d":"9 liters","e":"6 liters"},"options_float":{"a":5.0,"b":10.0,"c":15.0,"d":9.0,"e":6.0},"annotated_formula":"subtract(27, divide(multiply(27, 40), 60))","linear_formula":"multiply(n0,n1)|divide(#0,n2)|subtract(n0,#1)|","chain":"27 * 40<\/gadget>\n1_080<\/output>\n1_080 \/ 60<\/gadget>\n18<\/output>\n27 - 18<\/gadget>\n9<\/output>\n9<\/result>","index":1386} +{"problem":"marts income is 30 percent more than tims income and tims income is 40 percent less than juans income . what percentage of juans income is marts income","rationale":"\"m = ( 130 \/ 100 ) t t = ( 60 \/ 100 ) j = > m = ( 78 \/ 100 ) j answer c .\"","correct":"c","options":{"a":"124 % ","b":"120 % ","c":"78 % ","d":"80 %","e":"64 %"},"options_float":{"a":124.0,"b":120.0,"c":78.0,"d":80.0,"e":64.0},"annotated_formula":"multiply(divide(add(const_100, 30), multiply(divide(const_100, subtract(const_100, 40)), const_100)), const_100)","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|divide(const_100,#1)|multiply(#2,const_100)|divide(#0,#3)|multiply(#4,const_100)|","chain":"100 + 30<\/gadget>\n130<\/output>\n100 - 40<\/gadget>\n60<\/output>\n100 \/ 60<\/gadget>\n5\/3 = around 1.666667<\/output>\n(5\/3) * 100<\/gadget>\n500\/3 = around 166.666667<\/output>\n130 \/ (500\/3)<\/gadget>\n39\/50 = around 0.78<\/output>\n(39\/50) * 100<\/gadget>\n78<\/output>\n78<\/result>","index":1387} +{"problem":"there is a 5 - digit number . the fifth digit is one fourth of the third digit and one half of the fourth digit . third digit is one half of the first digit . second digit is 5 more than the fifth digit . what is that 5 - digit number ?","rationale":"we can prove it as follows suppose number is pqrst , now t = ( 1 \/ 4 ) r = ( 1 \/ 2 ) s . . . . . . ( 1 ) , r = ( 1 \/ 2 ) p . . . . . ( 2 ) , q = t + 5 . . . . ( 3 ) now from ( 1 ) 4 t = r = 2 s = k ( suppose ) . . . . ( 4 ) from ( 2 ) 2 r = p . . . ( 5 ) now ( 4 ) * 2 then we get 8 t = 2 r = 4 s = k ( again some integer ) . . . . ( 6 ) from ( 5 ) and ( 6 ) 8 t = 2 r = 4 s = p = k . . . . . . . . . ( 7 ) now we have to choose the value of k 1 ) if choose k = 0 then it is worthless number will be 00000 ( not a answer ) 2 ) for k = 1 to 7 and for k = 9 digits will be in fractions 3 ) for k = 8 we get p = 8 , s = 2 , r = 4 , t = 1 and q = 6 and number is pqrst = 86421 answer : a","correct":"a","options":{"a":"86421 ","b":"86424 ","c":"89421 ","d":"96421","e":"76421"},"options_float":{"a":86421.0,"b":86424.0,"c":89421.0,"d":96421.0,"e":76421.0},"annotated_formula":"add(add(add(add(multiply(multiply(const_4, const_2), multiply(const_1000, const_10)), multiply(5, const_1000)), multiply(const_4, const_100)), multiply(const_2, const_10)), const_1)","linear_formula":"multiply(const_2,const_4)|multiply(const_10,const_1000)|multiply(n0,const_1000)|multiply(const_100,const_4)|multiply(const_10,const_2)|multiply(#0,#1)|add(#5,#2)|add(#6,#3)|add(#7,#4)|add(#8,const_1)","chain":"4 * 2<\/gadget>\n8<\/output>\n1_000 * 10<\/gadget>\n10_000<\/output>\n8 * 10_000<\/gadget>\n80_000<\/output>\n5 * 1_000<\/gadget>\n5_000<\/output>\n80_000 + 5_000<\/gadget>\n85_000<\/output>\n4 * 100<\/gadget>\n400<\/output>\n85_000 + 400<\/gadget>\n85_400<\/output>\n2 * 10<\/gadget>\n20<\/output>\n85_400 + 20<\/gadget>\n85_420<\/output>\n85_420 + 1<\/gadget>\n85_421<\/output>\n85_421<\/result>","index":1388} +{"problem":"a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 840 . find the sum ?","rationale":"\"let the sum be rs . x . ( x * 18 * 2 ) \/ 100 - ( x * 12 * 2 ) \/ 100 = 840 = > 36 x \/ 100 - 24 x \/ 100 = 840 = > 12 x \/ 100 = 840 = > x = 7000 . answer : a\"","correct":"a","options":{"a":"7000 ","b":"2787 ","c":"2799 ","d":"2699","e":"2790"},"options_float":{"a":7000.0,"b":2787.0,"c":2799.0,"d":2699.0,"e":2790.0},"annotated_formula":"divide(840, divide(multiply(subtract(18, 12), const_2), const_100))","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)|","chain":"18 - 12<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n840 \/ (3\/25)<\/gadget>\n7_000<\/output>\n7_000<\/result>","index":1390} +{"problem":"for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.15 – 6.15 * ?","rationale":"\"since z * is defined as the greatest positive even integer less than or equal to z , then 6.15 * = 6 ( the greatest positive even integer less than or equal to 6.15 is 6 ) . hence , 6.15 – 6.15 * = 6.15 - 6 = 0.15 answer : b .\"","correct":"b","options":{"a":"1.5 ","b":"0.15 ","c":"6.25 ","d":"0.25","e":"6.0"},"options_float":{"a":1.5,"b":0.15,"c":6.25,"d":0.25,"e":6.0},"annotated_formula":"subtract(6.15, floor(6.15))","linear_formula":"floor(n0)|subtract(n0,#0)|","chain":"floor(6.15)<\/gadget>\n6<\/output>\n6.15 - 6<\/gadget>\n0.15<\/output>\n0.15<\/result>","index":1391} +{"problem":"when positive integer x is divided by positive integer y , the remainder is 9 . if x \/ y = 96.15 , what is the value of y ?","rationale":"\"guys , one more simple funda . 5 \/ 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 \/ 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 \/ 5 = 6.4 now . 4 x 5 = 2 is the remainder given x \/ y = 96.15 and remainder is 9 so . 15 x y = 9 hence y = 60 ans b\"","correct":"b","options":{"a":"96 ","b":"60 ","c":"48 ","d":"25","e":"12"},"options_float":{"a":96.0,"b":60.0,"c":48.0,"d":25.0,"e":12.0},"annotated_formula":"divide(9, subtract(96.15, floor(96.15)))","linear_formula":"floor(n1)|subtract(n1,#0)|divide(n0,#1)|","chain":"floor(96.15)<\/gadget>\n96<\/output>\n96.15 - 96<\/gadget>\n0.15<\/output>\n9 \/ 0.15<\/gadget>\n60<\/output>\n60<\/result>","index":1392} +{"problem":"the cost of the paint is rs . 40.00 per kg . if 1 kg of paint covers 20 sq . ft , how much will it cost to paint outside of a cube having 10 feet each side","rationale":"\"explanation : surface area of a cube = 6 x 10 ^ 2 = 600 s q . ft quantity of paint required = ( 600 \/ 20 ) = 30 kg cost of painting = 40 x 30 = rs . 1200 answer : a\"","correct":"a","options":{"a":"rs . 1200 ","b":"rs . 672 ","c":"rs . 546 ","d":"rs . 876","e":"none of these"},"options_float":{"a":1200.0,"b":672.0,"c":546.0,"d":876.0,"e":null},"annotated_formula":"multiply(divide(surface_cube(10), 20), 40.00)","linear_formula":"surface_cube(n3)|divide(#0,n2)|multiply(n0,#1)|","chain":"6 * (10 ** 2)<\/gadget>\n600<\/output>\n600 \/ 20<\/gadget>\n30<\/output>\n30 * 40<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":1393} +{"problem":"what is the perimeter of a rectangular field whose diagonal is 10 m and length is 3 m ?","rationale":"sol : breadth of the rectangular plot is = 10 ^ 2 - 3 ^ 2 = 1 m therefore , perimeter of the rectangular plot = 2 ( 3 + 1 ) = 8 m d ) 8 m","correct":"d","options":{"a":"12 m ","b":"5 m ","c":"10 m ","d":"8 m","e":"6 m"},"options_float":{"a":12.0,"b":5.0,"c":10.0,"d":8.0,"e":6.0},"annotated_formula":"divide(add(add(sqrt(subtract(power(10, const_2), power(3, const_2))), 3), add(sqrt(subtract(power(10, const_2), power(3, const_2))), 3)), const_3)","linear_formula":"power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|add(n1,#3)|add(#4,#4)|divide(#5,const_3)","chain":"10 ** 2<\/gadget>\n100<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n100 - 9<\/gadget>\n91<\/output>\n91 ** (1\/2)<\/gadget>\nsqrt(91) = around 9.539392<\/output>\n(sqrt(91)) + 3<\/gadget>\n3 + sqrt(91) = around 12.539392<\/output>\n(3 + sqrt(91)) + (3 + sqrt(91))<\/gadget>\n6 + 2*sqrt(91) = around 25.078784<\/output>\n(6 + 2*sqrt(91)) \/ 3<\/gadget>\n2 + 2*sqrt(91)\/3 = around 8.359595<\/output>\n2 + 2*sqrt(91)\/3 = around 8.359595<\/result>","index":1394} +{"problem":"cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 3 % sugar , what should be the ratio of cereal a to cereal b , by weight ?","rationale":"2 % is 1 % - point below 3 % and 10 % is 7 % - points above 3 % . the ratio of a : b should be 1 : 7 . the answer is c .","correct":"c","options":{"a":"2 : 9 ","b":"2 : 7 ","c":"1 : 7 ","d":"1 : 4","e":"1 : 3"},"options_float":{"a":0.2222222222,"b":0.2857142857,"c":0.1428571429,"d":0.25,"e":0.3333333333},"annotated_formula":"divide(subtract(3, 2), subtract(10, 3))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)","chain":"3 - 2<\/gadget>\n1<\/output>\n10 - 3<\/gadget>\n7<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1\/7 = around 0.142857<\/result>","index":1395} +{"problem":"a 120.00001 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ?","rationale":"\"explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 \/ 18 ) = 500 \/ 9 m \/ sec let the length of other train is x meter then x + 120 \/ 9 = 500 \/ 9 = > x + 120 = 500 = > x = 380 so the length of the train is 380 meters option c\"","correct":"c","options":{"a":"220 meter ","b":"225 meter ","c":"380 meter ","d":"435 meter","e":"none of these"},"options_float":{"a":220.0,"b":225.0,"c":380.0,"d":435.0,"e":null},"annotated_formula":"subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 120.00001)","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)|","chain":"120 + 80<\/gadget>\n200<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n200 * (5\/18)<\/gadget>\n500\/9 = around 55.555556<\/output>\n(500\/9) * 9<\/gadget>\n500<\/output>\n500 - 120.00001<\/gadget>\n379.99999<\/output>\n379.99999<\/result>","index":1397} +{"problem":"the ratio between the length and the breadth of a rectangular plot is 7 : 5 . if the perimeter of the plot is 288 metres , what is its area ?","rationale":"let the length and breadth be 7 x and 5 x respectively . then , p = 2 ( 7 x + 5 x ) = 288 â ‡ ’ x = 12 area = 7 ã — 12 ã — 5 ã — 12 = 5040 sq . m . answer d","correct":"d","options":{"a":"1320 sq . metres ","b":"3260 sq . metres ","c":"4280 sq . metres ","d":"5040 sq . metres","e":"none of these"},"options_float":{"a":1320.0,"b":3260.0,"c":4280.0,"d":5040.0,"e":null},"annotated_formula":"rectangle_area(divide(multiply(7, divide(288, multiply(const_2, add(const_1, divide(7, 5))))), 5), divide(288, multiply(const_2, add(const_1, divide(7, 5)))))","linear_formula":"divide(n0,n1)|add(#0,const_1)|multiply(#1,const_2)|divide(n2,#2)|multiply(n0,#3)|divide(#4,n1)|rectangle_area(#5,#3)","chain":"7 \/ 5<\/gadget>\n7\/5 = around 1.4<\/output>\n1 + (7\/5)<\/gadget>\n12\/5 = around 2.4<\/output>\n2 * (12\/5)<\/gadget>\n24\/5 = around 4.8<\/output>\n288 \/ (24\/5)<\/gadget>\n60<\/output>\n7 * 60<\/gadget>\n420<\/output>\n420 \/ 5<\/gadget>\n84<\/output>\n84 * 60<\/gadget>\n5_040<\/output>\n5_040<\/result>","index":1398} +{"problem":"during the first two weeks of january , the total rainfall in springdale was 20 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall during the second week of january ?","rationale":"\"total rainfall in 2 weeks = 20 inches . assume the rainfall in second week = 1.5 x rainfall in first week = x total rainfall = 2.5 x = 20 inches x = 8 and 1.5 x = 12 rainfall during second week = 12 inches option e\"","correct":"e","options":{"a":"5 inches ","b":"6 inches ","c":"9 inches ","d":"10 inches","e":"12 inches"},"options_float":{"a":5.0,"b":6.0,"c":9.0,"d":10.0,"e":12.0},"annotated_formula":"multiply(20, divide(1.5, add(const_1, 1.5)))","linear_formula":"add(n1,const_1)|divide(n1,#0)|multiply(n0,#1)|","chain":"1 + 1.5<\/gadget>\n2.5<\/output>\n1.5 \/ 2.5<\/gadget>\n0.6<\/output>\n20 * 0.6<\/gadget>\n12<\/output>\n12<\/result>","index":1399} +{"problem":"the set s consists of 5 numbers : { 1 , 2,3 , 4,5 } . if all possible subsets including the null set are created and one subset is chosen at random , what is the probability that the subset has 3 or 4 as its largest number ?","rationale":"the set s has 2 ^ 5 = 32 subsets . the number 5 is in half of these subsets . of the remaining 16 subsets , 4 is an element in 8 of them . thus 4 is the largest number in 8 subsets of s . of the remaining 8 subsets , 3 is an element in 4 of them . thus 3 is the largest number in 4 subsets of s . the probability that 3 or 4 is the largest number is 12 \/ 32 = 3 \/ 8 . the answer is c .","correct":"b","options":{"a":"3 \/ 4 ","b":"3 \/ 8 ","c":"3 \/ 16 ","d":"5 \/ 16","e":"3 \/ 32"},"options_float":{"a":0.75,"b":0.375,"c":0.1875,"d":0.3125,"e":0.09375},"annotated_formula":"divide(multiply(4, 3), power(const_2, 5))","linear_formula":"multiply(n4,n5)|power(const_2,n0)|divide(#0,#1)","chain":"4 * 3<\/gadget>\n12<\/output>\n2 ** 5<\/gadget>\n32<\/output>\n12 \/ 32<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":1400} +{"problem":"in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 \/ 4 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?","rationale":"\"the probability of survival for each of the first 3 months of life is 1 - 1 \/ 4 = 3 \/ 4 , so of 200 newborn 200 * 3 \/ 4 * 3 \/ 4 * 3 \/ 4 = ~ 85 is expected to survive . answer : e .\"","correct":"e","options":{"a":"40 ","b":"46 ","c":"52 ","d":"62","e":"85"},"options_float":{"a":40.0,"b":46.0,"c":52.0,"d":62.0,"e":85.0},"annotated_formula":"multiply(multiply(multiply(200, subtract(1, divide(1, 4))), subtract(1, divide(1, 4))), subtract(1, divide(1, 4)))","linear_formula":"divide(n1,n2)|subtract(n1,#0)|multiply(n3,#1)|multiply(#2,#1)|multiply(#3,#1)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n200 * (3\/4)<\/gadget>\n150<\/output>\n150 * (3\/4)<\/gadget>\n225\/2 = around 112.5<\/output>\n(225\/2) * (3\/4)<\/gadget>\n675\/8 = around 84.375<\/output>\n675\/8 = around 84.375<\/result>","index":1402} +{"problem":"a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 390 kilograms of pears that day , how many kilograms did he sell in the afternoon ?","rationale":"\"3 x = 390 x = 130 therefore , the salesman sold 130 kg in the morning and 2 ⋅ 130 = 260 kg in the afternoon . so answer is c .\"","correct":"c","options":{"a":"120 ","b":"180 ","c":"260 ","d":"280","e":"320"},"options_float":{"a":120.0,"b":180.0,"c":260.0,"d":280.0,"e":320.0},"annotated_formula":"multiply(divide(390, const_3), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_2)|","chain":"390 \/ 3<\/gadget>\n130<\/output>\n130 * 2<\/gadget>\n260<\/output>\n260<\/result>","index":1404} +{"problem":"if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability q that this product will be a multiple of 45 ?","rationale":"\"there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability q = 1 \/ 24 . answer c .\"","correct":"c","options":{"a":"1 \/ 32 ","b":"1 \/ 28 ","c":"1 \/ 24 ","d":"1 \/ 16","e":"1 \/ 14"},"options_float":{"a":0.03125,"b":0.0357142857,"c":0.0416666667,"d":0.0625,"e":0.0714285714},"annotated_formula":"divide(const_1, add(20, const_4))","linear_formula":"add(n1,const_4)|divide(const_1,#0)|","chain":"20 + 4<\/gadget>\n24<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n1\/24 = around 0.041667<\/result>","index":1405} +{"problem":"a certain high school has 500 students . of these students , 20 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ?","rationale":"\"we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 20 y = 20 z = 10 so , we have 20 students taking music , 20 taking art and 10 taking both music and art . that 10 person has been counted twice though ( once in the music ' group ' and once in the art ' group ' ) , so what we really have is . . . 10 student taking just music 10 student taking just art 10 student taking both music and art total = 30 students we ' re asked for the total number of students who are taking neither course . that is 5000 - 30 = 470 . e\"","correct":"e","options":{"a":"430 ","b":"440 ","c":"450 ","d":"460","e":"470"},"options_float":{"a":430.0,"b":440.0,"c":450.0,"d":460.0,"e":470.0},"annotated_formula":"subtract(500, subtract(add(20, 20), 10))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|","chain":"20 + 20<\/gadget>\n40<\/output>\n40 - 10<\/gadget>\n30<\/output>\n500 - 30<\/gadget>\n470<\/output>\n470<\/result>","index":1406} +{"problem":"a courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof . what could be the weight in grams of a package for which the charge is $ 2.15 ?","rationale":"\"the charge is 65 cents for the first 250 grams . this leaves a charge of $ 2.15 - $ 0.65 = $ 1.50 the charge for the next 1400 grams is $ 1.40 which leaves a charge of $ 0.10 the weight is somewhere between 1650 and 1750 . the answer is e .\"","correct":"e","options":{"a":"980 ","b":"1120 ","c":"1260 ","d":"1420","e":"1670"},"options_float":{"a":980.0,"b":1120.0,"c":1260.0,"d":1420.0,"e":1670.0},"annotated_formula":"divide(add(subtract(2.15, divide(65, 100)), multiply(divide(divide(10, 100), 100), 250)), divide(divide(10, 100), 100))","linear_formula":"divide(n2,n3)|divide(n0,n3)|divide(#0,n3)|subtract(n4,#1)|multiply(n1,#2)|add(#4,#3)|divide(#5,#2)|","chain":"65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n2.15 - (13\/20)<\/gadget>\n1.5<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) \/ 100<\/gadget>\n1\/1_000 = around 0.001<\/output>\n(1\/1_000) * 250<\/gadget>\n1\/4 = around 0.25<\/output>\n1.5 + (1\/4)<\/gadget>\n1.75<\/output>\n1.75 \/ (1\/1_000)<\/gadget>\n1_750<\/output>\n1_750<\/result>","index":1407} +{"problem":"if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 1 year the amount to be paid is ?","rationale":"\"a = 7500 ( 26 \/ 25 ) ^ 1 = 7800 answer : d\"","correct":"d","options":{"a":"3377 ","b":"2688 ","c":"2688 ","d":"7800","e":"1268"},"options_float":{"a":3377.0,"b":2688.0,"c":2688.0,"d":7800.0,"e":1268.0},"annotated_formula":"multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100)))","linear_formula":"add(n1,const_100)|divide(#0,const_100)|multiply(#1,#1)|multiply(n0,#2)|","chain":"100 + 4<\/gadget>\n104<\/output>\n104 \/ 100<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) * (26\/25)<\/gadget>\n676\/625 = around 1.0816<\/output>\n7_500 * (676\/625)<\/gadget>\n8_112<\/output>\n8_112<\/result>","index":1408} +{"problem":"a car gets 24 miles to the gallon . if it is modified to use a solar panel , it will use only 75 percent as much fuel as it does now . if the fuel tank holds 12 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ?","rationale":"\"originally , the distance the car could go on a full tank was 12 * 24 = 288 miles . after it has been modified , the car can go 24 \/ 0.75 = 32 miles per gallon . on a full tank , the car can go 32 * 12 = 384 miles , thus 96 miles more . the answer is a .\"","correct":"a","options":{"a":"96 ","b":"100 ","c":"104 ","d":"108","e":"112"},"options_float":{"a":96.0,"b":100.0,"c":104.0,"d":108.0,"e":112.0},"annotated_formula":"subtract(multiply(multiply(divide(24, 75), const_100), 12), multiply(24, 12))","linear_formula":"divide(n0,n1)|multiply(n0,n2)|multiply(#0,const_100)|multiply(n2,#2)|subtract(#3,#1)|","chain":"24 \/ 75<\/gadget>\n8\/25 = around 0.32<\/output>\n(8\/25) * 100<\/gadget>\n32<\/output>\n32 * 12<\/gadget>\n384<\/output>\n24 * 12<\/gadget>\n288<\/output>\n384 - 288<\/gadget>\n96<\/output>\n96<\/result>","index":1411} +{"problem":"the h . c . f . of two numbers is 15 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is :","rationale":"\"the numbers are ( 15 x 11 ) and ( 15 x 15 ) . larger number = ( 15 x 15 ) = 225 . answer : a\"","correct":"a","options":{"a":"225 ","b":"300 ","c":"299 ","d":"322","e":"345"},"options_float":{"a":225.0,"b":300.0,"c":299.0,"d":322.0,"e":345.0},"annotated_formula":"multiply(15, 15)","linear_formula":"multiply(n0,n2)|","chain":"15 * 15<\/gadget>\n225<\/output>\n225<\/result>","index":1412} +{"problem":"a metallic sheet is of rectangular shape with dimensions 46 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is :","rationale":"\"l = ( 46 - 16 ) m = 30 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 30 x 20 x 8 ) m 3 = 4800 m 3 . answer : option d\"","correct":"d","options":{"a":"4830 ","b":"5120 ","c":"6420 ","d":"4800","e":"8960"},"options_float":{"a":4830.0,"b":5120.0,"c":6420.0,"d":4800.0,"e":8960.0},"annotated_formula":"volume_rectangular_prism(subtract(46, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8)","linear_formula":"multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|","chain":"8 * 2<\/gadget>\n16<\/output>\n46 - 16<\/gadget>\n30<\/output>\n36 - 16<\/gadget>\n20<\/output>\n30 * 20 * 8<\/gadget>\n4_800<\/output>\n4_800<\/result>","index":1414} +{"problem":"in an election between two candidates , one got 60 % of the total valid votes , 30 % of the votes were invalid . if the total number of votes was 9000 , the number of valid votes that the other candidate got , was :","rationale":"\"explanation : number of valid votes = 70 % of 9000 = 6300 . valid votes polled by other candidates = 40 % of 6300 ( 40 \/ 1006300 ) = 2520 . harsh mishra a year ago 0 upvotes answer : b\"","correct":"b","options":{"a":"2500 ","b":"2520 ","c":"3000 ","d":"3100","e":"none of these"},"options_float":{"a":2500.0,"b":2520.0,"c":3000.0,"d":3100.0,"e":null},"annotated_formula":"multiply(multiply(subtract(const_1, divide(30, const_100)), subtract(const_1, divide(60, const_100))), 9000)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(7\/10) * (2\/5)<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) * 9_000<\/gadget>\n2_520<\/output>\n2_520<\/result>","index":1415} +{"problem":"on a certain road , 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ?","rationale":"suppose there are x motorists . 10 % of them exceeded the speed limit and received the ticket , i . e . x \/ 10 . again , suppose total no . of motorists who exceeded the speed limit are y . 20 % of y exceeded the speed limit but did n ' t received the ticket , i . e . y \/ 5 . it means 4 y \/ 5 received the ticket . hence , 4 y \/ 5 = x \/ 10 or y \/ x = 1 \/ 8 or y \/ x * 100 = 1 \/ 8 * 100 = 12.5 % answer : b","correct":"b","options":{"a":"10.5 % ","b":"12.5 % ","c":"15 % ","d":"22 %","e":"30 %"},"options_float":{"a":10.5,"b":12.5,"c":15.0,"d":22.0,"e":30.0},"annotated_formula":"multiply(divide(10, subtract(const_100, 20)), const_100)","linear_formula":"subtract(const_100,n1)|divide(n0,#0)|multiply(#1,const_100)","chain":"100 - 20<\/gadget>\n80<\/output>\n10 \/ 80<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":1416} +{"problem":"x and y are two towns . ganesh covers the distance from x to y at an average speed of 60 km \/ hr . however , he covers the distance from y to x at an average speed of 30 km \/ hr . his average speed during the whole journey in km \/ hr . is :","rationale":"\"solution : average speed = 2 xy \/ x + y = 2 * 60 * 30 \/ 60 + 30 = 40 answer : c\"","correct":"c","options":{"a":"34 ","b":"43 ","c":"40 ","d":"38","e":"29"},"options_float":{"a":34.0,"b":43.0,"c":40.0,"d":38.0,"e":29.0},"annotated_formula":"divide(add(add(subtract(60, const_10), const_100), add(subtract(60, const_10), const_100)), add(divide(add(subtract(60, const_10), const_100), 60), divide(add(subtract(60, const_10), const_100), 30)))","linear_formula":"subtract(n0,const_10)|add(#0,const_100)|add(#1,#1)|divide(#1,n0)|divide(#1,n1)|add(#3,#4)|divide(#2,#5)|","chain":"60 - 10<\/gadget>\n50<\/output>\n50 + 100<\/gadget>\n150<\/output>\n150 + 150<\/gadget>\n300<\/output>\n150 \/ 60<\/gadget>\n5\/2 = around 2.5<\/output>\n150 \/ 30<\/gadget>\n5<\/output>\n(5\/2) + 5<\/gadget>\n15\/2 = around 7.5<\/output>\n300 \/ (15\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":1417} +{"problem":"a drink vendor has 20 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ?","rationale":"\"the number of liters in each can = hcf of 20 , 144 and 368 = 4 liters . number of cans of maaza = 20 \/ 4 = 5 number of cans of pepsi = 144 \/ 4 = 36 number of cans of sprite = 368 \/ 4 = 92 the total number of cans required = 5 + 36 + 92 = 133 cans . answer : e\"","correct":"e","options":{"a":"135 ","b":"137 ","c":"142 ","d":"130","e":"133"},"options_float":{"a":135.0,"b":137.0,"c":142.0,"d":130.0,"e":133.0},"annotated_formula":"add(divide(368, gcd(gcd(20, 144), 368)), add(divide(20, gcd(gcd(20, 144), 368)), divide(144, gcd(gcd(20, 144), 368))))","linear_formula":"gcd(n0,n1)|gcd(n2,#0)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)|","chain":"gcd(20, 144)<\/gadget>\n4<\/output>\ngcd(4, 368)<\/gadget>\n4<\/output>\n368 \/ 4<\/gadget>\n92<\/output>\n20 \/ 4<\/gadget>\n5<\/output>\n144 \/ 4<\/gadget>\n36<\/output>\n5 + 36<\/gadget>\n41<\/output>\n92 + 41<\/gadget>\n133<\/output>\n133<\/result>","index":1418} +{"problem":"jonathan can type a 40 page document in 40 minutes , susan can type it in 30 minutes , and jack can type it in 24 minutes . working together , how much time will it take them to type the same document ?","rationale":"\"you may set up common equation like this : job \/ a + job \/ b + job \/ c = job \/ x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 40 \/ 40 + 40 \/ 30 + 40 \/ 24 = 40 \/ x if you solve this equation , you get the same answer b ( 10 )\"","correct":"b","options":{"a":"5 minutes ","b":"10 minutes ","c":"15 minutes ","d":"18 minutes","e":"20 minutes"},"options_float":{"a":5.0,"b":10.0,"c":15.0,"d":18.0,"e":20.0},"annotated_formula":"divide(40, add(divide(40, 24), add(divide(40, 40), divide(40, 30))))","linear_formula":"divide(n0,n1)|divide(n0,n2)|divide(n0,n3)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|","chain":"40 \/ 24<\/gadget>\n5\/3 = around 1.666667<\/output>\n40 \/ 40<\/gadget>\n1<\/output>\n40 \/ 30<\/gadget>\n4\/3 = around 1.333333<\/output>\n1 + (4\/3)<\/gadget>\n7\/3 = around 2.333333<\/output>\n(5\/3) + (7\/3)<\/gadget>\n4<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n10<\/result>","index":1419} +{"problem":"find the least number must be subtracted from 62575 so that remaining number is divisible by 99 .","rationale":"on dividing 62571 by 99 we get the remainder 7 , so 7 should be subtracted . answer is b","correct":"b","options":{"a":"93 ","b":"7 ","c":"54 ","d":"9","e":"6"},"options_float":{"a":93.0,"b":7.0,"c":54.0,"d":9.0,"e":6.0},"annotated_formula":"subtract(62575, multiply(floor(divide(62575, 99)), 99))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)","chain":"62_575 \/ 99<\/gadget>\n62_575\/99 = around 632.070707<\/output>\nfloor(62_575\/99)<\/gadget>\n632<\/output>\n632 * 99<\/gadget>\n62_568<\/output>\n62_575 - 62_568<\/gadget>\n7<\/output>\n7<\/result>","index":1420} +{"problem":"a man is 12 years older than his brother . in two years , his age will be twice the age of his brother . the present age of his brother is :","rationale":"let the brother ' s present age be x years . then , man ' s present age = ( x + 12 ) years . ( x + 12 ) + 2 = 2 ( x + 2 ) x + 14 = 2 x + 4 x = 10 . answer : a","correct":"a","options":{"a":"10 ","b":"21 ","c":"22 ","d":"23","e":"24"},"options_float":{"a":10.0,"b":21.0,"c":22.0,"d":23.0,"e":24.0},"annotated_formula":"subtract(add(12, const_2), multiply(const_2, const_2))","linear_formula":"add(n0,const_2)|multiply(const_2,const_2)|subtract(#0,#1)","chain":"12 + 2<\/gadget>\n14<\/output>\n2 * 2<\/gadget>\n4<\/output>\n14 - 4<\/gadget>\n10<\/output>\n10<\/result>","index":1421} +{"problem":"a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 280 pounds of the mixture ?","rationale":"\"almonds : walnuts = 5 : 2 total mixture has 7 parts in a 280 pound mixture , almonds are 5 \/ 7 ( total mixture ) = 5 \/ 7 * 280 = 200 pounds answer ( a )\"","correct":"a","options":{"a":"200 ","b":"84 ","c":"40 ","d":"28","e":"20"},"options_float":{"a":200.0,"b":84.0,"c":40.0,"d":28.0,"e":20.0},"annotated_formula":"divide(multiply(5, 280), add(5, 2))","linear_formula":"add(n0,n1)|multiply(n0,n2)|divide(#1,#0)|","chain":"5 * 280<\/gadget>\n1_400<\/output>\n5 + 2<\/gadget>\n7<\/output>\n1_400 \/ 7<\/gadget>\n200<\/output>\n200<\/result>","index":1422} +{"problem":"how many 4 - digit numbers that do not contain the digits 2 , 3 , 4 , 5 , 6 , or 7 are there ?","rationale":"the 1 st digit can be filled up by the numbers : { 1 , 8,9 } = 3 ways the 2 nd digit can be filled up by the numbers : { 0,1 , 8,9 } = 4 ways the 3 rd digit can be filled up by the numbers : { 0,1 , 8,9 } = 4 ways the 4 th digit can be filled up by the numbers : { 0,1 , 8,9 } = 4 ways the total number of such four - digit numbers is 3 * 4 * 4 * 4 = 192 the answer is d .","correct":"d","options":{"a":"168 ","b":"176 ","c":"184 ","d":"192","e":"200"},"options_float":{"a":168.0,"b":176.0,"c":184.0,"d":192.0,"e":200.0},"annotated_formula":"multiply(multiply(multiply(3, 4), 4), 4)","linear_formula":"multiply(n0,n2)|multiply(n0,#0)|multiply(n0,#1)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 * 4<\/gadget>\n48<\/output>\n48 * 4<\/gadget>\n192<\/output>\n192<\/result>","index":1423} +{"problem":"how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ?","rationale":"\"solution number of bricks = volume of the wall \/ volume of 1 brick = ( 800 x 600 x 22.5 \/ 25 x 11.25 x 6 ) = 6400 . answer c\"","correct":"c","options":{"a":"5600 ","b":"6000 ","c":"6400 ","d":"7200","e":"none"},"options_float":{"a":5600.0,"b":6000.0,"c":6400.0,"d":7200.0,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6))","linear_formula":"multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|","chain":"8 * 100<\/gadget>\n800<\/output>\n6 * 100<\/gadget>\n600<\/output>\n800 * 600<\/gadget>\n480_000<\/output>\n480_000 * 22.5<\/gadget>\n10_800_000<\/output>\n25 * 11.25<\/gadget>\n281.25<\/output>\n281.25 * 6<\/gadget>\n1_687.5<\/output>\n10_800_000 \/ 1_687.5<\/gadget>\n6_400<\/output>\n6_400<\/result>","index":1424} +{"problem":"the average weight of 6 person ' s increases by 1.8 kg when a new person comes in place of one of them weighing 69 kg . what might be the weight of the new person ?","rationale":"\"total weight increased = ( 6 x 1.8 ) kg = 10.8 kg . weight of new person = ( 69 + 10.8 ) kg = 79.8 kg . answer : option b\"","correct":"b","options":{"a":"73 kg ","b":"79.8 kg ","c":"86 kg ","d":"data inadequate","e":"none of these"},"options_float":{"a":73.0,"b":79.8,"c":86.0,"d":null,"e":null},"annotated_formula":"add(multiply(6, 1.8), 69)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"6 * 1.8<\/gadget>\n10.8<\/output>\n10.8 + 69<\/gadget>\n79.8<\/output>\n79.8<\/result>","index":1425} +{"problem":"a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 7200 interest . the principal amount borrowed by him was .","rationale":"\"explanation : s . i . = p â ˆ — r â ˆ — t \/ 100 = > p = s . i . â ˆ — 100 \/ r â ˆ — t = > p = 7200 â ˆ — 100 \/ 12 â ˆ — 3 = rs 20000 option d\"","correct":"d","options":{"a":"rs 14000 ","b":"rs 15000 ","c":"rs 16000 ","d":"rs 20000","e":"none of these"},"options_float":{"a":14000.0,"b":15000.0,"c":16000.0,"d":20000.0,"e":null},"annotated_formula":"divide(multiply(7200, const_100), multiply(12, 3))","linear_formula":"multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|","chain":"7_200 * 100<\/gadget>\n720_000<\/output>\n12 * 3<\/gadget>\n36<\/output>\n720_000 \/ 36<\/gadget>\n20_000<\/output>\n20_000<\/result>","index":1426} +{"problem":"a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1986 on her 30 nd birthday first be eligible to retire under this provision ?","rationale":"\"she must gain at least 70 points , now she has 30 and every year gives her two more points : one for age and one for additional year of employment , so 30 + 2 * ( # of years ) = 70 - - > ( # of years ) = 20 - - > 1986 + 20 = 2006 . answer : d .\"","correct":"d","options":{"a":"2003 ","b":"2004 ","c":"2005 ","d":"2006","e":"2007"},"options_float":{"a":2003.0,"b":2004.0,"c":2005.0,"d":2006.0,"e":2007.0},"annotated_formula":"add(1986, divide(subtract(70, 30), const_2))","linear_formula":"subtract(n0,n3)|divide(#0,const_2)|add(n2,#1)|","chain":"70 - 30<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n1_986 + 20<\/gadget>\n2_006<\/output>\n2_006<\/result>","index":1427} +{"problem":"i had rs . 10000 with me . out of this , money i lent some money to a for 2 years at 15 % simple interest . i lent the remaining money to b for an equal number of years at 18 % simple interest . after 2 years , i found that a had given me rs . 360 more as interest as compared to ( b ) the amount of money which i had lent to b must have been","rationale":"explanation : total amount = rs . 10000 let money lent a = rs . x and money lent to b = rs . ( 10000 x ) x = rs . 6000 . so the amount of money lent to b = rs . 4000 answer : c","correct":"c","options":{"a":"2000 ","b":"3000 ","c":"4000 ","d":"5000","e":"6000"},"options_float":{"a":2000.0,"b":3000.0,"c":4000.0,"d":5000.0,"e":6000.0},"annotated_formula":"divide(subtract(multiply(10000, divide(multiply(2, 15), const_100)), 360), add(divide(multiply(2, 15), const_100), divide(multiply(2, 18), const_100)))","linear_formula":"multiply(n1,n2)|multiply(n1,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|multiply(n0,#2)|subtract(#5,n5)|divide(#6,#4)","chain":"2 * 15<\/gadget>\n30<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n10_000 * (3\/10)<\/gadget>\n3_000<\/output>\n3_000 - 360<\/gadget>\n2_640<\/output>\n2 * 18<\/gadget>\n36<\/output>\n36 \/ 100<\/gadget>\n9\/25 = around 0.36<\/output>\n(3\/10) + (9\/25)<\/gadget>\n33\/50 = around 0.66<\/output>\n2_640 \/ (33\/50)<\/gadget>\n4_000<\/output>\n4_000<\/result>","index":1428} +{"problem":"to be considered for “ movie of the year , ” a film must appear in at least 1 \/ 4 of the top - 10 - movies lists submitted by the cinematic academy ’ s 785 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for “ movie of the year ” ?","rationale":"\"total movies submitted are 785 . as per question we need to take 1 \/ 4 of 785 to be considered for top 10 movies = 196.25 approximate the value we 197 . imo option b is the correct answer . . .\"","correct":"b","options":{"a":"191 ","b":"197 ","c":"193 ","d":"212","e":"213"},"options_float":{"a":191.0,"b":197.0,"c":193.0,"d":212.0,"e":213.0},"annotated_formula":"divide(785, 4)","linear_formula":"divide(n3,n1)|","chain":"785 \/ 4<\/gadget>\n785\/4 = around 196.25<\/output>\n785\/4 = around 196.25<\/result>","index":1430} +{"problem":"the number 0.5 is how much greater than 1 \/ 3 ?","rationale":"\"let x be the difference then . 5 - 1 \/ 3 = x 5 \/ 10 - 1 \/ 3 = x x = 1 \/ 6 ans a\"","correct":"a","options":{"a":"1 \/ 6 ","b":"2 \/ 10 ","c":"1 \/ 50 ","d":"1 \/ 500","e":"2 \/ 500"},"options_float":{"a":0.1666666667,"b":0.2,"c":0.02,"d":0.002,"e":0.004},"annotated_formula":"subtract(0.5, divide(1, 3))","linear_formula":"divide(n1,n2)|subtract(n0,#0)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n0.5 - (1\/3)<\/gadget>\n0.166667<\/output>\n0.166667<\/result>","index":1431} +{"problem":"the concentration of spirit in 3 different vessels a , b and c are 45 % , 30 % and 10 % respectively . if 4 litres from vessel a , 5 litres from vessel b and 6 litres from vessel c are mixed , find the concentration of spirit in the resultant solution .","rationale":"quantity of spirit in new solution = ( 4 * 45 \/ 100 ) + ( 5 * 30 \/ 100 ) + ( 6 * 10 \/ 100 ) = 1.8 + 1.5 + 0.6 = 3.9 liters so % concentration of spirit = 100 * 3.9 \/ ( 4 + 5 + 6 ) = 26 % answer : a","correct":"a","options":{"a":"26 % ","b":"33 % ","c":"34 % ","d":"35 %","e":"24 %"},"options_float":{"a":26.0,"b":33.0,"c":34.0,"d":35.0,"e":24.0},"annotated_formula":"divide(multiply(add(add(divide(multiply(4, 45), const_100), divide(multiply(30, 5), const_100)), divide(multiply(10, 6), const_100)), const_100), add(10, 5))","linear_formula":"add(n3,n5)|multiply(n1,n4)|multiply(n2,n5)|multiply(n3,n6)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|add(#4,#5)|add(#7,#6)|multiply(#8,const_100)|divide(#9,#0)","chain":"4 * 45<\/gadget>\n180<\/output>\n180 \/ 100<\/gadget>\n9\/5 = around 1.8<\/output>\n30 * 5<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n(9\/5) + (3\/2)<\/gadget>\n33\/10 = around 3.3<\/output>\n10 * 6<\/gadget>\n60<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(33\/10) + (3\/5)<\/gadget>\n39\/10 = around 3.9<\/output>\n(39\/10) * 100<\/gadget>\n390<\/output>\n10 + 5<\/gadget>\n15<\/output>\n390 \/ 15<\/gadget>\n26<\/output>\n26<\/result>","index":1432} +{"problem":"there are 2 sections a and b in a class , consisting of 50 and 70 students respectively . if the average weight of section a is 50 kg and that of section b is 70 kg , find the average of the whole class ?","rationale":"\"total weight of 50 + 70 students = 50 * 50 + 70 * 70 = 7400 average weight of the class is = 7400 \/ 120 = 61.67 kg answer is c\"","correct":"c","options":{"a":"50.78 kg ","b":"49.32 kg ","c":"61.67 kg ","d":"69.15 kg","e":"70.89 kg"},"options_float":{"a":50.78,"b":49.32,"c":61.67,"d":69.15,"e":70.89},"annotated_formula":"divide(add(multiply(50, 50), multiply(70, 70)), add(50, 70))","linear_formula":"add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)|","chain":"50 * 50<\/gadget>\n2_500<\/output>\n70 * 70<\/gadget>\n4_900<\/output>\n2_500 + 4_900<\/gadget>\n7_400<\/output>\n50 + 70<\/gadget>\n120<\/output>\n7_400 \/ 120<\/gadget>\n185\/3 = around 61.666667<\/output>\n185\/3 = around 61.666667<\/result>","index":1433} +{"problem":"if x ^ 2 – 3 x = 18 , then one possible value of x – 4 =","rationale":"x ^ 2 – 3 x = 18 i . e . x ^ 2 – 3 x - 18 = 0 i . e . x ^ 2 – 6 x + 3 x - 18 = 0 i . e . ( x - 6 ) ( x + 3 ) = 0 i . e . x = 6 or - 3 i . e . x - 4 = 6 - 4 or - 3 - 4 i . e . x - 4 = 2 or - 7 answer : option e","correct":"e","options":{"a":"- 9 ","b":"- 5 ","c":"- 3 ","d":"- 1","e":"2"},"options_float":{"a":-9.0,"b":-5.0,"c":-3.0,"d":-1.0,"e":2.0},"annotated_formula":"subtract(3, power(const_1, 2))","linear_formula":"power(const_1,n0)|subtract(n1,#0)","chain":"1 ** 2<\/gadget>\n1<\/output>\n3 - 1<\/gadget>\n2<\/output>\n2<\/result>","index":1435} +{"problem":"if 4 x + y = 8 and 3 x - 4 y = 5 , then what is the value of 7 x - 3 y ?","rationale":"4 x + y = 8 . . . equation 1 3 x - 4 y = 5 . . . equation 2 adding both the equations 7 x - 3 y = 23 correct answer option a","correct":"a","options":{"a":"23 ","b":"3 ","c":"15 ","d":"52 \/ 7","e":"60 \/ 7"},"options_float":{"a":23.0,"b":3.0,"c":15.0,"d":7.4285714286,"e":8.5714285714},"annotated_formula":"add(add(8, 5), const_10)","linear_formula":"add(n1,n4)|add(#0,const_10)","chain":"8 + 5<\/gadget>\n13<\/output>\n13 + 10<\/gadget>\n23<\/output>\n23<\/result>","index":1436} +{"problem":"working alone , a can complete a certain kind of job in 3 hours . a and d , working together at their respective rates , can complete one of these jobs in 2 hours . in how many hours can d , working alone , complete one of these jobs ?","rationale":"let total time taken by d to complete the job = d total time taken by a to complete the job = 3 work done by a in an hour 1 \/ a = 1 \/ 3 working together a and d can complete the job in 2 hours 1 \/ a + 1 \/ d = 1 \/ 2 = > 1 \/ d = 1 \/ 2 - 1 \/ a = 1 \/ 2 - 1 \/ 3 = 1 \/ 6 = > d = 6 hours answer c","correct":"c","options":{"a":"2 ","b":"4 ","c":"6 ","d":"8","e":"10"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":8.0,"e":10.0},"annotated_formula":"inverse(subtract(inverse(2), inverse(3)))","linear_formula":"inverse(n1)|inverse(n0)|subtract(#0,#1)|inverse(#2)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/2) - (1\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ (1\/6)<\/gadget>\n6<\/output>\n6<\/result>","index":1438} +{"problem":"a class of boys stands in a single line , one boy is 19 th in order from both the ends , how many boys are there in the class ?","rationale":"explanation : number of boys in the class = ( 18 + 1 + 18 ) = 37 answer : a ) 37","correct":"a","options":{"a":"37 ","b":"38 ","c":"27 ","d":"29","e":"11"},"options_float":{"a":37.0,"b":38.0,"c":27.0,"d":29.0,"e":11.0},"annotated_formula":"subtract(multiply(19, const_2), const_1)","linear_formula":"multiply(n0,const_2)|subtract(#0,const_1)","chain":"19 * 2<\/gadget>\n38<\/output>\n38 - 1<\/gadget>\n37<\/output>\n37<\/result>","index":1439} +{"problem":"what number must be added to 6 , 16 and 8 to get an average of 13 ?","rationale":"b 22","correct":"b","options":{"a":"23 ","b":"22 ","c":"44 ","d":"33","e":"55"},"options_float":{"a":23.0,"b":22.0,"c":44.0,"d":33.0,"e":55.0},"annotated_formula":"subtract(multiply(13, const_4), add(8, add(6, 16)))","linear_formula":"add(n0,n1)|multiply(n3,const_4)|add(n2,#0)|subtract(#1,#2)","chain":"13 * 4<\/gadget>\n52<\/output>\n6 + 16<\/gadget>\n22<\/output>\n8 + 22<\/gadget>\n30<\/output>\n52 - 30<\/gadget>\n22<\/output>\n22<\/result>","index":1440} +{"problem":"the average age of 30 students in a class is 15 years . if the age of teacher is also included , the average becomes 16 years , find the age of the teacher .","rationale":"\"explanation : if teacher ' s age is 15 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 16 . age of teacher = average age of all + total increase in age = 16 + ( 1 x 30 ) = 46 years answer : a\"","correct":"a","options":{"a":"46 ","b":"55 ","c":"43 ","d":"39","e":"54"},"options_float":{"a":46.0,"b":55.0,"c":43.0,"d":39.0,"e":54.0},"annotated_formula":"subtract(add(add(multiply(30, 15), 16), 30), multiply(30, 15))","linear_formula":"multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)|","chain":"30 * 15<\/gadget>\n450<\/output>\n450 + 16<\/gadget>\n466<\/output>\n466 + 30<\/gadget>\n496<\/output>\n496 - 450<\/gadget>\n46<\/output>\n46<\/result>","index":1441} +{"problem":"if x = the product of 7 distinct prime numbers , how many factors does x have besides 1 and itself ?","rationale":"since x has 7 distinct prime factors , x has a total of 2 ^ 7 = 128 factors . besides 1 and itself , x has 126 factors . the answer is d .","correct":"d","options":{"a":"108 ","b":"114 ","c":"120 ","d":"126","e":"132"},"options_float":{"a":108.0,"b":114.0,"c":120.0,"d":126.0,"e":132.0},"annotated_formula":"multiply(multiply(multiply(multiply(multiply(multiply(add(1, 1), add(1, 1)), add(1, 1)), add(1, 1)), add(1, 1)), add(1, 1)), add(1, 1))","linear_formula":"add(n1,n1)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)|multiply(#0,#3)|multiply(#0,#4)|multiply(#0,#5)","chain":"1 + 1<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16 * 2<\/gadget>\n32<\/output>\n32 * 2<\/gadget>\n64<\/output>\n64 * 2<\/gadget>\n128<\/output>\n128<\/result>","index":1442} +{"problem":"how many integers n are prime numbers in the range 200 < n < 220 ?","rationale":"to test a number n on whether it is a prime number , take square root ( n ) . consider as it is if it is a natural number . otherwise , increase the sq . root to the next natural no . then divide given no . by all prime numbers below the sq . root obtained . if the no . is divisible by any of the prime numbers , then it is not a prime number ; else it is . square root of 200 or 220 is less than 15 so , lets check if any number between 200 and 220 is not divisible by prime numbers 2 , 35 , 711 ( < 15 ) . that ' s it . out : 201 , 202 , 203 , 204 , 205 , 206 , 207 , 208 , 209 , 210 , 212 , 213 , 214 , 215 , 216 , 217 , 218 , 219 only 211 is a prime number in the given range correct option : a","correct":"a","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"divide(200, 200)","linear_formula":"divide(n0,n0)","chain":"200 \/ 200<\/gadget>\n1<\/output>\n1<\/result>","index":1443} +{"problem":"a sum of $ 800 amounts to $ 920 in 3 years at simple interest . if the interest rate is increased by 3 % it would amount to how much ?","rationale":"s . i = 920 - 800 = 120 p = $ 800 t = 3 years r = 100 * 120 \/ 800 * 3 = 5 % new rate = 5 + 3 = 8 % new s . i . = 800 * 8 * 3 \/ 100 = $ 192 new amount = 800 + 192 = $ 992 answer is c","correct":"c","options":{"a":"$ 506 ","b":"$ 774 ","c":"$ 992 ","d":"$ 854","e":"$ 1021"},"options_float":{"a":506.0,"b":774.0,"c":992.0,"d":854.0,"e":1021.0},"annotated_formula":"add(800, multiply(multiply(800, divide(add(divide(multiply(divide(subtract(920, 800), 3), const_100), 800), 3), const_100)), 3))","linear_formula":"subtract(n1,n0)|divide(#0,n2)|multiply(#1,const_100)|divide(#2,n0)|add(n2,#3)|divide(#4,const_100)|multiply(n0,#5)|multiply(n2,#6)|add(n0,#7)","chain":"920 - 800<\/gadget>\n120<\/output>\n120 \/ 3<\/gadget>\n40<\/output>\n40 * 100<\/gadget>\n4_000<\/output>\n4_000 \/ 800<\/gadget>\n5<\/output>\n5 + 3<\/gadget>\n8<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n800 * (2\/25)<\/gadget>\n64<\/output>\n64 * 3<\/gadget>\n192<\/output>\n800 + 192<\/gadget>\n992<\/output>\n992<\/result>","index":1444} +{"problem":"a copy machine , working at a constant rate , makes 35 copies per minute . a second copy machine , working at a constant rate , makes 65 copies per minute . working together at their respective rates , how many copies do the two machines make in half an hour ?","rationale":"\"together the two machines make 35 + 65 = 100 copies per minute . so , in half an hour they will make 100 * 30 = 3,000 copies . answer : a .\"","correct":"a","options":{"a":"3,000 ","b":"2,700 ","c":"4,500 ","d":"5,400","e":"324,000"},"options_float":{"a":3000.0,"b":2700.0,"c":4500.0,"d":5400.0,"e":324000.0},"annotated_formula":"divide(multiply(add(35, 65), const_60), const_2)","linear_formula":"add(n0,n1)|multiply(#0,const_60)|divide(#1,const_2)|","chain":"35 + 65<\/gadget>\n100<\/output>\n100 * 60<\/gadget>\n6_000<\/output>\n6_000 \/ 2<\/gadget>\n3_000<\/output>\n3_000<\/result>","index":1445} +{"problem":"10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 4554 , how much was it in the beginning ?","rationale":"\"x * ( 90 \/ 100 ) * ( 80 \/ 100 ) = 4554 x = 6325 answer : c\"","correct":"c","options":{"a":"7349 ","b":"6300 ","c":"6325 ","d":"6939","e":"2989"},"options_float":{"a":7349.0,"b":6300.0,"c":6325.0,"d":6939.0,"e":2989.0},"annotated_formula":"floor(divide(4554, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))))","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(9\/10) * (4\/5)<\/gadget>\n18\/25 = around 0.72<\/output>\n4_554 \/ (18\/25)<\/gadget>\n6_325<\/output>\nfloor(6_325)<\/gadget>\n6_325<\/output>\n6_325<\/result>","index":1448} +{"problem":"cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost w is what percent of the original cost ?","rationale":"\"original cost w , c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600\"","correct":"d","options":{"a":"w = 200 ","b":"w = 600 ","c":"w = 800 ","d":"w = 1600","e":"w = 50"},"options_float":{"a":200.0,"b":600.0,"c":800.0,"d":1600.0,"e":50.0},"annotated_formula":"multiply(power(const_2, 4), const_100)","linear_formula":"power(const_2,n0)|multiply(#0,const_100)|","chain":"2 ** 4<\/gadget>\n16<\/output>\n16 * 100<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":1449} +{"problem":"if a car went the first third of the distance at 80 kmh , the second third at 30 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ?","rationale":"assume d \/ 3 = 240 ( this number is convenient because it is divisible by 80 , 30 and 48 ) so : 240 = 80 * t 1 = 3 hrs 240 = 30 * t 2 = 8 hrs 240 = 48 * t 3 = 5 hrs t = t 1 + t 2 + t 3 = 16 hrs d = rt ( 240 * 3 ) = r * 16 r = 45 answer : d","correct":"d","options":{"a":"36 kmh ","b":"40 kmh ","c":"42 kmh ","d":"45 kmh","e":"50 kmh"},"options_float":{"a":36.0,"b":40.0,"c":42.0,"d":45.0,"e":50.0},"annotated_formula":"divide(const_3, add(add(divide(const_1, 80), divide(const_1, 30)), divide(const_1, 48)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_3,#4)","chain":"1 \/ 80<\/gadget>\n1\/80 = around 0.0125<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/80) + (1\/30)<\/gadget>\n11\/240 = around 0.045833<\/output>\n1 \/ 48<\/gadget>\n1\/48 = around 0.020833<\/output>\n(11\/240) + (1\/48)<\/gadget>\n1\/15 = around 0.066667<\/output>\n3 \/ (1\/15)<\/gadget>\n45<\/output>\n45<\/result>","index":1450} +{"problem":"a cement mixture is composed of 3 elements . by weight , 1 \/ 2 of the mixture is sand , 1 \/ 5 of the mixture is water , and the remaining 15 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ?","rationale":"\"let the total weight be x . sand content = ( 1 \/ 2 ) x water content = ( 1 \/ 5 ) x gravel = x - ( 1 \/ 2 ) x - ( 1 \/ 5 ) x = ( 3 \/ 10 ) x = 15 x = 50 then answer will be c = 50\"","correct":"c","options":{"a":"30 ","b":"40 ","c":"50 ","d":"60","e":"80"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":80.0},"annotated_formula":"divide(15, subtract(1, add(divide(1, 2), divide(1, 5))))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)|divide(n5,#3)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/2) + (1\/5)<\/gadget>\n7\/10 = around 0.7<\/output>\n1 - (7\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n15 \/ (3\/10)<\/gadget>\n50<\/output>\n50<\/result>","index":1451} +{"problem":"reema took a loan of rs 900 with simple interest for as many years as the rate of interest . if she paid rs . 729 as interest at the end of the loan period , what was the rate of interest .","rationale":"\"explanation : let rate = r % then time = r years . = > 900 ∗ r ∗ r \/ 100 = 729 = > r 2 = 81 = > r = 9 % option e\"","correct":"e","options":{"a":"5 % ","b":"6 % ","c":"7 % ","d":"8 %","e":"9 %"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"sqrt(divide(729, divide(900, const_100)))","linear_formula":"divide(n0,const_100)|divide(n1,#0)|sqrt(#1)|","chain":"900 \/ 100<\/gadget>\n9<\/output>\n729 \/ 9<\/gadget>\n81<\/output>\n81 ** (1\/2)<\/gadget>\n9<\/output>\n9<\/result>","index":1454} +{"problem":"a student committee on academic integrity has 132 ways to select a president and vice president from a group of candidates . the same person can not be both president and vice president . how many candidates are there ?","rationale":"\"xc 1 * ( x - 1 ) c 1 = 132 x ^ 2 - x - 132 = 0 ( x - 12 ) ( x + 11 ) = 0 x = 12 , - 11 - 11 ca n ' t possible . hence 12 should be the answer e\"","correct":"e","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"12"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":12.0},"annotated_formula":"divide(add(const_1, sqrt(add(multiply(const_4, 132), power(negate(const_1), const_2)))), const_2)","linear_formula":"multiply(n0,const_4)|negate(const_1)|power(#1,const_2)|add(#0,#2)|sqrt(#3)|add(#4,const_1)|divide(#5,const_2)|","chain":"4 * 132<\/gadget>\n528<\/output>\n-1<\/gadget>\n-1<\/output>\n(-1) ** 2<\/gadget>\n1<\/output>\n528 + 1<\/gadget>\n529<\/output>\n529 ** (1\/2)<\/gadget>\n23<\/output>\n1 + 23<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12<\/result>","index":1458} +{"problem":"a grocery store bought some mangoes at a rate of 5 for a dollar . they were separated into two stacks , one of which was sold at a rate of 3 for a dollar and the other at a rate of for a dollar . what was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes ?","rationale":"the cost price of a mango = 1 \/ 5 dollars . the selling price of a mango from the first stack = 1 \/ 3 dollars - - > the profit from one mango = 1 \/ 3 - 1 \/ 5 = 2 \/ 15 = 4 \/ 30 dollars . the selling price of a mango from the second stack = 1 \/ 8 dollars - - > the loss from one mango = 1 \/ 5 - 1 \/ 8 = 3 \/ 40 dollars . the profit from one mango from the first stack is 4 times the loss from one mango from the second stack . the ratio is 4 \/ 30 * 40 \/ 3 = 16 \/ 9 b","correct":"b","options":{"a":"1 : 4 ","b":"16 : 9 ","c":"2 : 3 ","d":"1 : 2","e":"2 : 5"},"options_float":{"a":0.25,"b":1.7777777778,"c":0.6666666667,"d":0.5,"e":0.4},"annotated_formula":"divide(subtract(divide(const_1, 3), divide(const_1, 5)), subtract(divide(const_1, 5), divide(const_1, multiply(const_2, const_4))))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|multiply(const_2,const_4)|divide(const_1,#2)|subtract(#0,#1)|subtract(#1,#3)|divide(#4,#5)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/3) - (1\/5)<\/gadget>\n2\/15 = around 0.133333<\/output>\n2 * 4<\/gadget>\n8<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/5) - (1\/8)<\/gadget>\n3\/40 = around 0.075<\/output>\n(2\/15) \/ (3\/40)<\/gadget>\n16\/9 = around 1.777778<\/output>\n16\/9 = around 1.777778<\/result>","index":1459} +{"problem":"if an investor puts $ 800 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ?","rationale":"\"1.05 * 1.05 * 800 = $ 882 the answer is c .\"","correct":"c","options":{"a":"$ 878 ","b":"$ 880 ","c":"$ 882 ","d":"$ 884","e":"$ 886"},"options_float":{"a":878.0,"b":880.0,"c":882.0,"d":884.0,"e":886.0},"annotated_formula":"multiply(800, power(add(const_1, divide(divide(10, const_2), const_100)), const_2))","linear_formula":"divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)|","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 + (1\/20)<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n800 * (441\/400)<\/gadget>\n882<\/output>\n882<\/result>","index":1460} +{"problem":"in an election between two candidates , one got 55 % of the total valid votes . 20 % of the votes were invalid . if the total number of votes was 7500 , what was the number of valid votes that the other candidate got ?","rationale":"\"total no of votes = 7500 20 % invalid = > 80 % valid tot valid votes = 7500 * 80 \/ 100 = > valid for 2 nd candidate got 7500 * 80 \/ 100 * 45 \/ 100 = 2700 answer b\"","correct":"b","options":{"a":"2800 ","b":"2700 ","c":"2100 ","d":"2500","e":"3000"},"options_float":{"a":2800.0,"b":2700.0,"c":2100.0,"d":2500.0,"e":3000.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n1 - (11\/20)<\/gadget>\n9\/20 = around 0.45<\/output>\n(4\/5) * (9\/20)<\/gadget>\n9\/25 = around 0.36<\/output>\n(9\/25) * 7_500<\/gadget>\n2_700<\/output>\n2_700<\/result>","index":1461} +{"problem":"indu gave bindu rs . 625 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ?","rationale":"\"625 = d ( 100 \/ 4 ) 2 d = 1 answer : a\"","correct":"a","options":{"a":"1 ","b":"2 ","c":"7 ","d":"4","e":"9"},"options_float":{"a":1.0,"b":2.0,"c":7.0,"d":4.0,"e":9.0},"annotated_formula":"subtract(subtract(multiply(625, power(add(const_1, divide(4, const_100)), 2)), 625), multiply(multiply(625, divide(4, const_100)), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n625 * (676\/625)<\/gadget>\n676<\/output>\n676 - 625<\/gadget>\n51<\/output>\n625 * (1\/25)<\/gadget>\n25<\/output>\n25 * 2<\/gadget>\n50<\/output>\n51 - 50<\/gadget>\n1<\/output>\n1<\/result>","index":1463} +{"problem":"a 1400 m long train crosses a tree in 100 sec , how much time will it take to pass a platform 700 m long ?","rationale":"\"l = s * t s = 1400 \/ 100 s = 14 m \/ sec . total length ( d ) = 2100 m t = d \/ s t = 2100 \/ 14 t = 150 sec answer : c\"","correct":"c","options":{"a":"288 ","b":"190 ","c":"150 ","d":"188","e":"12"},"options_float":{"a":288.0,"b":190.0,"c":150.0,"d":188.0,"e":12.0},"annotated_formula":"divide(add(1400, 700), divide(1400, 100))","linear_formula":"add(n0,n2)|divide(n0,n1)|divide(#0,#1)|","chain":"1_400 + 700<\/gadget>\n2_100<\/output>\n1_400 \/ 100<\/gadget>\n14<\/output>\n2_100 \/ 14<\/gadget>\n150<\/output>\n150<\/result>","index":1467} +{"problem":"during the last 4 years , a large truck company purchased diesel at rates of $ 1.2 , $ 1.3 , $ 1.8 and $ 2.1 per gallon . what was the mean cost of diesel over the 4 - year period if the company spends the same amount of dollars on diesel each year ?","rationale":"finding the harmonic mean here . hm = 4 \/ ( 1 \/ 12 + 1 \/ 13 + 1 \/ 18 + 1 \/ 21 ) = 15.18 c","correct":"c","options":{"a":"$ 2.5 ","b":"$ 2.55 ","c":"$ 1.52 ","d":"$ 3.25","e":"$ 3.0"},"options_float":{"a":2.5,"b":2.55,"c":1.52,"d":3.25,"e":3.0},"annotated_formula":"subtract(divide(add(add(1.2, 1.3), add(1.8, 2.1)), 4), divide(add(4, 4), const_100))","linear_formula":"add(n1,n2)|add(n3,n4)|add(n0,n0)|add(#0,#1)|divide(#2,const_100)|divide(#3,n0)|subtract(#5,#4)","chain":"1.2 + 1.3<\/gadget>\n2.5<\/output>\n1.8 + 2.1<\/gadget>\n3.9<\/output>\n2.5 + 3.9<\/gadget>\n6.4<\/output>\n6.4 \/ 4<\/gadget>\n1.6<\/output>\n4 + 4<\/gadget>\n8<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n1.6 - (2\/25)<\/gadget>\n1.52<\/output>\n1.52<\/result>","index":1469} +{"problem":"after spending rs . 5000 on rent , rs . 1500 on milk , rs . 4500 on groceries , rs . 2500 on childrens education rs . 2000 on petrol and rs . 5200 on miscellaneous expenses , mr . kishore saved 10 % of his monthly salary . how much did he save in rs . ?","rationale":"\"explanation : total exp = 5000 + 1500 + 4500 + 2500 + 2000 + 5200 = 20700 exp in % = 100 - 10 = 90 % , 20700 = 90 % saving = 10 % = 20700 x 10 \/ 90 = rs . 2300 answer : d\"","correct":"d","options":{"a":"2160 ","b":"2350 ","c":"2000 ","d":"2300","e":"none of these"},"options_float":{"a":2160.0,"b":2350.0,"c":2000.0,"d":2300.0,"e":null},"annotated_formula":"subtract(divide(multiply(add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 5200), const_100), subtract(const_100, 10)), add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 5200))","linear_formula":"add(n0,n1)|subtract(const_100,n6)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|multiply(#5,const_100)|divide(#6,#1)|subtract(#7,#5)|","chain":"5_000 + 1_500<\/gadget>\n6_500<\/output>\n6_500 + 4_500<\/gadget>\n11_000<\/output>\n11_000 + 2_500<\/gadget>\n13_500<\/output>\n13_500 + 2_000<\/gadget>\n15_500<\/output>\n15_500 + 5_200<\/gadget>\n20_700<\/output>\n20_700 * 100<\/gadget>\n2_070_000<\/output>\n100 - 10<\/gadget>\n90<\/output>\n2_070_000 \/ 90<\/gadget>\n23_000<\/output>\n23_000 - 20_700<\/gadget>\n2_300<\/output>\n2_300<\/result>","index":1470} +{"problem":"shekar scored 76 , 65 , 82 , 47 and 85 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ?","rationale":"\"explanation : average = ( 76 + 65 + 82 + 47 + 85 ) \/ 5 = 355 \/ 5 = 71 hence average = 71 answer : a\"","correct":"a","options":{"a":"71 ","b":"69 ","c":"75 ","d":"85","e":"90"},"options_float":{"a":71.0,"b":69.0,"c":75.0,"d":85.0,"e":90.0},"annotated_formula":"divide(add(add(add(add(76, 65), 82), 47), 85), add(const_1, const_4))","linear_formula":"add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|","chain":"76 + 65<\/gadget>\n141<\/output>\n141 + 82<\/gadget>\n223<\/output>\n223 + 47<\/gadget>\n270<\/output>\n270 + 85<\/gadget>\n355<\/output>\n1 + 4<\/gadget>\n5<\/output>\n355 \/ 5<\/gadget>\n71<\/output>\n71<\/result>","index":1471} +{"problem":"how many multiples of 4 are there between 12 and 104 , inclusive ?","rationale":"\"the multiples of 4 are from 4 * 3 up to 4 * 26 . 26 - 3 + 1 = 24 . the answer is d .\"","correct":"d","options":{"a":"21 ","b":"22 ","c":"23 ","d":"24","e":"25"},"options_float":{"a":21.0,"b":22.0,"c":23.0,"d":24.0,"e":25.0},"annotated_formula":"add(divide(subtract(104, 12), 4), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"104 - 12<\/gadget>\n92<\/output>\n92 \/ 4<\/gadget>\n23<\/output>\n23 + 1<\/gadget>\n24<\/output>\n24<\/result>","index":1473} +{"problem":"how many kg of rice at rs . 6.60 per kg . be mixed with 56 kg of rice at rs . 9.60 per kg to get a mixture worth rs . 8.20 per kg","rationale":"explanation : by the rule of alligation , we have cost of 1 kg of 1 st kind rice cost of 1 kg of 2 nd kind rice 6.6 9.6 price of 1 kg of the mixture 8.2 9.6 - 8.2 = 1.4 8.2 - 6.6 = 1.6 quantity of 1 st kind rice : quantity of 2 nd kind rice = 1.4 : 1.6 = 7 : 8 quantity of 1 st kind rice : 56 = 7 : 8 = > quantity of 1 st kind rice = 56 × 78 = 49 answer : option d","correct":"d","options":{"a":"56 kg ","b":"52 kg ","c":"44 kg ","d":"49 kg","e":"none of these"},"options_float":{"a":56.0,"b":52.0,"c":44.0,"d":49.0,"e":null},"annotated_formula":"divide(subtract(multiply(9.6, 56), multiply(56, 8.2)), subtract(8.2, 6.6))","linear_formula":"multiply(n1,n2)|multiply(n1,n3)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)","chain":"9.6 * 56<\/gadget>\n537.6<\/output>\n56 * 8.2<\/gadget>\n459.2<\/output>\n537.6 - 459.2<\/gadget>\n78.4<\/output>\n8.2 - 6.6<\/gadget>\n1.6<\/output>\n78.4 \/ 1.6<\/gadget>\n49<\/output>\n49<\/result>","index":1475} +{"problem":"a box measuring 36 inches long by 45 inches wide by 18 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ?","rationale":"\"least number of cubes will be required when the cubes that could fit in are biggest . 9 is the biggest number that could divide all three , 36 , 45 and 18 . thus side of cube must be 9 , and total number of cubes = 36 \/ 9 * 45 \/ 9 * 18 \/ 9 = 40 ans d .\"","correct":"d","options":{"a":"32 ","b":"34 ","c":"38 ","d":"40","e":"42"},"options_float":{"a":32.0,"b":34.0,"c":38.0,"d":40.0,"e":42.0},"annotated_formula":"divide(multiply(multiply(36, 45), 18), volume_cube(divide(18, const_2)))","linear_formula":"divide(n2,const_2)|multiply(n0,n1)|multiply(n2,#1)|volume_cube(#0)|divide(#2,#3)|","chain":"36 * 45<\/gadget>\n1_620<\/output>\n1_620 * 18<\/gadget>\n29_160<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n9 ** 3<\/gadget>\n729<\/output>\n29_160 \/ 729<\/gadget>\n40<\/output>\n40<\/result>","index":1476} +{"problem":"tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?","rationale":"\"circumference of a = 2 ( pie ) r = 8 so r = 4 \/ pie volume = pie ( 4 \/ pie ) ^ 2 * 10 = 160 \/ pie circumference of b = 2 ( pie ) r = 10 so r = 5 \/ pie volume = pie ( 5 \/ pie ) ^ 2 * 8 = 200 \/ pie so ratio of capacities = 160 \/ 200 = 0.8 so capacity of a will be 80 % of the capacity of b answer : e\"","correct":"e","options":{"a":"90 % ","b":"50 % ","c":"60 % ","d":"70 %","e":"80 %"},"options_float":{"a":90.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(multiply(power(divide(8, 10), const_2), divide(10, 8)), const_100)","linear_formula":"divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)|","chain":"8 \/ 10<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) ** 2<\/gadget>\n16\/25 = around 0.64<\/output>\n10 \/ 8<\/gadget>\n5\/4 = around 1.25<\/output>\n(16\/25) * (5\/4)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":1478} +{"problem":"at joes steakhouse the hourly wage for a chef is 20 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 6.50 per hour , how much less than a manager does a chef earn each hour ?","rationale":"manager wages per hour = $ 6.50 dishwasher wages per hour = half of manager ' s wages . = 1 \/ 2 ( $ 6.50 ) = = > $ 3.25 chef wages per hour = 20 % greater than dishwasher wages - - > 20 % of $ 3.25 = ( 20 * ( $ 3.25 ) ) \/ 100 - - > ( $ 65 ) \/ 100 - - > $ 0.65 therefore , chef wages per hour = $ 3.25 + $ 0.65 = = > $ 3.9 difference of wages between manager and chef = $ 6.50 - $ 3.9 = = > $ 2.6 answer : a","correct":"a","options":{"a":"$ 2.6 ","b":"$ 2.40 ","c":"$ 3.40 ","d":"$ 4.40","e":"$ 5.40"},"options_float":{"a":2.6,"b":2.4,"c":3.4,"d":4.4,"e":5.4},"annotated_formula":"multiply(subtract(const_1, multiply(divide(add(const_100, 20), const_100), divide(const_1, const_2))), 6.5)","linear_formula":"add(n0,const_100)|divide(const_1,const_2)|divide(#0,const_100)|multiply(#2,#1)|subtract(const_1,#3)|multiply(n1,#4)","chain":"100 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(6\/5) * (1\/2)<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 6.5<\/gadget>\n2.6<\/output>\n2.6<\/result>","index":1483} +{"problem":"the cost of paint is 60 per kilograme . a kilogram paint covers 20 square feet . how much will it cost to paint the outside of a cube having each side 10 feet ?","rationale":"area of cube = 6 × ( side ) 2 = 6 × 10 × 10 = 600 square feet . cost to paint outside of the cube = 600 ⁄ 20 × 60 = 1800 answer c","correct":"c","options":{"a":"3000 ","b":"900 ","c":"1800 ","d":"360","e":"none of these"},"options_float":{"a":3000.0,"b":900.0,"c":1800.0,"d":360.0,"e":null},"annotated_formula":"multiply(divide(multiply(multiply(const_2, const_3), power(10, const_2)), 20), 60)","linear_formula":"multiply(const_2,const_3)|power(n2,const_2)|multiply(#0,#1)|divide(#2,n1)|multiply(n0,#3)","chain":"2 * 3<\/gadget>\n6<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600 \/ 20<\/gadget>\n30<\/output>\n30 * 60<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":1484} +{"problem":"the original price of a suit is $ 160 . the price increased 25 % , and after this increase , the store published a 25 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 25 % off the increased price , how much did these consumers pay for the suit ?","rationale":"0.75 * ( 1.25 * 160 ) = $ 150 the answer is c .","correct":"c","options":{"a":"$ 140 ","b":"$ 145 ","c":"$ 150 ","d":"$ 155","e":"$ 160"},"options_float":{"a":140.0,"b":145.0,"c":150.0,"d":155.0,"e":160.0},"annotated_formula":"subtract(add(160, divide(multiply(160, 25), const_100)), divide(multiply(add(160, divide(multiply(160, 25), const_100)), 25), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|subtract(#2,#4)","chain":"160 * 25<\/gadget>\n4_000<\/output>\n4_000 \/ 100<\/gadget>\n40<\/output>\n160 + 40<\/gadget>\n200<\/output>\n200 * 25<\/gadget>\n5_000<\/output>\n5_000 \/ 100<\/gadget>\n50<\/output>\n200 - 50<\/gadget>\n150<\/output>\n150<\/result>","index":1486} +{"problem":"in a throw of dice what is the probability of getting number greater than 5","rationale":"\"number greater than 5 is 6 , so only 1 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 1 \/ 6 b )\"","correct":"b","options":{"a":"1 \/ 7 ","b":"1 \/ 6 ","c":"1 \/ 11 ","d":"1 \/ 13","e":"2 \/ 9"},"options_float":{"a":0.1428571429,"b":0.1666666667,"c":0.0909090909,"d":0.0769230769,"e":0.2222222222},"annotated_formula":"divide(subtract(const_6, 5), const_6)","linear_formula":"subtract(const_6,n0)|divide(#0,const_6)|","chain":"6 - 5<\/gadget>\n1<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":1489} +{"problem":"of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 19 boys and 41 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ?","rationale":"\"venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy \/ sad \/ neither . when you have two or more variables such as here where you have gender - boy \/ girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you will use logic : there are 6 happy boys . there are 4 sad girls but total 10 sad children . so rest 6 sad children must be sad boys . we have 6 happy boys and 6 sad boys . total we have 19 boys . so 19 - 6 - 6 = 7 boys must be neither happy nor sad . answer ( b )\"","correct":"b","options":{"a":"2 ","b":"7 ","c":"6 ","d":"8","e":"10"},"options_float":{"a":2.0,"b":7.0,"c":6.0,"d":8.0,"e":10.0},"annotated_formula":"subtract(subtract(19, subtract(10, 4)), 6)","linear_formula":"subtract(n2,n7)|subtract(n4,#0)|subtract(#1,n6)|","chain":"10 - 4<\/gadget>\n6<\/output>\n19 - 6<\/gadget>\n13<\/output>\n13 - 6<\/gadget>\n7<\/output>\n7<\/result>","index":1492} +{"problem":"a store sells chairs and tables . if the price of 2 chairs and 1 table is 60 % of the price of 1 chair and 2 tables , and the price of 1 table and 1 chair is $ 64 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . )","rationale":"\"let c = chair ; t = table 2 c + 1 t = 0.6 ( 1 c + 2 t ) or c ( 2 - 0.6 ) = 1.2 t - 1 t or 1.4 c = 0.2 t therefore c = 0.2 \/ 1.4 t = 1 \/ 7 t ic + 1 t = 64 or 1 \/ 7 t + 1 t = 64 therefore t = 64 * 7 \/ 8 = 56 e\"","correct":"e","options":{"a":"48 ","b":"52 ","c":"54 ","d":"58","e":"56"},"options_float":{"a":48.0,"b":52.0,"c":54.0,"d":58.0,"e":56.0},"annotated_formula":"divide(subtract(multiply(64, 2), multiply(64, divide(60, const_100))), add(divide(60, const_100), 1))","linear_formula":"divide(n2,const_100)|multiply(n7,n0)|add(#0,n1)|multiply(n7,#0)|subtract(#1,#3)|divide(#4,#2)|","chain":"64 * 2<\/gadget>\n128<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n64 * (3\/5)<\/gadget>\n192\/5 = around 38.4<\/output>\n128 - (192\/5)<\/gadget>\n448\/5 = around 89.6<\/output>\n(3\/5) + 1<\/gadget>\n8\/5 = around 1.6<\/output>\n(448\/5) \/ (8\/5)<\/gadget>\n56<\/output>\n56<\/result>","index":1493} +{"problem":"if the radius of a circle that centers at the origin is 5 , how many points q on the circle have integer coordinates ?","rationale":"\"i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + \/ - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + \/ - ( 3,4 or 4,3 ) - 8 possibilities . ans : q = c\"","correct":"c","options":{"a":"4 ","b":"8 ","c":"12 ","d":"15","e":"20"},"options_float":{"a":4.0,"b":8.0,"c":12.0,"d":15.0,"e":20.0},"annotated_formula":"subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2))","linear_formula":"add(n0,const_4)|multiply(n0,n0)|power(const_2,const_2)|subtract(#1,#0)|subtract(#3,#2)|","chain":"5 * 5<\/gadget>\n25<\/output>\n5 + 4<\/gadget>\n9<\/output>\n25 - 9<\/gadget>\n16<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n16 - 4<\/gadget>\n12<\/output>\n12<\/result>","index":1494} +{"problem":"the ratio of the radius of two circles is 1 : 5 , and then the ratio of their areas is ?","rationale":"\"r 1 : r 2 = 1 : 5 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 1 : 25 answer : c\"","correct":"c","options":{"a":"1 : 7 ","b":"2 : 9 ","c":"1 : 25 ","d":"3 : 7","e":"3 : 4"},"options_float":{"a":0.1428571429,"b":0.2222222222,"c":0.04,"d":0.4285714286,"e":0.75},"annotated_formula":"divide(circle_area(1), circle_area(5))","linear_formula":"circle_area(n0)|circle_area(n1)|divide(#0,#1)|","chain":"pi * (1 ** 2)<\/gadget>\npi = around 3.141593<\/output>\npi * (5 ** 2)<\/gadget>\n25*pi = around 78.539816<\/output>\npi \/ (25*pi)<\/gadget>\n1\/25 = around 0.04<\/output>\n1\/25 = around 0.04<\/result>","index":1495} +{"problem":"the average of 11 numbers is 60 . out of 11 numbers the average of first 6 no . is 98 , and last 6 numbers is 65 then find 6 th number ?","rationale":"6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 98 + 6 * 65 - 11 * 60 = 318 answer is a","correct":"a","options":{"a":"318 ","b":"65 ","c":"58 ","d":"62","e":"48"},"options_float":{"a":318.0,"b":65.0,"c":58.0,"d":62.0,"e":48.0},"annotated_formula":"subtract(add(multiply(6, 98), multiply(6, 65)), multiply(11, 60))","linear_formula":"multiply(n3,n4)|multiply(n3,n6)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)","chain":"6 * 98<\/gadget>\n588<\/output>\n6 * 65<\/gadget>\n390<\/output>\n588 + 390<\/gadget>\n978<\/output>\n11 * 60<\/gadget>\n660<\/output>\n978 - 660<\/gadget>\n318<\/output>\n318<\/result>","index":1496} +{"problem":"a man bought 40 shares of rs . 60 at 8 discount , the rate of dividend being 1212 % the rate of interest obtained is","rationale":"\"explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 8 = rs . 52 number of shares = 40 dividend = 12 1 \/ 2 % = 25 \/ 2 % dividend per share = 60 × 25 \/ 2 × 100 = rs . 7.5 total dividend = ( 40 × 7.5 ) ie , he got a dividend of ( 40 × 7.5 ) for an investment of rs . ( 40 × 52 ) interest obtained = 40 × 7.5 × 100 \/ 40 × 52 = 14.42 % answer : option d\"","correct":"d","options":{"a":"13.64 % ","b":"15.5 % ","c":"14 % ","d":"14.42 %","e":"14.95 %"},"options_float":{"a":13.64,"b":15.5,"c":14.0,"d":14.42,"e":14.95},"annotated_formula":"divide(multiply(multiply(40, 60), divide(1212, const_100)), multiply(40, subtract(60, 8)))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#1)|multiply(n0,#2)|divide(#3,#4)|","chain":"40 * 60<\/gadget>\n2_400<\/output>\n1_212 \/ 100<\/gadget>\n303\/25 = around 12.12<\/output>\n2_400 * (303\/25)<\/gadget>\n29_088<\/output>\n60 - 8<\/gadget>\n52<\/output>\n40 * 52<\/gadget>\n2_080<\/output>\n29_088 \/ 2_080<\/gadget>\n909\/65 = around 13.984615<\/output>\n909\/65 = around 13.984615<\/result>","index":1497} +{"problem":"the area of a triangle is 615 m 2 . if one of its sides is 123 metre , find the length of the perpendicular dropped on that side from opposite vertex .","rationale":"in a triangle , area = 1 ⁄ 2 × length of perpendicular × base or 615 = 1 ⁄ 2 × length of perpendicular × 123 ∴ length of perpendicular = 615 × 2 \/ 123 = 10 m answer c","correct":"c","options":{"a":"15 metres ","b":"12 metres ","c":"10 metres ","d":"9 metres","e":"none of these"},"options_float":{"a":15.0,"b":12.0,"c":10.0,"d":9.0,"e":null},"annotated_formula":"divide(multiply(615, const_2), 123)","linear_formula":"multiply(n0,const_2)|divide(#0,n2)","chain":"615 * 2<\/gadget>\n1_230<\/output>\n1_230 \/ 123<\/gadget>\n10<\/output>\n10<\/result>","index":1498} +{"problem":"in an election between the two candidates , the candidates who gets 60 % of votes polled is wined by 280 votes majority . what is the total number of votes polled ?","rationale":"note : majority ( 20 % ) = difference in votes polled to win ( 60 % ) & defeated candidates ( 40 % ) 20 % = 60 % - 40 % 20 % - - - - - > 280 ( 20 × 14 = 280 ) 100 % - - - - - > 1400 ( 100 × 14 = 1400 ) a","correct":"a","options":{"a":"1400 ","b":"1500 ","c":"1600 ","d":"1700","e":"1800"},"options_float":{"a":1400.0,"b":1500.0,"c":1600.0,"d":1700.0,"e":1800.0},"annotated_formula":"divide(multiply(const_100, 280), subtract(60, subtract(const_100, 60)))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n0)|subtract(n0,#1)|divide(#0,#2)","chain":"100 * 280<\/gadget>\n28_000<\/output>\n100 - 60<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n28_000 \/ 20<\/gadget>\n1_400<\/output>\n1_400<\/result>","index":1499} +{"problem":"what is the remainder if 7 ^ 12 is divided by 100 ?","rationale":"7 ^ 16 can be written as ( 7 ^ 4 ) ^ 3 if we divide 7 ^ 4 by 100 the reminder is 1 so , ( 7 ^ 4 ) ^ 3 by 100 , the reminder is 1 ^ 3 = 1 answer : a","correct":"a","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(divide(100, const_2), multiply(7, 7))","linear_formula":"divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)","chain":"100 \/ 2<\/gadget>\n50<\/output>\n7 * 7<\/gadget>\n49<\/output>\n50 - 49<\/gadget>\n1<\/output>\n1<\/result>","index":1501} +{"problem":"calculate the average of all the numbers between 10 and 86 which are divisible by 9 .","rationale":"\"explanation : numbers divisible by 9 are 18,27 , 36,45 , 54,63 , 72,81 , average = ( 18 + 27 + 36 + 45 + 54 + 63 + 72 + 81 , ) \/ 8 = 396 \/ 8 = 49.5 answer : a\"","correct":"a","options":{"a":"49.5 ","b":"41.5 ","c":"45.5 ","d":"40.5","e":"47.5"},"options_float":{"a":49.5,"b":41.5,"c":45.5,"d":40.5,"e":47.5},"annotated_formula":"multiply(divide(add(add(floor(divide(10, 9)), const_1), floor(divide(86, 9))), const_2), 9)","linear_formula":"divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6)|","chain":"10 \/ 9<\/gadget>\n10\/9 = around 1.111111<\/output>\nfloor(10\/9)<\/gadget>\n1<\/output>\n1 + 1<\/gadget>\n2<\/output>\n86 \/ 9<\/gadget>\n86\/9 = around 9.555556<\/output>\nfloor(86\/9)<\/gadget>\n9<\/output>\n2 + 9<\/gadget>\n11<\/output>\n11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\n(11\/2) * 9<\/gadget>\n99\/2 = around 49.5<\/output>\n99\/2 = around 49.5<\/result>","index":1502} +{"problem":"what is the length of a bridge ( in meters ) , which a train 156 meters long and travelling at 45 km \/ h can cross in 30 seconds ?","rationale":"speed = 45 km \/ h = 45000 m \/ 3600 s = 25 \/ 2 m \/ s in 30 seconds , the train can travel 25 \/ 2 * 30 = 375 meters 375 = length of train + length of bridge length of bridge = 375 - 156 = 219 meters the answer is c .","correct":"c","options":{"a":"203 ","b":"211 ","c":"219 ","d":"227","e":"235"},"options_float":{"a":203.0,"b":211.0,"c":219.0,"d":227.0,"e":235.0},"annotated_formula":"subtract(multiply(multiply(45, const_0_2778), 30), 156)","linear_formula":"multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n45 * (5\/18)<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 156<\/gadget>\n219<\/output>\n219<\/result>","index":1503} +{"problem":"the average of first 17 even numbers is ?","rationale":"\"sum of 17 even numbers = 17 * 18 = 306 average = 306 \/ 17 = 18 answer : c\"","correct":"c","options":{"a":"10 ","b":"11 ","c":"18 ","d":"20","e":"24"},"options_float":{"a":10.0,"b":11.0,"c":18.0,"d":20.0,"e":24.0},"annotated_formula":"add(17, const_1)","linear_formula":"add(n0,const_1)|","chain":"17 + 1<\/gadget>\n18<\/output>\n18<\/result>","index":1505} +{"problem":"a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 11 multiples of 14 , how many multiples of 7 are there in q ?","rationale":"\"halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 11 + 10 = 21 . the answer is c .\"","correct":"c","options":{"a":"19 ","b":"20 ","c":"21 ","d":"22","e":"23"},"options_float":{"a":19.0,"b":20.0,"c":21.0,"d":22.0,"e":23.0},"annotated_formula":"subtract(multiply(11, const_2), const_1)","linear_formula":"multiply(n1,const_2)|subtract(#0,const_1)|","chain":"11 * 2<\/gadget>\n22<\/output>\n22 - 1<\/gadget>\n21<\/output>\n21<\/result>","index":1507} +{"problem":"the ratio between the presents of the a and b is 2 : 3 . if b is 5 years older than a , what will be the ratio of the ages of a and b after 5 years ?","rationale":"\"let a and b ages will be 2 x and 3 x 3 x - 2 x = 5 x = 5 required ratio = 2 x + 5 : 3 x + 5 = 15 : 20 = 3 : 4 answer is a\"","correct":"a","options":{"a":"3 : 4 ","b":"1 : 2 ","c":"2 : 3 ","d":"5 : 6","e":"1 : 3"},"options_float":{"a":0.75,"b":0.5,"c":0.6666666667,"d":0.8333333333,"e":0.3333333333},"annotated_formula":"divide(add(subtract(multiply(3, 5), 5), 5), add(multiply(3, 5), 5))","linear_formula":"multiply(n1,n2)|add(n2,#0)|subtract(#0,n2)|add(n2,#2)|divide(#3,#1)|","chain":"3 * 5<\/gadget>\n15<\/output>\n15 - 5<\/gadget>\n10<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15 + 5<\/gadget>\n20<\/output>\n15 \/ 20<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":1508} +{"problem":"if x * y = xy – 2 ( x + y ) for all integers x and y , then 1 * ( – 3 ) =","rationale":"\"1 * ( - 3 ) = 1 * ( - 3 ) - 2 ( 1 + ( - 3 ) ) = - 3 + 4 = 1 option ( e )\"","correct":"e","options":{"a":"– 16 ","b":"– 11 ","c":"– 4 ","d":"4","e":"1"},"options_float":{"a":16.0,"b":11.0,"c":4.0,"d":4.0,"e":1.0},"annotated_formula":"add(negate(multiply(add(negate(3), 1), 2)), multiply(negate(3), 1))","linear_formula":"negate(n2)|add(n1,#0)|multiply(n1,#0)|multiply(#1,n0)|negate(#3)|add(#2,#4)|","chain":"-3<\/gadget>\n-3<\/output>\n(-3) + 1<\/gadget>\n-2<\/output>\n(-2) * 2<\/gadget>\n-4<\/output>\n-(-4)<\/gadget>\n4<\/output>\n(-3) * 1<\/gadget>\n-3<\/output>\n4 + (-3)<\/gadget>\n1<\/output>\n1<\/result>","index":1510} +{"problem":"on increasing the price of t . v . sets by 40 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ?","rationale":"\"explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 × 100 ) = rs . 10000 new sale value = rs . ( 140 × 80 ) = rs . 11200 increase % = 1200 \/ 10000 × 100 = 12 % answer : d\"","correct":"d","options":{"a":"4 ","b":"5 ","c":"56 ","d":"12","e":"7"},"options_float":{"a":4.0,"b":5.0,"c":56.0,"d":12.0,"e":7.0},"annotated_formula":"subtract(add(const_100, 40), add(divide(multiply(add(const_100, 40), 20), const_100), const_100))","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#2,const_100)|subtract(#0,#3)|","chain":"100 + 40<\/gadget>\n140<\/output>\n140 * 20<\/gadget>\n2_800<\/output>\n2_800 \/ 100<\/gadget>\n28<\/output>\n28 + 100<\/gadget>\n128<\/output>\n140 - 128<\/gadget>\n12<\/output>\n12<\/result>","index":1511} +{"problem":"a 10 % stock yields 8 % . the market value of the stock is :","rationale":"\"solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 10 , investment = rs . ( 100 \/ 8 x 10 ) = rs . 125 ∴ market value of rs . 100 stock = rs . 125 answer b\"","correct":"b","options":{"a":"rs . 72 ","b":"rs . 125 ","c":"rs . 112.50 ","d":"rs . 116.50","e":"none of these"},"options_float":{"a":72.0,"b":125.0,"c":112.5,"d":116.5,"e":null},"annotated_formula":"multiply(divide(const_100, 8), 10)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 8<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 10<\/gadget>\n125<\/output>\n125<\/result>","index":1513} +{"problem":"a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 300 . find the sum ?","rationale":"\"let the sum be rs . x . ( x * 18 * 2 ) \/ 100 - ( x * 12 * 2 ) \/ 100 = 300 = > 36 x \/ 100 - 24 x \/ 100 = 300 = > 12 x \/ 100 = 300 = > x = 2500 . answer : b\"","correct":"b","options":{"a":"7000 ","b":"2500 ","c":"2778 ","d":"2800","e":"2791"},"options_float":{"a":7000.0,"b":2500.0,"c":2778.0,"d":2800.0,"e":2791.0},"annotated_formula":"divide(300, divide(multiply(subtract(18, 12), const_2), const_100))","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)|","chain":"18 - 12<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n300 \/ (3\/25)<\/gadget>\n2_500<\/output>\n2_500<\/result>","index":1514} +{"problem":"at a monthly meeting , 3 \/ 5 of the attendees were males and 7 \/ 8 of the male attendees arrived on time . if 4 \/ 5 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ?","rationale":"\"males who did not arrive on time are 1 \/ 8 * 3 \/ 5 = 3 \/ 40 of the attendees . females who did not arrive on time are 1 \/ 5 * 2 \/ 5 = 2 \/ 25 of the attendees . the fraction of all attendees who did not arrive on time is 3 \/ 40 + 2 \/ 25 = 31 \/ 200 the answer is b .\"","correct":"b","options":{"a":"47 \/ 200 ","b":"31 \/ 200 ","c":"17 \/ 100 ","d":"11 \/ 100","e":"7 \/ 50"},"options_float":{"a":0.235,"b":0.155,"c":0.17,"d":0.11,"e":0.14},"annotated_formula":"add(multiply(subtract(const_1, divide(4, 5)), subtract(const_1, divide(3, 5))), multiply(subtract(const_1, divide(7, 8)), divide(3, 5)))","linear_formula":"divide(n4,n5)|divide(n0,n1)|divide(n2,n3)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#1,#5)|add(#6,#7)|","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/5) * (2\/5)<\/gadget>\n2\/25 = around 0.08<\/output>\n7 \/ 8<\/gadget>\n7\/8 = around 0.875<\/output>\n1 - (7\/8)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * (3\/5)<\/gadget>\n3\/40 = around 0.075<\/output>\n(2\/25) + (3\/40)<\/gadget>\n31\/200 = around 0.155<\/output>\n31\/200 = around 0.155<\/result>","index":1515} +{"problem":"how many quarters are equal to 4 dollars ?","rationale":"\"4 * 4 = 16 quarters answer : a\"","correct":"a","options":{"a":"16 ","b":"8 ","c":"12 ","d":"9","e":"7"},"options_float":{"a":16.0,"b":8.0,"c":12.0,"d":9.0,"e":7.0},"annotated_formula":"multiply(4, const_4)","linear_formula":"multiply(n0,const_4)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16<\/result>","index":1516} +{"problem":"danny and steve are running towards each other , each one from his own house . danny can reach steve ' s house in 33 minutes of running , which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time , how much time longer will it take steve to reach the halfway point between their houses than danny to reach the halfway point between their houses ?","rationale":"\"danny ' s time is 33 minutes . the time to reach halfway is 16.5 minutes . steve ' s time is 66 minutes . the time to reach halfway is 33 minutes . the time difference is 33 - 16.5 = 16.5 minutes the answer is a .\"","correct":"a","options":{"a":"16.5 minutes ","b":"33 minutes ","c":"49.5 minutes ","d":"66 minutes","e":"99 minutes"},"options_float":{"a":16.5,"b":33.0,"c":49.5,"d":66.0,"e":99.0},"annotated_formula":"divide(33, const_2)","linear_formula":"divide(n0,const_2)|","chain":"33 \/ 2<\/gadget>\n33\/2 = around 16.5<\/output>\n33\/2 = around 16.5<\/result>","index":1518} +{"problem":"a fair 2 sided coin is flipped 4 times . what is the probability that tails will be the result at least twice , but not more than 4 times ?","rationale":"\"at least twice , but not more than 4 timesmeans exactly 2 times , 3 times , 4 times the probability of getting exactly k results out of n flips is nck \/ 2 ^ n 4 c 2 \/ 2 ^ 4 + 4 c 3 \/ 2 ^ 4 = 10 \/ 16 = 5 \/ 8 option : c\"","correct":"c","options":{"a":"1 \/ 8 ","b":"2 \/ 3 ","c":"5 \/ 8 ","d":"4 \/ 7","e":"3 \/ 7"},"options_float":{"a":0.125,"b":0.6666666667,"c":0.625,"d":0.5714285714,"e":0.4285714286},"annotated_formula":"subtract(const_1, add(multiply(inverse(power(2, 4)), 4), add(inverse(power(2, 4)), inverse(power(2, 4)))))","linear_formula":"power(n0,n1)|inverse(#0)|add(#1,#1)|multiply(n1,#1)|add(#2,#3)|subtract(const_1,#4)|","chain":"2 ** 4<\/gadget>\n16<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/16) * 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/16) + (1\/16)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/4) + (1\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n1 - (3\/8)<\/gadget>\n5\/8 = around 0.625<\/output>\n5\/8 = around 0.625<\/result>","index":1520} +{"problem":"find the sum of divisors of 544 which are perfect squares .","rationale":"\"544 = 25 + 171 = ( 2 ) 0 + ( 2 ) 1 + ( 2 ) 2 + ( 2 ) 3 + ( 2 ) 4 + ( 2 ) 5 x ( 17 ) 0 + ( 17 ) 1 . here 20 = 1 = 11 a perfect square , 22 and 24 = 16 = 44 are perfect squares , hence , 1 + 4 + 16 = 21 . correct answer is : d\"","correct":"d","options":{"a":"32 ","b":"64 ","c":"42 ","d":"21","e":"none of these"},"options_float":{"a":32.0,"b":64.0,"c":42.0,"d":21.0,"e":null},"annotated_formula":"add(divide(divide(544, divide(divide(divide(divide(divide(544, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(544, divide(divide(divide(divide(divide(544, const_2), const_2), const_2), const_2), const_2)), const_2))))","linear_formula":"divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)|","chain":"544 \/ 2<\/gadget>\n272<\/output>\n272 \/ 2<\/gadget>\n136<\/output>\n136 \/ 2<\/gadget>\n68<\/output>\n68 \/ 2<\/gadget>\n34<\/output>\n34 \/ 2<\/gadget>\n17<\/output>\n544 \/ 17<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16 ** (1\/2)<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n16 + 5<\/gadget>\n21<\/output>\n21<\/result>","index":1524} +{"problem":"a glass was filled with 10 ounces of water , and 0.03 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ?","rationale":"\"we are given that 0.03 ounces of water evaporated each day . furthermore , we know that this process happened over a 20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply 0.01 by 20 . this gives us : 0.03 x 20 = 0.6 ounces finally , we are asked for “ what percent ” of the original amount of water evaporated during this period . to determine this percentage , we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated \/ original amount ) x 100 % ( 0.6 \/ 10 ) x 100 % ( 6 \/ 100 ) x 100 % = 6 % answer d\"","correct":"d","options":{"a":"0.006 % ","b":"0.06 % ","c":"0.6 % ","d":"6 %","e":"60 %"},"options_float":{"a":0.006,"b":0.06,"c":0.6,"d":6.0,"e":60.0},"annotated_formula":"multiply(divide(multiply(0.03, 20), 10), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"0.03 * 20<\/gadget>\n0.6<\/output>\n0.6 \/ 10<\/gadget>\n0.06<\/output>\n0.06 * 100<\/gadget>\n6<\/output>\n6<\/result>","index":1525} +{"problem":"a cistern is normally filled in 6 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ?","rationale":"1 \/ 6 - 1 \/ x = 1 \/ 8 x = 24 answer : c","correct":"c","options":{"a":"33 ","b":"88 ","c":"24 ","d":"99","e":"11"},"options_float":{"a":33.0,"b":88.0,"c":24.0,"d":99.0,"e":11.0},"annotated_formula":"inverse(subtract(divide(const_1, 6), divide(const_1, add(6, const_2))))","linear_formula":"add(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n6 + 2<\/gadget>\n8<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/6) - (1\/8)<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ (1\/24)<\/gadget>\n24<\/output>\n24<\/result>","index":1526} +{"problem":"a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 50 percent .","rationale":"1 hr = 20 percent . thus 15 min = 5 percent . now to charge 50 percent 150 min . answer : b","correct":"b","options":{"a":"145 ","b":"150 ","c":"175 ","d":"160","e":"130"},"options_float":{"a":145.0,"b":150.0,"c":175.0,"d":160.0,"e":130.0},"annotated_formula":"multiply(divide(50, 20), const_60)","linear_formula":"divide(n2,n1)|multiply(#0,const_60)|","chain":"50 \/ 20<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) * 60<\/gadget>\n150<\/output>\n150<\/result>","index":1527} +{"problem":"the diagonals of a rhombus are 10 cm and 12 cm . find its area ?","rationale":"\"1 \/ 2 * 10 * 12 = 60 answer : b\"","correct":"b","options":{"a":"158 ","b":"60 ","c":"150 ","d":"123","e":"117"},"options_float":{"a":158.0,"b":60.0,"c":150.0,"d":123.0,"e":117.0},"annotated_formula":"rhombus_area(10, 12)","linear_formula":"rhombus_area(n0,n1)|","chain":"(10 * 12) \/ 2<\/gadget>\n60<\/output>\n60<\/result>","index":1528} +{"problem":"a hollow iron pipe is 21 cm long and its external diameter is 8 cm . if the thickness of the pipe is 1 cm and iron weighs 8 g \/ cm 3 , then the weight of the pipe is :","rationale":"\"solution external radius = 4 cm , internal radius = 3 cm . volume of iron = ( 22 \/ 7 × ( 42 ) - ( 32 ) × 21 ) cm 3 = ( 22 \/ 7 × 7 × 1 × 21 ) cm 3 = 462 cm 3 . weight of iron = ( 462 x 8 ) gm = 3696 gm = 3.696 kg . answer b\"","correct":"b","options":{"a":"3.6 kg ","b":"3.696 kg ","c":"36 kg ","d":"36.9 kg","e":"none"},"options_float":{"a":3.6,"b":3.696,"c":36.0,"d":36.9,"e":null},"annotated_formula":"divide(multiply(subtract(volume_cylinder(divide(8, const_2), 21), volume_cylinder(subtract(divide(8, const_2), 1), 21)), 8), const_1000)","linear_formula":"divide(n1,const_2)|subtract(#0,n2)|volume_cylinder(#0,n0)|volume_cylinder(#1,n0)|subtract(#2,#3)|multiply(n1,#4)|divide(#5,const_1000)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\npi * (4 ** 2) * 21<\/gadget>\n336*pi = around 1_055.575132<\/output>\n4 - 1<\/gadget>\n3<\/output>\npi * (3 ** 2) * 21<\/gadget>\n189*pi = around 593.761012<\/output>\n(336*pi) - (189*pi)<\/gadget>\n147*pi = around 461.81412<\/output>\n(147*pi) * 8<\/gadget>\n1176*pi = around 3_694.512961<\/output>\n(1176*pi) \/ 1_000<\/gadget>\n147*pi\/125 = around 3.694513<\/output>\n147*pi\/125 = around 3.694513<\/result>","index":1529} +{"problem":"in the forestside summer camp there are 50 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ?","rationale":"given there are 50 students in the forestside summer camp , 90 % of 50 = 45 boys and remaining 5 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 5 % . . if we add 50 to existing 45 then the count will be 95 and the girls number will be 5 as it . now boys are 95 % and girls are 5 % . ( out of 100 students = 95 boys + 5 girls ) . imo option a is correct .","correct":"a","options":{"a":"50 . ","b":"45 . ","c":"40 . ","d":"30 .","e":"25 ."},"options_float":{"a":50.0,"b":45.0,"c":40.0,"d":30.0,"e":25.0},"annotated_formula":"subtract(divide(multiply(50, subtract(const_1, divide(90, const_100))), divide(5, const_100)), 50)","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#2)|divide(#3,#1)|subtract(#4,n0)","chain":"90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n50 * (1\/10)<\/gadget>\n5<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n5 \/ (1\/20)<\/gadget>\n100<\/output>\n100 - 50<\/gadget>\n50<\/output>\n50<\/result>","index":1530} +{"problem":"on dividing 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?","rationale":"\"d = ( d - r ) \/ q = ( 109 - 1 ) \/ 9 = 108 \/ 9 = 12 b )\"","correct":"b","options":{"a":"10 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"floor(divide(109, 9))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"109 \/ 9<\/gadget>\n109\/9 = around 12.111111<\/output>\nfloor(109\/9)<\/gadget>\n12<\/output>\n12<\/result>","index":1531} +{"problem":"if 0.75 : x : : 10 : 8 , then x is equal to :","rationale":"\"( x * 10 ) = ( 0.75 * 8 ) x = 6 \/ 10 x = 0.6 answer = a\"","correct":"a","options":{"a":"0.6 ","b":"0.8 ","c":"0.85 ","d":"0.5","e":"none"},"options_float":{"a":0.6,"b":0.8,"c":0.85,"d":0.5,"e":null},"annotated_formula":"divide(multiply(0.75, 8), 10)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"0.75 * 8<\/gadget>\n6<\/output>\n6 \/ 10<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":1533} +{"problem":"there are 2 sections a and b in a class , consisting of 36 and 44 students respectively . if the average weight of section a is 40 kg and that of section b is 35 kg , find the average of the whole class ?","rationale":"total weight of 36 + 44 students = 36 * 44 + 44 * 35 = 2980 average weight of the class is = 2980 \/ 80 = 37.25 kg answer is b","correct":"b","options":{"a":"35.29 kg ","b":"37.25 kg ","c":"42.45 kg ","d":"55.12 kg","e":"29.78 kg"},"options_float":{"a":35.29,"b":37.25,"c":42.45,"d":55.12,"e":29.78},"annotated_formula":"divide(add(multiply(36, 40), multiply(44, 35)), add(36, 44))","linear_formula":"add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)","chain":"36 * 40<\/gadget>\n1_440<\/output>\n44 * 35<\/gadget>\n1_540<\/output>\n1_440 + 1_540<\/gadget>\n2_980<\/output>\n36 + 44<\/gadget>\n80<\/output>\n2_980 \/ 80<\/gadget>\n149\/4 = around 37.25<\/output>\n149\/4 = around 37.25<\/result>","index":1534} +{"problem":"a box contains 100 pairs of shoes ( 200 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ?","rationale":"\"the problem with your solution is that we do n ' t choose 1 shoe from 200 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 \/ 1 * 1 \/ 199 ( as after taking one at random there are 199 shoes left and only one is the pair of the first one ) = 1 \/ 199 answer : c .\"","correct":"c","options":{"a":"1 \/ 190 ","b":"1 \/ 20 ","c":"1 \/ 199 ","d":"1 \/ 10","e":"1 \/ 9"},"options_float":{"a":0.0052631579,"b":0.05,"c":0.0050251256,"d":0.1,"e":0.1111111111},"annotated_formula":"divide(const_1, subtract(200, const_1))","linear_formula":"subtract(n1,const_1)|divide(const_1,#0)|","chain":"200 - 1<\/gadget>\n199<\/output>\n1 \/ 199<\/gadget>\n1\/199 = around 0.005025<\/output>\n1\/199 = around 0.005025<\/result>","index":1535} +{"problem":"the nasdaq index has decreased by 70 % since year 2000 . how much should the index increase in order to gain the same value as in year 2000 ?","rationale":"let x be the index in the year 2000 . after the decrease it becomes 0.3 x now for 0.3 x to reach 1 x it has to raise by 0.7 x . the increase needs to be 0.7 x \/ 0.3 x * 100 = 233.3 % answer c","correct":"c","options":{"a":"400 % ","b":"80 % ","c":"233 % ","d":"60 %","e":"none of the above"},"options_float":{"a":400.0,"b":80.0,"c":233.0,"d":60.0,"e":null},"annotated_formula":"multiply(divide(divide(70, const_100), subtract(const_1, divide(70, const_100))), const_100)","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(#0,#1)|multiply(#2,const_100)","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n1 - (7\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n(7\/10) \/ (3\/10)<\/gadget>\n7\/3 = around 2.333333<\/output>\n(7\/3) * 100<\/gadget>\n700\/3 = around 233.333333<\/output>\n700\/3 = around 233.333333<\/result>","index":1536} +{"problem":"if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 52 , what is the compound interest on the same sum at the rate and for the same time ?","rationale":"explanation : sum = ( 52 * 100 ) \/ ( 2 * 5 ) = rs . 520 amount = [ 520 * ( 1 + 5 \/ 100 ) 2 ] = rs . 573.3 c . i . = ( 573.3 - 520 ) = rs . 53.30 . answer : a","correct":"a","options":{"a":"s . 53.30 ","b":"s . 53.22 ","c":"s . 51.219 ","d":"s . 53.18","e":"s . 51.10"},"options_float":{"a":53.3,"b":53.22,"c":51.219,"d":53.18,"e":51.1},"annotated_formula":"subtract(add(add(divide(multiply(divide(52, multiply(divide(5, const_100), 2)), 5), const_100), divide(52, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(52, multiply(divide(5, const_100), 2)), 5), const_100), divide(52, multiply(divide(5, const_100), 2))), 5), const_100)), divide(52, multiply(divide(5, const_100), 2)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 2<\/gadget>\n1\/10 = around 0.1<\/output>\n52 \/ (1\/10)<\/gadget>\n520<\/output>\n520 * 5<\/gadget>\n2_600<\/output>\n2_600 \/ 100<\/gadget>\n26<\/output>\n26 + 520<\/gadget>\n546<\/output>\n546 * 5<\/gadget>\n2_730<\/output>\n2_730 \/ 100<\/gadget>\n273\/10 = around 27.3<\/output>\n546 + (273\/10)<\/gadget>\n5_733\/10 = around 573.3<\/output>\n(5_733\/10) - 520<\/gadget>\n533\/10 = around 53.3<\/output>\n533\/10 = around 53.3<\/result>","index":1539} +{"problem":"29 persons can repair a road in 16 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ?","rationale":"\"let the required number of days be x more persons , less days ( indirect proportion ) more hours , less days ( indirect proportion ) hence we can write as ( persons ) 29 : 30 } : : x : 16 ( hours ) 5 : 6 29 * 5 * 16 = 30 * 6 * x x = 12.88 answer : a\"","correct":"a","options":{"a":"12.88 ","b":"3.88 ","c":"13.88 ","d":"15","e":"18.8"},"options_float":{"a":12.88,"b":3.88,"c":13.88,"d":15.0,"e":18.8},"annotated_formula":"divide(multiply(multiply(29, 16), 5), multiply(30, 6))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|","chain":"29 * 16<\/gadget>\n464<\/output>\n464 * 5<\/gadget>\n2_320<\/output>\n30 * 6<\/gadget>\n180<\/output>\n2_320 \/ 180<\/gadget>\n116\/9 = around 12.888889<\/output>\n116\/9 = around 12.888889<\/result>","index":1540} +{"problem":"the income tax in country x is 8 % of any income up to $ 5000 . for incomes over $ 5000 , an 8 % tax is imposed on the first $ 5000 , and a 10 % tax is imposed on the portion of the total income in excess of $ 5000 . if perry paid $ 950 in income tax last month , then what was perry ' s income ?","rationale":"given : for incomes over $ 5,000 , an 8 % tax is imposed on the first $ 5,000 , and a 10 % tax is imposed on the portion of the total income in excess of $ 5000 . if perry paid $ 950 in income tax last month , then what was perry ' s income ? for first 5000 , 8 % tax then we get 400 $ as tax . now perry paid 950 $ tax . . . we need to check which option is going to yield 550 $ on the remaining amount . a . $ 10,500 - 5000 = 5500 , then we get 10 % tax , we get 550 $ . . . ( $ 10,500 - 5000 - here already for 5000 $ we have calculated the tax and we need to tally for the remaining amount ) 550 + 400 = 950 $ . . . hence this is the option . answer : option a .","correct":"a","options":{"a":"$ 10500 ","b":"$ 11000 ","c":"$ 13000 ","d":"$ 13500","e":"$ 14,000"},"options_float":{"a":10500.0,"b":11000.0,"c":13000.0,"d":13500.0,"e":14000.0},"annotated_formula":"add(add(5000, subtract(950, multiply(5000, divide(8, const_100)))), 5000)","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|subtract(n7,#1)|add(n1,#2)|add(n1,#3)","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n5_000 * (2\/25)<\/gadget>\n400<\/output>\n950 - 400<\/gadget>\n550<\/output>\n5_000 + 550<\/gadget>\n5_550<\/output>\n5_550 + 5_000<\/gadget>\n10_550<\/output>\n10_550<\/result>","index":1541} +{"problem":"a shopkeeper sold an book offering a discount of 5 % and earned a profit of 23.5 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"c 130 % let c . p . be $ 100 . then , s . p . = $ 123.50 let marked price be $ x . then , 95 \/ 100 x = 123.50 x = 12350 \/ 95 = $ 130 now , s . p . = $ 130 , c . p . = $ 100 profit % = 30 % .","correct":"c","options":{"a":"140 ","b":"120 ","c":"130 ","d":"110","e":"150"},"options_float":{"a":140.0,"b":120.0,"c":130.0,"d":110.0,"e":150.0},"annotated_formula":"multiply(const_100, divide(add(const_100, 23.5), subtract(const_100, 5)))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)","chain":"100 + 23.5<\/gadget>\n123.5<\/output>\n100 - 5<\/gadget>\n95<\/output>\n123.5 \/ 95<\/gadget>\n1.3<\/output>\n100 * 1.3<\/gadget>\n130<\/output>\n130<\/result>","index":1542} +{"problem":"if n = 2 ^ 0.15 and n ^ b = 8 , b must equal","rationale":"\"15 \/ 100 = 3 \/ 20 n = 2 ^ 3 \/ 20 n ^ b = 2 ^ 3 ( 2 ^ 3 \/ 20 ) ^ b = 2 ^ 3 b = 20 answer : c\"","correct":"c","options":{"a":"3 \/ 80 ","b":"3 \/ 5 ","c":"20 ","d":"5 \/ 3","e":"80 \/ 3"},"options_float":{"a":0.0375,"b":0.6,"c":20.0,"d":1.6666666667,"e":26.6666666667},"annotated_formula":"divide(log(8), log(power(2, 0.15)))","linear_formula":"log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)|","chain":"log(8)<\/gadget>\nlog(8) = around 2.079442<\/output>\n2 ** 0.15<\/gadget>\n1.109569<\/output>\nlog(1.109569)<\/gadget>\n0.103972<\/output>\nlog(8) \/ 0.103972<\/gadget>\n9.61797406994191*log(8) = around 20.000015<\/output>\n9.61797406994191*log(8) = around 20.000015<\/result>","index":1543} +{"problem":"last year sandy saved 10 % of her annual salary . this year , she made 10 % more money than last year , and she saved 6 % of her salary . the amount saved this year was what percent of the amount she saved last year ?","rationale":"\"let last year ' s salary be x . last year , sandy save 0.1 x this year , sandy saved 0.06 * 1.1 x = 0.066 x 0.066 x \/ 0.1 x = 0.66 = 66 % the answer is a .\"","correct":"a","options":{"a":"66 % ","b":"80 % ","c":"112 % ","d":"136 %","e":"148 %"},"options_float":{"a":66.0,"b":80.0,"c":112.0,"d":136.0,"e":148.0},"annotated_formula":"multiply(divide(multiply(divide(6, const_100), add(const_100, 10)), divide(multiply(10, const_100), const_100)), const_100)","linear_formula":"add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)|","chain":"6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n100 + 10<\/gadget>\n110<\/output>\n(3\/50) * 110<\/gadget>\n33\/5 = around 6.6<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n1_000 \/ 100<\/gadget>\n10<\/output>\n(33\/5) \/ 10<\/gadget>\n33\/50 = around 0.66<\/output>\n(33\/50) * 100<\/gadget>\n66<\/output>\n66<\/result>","index":1544} +{"problem":"the value of a scooter depreciates in such a way that its value of the end of each year is 3 \/ 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40000 , what is the value at the end of 2 years ?","rationale":"explanation : 40,000 * ( 3 \/ 4 ) ^ 2 answer : c","correct":"c","options":{"a":"3277 ","b":"2977 ","c":"22500 ","d":"6077","e":"17112"},"options_float":{"a":3277.0,"b":2977.0,"c":22500.0,"d":6077.0,"e":17112.0},"annotated_formula":"multiply(40000, power(divide(3, 4), 2))","linear_formula":"divide(n0,n1)|power(#0,n3)|multiply(n2,#1)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) ** 2<\/gadget>\n9\/16 = around 0.5625<\/output>\n40_000 * (9\/16)<\/gadget>\n22_500<\/output>\n22_500<\/result>","index":1545} +{"problem":"5 years ago , paula was 3 times as old as karl . in 6 years , paula will be twice as old as karl . what is the sum of their ages now ?","rationale":"p - 5 = 3 ( k - 5 ) and so p = 3 k - 10 p + 6 = 2 ( k + 6 ) ( 3 k - 10 ) + 6 = 2 k + 12 k = 16 p = 38 p + k = 54 the answer is d .","correct":"d","options":{"a":"42 ","b":"46 ","c":"50 ","d":"54","e":"58"},"options_float":{"a":42.0,"b":46.0,"c":50.0,"d":54.0,"e":58.0},"annotated_formula":"add(subtract(multiply(add(negate(subtract(6, multiply(6, const_2))), subtract(multiply(5, 3), 5)), const_3), subtract(multiply(5, 3), 5)), add(negate(subtract(6, multiply(6, const_2))), subtract(multiply(5, 3), 5)))","linear_formula":"multiply(n2,const_2)|multiply(n0,n1)|subtract(n2,#0)|subtract(#1,n0)|negate(#2)|add(#4,#3)|multiply(#5,const_3)|subtract(#6,#3)|add(#5,#7)","chain":"6 * 2<\/gadget>\n12<\/output>\n6 - 12<\/gadget>\n-6<\/output>\n-(-6)<\/gadget>\n6<\/output>\n5 * 3<\/gadget>\n15<\/output>\n15 - 5<\/gadget>\n10<\/output>\n6 + 10<\/gadget>\n16<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48 - 10<\/gadget>\n38<\/output>\n38 + 16<\/gadget>\n54<\/output>\n54<\/result>","index":1546} +{"problem":"one fourth of a solution that was 8 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ?","rationale":"\"say the second solution ( which was 1 \/ 4 th of total ) was x % sugar , then 3 \/ 4 * 0.08 + 1 \/ 4 * x = 1 * 0.16 - - > x = 0.4 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.08 + 25 * x = 100 * 0.16 - - > x = 0.4 . answer : c .\"","correct":"c","options":{"a":"74 % ","b":"52 % ","c":"40 % ","d":"18 %","e":"8.5 %"},"options_float":{"a":74.0,"b":52.0,"c":40.0,"d":18.0,"e":8.5},"annotated_formula":"multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(8, const_100))), multiply(divide(const_1, const_4), const_100)), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100)|","chain":"16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n100 * (4\/25)<\/gadget>\n16<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n100 - 25<\/gadget>\n75<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n75 * (2\/25)<\/gadget>\n6<\/output>\n16 - 6<\/gadget>\n10<\/output>\n10 \/ 25<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":1547} +{"problem":"a and b invests rs . 2000 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?","rationale":"\"( 2 * 6 + 4 * 6 ) : ( 4 * 12 ) 36 : 48 = > 3 : 4 answer : b\"","correct":"b","options":{"a":"3 : 5 ","b":"3 : 4 ","c":"9 : 2 ","d":"9 : 9","e":"3 : 1"},"options_float":{"a":0.6,"b":0.75,"c":4.5,"d":1.0,"e":3.0},"annotated_formula":"divide(add(multiply(2000, 6), multiply(multiply(2000, const_2), 6)), multiply(4000, add(6, 6)))","linear_formula":"add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)|","chain":"2_000 * 6<\/gadget>\n12_000<\/output>\n2_000 * 2<\/gadget>\n4_000<\/output>\n4_000 * 6<\/gadget>\n24_000<\/output>\n12_000 + 24_000<\/gadget>\n36_000<\/output>\n6 + 6<\/gadget>\n12<\/output>\n4_000 * 12<\/gadget>\n48_000<\/output>\n36_000 \/ 48_000<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":1548} +{"problem":"two trains of equal lengths take 8 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ?","rationale":"\"speed of the first train = 120 \/ 8 = 15 m \/ sec . speed of the second train = 120 \/ 15 = 8 m \/ sec . relative speed = 15 + 8 = 23 m \/ sec . required time = ( 120 + 120 ) \/ 23 = 10 sec . answer : d\"","correct":"d","options":{"a":"22 ","b":"12 ","c":"77 ","d":"10","e":"21"},"options_float":{"a":22.0,"b":12.0,"c":77.0,"d":10.0,"e":21.0},"annotated_formula":"divide(multiply(120, const_2), add(speed(120, 15), speed(120, 8)))","linear_formula":"multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)|","chain":"120 * 2<\/gadget>\n240<\/output>\n120 \/ 15<\/gadget>\n8<\/output>\n120 \/ 8<\/gadget>\n15<\/output>\n8 + 15<\/gadget>\n23<\/output>\n240 \/ 23<\/gadget>\n240\/23 = around 10.434783<\/output>\n240\/23 = around 10.434783<\/result>","index":1549} +{"problem":"a certain list consists of 21 different numbers . if n is in the list and n is 5 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ?","rationale":"\"series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 5 * x hence , a 1 + a 2 + . . + a 20 + n = 25 x so , the fraction asked = 5 x \/ 25 x = 1 \/ 5 answer c\"","correct":"c","options":{"a":"1 \/ 20 ","b":"1 \/ 6 ","c":"1 \/ 5 ","d":"4 \/ 21","e":"5 \/ 21"},"options_float":{"a":0.05,"b":0.1666666667,"c":0.2,"d":0.1904761905,"e":0.2380952381},"annotated_formula":"divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, const_4.0), 21), const_4.0), const_2), 5), const_3))","linear_formula":"divide(n2,const_4.0)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,const_4.0)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)|","chain":"1 * 1<\/gadget>\n1<\/output>\n20 \/ 4<\/gadget>\n5<\/output>\n5 + 21<\/gadget>\n26<\/output>\n26 \/ 4<\/gadget>\n13\/2 = around 6.5<\/output>\n(13\/2) * 2<\/gadget>\n13<\/output>\n13 - 5<\/gadget>\n8<\/output>\n8 - 3<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":1550} +{"problem":"danny and steve are running towards each other , each one from his own house . danny can reach steve ' s house in 27 minutes of running , which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time , how much time longer will it take steve to reach the halfway point between their houses than danny to reach the halfway point between their houses ?","rationale":"\"danny ' s time is 27 minutes . the time to reach halfway is 13.5 minutes . steve ' s time is 54 minutes . the time to reach halfway is 27 minutes . the time difference is 27 - 13.5 = 13.5 minutes the answer is a .\"","correct":"a","options":{"a":"13.5 minutes ","b":"27 minutes ","c":"35 minutes ","d":"54 minutes","e":"81 minutes"},"options_float":{"a":13.5,"b":27.0,"c":35.0,"d":54.0,"e":81.0},"annotated_formula":"divide(27, const_2)","linear_formula":"divide(n0,const_2)|","chain":"27 \/ 2<\/gadget>\n27\/2 = around 13.5<\/output>\n27\/2 = around 13.5<\/result>","index":1551} +{"problem":"a number exceeds by 40 from its 3 \/ 8 part . then the number is ?","rationale":"\"x – 3 \/ 8 x = 40 x = 64 answer : c\"","correct":"c","options":{"a":"32 ","b":"24 ","c":"64 ","d":"40","e":"45"},"options_float":{"a":32.0,"b":24.0,"c":64.0,"d":40.0,"e":45.0},"annotated_formula":"divide(multiply(40, 8), subtract(8, 3))","linear_formula":"multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)|","chain":"40 * 8<\/gadget>\n320<\/output>\n8 - 3<\/gadget>\n5<\/output>\n320 \/ 5<\/gadget>\n64<\/output>\n64<\/result>","index":1553} +{"problem":"amar takes as much time in running 22 meters as a car takes in covering 58 meters . what will be the distance covered by amar during the time the car covers 2.6 km ?","rationale":"\"distance covered by amar = 22 \/ 58 ( 2.6 km ) = 11 \/ 29 ( 2600 ) = 986 m answer : d\"","correct":"d","options":{"a":"700 m ","b":"978 m ","c":"970 m ","d":"986 m","e":"640 m"},"options_float":{"a":700.0,"b":978.0,"c":970.0,"d":986.0,"e":640.0},"annotated_formula":"divide(multiply(22, multiply(2.6, const_1000)), 58)","linear_formula":"multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)|","chain":"2.6 * 1_000<\/gadget>\n2_600<\/output>\n22 * 2_600<\/gadget>\n57_200<\/output>\n57_200 \/ 58<\/gadget>\n28_600\/29 = around 986.206897<\/output>\n28_600\/29 = around 986.206897<\/result>","index":1554} +{"problem":"a 340 - liter solution of kola is made from 88 % water , 5 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 10 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ?","rationale":"\"denominator : 340 + 10 + 3.2 + 6.8 = 360 numerator : 340 ( 1 - . 88 - . 05 ) + 3.2 340 ( 0.07 ) + 3.2 23.8 + 3.2 27 ratio : 27 \/ 360 = 3 \/ 40 answer : b\"","correct":"b","options":{"a":"6 % . ","b":"7.5 % . ","c":"9.2 % . ","d":"10.5 % .","e":"11 % ."},"options_float":{"a":6.0,"b":7.5,"c":9.2,"d":10.5,"e":11.0},"annotated_formula":"multiply(divide(add(subtract(subtract(340, multiply(340, divide(88, const_100))), multiply(340, divide(5, const_100))), 3.2), add(add(add(340, 3.2), 10), 6.8)), const_100)","linear_formula":"add(n0,n3)|divide(n1,const_100)|divide(n2,const_100)|add(n4,#0)|multiply(n0,#1)|multiply(n0,#2)|add(n5,#3)|subtract(n0,#4)|subtract(#7,#5)|add(n3,#8)|divide(#9,#6)|multiply(#10,const_100)|","chain":"88 \/ 100<\/gadget>\n22\/25 = around 0.88<\/output>\n340 * (22\/25)<\/gadget>\n1_496\/5 = around 299.2<\/output>\n340 - (1_496\/5)<\/gadget>\n204\/5 = around 40.8<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n340 * (1\/20)<\/gadget>\n17<\/output>\n(204\/5) - 17<\/gadget>\n119\/5 = around 23.8<\/output>\n(119\/5) + 3.2<\/gadget>\n27<\/output>\n340 + 3.2<\/gadget>\n343.2<\/output>\n343.2 + 10<\/gadget>\n353.2<\/output>\n353.2 + 6.8<\/gadget>\n360<\/output>\n27 \/ 360<\/gadget>\n3\/40 = around 0.075<\/output>\n(3\/40) * 100<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":1555} +{"problem":"when the perimeter of a regular polygon is divided by 5 , the length of a side is 25 . what is the name of the polygon ? what is the perimeter ?","rationale":"regular polygon . a polygon with equal sides and equal sides . divided by 5 to get the length of a side . it is the pentagon since it has 5 sides . so p = 5 × s to get the perimeter , just multiply a side by 5 . since 25 × 5 = 125 , the perimeter is 125 . answer a","correct":"a","options":{"a":"125 ","b":"126 ","c":"127 ","d":"128","e":"129"},"options_float":{"a":125.0,"b":126.0,"c":127.0,"d":128.0,"e":129.0},"annotated_formula":"multiply(25, divide(25, 5))","linear_formula":"divide(n1,n0)|multiply(n1,#0)","chain":"25 \/ 5<\/gadget>\n5<\/output>\n25 * 5<\/gadget>\n125<\/output>\n125<\/result>","index":1558} +{"problem":"a vessel of capacity 60 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ?","rationale":"explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) \/ t ] ^ n * t for the given problem , t = 60 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 60 - 9 ) \/ 60 ] ^ 2 ( 60 ) = 43.35 litres . answer : option d","correct":"d","options":{"a":"23.89 ","b":"72.9 ","c":"38.3 ","d":"43.35","e":"79.3"},"options_float":{"a":23.89,"b":72.9,"c":38.3,"d":43.35,"e":79.3},"annotated_formula":"subtract(subtract(60, multiply(const_3, const_3)), multiply(divide(subtract(60, multiply(const_3, const_3)), 60), multiply(const_3, const_3)))","linear_formula":"multiply(const_3,const_3)|subtract(n0,#0)|divide(#1,n0)|multiply(#2,#0)|subtract(#1,#3)","chain":"3 * 3<\/gadget>\n9<\/output>\n60 - 9<\/gadget>\n51<\/output>\n51 \/ 60<\/gadget>\n17\/20 = around 0.85<\/output>\n(17\/20) * 9<\/gadget>\n153\/20 = around 7.65<\/output>\n51 - (153\/20)<\/gadget>\n867\/20 = around 43.35<\/output>\n867\/20 = around 43.35<\/result>","index":1559} +{"problem":"an empty wooden vessel weighs 8 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled .","rationale":"\"an empty wooden vessel weighs 8 % of its total weight when filled with paint : vessel = 0.08 ( vessel + paint ) ; 8 v = v + p ( so the weight of completely filled vessel is 8 v ) p = 7 v ( so the weight of the paint when the vessels is completely filled is 7 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 \/ 2 * 8 v ; p ' = 3 v ( so the weight of the paint when the vessels is partially filled is 3 v ) . what fraction of the vessel is filled ? so , we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled : p ' \/ p = 3 v \/ 7 v = 3 \/ 7 . answer : a .\"","correct":"a","options":{"a":"3 \/ 7 ","b":"5 \/ 9 ","c":"1 \/ 24 ","d":"4 \/ 9","e":"2 \/ 5"},"options_float":{"a":0.4285714286,"b":0.5555555556,"c":0.0416666667,"d":0.4444444444,"e":0.4},"annotated_formula":"divide(subtract(divide(8, const_2), const_1), subtract(8, const_1))","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|subtract(#0,const_1)|divide(#2,#1)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\n4 - 1<\/gadget>\n3<\/output>\n8 - 1<\/gadget>\n7<\/output>\n3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":1560} +{"problem":"in a government institute 20 nos . staffs are employed . in a festival season their boss give festival allowance to all . he says his accountant to calculate for 30 days @ 100 . also he says balance payment will give to driver and cleaning staff and to be adjust from petty cash . he had given rs . 65000 \/ - to the accountant . how much amount will taken from petty cash ?","rationale":"for 20 staffs : 300 * 100 * 20 = 60000 for driver and cleaning staff : 300 * 100 * 2 = 6000 from petty cash : 1000 answer is a","correct":"a","options":{"a":"1000 ","b":"3000 ","c":"3500 ","d":"2500","e":"2250"},"options_float":{"a":1000.0,"b":3000.0,"c":3500.0,"d":2500.0,"e":2250.0},"annotated_formula":"divide(65000, subtract(100, add(20, divide(30, const_2))))","linear_formula":"divide(n1,const_2)|add(n0,#0)|subtract(n2,#1)|divide(n3,#2)","chain":"30 \/ 2<\/gadget>\n15<\/output>\n20 + 15<\/gadget>\n35<\/output>\n100 - 35<\/gadget>\n65<\/output>\n65_000 \/ 65<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1561} +{"problem":"if $ 5,000 is invested in an account that earns 4 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ?","rationale":"\"solution amount ( ci ) = p + ( 1 + r \/ n ) ^ nt = 5000 + ( 1 + 0.04 \/ 2 ) ^ 2 = 5404 amount ( si ) = p + ptr \/ 100 = 5000 + ( 5000 * 1 * 4 \/ 100 ) = 5400 difference = 5404 - 5400 = 4 $ a\"","correct":"a","options":{"a":"$ 4 ","b":"$ 8 ","c":"$ 12 ","d":"$ 16","e":"$ 432"},"options_float":{"a":4.0,"b":8.0,"c":12.0,"d":16.0,"e":432.0},"annotated_formula":"multiply(4, const_1)","linear_formula":"multiply(n1,const_1)|","chain":"4 * 1<\/gadget>\n4<\/output>\n4<\/result>","index":1562} +{"problem":"an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 7 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ?","rationale":"\"explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 7 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 2100 . hence , ( r - p ) = ( 2100 - 900 ) = 1200 answer : a\"","correct":"a","options":{"a":"rs 1200 ","b":"rs 1500 ","c":"rs 1600 ","d":"rs 1900","e":"rs 1700"},"options_float":{"a":1200.0,"b":1500.0,"c":1600.0,"d":1900.0,"e":1700.0},"annotated_formula":"multiply(divide(1500, 5), subtract(7, 3))","linear_formula":"divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)|","chain":"1_500 \/ 5<\/gadget>\n300<\/output>\n7 - 3<\/gadget>\n4<\/output>\n300 * 4<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":1563} +{"problem":"if a sum of money doubles itself in 25 years at simple interest , the ratepercent per annum is","rationale":"\"explanation : let sum = x then simple interest = x rate = ( 100 * x ) \/ ( x * 25 ) = 4 option e\"","correct":"e","options":{"a":"12 ","b":"12.5 ","c":"13 ","d":"13.5","e":"4"},"options_float":{"a":12.0,"b":12.5,"c":13.0,"d":13.5,"e":4.0},"annotated_formula":"divide(divide(const_2, divide(25, const_100)), const_2)","linear_formula":"divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n2 \/ (1\/4)<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":1566} +{"problem":"a company produces 50000 bottles of water everyday . if a case can hold 25 bottles of water . how many cases are required by the company to hold its one day production","rationale":"\"number of bottles that can be held in a case = 25 number of cases required to hold 50000 bottles = 50000 \/ 25 = 2000 cases . so the answer is a = 2000\"","correct":"a","options":{"a":"2000 ","b":"4500 ","c":"5000 ","d":"8000","e":"9000"},"options_float":{"a":2000.0,"b":4500.0,"c":5000.0,"d":8000.0,"e":9000.0},"annotated_formula":"divide(50000, 25)","linear_formula":"divide(n0,n1)|","chain":"50_000 \/ 25<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":1567} +{"problem":"a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 25 gallons of gasoline ?","rationale":"\"to maximize the distance that suv could be driven on 25 gallons of gasoline , we need consider only highway driving . max distance = 25 * 12.2 = 305 answer d\"","correct":"d","options":{"a":"190 ","b":"284.6 ","c":"300 ","d":"305","e":"312"},"options_float":{"a":190.0,"b":284.6,"c":300.0,"d":305.0,"e":312.0},"annotated_formula":"multiply(12.2, 25)","linear_formula":"multiply(n0,n2)|","chain":"12.2 * 25<\/gadget>\n305<\/output>\n305<\/result>","index":1571} +{"problem":"5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 120 only . men â € ™ s wages are ?","rationale":"\"5 m = xw = 8 b 5 m + xw + 8 b - - - - - 120 rs . 5 m + 5 m + 5 m - - - - - 120 rs . 15 m - - - - - - 120 rs . = > 1 m = 8 rs . answer : e\"","correct":"e","options":{"a":"6 rs ","b":"2 rs ","c":"4 rs ","d":"9 rs","e":"8 rs"},"options_float":{"a":6.0,"b":2.0,"c":4.0,"d":9.0,"e":8.0},"annotated_formula":"divide(120, multiply(const_3, 5))","linear_formula":"multiply(n0,const_3)|divide(n2,#0)|","chain":"3 * 5<\/gadget>\n15<\/output>\n120 \/ 15<\/gadget>\n8<\/output>\n8<\/result>","index":1572} +{"problem":"a and b are two partially filled buckets of water . if 5 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 4 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ?","rationale":"\"let a contains a , b contains b liters so , ( a - 5 ) \/ ( b + 5 ) = 1 \/ 3 . . . . . . . ( 1 ) again , ( b - 4 ) \/ ( a + 4 ) = 1 \/ 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 52 \/ 5 ans : ( b )\"","correct":"b","options":{"a":"11 ","b":"52 \/ 5 ","c":"17 ","d":"21","e":"23"},"options_float":{"a":11.0,"b":10.4,"c":17.0,"d":21.0,"e":23.0},"annotated_formula":"divide(multiply(add(add(5, const_3), const_2), divide(5, const_2)), add(const_2, divide(const_1, const_2)))","linear_formula":"add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)|","chain":"5 + 3<\/gadget>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n5 \/ 2<\/gadget>\n5\/2 = around 2.5<\/output>\n10 * (5\/2)<\/gadget>\n25<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n2 + (1\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n25 \/ (5\/2)<\/gadget>\n10<\/output>\n10<\/result>","index":1574} +{"problem":"a company pays 12.5 % dividend to its investors . if an investor buys rs . 60 shares and gets 25 % on investment , at what price did the investor buy the shares ?","rationale":"explanation : dividend on 1 share = ( 12.5 * 60 ) \/ 100 = rs . 7.5 rs . 25 is income on an investment of rs . 100 rs . 7.5 is income on an investment of rs . ( 7.5 * 100 ) \/ 25 = rs . 30 answer : c","correct":"c","options":{"a":"25 ","b":"66 ","c":"30 ","d":"19","e":"01"},"options_float":{"a":25.0,"b":66.0,"c":30.0,"d":19.0,"e":1.0},"annotated_formula":"divide(multiply(divide(multiply(12.5, 60), const_100), const_100), 25)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)","chain":"12.5 * 60<\/gadget>\n750<\/output>\n750 \/ 100<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) * 100<\/gadget>\n750<\/output>\n750 \/ 25<\/gadget>\n30<\/output>\n30<\/result>","index":1575} +{"problem":"what is the 10 digit to the right of the decimal point , in the decimal expansion of ( 1 \/ 2 ) ^ 10 ?","rationale":"step 1 convert fraction to decimal with two decimal places 1 \/ 2 = 0.50 then convert to scientific number = 5 * 10 ^ - 1 step 2 multiply 5 ^ 10 by 10 ^ - 10 step 3 5 has the sequence 5 , 25 , 125 answer is e","correct":"e","options":{"a":"9 ","b":"4 ","c":"2 ","d":"1","e":"5"},"options_float":{"a":9.0,"b":4.0,"c":2.0,"d":1.0,"e":5.0},"annotated_formula":"add(multiply(2, 2), 1)","linear_formula":"multiply(n2,n2)|add(n1,#0)","chain":"2 * 2<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5<\/result>","index":1577} +{"problem":"find the curved surface area , if the radius of a cone is 35 m and slant height is 30 m ?","rationale":"\"cone curved surface area = ï € rl 22 \/ 7 ã — 35 ã — 30 = 110 ã — 30 = 3300 m ( power 2 ) answer is c .\"","correct":"c","options":{"a":"1100 ","b":"2200 ","c":"3300 ","d":"4400","e":"5500"},"options_float":{"a":1100.0,"b":2200.0,"c":3300.0,"d":4400.0,"e":5500.0},"annotated_formula":"multiply(const_pi, multiply(35, 30))","linear_formula":"multiply(n0,n1)|multiply(#0,const_pi)|","chain":"35 * 30<\/gadget>\n1_050<\/output>\npi * 1_050<\/gadget>\n1050*pi = around 3_298.672286<\/output>\n1050*pi = around 3_298.672286<\/result>","index":1578} +{"problem":"a man have 8 t - shirt and some pant . he can dress in 72 ways . then tell me the no . of pant that man have .","rationale":"let he have x pant . by fundamental principle total no . of ways in which he can dress = 8 x 8 x = 72 x = 9 pants answer b","correct":"b","options":{"a":"8 ","b":"9 ","c":"10 ","d":"11","e":"12"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":12.0},"annotated_formula":"divide(72, 8)","linear_formula":"divide(n1,n0)","chain":"72 \/ 8<\/gadget>\n9<\/output>\n9<\/result>","index":1580} +{"problem":"on average , machine a will produce a completed item once every 4 minutes , and machine b will produce a completed item once every 5 minutes . the number of items machine a produces in a 24 - hour period is approximately what percent greater than the number of items machine b would produce in that same period ?","rationale":"machine a : 1 item in 4 minutes - - > 1 \/ 4 items in 1 minute ; machine b : 1 item in 5 minutes - - > 1 \/ 5 items in 1 minute . in the same time period , machine a produces ( 1 \/ 4 - 1 \/ 5 ) \/ ( 1 \/ 5 ) * 100 = ~ 25 % more items than machine b . answer : a .","correct":"a","options":{"a":"25 % ","b":"15 % ","c":"45 % ","d":"33 %","e":"55 %"},"options_float":{"a":25.0,"b":15.0,"c":45.0,"d":33.0,"e":55.0},"annotated_formula":"subtract(multiply(divide(multiply(divide(const_60, 4), 24), multiply(divide(const_60, 5), 24)), const_100), const_100)","linear_formula":"divide(const_60,n0)|divide(const_60,n1)|multiply(n2,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100)","chain":"60 \/ 4<\/gadget>\n15<\/output>\n15 * 24<\/gadget>\n360<\/output>\n60 \/ 5<\/gadget>\n12<\/output>\n12 * 24<\/gadget>\n288<\/output>\n360 \/ 288<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 100<\/gadget>\n125<\/output>\n125 - 100<\/gadget>\n25<\/output>\n25<\/result>","index":1581} +{"problem":"what is the divisor ? the dividend is 76 , the remainder is 8 , the quotient is 4","rationale":"divisor = dividend - remainder divided by the quotient divisor = 76 - 8 divided by 4 divisor = 68 \/ 4 divisor = 17","correct":"d","options":{"a":"32 ","b":"48 ","c":"67 ","d":"17","e":"26"},"options_float":{"a":32.0,"b":48.0,"c":67.0,"d":17.0,"e":26.0},"annotated_formula":"divide(subtract(76, 8), 4)","linear_formula":"subtract(n0,n1)|divide(#0,n2)","chain":"76 - 8<\/gadget>\n68<\/output>\n68 \/ 4<\/gadget>\n17<\/output>\n17<\/result>","index":1582} +{"problem":"at what rate percent per annum will a sum of money double in 6 years .","rationale":"\"let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) \/ ( p x 6 ) ] % = 16.6 % per annum . answer : e\"","correct":"e","options":{"a":"12.5 % ","b":"13.5 % ","c":"11.5 % ","d":"14.5 %","e":"16.6 %"},"options_float":{"a":12.5,"b":13.5,"c":11.5,"d":14.5,"e":16.6},"annotated_formula":"divide(const_100, 6)","linear_formula":"divide(const_100,n0)|","chain":"100 \/ 6<\/gadget>\n50\/3 = around 16.666667<\/output>\n50\/3 = around 16.666667<\/result>","index":1584} +{"problem":"if 8 gallons of gasoline are added to a tank that is already filled to 3 \/ 4 of its capacity , the tank is then filled to 9 \/ 10 of its capacity . how many gallons does the tank hold ?","rationale":"\"let the capacity of the tank = c ( 3 \/ 4 ) c + 8 = ( 9 \/ 10 ) c = > ( 9 \/ 10 ) c - ( 3 \/ 4 ) c = 8 = > ( 3 \/ 20 ) c = 8 = > c = ( 8 * 20 ) \/ 3 = 53.33 number of gallons of gasoline that the tank currently holds = 3 \/ 4 * c + 8 = 39.99 + 8 = 47.99 = 48 answer d\"","correct":"d","options":{"a":"20 ","b":"24 ","c":"36 ","d":"48","e":"60"},"options_float":{"a":20.0,"b":24.0,"c":36.0,"d":48.0,"e":60.0},"annotated_formula":"add(multiply(divide(8, subtract(divide(9, 10), divide(3, 4))), divide(3, 4)), 8)","linear_formula":"divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(n0,#2)|multiply(#3,#1)|add(n0,#4)|","chain":"9 \/ 10<\/gadget>\n9\/10 = around 0.9<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(9\/10) - (3\/4)<\/gadget>\n3\/20 = around 0.15<\/output>\n8 \/ (3\/20)<\/gadget>\n160\/3 = around 53.333333<\/output>\n(160\/3) * (3\/4)<\/gadget>\n40<\/output>\n40 + 8<\/gadget>\n48<\/output>\n48<\/result>","index":1585} +{"problem":"a certain company ’ s profit in 1996 was 15 percent greater than its profit in 1995 , and its profit in 1997 was 20 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ?","rationale":"\"soln : - profit in 1995 - 100 profit in 1996 - 115 % increase profit in 1997 in comparison to 1995 = 15 + 115 * 20 % = 38 % answer : e\"","correct":"e","options":{"a":"5 % ","b":"18 % ","c":"33 % ","d":"35 %","e":"38 %"},"options_float":{"a":5.0,"b":18.0,"c":33.0,"d":35.0,"e":38.0},"annotated_formula":"multiply(subtract(multiply(add(divide(20, const_100), const_1), add(const_1, divide(15, const_100))), const_1), const_100)","linear_formula":"divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n(6\/5) * (23\/20)<\/gadget>\n69\/50 = around 1.38<\/output>\n(69\/50) - 1<\/gadget>\n19\/50 = around 0.38<\/output>\n(19\/50) * 100<\/gadget>\n38<\/output>\n38<\/result>","index":1586} +{"problem":"i chose a number and divide it by 2 . then i subtracted 100 from the result and got 4 . what was the number i chose ?","rationale":"\"let x be the number i chose , then x \/ 2 − 100 = 4 x \/ 2 = 104 x = 208 answer is b .\"","correct":"b","options":{"a":"200 ","b":"208 ","c":"400 ","d":"500","e":"100"},"options_float":{"a":200.0,"b":208.0,"c":400.0,"d":500.0,"e":100.0},"annotated_formula":"multiply(add(100, 4), 2)","linear_formula":"add(n1,n2)|multiply(n0,#0)|","chain":"100 + 4<\/gadget>\n104<\/output>\n104 * 2<\/gadget>\n208<\/output>\n208<\/result>","index":1588} +{"problem":"the perimeter of an isosceles right triangle is 12 + 12 sq rt 2 . what is the length of the hypotenuse of the triangle ?","rationale":"side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 12 + 12 sqrt 2 or , a . ( 2 + sqrt 2 ) = 12 ( 1 + sqrt 2 ) , a = 12 * ( 1 + sqrt 2 ) \/ 2 + sqrt 2 = 12 * 2.414 \/ 3.414 = then hypotenuse = 8.485 d","correct":"d","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8.485","e":"8"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.485,"e":8.0},"annotated_formula":"divide(multiply(12, sqrt(2)), const_2)","linear_formula":"sqrt(n2)|multiply(n0,#0)|divide(#1,const_2)","chain":"2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n12 * (sqrt(2))<\/gadget>\n12*sqrt(2) = around 16.970563<\/output>\n(12*sqrt(2)) \/ 2<\/gadget>\n6*sqrt(2) = around 8.485281<\/output>\n6*sqrt(2) = around 8.485281<\/result>","index":1589} +{"problem":"a certain protective additive increases from 50 days to 60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ?","rationale":"\"general formula for percent increase or decrease , ( percent change ) : percent = change \/ original ∗ 100 so , the time between maintenance checks increased by 60 − 50 \/ 50 ∗ 100 = 20 answer : e .\"","correct":"e","options":{"a":"25 % ","b":"33 1 \/ 3 % ","c":"50 % ","d":"66 2 \/ 3 %","e":"20 %"},"options_float":{"a":25.0,"b":33.0,"c":50.0,"d":66.0,"e":20.0},"annotated_formula":"multiply(divide(subtract(60, 50), 50), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"60 - 50<\/gadget>\n10<\/output>\n10 \/ 50<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":1591} +{"problem":"given that 268 * 74 = 19832 , find the values of 2.68 * 0.74 .","rationale":"sum of decimal places = ( 2 + 2 ) = 4 2.68 * 0.74 = 1.9832 . answer is b .","correct":"b","options":{"a":"1.9328 ","b":"1.9832 ","c":"1.9823 ","d":"1.9382","e":"none of them"},"options_float":{"a":1.9328,"b":1.9832,"c":1.9823,"d":1.9382,"e":null},"annotated_formula":"multiply(2.68, 0.74)","linear_formula":"multiply(n3,n4)","chain":"2.68 * 0.74<\/gadget>\n1.9832<\/output>\n1.9832<\/result>","index":1592} +{"problem":"the probability of two events a and b are 0.25 and 0.40 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________","rationale":"\"we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 40 - . 15 = . 50 but the probability of neither a nor b = 1 - . 50 = 0.50 answer : c\"","correct":"c","options":{"a":"0.45 ","b":"0.4 ","c":"0.5 ","d":"0.05","e":"0.6"},"options_float":{"a":0.45,"b":0.4,"c":0.5,"d":0.05,"e":0.6},"annotated_formula":"subtract(const_1, subtract(add(0.25, 0.40), 0.15))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)|","chain":"0.25 + 0.4<\/gadget>\n0.65<\/output>\n0.65 - 0.15<\/gadget>\n0.5<\/output>\n1 - 0.5<\/gadget>\n0.5<\/output>\n0.5<\/result>","index":1593} +{"problem":"if henry were to add 9 gallons of water to a tank that is already 3 \/ 4 full of water , the tank would be 7 \/ 8 full . how many gallons of water would the tank hold if it were full ?","rationale":"\"7 \/ 8 x - 3 \/ 4 x = 9 galls 1 \/ 8 * x = 9 gallons x = 72 gallons answer d\"","correct":"d","options":{"a":"25 ","b":"40 ","c":"64 ","d":"72","e":"96"},"options_float":{"a":25.0,"b":40.0,"c":64.0,"d":72.0,"e":96.0},"annotated_formula":"multiply(9, divide(const_1, subtract(divide(7, 8), divide(3, 4))))","linear_formula":"divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(const_1,#2)|multiply(n0,#3)|","chain":"7 \/ 8<\/gadget>\n7\/8 = around 0.875<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(7\/8) - (3\/4)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n9 * 8<\/gadget>\n72<\/output>\n72<\/result>","index":1595} +{"problem":"bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne ’ s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many w hours does it currently take anne to clean the house on her own ?","rationale":"\"lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 \/ a + 1 \/ b portion of the work which equals 1 \/ 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 \/ a + 2 \/ b which is w equal to 1 \/ 3 ( as the work is completed in w = 3 hours ) solving these 2 equations we can find b as 12 so , d\"","correct":"d","options":{"a":"6 ","b":"7 ","c":"8 ","d":"12","e":"14"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":12.0,"e":14.0},"annotated_formula":"inverse(subtract(divide(const_1, 3), divide(const_1, 4)))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/3) - (1\/4)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":1596} +{"problem":"the price of a book is increased from $ 300 to $ 330 . what is the % of increase in its price ?","rationale":"explanation : change in the price = rs 330 – rs 300 = rs 30 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 30 \/ 300 ) * 100 = 10 % a","correct":"a","options":{"a":"10 % ","b":"20 % ","c":"30 % ","d":"35 %","e":"45 %"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":35.0,"e":45.0},"annotated_formula":"multiply(divide(subtract(330, 300), 300), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)","chain":"330 - 300<\/gadget>\n30<\/output>\n30 \/ 300<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":1598} +{"problem":"a sum of 14,800 amounts to 19,800 in 3 years at the rate of simple interest . what is the rate of interest ?","rationale":"\"b 11 % s . i . = ( 19800 - 14800 ) = 5000 . rate = ( 100 x 5000 ) \/ ( 14800 x 3 ) % = 11 %\"","correct":"b","options":{"a":"12 % ","b":"11 % ","c":"16 % ","d":"15 %","e":"18 %"},"options_float":{"a":12.0,"b":11.0,"c":16.0,"d":15.0,"e":18.0},"annotated_formula":"multiply(divide(divide(const_3, 3), add(multiply(const_3, 3), add(const_0_25, const_0_25))), const_100)","linear_formula":"add(const_0_25,const_0_25)|divide(const_3,n2)|multiply(const_3,n2)|add(#0,#2)|divide(#1,#3)|multiply(#4,const_100)|","chain":"3 \/ 3<\/gadget>\n1<\/output>\n3 * 3<\/gadget>\n9<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n9 + (1\/2)<\/gadget>\n19\/2 = around 9.5<\/output>\n1 \/ (19\/2)<\/gadget>\n2\/19 = around 0.105263<\/output>\n(2\/19) * 100<\/gadget>\n200\/19 = around 10.526316<\/output>\n200\/19 = around 10.526316<\/result>","index":1599} +{"problem":"from the sale of sleeping bags , a retailer made a gross profit of 16 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ?","rationale":"\"cost price * 1.16 = selling price - - > cost price * 1.16 = $ 28 - - > cost price = $ 24.13 . answer : c .\"","correct":"c","options":{"a":"3.0 ","b":"3.36 ","c":"24.13 ","d":"25.0","e":"31.36"},"options_float":{"a":3.0,"b":3.36,"c":24.13,"d":25.0,"e":31.36},"annotated_formula":"divide(multiply(28, const_100), add(const_100, 16))","linear_formula":"add(n0,const_100)|multiply(n1,const_100)|divide(#1,#0)|","chain":"28 * 100<\/gadget>\n2_800<\/output>\n100 + 16<\/gadget>\n116<\/output>\n2_800 \/ 116<\/gadget>\n700\/29 = around 24.137931<\/output>\n700\/29 = around 24.137931<\/result>","index":1602} +{"problem":"a number is doubled and 5 is added . if the resultant is trebled , it becomes 129 . what is that number ?","rationale":"explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 129 6 x + 15 = 129 6 x = 114 x = 19 answer : d","correct":"d","options":{"a":"12 ","b":"29 ","c":"27 ","d":"19","e":"99"},"options_float":{"a":12.0,"b":29.0,"c":27.0,"d":19.0,"e":99.0},"annotated_formula":"divide(subtract(129, multiply(const_3, 5)), multiply(const_3, const_2))","linear_formula":"multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)","chain":"3 * 5<\/gadget>\n15<\/output>\n129 - 15<\/gadget>\n114<\/output>\n3 * 2<\/gadget>\n6<\/output>\n114 \/ 6<\/gadget>\n19<\/output>\n19<\/result>","index":1603} +{"problem":"the average temperature for tuesday , wednesday and thursday was 42 ° c . the average temperature for wednesday , thursday and friday was 44 ° c . if the temperature on friday be 43 ° c , what was the temperature on tuesday ?","rationale":"\"explanation : t + w + t = 42 × 3 = 126 ° c w + t + f = 44 × 3 = 132 ° c also , temperature on friday = 43 ° c temperature on tuesday = 126 + 43 - 132 = 37 ° c answer : option c\"","correct":"c","options":{"a":"39 ° c ","b":"44 ° c ","c":"37 ° c ","d":"42 ° c","e":"none of these"},"options_float":{"a":39.0,"b":44.0,"c":37.0,"d":42.0,"e":null},"annotated_formula":"subtract(multiply(42, const_3), subtract(multiply(44, const_3), 43))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_3)|subtract(#1,n2)|subtract(#0,#2)|","chain":"42 * 3<\/gadget>\n126<\/output>\n44 * 3<\/gadget>\n132<\/output>\n132 - 43<\/gadget>\n89<\/output>\n126 - 89<\/gadget>\n37<\/output>\n37<\/result>","index":1604} +{"problem":"if a , b , and c are positive real numbers such that a ( b + c ) = 152 , b ( c + a ) = 162 , and c ( a + b ) = 170 , then abc is","rationale":"ab + bc = 152 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 ) bc + ba = 162 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ) ac + ab = 170 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 ) re - writing equation 3 as follows : ac + cb = 162 + 8 ac + cb = bc + ba + 8 ac = ba + 8 . . . . . . . . . . . . . . . ( 4 ) adding ( 1 ) ( 4 ) 2 ac = 160 ac = 80 abc has to be multiple of 80 , only 720 fits in answer = d","correct":"d","options":{"a":"672 ","b":"688 ","c":"704 ","d":"720","e":"750"},"options_float":{"a":672.0,"b":688.0,"c":704.0,"d":720.0,"e":750.0},"annotated_formula":"add(add(add(const_10, const_10), add(const_4, add(170, add(152, add(170, 162))))), add(const_10, const_10))","linear_formula":"add(const_10,const_10)|add(n1,n2)|add(n0,#1)|add(n2,#2)|add(#3,const_4)|add(#0,#4)|add(#5,#0)","chain":"10 + 10<\/gadget>\n20<\/output>\n170 + 162<\/gadget>\n332<\/output>\n152 + 332<\/gadget>\n484<\/output>\n170 + 484<\/gadget>\n654<\/output>\n4 + 654<\/gadget>\n658<\/output>\n20 + 658<\/gadget>\n678<\/output>\n678 + 20<\/gadget>\n698<\/output>\n698<\/result>","index":1607} +{"problem":"s is a set of 50 consecutive multiples of 2 . if the smallest number in s is 56 , then the greatest number in s is","rationale":"\"last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 50 consecutive multiples of 2 . if the smallest number in s is 56 , then the greatest number in s is first term = 56 ; total terms = 50 ; difference = 2 56 + ( 49 ) 2 = 154 ans b\"","correct":"b","options":{"a":"198 ","b":"154 ","c":"199 ","d":"102","e":"105"},"options_float":{"a":198.0,"b":154.0,"c":199.0,"d":102.0,"e":105.0},"annotated_formula":"add(56, multiply(subtract(50, const_1), 2))","linear_formula":"subtract(n0,const_1)|multiply(n1,#0)|add(n2,#1)|","chain":"50 - 1<\/gadget>\n49<\/output>\n49 * 2<\/gadget>\n98<\/output>\n56 + 98<\/gadget>\n154<\/output>\n154<\/result>","index":1608} +{"problem":"the total number of digit used in numbering the pages of a book having 366 pages , is :","rationale":"solution total number of digits = ( no . of digits in 1 - digits page nos . + no . of digits in 2 - digits page nos . + no . of digits i 3 - digits page nos . ) = ( 1 * 9 + 2 * 90 + 3 * 267 ) = ( 9 + 180 + 801 ) = 990 . answer b","correct":"b","options":{"a":"732 ","b":"990 ","c":"1098 ","d":"1305","e":"none of these"},"options_float":{"a":732.0,"b":990.0,"c":1098.0,"d":1305.0,"e":null},"annotated_formula":"add(add(subtract(const_10, const_1), multiply(const_2, multiply(const_10, subtract(const_10, const_1)))), multiply(const_3, subtract(366, subtract(const_100, const_1))))","linear_formula":"subtract(const_10,const_1)|subtract(const_100,const_1)|multiply(#0,const_10)|subtract(n0,#1)|multiply(#2,const_2)|multiply(#3,const_3)|add(#4,#0)|add(#6,#5)","chain":"10 - 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275 = 75 answer : b","correct":"b","options":{"a":"22 ","b":"75 ","c":"25 ","d":"70","e":"11"},"options_float":{"a":22.0,"b":75.0,"c":25.0,"d":70.0,"e":11.0},"annotated_formula":"divide(subtract(subtract(350, 25), 25), 4)","linear_formula":"subtract(n8,n5)|subtract(#0,n5)|divide(#1,n7)","chain":"350 - 25<\/gadget>\n325<\/output>\n325 - 25<\/gadget>\n300<\/output>\n300 \/ 4<\/gadget>\n75<\/output>\n75<\/result>","index":1612} +{"problem":"veena ranks 73 rd from the top in a class of 198 . what is her rank from the bottom if 22 students have failed the examination ?","rationale":"\"total student = 198 failed = 22 paasd student = 198 - 22 = 176 from bottom her rank is = 176 - 73 + 1 = 104 answer : b\"","correct":"b","options":{"a":"88 ","b":"104 ","c":"110 ","d":"90","e":"93"},"options_float":{"a":88.0,"b":104.0,"c":110.0,"d":90.0,"e":93.0},"annotated_formula":"subtract(subtract(subtract(198, 22), subtract(73, const_1)), const_4)","linear_formula":"subtract(n1,n2)|subtract(n0,const_1)|subtract(#0,#1)|subtract(#2,const_4)|","chain":"198 - 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1<\/gadget>\n99<\/output>\n60 * 99<\/gadget>\n5_940<\/output>\n46 * 100<\/gadget>\n4_600<\/output>\n5_940 - 4_600<\/gadget>\n1_340<\/output>\n1_340 \/ 4_600<\/gadget>\n67\/230 = around 0.291304<\/output>\n100 * (67\/230)<\/gadget>\n670\/23 = around 29.130435<\/output>\n670\/23 = around 29.130435<\/result>","index":1618} +{"problem":"a sells a bicycle to b and makes a profit of 25 % . b sells the same bicycle to c at a profit of 25 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a .","rationale":"\"explanation : let cp be 100 a sells at 25 % profit so sp = 125 b sells at 25 % profit = 125 x ( 1 + 25 \/ 100 ) = 156.25 cp sp 100 - - - 156.25 x - - - 225 cp = 225 x 100 \/ 156.25 = 144 answer : e\"","correct":"e","options":{"a":"237 ","b":"126 ","c":"971 ","d":"611","e":"144"},"options_float":{"a":237.0,"b":126.0,"c":971.0,"d":611.0,"e":144.0},"annotated_formula":"divide(divide(225, divide(add(const_100, 25), const_100)), divide(add(25, const_100), const_100))","linear_formula":"add(n1,const_100)|add(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#2)|divide(#4,#3)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n225 \/ (5\/4)<\/gadget>\n180<\/output>\n25 + 100<\/gadget>\n125<\/output>\n180 \/ (5\/4)<\/gadget>\n144<\/output>\n144<\/result>","index":1619} +{"problem":"what is the area of a regular hexagon with sides 1 in . long ?","rationale":"\"a regular hexagon of side 1 inch can be divided in to 6 equilateral triangle of side 1 inch . area of each triangle = ( root 3 \/ 4 ) * side ^ 2 = ( root 3 \/ 4 ) area of 6 triangles = 6 * ( root 3 \/ 4 ) = 1.5 root 3 = 2.59 square inch answer : b\"","correct":"b","options":{"a":"1.59 square inch ","b":"2.59 square inch ","c":"3.59 square inch ","d":"4.59 square inch","e":"2.69 square inch"},"options_float":{"a":1.59,"b":2.59,"c":3.59,"d":4.59,"e":2.69},"annotated_formula":"multiply(multiply(multiply(divide(divide(multiply(sqrt(1), const_4), const_pi), const_2), divide(divide(multiply(sqrt(1), const_4), const_pi), const_2)), const_pi), const_2)","linear_formula":"sqrt(n0)|multiply(#0,const_4)|divide(#1,const_pi)|divide(#2,const_2)|multiply(#3,#3)|multiply(#4,const_pi)|multiply(#5,const_2)|","chain":"1 ** (1\/2)<\/gadget>\n1<\/output>\n1 * 4<\/gadget>\n4<\/output>\n4 \/ pi<\/gadget>\n4\/pi = around 1.27324<\/output>\n(4\/pi) \/ 2<\/gadget>\n2\/pi = around 0.63662<\/output>\n(2\/pi) * (2\/pi)<\/gadget>\n4\/pi**2 = around 0.405285<\/output>\n(4\/pi**2) * pi<\/gadget>\n4\/pi = around 1.27324<\/output>\n(4\/pi) * 2<\/gadget>\n8\/pi = around 2.546479<\/output>\n8\/pi = around 2.546479<\/result>","index":1620} +{"problem":"ratio between rahul and deepak is 4 : 2 , after 10 years rahul age will be 26 years . what is deepak present age","rationale":"\"explanation : present age is 4 x and 3 x , = > 4 x + 10 = 26 = > x = 4 so deepak age is = 2 ( 4 ) = 8 answer : option e\"","correct":"e","options":{"a":"14 ","b":"15 ","c":"12 ","d":"22","e":"8"},"options_float":{"a":14.0,"b":15.0,"c":12.0,"d":22.0,"e":8.0},"annotated_formula":"divide(multiply(subtract(26, 10), 2), 4)","linear_formula":"subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|","chain":"26 - 10<\/gadget>\n16<\/output>\n16 * 2<\/gadget>\n32<\/output>\n32 \/ 4<\/gadget>\n8<\/output>\n8<\/result>","index":1621} +{"problem":"what is the rate percent when the simple interest on rs . 800 amount to rs . 160 in 5 years ?","rationale":"\"160 = ( 800 * 5 * r ) \/ 100 r = 4 % answer : e\"","correct":"e","options":{"a":"5 % ","b":"7 % ","c":"9 % ","d":"2 %","e":"4 %"},"options_float":{"a":5.0,"b":7.0,"c":9.0,"d":2.0,"e":4.0},"annotated_formula":"divide(multiply(const_100, 160), multiply(800, 5))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 160<\/gadget>\n16_000<\/output>\n800 * 5<\/gadget>\n4_000<\/output>\n16_000 \/ 4_000<\/gadget>\n4<\/output>\n4<\/result>","index":1623} +{"problem":"a plant manager must assign 12 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 3 new workers . how many different ways can she assign the new workers ?","rationale":"\"whatever : my take selecting team of 3 out of 12 to assign to the shifts = 12 c 3 = 220 ways . now 3 out of 12 means total of 4 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 4 * 3 * 2 * 1 = 24 here alt and alt are the same : so 24 \/ 2 = 12 ways . total ways of selecting = ( selecting 3 out of 12 ) * arranging those teams in shifts = 220 * 24 = 2640 ans : b\"","correct":"b","options":{"a":"2430 ","b":"2640 ","c":"3300 ","d":"4860","e":"5400"},"options_float":{"a":2430.0,"b":2640.0,"c":3300.0,"d":4860.0,"e":5400.0},"annotated_formula":"multiply(divide(factorial(divide(12, 3)), const_2), divide(factorial(12), multiply(factorial(subtract(12, 3)), factorial(3))))","linear_formula":"divide(n0,n1)|factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#0)|factorial(#3)|divide(#4,const_2)|multiply(#5,#2)|divide(#1,#7)|multiply(#6,#8)|","chain":"12 \/ 3<\/gadget>\n4<\/output>\nfactorial(4)<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\nfactorial(12)<\/gadget>\n479_001_600<\/output>\n12 - 3<\/gadget>\n9<\/output>\nfactorial(9)<\/gadget>\n362_880<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n362_880 * 6<\/gadget>\n2_177_280<\/output>\n479_001_600 \/ 2_177_280<\/gadget>\n220<\/output>\n12 * 220<\/gadget>\n2_640<\/output>\n2_640<\/result>","index":1627} +{"problem":"a and b invests rs . 5000 and rs . 6000 in a business . after 6 months , a withdraws two - fifth of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ?","rationale":"\"a : b ( 5000 * 6 ) + ( 3000 * 6 ) : ( 6000 * 8 ) + ( 4000 * 4 ) 48000 : 64000 3 : 4 answer : d\"","correct":"d","options":{"a":"3 : 5 ","b":"7 : 8 ","c":"5 : 6 ","d":"3 : 4","e":"7 : 9"},"options_float":{"a":0.6,"b":0.875,"c":0.8333333333,"d":0.75,"e":0.7777777778},"annotated_formula":"divide(add(multiply(5000, const_4.0), multiply(divide(6000, const_3), multiply(2, 6))), add(multiply(6000, multiply(2, const_3)), multiply(subtract(6000, divide(6000, const_3)), multiply(2, const_3))))","linear_formula":"divide(n1,const_3)|multiply(n0,const_4.0)|multiply(n3,n2)|multiply(n3,const_3)|multiply(#0,#2)|multiply(n1,#3)|subtract(n1,#0)|add(#1,#4)|multiply(#3,#6)|add(#5,#8)|divide(#7,#9)|","chain":"5_000 * 4<\/gadget>\n20_000<\/output>\n6_000 \/ 3<\/gadget>\n2_000<\/output>\n2 * 6<\/gadget>\n12<\/output>\n2_000 * 12<\/gadget>\n24_000<\/output>\n20_000 + 24_000<\/gadget>\n44_000<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6_000 * 6<\/gadget>\n36_000<\/output>\n6_000 - 2_000<\/gadget>\n4_000<\/output>\n4_000 * 6<\/gadget>\n24_000<\/output>\n36_000 + 24_000<\/gadget>\n60_000<\/output>\n44_000 \/ 60_000<\/gadget>\n11\/15 = around 0.733333<\/output>\n11\/15 = around 0.733333<\/result>","index":1628} +{"problem":"in a maths test , students were asked to find 5 \/ 16 of a certain number . one of the students by mistake found 5 \/ 6 th of that number and his answer was 150 more than the correct answer . find the number .","rationale":"\"explanation : let the number be x . 5 * x \/ 6 = 5 * x \/ 16 + 150 25 * x \/ 48 = 150 x = 288 answer b\"","correct":"b","options":{"a":"125 ","b":"288 ","c":"384 ","d":"400","e":"500"},"options_float":{"a":125.0,"b":288.0,"c":384.0,"d":400.0,"e":500.0},"annotated_formula":"divide(multiply(multiply(150, 16), 6), subtract(multiply(5, 16), multiply(5, 6)))","linear_formula":"multiply(n1,n4)|multiply(n0,n1)|multiply(n0,n3)|multiply(n3,#0)|subtract(#1,#2)|divide(#3,#4)|","chain":"150 * 16<\/gadget>\n2_400<\/output>\n2_400 * 6<\/gadget>\n14_400<\/output>\n5 * 16<\/gadget>\n80<\/output>\n5 * 6<\/gadget>\n30<\/output>\n80 - 30<\/gadget>\n50<\/output>\n14_400 \/ 50<\/gadget>\n288<\/output>\n288<\/result>","index":1629} +{"problem":"a , b and c enter into a partnership by investing $ 7000 , $ 11000 and $ 18000 respectively . at the end of 8 months , b receives $ 880 as his share . find the share of a .","rationale":"\"the ratio of capital of a , b and c = 7000 : 11000 : 18000 = 7 : 11 : 18 a ' s share = ( 7 \/ 11 ) * 880 = $ 560 the answer is c .\"","correct":"c","options":{"a":"$ 320 ","b":"$ 440 ","c":"$ 560 ","d":"$ 670","e":"$ 780"},"options_float":{"a":320.0,"b":440.0,"c":560.0,"d":670.0,"e":780.0},"annotated_formula":"divide(multiply(multiply(7000, divide(880, divide(multiply(11000, 8), const_100))), 8), const_100)","linear_formula":"multiply(n1,n3)|divide(#0,const_100)|divide(n4,#1)|multiply(n0,#2)|multiply(n3,#3)|divide(#4,const_100)|","chain":"11_000 * 8<\/gadget>\n88_000<\/output>\n88_000 \/ 100<\/gadget>\n880<\/output>\n880 \/ 880<\/gadget>\n1<\/output>\n7_000 * 1<\/gadget>\n7_000<\/output>\n7_000 * 8<\/gadget>\n56_000<\/output>\n56_000 \/ 100<\/gadget>\n560<\/output>\n560<\/result>","index":1630} +{"problem":"bill downloads the movierevenge of the avengersto his computer in 2.5 hours , using a download manager that downloads from 3 sources marked a , b and c . each source provides download at a constant rate , but the rates of different sources are not necessarily identical . if the movie was downloaded from sources a and c alone , it would take 4 hours to complete the download . the next day , source b is available , but the other sources are inactive . how long will it take to download the trailer of the movie , a file that is 10 times smaller , from source b alone ?","rationale":"let the movie size be 400 u . given , a + c = 4 hrs . a + c = 100 u \/ hr and a + b + c = 2.5 hrs or 400 \/ 2.5 = 160 u \/ hr b alone = 160 - 100 = 60 u \/ hr trailer = 40 times smaller or 400 \/ 10 = 40 u b will take 40 \/ 60 hrs or 40 minutes . ans a","correct":"a","options":{"a":"40 minutes ","b":"15 minutes ","c":"12 minutes ","d":"10 minutes","e":"3 minutes"},"options_float":{"a":40.0,"b":15.0,"c":12.0,"d":10.0,"e":3.0},"annotated_formula":"subtract(subtract(multiply(multiply(3, 4), 4), 4), 4)","linear_formula":"multiply(n1,n2)|multiply(n2,#0)|subtract(#1,n2)|subtract(#2,n2)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 * 4<\/gadget>\n48<\/output>\n48 - 4<\/gadget>\n44<\/output>\n44 - 4<\/gadget>\n40<\/output>\n40<\/result>","index":1631} +{"problem":"a bowl of fruit contains 14 apples and 21 oranges . how many oranges must be removed so that 70 % of the pieces of fruit in the bowl will be apples ?","rationale":"\"number of apples = 14 number of oranges = 23 let number of oranges that must be removed so that 70 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 14 + ( 21 - x ) = 35 - x 14 \/ ( 35 - x ) = 7 \/ 10 = > 20 = 35 - x = > x = 15 answer b\"","correct":"b","options":{"a":"3 ","b":"15 ","c":"14 ","d":"17","e":"20"},"options_float":{"a":3.0,"b":15.0,"c":14.0,"d":17.0,"e":20.0},"annotated_formula":"divide(subtract(multiply(add(14, 21), divide(70, const_100)), 14), divide(70, const_100))","linear_formula":"add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1)|","chain":"14 + 21<\/gadget>\n35<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n35 * (7\/10)<\/gadget>\n49\/2 = around 24.5<\/output>\n(49\/2) - 14<\/gadget>\n21\/2 = around 10.5<\/output>\n(21\/2) \/ (7\/10)<\/gadget>\n15<\/output>\n15<\/result>","index":1632} +{"problem":"the present worth of rs . 2310 due 2 years hence , the rate of interest being 15 % per annum , is :","rationale":"\"p . w . = rs . 100 x 2310 = rs . 1680 . 100 + 15 x 5 2 answer : b\"","correct":"b","options":{"a":"1689 ","b":"1680 ","c":"1682 ","d":"1681","e":"1686"},"options_float":{"a":1689.0,"b":1680.0,"c":1682.0,"d":1681.0,"e":1686.0},"annotated_formula":"divide(2310, power(add(divide(15, const_100), const_1), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) + 1<\/gadget>\n23\/20 = around 1.15<\/output>\n(23\/20) ** 2<\/gadget>\n529\/400 = around 1.3225<\/output>\n2_310 \/ (529\/400)<\/gadget>\n924_000\/529 = around 1_746.691871<\/output>\n924_000\/529 = around 1_746.691871<\/result>","index":1633} +{"problem":"after decreasing 24 % in the price of an article costs rs . 684 . find the actual cost of an article ?","rationale":"\"cp * ( 76 \/ 100 ) = 684 cp = 9 * 100 = > cp = 900 answer : c\"","correct":"c","options":{"a":"218 ","b":"777 ","c":"900 ","d":"2688","e":"1991"},"options_float":{"a":218.0,"b":777.0,"c":900.0,"d":2688.0,"e":1991.0},"annotated_formula":"divide(684, subtract(const_1, divide(24, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|","chain":"24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n1 - (6\/25)<\/gadget>\n19\/25 = around 0.76<\/output>\n684 \/ (19\/25)<\/gadget>\n900<\/output>\n900<\/result>","index":1634} +{"problem":"a man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream . the ratio of the speed of the boat in still water and stream is","rationale":"\"explanation : let speed downstream = x kmph then speed upstream = 2 x kmph so ratio will be , ( 2 x + x ) \/ 2 : ( 2 x - x ) \/ 2 = > 3 x \/ 2 : x \/ 2 = > 3 : 1 option a\"","correct":"a","options":{"a":"3 : 1 ","b":"1 : 3 ","c":"2 : 4 ","d":"4 : 2","e":"none of these"},"options_float":{"a":3.0,"b":0.3333333333,"c":0.5,"d":2.0,"e":null},"annotated_formula":"divide(add(const_1, const_2), subtract(const_2, const_1))","linear_formula":"add(const_1,const_2)|subtract(const_2,const_1)|divide(#0,#1)|","chain":"1 + 2<\/gadget>\n3<\/output>\n2 - 1<\/gadget>\n1<\/output>\n3 \/ 1<\/gadget>\n3<\/output>\n3<\/result>","index":1635} +{"problem":"ritesh and co . generated revenue of rs . 1,600 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ?","rationale":"explanation : given , ritesh and co . generated revenue of rs . 1,600 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1600 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 \/ 12.5 ) × 1600 . = > 12,800 . hence , the total revenue by end of 2007 is rs . 12,800 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 \/ 12800 ) × 100 . = > 19.53 % . answer : b","correct":"b","options":{"a":"12.5 % ","b":"19.53 % ","c":"25 % ","d":"50 %","e":"none of these"},"options_float":{"a":12.5,"b":19.53,"c":25.0,"d":50.0,"e":null},"annotated_formula":"multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100)","linear_formula":"add(const_1,const_4)|add(const_2,const_4)|divide(n2,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#3,const_100)|add(#6,#5)|multiply(#6,const_2)|add(#8,#4)|divide(#7,#2)|divide(#9,#10)|multiply(#11,const_100)","chain":"1 + 4<\/gadget>\n5<\/output>\n2 * 5<\/gadget>\n10<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n2 * 1_000<\/gadget>\n2_000<\/output>\n5 * 100<\/gadget>\n500<\/output>\n2_000 + 500<\/gadget>\n2_500<\/output>\n2 + 4<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n1_000 + 600<\/gadget>\n1_600<\/output>\n12.5 \/ 100<\/gadget>\n0.125<\/output>\n1_600 \/ 0.125<\/gadget>\n12_800<\/output>\n2_500 \/ 12_800<\/gadget>\n25\/128 = around 0.195312<\/output>\n(25\/128) * 100<\/gadget>\n625\/32 = around 19.53125<\/output>\n625\/32 = around 19.53125<\/result>","index":1636} +{"problem":"the average ( arithmetic mean ) of the integers from 100 to 400 , inclusive , is how much greater than the average of the integers from 50 to 250 , inclusive ?","rationale":"for an ap the mean or average of series is average of first and last term . so , average of numbers between 100 to 400 , inclusive = ( 100 + 400 ) \/ 2 = 250 average of numbers between 50 to 250 , inclusive = ( 50 + 250 ) \/ 2 = 150 difference = 250 - 150 = 100 answer is a","correct":"a","options":{"a":"100 ","b":"175 ","c":"200 ","d":"125","e":"300"},"options_float":{"a":100.0,"b":175.0,"c":200.0,"d":125.0,"e":300.0},"annotated_formula":"subtract(divide(add(100, 400), const_2), divide(add(50, 250), const_2))","linear_formula":"add(n0,n1)|add(n2,n3)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)","chain":"100 + 400<\/gadget>\n500<\/output>\n500 \/ 2<\/gadget>\n250<\/output>\n50 + 250<\/gadget>\n300<\/output>\n300 \/ 2<\/gadget>\n150<\/output>\n250 - 150<\/gadget>\n100<\/output>\n100<\/result>","index":1637} +{"problem":"a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 11 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ?","rationale":"\"correct answer : e the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.11 ( $ 1120 ) = $ 123.20 . the correct answer is e .\"","correct":"e","options":{"a":"$ 54.00 ","b":"$ 64.80 ","c":"$ 90.00 ","d":"$ 100.80","e":"$ 123.20"},"options_float":{"a":54.0,"b":64.8,"c":90.0,"d":100.8,"e":123.2},"annotated_formula":"divide(multiply(subtract(1720, 600), 11), const_100)","linear_formula":"subtract(n3,n0)|multiply(n1,#0)|divide(#1,const_100)|","chain":"1_720 - 600<\/gadget>\n1_120<\/output>\n1_120 * 11<\/gadget>\n12_320<\/output>\n12_320 \/ 100<\/gadget>\n616\/5 = around 123.2<\/output>\n616\/5 = around 123.2<\/result>","index":1638} +{"problem":"how much is 60 % of 40 is greater than 4 \/ 5 of 30 ?","rationale":"\"( 60 \/ 100 ) * 40 â € “ ( 4 \/ 5 ) * 30 24 - 24 = 0 answer : d\"","correct":"d","options":{"a":"8 ","b":"3 ","c":"7 ","d":"0","e":"1"},"options_float":{"a":8.0,"b":3.0,"c":7.0,"d":0.0,"e":1.0},"annotated_formula":"subtract(multiply(40, divide(60, const_100)), multiply(divide(4, 5), 30))","linear_formula":"divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n40 * (3\/5)<\/gadget>\n24<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 30<\/gadget>\n24<\/output>\n24 - 24<\/gadget>\n0<\/output>\n0<\/result>","index":1639} +{"problem":"find the greatest number which on dividing 6215 and 7373 , leaves a reminder of 23 and 29 respectively","rationale":"\"explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 6215 - 23 ) and ( 7373 - 29 ) = hcf ( 6192 , 7344 ) = 144 option e\"","correct":"e","options":{"a":"125 ","b":"127 ","c":"129 ","d":"131","e":"144"},"options_float":{"a":125.0,"b":127.0,"c":129.0,"d":131.0,"e":144.0},"annotated_formula":"gcd(subtract(6215, 23), subtract(7373, 29))","linear_formula":"subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)|","chain":"6_215 - 23<\/gadget>\n6_192<\/output>\n7_373 - 29<\/gadget>\n7_344<\/output>\ngcd(6_192, 7_344)<\/gadget>\n144<\/output>\n144<\/result>","index":1641} +{"problem":"the pinedale bus line travels at an average speed of 60 km \/ h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 8 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ?","rationale":"\"number of stops in an hour : 60 \/ 5 = 12 distance between stops : 60 \/ 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 8 = 40 km imo , correct answer is ` ` c . ' '\"","correct":"c","options":{"a":"20 km ","b":"30 km ","c":"40 km ","d":"50 km","e":"60 km"},"options_float":{"a":20.0,"b":30.0,"c":40.0,"d":50.0,"e":60.0},"annotated_formula":"multiply(60, divide(multiply(5, 8), 60))","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|","chain":"5 * 8<\/gadget>\n40<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n60 * (2\/3)<\/gadget>\n40<\/output>\n40<\/result>","index":1643} +{"problem":"the sum of ages of 5 children born at the intervals of 2 years each is 55 years . what is the age of the youngest child ?","rationale":"\"let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) and ( x + 8 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 50 5 x = 55 - 20 = 35 x = 7 . age of the youngest child = x = 7 years . answer : b\"","correct":"b","options":{"a":"4 ","b":"7 ","c":"6 ","d":"9","e":"8"},"options_float":{"a":4.0,"b":7.0,"c":6.0,"d":9.0,"e":8.0},"annotated_formula":"subtract(subtract(divide(55, 5), 2), 2)","linear_formula":"divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|","chain":"55 \/ 5<\/gadget>\n11<\/output>\n11 - 2<\/gadget>\n9<\/output>\n9 - 2<\/gadget>\n7<\/output>\n7<\/result>","index":1644} +{"problem":"two goods trains each 820 m long are running in opposite directions on parallel tracks . their speeds are 45 km \/ hr and 30 km \/ hr respectively . find the time taken by the slower train to pass the driver of the faster one ?","rationale":"relative speed = 45 + 30 = 75 km \/ hr . 75 * 5 \/ 18 = 125 \/ 6 m \/ sec . distance covered = 820 + 820 = 1640 m . required time = 1640 * 6 \/ 125 = 78.72 sec . answer : d","correct":"d","options":{"a":"228 ","b":"27.32 ","c":"76.6 ","d":"78.72","e":"21"},"options_float":{"a":228.0,"b":27.32,"c":76.6,"d":78.72,"e":21.0},"annotated_formula":"add(45, 30)","linear_formula":"add(n1,n2)","chain":"45 + 30<\/gadget>\n75<\/output>\n75<\/result>","index":1647} +{"problem":"20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours c will it take 12 beavers that work at the same pace , to build the same dam ?","rationale":"\"c . 5 hrs if there were 10 beavers it qould have taken double c = 6 hrs . . so closest to that option is 5 .\"","correct":"c","options":{"a":"2 . ","b":"4 . ","c":"c = 5 . ","d":"c = 6 .","e":"8 ."},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(multiply(3, 20), 12)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"3 * 20<\/gadget>\n60<\/output>\n60 \/ 12<\/gadget>\n5<\/output>\n5<\/result>","index":1648} +{"problem":"indu gave bindu rs . 1875 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ?","rationale":"\"1875 = d ( 100 \/ 4 ) 2 d = 3 answer : d\"","correct":"d","options":{"a":"s . 10 ","b":"s . 2 ","c":"s . 5 ","d":"s . 3","e":"s . 4"},"options_float":{"a":10.0,"b":2.0,"c":5.0,"d":3.0,"e":4.0},"annotated_formula":"subtract(subtract(multiply(1875, power(add(const_1, divide(4, const_100)), 2)), 1875), multiply(multiply(1875, divide(4, const_100)), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n1_875 * (676\/625)<\/gadget>\n2_028<\/output>\n2_028 - 1_875<\/gadget>\n153<\/output>\n1_875 * (1\/25)<\/gadget>\n75<\/output>\n75 * 2<\/gadget>\n150<\/output>\n153 - 150<\/gadget>\n3<\/output>\n3<\/result>","index":1649} +{"problem":"how many integers are between 7 and 57 \/ 4 , inclusive ?","rationale":"\"57 \/ 4 = 14 . xx we are not concerned about the exact value of 57 \/ 4 as we just need the integers . since the values are small , we can write down the integers . the different integers between 7 and 57 \/ 4 would be 7 , 8 , 9 , 10 , 11 , 12 , 13,14 total number of integers = 8 option d\"","correct":"d","options":{"a":"4 ","b":"5 ","c":"7 ","d":"8","e":"10"},"options_float":{"a":4.0,"b":5.0,"c":7.0,"d":8.0,"e":10.0},"annotated_formula":"add(subtract(divide(57, 4), 7), const_1)","linear_formula":"divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)|","chain":"57 \/ 4<\/gadget>\n57\/4 = around 14.25<\/output>\n(57\/4) - 7<\/gadget>\n29\/4 = around 7.25<\/output>\n(29\/4) + 1<\/gadget>\n33\/4 = around 8.25<\/output>\n33\/4 = around 8.25<\/result>","index":1650} +{"problem":"3 \/ 5 * [ ( 2 \/ 3 + 3 \/ 8 ) \/ 2 ] - 1 \/ 16 =","rationale":"we need to simplify the expression using pemdas . parentheses exponents multiplication anddivision addition andsubtraction the calculation of an equation or expression must be carried out in the following order : 1 ) operations within parentheses ( or absolute value bars or radicals ) 2 ) exponents 3 ) multiplication and division from left to right 4 ) addition and subtraction from left to right the answer is a .","correct":"a","options":{"a":"1 \/ 4 ","b":"19 \/ 16 ","c":"15 \/ 16 ","d":"9 \/ 13","e":"0"},"options_float":{"a":0.25,"b":1.1875,"c":0.9375,"d":0.6923076923,"e":0.0},"annotated_formula":"subtract(multiply(divide(3, 5), divide(add(divide(2, 3), divide(3, 8)), 2)), divide(1, 16))","linear_formula":"divide(n0,n1)|divide(n2,n0)|divide(n0,n5)|divide(n7,n8)|add(#1,#2)|divide(#4,n2)|multiply(#0,#5)|subtract(#6,#3)","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n3 \/ 8<\/gadget>\n3\/8 = around 0.375<\/output>\n(2\/3) + (3\/8)<\/gadget>\n25\/24 = around 1.041667<\/output>\n(25\/24) \/ 2<\/gadget>\n25\/48 = around 0.520833<\/output>\n(3\/5) * (25\/48)<\/gadget>\n5\/16 = around 0.3125<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n(5\/16) - (1\/16)<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":1652} +{"problem":"each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 93 % water by weight . what is the new weight of the cucumbers , in pounds ?","rationale":"\"out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 93 % water and 7 % of non - water , so now 1 pound of non - water composes 7 % of cucucmbers , which means that the new weight of cucumbers is 1 \/ 0.07 = 15 pounds . answer : b .\"","correct":"b","options":{"a":"2 ","b":"15 ","c":"92 ","d":"96","e":"98"},"options_float":{"a":2.0,"b":15.0,"c":92.0,"d":96.0,"e":98.0},"annotated_formula":"multiply(divide(subtract(100, 99), subtract(100, 93)), 100)","linear_formula":"subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|","chain":"100 - 99<\/gadget>\n1<\/output>\n100 - 93<\/gadget>\n7<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/7) * 100<\/gadget>\n100\/7 = around 14.285714<\/output>\n100\/7 = around 14.285714<\/result>","index":1654} +{"problem":"the cost of the paint is rs . 50 per kg . if 1 kg of paint covers 20 sq . ft , how much will it cost to paint outside of a cube having 20 feet each side","rationale":"\"explanation : surface area of a cube = 6 x 20 ^ 2 = 2400 sq . ft quantity of paint required = ( 2400 \/ 20 ) = 120 kg cost of painting = 50 x 120 = rs . 6000 answer : b\"","correct":"b","options":{"a":"rs . 962 ","b":"rs . 6000 ","c":"rs . 546 ","d":"rs . 876","e":"none of these"},"options_float":{"a":962.0,"b":6000.0,"c":546.0,"d":876.0,"e":null},"annotated_formula":"multiply(divide(surface_cube(20), 20), 50)","linear_formula":"surface_cube(n3)|divide(#0,n2)|multiply(n0,#1)|","chain":"6 * (20 ** 2)<\/gadget>\n2_400<\/output>\n2_400 \/ 20<\/gadget>\n120<\/output>\n120 * 50<\/gadget>\n6_000<\/output>\n6_000<\/result>","index":1655} +{"problem":"a train is 360 meter long is running at a speed of 45 km \/ hour . in what time will it pass a bridge of 160 meter length ?","rationale":"\"speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 360 + 160 = 520 meter time = distance \/ speed = 520 * ( 2 \/ 25 ) = 41.6 seconds answer : b\"","correct":"b","options":{"a":"11 seconds ","b":"41.6 seconds ","c":"40 seconds ","d":"88 seconds","e":"19 seconds"},"options_float":{"a":11.0,"b":41.6,"c":40.0,"d":88.0,"e":19.0},"annotated_formula":"divide(add(360, 160), divide(multiply(45, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"360 + 160<\/gadget>\n520<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n520 \/ (25\/2)<\/gadget>\n208\/5 = around 41.6<\/output>\n208\/5 = around 41.6<\/result>","index":1657} +{"problem":"the ratio of three numbers is 1 : 2 : 3 and the sum of their squares is 2016 . the sum of the numbers is ?","rationale":"\"let the numbers be x , 2 x , 3 x then , x ^ 2 + 4 x ^ 2 + 9 x ^ 2 = 2016 14 x ^ 2 = 2016 x ^ 2 = 144 x = 12 answer is b\"","correct":"b","options":{"a":"10 ","b":"12 ","c":"15 ","d":"14","e":"9"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":14.0,"e":9.0},"annotated_formula":"sqrt(divide(2016, add(power(3, 2), add(power(1, 2), power(2, 2)))))","linear_formula":"power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n1 ** 2<\/gadget>\n1<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n9 + 5<\/gadget>\n14<\/output>\n2_016 \/ 14<\/gadget>\n144<\/output>\n144 ** (1\/2)<\/gadget>\n12<\/output>\n12<\/result>","index":1658} +{"problem":"a certain airline ' s fleet consisted of 90 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ?","rationale":"\"let x be the number of years . 4 x > 90 - 3 x 7 x > 90 x > 12 + 6 \/ 7 the answer is c .\"","correct":"c","options":{"a":"11 ","b":"12 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"divide(90, add(3, 4))","linear_formula":"add(n3,n4)|divide(n0,#0)|","chain":"3 + 4<\/gadget>\n7<\/output>\n90 \/ 7<\/gadget>\n90\/7 = around 12.857143<\/output>\n90\/7 = around 12.857143<\/result>","index":1659} +{"problem":"john makes $ 60 a week from his job . he earns a raise andnow makes $ 78 a week . what is the % increase ?","rationale":"\"increase = ( 18 \/ 60 ) * 100 = ( 3 \/ 10 ) * 100 = 30 % . c\"","correct":"c","options":{"a":"16.12 % ","b":"16.66 % ","c":"30 % ","d":"17.66 %","e":"18.1 %"},"options_float":{"a":16.12,"b":16.66,"c":30.0,"d":17.66,"e":18.1},"annotated_formula":"multiply(divide(subtract(78, 60), 60), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"78 - 60<\/gadget>\n18<\/output>\n18 \/ 60<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":1660} +{"problem":"a jogger running at 9 km \/ hr along side a railway track is 250 m ahead of the engine of a 120 m long train running at 45 km \/ hr in the same direction . in how much time will the train pass the jogger ?","rationale":"\"speed of train relative to jogger = 45 - 9 = 36 km \/ hr . = 36 * 5 \/ 18 = 10 m \/ sec . distance to be covered = 250 + 120 = 370 m . time taken = 370 \/ 10 = 37 sec . answer : a\"","correct":"a","options":{"a":"37 sec ","b":"89 sec ","c":"36 sec ","d":"87 sec","e":"45 sec"},"options_float":{"a":37.0,"b":89.0,"c":36.0,"d":87.0,"e":45.0},"annotated_formula":"divide(add(250, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))","linear_formula":"add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|","chain":"250 + 120<\/gadget>\n370<\/output>\n45 - 9<\/gadget>\n36<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n5 \/ 18<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n370 \/ 10<\/gadget>\n37<\/output>\n37<\/result>","index":1662} +{"problem":"two pipes a and b can separately fill a cistern in 60 minutes and 75 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 50 minutes . in how much time , the third pipe alone can empty the cistern ?","rationale":"\"1 \/ 50 - ( 1 \/ 60 + 1 \/ 75 ) = - 1 \/ 100 third pipe can empty in 100 minutes answer : b\"","correct":"b","options":{"a":"90 min ","b":"100 min ","c":"110 min ","d":"120 min","e":"130 min"},"options_float":{"a":90.0,"b":100.0,"c":110.0,"d":120.0,"e":130.0},"annotated_formula":"inverse(subtract(add(inverse(60), inverse(75)), inverse(50)))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)|","chain":"1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n1 \/ 75<\/gadget>\n1\/75 = around 0.013333<\/output>\n(1\/60) + (1\/75)<\/gadget>\n3\/100 = around 0.03<\/output>\n1 \/ 50<\/gadget>\n1\/50 = around 0.02<\/output>\n(3\/100) - (1\/50)<\/gadget>\n1\/100 = around 0.01<\/output>\n1 \/ (1\/100)<\/gadget>\n100<\/output>\n100<\/result>","index":1664} +{"problem":"of the 20,210 employees of the anvil factory , 2 \/ 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ?","rationale":"the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 \/ 7 portion represents ( 14210 \/ 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 \/ 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 \/ 7 of the total workforce . if 1 \/ 7 of the workforce is journeymen , and 1 \/ 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 \/ 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 \/ 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 \/ 7 of it because 1 \/ 7 was eliminated . the remaining workers now account for 1 \/ 7 out of 6 \/ 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 49.67 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 \/ 12,180 is exactly 1 \/ 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you ’ re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . e","correct":"e","options":{"a":"14.3 % ","b":"16.67 % ","c":"33 % ","d":"28.6 %","e":"49.67 %"},"options_float":{"a":14.3,"b":16.67,"c":33.0,"d":28.6,"e":49.67},"annotated_formula":"multiply(multiply(divide(divide(divide(2, 7), const_2), add(divide(divide(2, 7), const_2), subtract(const_1, divide(2, 7)))), const_100), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_2)|subtract(const_1,#0)|add(#1,#2)|divide(#1,#3)|multiply(#4,const_100)|multiply(#5,const_3)","chain":"2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n(2\/7) \/ 2<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 - (2\/7)<\/gadget>\n5\/7 = around 0.714286<\/output>\n(1\/7) + (5\/7)<\/gadget>\n6\/7 = around 0.857143<\/output>\n(1\/7) \/ (6\/7)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 100<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 3<\/gadget>\n50<\/output>\n50<\/result>","index":1665} +{"problem":"if jimmy drinks 8 ounces of water each time for 8 times a day . how much gallons of water he would have to prepare ahead to drink for 5 days ? ( 1 ounce = 0.0078125 gallon )","rationale":"8 ounces * 8 times \/ day = 64 ounces of water = 64 * 0.0078125 = 0.5 gallon . 5 days need 0.5 * 5 = 2.5 gallons . correct option : 2.5 answer : d","correct":"d","options":{"a":"1 ","b":"1.5 ","c":"2 ","d":"2.5","e":"3"},"options_float":{"a":1.0,"b":1.5,"c":2.0,"d":2.5,"e":3.0},"annotated_formula":"multiply(multiply(multiply(8, 8), 0.0078125), 5)","linear_formula":"multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)","chain":"8 * 8<\/gadget>\n64<\/output>\n64 * 0.007812<\/gadget>\n0.499968<\/output>\n0.499968 * 5<\/gadget>\n2.49984<\/output>\n2.49984<\/result>","index":1666} +{"problem":"100 people are attending a newspaper conference . 45 of them are writers and more than 36 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ?","rationale":"\"{ total } = { writers } + { editors } - { both } + { neither } . { total } = 100 ; { writers } = 45 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 100 = 45 + { editors } - x + 2 x - - > x = 55 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 37 , thus x = { both } = 55 - 37 = 18 . answer : c .\"","correct":"c","options":{"a":"26 ","b":"24 ","c":"18 ","d":"20","e":"22"},"options_float":{"a":26.0,"b":24.0,"c":18.0,"d":20.0,"e":22.0},"annotated_formula":"subtract(100, add(add(36, const_1), 45))","linear_formula":"add(n2,const_1)|add(n1,#0)|subtract(n0,#1)|","chain":"36 + 1<\/gadget>\n37<\/output>\n37 + 45<\/gadget>\n82<\/output>\n100 - 82<\/gadget>\n18<\/output>\n18<\/result>","index":1667} +{"problem":"two trains , a and b , started simultaneously from opposite ends of a 120 - mile route and traveled toward each other on parallel tracks . train a , traveling at a constant rate , completed the 120 - mile trip in 6 hours ; train b , traveling at a constant rate , completed the 120 - mile trip in 4 hours . how many miles had train a traveled when it met train b ?","rationale":"\"as the ratio of the rates of a and b is 4 to 6 then the distance covered at the time of the meeting ( so after traveling the same time interval ) would also be in that ratio , which means that x would cover 4 \/ ( 4 + 6 ) = 4 \/ 10 of 120 miles : 120 * 4 \/ 10 = 48 miles . answer : c .\"","correct":"c","options":{"a":"28 ","b":"36 ","c":"48 ","d":"50","e":"54"},"options_float":{"a":28.0,"b":36.0,"c":48.0,"d":50.0,"e":54.0},"annotated_formula":"multiply(divide(4, add(6, 4)), 120)","linear_formula":"add(n2,n4)|divide(n4,#0)|multiply(n0,#1)|","chain":"6 + 4<\/gadget>\n10<\/output>\n4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 120<\/gadget>\n48<\/output>\n48<\/result>","index":1668} +{"problem":"if a speaks the truth 55 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time","rationale":"\"explanation : probability that a speaks truth is 55 \/ 100 = 0.55 probability that b speaks truth is 60 \/ 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) × p ( b ) = 0.55 × 0.6 = 0.33 answer : b\"","correct":"b","options":{"a":"0.29 ","b":"0.33 ","c":"0.41 ","d":"0.482","e":"0.411"},"options_float":{"a":0.29,"b":0.33,"c":0.41,"d":0.482,"e":0.411},"annotated_formula":"multiply(divide(55, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5)))","linear_formula":"multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)|","chain":"4 * 5<\/gadget>\n20<\/output>\n20 * 5<\/gadget>\n100<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(11\/20) * (3\/5)<\/gadget>\n33\/100 = around 0.33<\/output>\n33\/100 = around 0.33<\/result>","index":1669} +{"problem":"the closest approximation of ( 69.28 × 0.004 ) \/ 0.03 is","rationale":"\"( 69.28 × 0.004 ) \/ 0.03 1 . 0.004 = 4 × 10 ^ ( - 3 ) 2 . 0.03 = 3 × 10 ^ ( - 2 ) 3 . ( a × b ) \/ c = a × ( b \/ c ) 4 . 0.004 \/ 0.03 = 4 × 10 ^ ( - 3 ) \/ ( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) ) \/ 3 = 4 × 10 ^ ( - 1 ) \/ 3 = ( 4 \/ 3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ) therefore , ( 69.28 × 0.004 ) \/ 0.03 = 69.28 × ( 0.004 \/ 0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33 \/ 10 = 6.928 * 1.33 now , 7 × 2 = 14 7 × 1 = 7 or better : 6.9 × 1 = 6.9 6.9 × 2 = 13.8 hence , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only answer that satisfies this condition . answer : c\"","correct":"c","options":{"a":"0.092 ","b":"0.92 ","c":"9.2 ","d":"92","e":"920"},"options_float":{"a":0.092,"b":0.92,"c":9.2,"d":92.0,"e":920.0},"annotated_formula":"multiply(divide(0.004, 0.03), 69.28)","linear_formula":"divide(n1,n2)|multiply(n0,#0)|","chain":"0.004 \/ 0.03<\/gadget>\n0.133333<\/output>\n0.133333 * 69.28<\/gadget>\n9.23731<\/output>\n9.23731<\/result>","index":1670} +{"problem":"to complete a work a and b takes 8 days , b and c takes 12 days , a , b and c takes 6 days . how much time a and c will take","rationale":"explanation : a + b 1 day work = 1 \/ 8 b + c 1 day work = 1 \/ 12 a + b + c 1 day work = 1 \/ 6 we can get a work by ( a + b + c ) - ( b + c ) and c by ( a + b + c ) - ( a + b ) so a 1 day work = 1 \/ 6 − 1 \/ 12 = 1 \/ 12 similarly c 1 day work = 1 \/ 6 − 1 \/ 8 = 4 − 3 \/ 24 = 1 \/ 24 so a and c 1 day work = 1 \/ 12 + 1 \/ 24 = 3 \/ 24 = 1 \/ 8 so a and c can together do this work in 8 days option d","correct":"d","options":{"a":"24 days ","b":"16 days ","c":"12 days ","d":"8 days","e":"none of these"},"options_float":{"a":24.0,"b":16.0,"c":12.0,"d":8.0,"e":null},"annotated_formula":"divide(multiply(8, 6), add(subtract(8, 6), subtract(8, const_4)))","linear_formula":"multiply(n0,n2)|subtract(n0,n2)|subtract(n0,const_4)|add(#1,#2)|divide(#0,#3)","chain":"8 * 6<\/gadget>\n48<\/output>\n8 - 6<\/gadget>\n2<\/output>\n8 - 4<\/gadget>\n4<\/output>\n2 + 4<\/gadget>\n6<\/output>\n48 \/ 6<\/gadget>\n8<\/output>\n8<\/result>","index":1671} +{"problem":"maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 34 kilometers , maxwell ' s walking speed is 4 km \/ h , and brad ' s running speed is 6 km \/ h . what is the total time it takes maxwell before he meets up with brad ?","rationale":"\"total distance = 34 kms maxwell speed = 4 kms \/ hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance \/ relative speed total distance after brad started = 30 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms \/ hr time taken to meet brad after brad started = 30 \/ 10 = 3 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 3 = 12 + 4 = 16 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell \/ maxwell ' s speed = 16 \/ 4 = 4 hrs . . . answer b\"","correct":"b","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(add(34, 6), add(4, 6))","linear_formula":"add(n0,n2)|add(n1,n2)|divide(#0,#1)|","chain":"34 + 6<\/gadget>\n40<\/output>\n4 + 6<\/gadget>\n10<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n4<\/result>","index":1673} +{"problem":"the salary of a person was reduced by 25 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ?","rationale":"\"let the original salary be $ 100 new salary = $ 75 increase on 75 = 25 increase on 100 = [ ( 25 \/ 75 ) 100 ] % = 33.33 % answer is b\"","correct":"b","options":{"a":"10 % ","b":"33.33 % ","c":"20 % ","d":"25 %","e":"50 %"},"options_float":{"a":10.0,"b":33.33,"c":20.0,"d":25.0,"e":50.0},"annotated_formula":"multiply(divide(multiply(const_100, divide(25, const_100)), subtract(const_100, multiply(const_100, divide(25, const_100)))), const_100)","linear_formula":"divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n100 * (1\/4)<\/gadget>\n25<\/output>\n100 - 25<\/gadget>\n75<\/output>\n25 \/ 75<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":1675} +{"problem":"if ( x + 1 ) ( x + 3 ) = x ^ 2 + x , what is a possible value of x ?","rationale":"rearranging the given expression : ( x + 1 ) ( x + 3 ) = x ^ 2 + x i . e . x ^ 2 + 4 x + 3 = x ^ 2 + x i . e . 3 x = - 3 i . e . x = - 3 \/ 3 = - 1 answer : option e had it been a difficult expression to solve further , then we could have used options to check which one satisfies the expression after solving it as further as possible .","correct":"e","options":{"a":"2 ","b":"3 ","c":"- 1 \/ 3 ","d":"- 3","e":"- 1"},"options_float":{"a":2.0,"b":3.0,"c":-0.3333333333,"d":-3.0,"e":-1.0},"annotated_formula":"divide(negate(3), 3)","linear_formula":"negate(n1)|divide(#0,n1)","chain":"-3<\/gadget>\n-3<\/output>\n(-3) \/ 3<\/gadget>\n-1<\/output>\n-1<\/result>","index":1677} +{"problem":"as a treat for her two children , a mother runs to the freezer in which she has 3 cherry ice pops , 4 orange ice pops , and 4 lemon ice pops . she chooses two at random to bring outside to the children , but realizes as she runs out the door that she can not bring them different flavors without one being jealous of the other and getting even more upset . what is the probability that she has to return to the freezer to make sure that they each receive the same flavor ?","rationale":"p ( 2 cherry ) = 3 \/ 11 * 2 \/ 10 = 6 \/ 110 p ( 2 orange ) = 4 \/ 11 * 3 \/ 10 = 12 \/ 110 p ( 2 lemon ) = 4 \/ 11 * 3 \/ 10 = 12 \/ 110 p ( 2 of the same flavor ) = 30 \/ 110 p ( 2 different flavors ) = 1 - 30 \/ 110 = 80 \/ 110 = 8 \/ 11 the answer is c .","correct":"c","options":{"a":"3 \/ 5 ","b":"5 \/ 7 ","c":"8 \/ 11 ","d":"11 \/ 15","e":"13 \/ 20"},"options_float":{"a":0.6,"b":0.7142857143,"c":0.7272727273,"d":0.7333333333,"e":0.65},"annotated_formula":"subtract(const_1, divide(add(add(3, multiply(const_2, 3)), multiply(const_2, 3)), add(multiply(add(add(3, multiply(const_2, 3)), multiply(const_2, 3)), 3), const_10)))","linear_formula":"multiply(n0,const_2)|add(n0,#0)|add(#1,#0)|multiply(n0,#2)|add(#3,const_10)|divide(#2,#4)|subtract(const_1,#5)","chain":"2 * 3<\/gadget>\n6<\/output>\n3 + 6<\/gadget>\n9<\/output>\n9 + 6<\/gadget>\n15<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 + 10<\/gadget>\n55<\/output>\n15 \/ 55<\/gadget>\n3\/11 = around 0.272727<\/output>\n1 - (3\/11)<\/gadget>\n8\/11 = around 0.727273<\/output>\n8\/11 = around 0.727273<\/result>","index":1679} +{"problem":"a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 10 metres apart , how many poles will be needed ?","rationale":"\"solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 \/ 10 ] = 28 m answer a\"","correct":"a","options":{"a":"28 ","b":"56 ","c":"57 ","d":"58","e":"none of these"},"options_float":{"a":28.0,"b":56.0,"c":57.0,"d":58.0,"e":null},"annotated_formula":"divide(rectangle_perimeter(90, 50), 10)","linear_formula":"rectangle_perimeter(n0,n1)|divide(#0,n2)|","chain":"2 * (90 + 50)<\/gadget>\n280<\/output>\n280 \/ 10<\/gadget>\n28<\/output>\n28<\/result>","index":1681} +{"problem":"if in a race of 80 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by :","rationale":"\"explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 80 * 5 ) \/ 25 = 16 m hence , a beats b by 16 m . answer : a\"","correct":"a","options":{"a":"16 m ","b":"18 m ","c":"15 m ","d":"14 m","e":"11 m"},"options_float":{"a":16.0,"b":18.0,"c":15.0,"d":14.0,"e":11.0},"annotated_formula":"multiply(divide(subtract(25, 20), 25), 80)","linear_formula":"subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)|","chain":"25 - 20<\/gadget>\n5<\/output>\n5 \/ 25<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 80<\/gadget>\n16<\/output>\n16<\/result>","index":1682} +{"problem":"when two dice are rolled , what is the probability that the sum of the two numbers is 6 ?","rationale":"we have 5 outcomes that satisfy the question : 1 - 5 2 - 4 3 - 3 4 - 2 5 - 1 the probability is ( 1 \/ 6 * 1 \/ 6 ) * 5 = 5 \/ 36 . i think the answer is b","correct":"b","options":{"a":"1 \/ 9 ","b":"5 \/ 36 ","c":"1 \/ 3 ","d":"2 \/ 36","e":"none of the above"},"options_float":{"a":0.1111111111,"b":0.1388888889,"c":0.3333333333,"d":0.0555555556,"e":null},"annotated_formula":"divide(subtract(6, const_1), multiply(6, 6))","linear_formula":"multiply(n0,n0)|subtract(n0,const_1)|divide(#1,#0)","chain":"6 - 1<\/gadget>\n5<\/output>\n6 * 6<\/gadget>\n36<\/output>\n5 \/ 36<\/gadget>\n5\/36 = around 0.138889<\/output>\n5\/36 = around 0.138889<\/result>","index":1683} +{"problem":"two trains start from a & b and travel towards each other at speed of 50 kmph and 60 kmph resp . at the time of the meeting the second train has traveled 100 km more than the first . the distance between them .","rationale":"let the distance traveled by the first train be x km then distance covered by the second train is x + 100 km x \/ 50 = x + 100 \/ 60 x = 500 so the distance between a & b is x + x + 100 = 1100 km answer is d .","correct":"d","options":{"a":"800 km ","b":"900 km ","c":"1000 km ","d":"1100 km","e":"1200 km"},"options_float":{"a":800.0,"b":900.0,"c":1000.0,"d":1100.0,"e":1200.0},"annotated_formula":"add(multiply(50, divide(100, subtract(60, 50))), multiply(60, divide(100, subtract(60, 50))))","linear_formula":"subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)","chain":"60 - 50<\/gadget>\n10<\/output>\n100 \/ 10<\/gadget>\n10<\/output>\n50 * 10<\/gadget>\n500<\/output>\n60 * 10<\/gadget>\n600<\/output>\n500 + 600<\/gadget>\n1_100<\/output>\n1_100<\/result>","index":1684} +{"problem":"a dog takes 5 leaps for every 7 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is :","rationale":"explanation : dog : hare = ( 5 * 3 ) leaps of hare : 7 leaps of hare = 15 : 7 answer : b","correct":"b","options":{"a":"16 : 7 ","b":"15 : 7 ","c":"20 : 7 ","d":"7 : 15","e":"14 : 7"},"options_float":{"a":2.2857142857,"b":2.1428571429,"c":2.8571428571,"d":0.4666666667,"e":2.0},"annotated_formula":"divide(multiply(5, 3), 7)","linear_formula":"multiply(n0,n2)|divide(#0,n1)","chain":"5 * 3<\/gadget>\n15<\/output>\n15 \/ 7<\/gadget>\n15\/7 = around 2.142857<\/output>\n15\/7 = around 2.142857<\/result>","index":1685} +{"problem":"if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value a of z - x ?","rationale":"\"x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] a = z - x = x + 9 - x = 9 ans : d\"","correct":"d","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"add(add(5, const_2), const_2)","linear_formula":"add(n0,const_2)|add(#0,const_2)|","chain":"5 + 2<\/gadget>\n7<\/output>\n7 + 2<\/gadget>\n9<\/output>\n9<\/result>","index":1686} +{"problem":"a train 360 m long runs with a speed of 45 km \/ hr . what time will it take to pass a platform of 140 m long ?","rationale":"\"train length = 360 m platform length = 140 m time = ? length = speed * time total length = 360 + 140 = 500 m speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) ( m \/ sec ) = 12.5 m \/ sec time = length \/ speed = 500 \/ 12.5 = 40 sec answer : d\"","correct":"d","options":{"a":"38 sec ","b":"35 sec ","c":"44 sec ","d":"40 sec","e":"48 sec"},"options_float":{"a":38.0,"b":35.0,"c":44.0,"d":40.0,"e":48.0},"annotated_formula":"multiply(divide(add(divide(140, const_1000), divide(360, const_1000)), 45), const_3600)","linear_formula":"divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)|","chain":"140 \/ 1_000<\/gadget>\n7\/50 = around 0.14<\/output>\n360 \/ 1_000<\/gadget>\n9\/25 = around 0.36<\/output>\n(7\/50) + (9\/25)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) \/ 45<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/90) * 3_600<\/gadget>\n40<\/output>\n40<\/result>","index":1688} +{"problem":"two guys a and b leave point a and point b simultaneously and travel towards point b and point a on the same route at their respective constant speeds . they meet along the route and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively . how long will b take to cover the entire journey between point b and point a ?","rationale":"let x per minute be the speed of a and y per minute be the speed of b . after meeting at a point , a travels for 32 mins and b travels for 50 mins . so distance covered by each of them post point of crossing a = 32 x and b = 50 y the distance covered by a and b before they cross each would be distance covered by b and a post crossing respectively . therefore distance covered by b before he meets a = 32 x time taken by b cover 32 x distance = 32 x \/ y mins therefore total time taken by b = 32 x \/ y + 50 mins . . . . . . . . . . . . . . . . . i we need to find value of x in terms of y to arrive at final answer . total distance = 32 x + 50 y combined speed of a and b = x + y therefore time taken before a and b meet en - route = ( 32 x + 50 y ) \/ ( x + y ) time taken by b reach destination after meeting a = 50 mins total travel time for b = [ ( 32 x + 50 y ) \/ ( x + y ) ] + 50 mins . . . . . . . . . . . . . . . . . . . ii equate i and ii 32 x \/ y + 50 = [ ( 32 x + 50 y ) \/ ( x + y ) ] + 50 ( 32 x + 50 y ) \/ y = ( 82 x + 100 y ) \/ ( x + y ) 32 x ^ 2 + 50 xy + 32 xy + 50 y ^ 2 = 82 xy + 100 y ^ 2 32 x ^ 2 + 82 xy - 82 xy + 50 y ^ 2 - 100 y ^ 2 = 0 32 x ^ 2 - 50 y ^ 2 = 0 32 x ^ 2 = 50 y ^ 2 16 x ^ 2 = 25 y ^ 2 taking square root . . ( since x and y denote speed , square root ca n ' t be negative ) 4 x = 5 y y = 4 x \/ 5 . . . . . . . . . . . . iii substitute in i = 32 x \/ ( 4 x \/ 5 ) + 50 = 32 x * 5 \/ 4 x + 50 = 40 + 50 = 90 mins b","correct":"b","options":{"a":"65 ","b":"90 ","c":"70 ","d":"75","e":"80"},"options_float":{"a":65.0,"b":90.0,"c":70.0,"d":75.0,"e":80.0},"annotated_formula":"add(50, multiply(add(const_4, const_1), add(const_4, const_4)))","linear_formula":"add(const_1,const_4)|add(const_4,const_4)|multiply(#0,#1)|add(n1,#2)","chain":"4 + 1<\/gadget>\n5<\/output>\n4 + 4<\/gadget>\n8<\/output>\n5 * 8<\/gadget>\n40<\/output>\n50 + 40<\/gadget>\n90<\/output>\n90<\/result>","index":1690} +{"problem":"if 63 percent of a class answered the first question on a certain test correctly , 50 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ?","rationale":"\"63 % answered the first question correctly and 20 % answered neither correctly . then 17 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 50 % - 17 % = 33 % . the answer is d .\"","correct":"d","options":{"a":"24 % ","b":"27 % ","c":"30 % ","d":"33 %","e":"36 %"},"options_float":{"a":24.0,"b":27.0,"c":30.0,"d":33.0,"e":36.0},"annotated_formula":"subtract(add(add(63, 50), 20), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,const_100)|","chain":"63 + 50<\/gadget>\n113<\/output>\n113 + 20<\/gadget>\n133<\/output>\n133 - 100<\/gadget>\n33<\/output>\n33<\/result>","index":1692} +{"problem":"for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 95 units raises the average to 55 units per day , what is the value of n ?","rationale":"\"the average daily production was raised by 5 units for n days , which has a weighting of 5 n . 5 n = 95 - 55 = 40 n = 8 the answer is c .\"","correct":"c","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"subtract(divide(subtract(95, 50), subtract(55, 50)), const_1)","linear_formula":"subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)|","chain":"95 - 50<\/gadget>\n45<\/output>\n55 - 50<\/gadget>\n5<\/output>\n45 \/ 5<\/gadget>\n9<\/output>\n9 - 1<\/gadget>\n8<\/output>\n8<\/result>","index":1694} +{"problem":"how many seconds will a 300 m long train take to cross a man walking with a speed of 3 km \/ hr in the direction of the moving train if the speed of the train is 63 km \/ hr ?","rationale":"\"speed of train relative to man = 63 - 3 = 60 km \/ hr . = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec . time taken to pass the man = 300 * 3 \/ 50 = 18 sec . answer : e\"","correct":"e","options":{"a":"12 sec ","b":"30 sec ","c":"86 sec ","d":"15 sec","e":"18 sec"},"options_float":{"a":12.0,"b":30.0,"c":86.0,"d":15.0,"e":18.0},"annotated_formula":"divide(300, multiply(subtract(63, 3), const_0_2778))","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"63 - 3<\/gadget>\n60<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n300 \/ (50\/3)<\/gadget>\n18<\/output>\n18<\/result>","index":1695} +{"problem":"abcd is a parallelogram . bd = 2 . the angles of triangle bcd are all equal . what is the perimeter of the parallelogram ?","rationale":"so , bc = bd = cd = ba = ad = 2 the parallelogram has 4 equal sides bc = cd = ba = ad = 2 so the perimeter of the parallelogram will be 2 * 4 = 8 hence answer will be ( d ) 2","correct":"d","options":{"a":"12 ","b":"9 √ 3 ","c":"9 ","d":"8","e":"3 √ 3"},"options_float":{"a":12.0,"b":9.0,"c":9.0,"d":8.0,"e":3.0},"annotated_formula":"multiply(2, add(2, 2))","linear_formula":"add(n0,n0)|multiply(n0,#0)","chain":"2 + 2<\/gadget>\n4<\/output>\n2 * 4<\/gadget>\n8<\/output>\n8<\/result>","index":1697} +{"problem":"if the price of petrol increases by 40 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ?","rationale":"\"explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 40 ) i . e 140 . now , his consumption should be reduced to : - = ( 140 − 100 ) \/ 140 ∗ 100 . hence , the consumption should be reduced to 28.6 % . answer : d\"","correct":"d","options":{"a":"25 % ","b":"20 % ","c":"16.67 % ","d":"28.6 %","e":"none of these"},"options_float":{"a":25.0,"b":20.0,"c":16.67,"d":28.6,"e":null},"annotated_formula":"multiply(subtract(const_1, divide(const_100, add(const_100, 40))), const_100)","linear_formula":"add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"100 + 40<\/gadget>\n140<\/output>\n100 \/ 140<\/gadget>\n5\/7 = around 0.714286<\/output>\n1 - (5\/7)<\/gadget>\n2\/7 = around 0.285714<\/output>\n(2\/7) * 100<\/gadget>\n200\/7 = around 28.571429<\/output>\n200\/7 = around 28.571429<\/result>","index":1699} +{"problem":"the average of 6 no . ' s is 3.95 . the average of 2 of them is 3.8 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ?","rationale":"\"sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 3.8 * 2 ) + ( 3.85 * 2 ) ] = 8.4 . required average = ( 8.4 \/ 2 ) = 4.2 . a\"","correct":"a","options":{"a":"4.2 ","b":"4.4 ","c":"4.6 ","d":"5.6","e":"5.7"},"options_float":{"a":4.2,"b":4.4,"c":4.6,"d":5.6,"e":5.7},"annotated_formula":"divide(subtract(multiply(6, 3.95), add(multiply(2, 3.8), multiply(2, 3.85))), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2)|","chain":"6 * 3.95<\/gadget>\n23.7<\/output>\n2 * 3.8<\/gadget>\n7.6<\/output>\n2 * 3.85<\/gadget>\n7.7<\/output>\n7.6 + 7.7<\/gadget>\n15.3<\/output>\n23.7 - 15.3<\/gadget>\n8.4<\/output>\n8.4 \/ 2<\/gadget>\n4.2<\/output>\n4.2<\/result>","index":1701} +{"problem":"the average of 7 numbers is 24 . if each number be multiplied by 5 . find the average of new set of numbers ?","rationale":"\"explanation : average of new numbers = 24 * 5 = 120 answer : option c\"","correct":"c","options":{"a":"110 ","b":"122 ","c":"120 ","d":"125","e":"145"},"options_float":{"a":110.0,"b":122.0,"c":120.0,"d":125.0,"e":145.0},"annotated_formula":"multiply(24, 5)","linear_formula":"multiply(n1,n2)|","chain":"24 * 5<\/gadget>\n120<\/output>\n120<\/result>","index":1703} +{"problem":"a cycle is bought for rs . 900 and sold for rs . 1080 , find the gain percent ?","rationale":"\"900 - - - - 180 100 - - - - ? = > 20 % answer : b\"","correct":"b","options":{"a":"27 % ","b":"20 % ","c":"25 % ","d":"40 %","e":"28 %"},"options_float":{"a":27.0,"b":20.0,"c":25.0,"d":40.0,"e":28.0},"annotated_formula":"multiply(divide(subtract(1080, 900), 900), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_080 - 900<\/gadget>\n180<\/output>\n180 \/ 900<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":1704} +{"problem":"excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 30 kmph . of how many minutes does the train stop per hour ?","rationale":"\"explanation : t = 15 \/ 45 * 60 = 20 answer : option c\"","correct":"c","options":{"a":"5 ","b":"10 ","c":"20 ","d":"30","e":"40"},"options_float":{"a":5.0,"b":10.0,"c":20.0,"d":30.0,"e":40.0},"annotated_formula":"subtract(const_60, multiply(const_60, divide(30, 45)))","linear_formula":"divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|","chain":"30 \/ 45<\/gadget>\n2\/3 = around 0.666667<\/output>\n60 * (2\/3)<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20<\/result>","index":1705} +{"problem":"if rs . 544 be divided among a , b , c in such a way that a gets 2 \/ 3 of what b gets and b gets 1 \/ 4 of what c gets , then their shares are respectively ?","rationale":"\"( a = 2 \/ 3 b and b = 1 \/ 4 c ) = a \/ b = 2 \/ 3 and b \/ c = 1 \/ 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 544 * 2 \/ 17 = rs . 64 b ' s share = 544 * 3 \/ 17 = rs . 96 c ' s share = 544 * 12 \/ 17 = rs . 384 . answer : d\"","correct":"d","options":{"a":"s . 300 ","b":"s . 360 ","c":"s . 389 ","d":"s . 384","e":"s . 323"},"options_float":{"a":300.0,"b":360.0,"c":389.0,"d":384.0,"e":323.0},"annotated_formula":"divide(544, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1))","linear_formula":"divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(2\/3) * (1\/4)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) + (1\/4)<\/gadget>\n5\/12 = around 0.416667<\/output>\n(5\/12) + 1<\/gadget>\n17\/12 = around 1.416667<\/output>\n544 \/ (17\/12)<\/gadget>\n384<\/output>\n384<\/result>","index":1706} +{"problem":"m = { - 6 , - 5 , - 4 , - 3 , - 2 } t = { - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above , what is the probability that the product of the two integers will be negative ?","rationale":"we will have a negative product only if 1 , 2 , 3 , or 4 are selected from set t . p ( negative product ) = 4 \/ 8 = 1 \/ 2 the answer is d .","correct":"d","options":{"a":"0 ","b":"1 \/ 3 ","c":"2 \/ 5 ","d":"1 \/ 2","e":"3 \/ 5"},"options_float":{"a":0.0,"b":0.3333333333,"c":0.4,"d":0.5,"e":0.6},"annotated_formula":"divide(multiply(5, 4), multiply(5, multiply(4, 2)))","linear_formula":"multiply(n1,n2)|multiply(n2,n4)|multiply(n1,#1)|divide(#0,#2)","chain":"5 * 4<\/gadget>\n20<\/output>\n4 * 2<\/gadget>\n8<\/output>\n5 * 8<\/gadget>\n40<\/output>\n20 \/ 40<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":1707} +{"problem":"a 300 m long train crosses a platform in 48 sec while it crosses a signal pole in 18 sec . what is the length of the platform ?","rationale":"\"speed = 300 \/ 18 = 50 \/ 3 m \/ sec . let the length of the platform be x meters . then , ( x + 300 ) \/ 489 = 50 \/ 3 3 x + 900 = 2400 = > x = 500 m . answer : a\"","correct":"a","options":{"a":"500 ","b":"350 ","c":"828 ","d":"277","e":"122"},"options_float":{"a":500.0,"b":350.0,"c":828.0,"d":277.0,"e":122.0},"annotated_formula":"subtract(multiply(speed(300, 18), 48), 300)","linear_formula":"speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|","chain":"300 \/ 18<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 48<\/gadget>\n800<\/output>\n800 - 300<\/gadget>\n500<\/output>\n500<\/result>","index":1708} +{"problem":"the average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 22 years . find the average age of the women ?","rationale":"\"20 + 22 + 8 * 2 = 58 \/ 2 = 29 answer : d\"","correct":"d","options":{"a":"87 ","b":"98 ","c":"30 ","d":"29","e":"37"},"options_float":{"a":87.0,"b":98.0,"c":30.0,"d":29.0,"e":37.0},"annotated_formula":"divide(add(add(20, 22), multiply(8, 2)), const_2)","linear_formula":"add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)|","chain":"20 + 22<\/gadget>\n42<\/output>\n8 * 2<\/gadget>\n16<\/output>\n42 + 16<\/gadget>\n58<\/output>\n58 \/ 2<\/gadget>\n29<\/output>\n29<\/result>","index":1709} +{"problem":"if 20 % of x is 15 less than 15 % of 1500 , then x is ?","rationale":"\"20 % of x = x \/ 5 ; 15 % of 1500 = 15 \/ 100 * 1500 = 225 given that , x \/ 5 = 225 - 15 = > x \/ 5 = 210 = > x = 1050 . answer : e\"","correct":"e","options":{"a":"872 ","b":"738 ","c":"837 ","d":"840","e":"1050"},"options_float":{"a":872.0,"b":738.0,"c":837.0,"d":840.0,"e":1050.0},"annotated_formula":"divide(subtract(multiply(1500, divide(15, const_100)), 15), divide(20, const_100))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1_500 * (3\/20)<\/gadget>\n225<\/output>\n225 - 15<\/gadget>\n210<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n210 \/ (1\/5)<\/gadget>\n1_050<\/output>\n1_050<\/result>","index":1711} +{"problem":"if the average of 5 positive integers is 70 and the difference between the largest and the smallest of these 5 numbers is 10 , what is the maximum value possible for the largest of these 5 integers ?","rationale":"\"sum of 5 integer ( a , b , c , d , e ) = 5 * 70 = 350 e - a = 10 i . e . e = a + 10 for e to be maximum remaining 4 must be as small as possible since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers i . e . a + a + a + a + ( a + 10 ) = 350 i . e . 5 a = 340 i . e . a = 68 i . e . largest e = 68 + 10 = 78 answer : option d\"","correct":"d","options":{"a":"50 ","b":"52 ","c":"69 ","d":"78","e":"84"},"options_float":{"a":50.0,"b":52.0,"c":69.0,"d":78.0,"e":84.0},"annotated_formula":"add(divide(subtract(multiply(70, 5), 10), 5), 10)","linear_formula":"multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)|","chain":"70 * 5<\/gadget>\n350<\/output>\n350 - 10<\/gadget>\n340<\/output>\n340 \/ 5<\/gadget>\n68<\/output>\n68 + 10<\/gadget>\n78<\/output>\n78<\/result>","index":1715} +{"problem":"what is the normal price of an article sold at $ 72 after two successive discounts of 10 % and 20 % ?","rationale":"0.8 * 0.9 * cost price = $ 72 cost price = $ 100 the answer is c .","correct":"c","options":{"a":"$ 92 ","b":"$ 96 ","c":"$ 100 ","d":"$ 104","e":"$ 108"},"options_float":{"a":92.0,"b":96.0,"c":100.0,"d":104.0,"e":108.0},"annotated_formula":"divide(72, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100)))","linear_formula":"subtract(const_100,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n0,#4)","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(9\/10) * (4\/5)<\/gadget>\n18\/25 = around 0.72<\/output>\n72 \/ (18\/25)<\/gadget>\n100<\/output>\n100<\/result>","index":1723} +{"problem":"of the 200 stamps in a collection , 80 are foreign and 50 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ?","rationale":"\"20 stamps are both foreign and more than 10 years old . 60 stamps are foreign only . 30 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 60 + 30 ) = 90 the answer is b .\"","correct":"b","options":{"a":"60 ","b":"90 ","c":"100 ","d":"130","e":"150"},"options_float":{"a":60.0,"b":90.0,"c":100.0,"d":130.0,"e":150.0},"annotated_formula":"subtract(200, subtract(add(80, 50), 20))","linear_formula":"add(n1,n2)|subtract(#0,n4)|subtract(n0,#1)|","chain":"80 + 50<\/gadget>\n130<\/output>\n130 - 20<\/gadget>\n110<\/output>\n200 - 110<\/gadget>\n90<\/output>\n90<\/result>","index":1725} +{"problem":"joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 \/ 4 of all the paint . during the second week , he uses 1 \/ 3 of the remaining paint . how many gallons of paint has joe used ?","rationale":"\"total paint initially = 360 gallons paint used in the first week = ( 1 \/ 4 ) * 360 = 90 gallons . remaning paint = 270 gallons paint used in the second week = ( 1 \/ 3 ) * 270 = 90 gallons total paint used = 180 gallons . option b\"","correct":"b","options":{"a":"18 ","b":"180 ","c":"175 ","d":"216","e":"250"},"options_float":{"a":18.0,"b":180.0,"c":175.0,"d":216.0,"e":250.0},"annotated_formula":"add(multiply(divide(360, 4), 1), divide(subtract(360, multiply(divide(360, 4), 1)), 3))","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)|","chain":"360 \/ 4<\/gadget>\n90<\/output>\n90 * 1<\/gadget>\n90<\/output>\n360 - 90<\/gadget>\n270<\/output>\n270 \/ 3<\/gadget>\n90<\/output>\n90 + 90<\/gadget>\n180<\/output>\n180<\/result>","index":1727} +{"problem":"if a truck is traveling at a constant rate of 108 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters )","rationale":"speed = 108 km \/ hr = > 108,000 m \/ hr in one minute = > 108000 \/ 60 = 1800 meters in one sec = > 1800 \/ 60 = 30 meters time = total distance need to be covered \/ avg . speed = > 600 \/ 30 = 20 and hence the answer : a","correct":"a","options":{"a":"20 ","b":"24 ","c":"30 ","d":"36","e":"48"},"options_float":{"a":20.0,"b":24.0,"c":30.0,"d":36.0,"e":48.0},"annotated_formula":"multiply(divide(divide(600, 1000), 108), const_3600)","linear_formula":"divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)","chain":"600 \/ 1_000<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) \/ 108<\/gadget>\n1\/180 = around 0.005556<\/output>\n(1\/180) * 3_600<\/gadget>\n20<\/output>\n20<\/result>","index":1729} +{"problem":"what is the least number that should be added to 1054 , so the sum of the number is divisible by 23 ?","rationale":"\"( 1054 \/ 23 ) gives a remainder 19 so we need to add 4 . the answer is d .\"","correct":"d","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"multiply(subtract(add(const_1, floor(divide(1054, 23))), divide(1054, 23)), 23)","linear_formula":"divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)|","chain":"1_054 \/ 23<\/gadget>\n1_054\/23 = around 45.826087<\/output>\nfloor(1_054\/23)<\/gadget>\n45<\/output>\n1 + 45<\/gadget>\n46<\/output>\n46 - (1_054\/23)<\/gadget>\n4\/23 = around 0.173913<\/output>\n(4\/23) * 23<\/gadget>\n4<\/output>\n4<\/result>","index":1731} +{"problem":"two bullet trains of equal lengths take 10 seconds and 12 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?","rationale":"\"speed of the first bullet train = 120 \/ 10 m \/ sec = 12 m \/ sec . speed of the second bullet train = 120 \/ 12 m \/ sec = 10 m \/ sec . relative speed = ( 12 + 10 ) = 22 m \/ sec . required time = ( 120 + 120 ) \/ 22 sec = 11 sec . a\"","correct":"a","options":{"a":"11 sec . ","b":"15 sec . ","c":"12 sec . ","d":"17 sec .","e":"19 sec ."},"options_float":{"a":11.0,"b":15.0,"c":12.0,"d":17.0,"e":19.0},"annotated_formula":"divide(add(120, 120), add(speed(120, 10), speed(120, 12)))","linear_formula":"add(n2,n2)|speed(n2,n0)|speed(n2,n1)|add(#1,#2)|divide(#0,#3)|","chain":"120 + 120<\/gadget>\n240<\/output>\n120 \/ 10<\/gadget>\n12<\/output>\n120 \/ 12<\/gadget>\n10<\/output>\n12 + 10<\/gadget>\n22<\/output>\n240 \/ 22<\/gadget>\n120\/11 = around 10.909091<\/output>\n120\/11 = around 10.909091<\/result>","index":1732} +{"problem":"70 % of x is greater than 1 \/ 3 rd of x by 110 . what is x ?","rationale":"\"7 x \/ 10 - x \/ 3 = 110 11 x \/ 30 = 110 x = 300 answer : c\"","correct":"c","options":{"a":"100 ","b":"200 ","c":"300 ","d":"350","e":"400"},"options_float":{"a":100.0,"b":200.0,"c":300.0,"d":350.0,"e":400.0},"annotated_formula":"divide(110, subtract(divide(70, const_100), divide(1, 3)))","linear_formula":"divide(n0,const_100)|divide(n1,n2)|subtract(#0,#1)|divide(n3,#2)|","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(7\/10) - (1\/3)<\/gadget>\n11\/30 = around 0.366667<\/output>\n110 \/ (11\/30)<\/gadget>\n300<\/output>\n300<\/result>","index":1734} +{"problem":"in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 80 more candidates got selected than a . what was the number of candidates appeared from each state ?","rationale":"\"state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 80 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 80 = > total appeared candidates in state b = 80 x 100 = 8000 = > total appeared candidates in state a = total appeared candidates in state b = 8000\"","correct":"b","options":{"a":"7000 ","b":"8000 ","c":"8800 ","d":"8900","e":"9000"},"options_float":{"a":7000.0,"b":8000.0,"c":8800.0,"d":8900.0,"e":9000.0},"annotated_formula":"divide(80, divide(subtract(7, 6), const_100))","linear_formula":"subtract(n1,n0)|divide(#0,const_100)|divide(n2,#1)|","chain":"7 - 6<\/gadget>\n1<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n80 \/ (1\/100)<\/gadget>\n8_000<\/output>\n8_000<\/result>","index":1736} +{"problem":"the sector of a circle has perimeter of 144 cm and central angle 180 o . find its radius ?","rationale":"\"let the radius is x cm perimeter of the sector = length of the arc + 2 ( radius ) 144 = ( 180 \/ 360 * 2 * 22 \/ 7 * x ) + 2 ( x ) 144 = 22 x \/ 7 + 2 x 144 = 36 x \/ 7 36 x = 1008 x = 28 answer : d\"","correct":"d","options":{"a":"25 cm ","b":"32 cm ","c":"14 cm ","d":"28 cm","e":"21 cm"},"options_float":{"a":25.0,"b":32.0,"c":14.0,"d":28.0,"e":21.0},"annotated_formula":"divide(144, add(divide(multiply(multiply(const_2, divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), 180), divide(const_3600, const_10)), const_2))","linear_formula":"add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|add(#2,const_2)|divide(#3,#0)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,#1)|add(#7,const_2)|divide(n0,#8)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 2<\/gadget>\n22<\/output>\n4 + 3<\/gadget>\n7<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n2 * (22\/7)<\/gadget>\n44\/7 = around 6.285714<\/output>\n(44\/7) * 180<\/gadget>\n7_920\/7 = around 1_131.428571<\/output>\n3_600 \/ 10<\/gadget>\n360<\/output>\n(7_920\/7) \/ 360<\/gadget>\n22\/7 = around 3.142857<\/output>\n(22\/7) + 2<\/gadget>\n36\/7 = around 5.142857<\/output>\n144 \/ (36\/7)<\/gadget>\n28<\/output>\n28<\/result>","index":1737} +{"problem":"the dimensions of a room are 10 m x 7 m x 5 m . there are 2 doors and 3 windows in the room . the dimensions of the doors are 1 m x 3 m . one window is of size 2 m x 1.5 m and the other 2 windows are of size 1 m x 1.5 m . the cost of painting the walls at rs . 3 per sq m is ?","rationale":"\"area of 4 walls = 2 ( l + b ) h = 2 ( 10 + 7 ) x 5 = 170 sq m area of 2 doors and 3 windows = 2 ( 1 x 3 ) + ( 2 x 1.5 ) + 2 ( 1 x 1.5 ) = 12 sq m area to be planted = 170 - 12 = 158 sq m cost of painting = rs . 158 x 3 = rs . 474 answer : d\"","correct":"d","options":{"a":"rs . 229 ","b":"rs . 429 ","c":"rs . 129 ","d":"rs . 474","e":"rs . 111"},"options_float":{"a":229.0,"b":429.0,"c":129.0,"d":474.0,"e":111.0},"annotated_formula":"multiply(subtract(multiply(multiply(2, add(10, 7)), 5), add(add(multiply(2, 3), multiply(2, 1.5)), multiply(2, 1.5))), 3)","linear_formula":"add(n0,n1)|multiply(n3,n4)|multiply(n3,n8)|add(#1,#2)|multiply(n3,#0)|add(#3,#2)|multiply(n2,#4)|subtract(#6,#5)|multiply(n4,#7)|","chain":"10 + 7<\/gadget>\n17<\/output>\n2 * 17<\/gadget>\n34<\/output>\n34 * 5<\/gadget>\n170<\/output>\n2 * 3<\/gadget>\n6<\/output>\n2 * 1.5<\/gadget>\n3<\/output>\n6 + 3<\/gadget>\n9<\/output>\n9 + 3<\/gadget>\n12<\/output>\n170 - 12<\/gadget>\n158<\/output>\n158 * 3<\/gadget>\n474<\/output>\n474<\/result>","index":1738} +{"problem":"ravi and kavi start a business by investing â ‚ ¹ 2000 and â ‚ ¹ 72000 , respectively . find the ratio of their profits at the end of year .","rationale":"\"ratio of profit = ratio of investments = 2000 : 72000 = 1 : 36 answer : d\"","correct":"d","options":{"a":"2 : 36 ","b":"5 : 36 ","c":"7 : 36 ","d":"1 : 36","e":"3 : 36"},"options_float":{"a":0.0555555556,"b":0.1388888889,"c":0.1944444444,"d":0.0277777778,"e":0.0833333333},"annotated_formula":"divide(2000, 72000)","linear_formula":"divide(n0,n1)|","chain":"2_000 \/ 72_000<\/gadget>\n1\/36 = around 0.027778<\/output>\n1\/36 = around 0.027778<\/result>","index":1739} +{"problem":"the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 1000 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 1000 ( 100 \/ 125 ) = rs . 800 . answer : e\"","correct":"e","options":{"a":"620 ","b":"400 ","c":"45 ","d":"500","e":"800"},"options_float":{"a":620.0,"b":400.0,"c":45.0,"d":500.0,"e":800.0},"annotated_formula":"divide(1000, add(const_1, divide(25, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n1_000 \/ (5\/4)<\/gadget>\n800<\/output>\n800<\/result>","index":1740} +{"problem":"john makes $ 60 a week from his job . he earns a raise and now makes $ 100 a week . what is the % increase ?","rationale":"\"increase = ( 40 \/ 60 ) * 100 = ( 4 \/ 6 ) * 100 = 66.67 % . b\"","correct":"b","options":{"a":"16 % ","b":"66.67 % ","c":"17 % ","d":"17.61 %","e":"17.56 %"},"options_float":{"a":16.0,"b":66.67,"c":17.0,"d":17.61,"e":17.56},"annotated_formula":"multiply(divide(subtract(100, 60), 60), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"100 - 60<\/gadget>\n40<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 100<\/gadget>\n200\/3 = around 66.666667<\/output>\n200\/3 = around 66.666667<\/result>","index":1742} +{"problem":"a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 153 , while 3 \/ 4 of all his trees are pure fuji . how many of his trees are pure gala ?","rationale":"\"let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x \/ 4 = f and c + f = 153 = > . 1 x + 3 \/ 4 x = 153 = > x = 180 180 - 153 = pure gala = 27 . d\"","correct":"d","options":{"a":"22 ","b":"33 ","c":"55 ","d":"27","e":"88"},"options_float":{"a":22.0,"b":33.0,"c":55.0,"d":27.0,"e":88.0},"annotated_formula":"subtract(divide(153, add(divide(10, const_100), divide(3, 4))), 153)","linear_formula":"divide(n0,const_100)|divide(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,n1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(1\/10) + (3\/4)<\/gadget>\n17\/20 = around 0.85<\/output>\n153 \/ (17\/20)<\/gadget>\n180<\/output>\n180 - 153<\/gadget>\n27<\/output>\n27<\/result>","index":1744} +{"problem":"a cistern 5 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is :","rationale":"\"area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 5 x 1.25 ) + 5 x 4 ] m 2 = 42.5 m 2 . answer : option b\"","correct":"b","options":{"a":"49 m 2 ","b":"42.5 m 2 ","c":"53.5 m 2 ","d":"55 m 2","e":"57 m 2"},"options_float":{"a":49.0,"b":42.5,"c":53.5,"d":55.0,"e":57.0},"annotated_formula":"add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 5))), multiply(4, 5))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 4<\/gadget>\n5<\/output>\n(5\/4) * 5<\/gadget>\n25\/4 = around 6.25<\/output>\n5 + (25\/4)<\/gadget>\n45\/4 = around 11.25<\/output>\n2 * (45\/4)<\/gadget>\n45\/2 = around 22.5<\/output>\n4 * 5<\/gadget>\n20<\/output>\n(45\/2) + 20<\/gadget>\n85\/2 = around 42.5<\/output>\n85\/2 = around 42.5<\/result>","index":1746} +{"problem":"a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 63 , what is the total amount ?","rationale":"\"x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 63 39 - - - ? = > 273 answer : b\"","correct":"b","options":{"a":"115 ","b":"273 ","c":"117 ","d":"118","e":"119"},"options_float":{"a":115.0,"b":273.0,"c":117.0,"d":118.0,"e":119.0},"annotated_formula":"add(add(multiply(divide(const_100, 45), 63), multiply(divide(50, 45), 63)), 63)","linear_formula":"divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)|","chain":"100 \/ 45<\/gadget>\n20\/9 = around 2.222222<\/output>\n(20\/9) * 63<\/gadget>\n140<\/output>\n50 \/ 45<\/gadget>\n10\/9 = around 1.111111<\/output>\n(10\/9) * 63<\/gadget>\n70<\/output>\n140 + 70<\/gadget>\n210<\/output>\n210 + 63<\/gadget>\n273<\/output>\n273<\/result>","index":1747} +{"problem":"find the value of 1 \/ ( 3 + 1 \/ ( 3 + 1 \/ ( 3 - 1 \/ 3 ) ) )","rationale":"1 \/ [ 3 + ( 1 \/ ( 3 + 1 \/ ( 3 - 1 \/ 3 ) ) ) ] = > 1 \/ [ 3 + 1 \/ ( 3 + 1 \/ ( 8 \/ 3 ) ) ] = > 1 \/ [ 3 + 1 \/ ( 3 + 3 \/ 8 ) ] = > 1 \/ [ 3 + 8 \/ 27 ] = > 1 \/ ( 89 \/ 27 ) = > 27 \/ 89 b","correct":"b","options":{"a":"21 \/ 29 ","b":"27 \/ 89 ","c":"89 \/ 21 ","d":"29 \/ 21","e":"none"},"options_float":{"a":0.724137931,"b":0.3033707865,"c":4.2380952381,"d":1.380952381,"e":null},"annotated_formula":"divide(1, add(3, divide(1, add(3, divide(1, subtract(3, divide(1, 3)))))))","linear_formula":"divide(n0,n1)|subtract(n1,#0)|divide(n0,#1)|add(n1,#2)|divide(n0,#3)|add(n1,#4)|divide(n0,#5)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 - (1\/3)<\/gadget>\n8\/3 = around 2.666667<\/output>\n1 \/ (8\/3)<\/gadget>\n3\/8 = around 0.375<\/output>\n3 + (3\/8)<\/gadget>\n27\/8 = around 3.375<\/output>\n1 \/ (27\/8)<\/gadget>\n8\/27 = around 0.296296<\/output>\n3 + (8\/27)<\/gadget>\n89\/27 = around 3.296296<\/output>\n1 \/ (89\/27)<\/gadget>\n27\/89 = around 0.303371<\/output>\n27\/89 = around 0.303371<\/result>","index":1748} +{"problem":"if a - b = 5 and a 2 + b 2 = 31 , find the value of ab .","rationale":"\"2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 31 - 25 = 6 ab = 3 . answer : c\"","correct":"c","options":{"a":"9 ","b":"2 ","c":"3 ","d":"7","e":"12"},"options_float":{"a":9.0,"b":2.0,"c":3.0,"d":7.0,"e":12.0},"annotated_formula":"divide(subtract(31, power(5, 2)), 2)","linear_formula":"power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|","chain":"5 ** 2<\/gadget>\n25<\/output>\n31 - 25<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n3<\/result>","index":1749} +{"problem":"if circles x and y have the same area and circle x has a circumference of 18 π , half of the radius of circle y is :","rationale":"\"x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 18 * pi x = 9 thus y = 9 y \/ 2 = 4.5 ans : b\"","correct":"b","options":{"a":"16 ","b":"4.5 ","c":"4 ","d":"2","e":"1"},"options_float":{"a":16.0,"b":4.5,"c":4.0,"d":2.0,"e":1.0},"annotated_formula":"divide(divide(18, const_2), const_2)","linear_formula":"divide(n0,const_2)|divide(#0,const_2)|","chain":"18 \/ 2<\/gadget>\n9<\/output>\n9 \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":1750} +{"problem":"the present population of a town is 2000 . population increase rate is 20 % p . a . find the population of town after 1 years ?","rationale":"\"p = 2000 r = 20 % required population of town = p ( 1 + r \/ 100 ) ^ t = 2000 ( 1 + 20 \/ 100 ) = 2000 ( 6 \/ 5 ) = 2400 answer is b\"","correct":"b","options":{"a":"1000 ","b":"2400 ","c":"1500 ","d":"1600","e":"1250"},"options_float":{"a":1000.0,"b":2400.0,"c":1500.0,"d":1600.0,"e":1250.0},"annotated_formula":"add(2000, divide(multiply(2000, 20), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|","chain":"2_000 * 20<\/gadget>\n40_000<\/output>\n40_000 \/ 100<\/gadget>\n400<\/output>\n2_000 + 400<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":1751} +{"problem":"the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 490 , krishan has ?","rationale":"\"ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 490 : n & there n = 289 x 490 \/ 49 = rs . 2890 answer : a\"","correct":"a","options":{"a":"rs . 2890 ","b":"rs . 2330 ","c":"rs . 1190 ","d":"rs . 1620","e":"rs . 2680"},"options_float":{"a":2890.0,"b":2330.0,"c":1190.0,"d":1620.0,"e":2680.0},"annotated_formula":"multiply(divide(multiply(divide(490, 7), 17), 7), 17)","linear_formula":"divide(n4,n0)|multiply(n1,#0)|divide(#1,n0)|multiply(n1,#2)|","chain":"490 \/ 7<\/gadget>\n70<\/output>\n70 * 17<\/gadget>\n1_190<\/output>\n1_190 \/ 7<\/gadget>\n170<\/output>\n170 * 17<\/gadget>\n2_890<\/output>\n2_890<\/result>","index":1752} +{"problem":"by selling an article for $ 195 , a person gains $ 45 . what is the gain % ?","rationale":"s . p . = $ 195 gain = $ 45 c . p . = 195 - 45 = 150 gain % = 45 \/ 150 * 100 % = 30 % answer is b","correct":"b","options":{"a":"25 % ","b":"30 % ","c":"50 % ","d":"20 %","e":"10 %"},"options_float":{"a":25.0,"b":30.0,"c":50.0,"d":20.0,"e":10.0},"annotated_formula":"divide(multiply(45, const_100), subtract(195, 45))","linear_formula":"multiply(n1,const_100)|subtract(n0,n1)|divide(#0,#1)","chain":"45 * 100<\/gadget>\n4_500<\/output>\n195 - 45<\/gadget>\n150<\/output>\n4_500 \/ 150<\/gadget>\n30<\/output>\n30<\/result>","index":1753} +{"problem":"by selling an article at rs . 600 , a profit of 60 % is made . find its cost price ?","rationale":"\"sp = 600 cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 600 * [ 100 \/ ( 100 + 60 ) ] = 600 * [ 100 \/ 160 ] = rs . 375 answer : c\"","correct":"c","options":{"a":"228 ","b":"267 ","c":"375 ","d":"480","e":"811"},"options_float":{"a":228.0,"b":267.0,"c":375.0,"d":480.0,"e":811.0},"annotated_formula":"divide(multiply(600, const_100), add(const_100, 60))","linear_formula":"add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|","chain":"600 * 100<\/gadget>\n60_000<\/output>\n100 + 60<\/gadget>\n160<\/output>\n60_000 \/ 160<\/gadget>\n375<\/output>\n375<\/result>","index":1754} +{"problem":"ray writes a two digit number . he sees that the number exceeds 4 times the sum of its digits by 3 . if the number is increased by 18 , the result is the same as the number formed by reversing the digits . find the number .","rationale":"solution : let the two digit number be xy . 4 ( x + y ) + 3 = 10 x + y . . . . . . . ( 1 ) 10 x + y + 18 = 10 y + x . . . . ( 2 ) solving 1 st equation we get 2 x - y = 1 . . . . . ( 3 ) solving 2 nd equation we get y - x = 2 . . . . . ( 4 ) solving 3 and 4 , we get x = 3 and y = 5 answer : b","correct":"b","options":{"a":"2 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":2.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"divide(subtract(18, 3), 3)","linear_formula":"subtract(n2,n1)|divide(#0,n1)","chain":"18 - 3<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n5<\/result>","index":1755} +{"problem":"how many 3 - digit integers between 311 and 401 , exclusive , are divisible by 3 when the tens digit and the hundered digit are switched ?","rationale":"answer is 30 no of digits between 311 and 401 ( exclusive ) = 401 - 311 - 1 = 89 no intergers divisble by 3 = ~ 89 \/ 3 ~ 30 divsiblivity rule for 3 is sum of all digits must be divisble by 3 . it does n ' t matter if hundred digit and tens digits are switched . e . g 372 is divisble by 3 ( becae 3 + 7 + 2 = 12 ( divisble by 3 ) ) switch digits 732 also divisble by 3 ( becae 7 + 3 + 2 = 12 ( divisble by 3 ) ) divsiblivity rule for 3 is sum of all digits must be divisble by 3 . it does n ' t matter if hundred digit and tens digits are switched . you have right . so it is enough to proceed like usual to find the numbers in a given range that are divisible by 3 . so 399 - 312 \/ 3 + 1 = 30 . answer : a","correct":"a","options":{"a":"30 ","b":"19 ","c":"22 ","d":"3","e":"90"},"options_float":{"a":30.0,"b":19.0,"c":22.0,"d":3.0,"e":90.0},"annotated_formula":"divide(subtract(401, 311), const_3)","linear_formula":"subtract(n2,n1)|divide(#0,const_3)","chain":"401 - 311<\/gadget>\n90<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n30<\/result>","index":1756} +{"problem":"the length of a rectangle is reduced by 22 % . by what % would the width have to be increased to maintain the original area ?","rationale":"\"sol . required change = ( 22 * 100 ) \/ ( 100 - 22 ) = 28.2 % c\"","correct":"c","options":{"a":"14 % ","b":"15 % ","c":"28 % ","d":"30 %","e":"35 %"},"options_float":{"a":14.0,"b":15.0,"c":28.0,"d":30.0,"e":35.0},"annotated_formula":"multiply(divide(subtract(const_1, divide(subtract(const_100, 22), const_100)), divide(subtract(const_100, 22), const_100)), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 - 22<\/gadget>\n78<\/output>\n78 \/ 100<\/gadget>\n39\/50 = around 0.78<\/output>\n1 - (39\/50)<\/gadget>\n11\/50 = around 0.22<\/output>\n(11\/50) \/ (39\/50)<\/gadget>\n11\/39 = around 0.282051<\/output>\n(11\/39) * 100<\/gadget>\n1_100\/39 = around 28.205128<\/output>\n1_100\/39 = around 28.205128<\/result>","index":1757} +{"problem":"a car drives 60 miles on local roads at 20 mph , and 195 miles on the highway at 65 mph , what is the average speed of the entire trip ?","rationale":"\"so the answer is plainly d . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 60 miles , and v = 20 mph so we can obtain time : 20 * t = 60 or t = 60 \/ 20 = 3 hours the needed time to cover 60 miles in the same way we should divide 195 to 65 to find the needed time to cover 195 miles , so t = 3 hours so the total time for covering total distance would be 3 + 3 = 6 hours and total distance is 60 + 195 = 255 miles final stage : average speed is total distance divide to total time : 255 \/ 6 = 42.5 miles per hour rounded up to 43 mph . . . .\"","correct":"d","options":{"a":"36 mph ","b":"40 mph ","c":"44 mph ","d":"43 mph","e":"58 mph"},"options_float":{"a":36.0,"b":40.0,"c":44.0,"d":43.0,"e":58.0},"annotated_formula":"divide(add(60, 195), add(divide(60, 20), divide(195, 65)))","linear_formula":"add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)|","chain":"60 + 195<\/gadget>\n255<\/output>\n60 \/ 20<\/gadget>\n3<\/output>\n195 \/ 65<\/gadget>\n3<\/output>\n3 + 3<\/gadget>\n6<\/output>\n255 \/ 6<\/gadget>\n85\/2 = around 42.5<\/output>\n85\/2 = around 42.5<\/result>","index":1762} +{"problem":"if x > 3000 , then the value of ( 1.2 x ) \/ ( 0.5 x - 406 ) is closest to ?","rationale":"assume x = 3002 ( 1.2 x ) \/ ( 0.5 x - 406 ) = 3602.4 \/ ( 3002 * 0.5 - 406 ) = 3602.4 \/ 1095 = = 3 \/ 1 e","correct":"e","options":{"a":"1 \/ 6 ","b":"1 \/ 3 ","c":"10 \/ 21 ","d":"1 \/ 2","e":"3 \/ 1"},"options_float":{"a":0.1666666667,"b":0.3333333333,"c":0.4761904762,"d":0.5,"e":3.0},"annotated_formula":"multiply(subtract(const_2, 0.5), const_2)","linear_formula":"subtract(const_2,n2)|multiply(#0,const_2)","chain":"2 - 0.5<\/gadget>\n1.5<\/output>\n1.5 * 2<\/gadget>\n3<\/output>\n3<\/result>","index":1764} +{"problem":"a train passes a platform in 32 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km \/ hr , the length of the platform is","rationale":"\"explanation : speed of the train = 54 km \/ hr = ( 54 × 10 ) \/ 32 m \/ s = 15 m \/ s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) \/ 15 = > ( 300 + l ) \/ 15 = 32 = > 300 + l = 15 × 32 = 480 = > l = 480 - 300 = 180 meter answer : option a\"","correct":"a","options":{"a":"180 meter ","b":"240 meter ","c":"200 meter ","d":"260 meter","e":"none of these"},"options_float":{"a":180.0,"b":240.0,"c":200.0,"d":260.0,"e":null},"annotated_formula":"multiply(multiply(const_0_2778, 54), subtract(32, 20))","linear_formula":"multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 54<\/gadget>\n15<\/output>\n32 - 20<\/gadget>\n12<\/output>\n15 * 12<\/gadget>\n180<\/output>\n180<\/result>","index":1766} +{"problem":"the cost price of an book is 64 % of the marked price . calculate the gain percent after allowing a discount of 12 % ?","rationale":"c 37.5 % let marked price = $ 100 . then , c . p . = $ 64 , s . p . = $ 88 gain % = 24 \/ 64 * 100 = 37.5 % .","correct":"c","options":{"a":"33 % ","b":"28.02 % ","c":"37.5 % ","d":"30.5 %","e":"22 %"},"options_float":{"a":33.0,"b":28.02,"c":37.5,"d":30.5,"e":22.0},"annotated_formula":"multiply(subtract(multiply(divide(const_100, 64), divide(subtract(const_100, 12), const_100)), const_1), const_100)","linear_formula":"divide(const_100,n0)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_1)|multiply(#4,const_100)","chain":"100 \/ 64<\/gadget>\n25\/16 = around 1.5625<\/output>\n100 - 12<\/gadget>\n88<\/output>\n88 \/ 100<\/gadget>\n22\/25 = around 0.88<\/output>\n(25\/16) * (22\/25)<\/gadget>\n11\/8 = around 1.375<\/output>\n(11\/8) - 1<\/gadget>\n3\/8 = around 0.375<\/output>\n(3\/8) * 100<\/gadget>\n75\/2 = around 37.5<\/output>\n75\/2 = around 37.5<\/result>","index":1767} +{"problem":"3 numbers are in the ratio 3 : 5 : 7 . the largest number value is 70 . find difference between smallest & largest number is ?","rationale":"= = 3 : 5 : 7 total parts = 15 = the largest number value is 70 = the largest number is = 7 = then 7 parts - - - - - > 70 ( 7 * 10 = 70 ) = smallest number = 3 & largest number = 7 = difference between smallest number & largest number is = 7 - 3 = 4 = then 4 parts - - - - - > 40 ( 4 * 10 = 40 ) a","correct":"a","options":{"a":"40 ","b":"45 ","c":"35 ","d":"30","e":"50"},"options_float":{"a":40.0,"b":45.0,"c":35.0,"d":30.0,"e":50.0},"annotated_formula":"subtract(70, multiply(divide(70, 7), 3))","linear_formula":"divide(n4,n3)|multiply(n0,#0)|subtract(n4,#1)","chain":"70 \/ 7<\/gadget>\n10<\/output>\n10 * 3<\/gadget>\n30<\/output>\n70 - 30<\/gadget>\n40<\/output>\n40<\/result>","index":1770} +{"problem":"in a 80 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is :","rationale":"\"b runs 56 m in 7 sec . = > b runs 80 m in 7 \/ 56 * 80 = 10 seconds since a beats b by 7 seconds , a runs 80 m in ( 10 - 7 ) = 3 seconds hence , a ' s time over the course = 3 seconds answer : a\"","correct":"a","options":{"a":"3 seconds ","b":"12 seconds ","c":"10 seconds ","d":"18 seconds","e":"28 seconds"},"options_float":{"a":3.0,"b":12.0,"c":10.0,"d":18.0,"e":28.0},"annotated_formula":"subtract(multiply(divide(7, 56), 80), 7)","linear_formula":"divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)|","chain":"7 \/ 56<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 80<\/gadget>\n10<\/output>\n10 - 7<\/gadget>\n3<\/output>\n3<\/result>","index":1771} +{"problem":"there are 5 people who can build 5 houses in just 5 days . increase the amount of people to 100 . how long would it take these 100 people to build 100 houses ?","rationale":"it will take 5 days for 100 people to build 100 houses . 1 person can build 1 house in 5 days . so , 100 people can build 100 houses in 5 days . answer d","correct":"d","options":{"a":"2 days ","b":"3 days ","c":"4 days ","d":"5 days","e":"1 days"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":1.0},"annotated_formula":"multiply(divide(5, 5), 5)","linear_formula":"divide(n0,n0)|multiply(n0,#0)","chain":"5 \/ 5<\/gadget>\n1<\/output>\n1 * 5<\/gadget>\n5<\/output>\n5<\/result>","index":1772} +{"problem":"the smallest number which when diminished by 2 , is divisible by 12 , 16 , 18 , 21 and 28 is","rationale":"required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 2 = 1008 + 2 = 1010 answer : b","correct":"b","options":{"a":"1008 ","b":"1010 ","c":"1022 ","d":"1032","e":"1043"},"options_float":{"a":1008.0,"b":1010.0,"c":1022.0,"d":1032.0,"e":1043.0},"annotated_formula":"add(multiply(multiply(power(const_3, 2), power(2, const_4)), add(const_3, const_4)), const_2)","linear_formula":"add(const_3,const_4)|power(const_3,n0)|power(n0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,const_2)","chain":"3 ** 2<\/gadget>\n9<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n9 * 16<\/gadget>\n144<\/output>\n3 + 4<\/gadget>\n7<\/output>\n144 * 7<\/gadget>\n1_008<\/output>\n1_008 + 2<\/gadget>\n1_010<\/output>\n1_010<\/result>","index":1773} +{"problem":"45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 55 meters deep by working for 6 hours ?","rationale":"\"45 workers * 8 hours \/ 30 meters = x * 6 \/ 55 x = 110 total workers 110 - 45 = 65 new workers the answer is d .\"","correct":"d","options":{"a":"50 ","b":"55 ","c":"60 ","d":"65","e":"70"},"options_float":{"a":50.0,"b":55.0,"c":60.0,"d":65.0,"e":70.0},"annotated_formula":"subtract(multiply(multiply(45, divide(8, 6)), divide(55, 30)), 45)","linear_formula":"divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0)|","chain":"8 \/ 6<\/gadget>\n4\/3 = around 1.333333<\/output>\n45 * (4\/3)<\/gadget>\n60<\/output>\n55 \/ 30<\/gadget>\n11\/6 = around 1.833333<\/output>\n60 * (11\/6)<\/gadget>\n110<\/output>\n110 - 45<\/gadget>\n65<\/output>\n65<\/result>","index":1774} +{"problem":"in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 12 feet between each tree . how many trees can be planted if the road is 157 feet long ?","rationale":"\"let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 12 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number of 12 feet spaces will be t - 1 . then , the length of sidewalk required for 12 feet spaces will be 12 * ( t - 1 ) it is given that total lenght of sidewalk is 157 feet . or 12 ( t - 1 ) + t = 157 or 12 t - 12 + t = 157 or 13 t = 169 or t = 13 answer : - e\"","correct":"e","options":{"a":"8 ","b":"9 ","c":"10 ","d":"11","e":"13"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":13.0},"annotated_formula":"add(divide(subtract(157, const_1), add(12, const_1)), const_1)","linear_formula":"add(n0,const_1)|subtract(n1,const_1)|divide(#1,#0)|add(#2,const_1)|","chain":"157 - 1<\/gadget>\n156<\/output>\n12 + 1<\/gadget>\n13<\/output>\n156 \/ 13<\/gadget>\n12<\/output>\n13<\/result>","index":1777} +{"problem":"a train 360 m long runs with a speed of 45 km \/ hr . what time will it take to pass a platform of 150 m long ?","rationale":"\"speed = 45 km \/ hr = 45 ã — ( 5 \/ 18 ) m \/ s = 150 \/ 12 = 50 \/ 4 = 25 \/ 2 m \/ s total distance = length of the train + length of the platform = 360 + 150 = 510 meter time taken to cross the platform = 510 \/ ( 25 \/ 2 ) = 510 ã — 2 \/ 25 = 40.8 seconds answer : d\"","correct":"d","options":{"a":"38 sec ","b":"35 sec ","c":"44 sec ","d":"40.8 sec","e":"none of these"},"options_float":{"a":38.0,"b":35.0,"c":44.0,"d":40.8,"e":null},"annotated_formula":"multiply(divide(add(divide(150, const_1000), divide(360, const_1000)), 45), const_3600)","linear_formula":"divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)|","chain":"150 \/ 1_000<\/gadget>\n3\/20 = around 0.15<\/output>\n360 \/ 1_000<\/gadget>\n9\/25 = around 0.36<\/output>\n(3\/20) + (9\/25)<\/gadget>\n51\/100 = around 0.51<\/output>\n(51\/100) \/ 45<\/gadget>\n17\/1_500 = around 0.011333<\/output>\n(17\/1_500) * 3_600<\/gadget>\n204\/5 = around 40.8<\/output>\n204\/5 = around 40.8<\/result>","index":1779} +{"problem":"two brother x and y appeared for an exam . the probability of selection of x is 1 \/ 3 and that of b is 2 \/ 7 . find the probability that both of them are selected .","rationale":"explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 \/ 3 , p ( b ) = 2 \/ 7 . let c be the event that both are selected . p ( c ) = p ( a ) ã — p ( b ) as a and b are independent events : = ( 1 \/ 3 ) ã — ( 2 \/ 7 ) = 2 \/ 21 answer : b ) 2 \/ 21","correct":"b","options":{"a":"2 \/ 69 ","b":"2 \/ 21 ","c":"2 \/ 23 ","d":"2 \/ 29","e":"2 \/ 10"},"options_float":{"a":0.0289855072,"b":0.0952380952,"c":0.0869565217,"d":0.0689655172,"e":0.2},"annotated_formula":"multiply(divide(1, 3), divide(2, 7))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n(1\/3) * (2\/7)<\/gadget>\n2\/21 = around 0.095238<\/output>\n2\/21 = around 0.095238<\/result>","index":1781} +{"problem":"a box contains 22 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is","rationale":"\"solution p ( none is defective ) = 18 c 2 \/ 22 c 2 = 51 \/ 77 p ( at least one is defective ) = ( 1 - 51 \/ 77 ) = 26 \/ 77 . answer e\"","correct":"e","options":{"a":"4 \/ 19 ","b":"7 \/ 19 ","c":"12 \/ 19 ","d":"21 \/ 99","e":"26 \/ 77"},"options_float":{"a":0.2105263158,"b":0.3684210526,"c":0.6315789474,"d":0.2121212121,"e":0.3376623377},"annotated_formula":"subtract(const_1, divide(choose(subtract(22, 4), const_2), choose(22, const_2)))","linear_formula":"choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)|","chain":"22 - 4<\/gadget>\n18<\/output>\nbinomial(18, 2)<\/gadget>\n153<\/output>\nbinomial(22, 2)<\/gadget>\n231<\/output>\n153 \/ 231<\/gadget>\n51\/77 = around 0.662338<\/output>\n1 - (51\/77)<\/gadget>\n26\/77 = around 0.337662<\/output>\n26\/77 = around 0.337662<\/result>","index":1784} +{"problem":"the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 8 times the probability that both events a and b occur . what is the probability that event a occurs ?","rationale":"\"let us say probability of a occurring is a . let us say probability of b occurring is b . a = 2 b probability ( either a or b or both ) = 8 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 8 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 8 * 2 b * b 3 b - 2 b ^ 2 = 16 b ^ 2 3 b = 18 b ^ 2 b = 3 \/ 18 = 1 \/ 6 so , a = 2 b = 1 \/ 3 answer : d\"","correct":"d","options":{"a":"1 \/ 12 ","b":"1 \/ 8 ","c":"1 \/ 6 ","d":"1 \/ 3","e":"2 \/ 3"},"options_float":{"a":0.0833333333,"b":0.125,"c":0.1666666667,"d":0.3333333333,"e":0.6666666667},"annotated_formula":"multiply(divide(add(const_2, const_1), add(multiply(8, const_2), const_2)), const_2)","linear_formula":"add(const_1,const_2)|multiply(n1,const_2)|add(#1,const_2)|divide(#0,#2)|multiply(#3,const_2)|","chain":"2 + 1<\/gadget>\n3<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16 + 2<\/gadget>\n18<\/output>\n3 \/ 18<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 2<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":1785} +{"problem":"if - 2 and - 3 are negative integers , then - 2 * - 3 + 2 is","rationale":"answer : b","correct":"b","options":{"a":"- 5 ","b":"8 ","c":"- 6 ","d":"- 3","e":"- 8"},"options_float":{"a":-5.0,"b":8.0,"c":-6.0,"d":-3.0,"e":-8.0},"annotated_formula":"add(multiply(negate(2), negate(3)), 2)","linear_formula":"negate(n0)|negate(n1)|multiply(#0,#1)|add(n4,#2)|","chain":"-2<\/gadget>\n-2<\/output>\n-3<\/gadget>\n-3<\/output>\n(-2) * (-3)<\/gadget>\n6<\/output>\n6 + 2<\/gadget>\n8<\/output>\n8<\/result>","index":1786} +{"problem":"a can give b 50 meters start and c 100 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"a runs 1000 m while b runs 950 m and c runs 900 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 900 ) \/ 950 = 947.37 m . b can give c = 1000 - 947.37 = 52.63 m . answer : d\"","correct":"d","options":{"a":"111.12 ","b":"111.67 ","c":"111.64 ","d":"52.63","e":"101.12"},"options_float":{"a":111.12,"b":111.67,"c":111.64,"d":52.63,"e":101.12},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 100)), subtract(multiply(const_100, const_10), 50)))","linear_formula":"multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n1_000 * 900<\/gadget>\n900_000<\/output>\n1_000 - 50<\/gadget>\n950<\/output>\n900_000 \/ 950<\/gadget>\n18_000\/19 = around 947.368421<\/output>\n1_000 - (18_000\/19)<\/gadget>\n1_000\/19 = around 52.631579<\/output>\n1_000\/19 = around 52.631579<\/result>","index":1787} +{"problem":"how many seconds will a 800 meter long train moving with a speed of 63 km \/ hr take to cross a man walking with a speed of 3 km \/ hr in the direction of the train ?","rationale":"\"explanation : here distance d = 800 mts speed s = 63 - 3 = 60 kmph = 60 x 5 \/ 18 m \/ s time t = 800 x 18 \/ 60 x 5 = 48 sec . answer is a\"","correct":"a","options":{"a":"48 ","b":"52 ","c":"38 ","d":"42","e":"45"},"options_float":{"a":48.0,"b":52.0,"c":38.0,"d":42.0,"e":45.0},"annotated_formula":"divide(800, multiply(subtract(63, 3), const_0_2778))","linear_formula":"subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"63 - 3<\/gadget>\n60<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n800 \/ (50\/3)<\/gadget>\n48<\/output>\n48<\/result>","index":1795} +{"problem":"if a - b = 8 and a ^ 2 + b ^ 2 = 164 , find the value of ab","rationale":"2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 164 - 64 = 100 = > ab = 50 answer : c","correct":"c","options":{"a":"10 ","b":"12 ","c":"50 ","d":"18","e":"19"},"options_float":{"a":10.0,"b":12.0,"c":50.0,"d":18.0,"e":19.0},"annotated_formula":"multiply(multiply(add(8, divide(subtract(sqrt(164), 8), 2)), divide(subtract(sqrt(164), 8), 2)), 2)","linear_formula":"sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)","chain":"164 ** (1\/2)<\/gadget>\n2*sqrt(41) = around 12.806248<\/output>\n(2*sqrt(41)) - 8<\/gadget>\n-8 + 2*sqrt(41) = around 4.806248<\/output>\n(-8 + 2*sqrt(41)) \/ 2<\/gadget>\n-4 + sqrt(41) = around 2.403124<\/output>\n8 + (-4 + sqrt(41))<\/gadget>\n4 + sqrt(41) = around 10.403124<\/output>\n(4 + sqrt(41)) * (-4 + sqrt(41))<\/gadget>\n(-4 + sqrt(41))*(4 + sqrt(41)) = around 25<\/output>\n((-4 + sqrt(41))*(4 + sqrt(41))) * 2<\/gadget>\n2*(-4 + sqrt(41))*(4 + sqrt(41)) = around 50<\/output>\n2*(-4 + sqrt(41))*(4 + sqrt(41)) = around 50<\/result>","index":1796} +{"problem":"a sum fetched a total simple interest of rs . 929.20 at the rate of 8 p . c . p . a . in 5 years . what is the sum ?","rationale":"\"explanation : si = rs . 929.20 p = ? t = 5 years r = 8 % p = 100 × si \/ rt = 100 × 929.20 \/ 8 × 5 = rs . 2323 answer : option a\"","correct":"a","options":{"a":"2323 ","b":"1223 ","c":"2563 ","d":"1853","e":"2353"},"options_float":{"a":2323.0,"b":1223.0,"c":2563.0,"d":1853.0,"e":2353.0},"annotated_formula":"divide(divide(multiply(929.20, const_100), 8), 5)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"929.2 * 100<\/gadget>\n92_920<\/output>\n92_920 \/ 8<\/gadget>\n11_615<\/output>\n11_615 \/ 5<\/gadget>\n2_323<\/output>\n2_323<\/result>","index":1797} +{"problem":"? % of 360 = 151.2","rationale":"\"? % of 360 = 151.2 or , ? = 151.2 × 100 \/ 360 = 42 answer a\"","correct":"a","options":{"a":"42 ","b":"36 ","c":"64 ","d":"72","e":"none of these"},"options_float":{"a":42.0,"b":36.0,"c":64.0,"d":72.0,"e":null},"annotated_formula":"divide(multiply(151.2, const_100), 360)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"151.2 * 100<\/gadget>\n15_120<\/output>\n15_120 \/ 360<\/gadget>\n42<\/output>\n42<\/result>","index":1798} +{"problem":"the radius of a wheel is 22.5 cm . what is the distance covered by the wheel in making 500 resolutions ?","rationale":"\"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 \/ 7 * 22.5 = 70714 cm = 707.14 m answer : b\"","correct":"b","options":{"a":"708 m ","b":"707.14 m ","c":"774 m ","d":"714 m","e":"744 m"},"options_float":{"a":708.0,"b":707.14,"c":774.0,"d":714.0,"e":744.0},"annotated_formula":"divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.5), const_2), 500), const_100)","linear_formula":"add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)|","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21 + 1<\/gadget>\n22<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n(22\/7) * 22.5<\/gadget>\n70.714286<\/output>\n70.714286 * 2<\/gadget>\n141.428572<\/output>\n141.428572 * 500<\/gadget>\n70_714.286<\/output>\n70_714.286 \/ 100<\/gadget>\n707.14286<\/output>\n707.14286<\/result>","index":1799} +{"problem":"how many cubes of 10 cm edge can be put in a cubical box of 1 m edge ?","rationale":"\"sol . number of cubes = [ 100 * 100 * 100 \/ 10 * 10 * 10 ] = 1000 . answer d\"","correct":"d","options":{"a":"1 ","b":"10 ","c":"100 ","d":"1000","e":"none"},"options_float":{"a":1.0,"b":10.0,"c":100.0,"d":1000.0,"e":null},"annotated_formula":"divide(volume_cube(1), volume_cube(divide(10, const_100)))","linear_formula":"divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)|","chain":"1 ** 3<\/gadget>\n1<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) ** 3<\/gadget>\n1\/1_000 = around 0.001<\/output>\n1 \/ (1\/1_000)<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1801} +{"problem":"12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required ?","rationale":"\"that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 15 × 8 ? ( i . e . no . of men required ) = 12 × 8 × 10 \/ 15 × 10 = 8 days . a )\"","correct":"a","options":{"a":"8 days ","b":"9 days ","c":"11 days ","d":"13 days","e":"15 days"},"options_float":{"a":8.0,"b":9.0,"c":11.0,"d":13.0,"e":15.0},"annotated_formula":"divide(multiply(multiply(12, 10), 8), multiply(8, 15))","linear_formula":"multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)|","chain":"12 * 10<\/gadget>\n120<\/output>\n120 * 8<\/gadget>\n960<\/output>\n8 * 15<\/gadget>\n120<\/output>\n960 \/ 120<\/gadget>\n8<\/output>\n8<\/result>","index":1802} +{"problem":"if a man lost 10 % by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44 % ?","rationale":"90 % - - - - 12 144 % - - - - ? 90 \/ 144 * 12 = 7.50 answer : b","correct":"b","options":{"a":"6.5 ","b":"7.5 ","c":"7 ","d":"8","e":"2"},"options_float":{"a":6.5,"b":7.5,"c":7.0,"d":8.0,"e":2.0},"annotated_formula":"divide(multiply(subtract(const_100, 10), 12), add(const_100, 44))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)","chain":"100 - 10<\/gadget>\n90<\/output>\n90 * 12<\/gadget>\n1_080<\/output>\n100 + 44<\/gadget>\n144<\/output>\n1_080 \/ 144<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":1803} +{"problem":"jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 5 to 2 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ?","rationale":"\"explanation : indian stamps are common to both ratios . multiply both ratios by factors such that the indian stamps are represented by the same number . us : indian = 5 : 2 , and indian : british = 5 : 1 . multiply the first by 5 , and the second by 2 . hence the two ratios can be combined and us : british = 25 : 2 now us : indian = 25 : 10 , and indian : british = 10 : 2 answer : d\"","correct":"d","options":{"a":"10 : 5 ","b":"15 : 2 ","c":"20 : 2 ","d":"25 : 2","e":"none of these"},"options_float":{"a":2.0,"b":7.5,"c":10.0,"d":12.5,"e":null},"annotated_formula":"divide(multiply(5, 5), multiply(1, 2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)|","chain":"5 * 5<\/gadget>\n25<\/output>\n1 * 2<\/gadget>\n2<\/output>\n25 \/ 2<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":1804} +{"problem":"a canteen requires 13 dozen bananas per day . how many bananas will it require for 9 weeks ?","rationale":"requirement of bananas for 1 day in the canteen = 13 dozens ∴ requirement of bananas for 9 weeks i . e . 63 days = 63 × 13 dozens = 63 × 13 × 12 = 9828 . answer d","correct":"d","options":{"a":"728 ","b":"9828 ","c":"1404 ","d":"9882","e":"none"},"options_float":{"a":728.0,"b":9828.0,"c":1404.0,"d":9882.0,"e":null},"annotated_formula":"multiply(multiply(13, multiply(9, add(const_3, const_4))), multiply(const_3, const_4))","linear_formula":"add(const_3,const_4)|multiply(const_3,const_4)|multiply(n1,#0)|multiply(n0,#2)|multiply(#3,#1)","chain":"3 + 4<\/gadget>\n7<\/output>\n9 * 7<\/gadget>\n63<\/output>\n13 * 63<\/gadget>\n819<\/output>\n3 * 4<\/gadget>\n12<\/output>\n819 * 12<\/gadget>\n9_828<\/output>\n9_828<\/result>","index":1805} +{"problem":"an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 11 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ?","rationale":"\"explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 11 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 3300 . hence , ( r - p ) = ( 3300 - 900 ) = 2400 answer : e\"","correct":"e","options":{"a":"s 1200 ","b":"s 1500 ","c":"s 1600 ","d":"s 1900","e":"s 2400"},"options_float":{"a":1200.0,"b":1500.0,"c":1600.0,"d":1900.0,"e":2400.0},"annotated_formula":"multiply(divide(1500, 5), subtract(11, 3))","linear_formula":"divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)|","chain":"1_500 \/ 5<\/gadget>\n300<\/output>\n11 - 3<\/gadget>\n8<\/output>\n300 * 8<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":1806} +{"problem":"a shipment of 230 smartphones contains 84 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ?","rationale":"\"probability of chosing one defective phone from a lot of 230 which ontains 84 defective phones is = ( 84 \/ 230 ) probability of chosing one defective phone from a lot of 229 ( we already picked one ) which ontains 83 ( we already picked one ) defective phones is = ( 83 \/ 229 ) combined probability of series of events = product of the probabilities = ( 84 \/ 230 ) * ( 83 \/ 229 ) 84 \/ 230 is close to ( 23 \/ 63 ) and ( 83 \/ 229 ) = ( 29 \/ 80 ) so answer is ( 23 \/ 63 ) * ( 29 \/ 80 ) = ( 9 \/ 68 ) so , answer will be d\"","correct":"d","options":{"a":"1 \/ 250 ","b":"1 \/ 84 ","c":"1 \/ 11 ","d":"9 \/ 68","e":"1 \/ 3"},"options_float":{"a":0.004,"b":0.0119047619,"c":0.0909090909,"d":0.1323529412,"e":0.3333333333},"annotated_formula":"divide(multiply(84, subtract(84, const_1)), multiply(230, subtract(230, const_1)))","linear_formula":"subtract(n1,const_1)|subtract(n0,const_1)|multiply(n1,#0)|multiply(n0,#1)|divide(#2,#3)|","chain":"84 - 1<\/gadget>\n83<\/output>\n84 * 83<\/gadget>\n6_972<\/output>\n230 - 1<\/gadget>\n229<\/output>\n230 * 229<\/gadget>\n52_670<\/output>\n6_972 \/ 52_670<\/gadget>\n3_486\/26_335 = around 0.132371<\/output>\n3_486\/26_335 = around 0.132371<\/result>","index":1808} +{"problem":"a large box contains 20 small boxes and each small box contains 32 chocolate bars . how many chocolate bars are in the large box ?","rationale":"\"the only thing you have to do is a simple multiply we have : 20 * 32 = 640 correct answer is : e ) 640\"","correct":"e","options":{"a":"220 ","b":"490 ","c":"380 ","d":"450","e":"640"},"options_float":{"a":220.0,"b":490.0,"c":380.0,"d":450.0,"e":640.0},"annotated_formula":"multiply(20, 32)","linear_formula":"multiply(n0,n1)|","chain":"20 * 32<\/gadget>\n640<\/output>\n640<\/result>","index":1809} +{"problem":"in a mayoral election , candidate x received 1 \/ 2 more votes than candidate y , and candidate y received 2 \/ 5 fewer votes than z . if z received 25000 votes how many votes did candidate x received ?","rationale":"z = 25 - - > y received 2 \/ 5 fewer votes than z - - > y = z - 2 \/ 5 * z = 15 ; x received 1 \/ 2 more votes than y - - > x = y + 1 \/ 2 * y = 22,5 . answer : b .","correct":"b","options":{"a":"18000 ","b":"22500 ","c":"24000 ","d":"26000","e":"32000"},"options_float":{"a":18000.0,"b":22500.0,"c":24000.0,"d":26000.0,"e":32000.0},"annotated_formula":"multiply(multiply(add(const_1, divide(1, 2)), subtract(const_1, divide(2, 5))), 25000)","linear_formula":"divide(n0,n1)|divide(n1,n3)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n4,#4)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/2) * (3\/5)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 25_000<\/gadget>\n22_500<\/output>\n22_500<\/result>","index":1812} +{"problem":"a train 800 m long can cross a pole in 10 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 800 \/ 10 s = 80 m \/ sec speed = 80 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 288 kmph answer : e\"","correct":"e","options":{"a":"280 ","b":"282 ","c":"284 ","d":"286","e":"288"},"options_float":{"a":280.0,"b":282.0,"c":284.0,"d":286.0,"e":288.0},"annotated_formula":"divide(divide(800, const_1000), divide(10, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"800 \/ 1_000<\/gadget>\n4\/5 = around 0.8<\/output>\n10 \/ 3_600<\/gadget>\n1\/360 = around 0.002778<\/output>\n(4\/5) \/ (1\/360)<\/gadget>\n288<\/output>\n288<\/result>","index":1813} +{"problem":"a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is two times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that time , what is the ratio of water to salt in the solution ?","rationale":"water : soap = 3 : 2 soap : salt = 6 : 2 = > for 6 soap , salt = 2 = > for 2 soap , salt = ( 2 \/ 6 ) * 2 = 2 \/ 3 so , water : soap : salt = 3 : 2 : 2 \/ 3 = 18 : 12 : 4 after open container , water : soap : salt = 9 : 18 : 4 so , water : salt = 9 : 4 = 18 : 8 e","correct":"e","options":{"a":"1 : 1 ","b":"2 : 3 ","c":"3 : 2 ","d":"9 : 4","e":"18 : 8"},"options_float":{"a":1.0,"b":0.6666666667,"c":1.5,"d":2.25,"e":2.25},"annotated_formula":"divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2))","linear_formula":"multiply(n0,n1)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n18 \/ 8<\/gadget>\n9\/4 = around 2.25<\/output>\n9\/4 = around 2.25<\/result>","index":1814} +{"problem":"a string of 30 light bulbs is wired in such a way that if any individual light bulb fails , the entire string fails . if for each individual light bulb the probability of failing during time period t is 0.1 , what is the probability that the string of light bulbs will fail during the time period t ?","rationale":"\"the string of light bulbs will fail if at least one light bulb fails . so , let ' s find the probability of the opposite event and subtract that value from 1 . the opposite event is when none of the 30 light bulbs fails , since the probability of each light bulb not to fail is 1 - 0.1 = 0.90 the the probability that none of the 30 light bulbs fails is 0.90 ^ 30 . hence , the probability that at least one light bulb fails is 1 - 0.90 ^ 30 . answer : e . now , you should have spotted that your reasoning was not right because of one simple thing , consider the case when we have 100 light bulbs instead of 30 , then according to your logic the probability that the string of light bulbs will fail would be 100 * 0.1 = 10 , which is not possible since the probability of an event can not be more than 1 ( 100 % ) .\"","correct":"e","options":{"a":"0.1 ","b":"( 0.1 ) ^ 30 ","c":"1 - ( 0.1 ) ^ 30 ","d":"( 0.93 ) ^ 30","e":"1 - ( 0.90 ) ^ 30"},"options_float":{"a":0.1,"b":0.1,"c":1.0,"d":0.93,"e":1.0},"annotated_formula":"subtract(const_1, power(subtract(const_1, 0.1), 30))","linear_formula":"subtract(const_1,n1)|power(#0,n0)|subtract(const_1,#1)|","chain":"1 - 0.1<\/gadget>\n0.9<\/output>\n0.9 ** 30<\/gadget>\n0.042391<\/output>\n1 - 0.042391<\/gadget>\n0.957609<\/output>\n0.957609<\/result>","index":1817} +{"problem":"for a certain art exhibit , a museum sold admission tickets to a group of 30 people every 5 minutes from 9 : 00 in the morning to 4 : 55 in the afternoon , inclusive . the price of a regular admission ticket was $ 10 and the price of a student ticket was $ 6 . if on one day 3 times as many regular admission tickets were sold as student tickets , what was the total revenue from ticket sales that day ?","rationale":"from 9 : 00 in the morning to 4 : 55 in the afternoon , inclusive there are 8 * 12 = 96 five - minute intervals , thus total of 96 * 30 tickets were sold . say x student and 3 x regular tickets were sold , then x + 3 x = 96 * 30 - - > x = 24 * 30 and 3 x = 3 * ( 24 * 30 ) = 24 * 90 . therefore , the total revenue from ticket sales that day was 24 * 30 * 6 + 24 * 90 * 10 = $ 25,920 . answer : b .","correct":"b","options":{"a":"$ 24960 ","b":"$ 25920 ","c":"$ 28080 ","d":"$ 28500","e":"$ 29160"},"options_float":{"a":24960.0,"b":25920.0,"c":28080.0,"d":28500.0,"e":29160.0},"annotated_formula":"add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, const_4), 9), const_1), const_12), 30), add(3, const_1)), 3), 10), multiply(divide(multiply(multiply(add(subtract(add(const_12, const_4), 9), const_1), const_12), 30), add(3, const_1)), 6))","linear_formula":"add(const_12,const_4)|add(n8,const_1)|subtract(#0,n2)|add(#2,const_1)|multiply(#3,const_12)|multiply(n0,#4)|divide(#5,#1)|multiply(n8,#6)|multiply(n7,#6)|multiply(n6,#7)|add(#9,#8)","chain":"12 + 4<\/gadget>\n16<\/output>\n16 - 9<\/gadget>\n7<\/output>\n7 + 1<\/gadget>\n8<\/output>\n8 * 12<\/gadget>\n96<\/output>\n96 * 30<\/gadget>\n2_880<\/output>\n3 + 1<\/gadget>\n4<\/output>\n2_880 \/ 4<\/gadget>\n720<\/output>\n720 * 3<\/gadget>\n2_160<\/output>\n2_160 * 10<\/gadget>\n21_600<\/output>\n720 * 6<\/gadget>\n4_320<\/output>\n21_600 + 4_320<\/gadget>\n25_920<\/output>\n25_920<\/result>","index":1821} +{"problem":"during a car trip , maria stopped to rest after she traveled 1 \/ 2 of the total distance to her destination . she stopped again after she traveled 1 \/ 4 of the distance remaining between her first stop and her destination , and then she drove the remaining 105 miles to her detination . what was the total distance , in miles from maria ' s starting point to her destination ?","rationale":"\"easy 280 is the answer . 3 \/ 4 ( x \/ 2 ) = 105 x = 105 * 8 \/ 3 = 280 . answer a\"","correct":"a","options":{"a":"280 ","b":"320 ","c":"360 ","d":"420","e":"480"},"options_float":{"a":280.0,"b":320.0,"c":360.0,"d":420.0,"e":480.0},"annotated_formula":"divide(105, subtract(1, add(divide(1, 2), multiply(divide(1, 2), divide(1, 4)))))","linear_formula":"divide(n0,n1)|divide(n0,n3)|multiply(#0,#1)|add(#0,#2)|subtract(n0,#3)|divide(n4,#4)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) * (1\/4)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/2) + (1\/8)<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n105 \/ (3\/8)<\/gadget>\n280<\/output>\n280<\/result>","index":1822} +{"problem":"if 2 \/ 3 rd of the contents of a container evaporated on the 1 st day . and 1 \/ 4 th of the remaining evaporated on the second day . what part of the contents of the container is left at the end of the second day ?","rationale":"after first day , 1 \/ 3 of the contents remain . after second day , ( 1 \/ 3 ) - ( 1 \/ 4 ( 2 \/ 3 ) ) = ( 1 \/ 3 ) - ( 1 \/ 6 ) = 1 \/ 6 of the contents remain answer : d","correct":"d","options":{"a":"1 \/ 4 ","b":"1 \/ 12 ","c":"1 \/ 18 ","d":"1 \/ 6","e":"1 \/ 2"},"options_float":{"a":0.25,"b":0.0833333333,"c":0.0555555556,"d":0.1666666667,"e":0.5},"annotated_formula":"multiply(divide(subtract(multiply(4, 1), multiply(3, 1)), multiply(4, 1)), divide(2, 3))","linear_formula":"divide(n0,n1)|multiply(n2,n4)|multiply(n1,n2)|subtract(#1,#2)|divide(#3,#1)|multiply(#4,#0)","chain":"4 * 1<\/gadget>\n4<\/output>\n3 * 1<\/gadget>\n3<\/output>\n4 - 3<\/gadget>\n1<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/4) * (2\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":1824} +{"problem":"the average of first 7 prime numbers is ?","rationale":"\"sum of 7 prime no . = 58 average = 58 \/ 7 = 8.29 answer : c\"","correct":"c","options":{"a":"10.11 ","b":"11.11 ","c":"8.29 ","d":"13.11","e":"14.11"},"options_float":{"a":10.11,"b":11.11,"c":8.29,"d":13.11,"e":14.11},"annotated_formula":"add(7, const_1)","linear_formula":"add(n0,const_1)|","chain":"7 + 1<\/gadget>\n8<\/output>\n8<\/result>","index":1825} +{"problem":"in a group of 28 junior high school students , 5 take french , 10 take spanish , and 4 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ?","rationale":"b 13 add 1 + 10 + 4 to get 15 . then subtract 21 from the total students ⇒ 28 – 15 = 13 . answer is b","correct":"b","options":{"a":"7 ","b":"13 ","c":"9 ","d":"12","e":"10"},"options_float":{"a":7.0,"b":13.0,"c":9.0,"d":12.0,"e":10.0},"annotated_formula":"subtract(28, add(5, 10))","linear_formula":"add(n1,n2)|subtract(n0,#0)","chain":"5 + 10<\/gadget>\n15<\/output>\n28 - 15<\/gadget>\n13<\/output>\n13<\/result>","index":1826} +{"problem":"a , b , c can complete a piece of work in 3,9 and 15 days respectively . working together , they will complete the same work in how many days ?","rationale":"( a + b + c ) ' s 1 day work = ( 1 \/ 3 ) + ( 1 \/ 9 ) + ( 1 \/ 115 ) = 23 \/ 45 a , b , c together will complete the work in 45 \/ 23 days answer is a","correct":"a","options":{"a":"45 \/ 23 ","b":"55 \/ 23 ","c":"61 \/ 57 ","d":"12 \/ 13","e":"22 \/ 23"},"options_float":{"a":1.9565217391,"b":2.3913043478,"c":1.0701754386,"d":0.9230769231,"e":0.9565217391},"annotated_formula":"inverse(add(inverse(15), add(inverse(const_3), inverse(add(add(const_4, const_4), const_1)))))","linear_formula":"add(const_4,const_4)|inverse(const_3)|inverse(n1)|add(#0,const_1)|inverse(#3)|add(#1,#4)|add(#5,#2)|inverse(#6)","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n4 + 4<\/gadget>\n8<\/output>\n8 + 1<\/gadget>\n9<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/3) + (1\/9)<\/gadget>\n4\/9 = around 0.444444<\/output>\n(1\/15) + (4\/9)<\/gadget>\n23\/45 = around 0.511111<\/output>\n1 \/ (23\/45)<\/gadget>\n45\/23 = around 1.956522<\/output>\n45\/23 = around 1.956522<\/result>","index":1828} +{"problem":"in a group of cows and chickens , the number of legs was 12 more than twice the number of heads . the number of cows was :","rationale":"\"let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 12 more than twice the number of heads . therefore , 2 × ( x + y ) + 14 = 4 x + 2 y . or , 2 x + 2 y + 12 = 4 x + 2 y . or , 2 x + 12 = 4 x [ subtracting 2 y from both sides ] . or , 12 = 4 x – 2 x [ subtracting 2 x from both sides ] . or , 12 = 2 x . or , x = 6 [ dividing by 2 on both sides ] . therefore , the number of cows = 6 . correct answer : a ) 6\"","correct":"a","options":{"a":"6 ","b":"7 ","c":"10 ","d":"12","e":"14"},"options_float":{"a":6.0,"b":7.0,"c":10.0,"d":12.0,"e":14.0},"annotated_formula":"divide(12, subtract(const_4, const_2))","linear_formula":"subtract(const_4,const_2)|divide(n0,#0)|","chain":"4 - 2<\/gadget>\n2<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6<\/result>","index":1829} +{"problem":"a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ?","rationale":"\"new value – old value ) \/ old value x 100 we are given : february sales this year = 385 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 385 million and the old value = 320 million . let ’ s plug them into our percent change formula . ( new value – old value ) \/ old value x 100 [ ( 385 – 320 ) \/ 320 ] x 100 65 \/ 320 x 100 13 \/ 64 x 100 ≈ 13 \/ 65 x 100 ≈ 1 \/ 5 x 100 ≈ 20 % . the answer is c .\"","correct":"c","options":{"a":"2 % ","b":"17 % ","c":"20 % ","d":"65 %","e":"83 %"},"options_float":{"a":2.0,"b":17.0,"c":20.0,"d":65.0,"e":83.0},"annotated_formula":"multiply(divide(subtract(385, 320), 320), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|","chain":"385 - 320<\/gadget>\n65<\/output>\n65 \/ 320<\/gadget>\n13\/64 = around 0.203125<\/output>\n(13\/64) * 100<\/gadget>\n325\/16 = around 20.3125<\/output>\n325\/16 = around 20.3125<\/result>","index":1830} +{"problem":"if a - b = 3 and a ^ 2 + b ^ 2 = 29 , find the value of ab .","rationale":"\"2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 29 - 9 = 20 ab = 10 . b )\"","correct":"b","options":{"a":"5 ","b":"10 ","c":"12 ","d":"14","e":"15"},"options_float":{"a":5.0,"b":10.0,"c":12.0,"d":14.0,"e":15.0},"annotated_formula":"multiply(multiply(add(3, divide(subtract(sqrt(29), 3), 2)), divide(subtract(sqrt(29), 3), 2)), 2)","linear_formula":"sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)|","chain":"29 ** (1\/2)<\/gadget>\nsqrt(29) = around 5.385165<\/output>\n(sqrt(29)) - 3<\/gadget>\n-3 + sqrt(29) = around 2.385165<\/output>\n(-3 + sqrt(29)) \/ 2<\/gadget>\n-3\/2 + sqrt(29)\/2 = around 1.192582<\/output>\n3 + (-3\/2 + sqrt(29)\/2)<\/gadget>\n3\/2 + sqrt(29)\/2 = around 4.192582<\/output>\n(3\/2 + sqrt(29)\/2) * (-3\/2 + sqrt(29)\/2)<\/gadget>\n(-3\/2 + sqrt(29)\/2)*(3\/2 + sqrt(29)\/2) = around 5<\/output>\n((-3\/2 + sqrt(29)\/2)*(3\/2 + sqrt(29)\/2)) * 2<\/gadget>\n2*(-3\/2 + sqrt(29)\/2)*(3\/2 + sqrt(29)\/2) = around 10<\/output>\n2*(-3\/2 + sqrt(29)\/2)*(3\/2 + sqrt(29)\/2) = around 10<\/result>","index":1831} +{"problem":"students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 182 cms . what was the ratio of men to women ?","rationale":"\"we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 182 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 182 m ) \/ ( w + m ) = 180 170 w + 182 m = 180 w + 180 m 2 m = 10 w m = 5 w m \/ w = 5 \/ 1 the ratio of men to women is 5 to 1 . b\"","correct":"b","options":{"a":"5 : 2 ","b":"5 : 1 ","c":"4 : 3 ","d":"4 : 1","e":"3 : 1"},"options_float":{"a":2.5,"b":5.0,"c":1.3333333333,"d":4.0,"e":3.0},"annotated_formula":"divide(subtract(180, 170), subtract(182, 180))","linear_formula":"subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)|","chain":"180 - 170<\/gadget>\n10<\/output>\n182 - 180<\/gadget>\n2<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":1835} +{"problem":"the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 400 , by what percent is the price to be reduced ?","rationale":"\"price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 400 % change = ( final value - initial value ) * 100 \/ initial value % reduction = ( reduction in price ) * 100 \/ initial value i . e . % reduction = ( 400 ) * 100 \/ 500 = 80 % answer : option e\"","correct":"e","options":{"a":"10 % ","b":"15 % ","c":"20 % ","d":"25 %","e":"80 %"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":80.0},"annotated_formula":"multiply(divide(400, 500), const_100)","linear_formula":"divide(n1,n0)|multiply(#0,const_100)|","chain":"400 \/ 500<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":1837} +{"problem":"jack has two dice , one has 6 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , and the other has 7 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , 7 . if jack rolls both dice what is the probability that both of the number ' s sum will be 13 ?","rationale":"both dice addition result is 13 = ( 6 + 7 ) probability that the number on first die is 6 = 1 \/ 6 probability that the number on second die is 7 = 1 \/ 7 probability that both dice addition result is 13 = 1 \/ 6 * 1 \/ 7 = 1 \/ 42 d","correct":"d","options":{"a":"3 \/ 14 ","b":"2 \/ 7 ","c":"1 \/ 3 ","d":"1 \/ 42","e":"12 \/ 21"},"options_float":{"a":0.2142857143,"b":0.2857142857,"c":0.3333333333,"d":0.0238095238,"e":0.5714285714},"annotated_formula":"multiply(divide(const_1, 6), divide(1, 7))","linear_formula":"divide(const_1,n0)|divide(n1,n7)|multiply(#0,#1)","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/6) * (1\/7)<\/gadget>\n1\/42 = around 0.02381<\/output>\n1\/42 = around 0.02381<\/result>","index":1839} +{"problem":"if a = 105 and a ^ 3 = 21 * 25 * 315 * b , what is the value of b ?","rationale":"first step will be to break down all the numbers into their prime factors . 105 = 3 * 5 * 7 21 = 7 * 3 25 = 5 * 5 315 = 3 * 3 * 5 * 7 so , ( 105 ) ^ 3 = 3 * 7 * 5 * 5 * 3 * 3 * 5 * 7 * b therefore ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 ^ 2 * b therefore , b = 3 ^ 3 * 5 ^ 3 * 7 ^ 3 \/ 3 ^ 3 * 5 ^ 3 * 7 ^ 2 b = 7 correct answer b .","correct":"b","options":{"a":"35 ","b":"7 ","c":"45 ","d":"49","e":"54"},"options_float":{"a":35.0,"b":7.0,"c":45.0,"d":49.0,"e":54.0},"annotated_formula":"divide(power(105, 3), multiply(multiply(21, 25), 315))","linear_formula":"multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)","chain":"105 ** 3<\/gadget>\n1_157_625<\/output>\n21 * 25<\/gadget>\n525<\/output>\n525 * 315<\/gadget>\n165_375<\/output>\n1_157_625 \/ 165_375<\/gadget>\n7<\/output>\n7<\/result>","index":1841} +{"problem":"boy sells a book for rs . 540 he gets a loss of 10 % , to gain 10 % , what should be the sp ?","rationale":"\"cost price = 540 \/ 90 x 100 = 600 to gain 10 % = 600 x 10 \/ 100 = 60 sp = cp + gain = 600 + 60 = 660 answer : e\"","correct":"e","options":{"a":"430 ","b":"450 ","c":"550 ","d":"590","e":"660"},"options_float":{"a":430.0,"b":450.0,"c":550.0,"d":590.0,"e":660.0},"annotated_formula":"add(divide(540, subtract(const_1, divide(10, const_100))), multiply(divide(540, subtract(const_1, divide(10, const_100))), divide(10, const_100)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n0,#2)|multiply(#3,#1)|add(#3,#4)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n540 \/ (9\/10)<\/gadget>\n600<\/output>\n600 * (1\/10)<\/gadget>\n60<\/output>\n600 + 60<\/gadget>\n660<\/output>\n660<\/result>","index":1842} +{"problem":"xavier , yvonne , and zelda each try independently to solve a problem . if their individual probabilities for success are 1 \/ 3 , 1 \/ 2 and 5 \/ 8 , respectively , what is the probability that xavier and yvonne , but not zelda , will solve the problem ?","rationale":"\"p ( xavier will solve ) = 1 \/ 3 p ( yvonne will solve ) = 1 \/ 2 p ( zelda will not solve ) = 1 - 5 \/ 8 = 3 \/ 8 . now , we need to multiply all this ps to find an answer : p = ( 1 \/ 3 ) * ( 1 \/ 2 ) * ( 3 \/ 8 ) = 1 \/ 16 . ans . a .\"","correct":"a","options":{"a":"1 \/ 16 ","b":"7 \/ 8 ","c":"9 \/ 64 ","d":"5 \/ 64","e":"3 \/ 64"},"options_float":{"a":0.0625,"b":0.875,"c":0.140625,"d":0.078125,"e":0.046875},"annotated_formula":"multiply(multiply(divide(1, 3), divide(1, 2)), subtract(1, divide(5, 8)))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|multiply(#0,#1)|subtract(n0,#2)|multiply(#3,#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/3) * (1\/2)<\/gadget>\n1\/6 = around 0.166667<\/output>\n5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n(1\/6) * (3\/8)<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":1843} +{"problem":"if 0.2 of a number is equal to 0.08 of another number , the ratio of the numbers i","rationale":"\"sol . 0.2 a = 0.08 b â ‡ ” a \/ b = 0.08 \/ 0.20 = 8 \/ 20 = 2 \/ 5 â ˆ ´ a : b = 2 : 5 . answer b\"","correct":"b","options":{"a":"2 : 3 ","b":"2 : 5 ","c":"3 : 20 ","d":"20 : 3","e":"none"},"options_float":{"a":0.6666666667,"b":0.4,"c":0.15,"d":6.6666666667,"e":null},"annotated_formula":"divide(multiply(0.08, const_100), multiply(0.2, const_100))","linear_formula":"multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)|","chain":"0.08 * 100<\/gadget>\n8<\/output>\n0.2 * 100<\/gadget>\n20<\/output>\n8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":1845} +{"problem":"a person want to give his money of $ 1800 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ?","rationale":"\"b ' s share = 1800 * 3 \/ 9 = $ 600 answer is b\"","correct":"b","options":{"a":"$ 700 ","b":"$ 600 ","c":"$ 900 ","d":"$ 500","e":"$ 400"},"options_float":{"a":700.0,"b":600.0,"c":900.0,"d":500.0,"e":400.0},"annotated_formula":"multiply(divide(3, add(add(2, 3), 4)), 1800)","linear_formula":"add(n1,n2)|add(n4,#0)|divide(n1,#1)|multiply(n0,#2)|","chain":"2 + 3<\/gadget>\n5<\/output>\n5 + 4<\/gadget>\n9<\/output>\n3 \/ 9<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 1_800<\/gadget>\n600<\/output>\n600<\/result>","index":1846} +{"problem":"in what time will a train 240 metres long cross an electric pole , if its speed be 126 km \/ hr ?","rationale":"\"solution speed = ( 126 x 5 \/ 18 ) m \/ sec = 35 m \/ sec time taken = ( 240 \/ 35 ) sec = 6.85 sec . answer b\"","correct":"b","options":{"a":"6.75 sec ","b":"6.85 sec ","c":"6.55 sec ","d":"6.44 sec","e":"7.85 sec"},"options_float":{"a":6.75,"b":6.85,"c":6.55,"d":6.44,"e":7.85},"annotated_formula":"divide(240, multiply(126, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n126 * (5\/18)<\/gadget>\n35<\/output>\n240 \/ 35<\/gadget>\n48\/7 = around 6.857143<\/output>\n48\/7 = around 6.857143<\/result>","index":1847} +{"problem":"a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 7 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ?","rationale":"correct answer : b the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.07 ( $ 1120 ) = $ 78.40 . the correct answer is b .","correct":"b","options":{"a":"$ 54.00 ","b":"$ 78.40 ","c":"$ 90.00 ","d":"$ 100.80","e":"$ 154.80"},"options_float":{"a":54.0,"b":78.4,"c":90.0,"d":100.8,"e":154.8},"annotated_formula":"divide(multiply(subtract(1720, 600), 7), const_100)","linear_formula":"subtract(n3,n0)|multiply(n1,#0)|divide(#1,const_100)","chain":"1_720 - 600<\/gadget>\n1_120<\/output>\n1_120 * 7<\/gadget>\n7_840<\/output>\n7_840 \/ 100<\/gadget>\n392\/5 = around 78.4<\/output>\n392\/5 = around 78.4<\/result>","index":1848} +{"problem":"how long does a train 110 m long travelling at 60 kmph takes to cross a bridge of 200 m in length ?","rationale":"\"a 18.6 sec d = 110 + 200 = 310 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 310 * 3 \/ 50 = 18.6 sec answer is a\"","correct":"a","options":{"a":"18.6 sec ","b":"14.9 sec ","c":"12.4 sec ","d":"16.8 sec","e":"11.8 sec"},"options_float":{"a":18.6,"b":14.9,"c":12.4,"d":16.8,"e":11.8},"annotated_formula":"divide(add(110, 200), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"110 + 200<\/gadget>\n310<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n310 \/ (50\/3)<\/gadget>\n93\/5 = around 18.6<\/output>\n93\/5 = around 18.6<\/result>","index":1849} +{"problem":"excepting ‘ seldom ’ which means – rarely , infrequently and not often , is the antonym .","rationale":"answer : d","correct":"d","options":{"a":"22 ","b":"278 ","c":"267 ","d":"176","e":"971"},"options_float":{"a":22.0,"b":278.0,"c":267.0,"d":176.0,"e":971.0},"annotated_formula":"add(const_100, add(multiply(add(const_3, const_4), const_10), add(const_3, const_3)))","linear_formula":"add(const_3,const_3)|add(const_3,const_4)|multiply(#1,const_10)|add(#0,#2)|add(#3,const_100)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 10<\/gadget>\n70<\/output>\n3 + 3<\/gadget>\n6<\/output>\n70 + 6<\/gadget>\n76<\/output>\n100 + 76<\/gadget>\n176<\/output>\n176<\/result>","index":1850} +{"problem":"two pipes a and b can fill a tank in 24 minutes and 32 minutes respectively . if both the pipes are opened simultaneously , after how much time should b be closed so that the tank is full in 18 minutes ?","rationale":"pipe a will work for 18 minutes . suppose pipe b is closed after x minutes than we have 18 \/ 24 + x \/ 32 = 1 x = 8 minute . answer : d","correct":"d","options":{"a":"7 minute ","b":"9 minute ","c":"10 minute ","d":"8 minute","e":"11 minute"},"options_float":{"a":7.0,"b":9.0,"c":10.0,"d":8.0,"e":11.0},"annotated_formula":"multiply(subtract(const_1, multiply(divide(const_1, 24), 18)), 32)","linear_formula":"divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#1)|multiply(n1,#2)","chain":"1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/24) * 18<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 32<\/gadget>\n8<\/output>\n8<\/result>","index":1853} +{"problem":"the average height of 35 boys in a class was calculated as 181 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ?","rationale":"\"calculated average height of 35 boys = 181 cm . wrong total height of 35 boys = 181 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 181 cm - ( 166 cm - 106 cm ) \/ 35 = 181 cm - ( 166 cm - 106 cm ) \/ 35 = 181 cm - 60 \/ 35 cm = 181 cm - 1.71 cm = 179.29 cm . answer : b\"","correct":"b","options":{"a":"187.89 cm ","b":"179.29 cm ","c":"123.98 cm ","d":"149.98 cm","e":"146.89 cm"},"options_float":{"a":187.89,"b":179.29,"c":123.98,"d":149.98,"e":146.89},"annotated_formula":"floor(divide(add(subtract(multiply(35, 181), 166), 106), 35))","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)|","chain":"35 * 181<\/gadget>\n6_335<\/output>\n6_335 - 166<\/gadget>\n6_169<\/output>\n6_169 + 106<\/gadget>\n6_275<\/output>\n6_275 \/ 35<\/gadget>\n1_255\/7 = around 179.285714<\/output>\nfloor(1_255\/7)<\/gadget>\n179<\/output>\n179<\/result>","index":1854} +{"problem":"in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 280 vote ’ s majority . what is the total number of votes polled ?","rationale":"\"explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 280 ( 40 * 7 = 280 ) 100 % - - - - - > 700 ( 100 * 7 = 700 ) answer : option b\"","correct":"b","options":{"a":"750 ","b":"700 ","c":"800 ","d":"850","e":"none of these"},"options_float":{"a":750.0,"b":700.0,"c":800.0,"d":850.0,"e":null},"annotated_formula":"divide(multiply(const_100, 280), subtract(70, subtract(const_100, 70)))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n0)|subtract(n0,#1)|divide(#0,#2)|","chain":"100 * 280<\/gadget>\n28_000<\/output>\n100 - 70<\/gadget>\n30<\/output>\n70 - 30<\/gadget>\n40<\/output>\n28_000 \/ 40<\/gadget>\n700<\/output>\n700<\/result>","index":1855} +{"problem":"a dishonest dealer professes to sell goods at the cost price but uses a weight of 990 grams per kg , what is his percent ?","rationale":"\"990 - - - 10 100 - - - ? = > 1.01 % answer : a\"","correct":"a","options":{"a":"1.01 % ","b":"25 % ","c":"22 % ","d":"29 %","e":"45 %"},"options_float":{"a":1.01,"b":25.0,"c":22.0,"d":29.0,"e":45.0},"annotated_formula":"subtract(multiply(divide(const_100, 990), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)","linear_formula":"add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|","chain":"100 \/ 990<\/gadget>\n10\/99 = around 0.10101<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n(10\/99) * 1_000<\/gadget>\n10_000\/99 = around 101.010101<\/output>\n(10_000\/99) - 100<\/gadget>\n100\/99 = around 1.010101<\/output>\n100\/99 = around 1.010101<\/result>","index":1856} +{"problem":"50 % of x is greater than 1 \/ 3 rd of x by 110 . what is x ?","rationale":"\"5 x \/ 10 - x \/ 3 = 110 1 x \/ 6 = 110 x = 660 answer : e\"","correct":"e","options":{"a":"100 ","b":"200 ","c":"300 ","d":"350","e":"660"},"options_float":{"a":100.0,"b":200.0,"c":300.0,"d":350.0,"e":660.0},"annotated_formula":"divide(110, subtract(divide(50, const_100), divide(1, 3)))","linear_formula":"divide(n0,const_100)|divide(n1,n2)|subtract(#0,#1)|divide(n3,#2)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/2) - (1\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n110 \/ (1\/6)<\/gadget>\n660<\/output>\n660<\/result>","index":1858} +{"problem":"the average age of a , b and c is 26 years . if the average age of a and c is 29 years , what is the age of b in years ?","rationale":"\"c 20 age of b = age of ( a + b + c ) – age of ( a + c ) = 26 × 3 – 29 × 2 = 78 – 58 = 20 years\"","correct":"c","options":{"a":"19 ","b":"35 ","c":"20 ","d":"32","e":"21"},"options_float":{"a":19.0,"b":35.0,"c":20.0,"d":32.0,"e":21.0},"annotated_formula":"subtract(multiply(26, const_3), multiply(29, const_2))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_2)|subtract(#0,#1)|","chain":"26 * 3<\/gadget>\n78<\/output>\n29 * 2<\/gadget>\n58<\/output>\n78 - 58<\/gadget>\n20<\/output>\n20<\/result>","index":1860} +{"problem":"a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 25 % lemonade syrup ?","rationale":"\"let the total solution is 150 l with 80 l water 70 l syrup . to make 25 % syrup solution , the result solution must have 112.5 l syrup and 37.5 l syrup . therefore we are taking 32.5 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 32.5 l syrup in 69.6 l solution we started by multiplying 10 now to get to the result we need to divide by 32.5 = > amount of solution to be replaced with water = ( 69.6 \/ 32.5 ) = 2.14 . correct option : c\"","correct":"c","options":{"a":"1.5 ","b":"1.75 ","c":"2.14 ","d":"2.34","e":"2.64"},"options_float":{"a":1.5,"b":1.75,"c":2.14,"d":2.34,"e":2.64},"annotated_formula":"multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7))","linear_formula":"add(n0,n1)|add(const_2,const_3)|divide(n1,#0)|divide(const_2,#1)|subtract(#2,#3)|divide(#4,#2)|multiply(#0,#5)|","chain":"8 + 7<\/gadget>\n15<\/output>\n7 \/ 15<\/gadget>\n7\/15 = around 0.466667<\/output>\n2 + 3<\/gadget>\n5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(7\/15) - (2\/5)<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/15) \/ (7\/15)<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/7) * 15<\/gadget>\n15\/7 = around 2.142857<\/output>\n15\/7 = around 2.142857<\/result>","index":1861} +{"problem":"in what time will a train 175 m long cross an electric pole , it its speed be 180 km \/ hr ?","rationale":"\"speed = 180 * 5 \/ 18 = 50 m \/ sec time taken = 175 \/ 50 = 3.5 sec . answer : e\"","correct":"e","options":{"a":"7.5 ","b":"6.5 ","c":"5.5 ","d":"4.5","e":"3.5"},"options_float":{"a":7.5,"b":6.5,"c":5.5,"d":4.5,"e":3.5},"annotated_formula":"divide(175, multiply(180, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n180 * (5\/18)<\/gadget>\n50<\/output>\n175 \/ 50<\/gadget>\n7\/2 = around 3.5<\/output>\n7\/2 = around 3.5<\/result>","index":1862} +{"problem":"80 % of the population of a village is 64000 . the total population of the village is ?","rationale":"\"x * ( 80 \/ 100 ) = 64000 x = 800 * 100 x = 80000 answer : c\"","correct":"c","options":{"a":"26799 ","b":"24000 ","c":"80000 ","d":"29973","e":"12312"},"options_float":{"a":26799.0,"b":24000.0,"c":80000.0,"d":29973.0,"e":12312.0},"annotated_formula":"multiply(divide(const_100, 80), 64000)","linear_formula":"divide(const_100,n0)|multiply(n1,#0)|","chain":"100 \/ 80<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 64_000<\/gadget>\n80_000<\/output>\n80_000<\/result>","index":1863} +{"problem":"the perimeter of a triangle is 42 cm and the inradius of the triangle is 5.0 cm . what is the area of the triangle ?","rationale":"\"area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 5.0 * 42 \/ 2 = 105 cm 2 answer : c\"","correct":"c","options":{"a":"128 cm 2 ","b":"36.0 cm 2 ","c":"105 cm 2 ","d":"25 cm 2","e":"135 cm 2"},"options_float":{"a":128.0,"b":36.0,"c":105.0,"d":25.0,"e":135.0},"annotated_formula":"triangle_area(5.0, 42)","linear_formula":"triangle_area(n0,n1)|","chain":"(5 * 42) \/ 2<\/gadget>\n105<\/output>\n105<\/result>","index":1865} +{"problem":"company c sells a line of 20 products with an average retail price of $ 1,200 . if none of these products sells for less than $ 400 , and exactly 10 of the products sell for less than $ 1,000 , what is the greatest possible selling price of the most expensive product ?","rationale":"the average price of 20 products is $ 1,200 means that the total price of 20 products is 20 * 1,200 = $ 24,000 . next , since exactly 10 of the products sell for less than $ 1,000 , then let ' s make these 10 items to be at $ 400 each ( min possible ) . now , the remaining 9 items can not be priced less than $ 1,000 , thus the minimum possible price of each of these 9 items is $ 1,000 . thus the minimum possible value of 19 products is 10 * 400 + 9 * 1,000 = $ 13,000 . therefore , the greatest possible selling price of the most expensive product is $ 24,000 - $ 13,000 = $ 11,000 . answer : b .","correct":"b","options":{"a":"12000 ","b":"11000 ","c":"10000 ","d":"13000","e":"15000"},"options_float":{"a":12000.0,"b":11000.0,"c":10000.0,"d":13000.0,"e":15000.0},"annotated_formula":"subtract(multiply(multiply(const_12, const_100), 20), add(multiply(400, 10), multiply(subtract(subtract(20, 10), const_1), const_1000)))","linear_formula":"multiply(const_100,const_12)|multiply(n2,n3)|subtract(n0,n3)|multiply(n0,#0)|subtract(#2,const_1)|multiply(#4,const_1000)|add(#1,#5)|subtract(#3,#6)","chain":"12 * 100<\/gadget>\n1_200<\/output>\n1_200 * 20<\/gadget>\n24_000<\/output>\n400 * 10<\/gadget>\n4_000<\/output>\n20 - 10<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 * 1_000<\/gadget>\n9_000<\/output>\n4_000 + 9_000<\/gadget>\n13_000<\/output>\n24_000 - 13_000<\/gadget>\n11_000<\/output>\n11_000<\/result>","index":1867} +{"problem":"the total age of a and b is 12 year more than the total age of b and c . c is how many years younger than a ?","rationale":"\"solution ( a + b ) - ( b + c ) = 12 ⇔ a - c = 12 . answer a\"","correct":"a","options":{"a":"12 ","b":"24 ","c":"data inadequate ","d":"none of these","e":"34"},"options_float":{"a":12.0,"b":24.0,"c":null,"d":null,"e":34.0},"annotated_formula":"multiply(12, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"12 * 1<\/gadget>\n12<\/output>\n12<\/result>","index":1868} +{"problem":"the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 867 sq m , then what is the breadth of the rectangular plot ?","rationale":"answer : option b explanation : let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 867 3 b 2 = 867 b 2 = 289 = 172 ( b > 0 ) b = 17 m . answer : option b","correct":"b","options":{"a":"19 m ","b":"17 m ","c":"10 m ","d":"12 m","e":"18 m"},"options_float":{"a":19.0,"b":17.0,"c":10.0,"d":12.0,"e":18.0},"annotated_formula":"sqrt(divide(867, const_3))","linear_formula":"divide(n0,const_3)|sqrt(#0)","chain":"867 \/ 3<\/gadget>\n289<\/output>\n289 ** (1\/2)<\/gadget>\n17<\/output>\n17<\/result>","index":1869} +{"problem":"in a fuel station the service costs $ 1.05 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty - ?","rationale":"\"service cost of 3 van and 2 truck = 1.05 * ( 3 + 2 ) = 5.25 fuel in 3 van = 3 * 65 = 195 litre fuel in 2 trucks = 2 * 65 ( 1 + 120 \/ 100 ) = 286 total fuel ( van + truck ) = 481 litre total fuel cost = 481 * 0.6 = 288.6 total cost = fuel + service = 288.6 + 5.25 = 293.85 answer is d\"","correct":"d","options":{"a":"122.6 $ ","b":"128.9 $ ","c":"243.7 $ ","d":"293.85 $","e":"312.12 $"},"options_float":{"a":122.6,"b":128.9,"c":243.7,"d":293.85,"e":312.12},"annotated_formula":"add(multiply(multiply(add(65, divide(multiply(65, 120), const_100)), 2), 0.6), multiply(multiply(65, 3), 0.6))","linear_formula":"multiply(n4,n5)|multiply(n2,n4)|divide(#0,const_100)|multiply(n1,#1)|add(n4,#2)|multiply(n3,#4)|multiply(n1,#5)|add(#6,#3)|","chain":"65 * 120<\/gadget>\n7_800<\/output>\n7_800 \/ 100<\/gadget>\n78<\/output>\n65 + 78<\/gadget>\n143<\/output>\n143 * 2<\/gadget>\n286<\/output>\n286 * 0.6<\/gadget>\n171.6<\/output>\n65 * 3<\/gadget>\n195<\/output>\n195 * 0.6<\/gadget>\n117<\/output>\n171.6 + 117<\/gadget>\n288.6<\/output>\n288.6<\/result>","index":1870} +{"problem":"how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 34 % iodine solution ?","rationale":"\"solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) \/ ( x + 35 ) = 34 \/ 100 40 x + 700 = 34 x + 1220 6 x = 820 x = 136 lts correct option : e\"","correct":"e","options":{"a":"35 ","b":"49 ","c":"100 ","d":"105","e":"136"},"options_float":{"a":35.0,"b":49.0,"c":100.0,"d":105.0,"e":136.0},"annotated_formula":"add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 34), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35)","linear_formula":"divide(n0,const_100)|multiply(n1,n2)|divide(#1,const_100)|multiply(n2,#0)|multiply(n3,#2)|multiply(n2,#2)|subtract(#3,#2)|subtract(#4,#5)|divide(#7,#6)|add(n1,#8)|","chain":"35 * 20<\/gadget>\n700<\/output>\n700 \/ 100<\/gadget>\n7<\/output>\n7 * 34<\/gadget>\n238<\/output>\n7 * 20<\/gadget>\n140<\/output>\n238 - 140<\/gadget>\n98<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n20 * (2\/5)<\/gadget>\n8<\/output>\n8 - 7<\/gadget>\n1<\/output>\n98 \/ 1<\/gadget>\n98<\/output>\n98 + 35<\/gadget>\n133<\/output>\n133<\/result>","index":1871} +{"problem":"a car runs 375 km in 3 hours . what ' s the car ' s speed ?","rationale":"solution : 375 ÷ 3 = 125 answer b","correct":"b","options":{"a":"124 ","b":"125 ","c":"126 ","d":"127","e":"none"},"options_float":{"a":124.0,"b":125.0,"c":126.0,"d":127.0,"e":null},"annotated_formula":"speed(375, 3)","linear_formula":"speed(n0,n1)","chain":"375 \/ 3<\/gadget>\n125<\/output>\n125<\/result>","index":1872} +{"problem":"two men a and b working together complete a piece of work which it would have taken them respectively 12 and 18 days to complete if they worked separately . they received rs . 149.25 as their payment . find the share of a in the amount they received ?","rationale":"lcm of 12 and 18 is 36 no of untits of a is 3 units no of units of b is 2 units then 5 - - > 149.25 3 - - - > ? ( 3 * 149.25 ) \/ 5 = 89.55 answer : a","correct":"a","options":{"a":"rs . 89.55 ","b":"rs . 59.70 ","c":"rs . 85.56 ","d":"rs . 88.56","e":"none of these"},"options_float":{"a":89.55,"b":59.7,"c":85.56,"d":88.56,"e":null},"annotated_formula":"multiply(divide(inverse(12), add(inverse(12), inverse(18))), 149.25)","linear_formula":"inverse(n0)|inverse(n1)|add(#0,#1)|divide(#0,#2)|multiply(n2,#3)","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/12) + (1\/18)<\/gadget>\n5\/36 = around 0.138889<\/output>\n(1\/12) \/ (5\/36)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 149.25<\/gadget>\n89.55<\/output>\n89.55<\/result>","index":1873} +{"problem":"a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 1 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ?","rationale":"\"the ratio of economy to luxury cars is 3 : 1 - - > e : l = 3 : 1 = 12 : 4 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 12 : 9 . thus , l : s = 4 : 9 . answer : b .\"","correct":"b","options":{"a":"9 : 8 ","b":"4 : 9 ","c":"3 : 2 ","d":"2 : 3","e":"1 : 2"},"options_float":{"a":1.125,"b":0.4444444444,"c":1.5,"d":0.6666666667,"e":0.5},"annotated_formula":"divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(1, const_4)))","linear_formula":"multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)|","chain":"4 * 3<\/gadget>\n12<\/output>\n3 * 3<\/gadget>\n9<\/output>\n12 \/ 9<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 * 4<\/gadget>\n12<\/output>\n1 * 4<\/gadget>\n4<\/output>\n12 \/ 4<\/gadget>\n3<\/output>\n(4\/3) \/ 3<\/gadget>\n4\/9 = around 0.444444<\/output>\n4\/9 = around 0.444444<\/result>","index":1874} +{"problem":"the compound ratio of 7 : 6 , 3 : 2 and 4 : 5 ?","rationale":"7 \/ 6 * 3 \/ 2 * 4 \/ 5 = 7 \/ 5 7 : 5 answer : a","correct":"a","options":{"a":"7 : 5 ","b":"1 : 87 ","c":"1 : 6 ","d":"1 : 9","e":"1 : 2"},"options_float":{"a":1.4,"b":0.0114942529,"c":0.1666666667,"d":0.1111111111,"e":0.5},"annotated_formula":"divide(divide(multiply(7, 3), multiply(6, 2)), divide(multiply(3, 4), multiply(2, 5)))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)","chain":"7 * 3<\/gadget>\n21<\/output>\n6 * 2<\/gadget>\n12<\/output>\n21 \/ 12<\/gadget>\n7\/4 = around 1.75<\/output>\n3 * 4<\/gadget>\n12<\/output>\n2 * 5<\/gadget>\n10<\/output>\n12 \/ 10<\/gadget>\n6\/5 = around 1.2<\/output>\n(7\/4) \/ (6\/5)<\/gadget>\n35\/24 = around 1.458333<\/output>\n35\/24 = around 1.458333<\/result>","index":1875} +{"problem":"there are 50 students in a class . if 14 % are absent on a particular day , find the number of students present in the class .","rationale":"\"number of students absent on a particular day = 14 % of 50 i . e . , 14 \/ 100 × 50 = 7 therefore , the number of students present = 50 - 7 = 43 students . answer : a\"","correct":"a","options":{"a":"43 ","b":"36 ","c":"28 ","d":"129","e":"11"},"options_float":{"a":43.0,"b":36.0,"c":28.0,"d":129.0,"e":11.0},"annotated_formula":"multiply(subtract(const_1, divide(14, const_100.0)), 50)","linear_formula":"divide(n1,const_100.0)|subtract(const_1,#0)|multiply(#1,n0)|","chain":"14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n1 - (7\/50)<\/gadget>\n43\/50 = around 0.86<\/output>\n(43\/50) * 50<\/gadget>\n43<\/output>\n43<\/result>","index":1877} +{"problem":"excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 31 kmph . of how many minutes does the train stop per hour ?","rationale":"\"explanation : t = 14 \/ 45 * 60 = 18.6 answer : option a\"","correct":"a","options":{"a":"18.6 ","b":"27 ","c":"12 ","d":"121","e":"28"},"options_float":{"a":18.6,"b":27.0,"c":12.0,"d":121.0,"e":28.0},"annotated_formula":"subtract(const_60, multiply(const_60, divide(31, 45)))","linear_formula":"divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|","chain":"31 \/ 45<\/gadget>\n31\/45 = around 0.688889<\/output>\n60 * (31\/45)<\/gadget>\n124\/3 = around 41.333333<\/output>\n60 - (124\/3)<\/gadget>\n56\/3 = around 18.666667<\/output>\n56\/3 = around 18.666667<\/result>","index":1878} +{"problem":"a swimmer can swim in still water at 4 km \/ h . if the speed of the water current is 1 km \/ h , how many hours will the swimmer take to swim against the current for 6 km ?","rationale":"the swimmer can swim against the current at a speed of 4 - 1 = 3 km \/ h . the time it will take is 6 \/ 3 = 2 hours . the answer is e .","correct":"e","options":{"a":"1.2 ","b":"1.4 ","c":"1.6 ","d":"1.8","e":"2"},"options_float":{"a":1.2,"b":1.4,"c":1.6,"d":1.8,"e":2.0},"annotated_formula":"divide(6, subtract(4, 1))","linear_formula":"subtract(n0,n1)|divide(n2,#0)","chain":"4 - 1<\/gadget>\n3<\/output>\n6 \/ 3<\/gadget>\n2<\/output>\n2<\/result>","index":1881} +{"problem":"calculate how long it will take a swimmer to swim a distance of 12 km against the current of a river which flows at 1.2 km \/ hr , given that he can swim in still water at 5 km \/ h","rationale":"\"swim in still water at = 5 speed of river = 1.2 us = 5 - 1.2 = 3.8 distance = 12 t = 12 \/ 3.8 = 3.16 answer : b\"","correct":"b","options":{"a":"3.26 ","b":"3.16 ","c":"2.34 ","d":"3.98","e":"3.17"},"options_float":{"a":3.26,"b":3.16,"c":2.34,"d":3.98,"e":3.17},"annotated_formula":"divide(12, subtract(5, 1.2))","linear_formula":"subtract(n2,n1)|divide(n0,#0)|","chain":"5 - 1.2<\/gadget>\n3.8<\/output>\n12 \/ 3.8<\/gadget>\n3.157895<\/output>\n3.157895<\/result>","index":1883} +{"problem":"find the value of ( 100 + 20 \/ 90 ) × 90","rationale":"\"( 9000 + 20 ) \/ 90 * 90 = 9020 answer : e\"","correct":"e","options":{"a":"4520 ","b":"9100 ","c":"9150 ","d":"8120","e":"9020"},"options_float":{"a":4520.0,"b":9100.0,"c":9150.0,"d":8120.0,"e":9020.0},"annotated_formula":"multiply(add(divide(20, 90), 100), 90)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|","chain":"20 \/ 90<\/gadget>\n2\/9 = around 0.222222<\/output>\n(2\/9) + 100<\/gadget>\n902\/9 = around 100.222222<\/output>\n(902\/9) * 90<\/gadget>\n9_020<\/output>\n9_020<\/result>","index":1885} +{"problem":"a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 112 . what was the number he chose ?","rationale":"\"let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 112 x = 125 answer : c\"","correct":"c","options":{"a":"123 ","b":"267 ","c":"125 ","d":"267","e":"120"},"options_float":{"a":123.0,"b":267.0,"c":125.0,"d":267.0,"e":120.0},"annotated_formula":"divide(add(112, 138), 2)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"112 + 138<\/gadget>\n250<\/output>\n250 \/ 2<\/gadget>\n125<\/output>\n125<\/result>","index":1886} +{"problem":"how many multiples of 3 are there between 15 and 246 ?","rationale":"\"it should be mentioned whether 15 and 246 are inclusive . if 15 and 246 are inclusive , then the answer is ( 246 - 15 ) \/ 3 + 1 = 78 . if 15 and 246 are not inclusive , then the answer is ( 243 - 18 ) \/ 3 + 1 = 76 . since oa is d , then we have not inclusive case .\"","correct":"d","options":{"a":"128 ","b":"94 ","c":"89 ","d":"76","e":"60"},"options_float":{"a":128.0,"b":94.0,"c":89.0,"d":76.0,"e":60.0},"annotated_formula":"add(divide(subtract(246, 15), 3), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"246 - 15<\/gadget>\n231<\/output>\n231 \/ 3<\/gadget>\n77<\/output>\n77 + 1<\/gadget>\n78<\/output>\n78<\/result>","index":1888} +{"problem":"the area of a square land is a square feet and the perimeter is p feet . if 5 a = 10 p + 45 , what is the perimeter of the land , in feet ?","rationale":"you can also solve this via using the given numbers in the answer choices ! of course you need to be aware of the basic properties as outlined by the other posts above ( a = x ^ 2 and p = 4 x ) starting with b you will notice that x = 14 is way too big for your area ( 14 ^ 2 ) and will not satisfy : 5 a = 10 p + 45 or a = 2 p + 9 - - > eliminate b and e now pick d ( its either too big , then its a , or too small then you know its c or it is b itsself ) and picking d indeed solves the problem ! ( 36 \/ 4 - - > 9 ; a = 9 ^ 2 = 81 and 81 = 2 x 36 + 9 )","correct":"d","options":{"a":"18 ","b":"56 ","c":"40 ","d":"36","e":"64"},"options_float":{"a":18.0,"b":56.0,"c":40.0,"d":36.0,"e":64.0},"annotated_formula":"multiply(divide(add(divide(multiply(10, const_4), 5), sqrt(add(power(divide(multiply(10, const_4), 5), const_2), multiply(divide(45, 5), const_4)))), const_2), const_4)","linear_formula":"divide(n2,n0)|multiply(n1,const_4)|divide(#1,n0)|multiply(#0,const_4)|power(#2,const_2)|add(#3,#4)|sqrt(#5)|add(#2,#6)|divide(#7,const_2)|multiply(#8,const_4)","chain":"10 * 4<\/gadget>\n40<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n45 \/ 5<\/gadget>\n9<\/output>\n9 * 4<\/gadget>\n36<\/output>\n64 + 36<\/gadget>\n100<\/output>\n100 ** (1\/2)<\/gadget>\n10<\/output>\n8 + 10<\/gadget>\n18<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n36<\/result>","index":1889} +{"problem":"a cube of edge 16 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ?","rationale":"\"increase in volume = volume of the cube = 16 * 16 * 16 cm ^ 3 rise in water level = volume \/ area = 16 * 16 * 16 \/ 20 * 15 = 13.65 cm answer is c\"","correct":"c","options":{"a":"6 cm ","b":"8.25 cm ","c":"13.65 cm ","d":"15.12 cm","e":"20.62 cm"},"options_float":{"a":6.0,"b":8.25,"c":13.65,"d":15.12,"e":20.62},"annotated_formula":"divide(volume_cube(16), multiply(20, 15))","linear_formula":"multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)|","chain":"16 ** 3<\/gadget>\n4_096<\/output>\n20 * 15<\/gadget>\n300<\/output>\n4_096 \/ 300<\/gadget>\n1_024\/75 = around 13.653333<\/output>\n1_024\/75 = around 13.653333<\/result>","index":1890} +{"problem":"find the value of ( 45 + 23 \/ 89 ) × 89","rationale":"\"= ( 45 + 23 \/ 89 ) × 89 = ( 4005 + 23 ) \/ 89 × 89 = 4028 \/ 89 × 89 = 4028 answer is a .\"","correct":"a","options":{"a":"4028 ","b":"4026 ","c":"4248 ","d":"4538","e":"9028"},"options_float":{"a":4028.0,"b":4026.0,"c":4248.0,"d":4538.0,"e":9028.0},"annotated_formula":"multiply(add(divide(23, 89), 45), 89)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|","chain":"23 \/ 89<\/gadget>\n23\/89 = around 0.258427<\/output>\n(23\/89) + 45<\/gadget>\n4_028\/89 = around 45.258427<\/output>\n(4_028\/89) * 89<\/gadget>\n4_028<\/output>\n4_028<\/result>","index":1891} +{"problem":"the average age of father and his two sons is 24 years . 5 years ago , the average age of the two sons was 15 years . if the difference between the ages of the two sons is 4 years , what is the present age of the father ?","rationale":"the total present age of father and two sons is 3 x 24 = 72 yrs the total present age of sons is ( 15 + 5 ) x 2 = 40 years so , present age of father is 72 â € “ 40 = 32 yrs answer b","correct":"b","options":{"a":"45 ","b":"32 ","c":"47 ","d":"48","e":"49"},"options_float":{"a":45.0,"b":32.0,"c":47.0,"d":48.0,"e":49.0},"annotated_formula":"subtract(multiply(24, const_3), add(multiply(15, const_2), multiply(5, const_2)))","linear_formula":"multiply(n0,const_3)|multiply(n2,const_2)|multiply(n1,const_2)|add(#1,#2)|subtract(#0,#3)","chain":"24 * 3<\/gadget>\n72<\/output>\n15 * 2<\/gadget>\n30<\/output>\n5 * 2<\/gadget>\n10<\/output>\n30 + 10<\/gadget>\n40<\/output>\n72 - 40<\/gadget>\n32<\/output>\n32<\/result>","index":1892} +{"problem":"a caterer ordered 225 ice - cream bars and 125 sundaes . if the total price was $ 200.00 and the price of each ice - cream bar was $ 0.60 , what was the price of each sundae ?","rationale":"let price of a sundae = s price of ice cream bar = . 6 $ 225 * . 6 + 125 * s = 200 = > 125 * s = 65 = > s = 0.52 answer a","correct":"a","options":{"a":"$ 0.52 ","b":"$ 0.80 ","c":"$ 1.00 ","d":"$ 1.20","e":"$ 1.60"},"options_float":{"a":0.52,"b":0.8,"c":1.0,"d":1.2,"e":1.6},"annotated_formula":"divide(subtract(200, multiply(225, 0.6)), 125)","linear_formula":"multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)","chain":"225 * 0.6<\/gadget>\n135<\/output>\n200 - 135<\/gadget>\n65<\/output>\n65 \/ 125<\/gadget>\n13\/25 = around 0.52<\/output>\n13\/25 = around 0.52<\/result>","index":1893} +{"problem":"a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 50 as remainder . find the no . is ?","rationale":"\"( 555 + 445 ) * 2 * 110 + 50 = 220000 + 50 = 220050 e\"","correct":"e","options":{"a":"145646 ","b":"236578 ","c":"645353 ","d":"456546","e":"220050"},"options_float":{"a":145646.0,"b":236578.0,"c":645353.0,"d":456546.0,"e":220050.0},"annotated_formula":"add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 50)","linear_formula":"add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)|","chain":"555 + 445<\/gadget>\n1_000<\/output>\n1_000 * 2<\/gadget>\n2_000<\/output>\n555 - 445<\/gadget>\n110<\/output>\n2_000 * 110<\/gadget>\n220_000<\/output>\n220_000 + 50<\/gadget>\n220_050<\/output>\n220_050<\/result>","index":1896} +{"problem":"if p and q are prime numbers , how many divisors does the product p ^ 4 * q ^ 5 have ?","rationale":"when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is a . 5 * 6 = 30","correct":"a","options":{"a":"30 ","b":"40 ","c":"45 ","d":"50","e":"56"},"options_float":{"a":30.0,"b":40.0,"c":45.0,"d":50.0,"e":56.0},"annotated_formula":"multiply(add(4, const_1), add(5, const_1))","linear_formula":"add(n0,const_1)|add(n1,const_1)|multiply(#0,#1)","chain":"4 + 1<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n5 * 6<\/gadget>\n30<\/output>\n30<\/result>","index":1897} +{"problem":"if 20 % of a number = 300 , then 120 % of that number will be ?","rationale":"\"let the number x . then , 20 % of x = 300 x = ( 300 * 100 ) \/ 20 = 1500 120 % of x = ( 120 \/ 100 * 1500 ) = 1800 . answer : d\"","correct":"d","options":{"a":"20 ","b":"120 ","c":"360 ","d":"1800","e":"1820"},"options_float":{"a":20.0,"b":120.0,"c":360.0,"d":1800.0,"e":1820.0},"annotated_formula":"divide(multiply(120, 300), 20)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"120 * 300<\/gadget>\n36_000<\/output>\n36_000 \/ 20<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":1898} +{"problem":"a contractor undertook to do a certain piece of work in 6 days . he employed certain number of men , but 5 of them being absent from the very first day , the rest could finish the work in 10 days . the number of men originally employed were :","rationale":"explanation : let there be x men at the beginning . less men , more days ( indirect proportion ) 12 : 6 : : x : ( x - 5 ) 12 ( x - 5 ) = 6 x < = > 6 x = 60 x = 10 answer : b","correct":"b","options":{"a":"12 ","b":"10 ","c":"13 ","d":"11","e":"none of these"},"options_float":{"a":12.0,"b":10.0,"c":13.0,"d":11.0,"e":null},"annotated_formula":"divide(multiply(5, 10), 5)","linear_formula":"multiply(n1,n2)|divide(#0,n1)","chain":"5 * 10<\/gadget>\n50<\/output>\n50 \/ 5<\/gadget>\n10<\/output>\n10<\/result>","index":1900} +{"problem":"one gear of pulley rotates at a speed of 3 rev \/ sec another gear roates at 5 rev \/ sec . if both start together after how many seconds will they be together again","rationale":"3 rotates 3 , 6,9 , 12,15 5 rotates 5 , 10,15 ans is 15 answer : c","correct":"c","options":{"a":"3 sec ","b":"5 sec ","c":"15 sec ","d":"20 sec","e":"25 sec"},"options_float":{"a":3.0,"b":5.0,"c":15.0,"d":20.0,"e":25.0},"annotated_formula":"multiply(3, 5)","linear_formula":"multiply(n0,n1)","chain":"3 * 5<\/gadget>\n15<\/output>\n15<\/result>","index":1902} +{"problem":"one type of liquid contains 10 % of water and the second type of liquid contains 15 % of water . a glass is filled with 5 parts of 1 st liquid and 2 parts of 2 nd liquid . the percentage of water in the new mixture in the glass is ?","rationale":"\"required percentage = ( 10 % of 5 + 15 % of 2 ) \/ 5 + 2 * 100 = 0.8 \/ 7 * 100 = 11.4 % answer is b\"","correct":"b","options":{"a":"5.6 % ","b":"11.4 % ","c":"15.6 % ","d":"21.3 %","e":"18.9 %"},"options_float":{"a":5.6,"b":11.4,"c":15.6,"d":21.3,"e":18.9},"annotated_formula":"multiply(divide(const_100, add(5, 2)), add(divide(multiply(10, 5), const_100), divide(multiply(15, 2), const_100)))","linear_formula":"add(n2,n4)|multiply(n0,n2)|multiply(n1,n4)|divide(#1,const_100)|divide(#2,const_100)|divide(const_100,#0)|add(#3,#4)|multiply(#6,#5)|","chain":"5 + 2<\/gadget>\n7<\/output>\n100 \/ 7<\/gadget>\n100\/7 = around 14.285714<\/output>\n10 * 5<\/gadget>\n50<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n15 * 2<\/gadget>\n30<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/2) + (3\/10)<\/gadget>\n4\/5 = around 0.8<\/output>\n(100\/7) * (4\/5)<\/gadget>\n80\/7 = around 11.428571<\/output>\n80\/7 = around 11.428571<\/result>","index":1904} +{"problem":"mala and usha shared their water from a bottle . if mala and usha both drank for the same amount of time , but usha only drank 2 \/ 10 of the total bottle , what was the ratio of mala and usha speed ?","rationale":"the answer is suppose to be c . 4 : 1 . it ' s from the gmatprep answer : option c","correct":"c","options":{"a":"1 : 4 ","b":"2 : 5 ","c":"4 : 1 ","d":"3 : 5","e":"2 : 3"},"options_float":{"a":0.25,"b":0.4,"c":4.0,"d":0.6,"e":0.6666666667},"annotated_formula":"divide(subtract(10, 2), 2)","linear_formula":"subtract(n1,n0)|divide(#0,n0)","chain":"10 - 2<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":1906} +{"problem":"the floor of a rectangular room is 18 m long and 12 m wide . the room is surrounded by a veranda of width 2 m on all its sides . the area of the veranda is :","rationale":"\"area of the outer rectangle = 22 ã — 16 = 352 m 2 area of the inner rectangle = 18 ã — 12 = 216 m 2 required area = ( 352 â € “ 216 ) = 136 m 2 answer c\"","correct":"c","options":{"a":"124 m 2 ","b":"126 m 2 ","c":"136 m 2 ","d":"158 m 2","e":"none of these"},"options_float":{"a":124.0,"b":126.0,"c":136.0,"d":158.0,"e":null},"annotated_formula":"subtract(rectangle_area(add(multiply(2, 2), 18), add(12, multiply(2, 2))), rectangle_area(18, 12))","linear_formula":"multiply(n2,n2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4 + 18<\/gadget>\n22<\/output>\n12 + 4<\/gadget>\n16<\/output>\n22 * 16<\/gadget>\n352<\/output>\n18 * 12<\/gadget>\n216<\/output>\n352 - 216<\/gadget>\n136<\/output>\n136<\/result>","index":1907} +{"problem":"the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 84 , and the average of the female students ’ grades was 92 , how many female students took the test ?","rationale":"\"total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) \/ ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m \/ 8 = 84 thus m = 672 - - - - - - - - - 2 also , f \/ f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 672 + 92 f ) \/ ( 8 + f ) = 90 solving this we get f = 24 ans : e\"","correct":"e","options":{"a":"8 ","b":"9 ","c":"10 ","d":"21","e":"24"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":21.0,"e":24.0},"annotated_formula":"divide(subtract(multiply(90, 8), multiply(84, 8)), subtract(92, 90))","linear_formula":"multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|","chain":"90 * 8<\/gadget>\n720<\/output>\n84 * 8<\/gadget>\n672<\/output>\n720 - 672<\/gadget>\n48<\/output>\n92 - 90<\/gadget>\n2<\/output>\n48 \/ 2<\/gadget>\n24<\/output>\n24<\/result>","index":1908} +{"problem":"a box contains 21 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is","rationale":"\"solution p ( none is defective ) = 17 c 2 \/ 21 c 2 = 57 \/ 88 . p ( at least one is defective ) = ( 1 - 57 \/ 88 ) = 31 \/ 88 . answer e\"","correct":"e","options":{"a":"4 \/ 19 ","b":"7 \/ 19 ","c":"12 \/ 19 ","d":"21 \/ 95","e":"31 \/ 88"},"options_float":{"a":0.2105263158,"b":0.3684210526,"c":0.6315789474,"d":0.2210526316,"e":0.3522727273},"annotated_formula":"subtract(const_1, divide(choose(subtract(21, 4), const_2), choose(21, const_2)))","linear_formula":"choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)|","chain":"21 - 4<\/gadget>\n17<\/output>\nbinomial(17, 2)<\/gadget>\n136<\/output>\nbinomial(21, 2)<\/gadget>\n210<\/output>\n136 \/ 210<\/gadget>\n68\/105 = around 0.647619<\/output>\n1 - (68\/105)<\/gadget>\n37\/105 = around 0.352381<\/output>\n37\/105 = around 0.352381<\/result>","index":1909} +{"problem":"p , q and r together earn rs . 1710 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ?","rationale":"\"explanation : amount earned by p , q and r in 1 day = 1710 \/ 9 = 190 - - - ( 1 ) amount earned by p and r in 1 day = 600 \/ 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 \/ 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 130 - 190 = 60 = > amount earned by r in 1 day = 60 answer : option b\"","correct":"b","options":{"a":"s . 40 ","b":"s . 60 ","c":"s . 90 ","d":"s . 100","e":"s . 120"},"options_float":{"a":40.0,"b":60.0,"c":90.0,"d":100.0,"e":120.0},"annotated_formula":"subtract(add(divide(600, 5), divide(910, 7)), divide(1710, 9))","linear_formula":"divide(n2,n3)|divide(n5,n4)|divide(n0,n1)|add(#0,#1)|subtract(#3,#2)|","chain":"600 \/ 5<\/gadget>\n120<\/output>\n910 \/ 7<\/gadget>\n130<\/output>\n120 + 130<\/gadget>\n250<\/output>\n1_710 \/ 9<\/gadget>\n190<\/output>\n250 - 190<\/gadget>\n60<\/output>\n60<\/result>","index":1910} +{"problem":"a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 32 . find the profit percentage ?","rationale":"\"l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 160 ( 32 * 5 ) profit percentage = ( 160 - 100 ) \/ 100 * 100 = 60 % answer : e\"","correct":"e","options":{"a":"80 % ","b":"50 % ","c":"59 % ","d":"40 %","e":"60 %"},"options_float":{"a":80.0,"b":50.0,"c":59.0,"d":40.0,"e":60.0},"annotated_formula":"subtract(multiply(32, add(const_4, const_1)), multiply(25, const_4))","linear_formula":"add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)|","chain":"4 + 1<\/gadget>\n5<\/output>\n32 * 5<\/gadget>\n160<\/output>\n25 * 4<\/gadget>\n100<\/output>\n160 - 100<\/gadget>\n60<\/output>\n60<\/result>","index":1912} +{"problem":"a fraudulent dairy guy declares to market his dairy at price range but he combined it with water and therefore benefits 25 % . the percentage of water in the combination is :","rationale":"explanation : - > if capability price of 1 l of milk = rs . 1 and s . price of 1 l of combination = rs . 1 benefits = 20 % capability price of 1 l of combination = [ ( 100 \/ 100 + 25 ) * 1 ] = rs . 100 \/ 125 = rs . 4 \/ 5 - > from the rule of allegation i . capability price of 1 l of water = 0 ii . capability price of 1 l of milk = 1 iii . ( p ) = 4 \/ 5 iv . d – m = 1 – 4 \/ 5 = 1 \/ 5 v . m – c = 4 \/ 5 – 0 = 4 \/ 5 ratio of milk to water = 4 \/ 5 : 1 \/ 5 = 4 : 1 % of water in the combination = ( 1 \/ 5 x 100 ) % = 20 % answer : a","correct":"a","options":{"a":"20 % ","b":"30 % ","c":"50 % ","d":"35 %","e":"60 %"},"options_float":{"a":20.0,"b":30.0,"c":50.0,"d":35.0,"e":60.0},"annotated_formula":"subtract(const_100, multiply(divide(const_100, add(const_100, 25)), const_100))","linear_formula":"add(n0,const_100)|divide(const_100,#0)|multiply(#1,const_100)|subtract(const_100,#2)","chain":"100 + 25<\/gadget>\n125<\/output>\n100 \/ 125<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n20<\/result>","index":1913} +{"problem":"49 ã — 49 = 7 ^ ?","rationale":"49 ã — 49 = 7 ? or , 7 ( 2 ) ã — 7 ( 2 ) = 7 ? or 7 ( 4 ) = 7 ? or , ? = 4 answer a","correct":"a","options":{"a":"4 ","b":"7 ","c":"8 ","d":"16","e":"none of these"},"options_float":{"a":4.0,"b":7.0,"c":8.0,"d":16.0,"e":null},"annotated_formula":"divide(add(log(49), log(49)), log(7))","linear_formula":"log(n0)|log(n2)|add(#0,#0)|divide(#2,#1)","chain":"log(49)<\/gadget>\nlog(49) = around 3.89182<\/output>\nlog(49) + log(49)<\/gadget>\n2*log(49) = around 7.783641<\/output>\nlog(7)<\/gadget>\nlog(7) = around 1.94591<\/output>\n(2*log(49)) \/ log(7)<\/gadget>\n2*log(49)\/log(7) = around 4<\/output>\n2*log(49)\/log(7) = around 4<\/result>","index":1914} +{"problem":"a batsman in his 12 th inning makes a score of 60 and their by increasing his average by 4 . what is his average after the 12 th inning ?","rationale":"\"11 x + 60 = 12 ( x + 4 ) x = 12 + 4 = 16 answer : b\"","correct":"b","options":{"a":"12 ","b":"16 ","c":"20 ","d":"24","e":"28"},"options_float":{"a":12.0,"b":16.0,"c":20.0,"d":24.0,"e":28.0},"annotated_formula":"add(subtract(60, multiply(4, 12)), 4)","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|","chain":"4 * 12<\/gadget>\n48<\/output>\n60 - 48<\/gadget>\n12<\/output>\n12 + 4<\/gadget>\n16<\/output>\n16<\/result>","index":1915} +{"problem":"find the ones digit of 73 ^ 351","rationale":"\"the units digit of 73 ^ 351 will be the same as the units digit of 3 ^ 351 3 ^ 1 = 3 - - > the units digit is 3 ; 3 ^ 2 = 9 - - > the units digit is 9 ; 3 ^ 3 = 27 - - > the units digit is 7 ; 3 ^ 4 = 81 - - > the units digit is 1 ; 3 ^ 5 = 243 - - > the units digit is 3 again ; . . . so , as you can see the units digit repeats in blocks of 4 : { 3 , 9 , 7 , 1 } , { 3 , 9 , 7 , 1 } , . . . now , since 351 = 348 + 3 = ( multiple of 4 ) + 3 , then the units digit of 3 ^ 351 will be the third number in the pattern thus 7 . answer : d .\"","correct":"d","options":{"a":"3 ","b":"5 ","c":"6 ","d":"7","e":"9"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":7.0,"e":9.0},"annotated_formula":"subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(351, multiply(floor(divide(351, const_4)), const_4))), multiply(const_2, const_10))","linear_formula":"add(const_3,const_4)|divide(n1,const_4)|multiply(const_10,const_2)|floor(#1)|multiply(#0,const_10)|multiply(#3,const_4)|subtract(n0,#4)|subtract(n1,#5)|power(#6,#7)|subtract(#8,#2)|","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 10<\/gadget>\n70<\/output>\n73 - 70<\/gadget>\n3<\/output>\n351 \/ 4<\/gadget>\n351\/4 = around 87.75<\/output>\nfloor(351\/4)<\/gadget>\n87<\/output>\n87 * 4<\/gadget>\n348<\/output>\n351 - 348<\/gadget>\n3<\/output>\n3 ** 3<\/gadget>\n27<\/output>\n2 * 10<\/gadget>\n20<\/output>\n27 - 20<\/gadget>\n7<\/output>\n7<\/result>","index":1916} +{"problem":"5 is added to a certain number , the sum is multiplied by 7 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 66 . what is the number ?","rationale":"let number is x . when 5 added to it , = ( x + 5 ) 7 multiplied to sum , = 7 * ( x + 5 ) now , = [ { 7 * ( x + 5 ) } \/ 5 ] and , = [ { 7 * ( x + 5 ) } \/ 5 ] - 5 according to question , [ { 7 * ( x + 10 ) } \/ 5 ] - 5 = half of 66 [ ( 7 x + 35 ) \/ 5 ) = 33 + 5 7 x + 35 = 38 * 5 x = 155 \/ 7 x = 22.14 = 22 so , required number is : 22 . answer : c","correct":"c","options":{"a":"21 ","b":"20 ","c":"22 ","d":"30","e":"45"},"options_float":{"a":21.0,"b":20.0,"c":22.0,"d":30.0,"e":45.0},"annotated_formula":"subtract(divide(multiply(add(divide(66, const_2), 5), 5), 7), 5)","linear_formula":"divide(n4,const_2)|add(n0,#0)|multiply(n0,#1)|divide(#2,n1)|subtract(#3,n0)","chain":"66 \/ 2<\/gadget>\n33<\/output>\n33 + 5<\/gadget>\n38<\/output>\n38 * 5<\/gadget>\n190<\/output>\n190 \/ 7<\/gadget>\n190\/7 = around 27.142857<\/output>\n(190\/7) - 5<\/gadget>\n155\/7 = around 22.142857<\/output>\n155\/7 = around 22.142857<\/result>","index":1917} +{"problem":"john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 17 , what was the price before the first discount ?","rationale":"\"let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 17 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 17 solve for x x = $ 30.22 correct answer b\"","correct":"b","options":{"a":"$ 45.10 ","b":"$ 30.22 ","c":"$ 28.44 ","d":"$ 67.54","e":"$ 65.23"},"options_float":{"a":45.1,"b":30.22,"c":28.44,"d":67.54,"e":65.23},"annotated_formula":"divide(multiply(multiply(const_100, const_100), 17), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25)))","linear_formula":"multiply(const_100,const_100)|subtract(const_100,n0)|multiply(n2,#0)|multiply(#1,const_100)|multiply(n0,#1)|subtract(#3,#4)|divide(#2,#5)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n10_000 * 17<\/gadget>\n170_000<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75 * 100<\/gadget>\n7_500<\/output>\n75 * 25<\/gadget>\n1_875<\/output>\n7_500 - 1_875<\/gadget>\n5_625<\/output>\n170_000 \/ 5_625<\/gadget>\n272\/9 = around 30.222222<\/output>\n272\/9 = around 30.222222<\/result>","index":1918} +{"problem":"if a trader sold two cars each at rs . 325475 and gains 15 % on the first and loses 15 % on the second , then his profit or loss percent on the whole is ?","rationale":"\"sp of each car is rs . 325475 , he gains 15 % on first car and losses 15 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] \/ 100 = ( 15 ) ( 15 ) \/ 100 % = 1.44 % answer : e\"","correct":"e","options":{"a":"1.44 % ","b":"1.74 % ","c":"1.84 % ","d":"1.97 %","e":"2.25 %"},"options_float":{"a":1.44,"b":1.74,"c":1.84,"d":1.97,"e":2.25},"annotated_formula":"divide(multiply(15, 15), const_100)","linear_formula":"multiply(n1,n1)|divide(#0,const_100)|","chain":"15 * 15<\/gadget>\n225<\/output>\n225 \/ 100<\/gadget>\n9\/4 = around 2.25<\/output>\n9\/4 = around 2.25<\/result>","index":1919} +{"problem":"in township k , 1 \/ 5 of the housing units are equipped with cable television . if 1 \/ 10 of the housing units , including 1 \/ 3 of those that are equipped with cable television , are equipped with videocassette recorders , what fraction of the housing units have neither cable television nor videocassette recorders ?","rationale":"let total housing units = 5 x . as per question , equipped with cable tv = x equipped with vcr = x \/ 2 , which includes x \/ 3 ctv . so , only vcr = x \/ 2 - x \/ 3 = x \/ 6 . so , cable tv + vcr = x + x \/ 6 = 7 x \/ 6 no cable tv and vcr = 5 x - 7 x \/ 6 = 23 x \/ 6 . fraction = 23 x \/ ( 6 * 5 x ) = 23 \/ 30 . hence , a","correct":"a","options":{"a":"23 \/ 30 ","b":"11 \/ 15 ","c":"7 \/ 10 ","d":"1 \/ 6","e":"2 \/ 15"},"options_float":{"a":0.7666666667,"b":0.7333333333,"c":0.7,"d":0.1666666667,"e":0.1333333333},"annotated_formula":"add(subtract(subtract(const_1, divide(1, 5)), divide(1, 10)), multiply(divide(1, 5), divide(1, 3)))","linear_formula":"divide(n0,n1)|divide(n0,n5)|divide(n0,n3)|multiply(#0,#1)|subtract(const_1,#0)|subtract(#4,#2)|add(#3,#5)","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(4\/5) - (1\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/5) * (1\/3)<\/gadget>\n1\/15 = around 0.066667<\/output>\n(7\/10) + (1\/15)<\/gadget>\n23\/30 = around 0.766667<\/output>\n23\/30 = around 0.766667<\/result>","index":1922} +{"problem":"it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 16 minutes , how much time will she spend doing yoga ? ( rounded to minutes )","rationale":"\"the ratio is 2 : 3 = gym : ride , so ( 16 ) ( 3 \/ 2 ) = 24 minutes at the gym , and 24 + 16 = 40 minutes exercise , so ( 2 \/ 3 ) ( 40 ) = 27 minutes yoga . answer : c\"","correct":"c","options":{"a":"10 min . ","b":"41 min . ","c":"27 min . ","d":"23 min .","e":"25 min ."},"options_float":{"a":10.0,"b":41.0,"c":27.0,"d":23.0,"e":25.0},"annotated_formula":"divide(multiply(16, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3))))","linear_formula":"add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)|","chain":"2 + 3<\/gadget>\n5<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n16 * (3\/5)<\/gadget>\n48\/5 = around 9.6<\/output>\n(3\/5) * (3\/5)<\/gadget>\n9\/25 = around 0.36<\/output>\n(48\/5) \/ (9\/25)<\/gadget>\n80\/3 = around 26.666667<\/output>\n80\/3 = around 26.666667<\/result>","index":1924} +{"problem":"a cistern which could be filled in 8 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ?","rationale":"\"1 \/ 8 - 1 \/ x = 1 \/ 10 = > 40 hrs answer : e\"","correct":"e","options":{"a":"76 hrs ","b":"99 hrs ","c":"55 hrs ","d":"90 hrs","e":"40 hrs"},"options_float":{"a":76.0,"b":99.0,"c":55.0,"d":90.0,"e":40.0},"annotated_formula":"inverse(subtract(divide(const_1, 8), divide(const_1, const_10)))","linear_formula":"divide(const_1,n0)|divide(const_1,const_10)|subtract(#0,#1)|inverse(#2)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/8) - (1\/10)<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":1925} +{"problem":"a sum fetched a total simple interest of rs . 4016.25 at the rate of 1 % p . a . in 9 years . what is the sum ?","rationale":"\"principal = ( 100 * 4016.25 ) \/ ( 1 * 9 ) = rs . 44625 . answer : c\"","correct":"c","options":{"a":"42315 ","b":"42877 ","c":"44625 ","d":"38925","e":"22887"},"options_float":{"a":42315.0,"b":42877.0,"c":44625.0,"d":38925.0,"e":22887.0},"annotated_formula":"divide(divide(multiply(4016.25, const_100), 1), 9)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"4_016.25 * 100<\/gadget>\n401_625<\/output>\n401_625 \/ 1<\/gadget>\n401_625<\/output>\n401_625 \/ 9<\/gadget>\n44_625<\/output>\n44_625<\/result>","index":1928} +{"problem":"average expenditure of a person for the first 3 days of a week is rs . 320 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is :","rationale":"explanation : assumed mean = rs . 320 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 \/ 7 = rs . 40 therefore , average expenditure for 7 days = rs . 320 + rs . 40 = rs . 360 correct option : d","correct":"d","options":{"a":"350 ","b":"370 ","c":"390 ","d":"360","e":"none"},"options_float":{"a":350.0,"b":370.0,"c":390.0,"d":360.0,"e":null},"annotated_formula":"add(320, divide(multiply(4, subtract(420, 320)), add(3, 4)))","linear_formula":"add(n0,n2)|subtract(n3,n1)|multiply(n2,#1)|divide(#2,#0)|add(n1,#3)","chain":"420 - 320<\/gadget>\n100<\/output>\n4 * 100<\/gadget>\n400<\/output>\n3 + 4<\/gadget>\n7<\/output>\n400 \/ 7<\/gadget>\n400\/7 = around 57.142857<\/output>\n320 + (400\/7)<\/gadget>\n2_640\/7 = around 377.142857<\/output>\n2_640\/7 = around 377.142857<\/result>","index":1929} +{"problem":"in a certain school , the ratio of boys to girls is 5 to 13 . if there are 64 more girls than boys , how many boys are there ?","rationale":"the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 64 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 64 . so 1 part corresponds to 64 \/ 8 = 8 and so 5 parts correspond to 5 * 8 = 40 . b","correct":"b","options":{"a":"27 ","b":"40 ","c":"45 ","d":"72","e":"117"},"options_float":{"a":27.0,"b":40.0,"c":45.0,"d":72.0,"e":117.0},"annotated_formula":"subtract(divide(64, subtract(const_1, divide(5, 13))), 64)","linear_formula":"divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|subtract(#2,n2)","chain":"5 \/ 13<\/gadget>\n5\/13 = around 0.384615<\/output>\n1 - (5\/13)<\/gadget>\n8\/13 = around 0.615385<\/output>\n64 \/ (8\/13)<\/gadget>\n104<\/output>\n104 - 64<\/gadget>\n40<\/output>\n40<\/result>","index":1932} +{"problem":"a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 3 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ?","rationale":"\"by the question , the equation would be 150 p - 83 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 193 . answer e for me .\"","correct":"e","options":{"a":"$ 28 ","b":"$ 82 ","c":"$ 110 ","d":"$ 138","e":"$ 193"},"options_float":{"a":28.0,"b":82.0,"c":110.0,"d":138.0,"e":193.0},"annotated_formula":"divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(3, const_100)), multiply(add(80, 3), 150)), 150)","linear_formula":"add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16 * 1_000<\/gadget>\n16_000<\/output>\n3 * 100<\/gadget>\n300<\/output>\n16_000 + 300<\/gadget>\n16_300<\/output>\n80 + 3<\/gadget>\n83<\/output>\n83 * 150<\/gadget>\n12_450<\/output>\n16_300 + 12_450<\/gadget>\n28_750<\/output>\n28_750 \/ 150<\/gadget>\n575\/3 = around 191.666667<\/output>\n575\/3 = around 191.666667<\/result>","index":1933} +{"problem":"two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m \/ s and 8 m \/ s . if the circumference of the circle is 630 meters , after what time will they meet at the starting point ?","rationale":"\"they meet every 630 \/ 7 + 8 = 42 sec answer is c\"","correct":"c","options":{"a":"20 sec ","b":"15 sec ","c":"42 sec ","d":"50 sec","e":"1 min"},"options_float":{"a":20.0,"b":15.0,"c":42.0,"d":50.0,"e":1.0},"annotated_formula":"divide(630, add(8, 7))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"8 + 7<\/gadget>\n15<\/output>\n630 \/ 15<\/gadget>\n42<\/output>\n42<\/result>","index":1934} +{"problem":"calculate the sum of first 60 natural numbers .","rationale":"\"solution we know that ( 1 + 2 + 3 + . . . . . + 60 ) = n ( n + 1 ) \/ 2 therefore ( 1 + 2 + 3 + . . . . + 60 ) = ( 60 × 61 \/ 2 ) = 1830 . answer c\"","correct":"c","options":{"a":"1839 ","b":"2830 ","c":"1830 ","d":"1831","e":"1835"},"options_float":{"a":1839.0,"b":2830.0,"c":1830.0,"d":1831.0,"e":1835.0},"annotated_formula":"multiply(add(60, const_1), divide(60, const_2))","linear_formula":"add(n0,const_1)|divide(n0,const_2)|multiply(#0,#1)|","chain":"60 + 1<\/gadget>\n61<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n61 * 30<\/gadget>\n1_830<\/output>\n1_830<\/result>","index":1935} +{"problem":"buffalo gives 4 liter milk , cow gives ( 1 \/ 2 ) liter milk and goat gives 1 \/ 4 liter milk . you have to get 20 liter milk by 20 animals . what is the number of goats ?","rationale":"assume number of respective animals are x , y , z . x + y + z = 20 - - - ( 1 ) as the total number of animal has to be 20 amt of milk will be 4 x + ( 1 \/ 2 ) y + ( 1 \/ 4 ) z = 20 - - - ( 2 ) solving equation 1 and 2 we get 15 x + y = 60 - - - - ( 3 ) since buffalo gives 4 litre and total milk is 20 , x < 5 but from eq 3 , x can not be more than 4 ; further if x = 1 or 2 ; y > 20 . . . not possible , since total animal is 20 thus , x = 3 , y = 15 , z = 2 2 goats answer : b","correct":"b","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(4, 2)","linear_formula":"subtract(n0,n2)","chain":"4 - 2<\/gadget>\n2<\/output>\n2<\/result>","index":1936} +{"problem":"a batsman makes a score of 85 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning .","rationale":"\"let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 85 = 11 x x = 85 - 50 = 35 answer is e\"","correct":"e","options":{"a":"40 ","b":"50 ","c":"62 ","d":"45","e":"35"},"options_float":{"a":40.0,"b":50.0,"c":62.0,"d":45.0,"e":35.0},"annotated_formula":"add(subtract(85, multiply(11, 5)), 5)","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|","chain":"11 * 5<\/gadget>\n55<\/output>\n85 - 55<\/gadget>\n30<\/output>\n30 + 5<\/gadget>\n35<\/output>\n35<\/result>","index":1937} +{"problem":"each of the three people individually can complete a certain job in 3 , 4 , and 6 hours , respectively . what is the lowest fraction of the job that can be done in 1 hour by 2 of the people working together at their respective rates ?","rationale":"the two slowest people work at rates of 1 \/ 4 and 1 \/ 6 of the job per hour . the sum of these rates is 1 \/ 4 + 1 \/ 6 = 5 \/ 12 of the job per hour . the answer is d .","correct":"d","options":{"a":"1 \/ 3 ","b":"9 \/ 20 ","c":"8 \/ 15 ","d":"5 \/ 12","e":"2 \/ 3"},"options_float":{"a":0.3333333333,"b":0.45,"c":0.5333333333,"d":0.4166666667,"e":0.6666666667},"annotated_formula":"add(divide(const_1, 4), divide(const_1, 6))","linear_formula":"divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/4) + (1\/6)<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":1938} +{"problem":"in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 80 per kg ?","rationale":"( 96 - 80 ) \/ ( 80 - 68 ) = 16 \/ 12 = 4 \/ 3 answer : c","correct":"c","options":{"a":"5 : 8 ","b":"4 : 7 ","c":"4 : 3 ","d":"9 : 5","e":"9 : 8"},"options_float":{"a":0.625,"b":0.5714285714,"c":1.3333333333,"d":1.8,"e":1.125},"annotated_formula":"divide(divide(subtract(96, 80), subtract(96, 68)), subtract(const_1, divide(subtract(96, 80), subtract(96, 68))))","linear_formula":"subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)","chain":"96 - 80<\/gadget>\n16<\/output>\n96 - 68<\/gadget>\n28<\/output>\n16 \/ 28<\/gadget>\n4\/7 = around 0.571429<\/output>\n1 - (4\/7)<\/gadget>\n3\/7 = around 0.428571<\/output>\n(4\/7) \/ (3\/7)<\/gadget>\n4\/3 = around 1.333333<\/output>\n4\/3 = around 1.333333<\/result>","index":1939} +{"problem":"if a man lost 4 % by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44 % ?","rationale":"\"96 % - - - - 12 144 % - - - - ? 96 \/ 144 * 12 = 8 answer a\"","correct":"a","options":{"a":"8 ","b":"7 ","c":"5 ","d":"2","e":"1"},"options_float":{"a":8.0,"b":7.0,"c":5.0,"d":2.0,"e":1.0},"annotated_formula":"divide(multiply(subtract(const_100, 4), 12), add(const_100, 44))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|","chain":"100 - 4<\/gadget>\n96<\/output>\n96 * 12<\/gadget>\n1_152<\/output>\n100 + 44<\/gadget>\n144<\/output>\n1_152 \/ 144<\/gadget>\n8<\/output>\n8<\/result>","index":1940} +{"problem":"the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 6 miles per gallon , will not get him there . by how many miles per gallon must the aircraft ’ s fuel efficiency be improved to make the flight to cleveland possible ?","rationale":"\"actual miles \/ gallon is = 480 \/ 40 = 12 miles \/ gallon . current engine miles \/ gallon is 6 miles \/ gallon . additional 6 miles \/ gallon is required to match the actual mileage . imo option d .\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"12 ","d":"6","e":"160"},"options_float":{"a":2.0,"b":4.0,"c":12.0,"d":6.0,"e":160.0},"annotated_formula":"subtract(divide(480, 40), 6)","linear_formula":"divide(n1,n0)|subtract(#0,n2)|","chain":"480 \/ 40<\/gadget>\n12<\/output>\n12 - 6<\/gadget>\n6<\/output>\n6<\/result>","index":1941} +{"problem":"every year an amount increases by 1 \/ 8 th of itself . how much will it be after two years if its present value is rs . 70400 ?","rationale":"70400 * 9 \/ 8 * 9 \/ 8 = 89100 . answer : a","correct":"a","options":{"a":"89100 ","b":"81028 ","c":"27772 ","d":"29889","e":"22312"},"options_float":{"a":89100.0,"b":81028.0,"c":27772.0,"d":29889.0,"e":22312.0},"annotated_formula":"add(add(70400, multiply(divide(1, 8), 70400)), multiply(divide(1, 8), add(70400, multiply(divide(1, 8), 70400))))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 70_400<\/gadget>\n8_800<\/output>\n70_400 + 8_800<\/gadget>\n79_200<\/output>\n(1\/8) * 79_200<\/gadget>\n9_900<\/output>\n79_200 + 9_900<\/gadget>\n89_100<\/output>\n89_100<\/result>","index":1942} +{"problem":"an empty fuel tank is filled with brand z gasoline . when the tank is 1 \/ 4 empty , it is filled with brand y gasoline . when the tank is half empty again , it is filled with brand z gasoline . when the tank is half empty again , it is filled with brand y gasoline . at this time , what percent of the gasoline in the tank is brand z ?","rationale":"work with fraction of brand z in the tank . 1 st step : brand z is 1 2 nd step : brand z is 1 \/ 4 3 rd step : brand z is ( 1 \/ 2 ) * ( 1 \/ 4 ) + 1 \/ 2 = 5 \/ 8 4 th step : brand z is ( 1 \/ 2 ) * ( 5 \/ 8 ) = 5 \/ 16 answer ( a )","correct":"a","options":{"a":"31.25 % ","b":"30 % ","c":"40 % ","d":"50 %","e":"45 %"},"options_float":{"a":31.25,"b":30.0,"c":40.0,"d":50.0,"e":45.0},"annotated_formula":"divide(subtract(const_100, divide(subtract(const_100, multiply(divide(1, 4), const_100)), const_2)), const_2)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#2,const_2)|subtract(const_100,#3)|divide(#4,const_2)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75 \/ 2<\/gadget>\n75\/2 = around 37.5<\/output>\n100 - (75\/2)<\/gadget>\n125\/2 = around 62.5<\/output>\n(125\/2) \/ 2<\/gadget>\n125\/4 = around 31.25<\/output>\n125\/4 = around 31.25<\/result>","index":1943} +{"problem":"if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 29 , find the value of ab .","rationale":"\"2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 29 - 9 = 20 ab = 10 . answer is d .\"","correct":"d","options":{"a":"5 ","b":"8 ","c":"4 ","d":"10","e":"3"},"options_float":{"a":5.0,"b":8.0,"c":4.0,"d":10.0,"e":3.0},"annotated_formula":"divide(subtract(29, power(3, 2)), 2)","linear_formula":"power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n29 - 9<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":1944} +{"problem":"in an election , candidate a got 55 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate ?","rationale":"\"total number of invalid votes = 15 % of 560000 = 15 \/ 100 × 560000 = 8400000 \/ 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 55 % therefore , the number of valid votes polled in favour of candidate a = 55 % of 476000 = 55 \/ 100 × 476000 = 26180000 \/ 100 = 261800 a )\"","correct":"a","options":{"a":"261800 ","b":"355800 ","c":"356500 ","d":"356800","e":"357000"},"options_float":{"a":261800.0,"b":355800.0,"c":356500.0,"d":356800.0,"e":357000.0},"annotated_formula":"multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(55, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 - (3\/20)<\/gadget>\n17\/20 = around 0.85<\/output>\n560_000 * (17\/20)<\/gadget>\n476_000<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n476_000 * (11\/20)<\/gadget>\n261_800<\/output>\n261_800<\/result>","index":1945} +{"problem":"a certain number when divided by 39 leaves a remainder 20 , what is the remainder when the same number is divided by 13 ?","rationale":"\"39 + 20 = 59 \/ 13 = 7 ( remainder ) answer : a\"","correct":"a","options":{"a":"7 ","b":"8 ","c":"9 ","d":"0","e":"1"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":0.0,"e":1.0},"annotated_formula":"reminder(20, 13)","linear_formula":"reminder(n1,n2)|","chain":"20 % 13<\/gadget>\n7<\/output>\n7<\/result>","index":1947} +{"problem":"a sum fetched a total simple interest of $ 4196.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?","rationale":"\"e 8929 principal = $ 100 x 4196.25 \/ 9 x 5 = $ 419625 \/ 45 = $ 8929 .\"","correct":"e","options":{"a":"$ 8829 ","b":"$ 2840 ","c":"$ 6578 ","d":"$ 7782","e":"$ 8929"},"options_float":{"a":8829.0,"b":2840.0,"c":6578.0,"d":7782.0,"e":8929.0},"annotated_formula":"divide(divide(multiply(4196.25, const_100), 9), 5)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"4_196.25 * 100<\/gadget>\n419_625<\/output>\n419_625 \/ 9<\/gadget>\n46_625<\/output>\n46_625 \/ 5<\/gadget>\n9_325<\/output>\n9_325<\/result>","index":1948} +{"problem":"if the sum of 5 consecutive even positive integers is a , then the sum of the next 5 even consecutive integers in terms of a is :","rationale":"in case of consecutive integers or integers in arithmetic progression mean = median i . e . median = a \/ 5 = mean = third integer first integer = a \/ 5 - 4 second integer = a \/ 5 - 2 third integer = a \/ 5 fourth integer = a \/ 5 + 2 fifth integer = a \/ 5 + 4 i . e . sixth integer = a \/ 5 + 6 seventh integer = a \/ 5 + 8 eighth integer = a \/ 5 + 10 ninth integer = a \/ 5 + 12 tenth integer = a \/ 5 + 14 now mean of next 5 integers = median = a \/ 5 + 5 i . e . sum of next 5 integers = ( a \/ 5 + 10 ) * 5 = a + 50 answer : option d","correct":"d","options":{"a":"a + 5 ","b":"a + 10 ","c":"a + 25 ","d":"a + 50","e":"2 a + 5"},"options_float":{"a":5.0,"b":10.0,"c":25.0,"d":50.0,"e":2.0},"annotated_formula":"multiply(5, const_10)","linear_formula":"multiply(n0,const_10)","chain":"5 * 10<\/gadget>\n50<\/output>\n50<\/result>","index":1949} +{"problem":"an escalator moves towards the top level at the rate of 12 ft . sec and its length is 196 feet . if a person walks on the moving escalator at the rate of 2 feet per second towards the top level , how much time does he take to cover the entire length .","rationale":"\"time taken to cover the entire length = tot . dist \/ resultant speed = 196 \/ ( 12 + 2 ) = 14 sec answer : a\"","correct":"a","options":{"a":"14 sec ","b":"10 sec ","c":"12 sec ","d":"8 sec","e":"9 sec"},"options_float":{"a":14.0,"b":10.0,"c":12.0,"d":8.0,"e":9.0},"annotated_formula":"divide(196, add(12, 2))","linear_formula":"add(n0,n2)|divide(n1,#0)|","chain":"12 + 2<\/gadget>\n14<\/output>\n196 \/ 14<\/gadget>\n14<\/output>\n14<\/result>","index":1950} +{"problem":"pipe a fills a tank in 12 minutes . pipe b can fill the same tank 3 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ?","rationale":"\"a ' s rate is 1 \/ 12 and b ' s rate is 1 \/ 4 . the combined rate is 1 \/ 12 + 1 \/ 4 = 1 \/ 3 the pipes will fill the tank in 3 minutes . the answer is d .\"","correct":"d","options":{"a":"4 mins ","b":"6 mins ","c":"5 mins ","d":"3 mins","e":"7 mins"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":3.0,"e":7.0},"annotated_formula":"inverse(add(divide(const_1, 12), divide(3, 12)))","linear_formula":"divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n3 \/ 12<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/12) + (1\/4)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ (1\/3)<\/gadget>\n3<\/output>\n3<\/result>","index":1951} +{"problem":"a sum of rs . 2730 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ?","rationale":"( x * 8 * 3 ) \/ 100 = ( ( 2730 - x ) * 3 * 5 ) \/ 100 24 x \/ 100 = 40950 \/ 100 - 15 x \/ 100 39 x = 40950 = > x = 1050 second sum = 2730 â € “ 1050 = 1680 answer : b","correct":"b","options":{"a":"1629 ","b":"1680 ","c":"1677 ","d":"1698","e":"1679"},"options_float":{"a":1629.0,"b":1680.0,"c":1677.0,"d":1698.0,"e":1679.0},"annotated_formula":"subtract(2730, divide(multiply(multiply(3, 5), 2730), add(multiply(3, 5), multiply(8, 3))))","linear_formula":"multiply(n2,n4)|multiply(n1,n2)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|subtract(n0,#4)","chain":"3 * 5<\/gadget>\n15<\/output>\n15 * 2_730<\/gadget>\n40_950<\/output>\n8 * 3<\/gadget>\n24<\/output>\n15 + 24<\/gadget>\n39<\/output>\n40_950 \/ 39<\/gadget>\n1_050<\/output>\n2_730 - 1_050<\/gadget>\n1_680<\/output>\n1_680<\/result>","index":1952} +{"problem":"the speed at which a man can row a boat in still water is 22 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 80 metres ?","rationale":"speed of the boat downstream = 22 + 3 = 25 kmph = 25 * 5 \/ 18 = 6.94 m \/ s hence time taken to cover 80 m = 80 \/ 6.94 = 11.53 seconds . answer : b","correct":"b","options":{"a":"12.59 ","b":"11.53 ","c":"15.69 ","d":"14.56","e":"18.98"},"options_float":{"a":12.59,"b":11.53,"c":15.69,"d":14.56,"e":18.98},"annotated_formula":"divide(80, multiply(add(22, 3), const_0_2778))","linear_formula":"add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|","chain":"22 + 3<\/gadget>\n25<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n25 * (5\/18)<\/gadget>\n125\/18 = around 6.944444<\/output>\n80 \/ (125\/18)<\/gadget>\n288\/25 = around 11.52<\/output>\n288\/25 = around 11.52<\/result>","index":1954} +{"problem":"on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.70 per glass on the first day , what was the price per glass on the second day ?","rationale":"\"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.7 = 3 * x - - > x = $ 0.46 . answer : e .\"","correct":"e","options":{"a":"$ 0.15 ","b":"$ 0.20 ","c":"$ 0.30 ","d":"$ 0.40","e":"$ 0.46"},"options_float":{"a":0.15,"b":0.2,"c":0.3,"d":0.4,"e":0.46},"annotated_formula":"divide(multiply(add(const_1, const_1), 0.70), add(const_1, const_2))","linear_formula":"add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 * 0.7<\/gadget>\n1.4<\/output>\n1 + 2<\/gadget>\n3<\/output>\n1.4 \/ 3<\/gadget>\n0.466667<\/output>\n0.466667<\/result>","index":1955} +{"problem":"if one - third of one - fourth of a number is 18 , then 3 - tenth of that number is : a . 35","rationale":"explanation : the number is 1 \/ 3 of 1 \/ 4 is = 15 then 1 \/ 3 ã — 1 \/ 4 = 18 number is 216 then 216 ã — 3 \/ 10 = 64.8 answer : d","correct":"d","options":{"a":"23 ","b":"87 ","c":"26 ","d":"64.8","e":"01"},"options_float":{"a":23.0,"b":87.0,"c":26.0,"d":64.8,"e":1.0},"annotated_formula":"multiply(divide(const_3, const_10), multiply(multiply(3, const_4), 18))","linear_formula":"divide(const_3,const_10)|multiply(n1,const_4)|multiply(n0,#1)|multiply(#0,#2)","chain":"3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 * 18<\/gadget>\n216<\/output>\n(3\/10) * 216<\/gadget>\n324\/5 = around 64.8<\/output>\n324\/5 = around 64.8<\/result>","index":1956} +{"problem":"during a certain season , a team won 75 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ?","rationale":"we are first given that a team won 75 percent of its first 100 games . this means the team won 0.75 x 100 = 75 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 equals the number of remaining games in the season . thus we can say : 0.5 ( t – 100 ) = number of wins for remaining games 0.5 t – 50 = number of wins for remaining games lastly , we are given that team won 70 percent of all games played in the season . that is , they won 0.7 t games in the entire season . with this we can set up the equation : number of first 100 games won + number of games won for remaining games = total number of games won in the entire season 75 + 0.5 t – 50 = 0.7 t 25 = 0.2 t 250 = 2 t 125 = t answer is e .","correct":"e","options":{"a":"180 ","b":"170 ","c":"156 ","d":"150","e":"125"},"options_float":{"a":180.0,"b":170.0,"c":156.0,"d":150.0,"e":125.0},"annotated_formula":"add(100, divide(subtract(75, 70), subtract(divide(70, const_100), divide(50, const_100))))","linear_formula":"divide(n3,const_100)|divide(n2,const_100)|subtract(n0,n3)|subtract(#0,#1)|divide(#2,#3)|add(n1,#4)","chain":"75 - 70<\/gadget>\n5<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(7\/10) - (1\/2)<\/gadget>\n1\/5 = around 0.2<\/output>\n5 \/ (1\/5)<\/gadget>\n25<\/output>\n100 + 25<\/gadget>\n125<\/output>\n125<\/result>","index":1957} +{"problem":"if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability e that n is divisible by 3 ?","rationale":"\"prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 \/ 16 . ans d . took me 3 : 20 mins .\"","correct":"d","options":{"a":"1 \/ 3 ","b":"¼ ","c":"9 \/ 25 ","d":"5 \/ 16","e":"0"},"options_float":{"a":0.3333333333,"b":null,"c":0.36,"d":0.3125,"e":0.0},"annotated_formula":"divide(add(3, const_2), multiply(const_4, const_4))","linear_formula":"add(n0,const_2)|multiply(const_4,const_4)|divide(#0,#1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n4 * 4<\/gadget>\n16<\/output>\n5 \/ 16<\/gadget>\n5\/16 = around 0.3125<\/output>\n5\/16 = around 0.3125<\/result>","index":1958} +{"problem":"p , q and r together earn rs . 1890 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ?","rationale":"\"explanation : amount earned by p , q and r in 1 day = 1890 \/ 9 = 210 - - - ( 1 ) amount earned by p and r in 1 day = 600 \/ 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 \/ 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 130 - 210 = 40 = > amount earned by r in 1 day = 40 answer : option a\"","correct":"a","options":{"a":"s . 40 ","b":"s . 70 ","c":"s . 90 ","d":"s . 100","e":"s . 120"},"options_float":{"a":40.0,"b":70.0,"c":90.0,"d":100.0,"e":120.0},"annotated_formula":"subtract(add(divide(600, 5), divide(910, 7)), divide(1890, 9))","linear_formula":"divide(n2,n3)|divide(n5,n4)|divide(n0,n1)|add(#0,#1)|subtract(#3,#2)|","chain":"600 \/ 5<\/gadget>\n120<\/output>\n910 \/ 7<\/gadget>\n130<\/output>\n120 + 130<\/gadget>\n250<\/output>\n1_890 \/ 9<\/gadget>\n210<\/output>\n250 - 210<\/gadget>\n40<\/output>\n40<\/result>","index":1959} +{"problem":"the greatest number that divides 690 and 875 leaving remainders 10 and 25 respectively is :","rationale":"\"explanation : 690 - 10 = 680 , 875 – 25 = 850 highest number that can divide 680 and 850 exactly is hcm of the numbers . hcf of 680 and 850 = 170 answer : a\"","correct":"a","options":{"a":"170 ","b":"130 ","c":"150 ","d":"120","e":"180"},"options_float":{"a":170.0,"b":130.0,"c":150.0,"d":120.0,"e":180.0},"annotated_formula":"divide(subtract(875, 25), gcd(690, 875))","linear_formula":"gcd(n0,n1)|subtract(n1,n3)|divide(#1,#0)|","chain":"875 - 25<\/gadget>\n850<\/output>\ngcd(690, 875)<\/gadget>\n5<\/output>\n850 \/ 5<\/gadget>\n170<\/output>\n170<\/result>","index":1960} +{"problem":"a group of students was interviewed for that if it was asked whether or not they speak french and \/ or english . among those who speak french , 25 speak english well , while 65 of them do not speak english . if 55 % of students do not speak french , how many students were surveyed ?","rationale":"number of students who speak french are 65 + 25 = 90 of total students , the percentage of students who do not speak french was 55 % - - > percentage of who do is 45 % 90 - - - - - - - 45 % x - - - - - - - 100 % x = 90 * 100 \/ 45 = 200 = number of all students answer is d","correct":"d","options":{"a":"190 ","b":"175 ","c":"225 ","d":"200","e":"250"},"options_float":{"a":190.0,"b":175.0,"c":225.0,"d":200.0,"e":250.0},"annotated_formula":"divide(add(25, 65), divide(subtract(const_100, 55), const_100))","linear_formula":"add(n0,n1)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,#2)","chain":"25 + 65<\/gadget>\n90<\/output>\n100 - 55<\/gadget>\n45<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n90 \/ (9\/20)<\/gadget>\n200<\/output>\n200<\/result>","index":1961} +{"problem":"a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 510 kilograms of pears that day , how many kilograms did he sell in the afternoon ?","rationale":"\"3 x = 510 x = 170 therefore , the salesman sold 170 kg in the morning and 2 ⋅ 170 = 340 kg in the afternoon . so answer is c .\"","correct":"c","options":{"a":"120 ","b":"180 ","c":"340 ","d":"280","e":"320"},"options_float":{"a":120.0,"b":180.0,"c":340.0,"d":280.0,"e":320.0},"annotated_formula":"multiply(divide(510, const_3), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_2)|","chain":"510 \/ 3<\/gadget>\n170<\/output>\n170 * 2<\/gadget>\n340<\/output>\n340<\/result>","index":1962} +{"problem":"find the sum of all 3 digit natural numbers , which are divisible by 8 .","rationale":"the three digit natural numbers divisible by 8 are 104 , 112 , 120 , … . 992 . let sndenote their sum . that is , sn = 104 112 120 128 , 992 g + + + + + . now , the sequence 104 , 112 , 120 , g , 992 forms an a . p . a = 104 , d = 8 , l = 992 n = l - a \/ d n = 112 s 112 = n \/ 2 ( a + l ) = 61376 option a","correct":"a","options":{"a":"61376 ","b":"54411 ","c":"612314 ","d":"64170","e":"64171"},"options_float":{"a":61376.0,"b":54411.0,"c":612314.0,"d":64170.0,"e":64171.0},"annotated_formula":"multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 8)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 8), add(const_100, const_4)), 8), const_1), const_2))","linear_formula":"add(const_100,const_4)|multiply(const_10,const_100)|subtract(#1,n1)|add(#0,#2)|subtract(#2,#0)|divide(#4,n1)|add(#5,const_1)|divide(#6,const_2)|multiply(#3,#7)","chain":"100 + 4<\/gadget>\n104<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 8<\/gadget>\n992<\/output>\n104 + 992<\/gadget>\n1_096<\/output>\n992 - 104<\/gadget>\n888<\/output>\n888 \/ 8<\/gadget>\n111<\/output>\n111 + 1<\/gadget>\n112<\/output>\n112 \/ 2<\/gadget>\n56<\/output>\n1_096 * 56<\/gadget>\n61_376<\/output>\n61_376<\/result>","index":1963} +{"problem":"each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.65 % , what fraction of the products was checked by worker y ?","rationale":"\"x : 0.5 % is 0.15 % - points from 0.65 % . y : 0.8 % is 0.15 % - points from 0.65 % . therefore the ratio of products checked by y : x is 1 : 1 . thus , worker y checked 1 \/ 2 of the products . the answer is b .\"","correct":"b","options":{"a":"1 \/ 5 ","b":"1 \/ 2 ","c":"2 \/ 3 ","d":"1 \/ 4","e":"2 \/ 5"},"options_float":{"a":0.2,"b":0.5,"c":0.6666666667,"d":0.25,"e":0.4},"annotated_formula":"divide(subtract(0.65, 0.5), subtract(0.8, 0.5))","linear_formula":"subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)|","chain":"0.65 - 0.5<\/gadget>\n0.15<\/output>\n0.8 - 0.5<\/gadget>\n0.3<\/output>\n0.15 \/ 0.3<\/gadget>\n0.5<\/output>\n0.5<\/result>","index":1964} +{"problem":"the radius of the two circular fields is in the ratio 4 : 10 the area of the second field is what percent greater than the area of the first ?","rationale":"r = 4 ï € r 2 = 16 r = 10 ï € r 2 = 100 100 ï € â € “ 84 ï € 100 - - - - ? = > 84 % . answer : d","correct":"d","options":{"a":"72 % ","b":"96 % ","c":"94 % ","d":"84 %","e":"82 %"},"options_float":{"a":72.0,"b":96.0,"c":94.0,"d":84.0,"e":82.0},"annotated_formula":"subtract(power(10, const_2), power(4, const_2))","linear_formula":"power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)","chain":"10 ** 2<\/gadget>\n100<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n100 - 16<\/gadget>\n84<\/output>\n84<\/result>","index":1965} +{"problem":"a river 2 m deep and 45 m wide is flowing at the rate of 5 kmph the amount of water that runs into the sea per minute is ?","rationale":"\"explanation : ( 5000 * 2 * 45 ) \/ 60 = 7500 m 3 answer : option c\"","correct":"c","options":{"a":"4500 m 3 ","b":"4580 m 3 ","c":"7500 m 3 ","d":"4900 m 3","e":"4700 m 3"},"options_float":{"a":4500.0,"b":4580.0,"c":7500.0,"d":4900.0,"e":4700.0},"annotated_formula":"divide(multiply(multiply(2, 45), multiply(5, const_1000)), multiply(const_1, const_60))","linear_formula":"multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)|","chain":"2 * 45<\/gadget>\n90<\/output>\n5 * 1_000<\/gadget>\n5_000<\/output>\n90 * 5_000<\/gadget>\n450_000<\/output>\n1 * 60<\/gadget>\n60<\/output>\n450_000 \/ 60<\/gadget>\n7_500<\/output>\n7_500<\/result>","index":1966} +{"problem":"if the range q of the 6 numbers 4 , 314 , 710 and x is 12 , what is the difference between the greatest possible value of x and least possible value of x ?","rationale":"the range q of a set is the difference between the largest and smallest elements of a set . without x , the difference between the largest and smallest elements of a set is 14 - 3 = 11 < 12 , which means that in order 12 to be the range of the set x must be either the smallest element so that 14 - x = 12 - - - > x = 2 or x must the largest element so that x - 3 = 12 - - > x = 15 . the the difference between the greatest possible value of x and least possible value of x is 15 - 2 = 13 . answer : d .","correct":"d","options":{"a":"0 ","b":"2 ","c":"12 ","d":"13","e":"15"},"options_float":{"a":0.0,"b":2.0,"c":12.0,"d":13.0,"e":15.0},"annotated_formula":"divide(subtract(add(add(add(add(add(4, const_3), add(const_12, const_2)), add(const_4, const_3)), add(4, 6)), 12), add(const_10, const_1)), const_3)","linear_formula":"add(n1,const_3)|add(const_12,const_2)|add(const_3,const_4)|add(n0,n1)|add(const_1,const_10)|add(#0,#1)|add(#5,#2)|add(#6,#3)|add(n4,#7)|subtract(#8,#4)|divide(#9,const_3)","chain":"4 + 3<\/gadget>\n7<\/output>\n12 + 2<\/gadget>\n14<\/output>\n7 + 14<\/gadget>\n21<\/output>\n21 + 7<\/gadget>\n28<\/output>\n4 + 6<\/gadget>\n10<\/output>\n28 + 10<\/gadget>\n38<\/output>\n38 + 12<\/gadget>\n50<\/output>\n10 + 1<\/gadget>\n11<\/output>\n50 - 11<\/gadget>\n39<\/output>\n39 \/ 3<\/gadget>\n13<\/output>\n13<\/result>","index":1967} +{"problem":"if 80 percent of a class answered the first question on a certain test correctly , 55 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ?","rationale":"\"i did n ' t understand how to connect 20 percent with the whole equation . why 20 is out of the percentile diagram ( circles ) but inside of 100 ? 20 % appeared in the exam ; however did not answer question aquestion b correctly so they are out of the two circles , however as they have appeared in the exam , they have to be taken into consideration for calculation purpose i am doing it this way : suppose the total class size is 100 , then 20 percent of the class answered incorrectly and rest 80 percent answered correctly in question 1 or 2 or both . now , 100 - [ ( 80 - x ) + x + ( 55 - x ) ] = 20 . . . . . . . . . . values are in percent x = 20 - 100 + 135 x = 55 ans : e\"","correct":"e","options":{"a":"10 % ","b":"( 8 ) 20 % ","c":"30 % ","d":"50 %","e":"55 %"},"options_float":{"a":10.0,"b":8.0,"c":30.0,"d":50.0,"e":55.0},"annotated_formula":"subtract(add(add(80, 55), 20), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,const_100)|","chain":"80 + 55<\/gadget>\n135<\/output>\n135 + 20<\/gadget>\n155<\/output>\n155 - 100<\/gadget>\n55<\/output>\n55<\/result>","index":1969} +{"problem":"a train 300 m long can cross an electric pole in 20 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 300 \/ 20 s = 15 m \/ sec speed = 15 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 54 kmph answer : c\"","correct":"c","options":{"a":"72 ","b":"67 ","c":"54 ","d":"26","e":"27"},"options_float":{"a":72.0,"b":67.0,"c":54.0,"d":26.0,"e":27.0},"annotated_formula":"divide(divide(300, const_1000), divide(20, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"300 \/ 1_000<\/gadget>\n3\/10 = around 0.3<\/output>\n20 \/ 3_600<\/gadget>\n1\/180 = around 0.005556<\/output>\n(3\/10) \/ (1\/180)<\/gadget>\n54<\/output>\n54<\/result>","index":1970} +{"problem":"a train is 385 meter long is running at a speed of 45 km \/ hour . in what time will it pass a bridge of 140 meter length","rationale":"\"explanation : speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 385 + 140 = 525 meter time = distance \/ speed = 525 ∗ 2 \/ 25 = 42 seconds option a\"","correct":"a","options":{"a":"42 seconds ","b":"30 seconds ","c":"40 seconds ","d":"50 seconds","e":"none of these"},"options_float":{"a":42.0,"b":30.0,"c":40.0,"d":50.0,"e":null},"annotated_formula":"divide(add(385, 140), divide(multiply(45, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"385 + 140<\/gadget>\n525<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n525 \/ (25\/2)<\/gadget>\n42<\/output>\n42<\/result>","index":1971} +{"problem":"the distance between two cities a and b is 330 km . a train starts from a at 8 a . m . and travel towards b at 60 km \/ hr . another train starts from b at 9 a . m and travels towards a at 75 km \/ hr . at what time do they meet ?","rationale":"explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 330 . therefore , 60 x + 75 ( x - 1 ) = 330 . = > x = 3 . so , they meet at ( 8 + 3 ) i . e , 11 a . m . answer : c","correct":"c","options":{"a":"10 a . m ","b":"10.30 a . m ","c":"11 a . m ","d":"11.30 a . m","e":"none of these"},"options_float":{"a":10.0,"b":10.3,"c":11.0,"d":11.3,"e":null},"annotated_formula":"add(divide(add(330, 75), add(60, 75)), 8)","linear_formula":"add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)","chain":"330 + 75<\/gadget>\n405<\/output>\n60 + 75<\/gadget>\n135<\/output>\n405 \/ 135<\/gadget>\n3<\/output>\n3 + 8<\/gadget>\n11<\/output>\n11<\/result>","index":1973} +{"problem":"there are 10 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ?","rationale":"\"the total number of stations = 12 from 12 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 12 p ₂ ways . ² ⁰ p ₂ = 12 * 11 = 132 . answer : c\"","correct":"c","options":{"a":"156 ","b":"167 ","c":"132 ","d":"352","e":"380"},"options_float":{"a":156.0,"b":167.0,"c":132.0,"d":352.0,"e":380.0},"annotated_formula":"multiply(add(10, const_1), add(add(10, const_1), const_1))","linear_formula":"add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)|","chain":"10 + 1<\/gadget>\n11<\/output>\n11 + 1<\/gadget>\n12<\/output>\n11 * 12<\/gadget>\n132<\/output>\n132<\/result>","index":1975} +{"problem":"total dinning bill of 9 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared .","rationale":"\"dinner bill of 9 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) \/ 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 \/ 9 = $ 16.98 answer : c\"","correct":"c","options":{"a":"21.84 ","b":"12.84 ","c":"16.98 ","d":"24.84","e":"15.84"},"options_float":{"a":21.84,"b":12.84,"c":16.98,"d":24.84,"e":15.84},"annotated_formula":"divide(multiply(139.00, add(divide(const_1, 10), const_1)), 9)","linear_formula":"divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)|","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n139 * (11\/10)<\/gadget>\n1_529\/10 = around 152.9<\/output>\n(1_529\/10) \/ 9<\/gadget>\n1_529\/90 = around 16.988889<\/output>\n1_529\/90 = around 16.988889<\/result>","index":1976} +{"problem":"find the length of the longest pole that can be placed in a room 12 m long 8 m broad and 9 m high .","rationale":"\"length of longest pole = length of the diagonal of the room = sqrt ( 12 ^ 2 + 8 ^ 2 + 9 ^ 2 = sqrt ( 289 ) = 17 m . answer is e .\"","correct":"e","options":{"a":"13 ","b":"14 ","c":"15 ","d":"16","e":"17"},"options_float":{"a":13.0,"b":14.0,"c":15.0,"d":16.0,"e":17.0},"annotated_formula":"sqrt(add(power(9, const_2), add(power(12, const_2), power(8, const_2))))","linear_formula":"power(n0,const_2)|power(n1,const_2)|power(n2,const_2)|add(#0,#1)|add(#3,#2)|sqrt(#4)|","chain":"9 ** 2<\/gadget>\n81<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n144 + 64<\/gadget>\n208<\/output>\n81 + 208<\/gadget>\n289<\/output>\n289 ** (1\/2)<\/gadget>\n17<\/output>\n17<\/result>","index":1977} +{"problem":"if 9 ^ y = 3 ^ 14 , what is y ?","rationale":"\"9 ^ y = 3 ^ 2 y = 3 ^ 14 2 y = 14 y = 7 the answer is e .\"","correct":"e","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"7"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":7.0},"annotated_formula":"divide(14, const_2)","linear_formula":"divide(n2,const_2)|","chain":"14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":1979} +{"problem":"how many zeroes are there at the end of the number n , if n = 90 ! + 180 ! ?","rationale":"\"the number of zeroes at the end of 90 ! will be less than the number of zeroes at the end of 180 ! hence it is sufficient to calculate the number of zeroes at the end of 90 ! the number of zeroes = [ 90 \/ 5 ] + [ 90 \/ 25 ] + [ 90 \/ 125 ] = 18 + 3 + 0 = 21 the answer is c .\"","correct":"c","options":{"a":"15 ","b":"18 ","c":"21 ","d":"24","e":"27"},"options_float":{"a":15.0,"b":18.0,"c":21.0,"d":24.0,"e":27.0},"annotated_formula":"add(divide(90, add(const_4, const_1)), const_2)","linear_formula":"add(const_1,const_4)|divide(n0,#0)|add(#1,const_2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n90 \/ 5<\/gadget>\n18<\/output>\n18 + 2<\/gadget>\n20<\/output>\n20<\/result>","index":1980} +{"problem":"ms . harris invested in fund x and fund b . the total amount she invested , in both funds combined , was $ 100000 . in one year , fund x paid 23 % and fund b paid 17 % . the interest earned in fund b was exactly $ 200 greater than the interest earned in fund x . how much did ms . morris invest in fund x ?","rationale":"0.17 b - 0.23 x = 200 | * 100 17 b - 23 x = 20,000 x + b = 100,000 b = 100,000 - x 17 ( 100,000 - x ) - 23 x = 20,000 1 , 700,000 - 17 x - 23 x = 20,000 1 , 680,000 = 40 x divide by 40 first two digits : 42 . . . so answer is d .","correct":"d","options":{"a":"$ 32000 ","b":"$ 36000 ","c":"$ 40000 ","d":"$ 42000","e":"$ 45,000"},"options_float":{"a":32000.0,"b":36000.0,"c":40000.0,"d":42000.0,"e":45000.0},"annotated_formula":"subtract(100000, divide(add(multiply(23, 100000), 200), add(23, 17)))","linear_formula":"add(n1,n2)|multiply(n0,n1)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3)","chain":"23 * 100_000<\/gadget>\n2_300_000<\/output>\n2_300_000 + 200<\/gadget>\n2_300_200<\/output>\n23 + 17<\/gadget>\n40<\/output>\n2_300_200 \/ 40<\/gadget>\n57_505<\/output>\n100_000 - 57_505<\/gadget>\n42_495<\/output>\n42_495<\/result>","index":1981} +{"problem":"if the price of a certain computer increased 30 percent from b dollars to 351 dollars , then 2 b =","rationale":"\"before price increase price = b after 30 % price increase price = b + ( 30 \/ 100 ) * b = 1.3 b = 351 ( given ) i . e . b = 351 \/ 1.3 = $ 270 i . e . 2 b = 2 * 270 = 540 answer : option a\"","correct":"a","options":{"a":"540 ","b":"570 ","c":"619 ","d":"649","e":"700"},"options_float":{"a":540.0,"b":570.0,"c":619.0,"d":649.0,"e":700.0},"annotated_formula":"multiply(divide(351, divide(add(const_100, 30), const_100)), 2)","linear_formula":"add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)|","chain":"100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n351 \/ (13\/10)<\/gadget>\n270<\/output>\n270 * 2<\/gadget>\n540<\/output>\n540<\/result>","index":1984} +{"problem":"vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 5780 , how much amount did raghu invest ?","rationale":"\"let money invested by raghu = rs . x money invested by trishul = 9 \/ 10 x = 0.9 x money invested by vishal = 9 \/ 10 x * 110 \/ 100 = 0.99 x also , x + 0.9 x + 0.99 x = 5780 = x = 5780 \/ 2.89 = 2000 therefore , amount invested by raghu is rs . 2000 . answer : c\"","correct":"c","options":{"a":"1287 ","b":"2887 ","c":"2000 ","d":"1129","e":"1192"},"options_float":{"a":1287.0,"b":2887.0,"c":2000.0,"d":1129.0,"e":1192.0},"annotated_formula":"divide(5780, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|add(#3,#2)|add(#4,const_1)|divide(n2,#5)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(9\/10) * (11\/10)<\/gadget>\n99\/100 = around 0.99<\/output>\n(99\/100) + (9\/10)<\/gadget>\n189\/100 = around 1.89<\/output>\n(189\/100) + 1<\/gadget>\n289\/100 = around 2.89<\/output>\n5_780 \/ (289\/100)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":1985} +{"problem":"how many seconds will a train 100 meters long take to cross a bridge 170 meters long if the speed of the train is 36 kmph ?","rationale":"\"d = 100 + 170 = 270 s = 36 * 5 \/ 18 = 10 mps t = 270 \/ 10 = 27 sec answer : e\"","correct":"e","options":{"a":"54 sec ","b":"65 sec ","c":"25 sec ","d":"45 sec","e":"27 sec"},"options_float":{"a":54.0,"b":65.0,"c":25.0,"d":45.0,"e":27.0},"annotated_formula":"divide(add(170, 100), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"170 + 100<\/gadget>\n270<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n270 \/ 10<\/gadget>\n27<\/output>\n27<\/result>","index":1986} +{"problem":"in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 12 people all together . how many people can sit in a bus ?","rationale":"\"right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 12 total people can seat = 81 + 12 = 93 answer : e\"","correct":"e","options":{"a":"52 ","b":"49 ","c":"95 ","d":"88","e":"93"},"options_float":{"a":52.0,"b":49.0,"c":95.0,"d":88.0,"e":93.0},"annotated_formula":"add(multiply(add(15, subtract(15, 3)), 3), 12)","linear_formula":"subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)|","chain":"15 - 3<\/gadget>\n12<\/output>\n15 + 12<\/gadget>\n27<\/output>\n27 * 3<\/gadget>\n81<\/output>\n81 + 12<\/gadget>\n93<\/output>\n93<\/result>","index":1987} +{"problem":"some persons can do a piece of work in 16 days . two times the number of these people will do half of that work in ?","rationale":"\"16 \/ ( 2 * 2 ) = 4 days answer : b\"","correct":"b","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"multiply(multiply(16, divide(const_1, const_2)), divide(const_1, const_2))","linear_formula":"divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n16 * (1\/2)<\/gadget>\n8<\/output>\n8 * (1\/2)<\/gadget>\n4<\/output>\n4<\/result>","index":1988} +{"problem":"if an average hard drive had a capacity of 0.4 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ?","rationale":"\"0.4 * 2 ^ 10 = 0.4 * 1024 = 409.6 the answer is c .\"","correct":"c","options":{"a":"256 ","b":"512 ","c":"4096 ","d":"1024","e":"1536"},"options_float":{"a":256.0,"b":512.0,"c":4096.0,"d":1024.0,"e":1536.0},"annotated_formula":"multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.4, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.4), const_10)","linear_formula":"multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|multiply(#4,const_2)|multiply(#5,const_2)|multiply(#6,const_2)|multiply(#7,const_2)|multiply(#8,const_2)|add(n0,#9)|multiply(#10,const_10)|","chain":"0.4 * 2<\/gadget>\n0.8<\/output>\n0.8 * 2<\/gadget>\n1.6<\/output>\n1.6 * 2<\/gadget>\n3.2<\/output>\n3.2 * 2<\/gadget>\n6.4<\/output>\n6.4 * 2<\/gadget>\n12.8<\/output>\n12.8 * 2<\/gadget>\n25.6<\/output>\n25.6 * 2<\/gadget>\n51.2<\/output>\n51.2 * 2<\/gadget>\n102.4<\/output>\n102.4 * 2<\/gadget>\n204.8<\/output>\n204.8 * 2<\/gadget>\n409.6<\/output>\n409.6 + 0.4<\/gadget>\n410<\/output>\n410 * 10<\/gadget>\n4_100<\/output>\n4_100<\/result>","index":1990} +{"problem":"find the remainder when 6 ^ 50 is divided by 215","rationale":"\"( 6 ^ 3 ) ^ 16 * 6 ^ 2 \/ 215 ( 215 + 1 ) ^ 16 * 6 ^ 2 \/ 215 1 ^ 16 * 6 ^ 2 \/ 215 = 36 answer : c\"","correct":"c","options":{"a":"34 ","b":"35 ","c":"36 ","d":"37","e":"38"},"options_float":{"a":34.0,"b":35.0,"c":36.0,"d":37.0,"e":38.0},"annotated_formula":"reminder(power(6, 50), 215)","linear_formula":"power(n0,n1)|reminder(#0,n2)|","chain":"6 ** 50<\/gadget>\n808_281_277_464_764_060_643_139_600_456_536_293_376<\/output>\n808_281_277_464_764_060_643_139_600_456_536_293_376 % 215<\/gadget>\n36<\/output>\n36<\/result>","index":1994} +{"problem":"a train 140 m long is running at 60 kmph . in how much time wiu it pass a platform 260 m long ?","rationale":"\"s = d \/ t d = 140 + 260 = 400 m s = 60 * 5 \/ 18 m \/ sec = 50 \/ 3 m \/ sec t = 400 \/ ( 50 \/ 3 ) = 1200 \/ 50 = 24 sec answer : d\"","correct":"d","options":{"a":"21 sec ","b":"22 sec ","c":"23 sec ","d":"24 sec","e":"26 sec"},"options_float":{"a":21.0,"b":22.0,"c":23.0,"d":24.0,"e":26.0},"annotated_formula":"divide(add(140, 260), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"140 + 260<\/gadget>\n400<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n400 \/ (50\/3)<\/gadget>\n24<\/output>\n24<\/result>","index":1995} +{"problem":"the total marks obtained by a student in mathematics and physics is 50 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .","rationale":"\"let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 50 and c - p = 20 m + c \/ 2 = [ ( m + p ) + ( c - p ) ] \/ 2 = ( 50 + 20 ) \/ 2 = 35 . answer : c\"","correct":"c","options":{"a":"15 ","b":"25 ","c":"35 ","d":"45","e":"55"},"options_float":{"a":15.0,"b":25.0,"c":35.0,"d":45.0,"e":55.0},"annotated_formula":"divide(add(50, 20), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"50 + 20<\/gadget>\n70<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n35<\/result>","index":1996} +{"problem":"a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 70 paise per sq . metre","rationale":"\"area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 70 = 59500 p in rs = 59500 \/ 100 = rs 595 answer : a\"","correct":"a","options":{"a":"s 595 ","b":"s 780 ","c":"s 880 ","d":"s 480","e":"s 980"},"options_float":{"a":595.0,"b":780.0,"c":880.0,"d":480.0,"e":980.0},"annotated_formula":"divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 70), const_100)","linear_formula":"multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)|","chain":"110 * 65<\/gadget>\n7_150<\/output>\n2.5 * 2<\/gadget>\n5<\/output>\n110 - 5<\/gadget>\n105<\/output>\n65 - 5<\/gadget>\n60<\/output>\n105 * 60<\/gadget>\n6_300<\/output>\n7_150 - 6_300<\/gadget>\n850<\/output>\n850 * 70<\/gadget>\n59_500<\/output>\n59_500 \/ 100<\/gadget>\n595<\/output>\n595<\/result>","index":1997} +{"problem":"rahul played weel in this season . his current batting averagge is 46 . if he score 78 runs in today match . his batting average will become 54 . how many matches had he played in this season .","rationale":"\"46 x + 78 = 54 ( x + 1 ) = > 8 x = 24 = > x = 3 answer : c\"","correct":"c","options":{"a":"8 ","b":"10 ","c":"3 ","d":"6","e":"5"},"options_float":{"a":8.0,"b":10.0,"c":3.0,"d":6.0,"e":5.0},"annotated_formula":"divide(subtract(78, 54), subtract(54, 46))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"78 - 54<\/gadget>\n24<\/output>\n54 - 46<\/gadget>\n8<\/output>\n24 \/ 8<\/gadget>\n3<\/output>\n3<\/result>","index":1998} +{"problem":"find the simple interest on rs . 69,000 at 16 2 \/ 3 % per year for 9 months .","rationale":"p = rs . 69000 , r = 50 \/ 3 % p . a and t = 9 \/ 12 years = 3 \/ 4 years . simple interest = ( p * r * t ) \/ 100 = rs . ( 69,000 * ( 50 \/ 3 ) * ( 3 \/ 4 ) * ( 1 \/ 100 ) ) = rs . 8625 answer is c .","correct":"c","options":{"a":"7500 ","b":"6500 ","c":"8625 ","d":"9500","e":"none of them"},"options_float":{"a":7500.0,"b":6500.0,"c":8625.0,"d":9500.0,"e":null},"annotated_formula":"multiply(multiply(multiply(add(multiply(multiply(multiply(const_2, const_3), const_100), const_100), multiply(multiply(multiply(const_3, const_3), const_100), multiply(add(const_3, const_2), const_2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(const_3, const_3), multiply(2, multiply(const_2, const_3)))), divide(const_1, const_100))","linear_formula":"add(const_2,const_3)|divide(const_1,const_100)|multiply(const_3,const_3)|multiply(const_2,const_3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,const_2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600 * 100<\/gadget>\n60_000<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 100<\/gadget>\n900<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n900 * 10<\/gadget>\n9_000<\/output>\n60_000 + 9_000<\/gadget>\n69_000<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48 + 2<\/gadget>\n50<\/output>\n50 \/ 3<\/gadget>\n50\/3 = around 16.666667<\/output>\n69_000 * (50\/3)<\/gadget>\n1_150_000<\/output>\n2 * 6<\/gadget>\n12<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n1_150_000 * (3\/4)<\/gadget>\n862_500<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n862_500 * (1\/100)<\/gadget>\n8_625<\/output>\n8_625<\/result>","index":1999} +{"problem":"malar and roja combined will complete a task in 35 days , but malar alone can complete same work in 60 days . calculate in how many days roja can complete this work ?","rationale":"malar - 35 days = > 1 \/ 35 roja - 60 days = > 1 \/ 60 malar one day work = 1 \/ 35 - 1 \/ 60 = 1 \/ 84 = = > 84 days answer d","correct":"d","options":{"a":"85 days ","b":"45 days ","c":"90 days ","d":"84 days","e":"95 days"},"options_float":{"a":85.0,"b":45.0,"c":90.0,"d":84.0,"e":95.0},"annotated_formula":"divide(multiply(35, 60), subtract(60, 35))","linear_formula":"multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)","chain":"35 * 60<\/gadget>\n2_100<\/output>\n60 - 35<\/gadget>\n25<\/output>\n2_100 \/ 25<\/gadget>\n84<\/output>\n84<\/result>","index":2001} +{"problem":"the measure of the side of a square is quadrupled . if x represents the perimeter of the original square , what is the value of the new perimeter ?","rationale":"original perimeter = x hence original side = x \/ 4 new side = 4 x \/ 4 new perimeter = 4 * 4 x \/ 4 = 4 x correct option : b","correct":"b","options":{"a":"3 x ","b":"4 x ","c":"9 x ","d":"12 x","e":"27 x"},"options_float":{"a":3.0,"b":4.0,"c":9.0,"d":12.0,"e":27.0},"annotated_formula":"multiply(const_4, const_1)","linear_formula":"multiply(const_1,const_4)","chain":"4 * 1<\/gadget>\n4<\/output>\n4<\/result>","index":2002} +{"problem":"tough and tricky questions : word problems . micheal , saren and donald decided to club together to buy a present . each of them gave equal amount of money . luckily saren negotiated a 20 % discount for the present so that each of them paid 4 dollars less . how much did they pay for a present ?","rationale":"answer c . we know that saren negotiated a discount of 20 % , so each of them paid $ 4 less . since there are three people , the 20 % of the original price amounts to $ 12 . 5 times 12 $ is 60 $ , so the original price , before saren negotiated the discount , had been $ 60 . they paid $ 12 less than the base price , so they spent $ 48 .","correct":"c","options":{"a":"20 ","b":"36 ","c":"48 ","d":"60","e":"72"},"options_float":{"a":20.0,"b":36.0,"c":48.0,"d":60.0,"e":72.0},"annotated_formula":"subtract(divide(multiply(4, const_3), divide(20, const_100)), multiply(4, const_3))","linear_formula":"divide(n0,const_100)|multiply(n1,const_3)|divide(#1,#0)|subtract(#2,#1)","chain":"4 * 3<\/gadget>\n12<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n12 \/ (1\/5)<\/gadget>\n60<\/output>\n60 - 12<\/gadget>\n48<\/output>\n48<\/result>","index":2003} +{"problem":"a club wants to mix 25 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ?","rationale":"\"let number of pounds of 5 $ candy to be used be w 6 = ( 25 * 8 + 5 * w ) \/ ( 25 + w ) = > 150 + 6 w = 200 + 5 w = > w = 50 answer c\"","correct":"c","options":{"a":"30 ","b":"40 ","c":"50 ","d":"60","e":"70"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":70.0},"annotated_formula":"subtract(multiply(25, 8.00), multiply(6.00, 25))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|subtract(#0,#1)|","chain":"25 * 8<\/gadget>\n200<\/output>\n6 * 25<\/gadget>\n150<\/output>\n200 - 150<\/gadget>\n50<\/output>\n50<\/result>","index":2005} +{"problem":"pradeep has to obtain 20 % of the total marks to pass . he got 390 marks and failed by 25 marks . the maximum marks are","rationale":"\"explanation : let their maximum marks be x . then , 20 % of x = 390 + 25 = > 20 \/ 100 x = 415 x = ( 41500 \/ 20 ) x = 2075 . answer : b\"","correct":"b","options":{"a":"300 ","b":"2075 ","c":"800 ","d":"1000","e":"3900"},"options_float":{"a":300.0,"b":2075.0,"c":800.0,"d":1000.0,"e":3900.0},"annotated_formula":"divide(add(390, 25), divide(20, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"390 + 25<\/gadget>\n415<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n415 \/ (1\/5)<\/gadget>\n2_075<\/output>\n2_075<\/result>","index":2008} +{"problem":"a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6500 ?","rationale":"\"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6500 * 6 ) – 30,130 ] = rs . ( 39000 – 30,130 ) = rs . 8870 answer c\"","correct":"c","options":{"a":"rs . 5870 ","b":"rs . 5991 ","c":"rs . 8870 ","d":"rs . 6850","e":"none of these"},"options_float":{"a":5870.0,"b":5991.0,"c":8870.0,"d":6850.0,"e":null},"annotated_formula":"subtract(multiply(add(5, const_1), 6500), add(add(add(add(5420, 5660), 6200), 6350), 6500))","linear_formula":"add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|","chain":"5 + 1<\/gadget>\n6<\/output>\n6 * 6_500<\/gadget>\n39_000<\/output>\n5_420 + 5_660<\/gadget>\n11_080<\/output>\n11_080 + 6_200<\/gadget>\n17_280<\/output>\n17_280 + 6_350<\/gadget>\n23_630<\/output>\n23_630 + 6_500<\/gadget>\n30_130<\/output>\n39_000 - 30_130<\/gadget>\n8_870<\/output>\n8_870<\/result>","index":2009} +{"problem":"if 30 men take 15 days to to complete a job , in how many days can 25 men finish that work ?","rationale":"ans . 18 days","correct":"a","options":{"a":"18 ","b":"17 ","c":"16 ","d":"15","e":"14"},"options_float":{"a":18.0,"b":17.0,"c":16.0,"d":15.0,"e":14.0},"annotated_formula":"divide(multiply(30, 15), 25)","linear_formula":"multiply(n0,n1)|divide(#0,n2)","chain":"30 * 15<\/gadget>\n450<\/output>\n450 \/ 25<\/gadget>\n18<\/output>\n18<\/result>","index":2010} +{"problem":"if 7 a = 8 b and ab ≠ 0 , what is the ratio of a \/ 8 to b \/ 7 ?","rationale":"\"a nice fast approach is the first find a pair of numbers that satisfy the given equation : 7 a = 8 b here ' s one pair : a = 8 and b = 7 what is the ratio of a \/ 8 to b \/ 7 ? in other words , what is the value of ( a \/ 8 ) \/ ( b \/ 7 ) ? plug in values to get : ( a \/ 8 ) \/ ( b \/ 7 ) = ( 8 \/ 8 ) \/ ( 7 \/ 7 ) = 1 \/ 1 = 1 c\"","correct":"c","options":{"a":"16 \/ 14 ","b":"8 \/ 7 ","c":"1 ","d":"7 \/ 8","e":"64 \/ 49"},"options_float":{"a":1.1428571429,"b":1.1428571429,"c":1.0,"d":0.875,"e":1.306122449},"annotated_formula":"divide(multiply(7, 8), multiply(8, 7))","linear_formula":"multiply(n0,n1)|divide(#0,#0)|","chain":"7 * 8<\/gadget>\n56<\/output>\n8 * 7<\/gadget>\n56<\/output>\n56 \/ 56<\/gadget>\n1<\/output>\n1<\/result>","index":2011} +{"problem":"an empty fuel tank is filled with brand z gasoline . when the tank is 3 \/ 4 empty , it is filled with brand y gasoline . when the tank is half empty again , it is filled with brand z gasoline . when the tank is half empty again , it is filled with brand y gasoline . at this time , what percent of the gasoline in the tank is brand z ?","rationale":"\"work with fraction of brand z in the tank . 1 st step : brand z is 1 2 nd step : brand z is 1 \/ 2 3 rd step : brand z is ( 3 \/ 4 ) * ( 1 \/ 2 ) + 1 \/ 2 = 7 \/ 8 4 th step : brand z is ( 1 \/ 2 ) * ( 7 \/ 8 ) = 7 \/ 16 = 43.75 % answer ( c )\"","correct":"c","options":{"a":"40 % ","b":"41 % ","c":"43.75 % ","d":"42 %","e":"44 %"},"options_float":{"a":40.0,"b":41.0,"c":43.75,"d":42.0,"e":44.0},"annotated_formula":"divide(subtract(const_100, divide(subtract(const_100, multiply(divide(3, 4), const_100)), const_2)), const_2)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#2,const_2)|subtract(const_100,#3)|divide(#4,const_2)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n100 - 75<\/gadget>\n25<\/output>\n25 \/ 2<\/gadget>\n25\/2 = around 12.5<\/output>\n100 - (25\/2)<\/gadget>\n175\/2 = around 87.5<\/output>\n(175\/2) \/ 2<\/gadget>\n175\/4 = around 43.75<\/output>\n175\/4 = around 43.75<\/result>","index":2012} +{"problem":"3 candidates contested in an election and received 4136 , 7636 , 11628 votes respectively . what is the percentage of the total votes did the winning candidate get ?","rationale":"total number of votes polled = 4136 + 7636 + 11628 = 23400 required percentage = 11628 * 100 \/ 23400 = 50 % answer is c","correct":"c","options":{"a":"45 % ","b":"49 % ","c":"50 % ","d":"51 %","e":"62 %"},"options_float":{"a":45.0,"b":49.0,"c":50.0,"d":51.0,"e":62.0},"annotated_formula":"multiply(divide(11628, add(add(4136, 7636), 11628)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)","chain":"4_136 + 7_636<\/gadget>\n11_772<\/output>\n11_772 + 11_628<\/gadget>\n23_400<\/output>\n11_628 \/ 23_400<\/gadget>\n323\/650 = around 0.496923<\/output>\n(323\/650) * 100<\/gadget>\n646\/13 = around 49.692308<\/output>\n646\/13 = around 49.692308<\/result>","index":2015} +{"problem":"in a hostel , the number of students decreased by 12 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ?","rationale":"\"cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 12 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.88 n ) * ( 1.2 p ) = > f ( new ) = f \/ ( 0.88 * 1.2 ) = > f ( new ) = 0.9469 f therefore the new cost of food must be 94.69 % of the old cost , or the cost of food must decrease by 5.3 % ( option e )\"","correct":"e","options":{"a":"19 % ","b":"15 % ","c":"25 % ","d":"40 %","e":"5.3 %"},"options_float":{"a":19.0,"b":15.0,"c":25.0,"d":40.0,"e":5.3},"annotated_formula":"multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 12), add(const_100, 20)))), const_100)","linear_formula":"add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n100 - 12<\/gadget>\n88<\/output>\n100 + 20<\/gadget>\n120<\/output>\n88 * 120<\/gadget>\n10_560<\/output>\n10_000 \/ 10_560<\/gadget>\n125\/132 = around 0.94697<\/output>\n1 - (125\/132)<\/gadget>\n7\/132 = around 0.05303<\/output>\n(7\/132) * 100<\/gadget>\n175\/33 = around 5.30303<\/output>\n175\/33 = around 5.30303<\/result>","index":2016} +{"problem":"the sum of number of boys and girls in a school is 1150 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ?","rationale":"\"we have x + x % of 1150 = 1150 x + x \/ 100 * 1150 = 1150 25 \/ 2 * x = 1150 x = 1150 * 2 \/ 25 = 92 answer is d\"","correct":"d","options":{"a":"50 ","b":"40 ","c":"60 ","d":"92","e":"70"},"options_float":{"a":50.0,"b":40.0,"c":60.0,"d":92.0,"e":70.0},"annotated_formula":"divide(1150, add(divide(1150, const_100), const_1))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n0,#1)|","chain":"1_150 \/ 100<\/gadget>\n23\/2 = around 11.5<\/output>\n(23\/2) + 1<\/gadget>\n25\/2 = around 12.5<\/output>\n1_150 \/ (25\/2)<\/gadget>\n92<\/output>\n92<\/result>","index":2018} +{"problem":"p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 35 days to complete the same work . then q alone can do it in","rationale":"\"work done by p and q in 1 day = 1 \/ 10 work done by r in 1 day = 1 \/ 35 work done by p , q and r in 1 day = 1 \/ 10 + 1 \/ 35 = 9 \/ 70 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day ã — 2 = 9 \/ 70 = > work done by p in 1 day = 9 \/ 140 = > work done by q and r in 1 day = 9 \/ 140 hence work done by q in 1 day = 9 \/ 140 â € “ 1 \/ 35 = 1 \/ 28 so q alone can do the work in 28 days answer is d .\"","correct":"d","options":{"a":"20 ","b":"22 ","c":"25 ","d":"28","e":"30"},"options_float":{"a":20.0,"b":22.0,"c":25.0,"d":28.0,"e":30.0},"annotated_formula":"divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 35)), const_2), divide(const_1, 35)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4)|","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 35<\/gadget>\n1\/35 = around 0.028571<\/output>\n(1\/10) + (1\/35)<\/gadget>\n9\/70 = around 0.128571<\/output>\n(9\/70) \/ 2<\/gadget>\n9\/140 = around 0.064286<\/output>\n(9\/140) - (1\/35)<\/gadget>\n1\/28 = around 0.035714<\/output>\n1 \/ (1\/28)<\/gadget>\n28<\/output>\n28<\/result>","index":2019} +{"problem":"kim has 3 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 6 shoes , what is the probability that she will select 2 shoes of the same color ?","rationale":"\"total pairs = 6 c 2 = 15 ; same color pairs = 3 c 1 * 1 c 1 = 3 ; prob = 1 \/ 5 ans b\"","correct":"b","options":{"a":"2 \/ 5 ","b":"1 \/ 5 ","c":"1 \/ 9 ","d":"1 \/ 10","e":"1 \/ 25"},"options_float":{"a":0.4,"b":0.2,"c":0.1111111111,"d":0.1,"e":0.04},"annotated_formula":"divide(3, choose(6, 2))","linear_formula":"choose(n2,n1)|divide(n0,#0)|","chain":"binomial(6, 2)<\/gadget>\n15<\/output>\n3 \/ 15<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":2020} +{"problem":"simplify : 3005 - 3000 + 10","rationale":"3005 - 3000 + 10 = 3005 - ( 3000 \/ 10 ) = 3005 - 300 = 2705 . answer is c","correct":"c","options":{"a":"2505 ","b":"2509 ","c":"2705 ","d":"2508","e":"none of them"},"options_float":{"a":2505.0,"b":2509.0,"c":2705.0,"d":2508.0,"e":null},"annotated_formula":"subtract(3005, divide(3000, 10))","linear_formula":"divide(n1,n2)|subtract(n0,#0)","chain":"3_000 \/ 10<\/gadget>\n300<\/output>\n3_005 - 300<\/gadget>\n2_705<\/output>\n2_705<\/result>","index":2021} +{"problem":"a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 50 extra yellow balls and making ratio of white balls to yellow balls 8 \/ 15 . how many tennis balls did the retailer order originally .","rationale":"white : yellow = x : ( x + 50 ) = 8 : 15 - - > 15 x = 8 x + 400 - - > x = 57.14 . the total # of balls originally x + x = 57.14 + 57.14 = 114.28 . answer : b .","correct":"b","options":{"a":"110 ","b":"114.28 ","c":"140 ","d":"160","e":"214"},"options_float":{"a":110.0,"b":114.28,"c":140.0,"d":160.0,"e":214.0},"annotated_formula":"multiply(divide(multiply(8, 50), subtract(15, 8)), const_2)","linear_formula":"multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)|multiply(#2,const_2)","chain":"8 * 50<\/gadget>\n400<\/output>\n15 - 8<\/gadget>\n7<\/output>\n400 \/ 7<\/gadget>\n400\/7 = around 57.142857<\/output>\n(400\/7) * 2<\/gadget>\n800\/7 = around 114.285714<\/output>\n800\/7 = around 114.285714<\/result>","index":2022} +{"problem":"how many numbers are divisible by both 2 and 3 up to 300 ?","rationale":"divide 300 by 6 , as the numbers which are divisible by both 2 and 3 are also divisible by 6 . the quotient obtained after dividing 300 by 6 is 50 , hence answer is 50 . option a","correct":"a","options":{"a":"50 ","b":"100 ","c":"150 ","d":"200","e":"250"},"options_float":{"a":50.0,"b":100.0,"c":150.0,"d":200.0,"e":250.0},"annotated_formula":"divide(300, multiply(2, 3))","linear_formula":"multiply(n0,n1)|divide(n2,#0)","chain":"2 * 3<\/gadget>\n6<\/output>\n300 \/ 6<\/gadget>\n50<\/output>\n50<\/result>","index":2024} +{"problem":"the result when a number subtracted from 150 is the same as the number added to 68 . what is the number ?","rationale":"answer let the number be x . then , x - 68 = 150 - x â ‡ ’ 2 x = 150 + 68 = 218 â ‡ ’ 2 x = 218 â ˆ ´ x = 109 correct option : c","correct":"c","options":{"a":"110 ","b":"100 ","c":"109 ","d":"218","e":"210"},"options_float":{"a":110.0,"b":100.0,"c":109.0,"d":218.0,"e":210.0},"annotated_formula":"add(divide(subtract(150, 68), const_2), 68)","linear_formula":"subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)","chain":"150 - 68<\/gadget>\n82<\/output>\n82 \/ 2<\/gadget>\n41<\/output>\n41 + 68<\/gadget>\n109<\/output>\n109<\/result>","index":2025} +{"problem":"how many numbers up to 300 and 700 are divisible by 2 , 3 and 7 both together ?","rationale":"( 700 – 300 ) \/ 42 = 9 22 \/ 42 = > 9 numbers answer : a","correct":"a","options":{"a":"9 ","b":"10 ","c":"11 ","d":"12","e":"13"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":12.0,"e":13.0},"annotated_formula":"subtract(divide(subtract(700, 300), lcm(lcm(2, 3), 7)), divide(add(2, multiply(const_2, const_10)), lcm(lcm(2, 3), 7)))","linear_formula":"lcm(n2,n3)|multiply(const_10,const_2)|subtract(n1,n0)|add(n2,#1)|lcm(n4,#0)|divide(#2,#4)|divide(#3,#4)|subtract(#5,#6)","chain":"700 - 300<\/gadget>\n400<\/output>\nlcm(2, 3)<\/gadget>\n6<\/output>\nlcm(6, 7)<\/gadget>\n42<\/output>\n400 \/ 42<\/gadget>\n200\/21 = around 9.52381<\/output>\n2 * 10<\/gadget>\n20<\/output>\n2 + 20<\/gadget>\n22<\/output>\n22 \/ 42<\/gadget>\n11\/21 = around 0.52381<\/output>\n(200\/21) - (11\/21)<\/gadget>\n9<\/output>\n9<\/result>","index":2027} +{"problem":"total 90 cows 30 cow gives each 2 liter milk 30 cow gives each 3 \/ 4 liter milk 30 cow gives each 1 \/ 4 liter milk this is split into 3 son per each 30 cows & 30 liter milk how ?","rationale":"\"30 cow 2 liter each = 60 liter 30 cow 3 \/ 4 liter each = 3 \/ 4 = 0.75 * 30 = 22.5 30 cow 1 \/ 4 liter each = 1 \/ 4 = 0.25 * 30 = 7.5 add 60 + 22.5 + 7.5 = 90 milk split into 3 son each 30 liter then 90 \/ 3 = 30 answer : e\"","correct":"e","options":{"a":"10 ","b":"12 ","c":"15 ","d":"16","e":"30"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":16.0,"e":30.0},"annotated_formula":"add(add(divide(3, 4), multiply(divide(3, 4), 30)), multiply(const_0_25, 30))","linear_formula":"divide(n4,n5)|multiply(n1,const_0_25)|multiply(n1,#0)|add(#0,#2)|add(#3,#1)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 30<\/gadget>\n45\/2 = around 22.5<\/output>\n(3\/4) + (45\/2)<\/gadget>\n93\/4 = around 23.25<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 30<\/gadget>\n15\/2 = around 7.5<\/output>\n(93\/4) + (15\/2)<\/gadget>\n123\/4 = around 30.75<\/output>\n123\/4 = around 30.75<\/result>","index":2029} +{"problem":"the ratio between the length and the breadth of a rectangular park is 1 : 3 . if a man cycling along the boundary of the park at the speed of 12 km \/ hr completes one round in 8 min , then the area of the park ( in sq . m ) is ?","rationale":"\"perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 1 x metres and breadth = 3 x metres . then , 2 ( 1 x + 3 x ) = 1600 or x = 200 . length = 200 m and breadth = 600 m . area = ( 600 x 200 ) m 2 = 120000 m e\"","correct":"e","options":{"a":"124545 m ","b":"134561 m ","c":"156787 m ","d":"15450 m","e":"120000 m"},"options_float":{"a":124545.0,"b":134561.0,"c":156787.0,"d":15450.0,"e":120000.0},"annotated_formula":"rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(1, 3)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(1, 3)), const_2), 3))","linear_formula":"add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|","chain":"3 * 2<\/gadget>\n6<\/output>\n10 * 6<\/gadget>\n60<\/output>\n12 \/ 60<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 8<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) * 1_000<\/gadget>\n1_600<\/output>\n1 + 3<\/gadget>\n4<\/output>\n1_600 \/ 4<\/gadget>\n400<\/output>\n400 \/ 2<\/gadget>\n200<\/output>\n200 * 3<\/gadget>\n600<\/output>\n200 * 600<\/gadget>\n120_000<\/output>\n120_000<\/result>","index":2030} +{"problem":"the ratio of two quantities is incorrect when written as 10 to 14 . how should this ratio be written ?","rationale":"a ratio should be reduced to the lowest common denominator . 10 : 14 means 10 x : 14 x which can be written as 5 ( 2 ) : 7 ( 2 ) . . . answer : d","correct":"d","options":{"a":"5 : 14 ","b":"10 : 7 ","c":"20 : 24 ","d":"5 : 7","e":"it can not be determined from the information given"},"options_float":{"a":0.3571428571,"b":1.4285714286,"c":0.8333333333,"d":0.7142857143,"e":null},"annotated_formula":"divide(10, 14)","linear_formula":"divide(n0,n1)","chain":"10 \/ 14<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":2031} +{"problem":"cereal a is 11 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 5 % sugar , what should be the ratio of cereal a to cereal b , by weight ?","rationale":"\"2 % is 3 % - points below 5 % and 11 % is 6 % - points above 5 % . the ratio of a : b should be 3 : 6 = 1 : 2 . the answer is a .\"","correct":"a","options":{"a":"1 : 2 ","b":"2 : 3 ","c":"1 : 3 ","d":"1 : 4","e":"3 : 4"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.3333333333,"d":0.25,"e":0.75},"annotated_formula":"divide(subtract(5, 2), subtract(11, 5))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"5 - 2<\/gadget>\n3<\/output>\n11 - 5<\/gadget>\n6<\/output>\n3 \/ 6<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2032} +{"problem":"in a family 16 people eat only vegetarian , 9 people eat only non veg . , 12 people eat both veg and non veg . . how many people eat veg in the family ?","rationale":"total people eat veg = only veg + both veg and non veg total = 16 + 12 = 28 answer = a","correct":"a","options":{"a":"28 ","b":"26 ","c":"22 ","d":"31","e":"21"},"options_float":{"a":28.0,"b":26.0,"c":22.0,"d":31.0,"e":21.0},"annotated_formula":"add(16, 12)","linear_formula":"add(n0,n2)","chain":"16 + 12<\/gadget>\n28<\/output>\n28<\/result>","index":2033} +{"problem":"if the price of sugar rises from rs . 10 per kg to rs . 11 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by","rationale":"\"sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x × 11 = x = 91 kg . ∴ reduction in consumption = 9 % . answer a\"","correct":"a","options":{"a":"9 % ","b":"20 % ","c":"25 % ","d":"30 %","e":"none"},"options_float":{"a":9.0,"b":20.0,"c":25.0,"d":30.0,"e":null},"annotated_formula":"multiply(subtract(const_1, divide(multiply(const_1, 10), 11)), const_100)","linear_formula":"multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"1 * 10<\/gadget>\n10<\/output>\n10 \/ 11<\/gadget>\n10\/11 = around 0.909091<\/output>\n1 - (10\/11)<\/gadget>\n1\/11 = around 0.090909<\/output>\n(1\/11) * 100<\/gadget>\n100\/11 = around 9.090909<\/output>\n100\/11 = around 9.090909<\/result>","index":2034} +{"problem":"the sum of the even numbers between 1 and n is 85 * 86 , where n is an odd number , then n = ?","rationale":"\"let n - 1 = 2 a . 2 + 4 + . . . + 2 a = 2 * ( 1 + 2 + . . . + a ) = 2 * ( a ) ( a + 1 ) \/ 2 = ( a ) ( a + 1 ) = 85 * 86 then a = 85 and n = 171 . the answer is b .\"","correct":"b","options":{"a":"163 ","b":"171 ","c":"179 ","d":"185","e":"191"},"options_float":{"a":163.0,"b":171.0,"c":179.0,"d":185.0,"e":191.0},"annotated_formula":"add(multiply(85, const_2), 1)","linear_formula":"multiply(n1,const_2)|add(#0,n0)|","chain":"85 * 2<\/gadget>\n170<\/output>\n170 + 1<\/gadget>\n171<\/output>\n171<\/result>","index":2035} +{"problem":"total dinning bill of 6 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared .","rationale":"dinner bill of 6 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) \/ 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 \/ 6 = $ 25.48 answer : e","correct":"e","options":{"a":"21.84 ","b":"22.84 ","c":"23.84 ","d":"24.84","e":"25.48"},"options_float":{"a":21.84,"b":22.84,"c":23.84,"d":24.84,"e":25.48},"annotated_formula":"divide(multiply(139, add(divide(const_1, 10), const_1)), 6)","linear_formula":"divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n139 * (11\/10)<\/gadget>\n1_529\/10 = around 152.9<\/output>\n(1_529\/10) \/ 6<\/gadget>\n1_529\/60 = around 25.483333<\/output>\n1_529\/60 = around 25.483333<\/result>","index":2036} +{"problem":"in a t . v . factory , an average of 63 tvs are produced per day for the fist 25 days of the months . a few workers fellill for the next 5 daysreducing the daily avg for the month to 58 sets \/ day . the average production per day for day last 5 days is ?","rationale":"production during these 5 days = total production in a month - production in first 25 days . = 30 x 58 - 25 x 63 = 165 â ˆ ´ average for last 5 days = 165 \/ 5 = 33 b","correct":"b","options":{"a":"35 ","b":"33 ","c":"48 ","d":"50","e":"52"},"options_float":{"a":35.0,"b":33.0,"c":48.0,"d":50.0,"e":52.0},"annotated_formula":"divide(subtract(multiply(add(25, 5), 58), multiply(25, 63)), 5)","linear_formula":"add(n1,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2)","chain":"25 + 5<\/gadget>\n30<\/output>\n30 * 58<\/gadget>\n1_740<\/output>\n25 * 63<\/gadget>\n1_575<\/output>\n1_740 - 1_575<\/gadget>\n165<\/output>\n165 \/ 5<\/gadget>\n33<\/output>\n33<\/result>","index":2037} +{"problem":"a part - time employee ’ s hourly wage was increased by 25 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ?","rationale":"\"let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 \/ hour and worked 100 hours \/ week so , the total weekly income was $ 100 \/ week after the 25 % wage increase , the employee makes $ 1.25 \/ hour we want the employee ' s income to remain at $ 100 \/ week . so , we want ( $ 1.25 \/ hour ) ( new # of hours ) = $ 100 divide both sides by 1.25 to get : new # of hours = 100 \/ 1.25 ≈ 80 hours so , the number of hours decreases from 100 hours to ( approximately ) 80 hours . this represents a 20 % decrease ( approximately ) . answer : c\"","correct":"c","options":{"a":"9 % ","b":"15 % ","c":"20 % ","d":"50 %","e":"100 %"},"options_float":{"a":9.0,"b":15.0,"c":20.0,"d":50.0,"e":100.0},"annotated_formula":"multiply(divide(divide(25, const_100), divide(add(25, const_100), const_100)), const_100)","linear_formula":"add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n25 + 100<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n(1\/4) \/ (5\/4)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":2038} +{"problem":"a is twice as good a workman as b and they took 10 days together to do the work b alone can do it in .","rationale":"wc = 2 : 1 2 x + x = 1 \/ 10 x = 1 \/ 30 = > 30 days answer : c","correct":"c","options":{"a":"25 days ","b":"88 days ","c":"30 days ","d":"11 days","e":"13 days"},"options_float":{"a":25.0,"b":88.0,"c":30.0,"d":11.0,"e":13.0},"annotated_formula":"multiply(divide(multiply(10, add(const_2, const_1)), const_2), const_2)","linear_formula":"add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)","chain":"2 + 1<\/gadget>\n3<\/output>\n10 * 3<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15 * 2<\/gadget>\n30<\/output>\n30<\/result>","index":2041} +{"problem":"which number can we add to 425897 to get a number exactly divisible by 456 ?","rationale":"\"425897 \/ 456 = 933 with a remainder of 449 . we need to add 456 - 449 = 7 the answer is c .\"","correct":"c","options":{"a":"3 ","b":"5 ","c":"7 ","d":"9","e":"11"},"options_float":{"a":3.0,"b":5.0,"c":7.0,"d":9.0,"e":11.0},"annotated_formula":"subtract(456, reminder(425897, 456))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|","chain":"425_897 % 456<\/gadget>\n449<\/output>\n456 - 449<\/gadget>\n7<\/output>\n7<\/result>","index":2042} +{"problem":"the speed at which a man can row a boat in still water is 9 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ?","rationale":"\"speed of the boat downstream = 9 + 3 = 12 kmph = 12 * 5 \/ 18 = 10 \/ 3 m \/ s hence time taken to cover 60 m = 60 * 3 \/ 10 = 18 seconds . answer : d\"","correct":"d","options":{"a":"16 seconds ","b":"34 seconds ","c":"14 seconds ","d":"18 seconds","e":"15 seconds"},"options_float":{"a":16.0,"b":34.0,"c":14.0,"d":18.0,"e":15.0},"annotated_formula":"divide(60, multiply(add(9, 3), const_0_2778))","linear_formula":"add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|","chain":"9 + 3<\/gadget>\n12<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n12 * (5\/18)<\/gadget>\n10\/3 = around 3.333333<\/output>\n60 \/ (10\/3)<\/gadget>\n18<\/output>\n18<\/result>","index":2043} +{"problem":"the roof of an apartment building is rectangular and its length is 3 times longer than its width . if the area of the roof is 588 feet squared , what is the difference between the length and the width of the roof ?","rationale":"\"answer is a : 28 let w be the width , so length is 3 w . therefore : w * 3 w = 588 , solving for , w = 14 , so 3 w - w = 2 w = 2 * 14 = 28\"","correct":"a","options":{"a":"28 . ","b":"40 . ","c":"42 . ","d":"44 .","e":"46 ."},"options_float":{"a":28.0,"b":40.0,"c":42.0,"d":44.0,"e":46.0},"annotated_formula":"subtract(multiply(sqrt(divide(588, 3)), 3), sqrt(divide(588, 3)))","linear_formula":"divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)|","chain":"588 \/ 3<\/gadget>\n196<\/output>\n196 ** (1\/2)<\/gadget>\n14<\/output>\n14 * 3<\/gadget>\n42<\/output>\n42 - 14<\/gadget>\n28<\/output>\n28<\/result>","index":2044} +{"problem":"how many numbers from 15 to 65 are exactly divisible by 5 ?","rationale":"\"option ' d ' 15 \/ 5 = 3 and 65 \/ 5 = 13 = = > 13 - 3 = 10 numbers\"","correct":"d","options":{"a":"5 ","b":"7 ","c":"9 ","d":"10","e":"12"},"options_float":{"a":5.0,"b":7.0,"c":9.0,"d":10.0,"e":12.0},"annotated_formula":"add(divide(subtract(multiply(floor(divide(65, 5)), 5), multiply(add(floor(divide(15, 5)), const_1), 5)), 5), const_1)","linear_formula":"divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)|","chain":"65 \/ 5<\/gadget>\n13<\/output>\nfloor(13)<\/gadget>\n13<\/output>\n13 * 5<\/gadget>\n65<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\nfloor(3)<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4 * 5<\/gadget>\n20<\/output>\n65 - 20<\/gadget>\n45<\/output>\n45 \/ 5<\/gadget>\n9<\/output>\n9 + 1<\/gadget>\n10<\/output>\n10<\/result>","index":2046} +{"problem":"two cards are drawn together from a pack of 52 cards the probability that one is the spade and one is a heart is ?","rationale":"\"let s be the sample space then n ( s ) = 52 c 2 = ( 52 * 51 ) \/ ( 2 * 1 ) = 1326 let e be the event of getting 1 spade kings and 1 heart let n ( e ) = number of ways of chossing 1 spade out of 13 and 1 heart out of 13 = ( 13 c 1 * 13 c 1 ) = 13 * 13 = 169 p ( e ) = n ( e ) \/ n ( s ) = 169 \/ 1326 = 13 \/ 102 answer ( d )\"","correct":"d","options":{"a":"85 \/ 852 ","b":"45 \/ 69 ","c":"18 \/ 456 ","d":"13 \/ 102","e":"23 \/ 125"},"options_float":{"a":0.0997652582,"b":0.652173913,"c":0.0394736842,"d":0.1274509804,"e":0.184},"annotated_formula":"multiply(divide(multiply(divide(52, const_4), divide(52, const_4)), multiply(52, 52)), const_2)","linear_formula":"divide(n0,const_4)|multiply(n0,n0)|multiply(#0,#0)|divide(#2,#1)|multiply(#3,const_2)|","chain":"52 \/ 4<\/gadget>\n13<\/output>\n13 * 13<\/gadget>\n169<\/output>\n52 * 52<\/gadget>\n2_704<\/output>\n169 \/ 2_704<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/16) * 2<\/gadget>\n1\/8 = around 0.125<\/output>\n1\/8 = around 0.125<\/result>","index":2048} +{"problem":"on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 \/ 3 rds of the transactions that anthony handled , and jade handled 15 more transactions than cal . how much transactions did jade handled ?","rationale":"solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 \/ 3 rds of the transactions than anthony handled cal = 2 \/ 3 × 99 = 66 jade handled 15 more transactions than cal . jade = 66 + 15 = 81 jade handled = 81 transactions . answer : b","correct":"b","options":{"a":"80 ","b":"81 ","c":"82 ","d":"83","e":"84"},"options_float":{"a":80.0,"b":81.0,"c":82.0,"d":83.0,"e":84.0},"annotated_formula":"add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), const_3), 15)","linear_formula":"add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,const_3)|add(n4,#4)","chain":"90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n10 + 100<\/gadget>\n110<\/output>\n(9\/10) * 110<\/gadget>\n99<\/output>\n99 * 2<\/gadget>\n198<\/output>\n198 \/ 3<\/gadget>\n66<\/output>\n66 + 15<\/gadget>\n81<\/output>\n81<\/result>","index":2049} +{"problem":"peter invests a sum of money and gets back an amount of $ 815 in 3 years . david invests an equal amount of money and gets an amount of $ 850 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ?","rationale":"\"since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 850 - 815 = 35 interest earned for 3 years = 35 * 3 = 105 amount invested = 815 - 105 = 710 answer : e\"","correct":"e","options":{"a":"670 ","b":"664 ","c":"698 ","d":"744","e":"710"},"options_float":{"a":670.0,"b":664.0,"c":698.0,"d":744.0,"e":710.0},"annotated_formula":"subtract(815, multiply(divide(subtract(850, 815), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100)))","linear_formula":"divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5)|","chain":"850 - 815<\/gadget>\n35<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n3 \/ 100<\/gadget>\n3\/100 = around 0.03<\/output>\n(1\/25) - (3\/100)<\/gadget>\n1\/100 = around 0.01<\/output>\n35 \/ (1\/100)<\/gadget>\n3_500<\/output>\n3_500 * (3\/100)<\/gadget>\n105<\/output>\n815 - 105<\/gadget>\n710<\/output>\n710<\/result>","index":2050} +{"problem":"the sector of a circle has perimeter 83 cm and central angle 225 o . find its radius ?","rationale":"let radius = x cm perimeter of the sector = length of the arc + 2 ( radius ) 83 = ( 225 \/ 360 * 2 * 22 \/ 7 * x ) + 2 ( x ) 83 = 55 x \/ 14 + 2 x 83 = 83 x \/ 14 1162 = 83 x x = 14 answer : c","correct":"c","options":{"a":"21 cm ","b":"15 cm ","c":"14 cm ","d":"24 cm","e":"28 cm"},"options_float":{"a":21.0,"b":15.0,"c":14.0,"d":24.0,"e":28.0},"annotated_formula":"divide(83, add(divide(multiply(multiply(const_2, divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), 225), divide(const_3600, const_10)), const_2))","linear_formula":"add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|add(#2,const_2)|divide(#3,#0)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,#1)|add(#7,const_2)|divide(n0,#8)","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 2<\/gadget>\n22<\/output>\n4 + 3<\/gadget>\n7<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n2 * (22\/7)<\/gadget>\n44\/7 = around 6.285714<\/output>\n(44\/7) * 225<\/gadget>\n9_900\/7 = around 1_414.285714<\/output>\n3_600 \/ 10<\/gadget>\n360<\/output>\n(9_900\/7) \/ 360<\/gadget>\n55\/14 = around 3.928571<\/output>\n(55\/14) + 2<\/gadget>\n83\/14 = around 5.928571<\/output>\n83 \/ (83\/14)<\/gadget>\n14<\/output>\n14<\/result>","index":2052} +{"problem":"at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses at the restaurant ? ( assume that all boxes are filled to capacity . )","rationale":"\"1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can just ` ` add 28 s ' ' until we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e\"","correct":"e","options":{"a":"96 ","b":"240 ","c":"56 ","d":"384","e":"480"},"options_float":{"a":96.0,"b":240.0,"c":56.0,"d":384.0,"e":480.0},"annotated_formula":"multiply(multiply(16, const_2), 15)","linear_formula":"multiply(n1,const_2)|multiply(n2,#0)|","chain":"16 * 2<\/gadget>\n32<\/output>\n32 * 15<\/gadget>\n480<\/output>\n480<\/result>","index":2054} +{"problem":"in a certain alphabet , 9 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ?","rationale":"\"we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line \/ no dot box = 0 ) . first we find the total # of letters with lines : 9 + 24 = 33 ; next , we find the total # of letters without line : 40 - 33 = 7 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 7 - 0 = 7 . a\"","correct":"a","options":{"a":"7 ","b":"8 ","c":"14 ","d":"20","e":"28"},"options_float":{"a":7.0,"b":8.0,"c":14.0,"d":20.0,"e":28.0},"annotated_formula":"subtract(40, add(9, 24))","linear_formula":"add(n0,n1)|subtract(n2,#0)|","chain":"9 + 24<\/gadget>\n33<\/output>\n40 - 33<\/gadget>\n7<\/output>\n7<\/result>","index":2055} +{"problem":"if a train , travelling at a speed of 90 kmph , crosses a pole in 4 sec , then the length of train is ?","rationale":"\"d = 90 * 5 \/ 18 * 4 = 100 m answer : d\"","correct":"d","options":{"a":"281 ","b":"125 ","c":"288 ","d":"100","e":"121"},"options_float":{"a":281.0,"b":125.0,"c":288.0,"d":100.0,"e":121.0},"annotated_formula":"multiply(multiply(90, const_0_2778), 4)","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n90 * (5\/18)<\/gadget>\n25<\/output>\n25 * 4<\/gadget>\n100<\/output>\n100<\/result>","index":2056} +{"problem":"john ' s bank ' s saving amount is decreased 20 % due to loan payment and current balance is rs . 24000 . find the actual balance before deduction ?","rationale":"\"20 % decreased 80 % balance = 24000 100 % = 24000 \/ 80 * 100 = 30000 answer : e\"","correct":"e","options":{"a":"8000 ","b":"8500 ","c":"9000 ","d":"9500","e":"30000"},"options_float":{"a":8000.0,"b":8500.0,"c":9000.0,"d":9500.0,"e":30000.0},"annotated_formula":"add(divide(multiply(24000, 20), const_100), 24000)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n1,#1)|","chain":"24_000 * 20<\/gadget>\n480_000<\/output>\n480_000 \/ 100<\/gadget>\n4_800<\/output>\n4_800 + 24_000<\/gadget>\n28_800<\/output>\n28_800<\/result>","index":2057} +{"problem":"two employees x and y are paid a total of rs . 700 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?","rationale":"\"let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 700 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 ∴ 12 y \/ 10 + y = 700 ⇒ y [ 12 \/ 10 + 1 ] = 700 ⇒ 22 y \/ 10 = 700 ⇒ 22 y = 7000 ⇒ y = 7000 \/ 22 = rs . 318.18 e )\"","correct":"e","options":{"a":"s . 200.18 ","b":"s . 240.18 ","c":"s . 290.18 ","d":"s . 300.18","e":"s . 318.18"},"options_float":{"a":200.18,"b":240.18,"c":290.18,"d":300.18,"e":318.18},"annotated_formula":"divide(multiply(700, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2))","linear_formula":"add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)|","chain":"1 + 4<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n700 * 10<\/gadget>\n7_000<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n7_000 \/ 22<\/gadget>\n3_500\/11 = around 318.181818<\/output>\n3_500\/11 = around 318.181818<\/result>","index":2059} +{"problem":"james took a 3 - hour bike ride . in the second hour he travlled 12 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ?","rationale":"\"let the distance travelled in the first hour be x . thus , 1.2 x = 12 , x = 10 . now , the distance travelled in the 3 rd hour = 12 + 1 \/ 4 â ˆ — 12 = 15 therefore 12 + 15 + 10 = 37 answer : d\"","correct":"d","options":{"a":"54.0 ","b":"54.9 ","c":"55.5 ","d":"37.0","e":"63.0"},"options_float":{"a":54.0,"b":54.9,"c":55.5,"d":37.0,"e":63.0},"annotated_formula":"add(add(divide(multiply(multiply(const_1, const_100), 12), add(multiply(const_1, const_100), 20)), divide(multiply(25, 12), 20)), 12)","linear_formula":"multiply(const_1,const_100)|multiply(n1,n3)|add(n2,#0)|divide(#1,n2)|multiply(n1,#0)|divide(#4,#2)|add(#5,#3)|add(n1,#6)|","chain":"1 * 100<\/gadget>\n100<\/output>\n100 * 12<\/gadget>\n1_200<\/output>\n100 + 20<\/gadget>\n120<\/output>\n1_200 \/ 120<\/gadget>\n10<\/output>\n25 * 12<\/gadget>\n300<\/output>\n300 \/ 20<\/gadget>\n15<\/output>\n10 + 15<\/gadget>\n25<\/output>\n25 + 12<\/gadget>\n37<\/output>\n37<\/result>","index":2060} +{"problem":"in a group of 100 cars , 47 cars do not have air conditioning . if at least 53 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ?","rationale":"\"lets assume ac = 53 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 53 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 53 . . we have to assign atleast 6 tocars with ac and racing stripes . hence ac = 53 - 6 = 47 . the answer is b\"","correct":"b","options":{"a":"45 ","b":"47 ","c":"49 ","d":"51","e":"53"},"options_float":{"a":45.0,"b":47.0,"c":49.0,"d":51.0,"e":53.0},"annotated_formula":"subtract(100, 53)","linear_formula":"subtract(n0,n2)|","chain":"100 - 53<\/gadget>\n47<\/output>\n47<\/result>","index":2061} +{"problem":"the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 600 resolutions ?","rationale":"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 600 resolutions . = 600 * 2 * 22 \/ 7 * 22.4 = 84480 cm = 844.8 m answer : a","correct":"a","options":{"a":"844.8 m ","b":"704 m ","c":"774 m ","d":"714 m","e":"744 m"},"options_float":{"a":844.8,"b":704.0,"c":774.0,"d":714.0,"e":744.0},"annotated_formula":"divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 600), const_100)","linear_formula":"add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21 + 1<\/gadget>\n22<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n(22\/7) * 22.4<\/gadget>\n70.4<\/output>\n70.4 * 2<\/gadget>\n140.8<\/output>\n140.8 * 600<\/gadget>\n84_480<\/output>\n84_480 \/ 100<\/gadget>\n4_224\/5 = around 844.8<\/output>\n4_224\/5 = around 844.8<\/result>","index":2064} +{"problem":"mr . depak has $ 449774 in his bank account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among her 6 best friends ? options :","rationale":"to find the least amount deepak should add to his saving account to split the money evenly among his 6 best friends , he needs to make the total divisible by 6 simply add the individual digits of the total = 4 + 4 + 9 + 7 + 7 + 4 = 35 if you add 1 , the number is divisible by 6 ( 35 + 1 ) correct option : a","correct":"a","options":{"a":"$ 1 ","b":"$ 2 ","c":"$ 3 ","d":"$ 4","e":"$ 6"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":6.0},"annotated_formula":"subtract(add(449774, const_1), 449774)","linear_formula":"add(n0,const_1)|subtract(#0,n0)","chain":"449_774 + 1<\/gadget>\n449_775<\/output>\n449_775 - 449_774<\/gadget>\n1<\/output>\n1<\/result>","index":2065} +{"problem":"the ages of two person differ by 12 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively","rationale":"\"let their ages be x and ( x + 12 ) years . then , 5 ( x - 5 ) = ( x + 12 - 5 ) = > 4 x = 32 = > x = 8 their present ages are 20 years and 8 year . answer : e\"","correct":"e","options":{"a":"30 , 10 ","b":"25 , 5 ","c":"29 , 9 ","d":"50 , 30","e":"20 , 8"},"options_float":{"a":30.0,"b":25.0,"c":29.0,"d":50.0,"e":20.0},"annotated_formula":"add(divide(add(multiply(5, 5), subtract(12, 5)), subtract(5, const_1)), 12)","linear_formula":"multiply(n1,n2)|subtract(n0,n1)|subtract(n2,const_1)|add(#0,#1)|divide(#3,#2)|add(n0,#4)|","chain":"5 * 5<\/gadget>\n25<\/output>\n12 - 5<\/gadget>\n7<\/output>\n25 + 7<\/gadget>\n32<\/output>\n5 - 1<\/gadget>\n4<\/output>\n32 \/ 4<\/gadget>\n8<\/output>\n8 + 12<\/gadget>\n20<\/output>\n20<\/result>","index":2066} +{"problem":"a cab driver 5 days income was $ 300 , $ 150 , $ 750 , $ 200 , $ 600 . then his average income is ?","rationale":"\"avg = sum of observations \/ number of observations avg income = ( 300 + 150 + 750 + 200 + 600 ) \/ 5 = 400 answer is c\"","correct":"c","options":{"a":"$ 350 ","b":"$ 375 ","c":"$ 400 ","d":"$ 425","e":"$ 450"},"options_float":{"a":350.0,"b":375.0,"c":400.0,"d":425.0,"e":450.0},"annotated_formula":"divide(add(add(add(add(300, 150), 750), 200), 600), 5)","linear_formula":"add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)|","chain":"300 + 150<\/gadget>\n450<\/output>\n450 + 750<\/gadget>\n1_200<\/output>\n1_200 + 200<\/gadget>\n1_400<\/output>\n1_400 + 600<\/gadget>\n2_000<\/output>\n2_000 \/ 5<\/gadget>\n400<\/output>\n400<\/result>","index":2067} +{"problem":"3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 4 men and 2 women will do that task .","rationale":"\"3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 4 m + 2 w = 10 w answer is 10 \/ 14 = 5 \/ 7 answer : a\"","correct":"a","options":{"a":"5 \/ 7 ","b":"3 \/ 10 ","c":"5 \/ 18 ","d":"5 \/ 16","e":"5 \/ 11"},"options_float":{"a":0.7142857143,"b":0.3,"c":0.2777777778,"d":0.3125,"e":0.4545454545},"annotated_formula":"divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 4), 2), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2))","linear_formula":"subtract(n1,n3)|subtract(n2,n0)|divide(#0,#1)|multiply(n4,#2)|multiply(n2,#2)|add(n5,#3)|add(n3,#4)|divide(#5,#6)|","chain":"8 - 2<\/gadget>\n6<\/output>\n6 - 3<\/gadget>\n3<\/output>\n6 \/ 3<\/gadget>\n2<\/output>\n2 * 4<\/gadget>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 + 2<\/gadget>\n14<\/output>\n10 \/ 14<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":2068} +{"problem":"a trained covered x km at 90 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km .","rationale":"total time taken = x \/ 90 + 2 x \/ 20 hours = x \/ 9 hours average speed = 3 x \/ ( x \/ 9 ) = 27 kmph answer : c","correct":"c","options":{"a":"22 ","b":"99 ","c":"27 ","d":"66","e":"887"},"options_float":{"a":22.0,"b":99.0,"c":27.0,"d":66.0,"e":887.0},"annotated_formula":"divide(multiply(90, const_3), add(divide(90, 90), divide(multiply(const_2, 90), 20)))","linear_formula":"divide(n0,n0)|multiply(n0,const_3)|multiply(n0,const_2)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)","chain":"90 * 3<\/gadget>\n270<\/output>\n90 \/ 90<\/gadget>\n1<\/output>\n2 * 90<\/gadget>\n180<\/output>\n180 \/ 20<\/gadget>\n9<\/output>\n1 + 9<\/gadget>\n10<\/output>\n270 \/ 10<\/gadget>\n27<\/output>\n27<\/result>","index":2069} +{"problem":"sachin is younger than rahul by 4 years . if the ratio of their ages is 7 : 9 , find the age of sachin","rationale":"\"if rahul age is x , then sachin age is x - 4 , so ( x - 4 ) \/ x = 7 \/ 9 = > 9 x - 36 = 7 x = > 2 x = 36 = > x = 18 so sachin age is 18 - 4 = 14 answer : e\"","correct":"e","options":{"a":"24.58 ","b":"24.5 ","c":"24.3 ","d":"24.9","e":"14"},"options_float":{"a":24.58,"b":24.5,"c":24.3,"d":24.9,"e":14.0},"annotated_formula":"multiply(divide(4, subtract(9, 7)), 7)","linear_formula":"subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)|","chain":"9 - 7<\/gadget>\n2<\/output>\n4 \/ 2<\/gadget>\n2<\/output>\n2 * 7<\/gadget>\n14<\/output>\n14<\/result>","index":2070} +{"problem":"a triangle and a parallelogram are constructed on the same base such that their areas are equal . if the altitude of the parallelogram is 100 m , then the altitude of the triangle is .","rationale":"solution let the altitude of the triangle be h ï and base of each be b . then , ( ½ × b × h 1 ) where h 2 = 100 m . = h 1 = 2 h 2 = ( 2 x 100 ) m = 200 m . answer d","correct":"d","options":{"a":"10 √ 2 m ","b":"100 m ","c":"100 √ 2 m ","d":"200 m","e":"none"},"options_float":{"a":10.0,"b":100.0,"c":100.0,"d":200.0,"e":null},"annotated_formula":"multiply(100, const_2)","linear_formula":"multiply(n0,const_2)","chain":"100 * 2<\/gadget>\n200<\/output>\n200<\/result>","index":2071} +{"problem":"a man buys an item at rs . 1200 and sells it at the loss of 20 percent . then what is the selling price of that item","rationale":"\"explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 1200 = 80 \/ 100 * 1200 = 960 option d\"","correct":"d","options":{"a":"rs . 660 ","b":"rs . 760 ","c":"rs . 860 ","d":"rs . 960","e":"none of these"},"options_float":{"a":660.0,"b":760.0,"c":860.0,"d":960.0,"e":null},"annotated_formula":"multiply(1200, subtract(const_1, divide(20, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|multiply(n0,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1_200 * (4\/5)<\/gadget>\n960<\/output>\n960<\/result>","index":2072} +{"problem":"a man rows his boat 110 km downstream and 90 km upstream , taking 3 hours each time . find the speed of the stream ?","rationale":"\"speed downstream = d \/ t = 110 \/ ( 3 ) = 37 kmph speed upstream = d \/ t = 90 \/ ( 3 ) = 30 kmph the speed of the stream = ( 37 - 30 ) \/ 2 = 3.5 kmph answer : d\"","correct":"d","options":{"a":"6 ","b":"5 ","c":"7.5 ","d":"3.5","e":"2.5"},"options_float":{"a":6.0,"b":5.0,"c":7.5,"d":3.5,"e":2.5},"annotated_formula":"divide(subtract(divide(110, 3), divide(90, 3)), const_2)","linear_formula":"divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)|","chain":"110 \/ 3<\/gadget>\n110\/3 = around 36.666667<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n(110\/3) - 30<\/gadget>\n20\/3 = around 6.666667<\/output>\n(20\/3) \/ 2<\/gadget>\n10\/3 = around 3.333333<\/output>\n10\/3 = around 3.333333<\/result>","index":2074} +{"problem":"five women can do a work in ten days . 7 men can complete the same work in 5 days . what is the ratio between the capacity of a man and a woman ?","rationale":"explanation : ( 50 ã — 10 ) women can complete the work in 1 day . â ˆ ´ 1 woman ' s 1 day ' s work = 1 \/ 50 ( 7 ã — 5 ) men can complete the work in 1 day . â ˆ ´ 1 man ' s 1 day ' s work = 1 \/ 35 so , required ratio = 1 \/ 50 : 1 \/ 35 = 10 : 7 answer : a","correct":"a","options":{"a":"10 : 7 ","b":"11 : 10 ","c":"2 : 3 ","d":"3 : 2","e":"none of these"},"options_float":{"a":1.4285714286,"b":1.1,"c":0.6666666667,"d":1.5,"e":null},"annotated_formula":"divide(divide(const_1, multiply(7, 5)), divide(const_1, multiply(5, const_10)))","linear_formula":"multiply(n0,n1)|multiply(n1,const_10)|divide(const_1,#0)|divide(const_1,#1)|divide(#2,#3)","chain":"7 * 5<\/gadget>\n35<\/output>\n1 \/ 35<\/gadget>\n1\/35 = around 0.028571<\/output>\n5 * 10<\/gadget>\n50<\/output>\n1 \/ 50<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/35) \/ (1\/50)<\/gadget>\n10\/7 = around 1.428571<\/output>\n10\/7 = around 1.428571<\/result>","index":2075} +{"problem":"a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 340 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ?","rationale":"\"orchestra seats - a balcony seats - b a + b = 340 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 340 = 3320 - 2720 = 600 i . e . a = 150 and b = 340 - 150 = 190 more seats in balcony than orchestra = b - a = 190 - 150 = 40 answer : option a\"","correct":"a","options":{"a":"40 ","b":"110 ","c":"120 ","d":"130","e":"220"},"options_float":{"a":40.0,"b":110.0,"c":120.0,"d":130.0,"e":220.0},"annotated_formula":"subtract(divide(subtract(multiply(12, 340), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(340, divide(subtract(multiply(12, 340), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8))))","linear_formula":"multiply(n0,n2)|multiply(const_1000,const_3)|multiply(const_100,const_3)|multiply(const_10,const_2)|subtract(n0,n1)|add(#1,#2)|add(#5,#3)|subtract(#0,#6)|divide(#7,#4)|subtract(n2,#8)|subtract(#8,#9)|","chain":"12 * 340<\/gadget>\n4_080<\/output>\n3 * 1_000<\/gadget>\n3_000<\/output>\n3 * 100<\/gadget>\n300<\/output>\n3_000 + 300<\/gadget>\n3_300<\/output>\n2 * 10<\/gadget>\n20<\/output>\n3_300 + 20<\/gadget>\n3_320<\/output>\n4_080 - 3_320<\/gadget>\n760<\/output>\n12 - 8<\/gadget>\n4<\/output>\n760 \/ 4<\/gadget>\n190<\/output>\n340 - 190<\/gadget>\n150<\/output>\n190 - 150<\/gadget>\n40<\/output>\n40<\/result>","index":2076} +{"problem":"right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 8 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions ?","rationale":"we have the rectangle with dimensions 9 * 8 ( 9 horizontal dots and 8 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 8 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot is already occupied by a ) ; choose the y coordinate for vertex b ( as x coordinate is fixed by a ) : 7 c 1 , ( 8 - 1 = 7 as 1 vertical dot is already occupied by a ) . 9 c 1 * 8 c * 8 c 1 * 7 c 1 = 4032 . answer : b .","correct":"b","options":{"a":"54 ","b":"4032 ","c":"2,160 ","d":"2,916","e":"148,824"},"options_float":{"a":54.0,"b":4032.0,"c":2160.0,"d":2916.0,"e":148824.0},"annotated_formula":"multiply(multiply(8, subtract(8, const_1)), multiply(9, 8))","linear_formula":"multiply(n0,n3)|subtract(n0,const_1)|multiply(n0,#1)|multiply(#2,#0)","chain":"8 - 1<\/gadget>\n7<\/output>\n8 * 7<\/gadget>\n56<\/output>\n9 * 8<\/gadget>\n72<\/output>\n56 * 72<\/gadget>\n4_032<\/output>\n4_032<\/result>","index":2077} +{"problem":"there are 8 stores in town that had a total of 22 visitors on a particular day . however , only 12 people went shopping that day ; some people visited more than one store . if 8 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visited ?","rationale":"8 people visited 2 stores each for 16 visits . to maximize the number of stores that one person visited , let ' s assume that 3 people visited 1 store each . the number of remaining visits is 22 - 16 - 3 = 3 , which is the maximum that one person could have visited . the answer is b .","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"subtract(subtract(22, multiply(8, const_2)), subtract(subtract(12, 8), const_1))","linear_formula":"multiply(n0,const_2)|subtract(n2,n0)|subtract(n1,#0)|subtract(#1,const_1)|subtract(#2,#3)","chain":"8 * 2<\/gadget>\n16<\/output>\n22 - 16<\/gadget>\n6<\/output>\n12 - 8<\/gadget>\n4<\/output>\n4 - 1<\/gadget>\n3<\/output>\n6 - 3<\/gadget>\n3<\/output>\n3<\/result>","index":2078} +{"problem":"it takes avery 3 hours to build a brick wall while tom can do it in 3 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ?","rationale":"\"avery takes 3 hours tom takes 3 hours efficiency of avery is 1 \/ 3 units \/ hr efficiency of tom is 1 \/ 3 units \/ hr combined efficiency of tom and avery is 1 \/ 3 + 1 \/ 3 = 2 \/ 3 units \/ hr since they worked for 1 hour they completed 2 \/ 3 units of work and 1 \/ 3 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 1 \/ 3 \/ 1 \/ 3 hours = > 1 * 60 = 60 minutes . . . answer will be ( b )\"","correct":"b","options":{"a":"20 ","b":"60 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":20.0,"b":60.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"multiply(divide(subtract(const_1, add(divide(const_1, 3), divide(const_1, 3))), divide(const_1, 3)), const_60)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#1)|multiply(#4,const_60)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) + (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) \/ (1\/3)<\/gadget>\n1<\/output>\n1 * 60<\/gadget>\n60<\/output>\n60<\/result>","index":2080} +{"problem":"a student got 70 % in one subject , 80 % in other . to get overall 75 % how much he should get in third subject .","rationale":"already average of two subjects is ( 70 + 80 ) \/ 2 = 75 % to maintain the percentage at 75 % , he shouls get 75 % marks in third subject answer : b","correct":"b","options":{"a":"65 % ","b":"75 % ","c":"85 % ","d":"95 %","e":"88 %"},"options_float":{"a":65.0,"b":75.0,"c":85.0,"d":95.0,"e":88.0},"annotated_formula":"divide(add(add(70, 80), 75), const_3)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(#1,const_3)","chain":"70 + 80<\/gadget>\n150<\/output>\n150 + 75<\/gadget>\n225<\/output>\n225 \/ 3<\/gadget>\n75<\/output>\n75<\/result>","index":2081} +{"problem":"4 men and 3 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other )","rationale":"the list should be wmwmw . hence , from women 3 ! and men 4 ! , we get ( 3 ! ) ( 4 ! ) = 144 . therefore , the correct answer is b .","correct":"b","options":{"a":"12 ","b":"144 ","c":"18 ","d":"21","e":"24"},"options_float":{"a":12.0,"b":144.0,"c":18.0,"d":21.0,"e":24.0},"annotated_formula":"multiply(factorial(4), factorial(3))","linear_formula":"factorial(n0)|factorial(n1)|multiply(#0,#1)|","chain":"factorial(4)<\/gadget>\n24<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n24 * 6<\/gadget>\n144<\/output>\n144<\/result>","index":2082} +{"problem":"in a class , 10 students like to play basketball and 8 like to play cricket . 4 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ?","rationale":"\"draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 10 + 8 - 4 = 14 e )\"","correct":"e","options":{"a":"12 ","b":"15 ","c":"16 ","d":"18","e":"14"},"options_float":{"a":12.0,"b":15.0,"c":16.0,"d":18.0,"e":14.0},"annotated_formula":"subtract(add(10, 8), 4)","linear_formula":"add(n0,n1)|subtract(#0,n2)|","chain":"10 + 8<\/gadget>\n18<\/output>\n18 - 4<\/gadget>\n14<\/output>\n14<\/result>","index":2083} +{"problem":"a box contains 9 apples , 8 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ?","rationale":"\"p ( choosing a red apple 5 times in a row ) = 8 \/ 9 * 7 \/ 8 * 6 \/ 7 * 5 \/ 6 * 4 \/ 5 = 4 \/ 9 < 0.5 the answer is c .\"","correct":"c","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"add(multiply(0.5, 8), const_1)","linear_formula":"multiply(n2,n1)|add(#0,const_1)|","chain":"0.5 * 8<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5<\/result>","index":2084} +{"problem":"the length of the bridge , which a train 130 metres long and travelling at 45 km \/ hr can cross in 30 seconds , is","rationale":"\"sol . speed = [ 45 x 5 \/ 18 ] m \/ sec = ( 25 \/ 2 ) m \/ sec ; time = 30 sec . let the length of bridge be x metres . then , 130 + x \/ 30 = 25 \/ 2 ⇔ 2 ( 130 + x ) = 750 ⇔ x = 245 m . answer c\"","correct":"c","options":{"a":"200 m ","b":"225 m ","c":"245 m ","d":"250 m","e":"none"},"options_float":{"a":200.0,"b":225.0,"c":245.0,"d":250.0,"e":null},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 130<\/gadget>\n245<\/output>\n245<\/result>","index":2088} +{"problem":"the sum of the even numbers between 1 and n is 89 * 90 , where n is an odd number , then n = ?","rationale":"\"let n - 1 = 2 a . 2 + 4 + . . . + 2 a = 2 * ( 1 + 2 + . . . + a ) = 2 * ( a ) ( a + 1 ) \/ 2 = ( a ) ( a + 1 ) = 89 * 90 then a = 89 and n = 179 . the answer is c .\"","correct":"c","options":{"a":"167 ","b":"173 ","c":"179 ","d":"185","e":"193"},"options_float":{"a":167.0,"b":173.0,"c":179.0,"d":185.0,"e":193.0},"annotated_formula":"add(multiply(89, const_2), 1)","linear_formula":"multiply(n1,const_2)|add(#0,n0)|","chain":"89 * 2<\/gadget>\n178<\/output>\n178 + 1<\/gadget>\n179<\/output>\n179<\/result>","index":2089} +{"problem":"a person distributed 10 % of his income to his 2 children each . he deposited 20 % of his income to his wife ' s account . he donated 10 % of remaining amount to an orphan house . finally he has $ 500 . find his total income ?","rationale":"\"2 children got = 2 * 10 % = 20 % wife got = 20 % orphan house = 10 % total = 20 + 20 + 10 = 50 % remaining = 100 - 50 = 50 % 50 % = 500 100 % = 500 * 100 \/ 50 = $ 1000 answer is b\"","correct":"b","options":{"a":"$ 900 ","b":"$ 1000 ","c":"$ 1200 ","d":"$ 1150","e":"$ 1500"},"options_float":{"a":900.0,"b":1000.0,"c":1200.0,"d":1150.0,"e":1500.0},"annotated_formula":"multiply(divide(500, subtract(const_100, add(add(multiply(10, 2), 20), 10))), const_100)","linear_formula":"multiply(n0,n1)|add(n2,#0)|add(n3,#1)|subtract(const_100,#2)|divide(n4,#3)|multiply(#4,const_100)|","chain":"10 * 2<\/gadget>\n20<\/output>\n20 + 20<\/gadget>\n40<\/output>\n40 + 10<\/gadget>\n50<\/output>\n100 - 50<\/gadget>\n50<\/output>\n500 \/ 50<\/gadget>\n10<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":2091} +{"problem":"the average weight of 8 persons increases by 5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ?","rationale":"\"total weight increased = ( 8 x 5 ) kg = 40 kg . weight of new person = ( 65 + 40 ) kg = 105 kg . answer : e\"","correct":"e","options":{"a":"75 kg ","b":"65 kg ","c":"55 kg ","d":"85 kg","e":"105 kg"},"options_float":{"a":75.0,"b":65.0,"c":55.0,"d":85.0,"e":105.0},"annotated_formula":"add(65, multiply(8, 5))","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"8 * 5<\/gadget>\n40<\/output>\n65 + 40<\/gadget>\n105<\/output>\n105<\/result>","index":2093} +{"problem":"two cubes of their volumes in the ratio 64 : 343 . the ratio of their surface area is :","rationale":"\"the ratio of their surface area is 64 : 343 4 : 7 answer is e .\"","correct":"e","options":{"a":"1.5 : 5 ","b":"2 : 5 ","c":"3 : 5 ","d":"1 : 5","e":"4 : 7"},"options_float":{"a":0.3,"b":0.4,"c":0.6,"d":0.2,"e":0.5714285714},"annotated_formula":"divide(power(64, const_0_33), power(343, const_0_33))","linear_formula":"power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n64 ** (1\/3)<\/gadget>\n4<\/output>\n343 ** (1\/3)<\/gadget>\n7<\/output>\n4 \/ 7<\/gadget>\n4\/7 = around 0.571429<\/output>\n4\/7 = around 0.571429<\/result>","index":2094} +{"problem":"if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 28 , then n =","rationale":"\"2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 28 = > 4 x 2 ^ 2 n = 4 ^ 28 = 2 ^ 56 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 56 = > 2 ^ ( 2 n + 2 ) = 2 ^ 56 = > 2 n + 2 = 56 = > n = 27 so . answer will be d .\"","correct":"d","options":{"a":"3 ","b":"6 ","c":"12 ","d":"27","e":"24"},"options_float":{"a":3.0,"b":6.0,"c":12.0,"d":27.0,"e":24.0},"annotated_formula":"divide(subtract(multiply(28, 2), 2), 2)","linear_formula":"multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)|","chain":"28 * 2<\/gadget>\n56<\/output>\n56 - 2<\/gadget>\n54<\/output>\n54 \/ 2<\/gadget>\n27<\/output>\n27<\/result>","index":2095} +{"problem":"a welder received an order to make a 1 million liter cube - shaped tank . if he has only 4 x 2 meter sheets of metal that can be cut , how many metal sheets will be required for this order y ? ( 1 cubic meter = 1000 liters )","rationale":"i get 75 . a cube with 1 million liters cube would be a cube with the dimensions of 100 * 100 * 100 . 4 * 2 covers 8 sq liters so 100 \/ 8 = 12.5 . y = 12.5 * 6 = 75 . e","correct":"e","options":{"a":"92 ","b":"90 ","c":"82 ","d":"78","e":"75"},"options_float":{"a":92.0,"b":90.0,"c":82.0,"d":78.0,"e":75.0},"annotated_formula":"divide(divide(multiply(multiply(power(1000, divide(1, const_3)), power(1000, divide(1, const_3))), const_12), const_2), multiply(4, 2))","linear_formula":"divide(n0,const_3)|multiply(n1,n2)|power(n4,#0)|multiply(#2,#2)|multiply(#3,const_12)|divide(#4,const_2)|divide(#5,#1)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1_000 ** (1\/3)<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n100 * 12<\/gadget>\n1_200<\/output>\n1_200 \/ 2<\/gadget>\n600<\/output>\n4 * 2<\/gadget>\n8<\/output>\n600 \/ 8<\/gadget>\n75<\/output>\n75<\/result>","index":2096} +{"problem":"if a \/ b = 4 \/ 3 , then ( 3 a + 2 b ) \/ ( 3 a - 2 b ) = ?","rationale":"answer dividing numerator as well as denominator by b , we get given exp . = ( 3 a + 2 b ) \/ ( 3 a - 2 b ) = ( 3 a \/ b + 2 ) \/ ( 3 a \/ b - 2 ) = [ ( 3 x 4 ) \/ 3 + 2 ] \/ [ ( 3 x 4 ) \/ 3 - 2 ] = ( 4 + 2 ) \/ ( 4 - 2 ) = 3 option : b","correct":"b","options":{"a":"6 ","b":"3 ","c":"5 ","d":"- 1","e":"none"},"options_float":{"a":6.0,"b":3.0,"c":5.0,"d":-1.0,"e":null},"annotated_formula":"divide(add(multiply(3, 4), multiply(2, 3)), subtract(multiply(3, 4), multiply(2, 3)))","linear_formula":"multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)|subtract(#0,#1)|divide(#2,#3)","chain":"3 * 4<\/gadget>\n12<\/output>\n2 * 3<\/gadget>\n6<\/output>\n12 + 6<\/gadget>\n18<\/output>\n12 - 6<\/gadget>\n6<\/output>\n18 \/ 6<\/gadget>\n3<\/output>\n3<\/result>","index":2097} +{"problem":"a man spend 1 \/ 5 of his salary on food , 1 \/ 10 of his salary on house rent and 3 \/ 5 salary on clothes . he still has $ 15000 left with him . find salary . .","rationale":"[ 1 \/ ( x 1 \/ y 1 + x 2 \/ y 2 + x 3 \/ y 3 ) ] * total amount = balance amount [ 1 - ( 1 \/ 5 + 1 \/ 10 + 3 \/ 5 ) } * total salary = $ 15000 , = [ 1 - 9 \/ 10 ] * total salary = $ 15000 , total salary = $ 15000 * 10 = $ 150000 , correct answer ( a )","correct":"a","options":{"a":"$ 150000 ","b":"$ 18000 ","c":"$ 180000 ","d":"$ 1800","e":"none"},"options_float":{"a":150000.0,"b":18000.0,"c":180000.0,"d":1800.0,"e":null},"annotated_formula":"divide(15000, subtract(const_1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5))))","linear_formula":"divide(n0,n1)|divide(n0,n3)|divide(n4,n1)|add(#0,#1)|add(#3,#2)|subtract(const_1,#4)|divide(n6,#5)","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/5) + (1\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/10) + (3\/5)<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n15_000 \/ (1\/10)<\/gadget>\n150_000<\/output>\n150_000<\/result>","index":2100} +{"problem":"how many cuboids of length 5 m , width 3 m and height 2 m can be farmed from a cuboid of 12 m length , 14 m width and 10 m height .","rationale":"\"( 12 ã — 14 ã — 10 ) \/ ( 5 ã — 3 ã — 2 ) = 56 answer is b .\"","correct":"b","options":{"a":"82 ","b":"56 ","c":"80 ","d":"56","e":"100"},"options_float":{"a":82.0,"b":56.0,"c":80.0,"d":56.0,"e":100.0},"annotated_formula":"divide(multiply(multiply(12, 14), 10), multiply(multiply(5, 3), 2))","linear_formula":"multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)|","chain":"12 * 14<\/gadget>\n168<\/output>\n168 * 10<\/gadget>\n1_680<\/output>\n5 * 3<\/gadget>\n15<\/output>\n15 * 2<\/gadget>\n30<\/output>\n1_680 \/ 30<\/gadget>\n56<\/output>\n56<\/result>","index":2102} +{"problem":"the length of the bridge , which a train 160 m long and traveling at 45 km \/ hr can cross in 30 sec is ?","rationale":"\"speed = 45 * 5 \/ 18 = 25 \/ 2 m \/ sec . time = 30 sec let the length of bridge be x meters . then , ( 160 + x ) \/ 30 = 25 \/ 2 x = 215 m . answer : d\"","correct":"d","options":{"a":"377 ","b":"367 ","c":"237 ","d":"215","e":"267"},"options_float":{"a":377.0,"b":367.0,"c":237.0,"d":215.0,"e":267.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 160<\/gadget>\n215<\/output>\n215<\/result>","index":2104} +{"problem":"car dealership x , the total profit from sales increased by 30 percent over the previous years , while the number of cars sold decreased by 30 percent over the previous years . approximately what was the average percent increase in profit per car over the previous year .","rationale":"let profit be x and cars sold be y originally profit \/ car = x \/ y now it is 1.3 x \/ 0.7 y = 13 \/ 7 ( x \/ y ) increase in profit per car = ( ( 6 \/ 7 ) ( x \/ y ) ) \/ ( x \/ y ) * 100 = 85.71 % answer = e","correct":"e","options":{"a":"18 % ","b":"30 % ","c":"52 % ","d":"63 %","e":"85.71 %"},"options_float":{"a":18.0,"b":30.0,"c":52.0,"d":63.0,"e":85.71},"annotated_formula":"multiply(divide(subtract(divide(add(const_100, 30), subtract(const_10, const_3)), divide(const_100, const_10)), divide(const_100, const_10)), const_100)","linear_formula":"add(n0,const_100)|divide(const_100,const_10)|subtract(const_10,const_3)|divide(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)","chain":"100 + 30<\/gadget>\n130<\/output>\n10 - 3<\/gadget>\n7<\/output>\n130 \/ 7<\/gadget>\n130\/7 = around 18.571429<\/output>\n100 \/ 10<\/gadget>\n10<\/output>\n(130\/7) - 10<\/gadget>\n60\/7 = around 8.571429<\/output>\n(60\/7) \/ 10<\/gadget>\n6\/7 = around 0.857143<\/output>\n(6\/7) * 100<\/gadget>\n600\/7 = around 85.714286<\/output>\n600\/7 = around 85.714286<\/result>","index":2105} +{"problem":"arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 30 days .","rationale":"\"they together completed 4 \/ 10 work in 4 days . balance 6 \/ 10 work will be completed by arun alone in 30 * 6 \/ 10 = 18 days . answer : c\"","correct":"c","options":{"a":"16 days . ","b":"17 days . ","c":"18 days . ","d":"19 days .","e":"20 days ."},"options_float":{"a":16.0,"b":17.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"subtract(30, multiply(divide(30, 10), 4))","linear_formula":"divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)|","chain":"30 \/ 10<\/gadget>\n3<\/output>\n3 * 4<\/gadget>\n12<\/output>\n30 - 12<\/gadget>\n18<\/output>\n18<\/result>","index":2106} +{"problem":"10 men and 15 women together can complete a work in 8 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ?","rationale":"\"1 man ' s 1 day work = 1 \/ 100 ( 10 men + 15 women ) ' s 1 day work = 1 \/ 8 15 women ' s 1 day work = ( 1 \/ 8 - 10 \/ 100 ) = 1 \/ 40 1 woman ' s 1 day work = 1 \/ 600 1 woman alone can complete the work in 600 days . answer : e\"","correct":"e","options":{"a":"127 days ","b":"667 days ","c":"177 days ","d":"187 days","e":"600 days"},"options_float":{"a":127.0,"b":667.0,"c":177.0,"d":187.0,"e":600.0},"annotated_formula":"multiply(divide(multiply(const_1, 100), subtract(multiply(const_1, 100), multiply(10, 8))), multiply(15, 8))","linear_formula":"multiply(n3,const_1)|multiply(n0,n2)|multiply(n1,n2)|subtract(#0,#1)|divide(#0,#3)|multiply(#4,#2)|","chain":"1 * 100<\/gadget>\n100<\/output>\n10 * 8<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n100 \/ 20<\/gadget>\n5<\/output>\n15 * 8<\/gadget>\n120<\/output>\n5 * 120<\/gadget>\n600<\/output>\n600<\/result>","index":2107} +{"problem":"if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 50 , what is the compound interest on the same sum at the rate and for the same time ?","rationale":"\"sum = ( 50 * 100 ) \/ ( 2 * 5 ) = rs . 500 amount = [ 500 * ( 1 + 5 \/ 100 ) 2 ] = rs . 551.25 c . i . = ( 551.25 - 500 ) = rs . 51.25 . answer : a\"","correct":"a","options":{"a":"rs . 51.25 ","b":"rs . 51.27 ","c":"rs . 51.79 ","d":"rs . 51.8","e":"rs . 51.93"},"options_float":{"a":51.25,"b":51.27,"c":51.79,"d":51.8,"e":51.93},"annotated_formula":"subtract(add(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), 5), const_100)), divide(50, multiply(divide(5, const_100), 2)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 2<\/gadget>\n1\/10 = around 0.1<\/output>\n50 \/ (1\/10)<\/gadget>\n500<\/output>\n500 * 5<\/gadget>\n2_500<\/output>\n2_500 \/ 100<\/gadget>\n25<\/output>\n25 + 500<\/gadget>\n525<\/output>\n525 * 5<\/gadget>\n2_625<\/output>\n2_625 \/ 100<\/gadget>\n105\/4 = around 26.25<\/output>\n525 + (105\/4)<\/gadget>\n2_205\/4 = around 551.25<\/output>\n(2_205\/4) - 500<\/gadget>\n205\/4 = around 51.25<\/output>\n205\/4 = around 51.25<\/result>","index":2110} +{"problem":"the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 6.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 \/ 2 ) - x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ?","rationale":"\"i tried using time \/ rate approach : - initial price difference = 6.80 - 3.20 = 3.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled so , 3.6 = t ( 5.4 x ) - > t = ( 3.6 ) \/ 5.4 x final price = 3.20 + 5 x * t - > 3.20 + 5 * 3.6 \/ 5.4 = 6.5 answer e .\"","correct":"e","options":{"a":"$ 4.50 ","b":"$ 5.10 ","c":"$ 5.30 ","d":"$ 5.50","e":"$ 6.50"},"options_float":{"a":4.5,"b":5.1,"c":5.3,"d":5.5,"e":6.5},"annotated_formula":"add(3.20, multiply(divide(subtract(6.80, 3.20), add(5, subtract(sqrt(2), 1))), 5))","linear_formula":"sqrt(n3)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5)|","chain":"6.8 - 3.2<\/gadget>\n3.6<\/output>\n2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n(sqrt(2)) - 1<\/gadget>\n-1 + sqrt(2) = around 0.414214<\/output>\n5 + (-1 + sqrt(2))<\/gadget>\nsqrt(2) + 4 = around 5.414214<\/output>\n3.6 \/ (sqrt(2) + 4)<\/gadget>\n3.6\/(sqrt(2) + 4) = around 0.664917<\/output>\n(3.6\/(sqrt(2) + 4)) * 5<\/gadget>\n18.0\/(sqrt(2) + 4) = around 3.324583<\/output>\n3.2 + (18.0\/(sqrt(2) + 4))<\/gadget>\n3.2 + 18.0\/(sqrt(2) + 4) = around 6.524583<\/output>\n3.2 + 18.0\/(sqrt(2) + 4) = around 6.524583<\/result>","index":2111} +{"problem":"a shopkeeper has 280 kg of apples . he sells 40 % of these at 30 % profit and remaining 60 % at 30 % profit . find his % profit on total .","rationale":"\"if the total quantity was 100 then 40 x 30 % + 60 x 30 % = 30 this profit will remain same for any total quantity unless the % of products remains the same . hence ' e ' is the answer\"","correct":"e","options":{"a":"24 % ","b":"25 % ","c":"26 % ","d":"28 %","e":"30 %"},"options_float":{"a":24.0,"b":25.0,"c":26.0,"d":28.0,"e":30.0},"annotated_formula":"divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 30), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280)","linear_formula":"add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)|","chain":"280 * 40<\/gadget>\n11_200<\/output>\n11_200 \/ 100<\/gadget>\n112<\/output>\n100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n112 * (13\/10)<\/gadget>\n728\/5 = around 145.6<\/output>\n280 * 60<\/gadget>\n16_800<\/output>\n16_800 \/ 100<\/gadget>\n168<\/output>\n168 * (13\/10)<\/gadget>\n1_092\/5 = around 218.4<\/output>\n(728\/5) + (1_092\/5)<\/gadget>\n364<\/output>\n364 - 280<\/gadget>\n84<\/output>\n84 * 100<\/gadget>\n8_400<\/output>\n8_400 \/ 280<\/gadget>\n30<\/output>\n30<\/result>","index":2114} +{"problem":"the speed of a boat in still water in 42 km \/ hr and the rate of current is 4 km \/ hr . the distance travelled downstream in 44 minutes is :","rationale":"\"speed downstream = ( 42 + 4 ) = 46 kmph time = 44 minutes = 44 \/ 60 hour = 11 \/ 15 hour distance travelled = time × speed = 11 \/ 15 × 46 = 33.7 km answer : c\"","correct":"c","options":{"a":"86.6 km ","b":"46.6 km ","c":"33.7 km ","d":"35.6 km","e":"26.6 km"},"options_float":{"a":86.6,"b":46.6,"c":33.7,"d":35.6,"e":26.6},"annotated_formula":"multiply(add(42, 4), divide(44, const_60))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"42 + 4<\/gadget>\n46<\/output>\n44 \/ 60<\/gadget>\n11\/15 = around 0.733333<\/output>\n46 * (11\/15)<\/gadget>\n506\/15 = around 33.733333<\/output>\n506\/15 = around 33.733333<\/result>","index":2115} +{"problem":"a rectangular field has a length 10 meters more than it is width . if the area of the field is 119 , what is the length ( in meters ) of the rectangular field ?","rationale":"\"area = l * w = ( l ) * ( l - 10 ) = 171 trial and error : 20 * 10 = 200 ( too high ) 19 * 9 = 171 ( too high ) 18 * 8 = 144 ( too high ) 17 * 7 = 119 the length is 17 meters . the answer is c .\"","correct":"c","options":{"a":"13 ","b":"15 ","c":"17 ","d":"19","e":"21"},"options_float":{"a":13.0,"b":15.0,"c":17.0,"d":19.0,"e":21.0},"annotated_formula":"add(10, add(const_0_25, add(const_0_33, divide(divide(119, 10), const_2))))","linear_formula":"divide(n1,n0)|divide(#0,const_2)|add(#1,const_0_33)|add(#2,const_0_25)|add(n0,#3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n119 \/ 10<\/gadget>\n119\/10 = around 11.9<\/output>\n(119\/10) \/ 2<\/gadget>\n119\/20 = around 5.95<\/output>\n(1\/3) + (119\/20)<\/gadget>\n377\/60 = around 6.283333<\/output>\n(1\/4) + (377\/60)<\/gadget>\n98\/15 = around 6.533333<\/output>\n10 + (98\/15)<\/gadget>\n248\/15 = around 16.533333<\/output>\n248\/15 = around 16.533333<\/result>","index":2117} +{"problem":"in country z , 12 % of the people do not have a university diploma but have the job of their choice , and 25 % of the people who do not have the job of their choice have a university diploma . if 40 % of the people have the job of their choice , what percent of the people have a university diploma ?","rationale":"\"setting up a matrix is how i solve this one . diploma no diploma totals job of choice w \/ diploma job of choice w \/ o diploma = 12 % job of choice total = 40 % not job of choice with diploma = . 25 x not job of choice w \/ o diploma = . 75 x total not job of choice = x total with diploma total without diploma total citizen = 100 if 40 % of people have their job of choice , then 60 % of people do not have their job of choice . 25 % of 60 % = 15 % . we can also see that 28 % of the people have their job of choice and a diploma ( 40 % - 12 % = 28 % ) . 28 % + 15 % = 43 % . therefore 43 % of the people in country z have a diploma . ans b\"","correct":"b","options":{"a":"35 % ","b":"43 % ","c":"55 % ","d":"65 %","e":"75 %"},"options_float":{"a":35.0,"b":43.0,"c":55.0,"d":65.0,"e":75.0},"annotated_formula":"add(divide(multiply(25, subtract(const_100, 40)), const_100), subtract(40, 12))","linear_formula":"subtract(const_100,n2)|subtract(n2,n0)|multiply(n1,#0)|divide(#2,const_100)|add(#3,#1)|","chain":"100 - 40<\/gadget>\n60<\/output>\n25 * 60<\/gadget>\n1_500<\/output>\n1_500 \/ 100<\/gadget>\n15<\/output>\n40 - 12<\/gadget>\n28<\/output>\n15 + 28<\/gadget>\n43<\/output>\n43<\/result>","index":2118} +{"problem":"in a village of 2,500 people , 800 people are over 70 years old and 850 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a male or younger than 70 years old ?","rationale":"\"to solve this we would need to know the number of people who are less than 70 years old , the number of males , and the number of males who are less than 70 years old . overlapping sets : quantity of a or b = a + b - ( a ∩ b ) = quantity of a + quantity of b - ( intersection of a and b ) number of males = 2500 - 850 = 1650 number of people below 70 years old = 2500 - 800 = 1700 number of males below 70 years old = 1700 - ( 850 * 0.4 ) = 1360 total number of people who are male or below 70 = 1650 + 1700 - 1360 = 1990 probability of male or below 70 = 1990 \/ 2500 = 199 \/ 250 answer : b\"","correct":"b","options":{"a":"221 \/ 250 ","b":"199 \/ 250 ","c":"33 \/ 50 ","d":"8 \/ 25","e":"51 \/ 250"},"options_float":{"a":0.884,"b":0.796,"c":0.66,"d":0.32,"e":0.204},"annotated_formula":"divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(850, subtract(const_1, divide(40, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25)))","linear_formula":"multiply(const_5,const_5)|divide(#0,const_0_25)|divide(n4,#1)|multiply(#1,const_26)|subtract(const_1,#2)|multiply(n3,#4)|subtract(#3,#5)|divide(#6,#3)|","chain":"5 * 5<\/gadget>\n25<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n25 \/ (1\/4)<\/gadget>\n100<\/output>\n26 * 100<\/gadget>\n2_600<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n850 * (3\/5)<\/gadget>\n510<\/output>\n2_600 - 510<\/gadget>\n2_090<\/output>\n2_090 \/ 2_600<\/gadget>\n209\/260 = around 0.803846<\/output>\n209\/260 = around 0.803846<\/result>","index":2119} +{"problem":"in 10 years , a will be twice as old as b was 10 years ago . if a is now 5 years older than b the present age of b is","rationale":"\"let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 5 = > 2 b - b - 5 = 30 b = 35 so the present age of b is 35 years answer : d\"","correct":"d","options":{"a":"37 ","b":"38 ","c":"39 ","d":"35","e":"41"},"options_float":{"a":37.0,"b":38.0,"c":39.0,"d":35.0,"e":41.0},"annotated_formula":"add(add(multiply(const_2, 10), 10), 5)","linear_formula":"multiply(n0,const_2)|add(n0,#0)|add(n2,#1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 10<\/gadget>\n30<\/output>\n30 + 5<\/gadget>\n35<\/output>\n35<\/result>","index":2122} +{"problem":"a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many ways can the test be completed if every question is unanswered ?","rationale":"\"5 choices for each of the 4 questions , thus total of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways . answer : c .\"","correct":"c","options":{"a":"24 ","b":"120 ","c":"625 ","d":"720","e":"1024"},"options_float":{"a":24.0,"b":120.0,"c":625.0,"d":720.0,"e":1024.0},"annotated_formula":"power(5, 4)","linear_formula":"power(n1,n0)|","chain":"5 ** 4<\/gadget>\n625<\/output>\n625<\/result>","index":2123} +{"problem":"the total number of digits used in numbering the pages of a book having 369 pages is","rationale":"total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 270 ) = ( 9 + 180 + 810 ) = 999 . answer : c","correct":"c","options":{"a":"732 ","b":"990 ","c":"999 ","d":"1305","e":"1405"},"options_float":{"a":732.0,"b":990.0,"c":999.0,"d":1305.0,"e":1405.0},"annotated_formula":"subtract(subtract(multiply(369, const_3), subtract(const_100, const_1)), subtract(const_10, const_1))","linear_formula":"multiply(n0,const_3)|subtract(const_100,const_1)|subtract(const_10,const_1)|subtract(#0,#1)|subtract(#3,#2)","chain":"369 * 3<\/gadget>\n1_107<\/output>\n100 - 1<\/gadget>\n99<\/output>\n1_107 - 99<\/gadget>\n1_008<\/output>\n10 - 1<\/gadget>\n9<\/output>\n1_008 - 9<\/gadget>\n999<\/output>\n999<\/result>","index":2129} +{"problem":"when 732 is subtracted form the square of a number , the answer that is obtained is 5280 . what is the number ?","rationale":"explanation : let the number be x x 2 - 732 = 5280 x 2 = 5280 + ( 70 + 3 ) 2 = 5280 + 4900 + 420 + 9 = 10609 = 10000 + 2 ( 100 ) ( 3 ) + 32 = ( 100 + 3 ) 2 x = 100 + 3 = 103 . answer : b","correct":"b","options":{"a":"228 ","b":"103 ","c":"286 ","d":"122","e":"102"},"options_float":{"a":228.0,"b":103.0,"c":286.0,"d":122.0,"e":102.0},"annotated_formula":"subtract(multiply(sqrt(732), const_4), const_4)","linear_formula":"sqrt(n0)|multiply(#0,const_4)|subtract(#1,const_4)","chain":"732 ** (1\/2)<\/gadget>\n2*sqrt(183) = around 27.055499<\/output>\n(2*sqrt(183)) * 4<\/gadget>\n8*sqrt(183) = around 108.221994<\/output>\n(8*sqrt(183)) - 4<\/gadget>\n-4 + 8*sqrt(183) = around 104.221994<\/output>\n-4 + 8*sqrt(183) = around 104.221994<\/result>","index":2130} +{"problem":"the average height of 35 boys in a class was calculated as 182 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ?","rationale":"\"calculated average height of 35 boys = 182 cm . wrong total height of 35 boys = 182 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 182 cm - ( 166 cm - 106 cm ) \/ 35 = 182 cm - ( 166 cm - 106 cm ) \/ 35 = 182 cm - 60 \/ 35 cm = 182 cm - 1.71 cm = 180.29 cm . answer : b\"","correct":"b","options":{"a":"187.89 cm ","b":"180.29 cm ","c":"123.98 cm ","d":"149.98 cm","e":"146.89 cm"},"options_float":{"a":187.89,"b":180.29,"c":123.98,"d":149.98,"e":146.89},"annotated_formula":"floor(divide(add(subtract(multiply(35, 182), 166), 106), 35))","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)|","chain":"35 * 182<\/gadget>\n6_370<\/output>\n6_370 - 166<\/gadget>\n6_204<\/output>\n6_204 + 106<\/gadget>\n6_310<\/output>\n6_310 \/ 35<\/gadget>\n1_262\/7 = around 180.285714<\/output>\nfloor(1_262\/7)<\/gadget>\n180<\/output>\n180<\/result>","index":2131} +{"problem":"the length of a train and that of a platform are equal . if with a speed of 108 k \/ hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ?","rationale":"\"speed = [ 108 * 5 \/ 18 ] m \/ sec = 30 m \/ sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x \/ 60 = 30 = > x = 30 * 60 \/ 2 = 900 answer : d\"","correct":"d","options":{"a":"299 ","b":"266 ","c":"299 ","d":"900","e":"261"},"options_float":{"a":299.0,"b":266.0,"c":299.0,"d":900.0,"e":261.0},"annotated_formula":"divide(divide(multiply(108, const_1000), divide(const_60, const_1)), const_2)","linear_formula":"divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)|","chain":"108 * 1_000<\/gadget>\n108_000<\/output>\n60 \/ 1<\/gadget>\n60<\/output>\n108_000 \/ 60<\/gadget>\n1_800<\/output>\n1_800 \/ 2<\/gadget>\n900<\/output>\n900<\/result>","index":2132} +{"problem":"ravi and sunil are partners in a business . ravi invests rs . 12,000 for 4 months and sunil invested rs . 10000 for 5 months then after one year ratio of their profits will be","rationale":"= ( 12000 * 4 ) : ( 10000 * 5 ) = 48000 : 50000 = 24 : 25 answer : c","correct":"c","options":{"a":"1 : 3 ","b":"2 : 3 ","c":"24 : 25 ","d":"20 : 22","e":"3 : 4"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":0.96,"d":0.9090909091,"e":0.75},"annotated_formula":"divide(multiply(multiply(multiply(multiply(const_4, const_3), multiply(5, const_2)), const_100), 4), multiply(10000, 5))","linear_formula":"multiply(const_3,const_4)|multiply(n3,const_2)|multiply(n2,n3)|multiply(#0,#1)|multiply(#3,const_100)|multiply(n1,#4)|divide(#5,#2)","chain":"4 * 3<\/gadget>\n12<\/output>\n5 * 2<\/gadget>\n10<\/output>\n12 * 10<\/gadget>\n120<\/output>\n120 * 100<\/gadget>\n12_000<\/output>\n12_000 * 4<\/gadget>\n48_000<\/output>\n10_000 * 5<\/gadget>\n50_000<\/output>\n48_000 \/ 50_000<\/gadget>\n24\/25 = around 0.96<\/output>\n24\/25 = around 0.96<\/result>","index":2134} +{"problem":"the value of “ a ” varies in inverse proportion as the square of “ b ” . if the value of “ a ” is equal to 40 when “ b ” is equal to 12 . what would be the value of “ a ” when “ b ” is equal to 24 ?","rationale":"explanation : we have , a α 1 \/ b 2 a * b 2 = k ( constant ) when b = 12 , a = 40 … . . given k = 40 * ( 12 ) 2 = 5760 now , b = 24 . hence , a = 5760 \/ 576 = 10 answer : a","correct":"a","options":{"a":"10 ","b":"88 ","c":"26 ","d":"27","e":"28"},"options_float":{"a":10.0,"b":88.0,"c":26.0,"d":27.0,"e":28.0},"annotated_formula":"divide(multiply(power(12, const_2), 40), power(24, const_2))","linear_formula":"power(n1,const_2)|power(n2,const_2)|multiply(n0,#0)|divide(#2,#1)","chain":"12 ** 2<\/gadget>\n144<\/output>\n144 * 40<\/gadget>\n5_760<\/output>\n24 ** 2<\/gadget>\n576<\/output>\n5_760 \/ 576<\/gadget>\n10<\/output>\n10<\/result>","index":2135} +{"problem":"the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 4 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ?","rationale":"\"area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 4 = rs . 3624 answer : option b\"","correct":"b","options":{"a":"4000 ","b":"3624 ","c":"5673 ","d":"4530","e":"4566"},"options_float":{"a":4000.0,"b":3624.0,"c":5673.0,"d":4530.0,"e":4566.0},"annotated_formula":"multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 4)","linear_formula":"add(n0,n1)|multiply(n2,const_2)|multiply(n4,n5)|multiply(n5,n6)|multiply(#0,#1)|multiply(n5,#3)|subtract(#4,#2)|subtract(#6,#5)|multiply(n3,#7)|","chain":"2 * 12<\/gadget>\n24<\/output>\n15 + 25<\/gadget>\n40<\/output>\n24 * 40<\/gadget>\n960<\/output>\n6 * 3<\/gadget>\n18<\/output>\n960 - 18<\/gadget>\n942<\/output>\n4 * 3<\/gadget>\n12<\/output>\n3 * 12<\/gadget>\n36<\/output>\n942 - 36<\/gadget>\n906<\/output>\n906 * 4<\/gadget>\n3_624<\/output>\n3_624<\/result>","index":2136} +{"problem":"of the 65 cars on a car lot , 45 have power steering , 25 have power windows , and 17 have both power steering and power windows . how many of the cars on the lot have neither power steering nor power windows ?","rationale":"total - neither = all power steering + all power windows - both or 65 - neither = 45 + 25 - 17 = 53 . = > neither = 12 , hence c . answer : c","correct":"c","options":{"a":"2 ","b":"8 ","c":"12 ","d":"15","e":"18"},"options_float":{"a":2.0,"b":8.0,"c":12.0,"d":15.0,"e":18.0},"annotated_formula":"subtract(65, subtract(add(45, 25), 17))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)","chain":"45 + 25<\/gadget>\n70<\/output>\n70 - 17<\/gadget>\n53<\/output>\n65 - 53<\/gadget>\n12<\/output>\n12<\/result>","index":2137} +{"problem":"convert 0.32 in to a vulgar fraction ?","rationale":"\"answer 0.32 = 32 \/ 100 = 8 \/ 25 correct option : a\"","correct":"a","options":{"a":"8 \/ 25 ","b":"16 \/ 50 ","c":"17 \/ 50 ","d":"19 \/ 50","e":"none"},"options_float":{"a":0.32,"b":0.32,"c":0.34,"d":0.38,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.32), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))","linear_formula":"add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n100 * 0.32<\/gadget>\n32<\/output>\n32 \/ 100<\/gadget>\n8\/25 = around 0.32<\/output>\n8\/25 = around 0.32<\/result>","index":2139} +{"problem":"the l . c . m . of two numbers is 60 . the numbers are in the ratio 2 : 3 . then sum of the number is :","rationale":"\"let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 60 or x = 10 . the numbers are 20 and 30 . hence , required sum = ( 20 + 30 ) = 50 . answer : option e\"","correct":"e","options":{"a":"28 ","b":"32 ","c":"40 ","d":"53","e":"50"},"options_float":{"a":28.0,"b":32.0,"c":40.0,"d":53.0,"e":50.0},"annotated_formula":"add(multiply(divide(60, multiply(2, 3)), 3), multiply(divide(60, multiply(2, 3)), 2))","linear_formula":"multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)|","chain":"2 * 3<\/gadget>\n6<\/output>\n60 \/ 6<\/gadget>\n10<\/output>\n10 * 3<\/gadget>\n30<\/output>\n10 * 2<\/gadget>\n20<\/output>\n30 + 20<\/gadget>\n50<\/output>\n50<\/result>","index":2140} +{"problem":"a cycle is bought for rs . 900 and sold for rs . 1170 , find the gain percent ?","rationale":"\"900 - - - - 270 100 - - - - ? = > 30 % answer : d\"","correct":"d","options":{"a":"27 % ","b":"20 % ","c":"80 % ","d":"30 %","e":"24 %"},"options_float":{"a":27.0,"b":20.0,"c":80.0,"d":30.0,"e":24.0},"annotated_formula":"multiply(divide(subtract(1170, 900), 900), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_170 - 900<\/gadget>\n270<\/output>\n270 \/ 900<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":2142} +{"problem":"rs 378 is divided among the 3 children such that 12 times of anusha is equal to 8 times of babu which is equal to 6 times of esha . how much the rupee anusha get ?","rationale":"if money with anusha = a , babu = b & esha = e , then a + b + e = 378 - - - ( i ) & 12 a = 8 b = 6 e , so b = 3 a \/ 2 & e = 2 a substituting these values of b & e in ( i ) , a + ( 3 a \/ 2 ) + 2 a = 378 , on solving , a = 84 so , b = 126 , e = 168 answer : b","correct":"b","options":{"a":"74 ","b":"84 ","c":"94 ","d":"104","e":"114"},"options_float":{"a":74.0,"b":84.0,"c":94.0,"d":104.0,"e":114.0},"annotated_formula":"divide(378, add(add(const_1, divide(12, 8)), divide(12, 6)))","linear_formula":"divide(n2,n3)|divide(n2,n4)|add(#0,const_1)|add(#2,#1)|divide(n0,#3)","chain":"12 \/ 8<\/gadget>\n3\/2 = around 1.5<\/output>\n1 + (3\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n12 \/ 6<\/gadget>\n2<\/output>\n(5\/2) + 2<\/gadget>\n9\/2 = around 4.5<\/output>\n378 \/ (9\/2)<\/gadget>\n84<\/output>\n84<\/result>","index":2143} +{"problem":"the price of 10 chairs is equal to that of 4 tables . the price of 15 chairs and 2 tables together is rs . 4000 . the total price of 12 chairs and 3 tables is ?","rationale":"let the cost of a chair and that of a table be rs . x and rs . y respectively . then , 10 x = 4 y or y = 5 x . 2 15 x + 2 y = 4000 15 x + 2 x 5 x = 4000 2 20 x = 4000 x = 200 . so , y = 5 x 200 = 500 . 2 hence , the cost of 12 chairs and 3 tables = 12 x + 3 y = rs . ( 2400 + 1500 ) = rs . 3900 . a )","correct":"a","options":{"a":"rs . 3900 ","b":"rs . 4900 ","c":"rs . 4980 ","d":"rs . 5000","e":"rs . 5900"},"options_float":{"a":3900.0,"b":4900.0,"c":4980.0,"d":5000.0,"e":5900.0},"annotated_formula":"add(multiply(divide(4000, add(multiply(15, divide(4, 10)), 2)), 3), multiply(multiply(divide(4, 10), divide(4000, add(multiply(15, divide(4, 10)), 2))), 12))","linear_formula":"divide(n1,n0)|multiply(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n6,#3)|multiply(#0,#3)|multiply(n5,#5)|add(#4,#6)","chain":"4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n15 * (2\/5)<\/gadget>\n6<\/output>\n6 + 2<\/gadget>\n8<\/output>\n4_000 \/ 8<\/gadget>\n500<\/output>\n500 * 3<\/gadget>\n1_500<\/output>\n(2\/5) * 500<\/gadget>\n200<\/output>\n200 * 12<\/gadget>\n2_400<\/output>\n1_500 + 2_400<\/gadget>\n3_900<\/output>\n3_900<\/result>","index":2144} +{"problem":"a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 12 multiples of 14 , how many multiples of 7 are there in q ?","rationale":"\"halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 12 + 11 = 23 . the answer is d .\"","correct":"d","options":{"a":"20 ","b":"21 ","c":"22 ","d":"23","e":"24"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":23.0,"e":24.0},"annotated_formula":"subtract(multiply(12, const_2), const_1)","linear_formula":"multiply(n1,const_2)|subtract(#0,const_1)|","chain":"12 * 2<\/gadget>\n24<\/output>\n24 - 1<\/gadget>\n23<\/output>\n23<\/result>","index":2145} +{"problem":"the average age of 15 students of a class is 15 years . out of these , the average age of 6 students is 14 years and that of the other 8 students is 16 years . the age of the 15 th student is","rationale":"\"solution age of the 15 th student = [ 15 x 15 - ( 14 x 6 + 16 x 8 ) ] = ( 225 - 212 ) = 13 years . answer b\"","correct":"b","options":{"a":"9 years ","b":"13 years ","c":"14 years ","d":"21 years","e":"25 years"},"options_float":{"a":9.0,"b":13.0,"c":14.0,"d":21.0,"e":25.0},"annotated_formula":"subtract(multiply(15, 15), add(multiply(6, 14), multiply(8, 16)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n6 * 14<\/gadget>\n84<\/output>\n8 * 16<\/gadget>\n128<\/output>\n84 + 128<\/gadget>\n212<\/output>\n225 - 212<\/gadget>\n13<\/output>\n13<\/result>","index":2146} +{"problem":"a can do a work in 6 days . b can do the same work in 12 days . if both a & b are working together in how many days they will finish the work ?","rationale":"\"a rate = 1 \/ 6 b rate = 1 \/ 12 ( a + b ) rate = ( 1 \/ 6 ) + ( 1 \/ 12 ) = 1 \/ 4 a & b finish the work in 4 days correct option is c\"","correct":"c","options":{"a":"3 ","b":"5 ","c":"4 ","d":"2","e":"6"},"options_float":{"a":3.0,"b":5.0,"c":4.0,"d":2.0,"e":6.0},"annotated_formula":"divide(multiply(6, 12), add(6, 12))","linear_formula":"add(n0,n1)|multiply(n0,n1)|divide(#1,#0)|","chain":"6 * 12<\/gadget>\n72<\/output>\n6 + 12<\/gadget>\n18<\/output>\n72 \/ 18<\/gadget>\n4<\/output>\n4<\/result>","index":2147} +{"problem":"in an organisation there are 1000 employees out of which 250 earn below 10 k $ and 500 earn between 10 k $ and 50 k $ . what is the total percentage of employees who earn less than 50 k $ ?","rationale":"total percentage = ( 250 + 500 ) \/ 1000 = 0.75 b","correct":"b","options":{"a":"66 % ","b":"75 % ","c":"70 % ","d":"50 %","e":"82 %"},"options_float":{"a":66.0,"b":75.0,"c":70.0,"d":50.0,"e":82.0},"annotated_formula":"multiply(divide(subtract(1000, 250), 1000), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)","chain":"1_000 - 250<\/gadget>\n750<\/output>\n750 \/ 1_000<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n75<\/result>","index":2148} +{"problem":"how many quarters are equal to 6 dollars ?","rationale":"6 * 4 = 24 quarters answer : e","correct":"e","options":{"a":"1 ","b":"8 ","c":"12 ","d":"9","e":"24"},"options_float":{"a":1.0,"b":8.0,"c":12.0,"d":9.0,"e":24.0},"annotated_formula":"multiply(6, const_4)","linear_formula":"multiply(n0,const_4)","chain":"6 * 4<\/gadget>\n24<\/output>\n24<\/result>","index":2149} +{"problem":"the price of 2 sarees and 4 shirts is rs . 1600 . with the same money one can buy 1 saree and 6 shirts . if one wants to buy 12 shirts , how much shall he have to pay ?","rationale":"let the price of a saree and a shirt be rs . x and rs . y respectively . then , 2 x + 4 y = 1600 . . . . ( i ) and x + 6 y = 1600 . . . . ( ii ) divide equation ( i ) by 2 , we get the below equation . = > x + 2 y = 800 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 6 y = 1600 ( - ) x + 2 y = 800 - - - - - - - - - - - - - - - - 4 y = 800 - - - - - - - - - - - - - - - - therefore , y = 200 . now apply value of y in ( iii ) = > x + 2 x 200 = 800 = > x + 400 = 800 therefore x = 400 solving ( i ) and ( ii ) we get x = 400 , y = 200 . cost of 12 shirts = rs . ( 12 x 200 ) = rs . 2400 . answer is e .","correct":"e","options":{"a":"2500 ","b":"2700 ","c":"2200 ","d":"2900","e":"2400"},"options_float":{"a":2500.0,"b":2700.0,"c":2200.0,"d":2900.0,"e":2400.0},"annotated_formula":"multiply(divide(1600, add(multiply(2, divide(subtract(6, 4), subtract(2, 1))), 4)), 12)","linear_formula":"subtract(n4,n1)|subtract(n0,n3)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)|divide(n2,#4)|multiply(n5,#5)","chain":"6 - 4<\/gadget>\n2<\/output>\n2 - 1<\/gadget>\n1<\/output>\n2 \/ 1<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 + 4<\/gadget>\n8<\/output>\n1_600 \/ 8<\/gadget>\n200<\/output>\n200 * 12<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":2151} +{"problem":"in a company with 48 employees , some part - time and some full - time , exactly ( 1 \/ 3 ) of the part - time employees and ( 1 \/ 4 ) of the full - time employees take the subway to work . what is the greatest possible number y of employees who take the subway to work ?","rationale":"\"p \/ 3 + f \/ 4 = p \/ 3 + ( 48 - p ) \/ 4 = 12 + p \/ 2 p \/ 3 + f \/ 3 = ( p + f ) \/ 3 = 48 \/ 3 = 16 p \/ 4 + f \/ 4 = 12 p \/ 3 + f \/ 3 > p \/ 3 + f \/ 4 > p \/ 4 + f \/ 4 - - > 16 > 12 + p \/ 12 > 12 greatest possible y : 12 + p \/ 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer\"","correct":"d","options":{"a":"12 ","b":"13 ","c":"14 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4))))","linear_formula":"multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)|","chain":"48 * 3<\/gadget>\n144<\/output>\n3 * 4<\/gadget>\n12<\/output>\n144 \/ 12<\/gadget>\n12<\/output>\n48 - 12<\/gadget>\n36<\/output>\n36 \/ 12<\/gadget>\n3<\/output>\n12 + 3<\/gadget>\n15<\/output>\n15<\/result>","index":2152} +{"problem":"a garden center sells a certain grass seed in 5 - pound bags at $ 13.80 per bag , 10 - pound bags at $ 20.43 per bag , and 25 - pound bags $ 32.25 per bag . if a customer is to buy at least 65 pounds of the grass seed , but no more than 80 pounds , what is the least possible cost of the grass seed that the customer will buy ?","rationale":"there can be 2 cases 1 ) 25 + 25 + 10 + 5 = $ 98.73 or 2 ) 25 + 25 + 25 = $ 96.75 a","correct":"a","options":{"a":"$ 98.73 ","b":"$ 96.75 ","c":"$ 98.78 ","d":"$ 102.07","e":"$ 105.3"},"options_float":{"a":98.73,"b":96.75,"c":98.78,"d":102.07,"e":105.3},"annotated_formula":"add(add(multiply(const_2, 32.25), multiply(const_1, 20.43)), multiply(const_1, 13.8))","linear_formula":"multiply(n5,const_2)|multiply(n3,const_1)|multiply(n1,const_1)|add(#0,#1)|add(#3,#2)","chain":"2 * 32.25<\/gadget>\n64.5<\/output>\n1 * 20.43<\/gadget>\n20.43<\/output>\n64.5 + 20.43<\/gadget>\n84.93<\/output>\n1 * 13.8<\/gadget>\n13.8<\/output>\n84.93 + 13.8<\/gadget>\n98.73<\/output>\n98.73<\/result>","index":2153} +{"problem":"in the manufacture of a certain product , 5 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ?","rationale":"\"percent of defective produced = 5 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 \/ 100 ) * ( 5 \/ 100 ) * 100 % = ( 20 \/ 10000 ) * 100 % = ( 20 \/ 100 ) % = . 2 % answer b\"","correct":"b","options":{"a":"0.125 % ","b":"0.2 % ","c":"0.8 % ","d":"1.25 %","e":"2.0 %"},"options_float":{"a":0.125,"b":0.2,"c":0.8,"d":1.25,"e":2.0},"annotated_formula":"multiply(5, divide(4, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n5 * (1\/25)<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":2154} +{"problem":"a student appeared in the mock cat . the test paper contained 3 sections namely qa , di and va . the percentage marks in all va was equal to the average of percentage marks in all the 3 sections . coincidentally , if we reverse the digit of the percentage marks of qa we get the percentage marks of di . the percentage marks in va scored by student could be :","rationale":"solution : let the percentage marks in qa = ( 10 a + b ) % let the percentage marks in di = ( 10 b + a ) % let the percentage marks in va = x % . now , according to the question , we have , [ ( 10 a + b ) + x + ( 10 b + a ) ] \/ 3 = x → 11 a + 11 b + x = 3 x . or , x = 11 ( a + b ) \/ 2 . clearly , we can see that the percentage of the va section will be a multiple of 11 . so , required answer will be 66 . answer : option c","correct":"c","options":{"a":"81 ","b":"48 ","c":"66 ","d":"69","e":"none of these"},"options_float":{"a":81.0,"b":48.0,"c":66.0,"d":69.0,"e":null},"annotated_formula":"add(add(add(add(add(add(multiply(3, 3), const_2), add(multiply(3, 3), const_2)), add(multiply(3, 3), const_2)), add(multiply(3, 3), const_2)), add(multiply(3, 3), const_2)), add(multiply(3, 3), const_2))","linear_formula":"multiply(n0,n0)|add(#0,const_2)|add(#1,#1)|add(#2,#1)|add(#3,#1)|add(#4,#1)|add(#5,#1)","chain":"3 * 3<\/gadget>\n9<\/output>\n9 + 2<\/gadget>\n11<\/output>\n11 + 11<\/gadget>\n22<\/output>\n22 + 11<\/gadget>\n33<\/output>\n33 + 11<\/gadget>\n44<\/output>\n44 + 11<\/gadget>\n55<\/output>\n55 + 11<\/gadget>\n66<\/output>\n66<\/result>","index":2155} +{"problem":"two trains of equal are running on parallel lines in the same direction at 45 km \/ hr and 36 km \/ hr . the faster train passes the slower train in 36 sec . the length of each train is ?","rationale":"\"let the length of each train be x m . then , distance covered = 2 x m . relative speed = 45 - 36 = 9 km \/ hr . = 9 * 5 \/ 18 = 5 \/ 2 m \/ sec . 2 x \/ 36 = 5 \/ 2 = > x = 45 . answer : a\"","correct":"a","options":{"a":"45 ","b":"88 ","c":"66 ","d":"55","e":"22"},"options_float":{"a":45.0,"b":88.0,"c":66.0,"d":55.0,"e":22.0},"annotated_formula":"divide(multiply(36, divide(multiply(subtract(45, 36), const_1000), const_3600)), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|","chain":"45 - 36<\/gadget>\n9<\/output>\n9 * 1_000<\/gadget>\n9_000<\/output>\n9_000 \/ 3_600<\/gadget>\n5\/2 = around 2.5<\/output>\n36 * (5\/2)<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n45<\/result>","index":2157} +{"problem":"a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 2 cm when a man gets on it . the mass of man is","rationale":"\"solution volume of water displaced = ( 3 x 2 x 0.02 ) m 3 = 0.12 m 3 . mass of man = volume of water displaced × density of water = ( 0.12 × 1000 ) kg = 120 kg . answer d\"","correct":"d","options":{"a":"12 kg ","b":"60 kg ","c":"72 kg ","d":"120 kg","e":"none"},"options_float":{"a":12.0,"b":60.0,"c":72.0,"d":120.0,"e":null},"annotated_formula":"multiply(multiply(multiply(3, 2), divide(2, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"3 * 2<\/gadget>\n6<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n6 * (1\/50)<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) * 1_000<\/gadget>\n120<\/output>\n120<\/result>","index":2159} +{"problem":"in what ratio must a grocer mix two varieties of pulses costing rs . 15 and rs . 20 per kg respectively so as to get a mixture worth rs . 16.50 kg ?","rationale":"\"required rate = 3.50 : 1.50 = 7 : 3 . answer a\"","correct":"a","options":{"a":"7 : 3 ","b":"3 : 7 ","c":"4 : 6 ","d":"5 : 7","e":"8 : 9"},"options_float":{"a":2.3333333333,"b":0.4285714286,"c":0.6666666667,"d":0.7142857143,"e":0.8888888889},"annotated_formula":"divide(subtract(20, 16.50), subtract(16.50, 15))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"20 - 16.5<\/gadget>\n3.5<\/output>\n16.5 - 15<\/gadget>\n1.5<\/output>\n3.5 \/ 1.5<\/gadget>\n2.333333<\/output>\n2.333333<\/result>","index":2161} +{"problem":"if the personal income tax rate is lowered from 46 % to 32 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 36000 ?","rationale":"\"saving = ( 46 - 32 ) % of 36000 = 5040 . answer : b\"","correct":"b","options":{"a":"$ 3500 ","b":"$ 5040 ","c":"$ 3150 ","d":"$ 7000","e":"$ 10000"},"options_float":{"a":3500.0,"b":5040.0,"c":3150.0,"d":7000.0,"e":10000.0},"annotated_formula":"multiply(divide(36000, const_100), subtract(46, 32))","linear_formula":"divide(n2,const_100)|subtract(n0,n1)|multiply(#0,#1)|","chain":"36_000 \/ 100<\/gadget>\n360<\/output>\n46 - 32<\/gadget>\n14<\/output>\n360 * 14<\/gadget>\n5_040<\/output>\n5_040<\/result>","index":2162} +{"problem":"how many unique positive odd integers less than 100 are equal to the product of a positive multiple of 5 and an odd number ?","rationale":"the question basically asks how many positive odd integers less than 100 are odd multiples of 5 so we have 5,15 , 25,35 , 45,55 , 65,75 , 85,95 = 10 ans b","correct":"b","options":{"a":"4 ","b":"10 ","c":"11 ","d":"12","e":"15"},"options_float":{"a":4.0,"b":10.0,"c":11.0,"d":12.0,"e":15.0},"annotated_formula":"divide(divide(100, 5), const_2)","linear_formula":"divide(n0,n1)|divide(#0,const_2)","chain":"100 \/ 5<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":2163} +{"problem":"how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 625 meters in length ?","rationale":"\"explanation : t = ( 625 + 165 ) \/ 54 * 18 \/ 5 t = 52.7 answer : option e\"","correct":"e","options":{"a":"62 ","b":"72 ","c":"55 ","d":"82","e":"52.7"},"options_float":{"a":62.0,"b":72.0,"c":55.0,"d":82.0,"e":52.7},"annotated_formula":"divide(add(165, 625), multiply(54, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"165 + 625<\/gadget>\n790<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n54 * (5\/18)<\/gadget>\n15<\/output>\n790 \/ 15<\/gadget>\n158\/3 = around 52.666667<\/output>\n158\/3 = around 52.666667<\/result>","index":2164} +{"problem":"the length of a rectangle is one fourth of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 784 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 5 units ?","rationale":"given that the area of the square = 784 sq . units = > side of square = â ˆ š 784 = 28 units the radius of the circle = side of the square = 28 units length of the rectangle = 1 \/ 4 * 28 = 7 units given that breadth = 5 units area of the rectangle = lb = 7 * 5 = 35 sq . units answer : a","correct":"a","options":{"a":"35 sq . units ","b":"45 sq . units ","c":"55 sq . units ","d":"50 sq . units","e":"40 sq . units"},"options_float":{"a":35.0,"b":45.0,"c":55.0,"d":50.0,"e":40.0},"annotated_formula":"rectangle_area(divide(sqrt(784), const_4), 5)","linear_formula":"sqrt(n0)|divide(#0,const_4)|rectangle_area(n1,#1)","chain":"784 ** (1\/2)<\/gadget>\n28<\/output>\n28 \/ 4<\/gadget>\n7<\/output>\n7 * 5<\/gadget>\n35<\/output>\n35<\/result>","index":2166} +{"problem":"33 1 \/ 3 % of 36 ?","rationale":"explanation : 33 1 \/ 3 % = 1 \/ 3 1 \/ 3 * 36 = 12 answer : option d","correct":"d","options":{"a":"16 ","b":"21 ","c":"18 ","d":"12 number","e":"18"},"options_float":{"a":16.0,"b":21.0,"c":18.0,"d":12.0,"e":18.0},"annotated_formula":"divide(multiply(add(33, divide(1, 3)), 36), const_100)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n33 + (1\/3)<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) * 36<\/gadget>\n1_200<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n12<\/result>","index":2167} +{"problem":"an engine moves at the speed of 30 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 9 coaches are attached speed decreases to 18 kmph . what will be the speed of train when 16 coaches are attached .","rationale":"\"1 . no . of coaches = 9 sqr root = 3 speed decreases by 12 12 = k * 3 k = 4 no . of coaches = 16 swr root = 4 decrease = 4 * 4 = 16 new speed = 30 - 16 = 14 a\"","correct":"a","options":{"a":"14 ","b":"16 ","c":"18 ","d":"20","e":"22"},"options_float":{"a":14.0,"b":16.0,"c":18.0,"d":20.0,"e":22.0},"annotated_formula":"subtract(30, multiply(sqrt(16), divide(subtract(30, 18), sqrt(9))))","linear_formula":"sqrt(n1)|sqrt(n3)|subtract(n0,n2)|divide(#2,#0)|multiply(#3,#1)|subtract(n0,#4)|","chain":"16 ** (1\/2)<\/gadget>\n4<\/output>\n30 - 18<\/gadget>\n12<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n4 * 4<\/gadget>\n16<\/output>\n30 - 16<\/gadget>\n14<\/output>\n14<\/result>","index":2168} +{"problem":"at joes steakhouse the hourly wage for a chef is 20 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 10.50 per hour , how much less than a manager does a chef earn each hour ?","rationale":"\"manager wages per hour = $ 10.50 dishwasher wages per hour = half of manager ' s wages . = 1 \/ 2 ( $ 10.50 ) = = > $ 5.25 chef wages per hour = 20 % greater than dishwasher wages - - > 20 % of $ 5.25 = ( 20 * ( $ 5.25 ) ) \/ 100 - - > ( $ 95 ) \/ 100 - - > $ 0.95 therefore , chef wages per hour = $ 5.25 + $ 0.95 = = > $ 6.20 difference of wages between manager and chef = $ 10.50 - $ 6.20 = = > $ 4.30 answer : e\"","correct":"e","options":{"a":"$ 1.40 ","b":"$ 2.40 ","c":"$ 3.40 ","d":"$ 4.40","e":"$ 4.30"},"options_float":{"a":1.4,"b":2.4,"c":3.4,"d":4.4,"e":4.3},"annotated_formula":"multiply(subtract(const_1, multiply(divide(add(const_100, 20), const_100), divide(const_1, const_2))), 10.50)","linear_formula":"add(n0,const_100)|divide(const_1,const_2)|divide(#0,const_100)|multiply(#2,#1)|subtract(const_1,#3)|multiply(n1,#4)|","chain":"100 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(6\/5) * (1\/2)<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 10.5<\/gadget>\n4.2<\/output>\n4.2<\/result>","index":2171} +{"problem":"if the height of a cone is increased by 140 % then its volume is increased by ?","rationale":"140 % answer : b","correct":"b","options":{"a":"100 ","b":"140 ","c":"998 ","d":"729","e":"2879"},"options_float":{"a":100.0,"b":140.0,"c":998.0,"d":729.0,"e":2879.0},"annotated_formula":"rectangle_area(140, const_1)","linear_formula":"rectangle_area(n0,const_1)","chain":"140 * 1<\/gadget>\n140<\/output>\n140<\/result>","index":2172} +{"problem":"1345 - ( 1002 \/ 20.04 ) = ?","rationale":"\"1345 - 50 = 1295 answer : d\"","correct":"d","options":{"a":"2984 ","b":"2983 ","c":"2982 ","d":"1295","e":"none of these"},"options_float":{"a":2984.0,"b":2983.0,"c":2982.0,"d":1295.0,"e":null},"annotated_formula":"subtract(1345, divide(1002, 20.04))","linear_formula":"divide(n1,n2)|subtract(n0,#0)|","chain":"1_002 \/ 20.04<\/gadget>\n50<\/output>\n1_345 - 50<\/gadget>\n1_295<\/output>\n1_295<\/result>","index":2173} +{"problem":"the guests at a football banquet consumed a total of 325 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ?","rationale":"\"to minimize one quantity maximize other . 162 * 2 ( max possible amount of food a guest could consume ) = 324 pounds , so there must be more than 162 guests , next integer is 163 . answer : d .\"","correct":"d","options":{"a":"160 ","b":"161 ","c":"162 ","d":"163","e":"164"},"options_float":{"a":160.0,"b":161.0,"c":162.0,"d":163.0,"e":164.0},"annotated_formula":"add(floor(divide(325, 2)), const_1)","linear_formula":"divide(n0,n1)|floor(#0)|add(#1,const_1)|","chain":"325 \/ 2<\/gadget>\n325\/2 = around 162.5<\/output>\nfloor(325\/2)<\/gadget>\n162<\/output>\n162 + 1<\/gadget>\n163<\/output>\n163<\/result>","index":2174} +{"problem":"how many 3 - digit numbers abc , in which a , b , and c are each digits , satisfy the equation 2 b = a + c ?","rationale":"possible solutions are diff of 0 - - 111222 , . . . . . . . . , 999 = 9 diff of 1 - - 123 , 234 , . . . . . , 789 = 7 x 2 = 14 diff of 2 - - 135 , 246 . . . . . . . 579 = 5 x 2 = 10 diff of 3 - - 147 , 258369 = 3 x 2 = 6 diff of 4 - - 159 = 1 x 2 = 2 last 210 , 420 , 630840 = 4 so total is 9 + 14 + 10 + 6 + 2 + 4 = 45 . answer : d","correct":"d","options":{"a":"33 ","b":"36 ","c":"41 ","d":"45","e":"50"},"options_float":{"a":33.0,"b":36.0,"c":41.0,"d":45.0,"e":50.0},"annotated_formula":"add(add(multiply(3, const_12), multiply(2, const_4)), const_1)","linear_formula":"multiply(n0,const_12)|multiply(n1,const_4)|add(#0,#1)|add(#2,const_1)","chain":"3 * 12<\/gadget>\n36<\/output>\n2 * 4<\/gadget>\n8<\/output>\n36 + 8<\/gadget>\n44<\/output>\n44 + 1<\/gadget>\n45<\/output>\n45<\/result>","index":2176} +{"problem":"in cliff ’ s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 2 \/ 3 are shiny and the rest are matte , while 1 \/ 5 of the sedimentary rocks are shiny . if there are 40 shiny igneous rocks , how many total rocks does cliff have ?","rationale":"we can start with the known quantity and then go on to find the others . shiny igneous ricks are 40 . these are ( 2 \/ 3 ) of total igneous rocks . ( 2 \/ 3 ) * total igneous rocks = 40 total igneous rocks = 40 * ( 3 \/ 2 ) = 60 total sedimentary rocks = 2 * total igneous rocks = 2 * 60 = 120 total number of rocks = 60 + 120 = 180 answer ( a )","correct":"a","options":{"a":"180 ","b":"45 ","c":"60 ","d":"90","e":"135"},"options_float":{"a":180.0,"b":45.0,"c":60.0,"d":90.0,"e":135.0},"annotated_formula":"divide(multiply(divide(multiply(multiply(2, 3), 40), 2), 3), 2)","linear_formula":"multiply(n0,n1)|multiply(n4,#0)|divide(#1,n0)|multiply(n1,#2)|divide(#3,n0)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 40<\/gadget>\n240<\/output>\n240 \/ 2<\/gadget>\n120<\/output>\n120 * 3<\/gadget>\n360<\/output>\n360 \/ 2<\/gadget>\n180<\/output>\n180<\/result>","index":2177} +{"problem":"3 sets of 40 , 50 and 60 students appeared for an examination and the pass percentage was 100 , 90 and 80 respectively . the pass percentage of the whole set is","rationale":"total passed student = 40 + 45 + 48 = 133 133 is 88.66 % of 150 answer : c","correct":"c","options":{"a":"88.33 ","b":"84.33 ","c":"88.66 ","d":"84.66","e":"none of these"},"options_float":{"a":88.33,"b":84.33,"c":88.66,"d":84.66,"e":null},"annotated_formula":"multiply(divide(add(add(40, multiply(50, divide(90, const_100))), multiply(60, divide(80, const_100))), add(add(40, 50), 60)), const_100)","linear_formula":"add(n1,n2)|divide(n5,const_100)|divide(n6,const_100)|add(n3,#0)|multiply(n2,#1)|multiply(n3,#2)|add(n1,#4)|add(#6,#5)|divide(#7,#3)|multiply(#8,const_100)","chain":"90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n50 * (9\/10)<\/gadget>\n45<\/output>\n40 + 45<\/gadget>\n85<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n60 * (4\/5)<\/gadget>\n48<\/output>\n85 + 48<\/gadget>\n133<\/output>\n40 + 50<\/gadget>\n90<\/output>\n90 + 60<\/gadget>\n150<\/output>\n133 \/ 150<\/gadget>\n133\/150 = around 0.886667<\/output>\n(133\/150) * 100<\/gadget>\n266\/3 = around 88.666667<\/output>\n266\/3 = around 88.666667<\/result>","index":2178} +{"problem":"ake can dig a well in 16 days . paul can dig the same well in 24 days . jake , paul and hari together dig the well in 8 days . hari alone can dig the well in","rationale":"simple one . let the total work to be done is 48 meters . now jake can dig 3 mts , paul can dig 2 mts a day . now all of them combined dug in 8 days so per day they dug 48 \/ 8 = 6 mts . so of these 8 mts , hari capacity is 1 mt . so he takes 48 \/ 1 = 48 days to complete the digging job . answer : a","correct":"a","options":{"a":"48 ","b":"39 ","c":"37 ","d":"27","e":"29"},"options_float":{"a":48.0,"b":39.0,"c":37.0,"d":27.0,"e":29.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 8), add(divide(const_1, 16), divide(const_1, 24))))","linear_formula":"divide(const_1,n2)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|subtract(#0,#3)|divide(const_1,#4)","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/16) + (1\/24)<\/gadget>\n5\/48 = around 0.104167<\/output>\n(1\/8) - (5\/48)<\/gadget>\n1\/48 = around 0.020833<\/output>\n1 \/ (1\/48)<\/gadget>\n48<\/output>\n48<\/result>","index":2180} +{"problem":"a man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream . the ratio of the speed of the boat ( in still water ) and the stream is","rationale":"\"sol . let man ’ s rate upstream be x kmph . then , his rate downstream = 2 x kmph . ∴ ( speed in still water ) : ( speed of stream ) = [ 2 x + x \/ 2 ] : [ 2 x - x \/ 2 ] 3 x \/ 2 : x \/ 2 = 3 : 1 . answer d\"","correct":"d","options":{"a":"1 : 2 ","b":"1 : 3 ","c":"2 : 1 ","d":"3 : 1","e":"none"},"options_float":{"a":0.5,"b":0.3333333333,"c":2.0,"d":3.0,"e":null},"annotated_formula":"divide(add(const_1, const_2), subtract(const_2, const_1))","linear_formula":"add(const_1,const_2)|subtract(const_2,const_1)|divide(#0,#1)|","chain":"1 + 2<\/gadget>\n3<\/output>\n2 - 1<\/gadget>\n1<\/output>\n3 \/ 1<\/gadget>\n3<\/output>\n3<\/result>","index":2181} +{"problem":"a can complete the job in 6 hours and b can complete the same job in 3 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete","rationale":"a = 1 \/ 9","correct":"a","options":{"a":"1 \/ 9 ","b":"3 \/ 10 ","c":"1 \/ 2 ","d":"5 \/ 6","e":"8 \/ 9"},"options_float":{"a":0.1111111111,"b":0.3,"c":0.5,"d":0.8333333333,"e":0.8888888889},"annotated_formula":"divide(subtract(subtract(6, 1), 3), multiply(6, 3))","linear_formula":"multiply(n0,n1)|subtract(n0,n2)|subtract(#1,n1)|divide(#2,#0)","chain":"6 - 1<\/gadget>\n5<\/output>\n5 - 3<\/gadget>\n2<\/output>\n6 * 3<\/gadget>\n18<\/output>\n2 \/ 18<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":2182} +{"problem":"list r contains 5 numbers that have an average value of 60 . if the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number , what is the smallest possible value in the list ?","rationale":"the middle number is 60 . let the smallest number be x . then the largest number is 2 x + 20 . to make x as small as possible , let ' s make the remaining numbers as large as possible . so the second largest = the largest = 2 x + 20 . the second smallest should be equal to the median . the numbers are x , 60 , 60 , 2 x + 20 , 2 x + 20 . x + 60 + 60 + 2 x + 20 + 2 x + 20 = 5 * 60 = 300 5 x = 140 x = 28 the answer is b .","correct":"b","options":{"a":"25 ","b":"28 ","c":"31 ","d":"34","e":"37"},"options_float":{"a":25.0,"b":28.0,"c":31.0,"d":34.0,"e":37.0},"annotated_formula":"divide(subtract(subtract(subtract(subtract(multiply(60, 5), 60), 60), 20), 20), 5)","linear_formula":"multiply(n0,n1)|subtract(#0,n1)|subtract(#1,n1)|subtract(#2,n2)|subtract(#3,n2)|divide(#4,n0)","chain":"60 * 5<\/gadget>\n300<\/output>\n300 - 60<\/gadget>\n240<\/output>\n240 - 60<\/gadget>\n180<\/output>\n180 - 20<\/gadget>\n160<\/output>\n160 - 20<\/gadget>\n140<\/output>\n140 \/ 5<\/gadget>\n28<\/output>\n28<\/result>","index":2183} +{"problem":"if a - b = 6 and a 2 + b 2 = 50 , find the value of ab .","rationale":"\"explanation : 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 50 - 36 = 14 ab = 7 . answer : a\"","correct":"a","options":{"a":"7 ","b":"12 ","c":"15 ","d":"18","e":"20"},"options_float":{"a":7.0,"b":12.0,"c":15.0,"d":18.0,"e":20.0},"annotated_formula":"divide(subtract(50, power(6, 2)), 2)","linear_formula":"power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|","chain":"6 ** 2<\/gadget>\n36<\/output>\n50 - 36<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":2185} +{"problem":"in 5 football games thus far this season , barry has run for 98 , 107 , 85 , 89 , and 91 yards . at a minimum , how many yards m will he need to gain this sunday if he wants to keep his season average above 100 yards ?","rationale":"sorry the answer m = 141 is d .","correct":"d","options":{"a":"101 ","b":"116 ","c":"m = 130 ","d":"m = 131","e":"m = 141"},"options_float":{"a":101.0,"b":116.0,"c":130.0,"d":131.0,"e":141.0},"annotated_formula":"add(subtract(multiply(100, add(5, const_1)), add(91, add(89, add(85, add(98, 107))))), const_1)","linear_formula":"add(n0,const_1)|add(n1,n2)|add(n3,#1)|multiply(n6,#0)|add(n4,#2)|add(n5,#4)|subtract(#3,#5)|add(#6,const_1)","chain":"5 + 1<\/gadget>\n6<\/output>\n100 * 6<\/gadget>\n600<\/output>\n98 + 107<\/gadget>\n205<\/output>\n85 + 205<\/gadget>\n290<\/output>\n89 + 290<\/gadget>\n379<\/output>\n91 + 379<\/gadget>\n470<\/output>\n600 - 470<\/gadget>\n130<\/output>\n130 + 1<\/gadget>\n131<\/output>\n131<\/result>","index":2186} +{"problem":"the price of 3 pants and 6 t - shirts is rs . 750 . with the same money one can buy 1 pant and 12 t - shirts . if one wants to buy 8 t - shirts , how much shall she have to pay ?","rationale":"\"let the price of a pant and a t - shirt be rs . x and rs . y respectively . then , 3 x + 6 y = 750 . . . . ( i ) and x + 12 y = 750 . . . . ( ii ) divide equation ( i ) by 3 , we get the below equation . = x + 2 y = 250 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 12 y = 750 ( - ) x + 2 y = 250 - - - - - - - - - - - - - - - - 10 y = 500 - - - - - - - - - - - - - - - - y = 50 cost of 8 t - shirts = 8 * 50 = 400 answer : c\"","correct":"c","options":{"a":"150 ","b":"250 ","c":"400 ","d":"450","e":"550"},"options_float":{"a":150.0,"b":250.0,"c":400.0,"d":450.0,"e":550.0},"annotated_formula":"multiply(divide(750, add(multiply(divide(subtract(12, 6), subtract(3, 1)), 3), 6)), 8)","linear_formula":"subtract(n4,n1)|subtract(n0,n3)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)|divide(n2,#4)|multiply(n5,#5)|","chain":"12 - 6<\/gadget>\n6<\/output>\n3 - 1<\/gadget>\n2<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 + 6<\/gadget>\n15<\/output>\n750 \/ 15<\/gadget>\n50<\/output>\n50 * 8<\/gadget>\n400<\/output>\n400<\/result>","index":2187} +{"problem":"a certain debt will be paid in 65 installments from january 1 to december 31 of a certain year . each of the first 20 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 20 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ?","rationale":"\"total number of installments = 65 payment per installment for the first 20 installments = 410 payment per installment for the remaining 45 installments = 410 + 65 = 475 average = ( 20 * 410 + 45 * 475 ) \/ 65 = 455 answer b\"","correct":"b","options":{"a":"443 ","b":"455 ","c":"465 ","d":"468","e":"475"},"options_float":{"a":443.0,"b":455.0,"c":465.0,"d":468.0,"e":475.0},"annotated_formula":"divide(add(multiply(20, 410), multiply(add(410, 65), subtract(65, 20))), 65)","linear_formula":"add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0)|","chain":"20 * 410<\/gadget>\n8_200<\/output>\n410 + 65<\/gadget>\n475<\/output>\n65 - 20<\/gadget>\n45<\/output>\n475 * 45<\/gadget>\n21_375<\/output>\n8_200 + 21_375<\/gadget>\n29_575<\/output>\n29_575 \/ 65<\/gadget>\n455<\/output>\n455<\/result>","index":2188} +{"problem":"in 7 given numbers , the average of first 4 numbers is 4 and that of last 4 numbers is also 4 . if the average of these 7 numbers is 3 , the fourth number is ?","rationale":"let 7 numbers are a , b , c , d , e , f , g . given avg of first 4 nums are 4 . . . i . e ( a + b + c + d ) \/ 4 = 4 a + b + c + d = 16 given avg of last 4 nums are also 4 . . i . e . ( d + e + f + g ) \/ 4 = 4 d + e + f + g = 16 e + f + g = 16 - d given avg of 7 nums are 3 . . . . i . e ( a + b + c + d + e + f + g ) \/ 7 = 3 a + b + c + d + e + f + g = 21 16 + 16 - d = 21 d = 32 - 21 d = 11 answer : d","correct":"d","options":{"a":"3 ","b":"4 ","c":"7 ","d":"11","e":"12"},"options_float":{"a":3.0,"b":4.0,"c":7.0,"d":11.0,"e":12.0},"annotated_formula":"subtract(add(multiply(4, 4), multiply(4, 4)), multiply(7, 3))","linear_formula":"multiply(n1,n1)|multiply(n0,n6)|add(#0,#0)|subtract(#2,#1)","chain":"4 * 4<\/gadget>\n16<\/output>\n16 + 16<\/gadget>\n32<\/output>\n7 * 3<\/gadget>\n21<\/output>\n32 - 21<\/gadget>\n11<\/output>\n11<\/result>","index":2189} +{"problem":"at the beginning of a year , the owner of a jewel shop raised the price of all the jewels in his shop by x % and lowered them by x % . the price of one jewel after this up and down cycle reduced by rs . 100 . the owner carried out the same procedure after a month . after this second up - down cycle , the price of that jewel was rs . 2304 . find the original price of that jewel ( in rs . )","rationale":"solution : let the original price = y , after first change , it becomes , y * ( 1 + ( x \/ 100 ) ) after second change , it becomes y * ( 1 + ( x \/ 100 ) ) ( 1 - ( x \/ 100 ) ) = y ( 1 - ( x \/ 100 ) ^ 2 ) thus , x 2 * y = 106 - - - - - - - ( 1 ) x 2 = 106 \/ y now , y ( 1 - ( 106 \/ 10000 y ) ) 2 = 2304 ( similar to above ) y ( 1 - 100 \/ y ) 2 = 2304 y = 2500 . answer : option a","correct":"a","options":{"a":"2500 ","b":"550 ","c":"2600 ","d":"2650","e":"none of these"},"options_float":{"a":2500.0,"b":550.0,"c":2600.0,"d":2650.0,"e":null},"annotated_formula":"add(multiply(const_2, 100), 2304)","linear_formula":"multiply(n0,const_2)|add(n1,#0)","chain":"2 * 100<\/gadget>\n200<\/output>\n200 + 2_304<\/gadget>\n2_504<\/output>\n2_504<\/result>","index":2191} +{"problem":"in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds in the nature reserve are not hawks , paddyfield - warblers , or kingfishers ?","rationale":"\"since the problem does n ' t depend on the number of birds in reserve you can suppose that there are 100 birds . from them 100 * 0.3 = 30 are hawks . so , there are 70 non - hawks . there are 0.4 * 70 = 28 paddyfield - warblers . therefore , there are 0.25 * 28 = 7 kingfishers . the total number of birds in the nature reserve that are not hawks , paddyfield - warblers , or kingfishers is 100 - 30 - 28 - 7 = 35 . the answer is b\"","correct":"b","options":{"a":"25 % ","b":"35 % ","c":"45 % ","d":"70 %","e":"80 %"},"options_float":{"a":25.0,"b":35.0,"c":45.0,"d":70.0,"e":80.0},"annotated_formula":"add(const_10, divide(add(25, 25), const_2))","linear_formula":"add(n2,n2)|divide(#0,const_2)|add(#1,const_10)|","chain":"25 + 25<\/gadget>\n50<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n10 + 25<\/gadget>\n35<\/output>\n35<\/result>","index":2192} +{"problem":"alfred buys an old scooter for $ 4700 and spends $ 600 on its repairs . if he sells the scooter for $ 5800 , his gain percent is ?","rationale":"c . p . = 4700 + 600 = $ 5300 s . p . = $ 5800 gain = 5800 - 5300 = $ 500 gain % = 500 \/ 5300 * 100 = 9.43 % answer is d","correct":"d","options":{"a":"5.45 % ","b":"6.23 % ","c":"7 % ","d":"9.43 %","e":"10 %"},"options_float":{"a":5.45,"b":6.23,"c":7.0,"d":9.43,"e":10.0},"annotated_formula":"multiply(divide(subtract(5800, add(4700, 600)), add(4700, 600)), const_100)","linear_formula":"add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)","chain":"4_700 + 600<\/gadget>\n5_300<\/output>\n5_800 - 5_300<\/gadget>\n500<\/output>\n500 \/ 5_300<\/gadget>\n5\/53 = around 0.09434<\/output>\n(5\/53) * 100<\/gadget>\n500\/53 = around 9.433962<\/output>\n500\/53 = around 9.433962<\/result>","index":2194} +{"problem":"a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord , in feet ?","rationale":"\"because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . pi * r = 30 . round pi = 3 , then r = 10 . chord is about 10 feet long answer : d\"","correct":"d","options":{"a":"30 ","b":"25 ","c":"15 ","d":"10","e":"5"},"options_float":{"a":30.0,"b":25.0,"c":15.0,"d":10.0,"e":5.0},"annotated_formula":"divide(30, const_3)","linear_formula":"divide(n0,const_3)|","chain":"30 \/ 3<\/gadget>\n10<\/output>\n10<\/result>","index":2196} +{"problem":"the sum of two numbers is 66 . how much should be added to the sum for it to be exactly divisible by 9 leaving 0 as the reminder ?","rationale":"the sum is 66 . adding 6 to 66 = 66 + 6 = 72 . 72 is exactly divisible by 9 leaving 0 as the reminder = 8 . so the answer is a = 6","correct":"a","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"1"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":1.0},"annotated_formula":"subtract(subtract(subtract(subtract(const_100, 9), 9), const_10), 66)","linear_formula":"subtract(const_100,n1)|subtract(#0,n1)|subtract(#1,const_10)|subtract(#2,n0)","chain":"100 - 9<\/gadget>\n91<\/output>\n91 - 9<\/gadget>\n82<\/output>\n82 - 10<\/gadget>\n72<\/output>\n72 - 66<\/gadget>\n6<\/output>\n6<\/result>","index":2197} +{"problem":"what is next no . 11 23 47 83 131 191 263 347 443 551 ? ? ?","rationale":"11 + 12 * 1 = 23 23 + 12 * 2 = 47 47 + 12 * 3 = 83 83 + 12 * 4 = 131 131 + 12 * 5 = 191 191 + 12 * 6 = 263 263 + 12 * 7 = 347 347 + 12 * 8 = 443 443 + 12 * 9 = 551 551 + 12 * 10 = 671 answer : e","correct":"e","options":{"a":"171 ","b":"181 ","c":"191 ","d":"201","e":"671"},"options_float":{"a":171.0,"b":181.0,"c":191.0,"d":201.0,"e":671.0},"annotated_formula":"add(551, multiply(const_60, const_2))","linear_formula":"multiply(const_2,const_60)|add(n9,#0)","chain":"60 * 2<\/gadget>\n120<\/output>\n551 + 120<\/gadget>\n671<\/output>\n671<\/result>","index":2198} +{"problem":"zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 31.5 kg of zinc has been consumed in it ?","rationale":"\"sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 31.5 kg zinc , mixture , melted = [ 20 \/ 9 x 31.5 ] kg = 70 kg . answer d\"","correct":"d","options":{"a":"58 kg ","b":"60 kg ","c":"64 kg ","d":"70 kg","e":"none"},"options_float":{"a":58.0,"b":60.0,"c":64.0,"d":70.0,"e":null},"annotated_formula":"add(multiply(divide(11, 9), 31.5), 31.5)","linear_formula":"divide(n1,n0)|multiply(n2,#0)|add(n2,#1)|","chain":"11 \/ 9<\/gadget>\n11\/9 = around 1.222222<\/output>\n(11\/9) * 31.5<\/gadget>\n38.5<\/output>\n38.5 + 31.5<\/gadget>\n70<\/output>\n70<\/result>","index":2199} +{"problem":"a man is 20 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is","rationale":"\"explanation : let the son ' s present age be x years . then , man ' s present age = ( x + 20 ) years = > ( x + 20 ) + 2 = 2 ( x + 2 ) = > x + 22 = 2 x + 4 so , x = 18 . answer : d\"","correct":"d","options":{"a":"16 ","b":"14 ","c":"15 ","d":"18","e":"20"},"options_float":{"a":16.0,"b":14.0,"c":15.0,"d":18.0,"e":20.0},"annotated_formula":"divide(subtract(20, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))","linear_formula":"multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4 - 2<\/gadget>\n2<\/output>\n20 - 2<\/gadget>\n18<\/output>\n2 - 1<\/gadget>\n1<\/output>\n18 \/ 1<\/gadget>\n18<\/output>\n18<\/result>","index":2200} +{"problem":"a person buys an article at rs . 400 . at what price should he sell the article so as to make a profit of 40 % ?","rationale":"\"cost price = rs . 400 profit = 40 % of 400 = rs . 160 selling price = cost price + profit = 400 + 160 = 560 answer : c\"","correct":"c","options":{"a":"600 ","b":"277 ","c":"560 ","d":"261","e":"281"},"options_float":{"a":600.0,"b":277.0,"c":560.0,"d":261.0,"e":281.0},"annotated_formula":"add(400, multiply(400, divide(40, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n400 * (2\/5)<\/gadget>\n160<\/output>\n400 + 160<\/gadget>\n560<\/output>\n560<\/result>","index":2201} +{"problem":"an order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth . what be the cost of a carpet whose length and breadth are 40 % more and 25 % more respectively than the first carpet . given that the ratio of carpet is rs . 45 per sq m ?","rationale":"\"length of the first carpet = ( 1.44 ) ( 6 ) = 8.64 cm area of the second carpet = 8.64 ( 1 + 40 \/ 100 ) 6 ( 1 + 25 \/ 100 ) = 51.84 ( 1.4 ) ( 5 \/ 4 ) sq m = ( 12.96 ) ( 7 ) sq m cost of the second carpet = ( 45 ) ( 12.96 * 7 ) = 315 ( 13 - 0.04 ) = 4095 - 12.6 = rs . 4082.40 answer : d\"","correct":"d","options":{"a":"rs . 3642.40 ","b":"rs . 3868.80 ","c":"rs . 4216.20 ","d":"rs . 4082.40","e":"none of these"},"options_float":{"a":3642.4,"b":3868.8,"c":4216.2,"d":4082.4,"e":null},"annotated_formula":"multiply(multiply(multiply(multiply(6, 1.44), divide(add(const_100, 40), const_100)), multiply(6, divide(add(const_100, 25), const_100))), 45)","linear_formula":"add(n2,const_100)|add(n3,const_100)|multiply(n0,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#3,#2)|multiply(n0,#4)|multiply(#5,#6)|multiply(n4,#7)|","chain":"6 * 1.44<\/gadget>\n8.64<\/output>\n100 + 40<\/gadget>\n140<\/output>\n140 \/ 100<\/gadget>\n7\/5 = around 1.4<\/output>\n8.64 * (7\/5)<\/gadget>\n12.096<\/output>\n100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n6 * (5\/4)<\/gadget>\n15\/2 = around 7.5<\/output>\n12.096 * (15\/2)<\/gadget>\n90.72<\/output>\n90.72 * 45<\/gadget>\n4_082.4<\/output>\n4_082.4<\/result>","index":2202} +{"problem":"jack , jill , and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are 1 \/ 6 , 1 \/ 7 , and 1 \/ 8 respectively , what is the probability that jill and sandy will make a basket but jack will miss ?","rationale":"the probability that jill and sandy will make a basket but jack will miss is 5 \/ 6 * 1 \/ 7 * 1 \/ 8 = 5 \/ 336 . the answer is d .","correct":"d","options":{"a":"3 \/ 168 ","b":"5 \/ 168 ","c":"7 \/ 168 ","d":"5 \/ 336","e":"7 \/ 336"},"options_float":{"a":0.0178571429,"b":0.0297619048,"c":0.0416666667,"d":0.0148809524,"e":0.0208333333},"annotated_formula":"multiply(multiply(divide(1, 7), divide(1, 8)), subtract(const_1, divide(1, 6)))","linear_formula":"divide(n0,n3)|divide(n0,n5)|divide(n0,n1)|multiply(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)","chain":"1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/7) * (1\/8)<\/gadget>\n1\/56 = around 0.017857<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 - (1\/6)<\/gadget>\n5\/6 = around 0.833333<\/output>\n(1\/56) * (5\/6)<\/gadget>\n5\/336 = around 0.014881<\/output>\n5\/336 = around 0.014881<\/result>","index":2203} +{"problem":"a 11 % stock yields 8 % . the market value of the stock is :","rationale":"\"solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 11 , investment = rs . ( 100 \/ 8 x 11 ) = rs . 137.50 ∴ market value of rs . 100 stock = rs . 137.50 answer b\"","correct":"b","options":{"a":"rs . 72 ","b":"rs . 137.50 ","c":"rs . 112.50 ","d":"rs . 116.50","e":"none of these"},"options_float":{"a":72.0,"b":137.5,"c":112.5,"d":116.5,"e":null},"annotated_formula":"multiply(divide(const_100, 8), 11)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 8<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 11<\/gadget>\n275\/2 = around 137.5<\/output>\n275\/2 = around 137.5<\/result>","index":2205} +{"problem":"the ratio of the present age of viju to that of aju is 7 : 2 . 4 years from now , the ratio of the ages of viju to aju will be 5 : 2 . what was viju age 5 years ago ? ( rbi assistant 2015 )","rationale":"let the present age of viju be 7 x years and that of aju be 2 x years . then , 4 years from now 7 x + 4 \/ 2 x + 4 = 5 \/ 2 or 4 x = 12 or x = 3 viju present age = 7 * 3 = 21 years viju age 5 years ago = 21 - 5 = 16 years answer : d","correct":"d","options":{"a":"24 ","b":"18 ","c":"12 ","d":"16","e":"13"},"options_float":{"a":24.0,"b":18.0,"c":12.0,"d":16.0,"e":13.0},"annotated_formula":"subtract(divide(multiply(7, divide(subtract(multiply(5, 4), multiply(2, 4)), subtract(7, 5))), 2), 5)","linear_formula":"multiply(n2,n3)|multiply(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)|multiply(n0,#4)|divide(#5,n1)|subtract(#6,n3)","chain":"5 * 4<\/gadget>\n20<\/output>\n2 * 4<\/gadget>\n8<\/output>\n20 - 8<\/gadget>\n12<\/output>\n7 - 5<\/gadget>\n2<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n7 * 6<\/gadget>\n42<\/output>\n42 \/ 2<\/gadget>\n21<\/output>\n21 - 5<\/gadget>\n16<\/output>\n16<\/result>","index":2206} +{"problem":"yesterday it took robert 3 hours to drive from city a to city b . today it took robert 2.5 hours to drive back from city в to city a along the same route . if he had saved 30 minutes in both trips , the speed for the round trip would be 80 miles per hour . what is the distance between city a and city b ?","rationale":"\"2 d \/ 80 = 4.5 ( because time = 3 + 2.5 - 1 hrs ) = > d = 180 answer - b\"","correct":"b","options":{"a":"90 ","b":"180 ","c":"150 ","d":"240","e":"300"},"options_float":{"a":90.0,"b":180.0,"c":150.0,"d":240.0,"e":300.0},"annotated_formula":"divide(multiply(80, subtract(add(3, 2.5), const_1)), const_2)","linear_formula":"add(n0,n1)|subtract(#0,const_1)|multiply(n3,#1)|divide(#2,const_2)|","chain":"3 + 2.5<\/gadget>\n5.5<\/output>\n5.5 - 1<\/gadget>\n4.5<\/output>\n80 * 4.5<\/gadget>\n360<\/output>\n360 \/ 2<\/gadget>\n180<\/output>\n180<\/result>","index":2207} +{"problem":"if abc = ab ^ 3 = 180 where a , b and c are positive integers , what is the value of ` c ’ ?","rationale":"abc = a ( b ^ 3 ) = 180 on solving abc = a ( b ^ 3 ) we get c = b ^ 2 and for a ( b ^ 3 ) = 180 , there is only one possibility a = 180 and b = 1 so c = 1 ^ 2 = 1 answer : b","correct":"b","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"power(divide(180, 180), const_2)","linear_formula":"divide(n1,n1)|power(#0,const_2)","chain":"180 \/ 180<\/gadget>\n1<\/output>\n1 ** 2<\/gadget>\n1<\/output>\n1<\/result>","index":2208} +{"problem":"each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 95 % water by weight . what is the new weight of the cucumbers , in pounds ?","rationale":"\"out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 95 % water and 5 % of non - water , so now 1 pound of non - water composes 5 % of cucucmbers , which means that the new weight of cucumbers is 1 \/ 0.05 = 20 pounds . answer : b .\"","correct":"b","options":{"a":"2 ","b":"20 ","c":"92 ","d":"96","e":"98"},"options_float":{"a":2.0,"b":20.0,"c":92.0,"d":96.0,"e":98.0},"annotated_formula":"multiply(divide(subtract(100, 99), subtract(100, 95)), 100)","linear_formula":"subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|","chain":"100 - 99<\/gadget>\n1<\/output>\n100 - 95<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":2209} +{"problem":"a customer went to a shop and paid a total of $ 40 , out of which $ 1.28 was for sales tax on taxable purchases . if the tax rate was 8 % , then what was the cost of the tax free items ?","rationale":"\"the total cost was $ 40 . the tax was $ 1.28 let the original price of the taxable items = x given that tax rate = 8 % 0.08 x = 1.28 x = $ 16 the cost of the tax free items was $ 40 - $ 16 - $ 1.28 = $ 22.72 the answer is d .\"","correct":"d","options":{"a":"$ 19.54 ","b":"$ 2076 ","c":"$ 21.36 ","d":"$ 22.72","e":"$ 23.58"},"options_float":{"a":19.54,"b":2076.0,"c":21.36,"d":22.72,"e":23.58},"annotated_formula":"subtract(subtract(40, 1.28), multiply(8, const_2))","linear_formula":"multiply(n2,const_2)|subtract(n0,n1)|subtract(#1,#0)|","chain":"40 - 1.28<\/gadget>\n38.72<\/output>\n8 * 2<\/gadget>\n16<\/output>\n38.72 - 16<\/gadget>\n22.72<\/output>\n22.72<\/result>","index":2210} +{"problem":"the average of first 12 even numbers is ?","rationale":"\"sum of 12 even numbers = 12 * 13 = 156 average = 156 \/ 12 = 13 answer : d\"","correct":"d","options":{"a":"19 ","b":"18 ","c":"16 ","d":"13","e":"17"},"options_float":{"a":19.0,"b":18.0,"c":16.0,"d":13.0,"e":17.0},"annotated_formula":"add(12, const_1)","linear_formula":"add(n0,const_1)|","chain":"12 + 1<\/gadget>\n13<\/output>\n13<\/result>","index":2212} +{"problem":"if all 6 are replaced by 9 , the algebric sum of all numbers from 1 to 100 ( both inclusive ) varies by ,","rationale":"6 , 16,26 , 36,46 , 56,76 , 86,96 replace by 9 then 3 * 9 = 27 60,61 , 62,63 , 64,65 , 66,67 , 68,69 then increase by 30 i . e , 30 * 10 = 300 + 3 99 - 66 = 33 - 30 it will be 330 answer : b","correct":"b","options":{"a":"980 ","b":"330 ","c":"660 ","d":"990","e":"300"},"options_float":{"a":980.0,"b":330.0,"c":660.0,"d":990.0,"e":300.0},"annotated_formula":"add(multiply(subtract(9, 6), const_10), multiply(subtract(9, 6), const_100))","linear_formula":"subtract(n1,n0)|multiply(#0,const_10)|multiply(#0,const_100)|add(#1,#2)","chain":"9 - 6<\/gadget>\n3<\/output>\n3 * 10<\/gadget>\n30<\/output>\n3 * 100<\/gadget>\n300<\/output>\n30 + 300<\/gadget>\n330<\/output>\n330<\/result>","index":2213} +{"problem":"in a company with 48 employees , some part - time and some full - time , exactly ( 1 \/ 3 ) of the part - time employees and ( 1 \/ 4 ) of the full - time employees take the subway to work . what is the greatest possible number h of employees who take the subway to work ?","rationale":"\"p \/ 3 + f \/ 4 = p \/ 3 + ( 48 - p ) \/ 4 = 12 + p \/ 2 p \/ 3 + f \/ 3 = ( p + f ) \/ 3 = 48 \/ 3 = 16 p \/ 4 + f \/ 4 = 12 p \/ 3 + f \/ 3 > p \/ 3 + f \/ 4 > p \/ 4 + f \/ 4 - - > 16 > 12 + p \/ 12 > 12 greatest possible h : 12 + p \/ 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer\"","correct":"d","options":{"a":"12 ","b":"13 ","c":"14 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4))))","linear_formula":"multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)|","chain":"48 * 3<\/gadget>\n144<\/output>\n3 * 4<\/gadget>\n12<\/output>\n144 \/ 12<\/gadget>\n12<\/output>\n48 - 12<\/gadget>\n36<\/output>\n36 \/ 12<\/gadget>\n3<\/output>\n12 + 3<\/gadget>\n15<\/output>\n15<\/result>","index":2216} +{"problem":"a batsman in his 12 th innings makes a score of 60 and thereby increases his average by 2 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ?","rationale":"\"let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 2 ) + 60 ∴ x = 38 answer a\"","correct":"a","options":{"a":"38 ","b":"43 ","c":"44 ","d":"45","e":"46"},"options_float":{"a":38.0,"b":43.0,"c":44.0,"d":45.0,"e":46.0},"annotated_formula":"add(subtract(60, multiply(12, 2)), 2)","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|","chain":"12 * 2<\/gadget>\n24<\/output>\n60 - 24<\/gadget>\n36<\/output>\n36 + 2<\/gadget>\n38<\/output>\n38<\/result>","index":2218} +{"problem":"john can complete a given task in 18 days . jane will take only 12 days to complete the same task . john and jane set out to complete the task by beginning to work together . however , jane was indisposed 6 days before the work got over . in how many days did the work get over from the time john and jane started to work on it together ?","rationale":"\"in such questions , you need to start from the end . last 4 days john works alone and completes 6 * ( 1 \/ 18 ) = 1 \/ 3 of the work . so 2 \/ 3 of the work should have been completed by the two of them together before jane left . their combined rate of work is 1 \/ 18 + 1 \/ 12 = 5 \/ 36 time taken to complete 2 \/ 3 of the work = ( 2 \/ 3 ) \/ ( 5 \/ 36 ) = 24 \/ 5 days . so total number of days taken to complete the work = 24 \/ 5 + 6 = 54 \/ 5 days . d\"","correct":"d","options":{"a":"11 ","b":"34 \/ 5 ","c":"44 \/ 5 ","d":"54 \/ 5","e":"12"},"options_float":{"a":11.0,"b":6.8,"c":8.8,"d":10.8,"e":12.0},"annotated_formula":"add(6, divide(subtract(multiply(18, 12), multiply(12, 6)), add(12, 18)))","linear_formula":"add(n0,n1)|multiply(n0,n1)|multiply(n1,n2)|subtract(#1,#2)|divide(#3,#0)|add(n2,#4)|","chain":"18 * 12<\/gadget>\n216<\/output>\n12 * 6<\/gadget>\n72<\/output>\n216 - 72<\/gadget>\n144<\/output>\n12 + 18<\/gadget>\n30<\/output>\n144 \/ 30<\/gadget>\n24\/5 = around 4.8<\/output>\n6 + (24\/5)<\/gadget>\n54\/5 = around 10.8<\/output>\n54\/5 = around 10.8<\/result>","index":2219} +{"problem":"a man walks at a rate of 10 mph . after every ten miles , he rests for 7 minutes . how much time does he take to walk 50 miles ?","rationale":"\"to cover 50 miles the man needs ( time ) = ( distance ) \/ ( rate ) = 50 \/ 10 = 5 hours = 300 minutes . he will also rest 4 times ( after 10 , 20 , 30 and 40 miles ) , so total resting time = 4 * 7 = 28 minutes . total time = 300 + 28 = 328 minutes . answer : d .\"","correct":"d","options":{"a":"300 ","b":"318 ","c":"322 ","d":"328","e":"330"},"options_float":{"a":300.0,"b":318.0,"c":322.0,"d":328.0,"e":330.0},"annotated_formula":"add(multiply(7, const_4), multiply(divide(50, 10), const_60))","linear_formula":"divide(n2,n0)|multiply(n1,const_4)|multiply(#0,const_60)|add(#1,#2)|","chain":"7 * 4<\/gadget>\n28<\/output>\n50 \/ 10<\/gadget>\n5<\/output>\n5 * 60<\/gadget>\n300<\/output>\n28 + 300<\/gadget>\n328<\/output>\n328<\/result>","index":2220} +{"problem":"by selling an article for $ 225 a person gains $ 75 . what is the gain % ?","rationale":"\"s . p . = $ 225 gain = $ 75 c . p . = 225 - 75 = 150 gain % = 75 \/ 150 * 100 = 50 % answer is c\"","correct":"c","options":{"a":"25 % ","b":"30 % ","c":"50 % ","d":"20 %","e":"10 %"},"options_float":{"a":25.0,"b":30.0,"c":50.0,"d":20.0,"e":10.0},"annotated_formula":"divide(multiply(75, const_100), subtract(225, 75))","linear_formula":"multiply(n1,const_100)|subtract(n0,n1)|divide(#0,#1)|","chain":"75 * 100<\/gadget>\n7_500<\/output>\n225 - 75<\/gadget>\n150<\/output>\n7_500 \/ 150<\/gadget>\n50<\/output>\n50<\/result>","index":2221} +{"problem":"a mixture of 150 liters of wine and water contains 15 % water . how much more water should be added so that water becomes 25 % of the new mixture ?","rationale":"\"number of liters of water in 150 liters of the mixture = 15 % of 150 = 15 \/ 100 * 150 = 22.5 liters . p liters of water added to the mixture to make water 25 % of the new mixture . total amount of water becomes ( 22.5 + p ) and total volume of mixture is ( 150 + p ) . ( 22.5 + p ) = 25 \/ 100 * ( 150 + p ) 90 + 4 p = 150 + p p = 20 liters . answer : c\"","correct":"c","options":{"a":"10 ","b":"30 ","c":"20 ","d":"15","e":"17"},"options_float":{"a":10.0,"b":30.0,"c":20.0,"d":15.0,"e":17.0},"annotated_formula":"divide(subtract(multiply(divide(25, const_100), 150), multiply(divide(15, const_100), 150)), subtract(const_1, divide(25, const_100)))","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 150<\/gadget>\n75\/2 = around 37.5<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 150<\/gadget>\n45\/2 = around 22.5<\/output>\n(75\/2) - (45\/2)<\/gadget>\n15<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n15 \/ (3\/4)<\/gadget>\n20<\/output>\n20<\/result>","index":2222} +{"problem":"it is only in the second option we find the sentence a in continuation with the opening sentence d . sentence a alone explains , why the fear has such enormous effect on the human beings .","rationale":"answer : b","correct":"b","options":{"a":"28 ","b":"99 ","c":"728 ","d":"27","e":"271"},"options_float":{"a":28.0,"b":99.0,"c":728.0,"d":27.0,"e":271.0},"annotated_formula":"subtract(const_100, const_1)","linear_formula":"subtract(const_100,const_1)","chain":"100 - 1<\/gadget>\n99<\/output>\n99<\/result>","index":2223} +{"problem":"on a sum of money , the simple interest for 2 years is rs . 327 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is","rationale":"\"explanation : the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r ã — si ) \/ ( 2 ã — 100 ) difference between the compound interest and simple interest = 340 - 327 = 13 ( r ã — si ) \/ ( 2 ã — 100 ) = 13 ( r ã — 327 ) \/ ( 2 ã — 100 ) = 13 r = 7.95 % answer : option c\"","correct":"c","options":{"a":"15 % ","b":"14.25 % ","c":"7.95 % ","d":"10.5 %","e":"11.5 %"},"options_float":{"a":15.0,"b":14.25,"c":7.95,"d":10.5,"e":11.5},"annotated_formula":"divide(multiply(const_100, subtract(subtract(340, divide(327, 2)), divide(327, 2))), divide(327, 2))","linear_formula":"divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)|","chain":"327 \/ 2<\/gadget>\n327\/2 = around 163.5<\/output>\n340 - (327\/2)<\/gadget>\n353\/2 = around 176.5<\/output>\n(353\/2) - (327\/2)<\/gadget>\n13<\/output>\n100 * 13<\/gadget>\n1_300<\/output>\n1_300 \/ (327\/2)<\/gadget>\n2_600\/327 = around 7.95107<\/output>\n2_600\/327 = around 7.95107<\/result>","index":2224} +{"problem":"my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 120 years . can you tell me my age in years ?","rationale":"\"b i am 72 years old . let m be my age in years . if s is my son ' s age in years , then my son is 52 s weeks old . if g is my grandson ' s age in years , then my grandson is 365 g days old . thus , 365 g = 52 s . since my grandson is 12 g months old , 12 g = m . since my grandson , my son and i together are 120 years , g + s + m = 120 . the above system of 3 equations in 3 unknowns ( g , s and m ) can be solved as follows . m \/ 12 + 365 m \/ ( 52 x 12 ) + m = 120 or 52 m + 365 m + 624 m = 624 x 120 or m = 624 x 120 \/ 1041 = 72 . so , i am 72 years old .\"","correct":"b","options":{"a":"65 ","b":"72 ","c":"44 ","d":"73","e":"89"},"options_float":{"a":65.0,"b":72.0,"c":44.0,"d":73.0,"e":89.0},"annotated_formula":"multiply(divide(120, add(const_10, const_10)), const_12)","linear_formula":"add(const_10,const_10)|divide(n0,#0)|multiply(#1,const_12)|","chain":"10 + 10<\/gadget>\n20<\/output>\n120 \/ 20<\/gadget>\n6<\/output>\n6 * 12<\/gadget>\n72<\/output>\n72<\/result>","index":2225} +{"problem":"the ratio between speed of the two trains is 7 : 8 . if the 2 nd train runs 400 km in 4 hrs , what is the speed of the 1 st train ?","rationale":"answer : b explanation : speed of 2 nd train = 400 \/ 4 = 100 kmph since the ratios are in 7 : 8 speed of first train = 7 \/ 8 × 100 = 87.5 kmph answer : b","correct":"b","options":{"a":"87.9 ","b":"87.5 ","c":"87.7 ","d":"87.6","e":"43.3"},"options_float":{"a":87.9,"b":87.5,"c":87.7,"d":87.6,"e":43.3},"annotated_formula":"multiply(divide(400, 4), divide(7, 8))","linear_formula":"divide(n3,n4)|divide(n0,n1)|multiply(#0,#1)","chain":"400 \/ 4<\/gadget>\n100<\/output>\n7 \/ 8<\/gadget>\n7\/8 = around 0.875<\/output>\n100 * (7\/8)<\/gadget>\n175\/2 = around 87.5<\/output>\n175\/2 = around 87.5<\/result>","index":2228} +{"problem":"the circulation for magazine p in 1961 was 4 times the average ( arithmetic mean ) yearly circulation for magazine p for the years 1962 - 1970 . what is the ratio of the circulation in 1961 to the total circulation during 1961 - 1970 for magazine p ?","rationale":"\"there are 9 years from 1962 - 1970 , inclusive . let ' s say the average circulation every year between 1962 - 1970 inclusive is x . so the total circulation is 9 x from 1962 - 1970 , inclusive . in 1961 , the circulation is 4 x . so total circulation for 1961 - 1970 is 4 x + 9 x = 13 x . ratio of circulation in 1961 to total circulation during 1961 - 1970 is 4 x to 14 x or 4 \/ 14 = 2 \/ 7 answer : a\"","correct":"a","options":{"a":"2 \/ 7 ","b":"5 \/ 7 ","c":"3 \/ 7 ","d":"4 \/ 7","e":"1 \/ 7"},"options_float":{"a":0.2857142857,"b":0.7142857143,"c":0.4285714286,"d":0.5714285714,"e":0.1428571429},"annotated_formula":"divide(4, add(add(add(subtract(1970, 1962), const_1), 4), const_1))","linear_formula":"subtract(n3,n2)|add(#0,const_1)|add(n1,#1)|add(#2,const_1)|divide(n1,#3)|","chain":"1_970 - 1_962<\/gadget>\n8<\/output>\n8 + 1<\/gadget>\n9<\/output>\n9 + 4<\/gadget>\n13<\/output>\n13 + 1<\/gadget>\n14<\/output>\n4 \/ 14<\/gadget>\n2\/7 = around 0.285714<\/output>\n2\/7 = around 0.285714<\/result>","index":2229} +{"problem":"in how many years rs 100 will produce the same interest at 5 % as rs . 200 produce in 2 years at 10 %","rationale":"\"explanation : clue : firstly we need to calculate the si with prinical 200 , time 2 years and rate 10 % , it will be rs . 40 then we can get the time as time = ( 100 * 40 ) \/ ( 100 * 5 ) = 8 option a\"","correct":"a","options":{"a":"8 ","b":"10 ","c":"9 ","d":"12","e":"11"},"options_float":{"a":8.0,"b":10.0,"c":9.0,"d":12.0,"e":11.0},"annotated_formula":"divide(multiply(divide(multiply(200, 10), const_100), 2), multiply(divide(5, const_100), 100))","linear_formula":"divide(n1,const_100)|multiply(n2,n4)|divide(#1,const_100)|multiply(n0,#0)|multiply(n3,#2)|divide(#4,#3)|","chain":"200 * 10<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8<\/result>","index":2230} +{"problem":"the profit obtained by selling an article for rs . 54 is the same as the loss obtained by selling it for rs . 40 . what is the cost price of the article ?","rationale":"\"s . p 1 - c . p = c . p – s . p 2 54 - c . p = c . p - 40 2 c . p = 54 + 40 ; c . p = 94 \/ 2 = 47 answer : e\"","correct":"e","options":{"a":"rs . 40 ","b":"rs . 50 ","c":"rs . 49 ","d":"rs . 59","e":"rs . 47"},"options_float":{"a":40.0,"b":50.0,"c":49.0,"d":59.0,"e":47.0},"annotated_formula":"divide(add(54, 40), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"54 + 40<\/gadget>\n94<\/output>\n94 \/ 2<\/gadget>\n47<\/output>\n47<\/result>","index":2231} +{"problem":"mary , peter , and lucy were picking chestnuts . mary picked twice as much chestnuts than peter . lucy picked 2 kg more than peter . together the 3 of them picked 26 kg of chestnuts . how many kilograms did mary pick ?","rationale":"let x be the amount peter picked . then mary and lucy picked 2 x and x + 2 x , respectively . so x + 2 x + x + 2 = 26 4 x = 24 x = 6 so 2 x = 2 * 6 = 12 correct answer is a ) 12","correct":"a","options":{"a":"12 ","b":"24 ","c":"48 ","d":"52","e":"37"},"options_float":{"a":12.0,"b":24.0,"c":48.0,"d":52.0,"e":37.0},"annotated_formula":"multiply(divide(subtract(26, 2), add(2, 2)), 2)","linear_formula":"add(n0,n0)|subtract(n2,n0)|divide(#1,#0)|multiply(n0,#2)","chain":"26 - 2<\/gadget>\n24<\/output>\n2 + 2<\/gadget>\n4<\/output>\n24 \/ 4<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12<\/result>","index":2232} +{"problem":"a train traveling at 72 kmph crosses a platform in 35 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ?","rationale":"\"speed of train = 72 * ( 5 \/ 18 ) = 20 m \/ s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 35 = 700 m . so , length of platform = 700 - 360 = 340 m imo , answer a\"","correct":"a","options":{"a":"340 meters ","b":"360 meters ","c":"420 meters ","d":"600 meters","e":"can not be determined"},"options_float":{"a":340.0,"b":360.0,"c":420.0,"d":600.0,"e":null},"annotated_formula":"subtract(multiply(divide(multiply(72, const_1000), const_3600), 35), multiply(divide(multiply(72, const_1000), const_3600), 18))","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)|","chain":"72 * 1_000<\/gadget>\n72_000<\/output>\n72_000 \/ 3_600<\/gadget>\n20<\/output>\n20 * 35<\/gadget>\n700<\/output>\n20 * 18<\/gadget>\n360<\/output>\n700 - 360<\/gadget>\n340<\/output>\n340<\/result>","index":2233} +{"problem":"a train of 30 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ?","rationale":"\"d = 30 * 60 + 1500 = 3300 m t = 3300 \/ 60 * 18 \/ 5 = 198 sec = 3.3 mins answer : b\"","correct":"b","options":{"a":"6 ","b":"3.3 ","c":"4 ","d":"9","e":"2"},"options_float":{"a":6.0,"b":3.3,"c":4.0,"d":9.0,"e":2.0},"annotated_formula":"add(divide(multiply(add(30, const_1), 60), const_1000), 1.5)","linear_formula":"add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)|","chain":"30 + 1<\/gadget>\n31<\/output>\n31 * 60<\/gadget>\n1_860<\/output>\n1_860 \/ 1_000<\/gadget>\n93\/50 = around 1.86<\/output>\n(93\/50) + 1.5<\/gadget>\n3.36<\/output>\n3.36<\/result>","index":2235} +{"problem":"in the manufacture of a certain product , 8 percent of the units produced are defective and 5 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ?","rationale":"\"0.08 * 0.05 = 0.004 = 0.4 % the answer is b .\"","correct":"b","options":{"a":"0.2 % ","b":"0.4 % ","c":"0.8 % ","d":"1.2 %","e":"2.0 %"},"options_float":{"a":0.2,"b":0.4,"c":0.8,"d":1.2,"e":2.0},"annotated_formula":"multiply(8, divide(5, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n8 * (1\/20)<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":2236} +{"problem":"a number when divided by 779 gives a remainder 47 . by dividing the same number by 19 , what would be the remainder ?","rationale":"\"number = ( 779 x a ) + 47 , where ` ` a ' ' is the quotient = ( 19 x 41 x a ) + ( 19 x 2 ) + 9 = 19 x ( 41 a + 2 ) + 9 = 19 x ( new quotient ) + 9 \\ inline \\ fn _ jvn \\ therefore required remainder = 9 answer : a\"","correct":"a","options":{"a":"9 ","b":"8 ","c":"7 ","d":"6","e":"5"},"options_float":{"a":9.0,"b":8.0,"c":7.0,"d":6.0,"e":5.0},"annotated_formula":"subtract(47, multiply(19, const_2))","linear_formula":"multiply(n2,const_2)|subtract(n1,#0)|","chain":"19 * 2<\/gadget>\n38<\/output>\n47 - 38<\/gadget>\n9<\/output>\n9<\/result>","index":2237} +{"problem":"the cost price of 30 articles is the same as the selling price of x articles . if the profit is 20 % , what is x ?","rationale":"\"let the cost price = y the cost price of 30 articles = 30 y the selling price of x articles = 1.20 y * x 1.20 y * x = 30 y x = 30 \/ 1.2 = 25 the answer is d .\"","correct":"d","options":{"a":"20 ","b":"21 ","c":"23 ","d":"25","e":"27"},"options_float":{"a":20.0,"b":21.0,"c":23.0,"d":25.0,"e":27.0},"annotated_formula":"divide(multiply(30, const_4), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|","chain":"30 * 4<\/gadget>\n120<\/output>\n4 + 1<\/gadget>\n5<\/output>\n120 \/ 5<\/gadget>\n24<\/output>\n24<\/result>","index":2243} +{"problem":"if 32.5 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ?","rationale":"\"students enrolled in biology are 32.5 % and therefore not enrolled are 67.5 % . so of 880 is 880 * . 675 = 594 answer is d 594\"","correct":"d","options":{"a":"110 ","b":"330 ","c":"550 ","d":"594","e":"880"},"options_float":{"a":110.0,"b":330.0,"c":550.0,"d":594.0,"e":880.0},"annotated_formula":"multiply(divide(880, const_100), subtract(const_100, 32.5))","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|","chain":"880 \/ 100<\/gadget>\n44\/5 = around 8.8<\/output>\n100 - 32.5<\/gadget>\n67.5<\/output>\n(44\/5) * 67.5<\/gadget>\n594<\/output>\n594<\/result>","index":2244} +{"problem":"a sum fetched a total simple interest of $ 4151.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?","rationale":"\"e 8928 principal = $ 100 x 4151.25 \/ 9 x 5 = $ 415125 \/ 45 = $ 8928 .\"","correct":"e","options":{"a":"$ 8829 ","b":"$ 2840 ","c":"$ 6578 ","d":"$ 7782","e":"$ 8928"},"options_float":{"a":8829.0,"b":2840.0,"c":6578.0,"d":7782.0,"e":8928.0},"annotated_formula":"divide(divide(multiply(4151.25, const_100), 9), 5)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"4_151.25 * 100<\/gadget>\n415_125<\/output>\n415_125 \/ 9<\/gadget>\n46_125<\/output>\n46_125 \/ 5<\/gadget>\n9_225<\/output>\n9_225<\/result>","index":2246} +{"problem":"a car averages 30 miles per hour for the first 5 hours of a trip and averages 42 miles per hour for the rest of the trip . if the average speed for the entire trip is 34 miles per hour , how many hours long is the trip ?","rationale":"let t be the total time of the trip . 30 * 5 + 42 ( t - 5 ) = 34 t 8 t = 210 - 150 t = 7.5 the answer is b .","correct":"b","options":{"a":"6.5 ","b":"7.5 ","c":"8.5 ","d":"9.5","e":"10.5"},"options_float":{"a":6.5,"b":7.5,"c":8.5,"d":9.5,"e":10.5},"annotated_formula":"add(divide(subtract(multiply(34, 5), multiply(30, 5)), subtract(42, 34)), 5)","linear_formula":"multiply(n1,n3)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)","chain":"34 * 5<\/gadget>\n170<\/output>\n30 * 5<\/gadget>\n150<\/output>\n170 - 150<\/gadget>\n20<\/output>\n42 - 34<\/gadget>\n8<\/output>\n20 \/ 8<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) + 5<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":2248} +{"problem":"if jack walked 6 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ?","rationale":"\"distance walked in 1 hour and 15 mins = 6 miles speed per hour = distance \/ time = 6 \/ ( 5 \/ 4 ) = 4.8 miles per hour answer b\"","correct":"b","options":{"a":"4 ","b":"4.8 ","c":"6 ","d":"6.25","e":"15"},"options_float":{"a":4.0,"b":4.8,"c":6.0,"d":6.25,"e":15.0},"annotated_formula":"divide(multiply(6, const_60), add(15, const_60))","linear_formula":"add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0)|","chain":"6 * 60<\/gadget>\n360<\/output>\n15 + 60<\/gadget>\n75<\/output>\n360 \/ 75<\/gadget>\n24\/5 = around 4.8<\/output>\n24\/5 = around 4.8<\/result>","index":2250} +{"problem":"in a group of cows and hens , the number of legs are 8 more than twice the number of heads . the number of cows is :","rationale":"let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 8 2 x = 8 x = 4 . answer : e","correct":"e","options":{"a":"5 ","b":"6 ","c":"7 ","d":"10","e":"4"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":10.0,"e":4.0},"annotated_formula":"subtract(8, const_4)","linear_formula":"subtract(n0,const_4)","chain":"8 - 4<\/gadget>\n4<\/output>\n4<\/result>","index":2251} +{"problem":"the number of positive integer solutions for the equation x + y + z + t = 30 is","rationale":"\"the number of positive integer solutions for the equatio fx 1 + x 2 + ⋯ + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 30 - 1 c 4 - 1 = 29 c 3 = 3654 answer : b\"","correct":"b","options":{"a":"3854 ","b":"3654 ","c":"3764 ","d":"3960","e":"4964"},"options_float":{"a":3854.0,"b":3654.0,"c":3764.0,"d":3960.0,"e":4964.0},"annotated_formula":"divide(factorial(subtract(30, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(30, const_1), subtract(const_4, const_1)))))","linear_formula":"subtract(n0,const_1)|subtract(const_4,const_1)|factorial(#0)|factorial(#1)|subtract(#0,#1)|factorial(#4)|multiply(#3,#5)|divide(#2,#6)|","chain":"30 - 1<\/gadget>\n29<\/output>\nfactorial(29)<\/gadget>\n8_841_761_993_739_701_954_543_616_000_000<\/output>\n4 - 1<\/gadget>\n3<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n29 - 3<\/gadget>\n26<\/output>\nfactorial(26)<\/gadget>\n403_291_461_126_605_635_584_000_000<\/output>\n6 * 403_291_461_126_605_635_584_000_000<\/gadget>\n2_419_748_766_759_633_813_504_000_000<\/output>\n8_841_761_993_739_701_954_543_616_000_000 \/ 2_419_748_766_759_633_813_504_000_000<\/gadget>\n3_654<\/output>\n3_654<\/result>","index":2252} +{"problem":"reeya obtained 55 , 67 , 76 , 82 and 85 out of 100 in different subjects , what will be the average","rationale":"\"explanation : ( 55 + 67 + 76 + 82 + 85 \/ 5 ) = 73 option c\"","correct":"c","options":{"a":"70 ","b":"75 ","c":"73 ","d":"85","e":"90"},"options_float":{"a":70.0,"b":75.0,"c":73.0,"d":85.0,"e":90.0},"annotated_formula":"divide(add(add(add(add(55, 67), 76), 82), 85), add(const_4, const_1))","linear_formula":"add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|","chain":"55 + 67<\/gadget>\n122<\/output>\n122 + 76<\/gadget>\n198<\/output>\n198 + 82<\/gadget>\n280<\/output>\n280 + 85<\/gadget>\n365<\/output>\n4 + 1<\/gadget>\n5<\/output>\n365 \/ 5<\/gadget>\n73<\/output>\n73<\/result>","index":2253} +{"problem":"one night 17 percent of the female officers on a police force were on duty . if 204 police officers were on duty that night and half of these were female officers , how many female officers were on the police force ?","rationale":"\"let x be the number of female police officers on the police force . the number of female police officers on duty was 102 . 0.17 x = 102 x = 600 the answer is d .\"","correct":"d","options":{"a":"420 ","b":"480 ","c":"540 ","d":"600","e":"660"},"options_float":{"a":420.0,"b":480.0,"c":540.0,"d":600.0,"e":660.0},"annotated_formula":"divide(divide(204, const_2), divide(17, const_100))","linear_formula":"divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)|","chain":"204 \/ 2<\/gadget>\n102<\/output>\n17 \/ 100<\/gadget>\n17\/100 = around 0.17<\/output>\n102 \/ (17\/100)<\/gadget>\n600<\/output>\n600<\/result>","index":2255} +{"problem":"the workforce of company x is 60 % female . the company hired 24 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ?","rationale":"\"let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 24 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 24 ) 0.05 x = 13.2 x = 264 - this is quantity of employees before adding 24 men so after adding it will be 288 answer is c\"","correct":"c","options":{"a":"160 ","b":"220 ","c":"288 ","d":"360","e":"420"},"options_float":{"a":160.0,"b":220.0,"c":288.0,"d":360.0,"e":420.0},"annotated_formula":"add(divide(multiply(divide(55, const_100), 24), subtract(divide(60, const_100), divide(55, const_100))), 24)","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)|","chain":"55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n(11\/20) * 24<\/gadget>\n66\/5 = around 13.2<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) - (11\/20)<\/gadget>\n1\/20 = around 0.05<\/output>\n(66\/5) \/ (1\/20)<\/gadget>\n264<\/output>\n264 + 24<\/gadget>\n288<\/output>\n288<\/result>","index":2257} +{"problem":"1370 , x , 1070 , - 180 , - 6430","rationale":"1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 answer : a .","correct":"a","options":{"a":"1320 ","b":"6530 ","c":"6630 ","d":"6730","e":"6830"},"options_float":{"a":1320.0,"b":6530.0,"c":6630.0,"d":6730.0,"e":6830.0},"annotated_formula":"add(add(multiply(const_10, const_2), subtract(1370, 180)), const_100)","linear_formula":"multiply(const_10,const_2)|subtract(n0,n2)|add(#0,#1)|add(#2,const_100)","chain":"10 * 2<\/gadget>\n20<\/output>\n1_370 - 180<\/gadget>\n1_190<\/output>\n20 + 1_190<\/gadget>\n1_210<\/output>\n1_210 + 100<\/gadget>\n1_310<\/output>\n1_310<\/result>","index":2261} +{"problem":"the perimeter of a square is equal to the perimeter of a rectangle of length 17 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places )","rationale":"\"let the side of the square be a cm . perimeter of the rectangle = 2 ( 17 + 14 ) = 62 cm perimeter of the square = 62 cm i . e . 4 a = 62 a = 15.5 diameter of the semicircle = 15.5 cm circumference of the semicircle = 1 \/ 2 ( ∏ ) ( 15.5 ) = 1 \/ 2 ( 22 \/ 7 ) ( 15.5 ) = 24.35 cm to two decimal places answer : d\"","correct":"d","options":{"a":"23.57 ","b":"23.54 ","c":"23.5 ","d":"24.35","e":"24.51"},"options_float":{"a":23.57,"b":23.54,"c":23.5,"d":24.35,"e":24.51},"annotated_formula":"floor(divide(circumface(divide(divide(multiply(const_2, add(14, 17)), const_4), const_2)), const_2))","linear_formula":"add(n0,n1)|multiply(#0,const_2)|divide(#1,const_4)|divide(#2,const_2)|circumface(#3)|divide(#4,const_2)|floor(#5)|","chain":"14 + 17<\/gadget>\n31<\/output>\n2 * 31<\/gadget>\n62<\/output>\n62 \/ 4<\/gadget>\n31\/2 = around 15.5<\/output>\n(31\/2) \/ 2<\/gadget>\n31\/4 = around 7.75<\/output>\n2 * pi * (31\/4)<\/gadget>\n31*pi\/2 = around 48.694686<\/output>\n(31*pi\/2) \/ 2<\/gadget>\n31*pi\/4 = around 24.347343<\/output>\nfloor(31*pi\/4)<\/gadget>\n24<\/output>\n24<\/result>","index":2263} +{"problem":"? % of 360 = 180","rationale":"\"? % of 360 = 180 or , ? = 180 × 100 \/ 360 = 50 answer c\"","correct":"c","options":{"a":"277 ","b":"36 ","c":"50 ","d":"72","e":"none of these"},"options_float":{"a":277.0,"b":36.0,"c":50.0,"d":72.0,"e":null},"annotated_formula":"divide(multiply(180, const_100), 360)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"180 * 100<\/gadget>\n18_000<\/output>\n18_000 \/ 360<\/gadget>\n50<\/output>\n50<\/result>","index":2264} +{"problem":"the average of runs of a cricket player of 10 innings was 33 . how many runs must he make in his next innings so as to increase his average of runs by 4 ?","rationale":"\"explanation : average = total runs \/ no . of innings = 33 so , total = average x no . of innings = 33 x 10 = 330 . now increase in avg = 4 runs . so , new avg = 33 + 4 = 37 runs total runs = new avg x new no . of innings = 37 x 11 = 407 runs made in the 11 th inning = 407 - 330 = 77 answer : a\"","correct":"a","options":{"a":"77 ","b":"79 ","c":"85 ","d":"87","e":"89"},"options_float":{"a":77.0,"b":79.0,"c":85.0,"d":87.0,"e":89.0},"annotated_formula":"subtract(multiply(add(10, const_1), add(4, 33)), multiply(10, 33))","linear_formula":"add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"10 + 1<\/gadget>\n11<\/output>\n4 + 33<\/gadget>\n37<\/output>\n11 * 37<\/gadget>\n407<\/output>\n10 * 33<\/gadget>\n330<\/output>\n407 - 330<\/gadget>\n77<\/output>\n77<\/result>","index":2265} +{"problem":"the average score of a cricketer in 2 matches is 60 and in other 3 matches is 50 . then find the average score in all the 5 matches ?","rationale":"\"average in 5 matches = ( 2 * 60 + 3 * 50 ) \/ 2 + 3 = 120 + 150 \/ 5 = 270 \/ 5 = 54 answer is a\"","correct":"a","options":{"a":"54 ","b":"27 ","c":"30 ","d":"35","e":"42"},"options_float":{"a":54.0,"b":27.0,"c":30.0,"d":35.0,"e":42.0},"annotated_formula":"divide(add(multiply(2, 60), multiply(3, 50)), 5)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)|","chain":"2 * 60<\/gadget>\n120<\/output>\n3 * 50<\/gadget>\n150<\/output>\n120 + 150<\/gadget>\n270<\/output>\n270 \/ 5<\/gadget>\n54<\/output>\n54<\/result>","index":2267} +{"problem":"albert invested rs . 8000 in a scheme for 2 years at compound interest rate 5 % p . a . how much amount will albert get on maturity of the fixed deposit ?","rationale":"\"amount = [ 8000 * ( 1 + 5 \/ 100 ) 2 ] = 8000 * 21 \/ 20 * 21 \/ 20 = rs . 8820 answer : e\"","correct":"e","options":{"a":"8840 ","b":"2776 ","c":"2998 ","d":"2662","e":"8820"},"options_float":{"a":8840.0,"b":2776.0,"c":2998.0,"d":2662.0,"e":8820.0},"annotated_formula":"multiply(8000, power(add(const_1, divide(5, const_100)), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 + (1\/20)<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n8_000 * (441\/400)<\/gadget>\n8_820<\/output>\n8_820<\/result>","index":2270} +{"problem":"a person buys an article at $ 600 . at what price should he sell the article so as to make a profit of 8 % ?","rationale":"\"c 648 cost price = $ 600 profit = 8 % of 600 = $ 48 selling price = cost price + profit = 600 + 48 = 648\"","correct":"c","options":{"a":"670 ","b":"689 ","c":"648 ","d":"740","e":"889"},"options_float":{"a":670.0,"b":689.0,"c":648.0,"d":740.0,"e":889.0},"annotated_formula":"add(600, multiply(600, divide(8, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n600 * (2\/25)<\/gadget>\n48<\/output>\n600 + 48<\/gadget>\n648<\/output>\n648<\/result>","index":2271} +{"problem":"what is the remainder when ( 63 ) ( 65 ) is divided by 8 ?","rationale":"\"( 63 ) ( 65 ) = ( 64 - 1 ) ( 64 + 1 ) = 64 ^ 2 - 1 which is 1 less than a multiple of 8 . then the remainder will be 7 . the answer is e .\"","correct":"e","options":{"a":"1 ","b":"2 ","c":"4 ","d":"5","e":"7"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":5.0,"e":7.0},"annotated_formula":"reminder(multiply(65, 63), 8)","linear_formula":"multiply(n0,n1)|reminder(#0,n2)|","chain":"65 * 63<\/gadget>\n4_095<\/output>\n4_095 % 8<\/gadget>\n7<\/output>\n7<\/result>","index":2272} +{"problem":"if shares of two persons in profits are rs . 1600 and rs . 1200 then ratio of their capitals is","rationale":"\"profit = time * capital so 1600 : 1200 = 4 : 3 answer : c\"","correct":"c","options":{"a":"3 : 4 ","b":"2 : 3 ","c":"4 : 3 ","d":"1 : 3","e":"1 : 2"},"options_float":{"a":0.75,"b":0.6666666667,"c":1.3333333333,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(1600, 1200)","linear_formula":"divide(n0,n1)|","chain":"1_600 \/ 1_200<\/gadget>\n4\/3 = around 1.333333<\/output>\n4\/3 = around 1.333333<\/result>","index":2273} +{"problem":"irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did ingrid bake ?","rationale":"9.18 x + 5.17 x + 2.05 x = 16.4 x = 148 cookies x = 148 \/ 16.4 = 9 ( approx ) so , ingrid baked 9 * 5.17 cookies or 47 cookies ( approx ) % share = 47 \/ 148 = 31.7 approx hence , answer is c .","correct":"c","options":{"a":"0.125 % ","b":"1.25 % ","c":"31.7 % ","d":"125 %","e":"0.152 %"},"options_float":{"a":0.125,"b":1.25,"c":31.7,"d":125.0,"e":0.152},"annotated_formula":"multiply(divide(divide(multiply(148, 5.17), add(add(9.18, 5.17), 2.05)), 148), const_100)","linear_formula":"add(n0,n1)|multiply(n1,n3)|add(n2,#0)|divide(#1,#2)|divide(#3,n3)|multiply(#4,const_100)","chain":"148 * 5.17<\/gadget>\n765.16<\/output>\n9.18 + 5.17<\/gadget>\n14.35<\/output>\n14.35 + 2.05<\/gadget>\n16.4<\/output>\n765.16 \/ 16.4<\/gadget>\n46.656098<\/output>\n46.656098 \/ 148<\/gadget>\n0.315244<\/output>\n0.315244 * 100<\/gadget>\n31.5244<\/output>\n31.5244<\/result>","index":2274} +{"problem":"two integers are in the ratio of 1 to 4 . if 6 is added to the smaller number , the ratio becomes 1 to 3 . find the larger integer .","rationale":"\"one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 4 , then 4 x = y , where x is the smaller integer . if adding 6 to the smaller integer makes the ratio 1 to 3 , then 3 ( x + 6 ) = y . substituting y = 4 x into the second equation yields 3 x + 18 = 4 x . so , x = 18 ( smaller integer ) and , y = 4 x = 72 ( larger integer ) so d is the correct answer . another option is to test the answer choices . a ) the larger number is 4 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 1 if 6 is added to the smaller number , the ratio becomes 1 to 3 if we add 6 to the smaller number ( 1 ) , we get 7 . so , the new ratio is 7 to 4 . no good . we want a resulting ratio of 1 to 3 eliminate a b ) the larger number is 24 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 6 if 6 is added to the smaller number , the ratio becomes 1 to 3 if we add 6 to the smaller number ( 6 ) , we get 12 so , the new ratio is 12 to 24 which is 1 to 2 . no good . we want a resulting ratio of 1 to 3 eliminate b c ) the larger number is 30 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 7.5 ( which is not an integer ) since both numbers are integers , the larger number can not be 30 . eliminate c d ) the larger number is 72 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 18 if 6 is added to the smaller number , the ratio becomes 1 to 3 if we add 6 to the smaller number ( 18 ) , we get 24 so , the new ratio is 24 to 72 . this is the same as the ratio of 1 to 3 therefore ( d ) is correct e ) the larger number is 80 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 20 if 6 is added to the smaller number , the ratio becomes 1 to 3 if we add 6 to the smaller number ( 20 ) , we get 26 so , the new ratio is 26 to 80 which is 13 to 40 . no good . we want a resulting ratio of 1 to 3 eliminate e therefore d is the only correct choice .\"","correct":"d","options":{"a":"4 ","b":"24 ","c":"30 ","d":"72","e":"80"},"options_float":{"a":4.0,"b":24.0,"c":30.0,"d":72.0,"e":80.0},"annotated_formula":"multiply(divide(multiply(3, 6), subtract(4, 3)), 4)","linear_formula":"multiply(n2,n4)|subtract(n1,n4)|divide(#0,#1)|multiply(n1,#2)|","chain":"3 * 6<\/gadget>\n18<\/output>\n4 - 3<\/gadget>\n1<\/output>\n18 \/ 1<\/gadget>\n18<\/output>\n18 * 4<\/gadget>\n72<\/output>\n72<\/result>","index":2275} +{"problem":"a , b and c are partners in a business . their capitals are respectively , rs . 5000 , rs . 6000 and rs . 4000 . a gets 30 % of the total profit for managing the business . the remaining profit is divided among 3 in the ratio of their capitals . in the end of the year , the profit of a is rs . 200 more than the sum of the profits of b and c . find the total profit ?","rationale":"a : b : c = 5 : 6 : 4 let the total profit = 100 - 30 = 70 5 \/ 15 * 70 = 70 \/ 3 a share = 70 \/ 3 + 30 = 160 \/ 3 b + c share = 100 - 160 \/ 3 = 140 \/ 3 a - ( b + c ) = 160 \/ 3 - 140 \/ 3 = 20 \/ 3 20 \/ 3 - - - - 200 100 - - - - ? = > 3000 answer : d","correct":"d","options":{"a":"3089 ","b":"3098 ","c":"3057 ","d":"3000","e":"3012"},"options_float":{"a":3089.0,"b":3098.0,"c":3057.0,"d":3000.0,"e":3012.0},"annotated_formula":"multiply(divide(add(add(5000, 6000), 4000), multiply(const_100, const_10)), 200)","linear_formula":"add(n0,n1)|multiply(const_10,const_100)|add(n2,#0)|divide(#2,#1)|multiply(n5,#3)","chain":"5_000 + 6_000<\/gadget>\n11_000<\/output>\n11_000 + 4_000<\/gadget>\n15_000<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n15_000 \/ 1_000<\/gadget>\n15<\/output>\n15 * 200<\/gadget>\n3_000<\/output>\n3_000<\/result>","index":2276} +{"problem":"$ 100 is divided amongst a , b and c so that a may get 1 \/ 4 as much as b and c together , b may get 3 \/ 5 as much as a and c together , then the share of a is","rationale":"\"a : ( b + c ) = 1 : 4 a ' s share = 100 * 1 \/ 5 = $ 20 answer is b\"","correct":"b","options":{"a":"$ 15 ","b":"$ 20 ","c":"$ 25 ","d":"$ 18","e":"$ 30"},"options_float":{"a":15.0,"b":20.0,"c":25.0,"d":18.0,"e":30.0},"annotated_formula":"multiply(divide(100, add(divide(1, 4), const_1)), divide(1, 4))","linear_formula":"divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n100 \/ (5\/4)<\/gadget>\n80<\/output>\n80 * (1\/4)<\/gadget>\n20<\/output>\n20<\/result>","index":2277} +{"problem":"an urn contains 5 red , 6 blue and 8 green balls . 4 balls are randomly selected from the urn , find the probability that the drawn ball are 1 blue and 3 red ?","rationale":"sample space = no . of ways 4 balls were drawn from urn = 19 c 4 = 3876 no . ways 1 blue balls and 3 red were drawn from bag = 6 c 1 * 5 c 3 = 60 probability = 60 \/ 3876 = 5 \/ 323 ans - d","correct":"d","options":{"a":"1 \/ 969 ","b":"2 \/ 969 ","c":"4 \/ 969 ","d":"5 \/ 323","e":"21 \/ 789"},"options_float":{"a":0.0010319917,"b":0.0020639835,"c":0.004127967,"d":0.0154798762,"e":0.0266159696},"annotated_formula":"divide(multiply(choose(6, 1), choose(5, 3)), choose(add(add(5, 6), 8), 4))","linear_formula":"add(n0,n1)|choose(n1,n4)|choose(n0,n5)|add(n2,#0)|multiply(#1,#2)|choose(#3,n3)|divide(#4,#5)","chain":"binomial(6, 1)<\/gadget>\n6<\/output>\nbinomial(5, 3)<\/gadget>\n10<\/output>\n6 * 10<\/gadget>\n60<\/output>\n5 + 6<\/gadget>\n11<\/output>\n11 + 8<\/gadget>\n19<\/output>\nbinomial(19, 4)<\/gadget>\n3_876<\/output>\n60 \/ 3_876<\/gadget>\n5\/323 = around 0.01548<\/output>\n5\/323 = around 0.01548<\/result>","index":2278} +{"problem":"how many positive even integers less than 100 contain digits 5 or 9 ?","rationale":"two digit numbers : 5 at tens place : 50 , 52,54 , 56,58 9 at tens place : 90 , 92,94 , 96,98 if 5 and 9 is at units place , the number cant be even total : 5 + 5 = 10 answer b","correct":"b","options":{"a":"16 ","b":"10 ","c":"18 ","d":"19","e":"20"},"options_float":{"a":16.0,"b":10.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"multiply(5, const_2)","linear_formula":"multiply(n1,const_2)","chain":"5 * 2<\/gadget>\n10<\/output>\n10<\/result>","index":2279} +{"problem":"farm tax is levied on the 45 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . willam . mr . willam paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :","rationale":"\"only trick n this question is to ignore 45 % information as farm tax is levied uniformly in the village and that includes mr william ' s land . what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 \/ 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 \/ 45 ) = 27.78 % and the answer is not there in the options . the correct answer is ( d ) .\"","correct":"d","options":{"a":"15 % ","b":"25 % ","c":"15.125 % ","d":"27.78 %","e":"none"},"options_float":{"a":15.0,"b":25.0,"c":15.125,"d":27.78,"e":null},"annotated_formula":"divide(multiply(multiply(divide(480, 3840), const_100), const_100), 45)","linear_formula":"divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|","chain":"480 \/ 3_840<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 100<\/gadget>\n1_250<\/output>\n1_250 \/ 45<\/gadget>\n250\/9 = around 27.777778<\/output>\n250\/9 = around 27.777778<\/result>","index":2280} +{"problem":"johnson completes a piece of work in 10 days , vincent completes the same work in 40 days . if both of them work together , then the number of days required to complete the work is","rationale":"if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y \/ x + y days . that is , the required no . of days = 10 × 40 \/ 50 = 8 days . b )","correct":"b","options":{"a":"7 days ","b":"8 days ","c":"9 days ","d":"10 days","e":"12 days"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":12.0},"annotated_formula":"inverse(add(inverse(10), inverse(40)))","linear_formula":"inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/10) + (1\/40)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n8<\/result>","index":2283} +{"problem":"in triangle pqr , the angle q = 90 degree , pq = 4 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz","rationale":"\"look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 4 = 8 : 10 - - > qy = 3.2 . answer : e\"","correct":"e","options":{"a":"1.2 cm ","b":"2.4 cm ","c":"4.8 cm ","d":"2.16 cm","e":"3.2 cm"},"options_float":{"a":1.2,"b":2.4,"c":4.8,"d":2.16,"e":3.2},"annotated_formula":"divide(multiply(4, 8), const_10)","linear_formula":"multiply(n1,n2)|divide(#0,const_10)|","chain":"4 * 8<\/gadget>\n32<\/output>\n32 \/ 10<\/gadget>\n16\/5 = around 3.2<\/output>\n16\/5 = around 3.2<\/result>","index":2284} +{"problem":"a local elementary school has 1049 students and 9 teachers , how many students would have to be added to the school to make sure each teacher has the same amount of students in their classroom ?","rationale":"using the rule on the divisibility of nine , we add the digits of 1049 together to get 14 . since we are doing addition , the next number that is divisible by nine would be 18 . therefore , our answer 4 ( option a )","correct":"a","options":{"a":"4 ","b":"6 ","c":"5 ","d":"1","e":"2"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":1.0,"e":2.0},"annotated_formula":"subtract(9, subtract(1049, multiply(9, floor(divide(1049, 9)))))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|subtract(n1,#3)","chain":"1_049 \/ 9<\/gadget>\n1_049\/9 = around 116.555556<\/output>\nfloor(1_049\/9)<\/gadget>\n116<\/output>\n9 * 116<\/gadget>\n1_044<\/output>\n1_049 - 1_044<\/gadget>\n5<\/output>\n9 - 5<\/gadget>\n4<\/output>\n4<\/result>","index":2285} +{"problem":"the average marks of 30 students in a class is 100 . but a student mark is wrongly noted as 70 instead of 10 then find the correct average marks ?","rationale":"\"correct avg marks = 100 + ( 10 - 70 ) \/ 30 avg = 100 - 2 = 98 answer is c\"","correct":"c","options":{"a":"78 ","b":"82 ","c":"98 ","d":"91","e":"85"},"options_float":{"a":78.0,"b":82.0,"c":98.0,"d":91.0,"e":85.0},"annotated_formula":"divide(add(subtract(multiply(100, 30), 70), 10), 30)","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|","chain":"100 * 30<\/gadget>\n3_000<\/output>\n3_000 - 70<\/gadget>\n2_930<\/output>\n2_930 + 10<\/gadget>\n2_940<\/output>\n2_940 \/ 30<\/gadget>\n98<\/output>\n98<\/result>","index":2286} +{"problem":"a 440 - liter solution of kola is made from 88 % water , 8 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 10 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ?","rationale":"\"denominator : 440 + 10 + 3.2 + 6.8 = 460 numerator : 440 ( 1 - . 88 - . 08 ) + 3.2 440 ( 0.04 ) + 3.2 17.6 + 3.2 20.8 ratio : 20.8 \/ 460 = 0.045 answer : a\"","correct":"a","options":{"a":"4.52 % . ","b":"7.5 % . ","c":"9.2 % . ","d":"10.5 % .","e":"11 % ."},"options_float":{"a":4.52,"b":7.5,"c":9.2,"d":10.5,"e":11.0},"annotated_formula":"multiply(divide(add(subtract(subtract(440, multiply(440, divide(88, const_100))), multiply(440, divide(8, const_100))), 3.2), add(add(add(440, 3.2), 10), 6.8)), const_100)","linear_formula":"add(n0,n3)|divide(n1,const_100)|divide(n2,const_100)|add(n4,#0)|multiply(n0,#1)|multiply(n0,#2)|add(n5,#3)|subtract(n0,#4)|subtract(#7,#5)|add(n3,#8)|divide(#9,#6)|multiply(#10,const_100)|","chain":"88 \/ 100<\/gadget>\n22\/25 = around 0.88<\/output>\n440 * (22\/25)<\/gadget>\n1_936\/5 = around 387.2<\/output>\n440 - (1_936\/5)<\/gadget>\n264\/5 = around 52.8<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n440 * (2\/25)<\/gadget>\n176\/5 = around 35.2<\/output>\n(264\/5) - (176\/5)<\/gadget>\n88\/5 = around 17.6<\/output>\n(88\/5) + 3.2<\/gadget>\n20.8<\/output>\n440 + 3.2<\/gadget>\n443.2<\/output>\n443.2 + 10<\/gadget>\n453.2<\/output>\n453.2 + 6.8<\/gadget>\n460<\/output>\n20.8 \/ 460<\/gadget>\n0.045217<\/output>\n0.045217 * 100<\/gadget>\n4.5217<\/output>\n4.5217<\/result>","index":2288} +{"problem":"a wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule , whose length and breadth are in the ratio of 6 : 5 . what is the area of the rectangle ?","rationale":"the circumference of the circle is equal to the permeter of the rectangle . let l = 6 x and b = 5 x 2 ( 6 x + 5 x ) = 2 * 22 \/ 7 * 3.5 = > x = 1 therefore l = 6 cm and b = 5 cm area of the rectangle = 6 * 5 = 30 cm 2 answer : b","correct":"b","options":{"a":"65 cm 2 ","b":"30 cm 2 ","c":"76 cm 2 ","d":"68 cm 2","e":"46 cm 2"},"options_float":{"a":65.0,"b":30.0,"c":76.0,"d":68.0,"e":46.0},"annotated_formula":"rectangle_area(multiply(6, divide(circumface(3.5), multiply(add(6, 5), const_2))), multiply(5, divide(circumface(3.5), multiply(add(6, 5), const_2))))","linear_formula":"add(n1,n2)|circumface(n0)|multiply(#0,const_2)|divide(#1,#2)|multiply(n1,#3)|multiply(n2,#3)|rectangle_area(#4,#5)","chain":"2 * pi * 3.5<\/gadget>\n7.0*pi = around 21.991149<\/output>\n6 + 5<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n(7.0*pi) \/ 22<\/gadget>\n0.318181818181818*pi = around 0.999598<\/output>\n6 * (0.318181818181818*pi)<\/gadget>\n1.90909090909091*pi = around 5.997586<\/output>\n5 * (0.318181818181818*pi)<\/gadget>\n1.59090909090909*pi = around 4.997988<\/output>\n(1.90909090909091*pi) * (1.59090909090909*pi)<\/gadget>\n3.03719008264463*pi**2 = around 29.975865<\/output>\n3.03719008264463*pi**2 = around 29.975865<\/result>","index":2289} +{"problem":"sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 50 marks . how many sums did sandy get correct ?","rationale":"\"let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 50 5 x = 110 x = 22 the answer is b .\"","correct":"b","options":{"a":"20 ","b":"22 ","c":"24 ","d":"26","e":"28"},"options_float":{"a":20.0,"b":22.0,"c":24.0,"d":26.0,"e":28.0},"annotated_formula":"divide(add(multiply(2, 30), 50), add(2, 3))","linear_formula":"add(n0,n1)|multiply(n1,n2)|add(n3,#1)|divide(#2,#0)|","chain":"2 * 30<\/gadget>\n60<\/output>\n60 + 50<\/gadget>\n110<\/output>\n2 + 3<\/gadget>\n5<\/output>\n110 \/ 5<\/gadget>\n22<\/output>\n22<\/result>","index":2290} +{"problem":"a person borrows 20000 for 6 years at 8 % p . a . simple interest . he immediately lends it to another person at 9 % p . a . for 6 years . find his gain in the transaction per year .","rationale":"gain in 6 years = [ ( 20000 ã — 9 ã — 6 \/ 100 ) â ˆ ’ ( 20000 ã — 6 ã — 8 \/ 100 ) ] = ( 10800 â € “ 9600 ) = 1200 â ˆ ´ gain in 6 year = ( 1200 \/ 6 ) = 200 answer a","correct":"a","options":{"a":"200 ","b":"250 ","c":"210 ","d":"190","e":"180"},"options_float":{"a":200.0,"b":250.0,"c":210.0,"d":190.0,"e":180.0},"annotated_formula":"multiply(20000, divide(const_1, const_100))","linear_formula":"divide(const_1,const_100)|multiply(n0,#0)","chain":"1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n20_000 * (1\/100)<\/gadget>\n200<\/output>\n200<\/result>","index":2292} +{"problem":"if m is a positive integer and m ^ 2 is divisible by 24 , then the largest positive integer that must divide m is ?","rationale":"\"m ^ 2 is divisible by 48 so m ^ 2 must be multiple of 48 . if the value of m is multiples of 8 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is c but if m = 48 then answer should be 16 . is the question right ? or am i missing some thing ? c\"","correct":"c","options":{"a":"3 ","b":"6 ","c":"8 ","d":"12","e":"16"},"options_float":{"a":3.0,"b":6.0,"c":8.0,"d":12.0,"e":16.0},"annotated_formula":"multiply(const_3, divide(divide(24, const_3), const_3))","linear_formula":"divide(n1,const_3)|divide(#0,const_3)|multiply(#1,const_3)|","chain":"24 \/ 3<\/gadget>\n8<\/output>\n8 \/ 3<\/gadget>\n8\/3 = around 2.666667<\/output>\n3 * (8\/3)<\/gadget>\n8<\/output>\n8<\/result>","index":2296} +{"problem":"a flagstaff 17.5 m high casts a shadow of length 40.25 m . the height of the building , which casts a shadow of length 28.75 m under similar conditions will be ?","rationale":"\"let the height of the building x metres . less lengthy shadow , less in the height ( direct proportion ) 40.25 : 28.75 : : 17.5 : x 40.25 x x = 28.75 x 17.5 x = 28.75 x 17.5 \/ 40.25 x = 12.5 answer : b\"","correct":"b","options":{"a":"10 m ","b":"12.5 m ","c":"12 m ","d":"13 m","e":"14 m"},"options_float":{"a":10.0,"b":12.5,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"multiply(28.75, divide(17.5, 40.25))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|","chain":"17.5 \/ 40.25<\/gadget>\n0.434783<\/output>\n28.75 * 0.434783<\/gadget>\n12.500011<\/output>\n12.500011<\/result>","index":2297} +{"problem":"average monthly income of a family of 4 earning members was rs . 840 . one of the earning members died and therefore , the average income came down to rs 650 . the income of the deceased was ?","rationale":"answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 840 x 4 - 650 x 3 rs . = 1410 rs . correct option : d","correct":"d","options":{"a":"rs . 692.80 ","b":"rs . 820 ","c":"rs . 990 ","d":"rs . 1410","e":"none"},"options_float":{"a":692.8,"b":820.0,"c":990.0,"d":1410.0,"e":null},"annotated_formula":"subtract(multiply(840, 4), multiply(650, subtract(4, const_1)))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)","chain":"840 * 4<\/gadget>\n3_360<\/output>\n4 - 1<\/gadget>\n3<\/output>\n650 * 3<\/gadget>\n1_950<\/output>\n3_360 - 1_950<\/gadget>\n1_410<\/output>\n1_410<\/result>","index":2298} +{"problem":"x and y are two towns . ganesh covers the distance from x to y at an average speed of 43 km \/ hr . however , he covers the distance from y to x at an average speed of 34 km \/ hr . his average speed during the whole journey in km \/ hr . is :","rationale":"\"solution : average speed = 2 xy \/ x + y = 2 * 43 * 34 \/ 43 + 34 = 2924 \/ 77 = 37.974 . . . = 38 answer : d\"","correct":"d","options":{"a":"34 ","b":"43 ","c":"40 ","d":"38","e":"29"},"options_float":{"a":34.0,"b":43.0,"c":40.0,"d":38.0,"e":29.0},"annotated_formula":"divide(add(add(subtract(43, const_10), const_100), add(subtract(43, const_10), const_100)), add(divide(add(subtract(43, const_10), const_100), 43), divide(add(subtract(43, const_10), const_100), 34)))","linear_formula":"subtract(n0,const_10)|add(#0,const_100)|add(#1,#1)|divide(#1,n0)|divide(#1,n1)|add(#3,#4)|divide(#2,#5)|","chain":"43 - 10<\/gadget>\n33<\/output>\n33 + 100<\/gadget>\n133<\/output>\n133 + 133<\/gadget>\n266<\/output>\n133 \/ 43<\/gadget>\n133\/43 = around 3.093023<\/output>\n133 \/ 34<\/gadget>\n133\/34 = around 3.911765<\/output>\n(133\/43) + (133\/34)<\/gadget>\n10_241\/1_462 = around 7.004788<\/output>\n266 \/ (10_241\/1_462)<\/gadget>\n2_924\/77 = around 37.974026<\/output>\n2_924\/77 = around 37.974026<\/result>","index":2299} +{"problem":"what is the measure of the angle q made by the diagonals of the any adjacent sides of a cube .","rationale":"\"c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be q = 60 . c\"","correct":"c","options":{"a":"30 ","b":"45 ","c":"60 ","d":"75","e":"90"},"options_float":{"a":30.0,"b":45.0,"c":60.0,"d":75.0,"e":90.0},"annotated_formula":"divide(const_180, const_3)","linear_formula":"divide(const_180,const_3)|","chain":"180 \/ 3<\/gadget>\n60<\/output>\n60<\/result>","index":2300} +{"problem":"visitors to show were charged rs . 15 each on the first day . rs . 7.50 on the second day , rs . 2.50 on the third day and total attendance on the 3 days were in ratio 2 : 5 : 13 respectively . the average charge per person for the whole show is ?","rationale":"2 : 5 : 13 2 x 5 x 13 x 15 7.5 2.5 30 x + 37.5 x + 32.5 x = 100 x \/ 20 x average = 5 answer : c","correct":"c","options":{"a":"1 ","b":"4 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":1.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(add(add(multiply(15, 2), multiply(7.5, 5)), multiply(2.5, 13)), add(add(13, 5), 2))","linear_formula":"add(n5,n6)|multiply(n0,n4)|multiply(n1,n5)|multiply(n2,n6)|add(#1,#2)|add(n4,#0)|add(#4,#3)|divide(#6,#5)","chain":"15 * 2<\/gadget>\n30<\/output>\n7.5 * 5<\/gadget>\n37.5<\/output>\n30 + 37.5<\/gadget>\n67.5<\/output>\n2.5 * 13<\/gadget>\n32.5<\/output>\n67.5 + 32.5<\/gadget>\n100<\/output>\n13 + 5<\/gadget>\n18<\/output>\n18 + 2<\/gadget>\n20<\/output>\n100 \/ 20<\/gadget>\n5<\/output>\n5<\/result>","index":2301} +{"problem":"hillary and eddy are climbing to the summit of mt . everest from a base camp 4,900 ft from the summit . when they depart for the summit at 06 : 00 , hillary climbs at a rate of 800 ft \/ hr with eddy lagging behind at a slower rate of 500 ft \/ hr . if hillary stops 900 ft short of the summit and then descends at a rate of 1,000 ft \/ hr , at what time do hillary and eddy pass each other on her return trip ?","rationale":"\"solution : h stopped 900 ft before reaching the final point , time taken to reach 4000 ft = 4000 \/ 800 = 5 hrs . this means she reached there at 11 : 00 . speed difference between them is 800 - 500 = 300 ft \/ hr so by the time h stops they have 1500 ft of distance so now here we use relative speed formula they both are travelling toward each other with speed of 1000 and 500 total 1500 ft \/ hr and distance bwn them is 1500 ft so time taken to meet = 1 hr from 11 : 00 means 12 : 00 is the answer . d\"","correct":"d","options":{"a":"10.0 ","b":"11.0 ","c":"13.0 ","d":"12.0","e":"14.0"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":12.0,"e":14.0},"annotated_formula":"add(06, add(divide(multiply(divide(subtract(divide(500, const_100), const_1), divide(800, 1,000)), subtract(divide(800, 900), divide(500, 900))), add(divide(500, 900), const_1)), divide(subtract(divide(500, const_100), const_1), divide(800, 900))))","linear_formula":"divide(n4,const_100)|divide(n3,n6)|divide(n4,n5)|add(#2,const_1)|subtract(#0,const_1)|subtract(#1,#2)|divide(#4,#1)|multiply(#6,#5)|divide(#7,#3)|add(#8,#6)|add(n1,#9)|","chain":"500 \/ 100<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n800 \/ 1_000<\/gadget>\n4\/5 = around 0.8<\/output>\n4 \/ (4\/5)<\/gadget>\n5<\/output>\n800 \/ 900<\/gadget>\n8\/9 = around 0.888889<\/output>\n500 \/ 900<\/gadget>\n5\/9 = around 0.555556<\/output>\n(8\/9) - (5\/9)<\/gadget>\n1\/3 = around 0.333333<\/output>\n5 * (1\/3)<\/gadget>\n5\/3 = around 1.666667<\/output>\n(5\/9) + 1<\/gadget>\n14\/9 = around 1.555556<\/output>\n(5\/3) \/ (14\/9)<\/gadget>\n15\/14 = around 1.071429<\/output>\n4 \/ (8\/9)<\/gadget>\n9\/2 = around 4.5<\/output>\n(15\/14) + (9\/2)<\/gadget>\n39\/7 = around 5.571429<\/output>\n6 + (39\/7)<\/gadget>\n81\/7 = around 11.571429<\/output>\n81\/7 = around 11.571429<\/result>","index":2302} +{"problem":"a dishonest person wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing 33 $ per litre so as to make a profit of 70 % on cost when he sells the resulting milk and water mixture for 36 $ . in what ratio should he mix the water and milk ?","rationale":"\"first of all , let ' s consider 1 liter of the stuff he is going to sell - - - naive customers think it ' s pure milk , but we know it ' s some milk - water mixture . he is going to sell this liter of milk - water for $ 36 . this $ 36 should be a 20 % increase over cost . here , we need to think about percentage increases as multipliers . using multipliers ( cost ) * 1.70 = $ 36 cost = 36 \/ 1.7 = 360 \/ 12 = $ 30 if he wants a 70 % increase over cost on the sale of one liter of his milk - water , the cost has to be $ 21.17 well , a liter of milk costs $ 33 , so if he is going to use just $ 30 of milk in his mixture , that ' s 21.17 \/ 33 = 120 \/ 187 of a liter . if milk is 120 \/ 87 of the liter , then water is 67 \/ 187 of the liter , and the ratio of water to milk is 67 : 120 . answer choice ( b )\"","correct":"b","options":{"a":"1 : 120 ","b":"67 : 120 ","c":"67 : 8 ","d":"8 : 120","e":"1 : 2"},"options_float":{"a":0.0083333333,"b":0.5583333333,"c":8.375,"d":0.0666666667,"e":0.5},"annotated_formula":"divide(subtract(33, divide(36, divide(add(const_100, 70), const_100))), divide(36, divide(add(const_100, 70), const_100)))","linear_formula":"add(n1,const_100)|divide(#0,const_100)|divide(n2,#1)|subtract(n0,#2)|divide(#3,#2)|","chain":"100 + 70<\/gadget>\n170<\/output>\n170 \/ 100<\/gadget>\n17\/10 = around 1.7<\/output>\n36 \/ (17\/10)<\/gadget>\n360\/17 = around 21.176471<\/output>\n33 - (360\/17)<\/gadget>\n201\/17 = around 11.823529<\/output>\n(201\/17) \/ (360\/17)<\/gadget>\n67\/120 = around 0.558333<\/output>\n67\/120 = around 0.558333<\/result>","index":2303} +{"problem":"the price of a book is increased from $ 300 to $ 390 . what is the % of increase in its price ?","rationale":"\"explanation : change in the price = rs 390 – rs 300 = rs 90 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 90 \/ 300 ) * 100 = 30 % c\"","correct":"c","options":{"a":"10 % ","b":"20 % ","c":"30 % ","d":"35 %","e":"45 %"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":35.0,"e":45.0},"annotated_formula":"multiply(divide(subtract(390, 300), 300), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"390 - 300<\/gadget>\n90<\/output>\n90 \/ 300<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":2304} +{"problem":"a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 1170 , what will be b ' s share of the earnings ?","rationale":"\"explanatory answer a , b and c will share the amount of $ 1170 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 \/ 6 th of the work in a day . similarly , b will complete 1 \/ 8 th and c will complete 1 \/ 12 th of the work . so , the ratio of the work done by a : b : c when they work together will be equal to 1 \/ 6 : 1 \/ 8 : 1 \/ 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 \/ 6 : 24 \/ 8 : 24 \/ 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 1170 will be 4 : 3 : 2 . hence , b ' s share will be 3 * 1170 \/ 9 = 390 . correct choice is ( a )\"","correct":"a","options":{"a":"$ 390 ","b":"$ 520 ","c":"$ 1080 ","d":"$ 1170","e":"$ 630"},"options_float":{"a":390.0,"b":520.0,"c":1080.0,"d":1170.0,"e":630.0},"annotated_formula":"multiply(1170, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8)))))","linear_formula":"inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) + (1\/8)<\/gadget>\n7\/24 = around 0.291667<\/output>\n(1\/12) + (7\/24)<\/gadget>\n3\/8 = around 0.375<\/output>\n(1\/8) \/ (3\/8)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1_170 * (1\/3)<\/gadget>\n390<\/output>\n390<\/result>","index":2305} +{"problem":"if jack walked 8 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ?","rationale":"\"distance walked in 1 hour and 15 mins = 8 miles speed per hour = distance \/ time = 8 \/ ( 5 \/ 4 ) = 6.4 miles per hour answer d\"","correct":"d","options":{"a":"4 ","b":"4.5 ","c":"6 ","d":"6.4","e":"15"},"options_float":{"a":4.0,"b":4.5,"c":6.0,"d":6.4,"e":15.0},"annotated_formula":"divide(multiply(8, const_60), add(15, const_60))","linear_formula":"add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0)|","chain":"8 * 60<\/gadget>\n480<\/output>\n15 + 60<\/gadget>\n75<\/output>\n480 \/ 75<\/gadget>\n32\/5 = around 6.4<\/output>\n32\/5 = around 6.4<\/result>","index":2308} +{"problem":"a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 80 percent of books that were loaned out are returned and there are 67 books in the special collection at that time , how many books of the special collection were loaned out during that month ?","rationale":"\"there are 8 books less ( 75 - 68 ) which represents 20 % of the loaned books ( 100 - 80 ) so total loaned out books = 40 answer d\"","correct":"d","options":{"a":"20 ","b":"30 ","c":"35 ","d":"40","e":"55"},"options_float":{"a":20.0,"b":30.0,"c":35.0,"d":40.0,"e":55.0},"annotated_formula":"divide(subtract(75, 67), subtract(const_1, divide(80, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"75 - 67<\/gadget>\n8<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n8 \/ (1\/5)<\/gadget>\n40<\/output>\n40<\/result>","index":2309} +{"problem":"if a farmer wants to plough a farm field on time , he must plough 120 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ?","rationale":"let x be the number of days in the initial plan . therefore , the whole field is 120 ⋅ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 120 x = 85 ( x + 2 ) + 40 35 x = 210 x = 6 so the farmer planned to have the work done in 6 days , and the area of the farm field is 120 ⋅ 6 = 720 hectares correct answer c","correct":"c","options":{"a":"600 ","b":"490 ","c":"720 ","d":"435","e":"679"},"options_float":{"a":600.0,"b":490.0,"c":720.0,"d":435.0,"e":679.0},"annotated_formula":"multiply(divide(add(40, multiply(2, 85)), subtract(120, 85)), 120)","linear_formula":"multiply(n1,n2)|subtract(n0,n1)|add(n3,#0)|divide(#2,#1)|multiply(n0,#3)","chain":"2 * 85<\/gadget>\n170<\/output>\n40 + 170<\/gadget>\n210<\/output>\n120 - 85<\/gadget>\n35<\/output>\n210 \/ 35<\/gadget>\n6<\/output>\n6 * 120<\/gadget>\n720<\/output>\n720<\/result>","index":2310} +{"problem":"for each color copy , print shop x charges $ 1.20 and print shop y charges $ 1.70 . how much greater is the charge for 40 color copies at print shop y than at print shop x ?","rationale":"\"the difference in the two prices is $ 1.70 - $ 1.20 = $ 0.50 for each color copy . each color copy will cost an extra $ 0.50 at print shop y . 40 * $ 0.50 = $ 20 the answer is d .\"","correct":"d","options":{"a":"$ 14 ","b":"$ 16 ","c":"$ 18 ","d":"$ 20","e":"$ 22"},"options_float":{"a":14.0,"b":16.0,"c":18.0,"d":20.0,"e":22.0},"annotated_formula":"subtract(multiply(1.70, 40), multiply(1.20, 40))","linear_formula":"multiply(n1,n2)|multiply(n0,n2)|subtract(#0,#1)|","chain":"1.7 * 40<\/gadget>\n68<\/output>\n1.2 * 40<\/gadget>\n48<\/output>\n68 - 48<\/gadget>\n20<\/output>\n20<\/result>","index":2311} +{"problem":"in a school of 300 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?","rationale":"\"44 + 28 + 10 = 82 % 100 – 82 = 18 % 300 * 18 \/ 100 = 54 answer : a\"","correct":"a","options":{"a":"54 ","b":"63 ","c":"70 ","d":"42","e":"31"},"options_float":{"a":54.0,"b":63.0,"c":70.0,"d":42.0,"e":31.0},"annotated_formula":"divide(multiply(300, subtract(const_100, add(add(44, 28), 10))), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)|","chain":"44 + 28<\/gadget>\n72<\/output>\n72 + 10<\/gadget>\n82<\/output>\n100 - 82<\/gadget>\n18<\/output>\n300 * 18<\/gadget>\n5_400<\/output>\n5_400 \/ 100<\/gadget>\n54<\/output>\n54<\/result>","index":2312} +{"problem":"a number x is 12 times another number y . the percentage that y is less than x is","rationale":"\"say y = 1 and x = 12 . then y = 1 is less than x = 12 by ( 12 - 1 ) \/ 12 * 100 = 11 \/ 12 * 100 = 91.6 % . answer : a .\"","correct":"a","options":{"a":"91.6 % ","b":"87.5 % ","c":"80 % ","d":"11 %","e":"1 %"},"options_float":{"a":91.6,"b":87.5,"c":80.0,"d":11.0,"e":1.0},"annotated_formula":"multiply(divide(subtract(12, const_1), 12), const_100)","linear_formula":"subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"12 - 1<\/gadget>\n11<\/output>\n11 \/ 12<\/gadget>\n11\/12 = around 0.916667<\/output>\n(11\/12) * 100<\/gadget>\n275\/3 = around 91.666667<\/output>\n275\/3 = around 91.666667<\/result>","index":2313} +{"problem":"a train 90 m long is running with a speed of 60 km \/ hr . in what time will it pass a man who is running at 6 km \/ hr in the direction opposite to that in which the train is going ?","rationale":"\"speed of train relative to man = 60 + 6 = 66 km \/ hr . = 66 * 5 \/ 18 = 55 \/ 3 m \/ sec . time taken to pass the men = 90 * 3 \/ 55 = 5 sec . answer d\"","correct":"d","options":{"a":"7 ","b":"6 ","c":"8 ","d":"5","e":"4"},"options_float":{"a":7.0,"b":6.0,"c":8.0,"d":5.0,"e":4.0},"annotated_formula":"divide(90, multiply(add(60, 6), const_0_2778))","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"60 + 6<\/gadget>\n66<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n66 * (5\/18)<\/gadget>\n55\/3 = around 18.333333<\/output>\n90 \/ (55\/3)<\/gadget>\n54\/11 = around 4.909091<\/output>\n54\/11 = around 4.909091<\/result>","index":2314} +{"problem":"the average age of a family of 5 members is 20 years . if the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member ?","rationale":"solution : at present the total age of the family = 5 * 20 = 100 . the total age of the family at the time of the birth of the youngest member , = [ 100 − 10 − ( 10 * 4 ) ] = 50 . therefore , average age of the family at the time of birth of the youngest member , = 50 \/ 4 = 12.5 . answer : option d","correct":"d","options":{"a":"13.5 ","b":"14 ","c":"15 ","d":"12.5","e":"none"},"options_float":{"a":13.5,"b":14.0,"c":15.0,"d":12.5,"e":null},"annotated_formula":"subtract(divide(subtract(multiply(5, 20), 10), const_4), 10)","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_4)|subtract(#2,n2)","chain":"5 * 20<\/gadget>\n100<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 4<\/gadget>\n45\/2 = around 22.5<\/output>\n(45\/2) - 10<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":2315} +{"problem":"how many unique positive odd integers less than 90 are equal to the product of a positive multiple of 5 and an odd number ?","rationale":"\"the question basically asks how many positive odd integers less than 90 are odd multiples of 5 so we have 5 , 15,25 , 35,45 , 55,65 , 75,85 = 9 ans b\"","correct":"b","options":{"a":"4 ","b":"9 ","c":"11 ","d":"12","e":"15"},"options_float":{"a":4.0,"b":9.0,"c":11.0,"d":12.0,"e":15.0},"annotated_formula":"divide(divide(90, 5), const_2)","linear_formula":"divide(n0,n1)|divide(#0,const_2)|","chain":"90 \/ 5<\/gadget>\n18<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n9<\/result>","index":2317} +{"problem":"if # is an operation which results in adding the digits of integer until a single digit is left , what is the probability that a number picked up in first 90 positive integers will have the result of # as an odd digit ? ( example 99 = 9 + 9 = 18 = 1 + 8 = 9 , so # = 9 )","rationale":"first of all let us find out what is the maximum sum we will get by adding digits of numbers from 1 to 90 . that will be 17 ( 8 + 9 + 17 ) why did i calculate this ? i will come to that in a moment . so to get an odd digit by carrying out the operation # , the sum of the digits of the integer should be 1 , 3 , 5 , 7 , 9 , 10 , 12 , 14 , and 16 ( why not 18 ? because we know the maximum sum can be 17 ) number of integers , whose digits add to 1 : 1 and 10 = 2 number of integers , whose digits add to 3 : 3 , 12 , 21 , 30 = 4 number of integers , whose digits add to 5 : 5 , 14 , 23 , 32 , 41 , 50 = 6 number of integers , whose digits add to 7 : 7 , 16 , 25 , 34 , 43 , 52 , 61 , 70 = 8 number of integers , whose digits add to 9 : 10 multiples of 9 up to 90 = 10 number of integers , whose digits add to 10 : 19 , 28 , 37 , 46 , 55 , 64 , 73 , 82 = 8 number of integers , whose digits add to 12 : 39 , 48 , 57 , 66 , 75 , 84 = 6 number of integers , whose digits add to 14 : 59 , 68 , 77 , 86 = 4 number of integers , whose digits add to 16 : 79 , 88 = 2 so in total there are 2 + 4 + 6 + 8 + 10 + 8 + 6 + 4 + 2 = 50 such numbers probability = 50 \/ 90 = 5 \/ 9 correct answer e","correct":"e","options":{"a":"4 \/ 10 ","b":"4 \/ 9 ","c":"1 \/ 2 ","d":"6 \/ 10","e":"5 \/ 9"},"options_float":{"a":0.4,"b":0.4444444444,"c":0.5,"d":0.6,"e":0.5555555556},"annotated_formula":"divide(divide(add(9, 1), const_2), 9)","linear_formula":"add(n2,n5)|divide(#0,const_2)|divide(#1,n2)","chain":"9 + 1<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 \/ 9<\/gadget>\n5\/9 = around 0.555556<\/output>\n5\/9 = around 0.555556<\/result>","index":2318} +{"problem":"jerry and michelle play a card game . in the beginning of the game they have an equal number of cards . each player , at her turn , gives the other a third of her cards . michelle plays first , giving jerry a third of her cards . jerry plays next , and michelle follows . then the game ends . jerry ended up with 14 more cards than michelle . how many cards did each player have originally ?","rationale":"game michelle jerry michelle initially 2727 assume after game 1 1836 after game 2 3024 after game 3 2034 now jerry has 14 cards more than michelle . so the answer is a","correct":"a","options":{"a":"27 ","b":"30 ","c":"34 ","d":"36","e":"38"},"options_float":{"a":27.0,"b":30.0,"c":34.0,"d":36.0,"e":38.0},"annotated_formula":"subtract(multiply(14, const_3), 14)","linear_formula":"multiply(n0,const_3)|subtract(#0,n0)","chain":"14 * 3<\/gadget>\n42<\/output>\n42 - 14<\/gadget>\n28<\/output>\n28<\/result>","index":2320} +{"problem":"a circular swimming pool is surrounded by a concrete wall 4 feet wide . if the area of the wall is 11 \/ 25 of the area of the pool , then the radius of the pool in feet is ?","rationale":"let the radius of the pool be r . then area of the wall and pool = π ( r + 4 ) 2 π ( r + 4 ) 2 area of the pool = π ( r ) 2 π ( r ) 2 area of the wall = π ( r + 4 ) 2 − π ( r ) 2 π ( r + 4 ) 2 − π ( r ) 2 given π ( r + 4 ) 2 − π ( r ) 2 π ( r + 4 ) 2 − π ( r ) 2 = 1125 ( π r 2 ) 1125 ( π r 2 ) r 2 + 8 r + 16 − r 2 = 1125 r 2 r 2 + 8 r + 16 − r 2 = 1125 r 2 11 r 2 − 200 r − 400 = 011 r 2 − 200 r − 400 = 0 solving r = 20 answer : b","correct":"b","options":{"a":"22 ","b":"20 ","c":"9 ","d":"8","e":"1"},"options_float":{"a":22.0,"b":20.0,"c":9.0,"d":8.0,"e":1.0},"annotated_formula":"divide(divide(add(multiply(const_2, 4), sqrt(add(power(multiply(const_2, 4), const_2), multiply(multiply(divide(11, 25), const_4), power(4, const_2))))), divide(11, 25)), const_2)","linear_formula":"divide(n1,n2)|multiply(n0,const_2)|power(n0,const_2)|multiply(#0,const_4)|power(#1,const_2)|multiply(#3,#2)|add(#5,#4)|sqrt(#6)|add(#1,#7)|divide(#8,#0)|divide(#9,const_2)","chain":"2 * 4<\/gadget>\n8<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n11 \/ 25<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 4<\/gadget>\n44\/25 = around 1.76<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n(44\/25) * 16<\/gadget>\n704\/25 = around 28.16<\/output>\n64 + (704\/25)<\/gadget>\n2_304\/25 = around 92.16<\/output>\n(2_304\/25) ** (1\/2)<\/gadget>\n48\/5 = around 9.6<\/output>\n8 + (48\/5)<\/gadget>\n88\/5 = around 17.6<\/output>\n(88\/5) \/ (11\/25)<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":2322} +{"problem":"a truck travels 20 miles due north , 30 miles due east , and then 20 miles due north . how many miles is the truck from the starting point ?","rationale":"we have two right angle triangles with sides 20 and 15 the distance between starting and end point = sum of the hypotenuse of both the triangles . hypotenuse = [ 400 + 225 ] ^ ( 1 \/ 2 ) = 25 hence the total distance between end point and starting point = 25 + 25 = 50 correct option : c","correct":"c","options":{"a":"20.3 ","b":"44.7 ","c":"50 ","d":"70","e":"120"},"options_float":{"a":20.3,"b":44.7,"c":50.0,"d":70.0,"e":120.0},"annotated_formula":"add(sqrt(add(power(20, const_2), power(divide(30, const_2), const_2))), sqrt(add(power(20, const_2), power(divide(30, const_2), const_2))))","linear_formula":"divide(n1,const_2)|power(n0,const_2)|power(#0,const_2)|add(#1,#2)|sqrt(#3)|add(#4,#4)","chain":"20 ** 2<\/gadget>\n400<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15 ** 2<\/gadget>\n225<\/output>\n400 + 225<\/gadget>\n625<\/output>\n625 ** (1\/2)<\/gadget>\n25<\/output>\n25 + 25<\/gadget>\n50<\/output>\n50<\/result>","index":2324} +{"problem":"find the area of a parallelogram with base 22 cm and height 14 cm ?","rationale":"\"area of a parallelogram = base * height = 22 * 14 = 308 cm 2 answer : c\"","correct":"c","options":{"a":"298 cm 2 ","b":"384 cm 2 ","c":"308 cm 2 ","d":"286 cm 2","e":"276 cm 2"},"options_float":{"a":298.0,"b":384.0,"c":308.0,"d":286.0,"e":276.0},"annotated_formula":"multiply(22, 14)","linear_formula":"multiply(n0,n1)|","chain":"22 * 14<\/gadget>\n308<\/output>\n308<\/result>","index":2325} +{"problem":"the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 4180 . what was his original salary ?","rationale":"\"x * ( 110 \/ 100 ) * ( 95 \/ 100 ) = 4180 x * ( 11 \/ 10 ) * ( 1 \/ 100 ) = 44 x = 4000 answer : c\"","correct":"c","options":{"a":"2277 ","b":"2999 ","c":"4000 ","d":"2651","e":"1971"},"options_float":{"a":2277.0,"b":2999.0,"c":4000.0,"d":2651.0,"e":1971.0},"annotated_formula":"divide(4180, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (1\/20)<\/gadget>\n19\/20 = around 0.95<\/output>\n(11\/10) * (19\/20)<\/gadget>\n209\/200 = around 1.045<\/output>\n4_180 \/ (209\/200)<\/gadget>\n4_000<\/output>\n4_000<\/result>","index":2326} +{"problem":"the sum of all the digits of the integers from 18 to 21 inclusive is 24 ( 1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24 ) . what is the sum t of all the digits of the integers from 0 to 99 inclusive ?","rationale":"\"we want the sum of the digits from 0 to 99 , so i approximated : 0 - 9 - > 45 - > ( 9 + 0 ) * 10 \/ 2 40 - 49 - > 85 ( 13 + 4 ) * 10 \/ 2 90 - 99 - > 135 ( 18 + 9 ) * 10 \/ 2 we can see at a glance that theweightgoes up as the numbers go up ( meaning the difference between 85 and 45 is 40 , while 135 - 85 is 50 , this means that the second part of this sequence carries more weight for our result ) , so we know that the final answer has to be more than 850 ( 85 * 10 ) but close to it , and that ' s just t = 900 : the answer is c .\"","correct":"c","options":{"a":"450 ","b":"810 ","c":"900 ","d":"1000","e":"1100"},"options_float":{"a":450.0,"b":810.0,"c":900.0,"d":1000.0,"e":1100.0},"annotated_formula":"multiply(divide(multiply(9, add(9, 1)), 2), multiply(2, const_10))","linear_formula":"add(n6,n3)|multiply(const_10,n7)|multiply(n6,#0)|divide(#2,n7)|multiply(#3,#1)|","chain":"9 + 1<\/gadget>\n10<\/output>\n9 * 10<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n2 * 10<\/gadget>\n20<\/output>\n45 * 20<\/gadget>\n900<\/output>\n900<\/result>","index":2327} +{"problem":"in δ pqs above , if pq = 2 and ps = 3 , then","rationale":"there are two ways to calculate area of pqs . area remains same , so both are equal . 2 * 3 \/ 2 = pr * 4 \/ 2 pr = 3 \/ 2 a","correct":"a","options":{"a":"3 \/ 2 ","b":"12 \/ 5 ","c":"16 \/ 5 ","d":"15 \/ 4","e":"20 \/ 3"},"options_float":{"a":1.5,"b":2.4,"c":3.2,"d":3.75,"e":6.6666666667},"annotated_formula":"divide(3, 2)","linear_formula":"divide(n1,n0)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":2328} +{"problem":"a man whose speed is 8 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 3 kmph , find his average speed for the total journey ?","rationale":"\"explanation : speed of man = 8 kmph speed of stream = 3 kmph speed in downstream = 11 kmph speed in upstream = 5 kmph average speed = ( 2 x 11 x 5 ) \/ 16 = 6.8 kmph . answer : e\"","correct":"e","options":{"a":"1 kmph ","b":"5 kmph ","c":"6 kmph ","d":"4 kmph","e":"6.8 kmph"},"options_float":{"a":1.0,"b":5.0,"c":6.0,"d":4.0,"e":6.8},"annotated_formula":"divide(add(8, subtract(8, 3)), const_2)","linear_formula":"subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)|","chain":"8 - 3<\/gadget>\n5<\/output>\n8 + 5<\/gadget>\n13<\/output>\n13 \/ 2<\/gadget>\n13\/2 = around 6.5<\/output>\n13\/2 = around 6.5<\/result>","index":2330} +{"problem":"in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 9 feet between each tree . how many trees can be planted if the road is 151 feet long ?","rationale":"\"let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 14 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number of 9 feet spaces will be t - 1 . then , the length of sidewalk required for 9 feet spaces will be 9 * ( t - 1 ) it is given that total lenght of sidewalk is 151 feet . or 9 ( t - 1 ) + t = 151 or 9 t - 9 + t = 151 or 10 t = 160 or t = 16 answer : - e\"","correct":"e","options":{"a":"8 ","b":"9 ","c":"10 ","d":"11","e":"16"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":16.0},"annotated_formula":"add(divide(subtract(151, const_1), add(9, const_1)), const_1)","linear_formula":"add(n0,const_1)|subtract(n1,const_1)|divide(#1,#0)|add(#2,const_1)|","chain":"151 - 1<\/gadget>\n150<\/output>\n9 + 1<\/gadget>\n10<\/output>\n150 \/ 10<\/gadget>\n15<\/output>\n15 + 1<\/gadget>\n16<\/output>\n16<\/result>","index":2333} +{"problem":"the slant height of a cone is 35 cm and radius of the base is 14 cm , find the curved surface of the cone .","rationale":"π * 14 * 35 = 1540 answer : c","correct":"c","options":{"a":"4150 ","b":"1780 ","c":"1540 ","d":"1500","e":"6100"},"options_float":{"a":4150.0,"b":1780.0,"c":1540.0,"d":1500.0,"e":6100.0},"annotated_formula":"multiply(multiply(const_pi, 14), 35)","linear_formula":"multiply(n1,const_pi)|multiply(n0,#0)","chain":"pi * 14<\/gadget>\n14*pi = around 43.982297<\/output>\n(14*pi) * 35<\/gadget>\n490*pi = around 1_539.3804<\/output>\n490*pi = around 1_539.3804<\/result>","index":2334} +{"problem":"if a ã — b = 2 a - 3 b + ab , then 3 ã — 2 + 2 ã — 3 is equal to :","rationale":"\"explanation : 3 ã — 2 + 2 ã — 3 = ( 2 ã — 3 - 3 ã — 2 + 2 ã — 3 ) + ( 2 ã — 2 - 3 ã — 3 + 2 ã — 3 ) = ( 6 + 4 - 9 + 6 ) = 7 . answer : a\"","correct":"a","options":{"a":"7 ","b":"8 ","c":"9 ","d":"25","e":"26"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":25.0,"e":26.0},"annotated_formula":"add(add(subtract(multiply(2, 3), multiply(3, 2)), multiply(3, 2)), add(subtract(multiply(2, 2), multiply(3, 3)), multiply(2, 3)))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|multiply(n0,n0)|multiply(n1,n2)|subtract(#0,#1)|subtract(#2,#3)|add(#0,#4)|add(#0,#5)|add(#6,#7)|","chain":"2 * 3<\/gadget>\n6<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 - 6<\/gadget>\n0<\/output>\n0 + 6<\/gadget>\n6<\/output>\n2 * 2<\/gadget>\n4<\/output>\n3 * 3<\/gadget>\n9<\/output>\n4 - 9<\/gadget>\n-5<\/output>\n(-5) + 6<\/gadget>\n1<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7<\/result>","index":2335} +{"problem":"a 20 % stock yielding 12 % is quoted at :","rationale":"\"income of rs 12 on investment of rs 100 income of rs 20 on investment of ? = ( 20 * 100 ) \/ 12 = 167 answer : b\"","correct":"b","options":{"a":"rs . 83.33 ","b":"rs . 167 ","c":"rs . 112 ","d":"rs . 120","e":"rs . 140"},"options_float":{"a":83.33,"b":167.0,"c":112.0,"d":120.0,"e":140.0},"annotated_formula":"multiply(divide(const_100, 12), 20)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 12<\/gadget>\n25\/3 = around 8.333333<\/output>\n(25\/3) * 20<\/gadget>\n500\/3 = around 166.666667<\/output>\n500\/3 = around 166.666667<\/result>","index":2336} +{"problem":"if the length of the sides of two cubes are in the ratio 6 : 1 , what is the ratio of their total surface area ?","rationale":"\"let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 6 x ) ^ 2 = 216 x ^ 2 . the ratio of surface areas is 36 : 1 . the answer is d .\"","correct":"d","options":{"a":"6 : 1 ","b":"12 : 1 ","c":"24 : 1 ","d":"36 : 1","e":"72 : 1"},"options_float":{"a":6.0,"b":12.0,"c":24.0,"d":36.0,"e":72.0},"annotated_formula":"multiply(6, 6)","linear_formula":"multiply(n0,n0)|","chain":"6 * 6<\/gadget>\n36<\/output>\n36<\/result>","index":2337} +{"problem":"two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 12 seconds . the length of the faster train is","rationale":"\"solution relative speed = ( 36 + 45 ) km \/ hr = ( 81 x 5 \/ 18 ) m \/ sec = ( 45 \/ 2 ) m \/ sec length of the train = ( 45 \/ 2 x 12 ) m = 270 m . answer b\"","correct":"b","options":{"a":"80 m ","b":"270 m ","c":"120 m ","d":"180 m","e":"none"},"options_float":{"a":80.0,"b":270.0,"c":120.0,"d":180.0,"e":null},"annotated_formula":"multiply(multiply(add(36, 45), const_0_2778), 12)","linear_formula":"add(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)|","chain":"36 + 45<\/gadget>\n81<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n81 * (5\/18)<\/gadget>\n45\/2 = around 22.5<\/output>\n(45\/2) * 12<\/gadget>\n270<\/output>\n270<\/result>","index":2338} +{"problem":"if the perimeter of a rectangular garden is 1800 m , its length when its breadth is 400 m is ?","rationale":"2 ( l + 400 ) = 1800 = > l = 500 m answer : a","correct":"a","options":{"a":"500 ","b":"600 ","c":"700 ","d":"800","e":"900"},"options_float":{"a":500.0,"b":600.0,"c":700.0,"d":800.0,"e":900.0},"annotated_formula":"subtract(divide(1800, const_2), 400)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"1_800 \/ 2<\/gadget>\n900<\/output>\n900 - 400<\/gadget>\n500<\/output>\n500<\/result>","index":2340} +{"problem":"sarah is driving to the airport . after driving at 15 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 60 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did she drive in all ?","rationale":"after driving at 15 miles per hourfor one hour , this distance left to cover is d - 15 . say this distance is x miles . now , we know that the difference in time between covering this distance at 15 miles per hour and 60 miles per hour is 1 + 1 \/ 2 = 3 \/ 2 hours . so , we have that x \/ 15 - x \/ 60 = 3 \/ 2 - - > 4 x \/ 60 - x \/ 60 = 3 \/ 2 - - > 3 x \/ 60 = 3 \/ 2 - - > x = 30 . total distance = x + 15 = 45 miles . answer : a","correct":"a","options":{"a":"45 ","b":"175 ","c":"210 ","d":"245","e":"280"},"options_float":{"a":45.0,"b":175.0,"c":210.0,"d":245.0,"e":280.0},"annotated_formula":"subtract(30, subtract(15, 30))","linear_formula":"subtract(n0,n2)|subtract(n2,#0)","chain":"15 - 30<\/gadget>\n-15<\/output>\n30 - (-15)<\/gadget>\n45<\/output>\n45<\/result>","index":2342} +{"problem":"a shopkeeper sells 600 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ?","rationale":"sp per metre = 18000 \/ 600 = rs . 40 loss per metre = rs . 5 cp per metre = 40 + 5 = rs . 45 . answer : d","correct":"d","options":{"a":"12 ","b":"27 ","c":"29 ","d":"35","e":"21"},"options_float":{"a":12.0,"b":27.0,"c":29.0,"d":35.0,"e":21.0},"annotated_formula":"add(divide(18000, 600), 5)","linear_formula":"divide(n1,n0)|add(n2,#0)|","chain":"18_000 \/ 600<\/gadget>\n30<\/output>\n30 + 5<\/gadget>\n35<\/output>\n35<\/result>","index":2344} +{"problem":"a certain number of men can finish a piece of work in 100 days . if there were 10 men less , it would take 10 days more for the work to be finished . how many men were there originally ?","rationale":"originally let there be x men . less men , more days ( indirect proportion ) therefore ( x - 10 ) : x : : 100 : 110 = ( x - 10 ) x 110 - x ( 100 ) = 10 x = 1100 x = 110 answer is d .","correct":"d","options":{"a":"120 ","b":"105 ","c":"115 ","d":"110","e":"140"},"options_float":{"a":120.0,"b":105.0,"c":115.0,"d":110.0,"e":140.0},"annotated_formula":"divide(10, subtract(const_1, divide(100, add(100, 10))))","linear_formula":"add(n0,n1)|divide(n0,#0)|subtract(const_1,#1)|divide(n1,#2)","chain":"100 + 10<\/gadget>\n110<\/output>\n100 \/ 110<\/gadget>\n10\/11 = around 0.909091<\/output>\n1 - (10\/11)<\/gadget>\n1\/11 = around 0.090909<\/output>\n10 \/ (1\/11)<\/gadget>\n110<\/output>\n110<\/result>","index":2346} +{"problem":"find the cost of fencing around a circular field of diameter 16 m at the rate of rs . 3 a meter ?","rationale":"\"2 * 22 \/ 7 * 8 = 50.28 50.28 * 3 = rs . 150.84 answer : a\"","correct":"a","options":{"a":"150.84 ","b":"160.6 ","c":"150.45 ","d":"160.99","e":"180.4"},"options_float":{"a":150.84,"b":160.6,"c":150.45,"d":160.99,"e":180.4},"annotated_formula":"multiply(circumface(divide(16, const_2)), 3)","linear_formula":"divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|","chain":"16 \/ 2<\/gadget>\n8<\/output>\n2 * pi * 8<\/gadget>\n16*pi = around 50.265482<\/output>\n(16*pi) * 3<\/gadget>\n48*pi = around 150.796447<\/output>\n48*pi = around 150.796447<\/result>","index":2347} +{"problem":"a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 3200 . with the help of c , they completed the work in 3 days . how much is to be paid to c","rationale":"\"explanation : c ' s 1 day ' s work = 1 \/ 3 − ( 1 \/ 6 + 1 \/ 8 ) = ( 1 \/ 3 − 7 \/ 24 ) = 1 \/ 24 a : b : c = 1 \/ 6 : 1 \/ 8 : 1 \/ 24 = 4 : 3 : 1 c ′ sshare = 1 \/ 8 ∗ 3200 = 400 option b\"","correct":"b","options":{"a":"rs . 300 ","b":"rs . 400 ","c":"rs . 500 ","d":"rs . 600","e":"none of these"},"options_float":{"a":300.0,"b":400.0,"c":500.0,"d":600.0,"e":null},"annotated_formula":"multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 3200), 3)","linear_formula":"inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/8) + (1\/6)<\/gadget>\n7\/24 = around 0.291667<\/output>\n(1\/3) - (7\/24)<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/24) * 3_200<\/gadget>\n400\/3 = around 133.333333<\/output>\n(400\/3) * 3<\/gadget>\n400<\/output>\n400<\/result>","index":2348} +{"problem":"a worker can load one truck in 6 hours . a second worker can load the same truck in 4 hours . if both workers load one truck simultaneously while maintaining their constant rates , how long , in hours , will it take them to fill one truck ?","rationale":"the workers fill the truck at a rate of 1 \/ 6 + 1 \/ 4 = 10 \/ 24 = 5 \/ 12 of the truck per hour . then the time to fill one truck is 12 \/ 5 which is 2.4 hours . the answer is b .","correct":"b","options":{"a":"2.1 ","b":"2.4 ","c":"2.7 ","d":"3.0","e":"3.3"},"options_float":{"a":2.1,"b":2.4,"c":2.7,"d":3.0,"e":3.3},"annotated_formula":"inverse(add(divide(const_1, 6), divide(const_1, 4)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/6) + (1\/4)<\/gadget>\n5\/12 = around 0.416667<\/output>\n1 \/ (5\/12)<\/gadget>\n12\/5 = around 2.4<\/output>\n12\/5 = around 2.4<\/result>","index":2350} +{"problem":"40 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 3 like all . how many people like apple ?","rationale":"orange + mango - apple = 7 mango + apple - orange = 10 apple = 40 orange + mango + apple = 3 40 + 10 + 3 - 7 = 46 like apple answer : b","correct":"b","options":{"a":"47 ","b":"46 ","c":"54 ","d":"58","e":"62"},"options_float":{"a":47.0,"b":46.0,"c":54.0,"d":58.0,"e":62.0},"annotated_formula":"add(add(subtract(40, const_3), 7), subtract(10, 7))","linear_formula":"subtract(n0,const_3)|subtract(n2,n1)|add(n1,#0)|add(#2,#1)","chain":"40 - 3<\/gadget>\n37<\/output>\n37 + 7<\/gadget>\n44<\/output>\n10 - 7<\/gadget>\n3<\/output>\n44 + 3<\/gadget>\n47<\/output>\n47<\/result>","index":2351} +{"problem":"3 candidates in an election and received 1256 , 7636 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ?","rationale":"\"total number of votes polled = ( 1256 + 7636 + 11628 ) = 20520 so , required percentage = 11628 \/ 20520 * 100 = 56.7 % a\"","correct":"a","options":{"a":"56.7 % ","b":"55 % ","c":"57 % ","d":"60 %","e":"62 %"},"options_float":{"a":56.7,"b":55.0,"c":57.0,"d":60.0,"e":62.0},"annotated_formula":"multiply(divide(11628, add(add(1256, 7636), 11628)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)|","chain":"1_256 + 7_636<\/gadget>\n8_892<\/output>\n8_892 + 11_628<\/gadget>\n20_520<\/output>\n11_628 \/ 20_520<\/gadget>\n17\/30 = around 0.566667<\/output>\n(17\/30) * 100<\/gadget>\n170\/3 = around 56.666667<\/output>\n170\/3 = around 56.666667<\/result>","index":2354} +{"problem":"the ratio of the radius of two circles is 1 : 3 , and then the ratio of their areas is ?","rationale":"\"r 1 : r 2 = 1 : 3 π r 12 : π r 22 r 12 : r 22 = 1 : 9 answer : c\"","correct":"c","options":{"a":"1 : 5 ","b":"1 : 6 ","c":"1 : 9 ","d":"1 : 3","e":"1 : 2"},"options_float":{"a":0.2,"b":0.1666666667,"c":0.1111111111,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(circle_area(1), circle_area(3))","linear_formula":"circle_area(n0)|circle_area(n1)|divide(#0,#1)|","chain":"pi * (1 ** 2)<\/gadget>\npi = around 3.141593<\/output>\npi * (3 ** 2)<\/gadget>\n9*pi = around 28.274334<\/output>\npi \/ (9*pi)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":2356} +{"problem":"bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 63 more . what was his original selling price ?","rationale":"\"let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 63 0.07 = 63 p = 900 the original selling price was 900 * 1.1 = 990 . the correct answer is b .\"","correct":"b","options":{"a":"$ 850 ","b":"$ 990 ","c":"$ 1050 ","d":"$ 1180","e":"$ 1340"},"options_float":{"a":850.0,"b":990.0,"c":1050.0,"d":1180.0,"e":1340.0},"annotated_formula":"multiply(divide(63, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|subtract(#5,#2)|divide(n3,#6)|multiply(#2,#7)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n(9\/10) * (13\/10)<\/gadget>\n117\/100 = around 1.17<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(117\/100) - (11\/10)<\/gadget>\n7\/100 = around 0.07<\/output>\n63 \/ (7\/100)<\/gadget>\n900<\/output>\n900 * (11\/10)<\/gadget>\n990<\/output>\n990<\/result>","index":2360} +{"problem":"one - third of 1206 is what percent of 162 ?","rationale":"\"answer let one - third of 1206 is n % of 162 . ∵ 1206 \/ 3 = ( n x 162 ) \/ 100 ∴ n = ( 402 x 100 ) \/ 162 = 248.14 correct option : c\"","correct":"c","options":{"a":"324.11 ","b":"300.12 ","c":"248.14 ","d":"none of these","e":"can not be determined"},"options_float":{"a":324.11,"b":300.12,"c":248.14,"d":null,"e":null},"annotated_formula":"multiply(const_100, divide(divide(1206, const_3), 162))","linear_formula":"divide(n0,const_3)|divide(#0,n1)|multiply(#1,const_100)|","chain":"1_206 \/ 3<\/gadget>\n402<\/output>\n402 \/ 162<\/gadget>\n67\/27 = around 2.481481<\/output>\n100 * (67\/27)<\/gadget>\n6_700\/27 = around 248.148148<\/output>\n6_700\/27 = around 248.148148<\/result>","index":2361} +{"problem":"on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 50 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ?","rationale":"0.1 m = 0.50 e = > e \/ m = 1 \/ 5 * 100 = 20 % so answer is e . m - # of motorists e - # of motorists exceeding speed","correct":"e","options":{"a":"10.5 % ","b":"12.5 % ","c":"15 % ","d":"22 %","e":"20 %"},"options_float":{"a":10.5,"b":12.5,"c":15.0,"d":22.0,"e":20.0},"annotated_formula":"divide(const_100, multiply(multiply(divide(10, const_100), divide(50, const_100)), const_100))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/10) * (1\/2)<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20<\/result>","index":2363} +{"problem":"in a group of 7 boys and 8 girls , 5 members are to be selected . in how many different ways can they be selected such that at least 3 boys should be there ?","rationale":"no of boys = 7 no of girls = 8 therefore , no of ways at least 3 boy can be selected = 7 c 5 * 8 c 0 + 7 c 4 * 8 c 1 + 7 c 3 * 8 c 2 = 1281 ans - c","correct":"c","options":{"a":"1200 ","b":"1500 ","c":"1281 ","d":"1339","e":"1296"},"options_float":{"a":1200.0,"b":1500.0,"c":1281.0,"d":1339.0,"e":1296.0},"annotated_formula":"add(multiply(divide(multiply(factorial(7), factorial(7)), factorial(8)), const_2), multiply(3, 7))","linear_formula":"factorial(n0)|factorial(n1)|multiply(n0,n3)|multiply(#0,#0)|divide(#3,#1)|multiply(#4,const_2)|add(#5,#2)","chain":"factorial(7)<\/gadget>\n5_040<\/output>\n5_040 * 5_040<\/gadget>\n25_401_600<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n25_401_600 \/ 40_320<\/gadget>\n630<\/output>\n630 * 2<\/gadget>\n1_260<\/output>\n3 * 7<\/gadget>\n21<\/output>\n1_260 + 21<\/gadget>\n1_281<\/output>\n1_281<\/result>","index":2364} +{"problem":"{ - 10 , - 6 , - 5 , - 4 , - 2.5 , - 1 , 0 , 2.5 , 4 , 6 , 7 , 10 } a number is to be selected at random from the set above . what is the probability that the number selected will be a solution of the equation ( x + 5 ) ( x + 10 ) ( 2 x - 5 ) = 0 ?","rationale":"roots of the equation ( x - 5 ) ( x + 10 ) ( 2 x - 5 ) = 0 are x = - 5 , x = - 10 andx = 5 \/ 2 = 2.5 . so , three solutions present in our set of 12 distinct numbers , therefore p = 3 \/ 12 = 1 \/ 4 . answer : c .","correct":"c","options":{"a":"1 \/ 12 ","b":"1 \/ 6 ","c":"1 \/ 4 ","d":"1 \/ 3","e":"1 \/ 2"},"options_float":{"a":0.0833333333,"b":0.1666666667,"c":0.25,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(add(2, 1), add(10, 2))","linear_formula":"add(n5,n14)|add(n0,n14)|divide(#0,#1)","chain":"2 + 1<\/gadget>\n3<\/output>\n10 + 2<\/gadget>\n12<\/output>\n3 \/ 12<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":2366} +{"problem":"a towel , when bleached , was found to have lost 20 % of its length and 10 % of its breadth . the percentage of decrease in area is","rationale":"\"let original length = x and original breadth = y . decrease in area = xy - 80 x x 90 y 100 100 = xy - 18 xy 25 = 7 xy . 25 decrease % = 7 xy x 1 x 100 % = 28 % . 25 xy a )\"","correct":"a","options":{"a":"28 % ","b":"30 % ","c":"32 % ","d":"38 %","e":"40 %"},"options_float":{"a":28.0,"b":30.0,"c":32.0,"d":38.0,"e":40.0},"annotated_formula":"multiply(subtract(const_1, multiply(divide(subtract(const_100, 20), const_100), divide(subtract(const_100, 10), const_100))), const_100)","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n(4\/5) * (9\/10)<\/gadget>\n18\/25 = around 0.72<\/output>\n1 - (18\/25)<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) * 100<\/gadget>\n28<\/output>\n28<\/result>","index":2367} +{"problem":"the average of 10 numbers is calculated as 18 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ?","rationale":"\"explanation : 10 * 18 + 36 – 26 = 190 = > 190 \/ 10 = 19 c )\"","correct":"c","options":{"a":"16 ","b":"18 ","c":"19 ","d":"22","e":"24"},"options_float":{"a":16.0,"b":18.0,"c":19.0,"d":22.0,"e":24.0},"annotated_formula":"add(18, divide(subtract(36, 26), 10))","linear_formula":"subtract(n2,n3)|divide(#0,n0)|add(n1,#1)|","chain":"36 - 26<\/gadget>\n10<\/output>\n10 \/ 10<\/gadget>\n1<\/output>\n18 + 1<\/gadget>\n19<\/output>\n19<\/result>","index":2369} +{"problem":"the speed of a boat in still water in 42 km \/ hr and the rate of current is 5 km \/ hr . the distance travelled downstream in 44 minutes is :","rationale":"\"speed downstream = ( 42 + 5 ) = 47 kmph time = 44 minutes = 44 \/ 60 hour = 11 \/ 15 hour distance travelled = time × speed = 11 \/ 15 × 47 = 34.5 km answer : c\"","correct":"c","options":{"a":"86.6 km ","b":"46.6 km ","c":"34.5 km ","d":"35.6 km","e":"26.6 km"},"options_float":{"a":86.6,"b":46.6,"c":34.5,"d":35.6,"e":26.6},"annotated_formula":"multiply(add(42, 5), divide(44, const_60))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"42 + 5<\/gadget>\n47<\/output>\n44 \/ 60<\/gadget>\n11\/15 = around 0.733333<\/output>\n47 * (11\/15)<\/gadget>\n517\/15 = around 34.466667<\/output>\n517\/15 = around 34.466667<\/result>","index":2370} +{"problem":"the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1240 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ?","rationale":"\"sum of square of n numbers is found using the formula [ n ( n + 1 ) ( 2 n + 1 ) ] \/ 6 we have sum of 1 st 15 numbers = 1240 we need to find sum of squares from 16 to 30 which is = sum of squares of 1 st 30 + ve integers - sum of squares of 1 st 15 + ve integers . we will get d - 8215 ; answer : d\"","correct":"d","options":{"a":"2480 ","b":"3490 ","c":"6785 ","d":"8215","e":"9255"},"options_float":{"a":2480.0,"b":3490.0,"c":6785.0,"d":8215.0,"e":9255.0},"annotated_formula":"subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1240)","linear_formula":"add(n17,n1)|add(n5,n5)|multiply(n17,n2)|add(#2,n1)|multiply(n17,#0)|multiply(#3,#4)|divide(#5,#1)|subtract(#6,n9)|","chain":"30 + 1<\/gadget>\n31<\/output>\n30 * 31<\/gadget>\n930<\/output>\n2 * 30<\/gadget>\n60<\/output>\n60 + 1<\/gadget>\n61<\/output>\n930 * 61<\/gadget>\n56_730<\/output>\n3 + 3<\/gadget>\n6<\/output>\n56_730 \/ 6<\/gadget>\n9_455<\/output>\n9_455 - 1_240<\/gadget>\n8_215<\/output>\n8_215<\/result>","index":2374} +{"problem":"in a department store prize box , 40 % of the notes give the winner a dreamy vacation ; the other notes are blank . what is the approximate probability that 3 out of 5 people that draw the notes one after the other , and immediately return their note into the box get a dreamy vacation ?","rationale":"the probability of winning is 40 % = 40 \/ 100 = 2 \/ 5 . the probability of not winning is 60 % = 3 \/ 5 . p ( wwwnn ) = 5 ! 3 ! 2 ! ∗ ( 25 ) 3 ∗ ( 35 ) 2 = 144625 ≈ 23 p ( wwwnn ) = 5 ! 3 ! 2 ! ∗ ( 25 ) 3 ∗ ( 35 ) 2 = 144625 ≈ 23 ( we multiply by 5 ! 3 ! 2 ! 5 ! 3 ! 2 ! , because wwwnn scenario can occur in several ways : wwwnn , wwnwn , wnwwn , nwwwn , . . . the # of such cases basically equals to the # of permutations of 5 letters wwwnn , which is 5 ! 3 ! 2 ! 5 ! 3 ! 2 ! ) . answer : b .","correct":"b","options":{"a":"0.12 ","b":"0.23 ","c":"0.35 ","d":"0.45","e":"0.65"},"options_float":{"a":0.12,"b":0.23,"c":0.35,"d":0.45,"e":0.65},"annotated_formula":"multiply(multiply(choose(5, 3), power(divide(const_3, 5), const_2)), power(divide(const_2, 5), const_3))","linear_formula":"choose(n2,n1)|divide(const_3,n2)|divide(const_2,n2)|power(#1,const_2)|power(#2,const_3)|multiply(#0,#3)|multiply(#5,#4)","chain":"binomial(5, 3)<\/gadget>\n10<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) ** 2<\/gadget>\n9\/25 = around 0.36<\/output>\n10 * (9\/25)<\/gadget>\n18\/5 = around 3.6<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) ** 3<\/gadget>\n8\/125 = around 0.064<\/output>\n(18\/5) * (8\/125)<\/gadget>\n144\/625 = around 0.2304<\/output>\n144\/625 = around 0.2304<\/result>","index":2375} +{"problem":"if the volume of the cube is 729 cm 3 , then the surface area of the cube will be","rationale":"explanation : a 3 = 729 ; a = 9 surface area = ( 6 x 9 x 9 ) = 486 cm 2 answer : a","correct":"a","options":{"a":"486 sq . cm ","b":"456 sq . cm ","c":"446 sq . cm ","d":"476 sq . cm","e":"none of these"},"options_float":{"a":486.0,"b":456.0,"c":446.0,"d":476.0,"e":null},"annotated_formula":"surface_cube(cube_edge_by_volume(729))","linear_formula":"cube_edge_by_volume(n0)|surface_cube(#0)","chain":"729 ** (1\/3)<\/gadget>\n9<\/output>\n6 * (9 ** 2)<\/gadget>\n486<\/output>\n486<\/result>","index":2376} +{"problem":"what is the rate percent when the simple interest on rs . 1600 amount to rs . 200 in 4 years ?","rationale":"\"interest for 1 year = 200 \/ 4 = 50 interest on rs 1600 p \/ a = 50 interest rate = 50 \/ 1600 * 100 = 3.125 % answer : b\"","correct":"b","options":{"a":"4 % ","b":"3.125 % ","c":"5 % ","d":"6 %","e":"7 %"},"options_float":{"a":4.0,"b":3.125,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(multiply(const_100, 200), multiply(1600, 4))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 200<\/gadget>\n20_000<\/output>\n1_600 * 4<\/gadget>\n6_400<\/output>\n20_000 \/ 6_400<\/gadget>\n25\/8 = around 3.125<\/output>\n25\/8 = around 3.125<\/result>","index":2377} +{"problem":"if the simple interest on a certain sum of money for 2 years is one – fifth of the sum , then the rate of interest per annum is","rationale":"\"explanation : let the principal ( p ) be x then , simple interest ( si ) = x \/ 5 time ( t ) = 2 years rate of interest per annum ( r ) = ( 100 × si ) \/ pt = ( 100 × ( x \/ 5 ) \/ ( x × 2 ) = 20 \/ 2 = 10 % answer : option c\"","correct":"c","options":{"a":"14 % ","b":"7 % ","c":"10 % ","d":"15 %","e":"3 %"},"options_float":{"a":14.0,"b":7.0,"c":10.0,"d":15.0,"e":3.0},"annotated_formula":"divide(divide(const_100, add(const_1, const_4)), 2)","linear_formula":"add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0)|","chain":"1 + 4<\/gadget>\n5<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":2378} +{"problem":"a , b , c , d , e , f , g , h sitting in a row what is the probability that a , b , c sitting together ?","rationale":"total number of arrangement is = 8 ! = 40320 favorable event i . e ) a , b and c can be arranged in 3 ! and the remaining can be arranged in 6 ! since ab can be in any of the four positions . so 3 ! * 6 ! \/ 8 ! = 3 \/ 28 answer : c","correct":"c","options":{"a":"3 \/ 14 ","b":"5 \/ 28 ","c":"3 \/ 28 ","d":"1 \/ 7","e":"9 \/ 28"},"options_float":{"a":0.2142857143,"b":0.1785714286,"c":0.1071428571,"d":0.1428571429,"e":0.3214285714},"annotated_formula":"divide(multiply(factorial(add(subtract(add(const_4, const_4), const_3), const_1)), factorial(const_3)), factorial(add(const_4, const_4)))","linear_formula":"add(const_4,const_4)|factorial(const_3)|factorial(#0)|subtract(#0,const_3)|add(#3,const_1)|factorial(#4)|multiply(#5,#1)|divide(#6,#2)","chain":"4 + 4<\/gadget>\n8<\/output>\n8 - 3<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n720 * 6<\/gadget>\n4_320<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n4_320 \/ 40_320<\/gadget>\n3\/28 = around 0.107143<\/output>\n3\/28 = around 0.107143<\/result>","index":2379} +{"problem":"there are 2 sections a and b in a class , consisting of 50 and 50 students respectively . if the average weight of section a is 60 kg and that of section b is 80 kg , find the average of the whole class ?","rationale":"\"total weight of 50 + 50 students = 50 * 60 + 50 * 80 = 3000 + 4000 average weight of the class is = 7000 \/ 100 = 70 kg answer is d\"","correct":"d","options":{"a":"50.78 kg ","b":"49.32 kg ","c":"61.67 kg ","d":"70 kg","e":"70.89 kg"},"options_float":{"a":50.78,"b":49.32,"c":61.67,"d":70.0,"e":70.89},"annotated_formula":"divide(add(multiply(50, 60), multiply(50, 80)), add(50, 50))","linear_formula":"add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)|","chain":"50 * 60<\/gadget>\n3_000<\/output>\n50 * 80<\/gadget>\n4_000<\/output>\n3_000 + 4_000<\/gadget>\n7_000<\/output>\n50 + 50<\/gadget>\n100<\/output>\n7_000 \/ 100<\/gadget>\n70<\/output>\n70<\/result>","index":2380} +{"problem":"in a school with 632 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is","rationale":"\"sol . let the number of grils be x . then , number of boys = ( 632 - x ) . then , ( 11 3 \/ 4 × 632 ) ⇔ 11 x + 12 ( 632 - x ) ⇔ x = 7584 - 7426 ⇔ 158 . answer d\"","correct":"d","options":{"a":"150 ","b":"200 ","c":"250 ","d":"158","e":"none"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":158.0,"e":null},"annotated_formula":"subtract(multiply(12, 632), multiply(add(11, divide(9, 12)), 632))","linear_formula":"divide(n4,n1)|multiply(n0,n1)|add(n2,#0)|multiply(n0,#2)|subtract(#1,#3)|","chain":"12 * 632<\/gadget>\n7_584<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n11 + (3\/4)<\/gadget>\n47\/4 = around 11.75<\/output>\n(47\/4) * 632<\/gadget>\n7_426<\/output>\n7_584 - 7_426<\/gadget>\n158<\/output>\n158<\/result>","index":2381} +{"problem":"the salary of a person was reduced by 28 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ?","rationale":"\"let the original salary be $ 100 new salary = $ 72 increase on 72 = 28 increase on 100 = [ ( 28 \/ 72 ) 100 ] % = 38.9 % answer is e\"","correct":"e","options":{"a":"10 % ","b":"15 % ","c":"20 % ","d":"25 %","e":"38.9 %"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":38.9},"annotated_formula":"multiply(divide(multiply(const_100, divide(28, const_100)), subtract(const_100, multiply(const_100, divide(28, const_100)))), const_100)","linear_formula":"divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)|","chain":"28 \/ 100<\/gadget>\n7\/25 = around 0.28<\/output>\n100 * (7\/25)<\/gadget>\n28<\/output>\n100 - 28<\/gadget>\n72<\/output>\n28 \/ 72<\/gadget>\n7\/18 = around 0.388889<\/output>\n(7\/18) * 100<\/gadget>\n350\/9 = around 38.888889<\/output>\n350\/9 = around 38.888889<\/result>","index":2384} +{"problem":"the average height of 35 boys in a class was calculated as 183 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ?","rationale":"\"calculated average height of 35 boys = 183 cm . wrong total height of 35 boys = 183 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 183 cm - ( 166 cm - 106 cm ) \/ 35 = 183 cm - 60 \/ 35 cm = 183 cm - 1.71 cm = 181.29 cm . answer : b\"","correct":"b","options":{"a":"187.89 cm ","b":"181.29 cm ","c":"123.98 cm ","d":"149.98 cm","e":"146.89 cm"},"options_float":{"a":187.89,"b":181.29,"c":123.98,"d":149.98,"e":146.89},"annotated_formula":"floor(divide(add(subtract(multiply(35, 183), 166), 106), 35))","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)|","chain":"35 * 183<\/gadget>\n6_405<\/output>\n6_405 - 166<\/gadget>\n6_239<\/output>\n6_239 + 106<\/gadget>\n6_345<\/output>\n6_345 \/ 35<\/gadget>\n1_269\/7 = around 181.285714<\/output>\nfloor(1_269\/7)<\/gadget>\n181<\/output>\n181<\/result>","index":2385} +{"problem":"a is half good a work man as b and together they finish a job in 14 days . in how many days working alone b finish the job ?","rationale":"\"c 21 wc = 1 : 2 2 x + x = 1 \/ 14 = > x = 1 \/ 42 2 x = 1 \/ 21 = > 21 days\"","correct":"c","options":{"a":"23 ","b":"22 ","c":"21 ","d":"36","e":"48"},"options_float":{"a":23.0,"b":22.0,"c":21.0,"d":36.0,"e":48.0},"annotated_formula":"multiply(14, divide(const_3, const_2))","linear_formula":"divide(const_3,const_2)|multiply(n0,#0)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n14 * (3\/2)<\/gadget>\n21<\/output>\n21<\/result>","index":2386} +{"problem":"each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.55 % , what fraction of the products was checked by worker y ?","rationale":"\"x : 0.5 % is 0.05 % - points from 0.55 % . y : 0.8 % is 0.25 % - points from 0.55 % . therefore the ratio of products checked by y : x is 1 : 5 . thus , worker y checked 1 \/ 6 of the products . the answer is d .\"","correct":"d","options":{"a":"1 \/ 3 ","b":"1 \/ 4 ","c":"1 \/ 5 ","d":"1 \/ 6","e":"1 \/ 8"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.2,"d":0.1666666667,"e":0.125},"annotated_formula":"divide(subtract(0.55, 0.5), subtract(0.8, 0.5))","linear_formula":"subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)|","chain":"0.55 - 0.5<\/gadget>\n0.05<\/output>\n0.8 - 0.5<\/gadget>\n0.3<\/output>\n0.05 \/ 0.3<\/gadget>\n0.166667<\/output>\n0.166667<\/result>","index":2387} +{"problem":"a baseball card decreased in value 40 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ?","rationale":"\"consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.6 = 60 after second year price = 60 * 0.9 = 54 final decrease = [ ( 100 - 54 ) \/ 100 ] * 100 = 46 % correct answer - d\"","correct":"d","options":{"a":"28 % ","b":"30 % ","c":"32 % ","d":"46 %","e":"72 %"},"options_float":{"a":28.0,"b":30.0,"c":32.0,"d":46.0,"e":72.0},"annotated_formula":"subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(40, const_100))), const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(9\/10) * (3\/5)<\/gadget>\n27\/50 = around 0.54<\/output>\n(27\/50) * 100<\/gadget>\n54<\/output>\n100 - 54<\/gadget>\n46<\/output>\n46<\/result>","index":2388} +{"problem":"from a group of 3 boys and 3 girls , 4 children are to be randomly selected . what is the probability that equal numbers of boys and girls will be selected ?","rationale":"prob reqd = 1 - probab of not having eq numbers prob of not having eq no ' s is = prob of having 3 boys and 1 girl or prob of having 3 girls and 1 boy = ( 1 * 3 \/ 15 ) + ( 3 \/ 15 * 1 ) = 2 \/ 5 reqd prob = 1 - 2 \/ 5 = 3 \/ 5 answer : d","correct":"d","options":{"a":"1 \/ 10 ","b":"4 \/ 9 ","c":"1 \/ 2 ","d":"3 \/ 5","e":"2 \/ 3"},"options_float":{"a":0.1,"b":0.4444444444,"c":0.5,"d":0.6,"e":0.6666666667},"annotated_formula":"divide(multiply(choose(3, const_2), choose(const_3, const_2)), choose(add(3, 3), const_4))","linear_formula":"add(n0,n0)|choose(n0,const_2)|choose(const_3,const_2)|choose(#0,const_4)|multiply(#1,#2)|divide(#4,#3)","chain":"binomial(3, 2)<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n3 + 3<\/gadget>\n6<\/output>\nbinomial(6, 4)<\/gadget>\n15<\/output>\n9 \/ 15<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":2389} +{"problem":"9 years ago i was 5 times as old as my eldest son . today i am 3 times his age . how old am i now ?","rationale":"x - 9 = 5 ( y - 9 ) & x = 3 y x - 9 = 5 x \/ 3 - 45 x = 54 answer : b","correct":"b","options":{"a":"45 ","b":"54 ","c":"34 ","d":"58","e":"56"},"options_float":{"a":45.0,"b":54.0,"c":34.0,"d":58.0,"e":56.0},"annotated_formula":"add(multiply(5, 9), 9)","linear_formula":"multiply(n0,n1)|add(n0,#0)","chain":"5 * 9<\/gadget>\n45<\/output>\n45 + 9<\/gadget>\n54<\/output>\n54<\/result>","index":2390} +{"problem":"find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg","rationale":"\"by the rule of alligation : cost of 1 kg rice of 1 st kind cost of 1 kg rice of 2 nd kind required ratio = 60 : 90 = 2 : 3 answer : b\"","correct":"b","options":{"a":"2 : 0 ","b":"2 : 3 ","c":"2 : 1 ","d":"2 : 2","e":"2 : 8"},"options_float":{"a":null,"b":0.6666666667,"c":2.0,"d":1.0,"e":0.25},"annotated_formula":"divide(subtract(6.30, 5.70), subtract(7.20, 6.30))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"6.3 - 5.7<\/gadget>\n0.6<\/output>\n7.2 - 6.3<\/gadget>\n0.9<\/output>\n0.6 \/ 0.9<\/gadget>\n0.666667<\/output>\n0.666667<\/result>","index":2391} +{"problem":"a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is 4 times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is quartered by water evaporation . at that time , what is the ratio of water to salt in the solution ?","rationale":"water : soap = 3 : 2 soap : salt = 12 : 8 = > for 12 soap , salt = 8 = > for 2 soap , salt = ( 8 \/ 12 ) * 2 = 16 \/ 12 = 4 \/ 3 so , water : soap : salt = 3 : 2 : 4 \/ 3 = 9 : 6 : 4 after open container , water : soap : salt = 2.25 : 6 : 4 so , water : salt = 2.25 : 4 = 9 : 16 answer : c","correct":"c","options":{"a":"12 : 8 ","b":"8 : 16 ","c":"9 : 16 ","d":"8 : 12","e":"16 : 9"},"options_float":{"a":1.5,"b":0.5,"c":0.5625,"d":0.6666666667,"e":1.7777777778},"annotated_formula":"divide(multiply(divide(multiply(3, 4), multiply(2, 4)), 3), multiply(2, 4))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,#1)","chain":"3 * 4<\/gadget>\n12<\/output>\n2 * 4<\/gadget>\n8<\/output>\n12 \/ 8<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 3<\/gadget>\n9\/2 = around 4.5<\/output>\n(9\/2) \/ 8<\/gadget>\n9\/16 = around 0.5625<\/output>\n9\/16 = around 0.5625<\/result>","index":2392} +{"problem":"initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many q women are currently in the room ?","rationale":"\"the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number q = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d\"","correct":"d","options":{"a":"12 ","b":"14 ","c":"15 ","d":"24","e":"36"},"options_float":{"a":12.0,"b":14.0,"c":15.0,"d":24.0,"e":36.0},"annotated_formula":"multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3))","linear_formula":"subtract(n4,n2)|multiply(n1,#0)|divide(#1,n0)|subtract(#2,n3)|multiply(n2,#3)|","chain":"14 - 2<\/gadget>\n12<\/output>\n5 * 12<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15 - 3<\/gadget>\n12<\/output>\n2 * 12<\/gadget>\n24<\/output>\n24<\/result>","index":2393} +{"problem":"a caterer ordered 125 ice - cream bars and 125 sundaes . if the total price was $ 225.00 and the price of each ice - cream bar was $ 0.60 , what was the price of each sundae ?","rationale":"\"let price of a sundae = s price of ice cream bar = . 6 $ 125 * . 6 + 125 * s = 225 = > 125 * s = 150 = > s = 1.2 answer d\"","correct":"d","options":{"a":"$ 0.60 ","b":"$ 0.80 ","c":"$ 1.00 ","d":"$ 1.20","e":"$ 1.60"},"options_float":{"a":0.6,"b":0.8,"c":1.0,"d":1.2,"e":1.6},"annotated_formula":"divide(subtract(225.00, multiply(125, 0.60)), 125)","linear_formula":"multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)|","chain":"125 * 0.6<\/gadget>\n75<\/output>\n225 - 75<\/gadget>\n150<\/output>\n150 \/ 125<\/gadget>\n6\/5 = around 1.2<\/output>\n6\/5 = around 1.2<\/result>","index":2394} +{"problem":"a and b invests rs . 3000 and rs . 5500 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?","rationale":"\"( 3 * 6 + 6 * 6 ) : ( 5.5 * 12 ) 54 : 66 = > 9 : 11 answer : e\"","correct":"e","options":{"a":"8 : 8 ","b":"9 : 8 ","c":"9 : 7 ","d":"9 : 5","e":"9 : 11"},"options_float":{"a":1.0,"b":1.125,"c":1.2857142857,"d":1.8,"e":0.8181818182},"annotated_formula":"divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(5500, add(6, 6)))","linear_formula":"add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)|","chain":"3_000 * 6<\/gadget>\n18_000<\/output>\n3_000 * 2<\/gadget>\n6_000<\/output>\n6_000 * 6<\/gadget>\n36_000<\/output>\n18_000 + 36_000<\/gadget>\n54_000<\/output>\n6 + 6<\/gadget>\n12<\/output>\n5_500 * 12<\/gadget>\n66_000<\/output>\n54_000 \/ 66_000<\/gadget>\n9\/11 = around 0.818182<\/output>\n9\/11 = around 0.818182<\/result>","index":2395} +{"problem":"if a lends rs . 3150 to b at 8 % per annum and b lends the same sum to c at 12.5 % per annum then the gain of b in a period of 2 years is ?","rationale":"\"( 3150 * 4.5 * 2 ) \/ 100 = > 283.5 answer : c\"","correct":"c","options":{"a":"280 ","b":"295 ","c":"283.5 ","d":"245","e":"200"},"options_float":{"a":280.0,"b":295.0,"c":283.5,"d":245.0,"e":200.0},"annotated_formula":"subtract(divide(multiply(multiply(3150, 12.5), 2), const_100), divide(multiply(multiply(3150, 8), 2), const_100))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|","chain":"3_150 * 12.5<\/gadget>\n39_375<\/output>\n39_375 * 2<\/gadget>\n78_750<\/output>\n78_750 \/ 100<\/gadget>\n1_575\/2 = around 787.5<\/output>\n3_150 * 8<\/gadget>\n25_200<\/output>\n25_200 * 2<\/gadget>\n50_400<\/output>\n50_400 \/ 100<\/gadget>\n504<\/output>\n(1_575\/2) - 504<\/gadget>\n567\/2 = around 283.5<\/output>\n567\/2 = around 283.5<\/result>","index":2396} +{"problem":"apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 663 and for 36 kgs of apples is 726 then the cost of first 10 kgs of apples is","rationale":"ans : by framing equations we get 30 l + 3 q = 663 30 l + 6 q = 726 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first then we get l = 20 cost of 10 kgs of apples = 10 x 20 = 200 answer : c","correct":"c","options":{"a":"100 ","b":"150 ","c":"200 ","d":"350","e":"450"},"options_float":{"a":100.0,"b":150.0,"c":200.0,"d":350.0,"e":450.0},"annotated_formula":"multiply(divide(subtract(663, multiply(subtract(33, 30), divide(subtract(726, 663), subtract(36, 33)))), 30), 10)","linear_formula":"subtract(n4,n2)|subtract(n3,n1)|subtract(n1,n0)|divide(#0,#1)|multiply(#3,#2)|subtract(n2,#4)|divide(#5,n0)|multiply(n5,#6)","chain":"33 - 30<\/gadget>\n3<\/output>\n726 - 663<\/gadget>\n63<\/output>\n36 - 33<\/gadget>\n3<\/output>\n63 \/ 3<\/gadget>\n21<\/output>\n3 * 21<\/gadget>\n63<\/output>\n663 - 63<\/gadget>\n600<\/output>\n600 \/ 30<\/gadget>\n20<\/output>\n20 * 10<\/gadget>\n200<\/output>\n200<\/result>","index":2397} +{"problem":"a rower can row 7 km \/ h in still water . when the river is running at 2 km \/ h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ?","rationale":"let x be the distance to big rock . time = x \/ 5 + x \/ 9 = 1 x = 45 \/ 14 = 3.21 km the answer is d .","correct":"d","options":{"a":"2.92 ","b":"3.04 ","c":"3.12 ","d":"3.21","e":"3.35"},"options_float":{"a":2.92,"b":3.04,"c":3.12,"d":3.21,"e":3.35},"annotated_formula":"multiply(divide(subtract(7, 2), add(add(7, 2), subtract(7, 2))), add(7, 2))","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)","chain":"7 - 2<\/gadget>\n5<\/output>\n7 + 2<\/gadget>\n9<\/output>\n9 + 5<\/gadget>\n14<\/output>\n5 \/ 14<\/gadget>\n5\/14 = around 0.357143<\/output>\n(5\/14) * 9<\/gadget>\n45\/14 = around 3.214286<\/output>\n45\/14 = around 3.214286<\/result>","index":2400} +{"problem":"a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a women alone in 18 days , how long will a boy take to complete the work ?","rationale":"\"explanation : ( 1 man + 1 woman + 1 boy ) ’ s 1 day ’ s work = 1 \/ 3 1 man ’ s 1 day work = 1 \/ 6 1 womans 1 day ’ s work = 1 \/ 18 ( 1 man + 1 woman ) ‘ s 1 day ’ s work = 1 \/ 6 + 1 \/ 18 = 2 \/ 9 therefore , 1 boys 1 day ’ s work = 1 \/ 3 – 2 \/ 9 = 3 - 2 \/ 9 = 1 \/ 9 therefore , the boy alone can finish the work in 9 days . answer : option c\"","correct":"c","options":{"a":"6 days ","b":"8 days ","c":"9 days ","d":"7 days","e":"5 days"},"options_float":{"a":6.0,"b":8.0,"c":9.0,"d":7.0,"e":5.0},"annotated_formula":"inverse(subtract(inverse(3), add(inverse(6), inverse(18))))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/6) + (1\/18)<\/gadget>\n2\/9 = around 0.222222<\/output>\n(1\/3) - (2\/9)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9<\/result>","index":2402} +{"problem":"the least number which should be added to 1789 so that the sum is exactly divisible by 5 , 6 , 4 and 3 is ?","rationale":"l . c . m . of 5 , 6 , 4 and 3 = 60 . on dividing 1789 by 60 , the remainder is 49 . number to be added = ( 60 - 49 ) = 11 b )","correct":"b","options":{"a":"1 ","b":"11 ","c":"21 ","d":"31","e":"41"},"options_float":{"a":1.0,"b":11.0,"c":21.0,"d":31.0,"e":41.0},"annotated_formula":"subtract(multiply(3, multiply(4, 5)), multiply(subtract(divide(1789, multiply(3, multiply(4, 5))), subtract(multiply(const_3, const_10), const_1)), multiply(3, multiply(4, 5))))","linear_formula":"multiply(n1,n3)|multiply(const_10,const_3)|multiply(n4,#0)|subtract(#1,const_1)|divide(n0,#2)|subtract(#4,#3)|multiply(#2,#5)|subtract(#2,#6)","chain":"4 * 5<\/gadget>\n20<\/output>\n3 * 20<\/gadget>\n60<\/output>\n1_789 \/ 60<\/gadget>\n1_789\/60 = around 29.816667<\/output>\n3 * 10<\/gadget>\n30<\/output>\n30 - 1<\/gadget>\n29<\/output>\n(1_789\/60) - 29<\/gadget>\n49\/60 = around 0.816667<\/output>\n(49\/60) * 60<\/gadget>\n49<\/output>\n60 - 49<\/gadget>\n11<\/output>\n11<\/result>","index":2403} +{"problem":"i chose a number and divide it by 8 . then i subtracted 160 from the result and got 12 . what was the number i chose ?","rationale":"\"let x be the number i chose , then x \/ 8 − 160 = 12 x \/ 8 = 172 x = 1376 answer is a .\"","correct":"a","options":{"a":"1376 ","b":"1800 ","c":"1400 ","d":"2500","e":"100"},"options_float":{"a":1376.0,"b":1800.0,"c":1400.0,"d":2500.0,"e":100.0},"annotated_formula":"multiply(add(160, 12), 8)","linear_formula":"add(n1,n2)|multiply(n0,#0)|","chain":"160 + 12<\/gadget>\n172<\/output>\n172 * 8<\/gadget>\n1_376<\/output>\n1_376<\/result>","index":2404} +{"problem":"a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 40 , how many horses are there ?","rationale":"legs 8 * 4 = 32 now half on their horses so remaining on the walk so 4 men 4 men has 8 legs so , 8 + 32 = 40 legs walking answer : a","correct":"a","options":{"a":"8 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":8.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"divide(40, add(const_4, divide(const_2, const_2)))","linear_formula":"divide(const_2,const_2)|add(#0,const_4)|divide(n0,#1)","chain":"2 \/ 2<\/gadget>\n1<\/output>\n4 + 1<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8<\/result>","index":2406} +{"problem":"workers at a campaign office have 3000 fliers to send out . if they send out 1 \/ 5 of them in the morning and 1 \/ 4 of the remaining ones out during the afternoon , how many are left for the next day ?","rationale":"\"( 1 \/ 5 ) * 3000 = 600 remaining = 3000 - 600 = 2400 ( 1 \/ 4 ) of remaining = ( 1 \/ 4 ) * 2400 = 600 remaining now = 2400 - 600 = 1800 answer : option a\"","correct":"a","options":{"a":"1800 ","b":"800 ","c":"1100 ","d":"1200","e":"1900"},"options_float":{"a":1800.0,"b":800.0,"c":1100.0,"d":1200.0,"e":1900.0},"annotated_formula":"subtract(subtract(3000, divide(3000, 5)), divide(subtract(3000, divide(3000, 5)), 4))","linear_formula":"divide(n0,n2)|subtract(n0,#0)|divide(#1,n4)|subtract(#1,#2)|","chain":"3_000 \/ 5<\/gadget>\n600<\/output>\n3_000 - 600<\/gadget>\n2_400<\/output>\n2_400 \/ 4<\/gadget>\n600<\/output>\n2_400 - 600<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":2407} +{"problem":"the diagonal of a rhombus are 70 m and 160 m . its area is :","rationale":"\"area of the rhombus = 1 \/ 2 d 1 d 2 = ( 1 \/ 2 ã — 70 ã — 160 ) cm ( power ) 2 = 70 ã — 80 = 5600 cm ( power ) 2 answer is c .\"","correct":"c","options":{"a":"1940 ","b":"4800 ","c":"5600 ","d":"5200","e":"5000"},"options_float":{"a":1940.0,"b":4800.0,"c":5600.0,"d":5200.0,"e":5000.0},"annotated_formula":"rhombus_area(70, 160)","linear_formula":"rhombus_area(n0,n1)|","chain":"(70 * 160) \/ 2<\/gadget>\n5_600<\/output>\n5_600<\/result>","index":2409} +{"problem":"50 % of a number is more than 40 % of 120 by 180 . find the number ?","rationale":"\"( 50 \/ 100 ) * x – ( 40 \/ 100 ) * 120 = 180 1 \/ 2 x = 228 x = 456 answer : a\"","correct":"a","options":{"a":"456 ","b":"480 ","c":"420 ","d":"445","e":"320"},"options_float":{"a":456.0,"b":480.0,"c":420.0,"d":445.0,"e":320.0},"annotated_formula":"divide(multiply(add(divide(multiply(120, 40), const_100), 180), const_100), 50)","linear_formula":"multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)|","chain":"120 * 40<\/gadget>\n4_800<\/output>\n4_800 \/ 100<\/gadget>\n48<\/output>\n48 + 180<\/gadget>\n228<\/output>\n228 * 100<\/gadget>\n22_800<\/output>\n22_800 \/ 50<\/gadget>\n456<\/output>\n456<\/result>","index":2410} +{"problem":"sheela deposits rs . 5000 in bank savings account . if this is 20 % of her monthly income . what is her monthly income in ?","rationale":"\"explanation : 20 % of income = rs . 5000 100 % of income = 5000 x 100 \/ 20 = rs . 25000 answer : c\"","correct":"c","options":{"a":"22000 ","b":"20000 ","c":"25000 ","d":"22235","e":"none of these"},"options_float":{"a":22000.0,"b":20000.0,"c":25000.0,"d":22235.0,"e":null},"annotated_formula":"divide(multiply(5000, const_100), 20)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|","chain":"5_000 * 100<\/gadget>\n500_000<\/output>\n500_000 \/ 20<\/gadget>\n25_000<\/output>\n25_000<\/result>","index":2411} +{"problem":"there are 4 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 4 other amateurs , what is the total number of chess games possible to be played in the tournament ?","rationale":"\"each person is one participant of 4 games . so there are in all 4 * 4 = 16 instances of one participant games . but each game has 2 participants so total number of games = 16 \/ 2 = 8 b\"","correct":"b","options":{"a":"10 ","b":"8 ","c":"40 ","d":"60","e":"120"},"options_float":{"a":10.0,"b":8.0,"c":40.0,"d":60.0,"e":120.0},"annotated_formula":"divide(multiply(4, 4), const_2)","linear_formula":"multiply(n0,n1)|divide(#0,const_2)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2412} +{"problem":"a 24 ' ' x 24 ' ' square metal plate needs to be fixed by a carpenter on to a wooden board . the carpenter uses nails all along the edges of the square such that there are 25 nails on each side of the square . each nail is at the same distance from the neighboring nails . how many nails does the carpenter use ?","rationale":"every corner will have one , 4 corners will have 4 + and for 4 sides 23 no of nails = 4 + ( 4 * 23 ) = 96 answer : c","correct":"c","options":{"a":"94 ","b":"95 ","c":"96 ","d":"97","e":"98"},"options_float":{"a":94.0,"b":95.0,"c":96.0,"d":97.0,"e":98.0},"annotated_formula":"add(multiply(subtract(25, const_2), const_4), const_4)","linear_formula":"subtract(n2,const_2)|multiply(#0,const_4)|add(#1,const_4)","chain":"25 - 2<\/gadget>\n23<\/output>\n23 * 4<\/gadget>\n92<\/output>\n92 + 4<\/gadget>\n96<\/output>\n96<\/result>","index":2413} +{"problem":"let the number which when multiplied by 15 is increased by 280 .","rationale":"\"solution let the number be x . then , 15 x - x = 280 ‹ = › 14 x = 280 x ‹ = › 20 . answer b\"","correct":"b","options":{"a":"14 ","b":"20 ","c":"26 ","d":"28","e":"30"},"options_float":{"a":14.0,"b":20.0,"c":26.0,"d":28.0,"e":30.0},"annotated_formula":"divide(280, subtract(15, const_1))","linear_formula":"subtract(n0,const_1)|divide(n1,#0)|","chain":"15 - 1<\/gadget>\n14<\/output>\n280 \/ 14<\/gadget>\n20<\/output>\n20<\/result>","index":2415} +{"problem":"in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 105 ?","rationale":"\"let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 105 5 x - 20 = 105 5 x = 125 x = 25 the answer is b .\"","correct":"b","options":{"a":"24 ","b":"25 ","c":"26 ","d":"27","e":"28"},"options_float":{"a":24.0,"b":25.0,"c":26.0,"d":27.0,"e":28.0},"annotated_formula":"add(divide(subtract(105, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2))","linear_formula":"add(n1,n1)|add(n1,#0)|add(n1,#1)|add(#1,#1)|add(#3,#2)|subtract(n2,#4)|divide(#5,n0)|add(#2,#6)|","chain":"2 + 2<\/gadget>\n4<\/output>\n2 + 4<\/gadget>\n6<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6 + 6<\/gadget>\n12<\/output>\n6 + 2<\/gadget>\n8<\/output>\n12 + 8<\/gadget>\n20<\/output>\n105 - 20<\/gadget>\n85<\/output>\n85 \/ 5<\/gadget>\n17<\/output>\n17 + 8<\/gadget>\n25<\/output>\n25<\/result>","index":2416} +{"problem":"4 dices are thrown at the same time . what is the probability of getting only 2 dices showing the same face ?","rationale":"probability of getting two same faces : 1 and 1 \/ 6 for other two faces , probabilities : 5 \/ 6 and 4 \/ 6 respectively . the dice combinations can be arranged in 4 ! \/ 2 ! ways total probability = 1 * 1 \/ 6 * 5 \/ 6 * 4 \/ 6 * 4 ! \/ 2 ! = 5 \/ 9 answer : b","correct":"b","options":{"a":"4 \/ 9 ","b":"5 \/ 9 ","c":"11 \/ 18 ","d":"7 \/ 9","e":"none of the above"},"options_float":{"a":0.4444444444,"b":0.5555555556,"c":0.6111111111,"d":0.7777777778,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(choose(4, 2), const_6), const_5), const_4), power(const_6, const_4))","linear_formula":"choose(n0,n1)|power(const_6,const_4)|multiply(#0,const_6)|multiply(#2,const_5)|multiply(#3,const_4)|divide(#4,#1)","chain":"binomial(4, 2)<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n36 * 5<\/gadget>\n180<\/output>\n180 * 4<\/gadget>\n720<\/output>\n6 ** 4<\/gadget>\n1_296<\/output>\n720 \/ 1_296<\/gadget>\n5\/9 = around 0.555556<\/output>\n5\/9 = around 0.555556<\/result>","index":2420} +{"problem":"a boat having a length 7 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :","rationale":"\"volume of water displaced = ( 7 x 3 x 0.01 ) m 3 = 0.21 m 3 . mass of man = volume of water displaced x density of water = ( 0.21 x 1000 ) kg = 210 kg . answer : e\"","correct":"e","options":{"a":"100 kg ","b":"120 kg ","c":"89 kg ","d":"80 kg","e":"210 kg"},"options_float":{"a":100.0,"b":120.0,"c":89.0,"d":80.0,"e":210.0},"annotated_formula":"multiply(multiply(multiply(7, 3), divide(1, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"7 * 3<\/gadget>\n21<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n21 * (1\/100)<\/gadget>\n21\/100 = around 0.21<\/output>\n(21\/100) * 1_000<\/gadget>\n210<\/output>\n210<\/result>","index":2423} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 50 overs to reach the target of 282 runs ?","rationale":"\"explanation : runs scored in the first 10 overs = 10 × 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 50 run rate needed = 250 \/ 50 = 5 answer : option d\"","correct":"d","options":{"a":"6.25 ","b":"5.5 ","c":"7.4 ","d":"5","e":"6"},"options_float":{"a":6.25,"b":5.5,"c":7.4,"d":5.0,"e":6.0},"annotated_formula":"divide(subtract(282, multiply(10, 3.2)), 50)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 3.2<\/gadget>\n32<\/output>\n282 - 32<\/gadget>\n250<\/output>\n250 \/ 50<\/gadget>\n5<\/output>\n5<\/result>","index":2424} +{"problem":"the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 0.50 + 0.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ?","rationale":"\"number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 \/ 4 = 4 total number of axles = 5 t = 0.50 + 0.50 ( x − 2 ) = 0.50 + . 5 * 3 = 0.5 + 1.5 = 2 $ answer a\"","correct":"a","options":{"a":"$ 2.00 ","b":"$ 3.00 ","c":"$ 3.50 ","d":"$ 4.00","e":"$ 5.00"},"options_float":{"a":2.0,"b":3.0,"c":3.5,"d":4.0,"e":5.0},"annotated_formula":"add(0.50, multiply(0.50, subtract(add(divide(subtract(18, 2), 4), const_1), 2)))","linear_formula":"subtract(n3,n2)|divide(#0,n5)|add(#1,const_1)|subtract(#2,n2)|multiply(n1,#3)|add(n0,#4)|","chain":"18 - 2<\/gadget>\n16<\/output>\n16 \/ 4<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 - 2<\/gadget>\n3<\/output>\n0.5 * 3<\/gadget>\n1.5<\/output>\n0.5 + 1.5<\/gadget>\n2<\/output>\n2<\/result>","index":2425} +{"problem":"if a : b : : 3 : 7 , then what is ( 5 a + 2 b ) : ( a - 2 b ) ?","rationale":"\"a \/ b = 3 \/ 7 dividing numerator & denominator of ' ( 5 a + 6 b ) \/ ( a - 2 b ) ' by b , [ 5 ( a \/ b ) + 2 ] \/ [ ( a \/ b ) - 2 ] = [ 5 * ( 3 \/ 7 ) + 2 ] \/ [ ( 3 \/ 7 ) - 2 ] = - 29 \/ 11 answer : d\"","correct":"d","options":{"a":"57 : 11 ","b":"- 57 : 11 ","c":"11 : 10 ","d":"- 29 : 11","e":"- 1 : 10"},"options_float":{"a":5.1818181818,"b":-5.1818181818,"c":1.1,"d":-2.6363636364,"e":-0.1},"annotated_formula":"divide(add(multiply(5, 3), multiply(7, 2)), subtract(3, multiply(2, 7)))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|subtract(n0,#2)|divide(#3,#4)|","chain":"5 * 3<\/gadget>\n15<\/output>\n7 * 2<\/gadget>\n14<\/output>\n15 + 14<\/gadget>\n29<\/output>\n2 * 7<\/gadget>\n14<\/output>\n3 - 14<\/gadget>\n-11<\/output>\n29 \/ (-11)<\/gadget>\n-29\/11 = around -2.636364<\/output>\n-29\/11 = around -2.636364<\/result>","index":2426} +{"problem":"in a group of 100 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ?","rationale":"number of people whose age is less than 20 is given by 100 - 90 = 10 probability p that a person selected at random from the group is less than 20 is given by 10 \/ 100 = 0.1 correct answer a","correct":"a","options":{"a":"0.1 ","b":"0.55 ","c":"0.65 ","d":"0.75","e":"0.85"},"options_float":{"a":0.1,"b":0.55,"c":0.65,"d":0.75,"e":0.85},"annotated_formula":"divide(subtract(100, 90), 100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)","chain":"100 - 90<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1\/10 = around 0.1<\/result>","index":2427} +{"problem":"an article is bought for rs . 1200 and sold for rs . 800 , find the loss percent ?","rationale":"\"1200 - - - - 400 100 - - - - ? = > 33 % answer : c\"","correct":"c","options":{"a":"16 % ","b":"35 % ","c":"33 % ","d":"18 %","e":"12 %"},"options_float":{"a":16.0,"b":35.0,"c":33.0,"d":18.0,"e":12.0},"annotated_formula":"subtract(const_100, divide(multiply(800, const_100), 1200))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|","chain":"800 * 100<\/gadget>\n80_000<\/output>\n80_000 \/ 1_200<\/gadget>\n200\/3 = around 66.666667<\/output>\n100 - (200\/3)<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":2428} +{"problem":"how much is 80 % of 45 is greater than 4 \/ 5 of 25 ?","rationale":"\"( 80 \/ 100 ) * 45 â € “ ( 4 \/ 5 ) * 25 36 - 20 = 16 answer : c\"","correct":"c","options":{"a":"15 ","b":"57 ","c":"16 ","d":"16","e":"73"},"options_float":{"a":15.0,"b":57.0,"c":16.0,"d":16.0,"e":73.0},"annotated_formula":"subtract(multiply(45, divide(80, const_100)), multiply(divide(4, 5), 25))","linear_formula":"divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n45 * (4\/5)<\/gadget>\n36<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 25<\/gadget>\n20<\/output>\n36 - 20<\/gadget>\n16<\/output>\n16<\/result>","index":2429} +{"problem":"if p ( a ) = 5 \/ 7 and p ( b ) = 2 \/ 5 , find p ( a n b ) if a and b are independent events .","rationale":"\"p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 5 \/ 7 . 2 \/ 5 p ( a n b ) = 2 \/ 7 . d\"","correct":"d","options":{"a":"7 \/ 25 ","b":"3 \/ 25 ","c":"8 \/ 25 ","d":"2 \/ 7","e":"3 \/ 17"},"options_float":{"a":0.28,"b":0.12,"c":0.32,"d":0.2857142857,"e":0.1764705882},"annotated_formula":"multiply(divide(5, 7), divide(2, 5))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|","chain":"5 \/ 7<\/gadget>\n5\/7 = around 0.714286<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(5\/7) * (2\/5)<\/gadget>\n2\/7 = around 0.285714<\/output>\n2\/7 = around 0.285714<\/result>","index":2431} +{"problem":"in digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be ?","rationale":"\"20 * 10 * 5 = 1000 answer : b\"","correct":"b","options":{"a":"100 cu . m ","b":"1000 cu . m ","c":"10000 cu . m ","d":"2000 cu . m","e":"4000 cu . m"},"options_float":{"a":100.0,"b":1000.0,"c":10000.0,"d":2000.0,"e":4000.0},"annotated_formula":"multiply(multiply(20, 10), 5)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"20 * 10<\/gadget>\n200<\/output>\n200 * 5<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":2433} +{"problem":"in traveling from a dormitory to a certain city , a student went 1 \/ 3 of the way by foot , 3 \/ 5 of the way by bus , and the remaining 2 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ?","rationale":"\"whole trip = distance by foot + distance by bus + distance by car x = 1 \/ 3 x + 3 \/ 5 x + 2 x - 1 \/ 3 x - 3 \/ 5 x = 2 x = 30 km option : a\"","correct":"a","options":{"a":"30 ","b":"45 ","c":"60 ","d":"90","e":"120"},"options_float":{"a":30.0,"b":45.0,"c":60.0,"d":90.0,"e":120.0},"annotated_formula":"multiply(2, inverse(subtract(1, add(divide(1, 3), divide(3, 5)))))","linear_formula":"divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(n0,#2)|inverse(#3)|multiply(n4,#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(1\/3) + (3\/5)<\/gadget>\n14\/15 = around 0.933333<\/output>\n1 - (14\/15)<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ (1\/15)<\/gadget>\n15<\/output>\n2 * 15<\/gadget>\n30<\/output>\n30<\/result>","index":2434} +{"problem":"a factory produces 6500 toys per week . if the workers at this factory work 5 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ?","rationale":"\"to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 5 days ) by 5 . 6500 \/ 5 = 1300 toys correct answer c\"","correct":"c","options":{"a":"1375 toys ","b":"2375 toys ","c":"1300 toys ","d":"1375 toys","e":"2375 toys"},"options_float":{"a":1375.0,"b":2375.0,"c":1300.0,"d":1375.0,"e":2375.0},"annotated_formula":"divide(6500, 5)","linear_formula":"divide(n0,n1)|","chain":"6_500 \/ 5<\/gadget>\n1_300<\/output>\n1_300<\/result>","index":2435} +{"problem":"the charge for a single room at hotel p is 20 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ?","rationale":"\"let rate in r = 100 x then p = 80 x g = 100 y p = 90 y thus 80 x = 90 y or x = 1.125 y ans r = 112.5 y so increase = 12.5 % answer : c\"","correct":"c","options":{"a":"15 % ","b":"20 % ","c":"12.5 % ","d":"50 %","e":"150 %"},"options_float":{"a":15.0,"b":20.0,"c":12.5,"d":50.0,"e":150.0},"annotated_formula":"multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 20), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 20), subtract(const_100, 10)), const_100)), const_100)","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n100 - 10<\/gadget>\n90<\/output>\n80 \/ 90<\/gadget>\n8\/9 = around 0.888889<\/output>\n(8\/9) * 100<\/gadget>\n800\/9 = around 88.888889<\/output>\n100 - (800\/9)<\/gadget>\n100\/9 = around 11.111111<\/output>\n(100\/9) \/ (800\/9)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":2437} +{"problem":"line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 8 ) and ( 144 ) . what is the slope of line m ?","rationale":"the midpoint of ( 2,8 ) and ( 14,4 ) is ( 8,6 ) . the slope of a line through ( 0 , - 2 ) and ( 8,6 ) is ( 6 - ( - 2 ) ) \/ ( 8 - 0 ) = 8 \/ 8 = 1 the answer is d .","correct":"d","options":{"a":"- 2 ","b":"- 1 ","c":"0 ","d":"1","e":"2"},"options_float":{"a":-2.0,"b":-1.0,"c":0.0,"d":1.0,"e":2.0},"annotated_formula":"divide(divide(add(8, reminder(144, const_10)), const_2), subtract(divide(add(2, divide(subtract(144, reminder(144, const_10)), const_10)), const_2), 2))","linear_formula":"reminder(n3,const_10)|add(n2,#0)|subtract(n3,#0)|divide(#1,const_2)|divide(#2,const_10)|add(n0,#4)|divide(#5,const_2)|subtract(#6,n0)|divide(#3,#7)","chain":"144 % 10<\/gadget>\n4<\/output>\n8 + 4<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n144 - 4<\/gadget>\n140<\/output>\n140 \/ 10<\/gadget>\n14<\/output>\n2 + 14<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8 - 2<\/gadget>\n6<\/output>\n6 \/ 6<\/gadget>\n1<\/output>\n1<\/result>","index":2442} +{"problem":"the average weight of a , b and c is 60 kg . if the average weight of a and b be 70 kg and that of b and c be 50 kg , then the weight of b is :","rationale":"\"explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 60 x 3 ) = 180 â € ¦ . ( i ) a + b = ( 70 x 2 ) = 140 â € ¦ . ( ii ) b + c = ( 50 x 2 ) = 100 â € ¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 240 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 60 . b â € ™ s weight = 60 kg . answer b\"","correct":"b","options":{"a":"50 kg ","b":"60 kg ","c":"55 kg ","d":"57 kg","e":"62 kg"},"options_float":{"a":50.0,"b":60.0,"c":55.0,"d":57.0,"e":62.0},"annotated_formula":"subtract(add(multiply(70, const_2), multiply(50, const_2)), multiply(60, const_3))","linear_formula":"multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|","chain":"70 * 2<\/gadget>\n140<\/output>\n50 * 2<\/gadget>\n100<\/output>\n140 + 100<\/gadget>\n240<\/output>\n60 * 3<\/gadget>\n180<\/output>\n240 - 180<\/gadget>\n60<\/output>\n60<\/result>","index":2443} +{"problem":"length of a rectangular plot is 22 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ?","rationale":"\"let breadth = x metres . then , length = ( x + 22 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 22 ) + x ] = 200 2 x + 22 = 100 2 x = 78 x = 39 . hence , length = x + 22 = 61 m b\"","correct":"b","options":{"a":"51 m ","b":"61 m ","c":"71 m ","d":"81 m","e":"91 m"},"options_float":{"a":51.0,"b":61.0,"c":71.0,"d":81.0,"e":91.0},"annotated_formula":"divide(add(divide(5300, 26.50), multiply(const_2, 22)), const_4)","linear_formula":"divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)|","chain":"5_300 \/ 26.5<\/gadget>\n200<\/output>\n2 * 22<\/gadget>\n44<\/output>\n200 + 44<\/gadget>\n244<\/output>\n244 \/ 4<\/gadget>\n61<\/output>\n61<\/result>","index":2444} +{"problem":"a reduction of 20 % in the price of oil enables a house wife to obtain 4 kgs more for rs . 600 , what is the reduced price for kg ?","rationale":"\"600 * ( 20 \/ 100 ) = 120 - - - - 4 ? - - - - 1 = > rs . 30 answer : d\"","correct":"d","options":{"a":"20 ","b":"40 ","c":"36 ","d":"30","e":"35"},"options_float":{"a":20.0,"b":40.0,"c":36.0,"d":30.0,"e":35.0},"annotated_formula":"divide(divide(multiply(600, 20), const_100), 4)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|","chain":"600 * 20<\/gadget>\n12_000<\/output>\n12_000 \/ 100<\/gadget>\n120<\/output>\n120 \/ 4<\/gadget>\n30<\/output>\n30<\/result>","index":2445} +{"problem":"after spending rs . 5000 on rent , rs . 1500 on milk , rs . 4500 on groceries , rs . 2500 on childrens education rs . 2000 on petrol and rs . 2500 on miscellaneous expenses , mr . kishore saved 10 % of his monthly salary . how much did he save in rs . ?","rationale":"\"explanation : total exp = 5000 + 1500 + 4500 + 2500 + 2000 + 2500 = 18000 exp in % = 100 - 10 = 90 % , 18000 = 90 % saving = 10 % = 18000 x 10 \/ 90 = rs . 2000 answer : c\"","correct":"c","options":{"a":"2160 ","b":"2350 ","c":"2000 ","d":"2300","e":"none of these"},"options_float":{"a":2160.0,"b":2350.0,"c":2000.0,"d":2300.0,"e":null},"annotated_formula":"subtract(divide(multiply(add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 2500), const_100), subtract(const_100, 10)), add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 2500))","linear_formula":"add(n0,n1)|subtract(const_100,n6)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|multiply(#5,const_100)|divide(#6,#1)|subtract(#7,#5)|","chain":"5_000 + 1_500<\/gadget>\n6_500<\/output>\n6_500 + 4_500<\/gadget>\n11_000<\/output>\n11_000 + 2_500<\/gadget>\n13_500<\/output>\n13_500 + 2_000<\/gadget>\n15_500<\/output>\n15_500 + 2_500<\/gadget>\n18_000<\/output>\n18_000 * 100<\/gadget>\n1_800_000<\/output>\n100 - 10<\/gadget>\n90<\/output>\n1_800_000 \/ 90<\/gadget>\n20_000<\/output>\n20_000 - 18_000<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2446} +{"problem":"jim drove 384 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ?","rationale":"\"the number of miles to drive to finish his journey is given by 1200 - 384 = 816 miles correct answer d\"","correct":"d","options":{"a":"113 miles ","b":"432 miles ","c":"456 miles ","d":"816 miles","e":"767 miles"},"options_float":{"a":113.0,"b":432.0,"c":456.0,"d":816.0,"e":767.0},"annotated_formula":"subtract(1200, 384)","linear_formula":"subtract(n1,n0)|","chain":"1_200 - 384<\/gadget>\n816<\/output>\n816<\/result>","index":2447} +{"problem":"find the product of the local value and absolute value of 4 in 564823","rationale":"explanation : place value = local value face value = absolute value the place value of 4 in 564823 is 4 x 1000 = 4000 the face value of 4 in 564823 is nothing but 4 . = > 4000 x 4 = 16000 answer : option c","correct":"c","options":{"a":"1600 ","b":"8000 ","c":"16000 ","d":"12000","e":"18000"},"options_float":{"a":1600.0,"b":8000.0,"c":16000.0,"d":12000.0,"e":18000.0},"annotated_formula":"multiply(multiply(4, const_1000), 4)","linear_formula":"multiply(n0,const_1000)|multiply(n0,#0)","chain":"4 * 1_000<\/gadget>\n4_000<\/output>\n4_000 * 4<\/gadget>\n16_000<\/output>\n16_000<\/result>","index":2448} +{"problem":"the banker â € ™ s gain on a sum due 3 years hence at 10 % per annum is rs . 60 . the banker â € ™ s discount is","rationale":"\"solution t . d = ( b . g x 100 \/ r x t ) = rs . ( 60 x 100 \/ 10 x 3 ) = rs . 200 . b . d = rs ( 200 + 60 ) = rs . 260 . answer e\"","correct":"e","options":{"a":"rs . 960 ","b":"rs . 840 ","c":"rs . 1020 ","d":"rs . 760","e":"rs . 260"},"options_float":{"a":960.0,"b":840.0,"c":1020.0,"d":760.0,"e":260.0},"annotated_formula":"add(divide(multiply(60, const_100), multiply(3, 10)), 60)","linear_formula":"multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)|","chain":"60 * 100<\/gadget>\n6_000<\/output>\n3 * 10<\/gadget>\n30<\/output>\n6_000 \/ 30<\/gadget>\n200<\/output>\n200 + 60<\/gadget>\n260<\/output>\n260<\/result>","index":2449} +{"problem":"robert ' s salary was decreased by 70 % and subsequently increased by 70 % . how much percentage does he lose ?","rationale":"\"let original salary be $ 100 salary after decreasing 70 % = 100 - 100 x 70 \/ 100 = $ 30 salary after increasing 70 % on $ 30 = 30 + 30 x 70 \/ 100 = $ 51 percentage of loss = 100 - 51 = 49 % answer : e\"","correct":"e","options":{"a":"10 % ","b":"20 % ","c":"25 % ","d":"30 %","e":"49 %"},"options_float":{"a":10.0,"b":20.0,"c":25.0,"d":30.0,"e":49.0},"annotated_formula":"multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 70), add(const_100, 70))), multiply(const_100, const_100)), const_100)","linear_formula":"add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#1)|multiply(#5,const_100)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n100 - 70<\/gadget>\n30<\/output>\n100 + 70<\/gadget>\n170<\/output>\n30 * 170<\/gadget>\n5_100<\/output>\n10_000 - 5_100<\/gadget>\n4_900<\/output>\n4_900 \/ 10_000<\/gadget>\n49\/100 = around 0.49<\/output>\n(49\/100) * 100<\/gadget>\n49<\/output>\n49<\/result>","index":2450} +{"problem":"the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number e ?","rationale":"\"( 1 + 1 \/ 8 ) x - ( 1 - 1 \/ 4 ) x = 30 ( 9 \/ 8 ) x - ( 3 \/ 4 ) x = 30 x = 80 = e answer : b\"","correct":"b","options":{"a":"60 ","b":"80 ","c":"40 ","d":"120","e":"160"},"options_float":{"a":60.0,"b":80.0,"c":40.0,"d":120.0,"e":160.0},"annotated_formula":"divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#2,#3)|divide(n2,#4)|","chain":"12.5 \/ 100<\/gadget>\n0.125<\/output>\n1 + 0.125<\/gadget>\n1.125<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n1.125 - (3\/4)<\/gadget>\n0.375<\/output>\n30 \/ 0.375<\/gadget>\n80<\/output>\n80<\/result>","index":2452} +{"problem":"what is the remainder when you divide 2 ^ 200 - 3 by 7 ?","rationale":"answer is a . this is very similar to the 32 ^ 32 ^ 32 - 3 divided by 7 question .","correct":"a","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"reminder(multiply(200, 2), 3)","linear_formula":"multiply(n0,n1)|reminder(#0,n2)|","chain":"200 * 2<\/gadget>\n400<\/output>\n400 % 3<\/gadget>\n1<\/output>\n1<\/result>","index":2453} +{"problem":"if ( log 5 5 ) ( log 4 9 ) ( log 3 2 ) is equal to","rationale":"solution given expression = log 9 \/ log 4 × log 2 \/ log 3 ‹ = › log 3 ² \/ log 2 ² × log 2 \/ log 3 = 2 log 3 \/ 2 log 2 × log 2 \/ log 3 = 1 . answer a","correct":"a","options":{"a":"1 ","b":"3 \/ 2 ","c":"2 ","d":"5","e":"7"},"options_float":{"a":1.0,"b":1.5,"c":2.0,"d":5.0,"e":7.0},"annotated_formula":"divide(log(5), log(5))","linear_formula":"log(n0)|divide(#0,#0)","chain":"log(5)<\/gadget>\nlog(5) = around 1.609438<\/output>\nlog(5) \/ log(5)<\/gadget>\n1<\/output>\n1<\/result>","index":2455} +{"problem":"the total marks obtained by a student in mathematics and physics is 32 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .","rationale":"\"let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 32 and c - p = 20 m + c \/ 2 = [ ( m + p ) + ( c - p ) ] \/ 2 = ( 32 + 20 ) \/ 2 = 26 . answer : b\"","correct":"b","options":{"a":"40 ","b":"26 ","c":"27 ","d":"28","e":"22"},"options_float":{"a":40.0,"b":26.0,"c":27.0,"d":28.0,"e":22.0},"annotated_formula":"divide(add(32, 20), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"32 + 20<\/gadget>\n52<\/output>\n52 \/ 2<\/gadget>\n26<\/output>\n26<\/result>","index":2456} +{"problem":"in the rectangular coordinate system , if the line x = 4 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ?","rationale":"\"x = 4 y + 5 , and thus y = x \/ 4 - 5 \/ 4 the slope is 1 \/ 4 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) \/ ( m + 2 - m ) = p \/ 2 p \/ 2 = 1 \/ 4 and thus p = 1 \/ 2 the answer is c .\"","correct":"c","options":{"a":"3 ","b":"2 ","c":"1 \/ 2 ","d":"1 \/ 3","e":"1 \/ 4"},"options_float":{"a":3.0,"b":2.0,"c":0.5,"d":0.3333333333,"e":0.25},"annotated_formula":"divide(2, 4)","linear_formula":"divide(n2,n0)|","chain":"2 \/ 4<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2457} +{"problem":"the average weight of 4 students decreases by 8 kg when one of them weighing 96 kg is replaced by a new student . the weight of the student is","rationale":"explanation : let the weight of student be x kg . given , difference in average weight = 8 kg = > ( 96 - x ) \/ 4 = 8 = > x = 64 answer : c","correct":"c","options":{"a":"62 kg ","b":"60 kg ","c":"64 kg ","d":"72 kg","e":"none of these"},"options_float":{"a":62.0,"b":60.0,"c":64.0,"d":72.0,"e":null},"annotated_formula":"subtract(96, multiply(4, 8))","linear_formula":"multiply(n0,n1)|subtract(n2,#0)","chain":"4 * 8<\/gadget>\n32<\/output>\n96 - 32<\/gadget>\n64<\/output>\n64<\/result>","index":2458} +{"problem":"if 4 men working 10 hours a day earn rs . 800 per week , then 9 men working 6 hours a day will earn how much per week ?","rationale":"\"explanation : ( men 4 : 9 ) : ( hrs \/ day 10 : 6 ) : : 800 : x hence 4 * 10 * x = 9 * 6 * 800 or x = 9 * 6 * 800 \/ 4 * 10 = 1080 answer : c\"","correct":"c","options":{"a":"rs 840 ","b":"rs 1320 ","c":"rs 1080 ","d":"rs 1680","e":"none of these"},"options_float":{"a":840.0,"b":1320.0,"c":1080.0,"d":1680.0,"e":null},"annotated_formula":"multiply(divide(multiply(9, 6), multiply(4, 10)), 800)","linear_formula":"multiply(n3,n4)|multiply(n0,n1)|divide(#0,#1)|multiply(n2,#2)|","chain":"9 * 6<\/gadget>\n54<\/output>\n4 * 10<\/gadget>\n40<\/output>\n54 \/ 40<\/gadget>\n27\/20 = around 1.35<\/output>\n(27\/20) * 800<\/gadget>\n1_080<\/output>\n1_080<\/result>","index":2459} +{"problem":"in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 4.5 ) , ( 8 , 0 ) , ( 0 , - 4.5 ) , ( - 8 , 0 ) ?","rationale":"\"ares of rhombus = 1 \/ 2 * d 1 * d 2 length of 1 st diagonal = 8 + 8 = 16 length of 2 nd diagonal = 4.5 + 4.5 = 9 area = 1 \/ 2 * 16 * 9 = 72 b is the answer\"","correct":"b","options":{"a":"56 ","b":"72 ","c":"112 ","d":"116","e":"120"},"options_float":{"a":56.0,"b":72.0,"c":112.0,"d":116.0,"e":120.0},"annotated_formula":"rhombus_area(multiply(8, const_2), multiply(4.5, const_2))","linear_formula":"multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)|","chain":"8 * 2<\/gadget>\n16<\/output>\n4.5 * 2<\/gadget>\n9<\/output>\n(16 * 9) \/ 2<\/gadget>\n72<\/output>\n72<\/result>","index":2460} +{"problem":"a theater box office sold an average ( arithmetic mean ) of 63 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 78 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ?","rationale":"deviation from the mean for the daytime staff = 78 - 63 = 15 . deviation from the mean for the evening staff = 63 - 60 = 3 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 3 : 15 = 1 : 5 . the answer is c .","correct":"c","options":{"a":"2 : 3 ","b":"1 : 4 ","c":"1 : 5 ","d":"6 : 11","e":"4 : 7"},"options_float":{"a":0.6666666667,"b":0.25,"c":0.2,"d":0.5454545455,"e":0.5714285714},"annotated_formula":"divide(subtract(63, 60), subtract(78, 63))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)","chain":"63 - 60<\/gadget>\n3<\/output>\n78 - 63<\/gadget>\n15<\/output>\n3 \/ 15<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":2461} +{"problem":"a train running at the speed of 100 km \/ hr crosses a pole in 9 sec . what is the length of the train ?","rationale":"\"speed = 100 * 5 \/ 18 = 250 \/ 9 m \/ sec length of the train = speed * time = 250 \/ 9 * 9 = 250 m answer : b\"","correct":"b","options":{"a":"298 m ","b":"250 m ","c":"208 m ","d":"988 m","e":"299 m"},"options_float":{"a":298.0,"b":250.0,"c":208.0,"d":988.0,"e":299.0},"annotated_formula":"multiply(divide(multiply(100, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"100 * 1_000<\/gadget>\n100_000<\/output>\n100_000 \/ 3_600<\/gadget>\n250\/9 = around 27.777778<\/output>\n(250\/9) * 9<\/gadget>\n250<\/output>\n250<\/result>","index":2463} +{"problem":"an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 45 meters . how many times did the ball bounce ?","rationale":"\"going down = 16 m going up = 8 - - > total = 24 going down = 8 - - > total = 32 going up = 4 - - > total = 36 going down = 4 - - > total = 40 going up = 2 - - > total = 42 going down = 2 - - > total = 44 going up = 1 - - > total = 45 ( caught ) no of bounce = 4 answer : b\"","correct":"b","options":{"a":"5 ","b":"4 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":4.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"divide(divide(16, const_2), const_2)","linear_formula":"divide(n0,const_2)|divide(#0,const_2)|","chain":"16 \/ 2<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":2464} +{"problem":"average between 2 sets of numbers is closer to the set with more no ' s ?","rationale":"\"if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 × 90 + 2 × 805 = 4305 = 86.3 × 90 + 2 × 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 times . so the average is closer to 90 than to 80 as we have just calculated . b\"","correct":"b","options":{"a":"70 ","b":"80 ","c":"90 ","d":"92","e":"94"},"options_float":{"a":70.0,"b":80.0,"c":90.0,"d":92.0,"e":94.0},"annotated_formula":"multiply(multiply(const_2, const_4), const_10)","linear_formula":"multiply(const_2,const_4)|multiply(#0,const_10)|","chain":"2 * 4<\/gadget>\n8<\/output>\n8 * 10<\/gadget>\n80<\/output>\n80<\/result>","index":2465} +{"problem":"a merchant marks his goods up by 30 % and then offers a discount of 10 % on the marked price . what % profit does the merchant make after the discount ?","rationale":"\"let the price be 100 . the price becomes 130 after a 30 % markup . now a discount of 10 % on 130 . profit = 117 - 100 17 % answer a\"","correct":"a","options":{"a":"17 % ","b":"25 % ","c":"69 % ","d":"31 %","e":"19 %"},"options_float":{"a":17.0,"b":25.0,"c":69.0,"d":31.0,"e":19.0},"annotated_formula":"subtract(subtract(add(30, const_100), divide(multiply(add(30, const_100), 10), const_100)), const_100)","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)|","chain":"30 + 100<\/gadget>\n130<\/output>\n130 * 10<\/gadget>\n1_300<\/output>\n1_300 \/ 100<\/gadget>\n13<\/output>\n130 - 13<\/gadget>\n117<\/output>\n117 - 100<\/gadget>\n17<\/output>\n17<\/result>","index":2466} +{"problem":"elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ 46 and elvin ' s total telephone bill for february was 76 $ . if elvin ' s charge for the calls he made in february was twice the charge for the calls he made in january , what is elvin ' s fixed monthly charge for internet service ?","rationale":"\"bill = fixed charge + charge of calls made in jan , bill = fixed charge ( let , y ) + charge of calls made in jan ( let , x ) = $ 46 in feb , bill = fixed charge ( let , y ) + charge of calls made in feb ( then , 2 x ) = $ 76 i . e . x + y = 46 and 2 x + y = 76 take the difference if two equations i . e . ( 2 x + y ) - ( x + y ) = 76 - 46 i . e . x = 30 i . e . fixed monthly charge , y = 16 answer : option d\"","correct":"d","options":{"a":"$ 5 ","b":"$ 10 ","c":"$ 14 ","d":"$ 16","e":"$ 28"},"options_float":{"a":5.0,"b":10.0,"c":14.0,"d":16.0,"e":28.0},"annotated_formula":"subtract(multiply(46, const_2), 76)","linear_formula":"multiply(n0,const_2)|subtract(#0,n1)|","chain":"46 * 2<\/gadget>\n92<\/output>\n92 - 76<\/gadget>\n16<\/output>\n16<\/result>","index":2467} +{"problem":"if a , b and c together can finish a piece of work in 4 days . a alone in 12 days and b in 18 days , then c alone can do it in ?","rationale":"\"c 9 days c = 1 \/ 4 - 1 \/ 12 – 1 \/ 18 = 1 \/ 9 = > 9 days\"","correct":"c","options":{"a":"7 days ","b":"3 days ","c":"9 days ","d":"8 days","e":"2 days"},"options_float":{"a":7.0,"b":3.0,"c":9.0,"d":8.0,"e":2.0},"annotated_formula":"divide(const_1, subtract(subtract(const_0_25, divide(const_1, 12)), divide(const_1, 18)))","linear_formula":"divide(const_1,n1)|divide(const_1,n2)|subtract(const_0_25,#0)|subtract(#2,#1)|divide(const_1,#3)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/4) - (1\/12)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/6) - (1\/18)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9<\/result>","index":2469} +{"problem":"on a certain day , tim invested $ 600 at 10 percent annual interest , compounded annually , and lana invested 800 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ s investment in the first 2 years ?","rationale":"compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % \/ 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 600 ( 1.10 ) ^ 2 = 600 ( 1.21 ) = 726 and lana would have 800 ( 1.05 ) ^ 2 = 800 ( 1.1025 ) = 882 thus , tim earned 126 dollars , while lana earned 82 dollars the difference is $ 44 and the answer is c .","correct":"c","options":{"a":"$ 5 ","b":"$ 15 ","c":"$ 44 ","d":"$ 100","e":"$ 105"},"options_float":{"a":5.0,"b":15.0,"c":44.0,"d":100.0,"e":105.0},"annotated_formula":"subtract(subtract(multiply(600, power(add(const_1, divide(10, const_100)), 2)), 600), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 800), 800))","linear_formula":"divide(n1,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|power(#2,n4)|power(#3,n4)|multiply(n0,#4)|multiply(n2,#5)|subtract(#6,n0)|subtract(#7,n2)|subtract(#8,#9)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n600 * (121\/100)<\/gadget>\n726<\/output>\n726 - 600<\/gadget>\n126<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 + (1\/20)<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n(441\/400) * 800<\/gadget>\n882<\/output>\n882 - 800<\/gadget>\n82<\/output>\n126 - 82<\/gadget>\n44<\/output>\n44<\/result>","index":2471} +{"problem":"today mary ’ s father is 4 times as old as mary . 3 years ago , he was 5 times as old . how old is mary today ?","rationale":"dad ' s age = 4 m 5 * ( m - 3 ) = 4 m - 3 m = 12 the answer is d .","correct":"d","options":{"a":"6 ","b":"8 ","c":"10 ","d":"12","e":"14"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":12.0,"e":14.0},"annotated_formula":"subtract(multiply(5, 3), 3)","linear_formula":"multiply(n1,n2)|subtract(#0,n1)","chain":"5 * 3<\/gadget>\n15<\/output>\n15 - 3<\/gadget>\n12<\/output>\n12<\/result>","index":2473} +{"problem":"a horse is tethered to one corner of a rectangular grassy field 42 m by 26 m with a rope 16 m long . over how much area of the field can it graze ?","rationale":"\"area of the shaded portion = 1 ⁄ 4 × π × ( 16 ) 2 = 201 m 2 answer c\"","correct":"c","options":{"a":"154 cm 2 ","b":"308 m 2 ","c":"201 m 2 ","d":"205 m 2","e":"none of these"},"options_float":{"a":154.0,"b":308.0,"c":201.0,"d":205.0,"e":null},"annotated_formula":"divide(multiply(power(16, const_2), const_pi), const_4)","linear_formula":"power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)|","chain":"16 ** 2<\/gadget>\n256<\/output>\n256 * pi<\/gadget>\n256*pi = around 804.247719<\/output>\n(256*pi) \/ 4<\/gadget>\n64*pi = around 201.06193<\/output>\n64*pi = around 201.06193<\/result>","index":2474} +{"problem":"a batsman in his 17 th innings makes a score of 85 and their by increasing his average by 3 . what is his average after the 17 thinnings ?","rationale":"\"16 x + 85 = 17 ( x + 3 ) x = 34 + 3 = 37 answer : d\"","correct":"d","options":{"a":"22 ","b":"77 ","c":"26 ","d":"37","e":"88"},"options_float":{"a":22.0,"b":77.0,"c":26.0,"d":37.0,"e":88.0},"annotated_formula":"add(subtract(85, multiply(17, 3)), 3)","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|","chain":"17 * 3<\/gadget>\n51<\/output>\n85 - 51<\/gadget>\n34<\/output>\n34 + 3<\/gadget>\n37<\/output>\n37<\/result>","index":2476} +{"problem":"the average ( arithmetic mean ) of 4 different integers is 76 . if the largest integer is 90 , what is the least possible value of the smallest integer ?","rationale":"\"total of integers = 76 * 4 = 3004 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 304 - ( 90 + 89 + 88 ) = 304 - 267 = 37 answer : e\"","correct":"e","options":{"a":"1 ","b":"19 ","c":"29 ","d":"30","e":"37"},"options_float":{"a":1.0,"b":19.0,"c":29.0,"d":30.0,"e":37.0},"annotated_formula":"subtract(subtract(subtract(multiply(76, 4), 90), subtract(90, const_1)), subtract(90, const_2))","linear_formula":"multiply(n0,n1)|subtract(n2,const_1)|subtract(n2,const_2)|subtract(#0,n2)|subtract(#3,#1)|subtract(#4,#2)|","chain":"76 * 4<\/gadget>\n304<\/output>\n304 - 90<\/gadget>\n214<\/output>\n90 - 1<\/gadget>\n89<\/output>\n214 - 89<\/gadget>\n125<\/output>\n90 - 2<\/gadget>\n88<\/output>\n125 - 88<\/gadget>\n37<\/output>\n37<\/result>","index":2477} +{"problem":"a certain elevator has a safe weight limit of 2,000 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 200 pounds and the average weight of the others being 230 pounds ?","rationale":"lets assume there are 2 x people . half of them have average weight of 200 and other half has 230 . maximum weight is = 2000 so 200 * x + 230 * x = 2000 = > 430 x = 2000 = > x is approximately equal to 5 . so total people is 2 * 5 = 10 we are not taking 10 as answer because say 10 th person has minimum of 200 weight then 200 * 5 + 230 * 5 = 2150 ( which is more than 2000 ) answer c .","correct":"c","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"multiply(divide(multiply(const_10, 200), add(200, 230)), const_2)","linear_formula":"add(n1,n2)|multiply(n1,const_10)|divide(#1,#0)|multiply(#2,const_2)","chain":"10 * 200<\/gadget>\n2_000<\/output>\n200 + 230<\/gadget>\n430<\/output>\n2_000 \/ 430<\/gadget>\n200\/43 = around 4.651163<\/output>\n(200\/43) * 2<\/gadget>\n400\/43 = around 9.302326<\/output>\n400\/43 = around 9.302326<\/result>","index":2478} +{"problem":"the average salary of all the workers in a workshop is $ 8000 . the average salary of 7 technicians is $ 12000 and the average salary of the rest is $ 6000 . the total number of workers in the shop is ?","rationale":"let the total number of workers be x 8000 x = 12000 * 7 + 6000 ( x - 7 ) x = 21 answer is c","correct":"c","options":{"a":"15 ","b":"18 ","c":"21 ","d":"25","e":"31"},"options_float":{"a":15.0,"b":18.0,"c":21.0,"d":25.0,"e":31.0},"annotated_formula":"divide(subtract(multiply(7, 12000), multiply(6000, 7)), subtract(8000, 6000))","linear_formula":"multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)","chain":"7 * 12_000<\/gadget>\n84_000<\/output>\n6_000 * 7<\/gadget>\n42_000<\/output>\n84_000 - 42_000<\/gadget>\n42_000<\/output>\n8_000 - 6_000<\/gadget>\n2_000<\/output>\n42_000 \/ 2_000<\/gadget>\n21<\/output>\n21<\/result>","index":2479} +{"problem":"a farmer grows broccoli in his farm that is in the shape of a square . each broccoli takes 1 square foot of area in his garden . this year , he has increased his output by 101 broccoli when compared to last year . the shape of the area used for growing the broccoli has remained a square in both these years . how many broccoli did he produce this year ?","rationale":"explanatory answer the shape of the area used for growing broccoli has remained a square in both the years . let the side of the square area used for growing broccoli this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing broccoli last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of broccoli grown has increased by 101 , the area would have increased by 101 sq ft because each broccoli takes 1 sq ft space . hence , x 2 - y 2 = 101 ( x + y ) ( x - y ) = 101 . 101 is a prime number and hence it will have only two factors . i . e . , 101 and 1 . therefore , 101 can be expressed as product of 2 numbers in only way = 101 * 1 i . e . , ( x + y ) ( x - y ) = 101 * 1 so , ( x + y ) should be 101 and ( x - y ) should be 1 . solving the two equations we get x = 51 and y = 50 . therefore , number of broccoli produced this year = x 2 = 512 = 2601 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice a , c and d . as 2400,1987 and 2000 are not the square of any number . check choice b : if this year ' s produce is 2601 , last year ' s produce would have been 2601 - 101 = 2500 2500 is the square of 50 . so , 2601 is the answer . choice b","correct":"b","options":{"a":"2400 ","b":"2601 ","c":"1987 ","d":"2000","e":"can not be determined"},"options_float":{"a":2400.0,"b":2601.0,"c":1987.0,"d":2000.0,"e":null},"annotated_formula":"power(divide(add(101, 1), const_2), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|power(#1,const_2)","chain":"101 + 1<\/gadget>\n102<\/output>\n102 \/ 2<\/gadget>\n51<\/output>\n51 ** 2<\/gadget>\n2_601<\/output>\n2_601<\/result>","index":2480} +{"problem":"a person can swim in still water at 10 km \/ h . if the speed of water 8 km \/ h , how many hours will the man take to swim back against the current for 16 km ?","rationale":"\"m = 10 s = 8 us = 10 - 8 = 2 d = 16 t = 16 \/ 2 = 8 answer : d\"","correct":"d","options":{"a":"3 ","b":"7 ","c":"5 ","d":"8","e":"6"},"options_float":{"a":3.0,"b":7.0,"c":5.0,"d":8.0,"e":6.0},"annotated_formula":"divide(16, subtract(10, 8))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"10 - 8<\/gadget>\n2<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2482} +{"problem":"90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 60 % of the total school population of x students , what is x ?","rationale":"\"90 = x \/ 100 * 60 \/ 100 * x = > x ^ 2 = 9 * 10000 \/ 6 = > x = 122 a\"","correct":"a","options":{"a":"122 ","b":"150 ","c":"225 ","d":"250","e":"500"},"options_float":{"a":122.0,"b":150.0,"c":225.0,"d":250.0,"e":500.0},"annotated_formula":"sqrt(divide(multiply(90, const_100), divide(60, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)|","chain":"90 * 100<\/gadget>\n9_000<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n9_000 \/ (3\/5)<\/gadget>\n15_000<\/output>\n15_000 ** (1\/2)<\/gadget>\n50*sqrt(6) = around 122.474487<\/output>\n50*sqrt(6) = around 122.474487<\/result>","index":2483} +{"problem":"maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 65 kilometers , maxwell ' s walking speed is 2 km \/ h , and brad ' s running speed is 3 km \/ h . what is the distance traveled by maxwell when they meet in the middle ?","rationale":"\"consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 2 xt = 65 - 3 xt ( i . e distance = speed x time ) = > 5 t = 65 hence t = 13 oa = 2 x 13 = 26 km answer : e\"","correct":"e","options":{"a":"25 ","b":"23 ","c":"22 ","d":"24","e":"26"},"options_float":{"a":25.0,"b":23.0,"c":22.0,"d":24.0,"e":26.0},"annotated_formula":"multiply(2, divide(65, add(2, 3)))","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|","chain":"2 + 3<\/gadget>\n5<\/output>\n65 \/ 5<\/gadget>\n13<\/output>\n2 * 13<\/gadget>\n26<\/output>\n26<\/result>","index":2484} +{"problem":"the odds in favour of an events are 3 : 5 . the probability of occurrence of the event is","rationale":"number of cases favourable to e = 3 total number of cases = 3 + 5 = 8 p ( e ) = 3 \/ 8 answer is a","correct":"a","options":{"a":"3 \/ 8 ","b":"2 \/ 7 ","c":"1 \/ 3 ","d":"2 \/ 5","e":"6 \/ 7"},"options_float":{"a":0.375,"b":0.2857142857,"c":0.3333333333,"d":0.4,"e":0.8571428571},"annotated_formula":"divide(3, add(3, 5))","linear_formula":"add(n0,n1)|divide(n0,#0)","chain":"3 + 5<\/gadget>\n8<\/output>\n3 \/ 8<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":2486} +{"problem":"if john makes a contribution to a charity fund at school , the average contribution size will increase by 50 % reaching $ 75 per person . if there were 3 other contributions made before john ' s , what is the size of his donation ?","rationale":"\"cavg = average contribution before john cavg * 1.5 = 75 , therefore the average cont is $ 50 before john . if he needs to increase the average contribution by $ 25 , he must put in $ 25 for each of the 3 people . so $ 75 . but , he also has to put in the average for himself ( the fourth person ) , so add $ 75 . so $ 150 is your answer . answer b\"","correct":"b","options":{"a":"$ 100 ","b":"$ 150 ","c":"$ 200 ","d":"$ 250","e":"$ 450"},"options_float":{"a":100.0,"b":150.0,"c":200.0,"d":250.0,"e":450.0},"annotated_formula":"add(subtract(multiply(add(3, const_1), 75), multiply(add(3, const_1), 50)), 50)","linear_formula":"add(n2,const_1)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|add(n0,#3)|","chain":"3 + 1<\/gadget>\n4<\/output>\n4 * 75<\/gadget>\n300<\/output>\n4 * 50<\/gadget>\n200<\/output>\n300 - 200<\/gadget>\n100<\/output>\n100 + 50<\/gadget>\n150<\/output>\n150<\/result>","index":2487} +{"problem":"in a class of 40 students 26 play football and play 20 long tennis , if 17 play above , many play neither ?","rationale":"26 + 20 - 17 = 29 40 - 29 = 11 play neither answer is d","correct":"d","options":{"a":"6 ","b":"8 ","c":"10 ","d":"11","e":"14"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":11.0,"e":14.0},"annotated_formula":"subtract(40, subtract(add(26, 20), 17))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)","chain":"26 + 20<\/gadget>\n46<\/output>\n46 - 17<\/gadget>\n29<\/output>\n40 - 29<\/gadget>\n11<\/output>\n11<\/result>","index":2488} +{"problem":"seller selling an apple for rs . 18 , a seller loses 1 \/ 6 th of what it costs him . the cp of the apple is ?","rationale":"\"sp = 18 loss = cp 19 loss = cp − sp = cp − 18 ⇒ cp 19 = cp − 18 ⇒ 18 cp 19 = 18 ⇒ cp 19 = 1 ⇒ cp = 19 c\"","correct":"c","options":{"a":"10 ","b":"12 ","c":"19 ","d":"18","e":"20"},"options_float":{"a":10.0,"b":12.0,"c":19.0,"d":18.0,"e":20.0},"annotated_formula":"add(18, 1)","linear_formula":"add(n0,n1)|","chain":"18 + 1<\/gadget>\n19<\/output>\n19<\/result>","index":2489} +{"problem":"if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 8 , what is the least possible value of n ?","rationale":"we need 3 factors of 2 , so n must be at least 4 . therefore , the answer is b .","correct":"b","options":{"a":"7 ","b":"4 ","c":"5 ","d":"3","e":"6"},"options_float":{"a":7.0,"b":4.0,"c":5.0,"d":3.0,"e":6.0},"annotated_formula":"divide(8, const_2)","linear_formula":"divide(n1,const_2)","chain":"8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":2490} +{"problem":"a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 250 per month . what is his monthly salary ?","rationale":"\"income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 \/ 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 250 = > 6250 answer : d\"","correct":"d","options":{"a":"5000 ","b":"2999 ","c":"2878 ","d":"6250","e":"2771"},"options_float":{"a":5000.0,"b":2999.0,"c":2878.0,"d":6250.0,"e":2771.0},"annotated_formula":"divide(multiply(250, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100)))))","linear_formula":"divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)|","chain":"250 * 100<\/gadget>\n25_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n80 * (1\/5)<\/gadget>\n16<\/output>\n80 + 16<\/gadget>\n96<\/output>\n100 - 96<\/gadget>\n4<\/output>\n25_000 \/ 4<\/gadget>\n6_250<\/output>\n6_250<\/result>","index":2494} +{"problem":"if 15 students in a class average 75 % on an exam and 10 students average 90 % on the same exam , what is the average in percent for all 25 students ?","rationale":"\"( 15 * 75 + 10 * 90 ) \/ 25 = 81 % the answer is b .\"","correct":"b","options":{"a":"80 % ","b":"81 % ","c":"82 % ","d":"83 %","e":"84 %"},"options_float":{"a":80.0,"b":81.0,"c":82.0,"d":83.0,"e":84.0},"annotated_formula":"divide(add(multiply(15, 75), multiply(10, 90)), 25)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)|","chain":"15 * 75<\/gadget>\n1_125<\/output>\n10 * 90<\/gadget>\n900<\/output>\n1_125 + 900<\/gadget>\n2_025<\/output>\n2_025 \/ 25<\/gadget>\n81<\/output>\n81<\/result>","index":2495} +{"problem":"in a 100 m race a beats b by 25 m b beats c by 4 then a beats c by","rationale":"a : b = 100 : 75 b : c = 100 : 96 . a : c = ( a \/ b x b \/ c ) = ( 100 \/ 75 x 100 \/ 96 ) = 100 \/ 72 = 100 : 72 . a beats c by ( 100 - 72 ) m = 28 m . answer : e","correct":"e","options":{"a":"24 m ","b":"25 m ","c":"26 m ","d":"27 m","e":"28 m"},"options_float":{"a":24.0,"b":25.0,"c":26.0,"d":27.0,"e":28.0},"annotated_formula":"subtract(100, divide(100, multiply(divide(100, subtract(100, 25)), divide(100, subtract(100, 4)))))","linear_formula":"subtract(n0,n1)|subtract(n0,n2)|divide(n0,#0)|divide(n0,#1)|multiply(#2,#3)|divide(n0,#4)|subtract(n0,#5)","chain":"100 - 25<\/gadget>\n75<\/output>\n100 \/ 75<\/gadget>\n4\/3 = around 1.333333<\/output>\n100 - 4<\/gadget>\n96<\/output>\n100 \/ 96<\/gadget>\n25\/24 = around 1.041667<\/output>\n(4\/3) * (25\/24)<\/gadget>\n25\/18 = around 1.388889<\/output>\n100 \/ (25\/18)<\/gadget>\n72<\/output>\n100 - 72<\/gadget>\n28<\/output>\n28<\/result>","index":2496} +{"problem":"for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 50 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amount of the sale ?","rationale":"\"total sales = 800 comission = ( 20 \/ 100 ) * 500 + ( 50 \/ 100 ) * 300 = 100 + 150 = 250 % comission = ( 250 \/ 800 ) * 100 = 31.3 ~ 31 % answer is e\"","correct":"e","options":{"a":"22 % ","b":"24 % ","c":"25 % ","d":"27 %","e":"31 %"},"options_float":{"a":22.0,"b":24.0,"c":25.0,"d":27.0,"e":31.0},"annotated_formula":"multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(50, const_100), subtract(800, 500))), 800), const_100)","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n1)|multiply(n1,#0)|multiply(#1,#2)|add(#3,#4)|divide(#5,n4)|multiply(#6,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 500<\/gadget>\n100<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n800 - 500<\/gadget>\n300<\/output>\n(1\/2) * 300<\/gadget>\n150<\/output>\n100 + 150<\/gadget>\n250<\/output>\n250 \/ 800<\/gadget>\n5\/16 = around 0.3125<\/output>\n(5\/16) * 100<\/gadget>\n125\/4 = around 31.25<\/output>\n125\/4 = around 31.25<\/result>","index":2497} +{"problem":"a man swims downstream 60 km and upstream 100 km taking 10 hours each time , what is the speed of the man in still water ?","rationale":"\"60 - - - 10 ds = 6 ? - - - - 1 100 - - - - 10 us = 10 ? - - - - 1 m = ? m = ( 10 + 6 ) \/ 2 = 8 answer : d\"","correct":"d","options":{"a":"24 ","b":"42 ","c":"6 ","d":"8","e":"12"},"options_float":{"a":24.0,"b":42.0,"c":6.0,"d":8.0,"e":12.0},"annotated_formula":"divide(add(divide(100, 10), divide(60, 10)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"100 \/ 10<\/gadget>\n10<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n10 + 6<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2499} +{"problem":"a light flashes every 30 seconds , how many times will it flash in ? of an hour ?","rationale":"1 flash = 30 sec for 1 min = 2 flashes so for 1 hour = 2 * 60 = 120 flashes . answer : e","correct":"e","options":{"a":"550 ","b":"600 ","c":"650 ","d":"700","e":"120"},"options_float":{"a":550.0,"b":600.0,"c":650.0,"d":700.0,"e":120.0},"annotated_formula":"divide(const_3600, 30)","linear_formula":"divide(const_3600,n0)","chain":"3_600 \/ 30<\/gadget>\n120<\/output>\n120<\/result>","index":2500} +{"problem":"tourist purchased a total of 30 travelers checks in $ 50 and $ 100 denominations . the total worth of the travelers checks is $ 1800 . how many checks of $ 50 denominations can he spend so that average amount ( arithmetic mean ) of the remaining travelers checks is $ 100 ?","rationale":"\"you could set - up a quick table and brute force the answer . a 4 * 50 200 1800 - 200 1600 26 61.54 b 12 * 50 600 1800 - 600 1200 18 66.67 c 15 * 50 750 1800 - 750 1050 15 70.00 d 20 * 50 1000 1800 - 1000 800 10 80.00 e 24 * 50 1200 1800 - 1200 600 6 100.00 answer is e\"","correct":"e","options":{"a":"4 ","b":"12 ","c":"15 ","d":"20","e":"24"},"options_float":{"a":4.0,"b":12.0,"c":15.0,"d":20.0,"e":24.0},"annotated_formula":"divide(subtract(multiply(100, 30), 1800), subtract(100, 50))","linear_formula":"multiply(n0,n5)|subtract(n5,n1)|subtract(#0,n3)|divide(#2,#1)|","chain":"100 * 30<\/gadget>\n3_000<\/output>\n3_000 - 1_800<\/gadget>\n1_200<\/output>\n100 - 50<\/gadget>\n50<\/output>\n1_200 \/ 50<\/gadget>\n24<\/output>\n24<\/result>","index":2503} +{"problem":"at an election 2 candidates are participated and a candidate got 20 % of votes and defeated by 500 . and 10 votes are invalid . find the total polled votes ?","rationale":"\"winner votes = 100 - 20 = 80 polled votes = [ ( 100 * 500 ) \/ 2 * 80 - 100 ] + 10 = 843 ( approximately ) answer is a\"","correct":"a","options":{"a":"843 ","b":"1023 ","c":"965 ","d":"413","e":"1245"},"options_float":{"a":843.0,"b":1023.0,"c":965.0,"d":413.0,"e":1245.0},"annotated_formula":"divide(add(500, 10), subtract(const_1, divide(multiply(20, 2), const_100)))","linear_formula":"add(n2,n3)|multiply(n0,n1)|divide(#1,const_100)|subtract(const_1,#2)|divide(#0,#3)|","chain":"500 + 10<\/gadget>\n510<\/output>\n20 * 2<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n510 \/ (3\/5)<\/gadget>\n850<\/output>\n850<\/result>","index":2504} +{"problem":"a cat chases a rat 6 hours after the rat runs . cat takes 4 hours to reach the rat . if the average speed of the cat is 90 kmph , what s the average speed of the rat ?","rationale":"cat take 10 hours and rat take 4 hours . . . then distance chased by them is 90 * 4 . so speed of rat is ( 90 * 4 ) \/ 10 = 36 kmph . answer is d","correct":"d","options":{"a":"32 kmph ","b":"26 kmph ","c":"35 kmph ","d":"36 kmph","e":"32 kmph"},"options_float":{"a":32.0,"b":26.0,"c":35.0,"d":36.0,"e":32.0},"annotated_formula":"divide(multiply(90, 4), add(4, 6))","linear_formula":"add(n0,n1)|multiply(n1,n2)|divide(#1,#0)","chain":"90 * 4<\/gadget>\n360<\/output>\n4 + 6<\/gadget>\n10<\/output>\n360 \/ 10<\/gadget>\n36<\/output>\n36<\/result>","index":2505} +{"problem":"average of 5 consecutive even numbers is 35 . find the greatest number in these 5 numbers ?","rationale":"39 option ' c '","correct":"c","options":{"a":"26 ","b":"29 ","c":"39 ","d":"42","e":"43"},"options_float":{"a":26.0,"b":29.0,"c":39.0,"d":42.0,"e":43.0},"annotated_formula":"divide(add(multiply(5, 35), add(add(add(add(const_2, const_2), const_2), const_2), add(add(add(const_2, const_2), const_2), add(add(const_2, const_2), const_2)))), 5)","linear_formula":"add(const_2,const_2)|multiply(n0,n1)|add(#0,const_2)|add(#2,const_2)|add(#2,#2)|add(#3,#4)|add(#5,#1)|divide(#6,n0)","chain":"5 * 35<\/gadget>\n175<\/output>\n2 + 2<\/gadget>\n4<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6 + 2<\/gadget>\n8<\/output>\n6 + 6<\/gadget>\n12<\/output>\n8 + 12<\/gadget>\n20<\/output>\n175 + 20<\/gadget>\n195<\/output>\n195 \/ 5<\/gadget>\n39<\/output>\n39<\/result>","index":2506} +{"problem":"a wall photo 2 inches wide is placed around a rectangular paper with dimensions 8 inches by 12 inches . what is the area of the wall photo , in square inches ?","rationale":"this question is an example of a ' punch out ' question - we have to find the area of everything , then ' punch out ' the part that we do n ' t want . we ' re told that a wall photo 2 inches wide is placed around a rectangular paper with dimensions 8 inches by 12 inches . we ' re asked for the area of the wall photo , in square inches . area of a rectangle = ( length ) ( width ) so the area of the wall photo is . . . ( 8 ) ( 12 ) = 96 the wall photo ' adds ' 2 inches to the top , bottom , left and right ' sides ' of the picture , so the area of everything is . . . ( 8 + 2 + 2 ) ( 12 + 2 + 2 ) = ( 12 ) ( 16 ) = 196 when we ' punch out ' the area of the paper , we ' ll be left with the area of the wall photo : 192 - 96 = 96 final answer : a","correct":"a","options":{"a":"96 ","b":"86 ","c":"108 ","d":"144","e":"118"},"options_float":{"a":96.0,"b":86.0,"c":108.0,"d":144.0,"e":118.0},"annotated_formula":"subtract(multiply(multiply(8, 2), 12), rectangle_area(8, 12))","linear_formula":"multiply(n0,n1)|rectangle_area(n1,n2)|multiply(n2,#0)|subtract(#2,#1)","chain":"8 * 2<\/gadget>\n16<\/output>\n16 * 12<\/gadget>\n192<\/output>\n8 * 12<\/gadget>\n96<\/output>\n192 - 96<\/gadget>\n96<\/output>\n96<\/result>","index":2507} +{"problem":"the perimeter of a triangle is 36 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle","rationale":"\"explanation : area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 36 \/ 2 = 45 cm 2 answer : option b\"","correct":"b","options":{"a":"72 ","b":"45 ","c":"729 ","d":"34","e":"35"},"options_float":{"a":72.0,"b":45.0,"c":729.0,"d":34.0,"e":35.0},"annotated_formula":"triangle_area(2.5, 36)","linear_formula":"triangle_area(n0,n1)|","chain":"(2.5 * 36) \/ 2<\/gadget>\n45<\/output>\n45<\/result>","index":2508} +{"problem":"john found that the average of 15 no . ' s is 40 . if 10 is added to each number then the mean of number is ?","rationale":"( x + x 1 + . . . x 14 ) \/ 15 = 40 50 option a","correct":"a","options":{"a":"50 ","b":"55 ","c":"65 ","d":"72","e":"78"},"options_float":{"a":50.0,"b":55.0,"c":65.0,"d":72.0,"e":78.0},"annotated_formula":"add(40, 10)","linear_formula":"add(n1,n2)","chain":"40 + 10<\/gadget>\n50<\/output>\n50<\/result>","index":2509} +{"problem":"a car salesman earns a base salary of $ 1000 per month plus a commission of $ 200 for each car he sells . if the car salesman earned $ 2000 in march , how many cars does he need to sell in april in order to double his march earnings ?","rationale":"1000 + 200 x = 4000 x = 15 cars . the answer is b .","correct":"b","options":{"a":"14 ","b":"15 ","c":"16 ","d":"17","e":"18"},"options_float":{"a":14.0,"b":15.0,"c":16.0,"d":17.0,"e":18.0},"annotated_formula":"divide(subtract(multiply(2000, const_2), 1000), 200)","linear_formula":"multiply(n2,const_2)|subtract(#0,n0)|divide(#1,n1)","chain":"2_000 * 2<\/gadget>\n4_000<\/output>\n4_000 - 1_000<\/gadget>\n3_000<\/output>\n3_000 \/ 200<\/gadget>\n15<\/output>\n15<\/result>","index":2510} +{"problem":"barbata invests $ 2200 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ?","rationale":"\"let the additional invested amount for 8 % interest be x ; equation will be ; 2200 + 0.05 * 2200 + x + 0.08 x = 2200 + x + 0.06 ( 2200 + x ) 0.05 * 2200 + 0.08 x = 0.06 x + 0.06 * 2200 0.02 x = 2200 ( 0.06 - 0.05 ) x = 2200 * 0.01 \/ 0.02 = 1100 ans : ` ` a ' '\"","correct":"a","options":{"a":"1100 ","b":"3000 ","c":"1000 ","d":"3600","e":"2400"},"options_float":{"a":1100.0,"b":3000.0,"c":1000.0,"d":3600.0,"e":2400.0},"annotated_formula":"divide(subtract(multiply(divide(6, const_100), 2200), multiply(2200, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100)))","linear_formula":"divide(n3,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)|","chain":"6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n(3\/50) * 2_200<\/gadget>\n132<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n2_200 * (1\/20)<\/gadget>\n110<\/output>\n132 - 110<\/gadget>\n22<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) - (3\/50)<\/gadget>\n1\/50 = around 0.02<\/output>\n22 \/ (1\/50)<\/gadget>\n1_100<\/output>\n1_100<\/result>","index":2511} +{"problem":"24 lbs of coffee p and 25 lbs of coffee v are mixed to make coffee x and y . the ratio of p to v in coffee x is 4 to 1 , in y is 1 to 5 . how much of p is contained in the mixture x ?","rationale":"let there be 4 x lbs of p in x and x lbs of v in x let there be y lbs of p in y and 5 y lbs of v in y now total p is 4 x + y = 24 and total q is x + 5 y = 25 solving the two equations , we get x = 5 and y = 4 mixture x has 4 x lbs of p = 4 x = 4 * 5 = 20 lbs answer : b","correct":"b","options":{"a":"10 ","b":"20 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"multiply(divide(subtract(multiply(24, 5), 25), subtract(24, 5)), 4)","linear_formula":"multiply(n0,n5)|subtract(n0,n5)|subtract(#0,n1)|divide(#2,#1)|multiply(n2,#3)","chain":"24 * 5<\/gadget>\n120<\/output>\n120 - 25<\/gadget>\n95<\/output>\n24 - 5<\/gadget>\n19<\/output>\n95 \/ 19<\/gadget>\n5<\/output>\n5 * 4<\/gadget>\n20<\/output>\n20<\/result>","index":2512} +{"problem":"the average of 10 numbers is 0 . in those 10 , at the least , how many may be greater than or equal to zero ?","rationale":"explanation : average of 10 numbers = 0 sum of 10 numbers = ( 0 x 10 ) = 0 . it is quite possible that 9 of these numbers may be negative and if there sum is - a , then 10 th number is ( a ) . answer : b","correct":"b","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"5"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":5.0},"annotated_formula":"add(0, const_1)","linear_formula":"add(n1,const_1)","chain":"0 + 1<\/gadget>\n1<\/output>\n1<\/result>","index":2513} +{"problem":"if x + y = 260 , x - y = 200 , for integers of x and y , y = ?","rationale":"\"x + y = 260 x - y = 200 2 x = 60 x = 30 y = 230 answer is b\"","correct":"b","options":{"a":"200 ","b":"230 ","c":"50 ","d":"115","e":"150"},"options_float":{"a":200.0,"b":230.0,"c":50.0,"d":115.0,"e":150.0},"annotated_formula":"divide(add(260, 200), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"260 + 200<\/gadget>\n460<\/output>\n460 \/ 2<\/gadget>\n230<\/output>\n230<\/result>","index":2516} +{"problem":"a man can row 24 kmph in still water . it takes him thrice as long to row up as to row down the river . find the rate of the stream ?","rationale":"\"let man ' s rate upsteam be x kmph then his rate of downstream = 3 x kmph rate still water = 1 \/ 2 ( 3 x + x ) = 2 x 2 x = 24 x = 12 rate of upstream = 12 rate of downstream = 36 rate of stream 1 \/ 2 ( 36 - 12 ) = 12 kmph answer is b .\"","correct":"b","options":{"a":"6 ","b":"12 ","c":"18 ","d":"15","e":"20"},"options_float":{"a":6.0,"b":12.0,"c":18.0,"d":15.0,"e":20.0},"annotated_formula":"divide(multiply(24, const_2), add(const_3, const_1))","linear_formula":"add(const_1,const_3)|multiply(n0,const_2)|divide(#1,#0)|","chain":"24 * 2<\/gadget>\n48<\/output>\n3 + 1<\/gadget>\n4<\/output>\n48 \/ 4<\/gadget>\n12<\/output>\n12<\/result>","index":2517} +{"problem":"a bookseller sells his books at a 20 % markup in price . if he sells a book for $ 36.00 , how much did he pay for it ?","rationale":"\"let the cost price of book = x selling price of book = 36 $ markup % = 20 ( 120 \/ 100 ) x = 36 = > x = 30 answer d\"","correct":"d","options":{"a":"$ 34.40 ","b":"$ 34.00 ","c":"$ 32.00 ","d":"$ 30.00","e":"$ 35.00"},"options_float":{"a":34.4,"b":34.0,"c":32.0,"d":30.0,"e":35.0},"annotated_formula":"subtract(36.00, multiply(divide(20, const_100), 36.00))","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 36<\/gadget>\n36\/5 = around 7.2<\/output>\n36 - (36\/5)<\/gadget>\n144\/5 = around 28.8<\/output>\n144\/5 = around 28.8<\/result>","index":2518} +{"problem":"the hcf of two numbers is 42 and the other two factors of their lcm are 10 and 20 . what is the largest number .","rationale":"\"explanation : hcf of the two numbers = 42 hcf will be always a factor of lcm 42 is factor of lcm other two factors are 10 & 20 then the numbers are ( 42 * 10 ) and ( 42 x 20 ) = 420 and 840 answer : option c\"","correct":"c","options":{"a":"462 ","b":"450 ","c":"840 ","d":"504","e":"555"},"options_float":{"a":462.0,"b":450.0,"c":840.0,"d":504.0,"e":555.0},"annotated_formula":"multiply(42, 20)","linear_formula":"multiply(n0,n2)|","chain":"42 * 20<\/gadget>\n840<\/output>\n840<\/result>","index":2519} +{"problem":"suraj has a certain average of runs for 16 innings . in the 17 th innings he scores 92 runs thereby increasing his average by 4 runs . what is his average after the 17 th innings ?","rationale":"to improve his average by 4 runs per innings he has to contribute 16 x 4 = 64 runs for the previous 16 innings . thus , the average after the 17 th innings = 92 - 64 = 28 . answer : c","correct":"c","options":{"a":"48 ","b":"64 ","c":"28 ","d":"72","e":"27"},"options_float":{"a":48.0,"b":64.0,"c":28.0,"d":72.0,"e":27.0},"annotated_formula":"divide(subtract(92, multiply(16, 4)), subtract(17, 16))","linear_formula":"multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)","chain":"16 * 4<\/gadget>\n64<\/output>\n92 - 64<\/gadget>\n28<\/output>\n17 - 16<\/gadget>\n1<\/output>\n28 \/ 1<\/gadget>\n28<\/output>\n28<\/result>","index":2520} +{"problem":"a dog have birth to 12 puppies , 2 female and 10 male . what is the ratio of the female puppies to the male ?","rationale":"if female is 2 and male is 10 , then 1 \/ 5 . so ratio of female to male is = 2 \/ 10 = 1 \/ 5 answer : a","correct":"a","options":{"a":"1 \/ 5 ","b":"1 \/ 6 ","c":"1 \/ 10 ","d":"2 \/ 12","e":"1 \/ 8"},"options_float":{"a":0.2,"b":0.1666666667,"c":0.1,"d":0.1666666667,"e":0.125},"annotated_formula":"divide(2, 10)","linear_formula":"divide(n1,n2)","chain":"2 \/ 10<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":2522} +{"problem":"a box contains 6 black , 8 red and 4 green marbles . 2 marbles are drawn from the box at random . what is the probability that both the marbles are of the same color ?","rationale":"\"explanation : total marbles in a box = 6 black + 8 red + 4 green marbles = 18 marbles 2 marbles are drawn from 18 marbles at random . therefore , n ( s ) = 18 c 2 = 153 ways let a be the event that 2 marbles drawn at random are of the same color . number of cases favorable to the event a is n ( a ) = 6 c 2 + 8 c 2 + 4 c 2 = 15 + 28 + 6 = 49 therefore , by definition of probability of event a , p ( a ) = n ( a ) \/ n ( s ) = 49 \/ 153 answer : a\"","correct":"a","options":{"a":"49 \/ 153 ","b":"18 \/ 153 ","c":"49 \/ 153 ","d":"40 \/ 153","e":"24 \/ 72"},"options_float":{"a":0.3202614379,"b":0.1176470588,"c":0.3202614379,"d":0.2614379085,"e":0.3333333333},"annotated_formula":"divide(add(add(const_3.0, 4), add(const_10, add(const_3.0, 2))), multiply(add(const_10, 8), 4))","linear_formula":"add(const_3.0,n2)|add(const_3.0,n3)|add(const_10,n1)|add(#1,const_10)|multiply(n2,#2)|add(#0,#3)|divide(#5,#4)|","chain":"3 + 4<\/gadget>\n7<\/output>\n3 + 2<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n7 + 15<\/gadget>\n22<\/output>\n10 + 8<\/gadget>\n18<\/output>\n18 * 4<\/gadget>\n72<\/output>\n22 \/ 72<\/gadget>\n11\/36 = around 0.305556<\/output>\n11\/36 = around 0.305556<\/result>","index":2523} +{"problem":"a man has some hens and cows . if the number of heads be 48 and the number of feet equals 136 , then the number of hens will be","rationale":"\"explanation : let number of hens = h and number of cows = c number of heads = 48 = > h + c = 48 - - - ( equation 1 ) number of feet = 136 = > 2 h + 4 c = 136 = > h + 2 c = 68 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 68 - 48 = > c = 20 substituting the value of c in equation 1 , we get h + 20 = 48 = > h = 48 - 20 = 28 i . e . , number of hens = 28 answer : e\"","correct":"e","options":{"a":"22 ","b":"24 ","c":"26 ","d":"20","e":"28"},"options_float":{"a":22.0,"b":24.0,"c":26.0,"d":20.0,"e":28.0},"annotated_formula":"divide(subtract(multiply(48, const_4), 136), const_2)","linear_formula":"multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|","chain":"48 * 4<\/gadget>\n192<\/output>\n192 - 136<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":2525} +{"problem":"the average of 25 results is 20 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ?","rationale":"\"solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 20 or , x = 128 . answer : option a\"","correct":"a","options":{"a":"128 ","b":"75 ","c":"69 ","d":"78","e":"45"},"options_float":{"a":128.0,"b":75.0,"c":69.0,"d":78.0,"e":45.0},"annotated_formula":"subtract(subtract(multiply(25, 20), multiply(12, 17)), multiply(12, 14))","linear_formula":"multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)|","chain":"25 * 20<\/gadget>\n500<\/output>\n12 * 17<\/gadget>\n204<\/output>\n500 - 204<\/gadget>\n296<\/output>\n12 * 14<\/gadget>\n168<\/output>\n296 - 168<\/gadget>\n128<\/output>\n128<\/result>","index":2527} +{"problem":"if d is the smallest positive integer such that 3,150 multiplied by d is the square of an integer , then d must be","rationale":"solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3,150 . 3,150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3,150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that the exponents of both 2 and 7 are not even numbers . this tells us that 3,150 itself is not a perfect square . ) we also are given that 3,150 multiplied by d is the square of an integer . we can write this as : 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 x d = square of an integer according to our rule , we need all unique prime factors ' exponents to be even numbers . thus , we need one more 2 and one more 7 . therefore , d = 7 x 2 = 14 answer is e .","correct":"e","options":{"a":"2 ","b":"5 ","c":"6 ","d":"7","e":"14"},"options_float":{"a":2.0,"b":5.0,"c":6.0,"d":7.0,"e":14.0},"annotated_formula":"multiply(divide(divide(divide(divide(add(add(multiply(const_3, const_1000), divide(const_100, const_2)), const_100), const_2), power(add(const_3, const_2), const_2)), const_3), const_3), const_2)","linear_formula":"add(const_2,const_3)|divide(const_100,const_2)|multiply(const_1000,const_3)|add(#1,#2)|power(#0,const_2)|add(#3,const_100)|divide(#5,const_2)|divide(#6,#4)|divide(#7,const_3)|divide(#8,const_3)|multiply(#9,const_2)","chain":"3 * 1_000<\/gadget>\n3_000<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n3_000 + 50<\/gadget>\n3_050<\/output>\n3_050 + 100<\/gadget>\n3_150<\/output>\n3_150 \/ 2<\/gadget>\n1_575<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n1_575 \/ 25<\/gadget>\n63<\/output>\n63 \/ 3<\/gadget>\n21<\/output>\n21 \/ 3<\/gadget>\n7<\/output>\n7 * 2<\/gadget>\n14<\/output>\n14<\/result>","index":2528} +{"problem":"x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 14190 ?","rationale":"\"ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 14190 ã — 16 \/ 55 = rs . 4128 . share of z in the profit is rs . 4128 . option d\"","correct":"d","options":{"a":"3200 ","b":"4000 ","c":"3250 ","d":"4128","e":"3985"},"options_float":{"a":3200.0,"b":4000.0,"c":3250.0,"d":4128.0,"e":3985.0},"annotated_formula":"multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14190, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4)))))","linear_formula":"multiply(const_3,n2)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,n2)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5)|","chain":"3 * 4<\/gadget>\n12<\/output>\n12 - 4<\/gadget>\n8<\/output>\n48_000 * 8<\/gadget>\n384_000<\/output>\n36_000 * 12<\/gadget>\n432_000<\/output>\n42_000 * 12<\/gadget>\n504_000<\/output>\n432_000 + 504_000<\/gadget>\n936_000<\/output>\n936_000 + 384_000<\/gadget>\n1_320_000<\/output>\n14_190 \/ 1_320_000<\/gadget>\n43\/4_000 = around 0.01075<\/output>\n384_000 * (43\/4_000)<\/gadget>\n4_128<\/output>\n4_128<\/result>","index":2529} +{"problem":"there are 250 doctors and nurses at a hospital . if the ratio of doctors to nurses is 2 to 3 , how many nurses are at the hospital ?","rationale":"\"the number of nurses at the hospital is ( 3 \/ 5 ) * 250 = 150 . the answer is e .\"","correct":"e","options":{"a":"134 ","b":"138 ","c":"142 ","d":"146","e":"150"},"options_float":{"a":134.0,"b":138.0,"c":142.0,"d":146.0,"e":150.0},"annotated_formula":"multiply(divide(250, add(2, 3)), 3)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"2 + 3<\/gadget>\n5<\/output>\n250 \/ 5<\/gadget>\n50<\/output>\n50 * 3<\/gadget>\n150<\/output>\n150<\/result>","index":2530} +{"problem":"in an examination , 300 students appeared . out of these students ; 26 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed .","rationale":"\"the number of students with first division = 26 % of 300 = 26 \/ 100 × 300 = 7800 \/ 100 = 78 and , the number of students with second division = 54 % of 300 = 54 \/ 100 × 300 = 16200 \/ 100 = 162 therefore , the number of students who just passed = 300 – ( 78 + 162 ) = 60 answer : d\"","correct":"d","options":{"a":"33 ","b":"37 ","c":"54 ","d":"60","e":"01"},"options_float":{"a":33.0,"b":37.0,"c":54.0,"d":60.0,"e":1.0},"annotated_formula":"divide(multiply(300, subtract(const_100, add(26, 54))), const_100)","linear_formula":"add(n1,n2)|subtract(const_100,#0)|multiply(n0,#1)|divide(#2,const_100)|","chain":"26 + 54<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n300 * 20<\/gadget>\n6_000<\/output>\n6_000 \/ 100<\/gadget>\n60<\/output>\n60<\/result>","index":2531} +{"problem":"on a certain transatlantic crossing , 30 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 60 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ?","rationale":"\"0.30 p = rt + c 0.6 ( rt ) = no c = > 0.40 ( rt ) had c 0.30 p = 0.40 ( rt ) rt \/ p = 3 \/ 4 = 75 % answer - e\"","correct":"e","options":{"a":"33.3 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"75 %"},"options_float":{"a":33.3,"b":40.0,"c":50.0,"d":60.0,"e":75.0},"annotated_formula":"divide(30, divide(subtract(const_100, 60), const_100))","linear_formula":"subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|","chain":"100 - 60<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n30 \/ (2\/5)<\/gadget>\n75<\/output>\n75<\/result>","index":2532} +{"problem":"students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 184 cms . what was the ratio d of men to women ?","rationale":"( a ) d = 184 x 5 + 170 x 2 = 1260 . a","correct":"a","options":{"a":"5 : 2 ","b":"5 : 1 ","c":"4 : 3 ","d":"4 : 1","e":"3 : 1"},"options_float":{"a":2.5,"b":5.0,"c":1.3333333333,"d":4.0,"e":3.0},"annotated_formula":"divide(subtract(180, 170), subtract(184, 180))","linear_formula":"subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)|","chain":"180 - 170<\/gadget>\n10<\/output>\n184 - 180<\/gadget>\n4<\/output>\n10 \/ 4<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":2538} +{"problem":"at what rate of compound interest per annum will a sum of $ 1200 becomes 1348.32 in 2 years ?","rationale":"\"let the rate be r % p . a . 1200 * ( 1 + r \/ 100 ) ^ 2 = 1348.32 ( 1 + r \/ 100 ) ^ 2 = 134832 \/ 120000 = 11236 \/ 10000 ( 1 + r \/ 100 ) ^ 2 = ( 106 \/ 100 ) ^ 2 1 + r \/ 100 = 106 \/ 100 r = 6 % answer is b\"","correct":"b","options":{"a":"5 % ","b":"6 % ","c":"7 % ","d":"8 %","e":"10 %"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":10.0},"annotated_formula":"subtract(sqrt(divide(multiply(1348.32, const_100), divide(1200, const_100))), sqrt(divide(multiply(1200, const_100), divide(1200, const_100))))","linear_formula":"divide(n0,const_100)|multiply(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|divide(#2,#0)|sqrt(#3)|sqrt(#4)|subtract(#5,#6)|","chain":"1_348.32 * 100<\/gadget>\n134_832<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n134_832 \/ 12<\/gadget>\n11_236<\/output>\n11_236 ** (1\/2)<\/gadget>\n106<\/output>\n1_200 * 100<\/gadget>\n120_000<\/output>\n120_000 \/ 12<\/gadget>\n10_000<\/output>\n10_000 ** (1\/2)<\/gadget>\n100<\/output>\n106 - 100<\/gadget>\n6<\/output>\n6<\/result>","index":2539} +{"problem":"the food in a camp lasts for 10 men for 50 days . if 10 more men join , how many days will the food last ?","rationale":"\"one man can consume the same food in 10 * 50 = 500 days . 10 more men join , the total number of men = 20 the number of days the food will last = 500 \/ 20 = 25 days . answer : b\"","correct":"b","options":{"a":"22 days ","b":"25 days ","c":"50 days ","d":"16 days","e":"17 days"},"options_float":{"a":22.0,"b":25.0,"c":50.0,"d":16.0,"e":17.0},"annotated_formula":"divide(multiply(10, 50), add(10, 10))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"10 * 50<\/gadget>\n500<\/output>\n10 + 10<\/gadget>\n20<\/output>\n500 \/ 20<\/gadget>\n25<\/output>\n25<\/result>","index":2541} +{"problem":"a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 6 miles per day and it will take them two days , how many miles long is the route down the mountain ?","rationale":"\"on the way down , the rate is 1.5 * 6 = 9 miles per day . the distance of the route down the mountain is 2 * 9 = 18 miles . the answer is c .\"","correct":"c","options":{"a":"16 ","b":"17 ","c":"18 ","d":"19","e":"20"},"options_float":{"a":16.0,"b":17.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"multiply(multiply(6, const_2), divide(const_3, const_2))","linear_formula":"divide(const_3,const_2)|multiply(n0,const_2)|multiply(#0,#1)|","chain":"6 * 2<\/gadget>\n12<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n12 * (3\/2)<\/gadget>\n18<\/output>\n18<\/result>","index":2542} +{"problem":"equal amount of water were poured into two empty jars of different capacities , which made one jar 1 \/ 3 full and other jar 1 \/ 2 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ?","rationale":"same amount of water made bigger jar 1 \/ 3 full , thenthe same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 \/ 3 + 1 \/ 3 = 2 \/ 3 full . answer : a .","correct":"a","options":{"a":"2 \/ 3 ","b":"2 \/ 7 ","c":"1 \/ 2 ","d":"7 \/ 12","e":"1 \/ 7"},"options_float":{"a":0.6666666667,"b":0.2857142857,"c":0.5,"d":0.5833333333,"e":0.1428571429},"annotated_formula":"divide(const_2, 3)","linear_formula":"divide(const_2,n1)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":2544} +{"problem":"a certain candy manufacturer reduced the weight of candy bar m by 5 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ?","rationale":"assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.95 oz new price of candy = 1 \/ 0.95 = 1.052 price increase 5.2 % d","correct":"d","options":{"a":"5.5 ","b":"5.4 ","c":"5.3 ","d":"5.2","e":"5.1"},"options_float":{"a":5.5,"b":5.4,"c":5.3,"d":5.2,"e":5.1},"annotated_formula":"subtract(multiply(divide(const_1, subtract(const_1, divide(5, const_100))), const_100), const_100)","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(const_1,#1)|multiply(#2,const_100)|subtract(#3,const_100)","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (1\/20)<\/gadget>\n19\/20 = around 0.95<\/output>\n1 \/ (19\/20)<\/gadget>\n20\/19 = around 1.052632<\/output>\n(20\/19) * 100<\/gadget>\n2_000\/19 = around 105.263158<\/output>\n(2_000\/19) - 100<\/gadget>\n100\/19 = around 5.263158<\/output>\n100\/19 = around 5.263158<\/result>","index":2545} +{"problem":"mary ' s income is 60 percent more than tim ' s income , and tim ' s income is 60 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ?","rationale":"\"juan ' s income = 100 ( assume ) ; tim ' s income = 40 ( 60 percent less than juan ' s income ) ; mary ' s income = 64 ( 60 percent more than tim ' s income ) . thus , mary ' s income ( 64 ) is 64 % of juan ' s income ( 100 ) . answer : a .\"","correct":"a","options":{"a":"64 % ","b":"120 % ","c":"96 % ","d":"80 %","e":"64 %"},"options_float":{"a":64.0,"b":120.0,"c":96.0,"d":80.0,"e":64.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(60, const_100)), add(const_1, divide(60, const_100))), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n(2\/5) * (8\/5)<\/gadget>\n16\/25 = around 0.64<\/output>\n(16\/25) * 100<\/gadget>\n64<\/output>\n64<\/result>","index":2547} +{"problem":"12 . what is the dividend . divisor 17 , the quotient is 9 and the remainder is 7 .","rationale":"d = d * q + r d = 17 * 9 + 7 d = 153 + 7 d = 160 answer e","correct":"e","options":{"a":"150 ","b":"152 ","c":"154 ","d":"158","e":"160"},"options_float":{"a":150.0,"b":152.0,"c":154.0,"d":158.0,"e":160.0},"annotated_formula":"add(multiply(17, 9), 7)","linear_formula":"multiply(n1,n2)|add(n3,#0)","chain":"17 * 9<\/gadget>\n153<\/output>\n153 + 7<\/gadget>\n160<\/output>\n160<\/result>","index":2548} +{"problem":"if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs 1920 less than the principal . what was the principal ?","rationale":"\"p - 1920 = ( p * 5 * 4 ) \/ 100 p = 2400 answer : e\"","correct":"e","options":{"a":"1500 ","b":"2500 ","c":"2507 ","d":"3200","e":"2400"},"options_float":{"a":1500.0,"b":2500.0,"c":2507.0,"d":3200.0,"e":2400.0},"annotated_formula":"divide(1920, subtract(const_1, divide(multiply(4, 5), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|","chain":"4 * 5<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1_920 \/ (4\/5)<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":2549} +{"problem":"the diagonals of a rhombus are 15 cm and 20 cm . find its area ?","rationale":"\"1 \/ 2 * 15 * 20 = 150 answer : c\"","correct":"c","options":{"a":"158 ","b":"129 ","c":"150 ","d":"123","e":"117"},"options_float":{"a":158.0,"b":129.0,"c":150.0,"d":123.0,"e":117.0},"annotated_formula":"rhombus_area(15, 20)","linear_formula":"rhombus_area(n0,n1)|","chain":"(15 * 20) \/ 2<\/gadget>\n150<\/output>\n150<\/result>","index":2550} +{"problem":"if 1 \/ ( x + 4 ) + 1 \/ ( x - 4 ) = 1 \/ ( x - 4 ) , what is the value of x ?","rationale":"\"if we solve the question , we get x = 4 . option : b\"","correct":"b","options":{"a":"5 ","b":"4 ","c":"1 ","d":"2","e":"3"},"options_float":{"a":5.0,"b":4.0,"c":1.0,"d":2.0,"e":3.0},"annotated_formula":"subtract(add(4, 1), 1)","linear_formula":"add(n1,n0)|subtract(#0,n0)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n4<\/result>","index":2552} +{"problem":"the average of 9 observations was 8 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ?","rationale":"explanation : 1 to 9 = 8 * 9 = 72 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 72 = 18 option a","correct":"a","options":{"a":"18 ","b":"12 ","c":"15 ","d":"17","e":"19"},"options_float":{"a":18.0,"b":12.0,"c":15.0,"d":17.0,"e":19.0},"annotated_formula":"subtract(add(multiply(10, 5), multiply(8, 5)), multiply(8, 9))","linear_formula":"multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)","chain":"10 * 5<\/gadget>\n50<\/output>\n8 * 5<\/gadget>\n40<\/output>\n50 + 40<\/gadget>\n90<\/output>\n8 * 9<\/gadget>\n72<\/output>\n90 - 72<\/gadget>\n18<\/output>\n18<\/result>","index":2556} +{"problem":"sophia finished 2 \/ 3 of a book . she calculated that she finished 30 more pages than she has yet to read . how long is her book ?","rationale":"\"let x be the total number of pages in the book , then she finished 2 \/ 3 * x pages . then she has x − 2 \/ 3 * x = 1 \/ 3 * x pages left . 2 \/ 3 * x − 1 \/ 3 * x = 30 1 \/ 3 * x = 30 x = 90 so the book is 90 pages long . answer is a .\"","correct":"a","options":{"a":"90 ","b":"150 ","c":"130 ","d":"190","e":"210"},"options_float":{"a":90.0,"b":150.0,"c":130.0,"d":190.0,"e":210.0},"annotated_formula":"divide(30, subtract(const_1, divide(2, 3)))","linear_formula":"divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n30 \/ (1\/3)<\/gadget>\n90<\/output>\n90<\/result>","index":2557} +{"problem":"which expression is the greatest ?","rationale":"let 3252 \/ 3257 be the original fraction . notice that it is a positive proper fraction . hence adding the same constant to both the numerator and the denominator will increase the fraction as the fraction will move closer to one . hence we just have to pick the number that adds the greatest value to both the numerator and the denominator . options can be re - written as ( x - 5 ) x = > 1 - ( 5 \/ x ) a ) 1 - ( 5 \/ 3257 ) b ) 1 - ( 5 \/ 3461 ) c ) 1 - ( 5 \/ 3596 ) d ) 1 - ( 5 \/ 3351 ) e ) 1 - ( 5 \/ 3458 ) answer : c","correct":"c","options":{"a":"3252 \/ 3257 ","b":"3456 \/ 3461 ","c":"3591 \/ 3596 ","d":"3346 \/ 3351","e":"3453 \/ 3458"},"options_float":{"a":0.9984648449,"b":0.9985553308,"c":0.9986095662,"d":0.9985079081,"e":0.9985540775},"annotated_formula":"divide(subtract(const_3600, multiply(const_3, const_3)), subtract(const_3600, const_4))","linear_formula":"multiply(const_3,const_3)|subtract(const_3600,const_4)|subtract(const_3600,#0)|divide(#2,#1)","chain":"3 * 3<\/gadget>\n9<\/output>\n3_600 - 9<\/gadget>\n3_591<\/output>\n3_600 - 4<\/gadget>\n3_596<\/output>\n3_591 \/ 3_596<\/gadget>\n3_591\/3_596 = around 0.99861<\/output>\n3_591\/3_596 = around 0.99861<\/result>","index":2559} +{"problem":"if there are 200 questions in a 3 hr examination . among these questions are 100 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ?","rationale":"\"x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 100 x + 100 * 2 x = 180 x = 180 \/ 300 x = 0.6 time taken for type a problem = 100 * 2 * 0.6 = 120 min answer : e\"","correct":"e","options":{"a":"72 min ","b":"62 min ","c":"70 min ","d":"74 min","e":"120 min"},"options_float":{"a":72.0,"b":62.0,"c":70.0,"d":74.0,"e":120.0},"annotated_formula":"multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 100), multiply(const_2, 100)))), 100)","linear_formula":"multiply(n1,const_60)|multiply(n2,const_2)|subtract(n0,n2)|add(#1,#2)|divide(#0,#3)|multiply(#4,const_2)|multiply(n2,#5)|","chain":"3 * 60<\/gadget>\n180<\/output>\n200 - 100<\/gadget>\n100<\/output>\n2 * 100<\/gadget>\n200<\/output>\n100 + 200<\/gadget>\n300<\/output>\n180 \/ 300<\/gadget>\n3\/5 = around 0.6<\/output>\n2 * (3\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 100<\/gadget>\n120<\/output>\n120<\/result>","index":2561} +{"problem":"a and b are 90 km apart they start to move each other simultaneously a at speed 10 and b at 5 km \/ hr if every hour they double their speed what is distance that a pass until he meet b ?","rationale":"in 2 hours a covers ( 10 + 20 ) = 30 km b covers ( 5 + 10 ) = 15 km , total covered = 45 in the next hour they have to cover ( 90 - 45 ) km with the speeds 40 km \/ hr and 20 km \/ hr respectively let them take another t hrs to meet so sum of the distance to be covered by both is 45 km ( 40 * t ) + ( 20 * t ) = 45 km gives t = ( 3 \/ 4 ) hrs we need to fid out the distance covered by a which equals to = 30 + ( 40 * ( 3 \/ 4 ) ) = 60 km answer : d","correct":"d","options":{"a":"30 km ","b":"40 km ","c":"50 km ","d":"60 km","e":"70 km"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":70.0},"annotated_formula":"add(add(10, 5), add(add(10, 5), multiply(add(10, 5), const_2)))","linear_formula":"add(n1,n2)|multiply(#0,const_2)|add(#0,#1)|add(#0,#2)","chain":"10 + 5<\/gadget>\n15<\/output>\n15 * 2<\/gadget>\n30<\/output>\n15 + 30<\/gadget>\n45<\/output>\n15 + 45<\/gadget>\n60<\/output>\n60<\/result>","index":2562} +{"problem":"a card board of size 34 * 14 has to be attached to a wooden box and a total of 35 pins are to be used on the each side of the card box . find the total number of pins used .","rationale":"total 35 pins are there and 4 sides of card board . . . so 35 x 4 = 140 now in the rectangle 4 vertices have 4 pins which is common to the sides . . . . so 140 - 4 = 136 answer : b","correct":"b","options":{"a":"135 ","b":"136 ","c":"137 ","d":"138","e":"139"},"options_float":{"a":135.0,"b":136.0,"c":137.0,"d":138.0,"e":139.0},"annotated_formula":"add(const_100, subtract(multiply(subtract(const_10, const_1), add(const_4, const_1)), const_10))","linear_formula":"add(const_1,const_4)|subtract(const_10,const_1)|multiply(#0,#1)|subtract(#2,const_10)|add(#3,const_100)","chain":"10 - 1<\/gadget>\n9<\/output>\n4 + 1<\/gadget>\n5<\/output>\n9 * 5<\/gadget>\n45<\/output>\n45 - 10<\/gadget>\n35<\/output>\n100 + 35<\/gadget>\n135<\/output>\n135<\/result>","index":2563} +{"problem":"a cistern can be filled by a tap in 2 hours while it can be emptied by another tap in 4 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ?","rationale":"\"net part filled in 1 hour 1 \/ 2 - 1 \/ 4 = 1 \/ 4 the cistern will be filled in 4 hr answer is e\"","correct":"e","options":{"a":"2 hr ","b":"5 hr ","c":"7 hr ","d":"1 hr","e":"4 hr"},"options_float":{"a":2.0,"b":5.0,"c":7.0,"d":1.0,"e":4.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 2), divide(const_1, 4)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/2) - (1\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ (1\/4)<\/gadget>\n4<\/output>\n4<\/result>","index":2564} +{"problem":"evaluate combination 98 c 96 = 98 ! \/ ( 96 ) ! ( 2 ) !","rationale":"explanation : ncr = n ! \/ ( r ) ! ( n − r ) ! 98 c 96 = 98 ! \/ ( 96 ) ! ( 2 ) ! = 98 ∗ 97 * 96 ! \/ ( 96 ) ! ( 2 ) ! = 98 * 97 \/ 2 ∗ 1 = 4753 option b","correct":"b","options":{"a":"1617 ","b":"4753 ","c":"1417 ","d":"4353","e":"none of these"},"options_float":{"a":1617.0,"b":4753.0,"c":1417.0,"d":4353.0,"e":null},"annotated_formula":"divide(multiply(98, subtract(98, const_1)), 2)","linear_formula":"subtract(n0,const_1)|multiply(n0,#0)|divide(#1,n4)","chain":"98 - 1<\/gadget>\n97<\/output>\n98 * 97<\/gadget>\n9_506<\/output>\n9_506 \/ 2<\/gadget>\n4_753<\/output>\n4_753<\/result>","index":2565} +{"problem":"a ranch has both horses and ponies . exactly 3 \/ 10 of the ponies have horseshoes , and exactly 5 \/ 8 of the ponies with horseshoes are from iceland . if there are 4 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ?","rationale":"3 \/ 10 * p are ponies with horseshoes , so p is a multiple of 10 . 5 \/ 8 * 3 \/ 10 * p = 3 \/ 16 * p are icelandic ponies with horseshoes , so p is a multiple of 16 . the minimum value of p is 80 . then h = p + 4 = 84 . the minimum number of horses and ponies is 164 . the answer is d .","correct":"d","options":{"a":"116 ","b":"132 ","c":"148 ","d":"164","e":"180"},"options_float":{"a":116.0,"b":132.0,"c":148.0,"d":164.0,"e":180.0},"annotated_formula":"add(add(lcm(10, multiply(divide(10, 5), 8)), 4), lcm(10, multiply(divide(10, 5), 8)))","linear_formula":"divide(n1,n2)|multiply(n3,#0)|lcm(n1,#1)|add(n4,#2)|add(#3,#2)","chain":"10 \/ 5<\/gadget>\n2<\/output>\n2 * 8<\/gadget>\n16<\/output>\nlcm(10, 16)<\/gadget>\n80<\/output>\n80 + 4<\/gadget>\n84<\/output>\n84 + 80<\/gadget>\n164<\/output>\n164<\/result>","index":2566} +{"problem":"there are 5 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 8 other amateurs , what is the total number of chess games possible to be played in the tournament ?","rationale":"\"method 2 : each person is one participant of 4 games . so there are in all 8 * 5 = 40 instances of one participant games . but each game has 2 participants so total number of games = 40 \/ 2 = 20 b\"","correct":"b","options":{"a":"10 ","b":"20 ","c":"40 ","d":"60","e":"120"},"options_float":{"a":10.0,"b":20.0,"c":40.0,"d":60.0,"e":120.0},"annotated_formula":"divide(multiply(5, 8), const_2)","linear_formula":"multiply(n0,n1)|divide(#0,const_2)|","chain":"5 * 8<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":2567} +{"problem":"the speed of a train is 90 kmph . what is the distance covered by it in 12 minutes ?","rationale":"90 * 12 \/ 60 = 18 kmph answer : b","correct":"b","options":{"a":"15 ","b":"18 ","c":"77 ","d":"52","e":"42"},"options_float":{"a":15.0,"b":18.0,"c":77.0,"d":52.0,"e":42.0},"annotated_formula":"multiply(divide(12, const_60), 90)","linear_formula":"divide(n1,const_60)|multiply(n0,#0)","chain":"12 \/ 60<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 90<\/gadget>\n18<\/output>\n18<\/result>","index":2568} +{"problem":"in a chess competition involving some men and women , every player needs to play exactly one game with every other player . it was found that in 45 games , both the players were women and in 190 games , both players were men . what is the number of games in which one person was a man and other person was a woman ?","rationale":"this is similar to handshake problem , if there are nn people , then number of handshakes = 1 + 2 + . . . n = n ∗ ( n − 1 ) \/ 2 no . of women be xx and given x ∗ ( x − 1 ) \/ 2 = 45 this gives x = 10 | | | y no . of men : solve y ∗ ( y − 1 ) \/ 2 = 190 . this gives y = 20 so total men vs . women = 10 * 20 = 200 . answer is d","correct":"d","options":{"a":"40 ","b":"120 ","c":"180 ","d":"200","e":"220"},"options_float":{"a":40.0,"b":120.0,"c":180.0,"d":200.0,"e":220.0},"annotated_formula":"multiply(divide(add(const_1, sqrt(add(const_1, multiply(multiply(45, const_2), const_4)))), const_2), divide(add(const_1, sqrt(add(const_1, multiply(multiply(190, const_2), const_4)))), const_2))","linear_formula":"multiply(n0,const_2)|multiply(n1,const_2)|multiply(#0,const_4)|multiply(#1,const_4)|add(#2,const_1)|add(#3,const_1)|sqrt(#4)|sqrt(#5)|add(#6,const_1)|add(#7,const_1)|divide(#8,const_2)|divide(#9,const_2)|multiply(#10,#11)","chain":"45 * 2<\/gadget>\n90<\/output>\n90 * 4<\/gadget>\n360<\/output>\n1 + 360<\/gadget>\n361<\/output>\n361 ** (1\/2)<\/gadget>\n19<\/output>\n1 + 19<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n190 * 2<\/gadget>\n380<\/output>\n380 * 4<\/gadget>\n1_520<\/output>\n1 + 1_520<\/gadget>\n1_521<\/output>\n1_521 ** (1\/2)<\/gadget>\n39<\/output>\n1 + 39<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n10 * 20<\/gadget>\n200<\/output>\n200<\/result>","index":2569} +{"problem":"11 is subtracted from 40 % of a number and results is 23 . what is the number ?","rationale":"\"( 40 \/ 100 ) * x â € “ 11 = 23 4 x = 340 x = 85 answer : d\"","correct":"d","options":{"a":"160 ","b":"92 ","c":"88 ","d":"85","e":"34"},"options_float":{"a":160.0,"b":92.0,"c":88.0,"d":85.0,"e":34.0},"annotated_formula":"divide(add(11, 23), divide(40, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"11 + 23<\/gadget>\n34<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n34 \/ (2\/5)<\/gadget>\n85<\/output>\n85<\/result>","index":2570} +{"problem":"in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required is :","rationale":"\"explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 15 : 8 : : 12 : x = > 8 x 15 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 \/ ( 8 x 15 ) = > x = 8 answer : d\"","correct":"d","options":{"a":"4 ","b":"5 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"divide(divide(multiply(multiply(12, 8), 10), 15), 8)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n4)|divide(#2,n1)|","chain":"12 * 8<\/gadget>\n96<\/output>\n96 * 10<\/gadget>\n960<\/output>\n960 \/ 15<\/gadget>\n64<\/output>\n64 \/ 8<\/gadget>\n8<\/output>\n8<\/result>","index":2573} +{"problem":"what is the maximum number of pieces of birthday cake of size 4 ” by 4 ” that can be cut from a cake 20 ” by 20 ” ?","rationale":"\"each length = 20 ' ' each piece = 4 ' ' cake could be divided to maximum of 25 pieces answer : e\"","correct":"e","options":{"a":"5 ","b":"10 ","c":"16 ","d":"20","e":"25"},"options_float":{"a":5.0,"b":10.0,"c":16.0,"d":20.0,"e":25.0},"annotated_formula":"divide(multiply(20, 20), multiply(4, 4))","linear_formula":"multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)|","chain":"20 * 20<\/gadget>\n400<\/output>\n4 * 4<\/gadget>\n16<\/output>\n400 \/ 16<\/gadget>\n25<\/output>\n25<\/result>","index":2574} +{"problem":"the present worth of rs . 845 due in 2 years at 4 % per annum compound interest is","rationale":"\"solution present worth = rs . [ 845 \/ ( 1 + 4 \/ 100 ) ² ] = rs . ( 845 x 25 \/ 26 x 25 \/ 26 ) = rs . 781.25 answer c\"","correct":"c","options":{"a":"rs . 150.50 ","b":"rs . 154.75 ","c":"rs . 781.25 ","d":"rs . 158","e":"none"},"options_float":{"a":150.5,"b":154.75,"c":781.25,"d":158.0,"e":null},"annotated_formula":"divide(845, power(add(divide(4, const_100), const_1), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) + 1<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n845 \/ (676\/625)<\/gadget>\n3_125\/4 = around 781.25<\/output>\n3_125\/4 = around 781.25<\/result>","index":2575} +{"problem":"on the first of the year , james invested x dollars at proudstar bank in an account that yields 4 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ?","rationale":"per quarter , interest = 4 % so for a year , interest = 16 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 16.25 % answer = 16.25 % = answer : d","correct":"d","options":{"a":"22.04 % ","b":"26.12 % ","c":"28 % ","d":"16.25 %","e":"30 %"},"options_float":{"a":22.04,"b":26.12,"c":28.0,"d":16.25,"e":30.0},"annotated_formula":"multiply(4, const_4)","linear_formula":"multiply(n0,const_4)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16<\/result>","index":2576} +{"problem":"if a store owner increases a product ’ s price by 20 percent and then increases this price by another 15 percent , what percent of the original price is the total price increase ?","rationale":"let the initial price be 100 1 st increase = 20 % the price now is 120 2 nd increase of 15 % on 120 = 18 final price = 138 , total increase = 38 % answer : c","correct":"c","options":{"a":"20 % ","b":"35 % ","c":"38 % ","d":"65 %","e":"135 %"},"options_float":{"a":20.0,"b":35.0,"c":38.0,"d":65.0,"e":135.0},"annotated_formula":"add(multiply(add(multiply(const_100, divide(20, const_100)), const_100), divide(15, const_100)), multiply(const_100, divide(20, const_100)))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(#0,const_100)|add(#2,const_100)|multiply(#3,#1)|add(#4,#2)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n100 * (1\/5)<\/gadget>\n20<\/output>\n20 + 100<\/gadget>\n120<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n120 * (3\/20)<\/gadget>\n18<\/output>\n18 + 20<\/gadget>\n38<\/output>\n38<\/result>","index":2577} +{"problem":"if 5 students are made to sit in a row , then 6 do n ' t have seats . if 12 students are made to sit in a row , then 3 rows are empty . how many students have to sit a row such that each row has equal number of students and all students are seated ?","rationale":"let there be a rows . then 5 a + 6 = 12 ( a - 3 ) [ equating the number of students ] = > a = 6 and total number of students = 36 therefore 6 students much sit in each row such that each row has an equal number of students and all students are seated . option ( e )","correct":"e","options":{"a":"0 ","b":"1 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":0.0,"b":1.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"divide(add(multiply(12, 3), 6), subtract(12, 5))","linear_formula":"multiply(n2,n3)|subtract(n2,n0)|add(n1,#0)|divide(#2,#1)","chain":"12 * 3<\/gadget>\n36<\/output>\n36 + 6<\/gadget>\n42<\/output>\n12 - 5<\/gadget>\n7<\/output>\n42 \/ 7<\/gadget>\n6<\/output>\n6<\/result>","index":2578} +{"problem":"a pupil ' s marks were wrongly entered as 85 instead of 33 . due to that the average marks for the class got increased by half . the number of pupils in the class is :","rationale":"\"let there be x pupils in the class . total increase in marks = ( x * 1 \/ 2 ) = x \/ 2 . x \/ 2 = ( 85 - 33 ) = > x \/ 2 = 52 = > x = 104 . answer : b\"","correct":"b","options":{"a":"30 ","b":"104 ","c":"20 ","d":"25","e":"26"},"options_float":{"a":30.0,"b":104.0,"c":20.0,"d":25.0,"e":26.0},"annotated_formula":"multiply(subtract(85, 33), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|","chain":"85 - 33<\/gadget>\n52<\/output>\n52 * 2<\/gadget>\n104<\/output>\n104<\/result>","index":2579} +{"problem":"if the ratio of two number is 6 : 7 and lcm of the number is 450 then what is the number .","rationale":"\"product of two no = lcm * hcf 6 x * 7 x = 450 * x x = 11 answer : a\"","correct":"a","options":{"a":"11 ","b":"20 ","c":"25 ","d":"30","e":"35"},"options_float":{"a":11.0,"b":20.0,"c":25.0,"d":30.0,"e":35.0},"annotated_formula":"divide(450, multiply(6, 7))","linear_formula":"multiply(n0,n1)|divide(n2,#0)|","chain":"6 * 7<\/gadget>\n42<\/output>\n450 \/ 42<\/gadget>\n75\/7 = around 10.714286<\/output>\n75\/7 = around 10.714286<\/result>","index":2580} +{"problem":"jim is able to sell a hand - carved statue for $ 620 which was a 25 % profit over his cost . how much did the statue originally cost him ?","rationale":"\"620 = 1.25 * x x = 620 \/ 1.25 = 496.00 which rounds to $ 496.00 , which is ( a ) .\"","correct":"a","options":{"a":"$ 496.00 ","b":"$ 512.40 ","c":"$ 555.40 ","d":"$ 574.90","e":"$ 588.20"},"options_float":{"a":496.0,"b":512.4,"c":555.4,"d":574.9,"e":588.2},"annotated_formula":"divide(620, add(divide(25, const_100), const_1))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n620 \/ (5\/4)<\/gadget>\n496<\/output>\n496<\/result>","index":2582} +{"problem":"praveen starts business with rs . 3780 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ?","rationale":"\"let hari ’ s capital be rs . x . then , 3780 * 12 \/ 7 x = 2 \/ 3 = > 14 x = 136080 = > x = 9720 . answer : d\"","correct":"d","options":{"a":"s . 7500 ","b":"s . 8000 ","c":"s . 8500 ","d":"s . 9720","e":"s . 6000"},"options_float":{"a":7500.0,"b":8000.0,"c":8500.0,"d":9720.0,"e":6000.0},"annotated_formula":"divide(divide(3780, subtract(const_1, divide(5, const_12))), divide(2, 3))","linear_formula":"divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)|","chain":"5 \/ 12<\/gadget>\n5\/12 = around 0.416667<\/output>\n1 - (5\/12)<\/gadget>\n7\/12 = around 0.583333<\/output>\n3_780 \/ (7\/12)<\/gadget>\n6_480<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n6_480 \/ (2\/3)<\/gadget>\n9_720<\/output>\n9_720<\/result>","index":2583} +{"problem":"a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum marks are ?","rationale":"\"let the maximum marks be x then , 33 % of x = 125 + 40 33 x \/ 100 = 165 x = 500 answer is c\"","correct":"c","options":{"a":"450 ","b":"300 ","c":"500 ","d":"610","e":"175"},"options_float":{"a":450.0,"b":300.0,"c":500.0,"d":610.0,"e":175.0},"annotated_formula":"divide(add(125, 40), divide(33, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"125 + 40<\/gadget>\n165<\/output>\n33 \/ 100<\/gadget>\n33\/100 = around 0.33<\/output>\n165 \/ (33\/100)<\/gadget>\n500<\/output>\n500<\/result>","index":2585} +{"problem":"the telephone company wants to add an area code composed of 2 letters to every phone number . in order to do so , the company chose a special sign language containing 524 different signs . if the company used 522 of the signs fully and two remained unused , how many additional area codes can be created if the company uses all 524 signs ?","rationale":"# of 2 - letter codes possible from 524 different signs = 524 * 524 . # of 2 - letter codes possible from 522 different signs = 522 * 522 . the difference = 524 ^ 2 - 522 ^ 2 = ( 524 - 522 ) ( 524 + 522 ) = 2092 . answer : c .","correct":"c","options":{"a":"46 ","b":"248 ","c":"2092 ","d":"15,128","e":"30,256"},"options_float":{"a":46.0,"b":248.0,"c":2092.0,"d":15128.0,"e":30256.0},"annotated_formula":"multiply(multiply(522, subtract(524, 522)), subtract(524, 522))","linear_formula":"subtract(n1,n2)|multiply(n2,#0)|multiply(#1,#0)","chain":"524 - 522<\/gadget>\n2<\/output>\n522 * 2<\/gadget>\n1_044<\/output>\n1_044 * 2<\/gadget>\n2_088<\/output>\n2_088<\/result>","index":2586} +{"problem":"a case contains c cartons . each carton contains b boxes , and each box contains 200 paper clips . how many paper clips are contained in 2 cases ?","rationale":"\"2 cases * c cartons \/ case * b boxes \/ carton * 200 clips \/ box = 400 bc paper clips the answer is c .\"","correct":"c","options":{"a":"200 bc ","b":"200 b \/ c ","c":"400 bc ","d":"400 b \/ c","e":"400 \/ bc"},"options_float":{"a":200.0,"b":200.0,"c":400.0,"d":400.0,"e":400.0},"annotated_formula":"multiply(2, 200)","linear_formula":"multiply(n0,n1)|","chain":"2 * 200<\/gadget>\n400<\/output>\n400<\/result>","index":2587} +{"problem":"what is the value of x ^ 2 yz − xyz ^ 2 , if x = − 2 , y = 1 , and z = 1 ?","rationale":"4 * 1 * 1 - ( - 2 * 1 * 1 ) = 4 + 2 = 6 ans : a","correct":"a","options":{"a":"6 ","b":"24 ","c":"30 ","d":"32","e":"48"},"options_float":{"a":6.0,"b":24.0,"c":30.0,"d":32.0,"e":48.0},"annotated_formula":"multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1))","linear_formula":"negate(n0)|multiply(n3,#0)|subtract(#0,n3)|multiply(n3,#1)|multiply(#3,#2)","chain":"-2<\/gadget>\n-2<\/output>\n(-2) * 1<\/gadget>\n-2<\/output>\n(-2) - 1<\/gadget>\n-3<\/output>\n(-2) * (-3)<\/gadget>\n6<\/output>\n6<\/result>","index":2588} +{"problem":"2.205 divided by 0.3 gives","rationale":"\"explanation : 2.205 \/ 0.3 = 22.05 \/ 3 = 7.35 . answer : option a\"","correct":"a","options":{"a":"7.35 ","b":"6.25 ","c":"8.15 ","d":"6.95","e":"none of these"},"options_float":{"a":7.35,"b":6.25,"c":8.15,"d":6.95,"e":null},"annotated_formula":"divide(2.205, 0.3)","linear_formula":"divide(n0,n1)|","chain":"2.205 \/ 0.3<\/gadget>\n7.35<\/output>\n7.35<\/result>","index":2589} +{"problem":"having received his weekly allowance , john spent 3 \/ 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.88 at the candy store . what is john ’ s weekly allowance ?","rationale":"\"x = 3 x \/ 5 + 1 \/ 3 * 2 x \/ 5 + 88 4 x \/ 15 = 88 x = 330 = $ 3.30 the answer is c .\"","correct":"c","options":{"a":"$ 2.70 ","b":"$ 3.00 ","c":"$ 3.30 ","d":"$ 3.60","e":"$ 3.90"},"options_float":{"a":2.7,"b":3.0,"c":3.3,"d":3.6,"e":3.9},"annotated_formula":"divide(0.88, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5))))))","linear_formula":"divide(n0,n1)|divide(const_1,n0)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(const_1,#4)|divide(n2,#5)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/3) * (2\/5)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(3\/5) + (2\/15)<\/gadget>\n11\/15 = around 0.733333<\/output>\n1 - (11\/15)<\/gadget>\n4\/15 = around 0.266667<\/output>\n0.88 \/ (4\/15)<\/gadget>\n3.3<\/output>\n3.3<\/result>","index":2590} +{"problem":"exactly 3 \/ 7 of the people in the room are under the age of 21 , and exactly 5 \/ 10 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ?","rationale":"\"the total number of the people in the room must be a multiple of both 7 and 10 ( in order 3 \/ 7 and 5 \/ 10 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 10 , which is 70 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 70 people in the room . therefore there are 3 \/ 7 * 70 = 30 people in the room under the age of 21 . answer : d .\"","correct":"d","options":{"a":"21 ","b":"35 ","c":"39 ","d":"30","e":"65"},"options_float":{"a":21.0,"b":35.0,"c":39.0,"d":30.0,"e":65.0},"annotated_formula":"divide(multiply(multiply(7, 10), 3), 7)","linear_formula":"multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)|","chain":"7 * 10<\/gadget>\n70<\/output>\n70 * 3<\/gadget>\n210<\/output>\n210 \/ 7<\/gadget>\n30<\/output>\n30<\/result>","index":2593} +{"problem":"a 5 - digit number is formed by the digits 2 , 45 , 68 ( each digit used exactly once ) . what is the probability that the number formed is divisible by 4 ?","rationale":"a no . is divisible by 4 if last two digit is divisible by 4 so numbers ending with 2428 , 4852 , 5664 , 6884 are divisible by 4 last two places are fixed by 2428 , 4852 , 5664 , 6884 so remaining 3 places can be filled in 3 ! ways for each total = 8 * 3 ! = 48 = n ( e ) n ( s ) = 5 c 4 = 120 p = 48 \/ 120 = 2 \/ 5 answer : a","correct":"a","options":{"a":"2 \/ 5 ","b":"3 \/ 5 ","c":"4 \/ 5 ","d":"6 \/ 5","e":"7 \/ 5"},"options_float":{"a":0.4,"b":0.6,"c":0.8,"d":1.2,"e":1.4},"annotated_formula":"divide(multiply(factorial(4), 2), factorial(5))","linear_formula":"factorial(n4)|factorial(n0)|multiply(n1,#0)|divide(#2,#1)","chain":"factorial(4)<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\nfactorial(5)<\/gadget>\n120<\/output>\n48 \/ 120<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":2594} +{"problem":"if x is 20 percent more than y and y is 60 percent less than z , then x is what percent of z ?","rationale":"\"z = 100 ; y = 40 so x = 48 x as % of z = 48 \/ 100 * 100 = > 48 % answer will be ( b )\"","correct":"b","options":{"a":"500 % ","b":"48 % ","c":"500 \/ 3 % ","d":"125 %","e":"60 %"},"options_float":{"a":500.0,"b":48.0,"c":166.6666666667,"d":125.0,"e":60.0},"annotated_formula":"multiply(add(const_1, divide(20, const_100)), subtract(const_100, 60))","linear_formula":"divide(n0,const_100)|subtract(const_100,n1)|add(#0,const_1)|multiply(#2,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n100 - 60<\/gadget>\n40<\/output>\n(6\/5) * 40<\/gadget>\n48<\/output>\n48<\/result>","index":2595} +{"problem":"from mumbai railway station , two trains a and b start simultaneously from stations 400 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 40 miles per hour and 40 miles per hour , respectively , how many miles will train a have traveled when the trains pass each other , to the nearest mile ?","rationale":"since we know the distance ( 400 ) and the combined rate ( 80 ) , we plug it into the formula : distance = rate * time 400 = 80 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 400 \/ 80 from dividing 80 on both sides to isolate time in the equation above . time will be 20 hours so now you can plug that in for train a ’ s distance . distance = rate * time distance = 40 * 20 distance = 180 according to answer choice e .","correct":"e","options":{"a":"112 ","b":"133 ","c":"150 ","d":"167","e":"180"},"options_float":{"a":112.0,"b":133.0,"c":150.0,"d":167.0,"e":180.0},"annotated_formula":"multiply(divide(400, add(add(40, multiply(const_2, add(const_4, const_1))), 40)), 40)","linear_formula":"add(const_1,const_4)|multiply(#0,const_2)|add(n1,#1)|add(n1,#2)|divide(n0,#3)|multiply(n1,#4)","chain":"4 + 1<\/gadget>\n5<\/output>\n2 * 5<\/gadget>\n10<\/output>\n40 + 10<\/gadget>\n50<\/output>\n50 + 40<\/gadget>\n90<\/output>\n400 \/ 90<\/gadget>\n40\/9 = around 4.444444<\/output>\n(40\/9) * 40<\/gadget>\n1_600\/9 = around 177.777778<\/output>\n1_600\/9 = around 177.777778<\/result>","index":2597} +{"problem":"there is a church tower 150 feet tall and another catholic tower at a distance of 350 feet from it which is 200 feet tall . there is one each bird sitting on top of both the towers . they fly at a constant speed and time to reach a grain in b \/ w the towers at the same time . at what distance from the church is the grain ?","rationale":"if x ft is the distance of grain from 200 ft tall tower , then 150 ^ 2 + ( 350 - x ) ^ 2 = 200 ^ 2 + x ^ 2 then x = 150 meters answer : c","correct":"c","options":{"a":"170 meters ","b":"160 meters ","c":"150 meters ","d":"140 meters","e":"130 meters"},"options_float":{"a":170.0,"b":160.0,"c":150.0,"d":140.0,"e":130.0},"annotated_formula":"divide(subtract(add(power(150, const_2), power(350, const_2)), power(200, const_2)), multiply(const_2, 350))","linear_formula":"multiply(n1,const_2)|power(n0,const_2)|power(n1,const_2)|power(n2,const_2)|add(#1,#2)|subtract(#4,#3)|divide(#5,#0)","chain":"150 ** 2<\/gadget>\n22_500<\/output>\n350 ** 2<\/gadget>\n122_500<\/output>\n22_500 + 122_500<\/gadget>\n145_000<\/output>\n200 ** 2<\/gadget>\n40_000<\/output>\n145_000 - 40_000<\/gadget>\n105_000<\/output>\n2 * 350<\/gadget>\n700<\/output>\n105_000 \/ 700<\/gadget>\n150<\/output>\n150<\/result>","index":2600} +{"problem":"for 1 rs 4 p interest wat will be for 5000 rs ?","rationale":"for 1 rs , 4 p interest for 5000 rs , x x = 5000 \/ 1 * 4 p = = > 20000 paise to express in rs , 20000 \/ 100 = 200 rs answer : c","correct":"c","options":{"a":"350 rs ","b":"450 rs ","c":"200 rs ","d":"30 rs","e":"448 rs"},"options_float":{"a":350.0,"b":450.0,"c":200.0,"d":30.0,"e":448.0},"annotated_formula":"multiply(5000, divide(4, const_100))","linear_formula":"divide(n1,const_100)|multiply(n2,#0)","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n5_000 * (1\/25)<\/gadget>\n200<\/output>\n200<\/result>","index":2601} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 15 seconds . find the length of the train .","rationale":"\"speed = 60 * ( 5 \/ 18 ) m \/ sec = 50 \/ 3 m \/ sec length of train ( distance ) = speed * time ( 50 \/ 3 ) * 15 = 250 meter answer : e\"","correct":"e","options":{"a":"150 m ","b":"145 m ","c":"240 m ","d":"135 m","e":"250 m"},"options_float":{"a":150.0,"b":145.0,"c":240.0,"d":135.0,"e":250.0},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 15)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"60 * 1_000<\/gadget>\n60_000<\/output>\n60_000 \/ 3_600<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 15<\/gadget>\n250<\/output>\n250<\/result>","index":2602} +{"problem":"a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 600 more . what was the sum ?","rationale":"\"explanation : at 5 % more rate , the increase in s . i for 10 years = rs . 600 ( given ) so , at 5 % more rate , the increase in si for 1 year = 600 \/ 10 = rs . 60 \/ - i . e . rs . 60 is 5 % of the invested sum so , 1 % of the invested sum = 60 \/ 5 therefore , the invested sum = 60 × 100 \/ 5 = rs . 1200 answer : a\"","correct":"a","options":{"a":"rs . 1200 ","b":"rs . 1300 ","c":"rs . 1400 ","d":"rs . 1500","e":"none of these"},"options_float":{"a":1200.0,"b":1300.0,"c":1400.0,"d":1500.0,"e":null},"annotated_formula":"divide(600, multiply(divide(5, const_100), 10))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 10<\/gadget>\n1\/2 = around 0.5<\/output>\n600 \/ (1\/2)<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":2603} +{"problem":"a shop owner sells 60 mtr of cloth and gains sp of 10 mtrs . find the gain % ?","rationale":"\"here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 60 m cloth ) – ( c . p . of 60 m cloth ) selling price of 50 m cloth = selling price of 60 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 50 m cloth = rs . 5000 and s . p . of 50 m cloth = rs . rs . 6000 profit % = 10 \/ 50 × 100 = 20 % 20 profit of 20 % was made by the merchant . a\"","correct":"a","options":{"a":"20 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"70 %"},"options_float":{"a":20.0,"b":40.0,"c":50.0,"d":60.0,"e":70.0},"annotated_formula":"multiply(divide(10, subtract(60, 10)), const_100)","linear_formula":"subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)|","chain":"60 - 10<\/gadget>\n50<\/output>\n10 \/ 50<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":2604} +{"problem":"in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 4 : 7 : 9 . if the shop carries 364 dogs and bunnies total in stock , how many dogs are there ?","rationale":"\"let us assume the number of dogs , cats and bunnies to be 4 x , 7 x and 9 x total dogs and bunnies = 13 x . and we are given that 13 x = 364 . hence x = 28 . dogs = 4 x = 4 * 28 = 112 ( option d )\"","correct":"d","options":{"a":"42 ","b":"66 ","c":"98 ","d":"112","e":"154"},"options_float":{"a":42.0,"b":66.0,"c":98.0,"d":112.0,"e":154.0},"annotated_formula":"multiply(divide(364, add(4, 9)), 4)","linear_formula":"add(n0,n2)|divide(n3,#0)|multiply(n0,#1)|","chain":"4 + 9<\/gadget>\n13<\/output>\n364 \/ 13<\/gadget>\n28<\/output>\n28 * 4<\/gadget>\n112<\/output>\n112<\/result>","index":2606} +{"problem":"albert invested rs . 6500 in a scheme for 2 years at compound interest rate 6.5 % p . a . how much amount will albert get on maturity of the fixed deposit ?","rationale":"amount = [ 6500 * ( 1 + 6.5 \/ 100 ) 2 ] = 6500 * 106.5 \/ 100 * 106.5 \/ 100 = rs . 7372.46 answer : e","correct":"e","options":{"a":"s . 8829 ","b":"s . 7200 ","c":"s . 7200.5 ","d":"s . 4000","e":"s . 7372.46"},"options_float":{"a":8829.0,"b":7200.0,"c":7200.5,"d":4000.0,"e":7372.46},"annotated_formula":"multiply(6500, power(add(const_1, divide(6.5, const_100)), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)","chain":"6.5 \/ 100<\/gadget>\n0.065<\/output>\n1 + 0.065<\/gadget>\n1.065<\/output>\n1.065 ** 2<\/gadget>\n1.134225<\/output>\n6_500 * 1.134225<\/gadget>\n7_372.4625<\/output>\n7_372.4625<\/result>","index":2607} +{"problem":"keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 30 % of the total number of books in the library . if prior to the purchase , only 20 % of the books in keats library were biographies , by what percent has the number of biographies in the library increased ?","rationale":"\"let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.3 ( 5 b + x ) = b + x 0.5 b = 0.7 x x = 0.71 b the answer is c .\"","correct":"c","options":{"a":"47 % ","b":"59 % ","c":"71 % ","d":"83 %","e":"95 %"},"options_float":{"a":47.0,"b":59.0,"c":71.0,"d":83.0,"e":95.0},"annotated_formula":"multiply(divide(divide(subtract(multiply(30, const_100), multiply(20, const_100)), subtract(const_100, 30)), 20), const_100)","linear_formula":"multiply(n0,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#3,#2)|divide(#4,n1)|multiply(#5,const_100)|","chain":"30 * 100<\/gadget>\n3_000<\/output>\n20 * 100<\/gadget>\n2_000<\/output>\n3_000 - 2_000<\/gadget>\n1_000<\/output>\n100 - 30<\/gadget>\n70<\/output>\n1_000 \/ 70<\/gadget>\n100\/7 = around 14.285714<\/output>\n(100\/7) \/ 20<\/gadget>\n5\/7 = around 0.714286<\/output>\n(5\/7) * 100<\/gadget>\n500\/7 = around 71.428571<\/output>\n500\/7 = around 71.428571<\/result>","index":2608} +{"problem":"if the average ( arithmetic mean ) of the 4 numbers 3 , 16 , 33 , and ( n + 1 ) is 20 , then n =","rationale":"3 + 16 + 33 + n + 1 = 20 x 4 = 80 = > n + 53 = 80 = > n = 27 answer d","correct":"d","options":{"a":"19 ","b":"20 ","c":"21 ","d":"27","e":"29"},"options_float":{"a":19.0,"b":20.0,"c":21.0,"d":27.0,"e":29.0},"annotated_formula":"subtract(multiply(20, 4), add(add(add(3, 16), 33), 1))","linear_formula":"add(n1,n2)|multiply(n0,n5)|add(n3,#0)|add(n4,#2)|subtract(#1,#3)","chain":"20 * 4<\/gadget>\n80<\/output>\n3 + 16<\/gadget>\n19<\/output>\n19 + 33<\/gadget>\n52<\/output>\n52 + 1<\/gadget>\n53<\/output>\n80 - 53<\/gadget>\n27<\/output>\n27<\/result>","index":2609} +{"problem":"the mean of 50 observations is 200 . but later he found that there is decrements of 6 from each observations . what is the the updated mean is ?","rationale":"194 answer is e","correct":"e","options":{"a":"165 ","b":"185 ","c":"190 ","d":"198","e":"194"},"options_float":{"a":165.0,"b":185.0,"c":190.0,"d":198.0,"e":194.0},"annotated_formula":"subtract(200, 6)","linear_formula":"subtract(n1,n2)","chain":"200 - 6<\/gadget>\n194<\/output>\n194<\/result>","index":2610} +{"problem":"the ` ` racing magic ' ' takes 120 seconds to circle the racing track once . the ` ` charging bull ' ' makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ?","rationale":"time taken by racing magic to make one circle = 120 seconds = 2 mins time taken by ` ` charging bull ' ' to make one circle = 60 mins \/ 40 = 1.5 mins = 90 seconds lcm of 90 and 120 seconds = 360 seconds = 6 mins time taken for them to meet at the starting point for the second time = 6 mins * 2 = 12 mins answer d","correct":"d","options":{"a":"3 ","b":"6 ","c":"9 ","d":"12","e":"15"},"options_float":{"a":3.0,"b":6.0,"c":9.0,"d":12.0,"e":15.0},"annotated_formula":"add(add(const_2, const_4), add(const_2, const_4))","linear_formula":"add(const_2,const_4)|add(#0,#0)","chain":"2 + 4<\/gadget>\n6<\/output>\n6 + 6<\/gadget>\n12<\/output>\n12<\/result>","index":2613} +{"problem":"find the constant k so that : - x 2 - ( k + 10 ) x - 8 = - ( x - 2 ) ( x - 4 )","rationale":"\"- x 2 - ( k + 10 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 10 ) x - 8 = - x 2 + 6 x - 8 - ( k + 10 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 16 : solve the above for k correct answer c\"","correct":"c","options":{"a":"11 ","b":"12 ","c":"16 ","d":"14","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":16.0,"d":14.0,"e":15.0},"annotated_formula":"add(10, add(4, 2))","linear_formula":"add(n0,n4)|add(n1,#0)|","chain":"4 + 2<\/gadget>\n6<\/output>\n10 + 6<\/gadget>\n16<\/output>\n16<\/result>","index":2614} +{"problem":"102 ã — 102 + 98 ã — 98 = ?","rationale":"\"explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1022 + 982 = ( 100 + 2 ) 2 + ( 100 â ˆ ’ 2 ) 2 = 2 ( 1002 + 22 ) = 2 ( 10000 + 4 ) = 20008 . answer : option e\"","correct":"e","options":{"a":"20018 ","b":"20028 ","c":"20108 ","d":"10008","e":"20008"},"options_float":{"a":20018.0,"b":20028.0,"c":20108.0,"d":10008.0,"e":20008.0},"annotated_formula":"add(subtract(power(102, const_2), 102), subtract(power(98, const_2), 98))","linear_formula":"power(n0,const_2)|power(n2,const_2)|subtract(#0,n0)|subtract(#1,n2)|add(#2,#3)|","chain":"102 ** 2<\/gadget>\n10_404<\/output>\n10_404 - 102<\/gadget>\n10_302<\/output>\n98 ** 2<\/gadget>\n9_604<\/output>\n9_604 - 98<\/gadget>\n9_506<\/output>\n10_302 + 9_506<\/gadget>\n19_808<\/output>\n19_808<\/result>","index":2615} +{"problem":"two trains 140 m and 190 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ?","rationale":"\"relative speed = 60 + 40 = 100 km \/ hr . = 100 * 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = 140 + 190 = 330 m . required time = 330 * 9 \/ 250 = 11.88 sec . answer : a :\"","correct":"a","options":{"a":"11.88 sec ","b":"13.8 sec ","c":"53.8 sec ","d":"10.8 sec","e":"10.4 sec"},"options_float":{"a":11.88,"b":13.8,"c":53.8,"d":10.8,"e":10.4},"annotated_formula":"divide(add(140, 190), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"140 + 190<\/gadget>\n330<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n330 \/ (250\/9)<\/gadget>\n297\/25 = around 11.88<\/output>\n297\/25 = around 11.88<\/result>","index":2616} +{"problem":"mary works in a restaurant a maximum of 70 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ?","rationale":"\"mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 50 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 50 ( 20 ) = $ 500 . the total amount is $ 160 + $ 500 = $ 660 answer c 660 .\"","correct":"c","options":{"a":"600 ","b":"620 ","c":"660 ","d":"320","e":"400"},"options_float":{"a":600.0,"b":620.0,"c":660.0,"d":320.0,"e":400.0},"annotated_formula":"add(multiply(70, 8), multiply(subtract(70, 20), multiply(8, divide(25, const_100))))","linear_formula":"divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4)|","chain":"70 * 8<\/gadget>\n560<\/output>\n70 - 20<\/gadget>\n50<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n8 * (1\/4)<\/gadget>\n2<\/output>\n50 * 2<\/gadget>\n100<\/output>\n560 + 100<\/gadget>\n660<\/output>\n660<\/result>","index":2617} +{"problem":"the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 4 per metre approximately","rationale":"\"explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 \/ 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 \/ 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 4 ) = rs . 5943 . answer : option c\"","correct":"c","options":{"a":"4457 ","b":"4567 ","c":"5943 ","d":"4547","e":"4675"},"options_float":{"a":4457.0,"b":4567.0,"c":5943.0,"d":4547.0,"e":4675.0},"annotated_formula":"multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 4)","linear_formula":"divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)|","chain":"17.56 \/ pi<\/gadget>\n17.56\/pi = around 5.589522<\/output>\n(17.56\/pi) ** (1\/2)<\/gadget>\n4.19046536795139\/sqrt(pi) = around 2.364217<\/output>\n(4.19046536795139\/sqrt(pi)) * 100<\/gadget>\n419.046536795139\/sqrt(pi) = around 236.421691<\/output>\n2 * pi * (419.046536795139\/sqrt(pi))<\/gadget>\n838.093073590278*sqrt(pi) = around 1_485.481296<\/output>\n(838.093073590278*sqrt(pi)) * 4<\/gadget>\n3352.37229436111*sqrt(pi) = around 5_941.925183<\/output>\n3352.37229436111*sqrt(pi) = around 5_941.925183<\/result>","index":2618} +{"problem":"30 pens and 75 pencils were purchased for 750 . if the average price of a pencil was 2.00 , find the average price of a pen .","rationale":"\"since average price of a pencil = 2 ∴ price of 75 pencils = 150 ∴ price of 30 pens = ( 750 – 150 ) = 600 ∴ average price of a pen = 600 ⁄ 60 = 20 answer a\"","correct":"a","options":{"a":"20 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":20.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"divide(subtract(750, multiply(75, 2.00)), 30)","linear_formula":"multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0)|","chain":"75 * 2<\/gadget>\n150<\/output>\n750 - 150<\/gadget>\n600<\/output>\n600 \/ 30<\/gadget>\n20<\/output>\n20<\/result>","index":2620} +{"problem":"the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 16 and 17 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 23 * 16 ) and ( 23 * 17 ) . larger number = ( 23 * 17 ) = 391 . answer : e\"","correct":"e","options":{"a":"338 ","b":"278 ","c":"322 ","d":"231","e":"391"},"options_float":{"a":338.0,"b":278.0,"c":322.0,"d":231.0,"e":391.0},"annotated_formula":"multiply(23, 17)","linear_formula":"multiply(n0,n2)|","chain":"23 * 17<\/gadget>\n391<\/output>\n391<\/result>","index":2621} +{"problem":"to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of bucket is reduced to two - fifth of its present ?","rationale":"explanation : let the capacity of 1 bucket = x . then , capacity of tank = 25 x . new capacity of bucket = 2 \/ 5 x . required no of buckets = 25 x \/ ( 2 x \/ 5 ) = 62.5 answer is b","correct":"b","options":{"a":"52.5 ","b":"62.5 ","c":"72.5 ","d":"82.5","e":"81.5"},"options_float":{"a":52.5,"b":62.5,"c":72.5,"d":82.5,"e":81.5},"annotated_formula":"divide(multiply(25, add(const_4, const_1)), const_2)","linear_formula":"add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)","chain":"4 + 1<\/gadget>\n5<\/output>\n25 * 5<\/gadget>\n125<\/output>\n125 \/ 2<\/gadget>\n125\/2 = around 62.5<\/output>\n125\/2 = around 62.5<\/result>","index":2622} +{"problem":"the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 110 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 \/ 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ?","rationale":"\"if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 110 group # 2 = ( like understood ) = 3 \/ 4 ( 1 \/ 4 ) n = 110 n = 440 answer = ( e )\"","correct":"e","options":{"a":"90 ","b":"120 ","c":"160 ","d":"360","e":"440"},"options_float":{"a":90.0,"b":120.0,"c":160.0,"d":360.0,"e":440.0},"annotated_formula":"divide(110, subtract(const_1, divide(3, 4)))","linear_formula":"divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n110 \/ (1\/4)<\/gadget>\n440<\/output>\n440<\/result>","index":2623} +{"problem":"a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area r covered by the two roads ?","rationale":"\"area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads r = 1000 + 600 - 25 = 1575 answer : option c\"","correct":"c","options":{"a":"400 ","b":"1550 ","c":"1575 ","d":"1600","e":"1625"},"options_float":{"a":400.0,"b":1550.0,"c":1575.0,"d":1600.0,"e":1625.0},"annotated_formula":"add(rectangle_area(200, 5), rectangle_area(120, 5))","linear_formula":"rectangle_area(n0,n2)|rectangle_area(n1,n2)|add(#0,#1)|","chain":"200 * 5<\/gadget>\n1_000<\/output>\n120 * 5<\/gadget>\n600<\/output>\n1_000 + 600<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":2624} +{"problem":"a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 21 days , a finished the remaining work . in how many days a alone can finish the job ?","rationale":"\"a + b 10 days work = 10 * 1 \/ 40 = 1 \/ 4 remaining work = 1 - 1 \/ 4 = 3 \/ 4 3 \/ 4 work is done by a in 21 days whole work will be done by a in 21 * 4 \/ 3 = 28 days answer is e\"","correct":"e","options":{"a":"10 ","b":"25 ","c":"60 ","d":"30","e":"28"},"options_float":{"a":10.0,"b":25.0,"c":60.0,"d":30.0,"e":28.0},"annotated_formula":"divide(multiply(21, 40), subtract(40, 10))","linear_formula":"multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)|","chain":"21 * 40<\/gadget>\n840<\/output>\n40 - 10<\/gadget>\n30<\/output>\n840 \/ 30<\/gadget>\n28<\/output>\n28<\/result>","index":2625} +{"problem":"what will be the cost of building to paint which area equal to 196 sq ft , if the price per foot of building is rs . 15 ?","rationale":"let the side of the square plot be a ft . a 2 = 196 = > a = 14 length of the fence = perimeter of the plot = 4 a = 56 ft . cost of building the fence = 56 * 15 = rs . 840 . answer : e","correct":"e","options":{"a":"120 ","b":"240 ","c":"360 ","d":"380","e":"840"},"options_float":{"a":120.0,"b":240.0,"c":360.0,"d":380.0,"e":840.0},"annotated_formula":"multiply(square_perimeter(sqrt(196)), 15)","linear_formula":"sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)","chain":"196 ** (1\/2)<\/gadget>\n14<\/output>\n4 * 14<\/gadget>\n56<\/output>\n56 * 15<\/gadget>\n840<\/output>\n840<\/result>","index":2628} +{"problem":"exactly 3 sides of a certain 10 - sided die are red . what is the probability that kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll ?","rationale":"total no . of sides = 10 sides that are red = 3 probability that the die lands with red side up = 3 \/ 10 therefore , probability that the die does not land with red side up = 1 - 3 \/ 10 = 7 \/ 10 probability that kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll = ( 1 st roll - non red face ) x ( 2 nd roll - non red face ) x ( 3 rd roll - red face ) = ( 7 \/ 10 ) x ( 7 \/ 10 ) x ( 3 \/ 10 ) = 128 \/ 1000 = 0.147 d","correct":"d","options":{"a":"0.184 ","b":"0.16 ","c":"0.18 ","d":"0.147","e":"0.148"},"options_float":{"a":0.184,"b":0.16,"c":0.18,"d":0.147,"e":0.148},"annotated_formula":"multiply(power(divide(subtract(10, 3), 10), const_2), divide(3, 10))","linear_formula":"divide(n0,n1)|subtract(n1,n0)|divide(#1,n1)|power(#2,const_2)|multiply(#0,#3)","chain":"10 - 3<\/gadget>\n7<\/output>\n7 \/ 10<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) ** 2<\/gadget>\n49\/100 = around 0.49<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(49\/100) * (3\/10)<\/gadget>\n147\/1_000 = around 0.147<\/output>\n147\/1_000 = around 0.147<\/result>","index":2631} +{"problem":"the average runs scored by a batsman in 25 matches is 45 . in the next 7 matches the batsman scored an average of 15 runs . find his average in all the 32 matches ?","rationale":"total score of the batsman in 25 matches = 1125 . total score of the batsman in the next 7 matches = 105 . total score of the batsman in the 32 matches = 1230 . average score of the batsman = 1230 \/ 32 = 38.44 . answer : b","correct":"b","options":{"a":"31355 ","b":"38.44 ","c":"29.55 ","d":"97.66","e":"45.66"},"options_float":{"a":31355.0,"b":38.44,"c":29.55,"d":97.66,"e":45.66},"annotated_formula":"divide(add(multiply(45, 25), multiply(15, 7)), add(25, 7))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"45 * 25<\/gadget>\n1_125<\/output>\n15 * 7<\/gadget>\n105<\/output>\n1_125 + 105<\/gadget>\n1_230<\/output>\n25 + 7<\/gadget>\n32<\/output>\n1_230 \/ 32<\/gadget>\n615\/16 = around 38.4375<\/output>\n615\/16 = around 38.4375<\/result>","index":2632} +{"problem":"carl is facing very difficult financial times and can only pay the interest on a $ 10,000 loan he has taken . the bank charges him a quarterly compound rate of 4 % . what is the approximate interest he pays annually ?","rationale":"\"an easy way too attempt this is by approximation : for the first quarter , he pays 4 % of 10,000 which is $ 400 . so for the four quarters in the year , he will approximately play 400 * 4 = 1600 over the entire year . because of compound interest it will be more than 1600 . approx . option a is correct . answer : a\"","correct":"a","options":{"a":"$ 1750 ","b":"$ 2000 ","c":"$ 2150 ","d":"$ 2500","e":"$ 12000"},"options_float":{"a":1750.0,"b":2000.0,"c":2150.0,"d":2500.0,"e":12000.0},"annotated_formula":"subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(4, const_100)), const_4)), multiply(const_100, const_100))","linear_formula":"divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 4<\/gadget>\n456_976\/390_625 = around 1.169859<\/output>\n10_000 * (456_976\/390_625)<\/gadget>\n7_311_616\/625 = around 11_698.5856<\/output>\n(7_311_616\/625) - 10_000<\/gadget>\n1_061_616\/625 = around 1_698.5856<\/output>\n1_061_616\/625 = around 1_698.5856<\/result>","index":2633} +{"problem":"jack and jill work at a hospital with 7 other workers . for an internal review , 2 of the 9 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ?","rationale":"\"1 \/ 9 c 2 = 1 \/ 36 . answer : e .\"","correct":"e","options":{"a":"1 \/ 3 ","b":"1 \/ 4 ","c":"1 \/ 15 ","d":"3 \/ 8","e":"1 \/ 36"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.0666666667,"d":0.375,"e":0.0277777778},"annotated_formula":"inverse(divide(factorial(9), multiply(factorial(2), factorial(7))))","linear_formula":"factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)|","chain":"factorial(9)<\/gadget>\n362_880<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\nfactorial(7)<\/gadget>\n5_040<\/output>\n2 * 5_040<\/gadget>\n10_080<\/output>\n362_880 \/ 10_080<\/gadget>\n36<\/output>\n1 \/ 36<\/gadget>\n1\/36 = around 0.027778<\/output>\n1\/36 = around 0.027778<\/result>","index":2634} +{"problem":"a student has to obtain 50 % of the total marks to pass . he got 200 marks and failed by 20 marks . the maximum marks are ?","rationale":"\"let the maximum marks be x then , 50 % of x = 200 + 20 50 x \/ 100 = 220 50 x = 220 * 100 = 22000 x = 440 answer is e\"","correct":"e","options":{"a":"510 ","b":"520 ","c":"530 ","d":"540","e":"440"},"options_float":{"a":510.0,"b":520.0,"c":530.0,"d":540.0,"e":440.0},"annotated_formula":"divide(add(200, 20), divide(50, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"200 + 20<\/gadget>\n220<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n220 \/ (1\/2)<\/gadget>\n440<\/output>\n440<\/result>","index":2635} +{"problem":"the cost of 3 pens and 5 pencils is rs . 150 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ?","rationale":"\"explanation : let the cost of one pen is ‘ 5 x ’ and pencil is ‘ x ’ 3 x 5 x + 5 x = rs . 150 15 x + 5 x = rs . 150 x = 150 \/ 20 = 7.50 : . cost of 1 pen = 5 x = 5 x 7.50 = 37.50 : . cost of 12 pens , i . e . ( one dozen ) = 37.50 x 12 = rs . 450 answer : option d\"","correct":"d","options":{"a":"rs . 200 ","b":"rs . 250 ","c":"rs . 300 ","d":"rs . 450","e":"none of these"},"options_float":{"a":200.0,"b":250.0,"c":300.0,"d":450.0,"e":null},"annotated_formula":"multiply(multiply(3, const_4), divide(150, add(3, 1)))","linear_formula":"add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)|","chain":"3 * 4<\/gadget>\n12<\/output>\n3 + 1<\/gadget>\n4<\/output>\n150 \/ 4<\/gadget>\n75\/2 = around 37.5<\/output>\n12 * (75\/2)<\/gadget>\n450<\/output>\n450<\/result>","index":2636} +{"problem":"the tax on a commodity is diminished by 20 % but its consumption is increased by 10 % . find the decrease percent in the revenue derived from it ?","rationale":"\"answer : option d 100 * 100 = 10000 80 * 110 = 8800 10000 - - - - - - - 1200 100 - - - - - - - ? = 12 %\"","correct":"d","options":{"a":"20 % ","b":"19 % ","c":"15 % ","d":"12 %","e":"24 %"},"options_float":{"a":20.0,"b":19.0,"c":15.0,"d":12.0,"e":24.0},"annotated_formula":"subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 20)), const_100))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)|","chain":"100 + 10<\/gadget>\n110<\/output>\n100 - 20<\/gadget>\n80<\/output>\n110 * 80<\/gadget>\n8_800<\/output>\n8_800 \/ 100<\/gadget>\n88<\/output>\n100 - 88<\/gadget>\n12<\/output>\n12<\/result>","index":2640} +{"problem":"a store sells chairs and tables . if the price of 2 chairs and 1 table is 60 % of the price of 1 chair and 2 tables , and the price of 1 table and 1 chair is $ 72 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . )","rationale":"\"let c = chair ; t = table 2 c + 1 t = 0.6 ( 1 c + 2 t ) or c ( 2 - 0.6 ) = 1.2 t - 1 t or 1.4 c = 0.2 t therefore c = 0.2 \/ 1.4 t = 1 \/ 7 t ic + 1 t = 72 or 1 \/ 7 t + 1 t = 60 therefore t = 72 * 7 \/ 8 = 63 c\"","correct":"c","options":{"a":"55 ","b":"84 ","c":"63 ","d":"75","e":"70"},"options_float":{"a":55.0,"b":84.0,"c":63.0,"d":75.0,"e":70.0},"annotated_formula":"divide(subtract(multiply(72, 2), multiply(72, divide(60, const_100))), add(divide(60, const_100), 1))","linear_formula":"divide(n2,const_100)|multiply(n7,n0)|add(#0,n1)|multiply(n7,#0)|subtract(#1,#3)|divide(#4,#2)|","chain":"72 * 2<\/gadget>\n144<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n72 * (3\/5)<\/gadget>\n216\/5 = around 43.2<\/output>\n144 - (216\/5)<\/gadget>\n504\/5 = around 100.8<\/output>\n(3\/5) + 1<\/gadget>\n8\/5 = around 1.6<\/output>\n(504\/5) \/ (8\/5)<\/gadget>\n63<\/output>\n63<\/result>","index":2641} +{"problem":"a small pool filled only with water will require an additional 300 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 300 gallons of water will increase the amount of water in the pool by 25 % , what is the total capacity of the pool in gallons ?","rationale":"\"300 gallons of water increases capacity by 25 % that means 25 % is 300 gallons , so 100 % would be = 300 * 100 \/ 25 = 1200 gallons now 1200 + 300 gallons is 80 % capacity of tank . so 100 % capacity would be = 1500 * 100 \/ 80 = 1875 a is the answer\"","correct":"a","options":{"a":"1875 ","b":"1250 ","c":"1300 ","d":"1600","e":"1625"},"options_float":{"a":1875.0,"b":1250.0,"c":1300.0,"d":1600.0,"e":1625.0},"annotated_formula":"divide(add(divide(multiply(300, const_100), 25), 300), divide(80, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,const_100)|divide(#1,n3)|add(n0,#2)|divide(#3,#0)|","chain":"300 * 100<\/gadget>\n30_000<\/output>\n30_000 \/ 25<\/gadget>\n1_200<\/output>\n1_200 + 300<\/gadget>\n1_500<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1_500 \/ (4\/5)<\/gadget>\n1_875<\/output>\n1_875<\/result>","index":2642} +{"problem":"at 3 : 40 , the hour hand and the minute hand of a clock form an angle of","rationale":"\"angle between hands of a clock when the minute hand is behind the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( h − m \/ 5 ) + m \/ 2 degree when the minute hand is ahead of the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( m \/ 5 − h ) − m \/ 2 degree here h = 3 , m = 40 and minute hand is ahead of the hour hand . hence the angle = 30 ( m \/ 5 − h ) − m \/ 2 = 30 ( 40 \/ 5 − 3 ) − 40 \/ 2 = 30 ( 8 − 3 ) − 20 = 30 × 5 − 20 = 130 ° answer is c .\"","correct":"c","options":{"a":"160 ° ","b":"180 ° ","c":"130 ° ","d":"110 °","e":"150 °"},"options_float":{"a":160.0,"b":180.0,"c":130.0,"d":110.0,"e":150.0},"annotated_formula":"subtract(multiply(multiply(multiply(const_60, 3), const_2), divide(40, const_60)), multiply(divide(divide(multiply(multiply(const_60, 3), const_2), 3), const_4), add(divide(40, const_60), 3)))","linear_formula":"divide(n1,const_60)|multiply(n0,const_60)|add(n0,#0)|multiply(#1,const_2)|divide(#3,n0)|multiply(#0,#3)|divide(#4,const_4)|multiply(#2,#6)|subtract(#5,#7)|","chain":"60 * 3<\/gadget>\n180<\/output>\n180 * 2<\/gadget>\n360<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n360 * (2\/3)<\/gadget>\n240<\/output>\n360 \/ 3<\/gadget>\n120<\/output>\n120 \/ 4<\/gadget>\n30<\/output>\n(2\/3) + 3<\/gadget>\n11\/3 = around 3.666667<\/output>\n30 * (11\/3)<\/gadget>\n110<\/output>\n240 - 110<\/gadget>\n130<\/output>\n130<\/result>","index":2643} +{"problem":"heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 20 litres of tomato juice ?","rationale":"\"answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . ⇒ ⇒ 10 % ( 20 ) = 80 % ( x ) ⇒ ⇒ x = 2.5 liters . answer : b\"","correct":"b","options":{"a":"2.8 liters . ","b":"2.5 liters . ","c":"8.5 liters . ","d":"2.6 liters .","e":"2.1 liters ."},"options_float":{"a":2.8,"b":2.5,"c":8.5,"d":2.6,"e":2.1},"annotated_formula":"divide(multiply(divide(subtract(const_100, 90), const_100), 20), divide(subtract(const_100, 20), const_100))","linear_formula":"subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)|","chain":"100 - 90<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 20<\/gadget>\n2<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n2 \/ (4\/5)<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":2644} +{"problem":"if 2 ^ 5 , 3 ^ 3 , and 14 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ?","rationale":"here 156 has three two ' s two three ' s and one 14 rest of them must be in w so w = 14 * 3 * 4 = 168 smash e","correct":"e","options":{"a":"26 ","b":"39 ","c":"42 ","d":"65","e":"168"},"options_float":{"a":26.0,"b":39.0,"c":42.0,"d":65.0,"e":168.0},"annotated_formula":"multiply(multiply(multiply(power(2, 2), 3), divide(14, 2)), 2)","linear_formula":"divide(n4,n0)|power(n0,n0)|multiply(n2,#1)|multiply(#0,#2)|multiply(n0,#3)","chain":"2 ** 2<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n12 * 7<\/gadget>\n84<\/output>\n84 * 2<\/gadget>\n168<\/output>\n168<\/result>","index":2645} +{"problem":"a and b can finish a work in 10 days while b and c can do it in 18 days . a started the work , worked for 5 days , then b worked for 10 days and the remaining work was finished by c in 15 days . in how many days could c alone have finished the whole work ?","rationale":"let c completes the work in x days . work done by ( a + b ) in 1 day = 1 ⁄ 10 work done by ( b + c ) in 1 day = 1 ⁄ 18 a ’ s 5 days ’ work + b ’ s 10 days ’ work + c ’ s 15 days ’ work = 1 or ( a + b ) ’ s 5 days ’ work + ( b + c ) ’ s 5 days ’ work + c ’ s 10 days ’ work = 1 or 5 ⁄ 10 + 5 ⁄ 18 + 10 ⁄ x = 1 or x = 45 days answer c","correct":"c","options":{"a":"30 days ","b":"15 days ","c":"45 days ","d":"24 days","e":"none of these"},"options_float":{"a":30.0,"b":15.0,"c":45.0,"d":24.0,"e":null},"annotated_formula":"divide(10, subtract(const_1, add(divide(5, 10), divide(5, 18))))","linear_formula":"divide(n2,n0)|divide(n2,n1)|add(#0,#1)|subtract(const_1,#2)|divide(n0,#3)","chain":"5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n5 \/ 18<\/gadget>\n5\/18 = around 0.277778<\/output>\n(1\/2) + (5\/18)<\/gadget>\n7\/9 = around 0.777778<\/output>\n1 - (7\/9)<\/gadget>\n2\/9 = around 0.222222<\/output>\n10 \/ (2\/9)<\/gadget>\n45<\/output>\n45<\/result>","index":2646} +{"problem":"a boat running upstream takes 8 hours 48 minutes to cover a certain distance , while it take 4 hours to cover the same distance running downstream . what is the ratio between the speed of the boat and speed of the water current respectively ?","rationale":"\"sol . let the man rate upstream be x kmph and that downstream be y kmph . then , distance coverd upstream in 8 hrs 48 min . = distance coverd downstream in 4 hrs . ⇒ [ x 8 4 \/ 5 ] = ( y * 4 ) ⇒ 44 \/ 5 x = 4 y ⇒ y = 11 \/ 5 x . ∴ required ratio = [ y + x \/ 2 ] : [ 6 x \/ 5 * 1 \/ 2 ] : [ 6 x \/ 5 * 1 \/ 2 ] = 8 \/ 5 : 3 \/ 5 = 8 : 3 . answer d\"","correct":"d","options":{"a":"1 : 3 ","b":"3 : 8 ","c":"6 : 3 ","d":"8 : 3","e":"none"},"options_float":{"a":0.3333333333,"b":0.375,"c":2.0,"d":2.6666666667,"e":null},"annotated_formula":"divide(divide(add(divide(add(divide(48, const_60), 8), 4), const_1), const_2), divide(subtract(divide(add(divide(48, const_60), 8), 4), const_1), const_2))","linear_formula":"divide(n1,const_60)|add(n0,#0)|divide(#1,n2)|add(#2,const_1)|subtract(#2,const_1)|divide(#3,const_2)|divide(#4,const_2)|divide(#5,#6)|","chain":"48 \/ 60<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) + 8<\/gadget>\n44\/5 = around 8.8<\/output>\n(44\/5) \/ 4<\/gadget>\n11\/5 = around 2.2<\/output>\n(11\/5) + 1<\/gadget>\n16\/5 = around 3.2<\/output>\n(16\/5) \/ 2<\/gadget>\n8\/5 = around 1.6<\/output>\n(11\/5) - 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) \/ 2<\/gadget>\n3\/5 = around 0.6<\/output>\n(8\/5) \/ (3\/5)<\/gadget>\n8\/3 = around 2.666667<\/output>\n8\/3 = around 2.666667<\/result>","index":2647} +{"problem":"if 2 ^ x + 2 ^ y = 10 ^ 0 + 90 ^ 0 , where x and y are non negative integers , what is the greatest possible value of | x â € “ y | ?","rationale":"| x â € “ y | for x and y 2 ^ x + 2 ^ y = 10 ^ 0 + 90 ^ 0 a . x = y x = y = 0 1 + 1 = 1 + 1 right b . 1 x = 2 , y = 1 4 + 2 = 1 + 1 wrong c . 2 x = 3 , y = 1 8 + 2 = 1 + 1 wrong d . 3 x = 4 , y = 1 16 + 2 = 1 + 1 wrong e . 4 x = 5 , y = 1 32 + 2 = 1 + 1 wrong my answer is a .","correct":"a","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"subtract(2, 2)","linear_formula":"subtract(n0,n0)","chain":"2 - 2<\/gadget>\n0<\/output>\n0<\/result>","index":2648} +{"problem":"a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1032 in total compensation , how many total hours did he work that week ?","rationale":"for 40 hrs = 40 * 16 = 640 excess = 1032 - 640 = 392 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 392 \/ 28 = 98 \/ 7 = 14 total hrs = 40 + 14 = 54 answer e 52","correct":"e","options":{"a":"36 ","b":"40 ","c":"44 ","d":"48","e":"54"},"options_float":{"a":36.0,"b":40.0,"c":44.0,"d":48.0,"e":54.0},"annotated_formula":"add(40, divide(subtract(1032, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100)))","linear_formula":"add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)","chain":"16 * 40<\/gadget>\n640<\/output>\n1_032 - 640<\/gadget>\n392<\/output>\n100 + 75<\/gadget>\n175<\/output>\n16 * 175<\/gadget>\n2_800<\/output>\n2_800 \/ 100<\/gadget>\n28<\/output>\n392 \/ 28<\/gadget>\n14<\/output>\n40 + 14<\/gadget>\n54<\/output>\n54<\/result>","index":2649} +{"problem":"find the length of the wire required to go 15 times round a square field containing 69696 m 2 ?","rationale":"\"a 2 = 69696 = > a = 264 4 a = 1056 1056 * 15 = 15840 answer : a\"","correct":"a","options":{"a":"15840 ","b":"19879 ","c":"27997 ","d":"15842","e":"15823"},"options_float":{"a":15840.0,"b":19879.0,"c":27997.0,"d":15842.0,"e":15823.0},"annotated_formula":"multiply(square_perimeter(square_edge_by_area(69696)), 15)","linear_formula":"square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)|","chain":"69_696 ** (1\/2)<\/gadget>\n264<\/output>\n4 * 264<\/gadget>\n1_056<\/output>\n1_056 * 15<\/gadget>\n15_840<\/output>\n15_840<\/result>","index":2650} +{"problem":"p , q and r have rs . 7000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ?","rationale":"\"let the amount with r be rs . r r = 2 \/ 3 ( total amount with p and q ) r = 2 \/ 3 ( 7000 - r ) = > 3 r = 14000 - 2 r = > 5 r = 14000 = > r = 2800 . answer : d\"","correct":"d","options":{"a":"rs . 3000 ","b":"rs . 3600 ","c":"rs . 2400 ","d":"rs . 2800","e":"none of these"},"options_float":{"a":3000.0,"b":3600.0,"c":2400.0,"d":2800.0,"e":null},"annotated_formula":"multiply(divide(7000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3))","linear_formula":"divide(const_2,const_3)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 + (2\/3)<\/gadget>\n5\/3 = around 1.666667<\/output>\n7_000 \/ (5\/3)<\/gadget>\n4_200<\/output>\n4_200 * (2\/3)<\/gadget>\n2_800<\/output>\n2_800<\/result>","index":2651} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 40 overs to reach a target of 282 runs ?","rationale":"\"total runs in the first 10 overs = 10 × 3.2 = 32 run rate required in the remaining 40 overs = 282 − 32 \/ 40 = 250 \/ 40 = 6.25 runs per over . answer a\"","correct":"a","options":{"a":"6.25 ","b":"6.5 ","c":"6.75 ","d":"7.0","e":"none of the above"},"options_float":{"a":6.25,"b":6.5,"c":6.75,"d":7.0,"e":null},"annotated_formula":"divide(subtract(282, multiply(10, 3.2)), 40)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 3.2<\/gadget>\n32<\/output>\n282 - 32<\/gadget>\n250<\/output>\n250 \/ 40<\/gadget>\n25\/4 = around 6.25<\/output>\n25\/4 = around 6.25<\/result>","index":2653} +{"problem":"a and b together can complete a piece of work in 10 days . if a alone can complete the same work in 15 days , in how many days , in how many days can b alone complete that work ?","rationale":"( a + b ) ' s 1 day ' s work = 1 \/ 10 a ' s 1 day ' s work = 1 \/ 15 b ' s 1 days ' s work = ( 1 \/ 10 ) - ( 1 \/ 15 ) = 1 \/ 30 hence , b alone can complete the work in 30 days . answer is b","correct":"b","options":{"a":"10 ","b":"30 ","c":"25 ","d":"15","e":"45"},"options_float":{"a":10.0,"b":30.0,"c":25.0,"d":15.0,"e":45.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 10), divide(const_1, 15)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/10) - (1\/15)<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ (1\/30)<\/gadget>\n30<\/output>\n30<\/result>","index":2654} +{"problem":"a garrison of 2000 men has provisions for 54 days . at the end of 18 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ?","rationale":"\"2000 - - - - 54 2000 - - - - 36 x - - - - - 20 x * 20 = 2000 * 36 x = 3600 2000 - - - - - - - 1600 answer : b\"","correct":"b","options":{"a":"1977 ","b":"1600 ","c":"1979 ","d":"1900","e":"1278"},"options_float":{"a":1977.0,"b":1600.0,"c":1979.0,"d":1900.0,"e":1278.0},"annotated_formula":"subtract(divide(subtract(multiply(2000, 54), multiply(2000, 18)), 20), 2000)","linear_formula":"multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)|","chain":"2_000 * 54<\/gadget>\n108_000<\/output>\n2_000 * 18<\/gadget>\n36_000<\/output>\n108_000 - 36_000<\/gadget>\n72_000<\/output>\n72_000 \/ 20<\/gadget>\n3_600<\/output>\n3_600 - 2_000<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":2657} +{"problem":"the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 75 kg . what might be the weight of the new person ?","rationale":"\"a 95 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 75 + 20 ) kg = 95 kg .\"","correct":"a","options":{"a":"95 kg ","b":"90 kg ","c":"85 kg ","d":"data inadequate","e":"none of these"},"options_float":{"a":95.0,"b":90.0,"c":85.0,"d":null,"e":null},"annotated_formula":"add(multiply(8, 2.5), 75)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"8 * 2.5<\/gadget>\n20<\/output>\n20 + 75<\/gadget>\n95<\/output>\n95<\/result>","index":2658} +{"problem":"when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 109.20 , what was the total value of the item ?","rationale":"\"let x be the value of the item . 0.07 * ( x - 1000 ) = 109.20 x = 2560 the answer is e .\"","correct":"e","options":{"a":"$ 1,600 ","b":"$ 1,850 ","c":"$ 2,250 ","d":"$ 2,440","e":"$ 2,560"},"options_float":{"a":1600.0,"b":1850.0,"c":2250.0,"d":2440.0,"e":2560.0},"annotated_formula":"add(1,000, divide(109.20, divide(7, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,#0)|add(#1,n1)|","chain":"7 \/ 100<\/gadget>\n7\/100 = around 0.07<\/output>\n109.2 \/ (7\/100)<\/gadget>\n1_560<\/output>\n1_000 + 1_560<\/gadget>\n2_560<\/output>\n2_560<\/result>","index":2659} +{"problem":"a can finish a piece of work in 4 days . b can do it in 10 days . they work together for two days and then a goes away . in how many days will b finish the work ?","rationale":"2 \/ 4 + ( 2 + x ) \/ 10 = 1 = > x = 3 days answer : a","correct":"a","options":{"a":"3 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 10), const_2))), divide(const_1, 10))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 2<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/2) + (1\/5)<\/gadget>\n7\/10 = around 0.7<\/output>\n1 - (7\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) \/ (1\/10)<\/gadget>\n3<\/output>\n3<\/result>","index":2662} +{"problem":"if the sides of a triangle are 26 cm , 24 cm and 15 cm , what is its area ?","rationale":"\"the triangle with sides 26 cm , 24 cm and 15 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 \/ 2 * 24 * 15 = 180 cm 2 answer : c\"","correct":"c","options":{"a":"120 ","b":"772 ","c":"180 ","d":"266","e":"2848"},"options_float":{"a":120.0,"b":772.0,"c":180.0,"d":266.0,"e":2848.0},"annotated_formula":"divide(multiply(24, 15), const_2)","linear_formula":"multiply(n1,n2)|divide(#0,const_2)|","chain":"24 * 15<\/gadget>\n360<\/output>\n360 \/ 2<\/gadget>\n180<\/output>\n180<\/result>","index":2663} +{"problem":"two trains , a and b , started simultaneously from opposite ends of a 125 - mile route and traveled toward each other on parallel tracks . train a , traveling at a constant rate , completed the 125 - mile trip in 12 hours ; train b , traveling at a constant rate , completed the 125 - mile trip in 8 hours . how many miles had train a traveled when it met train b ?","rationale":"as the ratio of the rates of a and b is 8 to 12 then the distance covered at the time of the meeting ( so after traveling the same time interval ) would also be in that ratio , which means that a would cover 8 \/ ( 8 + 12 ) = 8 \/ 20 of 125 miles : 125 * 8 \/ 20 = 50 miles . answer : e .","correct":"e","options":{"a":"25 ","b":"37.5 ","c":"40 ","d":"45.5","e":"50"},"options_float":{"a":25.0,"b":37.5,"c":40.0,"d":45.5,"e":50.0},"annotated_formula":"multiply(divide(8, add(12, 8)), 125)","linear_formula":"add(n2,n4)|divide(n4,#0)|multiply(n0,#1)","chain":"12 + 8<\/gadget>\n20<\/output>\n8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 125<\/gadget>\n50<\/output>\n50<\/result>","index":2664} +{"problem":"if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2400 less than the principal . what was the principal ?","rationale":"\"p - 2400 = ( p * 5 * 4 ) \/ 100 p = 3000 answer : d\"","correct":"d","options":{"a":"1500 ","b":"2500 ","c":"2507 ","d":"3000","e":"11500"},"options_float":{"a":1500.0,"b":2500.0,"c":2507.0,"d":3000.0,"e":11500.0},"annotated_formula":"divide(2400, subtract(const_1, divide(multiply(4, 5), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|","chain":"4 * 5<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n2_400 \/ (4\/5)<\/gadget>\n3_000<\/output>\n3_000<\/result>","index":2665} +{"problem":"out of 250 students of a college , 160 play football , 90 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ?","rationale":"explanation : solution : n ( a ) = 160 , n ( b ) = 90 , n ( a u b ) = 250 - 50 = 200 . required number = > n ( a ^ b ) = n ( a ) + n ( b ) - n ( a u b ) = 160 + 90 - 200 = 250 - 200 = 50 . answer : c","correct":"c","options":{"a":"30 ","b":"40 ","c":"50 ","d":"60","e":"70"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":70.0},"annotated_formula":"subtract(add(50, add(160, 90)), 250)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)","chain":"160 + 90<\/gadget>\n250<\/output>\n50 + 250<\/gadget>\n300<\/output>\n300 - 250<\/gadget>\n50<\/output>\n50<\/result>","index":2666} +{"problem":"what is the area a of the square with the following coordinates : ( x , y ) , ( 20 , 20 ) , ( 20 , 5 ) , ( x , 5 ) ?","rationale":"length of one side = 15 ( 20 - 5 ) since , its a square , the area will be a = 15 ^ 2 = 225 d is the answer","correct":"d","options":{"a":"60 . ","b":"85 . ","c":"125 . ","d":"225 .","e":"it can not be determined from the information given"},"options_float":{"a":60.0,"b":85.0,"c":125.0,"d":225.0,"e":null},"annotated_formula":"square_area(subtract(20, 5))","linear_formula":"subtract(n0,n3)|square_area(#0)","chain":"20 - 5<\/gadget>\n15<\/output>\n15 ** 2<\/gadget>\n225<\/output>\n225<\/result>","index":2667} +{"problem":"a van takes 5 hours to cover a distance of 495 km . what speed in kph should the van maintain to cover the same distance in 3 \/ 2 of the previous time ?","rationale":"\"( 3 \/ 2 ) * 5 = 7.5 hours 495 \/ 7.5 = 66 kph the answer is c .\"","correct":"c","options":{"a":"60 ","b":"63 ","c":"66 ","d":"69","e":"72"},"options_float":{"a":60.0,"b":63.0,"c":66.0,"d":69.0,"e":72.0},"annotated_formula":"divide(495, multiply(divide(3, 2), 5))","linear_formula":"divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 5<\/gadget>\n15\/2 = around 7.5<\/output>\n495 \/ (15\/2)<\/gadget>\n66<\/output>\n66<\/result>","index":2668} +{"problem":"find the compound interest on $ 1000 in 2 years at 4 % per annum , the interest being compounded half - yearly ?","rationale":"\"principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 1000 * ( 1 + 2 \/ 100 ) ^ 4 = 1000 * 51 \/ 50 * 51 \/ 50 * 51 \/ 50 * 51 \/ 50 = $ 1082.43 c . i . = 1082.43 - 1000 = $ 82.43 answer is b\"","correct":"b","options":{"a":"$ 645.56 ","b":"$ 82.43 ","c":"$ 954.26 ","d":"$ 745.69","e":"$ 1020.45"},"options_float":{"a":645.56,"b":82.43,"c":954.26,"d":745.69,"e":1020.45},"annotated_formula":"subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 1000), 1000)","linear_formula":"divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) \/ 2<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) + 1<\/gadget>\n51\/50 = around 1.02<\/output>\n(51\/50) ** 4<\/gadget>\n6_765_201\/6_250_000 = around 1.082432<\/output>\n(6_765_201\/6_250_000) * 1_000<\/gadget>\n6_765_201\/6_250 = around 1_082.43216<\/output>\n(6_765_201\/6_250) - 1_000<\/gadget>\n515_201\/6_250 = around 82.43216<\/output>\n515_201\/6_250 = around 82.43216<\/result>","index":2670} +{"problem":"a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 15 seconds . what is the length of the platform in meters ?","rationale":"speed of the train in metres \/ sec = 72000 \/ 3600 = 20 distance travelled by train to cross the platform = 30 * 20 = 600 = length of train + length of platform distance travelled by train to cross the man = 15 * 20 = 300 = length of train length of platform = 600 - 300 = 300 answer : c","correct":"c","options":{"a":"240 meters ","b":"360 meters ","c":"300 meters ","d":"600 meters","e":"can not be determined"},"options_float":{"a":240.0,"b":360.0,"c":300.0,"d":600.0,"e":null},"annotated_formula":"subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 15))","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)","chain":"72 * 1_000<\/gadget>\n72_000<\/output>\n72_000 \/ 3_600<\/gadget>\n20<\/output>\n20 * 30<\/gadget>\n600<\/output>\n20 * 15<\/gadget>\n300<\/output>\n600 - 300<\/gadget>\n300<\/output>\n300<\/result>","index":2675} +{"problem":"brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 50 meters . they next meet after sally has run 150 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ?","rationale":"\"nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 50 meters , thus second time she runs 2 * 50 = 100 meters . since second time ( when they run full circumference ) brenda runs 200 meters and sally runs 150 meters , thus the circumference is 100 + 150 = 250 meters . answer : a .\"","correct":"a","options":{"a":"250 ","b":"300 ","c":"350 ","d":"400","e":"500"},"options_float":{"a":250.0,"b":300.0,"c":350.0,"d":400.0,"e":500.0},"annotated_formula":"add(multiply(const_2, 50), 150)","linear_formula":"multiply(n0,const_2)|add(n1,#0)|","chain":"2 * 50<\/gadget>\n100<\/output>\n100 + 150<\/gadget>\n250<\/output>\n250<\/result>","index":2676} +{"problem":"a man buys a cycle for rs . 1900 and sells it at a loss of 18 % . what is the selling price of the cycle ?","rationale":"\"s . p . = 82 % of rs . 1900 = rs . 82 x 1900 \/ 100 = rs . 1558 answer : option c\"","correct":"c","options":{"a":"s . 1590 ","b":"s . 1560 ","c":"s . 1558 ","d":"s . 1252","e":"s . 1062"},"options_float":{"a":1590.0,"b":1560.0,"c":1558.0,"d":1252.0,"e":1062.0},"annotated_formula":"divide(multiply(subtract(const_100, 18), 1900), const_100)","linear_formula":"subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|","chain":"100 - 18<\/gadget>\n82<\/output>\n82 * 1_900<\/gadget>\n155_800<\/output>\n155_800 \/ 100<\/gadget>\n1_558<\/output>\n1_558<\/result>","index":2677} +{"problem":"at the end of the day , february 14 th , a florist had 120 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 20 of which were white and 10 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ?","rationale":"\"r + w + p = 120 s + l = 120 1 \/ 3 * 120 = 40 short - stemmed white = 20 short - stemmed pink = 10 = > short - stemmed red = 10 10 \/ p = 10 \/ r = > r = p so total long stemmed = 80 and long stemmed red + long stemmed pink = 80 so long stemmed red \/ long stemmed = ? total white = 20 ( as no long stemmed white ) = > r + r + 20 = 120 = > 2 r = 100 and r = 50 long stemmed r = 50 - 10 = 40 so long stemmed red \/ r = 40 \/ 80 = 50 % answer - b\"","correct":"b","options":{"a":"20 % ","b":"50 % ","c":"55 % ","d":"75 %","e":"80 %"},"options_float":{"a":20.0,"b":50.0,"c":55.0,"d":75.0,"e":80.0},"annotated_formula":"multiply(divide(divide(multiply(subtract(120, divide(120, const_3)), subtract(subtract(divide(120, const_3), 20), 10)), add(10, subtract(subtract(divide(120, const_3), 20), 10))), subtract(120, divide(120, const_3))), const_100)","linear_formula":"divide(n1,const_3)|subtract(n1,#0)|subtract(#0,n2)|subtract(#2,n3)|add(n3,#3)|multiply(#1,#3)|divide(#5,#4)|divide(#6,#1)|multiply(#7,const_100)|","chain":"120 \/ 3<\/gadget>\n40<\/output>\n120 - 40<\/gadget>\n80<\/output>\n40 - 20<\/gadget>\n20<\/output>\n20 - 10<\/gadget>\n10<\/output>\n80 * 10<\/gadget>\n800<\/output>\n10 + 10<\/gadget>\n20<\/output>\n800 \/ 20<\/gadget>\n40<\/output>\n40 \/ 80<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":2678} +{"problem":"the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 27 . what is the difference between the two digits of that number ?","rationale":"\"sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 27 ⇔ 9 ( x - y ) = 27 ⇔ x - y = 3 answer a\"","correct":"a","options":{"a":"3 ","b":"5 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"divide(27, subtract(const_10, const_1))","linear_formula":"subtract(const_10,const_1)|divide(n0,#0)|","chain":"10 - 1<\/gadget>\n9<\/output>\n27 \/ 9<\/gadget>\n3<\/output>\n3<\/result>","index":2679} +{"problem":"a fair price shopkeeper takes 10 % profit on his goods . he lost 60 % goods during theft . his loss percent is :","rationale":"\"explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 40 . s . p . of each item = rs . 1.10 total sale = 1.10 * 40 = rs . 44 hence , loss % = 56 \/ 100 * 100 = 56 % answer : c\"","correct":"c","options":{"a":"72 % ","b":"42 % ","c":"56 % ","d":"12 %","e":"22 %"},"options_float":{"a":72.0,"b":42.0,"c":56.0,"d":12.0,"e":22.0},"annotated_formula":"subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 60), const_100)))","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)|","chain":"100 + 10<\/gadget>\n110<\/output>\n110 * 60<\/gadget>\n6_600<\/output>\n6_600 \/ 100<\/gadget>\n66<\/output>\n110 - 66<\/gadget>\n44<\/output>\n100 - 44<\/gadget>\n56<\/output>\n56<\/result>","index":2680} +{"problem":"a corporation that had $ 115.19 billion in profits for the year paid out $ 230.10 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 )","rationale":"required answer = [ employee benefit \/ profit ] * 100 = [ ( 230.10 million ) \/ ( 115.19 billion ) ] * 100 = [ ( 230.10 * 10 ^ 6 ) \/ ( 115.19 * 10 ^ 9 ) ] * 100 = ( 2 \/ 1000 ) * 100 = 0.2 % so answer is ( e )","correct":"e","options":{"a":"50 % ","b":"20 % ","c":"5 % ","d":"2 %","e":"0.2 %"},"options_float":{"a":50.0,"b":20.0,"c":5.0,"d":2.0,"e":0.2},"annotated_formula":"multiply(divide(multiply(230.1, power(10, add(const_3, const_3))), multiply(115.19, power(10, 9))), const_100)","linear_formula":"add(const_3,const_3)|power(n3,n4)|multiply(n0,#1)|power(n3,#0)|multiply(n1,#3)|divide(#4,#2)|multiply(#5,const_100)","chain":"3 + 3<\/gadget>\n6<\/output>\n10 ** 6<\/gadget>\n1_000_000<\/output>\n230.1 * 1_000_000<\/gadget>\n230_100_000<\/output>\n10 ** 9<\/gadget>\n1_000_000_000<\/output>\n115.19 * 1_000_000_000<\/gadget>\n115_190_000_000<\/output>\n230_100_000 \/ 115_190_000_000<\/gadget>\n2_301\/1_151_900 = around 0.001998<\/output>\n(2_301\/1_151_900) * 100<\/gadget>\n2_301\/11_519 = around 0.199757<\/output>\n2_301\/11_519 = around 0.199757<\/result>","index":2681} +{"problem":"39 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ?","rationale":"\"explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 39 : : 18 : x [ please pay attention , we have written 27 : 39 rather than 39 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \\ color { blue } x = \\ frac { 39 \\ times 18 } { 27 } } x = 26 so 26 days will be required to get work done by 27 men . answer : c\"","correct":"c","options":{"a":"24 ","b":"77 ","c":"26 ","d":"25","e":"13"},"options_float":{"a":24.0,"b":77.0,"c":26.0,"d":25.0,"e":13.0},"annotated_formula":"divide(multiply(18, 39), 27)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"18 * 39<\/gadget>\n702<\/output>\n702 \/ 27<\/gadget>\n26<\/output>\n26<\/result>","index":2682} +{"problem":"a metallic sphere of radius 24 cm is melted and drawn into a wire , whose radius of cross section is 16 cm . what is the length of the wire ?","rationale":"\"volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 16 ) ^ 2 * h = ( 4 \/ 3 ) π ( 24 ) ^ 3 = > h = 72 cm answer : e\"","correct":"e","options":{"a":"26 cm ","b":"45 cm ","c":"58 cm ","d":"63 cm","e":"72 cm"},"options_float":{"a":26.0,"b":45.0,"c":58.0,"d":63.0,"e":72.0},"annotated_formula":"divide(multiply(const_4, divide(power(24, const_3), power(16, const_2))), const_3)","linear_formula":"power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)|","chain":"24 ** 3<\/gadget>\n13_824<\/output>\n16 ** 2<\/gadget>\n256<\/output>\n13_824 \/ 256<\/gadget>\n54<\/output>\n4 * 54<\/gadget>\n216<\/output>\n216 \/ 3<\/gadget>\n72<\/output>\n72<\/result>","index":2684} +{"problem":"how many multiples of 4 are there between 32 and 252 ?","rationale":"\"it should be mentioned whether 32 and 252 are inclusive . if 32 and 252 are inclusive , then the answer is ( 252 - 32 ) \/ 4 + 1 = 56 . if 32 and 252 are not inclusive , then the answer is ( 248 - 36 ) \/ 4 + 1 = 54 . since oa is c , then we have not inclusive case .\"","correct":"c","options":{"a":"67 ","b":"64 ","c":"54 ","d":"56","e":"60"},"options_float":{"a":67.0,"b":64.0,"c":54.0,"d":56.0,"e":60.0},"annotated_formula":"add(divide(subtract(252, 32), 4), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"252 - 32<\/gadget>\n220<\/output>\n220 \/ 4<\/gadget>\n55<\/output>\n55 + 1<\/gadget>\n56<\/output>\n56<\/result>","index":2686} +{"problem":"set # 1 = { a , b , c , d , e , f } set # 2 = { k , l , m , n , o , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ?","rationale":"so not a vowel in set - 1 : 4 \/ 6 and not a vowel in ser - 2 : 5 \/ 6 now , 4 \/ 6 ∗ 5 \/ 6 = 5 \/ 9 this is for not a vowel . then for at least one vowel will be = 1 − 5 \/ 9 = 4 \/ 9 answer will be e .","correct":"e","options":{"a":"1 \/ 6 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"2 \/ 3","e":"4 \/ 9"},"options_float":{"a":0.1666666667,"b":0.3333333333,"c":0.5,"d":0.6666666667,"e":0.4444444444},"annotated_formula":"subtract(1, multiply(divide(const_4, add(const_4, const_2)), divide(add(const_3, const_2), add(const_4, const_2))))","linear_formula":"add(const_2,const_4)|add(const_2,const_3)|divide(const_4,#0)|divide(#1,#0)|multiply(#2,#3)|subtract(n0,#4)","chain":"4 + 2<\/gadget>\n6<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(2\/3) * (5\/6)<\/gadget>\n5\/9 = around 0.555556<\/output>\n1 - (5\/9)<\/gadget>\n4\/9 = around 0.444444<\/output>\n4\/9 = around 0.444444<\/result>","index":2688} +{"problem":"if a no . when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing thesame no . by 31","rationale":"p ÷ 44 = 432 = > p = 432 * 44 = 19008 p \/ 31 = 19008 \/ 31 = 613 , remainder = 5 c","correct":"c","options":{"a":"2 ","b":"4 ","c":"5 ","d":"7","e":"8"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":7.0,"e":8.0},"annotated_formula":"reminder(multiply(432, 44), 31)","linear_formula":"multiply(n0,n1)|reminder(#0,n3)","chain":"432 * 44<\/gadget>\n19_008<\/output>\n19_008 % 31<\/gadget>\n5<\/output>\n5<\/result>","index":2689} +{"problem":"the age of somu is one - third his father ' s . 7 years back he was one - fifth of his father ' s age . what is his persent age ?","rationale":"\"explanation : let somu ' s age be x and that of his father be 3 x . so , x - 7 = 3 x - 7 \/ 5 = x = 14 answer : option c\"","correct":"c","options":{"a":"11 ","b":"13 ","c":"14 ","d":"12","e":"10"},"options_float":{"a":11.0,"b":13.0,"c":14.0,"d":12.0,"e":10.0},"annotated_formula":"divide(subtract(multiply(add(const_4, const_1), 7), 7), subtract(add(const_4, const_1), const_3))","linear_formula":"add(const_1,const_4)|multiply(n0,#0)|subtract(#0,const_3)|subtract(#1,n0)|divide(#3,#2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 7<\/gadget>\n35<\/output>\n35 - 7<\/gadget>\n28<\/output>\n5 - 3<\/gadget>\n2<\/output>\n28 \/ 2<\/gadget>\n14<\/output>\n14<\/result>","index":2690} +{"problem":"sam ’ s car was fined when he gave joe and peter a ride , so they decided to help sam pay the fine . joe paid $ 6 more than 1 \/ 4 of the fine and peter paid $ 6 less than 1 \/ 3 of the fine , leaving pay $ 10 less than 1 \/ 2 the fine to complete the payment . what fraction of the fine did sam pay ?","rationale":"\"call the fine f . joe paid ( 1 \/ 4 ) f + 4 and peter paid ( 1 \/ 3 ) f – 4 , leaving ( 1 \/ 2 ) f – 5 left . if we add those three up , they should add up to f . f = [ ( 1 \/ 4 ) f + 6 ] + [ ( 1 \/ 3 ) f – 6 ] + [ ( 1 \/ 2 ) f – 10 ] f = ( 1 \/ 4 ) f + ( 1 \/ 3 ) f + ( 1 \/ 2 ) f – 10 multiply all terms by 12 to clear the fractions . 12 f = 3 f + 4 f + 6 f – 120 12 f = 13 f – 120 – f = – 120 f = 120 well , if the fine cost $ 120 , then sam paid the part not covered by joe or peter . half the fine is $ 60 , and sam paid $ 10 less than this : $ 50 . answer = a\"","correct":"a","options":{"a":"$ 50 ","b":"$ 40 ","c":"$ 35 ","d":"$ 44","e":"$ 45"},"options_float":{"a":50.0,"b":40.0,"c":35.0,"d":44.0,"e":45.0},"annotated_formula":"add(multiply(add(multiply(add(divide(1, 2), add(divide(1, 4), divide(1, 3))), 10), 10), 2), const_10)","linear_formula":"divide(n1,n2)|divide(n1,n5)|divide(n1,n8)|add(#0,#1)|add(#3,#2)|multiply(n6,#4)|add(n6,#5)|multiply(n8,#6)|add(#7,const_10)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/4) + (1\/3)<\/gadget>\n7\/12 = around 0.583333<\/output>\n(1\/2) + (7\/12)<\/gadget>\n13\/12 = around 1.083333<\/output>\n(13\/12) * 10<\/gadget>\n65\/6 = around 10.833333<\/output>\n(65\/6) + 10<\/gadget>\n125\/6 = around 20.833333<\/output>\n(125\/6) * 2<\/gadget>\n125\/3 = around 41.666667<\/output>\n(125\/3) + 10<\/gadget>\n155\/3 = around 51.666667<\/output>\n155\/3 = around 51.666667<\/result>","index":2691} +{"problem":"the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 91 per day . what was his wage on the 8 th day ?","rationale":"\"average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 91 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 91 * 7 ) = 1350 - ( 609 + 637 ) = 1350 - 1246 = 104 answer e\"","correct":"e","options":{"a":"$ 83 ","b":"$ 90 ","c":"$ 92 ","d":"$ 97","e":"$ 104"},"options_float":{"a":83.0,"b":90.0,"c":92.0,"d":97.0,"e":104.0},"annotated_formula":"subtract(multiply(90, 15), add(multiply(87, 7), multiply(91, 7)))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"90 * 15<\/gadget>\n1_350<\/output>\n87 * 7<\/gadget>\n609<\/output>\n91 * 7<\/gadget>\n637<\/output>\n609 + 637<\/gadget>\n1_246<\/output>\n1_350 - 1_246<\/gadget>\n104<\/output>\n104<\/result>","index":2692} +{"problem":"in a class , there are 18 boys who are over 160 cm tall . if these constitute 3 - fourths of the boys and the total number of boys is two - thirds of the total number of students in the class , what is the number of girls in the class ?","rationale":"let the number of boys be x . then , ( 3 \/ 4 ) x = 18 or x = 18 x ( 4 \/ 3 ) = 24 . if total number of students is y , then ( 2 \/ 3 ) y = 24 or y = 24 x ( 3 \/ 2 ) = 36 . therefore number of girls in the class = ( 36 - 24 ) = 12 . answer : b","correct":"b","options":{"a":"11 ","b":"12 ","c":"28 ","d":"27","e":"19"},"options_float":{"a":11.0,"b":12.0,"c":28.0,"d":27.0,"e":19.0},"annotated_formula":"subtract(divide(divide(18, divide(3, const_4)), divide(const_2, 3)), divide(18, divide(3, const_4)))","linear_formula":"divide(n2,const_4)|divide(const_2,n2)|divide(n0,#0)|divide(#2,#1)|subtract(#3,#2)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n18 \/ (3\/4)<\/gadget>\n24<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n24 \/ (2\/3)<\/gadget>\n36<\/output>\n36 - 24<\/gadget>\n12<\/output>\n12<\/result>","index":2694} +{"problem":"the perimeter of a triangle is 40 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ?","rationale":"\"area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 40 \/ 2 = 50 cm 2 answer : e\"","correct":"e","options":{"a":"87 cm 2 ","b":"28 cm 2 ","c":"98 cm 2 ","d":"26 cm 2","e":"50 cm 2"},"options_float":{"a":87.0,"b":28.0,"c":98.0,"d":26.0,"e":50.0},"annotated_formula":"triangle_area(2.5, 40)","linear_formula":"triangle_area(n0,n1)|","chain":"(2.5 * 40) \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":2696} +{"problem":"if 75 percent of a class answered the first question on a certain test correctly , 70 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ?","rationale":"{ total } = { first } + { second } - { both } + { neither } 100 % = 75 % + 70 % - { both } + 20 % - - > { both } = 65 % . answer : e .","correct":"e","options":{"a":"10 % ","b":"( 8 ) 20 % ","c":"30 % ","d":"50 %","e":"65 %"},"options_float":{"a":10.0,"b":8.0,"c":30.0,"d":50.0,"e":65.0},"annotated_formula":"subtract(add(add(75, 70), 20), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,const_100)","chain":"75 + 70<\/gadget>\n145<\/output>\n145 + 20<\/gadget>\n165<\/output>\n165 - 100<\/gadget>\n65<\/output>\n65<\/result>","index":2697} +{"problem":"if a man lost 4 % by selling oranges at the rate of 72 a rupee at how many a rupee must he sell them to gain 44 % ?","rationale":"\"96 % - - - - 72 144 % - - - - ? 96 \/ 144 * 72 = 48 answer : b\"","correct":"b","options":{"a":"41 ","b":"48 ","c":"49 ","d":"44","e":"43"},"options_float":{"a":41.0,"b":48.0,"c":49.0,"d":44.0,"e":43.0},"annotated_formula":"divide(multiply(subtract(const_100, 4), 72), add(const_100, 44))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|","chain":"100 - 4<\/gadget>\n96<\/output>\n96 * 72<\/gadget>\n6_912<\/output>\n100 + 44<\/gadget>\n144<\/output>\n6_912 \/ 144<\/gadget>\n48<\/output>\n48<\/result>","index":2698} +{"problem":"| 5 - 2 | - | 6 - 9 | = ?","rationale":"\"| 5 - 2 | - | 6 - 9 | = | 3 | - | - 3 | = 3 - 3 = 0 correct answer d\"","correct":"d","options":{"a":"3 ","b":"2 ","c":"1 ","d":"0","e":"4"},"options_float":{"a":3.0,"b":2.0,"c":1.0,"d":0.0,"e":4.0},"annotated_formula":"subtract(subtract(5, 2), subtract(9, 6))","linear_formula":"subtract(n0,n1)|subtract(n3,n2)|subtract(#0,#1)|","chain":"5 - 2<\/gadget>\n3<\/output>\n9 - 6<\/gadget>\n3<\/output>\n3 - 3<\/gadget>\n0<\/output>\n0<\/result>","index":2700} +{"problem":"find value of x : 121 × 54 = x","rationale":"121 × 54 = 121 × ( 102 ) 4 = 121 × 1000016 = 7.5625 × 10000 = 75625 e","correct":"e","options":{"a":"34545 ","b":"46576 ","c":"67887 ","d":"57687","e":"75625"},"options_float":{"a":34545.0,"b":46576.0,"c":67887.0,"d":57687.0,"e":75625.0},"annotated_formula":"multiply(add(add(121, 121), divide(multiply(121, 121), const_2)), const_10)","linear_formula":"add(n0,n0)|multiply(n0,n0)|divide(#1,const_2)|add(#0,#2)|multiply(#3,const_10)","chain":"121 + 121<\/gadget>\n242<\/output>\n121 * 121<\/gadget>\n14_641<\/output>\n14_641 \/ 2<\/gadget>\n14_641\/2 = around 7_320.5<\/output>\n242 + (14_641\/2)<\/gadget>\n15_125\/2 = around 7_562.5<\/output>\n(15_125\/2) * 10<\/gadget>\n75_625<\/output>\n75_625<\/result>","index":2701} +{"problem":"aishwarya rai walks into a bank to cash out her check . by mistake the bank teller gives her dollar amount in change , and her cent amount in dollars . on the way home she spends 5 cent , and then suddenly she notices that she has twice the amount of her check . how much was her check amount ?","rationale":"e the check was for dollars 31.63 . the bank teller gave her dollars 63.31 she spent . 05 , and then she had dollars 63.26 , which is twice the check . let x be the dollars of the check , and y be the cent . the check was for 100 x + y cent he was given 100 y + x cent also 100 y + x - 5 = 2 ( 100 x + y ) expanding this out and rearranging , we find : 98 y = 199 x + 5 or 199 x ≡ - 5 ( mod 98 ) or 98 * 2 * x + 3 x ≡ - 5 ( mod 98 ) 3 x ≡ - 5 ≡ 93 ( mod 98 ) this quickly leads to x = 31","correct":"e","options":{"a":"21.55 ","b":"25.89 ","c":"30.56 ","d":"31.55","e":"31.63"},"options_float":{"a":21.55,"b":25.89,"c":30.56,"d":31.55,"e":31.63},"annotated_formula":"divide(multiply(subtract(const_1, divide(5, subtract(const_100, const_2))), subtract(const_100, const_2)), const_3)","linear_formula":"subtract(const_100,const_2)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,#0)|divide(#3,const_3)","chain":"100 - 2<\/gadget>\n98<\/output>\n5 \/ 98<\/gadget>\n5\/98 = around 0.05102<\/output>\n1 - (5\/98)<\/gadget>\n93\/98 = around 0.94898<\/output>\n(93\/98) * 98<\/gadget>\n93<\/output>\n93 \/ 3<\/gadget>\n31<\/output>\n31<\/result>","index":2704} +{"problem":"the length of a room is 9 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre .","rationale":"\"area = 9 × 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 9 × 4.75 × 900 = 9 × 4275 = rs . 38475 answer is e .\"","correct":"e","options":{"a":"25650 ","b":"25750 ","c":"26550 ","d":"30750","e":"38475"},"options_float":{"a":25650.0,"b":25750.0,"c":26550.0,"d":30750.0,"e":38475.0},"annotated_formula":"multiply(900, multiply(9, 4.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"9 * 4.75<\/gadget>\n42.75<\/output>\n900 * 42.75<\/gadget>\n38_475<\/output>\n38_475<\/result>","index":2705} +{"problem":"if 2 ^ 5 , 3 ^ 3 , and 13 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ?","rationale":"\"here 156 has three two ' s two three ' s and one 13 rest of them must be in w so w = 13 * 3 * 4 = 156 answer : e\"","correct":"e","options":{"a":"26 ","b":"39 ","c":"42 ","d":"65","e":"156"},"options_float":{"a":26.0,"b":39.0,"c":42.0,"d":65.0,"e":156.0},"annotated_formula":"multiply(multiply(multiply(power(2, 2), 3), divide(13, 2)), 2)","linear_formula":"divide(n4,n0)|power(n0,n0)|multiply(n2,#1)|multiply(#0,#2)|multiply(n0,#3)|","chain":"2 ** 2<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n13 \/ 2<\/gadget>\n13\/2 = around 6.5<\/output>\n12 * (13\/2)<\/gadget>\n78<\/output>\n78 * 2<\/gadget>\n156<\/output>\n156<\/result>","index":2706} +{"problem":"i chose a number and divide it by 12 . then i subtracted 240 from the result and got 8 . what was the number i chose ?","rationale":"\"let x be the number i chose , then x \/ 12 − 240 = 8 x \/ 12 = 248 x = 2976 answer is a .\"","correct":"a","options":{"a":"2976 ","b":"2989 ","c":"2789 ","d":"2888","e":"2098"},"options_float":{"a":2976.0,"b":2989.0,"c":2789.0,"d":2888.0,"e":2098.0},"annotated_formula":"multiply(add(240, 8), 12)","linear_formula":"add(n1,n2)|multiply(n0,#0)|","chain":"240 + 8<\/gadget>\n248<\/output>\n248 * 12<\/gadget>\n2_976<\/output>\n2_976<\/result>","index":2708} +{"problem":"in what time will a train 100 meters long cross an electric pole , if its speed is 72 km \/ hr","rationale":"\"first convert speed into m \/ sec speed = 72 * ( 5 \/ 18 ) = 20 m \/ sec time = distance \/ speed = 100 \/ 20 = 5 seconds answer : a\"","correct":"a","options":{"a":"5 seconds ","b":"2.8 seconds ","c":"3.5 seconds ","d":"2.5 seconds","e":"2.6 seconds"},"options_float":{"a":5.0,"b":2.8,"c":3.5,"d":2.5,"e":2.6},"annotated_formula":"divide(100, multiply(72, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n100 \/ 20<\/gadget>\n5<\/output>\n5<\/result>","index":2711} +{"problem":"at joel ’ s bookstore , the current inventory is 30 % historical fiction . of the historical fiction books , 40 % are new releases , while 50 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ?","rationale":"let there be 100 books in all historic fiction books = 30 % of total = 30 other books = 70 new historic fiction = 40 % of 30 = 12 other new books = 50 % of 70 = 35 total new books = 35 + 12 = 47 fraction = 12 \/ 47 ans : b","correct":"b","options":{"a":"4 \/ 25 ","b":"12 \/ 47 ","c":"2 \/ 5 ","d":"8 \/ 15","e":"2 \/ 3"},"options_float":{"a":0.16,"b":0.2553191489,"c":0.4,"d":0.5333333333,"e":0.6666666667},"annotated_formula":"divide(divide(multiply(30, 40), const_100), add(divide(multiply(30, 40), const_100), divide(multiply(50, subtract(const_100, 30)), const_100)))","linear_formula":"multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)","chain":"30 * 40<\/gadget>\n1_200<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n100 - 30<\/gadget>\n70<\/output>\n50 * 70<\/gadget>\n3_500<\/output>\n3_500 \/ 100<\/gadget>\n35<\/output>\n12 + 35<\/gadget>\n47<\/output>\n12 \/ 47<\/gadget>\n12\/47 = around 0.255319<\/output>\n12\/47 = around 0.255319<\/result>","index":2712} +{"problem":"if 45 % of a class averages 95 % on a test , 50 % of the class averages 78 % on the test , and the remainder of the class averages 60 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) .","rationale":"\"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 45 % - 50 % = 5 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.45 x 95 + 0.50 x 78 + 0.05 x 60 = 84.75 the class average ( rounded ) is 85 % final answer c ) 85 %\"","correct":"c","options":{"a":"83 % ","b":"84 % ","c":"85 % ","d":"86 %","e":"87 %"},"options_float":{"a":83.0,"b":84.0,"c":85.0,"d":86.0,"e":87.0},"annotated_formula":"divide(add(add(multiply(45, 95), multiply(50, 78)), multiply(subtract(const_100, add(45, 50)), 60)), const_100)","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)|","chain":"45 * 95<\/gadget>\n4_275<\/output>\n50 * 78<\/gadget>\n3_900<\/output>\n4_275 + 3_900<\/gadget>\n8_175<\/output>\n45 + 50<\/gadget>\n95<\/output>\n100 - 95<\/gadget>\n5<\/output>\n5 * 60<\/gadget>\n300<\/output>\n8_175 + 300<\/gadget>\n8_475<\/output>\n8_475 \/ 100<\/gadget>\n339\/4 = around 84.75<\/output>\n339\/4 = around 84.75<\/result>","index":2713} +{"problem":"on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.30 per glass on the first day , what was the price per glass on the second day ?","rationale":"\"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.3 = 3 * x - - > x = $ 0.2 . answer : b .\"","correct":"b","options":{"a":"$ 0.15 ","b":"$ 0.20 ","c":"$ 0.30 ","d":"$ 0.40","e":"$ 0.45"},"options_float":{"a":0.15,"b":0.2,"c":0.3,"d":0.4,"e":0.45},"annotated_formula":"divide(multiply(add(const_1, const_1), 0.30), add(const_1, const_2))","linear_formula":"add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 * 0.3<\/gadget>\n0.6<\/output>\n1 + 2<\/gadget>\n3<\/output>\n0.6 \/ 3<\/gadget>\n0.2<\/output>\n0.2<\/result>","index":2714} +{"problem":"calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 40 % and its breadth is decreased by 25 % ?","rationale":"\"let l and b be 100 each 100 * 100 = 10000 l increase by 40 % = 140 b decrease by 25 % = 75 140 * 75 = 10500 5 % increase answer : a\"","correct":"a","options":{"a":"5 % increase ","b":"7 % increase ","c":"6 % increase ","d":"4 % increase","e":"3 % increase"},"options_float":{"a":5.0,"b":7.0,"c":6.0,"d":4.0,"e":3.0},"annotated_formula":"multiply(divide(subtract(multiply(add(const_100, 40), subtract(const_100, 25)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100)","linear_formula":"add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)|","chain":"100 + 40<\/gadget>\n140<\/output>\n100 - 25<\/gadget>\n75<\/output>\n140 * 75<\/gadget>\n10_500<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n10_500 - 10_000<\/gadget>\n500<\/output>\n500 \/ 10_000<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n5<\/result>","index":2715} +{"problem":"if 3 \/ w + 3 \/ x = 3 \/ y and wx = y , then the average ( arithmetic mean ) of w and x is","rationale":"\"given : 3 \/ w + 3 \/ x = 3 \/ ywx = y find : ( w + x ) \/ 2 = ? 3 ( 1 \/ w + 1 \/ x ) = 3 ( 1 \/ y ) - divide both sides by 3 ( 1 \/ w + 1 \/ x ) = 1 \/ y ( x + w ) \/ wx = 1 \/ wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) \/ 2 = 1 \/ 2 ans : a\"","correct":"a","options":{"a":"1 \/ 2 ","b":"1 ","c":"2 ","d":"4","e":"8"},"options_float":{"a":0.5,"b":1.0,"c":2.0,"d":4.0,"e":8.0},"annotated_formula":"divide(3, add(3, 3))","linear_formula":"add(n0,n0)|divide(n0,#0)|","chain":"3 + 3<\/gadget>\n6<\/output>\n3 \/ 6<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2716} +{"problem":"- - - - - - - - - - - - - - - - yes - - - - - - - - - no - - - - unsure subject m - - - - 500 - - - - - - - - 200 - - - - - 100 subject r - - - - 400 - - - - - - - - 100 - - - - - 300 a total of 800 students were asked whether they found two subjects , m and r , interesting . each answer was either yes or no or unsure , and the numbers of students who gave these answers are listed in the table above . if 140 students answered yes only for subject m , how many of the students did not answer yes for either subject ?","rationale":"\"since 140 students answered yes only for subject m , then the remaining 360 students who answered yes for subject m , also answered yes for subject r . so , 360 students answered yes for both subjects . if 360 students answered yes for both subjects , then 400 - 360 = 40 students answered yes only for subject r . so , we have that : 200 students answered yes only for subject m ; 40 students answered yes only for subject r ; 300 students answered yes for both subjects ; therefore 800 - ( 200 + 40 + 300 ) = 260 students did not answer yes for either subject . answer : b .\"","correct":"b","options":{"a":"100 ","b":"260 ","c":"360 ","d":"400","e":"500"},"options_float":{"a":100.0,"b":260.0,"c":360.0,"d":400.0,"e":500.0},"annotated_formula":"subtract(800, add(add(140, subtract(500, 140)), subtract(400, subtract(500, 140))))","linear_formula":"subtract(n0,n7)|add(n7,#0)|subtract(n3,#0)|add(#1,#2)|subtract(n6,#3)|","chain":"500 - 140<\/gadget>\n360<\/output>\n140 + 360<\/gadget>\n500<\/output>\n400 - 360<\/gadget>\n40<\/output>\n500 + 40<\/gadget>\n540<\/output>\n800 - 540<\/gadget>\n260<\/output>\n260<\/result>","index":2717} +{"problem":"the area of a square field is 7201 sq m . how long will a lady take to cross the field diagonally at the rate of 2.4 km \/ hr ?","rationale":"area of a square field = 7201 sq m let the side of square = a a ^ 2 = 7201 = > a = 84.86 diagonal = ( 2 ) ^ ( 1 \/ 2 ) * a = 1.414 * 84.86 = 120 speed of lady = 2.4 km \/ hour = 2400 m \/ hour = 40 m \/ min time taken by lady to cross the field diagonally = 120 \/ 40 = 3 min answer b","correct":"b","options":{"a":"2 min ","b":"3 min ","c":"4 min ","d":"5 min","e":"6 min"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"divide(7201, multiply(2.4, const_1000))","linear_formula":"multiply(n1,const_1000)|divide(n0,#0)","chain":"2.4 * 1_000<\/gadget>\n2_400<\/output>\n7_201 \/ 2_400<\/gadget>\n7_201\/2_400 = around 3.000417<\/output>\n7_201\/2_400 = around 3.000417<\/result>","index":2718} +{"problem":"running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 8 such machines produce in 4 minutes ?","rationale":"\"solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 8 ã — 4 ã — 270 â ‡ ” x = 8 x 4 x 270 \/ 6 = 1440 . answer d\"","correct":"d","options":{"a":"648 ","b":"1800 ","c":"2700 ","d":"1440","e":"none of these"},"options_float":{"a":648.0,"b":1800.0,"c":2700.0,"d":1440.0,"e":null},"annotated_formula":"multiply(multiply(divide(270, 6), 4), 8)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"270 \/ 6<\/gadget>\n45<\/output>\n45 * 4<\/gadget>\n180<\/output>\n180 * 8<\/gadget>\n1_440<\/output>\n1_440<\/result>","index":2722} +{"problem":"a number when divided b the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder . the number is :","rationale":"required number = ( 555 + 445 ) * 2 * 110 + 30 = 220000 + 30 = 220030 . answer : d","correct":"d","options":{"a":"1220 ","b":"1250 ","c":"2030 ","d":"22030","e":"1350"},"options_float":{"a":1220.0,"b":1250.0,"c":2030.0,"d":22030.0,"e":1350.0},"annotated_formula":"add(multiply(multiply(subtract(555, 445), const_2), const_100), 30)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|multiply(#1,const_100)|add(n2,#2)","chain":"555 - 445<\/gadget>\n110<\/output>\n110 * 2<\/gadget>\n220<\/output>\n220 * 100<\/gadget>\n22_000<\/output>\n22_000 + 30<\/gadget>\n22_030<\/output>\n22_030<\/result>","index":2724} +{"problem":"compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 108 grams of compound x ?","rationale":"\"total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 108 gms of x have 108 * 10 \/ 12 = 90 gms of b and 108 - 90 = 18 gms of a . cross check : - a \/ b = 18 \/ 90 = 2 \/ 10 ( as given ) ans a\"","correct":"a","options":{"a":"90 ","b":"162 ","c":"250 ","d":"270","e":"322"},"options_float":{"a":90.0,"b":162.0,"c":250.0,"d":270.0,"e":322.0},"annotated_formula":"divide(multiply(108, 10), add(2, 10))","linear_formula":"add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|","chain":"108 * 10<\/gadget>\n1_080<\/output>\n2 + 10<\/gadget>\n12<\/output>\n1_080 \/ 12<\/gadget>\n90<\/output>\n90<\/result>","index":2726} +{"problem":"tough and tricky questions : ratios 4 \/ 5 of a certain class left on a field trip . 1 \/ 3 of the students who stayed behind did not want to go on the field trip ( all the others did want to go ) . when another vehicle was located , 1 \/ 2 of the students who did want to go on the field trip but had been left behind were able to join . what fraction of the class ended up going on the field trip ?","rationale":"let total no . of students be 30 students left on a field trip = 4 \/ 5 ( 30 ) = 24 thus 6 students left behind 1 \/ 3 ( 6 ) = 2 students did not want to go . thus 4 students want to go on the trip 1 \/ 2 ( 4 ) = 2 were able to join the other students on the field trip . thus total number of students that went on a field trip = 24 + 2 = 26 thus required fraction = 26 \/ 30 = 13 \/ 15 c","correct":"c","options":{"a":"1 \/ 2 ","b":"2 \/ 3 ","c":"13 \/ 15 ","d":"23 \/ 30","e":"4 \/ 5"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.8666666667,"d":0.7666666667,"e":0.8},"annotated_formula":"divide(add(multiply(divide(1, 2), subtract(subtract(lcm(lcm(3, 5), 2), multiply(divide(4, 5), lcm(lcm(3, 5), 2))), multiply(divide(1, 3), subtract(lcm(lcm(3, 5), 2), multiply(divide(4, 5), lcm(lcm(3, 5), 2)))))), multiply(divide(4, 5), lcm(lcm(3, 5), 2))), lcm(lcm(3, 5), 2))","linear_formula":"divide(n2,n5)|divide(n0,n1)|divide(n2,n3)|lcm(n1,n3)|lcm(n5,#3)|multiply(#1,#4)|subtract(#4,#5)|multiply(#2,#6)|subtract(#6,#7)|multiply(#0,#8)|add(#9,#5)|divide(#10,#4)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\nlcm(3, 5)<\/gadget>\n15<\/output>\nlcm(15, 2)<\/gadget>\n30<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 30<\/gadget>\n24<\/output>\n30 - 24<\/gadget>\n6<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 6<\/gadget>\n2<\/output>\n6 - 2<\/gadget>\n4<\/output>\n(1\/2) * 4<\/gadget>\n2<\/output>\n2 + 24<\/gadget>\n26<\/output>\n26 \/ 30<\/gadget>\n13\/15 = around 0.866667<\/output>\n13\/15 = around 0.866667<\/result>","index":2728} +{"problem":"if x ^ 2 + 9 \/ x ^ 2 = 10 , what is the value of x - 3 \/ x","rationale":"to find : x - 3 \/ x . let it be t . = > x - 3 \/ x = t = > ( x ^ 2 + 9 \/ x ^ 2 ) - 2 * x * 3 \/ x = t ^ 2 ( squaring both sides ) . = > ( 10 ) - 2 * 3 = 4 = > t ^ 2 = 4 . thus t = 2 or t = - 2 . answer c","correct":"c","options":{"a":"36 ","b":"25 ","c":"2 ","d":"5","e":"3"},"options_float":{"a":36.0,"b":25.0,"c":2.0,"d":5.0,"e":3.0},"annotated_formula":"sqrt(subtract(10, multiply(const_2, const_3)))","linear_formula":"multiply(const_2,const_3)|subtract(n3,#0)|sqrt(#1)","chain":"2 * 3<\/gadget>\n6<\/output>\n10 - 6<\/gadget>\n4<\/output>\n4 ** (1\/2)<\/gadget>\n2<\/output>\n2<\/result>","index":2729} +{"problem":"the average salary \/ head of all the workers in a workshop is rs . 1100 , if the average salary \/ head of 9 technician is rs . 1300 and the average salary \/ head of the rest is rs . 880 , the total no . of workers in the work - shop is ?","rationale":"let the total number of workers be y . so sum of salary for all workers = sum of salary of 9 technician + sum of salary for other y - 9 workers . 9 x 1300 + 880 ( y - 9 ) = 1100 y ⇒ 11700 + 880 y - 7920 = 1100 y ⇒ 220 y = 3780 ∴ y = 17 so total number of workers = 17 d","correct":"d","options":{"a":"18 ","b":"20 ","c":"22 ","d":"17","e":"26"},"options_float":{"a":18.0,"b":20.0,"c":22.0,"d":17.0,"e":26.0},"annotated_formula":"divide(subtract(multiply(9, 1300), multiply(880, 9)), subtract(1100, 880))","linear_formula":"multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)|","chain":"9 * 1_300<\/gadget>\n11_700<\/output>\n880 * 9<\/gadget>\n7_920<\/output>\n11_700 - 7_920<\/gadget>\n3_780<\/output>\n1_100 - 880<\/gadget>\n220<\/output>\n3_780 \/ 220<\/gadget>\n189\/11 = around 17.181818<\/output>\n189\/11 = around 17.181818<\/result>","index":2730} +{"problem":"joe invested a certain sum of money in a simple interest bond whose value grew to $ 310 at the end of 3 years and to $ 410 at the end of another 5 years . what was the rate of interest in which he invested his sum ?","rationale":"\"in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 310 - $ 60 = 250 . the interest rate is $ 20 \/ $ 250 = 8 % the answer is c .\"","correct":"c","options":{"a":"6 % ","b":"7 % ","c":"8 % ","d":"9 %","e":"10 %"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"multiply(divide(divide(subtract(410, 310), 5), subtract(310, multiply(divide(subtract(410, 310), 5), 3))), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)|","chain":"410 - 310<\/gadget>\n100<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20 * 3<\/gadget>\n60<\/output>\n310 - 60<\/gadget>\n250<\/output>\n20 \/ 250<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) * 100<\/gadget>\n8<\/output>\n8<\/result>","index":2732} +{"problem":"a rhombus has area 432 sq m , one diagonal is 36 cm . find other diagonal ?","rationale":"area of rhombus = 1 \/ 2 * d 1 * d 2 432 = 1 \/ 2 * d 1 * . 36 d 1 = 2400 m answer : a","correct":"a","options":{"a":"2400 m ","b":"2500 m ","c":"2600 m ","d":"2700 m","e":"2900 m"},"options_float":{"a":2400.0,"b":2500.0,"c":2600.0,"d":2700.0,"e":2900.0},"annotated_formula":"divide(multiply(432, const_2), divide(36, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,const_2)|divide(#1,#0)","chain":"432 * 2<\/gadget>\n864<\/output>\n36 \/ 100<\/gadget>\n9\/25 = around 0.36<\/output>\n864 \/ (9\/25)<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":2733} +{"problem":"if { x } is the product of all even integers from 1 to x inclusive , what is the greatest prime factor of { 12 } + { 10 } ?","rationale":"\"soln : { 12 } + { 10 } = 12 * { 10 } + { 10 } = 13 * { 10 } answer : c\"","correct":"c","options":{"a":"23 ","b":"20 ","c":"13 ","d":"5","e":"2"},"options_float":{"a":23.0,"b":20.0,"c":13.0,"d":5.0,"e":2.0},"annotated_formula":"add(divide(add(12, 10), const_2), multiply(1, const_2))","linear_formula":"add(n1,n2)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)|","chain":"12 + 10<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n1 * 2<\/gadget>\n2<\/output>\n11 + 2<\/gadget>\n13<\/output>\n13<\/result>","index":2734} +{"problem":"a , b and c invested rs . 2400 , rs . 7200 and rs . 9600 respectively , in a partnership business . find the share of a in profit of rs . 9000 after a year ?","rationale":"\"explanation : 2400 : 7200 : 9600 1 : 3 : 4 1 \/ 8 * 9000 = 1125 answer : a\"","correct":"a","options":{"a":"1125 ","b":"2400 ","c":"7200 ","d":"2500","e":"2511"},"options_float":{"a":1125.0,"b":2400.0,"c":7200.0,"d":2500.0,"e":2511.0},"annotated_formula":"multiply(divide(2400, add(add(2400, 7200), 9600)), 9000)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|","chain":"2_400 + 7_200<\/gadget>\n9_600<\/output>\n9_600 + 9_600<\/gadget>\n19_200<\/output>\n2_400 \/ 19_200<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 9_000<\/gadget>\n1_125<\/output>\n1_125<\/result>","index":2735} +{"problem":"a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 165 . find the unknown rate of two towels ?","rationale":"\"10 * 150 = 1650 3 * 100 + 5 * 150 = 1050 1650 – 1050 = 600 e\"","correct":"e","options":{"a":"400 ","b":"450 ","c":"500 ","d":"550","e":"600"},"options_float":{"a":400.0,"b":450.0,"c":500.0,"d":550.0,"e":600.0},"annotated_formula":"subtract(subtract(multiply(add(add(3, 5), const_2), 165), multiply(5, 150)), multiply(3, 100))","linear_formula":"add(n0,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#0,const_2)|multiply(n4,#3)|subtract(#4,#1)|subtract(#5,#2)|","chain":"3 + 5<\/gadget>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n10 * 165<\/gadget>\n1_650<\/output>\n5 * 150<\/gadget>\n750<\/output>\n1_650 - 750<\/gadget>\n900<\/output>\n3 * 100<\/gadget>\n300<\/output>\n900 - 300<\/gadget>\n600<\/output>\n600<\/result>","index":2736} +{"problem":"aaron will jog from home at 2 miles per hour and then walk back home by the same route at 4 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ?","rationale":"xyt \/ ( x + y ) x = 2 , y = 4 t = 3 2 * 4 * 3 \/ 2 + 4 = 18 \/ 6 = 3 answer : c","correct":"c","options":{"a":"5 ","b":"7 ","c":"3 ","d":"9","e":"10"},"options_float":{"a":5.0,"b":7.0,"c":3.0,"d":9.0,"e":10.0},"annotated_formula":"divide(multiply(multiply(2, 4), 3), multiply(2, 4))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,#0)","chain":"2 * 4<\/gadget>\n8<\/output>\n8 * 3<\/gadget>\n24<\/output>\n24 \/ 8<\/gadget>\n3<\/output>\n3<\/result>","index":2737} +{"problem":"every year an amount increases by 1 \/ 8 th of itself . how much will it be after two years if its present value is rs . 57600 ?","rationale":"57600 * 9 \/ 8 * 9 \/ 8 = 72900 . answer : d","correct":"d","options":{"a":"81000 ","b":"81028 ","c":"27772 ","d":"72900","e":"22312"},"options_float":{"a":81000.0,"b":81028.0,"c":27772.0,"d":72900.0,"e":22312.0},"annotated_formula":"add(add(57600, multiply(divide(1, 8), 57600)), multiply(divide(1, 8), add(57600, multiply(divide(1, 8), 57600))))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 57_600<\/gadget>\n7_200<\/output>\n57_600 + 7_200<\/gadget>\n64_800<\/output>\n(1\/8) * 64_800<\/gadget>\n8_100<\/output>\n64_800 + 8_100<\/gadget>\n72_900<\/output>\n72_900<\/result>","index":2738} +{"problem":"for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 35 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amount of the sale ?","rationale":"\"total sales = 800 comission = ( 20 \/ 100 ) * 500 + ( 35 \/ 100 ) * 300 = 100 + 105 = 205 % comission = ( 205 \/ 800 ) * 100 = 25.6 ~ 26 % answer is d\"","correct":"d","options":{"a":"22 % ","b":"24 % ","c":"25 % ","d":"26 %","e":"28 %"},"options_float":{"a":22.0,"b":24.0,"c":25.0,"d":26.0,"e":28.0},"annotated_formula":"multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(35, const_100), subtract(800, 500))), 800), const_100)","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n1)|multiply(n1,#0)|multiply(#1,#2)|add(#3,#4)|divide(#5,n4)|multiply(#6,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 500<\/gadget>\n100<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n800 - 500<\/gadget>\n300<\/output>\n(7\/20) * 300<\/gadget>\n105<\/output>\n100 + 105<\/gadget>\n205<\/output>\n205 \/ 800<\/gadget>\n41\/160 = around 0.25625<\/output>\n(41\/160) * 100<\/gadget>\n205\/8 = around 25.625<\/output>\n205\/8 = around 25.625<\/result>","index":2739} +{"problem":"a boat can travel with a speed of 22 km \/ hr in still water . if the speed of the stream is 5 km \/ hr , find the time taken by the boat to go 135 km downstream","rationale":"\"explanation : speed of the boat in still water = 22 km \/ hr speed of the stream = 5 km \/ hr speed downstream = ( 22 + 5 ) = 27 km \/ hr distance travelled downstream = 135 km time taken = distance \/ speed = 135 \/ 27 = 5 hours . answer : option a\"","correct":"a","options":{"a":"5 hours ","b":"4 hours ","c":"3 hours ","d":"2 hours","e":"none of these"},"options_float":{"a":5.0,"b":4.0,"c":3.0,"d":2.0,"e":null},"annotated_formula":"divide(135, add(22, 5))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"22 + 5<\/gadget>\n27<\/output>\n135 \/ 27<\/gadget>\n5<\/output>\n5<\/result>","index":2740} +{"problem":"a bag contains 5 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is","rationale":"\"drawing two balls of same color from five green balls can be done in 5 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 5 c ₂ \/ ¹ ⁵ c ₂ + ⁸ c ₂ \/ ¹ ⁵ c ₂ = 19 \/ 39 answer : e\"","correct":"e","options":{"a":"8 \/ 15 ","b":"2 \/ 5 ","c":"3 \/ 5 ","d":"11 \/ 15","e":"19 \/ 39"},"options_float":{"a":0.5333333333,"b":0.4,"c":0.6,"d":0.7333333333,"e":0.4871794872},"annotated_formula":"add(multiply(divide(8, add(5, 8)), divide(subtract(8, const_1), subtract(add(5, 8), const_1))), multiply(divide(5, add(5, 8)), divide(subtract(5, const_1), subtract(add(5, 8), const_1))))","linear_formula":"add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)|","chain":"5 + 8<\/gadget>\n13<\/output>\n8 \/ 13<\/gadget>\n8\/13 = around 0.615385<\/output>\n8 - 1<\/gadget>\n7<\/output>\n13 - 1<\/gadget>\n12<\/output>\n7 \/ 12<\/gadget>\n7\/12 = around 0.583333<\/output>\n(8\/13) * (7\/12)<\/gadget>\n14\/39 = around 0.358974<\/output>\n5 \/ 13<\/gadget>\n5\/13 = around 0.384615<\/output>\n5 - 1<\/gadget>\n4<\/output>\n4 \/ 12<\/gadget>\n1\/3 = around 0.333333<\/output>\n(5\/13) * (1\/3)<\/gadget>\n5\/39 = around 0.128205<\/output>\n(14\/39) + (5\/39)<\/gadget>\n19\/39 = around 0.487179<\/output>\n19\/39 = around 0.487179<\/result>","index":2741} +{"problem":"find the 10 th term of an arithmetic progression whose first term is 8 and the common difference is 2 .","rationale":"\"n th term of a . p = a + ( n - 1 ) * d = 8 + ( 10 - 1 ) * 2 , = 8 + 18 = 26 . answer : b\"","correct":"b","options":{"a":"45 ","b":"26 ","c":"44 ","d":"40","e":"46"},"options_float":{"a":45.0,"b":26.0,"c":44.0,"d":40.0,"e":46.0},"annotated_formula":"add(multiply(subtract(10, const_1), 2), 8)","linear_formula":"subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1)|","chain":"10 - 1<\/gadget>\n9<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18 + 8<\/gadget>\n26<\/output>\n26<\/result>","index":2742} +{"problem":"if ( 10 x + 2 ) \/ 4 - ( 3 x - 6 ) \/ 18 = ( 2 x + 4 ) \/ 3 , then what is the value of x ?","rationale":"( 10 x + 2 ) \/ 4 - ( 3 x - 6 ) \/ 18 = ( 2 x + 4 ) \/ 3 = > 18 ( 10 x + 2 ) \/ 4 - ( 3 x - 6 ) = 18 ( 2 x + 4 ) \/ 3 = > 4.5 ( 10 x + 2 ) - 3 x + 6 = 6 ( 2 x + 4 ) = > 45 x + 9 - 3 x + 6 = 12 x + 24 = > 45 x - 3 x - 12 x = 24 - 9 - 6 = > 30 x = 9 = > x = 9 \/ 30 = > x = 3 \/ 10 answer : d","correct":"d","options":{"a":"3 ","b":"- 3 ","c":"10 \/ 3 ","d":"3 \/ 10","e":"- 10 \/ 3"},"options_float":{"a":3.0,"b":-3.0,"c":3.3333333333,"d":0.3,"e":-3.3333333333},"annotated_formula":"divide(subtract(subtract(multiply(divide(lcm(lcm(4, 18), 3), 3), 4), multiply(divide(lcm(lcm(4, 18), 3), 4), 2)), multiply(divide(lcm(lcm(4, 18), 3), 18), 6)), subtract(subtract(multiply(divide(lcm(lcm(4, 18), 3), 4), 10), multiply(divide(lcm(lcm(4, 18), 3), 18), 3)), multiply(divide(lcm(lcm(4, 18), 3), 3), 2)))","linear_formula":"lcm(n2,n5)|lcm(n3,#0)|divide(#1,n3)|divide(#1,n2)|divide(#1,n5)|multiply(n2,#2)|multiply(n1,#3)|multiply(n4,#4)|multiply(n0,#3)|multiply(n3,#4)|multiply(n1,#2)|subtract(#5,#6)|subtract(#8,#9)|subtract(#11,#7)|subtract(#12,#10)|divide(#13,#14)","chain":"lcm(4, 18)<\/gadget>\n36<\/output>\nlcm(36, 3)<\/gadget>\n36<\/output>\n36 \/ 3<\/gadget>\n12<\/output>\n12 * 4<\/gadget>\n48<\/output>\n36 \/ 4<\/gadget>\n9<\/output>\n9 * 2<\/gadget>\n18<\/output>\n48 - 18<\/gadget>\n30<\/output>\n36 \/ 18<\/gadget>\n2<\/output>\n2 * 6<\/gadget>\n12<\/output>\n30 - 12<\/gadget>\n18<\/output>\n9 * 10<\/gadget>\n90<\/output>\n2 * 3<\/gadget>\n6<\/output>\n90 - 6<\/gadget>\n84<\/output>\n12 * 2<\/gadget>\n24<\/output>\n84 - 24<\/gadget>\n60<\/output>\n18 \/ 60<\/gadget>\n3\/10 = around 0.3<\/output>\n3\/10 = around 0.3<\/result>","index":2743} +{"problem":"the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.80 per metre .","rationale":"\"explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.80 ) = rs . 6336 . answer : b ) 6336\"","correct":"b","options":{"a":"2399 ","b":"6336 ","c":"2999 ","d":"5808","e":"2888"},"options_float":{"a":2399.0,"b":6336.0,"c":2999.0,"d":5808.0,"e":2888.0},"annotated_formula":"multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.80)","linear_formula":"divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)|","chain":"13.86 \/ pi<\/gadget>\n13.86\/pi = around 4.411775<\/output>\n(13.86\/pi) ** (1\/2)<\/gadget>\n3.72290209379726\/sqrt(pi) = around 2.100423<\/output>\n(3.72290209379726\/sqrt(pi)) * 100<\/gadget>\n372.290209379726\/sqrt(pi) = around 210.042258<\/output>\n2 * pi * (372.290209379726\/sqrt(pi))<\/gadget>\n744.580418759452*sqrt(pi) = around 1_319.734431<\/output>\n(744.580418759452*sqrt(pi)) * 4.8<\/gadget>\n3573.98601004537*sqrt(pi) = around 6_334.725267<\/output>\n3573.98601004537*sqrt(pi) = around 6_334.725267<\/result>","index":2745} +{"problem":"a train 150 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ?","rationale":"\"d 9 sec speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 \/ 18 ) m \/ sec = ( 50 \/ 3 ) m \/ sec time taken by the train to cross the man = time taken by it to cover 150 m at 50 \/ 3 m \/ sec = 150 * 3 \/ 50 sec = 9 sec\"","correct":"d","options":{"a":"2 sec ","b":"6 sec ","c":"8 sec ","d":"9 sec","e":"5 sec"},"options_float":{"a":2.0,"b":6.0,"c":8.0,"d":9.0,"e":5.0},"annotated_formula":"divide(150, multiply(subtract(68, 8), const_0_2778))","linear_formula":"subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"68 - 8<\/gadget>\n60<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n150 \/ (50\/3)<\/gadget>\n9<\/output>\n9<\/result>","index":2747} +{"problem":"annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 30 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ?","rationale":"annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 30 minutes , sam rides 6 km so sam will be 4 km ahead . it will take annie 80 minutes to catch sam . the answer is d .","correct":"d","options":{"a":"20 ","b":"40 ","c":"60 ","d":"80","e":"100"},"options_float":{"a":20.0,"b":40.0,"c":60.0,"d":80.0,"e":100.0},"annotated_formula":"multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60)","linear_formula":"divide(n2,const_60)|subtract(n0,n1)|multiply(#0,#1)|divide(n1,#2)|subtract(#3,#2)|divide(#4,#1)|multiply(#5,const_60)","chain":"15 - 12<\/gadget>\n3<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n3 * (2\/3)<\/gadget>\n2<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6 - 2<\/gadget>\n4<\/output>\n4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) * 60<\/gadget>\n80<\/output>\n80<\/result>","index":2750} +{"problem":"john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.60 more than jane , what was the original price of the dish ?","rationale":"\"the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.60 - - > 15 p - 13.5 p = 60 - - > p = 40 . answer : d .\"","correct":"d","options":{"a":"24 ","b":"34.8 ","c":"37.8 ","d":"40","e":"84"},"options_float":{"a":24.0,"b":34.8,"c":37.8,"d":40.0,"e":84.0},"annotated_formula":"divide(0.60, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100))))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#0,#3)|divide(n2,#4)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * (3\/20)<\/gadget>\n27\/200 = around 0.135<\/output>\n(3\/20) - (27\/200)<\/gadget>\n3\/200 = around 0.015<\/output>\n0.6 \/ (3\/200)<\/gadget>\n40<\/output>\n40<\/result>","index":2751} +{"problem":"a certain industrial loom weaves 0.126 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ?","rationale":"\"let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.126 : 15 : : 1 : x = > 0.126 * x = 15 = > x = 15 \/ 0.126 = > x = 119 answer : e\"","correct":"e","options":{"a":"114 ","b":"115 ","c":"116 ","d":"117","e":"119"},"options_float":{"a":114.0,"b":115.0,"c":116.0,"d":117.0,"e":119.0},"annotated_formula":"divide(15, 0.126)","linear_formula":"divide(n1,n0)|","chain":"15 \/ 0.126<\/gadget>\n119.047619<\/output>\n119.047619<\/result>","index":2752} +{"problem":"a cistern is normally filled in 8 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ?","rationale":"if leakage \/ hour = 1 \/ x , then 1 \/ 8 - 1 \/ x = 1 \/ 10 , solving 1 \/ x = 1 \/ 40 so in 40 hours full cistern will be empty . answer : d","correct":"d","options":{"a":"10 hours ","b":"20 hours ","c":"30 hours ","d":"40 hours","e":"50 hours"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"inverse(subtract(divide(const_1, 8), divide(const_1, add(8, 2))))","linear_formula":"add(n0,n1)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n8 + 2<\/gadget>\n10<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/8) - (1\/10)<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":2753} +{"problem":"oil cans x and y are right circular cylinders and the height and radius of y are each 4 times those of x . if the oil in can x filled to capacity sells for $ 2 , how much does the oil in y sell for if y is only half filled ?","rationale":"formula for vol of a cyl is pi * r ^ 2 * h so vy = 64 * vy y when half filled will cost 32 times x so ans is d","correct":"d","options":{"a":"$ 60 ","b":"$ 52 ","c":"$ 63 ","d":"$ 64","e":"$ 68"},"options_float":{"a":60.0,"b":52.0,"c":63.0,"d":64.0,"e":68.0},"annotated_formula":"multiply(power(4, 2), 4)","linear_formula":"power(n0,n1)|multiply(n0,#0)","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 * 4<\/gadget>\n64<\/output>\n64<\/result>","index":2754} +{"problem":"how many distinct integer values of n satisfy the inequality | | n - 3 | + 4 | ≤ 15 ?","rationale":"so i can write this as in - 3 i + 4 < = 15 or in - 3 i < = 11 so n can have - 8 to 14 = 13 true values . . . . e","correct":"e","options":{"a":"15 ","b":"16 ","c":"17 ","d":"18","e":"13"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":13.0},"annotated_formula":"add(multiply(const_3, const_4), const_1)","linear_formula":"multiply(const_3,const_4)|add(#0,const_1)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 + 1<\/gadget>\n13<\/output>\n13<\/result>","index":2757} +{"problem":"a man rows his boat 75 km downstream and 45 km upstream , taking 5 hours each time . find the speed of the stream ?","rationale":"\"speed downstream = d \/ t = 75 \/ ( 5 ) = 15 kmph speed upstream = d \/ t = 45 \/ ( 5 ) = 9 kmph the speed of the stream = ( 15 - 9 ) \/ 2 = 3 kmph answer : d\"","correct":"d","options":{"a":"6 kmph ","b":"5 kmph ","c":"2 kmph ","d":"3 kmph","e":"1 kmph"},"options_float":{"a":6.0,"b":5.0,"c":2.0,"d":3.0,"e":1.0},"annotated_formula":"divide(subtract(divide(75, 5), divide(45, 5)), const_2)","linear_formula":"divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)|","chain":"75 \/ 5<\/gadget>\n15<\/output>\n45 \/ 5<\/gadget>\n9<\/output>\n15 - 9<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n3<\/result>","index":2758} +{"problem":"an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 25 metre of cloth ?","rationale":"\"explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 25 : x = > x = 25 ∗ 1 \/ 0.128 = > x = 195.31 so time will be approx 195 seconds option b\"","correct":"b","options":{"a":"194 seconds ","b":"195 seconds ","c":"196 seconds ","d":"197 seconds","e":"none of these"},"options_float":{"a":194.0,"b":195.0,"c":196.0,"d":197.0,"e":null},"annotated_formula":"divide(25, 0.128)","linear_formula":"divide(n1,n0)|","chain":"25 \/ 0.128<\/gadget>\n195.3125<\/output>\n195.3125<\/result>","index":2759} +{"problem":"at what time between 5 and 6 o ' clock are the hands of a 3 minutes apart ?","rationale":"in this type of problems the formuae is ( 5 * x + or - t ) * 12 \/ 11 here x is replaced by the first interval of given time . here x is 5 . t is spaces apart case 1 : ( 5 * x + t ) * 12 \/ 11 ( 5 * 5 + 3 ) * 12 \/ 11 28 * 12 \/ 11 = 336 \/ 11 = min therefore the hands will be 3 min apart at 31 5 \/ 11 min past 5 . case 2 : ( 5 * x - t ) * 12 \/ 11 ( 5 * 5 - 3 ) * 12 \/ 11 22 * 12 \/ 11 = 24 min therefore the hands will be 3 min apart at 24 min past 5 answer : a ) 24 min","correct":"a","options":{"a":"24 ","b":"55 ","c":"66 ","d":"88","e":"21"},"options_float":{"a":24.0,"b":55.0,"c":66.0,"d":88.0,"e":21.0},"annotated_formula":"multiply(subtract(multiply(5, 5), 3), divide(add(6, 6), add(6, 5)))","linear_formula":"add(n1,n1)|add(n0,n1)|multiply(n0,n0)|divide(#0,#1)|subtract(#2,n2)|multiply(#3,#4)","chain":"5 * 5<\/gadget>\n25<\/output>\n25 - 3<\/gadget>\n22<\/output>\n6 + 6<\/gadget>\n12<\/output>\n6 + 5<\/gadget>\n11<\/output>\n12 \/ 11<\/gadget>\n12\/11 = around 1.090909<\/output>\n22 * (12\/11)<\/gadget>\n24<\/output>\n24<\/result>","index":2760} +{"problem":"the probability is 1 \/ 2 that a certain coin turns up heads on any given toss . if the coin is tossed 6 times , what is the probability that the coin turns up tails on at least one of the tosses ?","rationale":"p ( 6 heads ) = 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 = 1 \/ 64 . p ( at least one tail ) = 1 - 1 \/ 64 = 63 \/ 64 . the answer is d .","correct":"d","options":{"a":"15 \/ 16 ","b":"21 \/ 32 ","c":"31 \/ 32 ","d":"63 \/ 64","e":"51 \/ 64"},"options_float":{"a":0.9375,"b":0.65625,"c":0.96875,"d":0.984375,"e":0.796875},"annotated_formula":"subtract(1, power(divide(1, 2), 6))","linear_formula":"divide(n0,n1)|power(#0,n2)|subtract(n0,#1)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 6<\/gadget>\n1\/64 = around 0.015625<\/output>\n1 - (1\/64)<\/gadget>\n63\/64 = around 0.984375<\/output>\n63\/64 = around 0.984375<\/result>","index":2761} +{"problem":"a football field is 8400 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 3500 square yards ?","rationale":"\"answer a ) 4800 yards need 1200 lbs 1 yard will need 1200 \/ 8400 = 1 \/ 7 lbs 3600 yards will need 1 \/ 7 * 3500 yards = 500 lbs c\"","correct":"c","options":{"a":"450 ","b":"600 ","c":"500 ","d":"2400","e":"3200"},"options_float":{"a":450.0,"b":600.0,"c":500.0,"d":2400.0,"e":3200.0},"annotated_formula":"multiply(3500, divide(1200, 8400))","linear_formula":"divide(n1,n0)|multiply(n2,#0)|","chain":"1_200 \/ 8_400<\/gadget>\n1\/7 = around 0.142857<\/output>\n3_500 * (1\/7)<\/gadget>\n500<\/output>\n500<\/result>","index":2762} +{"problem":"the lowest number which should be added to 11390 so that the sum is exactly divisible by 7 , 9 , 10 and 12 is :","rationale":"l . c . m . of 7 , 9 , 10 and 12 = 1260 . on dividing 11390 by 1260 , the remainder is 50 . number to be added = ( 1260 - 50 ) = 1210 . answer : option ' d '","correct":"d","options":{"a":"1010 ","b":"2210 ","c":"1250 ","d":"1210","e":"1110"},"options_float":{"a":1010.0,"b":2210.0,"c":1250.0,"d":1210.0,"e":1110.0},"annotated_formula":"divide(11390, 9)","linear_formula":"divide(n0,n2)","chain":"11_390 \/ 9<\/gadget>\n11_390\/9 = around 1_265.555556<\/output>\n11_390\/9 = around 1_265.555556<\/result>","index":2763} +{"problem":"a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 3 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ?","rationale":"\"total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes only the product of 333 will be odd , the remaining 26 will have either 2 or 4 in them , which ensures that their product will be even . therefore the number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 1 = 26 . answer : b\"","correct":"b","options":{"a":"20 ","b":"26 ","c":"24 ","d":"22","e":"30"},"options_float":{"a":20.0,"b":26.0,"c":24.0,"d":22.0,"e":30.0},"annotated_formula":"subtract(power(3, 3), const_1)","linear_formula":"power(n0,n0)|subtract(#0,const_1)|","chain":"3 ** 3<\/gadget>\n27<\/output>\n27 - 1<\/gadget>\n26<\/output>\n26<\/result>","index":2764} +{"problem":"120 students represent x percent of the boys at a school . if the boys at the school make up 30 % of the total school population of x students , what is x ?","rationale":"\"let b be the number of boys in the school . 120 = xb \/ 100 b = 0.3 x 12000 = 0.3 x ^ 2 x ^ 2 = 40000 x = 200 the answer is d .\"","correct":"d","options":{"a":"120 ","b":"150 ","c":"180 ","d":"200","e":"250"},"options_float":{"a":120.0,"b":150.0,"c":180.0,"d":200.0,"e":250.0},"annotated_formula":"sqrt(divide(multiply(120, const_100), divide(30, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)|","chain":"120 * 100<\/gadget>\n12_000<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n12_000 \/ (3\/10)<\/gadget>\n40_000<\/output>\n40_000 ** (1\/2)<\/gadget>\n200<\/output>\n200<\/result>","index":2765} +{"problem":"of the diplomats attending a summit conference , 17 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ?","rationale":"\"{ total } = { french } + { russian } - { both } + { neither } { total } = 17 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 150 . answer : e .\"","correct":"e","options":{"a":"72 ","b":"96 ","c":"108 ","d":"120","e":"150"},"options_float":{"a":72.0,"b":96.0,"c":108.0,"d":120.0,"e":150.0},"annotated_formula":"divide(subtract(32, 17), subtract(divide(20, const_100), divide(10, const_100)))","linear_formula":"divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)|","chain":"32 - 17<\/gadget>\n15<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/5) - (1\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n15 \/ (1\/10)<\/gadget>\n150<\/output>\n150<\/result>","index":2768} +{"problem":"a lady grows broccoli in her garden that is in the shape of a square . each broccoli takes 1 square foot of area in her garden . this year , she has increased her output by 79 broccoli when compared to last year . the shape of the area used for growing the broccoli has remained a square in both these years . how many broccoli did she produce this year ?","rationale":"explanatory answer the shape of the area used for growing broccoli has remained a square in both the years . let the side of the square area used for growing broccoli this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing broccoli last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of broccoli grown has increased by 79 , the area would have increased by 79 sq ft because each broccoli takes 1 sq ft space . hence , x 2 - y 2 = 79 ( x + y ) ( x - y ) = 79 . 79 is a prime number and hence it will have only two factors . i . e . , 79 and 1 . therefore , 79 can be expressed as product of 2 numbers in only way = 79 * 1 i . e . , ( x + y ) ( x - y ) = 79 * 1 so , ( x + y ) should be 79 and ( x - y ) should be 1 . solving the two equations we get x = 40 and y = 39 . therefore , number of broccoli produced this year = x 2 = 402 = 1600 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice b , c and d . as 1500,1650 and 1700 are not the square of any number . check choice a : if this year ' s produce is 1600 , last year ' s produce would have been 1600 - 79 = 1521 1521 is the square of 39 . so , 1600 is the answer . choice a","correct":"a","options":{"a":"1600 ","b":"1500 ","c":"1650 ","d":"1700","e":"can not be determined"},"options_float":{"a":1600.0,"b":1500.0,"c":1650.0,"d":1700.0,"e":null},"annotated_formula":"power(divide(add(79, 1), const_2), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|power(#1,const_2)","chain":"79 + 1<\/gadget>\n80<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40 ** 2<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":2771} +{"problem":"a train 120 m long is running at 60 kmph . in how much time will it pass a platform 240 m long ?","rationale":"distance travelled = 120 + 240 m = 360 m speed = 60 * 5 \/ 8 = 50 \/ 3 m time = 360 * 3 \/ 50 = 21.6 seconds answer : c .","correct":"c","options":{"a":"15.6 seconds ","b":"24.6 seconds ","c":"21.6 seconds ","d":"30.6 seconds","e":"35.6 seconds"},"options_float":{"a":15.6,"b":24.6,"c":21.6,"d":30.6,"e":35.6},"annotated_formula":"divide(add(120, 240), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)","chain":"120 + 240<\/gadget>\n360<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n360 \/ (50\/3)<\/gadget>\n108\/5 = around 21.6<\/output>\n108\/5 = around 21.6<\/result>","index":2772} +{"problem":"a is twice as good a workman as b and together they complete a work in 12 days . in how many days a alone can do the work ?","rationale":"a ’ s one day ’ s work : b ’ s one day ’ s work = 2 : 1 ( a + b ) ’ s one day ’ s work = 1 \/ 12 b ’ s one day ’ s work = 1 \/ 12 * 1 \/ 3 = 1 \/ 36 b alone can do the work in 36 days answer : e","correct":"e","options":{"a":"32 ","b":"33 ","c":"34 ","d":"35","e":"36"},"options_float":{"a":32.0,"b":33.0,"c":34.0,"d":35.0,"e":36.0},"annotated_formula":"multiply(add(const_1, const_2), 12)","linear_formula":"add(const_1,const_2)|multiply(n0,#0)","chain":"1 + 2<\/gadget>\n3<\/output>\n3 * 12<\/gadget>\n36<\/output>\n36<\/result>","index":2773} +{"problem":"10 women can complete a work in 5 days and 10 children take 10 days to complete the work . how many days will 5 women and 10 children take to complete the work ?","rationale":"\"1 women ' s 1 day work = 1 \/ 50 1 child ' s 1 day work = 1 \/ 100 ( 5 women + 10 children ) ' s 1 day work = ( 5 \/ 50 + 10 \/ 100 ) = 1 \/ 5 5 women and 10 children will complete the work in 5 days . b\"","correct":"b","options":{"a":"4 ","b":"5 ","c":"7 ","d":"8","e":"2"},"options_float":{"a":4.0,"b":5.0,"c":7.0,"d":8.0,"e":2.0},"annotated_formula":"inverse(add(divide(5, multiply(10, 5)), divide(10, multiply(10, 10))))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)|","chain":"10 * 5<\/gadget>\n50<\/output>\n5 \/ 50<\/gadget>\n1\/10 = around 0.1<\/output>\n10 * 10<\/gadget>\n100<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + (1\/10)<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ (1\/5)<\/gadget>\n5<\/output>\n5<\/result>","index":2774} +{"problem":"if p # q denotes the least common multiple of p and q , then e = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ?","rationale":"there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) \/ ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 \/ 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 \/ 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , e = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) \/ 24 = 2 * 72 = 144 the correct answer is c .","correct":"c","options":{"a":"216 ","b":"180 ","c":"144 ","d":"108","e":"72"},"options_float":{"a":216.0,"b":180.0,"c":144.0,"d":108.0,"e":72.0},"annotated_formula":"add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24)","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2)|add(n3,#3)","chain":"18 * 24<\/gadget>\n432<\/output>\n12 * 16<\/gadget>\n192<\/output>\n432 - 192<\/gadget>\n240<\/output>\n240 \/ 2<\/gadget>\n120<\/output>\n120 + 24<\/gadget>\n144<\/output>\n144<\/result>","index":2775} +{"problem":"if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 6 - m ) then m = ?","rationale":"( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 6 - m ) = 4 ^ ( ( 6 - m ) \/ 2 ) therefore , m = ( 6 - m ) \/ 2 2 m = 6 - m m = 2 answer b","correct":"b","options":{"a":"1 ","b":"2 ","c":"3 ","d":"5","e":"6"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"divide(6, add(2, const_1))","linear_formula":"add(n0,const_1)|divide(n3,#0)","chain":"2 + 1<\/gadget>\n3<\/output>\n6 \/ 3<\/gadget>\n2<\/output>\n2<\/result>","index":2776} +{"problem":"a train passes a platform in 44 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km \/ hr , the length of the platform is","rationale":"\"speed of the train = 54 km \/ hr = ( 54 × 10 ) \/ 36 m \/ s = 15 m \/ s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) \/ 15 = > ( 300 + l ) \/ 15 = 36 = > 300 + l = 15 × 44 = 660 = > l = 660 - 300 = 360 meter answer is c .\"","correct":"c","options":{"a":"240 ","b":"250 ","c":"360 ","d":"230","e":"220"},"options_float":{"a":240.0,"b":250.0,"c":360.0,"d":230.0,"e":220.0},"annotated_formula":"multiply(multiply(const_0_2778, 54), subtract(44, 20))","linear_formula":"multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 54<\/gadget>\n15<\/output>\n44 - 20<\/gadget>\n24<\/output>\n15 * 24<\/gadget>\n360<\/output>\n360<\/result>","index":2777} +{"problem":"exactly 35 % of the reporters for a certain wire service cover local politics in country x . if 30 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ?","rationale":"\"you are correct . people who cover local politics are a subset of people who cover politics . 30 % of reporters who cover politics do not cover local politics so 70 % do cover local politics . reporters covering local politics = 70 % of reporters covering politics = 35 % ofall reporters reporters covering politics \/ all reporters = 35 \/ 70 = 1 \/ 2 reporters not covering politics \/ all reporters = 1 \/ 2 = 50 % = d\"","correct":"d","options":{"a":"20 % ","b":"42 % ","c":"44 % ","d":"50 %","e":"84 %"},"options_float":{"a":20.0,"b":42.0,"c":44.0,"d":50.0,"e":84.0},"annotated_formula":"multiply(subtract(const_1, divide(35, subtract(const_100, 30))), const_100)","linear_formula":"subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"100 - 30<\/gadget>\n70<\/output>\n35 \/ 70<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":2778} +{"problem":"if rs . 10 be allowed as true discount on a bill of rs . 110 due at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is :","rationale":"\"solution s . i . on rs . ( 110 - 10 ) for a certain time = rs . 10 . s . i . on rs . 100 for double the time = rs . 20 . t . d . on rs . 120 = rs . ( 120 - 100 ) = rs . 20 . t . d . on rs . 110 = rs . ( 20 \/ 120 x 120 ) = rs . 18.33 . answer d\"","correct":"d","options":{"a":"rs . 20 ","b":"rs . 21.81 ","c":"rs . 22 ","d":"rs . 18.33","e":"none"},"options_float":{"a":20.0,"b":21.81,"c":22.0,"d":18.33,"e":null},"annotated_formula":"multiply(110, divide(subtract(add(10, 110), const_100), add(10, 110)))","linear_formula":"add(n0,n1)|subtract(#0,const_100)|divide(#1,#0)|multiply(n1,#2)|","chain":"10 + 110<\/gadget>\n120<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20 \/ 120<\/gadget>\n1\/6 = around 0.166667<\/output>\n110 * (1\/6)<\/gadget>\n55\/3 = around 18.333333<\/output>\n55\/3 = around 18.333333<\/result>","index":2779} +{"problem":"a shopkeeper gave an additional 20 per cent concession on the reduced price after giving 30 per cent standard concession on an article . if arun bought that article for 1120 , what was the original price ?","rationale":"original price = 1120 × 100 ⁄ 70 × 100 ⁄ 80 = 2000 answer d","correct":"d","options":{"a":"3000 ","b":"4000 ","c":"2400 ","d":"2000","e":"none of these"},"options_float":{"a":3000.0,"b":4000.0,"c":2400.0,"d":2000.0,"e":null},"annotated_formula":"divide(divide(multiply(multiply(1120, const_100), const_100), subtract(const_100, 20)), subtract(const_100, 30))","linear_formula":"multiply(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#0,const_100)|divide(#3,#1)|divide(#4,#2)","chain":"1_120 * 100<\/gadget>\n112_000<\/output>\n112_000 * 100<\/gadget>\n11_200_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n11_200_000 \/ 80<\/gadget>\n140_000<\/output>\n100 - 30<\/gadget>\n70<\/output>\n140_000 \/ 70<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2781} +{"problem":"if x , y , and z are positive integers and 3 x = 4 y = 7 z , then the least possible value of x - y + z is","rationale":"x - y + z = - 3 x \/ 4 + x + 3 x \/ 7 = ( - 21 + 28 + 12 ) x \/ 28 = 19 x \/ 28 19 is not divisible by 28 ( it ' s a prime # ) , so for least value , x = 28 answer - d","correct":"d","options":{"a":"33 ","b":"40 ","c":"49 ","d":"19","e":"61"},"options_float":{"a":33.0,"b":40.0,"c":49.0,"d":19.0,"e":61.0},"annotated_formula":"add(subtract(divide(multiply(multiply(3, 4), 7), 3), divide(multiply(multiply(3, 4), 7), 4)), divide(multiply(multiply(3, 4), 7), 7))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n2)|divide(#1,n0)|divide(#1,n1)|subtract(#3,#4)|add(#2,#5)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 * 7<\/gadget>\n84<\/output>\n84 \/ 3<\/gadget>\n28<\/output>\n84 \/ 4<\/gadget>\n21<\/output>\n28 - 21<\/gadget>\n7<\/output>\n84 \/ 7<\/gadget>\n12<\/output>\n7 + 12<\/gadget>\n19<\/output>\n19<\/result>","index":2782} +{"problem":"a man buys an article for $ 100 . and sells it for $ 130 . find the gain percent ?","rationale":"\"c . p . = $ 100 s . p . = $ 130 gain = $ 30 gain % = 30 \/ 100 * 100 = 30 % answer is e\"","correct":"e","options":{"a":"10 % ","b":"15 % ","c":"25 % ","d":"20 %","e":"30 %"},"options_float":{"a":10.0,"b":15.0,"c":25.0,"d":20.0,"e":30.0},"annotated_formula":"subtract(divide(130, divide(100, const_100)), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)|","chain":"100 \/ 100<\/gadget>\n1<\/output>\n130 \/ 1<\/gadget>\n130<\/output>\n130 - 100<\/gadget>\n30<\/output>\n30<\/result>","index":2784} +{"problem":"how many cubes of 20 cm edge can be put in a cubical box of 1 m edge .","rationale":"\"number of cubes = 100 â ˆ — 100 â ˆ — 100 \/ 20 â ˆ — 20 â ˆ — 20 = 125 note : 1 m = 100 cm answer : b\"","correct":"b","options":{"a":"177 cm ","b":"125 cm ","c":"786 cm ","d":"617 cm","e":"187 cm"},"options_float":{"a":177.0,"b":125.0,"c":786.0,"d":617.0,"e":187.0},"annotated_formula":"divide(volume_cube(1), volume_cube(divide(20, const_100)))","linear_formula":"divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)|","chain":"1 ** 3<\/gadget>\n1<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) ** 3<\/gadget>\n1\/125 = around 0.008<\/output>\n1 \/ (1\/125)<\/gadget>\n125<\/output>\n125<\/result>","index":2785} +{"problem":"3 houses are available in a new flat . 3 persons apply for the houses . each applies for 1 house without consulting others . the probability that all the 3 apply for the same house is ?","rationale":"one person can select one house out of 3 = \\ inline { \\ color { black } 3 c _ { 1 } } ways = 3 . hence , three persons can select one house out of 3 in 3 x 3 x 3 = 9 . therefore , probability that all thre apply for the same house is 1 \/ 9 b","correct":"b","options":{"a":"1 \/ 7 ","b":"1 \/ 9 ","c":"2 \/ 7 ","d":"3 \/ 7","e":"2 \/ 9"},"options_float":{"a":0.1428571429,"b":0.1111111111,"c":0.2857142857,"d":0.4285714286,"e":0.2222222222},"annotated_formula":"divide(1, multiply(choose(3, 1), choose(3, 1)))","linear_formula":"choose(n0,n2)|multiply(#0,#0)|divide(n2,#1)","chain":"binomial(3, 1)<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":2786} +{"problem":"if log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + 1 , then x is equal to :","rationale":"log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + 1 log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + log 10 10 log 10 [ 5 ( 5 x + 1 ) ] = log 10 [ 10 ( x + 5 ) ] 5 ( 5 x + 1 ) = 10 ( x + 5 ) 5 x + 1 = 2 x + 10 3 x = 9 = > x = 3 answer c","correct":"c","options":{"a":"1 ","b":"5 ","c":"3 ","d":"2","e":"6"},"options_float":{"a":1.0,"b":5.0,"c":3.0,"d":2.0,"e":6.0},"annotated_formula":"divide(subtract(10, 1), subtract(5, const_2))","linear_formula":"subtract(n0,n4)|subtract(n1,const_2)|divide(#0,#1)","chain":"10 - 1<\/gadget>\n9<\/output>\n5 - 2<\/gadget>\n3<\/output>\n9 \/ 3<\/gadget>\n3<\/output>\n3<\/result>","index":2787} +{"problem":"the sum of 4 consecutive even numbers is 140 . what would be the smallest number ?","rationale":"let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x - 2 ) = 32 answer : a","correct":"a","options":{"a":"32 ","b":"88 ","c":"70 ","d":"123","e":"12"},"options_float":{"a":32.0,"b":88.0,"c":70.0,"d":123.0,"e":12.0},"annotated_formula":"subtract(subtract(subtract(multiply(divide(140, add(4, 4)), const_2), const_1), const_1), const_1)","linear_formula":"add(n0,n0)|divide(n1,#0)|multiply(#1,const_2)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)","chain":"4 + 4<\/gadget>\n8<\/output>\n140 \/ 8<\/gadget>\n35\/2 = around 17.5<\/output>\n(35\/2) * 2<\/gadget>\n35<\/output>\n35 - 1<\/gadget>\n34<\/output>\n34 - 1<\/gadget>\n33<\/output>\n33 - 1<\/gadget>\n32<\/output>\n32<\/result>","index":2788} +{"problem":"an article is sold at a certain price . by selling it at 2 \/ 3 of that price one loses 10 % . find the gain percent at original price ?","rationale":"let the original s . p . be $ x then , new s . p . = $ 2 \/ 3 x loss = 10 % c . p . = 100 \/ 90 * 2 \/ 3 x = 20 x \/ 27 c . p . = 20 x \/ 27 gain = x - 20 x \/ 27 = 7 x \/ 27 gain % = 7 x \/ 27 * 27 \/ 20 x * 100 = 35 % answer is c","correct":"c","options":{"a":"25 % ","b":"20 % ","c":"35 % ","d":"40 %","e":"50 %"},"options_float":{"a":25.0,"b":20.0,"c":35.0,"d":40.0,"e":50.0},"annotated_formula":"subtract(divide(subtract(const_100, 10), divide(2, 3)), const_100)","linear_formula":"divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)","chain":"100 - 10<\/gadget>\n90<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n90 \/ (2\/3)<\/gadget>\n135<\/output>\n135 - 100<\/gadget>\n35<\/output>\n35<\/result>","index":2790} +{"problem":"in a sports club with 30 members , 17 play badminton and 19 play tennis and 3 do not play either . how many members play both badminton and tennis ?","rationale":"\"let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 19 - x ) + x + 3 = 30 39 - 2 x + x = 30 39 - x = 30 x = 9 so 9 members play both badminton and tennis . c\"","correct":"c","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"subtract(add(add(17, 19), 3), 30)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)|","chain":"17 + 19<\/gadget>\n36<\/output>\n36 + 3<\/gadget>\n39<\/output>\n39 - 30<\/gadget>\n9<\/output>\n9<\/result>","index":2791} +{"problem":"the local recycling plant had a contract requiring anyone who obtained recycled bottles for storage for them , to bring them back to be recycled again . the plant could make one new bottle from every 7 bottles returned . one week , on monday , they got 343 bottles to recycle . assuming that everybody brought back all the empties , how many could they eventually remake from the 343 ?","rationale":"in 1 st level they will make 343 \/ 7 = 49 bottles . in 2 nd level they will make 49 \/ 7 = 7 bottles . in 3 rd level they will make 7 \/ 7 = 1 bottle . totally they can remake 49 + 7 + 1 = 57 bottles answer : e","correct":"e","options":{"a":"53 ","b":"54 ","c":"55 ","d":"56","e":"57"},"options_float":{"a":53.0,"b":54.0,"c":55.0,"d":56.0,"e":57.0},"annotated_formula":"add(divide(343, 7), 7)","linear_formula":"divide(n1,n0)|add(n0,#0)","chain":"343 \/ 7<\/gadget>\n49<\/output>\n49 + 7<\/gadget>\n56<\/output>\n56<\/result>","index":2792} +{"problem":"the difference between the heights of two trees is 20 cm . 10 years ago , the taller one was thrice as taller as the shorter one . the present height of the taller tree is","rationale":"let their heights of x cm and ( x + 20 ) cm respectively . then , ( x + 20 ) - 10 = 3 ( x - 10 ) ⇔ x + 10 = 3 x - 30 ⇔ 2 x = 40 ∴ present height of taller tree = ( 20 + 10 ) = 30 cm . answer : 30 cm","correct":"a","options":{"a":"300 cm ","b":"15 cm ","c":"20 cm ","d":"10 cm","e":"40 cm"},"options_float":{"a":300.0,"b":15.0,"c":20.0,"d":10.0,"e":40.0},"annotated_formula":"multiply(add(20, 10), 10)","linear_formula":"add(n0,n1)|multiply(n1,#0)","chain":"20 + 10<\/gadget>\n30<\/output>\n30 * 10<\/gadget>\n300<\/output>\n300<\/result>","index":2794} +{"problem":"in an examination , 35 % of total students failed in hindi , 45 % failed in english and 20 % in both . the percentage of these who passed in both the subjects is :","rationale":"\"pass percentage = 100 - ( 35 + 45 - 20 ) = 100 - 60 = 40 answer : d\"","correct":"d","options":{"a":"10 % ","b":"20 % ","c":"30 % ","d":"40 %","e":"50 %"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"subtract(const_100, subtract(add(35, 45), 20))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|","chain":"35 + 45<\/gadget>\n80<\/output>\n80 - 20<\/gadget>\n60<\/output>\n100 - 60<\/gadget>\n40<\/output>\n40<\/result>","index":2796} +{"problem":"8 people went to a restaurant for dinner . 7 of them spent $ 10 each on their meals and the eighth spent $ 7 more than the average expenditure of all 8 people . what was the total amount of money that the 8 people spent ?","rationale":"let t be the total amount of money spent . 7 * 10 + ( t \/ 8 + 7 ) = t 77 = 7 t \/ 8 t = ( 8 * 77 ) \/ 7 = 88 the answer is b .","correct":"b","options":{"a":"$ 85 ","b":"$ 88 ","c":"$ 91 ","d":"$ 94","e":"$ 97"},"options_float":{"a":85.0,"b":88.0,"c":91.0,"d":94.0,"e":97.0},"annotated_formula":"add(multiply(7, 10), add(divide(add(multiply(7, 10), 7), subtract(8, const_1)), 7))","linear_formula":"multiply(n1,n2)|subtract(n0,const_1)|add(n1,#0)|divide(#2,#1)|add(n1,#3)|add(#4,#0)","chain":"7 * 10<\/gadget>\n70<\/output>\n70 + 7<\/gadget>\n77<\/output>\n8 - 1<\/gadget>\n7<\/output>\n77 \/ 7<\/gadget>\n11<\/output>\n11 + 7<\/gadget>\n18<\/output>\n70 + 18<\/gadget>\n88<\/output>\n88<\/result>","index":2797} +{"problem":"a person crosses a 1080 m long street in 12 minutes . what is his speed in km per hour ?","rationale":"\"speed = 1080 \/ ( 12 x 60 ) m \/ sec = 1.5 m \/ sec . converting m \/ sec to km \/ hr = 1.5 x ( 18 \/ 5 ) km \/ hr = 5.4 km \/ hr . answer : d\"","correct":"d","options":{"a":"4.1 ","b":"4.5 ","c":"4.8 ","d":"5.4","e":"5.5"},"options_float":{"a":4.1,"b":4.5,"c":4.8,"d":5.4,"e":5.5},"annotated_formula":"divide(divide(1080, const_1000), divide(multiply(12, const_60), const_3600))","linear_formula":"divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)|","chain":"1_080 \/ 1_000<\/gadget>\n27\/25 = around 1.08<\/output>\n12 * 60<\/gadget>\n720<\/output>\n720 \/ 3_600<\/gadget>\n1\/5 = around 0.2<\/output>\n(27\/25) \/ (1\/5)<\/gadget>\n27\/5 = around 5.4<\/output>\n27\/5 = around 5.4<\/result>","index":2798} +{"problem":"there were two candidates in an election . winner candidate received 62 % of votes and won the election by 384 votes . find the number of votes casted to the winning candidate ?","rationale":"\"w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 384 62 % - - - - - - - - ? = > 992 answer : c\"","correct":"c","options":{"a":"288 ","b":"744 ","c":"992 ","d":"298","e":"177"},"options_float":{"a":288.0,"b":744.0,"c":992.0,"d":298.0,"e":177.0},"annotated_formula":"divide(multiply(divide(384, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100)","linear_formula":"subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)|","chain":"100 - 62<\/gadget>\n38<\/output>\n62 - 38<\/gadget>\n24<\/output>\n24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n384 \/ (6\/25)<\/gadget>\n1_600<\/output>\n1_600 * 62<\/gadget>\n99_200<\/output>\n99_200 \/ 100<\/gadget>\n992<\/output>\n992<\/result>","index":2799} +{"problem":"a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 10 inches , then what is the maximum number of soap boxes that can be placed in each carton ?","rationale":"\"however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 42 * 60 dimensions of the soap = 10 * 6 * 7 we get = 5 x 6 x 5 = 150 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( d )\"","correct":"d","options":{"a":"210 ","b":"252 ","c":"280 ","d":"150","e":"420"},"options_float":{"a":210.0,"b":252.0,"c":280.0,"d":150.0,"e":420.0},"annotated_formula":"divide(multiply(multiply(25, 42), 60), multiply(multiply(7, 6), 10))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3)|","chain":"25 * 42<\/gadget>\n1_050<\/output>\n1_050 * 60<\/gadget>\n63_000<\/output>\n7 * 6<\/gadget>\n42<\/output>\n42 * 10<\/gadget>\n420<\/output>\n63_000 \/ 420<\/gadget>\n150<\/output>\n150<\/result>","index":2802} +{"problem":"selling an kite for rs . 30 , a shop keeper gains 25 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ?","rationale":"\"explanation : marked price = rs . 30 c . p . = 100 \/ 125 * 30 = rs . 24 sale price = 90 % of rs . 30 = rs . 27 required gain % = 3 \/ 24 * 100 = 12.5 % . answer : d\"","correct":"d","options":{"a":"8 % ","b":"10 % ","c":"11 % ","d":"12.5 %","e":"20 %"},"options_float":{"a":8.0,"b":10.0,"c":11.0,"d":12.5,"e":20.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(25, const_100))), divide(multiply(30, const_100), add(25, const_100))), const_100)","linear_formula":"add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n100 - 10<\/gadget>\n90<\/output>\n(3\/10) * 90<\/gadget>\n27<\/output>\n30 * 100<\/gadget>\n3_000<\/output>\n25 + 100<\/gadget>\n125<\/output>\n3_000 \/ 125<\/gadget>\n24<\/output>\n27 - 24<\/gadget>\n3<\/output>\n3 \/ 24<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":2803} +{"problem":"in a university the average age in an arts class is 21 and average age in technical class is 18 . what is the age average of university if there are 8 arts classes and 5 technical classes ?","rationale":"average age of university = ( ( 8 x 21 ) + ( 5 x 18 ) ) \/ 13 = 19.85 answer : b","correct":"b","options":{"a":"18.75 ","b":"19.85 ","c":"20.32 ","d":"20.9","e":"21.0"},"options_float":{"a":18.75,"b":19.85,"c":20.32,"d":20.9,"e":21.0},"annotated_formula":"divide(add(multiply(8, 21), multiply(5, 18)), add(8, 5))","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|divide(#3,#0)","chain":"8 * 21<\/gadget>\n168<\/output>\n5 * 18<\/gadget>\n90<\/output>\n168 + 90<\/gadget>\n258<\/output>\n8 + 5<\/gadget>\n13<\/output>\n258 \/ 13<\/gadget>\n258\/13 = around 19.846154<\/output>\n258\/13 = around 19.846154<\/result>","index":2804} +{"problem":"a theater box office sold an average ( arithmetic mean ) of 65 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 75 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ?","rationale":"\"deviation from the mean for the daytime staff = 75 - 65 = 10 . deviation from the mean for the evening staff = 65 - 60 = 5 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 5 : 10 = 1 : 2 . the answer is a .\"","correct":"a","options":{"a":"1 : 2 ","b":"1 : 4 ","c":"5 : 11 ","d":"1 : 3","e":"13 : 15"},"options_float":{"a":0.5,"b":0.25,"c":0.4545454545,"d":0.3333333333,"e":0.8666666667},"annotated_formula":"divide(subtract(65, 60), subtract(75, 65))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|","chain":"65 - 60<\/gadget>\n5<\/output>\n75 - 65<\/gadget>\n10<\/output>\n5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2805} +{"problem":"in a kilometer race , if abhishek gives bharti a 40 m start , abhishek wins by 19 sec . but if abhishek gives bharti a 30 sec start , bharti wins by 40 m . find the time taken by bharti to run 5,000 m ?","rationale":"explanation : if abhishek takes x seconds and bharti takes y seconds to run 1 km , then : = > y = 150 sec and x = 125 sec = > x + 19 = 960 y \/ 1000 and = ( 960 x \/ 1000 ) + 30 = y = > y = 150 sec and x = 125 sec answer = ( 150 \/ 1000 ) × 500 = 750 sec answer : b","correct":"b","options":{"a":"150 sec ","b":"750 sec ","c":"450 sec ","d":"825 sec","e":"none of these"},"options_float":{"a":150.0,"b":750.0,"c":450.0,"d":825.0,"e":null},"annotated_formula":"multiply(add(multiply(const_1000, const_4), const_1000), divide(add(multiply(divide(subtract(multiply(30, divide(subtract(const_1000, 40), const_1000)), 19), subtract(const_1, power(divide(subtract(const_1000, 40), const_1000), const_2))), divide(subtract(const_1000, 40), const_1000)), 30), const_1000))","linear_formula":"multiply(const_1000,const_4)|subtract(const_1000,n0)|add(#0,const_1000)|divide(#1,const_1000)|multiply(n2,#3)|power(#3,const_2)|subtract(#4,n1)|subtract(const_1,#5)|divide(#6,#7)|multiply(#8,#3)|add(n2,#9)|divide(#10,const_1000)|multiply(#2,#11)","chain":"1_000 * 4<\/gadget>\n4_000<\/output>\n4_000 + 1_000<\/gadget>\n5_000<\/output>\n1_000 - 40<\/gadget>\n960<\/output>\n960 \/ 1_000<\/gadget>\n24\/25 = around 0.96<\/output>\n30 * (24\/25)<\/gadget>\n144\/5 = around 28.8<\/output>\n(144\/5) - 19<\/gadget>\n49\/5 = around 9.8<\/output>\n(24\/25) ** 2<\/gadget>\n576\/625 = around 0.9216<\/output>\n1 - (576\/625)<\/gadget>\n49\/625 = around 0.0784<\/output>\n(49\/5) \/ (49\/625)<\/gadget>\n125<\/output>\n125 * (24\/25)<\/gadget>\n120<\/output>\n120 + 30<\/gadget>\n150<\/output>\n150 \/ 1_000<\/gadget>\n3\/20 = around 0.15<\/output>\n5_000 * (3\/20)<\/gadget>\n750<\/output>\n750<\/result>","index":2808} +{"problem":"every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 144 , how many more green disks than blue disks are in the bag ?","rationale":"\"let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 144 - - > x = 8 . g - b = 8 x - 3 x = 5 x = 40 . the answer is e .\"","correct":"e","options":{"a":"25 ","b":"28 ","c":"30 ","d":"35","e":"40"},"options_float":{"a":25.0,"b":28.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"multiply(divide(144, add(add(3, 7), 8)), subtract(8, 3))","linear_formula":"add(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(n3,#2)|multiply(#3,#1)|","chain":"3 + 7<\/gadget>\n10<\/output>\n10 + 8<\/gadget>\n18<\/output>\n144 \/ 18<\/gadget>\n8<\/output>\n8 - 3<\/gadget>\n5<\/output>\n8 * 5<\/gadget>\n40<\/output>\n40<\/result>","index":2809} +{"problem":"if the sum of two numbers is 22 and the sum of their squares is 404 , then the product of the numbers is","rationale":"\"according to the given conditions x + y = 22 and x ^ 2 + y ^ 2 = 404 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 22 ^ 2 = 404 + 2 xy so xy = 80 \/ 2 = 40 answer : a\"","correct":"a","options":{"a":"40 ","b":"44 ","c":"80 ","d":"88","e":"48"},"options_float":{"a":40.0,"b":44.0,"c":80.0,"d":88.0,"e":48.0},"annotated_formula":"divide(subtract(power(22, const_2), 404), const_2)","linear_formula":"power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|","chain":"22 ** 2<\/gadget>\n484<\/output>\n484 - 404<\/gadget>\n80<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40<\/result>","index":2810} +{"problem":"there are 200 female managers in a certain company . find the total number w of female employees in the company , if 2 \/ 5 of all the employees are managers and 2 \/ 5 of all male employees are managers .","rationale":"\"{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 \/ 5 of all the employees , thus { managers } = 2 \/ 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 \/ 5 of all male employees are managers : { male managers } = 2 \/ 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 \/ 5 ( m + f ) = 200 + 2 \/ 5 * m - - > f = 500 . answer : c .\"","correct":"c","options":{"a":"300 ","b":"w = 400 ","c":"w = 500 ","d":"w = 600","e":"none of these"},"options_float":{"a":300.0,"b":400.0,"c":500.0,"d":600.0,"e":null},"annotated_formula":"divide(200, divide(2, 5))","linear_formula":"divide(n1,n2)|divide(n0,#0)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n200 \/ (2\/5)<\/gadget>\n500<\/output>\n500<\/result>","index":2811} +{"problem":"a store reduced the price of all items in the store by 10 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ?","rationale":"\"consider price of the all items as $ 100 after a initial reduction of 10 % price becomes = 0.9 * 100 = $ 90 after the final reduction of 10 % price becomes = 0.9 * 90 = $ 81 price of all items on second day is 81 % of price on first day correct answer option c\"","correct":"c","options":{"a":"80.0 ","b":"80.9 ","c":"81.0 ","d":"81.1","e":"81.9"},"options_float":{"a":80.0,"b":80.9,"c":81.0,"d":81.1,"e":81.9},"annotated_formula":"multiply(multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 10), const_100)), const_100)","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * (9\/10)<\/gadget>\n81\/100 = around 0.81<\/output>\n(81\/100) * 100<\/gadget>\n81<\/output>\n81<\/result>","index":2812} +{"problem":"on the first of the year , james invested x dollars at proudstar bank in an account that yields 0.5 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ?","rationale":"per quarter , interest = 0.5 % so for a year , interest = 2 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 2 % answer = 2.04 % = answer : a","correct":"a","options":{"a":"2.04 % ","b":"4.12 % ","c":"5 % ","d":"8.25 %","e":"10 %"},"options_float":{"a":2.04,"b":4.12,"c":5.0,"d":8.25,"e":10.0},"annotated_formula":"multiply(0.5, const_4)","linear_formula":"multiply(n0,const_4)","chain":"0.5 * 4<\/gadget>\n2<\/output>\n2<\/result>","index":2813} +{"problem":"the chance of rain on any given day in tel - aviv is 50 % . what is the probability w that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?","rationale":"chances of rain on exactly 4 days and not rain on 2 days = ( 1 \/ 2 ) ^ 4 * ( 1 \/ 2 ) ^ 2 = 1 \/ 64 chosing 4 days out of 6 = 6 ! \/ ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days w = 15 \/ 64 ans a it is .","correct":"a","options":{"a":"15 \/ 64 ","b":"30 \/ 64 ","c":"1 \/ 2 ","d":"1 \/ 4","e":"52 \/ 64"},"options_float":{"a":0.234375,"b":0.46875,"c":0.5,"d":0.25,"e":0.8125},"annotated_formula":"divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), const_4), power(divide(const_1, const_2), const_2))))","linear_formula":"choose(n2,n1)|divide(const_1,const_2)|power(#1,const_4)|power(#1,const_2)|multiply(#2,#3)|inverse(#4)|divide(#0,#5)","chain":"binomial(6, 4)<\/gadget>\n15<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 4<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/2) ** 2<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/16) * (1\/4)<\/gadget>\n1\/64 = around 0.015625<\/output>\n1 \/ (1\/64)<\/gadget>\n64<\/output>\n15 \/ 64<\/gadget>\n15\/64 = around 0.234375<\/output>\n15\/64 = around 0.234375<\/result>","index":2814} +{"problem":"the sum of two numbers is 77 . 5 times one number is equal to 6 times the other . the bigger of the two numbers is","rationale":"a : b 5 : 6 a + b = 77 5 x + 6 x = 77 x = 7 then two numbers are 35,42 . the bigger number is 42 answer : b","correct":"b","options":{"a":"10 ","b":"42 ","c":"14 ","d":"15","e":"60"},"options_float":{"a":10.0,"b":42.0,"c":14.0,"d":15.0,"e":60.0},"annotated_formula":"subtract(77, divide(77, add(divide(6, 5), const_1)))","linear_formula":"divide(n2,n1)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)","chain":"6 \/ 5<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) + 1<\/gadget>\n11\/5 = around 2.2<\/output>\n77 \/ (11\/5)<\/gadget>\n35<\/output>\n77 - 35<\/gadget>\n42<\/output>\n42<\/result>","index":2815} +{"problem":"income and expenditure of a person are in the ratio 7 : 6 . if the income of the person is rs . 14000 , then find his savings ?","rationale":"\"let the income and the expenditure of the person be rs . 7 x and rs . 6 x respectively . income , 7 x = 14000 = > x = 2000 savings = income - expenditure = 7 x - 6 x = x so , savings = rs . 2000 . answer : d\"","correct":"d","options":{"a":"500 ","b":"1000 ","c":"1500 ","d":"2000","e":"2500"},"options_float":{"a":500.0,"b":1000.0,"c":1500.0,"d":2000.0,"e":2500.0},"annotated_formula":"subtract(14000, multiply(divide(6, 7), 14000))","linear_formula":"divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)|","chain":"6 \/ 7<\/gadget>\n6\/7 = around 0.857143<\/output>\n(6\/7) * 14_000<\/gadget>\n12_000<\/output>\n14_000 - 12_000<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2816} +{"problem":"3 people took gmat practice tests in which the minimum and maximum possible scores are 200 and 800 , respectively . they each took a test 3 times , and no one scored below 400 or over 700 . if the individual ranges of the 3 people ' s scores in those 3 practice tests were 40 , 70 and 100 , what is the difference between the maximum and minimum possible ranges of all their scores put together .","rationale":"according to the information in the question the maximum range would be between 400 and 700 ( 300 ) . the minimum range would be the greatest range for any one individual , which is listed in the problem as 100 . so the difference between these would be 300 - 100 or 200 . answer : e","correct":"e","options":{"a":"70 ","b":"80 ","c":"100 ","d":"140","e":"200"},"options_float":{"a":70.0,"b":80.0,"c":100.0,"d":140.0,"e":200.0},"annotated_formula":"subtract(subtract(700, 400), 100)","linear_formula":"subtract(n5,n4)|subtract(#0,n10)","chain":"700 - 400<\/gadget>\n300<\/output>\n300 - 100<\/gadget>\n200<\/output>\n200<\/result>","index":2817} +{"problem":"if p \/ q = 3 \/ 11 , then 2 p + q = ?","rationale":"let p = 3 , q = 11 then 2 * 3 + 11 = 17 so 2 p + q = 17 . answer : c","correct":"c","options":{"a":"12 ","b":"14 ","c":"17 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":14.0,"c":17.0,"d":15.0,"e":16.0},"annotated_formula":"add(multiply(3, 2), 11)","linear_formula":"multiply(n0,n2)|add(n1,#0)","chain":"3 * 2<\/gadget>\n6<\/output>\n6 + 11<\/gadget>\n17<\/output>\n17<\/result>","index":2818} +{"problem":"what is the probability of rolling a total of 5 with a single roll of two fair 6 - sided dice , each with the distinct numbers 1 through 6 on each side ?","rationale":"totalsample spacefrom two fair six - sided dice is 36 . there are six possible ways to be 5 i . e . ( 1 + 4 ) , ( 2 + 3 ) , ( 3 + 2 ) , ( 4 + 1 ) so , total probability is 4 \/ 36 or 1 \/ 9 . answer is e .","correct":"e","options":{"a":"1 \/ 12 ","b":"1 \/ 6 ","c":"2 \/ 7 ","d":"1 \/ 3","e":"1 \/ 9"},"options_float":{"a":0.0833333333,"b":0.1666666667,"c":0.2857142857,"d":0.3333333333,"e":0.1111111111},"annotated_formula":"divide(const_4, multiply(6, 6))","linear_formula":"multiply(n1,n1)|divide(const_4,#0)","chain":"6 * 6<\/gadget>\n36<\/output>\n4 \/ 36<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":2819} +{"problem":"if the simple interest on a certain sum of money for 6 years is one – fifth of the sum , then the rate of interest per annum is","rationale":"\"explanation : let the principal ( p ) be x then , simple interest ( si ) = x \/ 5 time ( t ) = 6 years rate of interest per annum ( r ) = ( 100 × si ) \/ pt = ( 100 × ( x \/ 5 ) \/ ( x × 6 ) = 20 \/ 6 = 3.3 % answer : option a\"","correct":"a","options":{"a":"3.3 % ","b":"7.2 % ","c":"6.1 % ","d":"5.0 %","e":"3.1 %"},"options_float":{"a":3.3,"b":7.2,"c":6.1,"d":5.0,"e":3.1},"annotated_formula":"divide(divide(const_100, add(const_1, const_4)), 6)","linear_formula":"add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0)|","chain":"1 + 4<\/gadget>\n5<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20 \/ 6<\/gadget>\n10\/3 = around 3.333333<\/output>\n10\/3 = around 3.333333<\/result>","index":2820} +{"problem":"the population of a city is 415600 . it increased by 25 % in the first year and decreased by 30 % in the second year . what is the population of the city at the end of second year ?","rationale":"as per given data , p = 415600 , r 1 = 25 % incresed , r 2 = 30 % decreased population of the city at the end of the second year = p ( 1 + r 1 \/ 100 ) ( 1 - r 2 \/ 100 ) = 415600 ( 1 + 25 \/ 100 ) ( 1 - 30 \/ 100 ) = 363650 answer : c","correct":"c","options":{"a":"367300 ","b":"363580 ","c":"363650 ","d":"373650","e":"386750"},"options_float":{"a":367300.0,"b":363580.0,"c":363650.0,"d":373650.0,"e":386750.0},"annotated_formula":"multiply(divide(subtract(const_100, 30), const_100), multiply(divide(add(25, const_100), const_100), 415600))","linear_formula":"add(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,const_100)|multiply(n0,#3)|multiply(#2,#4)","chain":"100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n25 + 100<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 415_600<\/gadget>\n519_500<\/output>\n(7\/10) * 519_500<\/gadget>\n363_650<\/output>\n363_650<\/result>","index":2821} +{"problem":"a thief is spotted by a policeman from a distance of 150 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km \/ hr and that of the policeman 10 km \/ hr , how far the thief will have run before he is overtaken ?","rationale":"\"relative speed of the policeman = ( 10 - 8 ) km \/ hr = 2 km \/ hr . time taken by police man to cover ( 150 m \/ 1000 ) x 1 \/ 2 hr = 3 \/ 40 hr . in 3 \/ 40 hrs , the thief covers a distance of 8 x 3 \/ 40 km = 3 \/ 5 km = 600 m answer is d\"","correct":"d","options":{"a":"350 m ","b":"200 m ","c":"400 m ","d":"600 m","e":"none of them"},"options_float":{"a":350.0,"b":200.0,"c":400.0,"d":600.0,"e":null},"annotated_formula":"divide(multiply(150, 8), subtract(10, 8))","linear_formula":"multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)|","chain":"150 * 8<\/gadget>\n1_200<\/output>\n10 - 8<\/gadget>\n2<\/output>\n1_200 \/ 2<\/gadget>\n600<\/output>\n600<\/result>","index":2823} +{"problem":"car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 70 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?","rationale":"\"car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 70 t t = 1.2 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 1.2 = 42 miles answer : - d\"","correct":"d","options":{"a":"15 ","b":"20 ","c":"40 ","d":"42","e":"68"},"options_float":{"a":15.0,"b":20.0,"c":40.0,"d":42.0,"e":68.0},"annotated_formula":"multiply(35, divide(multiply(divide(72, const_60), 35), subtract(70, 35)))","linear_formula":"divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)|","chain":"72 \/ 60<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 35<\/gadget>\n42<\/output>\n70 - 35<\/gadget>\n35<\/output>\n42 \/ 35<\/gadget>\n6\/5 = around 1.2<\/output>\n35 * (6\/5)<\/gadget>\n42<\/output>\n42<\/result>","index":2824} +{"problem":"a fruit juice company earns $ 150 per 100 gallons of apple juice sold , $ 170 per 100 gallons of guava juice sold , $ 130 per 100 gallons of grape juice sold , $ 100 per 100 gallons of orange juice . if the company manages to sell 230 gallons of apple , 210 gallons of guava , 310 gallons of grape , 270 gallons of orange on a particular day . what is the average earning of the company per 100 gallon of the juice sold ?","rationale":"earning of the company from apple = 150 \/ 100 x 230 = 345 earning of the company from guava = 170 \/ 100 x 210 = 357 earning of the company from grape = 130 \/ 100 x 310 = 403 earning of the company from orange = 100 \/ 100 x 270 = 270 total earning of the company = 345 + 357 + 403 + 270 = 1375 total juice sold = 1020 average earning per 100 gallon of juice sold = ( 1375 \/ 1020 ) x 100 = $ 134.8 answer : a","correct":"a","options":{"a":"134.8 ","b":"141.2 ","c":"147.2 ","d":"152.3","e":"160.1"},"options_float":{"a":134.8,"b":141.2,"c":147.2,"d":152.3,"e":160.1},"annotated_formula":"add(add(130, const_4), divide(add(const_4, const_4), const_10))","linear_formula":"add(n4,const_4)|add(const_4,const_4)|divide(#1,const_10)|add(#0,#2)","chain":"130 + 4<\/gadget>\n134<\/output>\n4 + 4<\/gadget>\n8<\/output>\n8 \/ 10<\/gadget>\n4\/5 = around 0.8<\/output>\n134 + (4\/5)<\/gadget>\n674\/5 = around 134.8<\/output>\n674\/5 = around 134.8<\/result>","index":2825} +{"problem":"if a mixture is 1 ⁄ 7 alcohol by volume and 2 ⁄ 7 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ?","rationale":"\"should be a sub - 600 level q . . volume = { 1 \/ 7 } \/ { 2 \/ 7 } = 1 \/ 2 c\"","correct":"c","options":{"a":"4 \/ 2 ","b":"3 \/ 2 ","c":"1 \/ 2 ","d":"5 \/ 2","e":"7 \/ 2"},"options_float":{"a":2.0,"b":1.5,"c":0.5,"d":2.5,"e":3.5},"annotated_formula":"divide(divide(1, 7), divide(2, 7))","linear_formula":"divide(n0,n1)|divide(n2,n1)|divide(#0,#1)|","chain":"1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n(1\/7) \/ (2\/7)<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2826} +{"problem":"the average of 10 numbers is 7 . if each number is multiplied by 12 , then the average of new set of numbers is","rationale":"explanation : if each number is multiplied by k , then the new average increases by k times . so new average = 84 answer : d","correct":"d","options":{"a":"28 ","b":"236 ","c":"288 ","d":"84","e":"26"},"options_float":{"a":28.0,"b":236.0,"c":288.0,"d":84.0,"e":26.0},"annotated_formula":"multiply(7, 12)","linear_formula":"multiply(n1,n2)","chain":"7 * 12<\/gadget>\n84<\/output>\n84<\/result>","index":2828} +{"problem":"a train 500 m long can cross an electric pole in 50 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 500 \/ 50 s = 10 m \/ sec speed = 10 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 36 kmph answer : e\"","correct":"e","options":{"a":"76 kmph ","b":"54 kmph ","c":"72 kmph ","d":"34 kmph","e":"36 kmph"},"options_float":{"a":76.0,"b":54.0,"c":72.0,"d":34.0,"e":36.0},"annotated_formula":"divide(divide(500, const_1000), divide(50, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"500 \/ 1_000<\/gadget>\n1\/2 = around 0.5<\/output>\n50 \/ 3_600<\/gadget>\n1\/72 = around 0.013889<\/output>\n(1\/2) \/ (1\/72)<\/gadget>\n36<\/output>\n36<\/result>","index":2829} +{"problem":"the length of a rectangle is increased by 40 % while its width is halved . what is the % change in area ?","rationale":"the original area is l * w the new area is 1.4 l * 0.5 w = 0.7 * l * w = l * w - 0.3 * l * w the area decreased by 30 % . the answer is b .","correct":"b","options":{"a":"25 % ","b":"30 % ","c":"50 % ","d":"65 %","e":"70 %"},"options_float":{"a":25.0,"b":30.0,"c":50.0,"d":65.0,"e":70.0},"annotated_formula":"multiply(subtract(const_1, multiply(divide(const_1, const_2), add(divide(40, const_100), const_1))), const_100)","linear_formula":"divide(n0,const_100)|divide(const_1,const_2)|add(#0,const_1)|multiply(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) + 1<\/gadget>\n7\/5 = around 1.4<\/output>\n(1\/2) * (7\/5)<\/gadget>\n7\/10 = around 0.7<\/output>\n1 - (7\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":2830} +{"problem":"if 5 x = 3 y and xy ≠ 0 , what is the ratio of 1 \/ 5 * x to 1 \/ 6 * y ?","rationale":"\"5 x = 3 y = > x \/ y = 3 \/ 5 1 \/ 5 * x to 1 \/ 6 * y = x \/ y * 6 \/ 5 = ( 3 \/ 5 ) * ( 6 \/ 5 ) = 18 \/ 25 ans : c\"","correct":"c","options":{"a":"25 \/ 6 ","b":"36 \/ 25 ","c":"18 \/ 25 ","d":"5 \/ 6","e":"25 \/ 36"},"options_float":{"a":4.1666666667,"b":1.44,"c":0.72,"d":0.8333333333,"e":0.6944444444},"annotated_formula":"multiply(divide(divide(1, 5), divide(1, 6)), divide(3, 5))","linear_formula":"divide(n3,n4)|divide(n3,n6)|divide(n1,n0)|divide(#0,#1)|multiply(#3,#2)|","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/5) \/ (1\/6)<\/gadget>\n6\/5 = around 1.2<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(6\/5) * (3\/5)<\/gadget>\n18\/25 = around 0.72<\/output>\n18\/25 = around 0.72<\/result>","index":2831} +{"problem":"for every integer n ≥ 3 , the function g ( n ) is defined as the product of all the odd integers from 1 to n , inclusive . what is the value of g ( 101 ) – g ( 99 ) ?","rationale":"\"g ( 101 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 g ( 99 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 g ( 101 ) - g ( 99 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 - 1 * 3 * 5 * 7 * 9 * . . . * 99 = 1 * 3 * 5 * 7 * 9 * . . . * 99 * ( 101 - 1 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 100 hence : d\"","correct":"d","options":{"a":"0 ","b":"98 ","c":"99 ","d":"100","e":"90"},"options_float":{"a":0.0,"b":98.0,"c":99.0,"d":100.0,"e":90.0},"annotated_formula":"subtract(101, 1)","linear_formula":"subtract(n2,n1)|","chain":"101 - 1<\/gadget>\n100<\/output>\n100<\/result>","index":2833} +{"problem":"two trains of equal length are running on parallel lines in the same directions at 46 km \/ hr . and 36 km \/ hr . the faster trains pass the slower train in 72 seconds . the length of each train is :","rationale":"\"explanation : the relative speed of train is 46 - 36 = 10 km \/ hr = ( 10 x 5 ) \/ 18 = 25 \/ 9 m \/ s 10 × 518 = 259 m \/ s in 72 secs the total distance traveled is 72 x 25 \/ 9 = 200 m . therefore the length of each train is = 200 \/ 2 = 100 m . answer d\"","correct":"d","options":{"a":"82 m ","b":"50 m ","c":"72 m ","d":"100 m","e":"none of these"},"options_float":{"a":82.0,"b":50.0,"c":72.0,"d":100.0,"e":null},"annotated_formula":"divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 72), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|","chain":"46 - 36<\/gadget>\n10<\/output>\n10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 3_600<\/gadget>\n25\/9 = around 2.777778<\/output>\n(25\/9) * 72<\/gadget>\n200<\/output>\n200 \/ 2<\/gadget>\n100<\/output>\n100<\/result>","index":2834} +{"problem":"a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 30 % ?","rationale":"c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 180 % of re . 1 = rs . 9 \/ 5 for rs . 9 \/ 5 , toffees sold = 6 . for re . 1 , toffees sold = 9 x 5 \/ 6 = 15 \/ 2 . answer e","correct":"e","options":{"a":"1 \/ 2 ","b":"2 \/ 6 ","c":"3 \/ 6 ","d":"9 \/ 6","e":"15 \/ 2"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.5,"d":1.5,"e":7.5},"annotated_formula":"multiply(6, add(const_1, divide(30, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n6 * (13\/10)<\/gadget>\n39\/5 = around 7.8<\/output>\n39\/5 = around 7.8<\/result>","index":2835} +{"problem":"how many seconds will a train 140 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?","rationale":"\"d = 140 + 150 = 290 s = 36 * 5 \/ 18 = 10 mps t = 290 \/ 10 = 29 sec b\"","correct":"b","options":{"a":"28 sec ","b":"29 sec ","c":"24 sec ","d":"25 sec","e":"26 sec"},"options_float":{"a":28.0,"b":29.0,"c":24.0,"d":25.0,"e":26.0},"annotated_formula":"divide(add(150, 140), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"150 + 140<\/gadget>\n290<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n290 \/ 10<\/gadget>\n29<\/output>\n29<\/result>","index":2836} +{"problem":"sophia finished 2323 of a book . she calculated that she finished 90 more pages than she has yet to read . how long is her book ?","rationale":"let x be the total number of pages in the book , then she finished 23 ⋅ x pages . then she has x − 23 ⋅ x = 13 ⋅ xpages left . 23 ⋅ x − 13 ⋅ x = 90 13 ⋅ x = 90 x = 270 so the book is 270 pages long . correct answer is d ) 270","correct":"d","options":{"a":"140 ","b":"320 ","c":"745 ","d":"270","e":"350"},"options_float":{"a":140.0,"b":320.0,"c":745.0,"d":270.0,"e":350.0},"annotated_formula":"divide(90, subtract(divide(const_2, const_3), subtract(const_1, divide(const_2, const_3))))","linear_formula":"divide(const_2,const_3)|subtract(const_1,#0)|subtract(#0,#1)|divide(n1,#2)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(2\/3) - (1\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n90 \/ (1\/3)<\/gadget>\n270<\/output>\n270<\/result>","index":2838} +{"problem":"on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 ⁄ 3 of the total distance and again after she traveled 1 ⁄ 4 of the distance remaining between her first stop and her destination . she then drove the remaining 300 miles and arrived safely at her destination . what was the total distance , in miles , from traci ’ s starting point to seal beach ?","rationale":"let d = total distance traci traveled 1 \/ 3 = d \/ 3 i . e . remaining distance = 2 d \/ 3 she traveled 1 \/ 4 th of 2 d \/ 3 = d \/ 6 thus : d = ( d \/ 3 ) + ( d \/ 6 ) + 300 d = 600 answer : a","correct":"a","options":{"a":"600 ","b":"300 ","c":"350 ","d":"400","e":"550"},"options_float":{"a":600.0,"b":300.0,"c":350.0,"d":400.0,"e":550.0},"annotated_formula":"multiply(300, const_2)","linear_formula":"multiply(n4,const_2)","chain":"300 * 2<\/gadget>\n600<\/output>\n600<\/result>","index":2839} +{"problem":"the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 4 . the marks obtained by vijay and abhishek are in the ratio of ?","rationale":"\"4 : 5 3 : 4 - - - - - - - 12 : 15 : 20 12 : 20 - - - - - - > 6 : 10 3 : 5 answer : b\"","correct":"b","options":{"a":"3 : 4 ","b":"3 : 5 ","c":"6 : 5 ","d":"3 : 6","e":"3 : 8"},"options_float":{"a":0.75,"b":0.6,"c":1.2,"d":0.5,"e":0.375},"annotated_formula":"divide(multiply(4, 3), multiply(5, 4))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)|","chain":"4 * 3<\/gadget>\n12<\/output>\n5 * 4<\/gadget>\n20<\/output>\n12 \/ 20<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":2841} +{"problem":"the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 1200 per sq . metre .","rationale":"\"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1200 x 20.625 ) = rs . 24,750 . answer b\"","correct":"b","options":{"a":"rs . 15,000 ","b":"rs . 24,750 ","c":"rs . 15,600 ","d":"rs . 16,500","e":"none"},"options_float":{"a":15000.0,"b":24750.0,"c":15600.0,"d":16500.0,"e":null},"annotated_formula":"multiply(1200, multiply(5.5, 3.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 3.75<\/gadget>\n20.625<\/output>\n1_200 * 20.625<\/gadget>\n24_750<\/output>\n24_750<\/result>","index":2842} +{"problem":"how many zeroes are there at the end of the number n , if n = 70 ! + 140 ! ?","rationale":"\"the number of zeroes at the end of 70 ! will be less than the number of zeroes at the end of 140 ! hence it is sufficient to calculate the number of zeroes at the end of 70 ! the number of zeroes = [ 70 \/ 5 ] + [ 70 \/ 25 ] + [ 70 \/ 125 ] = 14 + 2 + 0 = 16 the answer is c .\"","correct":"c","options":{"a":"8 ","b":"12 ","c":"16 ","d":"20","e":"24"},"options_float":{"a":8.0,"b":12.0,"c":16.0,"d":20.0,"e":24.0},"annotated_formula":"add(divide(70, add(const_4, const_1)), const_2)","linear_formula":"add(const_1,const_4)|divide(n0,#0)|add(#1,const_2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n70 \/ 5<\/gadget>\n14<\/output>\n14 + 2<\/gadget>\n16<\/output>\n16<\/result>","index":2845} +{"problem":"if 15 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ?","rationale":"\"let x be the number of students in the school . 0.15 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.15 x \/ 0.25 = 0.6 x students went to the camping which is 60 % . the answer is e\"","correct":"e","options":{"a":"95 ","b":"90 ","c":"85 ","d":"80","e":"60"},"options_float":{"a":95.0,"b":90.0,"c":85.0,"d":80.0,"e":60.0},"annotated_formula":"divide(15, divide(subtract(100, 75), 100))","linear_formula":"subtract(n1,n2)|divide(#0,n1)|divide(n0,#1)|","chain":"100 - 75<\/gadget>\n25<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n15 \/ (1\/4)<\/gadget>\n60<\/output>\n60<\/result>","index":2848} +{"problem":"an object thrown directly upward is at a height of h feet after t seconds , where h = - 14 ( t - 3 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?","rationale":"\"we see that h will be a maximum h = 150 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 14 ( 5 - 3 ) ^ 2 + 150 = - 14 ( 4 ) + 150 = 94 the answer is c .\"","correct":"c","options":{"a":"80 ","b":"88 ","c":"94 ","d":"112","e":"124"},"options_float":{"a":80.0,"b":88.0,"c":94.0,"d":112.0,"e":124.0},"annotated_formula":"add(multiply(negate(14), power(subtract(add(3, 2), 3), 2)), 150)","linear_formula":"add(n1,n2)|negate(n0)|subtract(#0,n1)|power(#2,n2)|multiply(#1,#3)|add(n3,#4)|","chain":"-14<\/gadget>\n-14<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 - 3<\/gadget>\n2<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n(-14) * 4<\/gadget>\n-56<\/output>\n(-56) + 150<\/gadget>\n94<\/output>\n94<\/result>","index":2849} +{"problem":"how many cones of radius 4 m , height 2 m can be formed from a cylinder of 12 m radius , 10 m height :","rationale":"( cylinder volume ) \/ ( cone volume ) ( π r ( power 2 ) h ) \/ ( 1 \/ 3 π r ( power 2 ) h ) = ( π × 12 × 12 × 10 ) \/ ( 1 \/ 3 π × 4 × 4 × 2 ) = 135 answer is a .","correct":"a","options":{"a":"135 ","b":"125 ","c":"145 ","d":"115","e":"155"},"options_float":{"a":135.0,"b":125.0,"c":145.0,"d":115.0,"e":155.0},"annotated_formula":"divide(volume_cylinder(12, 10), volume_cone(4, 2))","linear_formula":"volume_cone(n0,n1)|volume_cylinder(n2,n3)|divide(#1,#0)","chain":"pi * (12 ** 2) * 10<\/gadget>\n1440*pi = around 4_523.893421<\/output>\npi * (4 ** 2) * 2 \/ 3<\/gadget>\n32*pi\/3 = around 33.510322<\/output>\n(1440*pi) \/ (32*pi\/3)<\/gadget>\n135<\/output>\n135<\/result>","index":2851} +{"problem":"70 is increased by 50 % . find the final number .","rationale":"\"final number = initial number + 50 % ( original number ) = 70 + 50 % ( 70 ) = 70 + 35 = 105 . answer c\"","correct":"c","options":{"a":"100 ","b":"110 ","c":"105 ","d":"30","e":"160"},"options_float":{"a":100.0,"b":110.0,"c":105.0,"d":30.0,"e":160.0},"annotated_formula":"add(70, multiply(70, divide(50, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n70 * (1\/2)<\/gadget>\n35<\/output>\n70 + 35<\/gadget>\n105<\/output>\n105<\/result>","index":2854} +{"problem":"in a group of 8 semifinalists , all but 2 will advance to the final round . if in the final round only the top 3 will be awarded medals , then how many groups of medal winners are possible ?","rationale":"1 ) is it important here that how many go into finals . . . no . . . . what is important is how many get medals . . 3 finally these 3 can be any of the 8 - 8 c 3 = 8 ! 5 ! 3 ! = 8 ∗ 7 ∗ 63 ∗ 2 = 568 c 3 = 8 ! 5 ! 3 ! = 8 ∗ 7 ∗ 63 ∗ 2 = 56 2 ) is order important . . . no we are looking for groups only . . ans 56 answer : b","correct":"b","options":{"a":"20 ","b":"56 ","c":"120 ","d":"560","e":"720"},"options_float":{"a":20.0,"b":56.0,"c":120.0,"d":560.0,"e":720.0},"annotated_formula":"divide(multiply(divide(factorial(8), multiply(factorial(subtract(8, 2)), 2)), divide(factorial(subtract(8, 2)), multiply(factorial(3), factorial(3)))), const_10)","linear_formula":"factorial(n0)|factorial(n2)|subtract(n0,n1)|factorial(#2)|multiply(#1,#1)|divide(#3,#4)|multiply(n1,#3)|divide(#0,#6)|multiply(#7,#5)|divide(#8,const_10)","chain":"factorial(8)<\/gadget>\n40_320<\/output>\n8 - 2<\/gadget>\n6<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\n720 * 2<\/gadget>\n1_440<\/output>\n40_320 \/ 1_440<\/gadget>\n28<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n720 \/ 36<\/gadget>\n20<\/output>\n28 * 20<\/gadget>\n560<\/output>\n560 \/ 10<\/gadget>\n56<\/output>\n56<\/result>","index":2855} +{"problem":"martha has the unique talent of being able to guess other people ’ s height and weight . for every 6 people that martha meets , she consistently guesses the people ’ s correct height 5 times , and for every 8 people that she meets , she consistently guesses the people ’ s correct weight 6 times . if martha meets 3 people and her success rate remains constant , what is the probability that martha correctly guesses a person ’ s weight and height at least once ?","rationale":"probability that she guesses the height correctly p ( h ) = 5 \/ 6 probability that she guesses the weight correctlyp ( w ) = 6 \/ 8 probability that she guesses both weight and height correctly ( p ( h ) * p ( w ) ) = 5 \/ 6 * 6 \/ 8 = 30 \/ 48 now the q asks about the probability of this happening atleast once . we calculate it by finding the probability of not being able to guess in any of the three occasions . probability of not being able to guess any no . of times = 1 - 30 \/ 48 = 18 \/ 48 = 3 \/ 8 for all three occasions p ( a ) = 3 \/ 8 * 3 \/ 8 * 3 \/ 8 = 27 \/ 512 probability of the event happening atleast once = 1 - p ( a ) = 1 - 27 \/ 512 = 485 \/ 512 ans should be c","correct":"c","options":{"a":"8 \/ 27 ","b":"2 \/ 5 ","c":"485 \/ 512 ","d":"98 \/ 125","e":"625 \/ 952"},"options_float":{"a":0.2962962963,"b":0.4,"c":0.947265625,"d":0.784,"e":0.656512605},"annotated_formula":"subtract(const_1, power(subtract(const_1, multiply(divide(5, 6), divide(6, 8))), const_3))","linear_formula":"divide(n1,n0)|divide(n0,n2)|multiply(#0,#1)|subtract(const_1,#2)|power(#3,const_3)|subtract(const_1,#4)","chain":"5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n6 \/ 8<\/gadget>\n3\/4 = around 0.75<\/output>\n(5\/6) * (3\/4)<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n(3\/8) ** 3<\/gadget>\n27\/512 = around 0.052734<\/output>\n1 - (27\/512)<\/gadget>\n485\/512 = around 0.947266<\/output>\n485\/512 = around 0.947266<\/result>","index":2856} +{"problem":"the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 90 black cars in the lot , how many red cars are there in the lot ?","rationale":"b is correct r \/ b = 3 \/ 8 and b = 90 r = 90 * 3 \/ 8 = 33","correct":"b","options":{"a":"11 ","b":"33 ","c":"24 ","d":"27","e":"32"},"options_float":{"a":11.0,"b":33.0,"c":24.0,"d":27.0,"e":32.0},"annotated_formula":"floor(multiply(divide(3, 8), 90))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|floor(#1)","chain":"3 \/ 8<\/gadget>\n3\/8 = around 0.375<\/output>\n(3\/8) * 90<\/gadget>\n135\/4 = around 33.75<\/output>\nfloor(135\/4)<\/gadget>\n33<\/output>\n33<\/result>","index":2857} +{"problem":"a bowl was filled with 10 ounces of water , and 0.004 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ?","rationale":"\"total amount of water evaporated each day during a 50 - day period = . 004 * 50 = . 004 * 100 \/ 2 = . 4 \/ 2 = . 2 percent of the original amount of water evaporated during this period = ( . 2 \/ 10 ) * 100 % = 2 % answer d\"","correct":"d","options":{"a":"0.004 % ","b":"0.04 % ","c":"0.40 % ","d":"2 %","e":"40 %"},"options_float":{"a":0.004,"b":0.04,"c":0.4,"d":2.0,"e":40.0},"annotated_formula":"multiply(divide(multiply(50, 0.004), 10), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"50 * 0.004<\/gadget>\n0.2<\/output>\n0.2 \/ 10<\/gadget>\n0.02<\/output>\n0.02 * 100<\/gadget>\n2<\/output>\n2<\/result>","index":2860} +{"problem":"of the families in city x in 1994 , 35 percent owned a personal computer . the number of families in city x owning a computer in 1998 was 20 percent greater than it was in 1994 , and the total number of families in city x was 4 percent greater in 1998 than it was in 1994 . what percent of the families in city x owned a personal computer in 1998 ?","rationale":"\"say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 40 number of families owning computer in 1998 = 35 * 120 \/ 100 = 42 number of families in 1998 = 104 the percentage = 42 \/ 104 * 100 = 40.38 % . option : d\"","correct":"d","options":{"a":"50.12 % ","b":"40.52 % ","c":"40.56 % ","d":"40.38 %","e":"40.14 %"},"options_float":{"a":50.12,"b":40.52,"c":40.56,"d":40.38,"e":40.14},"annotated_formula":"multiply(const_100, divide(divide(multiply(add(20, const_100), 35), const_100), add(const_100, 4)))","linear_formula":"add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)|","chain":"20 + 100<\/gadget>\n120<\/output>\n120 * 35<\/gadget>\n4_200<\/output>\n4_200 \/ 100<\/gadget>\n42<\/output>\n100 + 4<\/gadget>\n104<\/output>\n42 \/ 104<\/gadget>\n21\/52 = around 0.403846<\/output>\n100 * (21\/52)<\/gadget>\n525\/13 = around 40.384615<\/output>\n525\/13 = around 40.384615<\/result>","index":2861} +{"problem":"the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each . the radius of the sphere is","rationale":"\"solution 4 π r 2 = 2 π 6 x 12 ⇒ r 2 = ( 6 x 12 \/ 2 ) ⇒ 36 ⇒ r = 6 cm . answer c\"","correct":"c","options":{"a":"3 cm ","b":"4 cm ","c":"6 cm ","d":"8 cm","e":"none"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":null},"annotated_formula":"sqrt(divide(multiply(multiply(const_pi, multiply(12, divide(12, const_2))), const_2), multiply(const_pi, const_4)))","linear_formula":"divide(n0,const_2)|multiply(const_4,const_pi)|multiply(n0,#0)|multiply(#2,const_pi)|multiply(#3,const_2)|divide(#4,#1)|sqrt(#5)|","chain":"12 \/ 2<\/gadget>\n6<\/output>\n12 * 6<\/gadget>\n72<\/output>\npi * 72<\/gadget>\n72*pi = around 226.194671<\/output>\n(72*pi) * 2<\/gadget>\n144*pi = around 452.389342<\/output>\npi * 4<\/gadget>\n4*pi = around 12.566371<\/output>\n(144*pi) \/ (4*pi)<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6<\/result>","index":2862} +{"problem":"the area of a parallelogram is 72 sq m and its base is 12 m . then the length of the height is ?","rationale":"b * h = 72 . h * 12 = 72 = > h = 6 m . answer : ( a )","correct":"a","options":{"a":"6 m ","b":"7 m ","c":"60 m ","d":"13 m","e":"none"},"options_float":{"a":6.0,"b":7.0,"c":60.0,"d":13.0,"e":null},"annotated_formula":"sqrt(divide(72, const_2))","linear_formula":"divide(n0,const_2)|sqrt(#0)|","chain":"72 \/ 2<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6<\/result>","index":2863} +{"problem":"a set of pictures of butterflies contains 10 pictures . jim bought 3 of the pictures . if 2 pictures are to be picked out from the 10 pictures , what is the probability that both pictures are not those that were already bought by jim ?","rationale":"probability of first pick not one of the jim ' s picture is = 7 \/ 10 ( except for 3 which jim purchased ) probability of second pick not one of the jim ' s picture is = 6 \/ 9 total = 7 \/ 10 ∗ 6 \/ 9 = 7 \/ 15 answer is d option d","correct":"d","options":{"a":"14 \/ 15 ","b":"3 \/ 5 ","c":"6 \/ 13 ","d":"7 \/ 15","e":"7 \/ 10"},"options_float":{"a":0.9333333333,"b":0.6,"c":0.4615384615,"d":0.4666666667,"e":0.7},"annotated_formula":"divide(divide(divide(factorial(subtract(10, 3)), factorial(subtract(subtract(10, 3), 2))), factorial(2)), divide(divide(factorial(10), factorial(subtract(10, 2))), factorial(2)))","linear_formula":"factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|divide(#1,#5)|factorial(#6)|divide(#4,#8)|divide(#7,#0)|divide(#9,#0)|divide(#11,#10)","chain":"10 - 3<\/gadget>\n7<\/output>\nfactorial(7)<\/gadget>\n5_040<\/output>\n7 - 2<\/gadget>\n5<\/output>\nfactorial(5)<\/gadget>\n120<\/output>\n5_040 \/ 120<\/gadget>\n42<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n42 \/ 2<\/gadget>\n21<\/output>\nfactorial(10)<\/gadget>\n3_628_800<\/output>\n10 - 2<\/gadget>\n8<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n3_628_800 \/ 40_320<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n21 \/ 45<\/gadget>\n7\/15 = around 0.466667<\/output>\n7\/15 = around 0.466667<\/result>","index":2866} +{"problem":"maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability q , that a book selected at random is either a hard cover book or a chemistry book ?","rationale":"first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 \/ 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , q = 9 provide the desired outcome , making the answer e .","correct":"e","options":{"a":"1 \/ 10 ","b":"3 \/ 20 ","c":"1 \/ 5 ","d":"1 \/ 4","e":"9 \/ 20"},"options_float":{"a":0.1,"b":0.15,"c":0.2,"d":0.25,"e":0.45},"annotated_formula":"subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))))","linear_formula":"add(n2,const_2)|add(#0,const_2)|add(n3,#1)|divide(const_2,#2)|divide(#1,#2)|add(#3,#4)|multiply(#3,#4)|subtract(#5,#6)","chain":"2 + 4<\/gadget>\n6<\/output>\n2 + 6<\/gadget>\n8<\/output>\n8 + 12<\/gadget>\n20<\/output>\n2 \/ 20<\/gadget>\n1\/10 = around 0.1<\/output>\n8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/10) + (2\/5)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/10) * (2\/5)<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/2) - (1\/25)<\/gadget>\n23\/50 = around 0.46<\/output>\n23\/50 = around 0.46<\/result>","index":2867} +{"problem":"a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 15 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m \/ s )","rationale":"\"the distance the sound traveled to the man is 15 * 330 = 4950 meters the answer is d .\"","correct":"d","options":{"a":"1650 ","b":"2750 ","c":"3850 ","d":"4950","e":"6050"},"options_float":{"a":1650.0,"b":2750.0,"c":3850.0,"d":4950.0,"e":6050.0},"annotated_formula":"multiply(330, 15)","linear_formula":"multiply(n1,n3)|","chain":"330 * 15<\/gadget>\n4_950<\/output>\n4_950<\/result>","index":2868} +{"problem":"eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 510 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy )","rationale":"distance traveled by eddy = 510 km time taken by eddy = 3 hours average speed of eddy = 510 \/ 3 = 170 km \/ hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 \/ 4 = 75 km \/ hour ratio of average speed of eddy to freddy = 170 \/ 75 = 34 \/ 15 answer e","correct":"e","options":{"a":"8 \/ 3 ","b":"3 \/ 8 ","c":"8 \/ 5 ","d":"5 \/ 8","e":"34 \/ 15"},"options_float":{"a":2.6666666667,"b":0.375,"c":1.6,"d":0.625,"e":2.2666666667},"annotated_formula":"divide(divide(510, 3), divide(300, 4))","linear_formula":"divide(n2,n0)|divide(n3,n1)|divide(#0,#1)","chain":"510 \/ 3<\/gadget>\n170<\/output>\n300 \/ 4<\/gadget>\n75<\/output>\n170 \/ 75<\/gadget>\n34\/15 = around 2.266667<\/output>\n34\/15 = around 2.266667<\/result>","index":2869} +{"problem":"two brothers ram and ravi appeared for an exam . the probability of selection of ram is 4 \/ 7 and that of ravi is 1 \/ 5 . find the probability that both of them are selected .","rationale":"let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 4 \/ 7 p ( b ) = 1 \/ 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 4 \/ 7 x 1 \/ 5 = 4 \/ 35 answer : a","correct":"a","options":{"a":"4 \/ 35 ","b":"2 \/ 3 ","c":"7 \/ 35 ","d":"5 \/ 7","e":"7 \/ 5"},"options_float":{"a":0.1142857143,"b":0.6666666667,"c":0.2,"d":0.7142857143,"e":1.4},"annotated_formula":"multiply(divide(4, 7), divide(1, 5))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)","chain":"4 \/ 7<\/gadget>\n4\/7 = around 0.571429<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(4\/7) * (1\/5)<\/gadget>\n4\/35 = around 0.114286<\/output>\n4\/35 = around 0.114286<\/result>","index":2870} +{"problem":"120 kg of an alloy a is mixed with 180 kg of alloy b . if alloy a has lead and tin in the ratio 2 : 3 and alloy b has tin and copper in the ratio 3 : 5 , then the amount of tin in the new alloy is ?","rationale":"\"quantity of tin in 120 kg of a = 120 * 3 \/ 5 = 72 kg quantity of tin in 180 kg of b = 180 * 3 \/ 8 = 67.5 kg quantity of tin in the new alloy = 72 + 67.5 = 139.5 kg answer is c\"","correct":"c","options":{"a":"100.6 kg ","b":"120.3 kg ","c":"139.5 kg ","d":"140.8 kg","e":"114.5 kg"},"options_float":{"a":100.6,"b":120.3,"c":139.5,"d":140.8,"e":114.5},"annotated_formula":"add(multiply(divide(120, add(2, 3)), 3), multiply(divide(180, add(3, 5)), 3))","linear_formula":"add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)|","chain":"2 + 3<\/gadget>\n5<\/output>\n120 \/ 5<\/gadget>\n24<\/output>\n24 * 3<\/gadget>\n72<\/output>\n3 + 5<\/gadget>\n8<\/output>\n180 \/ 8<\/gadget>\n45\/2 = around 22.5<\/output>\n(45\/2) * 3<\/gadget>\n135\/2 = around 67.5<\/output>\n72 + (135\/2)<\/gadget>\n279\/2 = around 139.5<\/output>\n279\/2 = around 139.5<\/result>","index":2872} +{"problem":"how many seconds will a train 120 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?","rationale":"\"d = 120 + 150 = 270 s = 36 * 5 \/ 18 = 10 mps t = 270 \/ 10 = 27 sec\"","correct":"d","options":{"a":"17 sec ","b":"21 sec ","c":"25 sec ","d":"27 sec","e":"29 sec"},"options_float":{"a":17.0,"b":21.0,"c":25.0,"d":27.0,"e":29.0},"annotated_formula":"divide(add(150, 120), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"150 + 120<\/gadget>\n270<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n270 \/ 10<\/gadget>\n27<\/output>\n27<\/result>","index":2873} +{"problem":"3 \/ 4 of a number is 19 less than the original number . the number is ?","rationale":"answer let the original number be n . so from question n - 3 n \/ 4 = 19 ⇒ 4 n - 3 n = 76 ∴ n = 76 option : c","correct":"c","options":{"a":"84 ","b":"64 ","c":"76 ","d":"72","e":"none"},"options_float":{"a":84.0,"b":64.0,"c":76.0,"d":72.0,"e":null},"annotated_formula":"multiply(19, 4)","linear_formula":"multiply(n1,n2)","chain":"19 * 4<\/gadget>\n76<\/output>\n76<\/result>","index":2874} +{"problem":"the cost price of 40 articles is the same as the selling price of x articles . if the profit is 25 % , what is x ?","rationale":"\"let the cost price = y the cost price of 40 articles = 40 y the selling price of x articles = 1.25 y * x 1.25 y * x = 40 y x = 40 \/ 1.25 = 32 the answer is d .\"","correct":"d","options":{"a":"26 ","b":"28 ","c":"30 ","d":"32","e":"34"},"options_float":{"a":26.0,"b":28.0,"c":30.0,"d":32.0,"e":34.0},"annotated_formula":"divide(multiply(40, const_4), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|","chain":"40 * 4<\/gadget>\n160<\/output>\n4 + 1<\/gadget>\n5<\/output>\n160 \/ 5<\/gadget>\n32<\/output>\n32<\/result>","index":2875} +{"problem":"a man can row 6 kmph in still water . when the river is running at 3 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?","rationale":"\"m = 6 s = 3 ds = 9 us = 3 x \/ 9 + x \/ 3 = 1 x = 2.25 d = 2.25 * 2 = 4.5 answer : b\"","correct":"b","options":{"a":"5.75 ","b":"4.5 ","c":"5.76 ","d":"5.74","e":"5.71"},"options_float":{"a":5.75,"b":4.5,"c":5.76,"d":5.74,"e":5.71},"annotated_formula":"multiply(divide(multiply(add(6, 3), subtract(6, 3)), add(add(6, 3), subtract(6, 3))), const_2)","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|","chain":"6 + 3<\/gadget>\n9<\/output>\n6 - 3<\/gadget>\n3<\/output>\n9 * 3<\/gadget>\n27<\/output>\n9 + 3<\/gadget>\n12<\/output>\n27 \/ 12<\/gadget>\n9\/4 = around 2.25<\/output>\n(9\/4) * 2<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":2876} +{"problem":"the interest on a certain deposit at 4.5 % p . a . is rs . 202.50 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a ?","rationale":"\"solution s . i = rs . 202.50 , r = 4.5 % , t = 1 year principal = rs ( 100 * 202.50 \/ 4.5 * 1 ) = rs . 4500 now p = rs . 4500 , r = 5 % , t = 1 year s . i = ( 4500 * 5 * 1 \/ 100 ) = rs . 225 different in interest = rs . ( 225 - 202.50 ) = rs . 22.50 answer b\"","correct":"b","options":{"a":"rs . 20.25 ","b":"rs . 22.50 ","c":"rs . 25 ","d":"rs . 42.75","e":"none"},"options_float":{"a":20.25,"b":22.5,"c":25.0,"d":42.75,"e":null},"annotated_formula":"subtract(divide(multiply(divide(202.50, divide(4.5, const_100)), 5), const_100), 202.50)","linear_formula":"divide(n0,const_100)|divide(n1,#0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1)|","chain":"4.5 \/ 100<\/gadget>\n0.045<\/output>\n202.5 \/ 0.045<\/gadget>\n4_500<\/output>\n4_500 * 5<\/gadget>\n22_500<\/output>\n22_500 \/ 100<\/gadget>\n225<\/output>\n225 - 202.5<\/gadget>\n22.5<\/output>\n22.5<\/result>","index":2877} +{"problem":"( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) \/ 5.0 = ?","rationale":"\"dove straight into calculation but quickly realized that the sum of two negatives is a negative so there is only one option . - 0.852 answer a\"","correct":"a","options":{"a":"- 0.852 ","b":"1.0 ","c":"1.07 ","d":"1.71","e":"2.71"},"options_float":{"a":-0.852,"b":1.0,"c":1.07,"d":1.71,"e":2.71},"annotated_formula":"divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 5.0)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|negate(#0)|subtract(#2,#1)|divide(#3,n4)|","chain":"1.9 * 0.6<\/gadget>\n1.14<\/output>\n-1.14<\/gadget>\n-1.14<\/output>\n2.6 * 1.2<\/gadget>\n3.12<\/output>\n(-1.14) - 3.12<\/gadget>\n-4.26<\/output>\n(-4.26) \/ 5<\/gadget>\n-0.852<\/output>\n-0.852<\/result>","index":2878} +{"problem":"if the wheel is 12 cm then the number of revolutions to cover a distance of 1056 cm is ?","rationale":"\"2 * 22 \/ 7 * 12 * x = 1056 = > x = 14 answer : c\"","correct":"c","options":{"a":"18 ","b":"26 ","c":"14 ","d":"12","e":"91"},"options_float":{"a":18.0,"b":26.0,"c":14.0,"d":12.0,"e":91.0},"annotated_formula":"divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 12))","linear_formula":"multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)|","chain":"3 * 100<\/gadget>\n300<\/output>\n1 * 10<\/gadget>\n10<\/output>\n300 + 10<\/gadget>\n310<\/output>\n310 + 4<\/gadget>\n314<\/output>\n314 \/ 100<\/gadget>\n157\/50 = around 3.14<\/output>\n2 * (157\/50)<\/gadget>\n157\/25 = around 6.28<\/output>\n(157\/25) * 12<\/gadget>\n1_884\/25 = around 75.36<\/output>\n1_056 \/ (1_884\/25)<\/gadget>\n2_200\/157 = around 14.012739<\/output>\n2_200\/157 = around 14.012739<\/result>","index":2880} +{"problem":"if the length and breadth of a rectangular room are each increased by 1 m , then the area of floor is increased by 21 sq . m . if the length is creased by 1 m and breadth is decreased by 1 m , then the area is decreased by 5 sq . m . the perimeter of the floor is :","rationale":"let length = x meters and breadth = y meters . then , ( x + 1 ) ( y + 1 ) - xy = 21 x + y = 20 ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ 1 and , xy - [ ( x + 1 ) ( y - 1 ) ] = 5 x - y = 6 ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ ã ¢ â ‚ ¬ â ¦ . . 2 solving ( i ) and ( ii ) , we get : x = 13 and y = 7 so , length = 13 m and breadth = 7 m . perimeter = [ 2 ( 13 + 7 ) ] m = 40 m . answer : c","correct":"c","options":{"a":"27 ","b":"35 ","c":"40 ","d":"49","e":"38"},"options_float":{"a":27.0,"b":35.0,"c":40.0,"d":49.0,"e":38.0},"annotated_formula":"rectangle_perimeter(divide(subtract(21, 5), const_2), subtract(subtract(21, 1), divide(subtract(21, 5), const_2)))","linear_formula":"subtract(n1,n4)|subtract(n1,n0)|divide(#0,const_2)|subtract(#1,#2)|rectangle_perimeter(#2,#3)","chain":"21 - 5<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n21 - 1<\/gadget>\n20<\/output>\n20 - 8<\/gadget>\n12<\/output>\n2 * (8 + 12)<\/gadget>\n40<\/output>\n40<\/result>","index":2881} +{"problem":"a car is 15 minutes late by running at a speed of 4 \/ 5 th of its actual speed . the actual time taken by the car to cover that distance is :","rationale":"explanation : new speed = 4 \/ 5 th of its actual speed new time = 5 \/ 4 th of its actual time = > 5 \/ 4 t - t = 15 \/ 60 = 1 \/ 4 = > 1 \/ 4 t = 1 \/ 4 = > t = 1 hour answer : b","correct":"b","options":{"a":"30 minutes ","b":"1 hour ","c":"2 hours ","d":"3 hours","e":"none of these"},"options_float":{"a":30.0,"b":1.0,"c":2.0,"d":3.0,"e":null},"annotated_formula":"subtract(5, 4)","linear_formula":"subtract(n2,n1)","chain":"5 - 4<\/gadget>\n1<\/output>\n1<\/result>","index":2882} +{"problem":"the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 30 runs . find his average in all the 30 matches ?","rationale":"\"total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 300 . total score of the batsman in the 30 matches = 1100 . average score of the batsman = 1100 \/ 30 = 36.67 . answer : b\"","correct":"b","options":{"a":"31 ","b":"36.67 ","c":"88 ","d":"13","e":"12"},"options_float":{"a":31.0,"b":36.67,"c":88.0,"d":13.0,"e":12.0},"annotated_formula":"divide(add(multiply(40, 20), multiply(30, 10)), add(20, 10))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"40 * 20<\/gadget>\n800<\/output>\n30 * 10<\/gadget>\n300<\/output>\n800 + 300<\/gadget>\n1_100<\/output>\n20 + 10<\/gadget>\n30<\/output>\n1_100 \/ 30<\/gadget>\n110\/3 = around 36.666667<\/output>\n110\/3 = around 36.666667<\/result>","index":2883} +{"problem":"the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability t that the next couple the maitre ' d seats will not order dessert ?","rationale":"\"could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . t = 2 \/ 10 of d are n ' t in d u c = so 8 \/ 10 of d are in duc which means = 60 = 8 \/ 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %\"","correct":"b","options":{"a":"20 % ","b":"25 % ","c":"40 % ","d":"60 %","e":"75 %"},"options_float":{"a":20.0,"b":25.0,"c":40.0,"d":60.0,"e":75.0},"annotated_formula":"multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5)))","linear_formula":"multiply(const_2,const_5)|multiply(#0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(const_1,#3)|divide(#2,#4)|subtract(const_1,#5)|multiply(#1,#6)|","chain":"2 * 5<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(3\/5) \/ (4\/5)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2884} +{"problem":"anne bought doughnuts for a class breakfast party . she bought 6 chocolate doughnuts , 8 coconut doughnuts , and 8 jam - filled doughnuts . how many doughnuts did anne buy in all ?","rationale":"\"add the numbers of doughnuts . 6 + 8 + 8 = 22 . answer is a .\"","correct":"a","options":{"a":"22 ","b":"36 ","c":"39 ","d":"21","e":"11"},"options_float":{"a":22.0,"b":36.0,"c":39.0,"d":21.0,"e":11.0},"annotated_formula":"add(add(6, 8), 8)","linear_formula":"add(n0,n1)|add(n2,#0)|","chain":"6 + 8<\/gadget>\n14<\/output>\n14 + 8<\/gadget>\n22<\/output>\n22<\/result>","index":2885} +{"problem":"i sold a book at a profit of 10 % . had i sold it for $ 110 more , 15 % would have been gained . find the cost price ?","rationale":"115 % of cost - 110 % of cost = $ 110 5 % of cost = $ 110 cost = 110 * 100 \/ 5 = $ 2200 answer is b","correct":"b","options":{"a":"$ 2000 ","b":"$ 2200 ","c":"$ 3000 ","d":"$ 3120","e":"$ 1540"},"options_float":{"a":2000.0,"b":2200.0,"c":3000.0,"d":3120.0,"e":1540.0},"annotated_formula":"divide(multiply(divide(multiply(110, 10), subtract(15, 10)), const_100), 10)","linear_formula":"multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)","chain":"110 * 10<\/gadget>\n1_100<\/output>\n15 - 10<\/gadget>\n5<\/output>\n1_100 \/ 5<\/gadget>\n220<\/output>\n220 * 100<\/gadget>\n22_000<\/output>\n22_000 \/ 10<\/gadget>\n2_200<\/output>\n2_200<\/result>","index":2886} +{"problem":"90 % of the population of a village is 36000 . the total population of the village is ?","rationale":"\"x * ( 90 \/ 100 ) = 36000 x = 400 * 100 x = 40000 answer : e\"","correct":"e","options":{"a":"26799 ","b":"24000 ","c":"26682 ","d":"29973","e":"40000"},"options_float":{"a":26799.0,"b":24000.0,"c":26682.0,"d":29973.0,"e":40000.0},"annotated_formula":"multiply(divide(const_100, 90), 36000)","linear_formula":"divide(const_100,n0)|multiply(n1,#0)|","chain":"100 \/ 90<\/gadget>\n10\/9 = around 1.111111<\/output>\n(10\/9) * 36_000<\/gadget>\n40_000<\/output>\n40_000<\/result>","index":2888} +{"problem":"( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) \/ 2.0 = ?","rationale":"\"dove straight into calculation ( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) \/ 2.0 = - 2.13 answer b\"","correct":"b","options":{"a":"- 0.71 ","b":"- 2.13 ","c":"1.07 ","d":"1.71","e":"2.71"},"options_float":{"a":-0.71,"b":-2.13,"c":1.07,"d":1.71,"e":2.71},"annotated_formula":"divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 2.0)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|negate(#0)|subtract(#2,#1)|divide(#3,n4)|","chain":"1.9 * 0.6<\/gadget>\n1.14<\/output>\n-1.14<\/gadget>\n-1.14<\/output>\n2.6 * 1.2<\/gadget>\n3.12<\/output>\n(-1.14) - 3.12<\/gadget>\n-4.26<\/output>\n(-4.26) \/ 2<\/gadget>\n-2.13<\/output>\n-2.13<\/result>","index":2889} +{"problem":"the probability that a number selected at random from the first 50 natural numbers is a composite number is","rationale":"the number of exhaustive events = ⁵ ⁰ c ₁ = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 \/ 50 = 17 \/ 25 . nswer : b","correct":"b","options":{"a":"17 \/ 26 ","b":"17 \/ 25 ","c":"17 \/ 20 ","d":"17 \/ 22","e":"17 \/ 21"},"options_float":{"a":0.6538461538,"b":0.68,"c":0.85,"d":0.7727272727,"e":0.8095238095},"annotated_formula":"divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50)","linear_formula":"multiply(const_3,const_6)|subtract(#0,const_1)|multiply(#1,const_2)|divide(#2,n0)","chain":"6 * 3<\/gadget>\n18<\/output>\n18 - 1<\/gadget>\n17<\/output>\n17 * 2<\/gadget>\n34<\/output>\n34 \/ 50<\/gadget>\n17\/25 = around 0.68<\/output>\n17\/25 = around 0.68<\/result>","index":2890} +{"problem":"a retailer buys 80 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ?","rationale":"\"let the market price of each pen be $ 1 then , cost price of 80 pens = $ 36 selling price of 80 pens = 99 % of $ 80 = $ 79.20 profit % = ( ( 43.20 * 100 ) \/ 36 ) % = 120 % answer e ) 120\"","correct":"e","options":{"a":"110 ","b":"115 ","c":"112 ","d":"118","e":"120"},"options_float":{"a":110.0,"b":115.0,"c":112.0,"d":118.0,"e":120.0},"annotated_formula":"multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 80)), divide(36, 80)))","linear_formula":"divide(n2,const_100)|divide(n1,n0)|subtract(n2,#0)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)|","chain":"1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n1 - (1\/100)<\/gadget>\n99\/100 = around 0.99<\/output>\n36 \/ 80<\/gadget>\n9\/20 = around 0.45<\/output>\n(99\/100) - (9\/20)<\/gadget>\n27\/50 = around 0.54<\/output>\n(27\/50) \/ (9\/20)<\/gadget>\n6\/5 = around 1.2<\/output>\n100 * (6\/5)<\/gadget>\n120<\/output>\n120<\/result>","index":2891} +{"problem":"if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 66 ?","rationale":"\"four wheeler = 16 * 4 = 64 ( max ) 2 wheel = 1 so no of 4 wheeler = 16 answer : d\"","correct":"d","options":{"a":"11 ","b":"12 ","c":"13 ","d":"16","e":"17"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":16.0,"e":17.0},"annotated_formula":"divide(subtract(66, 2), 4)","linear_formula":"subtract(n3,n0)|divide(#0,n1)|","chain":"66 - 2<\/gadget>\n64<\/output>\n64 \/ 4<\/gadget>\n16<\/output>\n16<\/result>","index":2892} +{"problem":"a train running at the speed of 27 km \/ hr crosses a post in 20 seconds . what is the length of the train ?","rationale":"\"speed = ( 27 x 5 \/ 18 ) = 7.5 m \/ sec . length of the train = ( speed x time ) . length of the train = 7.5 x 20 m = 150 m . answer : b\"","correct":"b","options":{"a":"200 ","b":"150 ","c":"300 ","d":"175","e":"250"},"options_float":{"a":200.0,"b":150.0,"c":300.0,"d":175.0,"e":250.0},"annotated_formula":"multiply(divide(multiply(27, const_1000), const_3600), 20)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"27 * 1_000<\/gadget>\n27_000<\/output>\n27_000 \/ 3_600<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) * 20<\/gadget>\n150<\/output>\n150<\/result>","index":2893} +{"problem":"what is the rate percent when the simple interest on rs . 800 amount to rs . 128 in 4 years ?","rationale":"\"128 = ( 800 * 4 * r ) \/ 100 r = 4 % answer : c\"","correct":"c","options":{"a":"5 % ","b":"3 % ","c":"4 % ","d":"9 %","e":"1 %"},"options_float":{"a":5.0,"b":3.0,"c":4.0,"d":9.0,"e":1.0},"annotated_formula":"divide(multiply(const_100, 128), multiply(800, 4))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 128<\/gadget>\n12_800<\/output>\n800 * 4<\/gadget>\n3_200<\/output>\n12_800 \/ 3_200<\/gadget>\n4<\/output>\n4<\/result>","index":2894} +{"problem":"an error 1 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is :","rationale":"\"explanation : 100 cm is read as 101 cm . a 1 = ( 100 × 100 ) cm 2 = 10000 and a 2 = ( 101 × 101 ) cm 2 = 10609 ( a 2 - a 1 ) = 10201 - 10000 = 201 = > 201 \/ 10000 * 100 = 2.01 answer : a\"","correct":"a","options":{"a":"2.01 ","b":"30.1 ","c":"3.01 ","d":"25.01","e":"4.05"},"options_float":{"a":2.01,"b":30.1,"c":3.01,"d":25.01,"e":4.05},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 1)), square_area(const_100)), const_100), square_area(const_100))","linear_formula":"add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|","chain":"100 + 1<\/gadget>\n101<\/output>\n101 ** 2<\/gadget>\n10_201<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n10_201 - 10_000<\/gadget>\n201<\/output>\n201 * 100<\/gadget>\n20_100<\/output>\n20_100 \/ 10_000<\/gadget>\n201\/100 = around 2.01<\/output>\n201\/100 = around 2.01<\/result>","index":2898} +{"problem":"the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 60 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam .","rationale":"\"let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 80 x - ( 5 * 60 ) = 90 ( x - 5 ) 150 = 10 x = > x = 15 . answer : c\"","correct":"c","options":{"a":"22 ","b":"27 ","c":"15 ","d":"99","e":"21"},"options_float":{"a":22.0,"b":27.0,"c":15.0,"d":99.0,"e":21.0},"annotated_formula":"divide(subtract(multiply(90, 5), multiply(5, 60)), subtract(90, 80))","linear_formula":"multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|","chain":"90 * 5<\/gadget>\n450<\/output>\n5 * 60<\/gadget>\n300<\/output>\n450 - 300<\/gadget>\n150<\/output>\n90 - 80<\/gadget>\n10<\/output>\n150 \/ 10<\/gadget>\n15<\/output>\n15<\/result>","index":2900} +{"problem":"ann and bob drive separately to a meeting . ann ' s average driving speed is greater than bob ' s avergae driving speed by one - third of bob ' s average driving speed , and ann drives twice as many miles as bob . what is the ratio w of the number of hours ann spends driving to the meeting to the number of hours bob spends driving to the meeting ?","rationale":"\"say the rate of bob is 3 mph and he covers 6 miles then he needs 6 \/ 3 = 2 hours to do that . now , in this case the rate of ann would be 3 + 3 * 1 \/ 3 = 4 mph and the distance she covers would be 6 * 2 = 12 miles , so she needs 12 \/ 4 = 3 hours for that . the ratio w of ann ' s time to bob ' s time is 3 : 2 . answer : b .\"","correct":"b","options":{"a":"8 : 3 ","b":"3 : 2 ","c":"4 : 3 ","d":"2 : 3","e":"3 : 8"},"options_float":{"a":2.6666666667,"b":1.5,"c":1.3333333333,"d":0.6666666667,"e":0.375},"annotated_formula":"divide(const_2, add(const_1, divide(const_1, const_3)))","linear_formula":"divide(const_1,const_3)|add(#0,const_1)|divide(const_2,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n2 \/ (4\/3)<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":2901} +{"problem":"find the value of 72515 x 9999 = m ?","rationale":"\"72515 x 9999 = 72515 x ( 10000 - 1 ) = 72515 x 10000 - 72515 x 1 = 725150000 - 72515 = 725077485 d\"","correct":"d","options":{"a":"345434667 ","b":"246465757 ","c":"465767867 ","d":"725077485","e":"645354643"},"options_float":{"a":345434667.0,"b":246465757.0,"c":465767867.0,"d":725077485.0,"e":645354643.0},"annotated_formula":"multiply(subtract(9999, const_4), 72515)","linear_formula":"subtract(n1,const_4)|multiply(#0,n0)|","chain":"9_999 - 4<\/gadget>\n9_995<\/output>\n9_995 * 72_515<\/gadget>\n724_787_425<\/output>\n724_787_425<\/result>","index":2903} +{"problem":"a can give b 50 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"a runs 1000 m while b runs 950 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) \/ 950 = 842.10 m . b can give c = 1000 - 842.10 = 157.89 m . answer : a\"","correct":"a","options":{"a":"157.89 ","b":"157.13 ","c":"157.22 ","d":"111.0","e":"111.12"},"options_float":{"a":157.89,"b":157.13,"c":157.22,"d":111.0,"e":111.12},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 50)))","linear_formula":"multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 200<\/gadget>\n800<\/output>\n1_000 * 800<\/gadget>\n800_000<\/output>\n1_000 - 50<\/gadget>\n950<\/output>\n800_000 \/ 950<\/gadget>\n16_000\/19 = around 842.105263<\/output>\n1_000 - (16_000\/19)<\/gadget>\n3_000\/19 = around 157.894737<\/output>\n3_000\/19 = around 157.894737<\/result>","index":2904} +{"problem":"the present population of a town is 1240 . population increase rate is 4 % p . a . find the population of town after 1 years ?","rationale":"\"p = 1240 r = 4 % required population of town = p * ( 1 + r \/ 100 ) ^ t = 1240 * ( 1 + 4 \/ 100 ) = 1240 * ( 26 \/ 25 ) = 1290 ( approximately ) answer is e\"","correct":"e","options":{"a":"990 ","b":"1215 ","c":"1345 ","d":"1142","e":"1290"},"options_float":{"a":990.0,"b":1215.0,"c":1345.0,"d":1142.0,"e":1290.0},"annotated_formula":"add(1240, divide(multiply(1240, 4), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|","chain":"1_240 * 4<\/gadget>\n4_960<\/output>\n4_960 \/ 100<\/gadget>\n248\/5 = around 49.6<\/output>\n1_240 + (248\/5)<\/gadget>\n6_448\/5 = around 1_289.6<\/output>\n6_448\/5 = around 1_289.6<\/result>","index":2905} +{"problem":"sandy bought 65 books for $ 1080 from one shop and 55 books for $ 840 from another shop . what is the average price that sandy paid per book ?","rationale":"\"average price per book = ( 1080 + 840 ) \/ ( 65 + 55 ) = 1920 \/ 120 = $ 16 the answer is b .\"","correct":"b","options":{"a":"$ 14 ","b":"$ 16 ","c":"$ 18 ","d":"$ 20","e":"$ 22"},"options_float":{"a":14.0,"b":16.0,"c":18.0,"d":20.0,"e":22.0},"annotated_formula":"divide(add(1080, 840), add(65, 55))","linear_formula":"add(n1,n3)|add(n0,n2)|divide(#0,#1)|","chain":"1_080 + 840<\/gadget>\n1_920<\/output>\n65 + 55<\/gadget>\n120<\/output>\n1_920 \/ 120<\/gadget>\n16<\/output>\n16<\/result>","index":2906} +{"problem":"a merchant marks goods up by 80 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 35 % . what % discount did the merchant offer ?","rationale":"\"let p be the original price of the goods and let x be the rate after the markup . ( 1.8 p ) * x = 1.35 p x = 1.35 \/ 1.8 = 0.75 which is a discount of 25 % . the answer is b .\"","correct":"b","options":{"a":"20 % ","b":"25 % ","c":"30 % ","d":"35 %","e":"40 %"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"multiply(divide(subtract(subtract(add(const_1, divide(80, const_100)), const_1), divide(35, const_100)), add(const_1, divide(80, const_100))), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(#2,const_1)|subtract(#3,#1)|divide(#4,#2)|multiply(#5,const_100)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 + (4\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n(9\/5) - 1<\/gadget>\n4\/5 = around 0.8<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n(4\/5) - (7\/20)<\/gadget>\n9\/20 = around 0.45<\/output>\n(9\/20) \/ (9\/5)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2907} +{"problem":"the average height of 20 students in a class was calculated as 175 cm . it has later found that the height of one of the students in the class was incorrectly written as 151 cm whereas the actual height was 136 cm . what was the actual average height of the students in the class ?","rationale":"the total height was 15 cm too much . the average height should be reduced by 15 cm \/ 20 = 0.75 cm the answer is b .","correct":"b","options":{"a":"174.75 cm ","b":"174.25 cm ","c":"173.75 cm ","d":"173.25 cm","e":"172.75 cm"},"options_float":{"a":174.75,"b":174.25,"c":173.75,"d":173.25,"e":172.75},"annotated_formula":"divide(subtract(multiply(20, 175), subtract(151, 136)), 20)","linear_formula":"multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)","chain":"20 * 175<\/gadget>\n3_500<\/output>\n151 - 136<\/gadget>\n15<\/output>\n3_500 - 15<\/gadget>\n3_485<\/output>\n3_485 \/ 20<\/gadget>\n697\/4 = around 174.25<\/output>\n697\/4 = around 174.25<\/result>","index":2908} +{"problem":"the cost price of a radio is rs . 3300 and it was sold for rs . 1230 , find the loss % ?","rationale":"\"3300 - - - - 2070 100 - - - - ? = > 62.7 % answer : a\"","correct":"a","options":{"a":"62.7 % ","b":"65 % ","c":"88 % ","d":"65 %","e":"62 %"},"options_float":{"a":62.7,"b":65.0,"c":88.0,"d":65.0,"e":62.0},"annotated_formula":"multiply(divide(subtract(3300, 1230), 3300), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"3_300 - 1_230<\/gadget>\n2_070<\/output>\n2_070 \/ 3_300<\/gadget>\n69\/110 = around 0.627273<\/output>\n(69\/110) * 100<\/gadget>\n690\/11 = around 62.727273<\/output>\n690\/11 = around 62.727273<\/result>","index":2909} +{"problem":"at what rate percent per annum will the simple interest on a sum of money be 3 \/ 5 of the amount in 10 years ?","rationale":"\"let sum = x . then , s . i . = 2 x \/ 5 , time = 10 years . rate = ( 100 * 3 x ) \/ ( x * 5 * 10 ) = 6 % answer : a\"","correct":"a","options":{"a":"6 % ","b":"7 % ","c":"9 % ","d":"3 %","e":"1 %"},"options_float":{"a":6.0,"b":7.0,"c":9.0,"d":3.0,"e":1.0},"annotated_formula":"divide(multiply(divide(3, 5), const_100), 10)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|divide(#1,n2)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n6<\/result>","index":2910} +{"problem":"working simultaneously and independently at an identical constant rate , 5 machines of a certain type can produce a total of x units of product p in 4 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 2 x units of product p in 2 days ?","rationale":"\"the rate of 5 machines is rate = job \/ time = x \/ 4 units per day - - > the rate of 1 machine 1 \/ 5 * ( x \/ 4 ) = x \/ 20 units per day ; now , again as { time } * { combined rate } = { job done } then 2 * ( m * x \/ 20 ) = 2 x - - > m = 20 . answer : e .\"","correct":"e","options":{"a":"11 ","b":"13 ","c":"15 ","d":"18","e":"20"},"options_float":{"a":11.0,"b":13.0,"c":15.0,"d":18.0,"e":20.0},"annotated_formula":"multiply(multiply(5, 2), divide(4, 2))","linear_formula":"divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)|","chain":"5 * 2<\/gadget>\n10<\/output>\n4 \/ 2<\/gadget>\n2<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":2912} +{"problem":"heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many litres of tomato puree will be obtained from 20 litres of tomato juice ?","rationale":"\"tomato juice has 90 % water and 10 % tomato tomato puree has 20 % water and 80 % tomato so 20 liter of tomato juice has 20 * 90 \/ 100 = 18 liters of water and 2 liters of tomato . as tomato puree has 80 % content of tomato , so 2 liters of tomato from 20 liter juice corresponds to 80 % so tomato purees that can be obtained = 100 * 2 \/ 80 = 2.5 answer : c\"","correct":"c","options":{"a":"2 litres ","b":"2.4 litres ","c":"2.5 litres ","d":"6 litres","e":"4 litres"},"options_float":{"a":2.0,"b":2.4,"c":2.5,"d":6.0,"e":4.0},"annotated_formula":"divide(multiply(divide(subtract(const_100, 90), const_100), 20), divide(subtract(const_100, 20), const_100))","linear_formula":"subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)|","chain":"100 - 90<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 20<\/gadget>\n2<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n2 \/ (4\/5)<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":2913} +{"problem":"a seller of used cars has 12 cars to sell and each of his clients selected 4 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ?","rationale":"\"ifno caris selected more than once then the number of clients = 12 \/ 4 = 3 but since every car is being selected three times so no . of clients must be thrice as well = 3 * 3 = 9 answer : option a\"","correct":"a","options":{"a":"9 ","b":"10 ","c":"12 ","d":"14","e":"16"},"options_float":{"a":9.0,"b":10.0,"c":12.0,"d":14.0,"e":16.0},"annotated_formula":"multiply(divide(12, 4), const_3)","linear_formula":"divide(n0,n1)|multiply(#0,const_3)|","chain":"12 \/ 4<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9<\/result>","index":2915} +{"problem":"set a contains all the even numbers between 2 and 50 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ?","rationale":"\"each term in set b is 60 more than the corresponding term in set a . the difference of the sums = 25 * 60 = 1500 . the answer is a .\"","correct":"a","options":{"a":"1500 ","b":"3100 ","c":"4400 ","d":"6200","e":"7400"},"options_float":{"a":1500.0,"b":3100.0,"c":4400.0,"d":6200.0,"e":7400.0},"annotated_formula":"multiply(subtract(62, 2), add(divide(subtract(50, 2), const_2), const_1))","linear_formula":"subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)|","chain":"62 - 2<\/gadget>\n60<\/output>\n50 - 2<\/gadget>\n48<\/output>\n48 \/ 2<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n60 * 25<\/gadget>\n1_500<\/output>\n1_500<\/result>","index":2917} +{"problem":"the cost price of 20 articles is the same as the selling price of x articles . if the profit is 25 % , then the value of x is :","rationale":"\"explanation : let the cost price of 1 article be z . so , the cost price of 20 article = 20 z - - - - - - - - - - - - - - - - - - - - - - - - - ( 1 ) selling price of 20 articles = 20 z + 25 % of 20 z = 25 z = > selling price of 1 article = 25 z \/ 20 = ( 5 \/ 4 ) * z = > selling price of x articles = ( 5 \/ 4 ) * z * x - - - - - - - - - - - - - - - - - - - - - - - - - ( 2 ) given that selling price ( s . p . ) of x articles = cost price ( c . p . ) of 20 articles = > ( 5 \/ 4 ) * z * x = 20 z = > x = 16 answer : b\"","correct":"b","options":{"a":"15 ","b":"16 ","c":"18 ","d":"25","e":"27"},"options_float":{"a":15.0,"b":16.0,"c":18.0,"d":25.0,"e":27.0},"annotated_formula":"divide(multiply(20, const_4), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|","chain":"20 * 4<\/gadget>\n80<\/output>\n4 + 1<\/gadget>\n5<\/output>\n80 \/ 5<\/gadget>\n16<\/output>\n16<\/result>","index":2918} +{"problem":"the ratio of male to female in a class is 2 : 3 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one fourth of the males and 3 \/ 4 th of the females in the class ?","rationale":"let the common ratio be x . . so m = 2 x and f = 3 x and total = 5 x 1 \/ 4 of m = 2 x \/ 4 and 3 \/ 4 of f = 9 \/ 4 x . . total preferring that carrer = x \/ 2 + 9 x \/ 4 = 11 x \/ 4 now 5 x = 360 , so x = 360 \/ 5 = 72 . . so 72 * 11 \/ 4 = 198 ans 198 b","correct":"b","options":{"a":"160 degree ","b":"198 degree ","c":"191 degree ","d":"192 degree","e":"204 degree"},"options_float":{"a":160.0,"b":198.0,"c":191.0,"d":192.0,"e":204.0},"annotated_formula":"add(divide(multiply(multiply(divide(const_360, add(2, 3)), 3), 3), 4), divide(multiply(divide(const_360, add(2, 3)), 2), const_4))","linear_formula":"add(n0,n1)|divide(const_360,#0)|multiply(n1,#1)|multiply(n0,#1)|divide(#3,const_4)|multiply(n1,#2)|divide(#5,n3)|add(#6,#4)","chain":"2 + 3<\/gadget>\n5<\/output>\n360 \/ 5<\/gadget>\n72<\/output>\n72 * 3<\/gadget>\n216<\/output>\n216 * 3<\/gadget>\n648<\/output>\n648 \/ 4<\/gadget>\n162<\/output>\n72 * 2<\/gadget>\n144<\/output>\n144 \/ 4<\/gadget>\n36<\/output>\n162 + 36<\/gadget>\n198<\/output>\n198<\/result>","index":2919} +{"problem":"at what price must an article costing rs . 47.50 be marked in order that after deducting 20 % from the list price . it may be sold at a profit of 25 % on the cost price ?","rationale":"cp = 47.50 sp = 47.50 * ( 125 \/ 100 ) = 59.375 mp * ( 80 \/ 100 ) = 59.375 mp = 74.2 answer : d","correct":"d","options":{"a":"62.5 ","b":"62.3 ","c":"62.7 ","d":"74.2","e":"62.9"},"options_float":{"a":62.5,"b":62.3,"c":62.7,"d":74.2,"e":62.9},"annotated_formula":"divide(multiply(add(47.5, divide(multiply(47.5, 25), const_100)), const_100), subtract(const_100, 20))","linear_formula":"multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)","chain":"47.5 * 25<\/gadget>\n1_187.5<\/output>\n1_187.5 \/ 100<\/gadget>\n11.875<\/output>\n47.5 + 11.875<\/gadget>\n59.375<\/output>\n59.375 * 100<\/gadget>\n5_937.5<\/output>\n100 - 20<\/gadget>\n80<\/output>\n5_937.5 \/ 80<\/gadget>\n74.21875<\/output>\n74.21875<\/result>","index":2920} +{"problem":"a mail handler needs 3 hours to sort the mail . his assistant takes 6 hours to sort the mail . how many hours will it take for both of them working together to sort the mail ?","rationale":"work hrs = ab \/ ( a + b ) = 18 \/ 9 = 2 answer is b","correct":"b","options":{"a":"3 ","b":"2 ","c":"1 ","d":"2 1 \/ 2","e":"1 1 \/ 5"},"options_float":{"a":3.0,"b":2.0,"c":1.0,"d":2.0,"e":1.0},"annotated_formula":"inverse(add(divide(const_1, 3), divide(const_1, 6)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/3) + (1\/6)<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ (1\/2)<\/gadget>\n2<\/output>\n2<\/result>","index":2921} +{"problem":"a bag contains 7 black and 8 white balls . if two balls are drawn simultaneously , what is the probability that both are of the same color is ?","rationale":"drawing two balls of same color from seven black balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = ⁷ c ₂ \/ ¹ ⁵ c ₂ + ⁸ c ₂ \/ ¹ ⁵ c ₂ = 7 \/ 15 answer is e","correct":"e","options":{"a":"1 \/ 2 ","b":"1 \/ 3 ","c":"2 \/ 3 ","d":"3 \/ 5","e":"7 \/ 15"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.6666666667,"d":0.6,"e":0.4666666667},"annotated_formula":"divide(add(divide(divide(factorial(8), factorial(subtract(8, const_2))), factorial(const_2)), divide(divide(factorial(7), factorial(subtract(7, const_2))), factorial(const_2))), divide(divide(factorial(add(7, 8)), factorial(subtract(add(7, 8), const_2))), factorial(const_2)))","linear_formula":"add(n0,n1)|factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|divide(#1,#6)|divide(#3,#7)|factorial(#9)|divide(#10,#2)|divide(#11,#2)|divide(#8,#12)|add(#13,#14)|divide(#15,#2)|divide(#16,#17)","chain":"factorial(8)<\/gadget>\n40_320<\/output>\n8 - 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fourth of the time . if together they take 18 days to complete the work , how much time shall b take to do it ?","rationale":"explanation : let ‘ x ’ be the number of days taken by b alone to finish the whole work then , a alone will finish the whole work in 3 x \/ 4 × 2 days = 3 x \/ 2 days . then , working together , 3 x \/ 2 × x \/ 3 x \/ 2 + x = 18 days 3 x × x \/ 5 x = 18 3 x \/ 5 = 18 therefore , the number of days taken by b = x = 18 × 5 \/ 3 = 30 days . answer : option a","correct":"a","options":{"a":"30 days ","b":"35 days ","c":"40 days ","d":"45 days","e":"50 days"},"options_float":{"a":30.0,"b":35.0,"c":40.0,"d":45.0,"e":50.0},"annotated_formula":"divide(multiply(18, add(3, const_2)), 3)","linear_formula":"add(n0,const_2)|multiply(n1,#0)|divide(#1,n0)","chain":"3 + 2<\/gadget>\n5<\/output>\n18 * 5<\/gadget>\n90<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n30<\/result>","index":2925} +{"problem":"a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 56 , find the sum of money ?","rationale":"\"a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 56 41 - - - - ? = > rs . 287 answer : b\"","correct":"b","options":{"a":"288 ","b":"287 ","c":"72 ","d":"205","e":"267"},"options_float":{"a":288.0,"b":287.0,"c":72.0,"d":205.0,"e":267.0},"annotated_formula":"multiply(divide(56, 40), add(add(const_100, 65), 40))","linear_formula":"add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)|","chain":"56 \/ 40<\/gadget>\n7\/5 = around 1.4<\/output>\n100 + 65<\/gadget>\n165<\/output>\n165 + 40<\/gadget>\n205<\/output>\n(7\/5) * 205<\/gadget>\n287<\/output>\n287<\/result>","index":2926} +{"problem":"in what time will a train 80 m long cross an electric pole , it its speed be 144 km \/ hr ?","rationale":"\"speed = 144 * 5 \/ 18 = 40 m \/ sec time taken = 80 \/ 40 = 2.0 sec . answer : a\"","correct":"a","options":{"a":"2.0 sec ","b":"2.8 sec ","c":"7.5 sec ","d":"2.3 sec","e":"1.5 sec"},"options_float":{"a":2.0,"b":2.8,"c":7.5,"d":2.3,"e":1.5},"annotated_formula":"divide(80, multiply(144, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n144 * (5\/18)<\/gadget>\n40<\/output>\n80 \/ 40<\/gadget>\n2<\/output>\n2<\/result>","index":2927} +{"problem":"john purchased 1375 large bottles at $ 1.75 per bottle and 690 small bottles at $ 1.35 per bottle . what was the approximate average price paid per bottle ?","rationale":"\"( 1375 * 1.75 + 690 * 1.35 ) \/ ( 1375 + 690 ) = ~ 1.62 option ( a )\"","correct":"a","options":{"a":"$ 1.62 ","b":"$ 1.64 ","c":"$ 1.68 ","d":"$ 1.72","e":"$ 1.76"},"options_float":{"a":1.62,"b":1.64,"c":1.68,"d":1.72,"e":1.76},"annotated_formula":"divide(add(multiply(1375, 1.75), multiply(690, 1.35)), add(1375, 690))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"1_375 * 1.75<\/gadget>\n2_406.25<\/output>\n690 * 1.35<\/gadget>\n931.5<\/output>\n2_406.25 + 931.5<\/gadget>\n3_337.75<\/output>\n1_375 + 690<\/gadget>\n2_065<\/output>\n3_337.75 \/ 2_065<\/gadget>\n1.616344<\/output>\n1.616344<\/result>","index":2928} +{"problem":"an investor deposited $ 5,000 to open a new savings account that earned 8 percent annual interest , compounded quarterly . if there were no other transactions in the account , what was the amount of money in the account 6 months after the account was opened ?","rationale":"\"the amount in the account after 6 months is 1.02 * 1.02 ( $ 5,000 ) = $ 5202 the answer is c .\"","correct":"c","options":{"a":"$ 5050 ","b":"$ 5101 ","c":"$ 5202 ","d":"$ 5303","e":"$ 5404"},"options_float":{"a":5050.0,"b":5101.0,"c":5202.0,"d":5303.0,"e":5404.0},"annotated_formula":"multiply(multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2)), power(add(divide(divide(8, const_4), const_100), const_1), const_2))","linear_formula":"add(const_2,const_3)|divide(n1,const_4)|divide(#1,const_100)|multiply(#0,const_100)|multiply(#0,const_2)|add(#2,const_1)|multiply(#3,#4)|power(#5,const_2)|multiply(#6,#7)|","chain":"2 + 3<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n5 * 2<\/gadget>\n10<\/output>\n500 * 10<\/gadget>\n5_000<\/output>\n8 \/ 4<\/gadget>\n2<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) + 1<\/gadget>\n51\/50 = around 1.02<\/output>\n(51\/50) ** 2<\/gadget>\n2_601\/2_500 = around 1.0404<\/output>\n5_000 * (2_601\/2_500)<\/gadget>\n5_202<\/output>\n5_202<\/result>","index":2929} +{"problem":"a pet store holds cats and dogs . if the difference between the number of cats and the number of dogs is 11 . what could be the ratio of cats to dogs in the pet store ?","rationale":"say theratioof cats to dogs is a \/ b . then thenumberof cats would be ax and thenumberof dogs bx , for some positive integer x . we are told that ax - bx = 11 - - > x ( a - b ) = 11 . since 11 is a prime number it could be broken into the product of two positive multiples only in one way : x ( a - b ) = 1 * 11 . the above implies that either x = 1 and ( a - b ) = 11 or x = 11 and ( a - b ) = 1 . therefore the correct answer should have the difference between numerator and denominator equal to 1 or 11 . for the original question only option which fits is e , 4 : 5 . cats = 11 * 4 = 44 and dogs = 11 * 5 = 55 . answer : d .","correct":"d","options":{"a":"1 : 3 ","b":"1 : 4 ","c":"1 : 5 ","d":"44 : 55","e":"4 : 6"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.2,"d":0.8,"e":0.6666666667},"annotated_formula":"divide(multiply(11, const_4), add(multiply(11, const_4), 11))","linear_formula":"multiply(n0,const_4)|add(n0,#0)|divide(#0,#1)","chain":"11 * 4<\/gadget>\n44<\/output>\n44 + 11<\/gadget>\n55<\/output>\n44 \/ 55<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":2931} +{"problem":"john makes $ 60 a week from his job . he earns a raise and now makes $ 90 a week . what is the % increase ?","rationale":"\"increase = ( 30 \/ 60 ) * 100 = ( 3 \/ 6 ) * 100 = 50 % . b\"","correct":"b","options":{"a":"16 % ","b":"50 % ","c":"17 % ","d":"17.61 %","e":"17.56 %"},"options_float":{"a":16.0,"b":50.0,"c":17.0,"d":17.61,"e":17.56},"annotated_formula":"multiply(divide(subtract(90, 60), 60), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"90 - 60<\/gadget>\n30<\/output>\n30 \/ 60<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":2932} +{"problem":"x can finish a work in 21 days . y can finish the same work in 15 days . yworked for 10 days and left the job . how many days does x alone need to finish the remaining work ?","rationale":"\"work done by x in 1 day = 1 \/ 21 work done by y in 1 day = 1 \/ 15 work done by y in 10 days = 10 \/ 15 = 2 \/ 3 remaining work = 1 – 2 \/ 3 = 1 \/ 3 number of days in which x can finish the remaining work = ( 1 \/ 3 ) \/ ( 1 \/ 21 ) = 7 c\"","correct":"c","options":{"a":"3 ","b":"5 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":3.0,"b":5.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"divide(subtract(const_1, multiply(10, divide(const_1, 15))), divide(const_1, 21))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n10 * (1\/15)<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 21<\/gadget>\n1\/21 = around 0.047619<\/output>\n(1\/3) \/ (1\/21)<\/gadget>\n7<\/output>\n7<\/result>","index":2934} +{"problem":"a sells a bicycle to b and makes a profit of 25 % . b sells the same bicycle to c at a profit of 50 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a .","rationale":"explanation : let cp be 100 a sells at 25 % profit so sp = 125 b sells at 50 % profit = 125 x ( 1 + 50 \/ 100 ) = 187.5 cp - - - sp 100 - - - 187.5 x - - - 225 cp = 225 x 100 \/ 187.5 = 120 answer : e","correct":"e","options":{"a":"237 ","b":"126 ","c":"971 ","d":"611","e":"120"},"options_float":{"a":237.0,"b":126.0,"c":971.0,"d":611.0,"e":120.0},"annotated_formula":"divide(divide(225, divide(add(const_100, 50), const_100)), divide(add(25, const_100), const_100))","linear_formula":"add(n1,const_100)|add(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#2)|divide(#4,#3)","chain":"100 + 50<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n225 \/ (3\/2)<\/gadget>\n150<\/output>\n25 + 100<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n150 \/ (5\/4)<\/gadget>\n120<\/output>\n120<\/result>","index":2935} +{"problem":"the average marks of a class of 12 students is 40 and that of another class of 28 students is 60 . find the average marks of all the students ?","rationale":"\"sum of the marks for the class of 12 students = 12 * 40 = 480 sum of the marks for the class of 28 students = 28 * 60 = 1680 sum of the marks for the class of 40 students = 480 + 1680 = 2160 average marks of all the students = 2160 \/ 40 = 54 answer : e\"","correct":"e","options":{"a":"52.2 ","b":"59.5 ","c":"52.8 ","d":"52.5","e":"54.0"},"options_float":{"a":52.2,"b":59.5,"c":52.8,"d":52.5,"e":54.0},"annotated_formula":"divide(add(multiply(12, 40), multiply(28, 60)), add(12, 28))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"12 * 40<\/gadget>\n480<\/output>\n28 * 60<\/gadget>\n1_680<\/output>\n480 + 1_680<\/gadget>\n2_160<\/output>\n12 + 28<\/gadget>\n40<\/output>\n2_160 \/ 40<\/gadget>\n54<\/output>\n54<\/result>","index":2937} +{"problem":"if 10 gallons of grape juice are added to 40 gallons of a mixture , which contains 20 percent grape juice then what percent of the resulting mixture is grape juice ?","rationale":"official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 40 × 0.20 = 8 gallons of grape juice . 40 × 0.80 = 32 gallons of other components . if we add 10 gallons of grape juice , we will end up with 18 gallons of grape juice and 32 gallons of other components , and we will have a total of 50 gallons of the mixture . so 18 \/ 50 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 18 \/ 50 = 36 \/ 100 = 36 % . the correct answer is choice ( d )","correct":"d","options":{"a":"14 % ","b":"25 % ","c":"28 % ","d":"36 %","e":"50 %"},"options_float":{"a":14.0,"b":25.0,"c":28.0,"d":36.0,"e":50.0},"annotated_formula":"multiply(divide(add(multiply(divide(20, const_100), 40), 10), add(40, 10)), const_100)","linear_formula":"add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|add(n0,#2)|divide(#3,#0)|multiply(#4,const_100)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 40<\/gadget>\n8<\/output>\n8 + 10<\/gadget>\n18<\/output>\n40 + 10<\/gadget>\n50<\/output>\n18 \/ 50<\/gadget>\n9\/25 = around 0.36<\/output>\n(9\/25) * 100<\/gadget>\n36<\/output>\n36<\/result>","index":2938} +{"problem":"15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 5 hrs . each would it take for 21 women if 3 women do as much work as 2 men ?","rationale":"\"let 1 man does 1 unit \/ hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 \/ 3 ) units \/ hr 21 w with 6 hrs a day will take ( 15 * 21 * 8 ) \/ ( 21 * 5 * ( 2 \/ 3 ) ) days = > 36 days answer : e\"","correct":"e","options":{"a":"30 ","b":"20 ","c":"16 ","d":"26","e":"36"},"options_float":{"a":30.0,"b":20.0,"c":16.0,"d":26.0,"e":36.0},"annotated_formula":"divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 5), divide(2, 3)))","linear_formula":"divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4)|","chain":"15 * 21<\/gadget>\n315<\/output>\n315 * 8<\/gadget>\n2_520<\/output>\n21 * 5<\/gadget>\n105<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n105 * (2\/3)<\/gadget>\n70<\/output>\n2_520 \/ 70<\/gadget>\n36<\/output>\n36<\/result>","index":2940} +{"problem":"ann and bob drive separately to a meeting . ann ' s average driving speed is greater than bob ' s avergae driving speed by one - third of bob ' s average driving speed , and ann drives twice as many miles as bob . what is the ratio e of the number of hours ann spends driving to the meeting to the number of hours bob spends driving to the meeting ?","rationale":"\"say the rate of bob is 3 mph and he covers 6 miles then he needs 6 \/ 3 = 2 hours to do that . now , in this case the rate of ann would be 3 + 3 * 1 \/ 3 = 4 mph and the distance she covers would be 6 * 2 = 12 miles , so she needs 12 \/ 4 = 3 hours for that . the ratio e of ann ' s time to bob ' s time is 3 : 2 . answer : b .\"","correct":"b","options":{"a":"8 : 3 ","b":"3 : 2 ","c":"4 : 3 ","d":"2 : 3","e":"3 : 8"},"options_float":{"a":2.6666666667,"b":1.5,"c":1.3333333333,"d":0.6666666667,"e":0.375},"annotated_formula":"divide(const_2, add(const_1, divide(const_1, const_3)))","linear_formula":"divide(const_1,const_3)|add(#0,const_1)|divide(const_2,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n2 \/ (4\/3)<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":2941} +{"problem":"as a treat for her two crying children , a mother runs to the freezer in which she has 4 cherry ice pops , 3 orange ice pops , and 4 lemon - lime ice pops . if she chooses two at random to bring outside to the children , but realizes as she runs out the door that she can not bring them different flavors without one invariably being jealous of the other and getting even more upset , what is the probability that she has to return to the freezer to make sure that they each receive the same flavor ?","rationale":"probability of not getting the same flavor - > fouvarable - > cherry - orange [ 4 c 1 * 3 c 1 [ or simply 4 * 3 ] or cherry - lemon [ 4 * 4 ] or orange - lemon [ 3 * 4 ] prob = ( 4 * 3 + 4 * 4 + 3 * 4 ) \/ 9 c 2 = 40 \/ 3960 = 1 \/ 99 answer - > a","correct":"a","options":{"a":"1 \/ 99 ","b":"1 \/ 6 ","c":"5 \/ 18 ","d":"13 \/ 18","e":"5 \/ 6"},"options_float":{"a":0.0101010101,"b":0.1666666667,"c":0.2777777778,"d":0.7222222222,"e":0.8333333333},"annotated_formula":"divide(subtract(4, 3), subtract(const_100, subtract(4, 3)))","linear_formula":"subtract(n0,n1)|subtract(const_100,#0)|divide(#0,#1)","chain":"4 - 3<\/gadget>\n1<\/output>\n100 - 1<\/gadget>\n99<\/output>\n1 \/ 99<\/gadget>\n1\/99 = around 0.010101<\/output>\n1\/99 = around 0.010101<\/result>","index":2942} +{"problem":"a and b together can complete a work in 12 days . a alone complete it in 20 days . if b does the work only for half a day daily , then in how many days a and b together will complete the work ?","rationale":"b ' s one day work = ( 1 \/ 12 - 1 \/ 20 ) = 1 \/ 30 . now , ( a + b ) ' s one day work = ( 1 \/ 20 + 1 \/ 60 ) = 1 \/ 15 . [ ∵ b work for half day only ] . so , a and b together will complete the work in 15 days . correct answer : ` ` c ' ' .","correct":"c","options":{"a":"110 days ","b":"11 days ","c":"15 days ","d":"20 days","e":"17 days"},"options_float":{"a":110.0,"b":11.0,"c":15.0,"d":20.0,"e":17.0},"annotated_formula":"add(12, const_3)","linear_formula":"add(n0,const_3)","chain":"12 + 3<\/gadget>\n15<\/output>\n15<\/result>","index":2943} +{"problem":"in a fuel station the service costs $ 1.65 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ?","rationale":"\"total cost = ( 1.65 * 12 ) + ( 0.65 * 12 * 55 ) = 19.8 + 429 = > 448.8 hence answer will be ( d )\"","correct":"d","options":{"a":"320 $ ","b":"380 $ ","c":"420 $ ","d":"448.8 $","e":"480 $"},"options_float":{"a":320.0,"b":380.0,"c":420.0,"d":448.8,"e":480.0},"annotated_formula":"multiply(multiply(0.65, 55), 12)","linear_formula":"multiply(n1,n3)|multiply(n2,#0)|","chain":"0.65 * 55<\/gadget>\n35.75<\/output>\n35.75 * 12<\/gadget>\n429<\/output>\n429<\/result>","index":2944} +{"problem":"a certain rectangular crate measures 6 feet by 8 feet by 10 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ?","rationale":"\"ans is ` ` b ' ' . for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . . therefore for max vol base should be 8 x 10 i . e . of radius 8 \/ 2 = 4 b\"","correct":"b","options":{"a":"4 ","b":"4 ","c":"6 ","d":"8","e":"10"},"options_float":{"a":4.0,"b":4.0,"c":6.0,"d":8.0,"e":10.0},"annotated_formula":"sqrt(divide(volume_cylinder(divide(8, const_2), 6), multiply(const_pi, 6)))","linear_formula":"divide(n1,const_2)|multiply(n0,const_pi)|volume_cylinder(#0,n0)|divide(#2,#1)|sqrt(#3)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\npi * (4 ** 2) * 6<\/gadget>\n96*pi = around 301.592895<\/output>\npi * 6<\/gadget>\n6*pi = around 18.849556<\/output>\n(96*pi) \/ (6*pi)<\/gadget>\n16<\/output>\n16 ** (1\/2)<\/gadget>\n4<\/output>\n4<\/result>","index":2945} +{"problem":"the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 950 per sq . metre .","rationale":"\"area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 950 ã — 22 ) = rs . 20900 answer : option c\"","correct":"c","options":{"a":"s . 20,550 ","b":"s . 15,600 ","c":"s . 20,900 ","d":"s . 17,600","e":"s . 17,900"},"options_float":{"a":20550.0,"b":15600.0,"c":20900.0,"d":17600.0,"e":17900.0},"annotated_formula":"multiply(950, multiply(5.5, 4))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 4<\/gadget>\n22<\/output>\n950 * 22<\/gadget>\n20_900<\/output>\n20_900<\/result>","index":2946} +{"problem":"how long does a train 110 m long running at the speed of 72 km \/ hr takes to cross a bridge 136 m length ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec total distance covered = 110 + 136 = 246 m . required time = 246 \/ 20 = 12.3 sec . answer : b\"","correct":"b","options":{"a":"82.1 sec . ","b":"12.3 sec ","c":"19.1 sec . ","d":"17.1 sec .","e":"42.1 sec ."},"options_float":{"a":82.1,"b":12.3,"c":19.1,"d":17.1,"e":42.1},"annotated_formula":"divide(add(110, 136), multiply(72, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"110 + 136<\/gadget>\n246<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n246 \/ 20<\/gadget>\n123\/10 = around 12.3<\/output>\n123\/10 = around 12.3<\/result>","index":2948} +{"problem":"two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability that jane will be on the same committee as albert ?","rationale":"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! \/ 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability = ( 8 \/ 20 ) * 100 = 40 % . ans : : d","correct":"d","options":{"a":"12 % ","b":"20 % ","c":"33 % ","d":"40 %","e":"50 %"},"options_float":{"a":12.0,"b":20.0,"c":33.0,"d":40.0,"e":50.0},"annotated_formula":"multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4))","linear_formula":"choose(const_4,const_1)|choose(n0,n1)|multiply(const_5,const_5)|multiply(#0,const_2)|multiply(#2,const_4)|divide(#3,#1)|multiply(#5,#4)","chain":"binomial(4, 1)<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\nbinomial(6, 3)<\/gadget>\n20<\/output>\n8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25 * 4<\/gadget>\n100<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":2950} +{"problem":"a group of 7 teenagers recently visited the japanese restaurant . the total bill for the meal , including 20 % gratuity , came to $ 840 . on average , how much did the meal of each individual cost before gratuity ?","rationale":"7 teenagers - so total 7 people the bill $ 840 includes 20 % gratuity . . . so the actual cost of dinner was $ 700 now , the cost per person will be $ 700 \/ 7 which is $ 10 option d","correct":"d","options":{"a":"$ 160 ","b":"$ 96 ","c":"$ 90 ","d":"$ 10","e":"$ 55"},"options_float":{"a":160.0,"b":96.0,"c":90.0,"d":10.0,"e":55.0},"annotated_formula":"subtract(subtract(subtract(divide(840, 20), 20), const_10), const_2)","linear_formula":"divide(n2,n1)|subtract(#0,n1)|subtract(#1,const_10)|subtract(#2,const_2)","chain":"840 \/ 20<\/gadget>\n42<\/output>\n42 - 20<\/gadget>\n22<\/output>\n22 - 10<\/gadget>\n12<\/output>\n12 - 2<\/gadget>\n10<\/output>\n10<\/result>","index":2951} +{"problem":"a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 68 books in the special collection at that time , how many books of the special collection were loaned out during that month ?","rationale":"\"total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 65 % = 35 % of books that were loaned out are not returned . now , there are 68 books , thus 76 - 68 = 7 books are not returned . { loaned out } * 0.35 = 7 - - > { loaned out } = 20 . answer : a .\"","correct":"a","options":{"a":"20 ","b":"30 ","c":"35 ","d":"40","e":"55"},"options_float":{"a":20.0,"b":30.0,"c":35.0,"d":40.0,"e":55.0},"annotated_formula":"divide(subtract(75, 68), subtract(const_1, divide(65, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"75 - 68<\/gadget>\n7<\/output>\n65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n1 - (13\/20)<\/gadget>\n7\/20 = around 0.35<\/output>\n7 \/ (7\/20)<\/gadget>\n20<\/output>\n20<\/result>","index":2952} +{"problem":"if 7 ^ q is a factor of the product of the integers from 1 to 100 , inclusive , what is the largest value of q ?","rationale":"so the question just means that we have to find all the multiples q of 7 between 1 to 100 so there are 14 multiples of 7 ( 7 - 98 ) but 49 and 98 contain two 7 ' s as factors so we have to add 14 + 2 = 16 e","correct":"e","options":{"a":"12 ","b":"13 ","c":"14 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"add(divide(100, 7), divide(divide(100, 7), 7))","linear_formula":"divide(n2,n0)|divide(#0,n0)|add(#0,#1)","chain":"100 \/ 7<\/gadget>\n100\/7 = around 14.285714<\/output>\n(100\/7) \/ 7<\/gadget>\n100\/49 = around 2.040816<\/output>\n(100\/7) + (100\/49)<\/gadget>\n800\/49 = around 16.326531<\/output>\n800\/49 = around 16.326531<\/result>","index":2954} +{"problem":"susan drove an average speed of 30 miles per hour for the first 60 miles of a tripthen at a average speed of 60 miles \/ hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles \/ hr for the entire trip","rationale":"\"avg . speed = total distance \/ total time total distance = 90 miles total time = 60 \/ 30 + 30 \/ 60 = 5 \/ 2 avg . speed = 36 answer - b\"","correct":"b","options":{"a":"35 ","b":"36 ","c":"45 ","d":"50","e":"55"},"options_float":{"a":35.0,"b":36.0,"c":45.0,"d":50.0,"e":55.0},"annotated_formula":"divide(add(60, 30), add(divide(60, 30), divide(30, 60)))","linear_formula":"add(n1,n3)|divide(n1,n0)|divide(n3,n2)|add(#1,#2)|divide(#0,#3)|","chain":"60 + 30<\/gadget>\n90<\/output>\n60 \/ 30<\/gadget>\n2<\/output>\n30 \/ 60<\/gadget>\n1\/2 = around 0.5<\/output>\n2 + (1\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n90 \/ (5\/2)<\/gadget>\n36<\/output>\n36<\/result>","index":2955} +{"problem":"a 415 m long train is running at 63 km \/ hr . in how much time will it cross a tunnel 285 m long ?","rationale":"\"speed of the train = ( 63 * 5 \/ 18 ) m \/ sec = 35 \/ 2 m \/ sec . distance travelled = ( speed x time ) time = ( distance travelled \/ speed ) distance travelled = length of the train + length of the tunnel = 415 + 285 = 700 m time = ( 700 \/ 35 \/ 2 ) = 40 sec answer : e\"","correct":"e","options":{"a":"25 ","b":"35 ","c":"45 ","d":"60","e":"40"},"options_float":{"a":25.0,"b":35.0,"c":45.0,"d":60.0,"e":40.0},"annotated_formula":"divide(add(415, 285), multiply(63, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"415 + 285<\/gadget>\n700<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n63 * (5\/18)<\/gadget>\n35\/2 = around 17.5<\/output>\n700 \/ (35\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":2956} +{"problem":"the bankers discount of a certain sum of money is rs . 288 and the true discount on the same sum for the same time is rs . 240 . the sum due is :","rationale":"\"sum = ( b . d * t . d ) \/ ( b . d - t . d ) ( 288 * 240 ) \/ 288 - 240 ; 1440 answer : c\"","correct":"c","options":{"a":"1400 ","b":"1420 ","c":"1440 ","d":"1460","e":"1480"},"options_float":{"a":1400.0,"b":1420.0,"c":1440.0,"d":1460.0,"e":1480.0},"annotated_formula":"divide(multiply(240, 288), subtract(288, 240))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"240 * 288<\/gadget>\n69_120<\/output>\n288 - 240<\/gadget>\n48<\/output>\n69_120 \/ 48<\/gadget>\n1_440<\/output>\n1_440<\/result>","index":2957} +{"problem":"a shopkeeper sold an article offering a discount of 4 % and earned a profit of 32 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"let c . p . be rs . 100 . then , s . p . = rs . 132 let marked price be rs . x . then , 96 \/ 100 x = 132 x = 13200 \/ 96 = rs . 137.5 now , s . p . = rs . 137.5 , c . p . = rs . 100 profit % = 37.5 % . answer : e","correct":"e","options":{"a":"24.5 ","b":"28.5 ","c":"30.5 ","d":"32.5","e":"37.5"},"options_float":{"a":24.5,"b":28.5,"c":30.5,"d":32.5,"e":37.5},"annotated_formula":"subtract(divide(multiply(add(const_100, 32), const_100), subtract(const_100, 4)), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)","chain":"100 + 32<\/gadget>\n132<\/output>\n132 * 100<\/gadget>\n13_200<\/output>\n100 - 4<\/gadget>\n96<\/output>\n13_200 \/ 96<\/gadget>\n275\/2 = around 137.5<\/output>\n(275\/2) - 100<\/gadget>\n75\/2 = around 37.5<\/output>\n75\/2 = around 37.5<\/result>","index":2958} +{"problem":"the volume of the sphere qq is ( dfrac { 37 } { 64 } % ) less than the volume of sphere pp and the volume of sphere rr is ( dfrac { 19 } { 27 } % ) less than that of sphere qq . by what is the surface areaof sphere rr less than the surfacearea of sphere pp ?","rationale":"let the volume of sphere pp be 64 parts . therefore volume of sphere qq = 64 − 3764 % = 64 − 3764 % of 6464 = 64 − 37 = 27 = 64 − 37 = 27 parts . the volume of rr = 27 − 1927 × 27 = 27 − 1927 × 27 = 27 − 19 = 8 = 27 − 19 = 8 parts . volume ratio : = p : q : r = 64 : 27 : 8 = p : q : r = 64 : 27 : 8 radius ratio : = p : q : r = 4 : 3 : 2 = p : q : r = 4 : 3 : 2 the surface area will be 16 : 9 : 516 : 9 : 5 surface area of rr is less than the surface area of sphere pp 16 k − 4 k = 12 k 16 k − 4 k = 12 k now , = 12 k 16 k × 100 = 12 k 16 k × 100 = 75 % = 75 % thus surface area of sphere rr is less than the surface area of sphere p by 75 % c","correct":"c","options":{"a":"50 % ","b":"60 % ","c":"75 % ","d":"80 %","e":"89 %"},"options_float":{"a":50.0,"b":60.0,"c":75.0,"d":80.0,"e":89.0},"annotated_formula":"subtract(const_100, multiply(multiply(multiply(divide(37, 64), divide(19, 27)), divide(37, 64)), const_100))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|multiply(#0,#2)|multiply(#3,const_100)|subtract(const_100,#4)","chain":"37 \/ 64<\/gadget>\n37\/64 = around 0.578125<\/output>\n19 \/ 27<\/gadget>\n19\/27 = around 0.703704<\/output>\n(37\/64) * (19\/27)<\/gadget>\n703\/1_728 = around 0.406829<\/output>\n(703\/1_728) * (37\/64)<\/gadget>\n26_011\/110_592 = around 0.235198<\/output>\n(26_011\/110_592) * 100<\/gadget>\n650_275\/27_648 = around 23.519784<\/output>\n100 - (650_275\/27_648)<\/gadget>\n2_114_525\/27_648 = around 76.480216<\/output>\n2_114_525\/27_648 = around 76.480216<\/result>","index":2959} +{"problem":"what will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of rs . 10 per square meter ?","rationale":"explanation : in this question , we are having perimeter . we know perimeter = 2 ( l + b ) , right so , 2 ( l + b ) = 340 as we have to make 1 meter boundary around this , so area of boundary = ( ( l + 2 ) + ( b + 2 ) - lb ) = 2 ( l + b ) + 4 = 340 + 4 = 344 so required cost will be = 344 * 10 = 3440 answer : option b","correct":"b","options":{"a":"rs . 3430 ","b":"rs . 3440 ","c":"rs . 3450 ","d":"rs . 3460","e":"rs . 3490"},"options_float":{"a":3430.0,"b":3440.0,"c":3450.0,"d":3460.0,"e":3490.0},"annotated_formula":"multiply(add(340, const_4), 10)","linear_formula":"add(n1,const_4)|multiply(n2,#0)","chain":"340 + 4<\/gadget>\n344<\/output>\n344 * 10<\/gadget>\n3_440<\/output>\n3_440<\/result>","index":2960} +{"problem":"a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 18 days . in how many days b alone can do the work ?","rationale":"\"explanation : a ’ s 5 day work = 5 * 1 \/ 15 = 1 \/ 3 remaining work = 1 - 1 \/ 3 = 2 \/ 3 b completes 2 \/ 3 work in 6 days b alone can do in x days 2 \/ 3 * x = 18 x = 27 days answer : option c\"","correct":"c","options":{"a":"5 days ","b":"7 days ","c":"27 days ","d":"9 days","e":"10 days"},"options_float":{"a":5.0,"b":7.0,"c":27.0,"d":9.0,"e":10.0},"annotated_formula":"inverse(multiply(inverse(18), subtract(const_1, multiply(5, inverse(15)))))","linear_formula":"inverse(n2)|inverse(n0)|multiply(n1,#1)|subtract(const_1,#2)|multiply(#0,#3)|inverse(#4)|","chain":"1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n5 * (1\/15)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/18) * (2\/3)<\/gadget>\n1\/27 = around 0.037037<\/output>\n1 \/ (1\/27)<\/gadget>\n27<\/output>\n27<\/result>","index":2961} +{"problem":"the h . c . f . of two numbers is 30 and the other two factors of their l . c . m . are 25 and 24 . the larger of the two numbers is","rationale":"\"solution clearly , the numbers are ( 30 x 25 ) and ( 30 x 24 ) . larger number = ( 30 x 25 ) = 750 . answer d\"","correct":"d","options":{"a":"276 ","b":"799 ","c":"712 ","d":"750","e":"none"},"options_float":{"a":276.0,"b":799.0,"c":712.0,"d":750.0,"e":null},"annotated_formula":"multiply(30, 24)","linear_formula":"multiply(n0,n2)|","chain":"30 * 24<\/gadget>\n720<\/output>\n720<\/result>","index":2962} +{"problem":"arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 15 % profit ?","rationale":"\"c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 115 % of rs . 630 = 115 \/ 100 * 630 = rs . 724.5 s . p . per kg = 724.5 \/ 50 = 14.49 a\"","correct":"a","options":{"a":"14.49 ","b":"16.33 ","c":"16.35 ","d":"16.3","e":"16.32"},"options_float":{"a":14.49,"b":16.33,"c":16.35,"d":16.3,"e":16.32},"annotated_formula":"divide(add(add(multiply(30, 11.50), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.50), multiply(20, 14.25)), const_100), 15)), add(30, 20))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,const_100)|multiply(n4,#4)|add(#3,#5)|divide(#6,#0)|","chain":"30 * 11.5<\/gadget>\n345<\/output>\n20 * 14.25<\/gadget>\n285<\/output>\n345 + 285<\/gadget>\n630<\/output>\n630 \/ 100<\/gadget>\n63\/10 = around 6.3<\/output>\n(63\/10) * 15<\/gadget>\n189\/2 = around 94.5<\/output>\n630 + (189\/2)<\/gadget>\n1_449\/2 = around 724.5<\/output>\n30 + 20<\/gadget>\n50<\/output>\n(1_449\/2) \/ 50<\/gadget>\n1_449\/100 = around 14.49<\/output>\n1_449\/100 = around 14.49<\/result>","index":2964} +{"problem":"one pipe can fill a tank four times as fast as another pipe . if together the two pipes can fill the tank in 36 minutes , then the slower pipe alone will be able to fill the tank in :","rationale":"explanation : let time taken by the faster pipe to fill the tank = x minutes . then , time taken by the slower pipe to fill the tank = 4 x minutes . time taken by both pipes together to fill the tank = x x 4 x \/ x + 4 x = 4 x \/ 5 4 x \/ 5 = 36 ⇒ x = 45 time taken by the slower pipe to fill the tank = 4 x = 180 minutes . answer : option a","correct":"a","options":{"a":"180 min ","b":"144 min . ","c":"126 min ","d":"114 min","e":"124 min"},"options_float":{"a":180.0,"b":144.0,"c":126.0,"d":114.0,"e":124.0},"annotated_formula":"multiply(add(const_1, const_4), 36)","linear_formula":"add(const_1,const_4)|multiply(n0,#0)","chain":"1 + 4<\/gadget>\n5<\/output>\n5 * 36<\/gadget>\n180<\/output>\n180<\/result>","index":2965} +{"problem":"carl is facing very difficult financial times and can only pay the interest on a $ 10,000 loan he has taken . the bank charges him a quarterly compound rate of 8 % . what is the approximate interest he pays annually ?","rationale":"\"an easy way too attempt this is by approximation : for the first quarter , he pays 8 % of 10,000 which is $ 800 . so for the four quarters in the year , he will approximately play 800 * 4 = 3200 over the entire year . because of compound interest it will be more than 3200 . approx . option d is correct . answer : d\"","correct":"d","options":{"a":"1500 ","b":"1750 ","c":"2000 ","d":"3500","e":"2500"},"options_float":{"a":1500.0,"b":1750.0,"c":2000.0,"d":3500.0,"e":2500.0},"annotated_formula":"subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(8, const_100)), const_4)), multiply(const_100, const_100))","linear_formula":"divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n1 + (2\/25)<\/gadget>\n27\/25 = around 1.08<\/output>\n(27\/25) ** 4<\/gadget>\n531_441\/390_625 = around 1.360489<\/output>\n10_000 * (531_441\/390_625)<\/gadget>\n8_503_056\/625 = around 13_604.8896<\/output>\n(8_503_056\/625) - 10_000<\/gadget>\n2_253_056\/625 = around 3_604.8896<\/output>\n2_253_056\/625 = around 3_604.8896<\/result>","index":2966} +{"problem":"a patient is given exactly 750 milliliters of a mixture of two medications daily . medication a contains 40 % pain killer and medication b contains 20 % pain killer . if the patient receives exactly 215 milliliters of pain killer daily , how many milliliters of medication b are in the mixture ?","rationale":"let there be a milliliters of medication a and b milliliters of medication b a + b = 750 - - 1 medication a contains 40 % pain killer and medication b contains 20 % pain killer . ( 40 \/ 100 ) * a + ( 20 \/ 100 ) * b = 215 = > ( 2 \/ 5 ) * a + ( 1 \/ 5 ) * b = 215 multiplying by 5 , we get 2 a + b = 1075 - - 2 equation 2 - equation 1 , we get a = 325 b = 425 answer d","correct":"d","options":{"a":"150 ","b":"225 ","c":"325 ","d":"425","e":"550"},"options_float":{"a":150.0,"b":225.0,"c":325.0,"d":425.0,"e":550.0},"annotated_formula":"divide(subtract(multiply(divide(40, const_100), 750), 215), subtract(divide(40, const_100), divide(20, const_100)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 750<\/gadget>\n300<\/output>\n300 - 215<\/gadget>\n85<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(2\/5) - (1\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n85 \/ (1\/5)<\/gadget>\n425<\/output>\n425<\/result>","index":2967} +{"problem":"let s is a set of all natural numbers less than 120 such that hcf of any element in s and 120 is 1 . find the sum of all elements in s ?","rationale":"s is a set of all natural numbers less than 120 such that hcf of any element in s and 120 is 1 . such numbers are , s = { 1 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 } sum of s = 1584 . answer : a","correct":"a","options":{"a":"1584 ","b":"7640 ","c":"5640 ","d":"7240","e":"6780"},"options_float":{"a":1584.0,"b":7640.0,"c":5640.0,"d":7240.0,"e":6780.0},"annotated_formula":"add(add(multiply(add(1, const_4), const_100), const_1000), add(multiply(multiply(const_2, const_4), const_10), const_4))","linear_formula":"add(n2,const_4)|multiply(const_2,const_4)|multiply(#0,const_100)|multiply(#1,const_10)|add(#2,const_1000)|add(#3,const_4)|add(#4,#5)","chain":"1 + 4<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n500 + 1_000<\/gadget>\n1_500<\/output>\n2 * 4<\/gadget>\n8<\/output>\n8 * 10<\/gadget>\n80<\/output>\n80 + 4<\/gadget>\n84<\/output>\n1_500 + 84<\/gadget>\n1_584<\/output>\n1_584<\/result>","index":2969} +{"problem":"a swimmer can swim in still water at 4 km \/ h . if the speed of the water current is 2 km \/ h , how many hours will the swimmer take to swim against the current for 5 km ?","rationale":"\"the swimmer can swim against the current at a speed of 4 - 2 = 2 km \/ h . the time it will take is 5 \/ 2 = 2.5 hours . the answer is b .\"","correct":"b","options":{"a":"2.0 ","b":"2.5 ","c":"3.0 ","d":"3.5","e":"4.0"},"options_float":{"a":2.0,"b":2.5,"c":3.0,"d":3.5,"e":4.0},"annotated_formula":"divide(5, subtract(4, 2))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"4 - 2<\/gadget>\n2<\/output>\n5 \/ 2<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":2970} +{"problem":"if a tire rotates at 400 revolutions per minute when the car is traveling 48 km \/ h , what is the circumference of the tire ?","rationale":"\"400 rev \/ minute = 400 * 60 rev \/ 60 minutes = 24,000 rev \/ hour 24,000 * c = 48,000 m : c is the circumference c = 2 meters correct answer d\"","correct":"d","options":{"a":"7 meters ","b":"9 meters ","c":"8 meters ","d":"2 meters","e":"3 meters"},"options_float":{"a":7.0,"b":9.0,"c":8.0,"d":2.0,"e":3.0},"annotated_formula":"multiply(divide(48, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000)","linear_formula":"multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n60 * 400<\/gadget>\n24_000<\/output>\n48 \/ 24_000<\/gadget>\n1\/500 = around 0.002<\/output>\n(1\/500) * 1_000<\/gadget>\n2<\/output>\n2<\/result>","index":2971} +{"problem":"a snooker tournament charges $ 45.00 for vip seats and $ 20.00 for general admission ( “ regular ” seats ) . on a certain night , a total of 320 tickets were sold , for a total cost of $ 7500 . how many fewer tickets were sold that night for vip seats than for general admission seats ?","rationale":"let no of sits in vip enclosure is x then x * 45 + 20 ( 320 - x ) = 7500 or 25 x = 7500 - 6400 , x = 1100 \/ 25 = 44 vip = 44 general 320 - 44 = 276 b","correct":"b","options":{"a":"200 ","b":"276 ","c":"230 ","d":"240","e":"250"},"options_float":{"a":200.0,"b":276.0,"c":230.0,"d":240.0,"e":250.0},"annotated_formula":"subtract(320, divide(subtract(divide(7500, 20), 320), subtract(divide(45, 20), const_1)))","linear_formula":"divide(n3,n1)|divide(n0,n1)|subtract(#0,n2)|subtract(#1,const_1)|divide(#2,#3)|subtract(n2,#4)","chain":"7_500 \/ 20<\/gadget>\n375<\/output>\n375 - 320<\/gadget>\n55<\/output>\n45 \/ 20<\/gadget>\n9\/4 = around 2.25<\/output>\n(9\/4) - 1<\/gadget>\n5\/4 = around 1.25<\/output>\n55 \/ (5\/4)<\/gadget>\n44<\/output>\n320 - 44<\/gadget>\n276<\/output>\n276<\/result>","index":2972} +{"problem":"find the principle on a certain sum of money at 5 % per annum for 3 1 \/ 5 years if the amount being rs . 1160 ?","rationale":"\"explanation : 1160 = p [ 1 + ( 5 * 16 \/ 5 ) \/ 100 ] p = 1000 answer : option a\"","correct":"a","options":{"a":"rs . 1000 ","b":"rs . 1100 ","c":"rs . 1010 ","d":"rs . 10000","e":"none of these"},"options_float":{"a":1000.0,"b":1100.0,"c":1010.0,"d":10000.0,"e":null},"annotated_formula":"divide(1160, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1))","linear_formula":"multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|","chain":"3 * 5<\/gadget>\n15<\/output>\n15 + 3<\/gadget>\n18<\/output>\n18 \/ 5<\/gadget>\n18\/5 = around 3.6<\/output>\n(18\/5) * 5<\/gadget>\n18<\/output>\n18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n(9\/50) + 1<\/gadget>\n59\/50 = around 1.18<\/output>\n1_160 \/ (59\/50)<\/gadget>\n58_000\/59 = around 983.050847<\/output>\n58_000\/59 = around 983.050847<\/result>","index":2973} +{"problem":"4 balls of different colors are to be placed in 3 different boxes such that any box contains at least 1 ball . what is the maximum number of different ways in which this can be done ?","rationale":"since the balls are all of different colors , let ' s permute them and then decide how many balls we put in each box . for example , arrange in a row the balls , then decide : two balls go into the first box , next two in the second box , and the last ball goes to the third box . since in each box there must be at least one ball , we have the possibilities of ( 2 , 2,1 ) , ( 2 , 1,2 ) , ( 1 , 2,2 ) or ( 3 , 1,1 ) , ( 1 , 3,1 ) , ( 1 , 1,3 ) balls in the three boxes . for the 2 , 2,1 type arrangements , we have [ 6 ! \/ ( 2 ! 2 ! 1 ! ) ] * 3 = 100 possibilities . inside a box , it does n ' t matter the order of the balls . for the 3 , 1,1 type arrangements , we have [ 6 ! \/ ( 3 ! 1 ! 1 ! ) ] * 3 = 80 possibilities . total of 100 + 80 = 180 possibilities . answer e .","correct":"e","options":{"a":"60 ","b":"90 ","c":"120 ","d":"150","e":"180"},"options_float":{"a":60.0,"b":90.0,"c":120.0,"d":150.0,"e":180.0},"annotated_formula":"add(multiply(multiply(multiply(4, 3), const_2), multiply(3, const_2)), multiply(multiply(3, const_2), multiply(3, const_2)))","linear_formula":"multiply(n0,n1)|multiply(n1,const_2)|multiply(#0,const_2)|multiply(#1,#1)|multiply(#2,#1)|add(#4,#3)","chain":"4 * 3<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n3 * 2<\/gadget>\n6<\/output>\n24 * 6<\/gadget>\n144<\/output>\n6 * 6<\/gadget>\n36<\/output>\n144 + 36<\/gadget>\n180<\/output>\n180<\/result>","index":2974} +{"problem":"the length of the bridge , which a train 140 metres long and travelling at 45 km \/ hr can cross in 30 seconds , is ?","rationale":"\"speed = [ 45 x 5 \/ 18 ] m \/ sec = [ 25 \/ 2 ] m \/ sec time = 30 sec let the length of bridge be x metres . then , ( 140 + x ) \/ 30 = 25 \/ 2 = > 2 ( 140 + x ) = 750 = > x = 235 m . answer : c\"","correct":"c","options":{"a":"255 ","b":"267 ","c":"235 ","d":"277","e":"211"},"options_float":{"a":255.0,"b":267.0,"c":235.0,"d":277.0,"e":211.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 140)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 30<\/gadget>\n375<\/output>\n375 - 140<\/gadget>\n235<\/output>\n235<\/result>","index":2979} +{"problem":"the average height of 35 boys in a class was calculated as 185 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ?","rationale":"\"calculated average height of 35 boys = 185 cm . wrong total height of 35 boys = 180 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 185 cm - ( 166 cm - 106 cm ) \/ 35 = 185 cm - 60 \/ 35 cm = 185 cm - 1.71 cm = 183.29 cm . answer : b\"","correct":"b","options":{"a":"187.89 cm ","b":"183.29 cm ","c":"123.98 cm ","d":"149.98 cm","e":"146.89 cm"},"options_float":{"a":187.89,"b":183.29,"c":123.98,"d":149.98,"e":146.89},"annotated_formula":"floor(divide(add(subtract(multiply(35, 185), 166), 106), 35))","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)|","chain":"35 * 185<\/gadget>\n6_475<\/output>\n6_475 - 166<\/gadget>\n6_309<\/output>\n6_309 + 106<\/gadget>\n6_415<\/output>\n6_415 \/ 35<\/gadget>\n1_283\/7 = around 183.285714<\/output>\nfloor(1_283\/7)<\/gadget>\n183<\/output>\n183<\/result>","index":2981} +{"problem":"whats the reminder when 65,985 , 241,545 , 898,754 , 582,556 , 898,522 , 454,889 is divided by 16","rationale":"a number ending in a 0 is divisible by 2 . a number ending in 2 zeroes is divisible by 4 . a number ending in 3 zeroes is divisible by 8 . a number ending in 4 zeroes in divisible by 16 . given the obscene number , you should immediately be convinced that you will need to focus on a very small part of it . 65,985 , 241,545 , 898,754 , 582,556 , 898,522 , 454,889 = 65,985 , 241,545 , 898,754 , 582,556 , 898,522 , 450,000 + 4,889 the first number is divisible by 16 . you just have to find the remainder when you divide 5287 by 16 . that will be the remainder when you divide the original number by 16 . 4889 \/ 16 gives remainder 9 . answer ( e )","correct":"e","options":{"a":"1 ","b":"7 . ","c":"2 . ","d":"3","e":"9"},"options_float":{"a":1.0,"b":7.0,"c":2.0,"d":3.0,"e":9.0},"annotated_formula":"add(multiply(const_4, const_2), reminder(add(add(multiply(subtract(const_10, const_1), const_100), multiply(multiply(add(const_3, const_2), const_100), const_10)), multiply(add(const_12, add(const_3, const_2)), add(const_3, const_2))), 16))","linear_formula":"add(const_2,const_3)|multiply(const_2,const_4)|subtract(const_10,const_1)|add(#0,const_12)|multiply(#2,const_100)|multiply(#0,const_100)|multiply(#5,const_10)|multiply(#3,#0)|add(#4,#6)|add(#8,#7)|reminder(#9,n6)|add(#1,#10)","chain":"4 * 2<\/gadget>\n8<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 * 100<\/gadget>\n900<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n500 * 10<\/gadget>\n5_000<\/output>\n900 + 5_000<\/gadget>\n5_900<\/output>\n12 + 5<\/gadget>\n17<\/output>\n17 * 5<\/gadget>\n85<\/output>\n5_900 + 85<\/gadget>\n5_985<\/output>\n5_985 % 16<\/gadget>\n1<\/output>\n8 + 1<\/gadget>\n9<\/output>\n9<\/result>","index":2982} +{"problem":"you enter a weight loss challenge game and manage to lose 12 % of your body weight . for the final weigh in you are forced to wear clothes that add 2 % to your weight . what percentage of weight loss is measured at the final weigh in ?","rationale":"\"( 100 % - 12 % ) * ( 100 % + 2 % ) = 0.88 * 1.02 = 10.24 % the weigh in records your weight loss at 10.24 % ! the answer is a\"","correct":"a","options":{"a":"10.24 % ","b":"9.22 % ","c":"9 % ","d":"14 %","e":"12 %"},"options_float":{"a":10.24,"b":9.22,"c":9.0,"d":14.0,"e":12.0},"annotated_formula":"multiply(subtract(const_1, multiply(subtract(const_1, divide(12, const_100)), add(const_1, divide(2, const_100)))), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100)|","chain":"12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n1 - (3\/25)<\/gadget>\n22\/25 = around 0.88<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n1 + (1\/50)<\/gadget>\n51\/50 = around 1.02<\/output>\n(22\/25) * (51\/50)<\/gadget>\n561\/625 = around 0.8976<\/output>\n1 - (561\/625)<\/gadget>\n64\/625 = around 0.1024<\/output>\n(64\/625) * 100<\/gadget>\n256\/25 = around 10.24<\/output>\n256\/25 = around 10.24<\/result>","index":2983} +{"problem":"if 20 men can build a water fountain 56 metres long in 14 days , what length of a similar water fountain can be built by 35 men in 3 days ?","rationale":"\"explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 14 : 3 : : 56 : x therefore ( 20 x 14 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) \/ 280 = 49 hence , the required length is 21 m . answer : a\"","correct":"a","options":{"a":"21 m ","b":"24 m ","c":"27 m ","d":"29 m","e":"30 m"},"options_float":{"a":21.0,"b":24.0,"c":27.0,"d":29.0,"e":30.0},"annotated_formula":"multiply(divide(56, multiply(20, 14)), multiply(35, 3))","linear_formula":"multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)|","chain":"20 * 14<\/gadget>\n280<\/output>\n56 \/ 280<\/gadget>\n1\/5 = around 0.2<\/output>\n35 * 3<\/gadget>\n105<\/output>\n(1\/5) * 105<\/gadget>\n21<\/output>\n21<\/result>","index":2984} +{"problem":"if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4650 , r will receive","rationale":"\"explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p \/ 3 r = 2 p \/ 5 p : q : r = p : 2 p \/ 3 : 2 p \/ 5 = 15 : 10 : 6 r ' s share = 4650 * ( 6 \/ 31 ) = 150 * 6 = 900 . answer : option d\"","correct":"d","options":{"a":"600 ","b":"700 ","c":"800 ","d":"900","e":"none of these"},"options_float":{"a":600.0,"b":700.0,"c":800.0,"d":900.0,"e":null},"annotated_formula":"multiply(4650, divide(6, add(add(add(10, add(4, const_1)), 10), 6)))","linear_formula":"add(n0,const_1)|add(n2,#0)|add(n2,#1)|add(n1,#2)|divide(n1,#3)|multiply(n3,#4)|","chain":"4 + 1<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15 + 10<\/gadget>\n25<\/output>\n25 + 6<\/gadget>\n31<\/output>\n6 \/ 31<\/gadget>\n6\/31 = around 0.193548<\/output>\n4_650 * (6\/31)<\/gadget>\n900<\/output>\n900<\/result>","index":2985} +{"problem":"the average age of 15 students of a class is 15 years . out of these , the average age of 8 students is 14 years and that of the other 6 students is 16 years , the age of the 15 th student is","rationale":"\"explanation : age of the 15 th student = [ 15 * 15 - ( 14 * 8 + 16 * 6 ) ] = 17 years . answer : a\"","correct":"a","options":{"a":"17 ","b":"12 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":17.0,"b":12.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"subtract(multiply(15, 15), add(multiply(8, 14), multiply(6, 16)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n8 * 14<\/gadget>\n112<\/output>\n6 * 16<\/gadget>\n96<\/output>\n112 + 96<\/gadget>\n208<\/output>\n225 - 208<\/gadget>\n17<\/output>\n17<\/result>","index":2986} +{"problem":"a committee is reviewing a total of 40 x black - and - white films and 10 y color films for a festival . if the committee selects y \/ x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ?","rationale":"\"say x = y = 10 . in this case we would have : 40 x = 400 black - and - white films ; 10 y = 100 color films . y \/ x % = 10 \/ 10 % = 1 % of the black - and - white films , so 4 black - and - white films and all 100 color films , thus total of 104 films were selected . color films thus compose 100 \/ 104 = 50 \/ 52 of the selected films . answer : a\"","correct":"a","options":{"a":"50 \/ 52 ","b":"55 \/ 52 ","c":"60 \/ 52 ","d":"40 \/ 52","e":"30 \/ 52"},"options_float":{"a":0.9615384615,"b":1.0576923077,"c":1.1538461538,"d":0.7692307692,"e":0.5769230769},"annotated_formula":"divide(10, add(divide(40, const_100), 10))","linear_formula":"divide(n0,const_100)|add(n1,#0)|divide(n1,#1)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) + 10<\/gadget>\n52\/5 = around 10.4<\/output>\n10 \/ (52\/5)<\/gadget>\n25\/26 = around 0.961538<\/output>\n25\/26 = around 0.961538<\/result>","index":2988} +{"problem":"if the cost price of 24 chocolates is equal to the selling price of 16 chocolates , the gain percent is :","rationale":"\"explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 16 chocolates = rs . 16 ; s . p . of 16 chocolates = rs . 24 . . ' . gain % = 8 * 100 \/ 16 = 50 % answer : c\"","correct":"c","options":{"a":"24 % ","b":"8 % ","c":"50 % ","d":"40 %","e":"none of these"},"options_float":{"a":24.0,"b":8.0,"c":50.0,"d":40.0,"e":null},"annotated_formula":"divide(const_100, divide(16, subtract(24, 16)))","linear_formula":"subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)|","chain":"24 - 16<\/gadget>\n8<\/output>\n16 \/ 8<\/gadget>\n2<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":2991} +{"problem":"a fort had provision of food for 150 men for 45 days . after 10 days , 25 men left the fort . find out the number of days for which the remaining food will last .","rationale":"explanation : given that fort had provision of food for 150 men for 45 days hence , after 10 days , the remaining food is sufficient for 150 men for 35 days remaining men after 10 days = 150 - 25 = 125 assume that after 10 days , the remaining food is sufficient for 125 men for x days more men , less days ( indirect proportion ) ( men ) 150 : 125 : : x : 35 ⇒ 150 × 35 = 125 x ⇒ 6 × 35 = 5 x ⇒ x = 6 × 7 = 42 i . e . , the remaining food is sufficient for 125 men for 42 days answer : option b","correct":"b","options":{"a":"44 ","b":"42 ","c":"40 ","d":"38","e":"36"},"options_float":{"a":44.0,"b":42.0,"c":40.0,"d":38.0,"e":36.0},"annotated_formula":"divide(multiply(150, add(25, 10)), subtract(150, 25))","linear_formula":"add(n2,n3)|subtract(n0,n3)|multiply(n0,#0)|divide(#2,#1)","chain":"25 + 10<\/gadget>\n35<\/output>\n150 * 35<\/gadget>\n5_250<\/output>\n150 - 25<\/gadget>\n125<\/output>\n5_250 \/ 125<\/gadget>\n42<\/output>\n42<\/result>","index":2992} +{"problem":"the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 14 and 15 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 23 * 14 ) and ( 23 * 15 ) . larger number = ( 23 * 15 ) = 345 . answer : c\"","correct":"c","options":{"a":"338 ","b":"278 ","c":"345 ","d":"231","e":"121"},"options_float":{"a":338.0,"b":278.0,"c":345.0,"d":231.0,"e":121.0},"annotated_formula":"multiply(23, 15)","linear_formula":"multiply(n0,n2)|","chain":"23 * 15<\/gadget>\n345<\/output>\n345<\/result>","index":2993} +{"problem":"during a given week a programmer spends 1 \/ 4 of his time preparing flow chart , 3 \/ 8 of his time coding and the rest of the time in debugging the programs . if he works 48 hours during the week , how many hours did he spend debugging the program .","rationale":"programmer takes 1 \/ 4 * 48 = 12 hrs a week to make flowchrt same way . . . . . 18 hrs a week for codin . . . so . . . . for debugging = 48 - ( 12 + 18 ) = 18 hrs answer : c","correct":"c","options":{"a":"15 hrs ","b":"16 hrs ","c":"18 hrs ","d":"19 hrs","e":"17 hrs"},"options_float":{"a":15.0,"b":16.0,"c":18.0,"d":19.0,"e":17.0},"annotated_formula":"divide(multiply(48, 3), 8)","linear_formula":"multiply(n2,n4)|divide(#0,n3)","chain":"48 * 3<\/gadget>\n144<\/output>\n144 \/ 8<\/gadget>\n18<\/output>\n18<\/result>","index":2994} +{"problem":"there is enough provisions for 600 men in an army camp for 25 days . if there were 250 men less , how long will the provision last ?","rationale":"\"exp : we have , m 1 d 1 = m 2 d 2 600 * 25 = 250 * d 2 d 2 = 600 * 25 \/ 250 = 60 days . answer : d\"","correct":"d","options":{"a":"10 days ","b":"40 days ","c":"50 days ","d":"60 days","e":"80 days"},"options_float":{"a":10.0,"b":40.0,"c":50.0,"d":60.0,"e":80.0},"annotated_formula":"divide(multiply(600, 25), 250)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"600 * 25<\/gadget>\n15_000<\/output>\n15_000 \/ 250<\/gadget>\n60<\/output>\n60<\/result>","index":2996} +{"problem":"in what time will a train 200 m long cross an electric pole , it its speed be 144 km \/ hr ?","rationale":"\"speed = 144 * 5 \/ 18 = 40 m \/ sec time taken = 200 \/ 40 = 5 sec . answer : e\"","correct":"e","options":{"a":"2.5 sec ","b":"1.9 sec ","c":"8.9 sec ","d":"6.9 sec","e":"5 sec"},"options_float":{"a":2.5,"b":1.9,"c":8.9,"d":6.9,"e":5.0},"annotated_formula":"divide(200, multiply(144, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n144 * (5\/18)<\/gadget>\n40<\/output>\n200 \/ 40<\/gadget>\n5<\/output>\n5<\/result>","index":2997} +{"problem":"in a group of 100 cars , 49 cars do not have air conditioning . if at least 51 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ?","rationale":"lets assume ac = 51 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 51 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 51 . . we have to assign atleast 2 tocars with ac and racing stripes . hence ac = 51 - 2 = 49 . the answer is a","correct":"a","options":{"a":"49 ","b":"47 ","c":"48 ","d":"51","e":"53"},"options_float":{"a":49.0,"b":47.0,"c":48.0,"d":51.0,"e":53.0},"annotated_formula":"subtract(100, 51)","linear_formula":"subtract(n0,n2)","chain":"100 - 51<\/gadget>\n49<\/output>\n49<\/result>","index":2998} +{"problem":"last year a certain bond price with a face value of 5000 yielded 10 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ?","rationale":"\"interest = 0.08 * 5000 = 0.065 * selling price - - > selling price = 0.1 * 5000 \/ 0.065 - - > selling price = ~ 7692 answer : a .\"","correct":"a","options":{"a":"7692 ","b":"5325 ","c":"5351 ","d":"6000","e":"6154"},"options_float":{"a":7692.0,"b":5325.0,"c":5351.0,"d":6000.0,"e":6154.0},"annotated_formula":"divide(multiply(5000, divide(10, const_100)), divide(6.5, const_100))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n5_000 * (1\/10)<\/gadget>\n500<\/output>\n6.5 \/ 100<\/gadget>\n0.065<\/output>\n500 \/ 0.065<\/gadget>\n7_692.307692<\/output>\n7_692.307692<\/result>","index":3000} +{"problem":"the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people c who purchased only book a ?","rationale":"\"this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b and this is 1 \/ 2 of those that purchased a . so , 1500 purchased a . less the 500 that purchased both , c = 1000 purchased a only . ( this is much simpler to solve using the venn diagram ) . correct answer is d . 1000\"","correct":"d","options":{"a":"250 ","b":"500 ","c":"750 ","d":"1000","e":"1500"},"options_float":{"a":250.0,"b":500.0,"c":750.0,"d":1000.0,"e":1500.0},"annotated_formula":"subtract(multiply(add(500, divide(500, const_2)), const_2), 500)","linear_formula":"divide(n0,const_2)|add(n0,#0)|multiply(#1,const_2)|subtract(#2,n0)|","chain":"500 \/ 2<\/gadget>\n250<\/output>\n500 + 250<\/gadget>\n750<\/output>\n750 * 2<\/gadget>\n1_500<\/output>\n1_500 - 500<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":3001} +{"problem":"in township k , 1 \/ 5 of the housing units are equiped with cable tv . if 1 \/ 10 of the housing units , including 1 \/ 6 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ?","rationale":"\"1 \/ 5 - - cable tv ( this includes some data from video cassette recorder ) 1 \/ 10 - - video cassette recorder including 1 \/ 6 ( equiped with cable tv ) i . e . 1 \/ 6 ( 1 \/ 5 ) = 1 \/ 30 therefore only video cassette recorder = 1 \/ 10 - 1 \/ 30 = 1 \/ 15 total = 1 \/ 5 + 1 \/ 15 + neither cable tv nor videocassette recorders 1 = 4 \/ 15 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = 1 - 4 \/ 15 = 11 \/ 15 hence b .\"","correct":"b","options":{"a":"23 \/ 30 ","b":"11 \/ 15 ","c":"7 \/ 10 ","d":"1 \/ 6","e":"2 \/ 15"},"options_float":{"a":0.7666666667,"b":0.7333333333,"c":0.7,"d":0.1666666667,"e":0.1333333333},"annotated_formula":"divide(add(10, 5), multiply(10, const_2))","linear_formula":"add(n1,n3)|multiply(n3,const_2)|divide(#0,#1)|","chain":"10 + 5<\/gadget>\n15<\/output>\n10 * 2<\/gadget>\n20<\/output>\n15 \/ 20<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":3003} +{"problem":"walking at 3 \/ 4 of her normal speed , a worker is 8 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is","rationale":"\"let v be her normal speed and let t be her normal time . d = ( 3 \/ 4 ) v * ( t + 8 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 3 \/ 4 ) v * ( t + 8 ) t \/ 4 = 6 t = 24 the answer is b .\"","correct":"b","options":{"a":"20 ","b":"24 ","c":"28 ","d":"32","e":"36"},"options_float":{"a":20.0,"b":24.0,"c":28.0,"d":32.0,"e":36.0},"annotated_formula":"multiply(3, 8)","linear_formula":"multiply(n0,n2)|","chain":"3 * 8<\/gadget>\n24<\/output>\n24<\/result>","index":3004} +{"problem":"the total age of a and b is 14 years more than the total age of b and c . c is how many years younger than a ?","rationale":"\"( a + b ) - ( b + c ) = 14 a - c = 14 . c is younger than a by 14 years . answer : a\"","correct":"a","options":{"a":"14 ","b":"27 ","c":"21 ","d":"18","e":"11"},"options_float":{"a":14.0,"b":27.0,"c":21.0,"d":18.0,"e":11.0},"annotated_formula":"multiply(14, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"14 * 1<\/gadget>\n14<\/output>\n14<\/result>","index":3005} +{"problem":"in a party every person shakes hands with every other person . if there were a total of 171 handshakes in the party then what is the number of persons present in the party ?","rationale":"explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 171 n ( n - 1 ) \/ 2 = 171 â ˆ ´ n = 19 answer : option e","correct":"e","options":{"a":"15 ","b":"16 ","c":"17 ","d":"18","e":"19"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":19.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(171, const_2), const_4), const_1)), const_1), const_2)","linear_formula":"multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)","chain":"171 * 2<\/gadget>\n342<\/output>\n342 * 4<\/gadget>\n1_368<\/output>\n1_368 + 1<\/gadget>\n1_369<\/output>\n1_369 ** (1\/2)<\/gadget>\n37<\/output>\n37 + 1<\/gadget>\n38<\/output>\n38 \/ 2<\/gadget>\n19<\/output>\n19<\/result>","index":3006} +{"problem":"a man performs 3 \/ 5 of the total journey by rail , 17 \/ 20 by bus and the remaining 6.5 km on foot . his total journey is","rationale":"explanation : let the journey be x km then , 3 x \/ 5 + 7 x \/ 20 + 6.5 = x 12 x + 7 x + 130 = 20 x x = 130 km answer : option d","correct":"d","options":{"a":"65 km ","b":"100 km ","c":"120 km ","d":"130 km","e":"140 km"},"options_float":{"a":65.0,"b":100.0,"c":120.0,"d":130.0,"e":140.0},"annotated_formula":"multiply(20, 6.5)","linear_formula":"multiply(n3,n4)","chain":"20 * 6.5<\/gadget>\n130<\/output>\n130<\/result>","index":3008} +{"problem":"two trains of length 100 meters and 200 meters are 450 meters apart . they are moving towards each other on parallel tracks , at speeds of 90 km \/ h and 72 km \/ h . after how many seconds will the trains meet ?","rationale":"\"the speeds are 90000 \/ 3600 = 25 m \/ s and 72000 \/ 3600 = 20 m \/ s the relative speed is 45 m \/ s . time = 450 \/ 45 = 10 seconds the answer is c .\"","correct":"c","options":{"a":"6 ","b":"8 ","c":"10 ","d":"12","e":"14"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":12.0,"e":14.0},"annotated_formula":"divide(450, multiply(add(90, 72), const_0_2778))","linear_formula":"add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)|","chain":"90 + 72<\/gadget>\n162<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n162 * (5\/18)<\/gadget>\n45<\/output>\n450 \/ 45<\/gadget>\n10<\/output>\n10<\/result>","index":3009} +{"problem":"a train is 360 meter long is running at a speed of 52 km \/ hour . in what time will it pass a bridge of 140 meter length .","rationale":"\"explanation : speed = 52 km \/ hr = 52 * ( 5 \/ 18 ) m \/ sec = 130 \/ 9 m \/ sec total distance = 360 + 140 = 500 meter time = distance \/ speed = 500 \/ 130 \/ 9 = 34.62 seconds answer : b\"","correct":"b","options":{"a":"24.62 seconds ","b":"34.62 seconds ","c":"44.62 seconds ","d":"54.62 seconds","e":"none of these"},"options_float":{"a":24.62,"b":34.62,"c":44.62,"d":54.62,"e":null},"annotated_formula":"divide(add(360, 140), divide(multiply(52, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"360 + 140<\/gadget>\n500<\/output>\n52 * 1_000<\/gadget>\n52_000<\/output>\n52_000 \/ 3_600<\/gadget>\n130\/9 = around 14.444444<\/output>\n500 \/ (130\/9)<\/gadget>\n450\/13 = around 34.615385<\/output>\n450\/13 = around 34.615385<\/result>","index":3011} +{"problem":"a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 540 ?","rationale":"\"slope of 2 and a y - intercept of 2 y - coordinate is 540 y = 2 x + 2 538 = 2 x x = 269 answer : d . 269\"","correct":"d","options":{"a":"249 ","b":"498 ","c":"676 ","d":"269","e":"1,002"},"options_float":{"a":249.0,"b":498.0,"c":676.0,"d":269.0,"e":1002.0},"annotated_formula":"divide(subtract(540, 2), 2)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|","chain":"540 - 2<\/gadget>\n538<\/output>\n538 \/ 2<\/gadget>\n269<\/output>\n269<\/result>","index":3012} +{"problem":"in what ratio must rice of rs . 16 per kg be mixed with rice of rs . 24 per kg so that cost of mixture is rs . 18 per kg ?","rationale":"\"( 18 - 24 ) \/ ( 16 - 18 ) = 6 \/ 2 = 3 : 1 answer : c\"","correct":"c","options":{"a":"4 : 1 ","b":"2 : 1 ","c":"3 : 1 ","d":"4 : 3","e":"4 : 5"},"options_float":{"a":4.0,"b":2.0,"c":3.0,"d":1.3333333333,"e":0.8},"annotated_formula":"divide(divide(subtract(18, 24), subtract(16, 24)), subtract(const_1, divide(subtract(18, 24), subtract(16, 24))))","linear_formula":"subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|","chain":"18 - 24<\/gadget>\n-6<\/output>\n16 - 24<\/gadget>\n-8<\/output>\n(-6) \/ (-8)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n(3\/4) \/ (1\/4)<\/gadget>\n3<\/output>\n3<\/result>","index":3013} +{"problem":"in a class of 25 students in an examination in maths 3 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ?","rationale":"\"explanation : total marks obtained by a class of 25 students = 3 * 95 + 3 * 0 + 19 * 45 = 1140 average marks of whole class = 1140 \/ 25 = 45.6 answer : option b\"","correct":"b","options":{"a":"47 ","b":"45.6 ","c":"44 ","d":"48","e":"49"},"options_float":{"a":47.0,"b":45.6,"c":44.0,"d":48.0,"e":49.0},"annotated_formula":"divide(add(multiply(45, subtract(25, add(3, 3))), multiply(95, 3)), 25)","linear_formula":"add(n1,n3)|multiply(n1,n2)|subtract(n0,#0)|multiply(n4,#2)|add(#3,#1)|divide(#4,n0)|","chain":"3 + 3<\/gadget>\n6<\/output>\n25 - 6<\/gadget>\n19<\/output>\n45 * 19<\/gadget>\n855<\/output>\n95 * 3<\/gadget>\n285<\/output>\n855 + 285<\/gadget>\n1_140<\/output>\n1_140 \/ 25<\/gadget>\n228\/5 = around 45.6<\/output>\n228\/5 = around 45.6<\/result>","index":3014} +{"problem":"ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 3 percent and the mobile phone at a profit of 10 percent . overall he make a .","rationale":"\"let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 3 \/ 100 ) = 15000 - 450 m = 8000 ( 1 + 10 \/ 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 450 + 800 = rs . 350 as this is positive , an overall profit of rs . 350 was made . answer : b\"","correct":"b","options":{"a":"228 ","b":"350 ","c":"27 ","d":"200","e":"881"},"options_float":{"a":228.0,"b":350.0,"c":27.0,"d":200.0,"e":881.0},"annotated_formula":"subtract(add(multiply(15000, subtract(const_1, divide(3, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0)|","chain":"3 \/ 100<\/gadget>\n3\/100 = around 0.03<\/output>\n1 - (3\/100)<\/gadget>\n97\/100 = around 0.97<\/output>\n15_000 * (97\/100)<\/gadget>\n14_550<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n8_000 * (11\/10)<\/gadget>\n8_800<\/output>\n14_550 + 8_800<\/gadget>\n23_350<\/output>\n15_000 + 8_000<\/gadget>\n23_000<\/output>\n23_350 - 23_000<\/gadget>\n350<\/output>\n350<\/result>","index":3015} +{"problem":"a man , a woman and a boy can together complete a piece of work in 3 days . if a woman alone can do it in 6 days and a boy alone in 18 days , how long will a man take to complete the work ?","rationale":"\"explanation : ( 1 man + 1 woman + 1 boy ) ’ s 1 day ’ s work = 1 \/ 3 1 wo man ’ s 1 day work = 1 \/ 6 1 boy ’ s 1 day ’ s work = 1 \/ 18 man alone can finish the work in 9 days . answer : option d\"","correct":"d","options":{"a":"5 days ","b":"8 days ","c":"7 days ","d":"9 days","e":"8 days"},"options_float":{"a":5.0,"b":8.0,"c":7.0,"d":9.0,"e":8.0},"annotated_formula":"inverse(subtract(inverse(3), add(inverse(6), inverse(18))))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/6) + (1\/18)<\/gadget>\n2\/9 = around 0.222222<\/output>\n(1\/3) - (2\/9)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9<\/result>","index":3017} +{"problem":"a train running at the speed of 70 km \/ hr crosses a pole in 9 sec . what is the length of the train ?","rationale":"\"speed = 70 * 5 \/ 18 = 175 \/ 9 m \/ sec length of the train = speed * time = 175 \/ 9 * 9 = 175 m answer : d\"","correct":"d","options":{"a":"227 m ","b":"150 m ","c":"167 m ","d":"175 m","e":"187 m"},"options_float":{"a":227.0,"b":150.0,"c":167.0,"d":175.0,"e":187.0},"annotated_formula":"multiply(divide(multiply(70, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"70 * 1_000<\/gadget>\n70_000<\/output>\n70_000 \/ 3_600<\/gadget>\n175\/9 = around 19.444444<\/output>\n(175\/9) * 9<\/gadget>\n175<\/output>\n175<\/result>","index":3019} +{"problem":"on my sister ' s birthday , she was 126 cm in height , having grown 5 % since the year before . how tall was she the previous year ?","rationale":"let the previous year ' s height be x . 1.05 x = 126 x = 120 the answer is b .","correct":"b","options":{"a":"124 cm ","b":"120 cm ","c":"116 cm ","d":"112 cm","e":"110 cm"},"options_float":{"a":124.0,"b":120.0,"c":116.0,"d":112.0,"e":110.0},"annotated_formula":"subtract(126, divide(multiply(126, 5), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)","chain":"126 * 5<\/gadget>\n630<\/output>\n630 \/ 100<\/gadget>\n63\/10 = around 6.3<\/output>\n126 - (63\/10)<\/gadget>\n1_197\/10 = around 119.7<\/output>\n1_197\/10 = around 119.7<\/result>","index":3020} +{"problem":"for a certain art exhibit , a museum sold admission tickets to a group of 50 people every 5 minutes from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive . the price of a regular admission ticket was $ 10 and the price of a student ticket was $ 6 . if on one day 3 times as many regular admission tickets were sold as student tickets , what was the total revenue from ticket sales that day ?","rationale":"\"from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive there are 9 * 12 = 108 five - minute intervals , thus total of 108 * 50 tickets were sold . say x student and 3 x regular tickets were sold , then x + 3 x = 108 * 50 - - > x = 27 * 50 and 3 x = 3 * ( 27 * 50 ) = 27 * 150 . therefore , the total revenue from ticket sales that day was 27 * 50 * 6 + 27 * 150 * 10 = $ 48,600 . answer : e .\"","correct":"e","options":{"a":"$ 24960 ","b":"$ 25920 ","c":"$ 28080 ","d":"$ 38500","e":"$ 48600"},"options_float":{"a":24960.0,"b":25920.0,"c":28080.0,"d":38500.0,"e":48600.0},"annotated_formula":"add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 50), add(3, const_1)), 3), 10), multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 50), add(3, const_1)), 6))","linear_formula":"add(const_12,n4)|add(n8,const_1)|subtract(#0,n2)|add(#2,const_1)|multiply(#3,const_12)|multiply(n0,#4)|divide(#5,#1)|multiply(n8,#6)|multiply(n7,#6)|multiply(n6,#7)|add(#9,#8)|","chain":"12 + 5<\/gadget>\n17<\/output>\n17 - 9<\/gadget>\n8<\/output>\n8 + 1<\/gadget>\n9<\/output>\n9 * 12<\/gadget>\n108<\/output>\n108 * 50<\/gadget>\n5_400<\/output>\n3 + 1<\/gadget>\n4<\/output>\n5_400 \/ 4<\/gadget>\n1_350<\/output>\n1_350 * 3<\/gadget>\n4_050<\/output>\n4_050 * 10<\/gadget>\n40_500<\/output>\n1_350 * 6<\/gadget>\n8_100<\/output>\n40_500 + 8_100<\/gadget>\n48_600<\/output>\n48_600<\/result>","index":3023} +{"problem":"sum of 3 numbers 264 , if the first number be twice then second and third number be one third of the first , then the second number is","rationale":"explanation : let the second number is x , then first is 2 x , and third is 1 \/ 3 ( 2 x ) = > 2 x + x + 2 x \/ 3 = 264 < = > 11 x \/ 3 = 264 = > x = 72 answer : option c","correct":"c","options":{"a":"70 ","b":"71 ","c":"72 ","d":"73","e":"74"},"options_float":{"a":70.0,"b":71.0,"c":72.0,"d":73.0,"e":74.0},"annotated_formula":"divide(264, add(add(const_2, const_1), divide(const_2, const_3)))","linear_formula":"add(const_1,const_2)|divide(const_2,const_3)|add(#0,#1)|divide(n1,#2)","chain":"2 + 1<\/gadget>\n3<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n3 + (2\/3)<\/gadget>\n11\/3 = around 3.666667<\/output>\n264 \/ (11\/3)<\/gadget>\n72<\/output>\n72<\/result>","index":3024} +{"problem":"a train moves with a speed of 216 kmph . its speed in metres per second is","rationale":"solution speed = 216 kmph = ( 216 x 5 \/ 18 ) m \/ sec = 60 m \/ sec . answer e","correct":"e","options":{"a":"10.8 ","b":"18 ","c":"30 ","d":"38.8","e":"60"},"options_float":{"a":10.8,"b":18.0,"c":30.0,"d":38.8,"e":60.0},"annotated_formula":"multiply(216, const_0_2778)","linear_formula":"multiply(n0,const_0_2778)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n216 * (5\/18)<\/gadget>\n60<\/output>\n60<\/result>","index":3027} +{"problem":"find the least number must be subtracted from 105829 so that remaining no . is divisible by 21 ?","rationale":"\"on dividing 105829 by 21 we get the remainder 10 , so 10 should be subtracted answer is b\"","correct":"b","options":{"a":"7 ","b":"10 ","c":"20 ","d":"32","e":"35"},"options_float":{"a":7.0,"b":10.0,"c":20.0,"d":32.0,"e":35.0},"annotated_formula":"subtract(105829, multiply(floor(divide(105829, 21)), 21))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"105_829 \/ 21<\/gadget>\n105_829\/21 = around 5_039.47619<\/output>\nfloor(105_829\/21)<\/gadget>\n5_039<\/output>\n5_039 * 21<\/gadget>\n105_819<\/output>\n105_829 - 105_819<\/gadget>\n10<\/output>\n10<\/result>","index":3028} +{"problem":"the average age of 50 students in a class is 14 years . if the age of teacher is also included , the average becomes 15 years , find the age of the teacher .","rationale":"\"explanation : if teacher ' s age is 14 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 15 . age of teacher = average age of all + total increase in age = 15 + ( 1 x 50 ) = 65 years answer : b\"","correct":"b","options":{"a":"30 ","b":"65 ","c":"60 ","d":"55","e":"56"},"options_float":{"a":30.0,"b":65.0,"c":60.0,"d":55.0,"e":56.0},"annotated_formula":"subtract(add(add(multiply(50, 14), 15), 50), multiply(50, 14))","linear_formula":"multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)|","chain":"50 * 14<\/gadget>\n700<\/output>\n700 + 15<\/gadget>\n715<\/output>\n715 + 50<\/gadget>\n765<\/output>\n765 - 700<\/gadget>\n65<\/output>\n65<\/result>","index":3030} +{"problem":"the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 4 people , the cost per person would have been $ 60 less . what was the total cost cost of the vacation ?","rationale":"\"c for cost . p price per person . c = 3 * p c = 4 * p - 240 substituting the value of p from the first equation onto the second we get p = 240 . plugging in the value of p in the first equation , we get c = 720 . which leads us to answer choice d\"","correct":"d","options":{"a":"$ 200 ","b":"$ 300 ","c":"$ 400 ","d":"$ 720","e":"$ 600"},"options_float":{"a":200.0,"b":300.0,"c":400.0,"d":720.0,"e":600.0},"annotated_formula":"multiply(multiply(4, 3), divide(60, subtract(4, 3)))","linear_formula":"multiply(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#2,#0)|","chain":"4 * 3<\/gadget>\n12<\/output>\n4 - 3<\/gadget>\n1<\/output>\n60 \/ 1<\/gadget>\n60<\/output>\n12 * 60<\/gadget>\n720<\/output>\n720<\/result>","index":3031} +{"problem":"twice the larger of two numbers is 3 more than 5 times the smaller , and the sum of 4 times the larger and 3 times the smaller is 71 . what are the numbers ?","rationale":"the larger number : x the smaller number : y twice the larger : 2 x three more than five times the smaller : 5 y + 3 relationship between ( ` ` is ' ' ) : 2 x = 5 y + 3 four times the larger : 4 x three times the smaller : 3 y relationship between ( ` ` sum of ' ' ) : 4 x + 3 y = 71 now i have two equations in two variables : 2 x = 5 y + 3 4 x + 3 y = 71 i will solve , say , the first equation for x : x = ( 5 \/ 2 ) y + ( 3 \/ 2 ) then i ' ll plug the right - hand side of this into the second equation in place of the ` ` x ' ' : 4 [ ( 5 \/ 2 ) y + ( 3 \/ 2 ) ] + 3 y = 71 10 y + 6 + 3 y = 71 13 y + 6 = 71 13 y = 65 y = 65 \/ 13 = 5 now that i have the value for y , i can solve for x : x = ( 5 \/ 2 ) y + ( 3 \/ 2 ) x = ( 5 \/ 2 ) ( 5 ) + ( 3 \/ 2 ) x = ( 25 \/ 2 ) + ( 3 \/ 2 ) x = 28 \/ 2 = 14 answer : e","correct":"e","options":{"a":"21 , 12 ","b":"9 , 0 ","c":"15 , 6 ","d":"10 , 1","e":"14 , 5"},"options_float":{"a":21.0,"b":9.0,"c":15.0,"d":10.0,"e":14.0},"annotated_formula":"divide(add(add(multiply(5, 5), 3), const_1), const_2)","linear_formula":"multiply(n1,n1)|add(n0,#0)|add(#1,const_1)|divide(#2,const_2)","chain":"5 * 5<\/gadget>\n25<\/output>\n25 + 3<\/gadget>\n28<\/output>\n28 + 1<\/gadget>\n29<\/output>\n29 \/ 2<\/gadget>\n29\/2 = around 14.5<\/output>\n29\/2 = around 14.5<\/result>","index":3032} +{"problem":"if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer w that must divide n is ?","rationale":"\"q : if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer w that must divide n is : a 6 , b 12 , c 24 , d 36 , e 48 n ^ 2 is divisible by 72 , but it must also be greater than 72 . if n is an integer , then n ^ 2 must be a perfect square . the factorization of 72 is ( 8 ) ( 9 ) , so if it is multiplied by 2 , it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144 , a perfect square . so n ^ 2 must be at least 144 or a multiple of 144 , which means that n must be 12 or a multiple of 12 . b\"","correct":"b","options":{"a":"6 ","b":"12 ","c":"24 ","d":"36","e":"48"},"options_float":{"a":6.0,"b":12.0,"c":24.0,"d":36.0,"e":48.0},"annotated_formula":"multiply(sqrt(divide(72, 2)), 2)","linear_formula":"divide(n1,n0)|sqrt(#0)|multiply(n0,#1)|","chain":"72 \/ 2<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12<\/result>","index":3033} +{"problem":"800 men have provisions for 20 days . if 200 more men join them , for how many days will the provisions last now ?","rationale":"\"800 * 20 = 1200 * x x = 16 . answer : c\"","correct":"c","options":{"a":"10 ","b":"8 ","c":"16 ","d":"11","e":"15"},"options_float":{"a":10.0,"b":8.0,"c":16.0,"d":11.0,"e":15.0},"annotated_formula":"divide(multiply(20, 800), add(800, 200))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"20 * 800<\/gadget>\n16_000<\/output>\n800 + 200<\/gadget>\n1_000<\/output>\n16_000 \/ 1_000<\/gadget>\n16<\/output>\n16<\/result>","index":3034} +{"problem":"the list price of an article is rs . 70 . a customer pays rs . 59.22 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?","rationale":"\"option e explanation : 70 * ( 90 \/ 100 ) * ( ( 100 - x ) \/ 100 ) = 59.22 x = 6 %\"","correct":"e","options":{"a":"8 % ","b":"7 % ","c":"10 % ","d":"12 %","e":"6 %"},"options_float":{"a":8.0,"b":7.0,"c":10.0,"d":12.0,"e":6.0},"annotated_formula":"multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 59.22), subtract(70, multiply(70, divide(10, const_100)))), const_100)","linear_formula":"divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n70 * (1\/10)<\/gadget>\n7<\/output>\n70 - 7<\/gadget>\n63<\/output>\n63 - 59.22<\/gadget>\n3.78<\/output>\n3.78 \/ 63<\/gadget>\n0.06<\/output>\n0.06 * 100<\/gadget>\n6<\/output>\n6<\/result>","index":3035} +{"problem":"a train crosses a platform of 200 m in 15 sec , same train crosses another platform of length 300 m in 20 sec . then find the length of the train ?","rationale":"\"length of the train be ‘ x ’ x + 200 \/ 15 = x + 300 \/ 20 4 x + 800 = 3 x + 900 x = 100 m answer : a\"","correct":"a","options":{"a":"100 m ","b":"180 m ","c":"159 m ","d":"250 m","e":"152 m"},"options_float":{"a":100.0,"b":180.0,"c":159.0,"d":250.0,"e":152.0},"annotated_formula":"subtract(multiply(300, divide(15, divide(15, const_3))), multiply(200, divide(20, divide(15, const_3))))","linear_formula":"divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)|","chain":"15 \/ 3<\/gadget>\n5<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\n300 * 3<\/gadget>\n900<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n200 * 4<\/gadget>\n800<\/output>\n900 - 800<\/gadget>\n100<\/output>\n100<\/result>","index":3036} +{"problem":"one man can do a piece of work in 10 days . one boys join him & he complete the work in 6 days . if he gets 50 rs for his work then what will be the share of the boy .","rationale":"1 day work of man = 1 \/ 10 , if boy ' s 1 day work = 1 \/ x , then ( 1 \/ 10 ) + ( 1 \/ x ) = 1 \/ 6 , x = 15 ratio of work of man & boy = 1 \/ 10 : 1 \/ 15 = 3 : 2 so out of 50 rs . boy ' s share = 2 * 50 \/ ( 3 + 2 ) = 20 answer : b","correct":"b","options":{"a":"10 ","b":"20 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(50, divide(10, const_4))","linear_formula":"divide(n0,const_4)|divide(n2,#0)","chain":"10 \/ 4<\/gadget>\n5\/2 = around 2.5<\/output>\n50 \/ (5\/2)<\/gadget>\n20<\/output>\n20<\/result>","index":3037} +{"problem":"evaluate : 10111 - 10 * 2 * 5 = ?","rationale":"according to order of operations , 10 ? 2 ? 5 ( division and multiplication ) is done first from left to right 10 \/ 2 = 5 * 2 = 10 hence 10111 - 10 * 2 * 5 = 10111 - 10 = 10101 correct answer a","correct":"a","options":{"a":"10101 ","b":"10110 ","c":"11100 ","d":"11001","e":"11011"},"options_float":{"a":10101.0,"b":10110.0,"c":11100.0,"d":11001.0,"e":11011.0},"annotated_formula":"subtract(10111, multiply(multiply(10, 2), 5))","linear_formula":"multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)","chain":"10 * 2<\/gadget>\n20<\/output>\n20 * 5<\/gadget>\n100<\/output>\n10_111 - 100<\/gadget>\n10_011<\/output>\n10_011<\/result>","index":3038} +{"problem":"seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 35 percent ryegrass , what percent of the weight of this mixture is x ?","rationale":"\"- - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 35 % 0.4 x + ( m - x ) 0.25 = 0.35 m 0.15 x = 0.10 m x = 0.6666 m x = 66.66 % of m e\"","correct":"e","options":{"a":"10 % ","b":"33.33 % ","c":"40 % ","d":"50 %","e":"66.66 %"},"options_float":{"a":10.0,"b":33.33,"c":40.0,"d":50.0,"e":66.66},"annotated_formula":"divide(subtract(35, 25), subtract(divide(40, const_100), divide(25, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n2)|subtract(#0,#1)|divide(#2,#3)|","chain":"35 - 25<\/gadget>\n10<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(2\/5) - (1\/4)<\/gadget>\n3\/20 = around 0.15<\/output>\n10 \/ (3\/20)<\/gadget>\n200\/3 = around 66.666667<\/output>\n200\/3 = around 66.666667<\/result>","index":3039} +{"problem":"a rectangular garden is to be twice as long as it is wide . if 150 yards of fencing , including the gate , will completely enclose the garden , what will be the length of the garden , in yards ?","rationale":"alternate approach backsolving ( using answer options to reach the correct answer ) can work wonders here if one is fast in calculations . given perimeter is 150 so , 2 ( l + b ) = 150 or , l + b = 75 now use the answer options ( given length ; breath will be half the length ) ( a ) 40 l = 40 ; b = 20 l + b = 60 ( b ) 50 l = 50 ; b = 25 l + b = 75 ( c ) 60 l = 60 ; b = 30 l + b = 90 ( d ) 70 l = 70 ; b = 35 l + b = 105 ( e ) 80 l = 80 ; b = 40 l + b = 120 thus you see no , need of any calculations , u can reach the correct option only by checking options ; correct answer will be ( b )","correct":"b","options":{"a":"40 ","b":"50 ","c":"60 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(divide(150, multiply(add(const_2, const_1), const_2)), const_2)","linear_formula":"add(const_1,const_2)|multiply(#0,const_2)|divide(n0,#1)|multiply(#2,const_2)","chain":"2 + 1<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n150 \/ 6<\/gadget>\n25<\/output>\n25 * 2<\/gadget>\n50<\/output>\n50<\/result>","index":3040} +{"problem":"a jar of 364 marbles is divided equally among a group of marble - players today . if 2 people joined the group in the future , each person would receive 1 marble less . how many people are there in the group today ?","rationale":"\"364 = 26 * 14 = 28 * 13 there are 26 people in the group today . the answer is d .\"","correct":"d","options":{"a":"20 ","b":"22 ","c":"24 ","d":"26","e":"28"},"options_float":{"a":20.0,"b":22.0,"c":24.0,"d":26.0,"e":28.0},"annotated_formula":"divide(subtract(sqrt(add(multiply(multiply(364, 2), const_4), power(2, 2))), 2), 2)","linear_formula":"multiply(n0,n1)|power(n1,n1)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n1)|divide(#5,n1)|","chain":"364 * 2<\/gadget>\n728<\/output>\n728 * 4<\/gadget>\n2_912<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n2_912 + 4<\/gadget>\n2_916<\/output>\n2_916 ** (1\/2)<\/gadget>\n54<\/output>\n54 - 2<\/gadget>\n52<\/output>\n52 \/ 2<\/gadget>\n26<\/output>\n26<\/result>","index":3041} +{"problem":"peter invested a certain sum of money in a simple interest bond whose value grew to $ 300 at the end of 3 years and to $ 400 at the end of another 5 years . what was the rate of interest in which he invested his sum ?","rationale":"\"now after a time of 3 years the principal p amounts to $ 300 and after a time of 8 years ( question says after another 5 years so 3 + 5 ) p becomes $ 400 . formulating the above data amount ( a 1 ) at end of 3 years a 1 = p ( 1 + 3 r \/ 100 ) = 300 amount ( a 2 ) at end of 8 years a 2 = p ( 1 + 8 r \/ 100 ) = 400 dividing a 2 by a 1 we get ( 1 + 8 r \/ 100 ) \/ ( 1 + 3 r \/ 100 ) = 4 \/ 3 after cross multiplication we are left with 12 r = 100 which gives r = 8.33 % answer : e\"","correct":"e","options":{"a":"12 % ","b":"12.5 % ","c":"6.67 % ","d":"6.25 %","e":"8.33 %"},"options_float":{"a":12.0,"b":12.5,"c":6.67,"d":6.25,"e":8.33},"annotated_formula":"multiply(divide(divide(subtract(400, 300), 5), subtract(300, multiply(divide(subtract(400, 300), 5), 3))), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)|","chain":"400 - 300<\/gadget>\n100<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20 * 3<\/gadget>\n60<\/output>\n300 - 60<\/gadget>\n240<\/output>\n20 \/ 240<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 100<\/gadget>\n25\/3 = around 8.333333<\/output>\n25\/3 = around 8.333333<\/result>","index":3044} +{"problem":"at what rate percent on simple interest will rs . 750 amount to rs . 900 in 5 years ?","rationale":"\"150 = ( 750 * 5 * r ) \/ 100 r = 4 % . answer : c\"","correct":"c","options":{"a":"6 ","b":"5 ","c":"4 ","d":"78","e":"2"},"options_float":{"a":6.0,"b":5.0,"c":4.0,"d":78.0,"e":2.0},"annotated_formula":"multiply(divide(divide(subtract(900, 750), 750), 5), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|","chain":"900 - 750<\/gadget>\n150<\/output>\n150 \/ 750<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ 5<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) * 100<\/gadget>\n4<\/output>\n4<\/result>","index":3045} +{"problem":"a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year .","rationale":"\"gain in 2 years = [ ( 5000 * 6 * 2 ) \/ 100 ] - [ ( 5000 * 4 * 2 ) \/ 100 ] 600 - 400 = 200 gain in 1 year = ( 200 \/ 2 ) = 100 rs answer : a\"","correct":"a","options":{"a":"100 rs ","b":"150 rs ","c":"160 rs ","d":"180 rs","e":"200 rs"},"options_float":{"a":100.0,"b":150.0,"c":160.0,"d":180.0,"e":200.0},"annotated_formula":"divide(subtract(divide(multiply(multiply(5000, 6), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2)","linear_formula":"multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)|","chain":"5_000 * 6<\/gadget>\n30_000<\/output>\n30_000 * 2<\/gadget>\n60_000<\/output>\n60_000 \/ 100<\/gadget>\n600<\/output>\n5_000 * 4<\/gadget>\n20_000<\/output>\n20_000 * 2<\/gadget>\n40_000<\/output>\n40_000 \/ 100<\/gadget>\n400<\/output>\n600 - 400<\/gadget>\n200<\/output>\n200 \/ 2<\/gadget>\n100<\/output>\n100<\/result>","index":3047} +{"problem":"if 1 \/ 5 th of a number decreased by 5 is 5 , then the number is","rationale":"explanation : let the number be a , then a \/ 5 − 5 = 5 ⇒ a \/ 5 = 10 ⇒ a = 50 correct option : b","correct":"b","options":{"a":"25 ","b":"50 ","c":"60 ","d":"75","e":"none"},"options_float":{"a":25.0,"b":50.0,"c":60.0,"d":75.0,"e":null},"annotated_formula":"divide(add(5, 5), divide(1, 5))","linear_formula":"add(n1,n1)|divide(n0,n1)|divide(#0,#1)","chain":"5 + 5<\/gadget>\n10<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n10 \/ (1\/5)<\/gadget>\n50<\/output>\n50<\/result>","index":3049} +{"problem":"earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 240 envelopes","rationale":"\"earl takes 1 min . for 36 envelopes . ellen takes 3 \/ 2 mins for the same . so ellen can stuff ( ( 36 ) \/ ( 3 \/ 2 ) ) in 1 min . i . e . , 24 envlpes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 240 envelopes they will take 240 \/ 60 mins . i . e . , 4 mins . answer : e\"","correct":"e","options":{"a":"6 minutes ","b":"5 minutes ","c":"7 minutes ","d":"3 minutes","e":"4 minutes"},"options_float":{"a":6.0,"b":5.0,"c":7.0,"d":3.0,"e":4.0},"annotated_formula":"divide(240, add(36, divide(36, divide(const_3, const_2))))","linear_formula":"divide(const_3,const_2)|divide(n0,#0)|add(n0,#1)|divide(n1,#2)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n36 \/ (3\/2)<\/gadget>\n24<\/output>\n36 + 24<\/gadget>\n60<\/output>\n240 \/ 60<\/gadget>\n4<\/output>\n4<\/result>","index":3050} +{"problem":"if ( x + 3 ) ^ 2 \/ ( 2 x + 15 ) = 3 , then the difference between the two possible values of x is :","rationale":"( x + 3 ) ^ 2 \/ ( 2 x + 15 ) = 3 ( x + 3 ) ^ 2 = 3 ( 2 x + 15 ) x ^ 2 + 6 x + 9 = 6 x + 45 x ^ 2 - 36 = 0 ( x - 6 ) ( x + 6 ) = 0 x = 6 or x = - 6 the answer is c .","correct":"c","options":{"a":"8 ","b":"10 ","c":"12 ","d":"14","e":"16"},"options_float":{"a":8.0,"b":10.0,"c":12.0,"d":14.0,"e":16.0},"annotated_formula":"multiply(sqrt(subtract(multiply(15, 3), multiply(const_3, 3))), const_2)","linear_formula":"multiply(n0,n3)|multiply(n0,const_3)|subtract(#0,#1)|sqrt(#2)|multiply(#3,const_2)","chain":"15 * 3<\/gadget>\n45<\/output>\n3 * 3<\/gadget>\n9<\/output>\n45 - 9<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12<\/result>","index":3051} +{"problem":"what is the measure of the angle y made by the diagonals of the any adjacent sides of a cube .","rationale":"\"c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be 60 . c\"","correct":"c","options":{"a":"30 ","b":"45 ","c":"60 ","d":"75","e":"90"},"options_float":{"a":30.0,"b":45.0,"c":60.0,"d":75.0,"e":90.0},"annotated_formula":"divide(const_180, const_3)","linear_formula":"divide(const_180,const_3)|","chain":"180 \/ 3<\/gadget>\n60<\/output>\n60<\/result>","index":3053} +{"problem":"a placed 3 sheets with two carbons to get two extra copies of the original . then he decided to get more carbon copies and folded the paper in such a way that the upper half of the sheets were on top of the lower half . then he typed . how many carbon copies did he get ?","rationale":"explanation : since the number of carbons is 2 , only two copies can be obtained . answer : b","correct":"b","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(3, const_1)","linear_formula":"subtract(n0,const_1)","chain":"3 - 1<\/gadget>\n2<\/output>\n2<\/result>","index":3054} +{"problem":"the average age of applicants for a new job is 30 , with a standard deviation of 8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the maximum number of different ages of the applicants ?","rationale":"\"minimum age = average - 1 standard deviation = 30 - 8 = 22 maximum age = average + 1 standard deviation = 30 + 8 = 38 maximum number of different ages of the applicants = 38 - 22 + 1 = 17 answer c\"","correct":"c","options":{"a":"8 ","b":"16 ","c":"17 ","d":"18","e":"34"},"options_float":{"a":8.0,"b":16.0,"c":17.0,"d":18.0,"e":34.0},"annotated_formula":"add(subtract(add(30, 8), subtract(30, 8)), const_1)","linear_formula":"add(n0,n1)|subtract(n0,n1)|subtract(#0,#1)|add(#2,const_1)|","chain":"30 + 8<\/gadget>\n38<\/output>\n30 - 8<\/gadget>\n22<\/output>\n38 - 22<\/gadget>\n16<\/output>\n16 + 1<\/gadget>\n17<\/output>\n17<\/result>","index":3056} +{"problem":"hcf and lcm two numbers are 16 and 396 respectively . if one of the numbers is 36 , then the other number is ?","rationale":"\"16 * 396 = 36 * x x = 176 answer : d\"","correct":"d","options":{"a":"36 ","b":"66 ","c":"132 ","d":"176","e":"364"},"options_float":{"a":36.0,"b":66.0,"c":132.0,"d":176.0,"e":364.0},"annotated_formula":"divide(multiply(16, 396), 36)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"16 * 396<\/gadget>\n6_336<\/output>\n6_336 \/ 36<\/gadget>\n176<\/output>\n176<\/result>","index":3057} +{"problem":"a 10000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ?","rationale":"\"in : we have : 1,000 \/ 2 min = 500 litres per minute out : we have : 1,000 \/ 4 + 1,000 \/ 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 5,000 by that net inflow to get the minutes ) - 60 min . answer e .\"","correct":"e","options":{"a":"8 ","b":"12 ","c":"18 ","d":"24","e":"60"},"options_float":{"a":8.0,"b":12.0,"c":18.0,"d":24.0,"e":60.0},"annotated_formula":"divide(multiply(add(add(6, 4), 2), divide(10000, 2)), const_1000)","linear_formula":"add(n4,n5)|divide(n0,n2)|add(n2,#0)|multiply(#2,#1)|divide(#3,const_1000)|","chain":"6 + 4<\/gadget>\n10<\/output>\n10 + 2<\/gadget>\n12<\/output>\n10_000 \/ 2<\/gadget>\n5_000<\/output>\n12 * 5_000<\/gadget>\n60_000<\/output>\n60_000 \/ 1_000<\/gadget>\n60<\/output>\n60<\/result>","index":3058} +{"problem":"in a village of 2,700 people , 900 people are over 70 years old and 1200 people are female . it is known that 60 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a male or younger than 70 years old ?","rationale":"\"the number of people younger than 70 years old is 2700 - 900 = 1800 the number of females older than 70 years old is 0.4 * 1200 = 480 the number of males older than 70 years old is 900 - 480 = 420 the number of people who are either male or younger than 70 is 1800 + 420 = 2220 . p ( a person is younger than 70 or male ) = 2220 \/ 2700 = 37 \/ 45 the answer is d .\"","correct":"d","options":{"a":"4 \/ 5 ","b":"5 \/ 7 ","c":"13 \/ 15 ","d":"37 \/ 45","e":"79 \/ 90"},"options_float":{"a":0.8,"b":0.7142857143,"c":0.8666666667,"d":0.8222222222,"e":0.8777777778},"annotated_formula":"divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(1200, subtract(const_1, divide(60, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25)))","linear_formula":"multiply(const_5,const_5)|divide(#0,const_0_25)|divide(n4,#1)|multiply(#1,const_26)|subtract(const_1,#2)|multiply(n3,#4)|subtract(#3,#5)|divide(#6,#3)|","chain":"5 * 5<\/gadget>\n25<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n25 \/ (1\/4)<\/gadget>\n100<\/output>\n26 * 100<\/gadget>\n2_600<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n1_200 * (2\/5)<\/gadget>\n480<\/output>\n2_600 - 480<\/gadget>\n2_120<\/output>\n2_120 \/ 2_600<\/gadget>\n53\/65 = around 0.815385<\/output>\n53\/65 = around 0.815385<\/result>","index":3060} +{"problem":"the cost of 8 dozen pencil and 2 dozen note book is 520 rupee . if cost of 3 pencil and 4 note book is 60 rupee , then sum of the cost of 1 pencil and 1 note book will be ( in rupee )","rationale":"let cost of pencil = rs x cost of notebook = rs y cost of 8 dozen pencils are 96 x cost of 2 dozen pencils are 24 y cost of 3 pencils are 3 x cost of 4 notebooks are 4 y equations are 96 x + 24 y = 520 - - - - - 1 3 x + 4 y = 60 - - - - - 2 multiplying 6 with equation ( 2 ) and then subtracting eqn ( 1 ) - eqn ( 2 ) , we get 96 x + 24 y = 520 18 x + 24 y = 360 after solving the equations , we get x = 2.05 y = 13.46 cost of 1 pencil and 1 notebook is 2.05 + 13.46 = rs 15.51 answer : b","correct":"b","options":{"a":"rs 13.51 ","b":"rs 15.51 ","c":"rs 16.51 ","d":"rs 14.51","e":"rs 17.51"},"options_float":{"a":13.51,"b":15.51,"c":16.51,"d":14.51,"e":17.51},"annotated_formula":"add(divide(subtract(520, multiply(60, add(const_3, const_3))), subtract(multiply(8, const_12), multiply(3, add(const_3, const_3)))), divide(subtract(60, multiply(divide(subtract(520, multiply(60, add(const_3, const_3))), subtract(multiply(8, const_12), multiply(3, add(const_3, const_3)))), 3)), 4))","linear_formula":"add(const_3,const_3)|multiply(n0,const_12)|multiply(n5,#0)|multiply(n3,#0)|subtract(n2,#2)|subtract(#1,#3)|divide(#4,#5)|multiply(n3,#6)|subtract(n5,#7)|divide(#8,n4)|add(#6,#9)","chain":"3 + 3<\/gadget>\n6<\/output>\n60 * 6<\/gadget>\n360<\/output>\n520 - 360<\/gadget>\n160<\/output>\n8 * 12<\/gadget>\n96<\/output>\n3 * 6<\/gadget>\n18<\/output>\n96 - 18<\/gadget>\n78<\/output>\n160 \/ 78<\/gadget>\n80\/39 = around 2.051282<\/output>\n(80\/39) * 3<\/gadget>\n80\/13 = around 6.153846<\/output>\n60 - (80\/13)<\/gadget>\n700\/13 = around 53.846154<\/output>\n(700\/13) \/ 4<\/gadget>\n175\/13 = around 13.461538<\/output>\n(80\/39) + (175\/13)<\/gadget>\n605\/39 = around 15.512821<\/output>\n605\/39 = around 15.512821<\/result>","index":3062} +{"problem":"8 years ago , p was half of q ' s age . if the ratio of their present ages is 3 : 4 what will be the total of their present ages ?","rationale":"let present age of p and q be 3 x 3 x and 4 x 4 x respectively . eight years ago , p was half of q ' s age â ‡ ’ ( 3 x â ˆ ’ 8 ) = 1 \/ 2 ( 4 x â ˆ ’ 8 ) â ‡ ’ 6 x â ˆ ’ 16 = 4 x â ˆ ’ 8 â ‡ ’ 2 x = 8 â ‡ ’ x = 4 total of their present ages = 3 x + 4 x = 7 x = 7 ã — 4 = 28 c","correct":"c","options":{"a":"33 ","b":"67 ","c":"28 ","d":"31","e":"35"},"options_float":{"a":33.0,"b":67.0,"c":28.0,"d":31.0,"e":35.0},"annotated_formula":"multiply(add(4, 3), divide(8, const_2))","linear_formula":"add(n1,n2)|divide(n0,const_2)|multiply(#0,#1)","chain":"4 + 3<\/gadget>\n7<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n7 * 4<\/gadget>\n28<\/output>\n28<\/result>","index":3063} +{"problem":"selling an kite for rs . 30 , a shop keeper gains 15 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ?","rationale":"\"explanation : marked price = rs . 30 c . p . = 100 \/ 115 * 30 = rs . 26.08 sale price = 90 % of rs . 30 = rs . 27 required gain % = 0.92 \/ 26.08 * 100 = 3.5 % . answer : a\"","correct":"a","options":{"a":"3.5 % ","b":"10 % ","c":"11 % ","d":"15 %","e":"20 %"},"options_float":{"a":3.5,"b":10.0,"c":11.0,"d":15.0,"e":20.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(15, const_100))), divide(multiply(30, const_100), add(15, const_100))), const_100)","linear_formula":"add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n100 - 10<\/gadget>\n90<\/output>\n(3\/10) * 90<\/gadget>\n27<\/output>\n30 * 100<\/gadget>\n3_000<\/output>\n15 + 100<\/gadget>\n115<\/output>\n3_000 \/ 115<\/gadget>\n600\/23 = around 26.086957<\/output>\n27 - (600\/23)<\/gadget>\n21\/23 = around 0.913043<\/output>\n(21\/23) \/ (600\/23)<\/gadget>\n7\/200 = around 0.035<\/output>\n(7\/200) * 100<\/gadget>\n7\/2 = around 3.5<\/output>\n7\/2 = around 3.5<\/result>","index":3064} +{"problem":"john makes $ 40 a week from his job . he earns a raise and now makes $ 80 a week . what is the % increase ?","rationale":"increase = ( 40 \/ 40 ) * 100 = ( 1 ) * 100 = 100 % . e","correct":"e","options":{"a":"16 % ","b":"16.66 % ","c":"17.9 % ","d":"18.12 %","e":"100 %"},"options_float":{"a":16.0,"b":16.66,"c":17.9,"d":18.12,"e":100.0},"annotated_formula":"multiply(divide(subtract(80, 40), 40), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)","chain":"80 - 40<\/gadget>\n40<\/output>\n40 \/ 40<\/gadget>\n1<\/output>\n1 * 100<\/gadget>\n100<\/output>\n100<\/result>","index":3065} +{"problem":"the average of 6 no . ' s is 4.60 . the average of 2 of them is 3.4 , while the average of the other 2 is 3.8 . what is the average of the remaining 2 no ' s ?","rationale":"\"sum of the remaining two numbers = ( 4.6 * 6 ) - [ ( 3.4 * 2 ) + ( 3.8 * 2 ) ] = 27.60 - ( 6.8 + 7.6 ) = 27.60 - 14.40 = 13.20 . required average = ( 13.2 \/ 2 ) = 6.6 . answer : b\"","correct":"b","options":{"a":"2.3 ","b":"6.6 ","c":"3.6 ","d":"4.5","e":"4.6"},"options_float":{"a":2.3,"b":6.6,"c":3.6,"d":4.5,"e":4.6},"annotated_formula":"divide(subtract(multiply(6, 4.60), add(multiply(2, 3.4), multiply(2, 3.8))), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2)|","chain":"6 * 4.6<\/gadget>\n27.6<\/output>\n2 * 3.4<\/gadget>\n6.8<\/output>\n2 * 3.8<\/gadget>\n7.6<\/output>\n6.8 + 7.6<\/gadget>\n14.4<\/output>\n27.6 - 14.4<\/gadget>\n13.2<\/output>\n13.2 \/ 2<\/gadget>\n6.6<\/output>\n6.6<\/result>","index":3067} +{"problem":"what least number must be subtracted from 13294 so that the remainder is exactly divisible by 97 ?","rationale":"\"sol . on dividing 13924 by 97 , we get remainder = 5 . ∴ required number to be subtracted = 5 . answer b\"","correct":"b","options":{"a":"3 ","b":"5 ","c":"7 ","d":"9","e":"10"},"options_float":{"a":3.0,"b":5.0,"c":7.0,"d":9.0,"e":10.0},"annotated_formula":"subtract(13294, multiply(floor(divide(13294, 97)), 97))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"13_294 \/ 97<\/gadget>\n13_294\/97 = around 137.051546<\/output>\nfloor(13_294\/97)<\/gadget>\n137<\/output>\n137 * 97<\/gadget>\n13_289<\/output>\n13_294 - 13_289<\/gadget>\n5<\/output>\n5<\/result>","index":3071} +{"problem":"amar takes as much time in running 18 meters as a car takes in covering 12 meters . what will be the distance covered by amar during the time the car covers 1.6 km ?","rationale":"\"b 2400 m distance covered by amar = 18 \/ 1.2 ( 1.6 km ) = 3 \/ 8 ( 1600 ) = 2400 m answer is b\"","correct":"b","options":{"a":"6200 m ","b":"2400 m ","c":"300 m ","d":"4300 m","e":"1200 m"},"options_float":{"a":6200.0,"b":2400.0,"c":300.0,"d":4300.0,"e":1200.0},"annotated_formula":"divide(multiply(18, multiply(1.6, const_1000)), 12)","linear_formula":"multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)|","chain":"1.6 * 1_000<\/gadget>\n1_600<\/output>\n18 * 1_600<\/gadget>\n28_800<\/output>\n28_800 \/ 12<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":3073} +{"problem":"with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity w of the pool ?","rationale":"\"d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) \/ ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or , 120 x = 80 ( x + 50 ) . or x = 100 . hence , the capacity of pool = 120 x = 12000 cubic meters .\"","correct":"d","options":{"a":"9000 cubic meters ","b":"10500 cubic meters ","c":"11750 cubic meters ","d":"12000 cubic meters","e":"12500 cubic meters"},"options_float":{"a":9000.0,"b":10500.0,"c":11750.0,"d":12000.0,"e":12500.0},"annotated_formula":"multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1))))","linear_formula":"add(const_1,const_4)|divide(n0,const_60)|multiply(#0,#1)|multiply(#2,const_60)|multiply(n2,#3)|","chain":"48 \/ 60<\/gadget>\n4\/5 = around 0.8<\/output>\n4 + 1<\/gadget>\n5<\/output>\n(4\/5) * 5<\/gadget>\n4<\/output>\n60 * 4<\/gadget>\n240<\/output>\n50 * 240<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":3074} +{"problem":"the average speed of a car is 1 4 \/ 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 4 hrs if the speed of the bike is twice speed of the tractor ?","rationale":"\"sol . average speed of a tractor = 25 km \/ h the speed of a bus in an hour = 25 × 2 = 50 km the speed of a car in an hour = 9 \/ 5 * 50 = 90 km so , the distance covered by car in 4 h is 90 × 4 = 360 km ans . ( c )\"","correct":"c","options":{"a":"300 km ","b":"320 km ","c":"360 km ","d":"390 km","e":"400 km"},"options_float":{"a":300.0,"b":320.0,"c":360.0,"d":390.0,"e":400.0},"annotated_formula":"multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 4)","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(n0,#0)|multiply(#1,const_2)|multiply(#2,#3)|multiply(n5,#4)|","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n1 + (4\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n575 \/ 23<\/gadget>\n25<\/output>\n2 * 25<\/gadget>\n50<\/output>\n(9\/5) * 50<\/gadget>\n90<\/output>\n90 * 4<\/gadget>\n360<\/output>\n360<\/result>","index":3076} +{"problem":"a man bought an article and sold it at a gain of 5 % . if he had bought it at 5 % less and sold it for re 5 less , he would have made a profit of 10 % . the c . p . of the article was","rationale":"\"explanation : let original cost price is x its selling price = ( 105 \/ 100 ) * x = 21 x \/ 20 new cost price = ( 95 \/ 100 ) * x = 19 x \/ 20 new selling price = ( 110 \/ 100 ) * ( 19 x \/ 20 ) = 209 x \/ 200 [ ( 21 x \/ 20 ) - ( 209 x \/ 200 ) ] = 1 = > x = 1000 answer : e ) rs 1000\"","correct":"e","options":{"a":"960 ","b":"980 ","c":"900 ","d":"970","e":"1000"},"options_float":{"a":960.0,"b":980.0,"c":900.0,"d":970.0,"e":1000.0},"annotated_formula":"multiply(multiply(multiply(5, const_4.0), 10), 5)","linear_formula":"multiply(n0,const_4.0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"5 * 4<\/gadget>\n20<\/output>\n20 * 10<\/gadget>\n200<\/output>\n200 * 5<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":3078} +{"problem":"the difference between a number and its 3 - fourth is 100 . what is 25 % of that number ?","rationale":"explanation : solution : let the number be x . then , x - 3 \/ 4 x = 100 = > x \/ 4 = 100 = > x = 400 . 25 % of x = 25 % of 400 = 25 \/ 100 * 400 = 100 . answer : b","correct":"b","options":{"a":"120 ","b":"100 ","c":"125 ","d":"150","e":"none of these"},"options_float":{"a":120.0,"b":100.0,"c":125.0,"d":150.0,"e":null},"annotated_formula":"multiply(divide(100, subtract(const_1, divide(3, const_4))), divide(25, 100))","linear_formula":"divide(n0,const_4)|divide(n2,n1)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n100 \/ (1\/4)<\/gadget>\n400<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n400 * (1\/4)<\/gadget>\n100<\/output>\n100<\/result>","index":3079} +{"problem":"i chose a number and divide it by 4 . then i subtracted 175 from the result and got 10 . what was the number i chose ?","rationale":"\"let x be the number i chose , then x \/ 4 − 175 = 10 x \/ 4 = 185 x = 740 answer is b .\"","correct":"b","options":{"a":"200 ","b":"740 ","c":"780 ","d":"500","e":"100"},"options_float":{"a":200.0,"b":740.0,"c":780.0,"d":500.0,"e":100.0},"annotated_formula":"multiply(add(175, 10), 4)","linear_formula":"add(n1,n2)|multiply(n0,#0)|","chain":"175 + 10<\/gadget>\n185<\/output>\n185 * 4<\/gadget>\n740<\/output>\n740<\/result>","index":3080} +{"problem":"ramu bought an old car for rs . 42000 . he spent rs . 15000 on repairs and sold it for rs . 64900 . what is his profit percent ?","rationale":"\"total cp = rs . 42000 + rs . 15000 = rs . 57000 and sp = rs . 64900 profit ( % ) = ( 64900 - 57000 ) \/ 57000 * 100 = 13.8 % answer : a\"","correct":"a","options":{"a":"13.8 ","b":"77 ","c":"18 ","d":"99","e":"88"},"options_float":{"a":13.8,"b":77.0,"c":18.0,"d":99.0,"e":88.0},"annotated_formula":"multiply(divide(subtract(64900, add(42000, 15000)), add(42000, 15000)), const_100)","linear_formula":"add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|","chain":"42_000 + 15_000<\/gadget>\n57_000<\/output>\n64_900 - 57_000<\/gadget>\n7_900<\/output>\n7_900 \/ 57_000<\/gadget>\n79\/570 = around 0.138596<\/output>\n(79\/570) * 100<\/gadget>\n790\/57 = around 13.859649<\/output>\n790\/57 = around 13.859649<\/result>","index":3081} +{"problem":"how many 3 digit numbers n are divisible by 78 or 91 ?","rationale":"the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as well n - - 13 * 7 * 6 = 91 * 6 = 546 so total number of multiples - - 11 + 9 - 1 = 19 . hence answer is 19 . b","correct":"b","options":{"a":"17 ","b":"19 ","c":"20 ","d":"21","e":"22"},"options_float":{"a":17.0,"b":19.0,"c":20.0,"d":21.0,"e":22.0},"annotated_formula":"add(subtract(91, 78), multiply(3, const_2))","linear_formula":"multiply(n0,const_2)|subtract(n2,n1)|add(#0,#1)","chain":"91 - 78<\/gadget>\n13<\/output>\n3 * 2<\/gadget>\n6<\/output>\n13 + 6<\/gadget>\n19<\/output>\n19<\/result>","index":3082} +{"problem":"the length and breadth of a rectangle is increased by 3 % and 6 % respectively . what is the increase in the area ?","rationale":"\"100 * 100 = 10000 103 * 106 = 10918 - - - - - - - - - - - 918 10000 - - - - - - - 918 100 - - - - - - - - - - 9.18 ans a\"","correct":"a","options":{"a":"9.18 ","b":"9.25 ","c":"9.5 ","d":"8.98","e":"8.88"},"options_float":{"a":9.18,"b":9.25,"c":9.5,"d":8.98,"e":8.88},"annotated_formula":"divide(multiply(subtract(rectangle_area(add(3, const_100), add(6, const_100)), rectangle_area(const_100, const_100)), const_100), rectangle_area(const_100, const_100))","linear_formula":"add(n0,const_100)|add(n1,const_100)|rectangle_area(const_100,const_100)|rectangle_area(#0,#1)|subtract(#3,#2)|multiply(#4,const_100)|divide(#5,#2)|","chain":"3 + 100<\/gadget>\n103<\/output>\n6 + 100<\/gadget>\n106<\/output>\n103 * 106<\/gadget>\n10_918<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n10_918 - 10_000<\/gadget>\n918<\/output>\n918 * 100<\/gadget>\n91_800<\/output>\n91_800 \/ 10_000<\/gadget>\n459\/50 = around 9.18<\/output>\n459\/50 = around 9.18<\/result>","index":3083} +{"problem":"a sells a bicycle to b and makes a profit of 60 % . b sells the same bicycle to c at a profit of 25 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a .","rationale":"\"explanation : let cp be 100 a sells at 60 % profit so sp = 160 b sells at 25 % profit = 160 x ( 1 + 25 \/ 100 ) = 200 cp sp 100 - - - 200 x - - - 225 cp = 225 x 100 \/ 200 = 112.5 answer : e\"","correct":"e","options":{"a":"237 ","b":"126 ","c":"971 ","d":"611","e":"112.5"},"options_float":{"a":237.0,"b":126.0,"c":971.0,"d":611.0,"e":112.5},"annotated_formula":"divide(divide(225, divide(add(const_100, 25), const_100)), divide(add(60, const_100), const_100))","linear_formula":"add(n1,const_100)|add(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#2)|divide(#4,#3)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n225 \/ (5\/4)<\/gadget>\n180<\/output>\n60 + 100<\/gadget>\n160<\/output>\n160 \/ 100<\/gadget>\n8\/5 = around 1.6<\/output>\n180 \/ (8\/5)<\/gadget>\n225\/2 = around 112.5<\/output>\n225\/2 = around 112.5<\/result>","index":3084} +{"problem":"if x is 50 percent greater than 88 , then x =","rationale":"\"x = 88 * 1.5 = 132 so the answer is b .\"","correct":"b","options":{"a":"68 ","b":"132 ","c":"86 ","d":"105.6","e":"108"},"options_float":{"a":68.0,"b":132.0,"c":86.0,"d":105.6,"e":108.0},"annotated_formula":"add(88, multiply(divide(50, const_100), 88))","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 88<\/gadget>\n44<\/output>\n88 + 44<\/gadget>\n132<\/output>\n132<\/result>","index":3087} +{"problem":"if x represents the product of the first 15 positive integers , then x is not a multiple of :","rationale":"a ) 99 = 9 * 11 b ) 84 = 4 * 7 * 3 c ) 72 = 9 * 11 d ) 65 = 5 * 13 e ) 57 = 19 * 3 since 19 is not there in first 15 positive numbers it is the only possibility d )","correct":"d","options":{"a":"99 ","b":"84 ","c":"72 ","d":"57","e":"67"},"options_float":{"a":99.0,"b":84.0,"c":72.0,"d":57.0,"e":67.0},"annotated_formula":"multiply(add(15, const_4), const_3)","linear_formula":"add(n0,const_4)|multiply(#0,const_3)","chain":"15 + 4<\/gadget>\n19<\/output>\n19 * 3<\/gadget>\n57<\/output>\n57<\/result>","index":3088} +{"problem":"a can complete a project in 20 days and b can complete the same project in 40 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ?","rationale":"\"let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 \/ 20 ) . the amount of work done by b is ( x ) * ( 1 \/ 40 ) . ( 1 \/ 20 ) * ( x - 10 ) + ( 1 \/ 40 ) * ( x ) = 1 ( x \/ 20 ) + ( x \/ 40 ) - ( 10 \/ 20 ) = 1 3 x \/ 40 = 3 \/ 2 x = 20 therefore , the answer is b : 20 .\"","correct":"b","options":{"a":"18 days ","b":"20 days ","c":"26.67 days ","d":"16 days","e":"12 days"},"options_float":{"a":18.0,"b":20.0,"c":26.67,"d":16.0,"e":12.0},"annotated_formula":"add(divide(subtract(const_1, multiply(divide(const_1, 40), 10)), add(divide(const_1, 20), divide(const_1, 40))), 10)","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)|","chain":"1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/40) * 10<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) + (1\/40)<\/gadget>\n3\/40 = around 0.075<\/output>\n(3\/4) \/ (3\/40)<\/gadget>\n10<\/output>\n10 + 10<\/gadget>\n20<\/output>\n20<\/result>","index":3089} +{"problem":"a can finish a work in 12 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ?","rationale":"\"explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 \/ 12 b ' s 1 day work = 1 \/ 6 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 \/ 12 + 1 \/ 6 ) = 1 \/ 4 so in one day 1 \/ 4 work will be done answer : a\"","correct":"a","options":{"a":"1 \/ 4 ","b":"1 \/ 6 ","c":"1 \/ 7 ","d":"1 \/ 8","e":"none of these"},"options_float":{"a":0.25,"b":0.1666666667,"c":0.1428571429,"d":0.125,"e":null},"annotated_formula":"add(divide(const_1, 12), multiply(divide(const_1, 12), const_2))","linear_formula":"divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 2<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/12) + (1\/6)<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":3092} +{"problem":"the sum of all the digits of the integers from 18 to 21 inclusive is 24 ( 1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24 ) . what is the sum e of all the digits of the integers from 0 to 99 inclusive ?","rationale":"we want the sum of the digits from 0 to 99 , so i approximated : 0 - 9 - > 45 - > ( 9 + 0 ) * 10 \/ 2 40 - 49 - > 85 ( 13 + 4 ) * 10 \/ 2 90 - 99 - > 135 ( 18 + 9 ) * 10 \/ 2 we can see at a glance that theweightgoes up as the numbers go up ( meaning the difference between 85 and 45 is 40 , while 135 - 85 is 50 , this means that the second part of this sequence carries more weight for our result ) , so we know that the final answer has to be more than 850 ( 85 * 10 ) but close to it , and that ' s just e = 900 : the answer is c .","correct":"c","options":{"a":"450 ","b":"810 ","c":"900 ","d":"1000","e":"1100"},"options_float":{"a":450.0,"b":810.0,"c":900.0,"d":1000.0,"e":1100.0},"annotated_formula":"multiply(divide(multiply(9, add(9, const_1)), const_2), multiply(const_2, const_10))","linear_formula":"add(n6,const_1)|multiply(const_10,const_2)|multiply(n6,#0)|divide(#2,const_2)|multiply(#3,#1)","chain":"9 + 1<\/gadget>\n10<\/output>\n9 * 10<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n2 * 10<\/gadget>\n20<\/output>\n45 * 20<\/gadget>\n900<\/output>\n900<\/result>","index":3094} +{"problem":"subtracting 3 % of a from a is equivalent to multiplying a by how much ?","rationale":"\"answer let a - 3 % of a = ab . ⇒ ( 97 x a ) \/ 100 = ab ∴ b = 0.97 correct option : a\"","correct":"a","options":{"a":"0.97 ","b":"9.4 ","c":"0.094 ","d":"94","e":"none"},"options_float":{"a":0.97,"b":9.4,"c":0.094,"d":94.0,"e":null},"annotated_formula":"divide(subtract(const_100, 3), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|","chain":"100 - 3<\/gadget>\n97<\/output>\n97 \/ 100<\/gadget>\n97\/100 = around 0.97<\/output>\n97\/100 = around 0.97<\/result>","index":3096} +{"problem":"sachin is older than rahul by 7 years . if the ratio of their ages is 11 : 9 , find the age of sachin","rationale":"explanation : if rahul age is x , then sachin age is x + 7 , so , 9 x + 63 = 11 x 2 x = 63 x = 31.5 so sachin age is 31.5 + 7 = 38.5 answer : c ) 38.5","correct":"c","options":{"a":"24.5 ","b":"24.8 ","c":"38.5 ","d":"24.88","e":"24.19"},"options_float":{"a":24.5,"b":24.8,"c":38.5,"d":24.88,"e":24.19},"annotated_formula":"divide(multiply(11, 7), subtract(11, 9))","linear_formula":"multiply(n0,n1)|subtract(n1,n2)|divide(#0,#1)","chain":"11 * 7<\/gadget>\n77<\/output>\n11 - 9<\/gadget>\n2<\/output>\n77 \/ 2<\/gadget>\n77\/2 = around 38.5<\/output>\n77\/2 = around 38.5<\/result>","index":3097} +{"problem":"when a certain number x is divided by 61 , the remainder is 24 . what is the remainder when x is divided by 5 ?","rationale":"\"when a certain number x is divided by 61 , the remainder is 24 . what is the remainder when x is divided by 5 ? putting a value say x = 24 we get remainder as 24 when divided by 61 . when 24 divided by 5 we get 4 as remainder . c is the answer .\"","correct":"c","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"reminder(24, 5)","linear_formula":"reminder(n1,n2)|","chain":"24 % 5<\/gadget>\n4<\/output>\n4<\/result>","index":3098} +{"problem":"solve below question 2 x + 1 = - 19","rationale":"\"2 x + 1 = - 19 x = - 10 e\"","correct":"e","options":{"a":"- 8 ","b":"- 9 ","c":"9 ","d":"8","e":"- 10"},"options_float":{"a":-8.0,"b":-9.0,"c":9.0,"d":8.0,"e":-10.0},"annotated_formula":"divide(negate(add(19, 1)), 2)","linear_formula":"add(n1,n2)|negate(#0)|divide(#1,n0)|","chain":"19 + 1<\/gadget>\n20<\/output>\n-20<\/gadget>\n-20<\/output>\n(-20) \/ 2<\/gadget>\n-10<\/output>\n-10<\/result>","index":3099} +{"problem":"a train 150 m long can cross an electric pole in 3 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 150 \/ 3 s = 50 m \/ sec speed = 50 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 180 kmph answer : a\"","correct":"a","options":{"a":"180 ","b":"120 ","c":"72 ","d":"27","e":"28"},"options_float":{"a":180.0,"b":120.0,"c":72.0,"d":27.0,"e":28.0},"annotated_formula":"divide(divide(150, const_1000), divide(3, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"150 \/ 1_000<\/gadget>\n3\/20 = around 0.15<\/output>\n3 \/ 3_600<\/gadget>\n1\/1_200 = around 0.000833<\/output>\n(3\/20) \/ (1\/1_200)<\/gadget>\n180<\/output>\n180<\/result>","index":3100} +{"problem":"approximately how many revolutions will be made by a car tire with a 10 - inch diameter if the car travels 1 ⁄ 2 mile ?","rationale":"\"diameter of car tire = 10 inches radius of car tire = 5 inches = 5 * 2.54 cm = 12.7 cm circumference of the car tire = 2 * pi * 12.7 = 2 * 3.14 * 12.7 = 79.76 cm distance traveled by car = . 5 mile = . 5 * 1.6 km = . 8 km = 800 m = 8 * 10 ^ 4 cm number of revolutions = distance traveled \/ circumference of tire = 8 * 10 ^ 4 \/ 79.76 = 1003 since the answer choices are far apart we should use estimation in this question . ideally , in the gmat the conversion between units would be provided . answer c\"","correct":"c","options":{"a":"120 ","b":"180 ","c":"1003 ","d":"1000","e":"1,440"},"options_float":{"a":120.0,"b":180.0,"c":1003.0,"d":1000.0,"e":1440.0},"annotated_formula":"divide(divide(1, 2), divide(multiply(add(divide(add(divide(add(divide(add(divide(multiply(const_3, 2), const_10), 1), const_10), const_4), const_10), 1), const_10), const_3), 10), multiply(add(add(multiply(add(multiply(add(multiply(multiply(const_3, 2), const_10), const_3), const_10), const_3), const_10), const_3), 2), const_10)))","linear_formula":"divide(n1,n2)|multiply(n2,const_3)|divide(#1,const_10)|multiply(#1,const_10)|add(#2,n1)|add(#3,const_3)|divide(#4,const_10)|multiply(#5,const_10)|add(#6,const_4)|add(#7,const_3)|divide(#8,const_10)|multiply(#9,const_10)|add(#10,n1)|add(#11,const_3)|add(#13,n2)|divide(#12,const_10)|add(#15,const_3)|multiply(#14,const_10)|multiply(n0,#16)|divide(#18,#17)|divide(#0,#19)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 \/ 10<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) + 1<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) \/ 10<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) + 4<\/gadget>\n104\/25 = around 4.16<\/output>\n(104\/25) \/ 10<\/gadget>\n52\/125 = around 0.416<\/output>\n(52\/125) + 1<\/gadget>\n177\/125 = around 1.416<\/output>\n(177\/125) \/ 10<\/gadget>\n177\/1_250 = around 0.1416<\/output>\n(177\/1_250) + 3<\/gadget>\n3_927\/1_250 = around 3.1416<\/output>\n(3_927\/1_250) * 10<\/gadget>\n3_927\/125 = around 31.416<\/output>\n6 * 10<\/gadget>\n60<\/output>\n60 + 3<\/gadget>\n63<\/output>\n63 * 10<\/gadget>\n630<\/output>\n630 + 3<\/gadget>\n633<\/output>\n633 * 10<\/gadget>\n6_330<\/output>\n6_330 + 3<\/gadget>\n6_333<\/output>\n6_333 + 2<\/gadget>\n6_335<\/output>\n6_335 * 10<\/gadget>\n63_350<\/output>\n(3_927\/125) \/ 63_350<\/gadget>\n561\/1_131_250 = around 0.000496<\/output>\n(1\/2) \/ (561\/1_131_250)<\/gadget>\n565_625\/561 = around 1_008.244207<\/output>\n565_625\/561 = around 1_008.244207<\/result>","index":3101} +{"problem":"calculate the share of z , if rs . 3720 is divided among x , y and z in the ratio 7 : 8 : 11 ?","rationale":"7 + 8 + 11 = 26 3720 \/ 26 = 143.1 so z ' s share = 11 * 143.1 answer : c","correct":"c","options":{"a":"1564.1 ","b":"1214.1 ","c":"1574.1 ","d":"1271.1","e":"1044.1"},"options_float":{"a":1564.1,"b":1214.1,"c":1574.1,"d":1271.1,"e":1044.1},"annotated_formula":"divide(add(3720, add(divide(3720, 8), divide(3720, 11))), const_3)","linear_formula":"divide(n0,n2)|divide(n0,n3)|add(#0,#1)|add(n0,#2)|divide(#3,const_3)","chain":"3_720 \/ 8<\/gadget>\n465<\/output>\n3_720 \/ 11<\/gadget>\n3_720\/11 = around 338.181818<\/output>\n465 + (3_720\/11)<\/gadget>\n8_835\/11 = around 803.181818<\/output>\n3_720 + (8_835\/11)<\/gadget>\n49_755\/11 = around 4_523.181818<\/output>\n(49_755\/11) \/ 3<\/gadget>\n16_585\/11 = around 1_507.727273<\/output>\n16_585\/11 = around 1_507.727273<\/result>","index":3102} +{"problem":"the area of a triangle is with base 18 m and height 6 m ?","rationale":"\"1 \/ 2 * 18 * 6 = 54 m 2 answer : d\"","correct":"d","options":{"a":"88 m 2 ","b":"10 m 2 ","c":"66 m 2 ","d":"54 m 2","e":"31 m 2"},"options_float":{"a":88.0,"b":10.0,"c":66.0,"d":54.0,"e":31.0},"annotated_formula":"triangle_area(18, 6)","linear_formula":"triangle_area(n0,n1)|","chain":"(18 * 6) \/ 2<\/gadget>\n54<\/output>\n54<\/result>","index":3103} +{"problem":"if a population of women in a town is 50 % of men . what is the population of men as a % of population of women ?","rationale":"\"we ' re told that the number of women in a town is equal to 50 % of the number of men in that town . if . . . . men = 10 women = 5 we ' re asked for the number of men , as a percentage of the number of women . m \/ w = 10 \/ 5 = 2 = 200 % answer is a\"","correct":"a","options":{"a":"200 % ","b":"100 % ","c":"50 % ","d":"65 %","e":"110 %"},"options_float":{"a":200.0,"b":100.0,"c":50.0,"d":65.0,"e":110.0},"annotated_formula":"multiply(divide(const_100, 50), const_100)","linear_formula":"divide(const_100,n0)|multiply(#0,const_100)|","chain":"100 \/ 50<\/gadget>\n2<\/output>\n2 * 100<\/gadget>\n200<\/output>\n200<\/result>","index":3104} +{"problem":"225 mtrs long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees ?","rationale":"26 trees have 25 gaps between them , required distance ( 225 \/ 25 ) = 10 a","correct":"a","options":{"a":"10 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"add(divide(225, subtract(26, const_1)), const_1)","linear_formula":"subtract(n1,const_1)|divide(n0,#0)|add(#1,const_1)","chain":"26 - 1<\/gadget>\n25<\/output>\n225 \/ 25<\/gadget>\n9<\/output>\n9 + 1<\/gadget>\n10<\/output>\n10<\/result>","index":3105} +{"problem":"in a circuit board factory , all circuit boards that pass a verification process are certified . every board that fails the verification process is indeed faulty , but 1 \/ 8 of those that pass are also faulty . approximately how many faulty circuit boards exist in a group of 3,200 circuit boards where 64 fail inspection ?","rationale":"total of 3,200 boards . all that fail verification are indeed faulty . so the 64 are indeed faulty . 1 \/ 8 those that pass are also faulty . from the 3,200 we know 64 fail . so 3,136 must pass . of these 1 \/ 8 are faulty . 3,136 divided by 8 gives you 392 . what one must do now is to add to the 392 which were not detected the actually detected faulty ones , namely the 64 . total faulty : 456 . answer : c","correct":"c","options":{"a":"72 ","b":"192 ","c":"456 ","d":"256","e":"264"},"options_float":{"a":72.0,"b":192.0,"c":456.0,"d":256.0,"e":264.0},"annotated_formula":"add(64, divide(subtract(multiply(multiply(add(divide(const_60, const_10), const_0_25), 8), 64), 64), 8))","linear_formula":"divide(const_60,const_10)|add(#0,const_0_25)|multiply(n1,#1)|multiply(n3,#2)|subtract(#3,n3)|divide(#4,n1)|add(n3,#5)","chain":"60 \/ 10<\/gadget>\n6<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n6 + (1\/4)<\/gadget>\n25\/4 = around 6.25<\/output>\n(25\/4) * 8<\/gadget>\n50<\/output>\n50 * 64<\/gadget>\n3_200<\/output>\n3_200 - 64<\/gadget>\n3_136<\/output>\n3_136 \/ 8<\/gadget>\n392<\/output>\n64 + 392<\/gadget>\n456<\/output>\n456<\/result>","index":3106} +{"problem":"the rate of spin of a certain gyroscope doubled every 15 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 400 meters per second , what was the speed , in meters per second , when the stopwatch was started ?","rationale":"\"let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 6 times . 2 ^ 6 * x = 400 x = ( 2 ^ 4 * 25 ) \/ 2 ^ 6 = 25 \/ 4 the answer is b .\"","correct":"b","options":{"a":"25 \/ 2 ","b":"25 \/ 4 ","c":"25 \/ 8 ","d":"25 \/ 16","e":"25 \/ 32"},"options_float":{"a":12.5,"b":6.25,"c":3.125,"d":1.5625,"e":0.78125},"annotated_formula":"divide(divide(400, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 15), const_1))), const_2)","linear_formula":"divide(const_60,const_2)|add(#0,const_60)|divide(#1,n0)|subtract(#2,const_1)|power(const_2,#3)|divide(n1,#4)|divide(#5,const_2)|","chain":"60 \/ 2<\/gadget>\n30<\/output>\n30 + 60<\/gadget>\n90<\/output>\n90 \/ 15<\/gadget>\n6<\/output>\n6 - 1<\/gadget>\n5<\/output>\n2 ** 5<\/gadget>\n32<\/output>\n400 \/ 32<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) \/ 2<\/gadget>\n25\/4 = around 6.25<\/output>\n25\/4 = around 6.25<\/result>","index":3107} +{"problem":"the area of a parallelogram is 288 sq m and its altitude is twice the corresponding base . then the length of the base is ?","rationale":"\"2 x * x = 288 = > x = 12 answer : a\"","correct":"a","options":{"a":"12 ","b":"9 ","c":"7 ","d":"6","e":"5"},"options_float":{"a":12.0,"b":9.0,"c":7.0,"d":6.0,"e":5.0},"annotated_formula":"sqrt(divide(288, const_2))","linear_formula":"divide(n0,const_2)|sqrt(#0)|","chain":"288 \/ 2<\/gadget>\n144<\/output>\n144 ** (1\/2)<\/gadget>\n12<\/output>\n12<\/result>","index":3108} +{"problem":"robert ' s salary was increased by 50 % and subsequently decreased by 10 % . how much percentage does he gain ?","rationale":"let original salary be $ 100 increased salary = 10 % of ( 150 % of $ 100 ) = ( 10 \/ 100 ) * ( 150 \/ 100 ) * 100 = $ 15 increase = 15 % correct option is a","correct":"a","options":{"a":"15 % ","b":"20 % ","c":"10 % ","d":"25 %","e":"30 %"},"options_float":{"a":15.0,"b":20.0,"c":10.0,"d":25.0,"e":30.0},"annotated_formula":"subtract(subtract(multiply(multiply(divide(add(const_100, 50), const_100), divide(subtract(const_100, const_10), const_100)), const_100), const_100), multiply(10, const_2))","linear_formula":"add(n0,const_100)|multiply(n1,const_2)|subtract(const_100,const_10)|divide(#0,const_100)|divide(#2,const_100)|multiply(#3,#4)|multiply(#5,const_100)|subtract(#6,const_100)|subtract(#7,#1)","chain":"100 + 50<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n(3\/2) * (9\/10)<\/gadget>\n27\/20 = around 1.35<\/output>\n(27\/20) * 100<\/gadget>\n135<\/output>\n135 - 100<\/gadget>\n35<\/output>\n10 * 2<\/gadget>\n20<\/output>\n35 - 20<\/gadget>\n15<\/output>\n15<\/result>","index":3109} +{"problem":"a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 29.16 , the number of the member is the group is :","rationale":"money collected = ( 29.16 x 100 ) paise = 29.16 paise numbers of members = 2916 squareroot = 54 answer b","correct":"b","options":{"a":"57 ","b":"54 ","c":"77 ","d":"87","e":"97"},"options_float":{"a":57.0,"b":54.0,"c":77.0,"d":87.0,"e":97.0},"annotated_formula":"sqrt(multiply(29.16, const_100))","linear_formula":"multiply(n0,const_100)|sqrt(#0)","chain":"29.16 * 100<\/gadget>\n2_916<\/output>\n2_916 ** (1\/2)<\/gadget>\n54<\/output>\n54<\/result>","index":3111} +{"problem":"a dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg , what is his percent ?","rationale":"\"800 - - - 200 100 - - - ? = > 25 % answer : b\"","correct":"b","options":{"a":"25 % ","b":"25 % ","c":"29 % ","d":"55 %","e":"45 %"},"options_float":{"a":25.0,"b":25.0,"c":29.0,"d":55.0,"e":45.0},"annotated_formula":"subtract(multiply(divide(const_100, 800), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)","linear_formula":"add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|","chain":"100 \/ 800<\/gadget>\n1\/8 = around 0.125<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n(1\/8) * 1_000<\/gadget>\n125<\/output>\n125 - 100<\/gadget>\n25<\/output>\n25<\/result>","index":3112} +{"problem":"the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 66 kg . what might be the weight of the new person ?","rationale":"\"c 86 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 66 + 20 ) kg = 86 kg .\"","correct":"c","options":{"a":"56 kg ","b":"90 kg ","c":"86 kg ","d":"data inadequate","e":"none of these"},"options_float":{"a":56.0,"b":90.0,"c":86.0,"d":null,"e":null},"annotated_formula":"add(multiply(8, 2.5), 66)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"8 * 2.5<\/gadget>\n20<\/output>\n20 + 66<\/gadget>\n86<\/output>\n86<\/result>","index":3113} +{"problem":"if ( 5 ^ 9 ) ( 9 ^ 5 ) = 3 ( 15 ^ x ) , what is the value of x ?","rationale":"\"( 5 ^ 9 ) ( 9 ^ 5 ) = 3 ( 15 ^ x ) = > 5 ^ 9 * 3 ^ 10 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 9 * 3 ^ 10 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 9 answer b\"","correct":"b","options":{"a":"7 ","b":"9 ","c":"11 ","d":"13","e":"15"},"options_float":{"a":7.0,"b":9.0,"c":11.0,"d":13.0,"e":15.0},"annotated_formula":"divide(log(divide(multiply(power(5, 9), power(9, 5)), 3)), log(15))","linear_formula":"log(n5)|power(n0,n1)|power(n2,n3)|multiply(#1,#2)|divide(#3,n4)|log(#4)|divide(#5,#0)|","chain":"5 ** 9<\/gadget>\n1_953_125<\/output>\n9 ** 5<\/gadget>\n59_049<\/output>\n1_953_125 * 59_049<\/gadget>\n115_330_078_125<\/output>\n115_330_078_125 \/ 3<\/gadget>\n38_443_359_375<\/output>\nlog(38_443_359_375)<\/gadget>\nlog(38443359375) = around 24.372452<\/output>\nlog(15)<\/gadget>\nlog(15) = around 2.70805<\/output>\nlog(38443359375) \/ log(15)<\/gadget>\nlog(38443359375)\/log(15) = around 9<\/output>\nlog(38443359375)\/log(15) = around 9<\/result>","index":3114} +{"problem":"a boat having a length 8 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :","rationale":"\"explanation : volume of water displaced = ( 8 x 2 x 0.01 ) m 3 = 0.16 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.16 x 1000 ) kg = 160 kg . answer : d\"","correct":"d","options":{"a":"12 kg ","b":"60 kg ","c":"72 kg ","d":"160 kg","e":"none of these"},"options_float":{"a":12.0,"b":60.0,"c":72.0,"d":160.0,"e":null},"annotated_formula":"multiply(multiply(multiply(8, 2), divide(1, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"8 * 2<\/gadget>\n16<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n16 * (1\/100)<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) * 1_000<\/gadget>\n160<\/output>\n160<\/result>","index":3115} +{"problem":"what is the sum between the place values of two 3 ' s in the numeral 63130","rationale":"required sum = 3000 + 30 = 3030 answer is d","correct":"d","options":{"a":"3111 ","b":"300 ","c":"3029 ","d":"3030","e":"3003"},"options_float":{"a":3111.0,"b":300.0,"c":3029.0,"d":3030.0,"e":3003.0},"annotated_formula":"add(multiply(3, const_10), multiply(3, const_1000))","linear_formula":"multiply(n0,const_10)|multiply(n0,const_1000)|add(#0,#1)","chain":"3 * 10<\/gadget>\n30<\/output>\n3 * 1_000<\/gadget>\n3_000<\/output>\n30 + 3_000<\/gadget>\n3_030<\/output>\n3_030<\/result>","index":3116} +{"problem":"lloyd normally works 7.5 hours per day and earns $ 4.00 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 1.5 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ?","rationale":"\"daily working hour * regular rate + overtime * increased rate 7.5 * 4.0 + 3 * 4.0 * 1.5 = 48 answer b\"","correct":"b","options":{"a":"$ 33.75 ","b":"$ 48.00 ","c":"$ 51.75 ","d":"$ 54.00","e":"$ 70.00"},"options_float":{"a":33.75,"b":48.0,"c":51.75,"d":54.0,"e":70.0},"annotated_formula":"add(multiply(7.5, 4.00), multiply(multiply(subtract(10.5, 7.5), 1.5), 4.00))","linear_formula":"multiply(n0,n1)|subtract(n4,n0)|multiply(n3,#1)|multiply(n1,#2)|add(#0,#3)|","chain":"7.5 * 4<\/gadget>\n30<\/output>\n10.5 - 7.5<\/gadget>\n3<\/output>\n3 * 1.5<\/gadget>\n4.5<\/output>\n4.5 * 4<\/gadget>\n18<\/output>\n30 + 18<\/gadget>\n48<\/output>\n48<\/result>","index":3117} +{"problem":"the average weight of 20 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class .","rationale":"\"explanation : average weight of 20 boys = 50.25 total weight of 20 boys = 50.25 × 20 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 20 ) + ( 45.15 × 8 ) total boys = 20 + 8 = 28 average weight of all the boys = ( 50.25 × 20 ) + ( 45.15 × 8 ) \/ 28 = answer : option a\"","correct":"a","options":{"a":"48.79286 ","b":"42.25 ","c":"50.87364 ","d":"51.25","e":"52.25474"},"options_float":{"a":48.79286,"b":42.25,"c":50.87364,"d":51.25,"e":52.25474},"annotated_formula":"divide(add(multiply(20, 50.25), multiply(8, 45.15)), add(20, 8))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"20 * 50.25<\/gadget>\n1_005<\/output>\n8 * 45.15<\/gadget>\n361.2<\/output>\n1_005 + 361.2<\/gadget>\n1_366.2<\/output>\n20 + 8<\/gadget>\n28<\/output>\n1_366.2 \/ 28<\/gadget>\n48.792857<\/output>\n48.792857<\/result>","index":3118} +{"problem":"what is the difference between the c . i . on rs . 9000 for 1 1 \/ 2 years at 4 % per annum compounded yearly and half - yearly ?","rationale":"c . i . when interest is compounded yearly = [ 9000 * ( 1 + 4 \/ 100 ) * ( 1 + ( 1 \/ 2 * 4 ) \/ 100 ] = 9000 * 26 \/ 25 * 51 \/ 50 = rs . 9547.2 c . i . when interest is compounded half - yearly = [ 9000 * ( 1 + 2 \/ 100 ) 2 ] = ( 9000 * 51 \/ 50 * 51 \/ 50 * 51 \/ 50 ) = rs . 9550.87 difference = ( 9550.87 - 9547.2 ) = rs . 3.67 . answer : c","correct":"c","options":{"a":"s . 2.04 ","b":"s . 2.08 ","c":"s . 3.67 ","d":"s . 2.83","e":"s . 2.42"},"options_float":{"a":2.04,"b":2.08,"c":3.67,"d":2.83,"e":2.42},"annotated_formula":"subtract(multiply(9000, multiply(multiply(add(const_1, divide(2, const_100)), add(const_1, divide(2, const_100))), add(const_1, divide(2, const_100)))), multiply(9000, multiply(add(const_1, divide(2, const_100)), add(const_1, divide(4, const_100)))))","linear_formula":"divide(n3,const_100)|divide(n4,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#2)|multiply(#2,#3)|multiply(#2,#4)|multiply(n0,#5)|multiply(n0,#6)|subtract(#8,#7)","chain":"2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n1 + (1\/50)<\/gadget>\n51\/50 = around 1.02<\/output>\n(51\/50) * (51\/50)<\/gadget>\n2_601\/2_500 = around 1.0404<\/output>\n(2_601\/2_500) * (51\/50)<\/gadget>\n132_651\/125_000 = around 1.061208<\/output>\n9_000 * (132_651\/125_000)<\/gadget>\n1_193_859\/125 = around 9_550.872<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(51\/50) * (26\/25)<\/gadget>\n663\/625 = around 1.0608<\/output>\n9_000 * (663\/625)<\/gadget>\n47_736\/5 = around 9_547.2<\/output>\n(1_193_859\/125) - (47_736\/5)<\/gadget>\n459\/125 = around 3.672<\/output>\n459\/125 = around 3.672<\/result>","index":3119} +{"problem":"a tank is filled by 3 pipes a , b , c in 3 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ?","rationale":"\"suppose pipe a alone take x hours to fill the tank then pipe b and c will take x \/ 2 and x \/ 4 hours respectively to fill the tank . 1 \/ x + 2 \/ x + 4 \/ x = 1 \/ 3 7 \/ x = 1 \/ 3 x = 21 hours answer is e\"","correct":"e","options":{"a":"25 hr ","b":"35 hr ","c":"40 hr ","d":"20 hr","e":"21 hr"},"options_float":{"a":25.0,"b":35.0,"c":40.0,"d":20.0,"e":21.0},"annotated_formula":"multiply(add(add(multiply(const_2, const_2), const_2), const_1), 3)","linear_formula":"multiply(const_2,const_2)|add(#0,const_2)|add(#1,const_1)|multiply(n1,#2)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21<\/result>","index":3120} +{"problem":"a girl multiplies 987 by a certain number and obtains 559981 as her answer . if in the answer , both 9 ’ s are wrong but the other digits are correct , then the correct answer will be","rationale":"987 = 3 x 7 x 47 . so , required number must be divisible by each one of 3 , 7 , 47 . none of the numbers in 553681 and 555181 are divisible by 3 . while 556581 is not divisible by 7 . correct answer is 555681 . answer : c","correct":"c","options":{"a":"553681 ","b":"555181 ","c":"555681 ","d":"556581","e":"none of these"},"options_float":{"a":553681.0,"b":555181.0,"c":555681.0,"d":556581.0,"e":null},"annotated_formula":"add(subtract(559981, multiply(add(multiply(9, const_10), 9), const_100)), multiply(add(multiply(const_2, const_3), multiply(const_10, add(const_1, const_4))), const_100))","linear_formula":"add(const_1,const_4)|multiply(const_2,const_3)|multiply(n2,const_10)|add(n2,#2)|multiply(#0,const_10)|add(#1,#4)|multiply(#3,const_100)|multiply(#5,const_100)|subtract(n1,#6)|add(#7,#8)","chain":"9 * 10<\/gadget>\n90<\/output>\n90 + 9<\/gadget>\n99<\/output>\n99 * 100<\/gadget>\n9_900<\/output>\n559_981 - 9_900<\/gadget>\n550_081<\/output>\n2 * 3<\/gadget>\n6<\/output>\n1 + 4<\/gadget>\n5<\/output>\n10 * 5<\/gadget>\n50<\/output>\n6 + 50<\/gadget>\n56<\/output>\n56 * 100<\/gadget>\n5_600<\/output>\n550_081 + 5_600<\/gadget>\n555_681<\/output>\n555_681<\/result>","index":3121} +{"problem":"a couple spent $ 198 in total while dining out and paid this amount using a credit card . the $ 198 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ?","rationale":"let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 198 x = $ 150 the correct answer is c .","correct":"c","options":{"a":"$ 130 ","b":"$ 140 ","c":"$ 150 ","d":"$ 160","e":"$ 170"},"options_float":{"a":130.0,"b":140.0,"c":150.0,"d":160.0,"e":170.0},"annotated_formula":"divide(multiply(divide(multiply(198, const_100), add(const_100, 20)), const_100), add(const_100, 10))","linear_formula":"add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1)","chain":"198 * 100<\/gadget>\n19_800<\/output>\n100 + 20<\/gadget>\n120<\/output>\n19_800 \/ 120<\/gadget>\n165<\/output>\n165 * 100<\/gadget>\n16_500<\/output>\n100 + 10<\/gadget>\n110<\/output>\n16_500 \/ 110<\/gadget>\n150<\/output>\n150<\/result>","index":3123} +{"problem":"in an election , candidate douglas won 58 percent of the total vote in counties x and y . he won 64 percent of the vote in county x . if the ratio of people who voted in county x to county y is 2 : 1 , what percent of the vote did candidate douglas win in county y ?","rationale":"\"given voters in ratio 2 : 1 let x has 200 votersy has 100 voters for x 64 % voted means 64 * 200 = 128 votes combined for xy has 300 voters and voted 58 % so total votes = 174 balance votes = 174 - 128 = 46 as y has 100 voters so 46 votes means 46 % of votes required ans e\"","correct":"e","options":{"a":"15 % ","b":"18 % ","c":"25 % ","d":"36 %","e":"46 %"},"options_float":{"a":15.0,"b":18.0,"c":25.0,"d":36.0,"e":46.0},"annotated_formula":"multiply(divide(subtract(divide(58, const_100), multiply(divide(64, const_100), divide(2, const_3))), divide(1, const_3)), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_3)|divide(n3,const_3)|multiply(#1,#2)|subtract(#0,#4)|divide(#5,#3)|multiply(#6,const_100)|","chain":"58 \/ 100<\/gadget>\n29\/50 = around 0.58<\/output>\n64 \/ 100<\/gadget>\n16\/25 = around 0.64<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(16\/25) * (2\/3)<\/gadget>\n32\/75 = around 0.426667<\/output>\n(29\/50) - (32\/75)<\/gadget>\n23\/150 = around 0.153333<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(23\/150) \/ (1\/3)<\/gadget>\n23\/50 = around 0.46<\/output>\n(23\/50) * 100<\/gadget>\n46<\/output>\n46<\/result>","index":3124} +{"problem":"if x is an integer and 2.134 × 10 ^ x is less than 21,000 , what is the greatest possible value for x ?","rationale":"\"x is an integer and 2.134 × 10 x is less than 21,000 , what is the greatest possible value for x ? for 2.134 × 10 x is less than 21,000 to remain true , the greatest number is 2,134 , which makes x = 3 e . 3\"","correct":"e","options":{"a":"7 ","b":"6 ","c":"5 ","d":"4","e":"3"},"options_float":{"a":7.0,"b":6.0,"c":5.0,"d":4.0,"e":3.0},"annotated_formula":"floor(divide(log(divide(21,000, 2.134)), log(10)))","linear_formula":"divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3)|","chain":"21_000 \/ 2.134<\/gadget>\n9_840.674789<\/output>\nlog(9_840.674789)<\/gadget>\n9.19428<\/output>\nlog(10)<\/gadget>\nlog(10) = around 2.302585<\/output>\n9.19428 \/ log(10)<\/gadget>\n9.19428\/log(10) = around 3.993025<\/output>\nfloor(9.19428\/log(10))<\/gadget>\n3<\/output>\n3<\/result>","index":3125} +{"problem":"a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 4 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? a . 480 liters","rationale":"1 \/ x - 1 \/ 20 = - 1 \/ 24 x = 120 120 * 4 = 480 answer : a","correct":"a","options":{"a":"480 ","b":"489 ","c":"486 ","d":"482","e":"483"},"options_float":{"a":480.0,"b":489.0,"c":486.0,"d":482.0,"e":483.0},"annotated_formula":"multiply(divide(multiply(24, 4), subtract(24, 20)), 20)","linear_formula":"multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2)","chain":"24 * 4<\/gadget>\n96<\/output>\n24 - 20<\/gadget>\n4<\/output>\n96 \/ 4<\/gadget>\n24<\/output>\n24 * 20<\/gadget>\n480<\/output>\n480<\/result>","index":3126} +{"problem":"the average weight of a group of persons increased from 48 kg to 51 kg , when two persons weighing 78 kg and 93 kg join the group . find the initial number of members in the group ?","rationale":"let the initial number of members in the group be n . initial total weight of all the members in the group = n ( 48 ) from the data , 48 n + 78 + 93 = 51 ( n + 2 ) = > 51 n - 48 n = 69 = > n = 23 therefore there were 23 members in the group initially . answer : c","correct":"c","options":{"a":"87 ","b":"67 ","c":"23 ","d":"75","e":"25"},"options_float":{"a":87.0,"b":67.0,"c":23.0,"d":75.0,"e":25.0},"annotated_formula":"divide(subtract(add(78, 93), multiply(51, const_2)), subtract(51, 48))","linear_formula":"add(n2,n3)|multiply(n1,const_2)|subtract(n1,n0)|subtract(#0,#1)|divide(#3,#2)","chain":"78 + 93<\/gadget>\n171<\/output>\n51 * 2<\/gadget>\n102<\/output>\n171 - 102<\/gadget>\n69<\/output>\n51 - 48<\/gadget>\n3<\/output>\n69 \/ 3<\/gadget>\n23<\/output>\n23<\/result>","index":3130} +{"problem":"convert 0.34 in to a vulgar fraction ?","rationale":"\"answer 0.34 = 34 \/ 100 = 17 \/ 50 correct option : c\"","correct":"c","options":{"a":"18 \/ 50 ","b":"16 \/ 50 ","c":"17 \/ 50 ","d":"19 \/ 50","e":"none"},"options_float":{"a":0.36,"b":0.32,"c":0.34,"d":0.38,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.34), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))","linear_formula":"add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n100 * 0.34<\/gadget>\n34<\/output>\n34 \/ 100<\/gadget>\n17\/50 = around 0.34<\/output>\n17\/50 = around 0.34<\/result>","index":3132} +{"problem":"dhoni bought 5 toys with the average cost of $ 10 . if david also buys the sixth toy with the price of $ 16 , what is the average ( arithmetic mean ) price of those 6 toys ?","rationale":"5 toys - > total cost = 5 * avg cost = 5 * 10 = 50 sixth toy cost = 16 total cost for 6 toys = 50 + 16 = 66 am = 66 \/ 6 = 11 hence e","correct":"e","options":{"a":"$ 10 ","b":"$ 7 ","c":"$ 8.5 ","d":"$ 9","e":"$ 11"},"options_float":{"a":10.0,"b":7.0,"c":8.5,"d":9.0,"e":11.0},"annotated_formula":"divide(add(10, const_12), const_2)","linear_formula":"add(n1,const_12)|divide(#0,const_2)","chain":"10 + 12<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n11<\/result>","index":3133} +{"problem":"a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds . what is the length of the train in meters ?","rationale":"\"relative speed of the train with respect to the motorbike should be the speed with which the length of the train is covered in 40 secs . calculate the relative speed = > 100 - 64 = 36 kmph and time is 40 sec = 40 \/ 3600 hrs distance ( in this case the length of the train ) = s * t = 36 * 40 \/ 3600 . 4 km or 400 meters . so answer ( a )\"","correct":"a","options":{"a":"400 meters ","b":"1111 meters ","c":"1777 meters ","d":"1822 meters","e":"none of these"},"options_float":{"a":400.0,"b":1111.0,"c":1777.0,"d":1822.0,"e":null},"annotated_formula":"multiply(multiply(subtract(100, 64), const_0_2778), 40)","linear_formula":"subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)|","chain":"100 - 64<\/gadget>\n36<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n10 * 40<\/gadget>\n400<\/output>\n400<\/result>","index":3134} +{"problem":"a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 350 m long platform ?","rationale":"\"let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 \/ 18 m = 120 m . time taken to cross the platform = ( 120 + 350 ) \/ 36 * 5 \/ 18 = 47 min . answer : d\"","correct":"d","options":{"a":"44 sec ","b":"49 sec ","c":"42 sec ","d":"47 sec","e":"none of these"},"options_float":{"a":44.0,"b":49.0,"c":42.0,"d":47.0,"e":null},"annotated_formula":"divide(add(350, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36))","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 36<\/gadget>\n10<\/output>\n10 * 12<\/gadget>\n120<\/output>\n350 + 120<\/gadget>\n470<\/output>\n470 \/ 10<\/gadget>\n47<\/output>\n47<\/result>","index":3135} +{"problem":"gold is 10 times as heavy as water and copper is 6 times as heavy as water . in what ratio should these be mixed to get an alloy 8 times as heavy as water ?","rationale":"\"g = 10 w c = 6 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 6 wx = x + 1 * 8 w 10 + 6 x = 8 ( x + 1 ) x = 1 \/ 1 ratio of gold with copper = 1 : 1 = 1 : 1 answer is b\"","correct":"b","options":{"a":"3 : 2 ","b":"1 : 1 ","c":"3 : 1 ","d":"5 : 2","e":"4 : 3"},"options_float":{"a":1.5,"b":1.0,"c":3.0,"d":2.5,"e":1.3333333333},"annotated_formula":"divide(subtract(8, 6), subtract(10, 8))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"8 - 6<\/gadget>\n2<\/output>\n10 - 8<\/gadget>\n2<\/output>\n2 \/ 2<\/gadget>\n1<\/output>\n1<\/result>","index":3136} +{"problem":"points a , b , and , c have xy - coordinates ( 2,0 ) , ( 8,12 ) , and ( 14,0 ) , respectively . points x , y , and z have xy - coordinates ( 6,0 ) , ( 8,4 ) , and ( 10,0 ) , respectively . what fraction q of the area of triangle abc is the area of triangle xyz ?","rationale":"\"if you notice , both triangles abc and xyz have a side on x axis . we can take these sides as bases for each triangle , therefore area of abc is 1 \/ 2 * 12 * 12 ( height of abc is the y coordinate of the third point ( 8,12 ) ) similarly area of xyz is 1 \/ 2 * 4 * 4 dividing area of xyz with that of abc gives q = 1 \/ 9 . a\"","correct":"a","options":{"a":"1 \/ 9 ","b":"1 \/ 8 ","c":"1 \/ 6 ","d":"1 \/ 5","e":"1 \/ 3"},"options_float":{"a":0.1111111111,"b":0.125,"c":0.1666666667,"d":0.2,"e":0.3333333333},"annotated_formula":"divide(divide(power(const_4, const_2), const_2), divide(power(add(const_10, const_2), const_2), const_2))","linear_formula":"add(const_10,const_2)|power(const_4,const_2)|divide(#1,const_2)|power(#0,const_2)|divide(#3,const_2)|divide(#2,#4)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n144 \/ 2<\/gadget>\n72<\/output>\n8 \/ 72<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":3137} +{"problem":"a waitress ' s income consists of her salary and tips . during one week , her tips were 3 \/ 4 of her salary . what fraction of her income for the week came from tips ?","rationale":"\"her tips were 3 \/ 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 3 \/ 4 ) ( $ 4 ) = $ 3 so , her total income = $ 4 + $ 3 = $ 7 what fraction of her income for the week came from tips $ 3 \/ $ 7 = 3 \/ 7 = c\"","correct":"c","options":{"a":"1 \/ 9 ","b":"1 \/ 6 ","c":"3 \/ 7 ","d":"4 \/ 9","e":"5 \/ 9"},"options_float":{"a":0.1111111111,"b":0.1666666667,"c":0.4285714286,"d":0.4444444444,"e":0.5555555556},"annotated_formula":"divide(3, add(3, 4))","linear_formula":"add(n0,n1)|divide(n0,#0)|","chain":"3 + 4<\/gadget>\n7<\/output>\n3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":3139} +{"problem":"the height of a triangle is decreased by 5 % and its base increased by 10 % . its area will","rationale":"explanation : let the initial height be 100 and the base be 200 units . â ˆ ´ area = 1 \/ 2 x 100 x 200 = 10000 sq units now , height = 95 units and base = 220 units â ˆ ´ new area = 1 \/ 2 x 95 x 220 = 10450 â ˆ ´ area is increased by ( 10450 - 10000 ) \/ 10000 x 10 = 4.5 % answer : option c","correct":"c","options":{"a":"decrease by 12.5 % ","b":"increase by 12.5 % ","c":"increase by 4.5 % ","d":"increase by 25 %","e":"none of these"},"options_float":{"a":12.5,"b":12.5,"c":4.5,"d":25.0,"e":null},"annotated_formula":"subtract(divide(multiply(subtract(const_100, 5), add(const_100, 10)), const_100), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)","chain":"100 - 5<\/gadget>\n95<\/output>\n100 + 10<\/gadget>\n110<\/output>\n95 * 110<\/gadget>\n10_450<\/output>\n10_450 \/ 100<\/gadget>\n209\/2 = around 104.5<\/output>\n(209\/2) - 100<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":3140} +{"problem":"the average age of 20 students of a class is 20 years . out of these , the average age of 9 students is 11 years and that of the other 10 students is 24 years . the age of the 20 th student is ?","rationale":"\"age of the 20 th student = 20 * 20 - ( 9 * 11 + 10 * 24 ) = 400 - 339 = 61 years answer is b\"","correct":"b","options":{"a":"65 ","b":"61 ","c":"59 ","d":"63","e":"64"},"options_float":{"a":65.0,"b":61.0,"c":59.0,"d":63.0,"e":64.0},"annotated_formula":"subtract(multiply(20, 20), add(multiply(9, 11), multiply(10, 24)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"20 * 20<\/gadget>\n400<\/output>\n9 * 11<\/gadget>\n99<\/output>\n10 * 24<\/gadget>\n240<\/output>\n99 + 240<\/gadget>\n339<\/output>\n400 - 339<\/gadget>\n61<\/output>\n61<\/result>","index":3142} +{"problem":"a man engaged a servant on the condition that he would pay him rs . 500 and a uniform after one year service . he served only for 9 months and received uniform and rs . 300 , find the price of the uniform ?","rationale":"\"9 \/ 12 = 3 \/ 4 * 500 = 375 300 - - - - - - - - - - - - - 75 1 \/ 4 - - - - - - - - 75 1 - - - - - - - - - ? = > rs . 300 answer : d\"","correct":"d","options":{"a":"s . 80 ","b":"s . 85 ","c":"s . 90 ","d":"s . 300","e":"s . 120"},"options_float":{"a":80.0,"b":85.0,"c":90.0,"d":300.0,"e":120.0},"annotated_formula":"multiply(divide(subtract(multiply(9, 500), multiply(multiply(const_3, const_4), 300)), multiply(multiply(const_3, const_4), const_1)), const_4)","linear_formula":"multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)|","chain":"9 * 500<\/gadget>\n4_500<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 * 300<\/gadget>\n3_600<\/output>\n4_500 - 3_600<\/gadget>\n900<\/output>\n12 * 1<\/gadget>\n12<\/output>\n900 \/ 12<\/gadget>\n75<\/output>\n75 * 4<\/gadget>\n300<\/output>\n300<\/result>","index":3143} +{"problem":"a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 800 sq . ft , how many feet of fencing will be required ?","rationale":"\"explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 800 = > breadth = 40 feet area to be fenced = 2 b + l = 2 * 40 + 20 = 100 feet answer : option a\"","correct":"a","options":{"a":"100 feet ","b":"120 feet ","c":"150 feet ","d":"180 feet","e":"186 feet"},"options_float":{"a":100.0,"b":120.0,"c":150.0,"d":180.0,"e":186.0},"annotated_formula":"add(multiply(divide(800, 20), const_2), 20)","linear_formula":"divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|","chain":"800 \/ 20<\/gadget>\n40<\/output>\n40 * 2<\/gadget>\n80<\/output>\n80 + 20<\/gadget>\n100<\/output>\n100<\/result>","index":3144} +{"problem":"every day daniel drives 60 miles back from work . on sunday , daniel drove all the way back from work at a constant speed of x miles per hour . on monday , daniel drove the first 32 miles back from work at ( 2 x ) miles per hour , and the rest of the way at ( x \/ 2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ?","rationale":"\"let ' s test x = 4 . . . . on sunday , daniel drove 60 miles at 4 miles \/ hour . d = ( r ) ( t ) 60 = ( 4 ) ( t ) 60 \/ 4 = 15 = t it takes 15 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles \/ hour and the rest of the way ( 28 miles ) at 4 \/ 2 = 2 miles \/ hour d = ( r ) ( t ) 32 = ( 8 ) ( t ) 32 \/ 8 = 4 = t it takes 4 hours for the first part d = ( r ) ( t ) 28 = ( 2 ) ( t ) 28 \/ 2 = 14 = t it takes 14 hours for the second part total time to drive home on monday = 4 + 14 = 18 hours we ' re asked by what percent 18 hours is greater than 15 hours . 18 \/ 15 = 1.2 , so it is 20 % greater . b\"","correct":"b","options":{"a":"10 % ","b":"20 % ","c":"30 % ","d":"40 %","e":"50 %"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"multiply(divide(subtract(add(divide(32, 2), multiply(subtract(60, 32), 2)), 60), 60), const_100)","linear_formula":"divide(n1,n2)|subtract(n0,n1)|multiply(#1,n2)|add(#0,#2)|subtract(#3,n0)|divide(#4,n0)|multiply(#5,const_100)|","chain":"32 \/ 2<\/gadget>\n16<\/output>\n60 - 32<\/gadget>\n28<\/output>\n28 * 2<\/gadget>\n56<\/output>\n16 + 56<\/gadget>\n72<\/output>\n72 - 60<\/gadget>\n12<\/output>\n12 \/ 60<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":3146} +{"problem":"the length of a rectangle is halved , while its breadth is tripled . watis the % change in area ?","rationale":"\"let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c\"","correct":"c","options":{"a":"43 % ","b":"47 % ","c":"50 % ","d":"65 %","e":"68 %"},"options_float":{"a":43.0,"b":47.0,"c":50.0,"d":65.0,"e":68.0},"annotated_formula":"multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100)","linear_formula":"divide(const_1,const_2)|multiply(#0,const_3)|subtract(#1,const_1)|divide(#2,const_1)|multiply(#3,const_100)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n3 * (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) - 1<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) \/ 1<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":3147} +{"problem":"the number of sailors on a ship is 85 % more than the number of officers . the ratio of sailors to officers would be","rationale":"\"sailor = 1.85 * officer sailor \/ officer = 1.85 \/ 1 = 185 \/ 100 = 37 \/ 20 answer will be a\"","correct":"a","options":{"a":"37 : 20 ","b":"47 : 20 ","c":"47 : 25 ","d":"22 : 25","e":"97 : 50"},"options_float":{"a":1.85,"b":2.35,"c":1.88,"d":0.88,"e":1.94},"annotated_formula":"divide(add(85, const_100), const_100)","linear_formula":"add(n0,const_100)|divide(#0,const_100)|","chain":"85 + 100<\/gadget>\n185<\/output>\n185 \/ 100<\/gadget>\n37\/20 = around 1.85<\/output>\n37\/20 = around 1.85<\/result>","index":3148} +{"problem":"a goods train runs at the speed of 72 km \/ hr and crosses a 270 m long platform in 26 sec . what is the length of the goods train ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec . time = 26 sec . let the length of the train be x meters . then , ( x + 270 ) \/ 26 = 20 x = 250 m . answer : d\"","correct":"d","options":{"a":"278 ","b":"166 ","c":"151 ","d":"250","e":"109"},"options_float":{"a":278.0,"b":166.0,"c":151.0,"d":250.0,"e":109.0},"annotated_formula":"subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 270)","linear_formula":"divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)|","chain":"72 \/ 3_600<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 1_000<\/gadget>\n20<\/output>\n20 * 26<\/gadget>\n520<\/output>\n520 - 270<\/gadget>\n250<\/output>\n250<\/result>","index":3149} +{"problem":"the volume of water inside a swimming pool doubles every hour . if the pool is filled to its full capacity within 8 hours , in how many hours was it filled to one quarter of its capacity ?","rationale":"d","correct":"d","options":{"a":"2 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"subtract(subtract(8, const_1), const_1)","linear_formula":"subtract(n0,const_1)|subtract(#0,const_1)","chain":"8 - 1<\/gadget>\n7<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6<\/result>","index":3150} +{"problem":"there is a rectangular prism made of 1 in cubes that has been covered in tin foil . there are exactly 128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these 128 cubes is twice the length and twice the height , what is the measure y in inches of the width of the foil covered prism ?","rationale":"\"if the width is w , then length and height would be w \/ 2 . so , w * w \/ 2 * w \/ 2 = 128 = > w ^ 3 = ( 2 ^ 3 ) * 64 = ( 2 ^ 3 ) * ( 4 ^ 3 ) = > w = 2 * 4 = 8 in . along the width of the cuboid , 8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = 8 + 2 = y = 10 ans e .\"","correct":"e","options":{"a":"4 ","b":"6 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":4.0,"b":6.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"add(multiply(power(divide(128, multiply(1, multiply(1, const_2))), inverse(const_3)), multiply(1, const_2)), multiply(1, const_2))","linear_formula":"inverse(const_3)|multiply(n0,const_2)|multiply(n0,#1)|divide(n1,#2)|power(#3,#0)|multiply(#1,#4)|add(#5,#1)|","chain":"1 * 2<\/gadget>\n2<\/output>\n128 \/ 2<\/gadget>\n64<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n64 ** (1\/3)<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n10<\/result>","index":3152} +{"problem":"a train 200 m long can cross an electric pole in 5 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 200 \/ 5 s = 40 m \/ sec speed = 40 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 144 kmph answer : d\"","correct":"d","options":{"a":"114 ","b":"124 ","c":"134 ","d":"144","e":"154"},"options_float":{"a":114.0,"b":124.0,"c":134.0,"d":144.0,"e":154.0},"annotated_formula":"divide(divide(200, const_1000), divide(5, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"200 \/ 1_000<\/gadget>\n1\/5 = around 0.2<\/output>\n5 \/ 3_600<\/gadget>\n1\/720 = around 0.001389<\/output>\n(1\/5) \/ (1\/720)<\/gadget>\n144<\/output>\n144<\/result>","index":3155} +{"problem":"water is leaking out from a cylinder container at the rate of 0.31 m ^ 3 per minute . after 10 minutes , the water level decreases 1 \/ 9 meters . what is value of the radius in meters ?","rationale":"\"10 * 0.31 = 3.1 = pi * r ^ 2 * h r ^ 2 = 3.1 \/ ( pi * 1 \/ 9 ) which is about 9 r = 3 the answer is b .\"","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"6","e":"9"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":6.0,"e":9.0},"annotated_formula":"divide(multiply(10, 0.31), 1)","linear_formula":"multiply(n0,n2)|divide(#0,n3)|","chain":"10 * 0.31<\/gadget>\n3.1<\/output>\n3.1 \/ 1<\/gadget>\n3.1<\/output>\n3.1<\/result>","index":3156} +{"problem":"a person crosses a 800 m long street in 5 minutes . what is his speed in km per hour ?","rationale":"\"distance = 800 meter time = 5 minutes = 5 x 60 seconds = 300 seconds speed = distance \/ time = 800 \/ 300 = 2.67 m \/ s = 2.67 ã — 18 \/ 5 km \/ hr = 9.61 km \/ hr answer : e\"","correct":"e","options":{"a":"1.23 ","b":"2.86 ","c":"4.36 ","d":"9.25","e":"9.61"},"options_float":{"a":1.23,"b":2.86,"c":4.36,"d":9.25,"e":9.61},"annotated_formula":"divide(divide(800, const_1000), divide(multiply(5, const_60), const_3600))","linear_formula":"divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)|","chain":"800 \/ 1_000<\/gadget>\n4\/5 = around 0.8<\/output>\n5 * 60<\/gadget>\n300<\/output>\n300 \/ 3_600<\/gadget>\n1\/12 = around 0.083333<\/output>\n(4\/5) \/ (1\/12)<\/gadget>\n48\/5 = around 9.6<\/output>\n48\/5 = around 9.6<\/result>","index":3157} +{"problem":"in a 100 m race , sam beats john by 5 seconds . on the contrary , if sam allowed john to start 15 m ahead of sam , then sam and john reach the finishing point at the same time . how long does sam take to run the 100 m race ?","rationale":"\"their difference is 5 second but this difference is 0 if john allows sam to start the race from 15 m ahead . that means jhon was 15 m away from finishing line when they started together . so he will cover 15 m in 5 seconds . so his speed = 15 \/ 5 = 3 metre \/ second . so time taken = 100 \/ 5 = 20 seconds . so sam took = 17 seconds . correct answer = e\"","correct":"e","options":{"a":"4 seconds ","b":"25 seconds ","c":"29 seconds ","d":"21 seconds","e":"17 seconds"},"options_float":{"a":4.0,"b":25.0,"c":29.0,"d":21.0,"e":17.0},"annotated_formula":"divide(subtract(100, 15), 5)","linear_formula":"subtract(n0,n2)|divide(#0,n1)|","chain":"100 - 15<\/gadget>\n85<\/output>\n85 \/ 5<\/gadget>\n17<\/output>\n17<\/result>","index":3159} +{"problem":"in an examination , 300 students appeared . out of these students ; 29 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed .","rationale":"\"the number of students with first division = 29 % of 300 = 29 \/ 100 × 300 = 8700 \/ 100 = 87 and , the number of students with second division = 54 % of 300 = 54 \/ 100 × 300 = 16200 \/ 100 = 162 therefore , the number of students who just passed = 300 – ( 87 + 162 ) = 51 answer : a\"","correct":"a","options":{"a":"51 ","b":"37 ","c":"54 ","d":"99","e":"01"},"options_float":{"a":51.0,"b":37.0,"c":54.0,"d":99.0,"e":1.0},"annotated_formula":"divide(multiply(300, subtract(const_100, add(29, 54))), const_100)","linear_formula":"add(n1,n2)|subtract(const_100,#0)|multiply(n0,#1)|divide(#2,const_100)|","chain":"29 + 54<\/gadget>\n83<\/output>\n100 - 83<\/gadget>\n17<\/output>\n300 * 17<\/gadget>\n5_100<\/output>\n5_100 \/ 100<\/gadget>\n51<\/output>\n51<\/result>","index":3161} +{"problem":"the length of rectangle is thrice its breadth and its perimeter is 104 m , find the area of the rectangle ?","rationale":"\"2 ( 3 x + x ) = 104 l = 39 b = 13 lb = 39 * 13 = 507 answer : a\"","correct":"a","options":{"a":"507 sq m ","b":"356 sq m ","c":"452 sq m ","d":"428 sq m","e":"525 sq m"},"options_float":{"a":507.0,"b":356.0,"c":452.0,"d":428.0,"e":525.0},"annotated_formula":"multiply(multiply(divide(104, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(104, add(multiply(const_3, const_2), multiply(const_1, const_2))))","linear_formula":"multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)|","chain":"3 * 2<\/gadget>\n6<\/output>\n1 * 2<\/gadget>\n2<\/output>\n6 + 2<\/gadget>\n8<\/output>\n104 \/ 8<\/gadget>\n13<\/output>\n13 * 3<\/gadget>\n39<\/output>\n39 * 13<\/gadget>\n507<\/output>\n507<\/result>","index":3162} +{"problem":"if x + y = 20 , x - y = 36 , for integers of x and y , x = ?","rationale":"\"x + y = 20 x - y = 36 2 x = 56 x = 28 answer is c\"","correct":"c","options":{"a":"20 ","b":"25 ","c":"28 ","d":"20","e":"42"},"options_float":{"a":20.0,"b":25.0,"c":28.0,"d":20.0,"e":42.0},"annotated_formula":"divide(add(20, 36), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"20 + 36<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":3163} +{"problem":"the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 14 % , then the cost price of the article is :","rationale":"\"110 % of s . p . = 616 s . p . = ( 616 * 100 ) \/ 110 = rs . 560 c . p = ( 110 * 560 ) \/ 114 = rs . 540 answer : option b\"","correct":"b","options":{"a":"500 ","b":"540 ","c":"555 ","d":"664","e":"5598"},"options_float":{"a":500.0,"b":540.0,"c":555.0,"d":664.0,"e":5598.0},"annotated_formula":"divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 14))","linear_formula":"add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)|","chain":"616 * 100<\/gadget>\n61_600<\/output>\n100 + 10<\/gadget>\n110<\/output>\n61_600 \/ 110<\/gadget>\n560<\/output>\n560 * 110<\/gadget>\n61_600<\/output>\n100 + 14<\/gadget>\n114<\/output>\n61_600 \/ 114<\/gadget>\n30_800\/57 = around 540.350877<\/output>\n30_800\/57 = around 540.350877<\/result>","index":3165} +{"problem":"the speed of a boat in still water is 14 km \/ hr and the rate of current is 5 km \/ hr . the distance travelled downstream in 8 minutes is","rationale":"\"explanation : speed downstreams = ( 14 + 5 ) kmph = 19 kmph . distance travelled = ( 19 x 8 \/ 60 ) km = 2.5 km option b\"","correct":"b","options":{"a":"1.6 km ","b":"2.5 km ","c":"3.6 km ","d":"4 km","e":"none of these"},"options_float":{"a":1.6,"b":2.5,"c":3.6,"d":4.0,"e":null},"annotated_formula":"multiply(divide(8, const_60), add(14, 5))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"8 \/ 60<\/gadget>\n2\/15 = around 0.133333<\/output>\n14 + 5<\/gadget>\n19<\/output>\n(2\/15) * 19<\/gadget>\n38\/15 = around 2.533333<\/output>\n38\/15 = around 2.533333<\/result>","index":3166} +{"problem":"what is the tens digit of 6 ^ 19 ?","rationale":"the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . . . 56 is 5 . the tens digit of 6 ^ 7 = . . . 36 is 3 again . etc . . . 19 has the form 5 n + 4 , so the tens digit of 6 ^ 19 is 9 . the answer is e .","correct":"e","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"floor(divide(reminder(power(6, reminder(19, add(const_4, const_1))), const_100), const_10))","linear_formula":"add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)","chain":"4 + 1<\/gadget>\n5<\/output>\n19 % 5<\/gadget>\n4<\/output>\n6 ** 4<\/gadget>\n1_296<\/output>\n1_296 % 100<\/gadget>\n96<\/output>\n96 \/ 10<\/gadget>\n48\/5 = around 9.6<\/output>\nfloor(48\/5)<\/gadget>\n9<\/output>\n9<\/result>","index":3167} +{"problem":"aa is two digit number then m times its cube of aa has 1 in its tens place then what is m .","rationale":"when aa = 11 7 * ( 11 ) ^ 3 = 7 * 1331 = 9317 m = 7 answer : e","correct":"e","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(multiply(power(add(const_10, 1), const_3), add(const_3, const_4)), power(add(const_10, 1), const_3))","linear_formula":"add(const_3,const_4)|add(n0,const_10)|power(#1,const_3)|multiply(#0,#2)|divide(#3,#2)","chain":"10 + 1<\/gadget>\n11<\/output>\n11 ** 3<\/gadget>\n1_331<\/output>\n3 + 4<\/gadget>\n7<\/output>\n1_331 * 7<\/gadget>\n9_317<\/output>\n9_317 \/ 1_331<\/gadget>\n7<\/output>\n7<\/result>","index":3168} +{"problem":"a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability w that out of the two , one is white and one is blue ?","rationale":"\"the required probability w = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) \/ 8 c 6 = 15 \/ 28 . d is thus the correct answer .\"","correct":"d","options":{"a":"15 \/ 56 ","b":"41 \/ 56 ","c":"13 \/ 28 ","d":"15 \/ 28","e":"5 \/ 14"},"options_float":{"a":0.2678571429,"b":0.7321428571,"c":0.4642857143,"d":0.5357142857,"e":0.3571428571},"annotated_formula":"divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2))","linear_formula":"add(n0,n1)|add(n1,const_2)|multiply(n0,n1)|multiply(#0,#1)|divide(#3,const_2)|divide(#2,#4)|","chain":"3 * 5<\/gadget>\n15<\/output>\n3 + 5<\/gadget>\n8<\/output>\n5 + 2<\/gadget>\n7<\/output>\n8 * 7<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n15 \/ 28<\/gadget>\n15\/28 = around 0.535714<\/output>\n15\/28 = around 0.535714<\/result>","index":3169} +{"problem":"the curved surface of a sphere is 64 π cm 2 . find its radius ?","rationale":"explanation : 4 π r 2 = 64 = > r = 4 answer : option c","correct":"c","options":{"a":"8 ","b":"3 ","c":"4 ","d":"9","e":"7"},"options_float":{"a":8.0,"b":3.0,"c":4.0,"d":9.0,"e":7.0},"annotated_formula":"sqrt(divide(64, const_4))","linear_formula":"divide(n0,const_4)|sqrt(#0)","chain":"64 \/ 4<\/gadget>\n16<\/output>\n16 ** (1\/2)<\/gadget>\n4<\/output>\n4<\/result>","index":3170} +{"problem":"dhoni spent 25 percent of his earning last month on rent and 10 percent less than what he spent on rent to purchase a new dishwasher . what percent of last month ' s earning did dhoni have left over ?","rationale":"\"say dhoni ' s earning last month was $ 100 . dhoni spent 30 percent of his earning last month on rent - - > $ 25 on rent ; 10 percent less than what he spent on rent to purchase a new dishwasher - - > $ 25 * 0.9 = $ 22.5 on the dishwasher . left over amount 100 - ( 25 + 22.5 ) = $ 52.5 answer : e\"","correct":"e","options":{"a":"50 % ","b":"55 % ","c":"60 % ","d":"62 %","e":"52.5 %"},"options_float":{"a":50.0,"b":55.0,"c":60.0,"d":62.0,"e":52.5},"annotated_formula":"subtract(const_100, add(25, multiply(25, subtract(const_1, divide(10, const_100)))))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|multiply(n0,#1)|add(n0,#2)|subtract(const_100,#3)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n25 * (9\/10)<\/gadget>\n45\/2 = around 22.5<\/output>\n25 + (45\/2)<\/gadget>\n95\/2 = around 47.5<\/output>\n100 - (95\/2)<\/gadget>\n105\/2 = around 52.5<\/output>\n105\/2 = around 52.5<\/result>","index":3171} +{"problem":"a present value of a machine is $ 800 . its value depletion rate is 10 % per annum then find the machine value after 2 years ?","rationale":"\"p = $ 800 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r \/ 100 ) ^ t ] = 800 * 9 \/ 10 * 9 \/ 10 = $ 648 answer is e\"","correct":"e","options":{"a":"$ 900 ","b":"$ 810 ","c":"$ 915 ","d":"$ 715","e":"$ 648"},"options_float":{"a":900.0,"b":810.0,"c":915.0,"d":715.0,"e":648.0},"annotated_formula":"multiply(800, power(subtract(const_1, divide(10, const_100)), 2))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) ** 2<\/gadget>\n81\/100 = around 0.81<\/output>\n800 * (81\/100)<\/gadget>\n648<\/output>\n648<\/result>","index":3172} +{"problem":"the effective annual rate of interest corresponding to a nominal rate of 16 % per annum payable half - yearly is ?","rationale":"\"amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 8 \/ 100 ) 2 ] = rs . 116.64 effective rate = ( 116.64 - 100 ) = 116.64 % answer : a\"","correct":"a","options":{"a":"16.64 % ","b":"16.07 % ","c":"16.08 % ","d":"16.09 %","e":"16.19 %"},"options_float":{"a":16.64,"b":16.07,"c":16.08,"d":16.09,"e":16.19},"annotated_formula":"add(add(divide(16, const_2), divide(16, const_2)), divide(multiply(divide(16, const_2), divide(16, const_2)), const_100))","linear_formula":"divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)|","chain":"16 \/ 2<\/gadget>\n8<\/output>\n8 + 8<\/gadget>\n16<\/output>\n8 * 8<\/gadget>\n64<\/output>\n64 \/ 100<\/gadget>\n16\/25 = around 0.64<\/output>\n16 + (16\/25)<\/gadget>\n416\/25 = around 16.64<\/output>\n416\/25 = around 16.64<\/result>","index":3173} +{"problem":"cost of 2 bats and 3 balls is rs . 1300 . the cost of 3 bats and 2 balls is rs . 1200 . the cost of each balls is more than that of each bat by ?","rationale":"2 bats + 3 ball = 1300 - - - ( 1 ) 3 bats + 3 ball = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - bats + ball = 100 = > ball - bats = 100 b","correct":"b","options":{"a":"50 ","b":"100 ","c":"110 ","d":"120","e":"130"},"options_float":{"a":50.0,"b":100.0,"c":110.0,"d":120.0,"e":130.0},"annotated_formula":"subtract(1300, 1200)","linear_formula":"subtract(n2,n5)","chain":"1_300 - 1_200<\/gadget>\n100<\/output>\n100<\/result>","index":3174} +{"problem":"m = { - 6 , - 5 , - 4 , - 3 , - 2 } t = { - 2 , - 1 , 1 , 2 , 3 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above , what is the probability that the product of the two integers will be negative ?","rationale":"answer e . total # of outcomes : 5 * 5 = 25 # of outcomes where product is - ve : ( - 6,1 ) , ( - 6,2 ) , ( - 6,3 ) . . . hence , total : 15 probability : 15 \/ 25 = 3 \/ 5","correct":"e","options":{"a":"0 ","b":"1 \/ 3 ","c":"2 \/ 5 ","d":"1 \/ 2","e":"3 \/ 5"},"options_float":{"a":0.0,"b":0.3333333333,"c":0.4,"d":0.5,"e":0.6},"annotated_formula":"divide(3, 5)","linear_formula":"divide(n3,n1)","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":3175} +{"problem":"what is the total cost of 2 sandwiches at $ 2.45 each and 4 sodas at $ 0.87 each ?","rationale":"\"answer = a 2 * 2.45 + 4 * 0.87 = 2 ( 2.50 - 0.05 ) + 4 ( 1.00 - 0.13 ) = 5 + 4 - 0.1 - 0.52 = 9 - 0.62 = 8.38\"","correct":"a","options":{"a":"$ 8.38 ","b":"$ 6.85 ","c":"$ 8.46 ","d":"$ 10.08","e":"$ 11.85"},"options_float":{"a":8.38,"b":6.85,"c":8.46,"d":10.08,"e":11.85},"annotated_formula":"add(multiply(2, 2.45), multiply(4, 0.87))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"2 * 2.45<\/gadget>\n4.9<\/output>\n4 * 0.87<\/gadget>\n3.48<\/output>\n4.9 + 3.48<\/gadget>\n8.38<\/output>\n8.38<\/result>","index":3177} +{"problem":"a woman complete a journey in 5 hours . she travels first half of the journey at the rate of 21 km \/ hr and second half at the rate of 24 km \/ hr . find the total journey in km .","rationale":"\"0.5 x \/ 21 + 0.5 x \/ 24 = 5 - - > x \/ 21 + x \/ 24 = 10 - - > x = 112 km . c\"","correct":"c","options":{"a":"334 km . ","b":"216 km . ","c":"112 km . ","d":"224 km .","e":"544 km ."},"options_float":{"a":334.0,"b":216.0,"c":112.0,"d":224.0,"e":544.0},"annotated_formula":"multiply(multiply(divide(multiply(5, 24), add(24, 21)), 21), const_2)","linear_formula":"add(n1,n2)|multiply(n0,n2)|divide(#1,#0)|multiply(n1,#2)|multiply(#3,const_2)|","chain":"5 * 24<\/gadget>\n120<\/output>\n24 + 21<\/gadget>\n45<\/output>\n120 \/ 45<\/gadget>\n8\/3 = around 2.666667<\/output>\n(8\/3) * 21<\/gadget>\n56<\/output>\n56 * 2<\/gadget>\n112<\/output>\n112<\/result>","index":3180} +{"problem":"a rectangular field is to be fenced on three sides leaving a side of 34 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ?","rationale":"\"given that length and area , so we can find the breadth . length x breadth = area 34 x breadth = 680 breadth = 20 feet area to be fenced = 2 b + l = 2 ( 20 ) + 34 = 88 feet answer : a ) 74 ft\"","correct":"a","options":{"a":"74 ft ","b":"88 ft ","c":"22 ft ","d":"11 ft","e":"66 ft"},"options_float":{"a":74.0,"b":88.0,"c":22.0,"d":11.0,"e":66.0},"annotated_formula":"add(multiply(divide(680, 34), const_2), 34)","linear_formula":"divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|","chain":"680 \/ 34<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n40 + 34<\/gadget>\n74<\/output>\n74<\/result>","index":3181} +{"problem":"how many shots of 1 cm radius can be prepared from a sphere of 6 cm radius ?","rationale":"\"4 \/ 3 ï € * 6 * 6 * 6 = 4 \/ 3 ï € * 1 * 1 * 1 * x x = 216 answer : c\"","correct":"c","options":{"a":"33 ","b":"88 ","c":"216 ","d":"288","e":"299"},"options_float":{"a":33.0,"b":88.0,"c":216.0,"d":288.0,"e":299.0},"annotated_formula":"multiply(power(6, const_2), 6)","linear_formula":"power(n1,const_2)|multiply(n1,#0)|","chain":"6 ** 2<\/gadget>\n36<\/output>\n36 * 6<\/gadget>\n216<\/output>\n216<\/result>","index":3182} +{"problem":"a man can do a piece of work in 5 days , but with the help of his son , he can finish it in 4 days . in what time can the son do it alone ?","rationale":"\"son ' s 1 day work = 1 \/ 4 - 1 \/ 5 = 1 \/ 20 son alone can do the work in 20 days = 20 days answer is e\"","correct":"e","options":{"a":"5 ","b":"5 1 \/ 2 ","c":"7 1 \/ 2 ","d":"6","e":"20"},"options_float":{"a":5.0,"b":5.0,"c":7.0,"d":6.0,"e":20.0},"annotated_formula":"divide(multiply(5, 4), subtract(5, 4))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"5 * 4<\/gadget>\n20<\/output>\n5 - 4<\/gadget>\n1<\/output>\n20 \/ 1<\/gadget>\n20<\/output>\n20<\/result>","index":3183} +{"problem":"a man & his wife appear in an interview for 2 vacancies in the same post . the probability of husband ' s selection is 1 \/ 7 & the probability of wife ' s selection is 1 \/ 5 . what is the probability that only 1 of them is selected ?","rationale":"let a = event that the husband is selected and b = event that the wife is selected . then , p ( a ) = 1 \/ 7 and p ( b ) = 1 \/ 5 ∴ p ( a ¯ ) = ( 1 - 17 \/ 5 ) = 6 \/ 7 and = ( 1 - 1 \/ 5 ) = 4 \/ 5 required probability = p [ ( a and not b ) or ( b and not a ) ] = p [ ( a and b ¯ ) or ( b and a ¯ ) ] = p ( a and b ¯ ) + p ( b and a ¯ ) = p ( a ) . p ( b ¯ ) + p ( b ) . p ( a ¯ ) = ( 1 \/ 7 * 4 \/ 5 ) + ( 1 \/ 5 * 6 \/ 7 ) = 10 \/ 35 = 2 \/ 7 c","correct":"c","options":{"a":"1 \/ 3 ","b":"2 \/ 3 ","c":"2 \/ 7 ","d":"3 \/ 11","e":"4 \/ 5"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":0.2857142857,"d":0.2727272727,"e":0.8},"annotated_formula":"add(multiply(divide(1, 5), subtract(1, divide(1, 7))), multiply(divide(1, 7), subtract(1, divide(1, 5))))","linear_formula":"divide(n1,n4)|divide(n1,n2)|subtract(n1,#1)|subtract(n1,#0)|multiply(#0,#2)|multiply(#1,#3)|add(#4,#5)","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 - (1\/7)<\/gadget>\n6\/7 = around 0.857143<\/output>\n(1\/5) * (6\/7)<\/gadget>\n6\/35 = around 0.171429<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(1\/7) * (4\/5)<\/gadget>\n4\/35 = around 0.114286<\/output>\n(6\/35) + (4\/35)<\/gadget>\n2\/7 = around 0.285714<\/output>\n2\/7 = around 0.285714<\/result>","index":3185} +{"problem":"find the length of the wire required to go 11 times round a square field containing 27889 m 2 .","rationale":"\"a 2 = 27889 = > a = 167 4 a = 668 668 * 11 = 7348 answer : a\"","correct":"a","options":{"a":"7348 ","b":"3388 ","c":"2667 ","d":"8766","e":"66711"},"options_float":{"a":7348.0,"b":3388.0,"c":2667.0,"d":8766.0,"e":66711.0},"annotated_formula":"multiply(square_perimeter(square_edge_by_area(27889)), 11)","linear_formula":"square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)|","chain":"27_889 ** (1\/2)<\/gadget>\n167<\/output>\n4 * 167<\/gadget>\n668<\/output>\n668 * 11<\/gadget>\n7_348<\/output>\n7_348<\/result>","index":3186} +{"problem":"there are 25 balls in a jar . you take out 5 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 \/ 5 . how many blue balls were there in the beginning ?","rationale":"\"9 = 5 blue balls + 20 \/ 5 answer : b\"","correct":"b","options":{"a":"12 . ","b":"9 . ","c":"8 . ","d":"7 .","e":"6 ."},"options_float":{"a":12.0,"b":9.0,"c":8.0,"d":7.0,"e":6.0},"annotated_formula":"add(divide(subtract(25, 5), 5), 5)","linear_formula":"subtract(n0,n1)|divide(#0,n3)|add(n1,#1)|","chain":"25 - 5<\/gadget>\n20<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n4 + 5<\/gadget>\n9<\/output>\n9<\/result>","index":3188} +{"problem":"a circular well with a diameter of 4 metres , is dug to a depth of 24 metres . what is the volume of the earth dug out ?","rationale":"\"solution volume = π r 2 h ‹ = › ( 22 \/ 7 × 2 × 2 × 24 ) m 3 ‹ = › 301.6 m 3 . answer e\"","correct":"e","options":{"a":"32 m 3 ","b":"36 m 3 ","c":"40 m 3 ","d":"44 m 3","e":"301.6"},"options_float":{"a":32.0,"b":36.0,"c":40.0,"d":44.0,"e":301.6},"annotated_formula":"volume_cylinder(divide(4, const_2), 24)","linear_formula":"divide(n0,const_2)|volume_cylinder(#0,n1)|","chain":"4 \/ 2<\/gadget>\n2<\/output>\npi * (2 ** 2) * 24<\/gadget>\n96*pi = around 301.592895<\/output>\n96*pi = around 301.592895<\/result>","index":3189} +{"problem":"priya and her father differ in their ages by 31 years . after 8 years . if the sum of their ages will be 69 , their present ages are","rationale":"a 11 , 42 let the age of priya = x years then , the age of her father = ( x + 31 ) years atq , ( x + 8 ) + ( x + 31 + 8 ) = 69 2 x + 47 = 69 2 x = 69 – 47 = 22 x = 11 the age of priya = 11 years the age of her father = 11 + 31 = 42 years","correct":"a","options":{"a":"11 , 42 ","b":"12 , 24 ","c":"15 , 42 ","d":"16 , 19","e":"33 , 44"},"options_float":{"a":11.0,"b":12.0,"c":15.0,"d":16.0,"e":33.0},"annotated_formula":"subtract(divide(add(subtract(subtract(69, 8), 8), 31), const_2), 31)","linear_formula":"subtract(n2,n1)|subtract(#0,n1)|add(n0,#1)|divide(#2,const_2)|subtract(#3,n0)","chain":"69 - 8<\/gadget>\n61<\/output>\n61 - 8<\/gadget>\n53<\/output>\n53 + 31<\/gadget>\n84<\/output>\n84 \/ 2<\/gadget>\n42<\/output>\n42 - 31<\/gadget>\n11<\/output>\n11<\/result>","index":3190} +{"problem":"if 7 spiders make 7 webs in 7 days , then how many days are needed for 1 spider to make 1 web ?","rationale":"\"explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 7 : 1 ( webs ) 1 : 7 } : : x : 7 ⇒ 7 × 1 × 7 = 1 × 7 × x ⇒ x = 7 answer : option b\"","correct":"b","options":{"a":"1 ","b":"7 ","c":"3 ","d":"14","e":"16"},"options_float":{"a":1.0,"b":7.0,"c":3.0,"d":14.0,"e":16.0},"annotated_formula":"multiply(1, 7)","linear_formula":"multiply(n0,n3)|","chain":"1 * 7<\/gadget>\n7<\/output>\n7<\/result>","index":3191} +{"problem":"the price of lunch for 13 people was $ 207 including a 15 % gratuity for service . what was the average price per person , excluding the gratuity ?","rationale":"\"clearly c is the answer i used poe here lets consider option ( c ) 13 * 13.85 = 180 now 180 ( 115 \/ 100 ) = 207 = > possible answer imo c\"","correct":"c","options":{"a":"11.73 ","b":"12 ","c":"13.85 ","d":"14","e":"15.87"},"options_float":{"a":11.73,"b":12.0,"c":13.85,"d":14.0,"e":15.87},"annotated_formula":"divide(divide(multiply(207, subtract(const_100, 15)), const_100), 13)","linear_formula":"subtract(const_100,n2)|multiply(n1,#0)|divide(#1,const_100)|divide(#2,n0)|","chain":"100 - 15<\/gadget>\n85<\/output>\n207 * 85<\/gadget>\n17_595<\/output>\n17_595 \/ 100<\/gadget>\n3_519\/20 = around 175.95<\/output>\n(3_519\/20) \/ 13<\/gadget>\n3_519\/260 = around 13.534615<\/output>\n3_519\/260 = around 13.534615<\/result>","index":3192} +{"problem":"find the principle on a certain sum of money at 5 % per annum for 2 years if the amount being rs . 1120 ?","rationale":"\"1120 = p [ 1 + ( 5 * 2 ) \/ 100 ] p = 1018 answer : a\"","correct":"a","options":{"a":"1018 ","b":"1067 ","c":"1977 ","d":"1056","e":"1097"},"options_float":{"a":1018.0,"b":1067.0,"c":1977.0,"d":1056.0,"e":1097.0},"annotated_formula":"divide(1120, add(const_1, divide(multiply(5, 2), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n2,#2)|","chain":"5 * 2<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n1_120 \/ (11\/10)<\/gadget>\n11_200\/11 = around 1_018.181818<\/output>\n11_200\/11 = around 1_018.181818<\/result>","index":3194} +{"problem":"a man swims downstream 24 km and upstream 16 km taking 4 hours each time , what is the speed of the man in still water ?","rationale":"\"24 - - - 4 ds = 6 ? - - - - 1 16 - - - - 4 us = 4 ? - - - - 1 m = ? m = ( 6 + 4 ) \/ 2 = 5 answer : b\"","correct":"b","options":{"a":"8 ","b":"5 ","c":"24 ","d":"12","e":"36"},"options_float":{"a":8.0,"b":5.0,"c":24.0,"d":12.0,"e":36.0},"annotated_formula":"divide(add(divide(16, 4), divide(24, 4)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"16 \/ 4<\/gadget>\n4<\/output>\n24 \/ 4<\/gadget>\n6<\/output>\n4 + 6<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":3195} +{"problem":"when positive integer k is divided by 5 , the remainder is 2 . when k is divided by 6 , the remainder is 5 . if k is less than 38 , what is the remainder when k is divided by 7 ?","rationale":"\"cant think of a straight approach but here is how i solved it : k is divided by 5 and remainder is 2 . this means k = 5 n + 2 ( n is an integer ) so the possible values of k = { 2 , 7 , 12 , 17 , 22 , 27 , 32 , 37 } ( less than 38 ) secondly , if k is divided by 6 , the remainder is 5 = > k = 6 m + 5 so the possible value set for k = { 5 , 11 , 17 , 23 , 29,35 } ( less than 38 ) 17 is the only common number in both the sets . hence k = 17 answer : c\"","correct":"c","options":{"a":"2 ","b":"4 ","c":"3 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":4.0,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"reminder(add(const_12, 5), 7)","linear_formula":"add(n0,const_12)|reminder(#0,n5)|","chain":"12 + 5<\/gadget>\n17<\/output>\n17 % 7<\/gadget>\n3<\/output>\n3<\/result>","index":3196} +{"problem":"a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 6000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ?","rationale":"\"c ' s 1 day ' s work = 1 \/ 3 - ( 1 \/ 6 + 1 \/ 8 ) = 1 \/ 3 - 7 \/ 24 = 1 \/ 24 a ' s wages : b ' s wages : c ' s wages = 1 \/ 6 : 1 \/ 8 : 1 \/ 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 \/ 24 * 6000 ) = rs . 750 answer = d\"","correct":"d","options":{"a":"s . 375 ","b":"s . 400 ","c":"s . 600 ","d":"s . 750","e":"s . 850"},"options_float":{"a":375.0,"b":400.0,"c":600.0,"d":750.0,"e":850.0},"annotated_formula":"multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 6000), 3)","linear_formula":"inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/8) + (1\/6)<\/gadget>\n7\/24 = around 0.291667<\/output>\n(1\/3) - (7\/24)<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/24) * 6_000<\/gadget>\n250<\/output>\n250 * 3<\/gadget>\n750<\/output>\n750<\/result>","index":3197} +{"problem":"if an organization were to sell n tickets for a theater production , the total revenue from ticket sales would be 20 percent greater than the total costs of the production . if the organization actually sold all but 5 percent of the n tickets , the total revenue from ticket sales was what percent greater than the total costs of the production ?","rationale":"let cost of prod for 1 ticket = 1 so cop for n tickets = n so for selling n tickets , sp = 1.2 * n = 1.2 n so sp of 1 ticket = 1.2 sp of 95 % tickets = 1.2 * 0.95 n so profit % = ( 1.2 * 0.95 n - n ) \/ n * 100 = 0.14 * 100 = 14 % answer - c","correct":"c","options":{"a":"4 % ","b":"10 % ","c":"14 % ","d":"15 %","e":"18 %"},"options_float":{"a":4.0,"b":10.0,"c":14.0,"d":15.0,"e":18.0},"annotated_formula":"subtract(add(const_12, const_4), const_2)","linear_formula":"add(const_12,const_4)|subtract(#0,const_2)","chain":"12 + 4<\/gadget>\n16<\/output>\n16 - 2<\/gadget>\n14<\/output>\n14<\/result>","index":3201} +{"problem":"virginia , adrienne , and dennis have taught history for a combined total of 96 years . if virginia has taught for 9 more years than adrienne and for 12 fewer years than dennis , for how many years has dennis taught ?","rationale":"let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 12 = > a = ( d - 12 ) - 9 = d - 21 d - 9 + d - 21 + d = 96 = > 3 d = 96 + 30 = 126 = > d = 42 answer e","correct":"e","options":{"a":"23 ","b":"32 ","c":"35 ","d":"41","e":"42"},"options_float":{"a":23.0,"b":32.0,"c":35.0,"d":41.0,"e":42.0},"annotated_formula":"add(divide(subtract(96, add(add(9, 12), 9)), const_3), add(9, 12))","linear_formula":"add(n1,n2)|add(n1,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)","chain":"9 + 12<\/gadget>\n21<\/output>\n21 + 9<\/gadget>\n30<\/output>\n96 - 30<\/gadget>\n66<\/output>\n66 \/ 3<\/gadget>\n22<\/output>\n22 + 21<\/gadget>\n43<\/output>\n43<\/result>","index":3202} +{"problem":"an aeroplane covers a certain distance of 670 kmph in 6 hours . to cover the same distance in 6 1 \/ 4 hours , it must travel at a speed of","rationale":"\"speed of aeroplane = 670 kmph distance travelled in 6 hours = 670 * 6 = 4020 km speed of aeroplane to acver 4020 km in 25 \/ 4 = 4020 * 4 \/ 25 = 643 km answer c .\"","correct":"c","options":{"a":"440 ","b":"540 ","c":"643 ","d":"740","e":"250"},"options_float":{"a":440.0,"b":540.0,"c":643.0,"d":740.0,"e":250.0},"annotated_formula":"divide(multiply(670, 6), divide(add(multiply(4, 6), 1), 4))","linear_formula":"multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3)|","chain":"670 * 6<\/gadget>\n4_020<\/output>\n4 * 6<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n25 \/ 4<\/gadget>\n25\/4 = around 6.25<\/output>\n4_020 \/ (25\/4)<\/gadget>\n3_216\/5 = around 643.2<\/output>\n3_216\/5 = around 643.2<\/result>","index":3204} +{"problem":"a cube is divided into 216 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ?","rationale":"each side of the cube has 6 x 6 = 36 cubelets . only the interior cubelets are painted one colour . on each side , 4 x 4 = 16 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 16 = 96 the answer is c .","correct":"c","options":{"a":"72 ","b":"84 ","c":"96 ","d":"108","e":"120"},"options_float":{"a":72.0,"b":84.0,"c":96.0,"d":108.0,"e":120.0},"annotated_formula":"divide(subtract(216, multiply(multiply(const_4, const_2), const_3)), const_2)","linear_formula":"multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)","chain":"4 * 2<\/gadget>\n8<\/output>\n8 * 3<\/gadget>\n24<\/output>\n216 - 24<\/gadget>\n192<\/output>\n192 \/ 2<\/gadget>\n96<\/output>\n96<\/result>","index":3205} +{"problem":"a sum fetched a total simple interest of $ 4061.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?","rationale":"\"e 8926 principal = $ 100 x 4016.25 \/ 9 x 5 = $ 406125 \/ 45 = $ 8926 .\"","correct":"e","options":{"a":"$ 8829 ","b":"$ 2840 ","c":"$ 6578 ","d":"$ 7782","e":"$ 8926"},"options_float":{"a":8829.0,"b":2840.0,"c":6578.0,"d":7782.0,"e":8926.0},"annotated_formula":"divide(divide(multiply(4061.25, const_100), 9), 5)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"4_061.25 * 100<\/gadget>\n406_125<\/output>\n406_125 \/ 9<\/gadget>\n45_125<\/output>\n45_125 \/ 5<\/gadget>\n9_025<\/output>\n9_025<\/result>","index":3206} +{"problem":"the average of 5 consecutive odd numbers is 55 . what is the difference between the highest and lowest numbers ?","rationale":"explanation : let the numbers be x , x + 2 , x + 4 , x + 6 and x + 8 . then , ( x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) ) \/ 5 = 55 5 x + 20 = 275 x = 255 \/ 5 = 51 . so , required difference = ( 51 + 8 ) 51 = 8 answer : c","correct":"c","options":{"a":"2 ","b":"5 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":2.0,"b":5.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"multiply(subtract(5, const_1), const_2)","linear_formula":"subtract(n0,const_1)|multiply(#0,const_2)","chain":"5 - 1<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":3207} +{"problem":"in a class of 25 students in an examination in maths 3 students scored 95 marks each , 5 get zero each and the average of the rest was 45 . what is the average of the whole class ?","rationale":"\"explanation : total marks obtained by a class of 25 students = 3 * 95 + 5 * 0 + 17 * 45 = 1050 average marks of whole class = 1050 \/ 25 = 42 answer : option c\"","correct":"c","options":{"a":"47 ","b":"45.6 ","c":"42 ","d":"48","e":"49"},"options_float":{"a":47.0,"b":45.6,"c":42.0,"d":48.0,"e":49.0},"annotated_formula":"divide(add(multiply(45, subtract(25, add(5, 3))), multiply(95, 3)), 25)","linear_formula":"add(n1,n3)|multiply(n1,n2)|subtract(n0,#0)|multiply(n4,#2)|add(#3,#1)|divide(#4,n0)|","chain":"5 + 3<\/gadget>\n8<\/output>\n25 - 8<\/gadget>\n17<\/output>\n45 * 17<\/gadget>\n765<\/output>\n95 * 3<\/gadget>\n285<\/output>\n765 + 285<\/gadget>\n1_050<\/output>\n1_050 \/ 25<\/gadget>\n42<\/output>\n42<\/result>","index":3208} +{"problem":"a school has 4 section of chemistry in class x having 40 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .","rationale":"\"required average marks = 40 × 50 + 35 × 60 + 45 × 55 + 42 × 45 \/ 40 + 35 + 45 + 42 = 2000 + 2100 + 2475 + 1890 \/ 162 = 8465 ⁄ 162 = 52.25 answer b\"","correct":"b","options":{"a":"50.25 ","b":"52.25 ","c":"51.25 ","d":"53.25","e":"none of the above"},"options_float":{"a":50.25,"b":52.25,"c":51.25,"d":53.25,"e":null},"annotated_formula":"divide(add(add(add(multiply(40, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(40, 35), 45), 42))","linear_formula":"add(n1,n2)|multiply(n1,n5)|multiply(n2,n6)|multiply(n3,n7)|multiply(n3,n4)|add(#1,#2)|add(n3,#0)|add(#5,#3)|add(n4,#6)|add(#7,#4)|divide(#9,#8)|","chain":"40 * 50<\/gadget>\n2_000<\/output>\n35 * 60<\/gadget>\n2_100<\/output>\n2_000 + 2_100<\/gadget>\n4_100<\/output>\n45 * 55<\/gadget>\n2_475<\/output>\n4_100 + 2_475<\/gadget>\n6_575<\/output>\n42 * 45<\/gadget>\n1_890<\/output>\n6_575 + 1_890<\/gadget>\n8_465<\/output>\n40 + 35<\/gadget>\n75<\/output>\n75 + 45<\/gadget>\n120<\/output>\n120 + 42<\/gadget>\n162<\/output>\n8_465 \/ 162<\/gadget>\n8_465\/162 = around 52.253086<\/output>\n8_465\/162 = around 52.253086<\/result>","index":3210} +{"problem":"ajay can walk 7 km in 1 hour . in how many hours he can walk 80 km ?","rationale":"\"1 hour he walk 8 km he walk 80 km in = 80 \/ 7 * 1 = 11.4 hours answer is b\"","correct":"b","options":{"a":"5 hrs ","b":"11.4 hrs ","c":"15.5 hrs ","d":"20.4 hrs","e":"30 hrs"},"options_float":{"a":5.0,"b":11.4,"c":15.5,"d":20.4,"e":30.0},"annotated_formula":"divide(80, 7)","linear_formula":"divide(n2,n0)|","chain":"80 \/ 7<\/gadget>\n80\/7 = around 11.428571<\/output>\n80\/7 = around 11.428571<\/result>","index":3211} +{"problem":"the percentage profit earned by selling an article for rs . 1320 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ?","rationale":"\"c . p . be rs . x . then , ( 1320 - x ) \/ x * 100 = ( x - 1280 ) \/ x * 100 1320 - x = x - 1280 2 x = 2600 = > x = 1300 required s . p . = 125 % of rs . 1300 = 125 \/ 100 * 1300 = rs . 1625 . anser d\"","correct":"d","options":{"a":"3000 ","b":"1230 ","c":"2000 ","d":"1625","e":"3400"},"options_float":{"a":3000.0,"b":1230.0,"c":2000.0,"d":1625.0,"e":3400.0},"annotated_formula":"multiply(divide(add(const_100, 25), const_100), divide(add(1320, 1280), const_2))","linear_formula":"add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n1_320 + 1_280<\/gadget>\n2_600<\/output>\n2_600 \/ 2<\/gadget>\n1_300<\/output>\n(5\/4) * 1_300<\/gadget>\n1_625<\/output>\n1_625<\/result>","index":3212} +{"problem":"130 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ?","rationale":"\"1 women ' s 1 day work = 1 \/ 910 1 child ' s 1 day work = 1 \/ 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 \/ 910 + 10 \/ 140 ) = 1 \/ 13 5 women and 10 children will complete the work in 13 days . answer : e\"","correct":"e","options":{"a":"8 days ","b":"6 days ","c":"7 days ","d":"9 days","e":"13 days"},"options_float":{"a":8.0,"b":6.0,"c":7.0,"d":9.0,"e":13.0},"annotated_formula":"inverse(add(divide(5, multiply(130, 7)), divide(130, multiply(130, 14))))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)|","chain":"130 * 7<\/gadget>\n910<\/output>\n5 \/ 910<\/gadget>\n1\/182 = around 0.005495<\/output>\n130 * 14<\/gadget>\n1_820<\/output>\n130 \/ 1_820<\/gadget>\n1\/14 = around 0.071429<\/output>\n(1\/182) + (1\/14)<\/gadget>\n1\/13 = around 0.076923<\/output>\n1 \/ (1\/13)<\/gadget>\n13<\/output>\n13<\/result>","index":3214} +{"problem":"a grocer has a sale of rs . 3435 , rs . 3920 , rs . 3855 , rs . 4230 and rs . 3560 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 3500 ?","rationale":"\"total sale for 5 months = rs . ( 3435 + 3920 + 3855 + 4230 + 3560 ) = rs . 19000 . required sale = rs . [ ( 3500 x 6 ) - 19000 ] = rs . ( 21000 - 19000 ) = rs . 2000 answer : option e\"","correct":"e","options":{"a":"500 ","b":"1000 ","c":"6000 ","d":"7000","e":"2000"},"options_float":{"a":500.0,"b":1000.0,"c":6000.0,"d":7000.0,"e":2000.0},"annotated_formula":"subtract(multiply(add(5, const_1), 3500), add(add(add(add(3435, 3920), 3855), 4230), 3560))","linear_formula":"add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|","chain":"5 + 1<\/gadget>\n6<\/output>\n6 * 3_500<\/gadget>\n21_000<\/output>\n3_435 + 3_920<\/gadget>\n7_355<\/output>\n7_355 + 3_855<\/gadget>\n11_210<\/output>\n11_210 + 4_230<\/gadget>\n15_440<\/output>\n15_440 + 3_560<\/gadget>\n19_000<\/output>\n21_000 - 19_000<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":3215} +{"problem":"a number of 55 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ?","rationale":"to maximize # of boxes we should minimize marbles per box : 17 * 3 + 1 * 4 + 1 * 5 = 50 - - > 17 + 1 + 1 = 19 . answer : e .","correct":"e","options":{"a":"10 ","b":"12 ","c":"15 ","d":"16","e":"19"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":16.0,"e":19.0},"annotated_formula":"divide(55, 3)","linear_formula":"divide(n0,n1)","chain":"55 \/ 3<\/gadget>\n55\/3 = around 18.333333<\/output>\n55\/3 = around 18.333333<\/result>","index":3216} +{"problem":"the length of a side of a hexagon is 7 inches . what is the perimeter ?","rationale":"\"hexagon . it means 6 equal sides . p = 6 ( 7 ) = 42 inches answer b\"","correct":"b","options":{"a":"48 ","b":"42 ","c":"44 ","d":"45","e":"40"},"options_float":{"a":48.0,"b":42.0,"c":44.0,"d":45.0,"e":40.0},"annotated_formula":"multiply(7, multiply(const_3, const_2))","linear_formula":"multiply(const_2,const_3)|multiply(n0,#0)|","chain":"3 * 2<\/gadget>\n6<\/output>\n7 * 6<\/gadget>\n42<\/output>\n42<\/result>","index":3217} +{"problem":"how many numbers from 45 to 110 are exactly divisible by 12 ?","rationale":"\"option ' c ' 45 \/ 12 = 3 and 110 \/ 12 = 9 = = > 9 - 3 = 6 numbers\"","correct":"c","options":{"a":"5 ","b":"7 ","c":"6 ","d":"11","e":"12"},"options_float":{"a":5.0,"b":7.0,"c":6.0,"d":11.0,"e":12.0},"annotated_formula":"add(divide(subtract(multiply(floor(divide(110, 12)), 12), multiply(add(floor(divide(45, 12)), const_1), 12)), 12), const_1)","linear_formula":"divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)|","chain":"110 \/ 12<\/gadget>\n55\/6 = around 9.166667<\/output>\nfloor(55\/6)<\/gadget>\n9<\/output>\n9 * 12<\/gadget>\n108<\/output>\n45 \/ 12<\/gadget>\n15\/4 = around 3.75<\/output>\nfloor(15\/4)<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4 * 12<\/gadget>\n48<\/output>\n108 - 48<\/gadget>\n60<\/output>\n60 \/ 12<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6<\/result>","index":3219} +{"problem":"40 is subtracted from 60 % of a number , the result is 50 . find the number ?","rationale":"\"( 60 \/ 100 ) * x – 40 = 50 6 x = 900 x = 150 answer : a\"","correct":"a","options":{"a":"150 ","b":"330 ","c":"277 ","d":"279","e":"712"},"options_float":{"a":150.0,"b":330.0,"c":277.0,"d":279.0,"e":712.0},"annotated_formula":"divide(add(40, 50), divide(60, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"40 + 50<\/gadget>\n90<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n90 \/ (3\/5)<\/gadget>\n150<\/output>\n150<\/result>","index":3220} +{"problem":"a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 108 . find the number of notes of rs . 50 denomination .","rationale":"\"total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 108 - x now , 50 * x + 500 * ( 108 - x ) = 10350 50 x + 54000 - 500 x = 10350 - 450 x = - 43650 x = 97 . no . of 50 rupees note = 97 . answer : option e\"","correct":"e","options":{"a":"15 ","b":"21 ","c":"17 ","d":"19","e":"97"},"options_float":{"a":15.0,"b":21.0,"c":17.0,"d":19.0,"e":97.0},"annotated_formula":"divide(subtract(multiply(500, 108), 10350), subtract(500, 50))","linear_formula":"multiply(n2,n3)|subtract(n2,n1)|subtract(#0,n0)|divide(#2,#1)|","chain":"500 * 108<\/gadget>\n54_000<\/output>\n54_000 - 10_350<\/gadget>\n43_650<\/output>\n500 - 50<\/gadget>\n450<\/output>\n43_650 \/ 450<\/gadget>\n97<\/output>\n97<\/result>","index":3221} +{"problem":"a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day 5 miles per hour for 6 hours . how many miles in total did she walk ?","rationale":"first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 5 miles per hour and 6 hours - 30 miles total 18 + 20 + 30 = 68 answer : option a .","correct":"a","options":{"a":"68 ","b":"44 ","c":"58 ","d":"60","e":"62"},"options_float":{"a":68.0,"b":44.0,"c":58.0,"d":60.0,"e":62.0},"annotated_formula":"add(add(18, multiply(5, const_4)), multiply(5, 6))","linear_formula":"multiply(n3,const_4)|multiply(n3,n4)|add(n1,#0)|add(#2,#1)","chain":"5 * 4<\/gadget>\n20<\/output>\n18 + 20<\/gadget>\n38<\/output>\n5 * 6<\/gadget>\n30<\/output>\n38 + 30<\/gadget>\n68<\/output>\n68<\/result>","index":3222} +{"problem":"after a storm deposits 120 billion gallons of water into the city reservoir , the reservoir is 85 % full . if the original contents of the reservoir totaled 220 billion gallons , the reservoir was approximately what percentage full before the storm ?","rationale":"\"when the storm deposited 120 billion gallons , volume of water in the reservoir = 220 + 120 = 340 billion gallons if this is only 85 % of the capacity of the reservoir , the total capacity of the reservoir = 340 \/ 0.85 = 400 billion gallons therefore percentage of reservoir that was full before the storm = ( 220 \/ 400 ) * 100 = 55 % option c\"","correct":"c","options":{"a":"45 % ","b":"48 % ","c":"55 % ","d":"58 %","e":"65 %"},"options_float":{"a":45.0,"b":48.0,"c":55.0,"d":58.0,"e":65.0},"annotated_formula":"multiply(divide(220, divide(add(120, 220), divide(85, const_100))), const_100)","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|divide(n2,#2)|multiply(#3,const_100)|","chain":"120 + 220<\/gadget>\n340<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n340 \/ (17\/20)<\/gadget>\n400<\/output>\n220 \/ 400<\/gadget>\n11\/20 = around 0.55<\/output>\n(11\/20) * 100<\/gadget>\n55<\/output>\n55<\/result>","index":3224} +{"problem":"find the value of 72517 x 9999 = m ?","rationale":"\"72518 x 9999 = 72517 x ( 10000 - 1 ) = 72517 x 10000 - 72517 x 1 = 725170000 - 72517 = 725097483 d\"","correct":"d","options":{"a":"345434667 ","b":"246465757 ","c":"465767867 ","d":"725097483","e":"645354643"},"options_float":{"a":345434667.0,"b":246465757.0,"c":465767867.0,"d":725097483.0,"e":645354643.0},"annotated_formula":"multiply(subtract(9999, const_4), 72517)","linear_formula":"subtract(n1,const_4)|multiply(#0,n0)|","chain":"9_999 - 4<\/gadget>\n9_995<\/output>\n9_995 * 72_517<\/gadget>\n724_807_415<\/output>\n724_807_415<\/result>","index":3225} +{"problem":"10 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 2907 , how much was it in the beginning ?","rationale":"\"x * ( 90 \/ 100 ) * ( 85 \/ 100 ) = 2907 x = 3800 answer : a\"","correct":"a","options":{"a":"3800 ","b":"4200 ","c":"4400 ","d":"4500","e":"4600"},"options_float":{"a":3800.0,"b":4200.0,"c":4400.0,"d":4500.0,"e":4600.0},"annotated_formula":"floor(divide(2907, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 15), const_100))))","linear_formula":"subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|","chain":"100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n(9\/10) * (17\/20)<\/gadget>\n153\/200 = around 0.765<\/output>\n2_907 \/ (153\/200)<\/gadget>\n3_800<\/output>\nfloor(3_800)<\/gadget>\n3_800<\/output>\n3_800<\/result>","index":3226} +{"problem":"average of 5 numbers is 20 , and the sum of 3 of the numbers is 48 , what is the average of the other 2 numbers ?","rationale":"let the five numbers be a , b , c , d , e . then their average is ( a + b + c + d + e ) \/ 5 = 20 a + b + c = 48 ( 48 + d + e ) \/ 5 = 20 48 + d + e = 100 d + e = 52 average = 52 \/ 2 = 26 ans e","correct":"e","options":{"a":"24 ","b":"34 ","c":"29 ","d":"35","e":"26"},"options_float":{"a":24.0,"b":34.0,"c":29.0,"d":35.0,"e":26.0},"annotated_formula":"divide(subtract(multiply(20, 5), 48), 2)","linear_formula":"multiply(n0,n1)|subtract(#0,n3)|divide(#1,n4)","chain":"20 * 5<\/gadget>\n100<\/output>\n100 - 48<\/gadget>\n52<\/output>\n52 \/ 2<\/gadget>\n26<\/output>\n26<\/result>","index":3227} +{"problem":"the population of a city increases by 4 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ?","rationale":"\"actual increase in population = 3 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 3 \/ 100 ) ^ 3 = 109.2727 ∴ required percentage = 9.27 % answer : b\"","correct":"b","options":{"a":"9 % ","b":"9.27 % ","c":"27 % ","d":"12 %","e":"none of these"},"options_float":{"a":9.0,"b":9.27,"c":27.0,"d":12.0,"e":null},"annotated_formula":"add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(4, 1)), subtract(4, 1)), const_100), subtract(4, 1))), subtract(4, 1)), const_100), add(divide(multiply(add(const_100, subtract(4, 1)), subtract(4, 1)), const_100), subtract(4, 1)))","linear_formula":"subtract(n0,n1)|add(#0,const_100)|multiply(#1,#0)|divide(#2,const_100)|add(#3,#0)|add(#4,const_100)|multiply(#5,#0)|divide(#6,const_100)|add(#4,#7)|","chain":"4 - 1<\/gadget>\n3<\/output>\n100 + 3<\/gadget>\n103<\/output>\n103 * 3<\/gadget>\n309<\/output>\n309 \/ 100<\/gadget>\n309\/100 = around 3.09<\/output>\n(309\/100) + 3<\/gadget>\n609\/100 = around 6.09<\/output>\n100 + (609\/100)<\/gadget>\n10_609\/100 = around 106.09<\/output>\n(10_609\/100) * 3<\/gadget>\n31_827\/100 = around 318.27<\/output>\n(31_827\/100) \/ 100<\/gadget>\n31_827\/10_000 = around 3.1827<\/output>\n(31_827\/10_000) + (609\/100)<\/gadget>\n92_727\/10_000 = around 9.2727<\/output>\n92_727\/10_000 = around 9.2727<\/result>","index":3228} +{"problem":"a rectangle measures 8 cm on length and its diagonal measures 17 cm . what is the perimeter of the rectangle ?","rationale":"second side = √ 17 power 2 - 8 power 2 = √ 289 - 64 = 15 cm perimeter = 2 ( l + b ) = 2 ( 8 + 5 ) cm = 2 ( 23 ) = 46 cm answer is c .","correct":"c","options":{"a":"44 cm ","b":"42 cm ","c":"46 cm ","d":"49 cm","e":"41 cm"},"options_float":{"a":44.0,"b":42.0,"c":46.0,"d":49.0,"e":41.0},"annotated_formula":"rectangle_perimeter(sqrt(subtract(power(17, const_2), power(8, const_2))), 8)","linear_formula":"power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_perimeter(n0,#3)","chain":"17 ** 2<\/gadget>\n289<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n289 - 64<\/gadget>\n225<\/output>\n225 ** (1\/2)<\/gadget>\n15<\/output>\n2 * (15 + 8)<\/gadget>\n46<\/output>\n46<\/result>","index":3229} +{"problem":"a train running at the speed of 48 km \/ hr crosses a pole in 9 seconds . what is the length of the train ?","rationale":"\"speed = ( 48 x 5 \/ 18 ) m \/ sec = ( 40 \/ 3 ) m \/ sec . length of the train = ( speed x time ) . length of the train = ( 40 \/ 3 x 9 ) m = 120 m . answer is c .\"","correct":"c","options":{"a":"140 ","b":"130 ","c":"120 ","d":"170","e":"160"},"options_float":{"a":140.0,"b":130.0,"c":120.0,"d":170.0,"e":160.0},"annotated_formula":"multiply(divide(multiply(48, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"48 * 1_000<\/gadget>\n48_000<\/output>\n48_000 \/ 3_600<\/gadget>\n40\/3 = around 13.333333<\/output>\n(40\/3) * 9<\/gadget>\n120<\/output>\n120<\/result>","index":3233} +{"problem":"the average height of 35 boys in a class was calculated as 185 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ?","rationale":"\"calculated average height of 35 boys = 185 cm . wrong total height of 35 boys = 185 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 185 cm - ( 166 cm - 106 cm ) \/ 35 = 185 cm - 60 \/ 35 cm = 185 cm - 1.71 cm = 183.29 cm . answer : b\"","correct":"b","options":{"a":"178.27 cm ","b":"183.29 cm ","c":"978.29 cm ","d":"178.89 cm","e":"176.29 cm"},"options_float":{"a":178.27,"b":183.29,"c":978.29,"d":178.89,"e":176.29},"annotated_formula":"floor(divide(add(subtract(multiply(35, 185), 166), 106), 35))","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)|","chain":"35 * 185<\/gadget>\n6_475<\/output>\n6_475 - 166<\/gadget>\n6_309<\/output>\n6_309 + 106<\/gadget>\n6_415<\/output>\n6_415 \/ 35<\/gadget>\n1_283\/7 = around 183.285714<\/output>\nfloor(1_283\/7)<\/gadget>\n183<\/output>\n183<\/result>","index":3235} +{"problem":"the cost to park a car in a certain parking garage is $ 10.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ?","rationale":"\"total cost of parking for 10 hours = 10 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 10 + 7 * 1.75 = 22.25 thus the average parking price = 22.25 \/ 9 = 2.47 $ d is the correct answer .\"","correct":"d","options":{"a":"$ 1.09 ","b":"$ 1.67 ","c":"$ 2.225 ","d":"$ 2.47","e":"$ 2.50"},"options_float":{"a":1.09,"b":1.67,"c":2.225,"d":2.47,"e":2.5},"annotated_formula":"divide(add(10.00, multiply(1.75, subtract(9, 2))), 9)","linear_formula":"subtract(n4,n1)|multiply(n2,#0)|add(n0,#1)|divide(#2,n4)|","chain":"9 - 2<\/gadget>\n7<\/output>\n1.75 * 7<\/gadget>\n12.25<\/output>\n10 + 12.25<\/gadget>\n22.25<\/output>\n22.25 \/ 9<\/gadget>\n2.472222<\/output>\n2.472222<\/result>","index":3236} +{"problem":"at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 205 ?","rationale":"\"i think it should be c . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 7 20 - pack for 3.05 * 7 = $ 21.35 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 7 = 2140\"","correct":"c","options":{"a":"1,108 ","b":"2,100 ","c":"2,140 ","d":"2,124","e":"2,256"},"options_float":{"a":1108.0,"b":2100.0,"c":2140.0,"d":2124.0,"e":2256.0},"annotated_formula":"multiply(divide(205, 22.95), 250)","linear_formula":"divide(n6,n5)|multiply(n4,#0)|","chain":"205 \/ 22.95<\/gadget>\n8.932462<\/output>\n8.932462 * 250<\/gadget>\n2_233.1155<\/output>\n2_233.1155<\/result>","index":3237} +{"problem":"a man can row upstream at 25 kmph and downstream at 45 kmph , and then find the speed of the man in still water ?","rationale":"\"us = 25 ds = 45 m = ( 45 + 25 ) \/ 2 = 35 answer : e\"","correct":"e","options":{"a":"11 ","b":"77 ","c":"30 ","d":"88","e":"35"},"options_float":{"a":11.0,"b":77.0,"c":30.0,"d":88.0,"e":35.0},"annotated_formula":"divide(add(25, 45), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"25 + 45<\/gadget>\n70<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n35<\/result>","index":3240} +{"problem":"the least number which when divided by 5 , 6 , 7 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder , is :","rationale":"\"explanation : l . c . m . of 5 , 6 , 7 , 8 = 840 . { \\ color { blue } \\ therefore } required number is of the form 840 k + 3 least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . { \\ color { blue } \\ therefore } required number = ( 840 x 2 + 3 ) = 1683 . answer : b ) 1683\"","correct":"b","options":{"a":"2377 ","b":"1683 ","c":"2677 ","d":"9972","e":"1611"},"options_float":{"a":2377.0,"b":1683.0,"c":2677.0,"d":9972.0,"e":1611.0},"annotated_formula":"multiply(lcm(lcm(lcm(5, 6), 7), 8), const_2)","linear_formula":"lcm(n0,n1)|lcm(n2,#0)|lcm(n3,#1)|multiply(#2,const_2)|","chain":"lcm(5, 6)<\/gadget>\n30<\/output>\nlcm(30, 7)<\/gadget>\n210<\/output>\nlcm(210, 8)<\/gadget>\n840<\/output>\n840 * 2<\/gadget>\n1_680<\/output>\n1_680<\/result>","index":3241} +{"problem":"1 men takes 37 $ to eat a pizza in restaurant such that it have discount multiple of 4 if u eat more pizza . discount increase by multiple of 4 every pizza u eat . for 1 pizza 0 % off . for 2 pizza % off for 3 pizza 8 % off . pizza without discount is 10 $ . how much that man eats the pizza .","rationale":"solution : b . for 1 pizza 10 $ . for 2 pizza 10 $ + 9.6 $ ( 4 % off ) = 19.6 , for 3 pizza 19.6 + 9.2 ( 8 % off ) = 28.8 , for 4 pizza 28.8 + 8.8 ( 12 % off ) = 37.6 . so he have 37 so he cant eat 4 th pizza .","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"subtract(multiply(10, 4), 37)","linear_formula":"multiply(n2,n9)|subtract(#0,n1)","chain":"10 * 4<\/gadget>\n40<\/output>\n40 - 37<\/gadget>\n3<\/output>\n3<\/result>","index":3242} +{"problem":"if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 46 ?","rationale":"\"four wheeler = 11 * 4 = 44 ( max ) 2 wheel = 1 so no of 4 wheeler = 11 answer : a\"","correct":"a","options":{"a":"11 ","b":"12 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"divide(subtract(46, 2), 4)","linear_formula":"subtract(n3,n0)|divide(#0,n1)|","chain":"46 - 2<\/gadget>\n44<\/output>\n44 \/ 4<\/gadget>\n11<\/output>\n11<\/result>","index":3244} +{"problem":"a train 250 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ?","rationale":"\"d = 72 * 5 \/ 18 = 25 = 500 – 250 = 250 m answer : c\"","correct":"c","options":{"a":"150 m ","b":"200 m ","c":"250 m ","d":"270 m","e":"300 m"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":270.0,"e":300.0},"annotated_formula":"subtract(multiply(25, multiply(72, const_0_2778)), 250)","linear_formula":"multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n25 * 20<\/gadget>\n500<\/output>\n500 - 250<\/gadget>\n250<\/output>\n250<\/result>","index":3246} +{"problem":"a 9 % stock yields 8 % . the market value of the stock is :","rationale":"\"solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 9 , investment = rs . ( 100 \/ 8 x 9 ) = rs . 112.50 ∴ market value of rs . 100 stock = rs . 112.50 answer c\"","correct":"c","options":{"a":"rs . 72 ","b":"rs . 92 ","c":"rs . 112.50 ","d":"rs . 116.50","e":"none of these"},"options_float":{"a":72.0,"b":92.0,"c":112.5,"d":116.5,"e":null},"annotated_formula":"multiply(divide(const_100, 8), 9)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 8<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 9<\/gadget>\n225\/2 = around 112.5<\/output>\n225\/2 = around 112.5<\/result>","index":3247} +{"problem":"one side of a rectangular field is 15 m and one of its diagonal is 18 m . find the area of the field .","rationale":"\"solution other side = √ ( 18 ) 2 - ( 15 ) 2 = √ 324 - 225 = √ 99 = 9.9 m . ∴ area = ( 15 x 9.9 ) m 2 = 148.5 m 2 . answer d\"","correct":"d","options":{"a":"100 ","b":"120 ","c":"150 ","d":"148.5","e":"none"},"options_float":{"a":100.0,"b":120.0,"c":150.0,"d":148.5,"e":null},"annotated_formula":"rectangle_area(15, sqrt(subtract(power(18, const_2), power(15, const_2))))","linear_formula":"power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)|","chain":"18 ** 2<\/gadget>\n324<\/output>\n15 ** 2<\/gadget>\n225<\/output>\n324 - 225<\/gadget>\n99<\/output>\n99 ** (1\/2)<\/gadget>\n3*sqrt(11) = around 9.949874<\/output>\n15 * (3*sqrt(11))<\/gadget>\n45*sqrt(11) = around 149.248116<\/output>\n45*sqrt(11) = around 149.248116<\/result>","index":3248} +{"problem":"the average of temperatures at noontime from monday to friday is 60 ; the lowest one is 40 , what is the possible maximum range of the temperatures ?","rationale":"\"average = 60 , sum of temperatures = 60 * 5 = 300 as the min temperature is 40 , max would be 300 - 4 * 40 = 140 - - > the range = 140 ( max ) - 40 ( min ) = 100 answer : a\"","correct":"a","options":{"a":"100 ","b":"25 ","c":"40 ","d":"45","e":"75"},"options_float":{"a":100.0,"b":25.0,"c":40.0,"d":45.0,"e":75.0},"annotated_formula":"subtract(subtract(multiply(60, add(const_2, const_3)), multiply(40, const_4)), 40)","linear_formula":"add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1)|","chain":"2 + 3<\/gadget>\n5<\/output>\n60 * 5<\/gadget>\n300<\/output>\n40 * 4<\/gadget>\n160<\/output>\n300 - 160<\/gadget>\n140<\/output>\n140 - 40<\/gadget>\n100<\/output>\n100<\/result>","index":3250} +{"problem":"find out the square of a number which when doubled exceeds its one sixth by 11 ?","rationale":"\"a let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p \/ 6 ) + 11 = > 12 p = p + 66 = > p = 6 p ^ 2 = 6 ^ 2 = 36 . answer : a\"","correct":"a","options":{"a":"36 ","b":"25 ","c":"19 ","d":"26","e":"17"},"options_float":{"a":36.0,"b":25.0,"c":19.0,"d":26.0,"e":17.0},"annotated_formula":"power(divide(11, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2)","linear_formula":"add(const_1,const_4)|divide(const_1,#0)|subtract(const_2,#1)|divide(n0,#2)|power(#3,const_2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n2 - (1\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n11 \/ (9\/5)<\/gadget>\n55\/9 = around 6.111111<\/output>\n(55\/9) ** 2<\/gadget>\n3_025\/81 = around 37.345679<\/output>\n3_025\/81 = around 37.345679<\/result>","index":3251} +{"problem":"a train overtakes two persons who are walking in the same direction in which the train is going , at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . find the length of train ?","rationale":"\"explanation : first person speed = 2 * ( 5 \/ 18 ) = 5 \/ 9 m \/ sec second person speed = 4 * ( 5 \/ 18 ) = 10 \/ 9 m \/ sec let the length of train is x metre and speed is y m \/ sec then , x \/ y − 5 \/ 9 = 9 = > 9 y − 5 = x = > 9 y − x = 5 . . . . . ( i ) also , x \/ y − 10 \/ 9 = 10 90 y − 9 x = 100 . . . . . ( ii ) from ( i ) and ( ii ) , we get , x = 50 so length of train is 50 metre option b\"","correct":"b","options":{"a":"45 m ","b":"50 m ","c":"55 m ","d":"60 m","e":"65 m"},"options_float":{"a":45.0,"b":50.0,"c":55.0,"d":60.0,"e":65.0},"annotated_formula":"multiply(multiply(subtract(divide(subtract(multiply(2, divide(9, const_3600)), multiply(4, divide(10, const_3600))), subtract(divide(9, const_3600), divide(10, const_3600))), 2), divide(9, const_3600)), const_1000)","linear_formula":"divide(n2,const_3600)|divide(n3,const_3600)|multiply(n0,#0)|multiply(n1,#1)|subtract(#0,#1)|subtract(#2,#3)|divide(#5,#4)|subtract(#6,n0)|multiply(#0,#7)|multiply(#8,const_1000)|","chain":"9 \/ 3_600<\/gadget>\n1\/400 = around 0.0025<\/output>\n2 * (1\/400)<\/gadget>\n1\/200 = around 0.005<\/output>\n10 \/ 3_600<\/gadget>\n1\/360 = around 0.002778<\/output>\n4 * (1\/360)<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/200) - (1\/90)<\/gadget>\n-11\/1_800 = around -0.006111<\/output>\n(1\/400) - (1\/360)<\/gadget>\n-1\/3_600 = around -0.000278<\/output>\n(-11\/1_800) \/ (-1\/3_600)<\/gadget>\n22<\/output>\n22 - 2<\/gadget>\n20<\/output>\n20 * (1\/400)<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 1_000<\/gadget>\n50<\/output>\n50<\/result>","index":3252} +{"problem":"a number is doubled and 9 is added . if resultant is trebled , it becomes 63 . what is that number","rationale":"\"explanation : = > 3 ( 2 x + 9 ) = 63 = > 2 x + 9 = 21 = > x = 6 answer : option a\"","correct":"a","options":{"a":"6 ","b":"10 ","c":"12 ","d":"14","e":"15"},"options_float":{"a":6.0,"b":10.0,"c":12.0,"d":14.0,"e":15.0},"annotated_formula":"divide(subtract(63, multiply(const_3, 9)), multiply(const_3, const_2))","linear_formula":"multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)|","chain":"3 * 9<\/gadget>\n27<\/output>\n63 - 27<\/gadget>\n36<\/output>\n3 * 2<\/gadget>\n6<\/output>\n36 \/ 6<\/gadget>\n6<\/output>\n6<\/result>","index":3253} +{"problem":"a 80 cm long wire is to be cut into two pieces so that one piece will be 3 \/ 5 th of the other , how many centimeters will the shorter piece be ?","rationale":"\"explanation : 1 : 3 \/ 5 = 5 : 3 3 \/ 8 * 80 = 30 answer : option a\"","correct":"a","options":{"a":"30 ","b":"20 ","c":"83 ","d":"21","e":"52"},"options_float":{"a":30.0,"b":20.0,"c":83.0,"d":21.0,"e":52.0},"annotated_formula":"subtract(80, divide(80, add(divide(3, 5), const_1)))","linear_formula":"divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) + 1<\/gadget>\n8\/5 = around 1.6<\/output>\n80 \/ (8\/5)<\/gadget>\n50<\/output>\n80 - 50<\/gadget>\n30<\/output>\n30<\/result>","index":3255} +{"problem":"a waiter ' s salary consists of his salary and tips . during one week his tips were 7 \/ 4 of his salary . what fraction of his income came from tips ?","rationale":"\"income = salary ( s ) + tips = s + s * 7 \/ 4 = s * 11 \/ 4 tips = s * 7 \/ 4 fraction of his income came from tips = ( s * 7 \/ 4 ) \/ ( s * 11 \/ 4 ) = 7 \/ 11 answer : e\"","correct":"e","options":{"a":"4 \/ 9 ","b":"5 \/ 4 ","c":"5 \/ 8 ","d":"5 \/ 9","e":"7 \/ 11"},"options_float":{"a":0.4444444444,"b":1.25,"c":0.625,"d":0.5555555556,"e":0.6363636364},"annotated_formula":"divide(divide(7, 4), add(divide(7, 4), const_1))","linear_formula":"divide(n0,n1)|add(#0,const_1)|divide(#0,#1)|","chain":"7 \/ 4<\/gadget>\n7\/4 = around 1.75<\/output>\n(7\/4) + 1<\/gadget>\n11\/4 = around 2.75<\/output>\n(7\/4) \/ (11\/4)<\/gadget>\n7\/11 = around 0.636364<\/output>\n7\/11 = around 0.636364<\/result>","index":3259} +{"problem":"the total age of a and b is 15 years more than the total age of b and c . c is how many year younger than","rationale":"\"given that a + b = 15 + b + c = > a â € “ c = 15 + b â € “ b = 15 = > c is younger than a by 15 years answer : e\"","correct":"e","options":{"a":"14 years ","b":"12 years ","c":"56 years ","d":"66 years","e":"15 years"},"options_float":{"a":14.0,"b":12.0,"c":56.0,"d":66.0,"e":15.0},"annotated_formula":"multiply(15, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"15 * 1<\/gadget>\n15<\/output>\n15<\/result>","index":3260} +{"problem":"a bag has 9 green balls and 23 pink balls . how many green balls should be added to the bag so that the ratio of green balls and pink balls in the bag becomes 1 : 1 ?","rationale":"suppose x green balls need to be added . then , the number of green balls becomes 9 + x , and the number of pink balls remains the same , 23 . but now the ratio of green and pink balls is 1 : 1 this means , ( 9 + x ) \/ 23 = 1 \/ 1 solving this we get x = 14 . ans e","correct":"e","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"subtract(23, 9)","linear_formula":"subtract(n1,n0)","chain":"23 - 9<\/gadget>\n14<\/output>\n14<\/result>","index":3262} +{"problem":"what is remainder of the division ( 1625 * 1627 * 1629 ) \/ 12 ?","rationale":"remainder will be number \/ 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : a","correct":"a","options":{"a":"3 ","b":"2 ","c":"1 ","d":"0","e":"4"},"options_float":{"a":3.0,"b":2.0,"c":1.0,"d":0.0,"e":4.0},"annotated_formula":"subtract(multiply(multiply(1625, 1627), 1629), subtract(multiply(multiply(1625, 1627), 1629), const_3))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)","chain":"1_625 * 1_627<\/gadget>\n2_643_875<\/output>\n2_643_875 * 1_629<\/gadget>\n4_306_872_375<\/output>\n4_306_872_375 - 3<\/gadget>\n4_306_872_372<\/output>\n4_306_872_375 - 4_306_872_372<\/gadget>\n3<\/output>\n3<\/result>","index":3265} +{"problem":"if henry were to add 8 gallons of water to a tank that is already 3 \/ 4 full of water , the tank would be 7 \/ 8 full . how many gallons of water would the tank hold if it were full ?","rationale":"\"7 \/ 8 x - 3 \/ 4 x = 8 galls 1 \/ 8 * x = 8 gallons x = 64 gallons answer c\"","correct":"c","options":{"a":"25 ","b":"40 ","c":"64 ","d":"80","e":"96"},"options_float":{"a":25.0,"b":40.0,"c":64.0,"d":80.0,"e":96.0},"annotated_formula":"multiply(8, divide(const_1, subtract(divide(7, 8), divide(3, 4))))","linear_formula":"divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(const_1,#2)|multiply(n0,#3)|","chain":"7 \/ 8<\/gadget>\n7\/8 = around 0.875<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(7\/8) - (3\/4)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n8 * 8<\/gadget>\n64<\/output>\n64<\/result>","index":3266} +{"problem":"find compound interest on $ 4000 at 15 % per annum for 2 years 4 months , compounded annually .","rationale":"\"time = 2 years 4 months = 2 ( 4 \/ 12 ) years = 2 ( 1 \/ 3 ) years . amount = $ [ 4000 x ( 1 + ­ ( 15 \/ 100 ) ) 2 x ( 1 + ( ( 1 \/ 3 ) * 15 ) \/ 100 ) ] = $ [ 4000 * ( 23 \/ 20 ) * ( 23 \/ 20 ) * ( 21 \/ 20 ) ] = $ 5554.50 . : . c . i . = rs . ( 5554.50 - 4000 ) = $ 1554.50 answer a .\"","correct":"a","options":{"a":"1554.5 ","b":"1509.5 ","c":"1108.5 ","d":"1107.5","e":"1100.5"},"options_float":{"a":1554.5,"b":1509.5,"c":1108.5,"d":1107.5,"e":1100.5},"annotated_formula":"add(add(multiply(4000, divide(15, const_100)), multiply(add(4000, multiply(4000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(4000, multiply(4000, divide(15, const_100))), multiply(add(4000, multiply(4000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const_3)))","linear_formula":"divide(n1,const_100)|divide(#0,const_3)|multiply(n0,#0)|add(n0,#2)|multiply(#3,#0)|add(#2,#4)|add(#3,#4)|multiply(#6,#1)|add(#5,#7)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n4_000 * (3\/20)<\/gadget>\n600<\/output>\n4_000 + 600<\/gadget>\n4_600<\/output>\n4_600 * (3\/20)<\/gadget>\n690<\/output>\n600 + 690<\/gadget>\n1_290<\/output>\n4_600 + 690<\/gadget>\n5_290<\/output>\n(3\/20) \/ 3<\/gadget>\n1\/20 = around 0.05<\/output>\n5_290 * (1\/20)<\/gadget>\n529\/2 = around 264.5<\/output>\n1_290 + (529\/2)<\/gadget>\n3_109\/2 = around 1_554.5<\/output>\n3_109\/2 = around 1_554.5<\/result>","index":3267} +{"problem":"a train is 485 meter long is running at a speed of 45 km \/ hour . in what time will it pass a bridge of 140 meter length","rationale":"\"explanation : speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 485 + 140 = 625 meter time = distance \/ speed = 625 ∗ 2 \/ 25 = 50 seconds option d\"","correct":"d","options":{"a":"20 seconds ","b":"30 seconds ","c":"40 seconds ","d":"50 seconds","e":"none of these"},"options_float":{"a":20.0,"b":30.0,"c":40.0,"d":50.0,"e":null},"annotated_formula":"divide(add(485, 140), divide(multiply(45, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"485 + 140<\/gadget>\n625<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n625 \/ (25\/2)<\/gadget>\n50<\/output>\n50<\/result>","index":3269} +{"problem":"if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 1110 will be :","rationale":"\"z share = z , y = 1.2 z x = 1.25 ã — 1.2 z , x + y + z = 111 ( 1.25 ã — 1.2 + 1.2 + 1 ) z = 1110 3.7 z = 1110 , z = 300 answer : . a\"","correct":"a","options":{"a":"rs . 300 ","b":"rs . 200 ","c":"rs . 240 ","d":"rs . 350","e":"none of these"},"options_float":{"a":300.0,"b":200.0,"c":240.0,"d":350.0,"e":null},"annotated_formula":"divide(1110, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,#2)|add(#2,#4)|add(#5,const_1)|divide(n2,#6)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n(5\/4) * (6\/5)<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) + (6\/5)<\/gadget>\n27\/10 = around 2.7<\/output>\n(27\/10) + 1<\/gadget>\n37\/10 = around 3.7<\/output>\n1_110 \/ (37\/10)<\/gadget>\n300<\/output>\n300<\/result>","index":3270} +{"problem":"a train moves with a speed of 135 kmph . its speed in metres per second is","rationale":"\"solution speed = 135 kmph = ( 135 x 5 \/ 18 ) m \/ sec = 37.5 m \/ sec . answer b\"","correct":"b","options":{"a":"10.8 ","b":"37.5 ","c":"30 ","d":"38.8","e":"none"},"options_float":{"a":10.8,"b":37.5,"c":30.0,"d":38.8,"e":null},"annotated_formula":"multiply(135, const_0_2778)","linear_formula":"multiply(n0,const_0_2778)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n135 * (5\/18)<\/gadget>\n75\/2 = around 37.5<\/output>\n75\/2 = around 37.5<\/result>","index":3271} +{"problem":"a 50 ml after - shave lotion at 30 % alcohol is mixed with 30 ml of pure water . what is the percentage of alcohol in the new solution ?","rationale":"the amount of the final mixture is given by 50 ml + 30 ml = 80 ml the amount of alcohol is equal to the amount of alcohol in pure water ( which is 0 ) plus the amount of alcohol in the 30 % solution . let x be the percentage of alcohol in the final solution . hence 0 + 30 % 50 ml = x ( 80 ) solve for x x = 0.1817 = 18.75 % answer c","correct":"c","options":{"a":"16.75 % ","b":"17.75 % ","c":"18.75 % ","d":"19.75 %","e":"none of these"},"options_float":{"a":16.75,"b":17.75,"c":18.75,"d":19.75,"e":null},"annotated_formula":"multiply(divide(divide(multiply(30, 50), const_100), add(50, 30)), const_100)","linear_formula":"add(n0,n1)|multiply(n0,n1)|divide(#1,const_100)|divide(#2,#0)|multiply(#3,const_100)","chain":"30 * 50<\/gadget>\n1_500<\/output>\n1_500 \/ 100<\/gadget>\n15<\/output>\n50 + 30<\/gadget>\n80<\/output>\n15 \/ 80<\/gadget>\n3\/16 = around 0.1875<\/output>\n(3\/16) * 100<\/gadget>\n75\/4 = around 18.75<\/output>\n75\/4 = around 18.75<\/result>","index":3272} +{"problem":"what is the value of 11 ^ x - 11 ^ ( x + 2 ) , where x is the largest integer such that 11 ^ x is a factor of 30030 ?","rationale":"the given expression is 11 x - 11 ( x ^ + 2 ) = 11 x ( 1 - x ) 1 x needs to be factor of 30030 and x needs to be the largest integer possible . this means 11 x needs to be the largest factor possible . the largest factor of a number is the number itself . the largest factor of 30030 is 30030 = ( 11 * 2730 ) x must be 2730 the value of 11 x ( 1 - x ) = – 1320 . answer : b","correct":"b","options":{"a":"- 1331 ","b":"- 1320 ","c":"- 121 ","d":"- 120","e":"- 1"},"options_float":{"a":-1331.0,"b":-1320.0,"c":-121.0,"d":-120.0,"e":-1.0},"annotated_formula":"subtract(11, power(11, const_3))","linear_formula":"power(n0,const_3)|subtract(n0,#0)","chain":"11 ** 3<\/gadget>\n1_331<\/output>\n11 - 1_331<\/gadget>\n-1_320<\/output>\n-1_320<\/result>","index":3274} +{"problem":"a ’ s speed is 17 \/ 14 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ?","rationale":"\"let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d \/ ( 17 v \/ 14 ) = xd \/ v ( 14 \/ 17 ) * d \/ v = x * d \/ v x = 14 \/ 17 b should have a head start of 3 \/ 17 of the full distance . the answer is d .\"","correct":"d","options":{"a":"1 \/ 14 ","b":"1 \/ 17 ","c":"3 \/ 14 ","d":"3 \/ 17","e":"1 \/ 31"},"options_float":{"a":0.0714285714,"b":0.0588235294,"c":0.2142857143,"d":0.1764705882,"e":0.0322580645},"annotated_formula":"divide(subtract(17, 14), 17)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|","chain":"17 - 14<\/gadget>\n3<\/output>\n3 \/ 17<\/gadget>\n3\/17 = around 0.176471<\/output>\n3\/17 = around 0.176471<\/result>","index":3276} +{"problem":"a invested $ 100 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find a shares ?","rationale":"\"a : b = 100 * 12 : 200 * 6 a : b = 1 : 1 a ' s share = 100 \/ 2 = $ 50 answer is e\"","correct":"e","options":{"a":"$ 100 ","b":"$ 75 ","c":"$ 20 ","d":"$ 120","e":"$ 50"},"options_float":{"a":100.0,"b":75.0,"c":20.0,"d":120.0,"e":50.0},"annotated_formula":"multiply(100, subtract(const_1, divide(divide(200, const_2), add(100, divide(200, const_2)))))","linear_formula":"divide(n2,const_2)|add(n0,#0)|divide(#0,#1)|subtract(const_1,#2)|multiply(n3,#3)|","chain":"200 \/ 2<\/gadget>\n100<\/output>\n100 + 100<\/gadget>\n200<\/output>\n100 \/ 200<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n100 * (1\/2)<\/gadget>\n50<\/output>\n50<\/result>","index":3277} +{"problem":"in a family 13 people eat only vegetarian , 8 people eat only non veg . , 6 people eat both veg and non veg . . how many people eat veg in the family ?","rationale":"\"total people eat veg = only veg + both veg and non veg total = 13 + 6 = 19 answer = c\"","correct":"c","options":{"a":"20 ","b":"11 ","c":"19 ","d":"31","e":"21"},"options_float":{"a":20.0,"b":11.0,"c":19.0,"d":31.0,"e":21.0},"annotated_formula":"add(13, 6)","linear_formula":"add(n0,n2)|","chain":"13 + 6<\/gadget>\n19<\/output>\n19<\/result>","index":3278} +{"problem":"a shopkeeper labeled the price of his articles so as to earn a profit of 40 % on the cost price . he then sold the articles by offering a discount of 5 % on the labeled price . what is the actual percent profit earned in the deal ?","rationale":"\"explanation : let the cp of the article = rs . 100 . then labeled price = rs . 140 . sp = rs . 140 - 5 % of 140 = rs . 140 - 7 = rs . 133 . gain = rs . 133 â € “ rs . 100 = rs . 33 therefore , gain \/ profit percent = 33 % . answer : option d\"","correct":"d","options":{"a":"18 % ","b":"20 % ","c":"17 % ","d":"33 %","e":"none of these"},"options_float":{"a":18.0,"b":20.0,"c":17.0,"d":33.0,"e":null},"annotated_formula":"subtract(subtract(add(const_100, 40), multiply(add(const_100, 40), divide(5, const_100))), const_100)","linear_formula":"add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(#0,#2)|subtract(#3,const_100)|","chain":"100 + 40<\/gadget>\n140<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n140 * (1\/20)<\/gadget>\n7<\/output>\n140 - 7<\/gadget>\n133<\/output>\n133 - 100<\/gadget>\n33<\/output>\n33<\/result>","index":3279} +{"problem":"the total of 336 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is","rationale":"\"explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 336 - x ) . 0.20 * ( x ) + 0.25 ( 336 - x ) = 71 = > x = 260 . . answer : d ) 260\"","correct":"d","options":{"a":"238 ","b":"277 ","c":"278 ","d":"260","e":"288"},"options_float":{"a":238.0,"b":277.0,"c":278.0,"d":260.0,"e":288.0},"annotated_formula":"divide(subtract(multiply(336, 25), multiply(71, const_100)), subtract(25, 20))","linear_formula":"multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|","chain":"336 * 25<\/gadget>\n8_400<\/output>\n71 * 100<\/gadget>\n7_100<\/output>\n8_400 - 7_100<\/gadget>\n1_300<\/output>\n25 - 20<\/gadget>\n5<\/output>\n1_300 \/ 5<\/gadget>\n260<\/output>\n260<\/result>","index":3280} +{"problem":"a factory produces 6400 toys per week . if the workers at this factory work 3 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ?","rationale":"\"to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 3 days ) by 3 . 6400 \/ 3 = 2133 toys correct answer b\"","correct":"b","options":{"a":"1375 toys ","b":"2133 toys ","c":"3375 toys ","d":"4375 toys","e":"5375 toys"},"options_float":{"a":1375.0,"b":2133.0,"c":3375.0,"d":4375.0,"e":5375.0},"annotated_formula":"divide(6400, 3)","linear_formula":"divide(n0,n1)|","chain":"6_400 \/ 3<\/gadget>\n6_400\/3 = around 2_133.333333<\/output>\n6_400\/3 = around 2_133.333333<\/result>","index":3281} +{"problem":"walking at 5 \/ 6 th of its usual speed a cab is 15 mnts late . find its usual time to cover the journey ?","rationale":"new speed = 5 \/ 6 th of usual speed new time = 6 \/ 5 th of usual time 6 \/ 5 ut - ut = 15 m ut \/ 5 = 15 m ut = 75 m answer is d","correct":"d","options":{"a":"25 m ","b":"45 m ","c":"32 m ","d":"75 m","e":"62 m"},"options_float":{"a":25.0,"b":45.0,"c":32.0,"d":75.0,"e":62.0},"annotated_formula":"multiply(5, 15)","linear_formula":"multiply(n0,n2)","chain":"5 * 15<\/gadget>\n75<\/output>\n75<\/result>","index":3282} +{"problem":"a person purchased a tv set for rs . 16000 and a dvd player for rs . 6250 . he sold both the items together for rs . 35600 . what percentage of profit did he make ?","rationale":"the total cp = rs . 16000 + rs . 6250 = rs . 22250 and sp = rs . 35600 profit ( % ) = ( 35600 - 22250 ) \/ 22250 * 100 = 60 % answer : d","correct":"d","options":{"a":"80 % ","b":"49 % ","c":"40 % ","d":"60 %","e":"90 %"},"options_float":{"a":80.0,"b":49.0,"c":40.0,"d":60.0,"e":90.0},"annotated_formula":"multiply(subtract(divide(35600, add(16000, 6250)), const_1), const_100)","linear_formula":"add(n0,n1)|divide(n2,#0)|subtract(#1,const_1)|multiply(#2,const_100)","chain":"16_000 + 6_250<\/gadget>\n22_250<\/output>\n35_600 \/ 22_250<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) - 1<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n60<\/result>","index":3284} +{"problem":"two trains each 180 m in length are running on the same parallel lines in opposite directions with the speed of each 80 kmph . in what time will they cross each other completely ?","rationale":"d = 180 m + 180 m = 360 m * 1 \/ 1000 = 0.36 kms rs = 80 + 80 = 160 kmph t = ( 0.36 \/ 160 ) * 3600 = 8.1 sec answer : b","correct":"b","options":{"a":"7.8 sec ","b":"8.1 sec ","c":"8.3 sec ","d":"8.2 sec","e":"8.4 sec"},"options_float":{"a":7.8,"b":8.1,"c":8.3,"d":8.2,"e":8.4},"annotated_formula":"divide(180, multiply(80, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n80 * (5\/18)<\/gadget>\n200\/9 = around 22.222222<\/output>\n180 \/ (200\/9)<\/gadget>\n81\/10 = around 8.1<\/output>\n81\/10 = around 8.1<\/result>","index":3285} +{"problem":"20 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ?","rationale":"\"capacity of the tank = 20 × 13.5 = 270 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 270 ⁄ 9 = 30 answer a\"","correct":"a","options":{"a":"30 ","b":"32 ","c":"60 ","d":"data inadequate","e":"none of these"},"options_float":{"a":30.0,"b":32.0,"c":60.0,"d":null,"e":null},"annotated_formula":"divide(multiply(13.5, 20), 9)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"13.5 * 20<\/gadget>\n270<\/output>\n270 \/ 9<\/gadget>\n30<\/output>\n30<\/result>","index":3286} +{"problem":"the number of the members of a club is more than 50 and less than 80 . when 4 people sit at a table , other people exactly divided to 8 - people groups ( 8 people sit at a table ) or 7 - people groups ( 7 people sit at a table ) . if the members are divided to 9 - people groups , how many people will be left ?","rationale":"\"the number of members is 7 k + 4 = 8 j + 4 the only number in this range which satisfies this is 60 . 60 \/ 9 = 6 ( 9 ) + 6 the answer is d .\"","correct":"d","options":{"a":"0 ","b":"2 ","c":"4 ","d":"6","e":"8"},"options_float":{"a":0.0,"b":2.0,"c":4.0,"d":6.0,"e":8.0},"annotated_formula":"reminder(add(lcm(8, 7), 4), 9)","linear_formula":"lcm(n3,n5)|add(n2,#0)|reminder(#1,n7)|","chain":"lcm(8, 7)<\/gadget>\n56<\/output>\n56 + 4<\/gadget>\n60<\/output>\n60 % 9<\/gadget>\n6<\/output>\n6<\/result>","index":3287} +{"problem":"a train running at the speed of 72 km \/ hr crosses a pole in 9 seconds . find the length of the train .","rationale":"\"speed = 72 * ( 5 \/ 18 ) m \/ sec = 20 m \/ sec length of train ( distance ) = speed * time ( 20 ) * 9 = 180 meter answer : c\"","correct":"c","options":{"a":"150 meter ","b":"286 meter ","c":"180 meter ","d":"288 meter","e":"265 meter"},"options_float":{"a":150.0,"b":286.0,"c":180.0,"d":288.0,"e":265.0},"annotated_formula":"multiply(divide(multiply(72, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"72 * 1_000<\/gadget>\n72_000<\/output>\n72_000 \/ 3_600<\/gadget>\n20<\/output>\n20 * 9<\/gadget>\n180<\/output>\n180<\/result>","index":3288} +{"problem":"two clocks are set correctly at 1 0 ' clock in the afternoon . one keeps correct time . the other gains 5 seconds per minute . that evening , when the fast clock shows 6 : 20 , what does the correct clock show ?","rationale":"faster clock gains 5 seconds per minute or 5 * 60 = 300 seconds or 5 minutes per hour . both clocks are set at 1 : 00 at 2 : 00 correct clock is at 2 : 00 faster clock is at 2 : 05 : 00 ( 1 hour + gained 5 minutes ) at 3 : 00 - correct clock is at 3 : 00 faster clock is at 3 : 10 : 00 ( 2 hours + gained 10 minutes ) . in 5 hours , faster clock would gain 20 minutes . it will show 6 : 20 . at that point , the correct clock would show 6 : 00 . answer is c","correct":"c","options":{"a":"4.0 ","b":"5.0 ","c":"6.0 ","d":"7.2","e":"3.5"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.2,"e":3.5},"annotated_formula":"add(1, divide(add(multiply(5, const_3600), multiply(20, const_60)), add(const_3600, multiply(5, const_60))))","linear_formula":"multiply(n2,const_3600)|multiply(n4,const_60)|multiply(n2,const_60)|add(#0,#1)|add(#2,const_3600)|divide(#3,#4)|add(n0,#5)","chain":"5 * 3_600<\/gadget>\n18_000<\/output>\n20 * 60<\/gadget>\n1_200<\/output>\n18_000 + 1_200<\/gadget>\n19_200<\/output>\n5 * 60<\/gadget>\n300<\/output>\n3_600 + 300<\/gadget>\n3_900<\/output>\n19_200 \/ 3_900<\/gadget>\n64\/13 = around 4.923077<\/output>\n1 + (64\/13)<\/gadget>\n77\/13 = around 5.923077<\/output>\n77\/13 = around 5.923077<\/result>","index":3289} +{"problem":"the triplicate ratio of 1 : 3 is ?","rationale":"\"1 ^ 3 : 3 ^ 3 = 1 : 27 answer : a\"","correct":"a","options":{"a":"1 : 27 ","b":"1 : 8 ","c":"1 : 3 ","d":"1 : 1","e":"1 : 2"},"options_float":{"a":0.037037037,"b":0.125,"c":0.3333333333,"d":1.0,"e":0.5},"annotated_formula":"divide(power(1, 3), power(3, 3))","linear_formula":"power(n0,n1)|power(n1,n1)|divide(#0,#1)|","chain":"1 ** 3<\/gadget>\n1<\/output>\n3 ** 3<\/gadget>\n27<\/output>\n1 \/ 27<\/gadget>\n1\/27 = around 0.037037<\/output>\n1\/27 = around 0.037037<\/result>","index":3290} +{"problem":"a shopkeeper sold an article offering a discount of 5 % and earned a profit of 22.55 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"\"let c . p . be rs . 100 . then , s . p . = rs . 122.55 let marked price be rs . x . then , 95 \/ 100 x = 122.55 x = 12255 \/ 95 = rs . 129 now , s . p . = rs . 129 , c . p . = rs . 100 profit % = 29 % . answer : b\"","correct":"b","options":{"a":"60 % ","b":"29 % ","c":"39 % ","d":"56 %","e":"73 %"},"options_float":{"a":60.0,"b":29.0,"c":39.0,"d":56.0,"e":73.0},"annotated_formula":"subtract(divide(multiply(add(const_100, 22.55), const_100), subtract(const_100, 5)), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)|","chain":"100 + 22.55<\/gadget>\n122.55<\/output>\n122.55 * 100<\/gadget>\n12_255<\/output>\n100 - 5<\/gadget>\n95<\/output>\n12_255 \/ 95<\/gadget>\n129<\/output>\n129 - 100<\/gadget>\n29<\/output>\n29<\/result>","index":3291} +{"problem":"5 men are equal to as many women as are equal to 8 boys . all of them earn $ 90 only . men ’ s wages are ?","rationale":"\"b $ 6 5 m = xw = 8 b 5 m + xw + 8 b - - - - - $ 90 rs . 5 m + 5 m + 5 m - - - - - $ 90 rs . 15 m - - - - - - $ 90 = > 1 m = $ 6 .\"","correct":"b","options":{"a":"$ 4 ","b":"$ 6 ","c":"$ 7 ","d":"$ 9","e":"$ 3"},"options_float":{"a":4.0,"b":6.0,"c":7.0,"d":9.0,"e":3.0},"annotated_formula":"divide(90, multiply(const_3, 5))","linear_formula":"multiply(n0,const_3)|divide(n2,#0)|","chain":"3 * 5<\/gadget>\n15<\/output>\n90 \/ 15<\/gadget>\n6<\/output>\n6<\/result>","index":3292} +{"problem":"two trains of equal are running on parallel lines in the same direction at 48 km \/ hr and 36 km \/ hr . the faster train passes the slower train in 36 sec . the length of each train is ?","rationale":"\"let the length of each train be x m . then , distance covered = 2 x m . relative speed = 48 - 36 = 12 km \/ hr . = 12 * 5 \/ 18 = 10 \/ 3 m \/ sec . 2 x \/ 36 = 10 \/ 3 = > x = 60 . answer : a\"","correct":"a","options":{"a":"60 m ","b":"72 m ","c":"80 m ","d":"82 m","e":"84 m"},"options_float":{"a":60.0,"b":72.0,"c":80.0,"d":82.0,"e":84.0},"annotated_formula":"divide(multiply(36, divide(multiply(subtract(48, 36), const_1000), const_3600)), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|","chain":"48 - 36<\/gadget>\n12<\/output>\n12 * 1_000<\/gadget>\n12_000<\/output>\n12_000 \/ 3_600<\/gadget>\n10\/3 = around 3.333333<\/output>\n36 * (10\/3)<\/gadget>\n120<\/output>\n120 \/ 2<\/gadget>\n60<\/output>\n60<\/result>","index":3293} +{"problem":"a number is said to be prime saturated if the product of all the different positive prime factors of t is less than the square root of t . what is the greatest two digit prime saturated integer ?","rationale":"\"t = 96 = 3 * 32 = 3 * 2 ^ 5 answer is d .\"","correct":"d","options":{"a":"99 ","b":"98 ","c":"97 ","d":"96","e":"95"},"options_float":{"a":99.0,"b":98.0,"c":97.0,"d":96.0,"e":95.0},"annotated_formula":"add(multiply(const_2, const_3), subtract(const_100, const_10))","linear_formula":"multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)|","chain":"2 * 3<\/gadget>\n6<\/output>\n100 - 10<\/gadget>\n90<\/output>\n6 + 90<\/gadget>\n96<\/output>\n96<\/result>","index":3295} +{"problem":"a cycle is bought for rs . 1500 and sold for rs . 1620 , find the gain percent ?","rationale":"\"1500 - - - - 120 100 - - - - ? = > 8 % answer : b\"","correct":"b","options":{"a":"11 ","b":"8 ","c":"10 ","d":"20","e":"12"},"options_float":{"a":11.0,"b":8.0,"c":10.0,"d":20.0,"e":12.0},"annotated_formula":"multiply(divide(subtract(1620, 1500), 1500), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_620 - 1_500<\/gadget>\n120<\/output>\n120 \/ 1_500<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) * 100<\/gadget>\n8<\/output>\n8<\/result>","index":3296} +{"problem":"a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 76 , how many questions did student a answer correctly ?","rationale":"\"let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 76 ( subtracting twice of incorrect from correct ) 3 x = 276 x = 92 answer : a\"","correct":"a","options":{"a":"92 ","b":"60 ","c":"73 ","d":"82","e":"91"},"options_float":{"a":92.0,"b":60.0,"c":73.0,"d":82.0,"e":91.0},"annotated_formula":"subtract(100, divide(subtract(100, 76), const_3))","linear_formula":"subtract(n0,n1)|divide(#0,const_3)|subtract(n0,#1)|","chain":"100 - 76<\/gadget>\n24<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n100 - 8<\/gadget>\n92<\/output>\n92<\/result>","index":3297} +{"problem":"ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 11 percent . overall he make a .","rationale":"\"let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 \/ 100 ) = 15000 - 600 m = 8000 ( 1 + 11 \/ 100 ) = 8000 + 880 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 880 = rs . 280 as this is positive , an overall profit of rs . 280 was made . answer : e\"","correct":"e","options":{"a":"228 ","b":"288 ","c":"27 ","d":"200","e":"280"},"options_float":{"a":228.0,"b":288.0,"c":27.0,"d":200.0,"e":280.0},"annotated_formula":"subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(11, const_100)))), add(15000, 8000))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 - (1\/25)<\/gadget>\n24\/25 = around 0.96<\/output>\n15_000 * (24\/25)<\/gadget>\n14_400<\/output>\n11 \/ 100<\/gadget>\n11\/100 = around 0.11<\/output>\n1 + (11\/100)<\/gadget>\n111\/100 = around 1.11<\/output>\n8_000 * (111\/100)<\/gadget>\n8_880<\/output>\n14_400 + 8_880<\/gadget>\n23_280<\/output>\n15_000 + 8_000<\/gadget>\n23_000<\/output>\n23_280 - 23_000<\/gadget>\n280<\/output>\n280<\/result>","index":3298} +{"problem":"in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 5 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ?","rationale":"\"since 5 % of the customers who sample the candyare notcaught , then 88 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.95 = 0.22 ; { % of customers who sample candy } = 0.231 . answer : b .\"","correct":"b","options":{"a":"22 % ","b":"23 % ","c":"24 % ","d":"25 %","e":"34 %"},"options_float":{"a":22.0,"b":23.0,"c":24.0,"d":25.0,"e":34.0},"annotated_formula":"divide(22, divide(subtract(const_100, 5), const_100))","linear_formula":"subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n22 \/ (19\/20)<\/gadget>\n440\/19 = around 23.157895<\/output>\n440\/19 = around 23.157895<\/result>","index":3299} +{"problem":"p , q and r have $ 4000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ?","rationale":"\"e 1600 let the amount with r be $ r r = 2 \/ 3 ( total amount with p and q ) r = 2 \/ 3 ( 4000 - r ) = > 3 r = 8000 - 2 r = > 5 r = 8000 = > r = 1600 .\"","correct":"e","options":{"a":"2400 ","b":"2403 ","c":"3998 ","d":"2539","e":"1600"},"options_float":{"a":2400.0,"b":2403.0,"c":3998.0,"d":2539.0,"e":1600.0},"annotated_formula":"divide(multiply(4000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3)))","linear_formula":"multiply(const_2,const_2)|multiply(const_1,const_3)|divide(#0,const_3)|multiply(n0,#0)|multiply(#2,const_3)|add(#4,#1)|add(#5,#1)|divide(#3,#6)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4_000 * 4<\/gadget>\n16_000<\/output>\n4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) * 3<\/gadget>\n4<\/output>\n1 * 3<\/gadget>\n3<\/output>\n4 + 3<\/gadget>\n7<\/output>\n7 + 3<\/gadget>\n10<\/output>\n16_000 \/ 10<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":3302} +{"problem":"rahul and sham together can complete a task in 35 days , but rahul alone can complete same work in 60 days . calculate in how many days sham can complete this work ?","rationale":"explanation : as rahul and sham together can finish work in 35 days . 1 days work of rahul and sham is 1 \/ 35 rahul can alone complete this work in 60 days , so , rahul one day work is 1 \/ 60 clearly , sham one day work will be = ( rahul and sham one day work ) - ( rahul one day work ) = 1 \/ 35 - 1 \/ 60 = 1 \/ 84 hence sham will complete the given work in 84 days . answer is a","correct":"a","options":{"a":"84 days ","b":"82 days ","c":"76 days ","d":"68 days","e":"60 days"},"options_float":{"a":84.0,"b":82.0,"c":76.0,"d":68.0,"e":60.0},"annotated_formula":"inverse(subtract(inverse(35), inverse(60)))","linear_formula":"inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)","chain":"1 \/ 35<\/gadget>\n1\/35 = around 0.028571<\/output>\n1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n(1\/35) - (1\/60)<\/gadget>\n1\/84 = around 0.011905<\/output>\n1 \/ (1\/84)<\/gadget>\n84<\/output>\n84<\/result>","index":3303} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 6 sec . what is the length of the train ?","rationale":"\"speed = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec length of the train = speed * time = 50 \/ 3 * 6 = 100 m answer : d\"","correct":"d","options":{"a":"356 ","b":"278 ","c":"127 ","d":"100","e":"150"},"options_float":{"a":356.0,"b":278.0,"c":127.0,"d":100.0,"e":150.0},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 6)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"60 * 1_000<\/gadget>\n60_000<\/output>\n60_000 \/ 3_600<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 6<\/gadget>\n100<\/output>\n100<\/result>","index":3304} +{"problem":"lucy bought a 1 - year , 10000 certificate of deposit that paid interest at an annual rate of 8 percent compunded semiannually . what was the total amount of interest paid on this certificate at maturity ?","rationale":"solution : it says , 8 percent per annum per compounded semi annually . therefore r = 8 \/ 2 = 4 % . so , p = 10000 r = 4 n = 2 net amount a = 10816 , i = 800 . answer : d","correct":"d","options":{"a":"10464 ","b":"864 ","c":"816 ","d":"800","e":"480"},"options_float":{"a":10464.0,"b":864.0,"c":816.0,"d":800.0,"e":480.0},"annotated_formula":"subtract(multiply(10000, power(add(const_1, divide(divide(8, const_100), const_2)), const_2)), 10000)","linear_formula":"divide(n2,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_2)|multiply(n1,#3)|subtract(#4,n1)","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) \/ 2<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n10_000 * (676\/625)<\/gadget>\n10_816<\/output>\n10_816 - 10_000<\/gadget>\n816<\/output>\n816<\/result>","index":3305} +{"problem":"a bag marked at $ 125 is sold for $ 120 . the rate of discount is ?","rationale":"rate of discount = 5 \/ 125 * 100 = 4 % answer is e","correct":"e","options":{"a":"10 % ","b":"25 % ","c":"20 % ","d":"50 %","e":"4 %"},"options_float":{"a":10.0,"b":25.0,"c":20.0,"d":50.0,"e":4.0},"annotated_formula":"multiply(divide(subtract(125, 120), 125), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)","chain":"125 - 120<\/gadget>\n5<\/output>\n5 \/ 125<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) * 100<\/gadget>\n4<\/output>\n4<\/result>","index":3307} +{"problem":"the radius of a cylindrical vessel is 8 cm and height is 3 cm . find the whole surface of the cylinder ?","rationale":"\"r = 8 h = 3 2 π r ( h + r ) = 2 * 22 \/ 7 * 8 ( 11 ) = 553.1 answer : d\"","correct":"d","options":{"a":"308 sq cm ","b":"220 sq cm ","c":"440 sq cm ","d":"553.1 sq cm","e":"138 sq cm"},"options_float":{"a":308.0,"b":220.0,"c":440.0,"d":553.1,"e":138.0},"annotated_formula":"surface_cylinder(8, 3)","linear_formula":"surface_cylinder(n0,n1)|","chain":"2 * pi * 8 * (8 + 3)<\/gadget>\n176*pi = around 552.920307<\/output>\n176*pi = around 552.920307<\/result>","index":3308} +{"problem":"the marks obtained by vijay and amith are in the ratio 6 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ?","rationale":"\"6 : 5 3 : 2 - - - - - - - 18 : 15 : 10 18 : 10 9 : 5 answer : c\"","correct":"c","options":{"a":"1 : 5 ","b":"4 : 5 ","c":"9 : 5 ","d":"9 : 8","e":"9 : 7"},"options_float":{"a":0.2,"b":0.8,"c":1.8,"d":1.125,"e":1.2857142857},"annotated_formula":"divide(multiply(6, 3), multiply(5, 2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)|","chain":"6 * 3<\/gadget>\n18<\/output>\n5 * 2<\/gadget>\n10<\/output>\n18 \/ 10<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":3309} +{"problem":"right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 66 years . what is molly ' s age right now ?","rationale":"\"now , sandy is 66 - 6 = 60 molly ' s age is ( 3 \/ 4 ) * 60 = 45 the answer is d .\"","correct":"d","options":{"a":"15 years ","b":"35 years ","c":"25 years ","d":"45 years","e":"55 years"},"options_float":{"a":15.0,"b":35.0,"c":25.0,"d":45.0,"e":55.0},"annotated_formula":"multiply(divide(subtract(66, 6), 4), 3)","linear_formula":"subtract(n3,n2)|divide(#0,n0)|multiply(n1,#1)|","chain":"66 - 6<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45<\/result>","index":3312} +{"problem":"( 935421 x 625 ) = ?","rationale":"\"explanation : 935421 × 625 = 935421 × 54 = 935421 × ( 10 \/ 2 ) 4 = ( 935421 × 10000 ) \/ 16 = 584638125 . answer : option d\"","correct":"d","options":{"a":"542622125 ","b":"584632125 ","c":"544638125 ","d":"584638125","e":"none of these"},"options_float":{"a":542622125.0,"b":584632125.0,"c":544638125.0,"d":584638125.0,"e":null},"annotated_formula":"multiply(935421, power(add(const_4, const_1), const_4))","linear_formula":"add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 ** 4<\/gadget>\n625<\/output>\n935_421 * 625<\/gadget>\n584_638_125<\/output>\n584_638_125<\/result>","index":3314} +{"problem":"bert and rebecca were looking at the price of a condominium . the price of the condominium was 100 % more than bert had in savings , and separately , the same price was also 60 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings .","rationale":"\"suppose bert had 100 so price becomes 200 , this 200 = 1.6 times r ' s saving . . so r ' s saving becomes 125 so required ratio is 100 : 125 = 4 : 5 answer : d\"","correct":"d","options":{"a":"1 : 5 ","b":"2 : 5 ","c":"3 : 5 ","d":"4 : 5","e":"1 : 1"},"options_float":{"a":0.2,"b":0.4,"c":0.6,"d":0.8,"e":1.0},"annotated_formula":"divide(divide(const_100, add(const_100, 100)), divide(const_100, add(const_100, 60)))","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(const_100,#0)|divide(const_100,#1)|divide(#2,#3)|","chain":"100 + 100<\/gadget>\n200<\/output>\n100 \/ 200<\/gadget>\n1\/2 = around 0.5<\/output>\n100 + 60<\/gadget>\n160<\/output>\n100 \/ 160<\/gadget>\n5\/8 = around 0.625<\/output>\n(1\/2) \/ (5\/8)<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":3317} +{"problem":"a man swims downstream 30 km and upstream 20 km taking 5 hours each time , what is the speed of the man in still water ?","rationale":"\"30 - - - 5 ds = 6 ? - - - - 1 20 - - - - 5 us = 4 ? - - - - 1 m = ? m = ( 6 + 4 ) \/ 2 = 5 answer : d\"","correct":"d","options":{"a":"7 ","b":"8 ","c":"2 ","d":"5","e":"4"},"options_float":{"a":7.0,"b":8.0,"c":2.0,"d":5.0,"e":4.0},"annotated_formula":"divide(add(divide(20, 5), divide(30, 5)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"20 \/ 5<\/gadget>\n4<\/output>\n30 \/ 5<\/gadget>\n6<\/output>\n4 + 6<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":3318} +{"problem":"andrew purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ?","rationale":"\"cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 560 + 490 = 1055 b\"","correct":"b","options":{"a":"1000 ","b":"1055 ","c":"1060 ","d":"1075","e":"1080"},"options_float":{"a":1000.0,"b":1055.0,"c":1060.0,"d":1075.0,"e":1080.0},"annotated_formula":"add(multiply(8, 70), multiply(9, 55))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"8 * 70<\/gadget>\n560<\/output>\n9 * 55<\/gadget>\n495<\/output>\n560 + 495<\/gadget>\n1_055<\/output>\n1_055<\/result>","index":3319} +{"problem":"how many integers v are prime numbers in the range 200 < v < 220 ?","rationale":"my friend , every single odd number greater than can be written either as 4 v + 1 or as 4 v + 3 . if you divide any odd number by 4 , you will get a remainder of either 1 or 3 . that ' s not a rule unique to prime numbers at all . the 6 v + 1 or 6 v - 1 rule is basically every odd number that is not divisible by three , so it narrows the search a little . here ' s how i thought about the problem . first , eliminate all the even numbers and the odd multiples of 5 in that range . that leaves us with : { 201 , 203 , 207 , 209 , 211 , 213 , 217 , 219 } eliminate the four multiples of 3 . notice that 21 is a multiple of 3 , so 210 is also a multiple of 3 . if we add or subtract 3 or 9 , we get more multiples of three . when we eliminate those , we are left with . { 203 , 209 , 211 , 217 } now , notice that a cool thing about this range is that 210 is also a multiple 7 ( again , because 21 is a multiple of 7 ) . this means that 210 - 7 = 203 210 + 7 = 217 those two numbers are also multiples of 7 , so eliminate them from the list . now , we are left with { 209 , 211 } . we ' ve already checked all the prime numbers less than 10 , so we know that neither of these numbers is divisible by anything less than 10 . we have to check 11 now . we know that 22 is a multiple of 11 , so 220 is also a multiple of 11 . this means that 220 - 11 = 209 is also a multiple of 11 . we can eliminate this from the list also . that leaves us with just 211 . there ' s no zero option in the question , so this must be a prime number . answer = ( a )","correct":"a","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(divide(subtract(subtract(220, 200), const_12), const_4), const_1)","linear_formula":"subtract(n1,n0)|subtract(#0,const_12)|divide(#1,const_4)|subtract(#2,const_1)","chain":"220 - 200<\/gadget>\n20<\/output>\n20 - 12<\/gadget>\n8<\/output>\n8 \/ 4<\/gadget>\n2<\/output>\n2 - 1<\/gadget>\n1<\/output>\n1<\/result>","index":3320} +{"problem":"if the average of 20 digits is 500 and the average of 14 of them is 390 , calculate the average of the remaining 6 numbers ?","rationale":"\"explanation : total of the 20 digits - 20 * 500 = 10000 total of the 14 digits - 14 * 390 = 5460 total of the remaining 6 digits - 10000 - 5460 = 4540 average of the remaining 6 numbers = 4540 \/ 6 = 756.7 answer : a\"","correct":"a","options":{"a":"756.7 ","b":"751.7 ","c":"753.7 ","d":"752.7","e":"722.7"},"options_float":{"a":756.7,"b":751.7,"c":753.7,"d":752.7,"e":722.7},"annotated_formula":"divide(subtract(multiply(20, 500), multiply(14, 390)), 6)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)|","chain":"20 * 500<\/gadget>\n10_000<\/output>\n14 * 390<\/gadget>\n5_460<\/output>\n10_000 - 5_460<\/gadget>\n4_540<\/output>\n4_540 \/ 6<\/gadget>\n2_270\/3 = around 756.666667<\/output>\n2_270\/3 = around 756.666667<\/result>","index":3321} +{"problem":"two bullet train s 140 m and 180 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is :","rationale":"\"relative speed = ( 60 + 40 ) km \/ hr = 100 x 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = ( 140 + 180 ) m = 320 m . required time = 320 x 9 \/ 250 = 11.52 sec . answer b\"","correct":"b","options":{"a":"15.8 sec . ","b":"11.52 sec . ","c":"11.8 sec . ","d":"10.8 sec .","e":"08.8 sec ."},"options_float":{"a":15.8,"b":11.52,"c":11.8,"d":10.8,"e":8.8},"annotated_formula":"divide(add(140, 180), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"140 + 180<\/gadget>\n320<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n320 \/ (250\/9)<\/gadget>\n288\/25 = around 11.52<\/output>\n288\/25 = around 11.52<\/result>","index":3322} +{"problem":"a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 25 per kg , then his gain percent is","rationale":"\"explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 25 ) = rs . 200 . profit = 200 - 150 = 50 so , profit % = ( 50 \/ 150 ) * 100 = 33 % option b\"","correct":"b","options":{"a":"12 % ","b":"33 % ","c":"14 % ","d":"15 %","e":"16 %"},"options_float":{"a":12.0,"b":33.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"divide(multiply(subtract(multiply(25, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20)))","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3)|","chain":"5 + 3<\/gadget>\n8<\/output>\n25 * 8<\/gadget>\n200<\/output>\n5 * 18<\/gadget>\n90<\/output>\n3 * 20<\/gadget>\n60<\/output>\n90 + 60<\/gadget>\n150<\/output>\n200 - 150<\/gadget>\n50<\/output>\n50 * 100<\/gadget>\n5_000<\/output>\n5_000 \/ 150<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":3323} +{"problem":"if p \/ q = 4 \/ 5 , then the value of 25 \/ 7 + { ( 2 q - p ) \/ ( 2 q + p ) } is ?","rationale":"\"answer given exp . = 25 \/ 7 + { ( 2 q - p ) \/ ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 25 \/ 7 + { 2 - p \/ q ) \/ ( 2 + p \/ q ) } = 25 \/ 7 + { ( 2 - 4 \/ 5 ) \/ ( 2 + 4 \/ 5 ) } = 25 \/ 7 + 6 \/ 14 = 25 \/ 7 + 3 \/ 7 = 28 \/ 7 = 4 correct option : a\"","correct":"a","options":{"a":"4 ","b":"34 ","c":"1 ","d":"2","e":"3"},"options_float":{"a":4.0,"b":34.0,"c":1.0,"d":2.0,"e":3.0},"annotated_formula":"add(divide(25, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5))))","linear_formula":"divide(n2,n3)|divide(n0,n1)|add(n4,#1)|subtract(n4,#1)|divide(#3,#2)|add(#0,#4)|","chain":"25 \/ 7<\/gadget>\n25\/7 = around 3.571429<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n2 - (4\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n2 + (4\/5)<\/gadget>\n14\/5 = around 2.8<\/output>\n(6\/5) \/ (14\/5)<\/gadget>\n3\/7 = around 0.428571<\/output>\n(25\/7) + (3\/7)<\/gadget>\n4<\/output>\n4<\/result>","index":3326} +{"problem":"a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price r of the machine ?","rationale":"\"since the wholesale price was $ 90 and the profit was 20 % of the wholesale price ( [ . 2 ] [ 90 ] = $ 18 ) , the retail price would have to be above $ 108 , but not that much greater than that . let ' s start by testing answer c : $ 120 . . . . if . . . . . retail price r = $ 120 10 % discount off = $ 120 - ( . 1 ) ( 120 ) = 120 - 12 = 108 20 % profit on wholesale price = 90 + ( . 2 ) ( 90 ) = 90 + 18 = 108 these two numbers match , so this must be the answer ! final answer : [ reveal ] spoiler : c\"","correct":"c","options":{"a":"81 ","b":"100 ","c":"120 ","d":"135","e":"160"},"options_float":{"a":81.0,"b":100.0,"c":120.0,"d":135.0,"e":160.0},"annotated_formula":"divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10))","linear_formula":"multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)|","chain":"90 * 20<\/gadget>\n1_800<\/output>\n1_800 \/ 100<\/gadget>\n18<\/output>\n90 + 18<\/gadget>\n108<\/output>\n108 * 100<\/gadget>\n10_800<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n10_800 \/ 90<\/gadget>\n120<\/output>\n120<\/result>","index":3327} +{"problem":"bottle r contains 350 capsules and costs $ 6.25 . bottle t contains 130 capsules and costs $ 2.99 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ?","rationale":"\"cost per capsule in r is 6.25 \/ 350 = 0.0178 cost per capsule in t is 2.99 \/ 130 = 0.023 the difference is 0.0052 the answer is c\"","correct":"c","options":{"a":"$ 0.25 ","b":"$ 0.12 ","c":"$ 0.0052 ","d":"$ 0.003","e":"$ 0.002"},"options_float":{"a":0.25,"b":0.12,"c":0.0052,"d":0.003,"e":0.002},"annotated_formula":"subtract(divide(2.99, 130), divide(6.25, 350))","linear_formula":"divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|","chain":"2.99 \/ 130<\/gadget>\n0.023<\/output>\n6.25 \/ 350<\/gadget>\n0.017857<\/output>\n0.023 - 0.017857<\/gadget>\n0.005143<\/output>\n0.005143<\/result>","index":3328} +{"problem":"the milk level in a rectangular box measuring 58 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons )","rationale":"\"6 inches = 1 \/ 2 feet ( there are 12 inches in a foot . ) , so 58 * 25 * 1 \/ 2 = 725 feet ^ 3 of milk must be removed , which equals to 725 * 7.5 = 5437.5 gallons . answer : d .\"","correct":"d","options":{"a":"100 ","b":"250 ","c":"750 ","d":"5437.5","e":"5635.5"},"options_float":{"a":100.0,"b":250.0,"c":750.0,"d":5437.5,"e":5635.5},"annotated_formula":"multiply(multiply(multiply(58, 25), divide(1, const_2)), 7.5)","linear_formula":"divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)|","chain":"58 * 25<\/gadget>\n1_450<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1_450 * (1\/2)<\/gadget>\n725<\/output>\n725 * 7.5<\/gadget>\n5_437.5<\/output>\n5_437.5<\/result>","index":3329} +{"problem":"a rectangular block 15 cm by 30 cm by 75 cm is cut into an exact number of equal cubes . find the least possible number of cubes ?","rationale":"volume of the block = 15 * 30 * 75 = 33750 cm ^ 3 side of the largest cube = h . c . f of 15 , 30,75 = 15 cm volume of the cube = 15 * 15 * 15 = 3375 cm ^ 3 number of cubes = 33750 \/ 3375 = 10 answer is d","correct":"d","options":{"a":"6 ","b":"40 ","c":"15 ","d":"10","e":"22"},"options_float":{"a":6.0,"b":40.0,"c":15.0,"d":10.0,"e":22.0},"annotated_formula":"divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(75, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(75, const_10))), const_10))","linear_formula":"multiply(const_1000,const_3)|multiply(n2,const_10)|multiply(const_2,const_3)|rectangle_area(const_1000,const_360)|add(#0,#1)|divide(#3,const_10)|multiply(#2,const_1000)|subtract(#5,#6)|add(#4,#7)|divide(#8,const_10)|divide(#8,#9)","chain":"360 * 1_000<\/gadget>\n360_000<\/output>\n360_000 \/ 10<\/gadget>\n36_000<\/output>\n3 * 2<\/gadget>\n6<\/output>\n1_000 * 6<\/gadget>\n6_000<\/output>\n36_000 - 6_000<\/gadget>\n30_000<\/output>\n3 * 1_000<\/gadget>\n3_000<\/output>\n75 * 10<\/gadget>\n750<\/output>\n3_000 + 750<\/gadget>\n3_750<\/output>\n30_000 + 3_750<\/gadget>\n33_750<\/output>\n33_750 \/ 10<\/gadget>\n3_375<\/output>\n33_750 \/ 3_375<\/gadget>\n10<\/output>\n10<\/result>","index":3332} +{"problem":"after replacing an old member by a new member , it was found that the average age of 5 members of a club is the same as it was 3 years ago . what is the difference between the ages of the replaced and the new member ?","rationale":"explanation : i ) let the ages of the five members at present be a , b , c , d & e years . and the age of the new member be f years . ii ) so the new average of five members ' age = ( a + b + c + d + f ) \/ 5 - - - - - - - ( 1 ) iii ) their corresponding ages 3 years ago = ( a - 3 ) , ( b - 3 ) , ( c - 3 ) , ( d - 3 ) & ( e - 3 ) years so their average age 3 years ago = ( a + b + c + d + e - 15 ) \/ 5 = x - - - - - ( 2 ) = = > a + b + c + d + e = 5 x + 15 = = > a + b + c + d = 5 x + 15 - e - - - - - - ( 3 ) iv ) substituting this value of a + b + c + d = 5 x + 15 - e in ( 1 ) above , the new average is : ( 5 x + 15 - e + f ) \/ 5 equating this to the average age of x years , 3 yrs , ago as in ( 2 ) above , ( 5 x + 15 - e + f ) \/ 5 = x = = > ( 5 x + 15 - e + f ) = 5 x solving e - f = 15 years . thus the difference of ages between replaced and new member = 15 years . answer : d","correct":"d","options":{"a":"12 ","b":"13 ","c":"14 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"multiply(5, 3)","linear_formula":"multiply(n0,n1)","chain":"5 * 3<\/gadget>\n15<\/output>\n15<\/result>","index":3333} +{"problem":"zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 26.1 kg of zinc has been consumed in it ?","rationale":"\"sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 26.1 kg zinc , mixture , melted = [ 20 \/ 9 x 26.1 ] kg = 58 kg . answer a\"","correct":"a","options":{"a":"58 kg ","b":"60 kg ","c":"64 kg ","d":"70 kg","e":"none"},"options_float":{"a":58.0,"b":60.0,"c":64.0,"d":70.0,"e":null},"annotated_formula":"add(multiply(divide(11, 9), 26.1), 26.1)","linear_formula":"divide(n1,n0)|multiply(n2,#0)|add(n2,#1)|","chain":"11 \/ 9<\/gadget>\n11\/9 = around 1.222222<\/output>\n(11\/9) * 26.1<\/gadget>\n31.9<\/output>\n31.9 + 26.1<\/gadget>\n58<\/output>\n58<\/result>","index":3334} +{"problem":"the ratio of boys to girls is 6 : 4.60 % of the boys and 40 % of girls take lunch in the canteen . what % of class takes lunch ?","rationale":"let total no of students 100 , boy 60 and girl 40 , so lunch taken by boy and girl respectively 60 * 60 \/ 100 = 36 and 40 * 40 \/ 100 = 16 so total 52 students takes lunch out of 100 students . . so required % is 52 % answer : b","correct":"b","options":{"a":"51 % ","b":"52 % ","c":"53 % ","d":"54 %","e":"55 %"},"options_float":{"a":51.0,"b":52.0,"c":53.0,"d":54.0,"e":55.0},"annotated_formula":"add(multiply(divide(6, const_10), subtract(const_100, 40)), multiply(divide(const_4, const_10), 40))","linear_formula":"divide(n0,const_10)|divide(const_4,const_10)|subtract(const_100,n2)|multiply(#0,#2)|multiply(n2,#1)|add(#3,#4)","chain":"6 \/ 10<\/gadget>\n3\/5 = around 0.6<\/output>\n100 - 40<\/gadget>\n60<\/output>\n(3\/5) * 60<\/gadget>\n36<\/output>\n4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 40<\/gadget>\n16<\/output>\n36 + 16<\/gadget>\n52<\/output>\n52<\/result>","index":3335} +{"problem":"a person x started at 3 hours earlier at 40 km \/ h from a place p , then another person y followed him at 60 km \/ h . started his journey at 3 o ' clock , afternoon . what is the diference in time when x was 30 km ahead of y and when y was 30 km ahead of x ?","rationale":"explanation : time ( when x was 30 km ahead of y ) = ( 120 - 30 ) \/ 20 = 4.5 h time ( when y was 30 km ahead of x ) = ( 120 + 30 ) \/ 20 = 7.5 h thus , required difference in time = 3 h answer : b","correct":"b","options":{"a":"8 h ","b":"3 h ","c":"6 h ","d":"3 h","e":"1 h"},"options_float":{"a":8.0,"b":3.0,"c":6.0,"d":3.0,"e":1.0},"annotated_formula":"subtract(divide(add(multiply(40, 3), 30), divide(40, const_2)), divide(subtract(multiply(40, 3), 30), divide(40, const_2)))","linear_formula":"divide(n1,const_2)|multiply(n0,n1)|add(n4,#1)|subtract(#1,n4)|divide(#2,#0)|divide(#3,#0)|subtract(#4,#5)","chain":"40 * 3<\/gadget>\n120<\/output>\n120 + 30<\/gadget>\n150<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n150 \/ 20<\/gadget>\n15\/2 = around 7.5<\/output>\n120 - 30<\/gadget>\n90<\/output>\n90 \/ 20<\/gadget>\n9\/2 = around 4.5<\/output>\n(15\/2) - (9\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":3336} +{"problem":"a room of 5 m 44 cm long and 3 m 74 cm broad is to be paved with squre tiles . find the least number of squre tiles required to cover the floor ?","rationale":"\"area of the room = 544 * 374 sq cm size of largest square tile = h . c . f of 544 cm and 374 cm = 34 cm area of 1 tile = 34 * 34 sq cm no . of tiles required = ( 544 * 374 ) \/ ( 34 * 34 ) = 176 answer : d\"","correct":"d","options":{"a":"178 ","b":"128 ","c":"127 ","d":"176","e":"171"},"options_float":{"a":178.0,"b":128.0,"c":127.0,"d":176.0,"e":171.0},"annotated_formula":"divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), power(divide(add(multiply(5, const_100), 44), power(const_2, const_4)), const_2))","linear_formula":"multiply(n0,const_100)|multiply(n2,const_100)|power(const_2,const_4)|add(n1,#0)|add(n3,#1)|divide(#3,#2)|multiply(#3,#4)|power(#5,const_2)|divide(#6,#7)|","chain":"5 * 100<\/gadget>\n500<\/output>\n500 + 44<\/gadget>\n544<\/output>\n3 * 100<\/gadget>\n300<\/output>\n300 + 74<\/gadget>\n374<\/output>\n544 * 374<\/gadget>\n203_456<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n544 \/ 16<\/gadget>\n34<\/output>\n34 ** 2<\/gadget>\n1_156<\/output>\n203_456 \/ 1_156<\/gadget>\n176<\/output>\n176<\/result>","index":3337} +{"problem":"if 35 % of a number is 24 less than 50 % of that number , then the number is ?","rationale":"explanation : let the number be n . then , 50 % of n - 35 % of n = 24 50 \/ 100 n - 35 \/ 100 n = 24 n = ( 24 x 100 ) \/ 15 = 160 . answer c","correct":"c","options":{"a":"180 ","b":"450 ","c":"160 ","d":"170","e":"185"},"options_float":{"a":180.0,"b":450.0,"c":160.0,"d":170.0,"e":185.0},"annotated_formula":"divide(24, divide(subtract(50, 35), const_100))","linear_formula":"subtract(n2,n0)|divide(#0,const_100)|divide(n1,#1)","chain":"50 - 35<\/gadget>\n15<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n24 \/ (3\/20)<\/gadget>\n160<\/output>\n160<\/result>","index":3339} +{"problem":"tom opened a shop investing rs . 3000 . jose joined him 2 months later , investing rs . 4500 . they earned a profit of rs . 5400 after completion of one year . what will be jose ' s share of profit ?","rationale":"\"sol = ~ s - so anju ’ s share = [ 5 \/ 9 ] x 5400 = 3000 a\"","correct":"a","options":{"a":"3000 ","b":"2200 ","c":"2309 ","d":"2400","e":"3200"},"options_float":{"a":3000.0,"b":2200.0,"c":2309.0,"d":2400.0,"e":3200.0},"annotated_formula":"multiply(5400, subtract(const_1, divide(multiply(3000, multiply(2, multiply(2, const_3))), add(multiply(4500, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(3000, multiply(2, multiply(2, const_3)))))))","linear_formula":"multiply(n1,const_3)|multiply(#0,n1)|multiply(n0,#1)|subtract(#1,n1)|multiply(n2,#3)|add(#4,#2)|divide(#2,#5)|subtract(const_1,#6)|multiply(n3,#7)|","chain":"2 * 3<\/gadget>\n6<\/output>\n2 * 6<\/gadget>\n12<\/output>\n3_000 * 12<\/gadget>\n36_000<\/output>\n12 - 2<\/gadget>\n10<\/output>\n4_500 * 10<\/gadget>\n45_000<\/output>\n45_000 + 36_000<\/gadget>\n81_000<\/output>\n36_000 \/ 81_000<\/gadget>\n4\/9 = around 0.444444<\/output>\n1 - (4\/9)<\/gadget>\n5\/9 = around 0.555556<\/output>\n5_400 * (5\/9)<\/gadget>\n3_000<\/output>\n3_000<\/result>","index":3340} +{"problem":"there are 5 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ?","rationale":"number of ways of selecting 2 fiction books = 5 c 2 number of ways of selecting 2 non fiction books = 6 c 2 5 c 2 * 6 c 2 = 10 * 15 = 150 answer : c","correct":"c","options":{"a":"90 ","b":"120 ","c":"150 ","d":"180","e":"200"},"options_float":{"a":90.0,"b":120.0,"c":150.0,"d":180.0,"e":200.0},"annotated_formula":"divide(multiply(multiply(5, const_4), multiply(6, 5)), power(factorial(const_2), const_2))","linear_formula":"factorial(const_2)|multiply(n0,const_4)|multiply(n0,n1)|multiply(#1,#2)|power(#0,const_2)|divide(#3,#4)","chain":"5 * 4<\/gadget>\n20<\/output>\n6 * 5<\/gadget>\n30<\/output>\n20 * 30<\/gadget>\n600<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n600 \/ 4<\/gadget>\n150<\/output>\n150<\/result>","index":3342} +{"problem":"the area of a triangle is with base 4 m and height 6 m ?","rationale":"\"1 \/ 2 * 4 * 6 = 12 m 2 answer : e\"","correct":"e","options":{"a":"11 m 2 ","b":"10 m 2 ","c":"18 m 2 ","d":"19 m 2","e":"12 m 2"},"options_float":{"a":11.0,"b":10.0,"c":18.0,"d":19.0,"e":12.0},"annotated_formula":"triangle_area(4, 6)","linear_formula":"triangle_area(n0,n1)|","chain":"(4 * 6) \/ 2<\/gadget>\n12<\/output>\n12<\/result>","index":3344} +{"problem":"if 40 sheep eat 40 bags of grass in 40 days . in how many days one sheep will eat one bag of grass ?","rationale":"explanation : let the required number of days be x . less sheep , more days ( indirect proportion ) less bags , less days ( direct proportion ) sheep 1 : 40 : : 40 : x bags 40 : 1 1 x 40 x x = 40 x 1 x 40 x = 40 . answer : c","correct":"c","options":{"a":"1 ","b":"20 ","c":"40 ","d":"80","e":"70"},"options_float":{"a":1.0,"b":20.0,"c":40.0,"d":80.0,"e":70.0},"annotated_formula":"divide(divide(multiply(multiply(40, 40), 40), 40), 40)","linear_formula":"multiply(n0,n0)|multiply(n0,#0)|divide(#1,n0)|divide(#2,n0)","chain":"40 * 40<\/gadget>\n1_600<\/output>\n1_600 * 40<\/gadget>\n64_000<\/output>\n64_000 \/ 40<\/gadget>\n1_600<\/output>\n1_600 \/ 40<\/gadget>\n40<\/output>\n40<\/result>","index":3345} +{"problem":"increasing the original price of an article by 20 percent and then increasing the new price by 20 percent is equivalent to increasing the original price by","rationale":"\"1.2 * 1.2 * x = 1.44 * x the answer is e .\"","correct":"e","options":{"a":"40 % ","b":"41 % ","c":"42 % ","d":"43 %","e":"44 %"},"options_float":{"a":40.0,"b":41.0,"c":42.0,"d":43.0,"e":44.0},"annotated_formula":"multiply(const_100, subtract(multiply(add(const_1, divide(20, const_100)), add(const_1, divide(20, const_100))), const_1))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * (6\/5)<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n100 * (11\/25)<\/gadget>\n44<\/output>\n44<\/result>","index":3346} +{"problem":"the age of somu is one - third his father ' s . 6 years back he was one - fifth of his father ' s age . what is his persent age ?","rationale":"\"explanation : let somu ' s age be x and that of his father be 3 x . so , x - 6 = 3 x - 6 \/ 5 = x = 12 answer : option d\"","correct":"d","options":{"a":"11 ","b":"13 ","c":"14 ","d":"12","e":"10"},"options_float":{"a":11.0,"b":13.0,"c":14.0,"d":12.0,"e":10.0},"annotated_formula":"divide(subtract(multiply(add(const_4, const_1), 6), 6), subtract(add(const_4, const_1), const_3))","linear_formula":"add(const_1,const_4)|multiply(n0,#0)|subtract(#0,const_3)|subtract(#1,n0)|divide(#3,#2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 6<\/gadget>\n30<\/output>\n30 - 6<\/gadget>\n24<\/output>\n5 - 3<\/gadget>\n2<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12<\/result>","index":3347} +{"problem":"john builds a square with a side of 12 cm from some wire . if from the same wire , he builds a rectangle with a width of 6 cm , what will be the length of the rectangle ?","rationale":"solution : if the side of the square is 12 cm , then its perimetre will be 48 cm . we build the rectangle from the same wire , so it will have the same perimetre . its width is 6 cm , so its length will be ( 48 - 12 ) ÷ 2 = 36 ÷ 2 = 18 cm answer a","correct":"a","options":{"a":"18 cm ","b":"19 cm ","c":"20 cm ","d":"21 cm","e":"none"},"options_float":{"a":18.0,"b":19.0,"c":20.0,"d":21.0,"e":null},"annotated_formula":"divide(subtract(square_perimeter(12), multiply(6, const_2)), const_2)","linear_formula":"multiply(n1,const_2)|square_perimeter(n0)|subtract(#1,#0)|divide(#2,const_2)","chain":"4 * 12<\/gadget>\n48<\/output>\n6 * 2<\/gadget>\n12<\/output>\n48 - 12<\/gadget>\n36<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n18<\/result>","index":3348} +{"problem":"a train 560 m long , running with a speed of 63 km \/ hr will pass a tree in ?","rationale":"\"speed = 63 * 5 \/ 18 = 35 \/ 2 m \/ sec time taken = 560 * 2 \/ 35 = 32 sec answer : option e\"","correct":"e","options":{"a":"15 ","b":"16 ","c":"17 ","d":"28","e":"32"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":28.0,"e":32.0},"annotated_formula":"multiply(divide(560, multiply(63, const_1000)), const_3600)","linear_formula":"multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|","chain":"63 * 1_000<\/gadget>\n63_000<\/output>\n560 \/ 63_000<\/gadget>\n2\/225 = around 0.008889<\/output>\n(2\/225) * 3_600<\/gadget>\n32<\/output>\n32<\/result>","index":3349} +{"problem":"the banker â € ™ s gain on a sum due 3 years hence at 12 % per annum is rs . 180 . the banker â € ™ s discount is","rationale":"solution t . d = ( b . g x 100 \/ r x t ) = rs . ( 180 x 100 \/ 12 x 3 ) = rs . 500 . b . d = rs ( 500 + 180 ) = rs . 680 . answer d","correct":"d","options":{"a":"rs . 960 ","b":"rs . 840 ","c":"rs . 1020 ","d":"rs . 680","e":"none"},"options_float":{"a":960.0,"b":840.0,"c":1020.0,"d":680.0,"e":null},"annotated_formula":"add(divide(multiply(180, const_100), multiply(3, 12)), 180)","linear_formula":"multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)","chain":"180 * 100<\/gadget>\n18_000<\/output>\n3 * 12<\/gadget>\n36<\/output>\n18_000 \/ 36<\/gadget>\n500<\/output>\n500 + 180<\/gadget>\n680<\/output>\n680<\/result>","index":3350} +{"problem":"12 . what is the dividend . divisor 17 , the quotient is 9 and the remainder is 10 .","rationale":"\"d = d * q + r d = 17 * 9 + 10 d = 153 + 10 d = 163 answer d\"","correct":"d","options":{"a":"150 ","b":"152 ","c":"154 ","d":"163","e":"144"},"options_float":{"a":150.0,"b":152.0,"c":154.0,"d":163.0,"e":144.0},"annotated_formula":"add(multiply(17, 9), 10)","linear_formula":"multiply(n1,n2)|add(n3,#0)|","chain":"17 * 9<\/gadget>\n153<\/output>\n153 + 10<\/gadget>\n163<\/output>\n163<\/result>","index":3351} +{"problem":"a lady builds 8 cm length , 15 cm width , and 5 cm height box using 10 cubic cm cubes . what is the minimum number of cubes required to build the box ?","rationale":"\"number of cubes required = volume of box \/ volume of cube = 8 * 15 * 5 \/ 10 = 60 cubes answer : b\"","correct":"b","options":{"a":"107 ","b":"60 ","c":"70 ","d":"89","e":"78"},"options_float":{"a":107.0,"b":60.0,"c":70.0,"d":89.0,"e":78.0},"annotated_formula":"divide(multiply(multiply(8, 15), 5), 10)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|","chain":"8 * 15<\/gadget>\n120<\/output>\n120 * 5<\/gadget>\n600<\/output>\n600 \/ 10<\/gadget>\n60<\/output>\n60<\/result>","index":3352} +{"problem":"there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total w of the book ?","rationale":"\"first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) w 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) .\"","correct":"a","options":{"a":"3003 ","b":"3027 ","c":"3024 ","d":"3021","e":"3018"},"options_float":{"a":3003.0,"b":3027.0,"c":3024.0,"d":3021.0,"e":3018.0},"annotated_formula":"add(3000, 3)","linear_formula":"add(n0,n5)|","chain":"3_000 + 3<\/gadget>\n3_003<\/output>\n3_003<\/result>","index":3354} +{"problem":"the population of a bacteria colony doubles every day . if it was started 8 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ?","rationale":"\"8 days ago - 2 7 days ago - 4 6 days ago - 8 5 days ago - 16 4 days ago - 32 3 days ago - 64 2 days ago - 128 yesterday - 256 today - 512 ans : a\"","correct":"a","options":{"a":"512 ","b":"768 ","c":"1024 ","d":"2048","e":"4096"},"options_float":{"a":512.0,"b":768.0,"c":1024.0,"d":2048.0,"e":4096.0},"annotated_formula":"subtract(power(2, add(8, const_1)), const_1)","linear_formula":"add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|","chain":"8 + 1<\/gadget>\n9<\/output>\n2 ** 9<\/gadget>\n512<\/output>\n512 - 1<\/gadget>\n511<\/output>\n511<\/result>","index":3356} +{"problem":"how many quarters are equal to 7 dollars ?","rationale":"\"7 * 4 = 28 quarters answer : b\"","correct":"b","options":{"a":"1 ","b":"28 ","c":"12 ","d":"9","e":"7"},"options_float":{"a":1.0,"b":28.0,"c":12.0,"d":9.0,"e":7.0},"annotated_formula":"multiply(7, const_4)","linear_formula":"multiply(n0,const_4)|","chain":"7 * 4<\/gadget>\n28<\/output>\n28<\/result>","index":3357} +{"problem":"what is 0.1 percent of 12,356 ?","rationale":"since , percent = 1 \/ 100 , what = something ( s ) , and is : = . we can write the question as s = 0.1 ( 1 \/ 100 ) 12,356 . the answer is 12.356 . hence , the correct answer is c .","correct":"c","options":{"a":"0.12356 ","b":"1.2356 ","c":"12.356 ","d":"0.012356","e":"0.0012356"},"options_float":{"a":0.12356,"b":1.2356,"c":12.356,"d":0.012356,"e":0.0012356},"annotated_formula":"divide(multiply(0.1, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)","linear_formula":"add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 * 100<\/gadget>\n1_200<\/output>\n10 * 1_200<\/gadget>\n12_000<\/output>\n3 + 4<\/gadget>\n7<\/output>\n7 * 5<\/gadget>\n35<\/output>\n35 * 10<\/gadget>\n350<\/output>\n12_000 + 350<\/gadget>\n12_350<\/output>\n3 + 3<\/gadget>\n6<\/output>\n12_350 + 6<\/gadget>\n12_356<\/output>\n0.1 * 12_356<\/gadget>\n1_235.6<\/output>\n1_235.6 \/ 100<\/gadget>\n12.356<\/output>\n12.356<\/result>","index":3358} +{"problem":"the difference between the value of a number increased by 25 % and the value of the original number decreased by 30 % is 22 . what is the original number ?","rationale":"\"let the original # be x then 1.25 x − 0.7 x = 22 - - > x = 40 answer : a .\"","correct":"a","options":{"a":"40 ","b":"50 ","c":"65 ","d":"70","e":"90"},"options_float":{"a":40.0,"b":50.0,"c":65.0,"d":70.0,"e":90.0},"annotated_formula":"divide(22, subtract(add(const_1, divide(25, const_100)), subtract(const_1, divide(30, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#2,#3)|divide(n2,#4)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n(5\/4) - (7\/10)<\/gadget>\n11\/20 = around 0.55<\/output>\n22 \/ (11\/20)<\/gadget>\n40<\/output>\n40<\/result>","index":3359} +{"problem":"if a - b = 3 and a ^ 2 + b ^ 2 = 33 , find the value of ab .","rationale":"\"2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 33 - 9 = 24 ab = 12 . answer is a .\"","correct":"a","options":{"a":"12 ","b":"15 ","c":"10 ","d":"18","e":"13"},"options_float":{"a":12.0,"b":15.0,"c":10.0,"d":18.0,"e":13.0},"annotated_formula":"multiply(multiply(add(3, divide(subtract(sqrt(33), 3), 2)), divide(subtract(sqrt(33), 3), 2)), 2)","linear_formula":"sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)|","chain":"33 ** (1\/2)<\/gadget>\nsqrt(33) = around 5.744563<\/output>\n(sqrt(33)) - 3<\/gadget>\n-3 + sqrt(33) = around 2.744563<\/output>\n(-3 + sqrt(33)) \/ 2<\/gadget>\n-3\/2 + sqrt(33)\/2 = around 1.372281<\/output>\n3 + (-3\/2 + sqrt(33)\/2)<\/gadget>\n3\/2 + sqrt(33)\/2 = around 4.372281<\/output>\n(3\/2 + sqrt(33)\/2) * (-3\/2 + sqrt(33)\/2)<\/gadget>\n(-3\/2 + sqrt(33)\/2)*(3\/2 + sqrt(33)\/2) = around 6<\/output>\n((-3\/2 + sqrt(33)\/2)*(3\/2 + sqrt(33)\/2)) * 2<\/gadget>\n2*(-3\/2 + sqrt(33)\/2)*(3\/2 + sqrt(33)\/2) = around 12<\/output>\n2*(-3\/2 + sqrt(33)\/2)*(3\/2 + sqrt(33)\/2) = around 12<\/result>","index":3360} +{"problem":"1 : 3 = 1 2 \/ 3 : x . the value of x is ?","rationale":"x * 1 = 3 * 5 \/ 3 x = 5 answer : d","correct":"d","options":{"a":"1 ","b":"6 ","c":"7 ","d":"5","e":"8"},"options_float":{"a":1.0,"b":6.0,"c":7.0,"d":5.0,"e":8.0},"annotated_formula":"divide(add(divide(2, 3), 1), divide(1, 3))","linear_formula":"divide(n3,n1)|divide(n0,n1)|add(n0,#0)|divide(#2,#1)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) + 1<\/gadget>\n5\/3 = around 1.666667<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(5\/3) \/ (1\/3)<\/gadget>\n5<\/output>\n5<\/result>","index":3361} +{"problem":"how many figures are required to number the pages the pages of a book containing 709 pages ?","rationale":"\"1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 709 = 610 * 3 = 1830 - - - - - - - - - - - 2019 answer : b\"","correct":"b","options":{"a":"2020 ","b":"2019 ","c":"1019 ","d":"2029","e":"2039"},"options_float":{"a":2020.0,"b":2019.0,"c":1019.0,"d":2029.0,"e":2039.0},"annotated_formula":"add(add(subtract(divide(divide(709, const_10), const_10), const_1), subtract(subtract(divide(709, const_10), const_1), subtract(divide(divide(709, const_10), const_10), const_1))), multiply(subtract(subtract(709, const_1), subtract(divide(709, const_10), const_1)), const_3))","linear_formula":"divide(n0,const_10)|subtract(n0,const_1)|divide(#0,const_10)|subtract(#0,const_1)|subtract(#2,const_1)|subtract(#1,#3)|multiply(#5,const_3)|subtract(#3,#4)|add(#4,#7)|add(#8,#6)|","chain":"709 \/ 10<\/gadget>\n709\/10 = around 70.9<\/output>\n(709\/10) \/ 10<\/gadget>\n709\/100 = around 7.09<\/output>\n(709\/100) - 1<\/gadget>\n609\/100 = around 6.09<\/output>\n(709\/10) - 1<\/gadget>\n699\/10 = around 69.9<\/output>\n(699\/10) - (609\/100)<\/gadget>\n6_381\/100 = around 63.81<\/output>\n(609\/100) + (6_381\/100)<\/gadget>\n699\/10 = around 69.9<\/output>\n709 - 1<\/gadget>\n708<\/output>\n708 - (699\/10)<\/gadget>\n6_381\/10 = around 638.1<\/output>\n(6_381\/10) * 3<\/gadget>\n19_143\/10 = around 1_914.3<\/output>\n(699\/10) + (19_143\/10)<\/gadget>\n9_921\/5 = around 1_984.2<\/output>\n9_921\/5 = around 1_984.2<\/result>","index":3362} +{"problem":"a cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm . the percentage of the wood wasted is :","rationale":"explanation : we will first subtract the cone volume from wood volume to get the wood wasted . then we can calculate its percentage . sphere volume = 4 \/ 3 π r 3 cone volume = 1 \/ 3 π r 2 h volume of wood wasted = ( 4 \/ 3 π ∗ 9 ∗ 9 ∗ 9 ) − ( 1 \/ 3 π ∗ 9 ∗ 9 ∗ 9 ) = π ∗ 9 ∗ 9 ∗ 9 cm 3 required percentage = π ∗ 9 ∗ 9 ∗ 9 \/ 4 \/ 3 π ∗ 9 ∗ 9 ∗ 9 ∗ 100 % = 3 \/ 4 ∗ 100 % = 75 % option d","correct":"d","options":{"a":"45 % ","b":"56 % ","c":"67 % ","d":"75 %","e":"none of these"},"options_float":{"a":45.0,"b":56.0,"c":67.0,"d":75.0,"e":null},"annotated_formula":"multiply(divide(subtract(volume_sphere(9), volume_cone(9, 9)), volume_sphere(9)), const_100)","linear_formula":"volume_cone(n0,n0)|volume_sphere(n0)|subtract(#1,#0)|divide(#2,#1)|multiply(#3,const_100)","chain":"4\/3 * pi * (9 ** 3)<\/gadget>\n972*pi = around 3_053.628059<\/output>\npi * (9 ** 2) * 9 \/ 3<\/gadget>\n243*pi = around 763.407015<\/output>\n(972*pi) - (243*pi)<\/gadget>\n729*pi = around 2_290.221044<\/output>\n(729*pi) \/ (972*pi)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n75<\/result>","index":3363} +{"problem":"of the votes cast on a certain proposal , 66 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . )","rationale":"\"let x be the total number of votes cast . 0.6 x = 0.4 x + 66 0.2 x = 66 x = 330 the answer is c .\"","correct":"c","options":{"a":"290 ","b":"310 ","c":"330 ","d":"350","e":"370"},"options_float":{"a":290.0,"b":310.0,"c":330.0,"d":350.0,"e":370.0},"annotated_formula":"divide(66, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n0,#2)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) - (2\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n66 \/ (1\/5)<\/gadget>\n330<\/output>\n330<\/result>","index":3364} +{"problem":"the average earning of amechanic for the first 4 days of a week is rs . 25 and for the last 4 days is rs . 22 . if heearns rs . 20 on the fourth day , his average earning forthe whole week is ?","rationale":"total earning for the week = sum of earning of first four days + sum of earning of last four days - earning of 4 th day = 4 x 25 + 4 x 22 - 20 = rs . 168 â ˆ ´ average earning = 168 \/ 7 = rs . 24 b","correct":"b","options":{"a":"s . 10 ","b":"s . 24 ","c":"s . 30 ","d":"s . 40","e":"s . 50"},"options_float":{"a":10.0,"b":24.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(subtract(add(multiply(4, 25), multiply(4, 22)), 20), add(const_3, const_4))","linear_formula":"add(const_3,const_4)|multiply(n0,n1)|multiply(n0,n3)|add(#1,#2)|subtract(#3,n4)|divide(#4,#0)","chain":"4 * 25<\/gadget>\n100<\/output>\n4 * 22<\/gadget>\n88<\/output>\n100 + 88<\/gadget>\n188<\/output>\n188 - 20<\/gadget>\n168<\/output>\n3 + 4<\/gadget>\n7<\/output>\n168 \/ 7<\/gadget>\n24<\/output>\n24<\/result>","index":3366} +{"problem":"a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 3025 in 3 years . find the rate percent ?","rationale":"\"explanation : 2420 - - - 605 100 - - - ? = > 25 % answer : option d\"","correct":"d","options":{"a":"11 ","b":"10 ","c":"28 ","d":"25","e":"82"},"options_float":{"a":11.0,"b":10.0,"c":28.0,"d":25.0,"e":82.0},"annotated_formula":"multiply(divide(subtract(3025, 2420), 2420), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"3_025 - 2_420<\/gadget>\n605<\/output>\n605 \/ 2_420<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":3367} +{"problem":"9 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ?","rationale":"\"( 9 * 8 ) \/ 30 = ( x * 6 ) \/ 50 = > x = 20 20 – 9 = 11 answer : d\"","correct":"d","options":{"a":"33 ","b":"66 ","c":"88 ","d":"11","e":"281"},"options_float":{"a":33.0,"b":66.0,"c":88.0,"d":11.0,"e":281.0},"annotated_formula":"subtract(divide(multiply(divide(multiply(9, 8), 30), 50), 6), 9)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)|","chain":"9 * 8<\/gadget>\n72<\/output>\n72 \/ 30<\/gadget>\n12\/5 = around 2.4<\/output>\n(12\/5) * 50<\/gadget>\n120<\/output>\n120 \/ 6<\/gadget>\n20<\/output>\n20 - 9<\/gadget>\n11<\/output>\n11<\/result>","index":3368} +{"problem":"a building contractor employs 20 male , 15 female and 5 child workers . to a male worker he pays rs . 25 per day , to a female worker rs . 20 per day and a child worker rs . 8 per day . the average wage per day paid by the contractor is ?","rationale":"explanation : 20 15 5 25 20 8 500 + 300 + 40 = 840 \/ 40 = 21 b","correct":"b","options":{"a":"19 ","b":"21 ","c":"23 ","d":"25","e":"27"},"options_float":{"a":19.0,"b":21.0,"c":23.0,"d":25.0,"e":27.0},"annotated_formula":"divide(add(add(multiply(20, 25), multiply(15, 20)), multiply(5, 8)), add(add(20, 15), 5))","linear_formula":"add(n0,n1)|multiply(n0,n3)|multiply(n0,n1)|multiply(n2,n5)|add(#1,#2)|add(n2,#0)|add(#4,#3)|divide(#6,#5)","chain":"20 * 25<\/gadget>\n500<\/output>\n15 * 20<\/gadget>\n300<\/output>\n500 + 300<\/gadget>\n800<\/output>\n5 * 8<\/gadget>\n40<\/output>\n800 + 40<\/gadget>\n840<\/output>\n20 + 15<\/gadget>\n35<\/output>\n35 + 5<\/gadget>\n40<\/output>\n840 \/ 40<\/gadget>\n21<\/output>\n21<\/result>","index":3369} +{"problem":"find ( 7 x + 2 y ) \/ ( x - 2 y ) if x \/ 2 y = 3 \/ 2 ?","rationale":"\"x \/ 2 y = 3 \/ 2 = > x = 6 y \/ 2 = 3 y = > ( 7 x + 2 y ) \/ ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 2 y ) \/ ( 3 y - 2 y ) = > 23 y \/ y = 23 answer : a\"","correct":"a","options":{"a":"23 ","b":"25 ","c":"26 ","d":"27","e":"29"},"options_float":{"a":23.0,"b":25.0,"c":26.0,"d":27.0,"e":29.0},"annotated_formula":"divide(add(multiply(multiply(divide(3, 2), 2), 7), 2), subtract(multiply(divide(3, 2), 2), 2))","linear_formula":"divide(n4,n2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#1,n2)|add(n1,#2)|divide(#4,#3)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 2<\/gadget>\n3<\/output>\n3 * 7<\/gadget>\n21<\/output>\n21 + 2<\/gadget>\n23<\/output>\n3 - 2<\/gadget>\n1<\/output>\n23 \/ 1<\/gadget>\n23<\/output>\n23<\/result>","index":3371} +{"problem":"john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 45 mph to reach his destination . how many miles did john drive ?","rationale":"\"the total distance d traveled by john is given by d = 45 * 2 + 3 * 45 = 225 miles . answer c\"","correct":"c","options":{"a":"235 miles . ","b":"245 miles . ","c":"225 miles . ","d":"265 miles .","e":"275 miles ."},"options_float":{"a":235.0,"b":245.0,"c":225.0,"d":265.0,"e":275.0},"annotated_formula":"add(multiply(45, 2), multiply(3, 45))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"45 * 2<\/gadget>\n90<\/output>\n3 * 45<\/gadget>\n135<\/output>\n90 + 135<\/gadget>\n225<\/output>\n225<\/result>","index":3372} +{"problem":"a statue is being carved by a sculptor . the original piece of marble weighed 190 kg . in the first week 25 percent is cut away . in the second week 15 percent of the remainder is cut away . in the third week the statue is completed when 10 percent of the remainder is cut away . what is the weight of the final statue ?","rationale":"\"c 109 kg 190 ã — 0.75 ã — 0.85 ã — 0.9 = 109 kg .\"","correct":"c","options":{"a":"105 kg ","b":"103 kg ","c":"109 kg ","d":"125 kg","e":"117 kg"},"options_float":{"a":105.0,"b":103.0,"c":109.0,"d":125.0,"e":117.0},"annotated_formula":"multiply(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(15, const_100)), multiply(190, subtract(const_1, divide(25, const_100)))))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(n0,#3)|multiply(#6,#4)|multiply(#7,#5)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 - (3\/20)<\/gadget>\n17\/20 = around 0.85<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n190 * (3\/4)<\/gadget>\n285\/2 = around 142.5<\/output>\n(17\/20) * (285\/2)<\/gadget>\n969\/8 = around 121.125<\/output>\n(9\/10) * (969\/8)<\/gadget>\n8_721\/80 = around 109.0125<\/output>\n8_721\/80 = around 109.0125<\/result>","index":3373} +{"problem":"the population of a town increased from 1 , 75,000 to 2 , 97,500 in a decade . the average percent increase of population per year is","rationale":"solution increase in 10 years = ( 297500 - 175000 ) = 122500 increase % = ( 122500 \/ 175000 ã — 100 ) % = 70 % . required average = ( 70 \/ 10 ) % = 7 % . answer e","correct":"e","options":{"a":"4.37 % ","b":"5 % ","c":"6 % ","d":"8.75 %","e":"7 %"},"options_float":{"a":4.37,"b":5.0,"c":6.0,"d":8.75,"e":7.0},"annotated_formula":"add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(const_2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(const_2, const_3)), const_1000)), const_1), const_10), const_100), const_4)","linear_formula":"add(const_2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,const_1)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 * 10<\/gadget>\n100_000<\/output>\n100_000 - 1_000<\/gadget>\n99_000<\/output>\n99_000 - 1_000<\/gadget>\n98_000<\/output>\n2 + 3<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n98_000 - 500<\/gadget>\n97_500<\/output>\n3 + 4<\/gadget>\n7<\/output>\n7 * 10<\/gadget>\n70<\/output>\n70 + 5<\/gadget>\n75<\/output>\n75 * 1_000<\/gadget>\n75_000<\/output>\n97_500 \/ 75_000<\/gadget>\n13\/10 = around 1.3<\/output>\n(13\/10) - 1<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) \/ 10<\/gadget>\n3\/100 = around 0.03<\/output>\n(3\/100) * 100<\/gadget>\n3<\/output>\n7<\/result>","index":3374} +{"problem":"in an exam , gibi scored 59 percent , jigi scored 55 percent , mike scored 99 percent and lizzy 67 percent . the maximum score awarded in the exam is 700 . find the average mark scored by all the 4 students ?","rationale":"average mark scored by all the three boys = [ 59 \/ 100 ( 700 ) + 55 \/ 100 ( 700 ) + 99 \/ 100 ( 700 ) + 67 \/ 100 ( 700 ) ] \/ 4 ( 413 + 385 + 693 + 469 ) \/ 4 = 490 answer : b","correct":"b","options":{"a":"471 ","b":"490 ","c":"480 ","d":"453","e":"456"},"options_float":{"a":471.0,"b":490.0,"c":480.0,"d":453.0,"e":456.0},"annotated_formula":"divide(divide(multiply(add(add(59, 55), add(99, 67)), 700), const_100), 4)","linear_formula":"add(n0,n1)|add(n2,n3)|add(#0,#1)|multiply(n4,#2)|divide(#3,const_100)|divide(#4,n5)","chain":"59 + 55<\/gadget>\n114<\/output>\n99 + 67<\/gadget>\n166<\/output>\n114 + 166<\/gadget>\n280<\/output>\n280 * 700<\/gadget>\n196_000<\/output>\n196_000 \/ 100<\/gadget>\n1_960<\/output>\n1_960 \/ 4<\/gadget>\n490<\/output>\n490<\/result>","index":3375} +{"problem":"in how many no . between 10 and 50 exactly two of the digits is 5 ?","rationale":"\"it ' s simple can be solved by elimination of answer choices . option b and d are too large , not possible . even ae are large to have correct choice . ans : c\"","correct":"c","options":{"a":"47 ","b":"65 ","c":"1 ","d":"35","e":"65"},"options_float":{"a":47.0,"b":65.0,"c":1.0,"d":35.0,"e":65.0},"annotated_formula":"divide(divide(50, 10), 5)","linear_formula":"divide(n1,n0)|divide(#0,n2)|","chain":"50 \/ 10<\/gadget>\n5<\/output>\n5 \/ 5<\/gadget>\n1<\/output>\n1<\/result>","index":3376} +{"problem":"john and david work at a hospital with 8 other workers . for an internal review , 2 of the 8 workers will be randomly chosen to be interviewed . what is the probability that john and david will both be chosen ?","rationale":"probability that john and david will both be chosen out of 8 workers = ( 2 \/ 8 ) * ( 1 \/ 7 ) = 1 \/ 28 answer d","correct":"d","options":{"a":"1 \/ 30 ","b":"1 \/ 25 ","c":"1 \/ 20 ","d":"1 \/ 28","e":"1 \/ 30"},"options_float":{"a":0.0333333333,"b":0.04,"c":0.05,"d":0.0357142857,"e":0.0333333333},"annotated_formula":"multiply(2, multiply(divide(const_1, subtract(8, const_1)), divide(const_1, 8)))","linear_formula":"divide(const_1,n0)|subtract(n0,const_1)|divide(const_1,#1)|multiply(#2,#0)|multiply(n1,#3)","chain":"8 - 1<\/gadget>\n7<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/7) * (1\/8)<\/gadget>\n1\/56 = around 0.017857<\/output>\n2 * (1\/56)<\/gadget>\n1\/28 = around 0.035714<\/output>\n1\/28 = around 0.035714<\/result>","index":3377} +{"problem":"by selling 8 pens for a rupee a woman loses 25 % . how many for a rupee should he sell in order to gain 25 % ?","rationale":"\"d 75 % - - - 8 125 % - - - ? 75 \/ 125 * 8 = 5\"","correct":"d","options":{"a":"2 ","b":"14 ","c":"3 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":14.0,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"inverse(add(divide(divide(const_100, subtract(const_100, 25)), 8), divide(multiply(divide(divide(const_100, subtract(const_100, 25)), 8), 25), const_100)))","linear_formula":"subtract(const_100,n1)|divide(const_100,#0)|divide(#1,n0)|multiply(n2,#2)|divide(#3,const_100)|add(#2,#4)|inverse(#5)|","chain":"100 - 25<\/gadget>\n75<\/output>\n100 \/ 75<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) \/ 8<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 25<\/gadget>\n25\/6 = around 4.166667<\/output>\n(25\/6) \/ 100<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/6) + (1\/24)<\/gadget>\n5\/24 = around 0.208333<\/output>\n1 \/ (5\/24)<\/gadget>\n24\/5 = around 4.8<\/output>\n24\/5 = around 4.8<\/result>","index":3378} +{"problem":"3.20 can be expressed in terms of percentage as","rationale":"explanation : while calculation in terms of percentage we need to multiply by 100 , so 3.20 * 100 = 320 answer : option c","correct":"c","options":{"a":"3.20 % ","b":"30.2 % ","c":"320.00 % ","d":"0.320 %","e":"none of these"},"options_float":{"a":3.2,"b":30.2,"c":320.0,"d":0.32,"e":null},"annotated_formula":"multiply(3.2, const_100)","linear_formula":"multiply(n0,const_100)","chain":"3.2 * 100<\/gadget>\n320<\/output>\n320<\/result>","index":3379} +{"problem":"a person was asked to subtract 25 % of a number from the original but he subtracted 25 from the number . his answer was 25 more than correct one . what was the original number ?","rationale":"25 + 25 = 50 25 % - - - - - - 50 100 % - - - - - ? = > 200 answer : d","correct":"d","options":{"a":"288 ","b":"266 ","c":"288 ","d":"200","e":"111"},"options_float":{"a":288.0,"b":266.0,"c":288.0,"d":200.0,"e":111.0},"annotated_formula":"divide(add(25, 25), subtract(const_1, subtract(const_1, divide(25, const_100))))","linear_formula":"add(n0,n0)|divide(n0,const_100)|subtract(const_1,#1)|subtract(const_1,#2)|divide(#0,#3)","chain":"25 + 25<\/gadget>\n50<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n50 \/ (1\/4)<\/gadget>\n200<\/output>\n200<\/result>","index":3380} +{"problem":"if x : y is 1 : 5 and y : z is 5 : 7 then x : z is equal to","rationale":"the two ratios given are having the same number 5 for y in both the ratios . hence - x : y = 1 : 5 y : z = 5 : 7 = > x : z = 1 : 7 answer d","correct":"d","options":{"a":"1 : 6 ","b":"1 : 5 ","c":"1 : 4 ","d":"1 : 7","e":"1 : 3"},"options_float":{"a":0.1666666667,"b":0.2,"c":0.25,"d":0.1428571429,"e":0.3333333333},"annotated_formula":"divide(1, 7)","linear_formula":"divide(n0,n3)","chain":"1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1\/7 = around 0.142857<\/result>","index":3381} +{"problem":"maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 72 kilometers , maxwell ' s walking speed is 6 km \/ h , and brad ' s running speed is 12 km \/ h . what is the distance traveled by maxwell when they meet in the middle ?","rationale":"\"consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 6 xt = 72 - 12 xt ( i . e distance = speed x time ) = > 18 t = 72 hence t = 4 oa = 6 x 4 = 24 km answer : b\"","correct":"b","options":{"a":"14 ","b":"24 ","c":"16 ","d":"18","e":"20"},"options_float":{"a":14.0,"b":24.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"multiply(6, divide(72, add(6, 12)))","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|","chain":"6 + 12<\/gadget>\n18<\/output>\n72 \/ 18<\/gadget>\n4<\/output>\n6 * 4<\/gadget>\n24<\/output>\n24<\/result>","index":3383} +{"problem":"carolyn bought 17 gumballs , lew bought 12 gumballs , and carey bought x gumballs . the average ( arithmetic mean ) number of gumballs the 3 bought is between 19 and 25 , inclusive . what is the difference between the greatest number and the smallest number of gumballs carey could have bought ?","rationale":"smallest gumballs = ( 19 - 17 ) + ( 19 - 12 ) + 19 = 28 largest gumballs = ( 25 - 17 ) + ( 25 - 12 ) + 25 = 46 difference = 46 - 28 = 18 a","correct":"a","options":{"a":"18 ","b":"20 ","c":"22 ","d":"24","e":"26"},"options_float":{"a":18.0,"b":20.0,"c":22.0,"d":24.0,"e":26.0},"annotated_formula":"subtract(subtract(multiply(25, 3), add(17, 12)), subtract(multiply(19, 3), add(17, 12)))","linear_formula":"add(n0,n1)|multiply(n2,n4)|multiply(n2,n3)|subtract(#1,#0)|subtract(#2,#0)|subtract(#3,#4)","chain":"25 * 3<\/gadget>\n75<\/output>\n17 + 12<\/gadget>\n29<\/output>\n75 - 29<\/gadget>\n46<\/output>\n19 * 3<\/gadget>\n57<\/output>\n57 - 29<\/gadget>\n28<\/output>\n46 - 28<\/gadget>\n18<\/output>\n18<\/result>","index":3384} +{"problem":"a train 250 m long passed a pole in 10 sec . how long will it take to pass a platform 1250 m long ?","rationale":"speed = 250 \/ 10 = 25 m \/ sec . required time = ( 250 + 1250 ) \/ 25 = 60 sec answer : a","correct":"a","options":{"a":"60 sec ","b":"90 sec ","c":"120 sec ","d":"150 sec","e":"180 sec"},"options_float":{"a":60.0,"b":90.0,"c":120.0,"d":150.0,"e":180.0},"annotated_formula":"divide(add(250, 1250), divide(250, 10))","linear_formula":"add(n0,n2)|divide(n0,n1)|divide(#0,#1)","chain":"250 + 1_250<\/gadget>\n1_500<\/output>\n250 \/ 10<\/gadget>\n25<\/output>\n1_500 \/ 25<\/gadget>\n60<\/output>\n60<\/result>","index":3385} +{"problem":"one ball will drop from a certain height . the height it will reach after rebounding from the floor is 50 percent of the previous height . the total travel is 225 cm when it touches the floor on third time . what is the value of the original height ?","rationale":"\"when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = h ( 1 + 1 + 0.5 ) = 225 2.5 h = 225 h = 90 . b is the answer .\"","correct":"b","options":{"a":"80 cm ","b":"90 cm ","c":"100 cm ","d":"120 cm","e":"130 cm"},"options_float":{"a":80.0,"b":90.0,"c":100.0,"d":120.0,"e":130.0},"annotated_formula":"divide(225, add(const_2, divide(50, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_2)|divide(n1,#1)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n2 + (1\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n225 \/ (5\/2)<\/gadget>\n90<\/output>\n90<\/result>","index":3386} +{"problem":"a man fell in a 30 meter deep well , in one day he climbs 4 meters up and slips 3 meters down . how many days would it take for him to come out of the well ?","rationale":"d i will take 27 days to come out of that well for that man , he climbs 4 meter every day and slips 3 meter down that means he climbs 1 meter in total each day , so like this on 26 th day he would have climbed 26 meter and on 27 th day he will climb 4 meeter again so total 30 meter he will climb in 27 days .","correct":"d","options":{"a":"22 ","b":"46 ","c":"17 ","d":"27","e":"33"},"options_float":{"a":22.0,"b":46.0,"c":17.0,"d":27.0,"e":33.0},"annotated_formula":"subtract(30, 3)","linear_formula":"subtract(n0,n2)","chain":"30 - 3<\/gadget>\n27<\/output>\n27<\/result>","index":3387} +{"problem":"if a mixture is 2 ⁄ 7 alcohol by volume and 3 ⁄ 7 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ?","rationale":"should be a sub - 600 level q . . volume = { 2 \/ 7 } \/ { 3 \/ 7 } = 2 \/ 3 c","correct":"c","options":{"a":"3 ","b":"1 \/ 3 ","c":"2 \/ 3 ","d":"3 \/ 5","e":"4 \/ 5"},"options_float":{"a":3.0,"b":0.3333333333,"c":0.6666666667,"d":0.6,"e":0.8},"annotated_formula":"divide(divide(2, 7), divide(3, 7))","linear_formula":"divide(n0,n1)|divide(n2,n1)|divide(#0,#1)","chain":"2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n(2\/7) \/ (3\/7)<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":3388} +{"problem":"20 percent of country y ' s yearly exports come from fruit exports . one - sixth of all fruit exports from country y are orange exports . if country y generates $ 4.25 million from its orange exports , how much money does it generate from its yearly total of exports ?","rationale":"2 \/ 10 * 1 \/ 6 * ( total ) = 4.25 1 \/ 30 * ( total ) = 4.25 ( total ) = 4.25 * 30 = 127.5 answer : d .","correct":"d","options":{"a":"$ 21.25 m ","b":"$ 25.5 m ","c":"$ 106.25 m ","d":"$ 127.5 m","e":"$ 153 m"},"options_float":{"a":21.25,"b":25.5,"c":106.25,"d":127.5,"e":153.0},"annotated_formula":"divide(multiply(divide(4.25, divide(const_1, multiply(const_2, const_3))), const_100), 20)","linear_formula":"multiply(const_2,const_3)|divide(const_1,#0)|divide(n1,#1)|multiply(#2,const_100)|divide(#3,n0)","chain":"2 * 3<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n4.25 \/ (1\/6)<\/gadget>\n25.5<\/output>\n25.5 * 100<\/gadget>\n2_550<\/output>\n2_550 \/ 20<\/gadget>\n255\/2 = around 127.5<\/output>\n255\/2 = around 127.5<\/result>","index":3389} +{"problem":"what is the value of n if the sum of the consecutive odd intergers w from 1 to n equals 169 ?","rationale":"\"before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n \/ 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is .\"","correct":"b","options":{"a":"47 ","b":"25 ","c":"37 ","d":"33","e":"29"},"options_float":{"a":47.0,"b":25.0,"c":37.0,"d":33.0,"e":29.0},"annotated_formula":"add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1)","linear_formula":"multiply(n0,const_2)|sqrt(n1)|multiply(#1,const_2)|subtract(#2,#0)|add(n0,#3)|","chain":"169 ** (1\/2)<\/gadget>\n13<\/output>\n13 * 2<\/gadget>\n26<\/output>\n2 * 1<\/gadget>\n2<\/output>\n26 - 2<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n25<\/result>","index":3390} +{"problem":"indu gave bindu rs . 7500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ?","rationale":"\"7500 = d ( 100 \/ 4 ) 2 d = 12 answer : e\"","correct":"e","options":{"a":"5 ","b":"2 ","c":"9 ","d":"5","e":"12"},"options_float":{"a":5.0,"b":2.0,"c":9.0,"d":5.0,"e":12.0},"annotated_formula":"subtract(subtract(multiply(7500, power(add(const_1, divide(4, const_100)), 2)), 7500), multiply(multiply(7500, divide(4, const_100)), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n7_500 * (676\/625)<\/gadget>\n8_112<\/output>\n8_112 - 7_500<\/gadget>\n612<\/output>\n7_500 * (1\/25)<\/gadget>\n300<\/output>\n300 * 2<\/gadget>\n600<\/output>\n612 - 600<\/gadget>\n12<\/output>\n12<\/result>","index":3391} +{"problem":"evaluate : 7899665 - 12 * 3 * 2 = ?","rationale":"\"according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 7899665 - 12 * 3 * 2 = 7899665 - 8 = 7899657 correct answer d\"","correct":"d","options":{"a":"2455835 ","b":"2789255 ","c":"6868575 ","d":"7899657","e":"1023666"},"options_float":{"a":2455835.0,"b":2789255.0,"c":6868575.0,"d":7899657.0,"e":1023666.0},"annotated_formula":"subtract(7899665, multiply(multiply(12, 3), 2))","linear_formula":"multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|","chain":"12 * 3<\/gadget>\n36<\/output>\n36 * 2<\/gadget>\n72<\/output>\n7_899_665 - 72<\/gadget>\n7_899_593<\/output>\n7_899_593<\/result>","index":3392} +{"problem":"a can do a half of certain work in 70 days and b one third of the same in 35 days . they together will do the whole work in ?","rationale":"a = 140 days b = 105 days 1 \/ 140 + 1 \/ 105 = 7 \/ 420 = 1 \/ 60 = > 60 days answer : d","correct":"d","options":{"a":"1 \/ 65 ","b":"1 \/ 62 ","c":"1 \/ 63 ","d":"1 \/ 60","e":"1 \/ 61"},"options_float":{"a":0.0153846154,"b":0.0161290323,"c":0.0158730159,"d":0.0166666667,"e":0.0163934426},"annotated_formula":"add(inverse(multiply(70, const_2)), inverse(multiply(35, const_3)))","linear_formula":"multiply(n0,const_2)|multiply(n1,const_3)|inverse(#0)|inverse(#1)|add(#2,#3)","chain":"70 * 2<\/gadget>\n140<\/output>\n1 \/ 140<\/gadget>\n1\/140 = around 0.007143<\/output>\n35 * 3<\/gadget>\n105<\/output>\n1 \/ 105<\/gadget>\n1\/105 = around 0.009524<\/output>\n(1\/140) + (1\/105)<\/gadget>\n1\/60 = around 0.016667<\/output>\n1\/60 = around 0.016667<\/result>","index":3393} +{"problem":"if 9 \/ ( 1 + 4 \/ x ) = 1 , then x =","rationale":"\"the expression 9 \/ ( 1 + 4 \/ x ) = 1 should have been equal to something . if 9 \/ ( 1 + 4 \/ x ) = 1 = 1 = > 9 x \/ ( x + 4 ) = 1 = > 9 x = x + 4 = > 8 x = 4 = > x = 1 \/ 2 correct option : b\"","correct":"b","options":{"a":"3 ","b":"1 \/ 2 ","c":"1 \/ 3 ","d":"- 1 \/ 3","e":"- 3"},"options_float":{"a":3.0,"b":0.5,"c":0.3333333333,"d":-0.3333333333,"e":-3.0},"annotated_formula":"divide(4, subtract(9, 1))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"9 - 1<\/gadget>\n8<\/output>\n4 \/ 8<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":3394} +{"problem":"an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 \/ 5 the price for each package in excess of 10 . during a certain week , the supplier sold 60 packages of gaskets . if 15 percent of the gaskets went to company x , 15 percent to company y , and the rest to company z , what was the total amount , in dollars , that the parts supplier received in payment for the gaskets ?","rationale":"$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 \/ 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 15 % of the gaskets to x = 9 gaskets = 25 * 9 = 225 2 . he sold 15 % of the gaskets to y = 9 gaskets = 25 * 9 = 225 3 . he sold remaining 70 % to z = 42 gaskets = 25 * 10 = 250 + 20 * 32 = 890 thus , total money earned 225 + 225 + 890 = 1340 answer is a","correct":"a","options":{"a":"1340 ","b":"1445 ","c":"1375 ","d":"1415","e":"1455"},"options_float":{"a":1340.0,"b":1445.0,"c":1375.0,"d":1415.0,"e":1455.0},"annotated_formula":"add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(60, multiply(60, divide(15, const_100))), multiply(60, divide(15, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(60, divide(15, const_100))), multiply(25, multiply(60, divide(15, const_100)))))","linear_formula":"divide(n2,n3)|divide(n6,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n5,#1)|multiply(n0,#4)|subtract(n5,#4)|add(#5,#5)|subtract(#6,#4)|subtract(#8,n1)|multiply(#3,#9)|add(#10,#2)|add(#11,#7)","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n25 * (4\/5)<\/gadget>\n20<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n60 * (3\/20)<\/gadget>\n9<\/output>\n60 - 9<\/gadget>\n51<\/output>\n51 - 9<\/gadget>\n42<\/output>\n42 - 10<\/gadget>\n32<\/output>\n20 * 32<\/gadget>\n640<\/output>\n25 * 10<\/gadget>\n250<\/output>\n640 + 250<\/gadget>\n890<\/output>\n25 * 9<\/gadget>\n225<\/output>\n225 + 225<\/gadget>\n450<\/output>\n890 + 450<\/gadget>\n1_340<\/output>\n1_340<\/result>","index":3402} +{"problem":"suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3982 . what is the difference between rohan ’ s and sudhir ’ s share in the profit ?","rationale":"\"suresh : rohan : sudhir ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8 = 6 : 3 : 2 the difference between rohan ’ s and sudhir ’ s share = 1 share : . i . e . = rs . 3982 × 1 \/ 11 = rs . 362 . a )\"","correct":"a","options":{"a":"rs . 362 ","b":"rs . 383 ","c":"rs . 398 ","d":"rs . 399","e":"rs . 400"},"options_float":{"a":362.0,"b":383.0,"c":398.0,"d":399.0,"e":400.0},"annotated_formula":"multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(3982, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4))))))","linear_formula":"multiply(n1,n2)|subtract(#0,n1)|subtract(#0,n2)|divide(#1,#0)|divide(#2,#0)|multiply(n3,#3)|multiply(n4,#4)|add(n0,#5)|subtract(#5,#6)|add(#7,#6)|divide(n5,#9)|multiply(#10,#8)|","chain":"3 * 4<\/gadget>\n12<\/output>\n12 - 3<\/gadget>\n9<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n12_000 * (3\/4)<\/gadget>\n9_000<\/output>\n12 - 4<\/gadget>\n8<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n9_000 * (2\/3)<\/gadget>\n6_000<\/output>\n9_000 - 6_000<\/gadget>\n3_000<\/output>\n18_000 + 9_000<\/gadget>\n27_000<\/output>\n27_000 + 6_000<\/gadget>\n33_000<\/output>\n3_982 \/ 33_000<\/gadget>\n181\/1_500 = around 0.120667<\/output>\n3_000 * (181\/1_500)<\/gadget>\n362<\/output>\n362<\/result>","index":3403} +{"problem":"a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 360 kilograms of pears that day , how many kilograms did he sell in the morning and how many in the afternoon ?","rationale":"\"let x be the number of kilograms he sold in the morning . then in the afternoon he sold 2 x kilograms . so , the total is x + 2 x = 3 x . this must be equal to 360 . 3 x = 360 x = 3603 x = 120 therefore , the salesman sold 120 kg in the morning and 2 ⋅ 120 = 240 kg in the afternoon . so answer is c .\"","correct":"c","options":{"a":"120 ","b":"180 ","c":"240 ","d":"280","e":"320"},"options_float":{"a":120.0,"b":180.0,"c":240.0,"d":280.0,"e":320.0},"annotated_formula":"multiply(divide(360, const_3), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_2)|","chain":"360 \/ 3<\/gadget>\n120<\/output>\n120 * 2<\/gadget>\n240<\/output>\n240<\/result>","index":3407} +{"problem":"the average of the marks of 25 students in a class is 70 . if the marks of each student are doubled , find the new average ?","rationale":"\"sum of the marks for the 25 students = 25 * 70 = 1750 . the marks of each student are doubled , the sum also will be doubled . the new sum = 1750 * 2 = 3500 . so , the new average = 3500 \/ 25 = 140 . answer : a\"","correct":"a","options":{"a":"140 ","b":"100 ","c":"85 ","d":"160","e":"190"},"options_float":{"a":140.0,"b":100.0,"c":85.0,"d":160.0,"e":190.0},"annotated_formula":"multiply(70, const_2)","linear_formula":"multiply(n1,const_2)|","chain":"70 * 2<\/gadget>\n140<\/output>\n140<\/result>","index":3408} +{"problem":"a train after traveling for 50 km meets with an accident and then proceeds at 3 \/ 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed w of the train .","rationale":"\"let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 \/ 4 of its former speed and arrives at its destination 35 minutes late so , y \/ ( 3 x \/ 4 ) - y \/ x = 35 \/ 60 4 y \/ 3 x - y \/ x = 7 \/ 12 y \/ x ( 4 \/ 3 - 1 ) = 7 \/ 12 y \/ x * 1 \/ 3 = 7 \/ 12 y \/ x = 7 \/ 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) \/ ( 3 x \/ 4 ) - ( y - 24 ) \/ x = 25 \/ 60 4 ( y - 24 ) \/ 3 x - ( y - 24 ) \/ x = 5 \/ 12 ( y - 24 ) \/ x ( 4 \/ 3 - 1 ) = 5 \/ 12 ( y - 24 ) \/ x * 1 \/ 3 = 5 \/ 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = w ans = c\"","correct":"c","options":{"a":"45 ","b":"33 ","c":"48 ","d":"55","e":"61"},"options_float":{"a":45.0,"b":33.0,"c":48.0,"d":55.0,"e":61.0},"annotated_formula":"multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25)))","linear_formula":"multiply(n2,n4)|subtract(n3,n5)|add(n0,#1)|divide(#0,n1)|divide(#2,#1)|subtract(#3,n4)|multiply(#4,#5)|","chain":"24 * 4<\/gadget>\n96<\/output>\n96 \/ 3<\/gadget>\n32<\/output>\n32 - 24<\/gadget>\n8<\/output>\n35 - 25<\/gadget>\n10<\/output>\n10 + 50<\/gadget>\n60<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n8 * 6<\/gadget>\n48<\/output>\n48<\/result>","index":3409} +{"problem":"amit and ananthu can do a work in 10 days and 20 days respectively . amit started the work and left after 2 days . ananthu took over and completed the work . in how many days was the total work completed ?","rationale":"amit ’ s one day ’ s work = 1 \/ 10 amit ’ s 2 day ’ s work = 1 \/ 10 * 2 = 1 \/ 5 work left = 1 - 1 \/ 5 = 4 \/ 5 ananthu ’ s one day ’ s work = 1 \/ 20 ananthu can do work in = 4 \/ 5 * 20 = 16 days so total days = 16 + 2 = 18 days answer : a","correct":"a","options":{"a":"18 days ","b":"20 days ","c":"23 days ","d":"25 days","e":"27 days"},"options_float":{"a":18.0,"b":20.0,"c":23.0,"d":25.0,"e":27.0},"annotated_formula":"add(divide(subtract(const_1, multiply(inverse(10), const_2)), inverse(20)), const_2)","linear_formula":"inverse(n0)|inverse(n1)|multiply(#0,const_2)|subtract(const_1,#2)|divide(#3,#1)|add(#4,const_2)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 2<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(4\/5) \/ (1\/20)<\/gadget>\n16<\/output>\n16 + 2<\/gadget>\n18<\/output>\n18<\/result>","index":3410} +{"problem":"cheese , bologna , and peanut butter sandwiches were made for a picnic in a ratio of 1 to 7 to 8 . if a total of 80 sandwiches were made , how many bologna sandwiches were made ?","rationale":"for deciding such task we should calculate all parts 1 + 7 + 8 = 16 parts and we should calculate how many sandwiches holds 1 part : 80 \/ 16 = 5 sandwiches in one part for bologna we have 7 parts so : 7 * 5 = 35 answer is c","correct":"c","options":{"a":"15 ","b":"30 ","c":"35 ","d":"42","e":"48"},"options_float":{"a":15.0,"b":30.0,"c":35.0,"d":42.0,"e":48.0},"annotated_formula":"multiply(divide(80, add(add(1, 7), 8)), 7)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n1,#2)","chain":"1 + 7<\/gadget>\n8<\/output>\n8 + 8<\/gadget>\n16<\/output>\n80 \/ 16<\/gadget>\n5<\/output>\n5 * 7<\/gadget>\n35<\/output>\n35<\/result>","index":3412} +{"problem":"for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 110 units raises the average to 55 units per day , what is the value of n ?","rationale":"\"the daily production was raised by 5 units for n days , which has a weighting of 5 n . 5 n = 110 - 55 = 55 n = 11 the answer is b .\"","correct":"b","options":{"a":"9 ","b":"11 ","c":"13 ","d":"15","e":"17"},"options_float":{"a":9.0,"b":11.0,"c":13.0,"d":15.0,"e":17.0},"annotated_formula":"subtract(divide(subtract(110, 50), subtract(55, 50)), const_1)","linear_formula":"subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)|","chain":"110 - 50<\/gadget>\n60<\/output>\n55 - 50<\/gadget>\n5<\/output>\n60 \/ 5<\/gadget>\n12<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11<\/result>","index":3413} +{"problem":"a train is 360 meter long is running at a speed of 50 km \/ hour . in what time will it pass a bridge of 140 meter length ?","rationale":"\"speed = 50 km \/ hr = 50 * ( 5 \/ 18 ) m \/ sec = 125 \/ 9 m \/ sec total distance = 360 + 140 = 500 meter time = distance \/ speed = 500 * ( 9 \/ 125 ) = 36 seconds answer : c\"","correct":"c","options":{"a":"87 ","b":"69 ","c":"36 ","d":"72","e":"21"},"options_float":{"a":87.0,"b":69.0,"c":36.0,"d":72.0,"e":21.0},"annotated_formula":"divide(add(360, 140), divide(multiply(50, const_1000), const_3600))","linear_formula":"add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|","chain":"360 + 140<\/gadget>\n500<\/output>\n50 * 1_000<\/gadget>\n50_000<\/output>\n50_000 \/ 3_600<\/gadget>\n125\/9 = around 13.888889<\/output>\n500 \/ (125\/9)<\/gadget>\n36<\/output>\n36<\/result>","index":3414} +{"problem":"the cash difference between the selling prices of an book at a profit of 12 % and 20 % is $ 3 . the ratio of the two selling prices is :","rationale":"\"let c . p . of the book be $ x . then , required ratio = 112 % of x \/ 120 % of x = 112 \/ 120 = 14 \/ 15 = 14 : 15 e\"","correct":"e","options":{"a":"55 : 23 ","b":"52 : 33 ","c":"52 : 53 ","d":"45 : 53","e":"14 : 15"},"options_float":{"a":2.3913043478,"b":1.5757575758,"c":0.9811320755,"d":0.8490566038,"e":0.9333333333},"annotated_formula":"divide(add(const_100, 12), add(const_100, 20))","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|","chain":"100 + 12<\/gadget>\n112<\/output>\n100 + 20<\/gadget>\n120<\/output>\n112 \/ 120<\/gadget>\n14\/15 = around 0.933333<\/output>\n14\/15 = around 0.933333<\/result>","index":3416} +{"problem":"mixture contains alcohol and water in the ratio 4 : 3 . if 5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture .","rationale":"\"let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 20 x = 4 ( 3 x + 5 ) 8 x = 20 x = 2.5 quantity of alcohol = ( 4 x 2.5 ) litres = 10 litres . answer : a\"","correct":"a","options":{"a":"10 ","b":"99 ","c":"27 ","d":"22","e":"29"},"options_float":{"a":10.0,"b":99.0,"c":27.0,"d":22.0,"e":29.0},"annotated_formula":"multiply(divide(multiply(4, 5), subtract(multiply(5, 4), multiply(3, 4))), 4)","linear_formula":"multiply(n0,n4)|multiply(n0,n2)|multiply(n0,n1)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)|","chain":"4 * 5<\/gadget>\n20<\/output>\n5 * 4<\/gadget>\n20<\/output>\n3 * 4<\/gadget>\n12<\/output>\n20 - 12<\/gadget>\n8<\/output>\n20 \/ 8<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) * 4<\/gadget>\n10<\/output>\n10<\/result>","index":3417} +{"problem":"two pipes a and b can fill a tank in 36 hours and 46 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ?","rationale":"\"a fill in one hour = 1 \/ 36 b fill in one hour = 1 \/ 46 both fill in one hour = 1 \/ 36 + 1 \/ 46 = 82 \/ 1656 time taken by both pipes = 1656 \/ 82 = 20.195 hrs answer : a\"","correct":"a","options":{"a":"20.195 hrs ","b":"21.195 hrs ","c":"22.195 hrs ","d":"23.195 hrs","e":"20.495 hrs"},"options_float":{"a":20.195,"b":21.195,"c":22.195,"d":23.195,"e":20.495},"annotated_formula":"divide(const_1, add(divide(const_1, 36), divide(const_1, 46)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 36<\/gadget>\n1\/36 = around 0.027778<\/output>\n1 \/ 46<\/gadget>\n1\/46 = around 0.021739<\/output>\n(1\/36) + (1\/46)<\/gadget>\n41\/828 = around 0.049517<\/output>\n1 \/ (41\/828)<\/gadget>\n828\/41 = around 20.195122<\/output>\n828\/41 = around 20.195122<\/result>","index":3420} +{"problem":"a grocer has a sale of rs . 5435 , rs . 5927 , rs . 5855 , rs . 6230 and rs . 5562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5500 ?","rationale":"\"total sale for 5 months = rs . ( 5435 + 5927 + 5855 + 6230 + 5562 ) = rs . 29009 . required sale = rs . [ ( 5500 x 6 ) - 29009 ] = rs . ( 33000 - 29009 ) = rs . 3991 answer : option c\"","correct":"c","options":{"a":"4991 ","b":"5991 ","c":"3991 ","d":"6542","e":"6995"},"options_float":{"a":4991.0,"b":5991.0,"c":3991.0,"d":6542.0,"e":6995.0},"annotated_formula":"subtract(multiply(add(5, const_1), 5500), add(add(add(add(5435, 5927), 5855), 6230), 5562))","linear_formula":"add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|","chain":"5 + 1<\/gadget>\n6<\/output>\n6 * 5_500<\/gadget>\n33_000<\/output>\n5_435 + 5_927<\/gadget>\n11_362<\/output>\n11_362 + 5_855<\/gadget>\n17_217<\/output>\n17_217 + 6_230<\/gadget>\n23_447<\/output>\n23_447 + 5_562<\/gadget>\n29_009<\/output>\n33_000 - 29_009<\/gadget>\n3_991<\/output>\n3_991<\/result>","index":3421} +{"problem":"find the result of equation of the ellipse whose foci are at ( 0 , - 5 ) and ( 0 , 5 ) and the length of its major axis is 14 .","rationale":"from the coordinates of the foci , c = 5 and the major axis is the y axis . from the length of the major axis , we obtain a = 7 . also b 2 = a 2 - c 2 = 24 . the equation of the ellipse is given by x 2 \/ 24 + y 2 \/ 49 = 1 correct answer is e ) 1","correct":"e","options":{"a":"5 ","b":"4 ","c":"3 ","d":"2","e":"1"},"options_float":{"a":5.0,"b":4.0,"c":3.0,"d":2.0,"e":1.0},"annotated_formula":"divide(add(add(5, 5), const_4), 14)","linear_formula":"add(n1,n1)|add(#0,const_4)|divide(#1,n4)","chain":"5 + 5<\/gadget>\n10<\/output>\n10 + 4<\/gadget>\n14<\/output>\n14 \/ 14<\/gadget>\n1<\/output>\n1<\/result>","index":3422} +{"problem":"if 12 machine can finish a job in 40 days , then how many more machines would be needed to finish the job in one - fourth less time ?","rationale":"you might think of this in a management context - we can use the principle of ' person - hours ' to solve any problem where we have identical workers . so , using simpler numbers , suppose you know that 6 identical employees , working simultaneously , would finish a job in 5 hours . then that job requires 6 * 5 = 30 total hours of person - work . if instead you wanted the job done in 3 hours , you ' d assign 30 \/ 3 = 10 employees to do the job , because you want to get a total of 30 hours of work from the employees . we can solve this problem identically . if 12 machines ( identical ones , i assume ) work simultaneously for 40 days , they will do a total of 12 * 40 machine - days of work . so the job requires 12 * 40 days of machine work in total . we instead want the job done in 1 \/ 4 less time , so in 30 days . so we ' ll need 12 * 40 \/ 30 = 16 machines , or 4 additional machines . a","correct":"a","options":{"a":"a . 4 ","b":"b . 8 ","c":"c . 10 ","d":"d . 12","e":"e . 16"},"options_float":{"a":4.0,"b":8.0,"c":10.0,"d":12.0,"e":16.0},"annotated_formula":"subtract(divide(multiply(12, 40), multiply(divide(const_3, const_4), 40)), 12)","linear_formula":"divide(const_3,const_4)|multiply(n0,n1)|multiply(n1,#0)|divide(#1,#2)|subtract(#3,n0)","chain":"12 * 40<\/gadget>\n480<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 40<\/gadget>\n30<\/output>\n480 \/ 30<\/gadget>\n16<\/output>\n16 - 12<\/gadget>\n4<\/output>\n4<\/result>","index":3424} +{"problem":"following an increase in prices , the price of a candy box was 20 pounds and the price of a can of soda was 6 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ?","rationale":"\"price of candy before price increase = 20 \/ 1.25 = 16 price of soda before price increase = 6 \/ 1.5 = 4 total price = 16 + 4 = 20 b is the answer\"","correct":"b","options":{"a":"11 . ","b":"20 . ","c":"13 . ","d":"14 .","e":"14.5"},"options_float":{"a":11.0,"b":20.0,"c":13.0,"d":14.0,"e":14.5},"annotated_formula":"add(divide(multiply(20, const_100), add(const_100, 25)), divide(multiply(6, const_100), add(const_100, 50)))","linear_formula":"add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|multiply(n1,const_100)|divide(#2,#0)|divide(#3,#1)|add(#4,#5)|","chain":"20 * 100<\/gadget>\n2_000<\/output>\n100 + 25<\/gadget>\n125<\/output>\n2_000 \/ 125<\/gadget>\n16<\/output>\n6 * 100<\/gadget>\n600<\/output>\n100 + 50<\/gadget>\n150<\/output>\n600 \/ 150<\/gadget>\n4<\/output>\n16 + 4<\/gadget>\n20<\/output>\n20<\/result>","index":3425} +{"problem":"a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 14 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ?","rationale":"\"explanation : l * 30 = 1200 è l = 40 40 + 30 + 50 = 120 120 * 14 = 1680 answer : option b\"","correct":"b","options":{"a":"1276 ","b":"1680 ","c":"2832 ","d":"1299","e":"1236"},"options_float":{"a":1276.0,"b":1680.0,"c":2832.0,"d":1299.0,"e":1236.0},"annotated_formula":"multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 14)","linear_formula":"divide(n1,n3)|power(n3,n2)|add(n3,#0)|power(#0,n2)|add(#1,#3)|sqrt(#4)|add(#2,#5)|multiply(n0,#6)|","chain":"1_200 \/ 30<\/gadget>\n40<\/output>\n30 + 40<\/gadget>\n70<\/output>\n30 ** 2<\/gadget>\n900<\/output>\n40 ** 2<\/gadget>\n1_600<\/output>\n900 + 1_600<\/gadget>\n2_500<\/output>\n2_500 ** (1\/2)<\/gadget>\n50<\/output>\n70 + 50<\/gadget>\n120<\/output>\n120 * 14<\/gadget>\n1_680<\/output>\n1_680<\/result>","index":3426} +{"problem":". a car covers a distance of 810 km in 5 hours . find its speed ?","rationale":"\"810 \/ 5 = 162 kmph answer : c\"","correct":"c","options":{"a":"104 ","b":"55 ","c":"162 ","d":"174","e":"2761"},"options_float":{"a":104.0,"b":55.0,"c":162.0,"d":174.0,"e":2761.0},"annotated_formula":"divide(810, 5)","linear_formula":"divide(n0,n1)|","chain":"810 \/ 5<\/gadget>\n162<\/output>\n162<\/result>","index":3427} +{"problem":"the compounded ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 2 ) is","rationale":"answer : b ) 2 : 1","correct":"b","options":{"a":"2 : 0 ","b":"2 : 1 ","c":"2 : 9 ","d":"2 : 2","e":"2 : 5"},"options_float":{"a":null,"b":2.0,"c":0.2222222222,"d":1.0,"e":0.4},"annotated_formula":"multiply(multiply(divide(2, 3), divide(6, 11)), divide(11, 2))","linear_formula":"divide(n3,n0)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n6 \/ 11<\/gadget>\n6\/11 = around 0.545455<\/output>\n(2\/3) * (6\/11)<\/gadget>\n4\/11 = around 0.363636<\/output>\n11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\n(4\/11) * (11\/2)<\/gadget>\n2<\/output>\n2<\/result>","index":3428} +{"problem":"in a photography exhibition , some photographs were taken by octavia the photographer , and some photographs were framed by jack the framer . jack had framed 24 photographs taken by octavia , and 12 photographs taken by other photographers . if 36 of the photographs in the exhibition were taken by octavia , how many photographs were either framed by jack or taken by octavia ?","rationale":"the number of photographs either framed by jack or taken by octavia = { jack } + { octavia } - { overlap } = 36 + 36 - 24 = 48 ( { overlap } is green box ) . answer : b","correct":"b","options":{"a":"36 ","b":"48 ","c":"72 ","d":"96","e":"108"},"options_float":{"a":36.0,"b":48.0,"c":72.0,"d":96.0,"e":108.0},"annotated_formula":"add(add(24, 12), subtract(36, 24))","linear_formula":"add(n0,n1)|subtract(n2,n0)|add(#0,#1)","chain":"24 + 12<\/gadget>\n36<\/output>\n36 - 24<\/gadget>\n12<\/output>\n36 + 12<\/gadget>\n48<\/output>\n48<\/result>","index":3429} +{"problem":"what is value of ( p + q ) \/ ( p - q ) if p \/ q = 7 ?","rationale":"explanation : ( p + q ) \/ ( p - q ) = ( p \/ q + 1 ) \/ ( p \/ q - 1 ) = ( 7 + 1 ) \/ ( 7 - 1 ) = 4 \/ 3 . answer : c","correct":"c","options":{"a":"1 \/ 3 ","b":"2 \/ 3 ","c":"4 \/ 3 ","d":"5 \/ 3","e":"none of these"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":1.3333333333,"d":1.6666666667,"e":null},"annotated_formula":"divide(add(7, const_1), subtract(7, const_1))","linear_formula":"add(n0,const_1)|subtract(n0,const_1)|divide(#0,#1)","chain":"7 + 1<\/gadget>\n8<\/output>\n7 - 1<\/gadget>\n6<\/output>\n8 \/ 6<\/gadget>\n4\/3 = around 1.333333<\/output>\n4\/3 = around 1.333333<\/result>","index":3431} +{"problem":"find the area of a cuboid of length 8 cm , breadth 6 cm . and height 9 cm .","rationale":"\"area of a cuboid = lxbxh = 8 cm x 6 cm x 9 cm = 432 cm cube answer : d\"","correct":"d","options":{"a":"400 cm cube ","b":"410 cm cube ","c":"420 cm cube ","d":"432 cm cube","e":"480 cm cube"},"options_float":{"a":400.0,"b":410.0,"c":420.0,"d":432.0,"e":480.0},"annotated_formula":"multiply(multiply(8, 6), 9)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"8 * 6<\/gadget>\n48<\/output>\n48 * 9<\/gadget>\n432<\/output>\n432<\/result>","index":3432} +{"problem":"a windmill is taking advantage of strong air currents in order to produce electrical energy . on a typical day the wind speed is around 20 mph and in that speed the windmill produces 400 kw \/ h ( kilowatts per hour ) . on a stormy day a windmill produces 20 % more energy . how much kw \/ h can 3 windmills produce in two hours on a stormy day ?","rationale":"normal day = 400 kw \/ h stormy day = 400 * 1.2 = 480 kw \/ h 3 windmills for 2 hours on stormy day 480 * 3 * 2 = 2880 answer : a","correct":"a","options":{"a":"2880 . ","b":"4860 . ","c":"5780 ","d":"5760","e":"6380 ."},"options_float":{"a":2880.0,"b":4860.0,"c":5780.0,"d":5760.0,"e":6380.0},"annotated_formula":"multiply(multiply(add(400, divide(multiply(400, 20), const_100)), 3), const_2)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n1,#1)|multiply(n3,#2)|multiply(#3,const_2)","chain":"400 * 20<\/gadget>\n8_000<\/output>\n8_000 \/ 100<\/gadget>\n80<\/output>\n400 + 80<\/gadget>\n480<\/output>\n480 * 3<\/gadget>\n1_440<\/output>\n1_440 * 2<\/gadget>\n2_880<\/output>\n2_880<\/result>","index":3434} +{"problem":"ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 18 % if no discount was offered ?","rationale":"\"price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 \/ 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \\ total price of tv set = 15625 + 375 = rs . 16000 the price at whichthe tv should be sold to get a profit of 18 % if no discount was offered = 16000 * 118 \/ 100 = rs . 18880 . answer : e\"","correct":"e","options":{"a":"17608 ","b":"17606 ","c":"17604 ","d":"17600","e":"18880"},"options_float":{"a":17608.0,"b":17606.0,"c":17604.0,"d":17600.0,"e":18880.0},"annotated_formula":"divide(multiply(add(const_100, 18), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100)","linear_formula":"add(n4,const_100)|add(n2,n3)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#2,#3)|add(#1,#4)|multiply(#0,#5)|divide(#6,const_100)|","chain":"100 + 18<\/gadget>\n118<\/output>\n12_500 * 100<\/gadget>\n1_250_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n1_250_000 \/ 80<\/gadget>\n15_625<\/output>\n125 + 250<\/gadget>\n375<\/output>\n15_625 + 375<\/gadget>\n16_000<\/output>\n118 * 16_000<\/gadget>\n1_888_000<\/output>\n1_888_000 \/ 100<\/gadget>\n18_880<\/output>\n18_880<\/result>","index":3438} +{"problem":"a hall is 15 meters long and 15 meters wide . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , what is the volume of the hall ( in cubic meters ) ?","rationale":"\"2 hl + 2 hw = 2 lw h = lw \/ ( l + w ) volume = lwh = ( lw ) ^ 2 \/ ( l + w ) = 1687.5 the answer is d .\"","correct":"d","options":{"a":"975.5 ","b":"1250.5 ","c":"1465.5 ","d":"1687.5","e":"1824.5"},"options_float":{"a":975.5,"b":1250.5,"c":1465.5,"d":1687.5,"e":1824.5},"annotated_formula":"volume_rectangular_prism(15, 15, divide(multiply(rectangle_area(15, 15), const_2), rectangle_perimeter(15, 15)))","linear_formula":"rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|multiply(#0,const_2)|divide(#2,#1)|volume_rectangular_prism(n0,n1,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n225 * 2<\/gadget>\n450<\/output>\n2 * (15 + 15)<\/gadget>\n60<\/output>\n450 \/ 60<\/gadget>\n15\/2 = around 7.5<\/output>\n15 * 15 * (15\/2)<\/gadget>\n3_375\/2 = around 1_687.5<\/output>\n3_375\/2 = around 1_687.5<\/result>","index":3439} +{"problem":"what is the total surface area in square meters of a rectangular solid whose length is 9 meters , width is 8 meters , and depth is 5 meters ?","rationale":"\"surface area of a cuboid = 2 ( lb + bh + lh ) = 2 ( 9 * 8 + 8 * 5 + 9 * 5 ) = 2 ( 72 + 40 + 45 ) = 2 * 157 = 314 m 2 answer : e\"","correct":"e","options":{"a":"152 m 2 ","b":"162 m 2 ","c":"180 m 2 ","d":"300 m 2","e":"314 m 2"},"options_float":{"a":152.0,"b":162.0,"c":180.0,"d":300.0,"e":314.0},"annotated_formula":"surface_rectangular_prism(8, 9, 5)","linear_formula":"surface_rectangular_prism(n0,n1,n2)|","chain":"2 * (8 * 9 + 9 * 5 + 8 * 5)<\/gadget>\n314<\/output>\n314<\/result>","index":3440} +{"problem":"a car traveling at a certain constant speed takes 4 seconds longer to travel 1 km than it would take to travel 1 km at 80 km \/ hour . at what speed , in km \/ hr , is the car traveling ?","rationale":"\"time to cover 1 kilometer at 80 kilometers per hour is 1 \/ 80 hours = 3,600 \/ 80 seconds = 45 seconds ; time to cover 1 kilometer at regular speed is 45 + 4 = 49 seconds = 49 \/ 3,600 hours = 1 \/ 74 hours ; so , we get that to cover 1 kilometer 1 \/ 74 hours is needed - - > regular speed 74 kilometers per hour ( rate is a reciprocal of time or rate = distance \/ time ) . answer : c\"","correct":"c","options":{"a":"70 ","b":"72 ","c":"74 ","d":"75","e":"78"},"options_float":{"a":70.0,"b":72.0,"c":74.0,"d":75.0,"e":78.0},"annotated_formula":"divide(1, divide(add(multiply(const_3600, divide(1, 80)), 4), const_3600))","linear_formula":"divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)|","chain":"1 \/ 80<\/gadget>\n1\/80 = around 0.0125<\/output>\n3_600 * (1\/80)<\/gadget>\n45<\/output>\n45 + 4<\/gadget>\n49<\/output>\n49 \/ 3_600<\/gadget>\n49\/3_600 = around 0.013611<\/output>\n1 \/ (49\/3_600)<\/gadget>\n3_600\/49 = around 73.469388<\/output>\n3_600\/49 = around 73.469388<\/result>","index":3442} +{"problem":"a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12500 after a year ?","rationale":"6300 : 4200 : 10500 3 : 2 : 5 3 \/ 10 * 12100 = 3750 . answer : d","correct":"d","options":{"a":"3630 ","b":"2881 ","c":"2887 ","d":"3750","e":"2212"},"options_float":{"a":3630.0,"b":2881.0,"c":2887.0,"d":3750.0,"e":2212.0},"annotated_formula":"multiply(divide(6300, add(add(6300, 4200), 10500)), 12500)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)","chain":"6_300 + 4_200<\/gadget>\n10_500<\/output>\n10_500 + 10_500<\/gadget>\n21_000<\/output>\n6_300 \/ 21_000<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 12_500<\/gadget>\n3_750<\/output>\n3_750<\/result>","index":3443} +{"problem":"a student needs 30 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 100 marks , find the maximum marks set for the test .","rationale":"\"30 % = 180 marks 1 % = 6 marks 100 % = 600 marks the answer is b .\"","correct":"b","options":{"a":"400 ","b":"600 ","c":"800 ","d":"1000","e":"1200"},"options_float":{"a":400.0,"b":600.0,"c":800.0,"d":1000.0,"e":1200.0},"annotated_formula":"divide(add(80, 100), divide(30, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"80 + 100<\/gadget>\n180<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n180 \/ (3\/10)<\/gadget>\n600<\/output>\n600<\/result>","index":3445} +{"problem":"to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 8 hours a day ?","rationale":"\"man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 8 = 160 therefore number of days = 5760 \/ 240 = 36 days ans = b\"","correct":"b","options":{"a":"40 ","b":"36 ","c":"42 ","d":"44","e":"46"},"options_float":{"a":40.0,"b":36.0,"c":42.0,"d":44.0,"e":46.0},"annotated_formula":"divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 8))","linear_formula":"multiply(n1,n2)|multiply(n4,n6)|multiply(n3,#0)|multiply(n5,#2)|divide(#3,#1)|","chain":"30 * 12<\/gadget>\n360<\/output>\n360 * 8<\/gadget>\n2_880<\/output>\n2_880 * 2<\/gadget>\n5_760<\/output>\n20 * 8<\/gadget>\n160<\/output>\n5_760 \/ 160<\/gadget>\n36<\/output>\n36<\/result>","index":3446} +{"problem":"if x and j are integers and 2 x – j = 11 , then 4 x + j can not be","rationale":"2 x - j = 11 . . . . j = 2 x - 11 4 x + j = 4 x + 2 x - 11 = 6 x - 11 6 x - 11 = - 5 . . . x = 1 6 x - 11 = 1 . . . x = 2 6 x - 11 = 13 . . . x = 4 6 x - 11 = 17 . . x is not integer 6 x - 11 = 551 . . x is not integer i think the choice e is 55 not 551 . otherwise both de can not be solution = d","correct":"d","options":{"a":"– 5 ","b":"1 ","c":"13 ","d":"17","e":"551"},"options_float":{"a":5.0,"b":1.0,"c":13.0,"d":17.0,"e":551.0},"annotated_formula":"subtract(multiply(11, 2), subtract(11, add(4, 2)))","linear_formula":"add(n0,n2)|multiply(n0,n1)|subtract(n1,#0)|subtract(#1,#2)","chain":"11 * 2<\/gadget>\n22<\/output>\n4 + 2<\/gadget>\n6<\/output>\n11 - 6<\/gadget>\n5<\/output>\n22 - 5<\/gadget>\n17<\/output>\n17<\/result>","index":3447} +{"problem":"an investment yields an interest payment of $ 228 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ?","rationale":"\"let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 \/ 12 ) = ( 3 \/ 4 ) % ( 3 \/ 4 ) * ( p \/ 100 ) = 228 = > p = ( 228 * 4 * 10 ^ 2 ) \/ 3 = 76 * 4 * 10 ^ 2 = 304 * 10 ^ 2 answer b\"","correct":"b","options":{"a":"$ 28,300 ","b":"$ 30,400 ","c":"$ 31,300 ","d":"$ 32,500","e":"$ 35,100"},"options_float":{"a":28300.0,"b":30400.0,"c":31300.0,"d":32500.0,"e":35100.0},"annotated_formula":"multiply(divide(228, divide(9, multiply(const_3, const_4))), const_100)","linear_formula":"multiply(const_3,const_4)|divide(n1,#0)|divide(n0,#1)|multiply(#2,const_100)|","chain":"3 * 4<\/gadget>\n12<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n228 \/ (3\/4)<\/gadget>\n304<\/output>\n304 * 100<\/gadget>\n30_400<\/output>\n30_400<\/result>","index":3450} +{"problem":"joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 45 . he was allowed to drop his lowest score . after doing so , his average test score improved to 50 . what is the lowest test score that was dropped ?","rationale":"\"the arithmetic mean of 4 equally weighted tests was 45 . so what we can assume is that we have 4 test scores , each 45 . he dropped his lowest score and the avg went to 50 . this means that the lowest score was not 45 and other three scores had given the lowest score 5 each to make it up to 45 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 45 . so the lowest score = 45 - 15 = 30 answer ( b )\"","correct":"b","options":{"a":"20 ","b":"30 ","c":"55 ","d":"65","e":"80"},"options_float":{"a":20.0,"b":30.0,"c":55.0,"d":65.0,"e":80.0},"annotated_formula":"subtract(multiply(45, 4), multiply(50, const_3))","linear_formula":"multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)|","chain":"45 * 4<\/gadget>\n180<\/output>\n50 * 3<\/gadget>\n150<\/output>\n180 - 150<\/gadget>\n30<\/output>\n30<\/result>","index":3452} +{"problem":"if y > 0 , ( 4 y ) \/ 20 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"can be reduced to 2 y \/ 10 + 3 y \/ 10 = y \/ 2 = 50 % answer b\"","correct":"b","options":{"a":"40 % ","b":"50 % ","c":"60 % ","d":"70 %","e":"80 %"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(const_100, add(divide(4, 20), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"4 \/ 20<\/gadget>\n1\/5 = around 0.2<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/5) + (3\/10)<\/gadget>\n1\/2 = around 0.5<\/output>\n100 * (1\/2)<\/gadget>\n50<\/output>\n50<\/result>","index":3453} +{"problem":"in how many years rs 100 will produce the same interest at 5 % as rs . 600 produce in 4 years at 10 %","rationale":"\"explanation : clue : firstly we need to calculate the si with prinical 600 , time 4 years and rate 10 % , it will be rs . 240 then we can get the time as time = ( 100 * 240 ) \/ ( 100 * 5 ) = 48 option e\"","correct":"e","options":{"a":"44 ","b":"52 ","c":"50 ","d":"46","e":"48"},"options_float":{"a":44.0,"b":52.0,"c":50.0,"d":46.0,"e":48.0},"annotated_formula":"divide(multiply(divide(multiply(600, 10), const_100), 4), multiply(divide(5, const_100), 100))","linear_formula":"divide(n1,const_100)|multiply(n2,n4)|divide(#1,const_100)|multiply(n0,#0)|multiply(n3,#2)|divide(#4,#3)|","chain":"600 * 10<\/gadget>\n6_000<\/output>\n6_000 \/ 100<\/gadget>\n60<\/output>\n60 * 4<\/gadget>\n240<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n240 \/ 5<\/gadget>\n48<\/output>\n48<\/result>","index":3454} +{"problem":"in the number 10,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 55 , what is the greatest possible value of b × a ?","rationale":"you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 15 ( next number wo n ' t have 110 as the first 3 digits so we have only two options 0 and 25 ) . answer : e . ! please post ps questions in the ps subforum : gmat - problem - solving - ps - 140 \/ please post ds questions in the ds subforum : gmat - data - sufficiency - ds - 141 \/ no posting of ps \/ ds questions is allowed in the mainmath forum . d","correct":"d","options":{"a":"0 ","b":"5 ","c":"10 ","d":"15","e":"25"},"options_float":{"a":0.0,"b":5.0,"c":10.0,"d":15.0,"e":25.0},"annotated_formula":"subtract(multiply(reminder(reminder(multiply(add(divide(add(power(const_100, const_2), const_1000), 55), const_1), 55), const_100), const_10), divide(subtract(reminder(multiply(add(divide(add(power(const_100, const_2), const_1000), 55), const_1), 55), const_100), reminder(reminder(multiply(add(divide(add(power(const_100, const_2), const_1000), 55), const_1), 55), const_100), const_10)), const_10)), const_10)","linear_formula":"power(const_100,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,const_100)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)","chain":"100 ** 2<\/gadget>\n10_000<\/output>\n10_000 + 1_000<\/gadget>\n11_000<\/output>\n11_000 \/ 55<\/gadget>\n200<\/output>\n200 + 1<\/gadget>\n201<\/output>\n201 * 55<\/gadget>\n11_055<\/output>\n11_055 % 100<\/gadget>\n55<\/output>\n55 % 10<\/gadget>\n5<\/output>\n55 - 5<\/gadget>\n50<\/output>\n50 \/ 10<\/gadget>\n5<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25 - 10<\/gadget>\n15<\/output>\n15<\/result>","index":3455} +{"problem":"if mn ≠ 0 and 25 percent of n equals 37 1 \/ 2 percent of m , what is the value of 12 n \/ m ?","rationale":"given : 25 percent of n equals 37 ( 1 \/ 2 ) percent of m = > ( 25 \/ 100 ) * n = ( 75 \/ 200 ) * m = > 2 n = 3 m . 12 n \/ m = 6 * 2 n \/ m = 6 * 3 m \/ m = 18 . hence a .","correct":"a","options":{"a":"18 ","b":"323323 ","c":"8 ","d":"3","e":"98"},"options_float":{"a":18.0,"b":323323.0,"c":8.0,"d":3.0,"e":98.0},"annotated_formula":"add(divide(multiply(multiply(divide(37, const_100), const_100), const_10), multiply(divide(25, const_100), const_100)), const_3)","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#2,const_10)|divide(#4,#3)|add(#5,const_3)","chain":"37 \/ 100<\/gadget>\n37\/100 = around 0.37<\/output>\n(37\/100) * 100<\/gadget>\n37<\/output>\n37 * 10<\/gadget>\n370<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n370 \/ 25<\/gadget>\n74\/5 = around 14.8<\/output>\n(74\/5) + 3<\/gadget>\n89\/5 = around 17.8<\/output>\n89\/5 = around 17.8<\/result>","index":3457} +{"problem":"if the cost price of 18 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ?","rationale":"\"explanation : let cost price of 1 article be re . 1 . therefore , cost price of 18 articles = rs . 18 . selling price of 16 articles = rs . 18 therefore , selling price of 18 articles is : - = > 18 \/ 16 ã — 18 = > 20.25 . therefore , profit = selling price - cost price . = > 20.25 â ˆ ’ 18 = 2.25 . hence , the percentage of profit = profit x 100 \/ c . p . = > 2.25 \/ 18 ã — 100 . = > 12.5 % profit . answer : b\"","correct":"b","options":{"a":"20 % loss ","b":"12.5 % profit ","c":"33.33 % loss ","d":"30.33 % loss","e":"none of these"},"options_float":{"a":20.0,"b":12.5,"c":33.33,"d":30.33,"e":null},"annotated_formula":"multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 18)), divide(multiply(const_100, 16), 18)))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 * 16<\/gadget>\n1_600<\/output>\n1_600 \/ 18<\/gadget>\n800\/9 = around 88.888889<\/output>\n100 - (800\/9)<\/gadget>\n100\/9 = around 11.111111<\/output>\n(100\/9) \/ (800\/9)<\/gadget>\n1\/8 = around 0.125<\/output>\n100 * (1\/8)<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":3458} +{"problem":"if the length of an edge of cube p is thrice the length of an edge of cube q , what is the ratio of the volume of cube q to the volume of cube p ?","rationale":"the length of cube q = 1 ; the length of cube p = 3 ; the ratio of the volume of cube q to the volume of cube p = 1 ^ 3 \/ 3 ^ 3 = 1 \/ 27 answer : d","correct":"d","options":{"a":"1 \/ 4 ","b":"1 \/ 5 ","c":"1 \/ 6 ","d":"1 \/ 27","e":"1 \/ 9"},"options_float":{"a":0.25,"b":0.2,"c":0.1666666667,"d":0.037037037,"e":0.1111111111},"annotated_formula":"power(inverse(const_3), const_3)","linear_formula":"inverse(const_3)|power(#0,const_3)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) ** 3<\/gadget>\n1\/27 = around 0.037037<\/output>\n1\/27 = around 0.037037<\/result>","index":3460} +{"problem":"si on a certain sum of money for 3 yrs at 8 percent \/ annum is 1 \/ 2 the ci on rs . 4000 for 2 yrs at 10 percent \/ annum . the sum placed on si is ?","rationale":"c . i . = rs . 4000 x 1 + 10 2 - 4000 100 = rs . 4000 x 11 x 11 - 4000 10 10 = rs . 840 . sum = rs . 420 x 100 = rs . 1750 . 3 x 8 c","correct":"c","options":{"a":"rs . 1700 ","b":"rs . 1720 ","c":"rs . 1750 ","d":"rs . 1850","e":"rs . 2000"},"options_float":{"a":1700.0,"b":1720.0,"c":1750.0,"d":1850.0,"e":2000.0},"annotated_formula":"divide(multiply(divide(subtract(multiply(4000, power(add(1, divide(10, const_100)), const_2)), 4000), const_2), const_100), multiply(3, 8))","linear_formula":"divide(n6,const_100)|multiply(n0,n1)|add(n2,#0)|power(#2,const_2)|multiply(n4,#3)|subtract(#4,n4)|divide(#5,const_2)|multiply(#6,const_100)|divide(#7,#1)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n4_000 * (121\/100)<\/gadget>\n4_840<\/output>\n4_840 - 4_000<\/gadget>\n840<\/output>\n840 \/ 2<\/gadget>\n420<\/output>\n420 * 100<\/gadget>\n42_000<\/output>\n3 * 8<\/gadget>\n24<\/output>\n42_000 \/ 24<\/gadget>\n1_750<\/output>\n1_750<\/result>","index":3461} +{"problem":"what is the sum of the first 23 natural numbers ( starting from 1 ) ?","rationale":"the sum of n natural numbers = n * ( n + 1 ) \/ 2 = 23 * 24 \/ 2 = 276 the answer is c .","correct":"c","options":{"a":"256 ","b":"266 ","c":"276 ","d":"286","e":"296"},"options_float":{"a":256.0,"b":266.0,"c":276.0,"d":286.0,"e":296.0},"annotated_formula":"multiply(divide(23, const_2), add(subtract(23, 1), const_2))","linear_formula":"divide(n0,const_2)|subtract(n0,n1)|add(#1,const_2)|multiply(#2,#0)","chain":"23 \/ 2<\/gadget>\n23\/2 = around 11.5<\/output>\n23 - 1<\/gadget>\n22<\/output>\n22 + 2<\/gadget>\n24<\/output>\n(23\/2) * 24<\/gadget>\n276<\/output>\n276<\/result>","index":3463} +{"problem":"in a garden , 26 trees are planted at equal distances along a yard 700 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ?","rationale":"\"26 trees have 25 gaps between them . length of each gap = 700 \/ 25 = 28 i . e . , distance between two consecutive trees = 28 answer is b .\"","correct":"b","options":{"a":"10 ","b":"28 ","c":"12 ","d":"14","e":"16"},"options_float":{"a":10.0,"b":28.0,"c":12.0,"d":14.0,"e":16.0},"annotated_formula":"divide(700, subtract(26, const_1))","linear_formula":"subtract(n0,const_1)|divide(n1,#0)|","chain":"26 - 1<\/gadget>\n25<\/output>\n700 \/ 25<\/gadget>\n28<\/output>\n28<\/result>","index":3464} +{"problem":"what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 20 % ?","rationale":"\"the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 20 % increase a surface = 6 * ( ( 1.2 a ) ^ 2 ) = 6 * 1.44 * a ^ 2 the increase in the surface area = ( 6 * 1.44 * a ^ 2 - 6 * a ^ 2 ) \/ 6 * a ^ 2 = ( 6 * a ^ 2 ( 1.44 - 1 ) ) \/ ( 6 * a ^ 2 ) = 1.44 - 1 = 0.44 = 44 % answer : b\"","correct":"b","options":{"a":"33 % ","b":"44 % ","c":"55 % ","d":"56 %","e":"73 %"},"options_float":{"a":33.0,"b":44.0,"c":55.0,"d":56.0,"e":73.0},"annotated_formula":"multiply(subtract(power(add(const_1, divide(20, const_100)), const_2), const_1), const_100)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 2<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 100<\/gadget>\n44<\/output>\n44<\/result>","index":3466} +{"problem":"a semicircular shaped window has diameter of 63 cm . its perimeter equals","rationale":"explanation : perimeter of window = { \\ color { blue } \\ pi } r + 2 r = [ { ( 22 \/ 7 ) * ( 63 \/ 2 ) } + 63 ] = 99 + 63 = 162 cm answer : a ) 162","correct":"a","options":{"a":"162 ","b":"766 ","c":"298 ","d":"277","e":"198"},"options_float":{"a":162.0,"b":766.0,"c":298.0,"d":277.0,"e":198.0},"annotated_formula":"add(divide(circumface(divide(63, const_2)), const_2), 63)","linear_formula":"divide(n0,const_2)|circumface(#0)|divide(#1,const_2)|add(n0,#2)","chain":"63 \/ 2<\/gadget>\n63\/2 = around 31.5<\/output>\n2 * pi * (63\/2)<\/gadget>\n63*pi = around 197.920337<\/output>\n(63*pi) \/ 2<\/gadget>\n63*pi\/2 = around 98.960169<\/output>\n(63*pi\/2) + 63<\/gadget>\n63 + 63*pi\/2 = around 161.960169<\/output>\n63 + 63*pi\/2 = around 161.960169<\/result>","index":3470} +{"problem":"by selling 100 pens , a trader gains the cost of 30 pens . find his gain percentage ?","rationale":"\"let the cp of each pen be rs . 1 . cp of 100 pens = rs . 100 profit = cost of 30 pens = rs . 30 profit % = 30 \/ 100 * 100 = 30 % answer : a\"","correct":"a","options":{"a":"30 % ","b":"33 % ","c":"31 % ","d":"32 %","e":"34 %"},"options_float":{"a":30.0,"b":33.0,"c":31.0,"d":32.0,"e":34.0},"annotated_formula":"multiply(divide(30, 100), const_100)","linear_formula":"divide(n1,n0)|multiply(#0,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":3471} +{"problem":"out of 3 consecutive odd numbers 11 times the first number is equal to addition of thrice the third number and adding 16 to 4 times the second . what is the first number ?","rationale":"description : = > 11 x = 4 ( x + 2 ) + 16 + 3 ( x + 4 ) = > 11 x = 7 x + 36 = > 4 x = 36 x = 9 answer e","correct":"e","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"subtract(16, subtract(11, 4))","linear_formula":"subtract(n1,n3)|subtract(n2,#0)","chain":"11 - 4<\/gadget>\n7<\/output>\n16 - 7<\/gadget>\n9<\/output>\n9<\/result>","index":3472} +{"problem":"working together at their respective constant rates , machine a and machine b can produce 1600 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 1600 unit total would it produce ?","rationale":"1 \/ a + 1 \/ b = 1 \/ t 1 \/ a + 1 \/ 12 = 1 \/ 8 ( 50 % more of 8 is 12 ) 1 \/ a = 1 \/ 24 machine a can produce 1600 units in 24 hrs , so it can produce 1600 * 8 \/ 24 = 533.33 units is 8 hrs . 533.33 is 33.33 % of 1600 . b is the answer","correct":"b","options":{"a":"25 ","b":"33 ","c":"50 ","d":"67","e":"75"},"options_float":{"a":25.0,"b":33.0,"c":50.0,"d":67.0,"e":75.0},"annotated_formula":"multiply(divide(multiply(subtract(divide(1600, 8), divide(1600, multiply(add(divide(50, const_100), const_1), 8))), 8), 1600), const_100)","linear_formula":"divide(n0,n1)|divide(n2,const_100)|add(#1,const_1)|multiply(n1,#2)|divide(n0,#3)|subtract(#0,#4)|multiply(n1,#5)|divide(#6,n0)|multiply(#7,const_100)","chain":"1_600 \/ 8<\/gadget>\n200<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) + 1<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 8<\/gadget>\n12<\/output>\n1_600 \/ 12<\/gadget>\n400\/3 = around 133.333333<\/output>\n200 - (400\/3)<\/gadget>\n200\/3 = around 66.666667<\/output>\n(200\/3) * 8<\/gadget>\n1_600\/3 = around 533.333333<\/output>\n(1_600\/3) \/ 1_600<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":3473} +{"problem":"according to the formula f = 9 \/ 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 26 , by how much does the temperature in degrees celsius ( c ) increase ?","rationale":"\"you can plug in values . c = 5 \/ 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 26 = 58 - - > c = 5 \/ 9 * 26 = 14.44 . increase = 14.44 degrees . answer : b .\"","correct":"b","options":{"a":"9 ","b":"14.44 ","c":"47 ","d":"48 3 \/ 5","e":"59"},"options_float":{"a":9.0,"b":14.44,"c":47.0,"d":48.0,"e":59.0},"annotated_formula":"divide(multiply(26, 5), 9)","linear_formula":"multiply(n1,n3)|divide(#0,n0)|","chain":"26 * 5<\/gadget>\n130<\/output>\n130 \/ 9<\/gadget>\n130\/9 = around 14.444444<\/output>\n130\/9 = around 14.444444<\/result>","index":3474} +{"problem":"if price of t . v set is reduced by 20 % , then its sale increases by 80 % , find net effect on sale value","rationale":"\"- a + b + ( ( - a ) ( b ) \/ 100 ) = - 20 + 80 + ( - 20 * 80 ) \/ 100 = - 20 + 80 - 16 = 44 answer : a\"","correct":"a","options":{"a":"44 ","b":"45 ","c":"46 ","d":"47","e":"48"},"options_float":{"a":44.0,"b":45.0,"c":46.0,"d":47.0,"e":48.0},"annotated_formula":"multiply(subtract(multiply(divide(subtract(const_100, 20), const_100), divide(add(const_100, 80), const_100)), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n100 + 80<\/gadget>\n180<\/output>\n180 \/ 100<\/gadget>\n9\/5 = around 1.8<\/output>\n(4\/5) * (9\/5)<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 100<\/gadget>\n44<\/output>\n44<\/result>","index":3475} +{"problem":"the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average of the team ?","rationale":"\"let the average of the whole team be x years . 11 x - ( 26 + 29 ) = 9 ( x - 1 ) = 11 x - 9 x = 46 = 2 x = 46 = > x = 23 so , average age of the team is 23 years . answer a\"","correct":"a","options":{"a":"23 ","b":"25 ","c":"30 ","d":"22","e":"40"},"options_float":{"a":23.0,"b":25.0,"c":30.0,"d":22.0,"e":40.0},"annotated_formula":"divide(subtract(add(26, add(26, 3)), multiply(3, 3)), const_2)","linear_formula":"add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)|","chain":"26 + 3<\/gadget>\n29<\/output>\n26 + 29<\/gadget>\n55<\/output>\n3 * 3<\/gadget>\n9<\/output>\n55 - 9<\/gadget>\n46<\/output>\n46 \/ 2<\/gadget>\n23<\/output>\n23<\/result>","index":3477} +{"problem":"tom and marry each have a bag of marbles , each of which contains 4 red marbles , 10 blue marbles , and 6 yellow marbles . if tom and marry each select one marble from their respective bags , what is the probability that either tom or marry select a blue marble ?","rationale":"probability that either selects a red marble , p ( either ) = 1 - probability that neither selects a blue marble , p ( neither ) probability that tom selects non blue = 10 c 1 \/ 20 c 1 = 1 \/ 2 probability that marry selects non blue = 10 c 1 \/ 20 c 1 = 1 \/ 2 = > p ( either ) = 1 - ( 1 \/ 2 * 1 \/ 2 ) = 1 - 1 \/ 4 = 3 \/ 4 . c )","correct":"c","options":{"a":"1 \/ 7 ","b":"2 \/ 4 ","c":"3 \/ 4 ","d":"1 \/ 8","e":"1 \/ 16"},"options_float":{"a":0.1428571429,"b":0.5,"c":0.75,"d":0.125,"e":0.0625},"annotated_formula":"subtract(const_1, multiply(divide(10, add(add(4, 6), 10)), divide(10, add(add(4, 6), 10))))","linear_formula":"add(n0,n2)|add(n1,#0)|divide(n1,#1)|multiply(#2,#2)|subtract(const_1,#3)","chain":"4 + 6<\/gadget>\n10<\/output>\n10 + 10<\/gadget>\n20<\/output>\n10 \/ 20<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * (1\/2)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":3478} +{"problem":"if 7 cats can kill 7 rats in 7 minutes , how long will it take 100 cats to kill 100 rats ?","rationale":"\"it will take 7 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 7 minutes , so 100 cats can kill 100 rats in 7 minutes answer c\"","correct":"c","options":{"a":"5 minutes ","b":"6 minutes ","c":"7 minutes ","d":"8 minutes","e":"9 minutes"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"multiply(7, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"7 * 1<\/gadget>\n7<\/output>\n7<\/result>","index":3479} +{"problem":"at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 20.95 . what is the greatest number of hot dogs you can buy at this store with $ 300 ?","rationale":"to maximize number of hot dogs with 300 $ total number of hot dogs bought in 250 - pack = 20.95 * 14 = 293.30 $ amount remaining = 300 - 293.30 = 6.7 $ total number of hot dogs bought in 20 - pack = 3.05 * 2 = 6.10 $ amount remaining = 6.70 - 6.10 = 0.6 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 300 $ = 250 * 14 + 20 * 2 = 3,540 answer d","correct":"d","options":{"a":"3,108 ","b":"3,100 ","c":"3,108 ","d":"3,540","e":"3,256"},"options_float":{"a":3108.0,"b":3100.0,"c":3108.0,"d":3540.0,"e":3256.0},"annotated_formula":"multiply(divide(300, 20.95), 250)","linear_formula":"divide(n6,n5)|multiply(n4,#0)","chain":"300 \/ 20.95<\/gadget>\n14.319809<\/output>\n14.319809 * 250<\/gadget>\n3_579.95225<\/output>\n3_579.95225<\/result>","index":3480} +{"problem":"a person spends 1 \/ 3 rd of the money with him on clothes , 1 \/ 5 th of the remaining on food and 1 \/ 4 th of the remaining on travel . now , he is left with rs 200 . how much did he have with him in the beginning ?","rationale":"suppose the amount in the beginning was rs ’ x ’ money spent on clothes = rs 1 x \/ 3 balance = rs 2 x \/ 3 money spent on food = 1 \/ 5 of 2 x \/ 3 = rs 2 x \/ 15 balance = 2 x \/ 3 - 2 x \/ 15 = rs 8 x \/ 15 money spent on travel = 1 \/ 4 of 8 x \/ 15 = rs 2 x \/ 15 = 8 x \/ 15 - 2 x \/ 15 = 6 x \/ 15 = rs 2 x \/ 5 therefore 2 x \/ 5 = 200 = 500 answer : b","correct":"b","options":{"a":"s 200 ","b":"s 500 ","c":"s 300 ","d":"s 450","e":"s 550"},"options_float":{"a":200.0,"b":500.0,"c":300.0,"d":450.0,"e":550.0},"annotated_formula":"divide(200, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4)))","linear_formula":"divide(n0,n1)|subtract(n0,#0)|divide(#1,n3)|subtract(#1,#2)|divide(#3,n5)|subtract(#3,#4)|divide(n6,#5)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) \/ 5<\/gadget>\n2\/15 = around 0.133333<\/output>\n(2\/3) - (2\/15)<\/gadget>\n8\/15 = around 0.533333<\/output>\n(8\/15) \/ 4<\/gadget>\n2\/15 = around 0.133333<\/output>\n(8\/15) - (2\/15)<\/gadget>\n2\/5 = around 0.4<\/output>\n200 \/ (2\/5)<\/gadget>\n500<\/output>\n500<\/result>","index":3481} +{"problem":"the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 76 % of the required amount . what is the net difference in the expenditure on that product ?","rationale":"quantity x rate = price 1 x 1 = 1 0.76 x 1.25 = 0.95 decrease in price = ( 0.05 \/ 1 ) × 100 = 5 % c )","correct":"c","options":{"a":"12.5 % ","b":"13 % ","c":"5 % ","d":"17 %","e":"19 %"},"options_float":{"a":12.5,"b":13.0,"c":5.0,"d":17.0,"e":19.0},"annotated_formula":"divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 76)), const_100), multiply(const_100, const_100))","linear_formula":"add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)","chain":"100 * 100<\/gadget>\n10_000<\/output>\n100 + 25<\/gadget>\n125<\/output>\n125 * 76<\/gadget>\n9_500<\/output>\n10_000 - 9_500<\/gadget>\n500<\/output>\n500 * 100<\/gadget>\n50_000<\/output>\n50_000 \/ 10_000<\/gadget>\n5<\/output>\n5<\/result>","index":3482} +{"problem":"if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability r that n is divisible by 3 ?","rationale":"\"prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 \/ 16 . ans d . took me 3 : 20 mins .\"","correct":"d","options":{"a":"1 \/ 3 ","b":"¼ ","c":"9 \/ 25 ","d":"5 \/ 16","e":"0"},"options_float":{"a":0.3333333333,"b":null,"c":0.36,"d":0.3125,"e":0.0},"annotated_formula":"divide(add(3, const_2), multiply(const_4, const_4))","linear_formula":"add(n0,const_2)|multiply(const_4,const_4)|divide(#0,#1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n4 * 4<\/gadget>\n16<\/output>\n5 \/ 16<\/gadget>\n5\/16 = around 0.3125<\/output>\n5\/16 = around 0.3125<\/result>","index":3483} +{"problem":"0 - - - : - | - - - : - - | - : - - : - | - - : - - - | - : - - - - 1 on the number line above , the segment from 0 to 1 has been divided into fifths , as indicated by the large tick marks , and also into sevenths , as indicated by the small tick marks . what is the least possible distance x between any two of the tick marks ?","rationale":"divisions are 0 1 \/ 7 2 \/ 7 3 \/ 7 4 \/ 7 5 \/ 7 6 \/ 7 1 1 \/ 5 2 \/ 5 3 \/ 5 4 \/ 5 5 \/ 5 expressing the same in terms of the lcm of 57 , i . e 35 0 7 \/ 35 14 \/ 35 21 \/ 35 28 \/ 35 35 \/ 35 5 \/ 35 10 \/ 35 15 \/ 35 20 \/ 35 25 \/ 35 30 \/ 35 by comparing the divisions with each other we can see 1 \/ 35 is the shortest possible distance between any two selected divisions . x = 1 \/ 35 answer is b .","correct":"b","options":{"a":"1 \/ 70 ","b":"1 \/ 35 ","c":"2 \/ 35 ","d":"1 \/ 12","e":"1 \/ 7"},"options_float":{"a":0.0142857143,"b":0.0285714286,"c":0.0571428571,"d":0.0833333333,"e":0.1428571429},"annotated_formula":"divide(1, multiply(add(const_2, const_3), add(const_3, const_4)))","linear_formula":"add(const_2,const_3)|add(const_3,const_4)|multiply(#0,#1)|divide(n1,#2)","chain":"2 + 3<\/gadget>\n5<\/output>\n3 + 4<\/gadget>\n7<\/output>\n5 * 7<\/gadget>\n35<\/output>\n1 \/ 35<\/gadget>\n1\/35 = around 0.028571<\/output>\n1\/35 = around 0.028571<\/result>","index":3484} +{"problem":"find the area of a parallelogram with base 21 cm and height 11 cm ?","rationale":"\"area of a parallelogram = base * height = 21 * 11 = 231 cm 2 answer : a\"","correct":"a","options":{"a":"231 cm 2 ","b":"384 cm 2 ","c":"672 cm 2 ","d":"267 cm 2","e":"286 cm 2"},"options_float":{"a":231.0,"b":384.0,"c":672.0,"d":267.0,"e":286.0},"annotated_formula":"multiply(21, 11)","linear_formula":"multiply(n0,n1)|","chain":"21 * 11<\/gadget>\n231<\/output>\n231<\/result>","index":3485} +{"problem":"a cycle is bought for rs . 1120 and sold for rs . 1200 , find the gain percent ?","rationale":"\"explanation : 1120 - - - - 80 100 - - - - ? = > 7 % answer : a\"","correct":"a","options":{"a":"7 % ","b":"2 % ","c":"6 % ","d":"1 %","e":"8 %"},"options_float":{"a":7.0,"b":2.0,"c":6.0,"d":1.0,"e":8.0},"annotated_formula":"multiply(divide(subtract(1200, 1120), 1120), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_200 - 1_120<\/gadget>\n80<\/output>\n80 \/ 1_120<\/gadget>\n1\/14 = around 0.071429<\/output>\n(1\/14) * 100<\/gadget>\n50\/7 = around 7.142857<\/output>\n50\/7 = around 7.142857<\/result>","index":3486} +{"problem":"a trader bought a car at 20 % discount on its original price . he sold it at a 70 % increase on the price he bought it . what percent of profit did he make on the original price ?","rationale":"\"original price = 100 cp = 80 s = 80 * ( 170 \/ 100 ) = 136 100 - 136 = 36 % answer : e\"","correct":"e","options":{"a":"17 % ","b":"72 % ","c":"12 % ","d":"82 %","e":"36 %"},"options_float":{"a":17.0,"b":72.0,"c":12.0,"d":82.0,"e":36.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 70)), const_100), const_100), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n100 + 70<\/gadget>\n170<\/output>\n80 * 170<\/gadget>\n13_600<\/output>\n13_600 \/ 100<\/gadget>\n136<\/output>\n136 \/ 100<\/gadget>\n34\/25 = around 1.36<\/output>\n(34\/25) - 1<\/gadget>\n9\/25 = around 0.36<\/output>\n(9\/25) * 100<\/gadget>\n36<\/output>\n36<\/result>","index":3487} +{"problem":"a boat having a length 9 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :","rationale":"\"volume of water displaced = ( 9 x 3 x 0.01 ) m 3 = 0.27 m 3 . mass of man = volume of water displaced x density of water = ( 0.27 x 1000 ) kg = 270 kg . answer : e\"","correct":"e","options":{"a":"100 kg ","b":"120 kg ","c":"189 kg ","d":"180 kg","e":"270 kg"},"options_float":{"a":100.0,"b":120.0,"c":189.0,"d":180.0,"e":270.0},"annotated_formula":"multiply(multiply(multiply(9, 3), divide(1, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"9 * 3<\/gadget>\n27<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n27 * (1\/100)<\/gadget>\n27\/100 = around 0.27<\/output>\n(27\/100) * 1_000<\/gadget>\n270<\/output>\n270<\/result>","index":3488} +{"problem":"what annual payment will discharge a debt of rs . 1030 due in 2 years at the rate of 5 % compound interest ?","rationale":"\"explanation : let each installment be rs . x . then , x \/ ( 1 + 5 \/ 100 ) + x \/ ( 1 + 5 \/ 100 ) 2 = 1030 820 x + 1030 * 441 x = 553.94 so , value of each installment = rs . 553.94 answer : option a\"","correct":"a","options":{"a":"553.94 ","b":"551.25 ","c":"534.33 ","d":"543.33","e":"646.33"},"options_float":{"a":553.94,"b":551.25,"c":534.33,"d":543.33,"e":646.33},"annotated_formula":"divide(multiply(power(add(divide(5, const_100), const_1), 2), 1030), 2)","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|divide(#3,n1)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) + 1<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n(441\/400) * 1_030<\/gadget>\n45_423\/40 = around 1_135.575<\/output>\n(45_423\/40) \/ 2<\/gadget>\n45_423\/80 = around 567.7875<\/output>\n45_423\/80 = around 567.7875<\/result>","index":3489} +{"problem":"in a factory , there are 90 % technicians and 10 % non - technicians . if the 90 % of the technicians and 10 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ?","rationale":"\"total = 100 t = 90 nt = 10 90 * ( 10 \/ 100 ) = 910 * ( 90 \/ 100 ) = 9 9 + 9 = 18 = > 100 - 18 = 82 % answer : d\"","correct":"d","options":{"a":"62 % ","b":"57 % ","c":"52 % ","d":"82 %","e":"42 %"},"options_float":{"a":62.0,"b":57.0,"c":52.0,"d":82.0,"e":42.0},"annotated_formula":"subtract(const_100, add(multiply(90, divide(10, const_100)), multiply(divide(90, const_100), 10)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(n0,#0)|multiply(n1,#1)|add(#2,#3)|subtract(const_100,#4)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n90 * (1\/10)<\/gadget>\n9<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 10<\/gadget>\n9<\/output>\n9 + 9<\/gadget>\n18<\/output>\n100 - 18<\/gadget>\n82<\/output>\n82<\/result>","index":3490} +{"problem":"how many seconds will a 450 metre long train take to cross a man running with a speed of 6 km \/ hr in the direction of the moving train if the speed of the train is 60 km \/ hr ?","rationale":"explanation : speed of train relatively to man = ( 60 - 6 ) km \/ hr = 54 km \/ hr = ( 54 x 5 \/ 18 ) m \/ sec = 15 m \/ sec time taken to pass the man = ( 450 \/ 15 ) sec = 30 sec . answer : b","correct":"b","options":{"a":"25 ","b":"30 ","c":"40 ","d":"45","e":"60"},"options_float":{"a":25.0,"b":30.0,"c":40.0,"d":45.0,"e":60.0},"annotated_formula":"divide(450, multiply(const_0_2778, subtract(60, 6)))","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 - 6<\/gadget>\n54<\/output>\n(5\/18) * 54<\/gadget>\n15<\/output>\n450 \/ 15<\/gadget>\n30<\/output>\n30<\/result>","index":3491} +{"problem":"90 kg of an alloy a is mixed with 140 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 4 and alloy b has tin and copper in the ratio 2 : 5 , then the amount of tin in the new alloy is ?","rationale":"\"quantity of tin in 90 kg of a = 90 * 4 \/ 7 = 51.4 kg quantity of tin in 140 kg of b = 140 * 2 \/ 7 = 40 kg quantity of tin in the new alloy = 51.4 + 40 = 91.4 kg answer is d\"","correct":"d","options":{"a":"100.6 kg ","b":"95.4 kg ","c":"117.5 kg ","d":"91.4 kg","e":"92 kg"},"options_float":{"a":100.6,"b":95.4,"c":117.5,"d":91.4,"e":92.0},"annotated_formula":"add(multiply(divide(90, add(3, 4)), 4), multiply(divide(140, add(2, 5)), 2))","linear_formula":"add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)|","chain":"3 + 4<\/gadget>\n7<\/output>\n90 \/ 7<\/gadget>\n90\/7 = around 12.857143<\/output>\n(90\/7) * 4<\/gadget>\n360\/7 = around 51.428571<\/output>\n2 + 5<\/gadget>\n7<\/output>\n140 \/ 7<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n(360\/7) + 40<\/gadget>\n640\/7 = around 91.428571<\/output>\n640\/7 = around 91.428571<\/result>","index":3493} +{"problem":"a rectangular garden is to be twice as long as it is wide . if 180 yards of fencing , including the gate , will completely enclose the garden , what will be the length of the garden , in yards ?","rationale":"\"alternate approach backsolving ( using answer options to reach the correct answer ) can work wonders here if one is fast in calculations . given perimeter is 180 so , 2 ( l + b ) = 180 or , l + b = 90 now use the answer options ( given length ; breath will be half the length ) ( a ) 60 l = 60 ; b = 30 l + b = 90 ( b ) 70 l = 70 ; b = 35 l + b = 105 ( c ) 80 l = 80 ; b = 40 l + b = 120 ( d ) 90 l = 90 ; b = 45 l + b = 135 ( e ) 100 l = 100 ; b = 50 l + b = 150 thus you see no , need of any calculations , u can reach the correct option only by checking options ; correct answer will be ( a )\"","correct":"a","options":{"a":"60 ","b":"70 ","c":"80 ","d":"90","e":"100"},"options_float":{"a":60.0,"b":70.0,"c":80.0,"d":90.0,"e":100.0},"annotated_formula":"multiply(divide(180, multiply(add(const_2, const_1), const_2)), const_2)","linear_formula":"add(const_1,const_2)|multiply(#0,const_2)|divide(n0,#1)|multiply(#2,const_2)|","chain":"2 + 1<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n180 \/ 6<\/gadget>\n30<\/output>\n30 * 2<\/gadget>\n60<\/output>\n60<\/result>","index":3494} +{"problem":"if the sides of a cube are in the ratio 9 : 7 . what is the ratio of their diagonals ?","rationale":"\"explanation : diagonal of a cube = a √ 3 where a is side a 1 : a 2 = 9 : 7 d 1 : d 2 = 9 : 7 where √ 3 cancelled both side answer : c\"","correct":"c","options":{"a":"4 : 3 ","b":"16 : 9 ","c":"9 : 7 ","d":"3 : 4","e":"3 : 8"},"options_float":{"a":1.3333333333,"b":1.7777777778,"c":1.2857142857,"d":0.75,"e":0.375},"annotated_formula":"divide(9, 7)","linear_formula":"divide(n0,n1)|","chain":"9 \/ 7<\/gadget>\n9\/7 = around 1.285714<\/output>\n9\/7 = around 1.285714<\/result>","index":3498} +{"problem":"last year elaine spent 20 % of her annual earnings on rent . this year she earned 20 % more than last year and she spent 30 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ?","rationale":"\"for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 20 of this on rent . this year she earned 20 % more , or $ 120 she would ' ve spent 30 % of this on rent , or $ 36 do $ 36 \/ $ 20 this will give you 180 % b is the correct answer .\"","correct":"b","options":{"a":"185.5 ","b":"180.0 ","c":"167.5 ","d":"172.5","e":"177.5"},"options_float":{"a":185.5,"b":180.0,"c":167.5,"d":172.5,"e":177.5},"annotated_formula":"multiply(divide(multiply(add(divide(20, const_100), const_1), 30), 20), const_100)","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n2,#1)|divide(#2,n0)|multiply(#3,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 30<\/gadget>\n36<\/output>\n36 \/ 20<\/gadget>\n9\/5 = around 1.8<\/output>\n(9\/5) * 100<\/gadget>\n180<\/output>\n180<\/result>","index":3499} +{"problem":"two bullet train s 140 m and 200 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is :","rationale":"relative speed = ( 60 + 40 ) km \/ hr = 100 x 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = ( 140 + 200 ) m = 340 m . required time = 340 x 9 \/ 250 = 54 \/ 5 = 12.24 sec . answer b","correct":"b","options":{"a":"15.8 sec . ","b":"12.24 sec . ","c":"11.8 sec . ","d":"10.8 sec .","e":"08.8 sec ."},"options_float":{"a":15.8,"b":12.24,"c":11.8,"d":10.8,"e":8.8},"annotated_formula":"divide(add(140, 200), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)","chain":"140 + 200<\/gadget>\n340<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n340 \/ (250\/9)<\/gadget>\n306\/25 = around 12.24<\/output>\n306\/25 = around 12.24<\/result>","index":3501} +{"problem":"in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.5 times as old as robert ?","rationale":"\"in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 ro = 10 , ri = 40 now for each year we can calculate : 2000 : ri = 60 , ro = 90 thus ans : e\"","correct":"e","options":{"a":"1960 ","b":"1965 ","c":"1970 ","d":"1975","e":"2000"},"options_float":{"a":1960.0,"b":1965.0,"c":1970.0,"d":1975.0,"e":2000.0},"annotated_formula":"add(1950, add(1.5, add(4, 3)))","linear_formula":"add(n3,n1)|add(n4,#0)|add(n0,#1)|","chain":"4 + 3<\/gadget>\n7<\/output>\n1.5 + 7<\/gadget>\n8.5<\/output>\n1_950 + 8.5<\/gadget>\n1_958.5<\/output>\n1_958.5<\/result>","index":3502} +{"problem":"in what ratio must tea at rs . 60 per kg be mixed with tea at rs . 70 per kg so that the mixture must be worth rs . 67.5 per kg ?","rationale":"\"required ratio = 250 : 750 = 1 : 3 answer d\"","correct":"d","options":{"a":"3 : 1 ","b":"3 : 2 ","c":"4 : 3 ","d":"1 : 3","e":"none"},"options_float":{"a":3.0,"b":1.5,"c":1.3333333333,"d":0.3333333333,"e":null},"annotated_formula":"divide(subtract(70, 67.5), subtract(67.5, 60))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"70 - 67.5<\/gadget>\n2.5<\/output>\n67.5 - 60<\/gadget>\n7.5<\/output>\n2.5 \/ 7.5<\/gadget>\n0.333333<\/output>\n0.333333<\/result>","index":3504} +{"problem":"in what time will a train 50 m long cross an electric pole , it its speed be 360 km \/ hr ?","rationale":"\"speed = 360 * 5 \/ 18 = 100 m \/ sec time taken = 50 \/ 100 = 0.5 sec . answer : b\"","correct":"b","options":{"a":"0.25 ","b":"0.5 ","c":"1.5 ","d":"2.5","e":"3.5"},"options_float":{"a":0.25,"b":0.5,"c":1.5,"d":2.5,"e":3.5},"annotated_formula":"divide(50, multiply(360, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n360 * (5\/18)<\/gadget>\n100<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":3505} +{"problem":"each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.74 % , what fraction of the products was checked by worker y ?","rationale":"\"x : 0.5 % is 0.24 % - points from 0.74 % . y : 0.8 % is 0.06 % - points from 0.74 % . therefore the ratio of products checked by y : x is 4 : 1 . thus , worker y checked 4 \/ 5 of the products . the answer is e .\"","correct":"e","options":{"a":"1 \/ 2 ","b":"2 \/ 3 ","c":"3 \/ 5 ","d":"3 \/ 4","e":"4 \/ 5"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.6,"d":0.75,"e":0.8},"annotated_formula":"divide(subtract(0.74, 0.5), subtract(0.8, 0.5))","linear_formula":"subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)|","chain":"0.74 - 0.5<\/gadget>\n0.24<\/output>\n0.8 - 0.5<\/gadget>\n0.3<\/output>\n0.24 \/ 0.3<\/gadget>\n0.8<\/output>\n0.8<\/result>","index":3506} +{"problem":"workers at a campaign office have 2000 fliers to send out . if they send out 1 \/ 4 of them in the morning and 1 \/ 5 of the remaining ones out during the afternoon , how many are left for the next day ?","rationale":"\"( 1 \/ 4 ) * 2000 = 500 remaining = 2000 - 500 = 1500 ( 1 \/ 5 ) of remaining = ( 1 \/ 5 ) * 1600 = 320 remaining now = 1500 - 320 = 1180 answer : option e\"","correct":"e","options":{"a":"300 ","b":"800 ","c":"1100 ","d":"1200","e":"1180"},"options_float":{"a":300.0,"b":800.0,"c":1100.0,"d":1200.0,"e":1180.0},"annotated_formula":"subtract(subtract(2000, divide(2000, 4)), divide(subtract(2000, divide(2000, 4)), 5))","linear_formula":"divide(n0,n2)|subtract(n0,#0)|divide(#1,n4)|subtract(#1,#2)|","chain":"2_000 \/ 4<\/gadget>\n500<\/output>\n2_000 - 500<\/gadget>\n1_500<\/output>\n1_500 \/ 5<\/gadget>\n300<\/output>\n1_500 - 300<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":3508} +{"problem":"abcd is a square where ab = â ˆ š 4032 . let x be a point on ab and y be a point on cd such that ax = cy . compute the area of trapezoid axyd .","rationale":"note that trapezoids axy d and bxy c are congruent , so the area of axy d is always 4032 \/ 2 = 2016 . correct answer d","correct":"d","options":{"a":"3008 ","b":"2002 ","c":"1008 ","d":"2016","e":"3000"},"options_float":{"a":3008.0,"b":2002.0,"c":1008.0,"d":2016.0,"e":3000.0},"annotated_formula":"divide(4032, const_2)","linear_formula":"divide(n0,const_2)","chain":"4_032 \/ 2<\/gadget>\n2_016<\/output>\n2_016<\/result>","index":3509} +{"problem":"a sports equipment store sold ping pong rackets for a total of $ 637 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ?","rationale":"\"average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 637 x = 65 pairs were sold . answer : c\"","correct":"c","options":{"a":"50 ","b":"100 ","c":"65 ","d":"500","e":"980"},"options_float":{"a":50.0,"b":100.0,"c":65.0,"d":500.0,"e":980.0},"annotated_formula":"divide(637, 9.8)","linear_formula":"divide(n0,n1)|","chain":"637 \/ 9.8<\/gadget>\n65<\/output>\n65<\/result>","index":3512} +{"problem":"a farmer has enough food to feed her 1000 cows for 50 days . after 10 days some of the cows run away and the food was now enough for the same period of 50 days as before . how many cows ran away ?","rationale":"1000 - - - - 50 1000 - - - - 40 x - - - - 50 x * 50 = 1000 * 40 x = 800 1000 - - - - - - - 200 answer = b","correct":"b","options":{"a":"300 ","b":"200 ","c":"500 ","d":"400","e":"250"},"options_float":{"a":300.0,"b":200.0,"c":500.0,"d":400.0,"e":250.0},"annotated_formula":"multiply(1000, subtract(const_1, multiply(subtract(50, 10), divide(const_1, 50))))","linear_formula":"divide(const_1,n1)|subtract(n1,n2)|multiply(#0,#1)|subtract(const_1,#2)|multiply(n0,#3)","chain":"50 - 10<\/gadget>\n40<\/output>\n1 \/ 50<\/gadget>\n1\/50 = around 0.02<\/output>\n40 * (1\/50)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n1_000 * (1\/5)<\/gadget>\n200<\/output>\n200<\/result>","index":3513} +{"problem":"a certain number of men can do a work in 60 days . if there were 8 men less it could be finished in 10 days more . how many men are there ?","rationale":"\"the original number of men = 8 ( 60 + 10 ) \/ 10 = 56 men answer is a\"","correct":"a","options":{"a":"56 ","b":"42 ","c":"60 ","d":"45","e":"36"},"options_float":{"a":56.0,"b":42.0,"c":60.0,"d":45.0,"e":36.0},"annotated_formula":"divide(multiply(add(60, 10), 8), 10)","linear_formula":"add(n0,n2)|multiply(n1,#0)|divide(#1,n2)|","chain":"60 + 10<\/gadget>\n70<\/output>\n70 * 8<\/gadget>\n560<\/output>\n560 \/ 10<\/gadget>\n56<\/output>\n56<\/result>","index":3514} +{"problem":"if - 3 x + 4 y = 28 and 3 x - 2 y = 8 , what is the product of x and y ?","rationale":"given - 3 x + 4 y = 28 - - - eq 1 3 x - 2 y = 8 - - eq 2 sum both eqns we get 2 y = 36 = > y = 18 sub 2 y in eq 2 = > 3 x - 36 = 8 . = > x = 44 \/ 3 now xy = 44 \/ 3 * 18 = 264 . option a is correct answer .","correct":"a","options":{"a":"264 . ","b":"428 ","c":"464 . ","d":"622","e":"642"},"options_float":{"a":264.0,"b":428.0,"c":464.0,"d":622.0,"e":642.0},"annotated_formula":"multiply(divide(add(8, multiply(divide(add(28, 8), 2), 2)), 3), divide(add(28, 8), 2))","linear_formula":"add(n2,n5)|divide(#0,n4)|multiply(n4,#1)|add(n5,#2)|divide(#3,n0)|multiply(#4,#1)","chain":"28 + 8<\/gadget>\n36<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n18 * 2<\/gadget>\n36<\/output>\n8 + 36<\/gadget>\n44<\/output>\n44 \/ 3<\/gadget>\n44\/3 = around 14.666667<\/output>\n(44\/3) * 18<\/gadget>\n264<\/output>\n264<\/result>","index":3515} +{"problem":"two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 25 hours and 81 hours respectively . the ratio of their speeds is :","rationale":"\"let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 81 : â ˆ š 25 = 9 : 5 . answer c\"","correct":"c","options":{"a":"2 : 3 ","b":"4 : 3 ","c":"9 : 5 ","d":"9 : 16","e":"none of these"},"options_float":{"a":0.6666666667,"b":1.3333333333,"c":1.8,"d":0.5625,"e":null},"annotated_formula":"divide(sqrt(81), sqrt(25))","linear_formula":"sqrt(n1)|sqrt(n0)|divide(#0,#1)|","chain":"81 ** (1\/2)<\/gadget>\n9<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n9 \/ 5<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":3517} +{"problem":"a train 100 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ?","rationale":"\"let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 \/ 18 m \/ sec 100 \/ [ ( x + 5 ) * 5 \/ 18 ] = 6 30 ( x + 5 ) = 1800 x = 55 kmph answer is b\"","correct":"b","options":{"a":"45 kmph ","b":"55 kmph ","c":"32 kmph ","d":"61 kmph","e":"70 kmph"},"options_float":{"a":45.0,"b":55.0,"c":32.0,"d":61.0,"e":70.0},"annotated_formula":"subtract(divide(divide(100, 6), const_0_2778), 5)","linear_formula":"divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2)|","chain":"100 \/ 6<\/gadget>\n50\/3 = around 16.666667<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(50\/3) \/ (5\/18)<\/gadget>\n60<\/output>\n60 - 5<\/gadget>\n55<\/output>\n55<\/result>","index":3518} +{"problem":"evaluate : 96385 + 12 * 3 * 25 = ?","rationale":"\"according to order of operations , 12 ? 3 ? 25 ( division and multiplication ) is done first from left to right 12 \/ 2 = 4 * 25 = 100 hence 96385 + 12 * 3 * 2 = 96385 + 100 = 96485 correct answer d\"","correct":"d","options":{"a":"96385 ","b":"96358 ","c":"96785 ","d":"96485","e":"96855"},"options_float":{"a":96385.0,"b":96358.0,"c":96785.0,"d":96485.0,"e":96855.0},"annotated_formula":"subtract(96385, multiply(multiply(12, 3), 25))","linear_formula":"multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|","chain":"12 * 3<\/gadget>\n36<\/output>\n36 * 25<\/gadget>\n900<\/output>\n96_385 - 900<\/gadget>\n95_485<\/output>\n95_485<\/result>","index":3519} +{"problem":"if y > 0 , ( 9 y ) \/ 20 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"can be reduced to 9 y \/ 20 + 6 y \/ 20 = 15 y \/ 20 = 75 % answer e\"","correct":"e","options":{"a":"40 % ","b":"50 % ","c":"60 % ","d":"70 %","e":"75 %"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":75.0},"annotated_formula":"multiply(const_100, add(divide(9, 20), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"9 \/ 20<\/gadget>\n9\/20 = around 0.45<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(9\/20) + (3\/10)<\/gadget>\n3\/4 = around 0.75<\/output>\n100 * (3\/4)<\/gadget>\n75<\/output>\n75<\/result>","index":3521} +{"problem":"all the students of class are told to sit in circle shape . here the boy at the 5 th position is exactly opposite to 20 th boy . total number of boys in the class ?","rationale":"\"as half the circle shape consist of 20 - 5 = 15 boys , so total number of boys in full circle = 2 * 15 = 30 answer : a\"","correct":"a","options":{"a":"30 ","b":"40 ","c":"45 ","d":"25","e":"20"},"options_float":{"a":30.0,"b":40.0,"c":45.0,"d":25.0,"e":20.0},"annotated_formula":"multiply(subtract(20, 5), const_2)","linear_formula":"subtract(n1,n0)|multiply(#0,const_2)|","chain":"20 - 5<\/gadget>\n15<\/output>\n15 * 2<\/gadget>\n30<\/output>\n30<\/result>","index":3522} +{"problem":"an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how many meters f long ?","rationale":"\"let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m , leaving 1 m of empty space . thus , i believe the answer is b ) 1\"","correct":"b","options":{"a":"0.5 ","b":"1 ","c":"1.5 ","d":"2","e":"3"},"options_float":{"a":0.5,"b":1.0,"c":1.5,"d":2.0,"e":3.0},"annotated_formula":"subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5)))","linear_formula":"multiply(n1,const_4)|multiply(n2,const_4)|add(#0,#1)|subtract(n0,#2)|","chain":"4 * 2<\/gadget>\n8<\/output>\n4 * 1.5<\/gadget>\n6<\/output>\n8 + 6<\/gadget>\n14<\/output>\n15 - 14<\/gadget>\n1<\/output>\n1<\/result>","index":3524} +{"problem":"for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 24 ) ?","rationale":"f ( 24 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 = 2 ^ 12 * ( 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 ) - - > the greatest prime factor is 11 . answer : e .","correct":"e","options":{"a":"23 ","b":"19 ","c":"17 ","d":"13","e":"11"},"options_float":{"a":23.0,"b":19.0,"c":17.0,"d":13.0,"e":11.0},"annotated_formula":"subtract(divide(24, 2), const_1)","linear_formula":"divide(n8,n0)|subtract(#0,const_1)","chain":"24 \/ 2<\/gadget>\n12<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11<\/result>","index":3526} +{"problem":"a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 80 percent of books that were loaned out are returned and there are 66 books in the special collection at that time , how many books of the special collection were loaned out during that month ?","rationale":"\"there are 9 books less ( 75 - 66 ) which represents 20 % of the loaned books ( 100 - 80 ) so total loaned out books = 45 answer d\"","correct":"d","options":{"a":"20 ","b":"30 ","c":"35 ","d":"45","e":"55"},"options_float":{"a":20.0,"b":30.0,"c":35.0,"d":45.0,"e":55.0},"annotated_formula":"divide(subtract(75, 66), subtract(const_1, divide(80, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"75 - 66<\/gadget>\n9<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n9 \/ (1\/5)<\/gadget>\n45<\/output>\n45<\/result>","index":3530} +{"problem":"find the simple interest on rs . 70,000 at 16 2 \/ 3 % per year for 9 months .","rationale":"\"p = rs . 70000 , r = 50 \/ 3 % p . a and t = 9 \/ 12 years = 3 \/ 4 years . simple interest = ( p * r * t ) \/ 100 = rs . ( 70,000 * ( 50 \/ 3 ) * ( 3 \/ 4 ) * ( 1 \/ 100 ) ) = rs . 8750 answer is c .\"","correct":"c","options":{"a":"7500 ","b":"6500 ","c":"8750 ","d":"9500","e":"none of them"},"options_float":{"a":7500.0,"b":6500.0,"c":8750.0,"d":9500.0,"e":null},"annotated_formula":"multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100))","linear_formula":"add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600 * 100<\/gadget>\n60_000<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 100<\/gadget>\n900<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n900 * 10<\/gadget>\n9_000<\/output>\n60_000 + 9_000<\/gadget>\n69_000<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48 + 2<\/gadget>\n50<\/output>\n50 \/ 3<\/gadget>\n50\/3 = around 16.666667<\/output>\n69_000 * (50\/3)<\/gadget>\n1_150_000<\/output>\n2 * 6<\/gadget>\n12<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n1_150_000 * (3\/4)<\/gadget>\n862_500<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n862_500 * (1\/100)<\/gadget>\n8_625<\/output>\n8_625<\/result>","index":3532} +{"problem":"a is twice as good a workman as b and they took 6 days together to do the work b alone can do it in .","rationale":"\"wc = 2 : 1 2 x + x = 1 \/ 6 x = 1 \/ 18 = > 18 days answer : a\"","correct":"a","options":{"a":"18 days ","b":"12 days ","c":"29 days ","d":"25 days","e":"27 days"},"options_float":{"a":18.0,"b":12.0,"c":29.0,"d":25.0,"e":27.0},"annotated_formula":"multiply(divide(multiply(6, add(const_2, const_1)), const_2), const_2)","linear_formula":"add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)|","chain":"2 + 1<\/gadget>\n3<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18<\/result>","index":3533} +{"problem":"donovan and michael are racing around a circular 400 - meter track . if donovan runs each lap in 45 seconds and michael runs each lap in 36 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ?","rationale":"\"one way of approaching this question is by relative speed method 1 . speed \/ rate of donovan = distance \/ time = > 400 \/ 45 = > 80 \/ 9 2 . speed \/ rate of michael = distance \/ time = > 400 \/ 36 = > 100 \/ 9 relative speed between them = 100 \/ 9 - 80 \/ 9 = > 20 \/ 9 ( we subtract the rates if moving in the same direction and add the rates if moving in the opposite direction ) in order to pass donovan - distance to be covered = 400 , relative rate = 20 \/ 9 total time taken by micheal to surpass donovan = distance \/ rate = > 400 * 9 \/ 20 = > 3600 \/ 10 = > 180 no . of laps taken by michael = total time \/ michael ' s rate = > 180 \/ 36 = > 5 hence correct answer is 5 laps . d\"","correct":"d","options":{"a":"8 ","b":"7 ","c":"6 ","d":"5","e":"9"},"options_float":{"a":8.0,"b":7.0,"c":6.0,"d":5.0,"e":9.0},"annotated_formula":"divide(divide(400, subtract(divide(400, 36), divide(400, 45))), 36)","linear_formula":"divide(n0,n2)|divide(n0,n1)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n2)|","chain":"400 \/ 36<\/gadget>\n100\/9 = around 11.111111<\/output>\n400 \/ 45<\/gadget>\n80\/9 = around 8.888889<\/output>\n(100\/9) - (80\/9)<\/gadget>\n20\/9 = around 2.222222<\/output>\n400 \/ (20\/9)<\/gadget>\n180<\/output>\n180 \/ 36<\/gadget>\n5<\/output>\n5<\/result>","index":3534} +{"problem":"if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 9 people to do 9 times of that particular work ?","rationale":"\"3 people can do the work one time in one day . 1 person can do 1 \/ 3 of the work in one day . 9 people can do 9 \/ 3 of the work in one day . 9 people can do 9 times the work in 3 days . the answer is c .\"","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"9","e":"12"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":9.0,"e":12.0},"annotated_formula":"multiply(3, divide(9, 9))","linear_formula":"divide(n3,n3)|multiply(n0,#0)|","chain":"9 \/ 9<\/gadget>\n1<\/output>\n3 * 1<\/gadget>\n3<\/output>\n3<\/result>","index":3535} +{"problem":"the average age of students of a class is 15.8 years . the average age of boys in the class is 16.5 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is :","rationale":"\"let the ratio be k : 1 . then , k * 16.5 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.5 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 \/ 0.7 = 4 \/ 7 required ratio = 4 \/ 7 : 1 = 4 : 7 . answer : e\"","correct":"e","options":{"a":"2 : 5 ","b":"2 : 3 ","c":"2 : 4 ","d":"2 : 1","e":"4 : 7"},"options_float":{"a":0.4,"b":0.6666666667,"c":0.5,"d":2.0,"e":0.5714285714},"annotated_formula":"divide(subtract(15.8, 15.4), subtract(16.5, 15.8))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|","chain":"15.8 - 15.4<\/gadget>\n0.4<\/output>\n16.5 - 15.8<\/gadget>\n0.7<\/output>\n0.4 \/ 0.7<\/gadget>\n0.571429<\/output>\n0.571429<\/result>","index":3537} +{"problem":"4 wolves , wolverine , man - hunter , blue - wolf , and jasmine , caught rabbits for their evening meal . if man - hunter caught 3 more rabbits than blue - wolf did , jasmine caught 3 less than wolverine did , and wolverine caught 1 more than man - hunter did , how many more fewer rabbits did blue - wolf catch than jasmine ?","rationale":"in this word problem , the first key is to turn the story into equations : man - hunter caught 3 more than blue - wolf did : m = 3 + b jasmine caught 3 less than wolverine did : j = w - 3 wolverine caught 3 more than man - hunter did : w = 1 + m then , in classic gmat style , the problem does n ' t ask you to solve for a single variable , but instead to solve for the difference between b and j ( how many more rabbits did blue - wolf catch than jasmine did ? ) . this means that your goal is to get the equations in terms of b and j so that you can solve for the expression b - j . taking the first equation and solving for b , you have b = m - 3 . taking the second equation and solving for j you have j = w - 3 , and then with the third equation you can replace w with 1 + m so that both b and j are in terms of m : j = ( 1 + m ) - 3 . so if b = m - 3 and j = m - 2 , then : b - j = ( m - 3 ) - ( m - 2 ) b - j = m - 3 - m + 2 ( remove the parentheses by distributing the multiplication ) b - j = - 1 , so the answer is e .","correct":"e","options":{"a":"- 3 ","b":"- 2 ","c":"- 5 ","d":"- 12","e":"- 1"},"options_float":{"a":-3.0,"b":-2.0,"c":-5.0,"d":-12.0,"e":-1.0},"annotated_formula":"subtract(add(4, 1), add(3, 3))","linear_formula":"add(n0,n3)|add(n1,n1)|subtract(#0,#1)","chain":"4 + 1<\/gadget>\n5<\/output>\n3 + 3<\/gadget>\n6<\/output>\n5 - 6<\/gadget>\n-1<\/output>\n-1<\/result>","index":3539} +{"problem":"a bullet train 160 m long is running with a speed of 70 kmph . in what time will it pass a man who is running at 8 kmph in the direction opposite to that in which the bullet train is going ?","rationale":"\"c 7.38 sec speed of the bullet train relative to man = ( 70 + 8 ) kmph = 78 * 5 \/ 18 m \/ sec = 65 \/ 3 m \/ sec . time taken by the bullet train to cross the man = time taken by it to cover 160 m at ( 65 \/ 3 ) m \/ sec = ( 160 * 3 \/ 65 ) sec = 7.38 sec\"","correct":"c","options":{"a":"23 sec ","b":"15 sec ","c":"7.38 sec ","d":"6.92 sec","e":"16 sec"},"options_float":{"a":23.0,"b":15.0,"c":7.38,"d":6.92,"e":16.0},"annotated_formula":"divide(160, divide(multiply(add(70, 8), const_1000), const_3600))","linear_formula":"add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|","chain":"70 + 8<\/gadget>\n78<\/output>\n78 * 1_000<\/gadget>\n78_000<\/output>\n78_000 \/ 3_600<\/gadget>\n65\/3 = around 21.666667<\/output>\n160 \/ (65\/3)<\/gadget>\n96\/13 = around 7.384615<\/output>\n96\/13 = around 7.384615<\/result>","index":3540} +{"problem":"a man can do a job in 5 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ?","rationale":"\"1 day work of the three persons = ( 1 \/ 5 + 1 \/ 20 + 1 \/ 25 ) = 29 \/ 100 so , all three together will complete the work in 100 \/ 29 = 3.5 days . answer : c\"","correct":"c","options":{"a":"6.3 ","b":"6.9 ","c":"3.5 ","d":"6.1","e":"6.2"},"options_float":{"a":6.3,"b":6.9,"c":3.5,"d":6.1,"e":6.2},"annotated_formula":"divide(const_1, add(divide(const_1, 25), add(divide(const_1, 5), divide(const_1, 20))))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4)|","chain":"1 \/ 25<\/gadget>\n1\/25 = around 0.04<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/5) + (1\/20)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/25) + (1\/4)<\/gadget>\n29\/100 = around 0.29<\/output>\n1 \/ (29\/100)<\/gadget>\n100\/29 = around 3.448276<\/output>\n100\/29 = around 3.448276<\/result>","index":3542} +{"problem":"in a kilometer race , a beats b by 52 meters or 18 seconds . what time does a take to complete the race ?","rationale":"\"time taken by b run 1000 meters = ( 1000 * 18 ) \/ 52 = 346 sec . time taken by a = 346 - 18 = 328 sec . answer : c\"","correct":"c","options":{"a":"277 sec ","b":"190 sec ","c":"328 sec ","d":"367 sec","e":"376 sec"},"options_float":{"a":277.0,"b":190.0,"c":328.0,"d":367.0,"e":376.0},"annotated_formula":"subtract(divide(multiply(const_1, const_1000), divide(52, 18)), 18)","linear_formula":"divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|","chain":"1 * 1_000<\/gadget>\n1_000<\/output>\n52 \/ 18<\/gadget>\n26\/9 = around 2.888889<\/output>\n1_000 \/ (26\/9)<\/gadget>\n4_500\/13 = around 346.153846<\/output>\n(4_500\/13) - 18<\/gadget>\n4_266\/13 = around 328.153846<\/output>\n4_266\/13 = around 328.153846<\/result>","index":3546} +{"problem":"a towel , when bleached , lost 30 % of its length and 25 % of its breadth . what is the percentage decrease in area ?","rationale":"\"percentage change in area = ( − 30 − 25 + ( 30 × 25 ) \/ 100 ) % = − 47.5 % i . e . , area is decreased by 47.5 % answer : e\"","correct":"e","options":{"a":"24 % ","b":"30 % ","c":"44 % ","d":"54 %","e":"47.5 %"},"options_float":{"a":24.0,"b":30.0,"c":44.0,"d":54.0,"e":47.5},"annotated_formula":"divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), subtract(const_100, 25))), const_100)","linear_formula":"multiply(const_100,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#1,#2)|subtract(#0,#3)|divide(#4,const_100)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n100 - 30<\/gadget>\n70<\/output>\n100 - 25<\/gadget>\n75<\/output>\n70 * 75<\/gadget>\n5_250<\/output>\n10_000 - 5_250<\/gadget>\n4_750<\/output>\n4_750 \/ 100<\/gadget>\n95\/2 = around 47.5<\/output>\n95\/2 = around 47.5<\/result>","index":3548} +{"problem":"a train 270 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in :","rationale":"\"relative speed = ( 25 + 2 ) km \/ hr = 27 km \/ hr = ( 27 × 5 \/ 18 ) m \/ sec = 15 \/ 2 m \/ sec . time taken by the train to pass the man = ( 270 × 2 \/ 15 ) sec = 36 sec answer : c\"","correct":"c","options":{"a":"30 sec ","b":"32 sec ","c":"36 sec ","d":"38 sec","e":"40 sec"},"options_float":{"a":30.0,"b":32.0,"c":36.0,"d":38.0,"e":40.0},"annotated_formula":"multiply(const_3600, divide(divide(270, const_1000), add(25, 2)))","linear_formula":"add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)|","chain":"270 \/ 1_000<\/gadget>\n27\/100 = around 0.27<\/output>\n25 + 2<\/gadget>\n27<\/output>\n(27\/100) \/ 27<\/gadget>\n1\/100 = around 0.01<\/output>\n3_600 * (1\/100)<\/gadget>\n36<\/output>\n36<\/result>","index":3554} +{"problem":"speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph . a man rows to a place at a distance of 4864 km and comes back to the starting point . the total time taken by him is :","rationale":"\"explanation : speed downstream = ( 14 + 1.2 ) = 15.2 kmph speed upstream = ( 14 - 1.2 ) = 12.8 kmph total time taken = 4864 \/ 15.2 + 4864 \/ 12.8 = 320 + 380 = 700 hours . answer : a\"","correct":"a","options":{"a":"700 ","b":"200 ","c":"540 ","d":"120","e":"635"},"options_float":{"a":700.0,"b":200.0,"c":540.0,"d":120.0,"e":635.0},"annotated_formula":"add(divide(4864, add(14, 1.2)), divide(4864, subtract(14, 1.2)))","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3)|","chain":"14 + 1.2<\/gadget>\n15.2<\/output>\n4_864 \/ 15.2<\/gadget>\n320<\/output>\n14 - 1.2<\/gadget>\n12.8<\/output>\n4_864 \/ 12.8<\/gadget>\n380<\/output>\n320 + 380<\/gadget>\n700<\/output>\n700<\/result>","index":3555} +{"problem":"you enter a weight loss challenge game and manage to lose 13 % of your body weight . for the final weigh in you are forced to wear clothes that add 2 % to your weight . what percentage of weight loss is measured at the final weigh in ?","rationale":"( 100 % - 13 % ) * ( 100 % + 2 % ) = 0.87 * 1.02 = 11.26 % the weigh in records your weight loss at 11.26 % ! the answer is c","correct":"c","options":{"a":"13 % ","b":"9.22 % ","c":"11.26 % ","d":"14 %","e":"12 %"},"options_float":{"a":13.0,"b":9.22,"c":11.26,"d":14.0,"e":12.0},"annotated_formula":"multiply(subtract(const_1, multiply(subtract(const_1, divide(13, const_100)), add(const_1, divide(2, const_100)))), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100)","chain":"13 \/ 100<\/gadget>\n13\/100 = around 0.13<\/output>\n1 - (13\/100)<\/gadget>\n87\/100 = around 0.87<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n1 + (1\/50)<\/gadget>\n51\/50 = around 1.02<\/output>\n(87\/100) * (51\/50)<\/gadget>\n4_437\/5_000 = around 0.8874<\/output>\n1 - (4_437\/5_000)<\/gadget>\n563\/5_000 = around 0.1126<\/output>\n(563\/5_000) * 100<\/gadget>\n563\/50 = around 11.26<\/output>\n563\/50 = around 11.26<\/result>","index":3556} +{"problem":"i bought two books ; for rs . 500 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ?","rationale":"\"x * ( 85 \/ 100 ) = ( 500 - x ) 119 \/ 100 x = 291.67 answer : e\"","correct":"e","options":{"a":"337 ","b":"280 ","c":"299.67 ","d":"266.67","e":"291.67"},"options_float":{"a":337.0,"b":280.0,"c":299.67,"d":266.67,"e":291.67},"annotated_formula":"divide(multiply(500, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19)))","linear_formula":"add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|","chain":"100 + 19<\/gadget>\n119<\/output>\n500 * 119<\/gadget>\n59_500<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 + 119<\/gadget>\n204<\/output>\n59_500 \/ 204<\/gadget>\n875\/3 = around 291.666667<\/output>\n875\/3 = around 291.666667<\/result>","index":3557} +{"problem":"in a group of ducks and cows , the total number of legs are 40 more than twice the number of heads . find the total number of cows .","rationale":"\"let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 40 more than twice the number of heads = > 2 ( d + 2 c ) = 40 + 2 ( c + d ) = > d + 2 c = 20 + c + d = > 2 c = 20 + c = > c = 20 i . e . , total number of cows = 20 answer is e .\"","correct":"e","options":{"a":"12 ","b":"14 ","c":"16 ","d":"18","e":"20"},"options_float":{"a":12.0,"b":14.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"divide(40, const_2)","linear_formula":"divide(n0,const_2)|","chain":"40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":3560} +{"problem":"a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 102 . what was the number he chose ?","rationale":"\"solution : let x be the number he chose , then 2 * x * 138 = 102 2 x = 240 x = 120 correct answer c\"","correct":"c","options":{"a":"90 ","b":"100 ","c":"120 ","d":"160","e":"200"},"options_float":{"a":90.0,"b":100.0,"c":120.0,"d":160.0,"e":200.0},"annotated_formula":"divide(add(102, 138), 2)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"102 + 138<\/gadget>\n240<\/output>\n240 \/ 2<\/gadget>\n120<\/output>\n120<\/result>","index":3561} +{"problem":"the product of two successive numbers is 4160 . which is the smaller of the two numbers ?","rationale":"\"b 4160 from the given alternatives , 64 × 65 = 4160 ∴ smaller number = 64\"","correct":"b","options":{"a":"78 ","b":"64 ","c":"88 ","d":"67","e":"37"},"options_float":{"a":78.0,"b":64.0,"c":88.0,"d":67.0,"e":37.0},"annotated_formula":"sqrt(4160)","linear_formula":"sqrt(n0)|","chain":"4_160 ** (1\/2)<\/gadget>\n8*sqrt(65) = around 64.498062<\/output>\n8*sqrt(65) = around 64.498062<\/result>","index":3562} +{"problem":"soymilk are sold in packages of two or 3 only . if hannah bought 17 soymilk exactly , what could be the number of large packs hannah bought ?","rationale":"let number of packs of three = t let number of packs of two = w 3 t + 2 w = 17 now , we need to test for values of w . since sum 17 is odd and 2 w will always be even , therefore w ca n ' t be even . now , we can test for values w = 1 , 3 and 5 2 * 4 + 3 * 3 = 8 + 9 = 17 answer c","correct":"c","options":{"a":"1 ","b":"4 ","c":"3 ","d":"5","e":"8"},"options_float":{"a":1.0,"b":4.0,"c":3.0,"d":5.0,"e":8.0},"annotated_formula":"divide(subtract(17, multiply(const_2, const_4)), 3)","linear_formula":"multiply(const_2,const_4)|subtract(n1,#0)|divide(#1,n0)","chain":"2 * 4<\/gadget>\n8<\/output>\n17 - 8<\/gadget>\n9<\/output>\n9 \/ 3<\/gadget>\n3<\/output>\n3<\/result>","index":3563} +{"problem":"what ratio must a shopkeeper mix peas and soybean of rs . 16 and rs . 25 \/ kg , as to obtain a mixture of rs . 21.50 ?","rationale":"\"correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 3.50 : 5.50 = 7 : 11\"","correct":"c","options":{"a":"10 : 7 ","b":"9 : 8 ","c":"7 : 11 ","d":"13 : 11","e":"14 : 8"},"options_float":{"a":1.4285714286,"b":1.125,"c":0.6363636364,"d":1.1818181818,"e":1.75},"annotated_formula":"divide(subtract(25, 21.50), subtract(21.50, 16))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"25 - 21.5<\/gadget>\n3.5<\/output>\n21.5 - 16<\/gadget>\n5.5<\/output>\n3.5 \/ 5.5<\/gadget>\n0.636364<\/output>\n0.636364<\/result>","index":3564} +{"problem":"a 300 meter long train crosses a platform in 33 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ?","rationale":"\"speed = [ 300 \/ 18 ] m \/ sec = 50 \/ 3 m \/ sec . let the length of the platform be x meters . then , x + 300 \/ 33 = 50 \/ 3 3 ( x + 300 ) = 1650 è x = 250 m . answer : a\"","correct":"a","options":{"a":"250 ","b":"266 ","c":"230 ","d":"288","e":"271"},"options_float":{"a":250.0,"b":266.0,"c":230.0,"d":288.0,"e":271.0},"annotated_formula":"subtract(multiply(divide(300, 18), 33), 300)","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|","chain":"300 \/ 18<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 33<\/gadget>\n550<\/output>\n550 - 300<\/gadget>\n250<\/output>\n250<\/result>","index":3565} +{"problem":"bucket p has thrice the capacity as bucket q . it takes 60 turns for bucket p to fill the empty drum . how many turns it will take for both the buckets p and q , having each turn together to fill the empty drum ?","rationale":"\"let capacity of p be x liters then capacity of q = x \/ 3 liters capacity of the drum = 60 x required number of turns = 60 x \/ ( x + x \/ 3 ) = 60 x * 3 \/ 4 x = 45 answer is b\"","correct":"b","options":{"a":"30 ","b":"45 ","c":"50 ","d":"25","e":"60"},"options_float":{"a":30.0,"b":45.0,"c":50.0,"d":25.0,"e":60.0},"annotated_formula":"divide(const_1, add(divide(const_1, 60), divide(const_1, multiply(60, const_3))))","linear_formula":"divide(const_1,n0)|multiply(n0,const_3)|divide(const_1,#1)|add(#0,#2)|divide(const_1,#3)|","chain":"1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n60 * 3<\/gadget>\n180<\/output>\n1 \/ 180<\/gadget>\n1\/180 = around 0.005556<\/output>\n(1\/60) + (1\/180)<\/gadget>\n1\/45 = around 0.022222<\/output>\n1 \/ (1\/45)<\/gadget>\n45<\/output>\n45<\/result>","index":3566} +{"problem":"if 4 \/ ( 1 + 3 \/ x ) = 1 , then x =","rationale":"\"the expression 4 \/ ( 1 + 3 \/ x ) = 1 should have been equal to something . if 4 \/ ( 1 + 3 \/ x ) = 1 = > 4 x \/ ( x + 3 ) = 1 = > 4 x = x + 3 = > 3 x = 3 = > x = 1 correct option : b\"","correct":"b","options":{"a":"3 ","b":"1 ","c":"1 \/ 3 ","d":"- 1 \/ 3","e":"- 3"},"options_float":{"a":3.0,"b":1.0,"c":0.3333333333,"d":-0.3333333333,"e":-3.0},"annotated_formula":"divide(3, subtract(4, 1))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"4 - 1<\/gadget>\n3<\/output>\n3 \/ 3<\/gadget>\n1<\/output>\n1<\/result>","index":3568} +{"problem":"a 16 % stock yielding 10 % is quoted at :","rationale":"\"solution to earn rs . 10 , money invested = rs . 100 . to earn rs . 16 , money invested = rs . ( 100 \/ 10 x 16 ) = rs . 160 . â ˆ ´ market value of rs . 100 stock = rs . 160 answer e\"","correct":"e","options":{"a":"83.33 ","b":"110 ","c":"112 ","d":"120","e":"160"},"options_float":{"a":83.33,"b":110.0,"c":112.0,"d":120.0,"e":160.0},"annotated_formula":"multiply(divide(const_100, 10), 16)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 10<\/gadget>\n10<\/output>\n10 * 16<\/gadget>\n160<\/output>\n160<\/result>","index":3569} +{"problem":"there are , in a certain league , 20 teams , and each team face another team for a total of 10 times . how many games are played in the season ?","rationale":"\"1 st team : 190 2 nd team : 180 3 rd team : 170 similarly the last team will play 0 therefore ( 190 + 180 + 170 . . . . 10 ) 10 ( 19 + 18 + 17 + 16 . . . . 1 ) 10 ( n ( n + 1 ) \/ 2 ) = 10 ( 19 * 20 \/ 2 ) = 1900 answer : a\"","correct":"a","options":{"a":"1900 ","b":"3600 ","c":"2400 ","d":"180","e":"5400"},"options_float":{"a":1900.0,"b":3600.0,"c":2400.0,"d":180.0,"e":5400.0},"annotated_formula":"multiply(multiply(subtract(20, const_1), 10), divide(20, const_2))","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|multiply(#1,n1)|multiply(#0,#2)|","chain":"20 - 1<\/gadget>\n19<\/output>\n19 * 10<\/gadget>\n190<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n190 * 10<\/gadget>\n1_900<\/output>\n1_900<\/result>","index":3570} +{"problem":"if integer k is equal to the sum of all even multiples of 15 between 390 and 615 , what is the greatest prime factor of k ?","rationale":"\"if we break down what the stem is asking what is the sum of all mult of 30 between 390 and 600 . using arithmetic progression to find n : 600 = 390 + ( n - 1 ) 30 210 + 30 = 30 n 240 = 3 n = > n = 8 the sum would be : 8 * mean mean = [ 600 + 390 ] \/ 2 = 495 8 * 495 c\"","correct":"c","options":{"a":"5 ","b":"7 ","c":"8 ","d":"13","e":"17"},"options_float":{"a":5.0,"b":7.0,"c":8.0,"d":13.0,"e":17.0},"annotated_formula":"divide(add(divide(subtract(multiply(floor(divide(615, 15)), 15), multiply(floor(divide(390, 15)), 15)), 15), const_1), const_2)","linear_formula":"divide(n2,n0)|divide(n1,n0)|floor(#0)|floor(#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,#5)|divide(#6,n0)|add(#7,const_1)|divide(#8,const_2)|","chain":"615 \/ 15<\/gadget>\n41<\/output>\nfloor(41)<\/gadget>\n41<\/output>\n41 * 15<\/gadget>\n615<\/output>\n390 \/ 15<\/gadget>\n26<\/output>\nfloor(26)<\/gadget>\n26<\/output>\n26 * 15<\/gadget>\n390<\/output>\n615 - 390<\/gadget>\n225<\/output>\n225 \/ 15<\/gadget>\n15<\/output>\n15 + 1<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":3571} +{"problem":"a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 54 , what is the total amount ?","rationale":"\"x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 54 35 - - - ? = > 210 answer : e\"","correct":"e","options":{"a":"166 ","b":"105 ","c":"178 ","d":"177","e":"210"},"options_float":{"a":166.0,"b":105.0,"c":178.0,"d":177.0,"e":210.0},"annotated_formula":"add(add(multiply(divide(const_100, 45), 54), multiply(divide(30, 45), 54)), 54)","linear_formula":"divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)|","chain":"100 \/ 45<\/gadget>\n20\/9 = around 2.222222<\/output>\n(20\/9) * 54<\/gadget>\n120<\/output>\n30 \/ 45<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 54<\/gadget>\n36<\/output>\n120 + 36<\/gadget>\n156<\/output>\n156 + 54<\/gadget>\n210<\/output>\n210<\/result>","index":3573} +{"problem":"the expenditure of a businessman increase consistently by 10 % per year . if his present expenditure is rs . 20000 then what will his expenditure be after 2 years ?","rationale":"explanation : expenditure = 20000 x 1.1 x 1.1 = 24200 answer : option d","correct":"d","options":{"a":"24500 ","b":"24400 ","c":"24300 ","d":"24200","e":"24100"},"options_float":{"a":24500.0,"b":24400.0,"c":24300.0,"d":24200.0,"e":24100.0},"annotated_formula":"multiply(20000, power(divide(add(const_100, 10), const_100), 2))","linear_formula":"add(n0,const_100)|divide(#0,const_100)|power(#1,n2)|multiply(n1,#2)","chain":"100 + 10<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n20_000 * (121\/100)<\/gadget>\n24_200<\/output>\n24_200<\/result>","index":3574} +{"problem":"farm tax is levied on the 25 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :","rationale":"\"this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 \/ 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 \/ 25 ) = 50 % the correct answer is ( c ) .\"","correct":"c","options":{"a":"15 % ","b":"25 % ","c":"50 % ","d":"0.2083 %","e":"none"},"options_float":{"a":15.0,"b":25.0,"c":50.0,"d":0.2083,"e":null},"annotated_formula":"divide(multiply(multiply(divide(480, 3840), const_100), const_100), 25)","linear_formula":"divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|","chain":"480 \/ 3_840<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 100<\/gadget>\n1_250<\/output>\n1_250 \/ 25<\/gadget>\n50<\/output>\n50<\/result>","index":3575} +{"problem":"a textile manufacturing firm employees 100 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 100 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is :","rationale":"\"explanation : profit = 5 , 00,000 â ˆ ’ ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 ã — ( 99 \/ 100 ) â ˆ ’ 150000 ã — ( 99 \/ 100 ) â ˆ ’ 75000 . = > rs 2 , 68,000 . decrease in profit = > 2 , 75,000 â ˆ ’ 2 , 71,500 = > rs . 3,500 . answer : d\"","correct":"d","options":{"a":"13000 ","b":"7000 ","c":"10000 ","d":"3500","e":"none of these"},"options_float":{"a":13000.0,"b":7000.0,"c":10000.0,"d":3500.0,"e":null},"annotated_formula":"subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(100, 1), 100), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(100, 1), 100))), 75000))","linear_formula":"multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10)|","chain":"5 * 1_000<\/gadget>\n5_000<\/output>\n5_000 * 100<\/gadget>\n500_000<\/output>\n75_000 * 2<\/gadget>\n150_000<\/output>\n150_000 + 75_000<\/gadget>\n225_000<\/output>\n500_000 - 225_000<\/gadget>\n275_000<\/output>\n100 - 1<\/gadget>\n99<\/output>\n99 \/ 100<\/gadget>\n99\/100 = around 0.99<\/output>\n(99\/100) * 500_000<\/gadget>\n495_000<\/output>\n150_000 * (99\/100)<\/gadget>\n148_500<\/output>\n495_000 - 148_500<\/gadget>\n346_500<\/output>\n346_500 - 75_000<\/gadget>\n271_500<\/output>\n275_000 - 271_500<\/gadget>\n3_500<\/output>\n3_500<\/result>","index":3576} +{"problem":"it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 18 minutes , how much time will she spend doing yoga ? ( rounded to minutes )","rationale":"\"the ratio is 2 : 3 = gym : ride , so ( 18 ) ( 3 \/ 2 ) = 27 minutes at the gym , and 27 + 18 = 45 minutes exercise , so ( 2 \/ 3 ) ( 45 ) = 30 minutes yoga . answer : a\"","correct":"a","options":{"a":"30 min . ","b":"41 min . ","c":"17 min . ","d":"23 min .","e":"25 min ."},"options_float":{"a":30.0,"b":41.0,"c":17.0,"d":23.0,"e":25.0},"annotated_formula":"divide(multiply(18, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3))))","linear_formula":"add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)|","chain":"2 + 3<\/gadget>\n5<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n18 * (3\/5)<\/gadget>\n54\/5 = around 10.8<\/output>\n(3\/5) * (3\/5)<\/gadget>\n9\/25 = around 0.36<\/output>\n(54\/5) \/ (9\/25)<\/gadget>\n30<\/output>\n30<\/result>","index":3577} +{"problem":"when positive integer x is divided by 11 , the quotient is y and the remainder is 4 . when 2 x is divided by 8 , the quotient is 3 y and the remainder is 1 . what is the value of 13 y – x ?","rationale":"( 1 ) x = 11 y + 4 ( 2 ) 2 x = 24 y + 1 let ' s subtract equation ( 1 ) from equation ( 2 ) . 13 y - 3 = x 13 y - x = 3 the answer is b .","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"subtract(multiply(13, divide(subtract(multiply(2, 4), 1), subtract(multiply(8, 3), multiply(2, 11)))), add(multiply(11, divide(subtract(multiply(2, 4), 1), subtract(multiply(8, 3), multiply(2, 11)))), 4))","linear_formula":"multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n2)|subtract(#0,n5)|subtract(#1,#2)|divide(#3,#4)|multiply(n6,#5)|multiply(n0,#5)|add(n1,#7)|subtract(#6,#8)","chain":"2 * 4<\/gadget>\n8<\/output>\n8 - 1<\/gadget>\n7<\/output>\n8 * 3<\/gadget>\n24<\/output>\n2 * 11<\/gadget>\n22<\/output>\n24 - 22<\/gadget>\n2<\/output>\n7 \/ 2<\/gadget>\n7\/2 = around 3.5<\/output>\n13 * (7\/2)<\/gadget>\n91\/2 = around 45.5<\/output>\n11 * (7\/2)<\/gadget>\n77\/2 = around 38.5<\/output>\n(77\/2) + 4<\/gadget>\n85\/2 = around 42.5<\/output>\n(91\/2) - (85\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":3578} +{"problem":"what is the least number of squares tiles required to pave the floor of a room 15.17 m long and 9.02 m broad ?","rationale":"explanation : length of largest tile = h . c . f . of 1517 cm and 902 cm = 41 cm . area of each tile = required number of tiles = = 814 answer : a ) 814","correct":"a","options":{"a":"814 ","b":"266 ","c":"287 ","d":"999","e":"661"},"options_float":{"a":814.0,"b":266.0,"c":287.0,"d":999.0,"e":661.0},"annotated_formula":"divide(rectangle_area(multiply(15.17, const_100), multiply(9.02, const_100)), square_area(add(multiply(const_4, const_10), const_1)))","linear_formula":"multiply(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_1)|rectangle_area(#0,#1)|square_area(#3)|divide(#4,#5)","chain":"15.17 * 100<\/gadget>\n1_517<\/output>\n9.02 * 100<\/gadget>\n902<\/output>\n1_517 * 902<\/gadget>\n1_368_334<\/output>\n4 * 10<\/gadget>\n40<\/output>\n40 + 1<\/gadget>\n41<\/output>\n41 ** 2<\/gadget>\n1_681<\/output>\n1_368_334 \/ 1_681<\/gadget>\n814<\/output>\n814<\/result>","index":3579} +{"problem":"if a drawing has an outer circle of diameter 30 and an inner circle of diameter 24 , then what fraction of outer circle ' s surface is not covered by the inner circle ?","rationale":"the questions asks us to find the surface which is not covered by the inner circle i . e . , area of the surface not covered . where as circumference is the length along the edge of the circle , 2 * pi * r implies the length of the curve pi * r ^ 2 implies area enclosed by that curve . . hence area of the circle is considered for this problem area of the inner circle = pi * r ^ 2 , where r = 12 = > pi * 144 area of the outer circle = pi * 225 surface not covered by the inner circle = pi * 225 - pi * 144 = pi * 81 fraction of outer circle ' s surface is not covered by the inner circle = pi * 81 \/ total area of the outer circle = > pi * 81 \/ pi * 225 = 9 \/ 25 answer is d","correct":"d","options":{"a":"3 \/ 6 ","b":"8 \/ 24 ","c":"14 \/ 16 ","d":"9 \/ 25","e":"2 \/ 3"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.875,"d":0.36,"e":0.6666666667},"annotated_formula":"divide(subtract(multiply(const_pi, power(divide(30, const_2), const_2)), multiply(const_pi, power(divide(24, const_2), const_2))), multiply(const_pi, power(divide(30, const_2), const_2)))","linear_formula":"divide(n0,const_2)|divide(n1,const_2)|power(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi)|multiply(#3,const_pi)|subtract(#4,#5)|divide(#6,#4)","chain":"30 \/ 2<\/gadget>\n15<\/output>\n15 ** 2<\/gadget>\n225<\/output>\npi * 225<\/gadget>\n225*pi = around 706.858347<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12 ** 2<\/gadget>\n144<\/output>\npi * 144<\/gadget>\n144*pi = around 452.389342<\/output>\n(225*pi) - (144*pi)<\/gadget>\n81*pi = around 254.469005<\/output>\n(81*pi) \/ (225*pi)<\/gadget>\n9\/25 = around 0.36<\/output>\n9\/25 = around 0.36<\/result>","index":3581} +{"problem":"x can finish a work in 24 days . y can finish the same work in 16 days . yworked for 10 days and left the job . how many days does x alone need to finish the remaining work ?","rationale":"\"work done by x in 1 day = 1 \/ 24 work done by y in 1 day = 1 \/ 16 work done by y in 10 days = 10 \/ 16 = 5 \/ 8 remaining work = 1 – 5 \/ 8 = 3 \/ 8 number of days in which x can finish the remaining work = ( 3 \/ 8 ) \/ ( 1 \/ 24 ) = 9 e\"","correct":"e","options":{"a":"3 ","b":"5 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"divide(subtract(const_1, multiply(10, divide(const_1, 16))), divide(const_1, 24))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)|","chain":"1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n10 * (1\/16)<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(3\/8) \/ (1\/24)<\/gadget>\n9<\/output>\n9<\/result>","index":3583} +{"problem":"some of 45 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 40 % . what fraction of the original paint was replaced ?","rationale":"\"let total paint = 1 let amount replaced = x 45 ( 1 - x ) + 25 x = 40 x = 1 \/ 4 answer : b\"","correct":"b","options":{"a":"2 \/ 5 ","b":"1 \/ 4 ","c":"1 \/ 5 ","d":"1 \/ 7","e":"2 \/ 3"},"options_float":{"a":0.4,"b":0.25,"c":0.2,"d":0.1428571429,"e":0.6666666667},"annotated_formula":"divide(subtract(divide(40, const_100), divide(45, const_100)), subtract(divide(25, const_100), divide(45, const_100)))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n(2\/5) - (9\/20)<\/gadget>\n-1\/20 = around -0.05<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) - (9\/20)<\/gadget>\n-1\/5 = around -0.2<\/output>\n(-1\/20) \/ (-1\/5)<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":3584} +{"problem":"what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 59 ?","rationale":"\"explanation : let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 59 = rs . 4012 . answer : option e\"","correct":"e","options":{"a":"3944 ","b":"920 ","c":"7290 ","d":"6928","e":"4012"},"options_float":{"a":3944.0,"b":920.0,"c":7290.0,"d":6928.0,"e":4012.0},"annotated_formula":"multiply(square_perimeter(sqrt(289)), 59)","linear_formula":"sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|","chain":"289 ** (1\/2)<\/gadget>\n17<\/output>\n4 * 17<\/gadget>\n68<\/output>\n68 * 59<\/gadget>\n4_012<\/output>\n4_012<\/result>","index":3586} +{"problem":"at jerri ’ s pet shop , 60 cups of bird seed are used every 5 days to feed 20 parakeets . how many cups of bird seed would be required to feed 10 parakeets for 5 days ?","rationale":"let ' s go step - by - step . 60 cups are used over a 5 day period , which means 12 cups a day . this feeds 20 parakeets , which means each parakeet needs . 60 of a cup every day . for 10 parakeets , we need 10 * . 60 cups = 6 cups a day . over 5 days , we need 30 cups . choice d .","correct":"d","options":{"a":"45.6 ","b":"20 ","c":"9 ","d":"30","e":"65"},"options_float":{"a":45.6,"b":20.0,"c":9.0,"d":30.0,"e":65.0},"annotated_formula":"divide(60, const_2)","linear_formula":"divide(n0,const_2)","chain":"60 \/ 2<\/gadget>\n30<\/output>\n30<\/result>","index":3588} +{"problem":"a jar of 180 marbles is divided equally among a group of marble - players today . if 2 people joined the group in the future , each person would receive 1 marble less . how many people are there in the group today ?","rationale":"\"180 = 18 * 10 = 20 * 9 there are 18 people in the group today . the answer is b .\"","correct":"b","options":{"a":"16 ","b":"18 ","c":"20 ","d":"22","e":"24"},"options_float":{"a":16.0,"b":18.0,"c":20.0,"d":22.0,"e":24.0},"annotated_formula":"divide(subtract(sqrt(add(multiply(multiply(180, 2), const_4), power(2, 2))), 2), 2)","linear_formula":"multiply(n0,n1)|power(n1,n1)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n1)|divide(#5,n1)|","chain":"180 * 2<\/gadget>\n360<\/output>\n360 * 4<\/gadget>\n1_440<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n1_440 + 4<\/gadget>\n1_444<\/output>\n1_444 ** (1\/2)<\/gadget>\n38<\/output>\n38 - 2<\/gadget>\n36<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n18<\/result>","index":3590} +{"problem":"a cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and there by decreases his average by 0.4 . the number age of the family now is :","rationale":"\"let the number of wickets taken till the last match be x . then , ( 12.4 x + 26 ) \/ ( x + 5 ) = 12 = 12.4 x + 26 = 12 x + 60 = 0.4 x = 34 = x = 340 \/ 4 = 85 . answer : d\"","correct":"d","options":{"a":"18 ","b":"17 ","c":"76 ","d":"85","e":"86"},"options_float":{"a":18.0,"b":17.0,"c":76.0,"d":85.0,"e":86.0},"annotated_formula":"divide(subtract(multiply(5, subtract(12.4, 0.4)), 26), 0.4)","linear_formula":"subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)|","chain":"12.4 - 0.4<\/gadget>\n12<\/output>\n5 * 12<\/gadget>\n60<\/output>\n60 - 26<\/gadget>\n34<\/output>\n34 \/ 0.4<\/gadget>\n85<\/output>\n85<\/result>","index":3591} +{"problem":"the average of first 5 multiples of 5 is","rationale":"solution average = 5 ( 1 + 2 + 3 + 4 + 5 ) \/ 5 = 75 \/ 5 . = 15 answer e","correct":"e","options":{"a":"3 ","b":"6 ","c":"9 ","d":"12","e":"15"},"options_float":{"a":3.0,"b":6.0,"c":9.0,"d":12.0,"e":15.0},"annotated_formula":"divide(add(add(add(add(5, multiply(5, const_2)), multiply(5, const_3)), multiply(5, const_4)), multiply(5, 5)), 5)","linear_formula":"multiply(n0,const_2)|multiply(n0,const_3)|multiply(n0,const_4)|multiply(n0,n0)|add(n0,#0)|add(#4,#1)|add(#5,#2)|add(#6,#3)|divide(#7,n0)","chain":"5 * 2<\/gadget>\n10<\/output>\n5 + 10<\/gadget>\n15<\/output>\n5 * 3<\/gadget>\n15<\/output>\n15 + 15<\/gadget>\n30<\/output>\n5 * 4<\/gadget>\n20<\/output>\n30 + 20<\/gadget>\n50<\/output>\n5 * 5<\/gadget>\n25<\/output>\n50 + 25<\/gadget>\n75<\/output>\n75 \/ 5<\/gadget>\n15<\/output>\n15<\/result>","index":3592} +{"problem":"a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 50 percent of the drive from the center , what percent of the round - trip has the technician completed ?","rationale":"\"round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 50 % of 50 % i . e 25 % . so he completed 50 + 25 = 75 % of total trip . d\"","correct":"d","options":{"a":"70 ","b":"65 ","c":"60 ","d":"75","e":"80"},"options_float":{"a":70.0,"b":65.0,"c":60.0,"d":75.0,"e":80.0},"annotated_formula":"add(divide(const_100, const_2), divide(multiply(50, divide(const_100, const_2)), const_100))","linear_formula":"divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)|","chain":"100 \/ 2<\/gadget>\n50<\/output>\n50 * 50<\/gadget>\n2_500<\/output>\n2_500 \/ 100<\/gadget>\n25<\/output>\n50 + 25<\/gadget>\n75<\/output>\n75<\/result>","index":3593} +{"problem":"the difference of two numbers is 1415 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ?","rationale":"let the smaller number be x . then larger number = ( x + 1415 ) . x + 1415 = 6 x + 15 5 x = 1400 x = 280 smaller number = 280 . b )","correct":"b","options":{"a":"270 ","b":"280 ","c":"290 ","d":"300","e":"310"},"options_float":{"a":270.0,"b":280.0,"c":290.0,"d":300.0,"e":310.0},"annotated_formula":"divide(add(1415, 15), subtract(6, const_1))","linear_formula":"add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)","chain":"1_415 + 15<\/gadget>\n1_430<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_430 \/ 5<\/gadget>\n286<\/output>\n286<\/result>","index":3594} +{"problem":"the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 25 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ?","rationale":"\"1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 56.25 4 ) now 56.25 has to be added byx % in order to equal the original price . 56.25 + ( x % ) 56.25 = 100 . solving this eq for x , we get x = 77.78 ans is b .\"","correct":"b","options":{"a":"32.5 ","b":"77.78 ","c":"48 ","d":"65","e":"67.5"},"options_float":{"a":32.5,"b":77.78,"c":48.0,"d":65.0,"e":67.5},"annotated_formula":"multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(25, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(25, const_100)))))","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)|","chain":"100 - 25<\/gadget>\n75<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n75 * (1\/4)<\/gadget>\n75\/4 = around 18.75<\/output>\n75 - (75\/4)<\/gadget>\n225\/4 = around 56.25<\/output>\n100 - (225\/4)<\/gadget>\n175\/4 = around 43.75<\/output>\n(175\/4) \/ (225\/4)<\/gadget>\n7\/9 = around 0.777778<\/output>\n100 * (7\/9)<\/gadget>\n700\/9 = around 77.777778<\/output>\n700\/9 = around 77.777778<\/result>","index":3595} +{"problem":"of the 180 people at a party , 70 were women , and 30 women tried the appetizer . if 50 people did not try the appetizer , what is the total number of men who tried the appetizer ?","rationale":"\"total people at party = 180 women = 70 so men 180 - 70 = 110 no . of pple who tried appetizer = 180 - 50 ( given info ) = 130 no of women who tried appetizer = 30 so remaining ppl ( men ) who tried the appetizer = 130 - 30 = 100 correct option e\"","correct":"e","options":{"a":"40 ","b":"50 ","c":"60 ","d":"70","e":"100"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":100.0},"annotated_formula":"subtract(subtract(180, 50), 30)","linear_formula":"subtract(n0,n3)|subtract(#0,n2)|","chain":"180 - 50<\/gadget>\n130<\/output>\n130 - 30<\/gadget>\n100<\/output>\n100<\/result>","index":3597} +{"problem":"danny and steve are running towards each other , each one from his own house . danny can reach steve ' s house in 25 minutes of running , which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time , how much time longer will it take steve to reach the halfway point between their houses than danny to reach the halfway point between their houses ?","rationale":"\"if the distance between the two houses is d . then speed of danny = ( d \/ 25 ) speed of steve = ( d \/ 50 ) [ steve takes 50 minutes to reach danny ' s house . double the time taken by danny ) obviously speed of danny is twice that of steve . now in order to cover distance ( d \/ 2 ) [ they meet in middle ] , steve will take : ( d \/ 2 ) \/ ( d \/ 50 ) = 25 minutes . distance is same and danny ' s speed is twice of steve . thus he will take 12.5 minutes to cover this same distance . steve take 25 - 12.5 = 12.5 minutes longer . a is the answer .\"","correct":"a","options":{"a":"12.5 minutes . ","b":"25 minutes ","c":"35 minutes . ","d":"35 minutes","e":"75 minutes"},"options_float":{"a":12.5,"b":25.0,"c":35.0,"d":35.0,"e":75.0},"annotated_formula":"divide(25, const_2)","linear_formula":"divide(n0,const_2)|","chain":"25 \/ 2<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":3598} +{"problem":"we run a business that rents out canoes and kayaks . a canoe rental costs $ 15 per day , and a kayak rental costs $ 18 dollars per day . one day , our business rents out 3 canoes for every 2 kayaks and receives a total of $ 405 in revenue . how many more canoes than kayaks were rented out ?","rationale":"\"let x be the number of canoes . then 2 x \/ 3 is the number of kayaks . 15 x + ( 2 x \/ 3 ) * 18 = 405 15 x + 12 x = 405 27 x = 405 x = 15 ( canoes ) 2 x \/ 3 = 10 ( kayaks ) there were 15 - 10 = 5 more canoes rented out . the answer is a .\"","correct":"a","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"multiply(subtract(3, 2), divide(405, add(multiply(3, 15), multiply(2, 18))))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|subtract(n2,n3)|add(#0,#1)|divide(n4,#3)|multiply(#4,#2)|","chain":"3 - 2<\/gadget>\n1<\/output>\n3 * 15<\/gadget>\n45<\/output>\n2 * 18<\/gadget>\n36<\/output>\n45 + 36<\/gadget>\n81<\/output>\n405 \/ 81<\/gadget>\n5<\/output>\n1 * 5<\/gadget>\n5<\/output>\n5<\/result>","index":3599} +{"problem":"in a company the manager wants to give some gifts to all of the workers . in each block there are about 100 workers are there . the total amount for giving the gifts for all the workers is $ 4000 . the worth of the gift is $ 4 . how many blocks are there in the company ?","rationale":"each employee will get a gift worth of = $ 4 total employees = 4000 \/ 4 = 1000 total blocks = 1000 \/ 100 = 10 correct option is b","correct":"b","options":{"a":"12 ","b":"10 ","c":"15 ","d":"17","e":"20"},"options_float":{"a":12.0,"b":10.0,"c":15.0,"d":17.0,"e":20.0},"annotated_formula":"divide(divide(4000, 4), 100)","linear_formula":"divide(n1,n2)|divide(#0,n0)","chain":"4_000 \/ 4<\/gadget>\n1_000<\/output>\n1_000 \/ 100<\/gadget>\n10<\/output>\n10<\/result>","index":3601} +{"problem":"in what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way ?","rationale":"\"t = 60 \/ 36 * 18 \/ 5 = 6 sec answer : d\"","correct":"d","options":{"a":"2 ","b":"7 ","c":"8 ","d":"6","e":"9"},"options_float":{"a":2.0,"b":7.0,"c":8.0,"d":6.0,"e":9.0},"annotated_formula":"divide(60, multiply(36, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n6<\/result>","index":3602} +{"problem":"a man buys 54 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ?","rationale":"\"explanation : let marked price be re . 1 each c . p . of 54 pens = rs . 46 s . p . of 54 pens = 99 % of rs . 54 = rs . 53.46 profit % = ( profit \/ c . p . ) x 100 profit % = ( 7.46 \/ 46 ) x 100 = 16.21 % answer : c\"","correct":"c","options":{"a":"7.6 % ","b":"7.7 % ","c":"16.21 % ","d":"13.6 %","e":"7.8 %"},"options_float":{"a":7.6,"b":7.7,"c":16.21,"d":13.6,"e":7.8},"annotated_formula":"multiply(const_100, divide(subtract(multiply(54, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100)))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n2)|multiply(n0,#1)|subtract(#2,#0)|divide(#3,#0)|multiply(#4,const_100)|","chain":"100 - 1<\/gadget>\n99<\/output>\n54 * 99<\/gadget>\n5_346<\/output>\n46 * 100<\/gadget>\n4_600<\/output>\n5_346 - 4_600<\/gadget>\n746<\/output>\n746 \/ 4_600<\/gadget>\n373\/2_300 = around 0.162174<\/output>\n100 * (373\/2_300)<\/gadget>\n373\/23 = around 16.217391<\/output>\n373\/23 = around 16.217391<\/result>","index":3604} +{"problem":"find the area of a cuboid of length 15 cm , breadth 10 cm . and height 16 cm .","rationale":"\"area of a cuboid = lxbxh = 15 cm x 10 cm x 16 cm = 2400 cm cube answer : a\"","correct":"a","options":{"a":"2400 cm cube ","b":"2410 cm cube ","c":"2420 cm cube ","d":"2430 cm cube","e":"2480 cm cube"},"options_float":{"a":2400.0,"b":2410.0,"c":2420.0,"d":2430.0,"e":2480.0},"annotated_formula":"multiply(multiply(15, 10), 16)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"15 * 10<\/gadget>\n150<\/output>\n150 * 16<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":3605} +{"problem":"a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 6 days . in how many days b alone can do the work ?","rationale":"\"explanation : a ’ s 5 day work = 5 * 1 \/ 15 = 1 \/ 3 remaining work = 1 - 1 \/ 3 = 2 \/ 3 b completes 2 \/ 3 work in 6 days b alone can do in x days 2 \/ 3 * x = 6 x = 9 days answer : option d\"","correct":"d","options":{"a":"5 days ","b":"7 days ","c":"12 days ","d":"9 days","e":"10 days"},"options_float":{"a":5.0,"b":7.0,"c":12.0,"d":9.0,"e":10.0},"annotated_formula":"inverse(multiply(inverse(6), subtract(const_1, multiply(5, inverse(15)))))","linear_formula":"inverse(n2)|inverse(n0)|multiply(n1,#1)|subtract(const_1,#2)|multiply(#0,#3)|inverse(#4)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n5 * (1\/15)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/6) * (2\/3)<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9<\/result>","index":3606} +{"problem":"the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 15 and 41 inclusive .","rationale":"\"we ' re dealing with a sequence of consecutive odd integers : 15 to 41 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 15 + 41 = 56 2 ) now look at the ' next smallest ' and the ' next biggest ' : 17 + 39 = 56 now we have proof that there is no middle term . we have 7 bunches of 56 . 7 ( 56 ) = 392 b\"","correct":"b","options":{"a":"351 ","b":"392 ","c":"410 ","d":"424","e":"450"},"options_float":{"a":351.0,"b":392.0,"c":410.0,"d":424.0,"e":450.0},"annotated_formula":"subtract(power(divide(add(41, const_1), 2), 2), power(divide(add(subtract(15, 2), const_1), 2), 2))","linear_formula":"add(n2,const_1)|subtract(n1,n0)|add(#1,const_1)|divide(#0,n0)|divide(#2,n0)|power(#3,n0)|power(#4,n0)|subtract(#5,#6)|","chain":"41 + 1<\/gadget>\n42<\/output>\n42 \/ 2<\/gadget>\n21<\/output>\n21 ** 2<\/gadget>\n441<\/output>\n15 - 2<\/gadget>\n13<\/output>\n13 + 1<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7 ** 2<\/gadget>\n49<\/output>\n441 - 49<\/gadget>\n392<\/output>\n392<\/result>","index":3607} +{"problem":"the cost price of a radio is rs . 1800 and it was sold for rs . 1430 , find the loss % ?","rationale":"\"1800 - - - - 370 100 - - - - ? = > 20.5 % answer : b\"","correct":"b","options":{"a":"18 % ","b":"20.5 % ","c":"30 % ","d":"45 %","e":"12 %"},"options_float":{"a":18.0,"b":20.5,"c":30.0,"d":45.0,"e":12.0},"annotated_formula":"multiply(divide(subtract(1800, 1430), 1800), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_800 - 1_430<\/gadget>\n370<\/output>\n370 \/ 1_800<\/gadget>\n37\/180 = around 0.205556<\/output>\n(37\/180) * 100<\/gadget>\n185\/9 = around 20.555556<\/output>\n185\/9 = around 20.555556<\/result>","index":3608} +{"problem":"what is the dividend . divisor 17 , the quotient is 8 and the remainder is 5","rationale":"\"d = d * q + r d = 17 * 8 + 5 d = 136 + 5 d = 141 answer : b\"","correct":"b","options":{"a":"145 ","b":"141 ","c":"150 ","d":"153","e":"158"},"options_float":{"a":145.0,"b":141.0,"c":150.0,"d":153.0,"e":158.0},"annotated_formula":"add(multiply(17, 8), 5)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"17 * 8<\/gadget>\n136<\/output>\n136 + 5<\/gadget>\n141<\/output>\n141<\/result>","index":3611} +{"problem":"a searchlight on top of the watch - tower makes 3 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 15 seconds ?","rationale":"\"3 revolutions per minute = 1 revolution every 20 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than 20 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least 15 seconds ) = 1 - p ( man in dark for max of 15 seconds ) = 1 - 15 \/ 20 = 1 - 3 \/ 4 = 1 \/ 4 or the other way would be : p ( man in dark for at least 5 seconds ) is like saying he can be in dark for 5 , 6,7 . . . all the way to 20 seconds because that is the max . in this approach it would be 5 \/ 20 seconds = 1 \/ 4 . answer is a\"","correct":"a","options":{"a":"1 \/ 4 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"2 \/ 3","e":"3 \/ 4"},"options_float":{"a":0.25,"b":0.3333333333,"c":0.5,"d":0.6666666667,"e":0.75},"annotated_formula":"subtract(const_1, divide(15, divide(multiply(const_60, const_1), 3)))","linear_formula":"multiply(const_1,const_60)|divide(#0,n0)|divide(n1,#1)|subtract(const_1,#2)|","chain":"60 * 1<\/gadget>\n60<\/output>\n60 \/ 3<\/gadget>\n20<\/output>\n15 \/ 20<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":3612} +{"problem":"john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of 3 , 4 , or 5 miles per hour and amanda runs at a constant rate of 3 , 4 , 5 , 6 , or 7 miles per hour . what is the probability that john has traveled farther than amanda by the time they meet ?","rationale":"john will run farther if he runs at 5 mph and amanda runs at 4 mph or 3 mph . in this case , p ( john runs farther ) = 1 \/ 3 * 2 \/ 5 = 2 \/ 15 john will run farther if he runs at 4 mph and amanda runs at 3 mph . in this case , p ( john runs farther ) = 1 \/ 3 * 1 \/ 5 = 1 \/ 15 p ( john runs farther ) = 2 \/ 15 + 1 \/ 15 = 3 \/ 15 = 1 \/ 5 the answer is c .","correct":"c","options":{"a":"1 \/ 2 ","b":"1 \/ 3 ","c":"1 \/ 5 ","d":"2 \/ 15","e":"4 \/ 15"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.2,"d":0.1333333333,"e":0.2666666667},"annotated_formula":"divide(const_3, multiply(3, 5))","linear_formula":"multiply(n0,n2)|divide(const_3,#0)","chain":"3 * 5<\/gadget>\n15<\/output>\n3 \/ 15<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":3613} +{"problem":"two bullet train s 140 m and 210 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is :","rationale":"\"relative speed = ( 60 + 40 ) km \/ hr = 100 x 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = ( 140 + 210 ) m = 350 m . required time = 350 x 9 \/ 250 = 54 \/ 5 = 12.6 sec . answer b\"","correct":"b","options":{"a":"15.8 sec . ","b":"12.6 sec . ","c":"11.8 sec . ","d":"10.8 sec .","e":"08.8 sec ."},"options_float":{"a":15.8,"b":12.6,"c":11.8,"d":10.8,"e":8.8},"annotated_formula":"divide(add(140, 210), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"140 + 210<\/gadget>\n350<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n350 \/ (250\/9)<\/gadget>\n63\/5 = around 12.6<\/output>\n63\/5 = around 12.6<\/result>","index":3615} +{"problem":"4 friends are pooling their money for pizza . emma has 8 dollars . daya has 25 % more money than emma , jeff has 2 \/ 5 of the money daya has , and brenda has 4 more dollars than jeff . how much money does brenda have ?","rationale":"d 8 emma has 8 dollars . daya has 25 % more dollars than emma . daya = 8 + 8 x 25 % = 8 + 8 x . 25 = 8 + 2 = 10 jeff has 2 \/ 5 of what daya has . = 10 x 2 \/ 5 = 4 brenda has 4 more dollars than jeff . = 4 + 4 = 8 brenda has 8 dollars . the correct answer is d .","correct":"d","options":{"a":"14 ","b":"6 ","c":"7 ","d":"8","e":"10"},"options_float":{"a":14.0,"b":6.0,"c":7.0,"d":8.0,"e":10.0},"annotated_formula":"add(divide(multiply(multiply(divide(add(25, const_100), const_100), 8), 2), 5), 4)","linear_formula":"add(n2,const_100)|divide(#0,const_100)|multiply(n1,#1)|multiply(n3,#2)|divide(#3,n4)|add(n0,#4)","chain":"25 + 100<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 8<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n4 + 4<\/gadget>\n8<\/output>\n8<\/result>","index":3616} +{"problem":"in one alloy there is 12 % chromium while in another alloy it is 8 % . 35 kg of the first alloy was melted together with 40 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy .","rationale":"\"the amount of chromium in the new 35 + 45 = 80 kg alloy is 0.12 * 35 + 0.08 * 45 = 7.8 kg , so the percentage is 7.8 \/ 80 * 100 = 9.75 % . answer : b .\"","correct":"b","options":{"a":"8.8 % ","b":"9.75 % ","c":"9.2 % ","d":"8.6 %","e":"8.4 %"},"options_float":{"a":8.8,"b":9.75,"c":9.2,"d":8.6,"e":8.4},"annotated_formula":"multiply(divide(add(divide(multiply(12, 35), const_100), divide(multiply(8, 40), const_100)), add(35, 40)), const_100)","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"12 * 35<\/gadget>\n420<\/output>\n420 \/ 100<\/gadget>\n21\/5 = around 4.2<\/output>\n8 * 40<\/gadget>\n320<\/output>\n320 \/ 100<\/gadget>\n16\/5 = around 3.2<\/output>\n(21\/5) + (16\/5)<\/gadget>\n37\/5 = around 7.4<\/output>\n35 + 40<\/gadget>\n75<\/output>\n(37\/5) \/ 75<\/gadget>\n37\/375 = around 0.098667<\/output>\n(37\/375) * 100<\/gadget>\n148\/15 = around 9.866667<\/output>\n148\/15 = around 9.866667<\/result>","index":3617} +{"problem":"mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 26 miles , how many miles was mike ' s ride ?","rationale":"\"the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 26 ) = $ 14 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 14 0.25 * x = 11.5 x = 46 miles the answer is d .\"","correct":"d","options":{"a":"20 ","b":"30 ","c":"36 ","d":"46","e":"54"},"options_float":{"a":20.0,"b":30.0,"c":36.0,"d":46.0,"e":54.0},"annotated_formula":"divide(subtract(add(add(2.50, 5.00), multiply(0.25, 26)), 2.50), 0.25)","linear_formula":"add(n0,n3)|multiply(n1,n5)|add(#0,#1)|subtract(#2,n0)|divide(#3,n1)|","chain":"2.5 + 5<\/gadget>\n7.5<\/output>\n0.25 * 26<\/gadget>\n6.5<\/output>\n7.5 + 6.5<\/gadget>\n14<\/output>\n14 - 2.5<\/gadget>\n11.5<\/output>\n11.5 \/ 0.25<\/gadget>\n46<\/output>\n46<\/result>","index":3618} +{"problem":"fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 8.91 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discount rates is 22 percent , what is the discount rate on pony jeans ?","rationale":"\"let x be the discount on pony jeans . then 0.22 - x is the discount on fox jeans . 3 ( 0.22 - x ) ( 15 ) + 2 x ( 18 ) = 8.91 9.9 - 45 x + 36 x = 8.91 9 x = 0.99 x = 0.11 the answer is c .\"","correct":"c","options":{"a":"9 % ","b":"10 % ","c":"11 % ","d":"12 %","e":"15 %"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":12.0,"e":15.0},"annotated_formula":"multiply(subtract(divide(22, const_100), divide(subtract(8.91, multiply(divide(22, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100)","linear_formula":"divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)|","chain":"22 \/ 100<\/gadget>\n11\/50 = around 0.22<\/output>\n18 * 2<\/gadget>\n36<\/output>\n(11\/50) * 36<\/gadget>\n198\/25 = around 7.92<\/output>\n8.91 - (198\/25)<\/gadget>\n0.99<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 - 36<\/gadget>\n9<\/output>\n0.99 \/ 9<\/gadget>\n0.11<\/output>\n(11\/50) - 0.11<\/gadget>\n0.11<\/output>\n0.11 * 100<\/gadget>\n11<\/output>\n11<\/result>","index":3619} +{"problem":"find the simple interest on $ 10000 at 4 % per annum for 12 months ?","rationale":"\"p = $ 10000 r = 4 % t = 12 \/ 12 years = 1 year s . i . = p * r * t \/ 100 = 10000 * 4 * 1 \/ 100 = $ 400 answer is a\"","correct":"a","options":{"a":"$ 400 ","b":"$ 500 ","c":"$ 650 ","d":"$ 710","e":"$ 1000"},"options_float":{"a":400.0,"b":500.0,"c":650.0,"d":710.0,"e":1000.0},"annotated_formula":"multiply(10000, divide(4, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n10_000 * (1\/25)<\/gadget>\n400<\/output>\n400<\/result>","index":3620} +{"problem":"a contractor undertakes to built a walls in 50 days . he employs 40 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?","rationale":"\"40 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 40 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 40 × 25 × 0.6 \/ 25 × 0.4 = 60 men answera\"","correct":"a","options":{"a":"60 ","b":"30 ","c":"35 ","d":"20","e":"none of these"},"options_float":{"a":60.0,"b":30.0,"c":35.0,"d":20.0,"e":null},"annotated_formula":"divide(multiply(40, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10))","linear_formula":"divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0)|","chain":"100 - 40<\/gadget>\n60<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n40 * (3\/5)<\/gadget>\n24<\/output>\n4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n24 \/ (2\/5)<\/gadget>\n60<\/output>\n60<\/result>","index":3622} +{"problem":"the average weight of 6 person ' s increases by 4.5 kg when a new person comes in place of one of them weighing 75 kg . what might be the weight of the new person ?","rationale":"\"total weight increased = ( 6 x 4.5 ) kg = 27 kg . weight of new person = ( 75 + 27 ) kg = 102 kg option d\"","correct":"d","options":{"a":"160 kg ","b":"175 kg ","c":"180 kg ","d":"102 kg","e":"190 kg"},"options_float":{"a":160.0,"b":175.0,"c":180.0,"d":102.0,"e":190.0},"annotated_formula":"add(multiply(6, 4.5), 75)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"6 * 4.5<\/gadget>\n27<\/output>\n27 + 75<\/gadget>\n102<\/output>\n102<\/result>","index":3623} +{"problem":"sandy buys an old scooter for $ 900 and spends $ 300 on its repairs . if sandy sells the scooter for $ 1500 , what is the gain percent ?","rationale":"\"selling price \/ total cost = 1500 \/ 1200 = 1.25 the gain percent is 25 % . the answer is d .\"","correct":"d","options":{"a":"10 % ","b":"15 % ","c":"20 % ","d":"25 %","e":"30 %"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"multiply(divide(subtract(1500, add(300, 900)), add(300, 900)), const_100)","linear_formula":"add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|","chain":"300 + 900<\/gadget>\n1_200<\/output>\n1_500 - 1_200<\/gadget>\n300<\/output>\n300 \/ 1_200<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":3625} +{"problem":"a , b , c , d , e are the 5 electronic shops in the naza market , which have 20 , 30 , 60 , 80 and 50 t . v . sets with them respectively , then the average number of t . v . sets in each shop is","rationale":"answer average number of t . v . sets = ( 20 + 30 + 60 + 80 + 50 ) \/ 5 = 48 correct option : b","correct":"b","options":{"a":"24 ","b":"48 ","c":"50 ","d":"60","e":"none"},"options_float":{"a":24.0,"b":48.0,"c":50.0,"d":60.0,"e":null},"annotated_formula":"divide(add(add(60, 80), add(50, add(20, 30))), 5)","linear_formula":"add(n3,n4)|add(n1,n2)|add(n5,#1)|add(#0,#2)|divide(#3,n0)","chain":"60 + 80<\/gadget>\n140<\/output>\n20 + 30<\/gadget>\n50<\/output>\n50 + 50<\/gadget>\n100<\/output>\n140 + 100<\/gadget>\n240<\/output>\n240 \/ 5<\/gadget>\n48<\/output>\n48<\/result>","index":3626} +{"problem":"a no . when divided by 281 gives a remainder 160 , what remainder will be obtainedby dividingthe same no . 21 ?","rationale":"\"281 + 160 = 441 \/ 21 = 21 ( remainder ) a\"","correct":"a","options":{"a":"21 ","b":"22 ","c":"23 ","d":"24","e":"25"},"options_float":{"a":21.0,"b":22.0,"c":23.0,"d":24.0,"e":25.0},"annotated_formula":"divide(add(281, 160), 21)","linear_formula":"add(n0,n1)|divide(#0,n2)|","chain":"281 + 160<\/gadget>\n441<\/output>\n441 \/ 21<\/gadget>\n21<\/output>\n21<\/result>","index":3627} +{"problem":"in the rectangular coordinate system , points ( 16 , 0 ) and ( – 16 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ?","rationale":"\"d it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 16 d\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"8 ","d":"16","e":"none of the above"},"options_float":{"a":2.0,"b":4.0,"c":8.0,"d":16.0,"e":null},"annotated_formula":"sqrt(power(16, const_2))","linear_formula":"power(n0,const_2)|sqrt(#0)|","chain":"16 ** 2<\/gadget>\n256<\/output>\n256 ** (1\/2)<\/gadget>\n16<\/output>\n16<\/result>","index":3628} +{"problem":"the speed of a railway engine is 74 km per hour when no compartment is attached , and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached , the maximum number of compartments that can be carried by the engine is :","rationale":"\"the reduction in speed is directly proportional to the square root of the number of compartments attached doesreductionmean amount subtracted ? or percentage decrease ? there are at least two interpretations , and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words , if s = speed , and n = number of compartments , then s = 74 - k * sqrt ( n ) wherekis a constant of the proportionality . in general , if a is directly proportional to b , we can write a = k * b and solve for k . if n = 9 , then s = 24 24 = 74 - k * sqrt ( 9 ) = 74 - 3 k k = 16.67 now , we need to know : what value of n makes s go to zero ? 0 = 74 - 16.67 * sqrt ( n ) 16.67 * sqrt ( n ) = 74 sqrt ( n ) = 4.44 n = 4.44 ^ 2 > 19 with 20 compartments , the train does not budge . therefore , it would budge if there were one fewer cars . thus , 19 is the maximum number of cars the engine can pull and still move . e\"","correct":"e","options":{"a":"49 ","b":"18 ","c":"16 ","d":"17","e":"19"},"options_float":{"a":49.0,"b":18.0,"c":16.0,"d":17.0,"e":19.0},"annotated_formula":"power(divide(74, divide(subtract(74, 24), sqrt(9))), const_2)","linear_formula":"sqrt(n2)|subtract(n0,n1)|divide(#1,#0)|divide(n0,#2)|power(#3,const_2)|","chain":"74 - 24<\/gadget>\n50<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n50 \/ 3<\/gadget>\n50\/3 = around 16.666667<\/output>\n74 \/ (50\/3)<\/gadget>\n111\/25 = around 4.44<\/output>\n(111\/25) ** 2<\/gadget>\n12_321\/625 = around 19.7136<\/output>\n12_321\/625 = around 19.7136<\/result>","index":3629} +{"problem":"the sum of even numbers between 1 and 31 is","rationale":"\"solution required numbers are 2 , 4 , 6 , . . 30 . this is an a . p containing 15 terms . required sum = n \/ 2 ( first term + last term ) ‹ = › 15 \/ 2 ( 2 + 30 ) ‹ = › 240 . answer c\"","correct":"c","options":{"a":"16 ","b":"128 ","c":"240 ","d":"512","e":"none"},"options_float":{"a":16.0,"b":128.0,"c":240.0,"d":512.0,"e":null},"annotated_formula":"multiply(add(31, 1), divide(floor(divide(31, const_2)), const_2))","linear_formula":"add(n0,n1)|divide(n1,const_2)|floor(#1)|divide(#2,const_2)|multiply(#0,#3)|","chain":"31 + 1<\/gadget>\n32<\/output>\n31 \/ 2<\/gadget>\n31\/2 = around 15.5<\/output>\nfloor(31\/2)<\/gadget>\n15<\/output>\n15 \/ 2<\/gadget>\n15\/2 = around 7.5<\/output>\n32 * (15\/2)<\/gadget>\n240<\/output>\n240<\/result>","index":3630} +{"problem":"if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 50 , what is the value of c â ˆ ’ a ?","rationale":"\"the arithmetic mean of a and b = ( a + b ) \/ 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 100 - - 2 subtracting 1 from 2 we have c - a = 10 ; answer : b\"","correct":"b","options":{"a":"25 ","b":"10 ","c":"90 ","d":"14","e":"it can not be determined from the information given"},"options_float":{"a":25.0,"b":10.0,"c":90.0,"d":14.0,"e":null},"annotated_formula":"subtract(multiply(50, const_2), multiply(45, const_2))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|","chain":"50 * 2<\/gadget>\n100<\/output>\n45 * 2<\/gadget>\n90<\/output>\n100 - 90<\/gadget>\n10<\/output>\n10<\/result>","index":3632} +{"problem":"a certain plant was 11 feet long on the day it was plantedthe length grows by a constant amount everyday ( inc . the first day it was planted ) . if this plant grew by 30 % b \/ w the 4 th day and 10 th day after it was planted , how many feet per day does the plant grow","rationale":"if the plant is 11 feet tall when it is planted , it will be 11 + k feet tall at the end of the day ( where k is the constant length it grows by every day ) . similarly , at the end of the 4 th day the plant must be 11 + 4 k feet tall and at the end of the 10 th day it must be 11 + 10 k feet tall . therefore 6 k = 0.3 ( 11 + 4 k ) = > k = 0.69 therefore the plant grows by 0.69 feet every day . option ( c )","correct":"c","options":{"a":"2 ","b":"4.5 ","c":"0.69 ","d":"5.5","e":"3"},"options_float":{"a":2.0,"b":4.5,"c":0.69,"d":5.5,"e":3.0},"annotated_formula":"divide(subtract(multiply(divide(add(const_100, 30), const_100), 11), 11), subtract(10, multiply(divide(add(const_100, 30), const_100), 4)))","linear_formula":"add(n1,const_100)|divide(#0,const_100)|multiply(n0,#1)|multiply(n2,#1)|subtract(#2,n0)|subtract(n3,#3)|divide(#4,#5)","chain":"100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n(13\/10) * 11<\/gadget>\n143\/10 = around 14.3<\/output>\n(143\/10) - 11<\/gadget>\n33\/10 = around 3.3<\/output>\n(13\/10) * 4<\/gadget>\n26\/5 = around 5.2<\/output>\n10 - (26\/5)<\/gadget>\n24\/5 = around 4.8<\/output>\n(33\/10) \/ (24\/5)<\/gadget>\n11\/16 = around 0.6875<\/output>\n11\/16 = around 0.6875<\/result>","index":3633} +{"problem":"john and steve are speed walkers in a race . john is 14 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m \/ s , while steve maintains a blistering 3.7 m \/ s speed . if john finishes the race 2 meters ahead of steve , how long was john ’ s final push ?","rationale":"\"let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.7 t + 14 + 2 - - - > 0.5 t = 16 - - - > t = 32 seconds . d is the correct answer .\"","correct":"d","options":{"a":"13 seconds ","b":"17 seconds ","c":"26 seconds ","d":"32 seconds","e":"51 seconds"},"options_float":{"a":13.0,"b":17.0,"c":26.0,"d":32.0,"e":51.0},"annotated_formula":"divide(add(divide(multiply(3.7, add(14, 2)), subtract(4.2, 3.7)), add(14, 2)), 4.2)","linear_formula":"add(n0,n3)|subtract(n1,n2)|multiply(n2,#0)|divide(#2,#1)|add(#0,#3)|divide(#4,n1)|","chain":"14 + 2<\/gadget>\n16<\/output>\n3.7 * 16<\/gadget>\n59.2<\/output>\n4.2 - 3.7<\/gadget>\n0.5<\/output>\n59.2 \/ 0.5<\/gadget>\n118.4<\/output>\n118.4 + 16<\/gadget>\n134.4<\/output>\n134.4 \/ 4.2<\/gadget>\n32<\/output>\n32<\/result>","index":3635} +{"problem":"the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . therefore decreasing theaverage by 4 year . find what was theoriginal strength of class ?","rationale":"\"let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 ⇒ 40 y + 384 = 36 y + 432 ⇒ 4 y = 48 ∴ y = 12 b\"","correct":"b","options":{"a":"11 ","b":"12 ","c":"17 ","d":"19","e":"21"},"options_float":{"a":11.0,"b":12.0,"c":17.0,"d":19.0,"e":21.0},"annotated_formula":"divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 32)), 4)","linear_formula":"multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)|","chain":"40 - 4<\/gadget>\n36<\/output>\n12 * 36<\/gadget>\n432<\/output>\n12 * 32<\/gadget>\n384<\/output>\n432 - 384<\/gadget>\n48<\/output>\n48 \/ 4<\/gadget>\n12<\/output>\n12<\/result>","index":3636} +{"problem":"running at the same constant rate , 5 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ?","rationale":"\"5 machines produce 270 bottles per minute ; 1 machine produces 270 \/ 5 = 54 bottles per minute ; 10 machines produce 54 * 10 = 540 bottles per minute ; in 4 minutes 10 machines produce 540 * 4 = 2,160 bottles . answer : c .\"","correct":"c","options":{"a":"648 ","b":"1,800 ","c":"2,160 ","d":"10,800","e":"64,800"},"options_float":{"a":648.0,"b":1800.0,"c":2160.0,"d":10800.0,"e":64800.0},"annotated_formula":"multiply(multiply(divide(270, 5), 4), 10)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"270 \/ 5<\/gadget>\n54<\/output>\n54 * 4<\/gadget>\n216<\/output>\n216 * 10<\/gadget>\n2_160<\/output>\n2_160<\/result>","index":3637} +{"problem":"the radius of the wheel of a bus is 100 cms and the speed of the bus is 66 km \/ h , then the r . p . m . ( revolutions per minutes ) of the wheel is","rationale":"\"radius of the wheel of bus = 100 cm . then , circumference of wheel = 2 ï € r = 200 ï € = 628.57 cm distance covered by bus in 1 minute = 66 ⠁ „ 60 ã — 1000 ã — 100 cms distance covered by one revolution of wheel = circumference of wheel = 440 cm â ˆ ´ revolutions per minute = 6600000 \/ 60 ã — 628.57 = 175 answer a\"","correct":"a","options":{"a":"175 ","b":"250 ","c":"300 ","d":"330","e":"none of these"},"options_float":{"a":175.0,"b":250.0,"c":300.0,"d":330.0,"e":null},"annotated_formula":"divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 100), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2)))))","linear_formula":"multiply(n1,const_1000)|multiply(const_1,const_60)|multiply(const_3,const_4)|multiply(const_2,const_4)|multiply(n0,const_2)|add(#2,const_2)|add(#3,const_2)|multiply(#0,const_100)|divide(#7,#1)|power(#6,const_2)|divide(#5,#9)|add(#10,const_3)|multiply(#11,#4)|divide(#8,#12)|","chain":"1_000 * 66<\/gadget>\n66_000<\/output>\n100 * 66_000<\/gadget>\n6_600_000<\/output>\n60 * 1<\/gadget>\n60<\/output>\n6_600_000 \/ 60<\/gadget>\n110_000<\/output>\n2 * 100<\/gadget>\n200<\/output>\n3 * 4<\/gadget>\n12<\/output>\n2 + 12<\/gadget>\n14<\/output>\n4 * 2<\/gadget>\n8<\/output>\n2 + 8<\/gadget>\n10<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n3 + (7\/50)<\/gadget>\n157\/50 = around 3.14<\/output>\n200 * (157\/50)<\/gadget>\n628<\/output>\n110_000 \/ 628<\/gadget>\n27_500\/157 = around 175.159236<\/output>\n27_500\/157 = around 175.159236<\/result>","index":3638} +{"problem":"if n is an integer and 101 n ^ 2 is less than or equal to 6400 , what is the greatest possible value of n ?","rationale":"101 * n ^ 2 < = 6400 n ^ 2 < = 6400 \/ 101 which will be less than 81 since 6400 \/ 100 = 64 which is the square of 9 next closest value of n where n ^ 2 < = 64 is 7 ans a","correct":"a","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"floor(sqrt(divide(6400, 101)))","linear_formula":"divide(n2,n0)|sqrt(#0)|floor(#1)","chain":"6_400 \/ 101<\/gadget>\n6_400\/101 = around 63.366337<\/output>\n(6_400\/101) ** (1\/2)<\/gadget>\n80*sqrt(101)\/101 = around 7.960298<\/output>\nfloor(80*sqrt(101)\/101)<\/gadget>\n7<\/output>\n7<\/result>","index":3640} +{"problem":"a train 360 m long runs with a speed of 45 km \/ hr . what time will it take to pass a platform of 390 m long ?","rationale":"\"explanation : speed = 45 km \/ hr = 45 × ( 10 \/ 36 ) m \/ s = 150 \/ 12 = 50 \/ 4 = 25 \/ 2 m \/ s total distance = length of the train + length of the platform = 360 + 390 = 750 meter time taken to cross the platform = 750 \/ ( 25 \/ 2 ) = 750 × 2 \/ 25 = 60 seconds answer : option e\"","correct":"e","options":{"a":"38 sec ","b":"35 sec ","c":"44 sec ","d":"40 sec","e":"60 sec"},"options_float":{"a":38.0,"b":35.0,"c":44.0,"d":40.0,"e":60.0},"annotated_formula":"multiply(divide(add(divide(390, const_1000), divide(360, const_1000)), 45), const_3600)","linear_formula":"divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)|","chain":"390 \/ 1_000<\/gadget>\n39\/100 = around 0.39<\/output>\n360 \/ 1_000<\/gadget>\n9\/25 = around 0.36<\/output>\n(39\/100) + (9\/25)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) \/ 45<\/gadget>\n1\/60 = around 0.016667<\/output>\n(1\/60) * 3_600<\/gadget>\n60<\/output>\n60<\/result>","index":3642} +{"problem":"a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 500 more than d , what is a ' s share ?","rationale":"sol . let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 500 ⇔ x = 500 . ∴ a ' s share = rs . 5 x = rs . ( 5 x 500 ) = rs . 2500 . answer a","correct":"a","options":{"a":"rs . 2500 ","b":"rs . 1000 ","c":"rs . 1500 ","d":"rs . 2000","e":"none"},"options_float":{"a":2500.0,"b":1000.0,"c":1500.0,"d":2000.0,"e":null},"annotated_formula":"multiply(divide(500, subtract(divide(4, add(add(add(5, 2), 4), 3)), divide(3, add(add(add(5, 2), 4), 3)))), divide(5, add(add(add(5, 2), 4), 3)))","linear_formula":"add(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n2,#2)|divide(n3,#2)|divide(n0,#2)|subtract(#3,#4)|divide(n4,#6)|multiply(#7,#5)","chain":"5 + 2<\/gadget>\n7<\/output>\n7 + 4<\/gadget>\n11<\/output>\n11 + 3<\/gadget>\n14<\/output>\n4 \/ 14<\/gadget>\n2\/7 = around 0.285714<\/output>\n3 \/ 14<\/gadget>\n3\/14 = around 0.214286<\/output>\n(2\/7) - (3\/14)<\/gadget>\n1\/14 = around 0.071429<\/output>\n500 \/ (1\/14)<\/gadget>\n7_000<\/output>\n5 \/ 14<\/gadget>\n5\/14 = around 0.357143<\/output>\n7_000 * (5\/14)<\/gadget>\n2_500<\/output>\n2_500<\/result>","index":3643} +{"problem":"a certain high school has 500 students . of these students , 50 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ?","rationale":"\"we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 50 y = 20 z = 10 so , we have 50 students taking music , 20 taking art and 10 taking both music and art . 40 student taking just music 10 student taking just art 10 student taking both music and art total = 60 students we ' re asked for the total number of students who are taking neither course . that is 500 - 60 = 440 . b\"","correct":"b","options":{"a":"430 ","b":"440 ","c":"450 ","d":"460","e":"470"},"options_float":{"a":430.0,"b":440.0,"c":450.0,"d":460.0,"e":470.0},"annotated_formula":"subtract(500, subtract(add(50, 20), 10))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|","chain":"50 + 20<\/gadget>\n70<\/output>\n70 - 10<\/gadget>\n60<\/output>\n500 - 60<\/gadget>\n440<\/output>\n440<\/result>","index":3644} +{"problem":"praveen starts business with rs . 3920 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ?","rationale":"\"let hari ’ s capital be rs . x . then , 3920 * 12 \/ 7 x = 2 \/ 3 = > 14 x = 141120 = > x = 10080 . answer : a\"","correct":"a","options":{"a":"s . 10080 ","b":"s . 8000 ","c":"s . 8500 ","d":"s . 9000","e":"s . 6000"},"options_float":{"a":10080.0,"b":8000.0,"c":8500.0,"d":9000.0,"e":6000.0},"annotated_formula":"divide(divide(3920, subtract(const_1, divide(5, const_12))), divide(2, 3))","linear_formula":"divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)|","chain":"5 \/ 12<\/gadget>\n5\/12 = around 0.416667<\/output>\n1 - (5\/12)<\/gadget>\n7\/12 = around 0.583333<\/output>\n3_920 \/ (7\/12)<\/gadget>\n6_720<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n6_720 \/ (2\/3)<\/gadget>\n10_080<\/output>\n10_080<\/result>","index":3645} +{"problem":"a woman can do a piece of work in 40 days . man is 25 % more efficient than woman . in how many days a man can do the same piece of work ?","rationale":"the ratio of the efficiencies of a woman and man = 100 : 125 = 4 : 5 the ratio of the days taken by woman and man to finish the work = 5 : 4 woman takes 32 days to finish the work . = > 5 parts = 40 the number of days taken by man to finish the work = 4 parts = 40 \/ 5 x 4 = 32 days a","correct":"a","options":{"a":"32 days ","b":"36 days ","c":"38 days ","d":"40 days","e":"53 days"},"options_float":{"a":32.0,"b":36.0,"c":38.0,"d":40.0,"e":53.0},"annotated_formula":"multiply(divide(40, add(const_100, 25)), const_100)","linear_formula":"add(n1,const_100)|divide(n0,#0)|multiply(#1,const_100)","chain":"100 + 25<\/gadget>\n125<\/output>\n40 \/ 125<\/gadget>\n8\/25 = around 0.32<\/output>\n(8\/25) * 100<\/gadget>\n32<\/output>\n32<\/result>","index":3648} +{"problem":"the ratio between the present ages of a and b is 3.5 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ?","rationale":"\"let the present ages of a and b be 3.5 x and 3 x years respectively . then , ( 3.5 x - 4 ) \/ ( 3 x + 4 ) = 1 \/ 1 0.5 x = 8 = > x = 16 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 60 : 44 = 15 : 11 . answer : c\"","correct":"c","options":{"a":"14 : 4 ","b":"13 : 0 ","c":"15 : 11 ","d":"13 : 12","e":"13 : 17"},"options_float":{"a":3.5,"b":null,"c":1.3636363636,"d":1.0833333333,"e":0.7647058824},"annotated_formula":"divide(add(multiply(3.5, divide(add(3.5, 3.5), subtract(3.5, 3))), 3.5), subtract(multiply(3, divide(add(3.5, 3.5), subtract(3.5, 3))), 3.5))","linear_formula":"add(n0,n0)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n0,#3)|subtract(#4,n0)|divide(#5,#6)|","chain":"3.5 + 3.5<\/gadget>\n7<\/output>\n3.5 - 3<\/gadget>\n0.5<\/output>\n7 \/ 0.5<\/gadget>\n14<\/output>\n3.5 * 14<\/gadget>\n49<\/output>\n49 + 3.5<\/gadget>\n52.5<\/output>\n3 * 14<\/gadget>\n42<\/output>\n42 - 3.5<\/gadget>\n38.5<\/output>\n52.5 \/ 38.5<\/gadget>\n1.363636<\/output>\n1.363636<\/result>","index":3650} +{"problem":"ram , who is half as efficient as krish , will take 21 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ?","rationale":"\"number of days taken by ram to complete task = 21 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 24 days is 21 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work done by krish and ram in a day = 3 w total amount of time needed by krish and ram to complete task = 21 w \/ 3 w = 7 days answer d\"","correct":"d","options":{"a":"16 days ","b":"12 days ","c":"8 days ","d":"7 days","e":"18 days"},"options_float":{"a":16.0,"b":12.0,"c":8.0,"d":7.0,"e":18.0},"annotated_formula":"inverse(add(divide(const_1, 21), divide(const_1, divide(21, const_2))))","linear_formula":"divide(const_1,n0)|divide(n0,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3)|","chain":"1 \/ 21<\/gadget>\n1\/21 = around 0.047619<\/output>\n21 \/ 2<\/gadget>\n21\/2 = around 10.5<\/output>\n1 \/ (21\/2)<\/gadget>\n2\/21 = around 0.095238<\/output>\n(1\/21) + (2\/21)<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ (1\/7)<\/gadget>\n7<\/output>\n7<\/result>","index":3651} +{"problem":"calculate the l . c . m of 7 \/ 10 , 8 \/ 9 , 3 \/ 8 , 5 \/ 12 is :","rationale":"required l . c . m = l . c . m . of 7 , 8 , 3 , 5 \/ h . c . f . of 10 , 9 , 8 , 12 = 840 \/ 1 = 840 answer is a","correct":"a","options":{"a":"840 ","b":"70 ","c":"740 ","d":"140","e":"940"},"options_float":{"a":840.0,"b":70.0,"c":740.0,"d":140.0,"e":940.0},"annotated_formula":"multiply(multiply(7, 8), multiply(3, 5))","linear_formula":"multiply(n0,n2)|multiply(n4,n6)|multiply(#0,#1)","chain":"7 * 8<\/gadget>\n56<\/output>\n3 * 5<\/gadget>\n15<\/output>\n56 * 15<\/gadget>\n840<\/output>\n840<\/result>","index":3652} +{"problem":"a circular mat with diameter 18 inches is placed on a square tabletop , each of whose sides is 24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ?","rationale":"so we are looking for the area of the cloth over the area of the table area of the cloth = ( pi ) ( r ) ^ 2 which is about ( 3 ) ( 9 ) ( 9 ) area of the table = ( 24 ) ( 24 ) so the quick way to estimate is looking at the fraction like this : ( 3 \/ 24 ) ( 81 \/ 24 ) i hope this is easy to follow , so with some simplification i get ( 1 \/ 8 ) ( 3 ) = ( 3 \/ 8 ) answer is d","correct":"d","options":{"a":"5 \/ 12 ","b":"3 \/ 5 ","c":"1 \/ 2 ","d":"3 \/ 8","e":"5 \/ 6"},"options_float":{"a":0.4166666667,"b":0.6,"c":0.5,"d":0.375,"e":0.8333333333},"annotated_formula":"divide(divide(18, const_2), 24)","linear_formula":"divide(n0,const_2)|divide(#0,n1)","chain":"18 \/ 2<\/gadget>\n9<\/output>\n9 \/ 24<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":3653} +{"problem":"a man swims downstream 36 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ?","rationale":"\"36 - - - 3 ds = 12 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 12 + 6 ) \/ 2 = 9 answer : c\"","correct":"c","options":{"a":"7 ","b":"8 ","c":"9 ","d":"2","e":"4"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":2.0,"e":4.0},"annotated_formula":"divide(add(divide(18, 3), divide(36, 3)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"18 \/ 3<\/gadget>\n6<\/output>\n36 \/ 3<\/gadget>\n12<\/output>\n6 + 12<\/gadget>\n18<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n9<\/result>","index":3654} +{"problem":"the cash realised on selling a 14 % stock is rs . 109.25 , brokerage being 1 \/ 4 % is ?","rationale":"\"cash realised = rs . ( 109.25 - 0.25 ) = rs . 109 . answer : d\"","correct":"d","options":{"a":"366 ","b":"106 ","c":"102 ","d":"109","e":"122"},"options_float":{"a":366.0,"b":106.0,"c":102.0,"d":109.0,"e":122.0},"annotated_formula":"subtract(109.25, divide(1, 4))","linear_formula":"divide(n2,n3)|subtract(n1,#0)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n109.25 - (1\/4)<\/gadget>\n109<\/output>\n109<\/result>","index":3655} +{"problem":"the price of maruti car rises by 30 percent while the sales of the car come down by 20 % . what is the percentage change in the total revenue ?","rationale":"solution : let initial price of maruti car be rs . 100 . as price increases 30 % , price of car will become , ( 100 + 30 % of 100 ) = rs . 130 . due to increase in price , sales is down by 20 % . it means , it is going make 20 % less revenue as expected after increment of price . so , new revenue = ( 130 - 20 % of 130 ) = rs . 104 . the initial revenue was rs . 100 which becomes rs . 104 at the end . it means there is 4 % increment in the total revenue . mind calculation method : 100 = = 30 % ↑ ( price effect ) = = > 130 = = 20 % ↓ ( sales effects ) = = > 104 . hence , 4 % rises . answer : option c","correct":"c","options":{"a":"- 4 % ","b":"- 2 % ","c":"+ 4 % ","d":"+ 2 %","e":"none of these"},"options_float":{"a":-4.0,"b":-2.0,"c":4.0,"d":2.0,"e":null},"annotated_formula":"subtract(subtract(add(30, const_100), divide(multiply(20, add(30, const_100)), const_100)), const_100)","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)","chain":"30 + 100<\/gadget>\n130<\/output>\n20 * 130<\/gadget>\n2_600<\/output>\n2_600 \/ 100<\/gadget>\n26<\/output>\n130 - 26<\/gadget>\n104<\/output>\n104 - 100<\/gadget>\n4<\/output>\n4<\/result>","index":3656} +{"problem":"if p and q are positive integers each greater than 1 , and 17 ( p + 1 ) = 21 ( q + 1 ) , what is the least possible value of p + q ?","rationale":"\"17 ( p + 1 ) = 21 ( q + 1 ) - - > ( p + 1 ) \/ ( q + 1 ) = 21 \/ 17 - - > the least positive value of p + 1 is 21 , so the least value of p is 20 and the least positive value of q + 1 is 17 , so the least value of q is 16 - - > the least value of p + q is 20 + 16 = 36 . answer : a .\"","correct":"a","options":{"a":"36 ","b":"42 ","c":"44 ","d":"46","e":"none"},"options_float":{"a":36.0,"b":42.0,"c":44.0,"d":46.0,"e":null},"annotated_formula":"subtract(add(21, 17), const_2)","linear_formula":"add(n1,n3)|subtract(#0,const_2)|","chain":"21 + 17<\/gadget>\n38<\/output>\n38 - 2<\/gadget>\n36<\/output>\n36<\/result>","index":3657} +{"problem":"the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 25 percent of a , and m equals b decreased by 60 percent of b , what is the value of m \/ x ?","rationale":"\"a \/ b = 4 \/ 5 m \/ x = ( 2 \/ 5 ) * 5 \/ ( 5 \/ 4 ) * 4 = 2 \/ 5 the answer is b .\"","correct":"b","options":{"a":"1 \/ 3 ","b":"2 \/ 5 ","c":"3 \/ 4 ","d":"3 \/ 2","e":"2 \/ 3"},"options_float":{"a":0.3333333333,"b":0.4,"c":0.75,"d":1.5,"e":0.6666666667},"annotated_formula":"multiply(divide(subtract(const_100, 60), add(const_100, 25)), divide(5, 4))","linear_formula":"add(n2,const_100)|divide(n1,n0)|subtract(const_100,n3)|divide(#2,#0)|multiply(#3,#1)|","chain":"100 - 60<\/gadget>\n40<\/output>\n100 + 25<\/gadget>\n125<\/output>\n40 \/ 125<\/gadget>\n8\/25 = around 0.32<\/output>\n5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n(8\/25) * (5\/4)<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":3658} +{"problem":"three rugs have a combined area of 212 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ?","rationale":"\"total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 212 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 24 . answer : c .\"","correct":"c","options":{"a":"18 square meters ","b":"20 square meters ","c":"24 square meters ","d":"28 square meters","e":"30 square meters"},"options_float":{"a":18.0,"b":20.0,"c":24.0,"d":28.0,"e":30.0},"annotated_formula":"divide(subtract(subtract(212, 140), 24), const_2)","linear_formula":"subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|","chain":"212 - 140<\/gadget>\n72<\/output>\n72 - 24<\/gadget>\n48<\/output>\n48 \/ 2<\/gadget>\n24<\/output>\n24<\/result>","index":3660} +{"problem":"when 2 \/ 9 of the votes on a certain resolution have been counted , 3 \/ 4 of those counted are in favor of the resolution . what fraction q of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution ?","rationale":"\"if we use variable for total votes there will be too many fractions to manipulate with , so pick some smart # : let set total # of votes is 18 . 2 \/ 9 of the votes on a certain resolution have been counted - - > 4 counted and 18 - 4 = 14 votes left to be counted ; 3 \/ 4 of those counted are in favor of the resolution - - > 3 in favor and 1 against ; ratio of those who voted against to those who voted for to be 2 to 1 there should be total of 18 * 2 \/ 3 = 12 people who voted against , so in the remaining 14 votes there should be 12 - 1 = 11 people who voted against . thus q = 11 \/ 14 of the remaining votes must be against . answer : a .\"","correct":"a","options":{"a":"11 \/ 14 ","b":"13 \/ 18 ","c":"4 \/ 7 ","d":"3 \/ 7","e":"3 \/ 14"},"options_float":{"a":0.7857142857,"b":0.7222222222,"c":0.5714285714,"d":0.4285714286,"e":0.2142857143},"annotated_formula":"divide(subtract(divide(2, add(1, 2)), subtract(divide(2, 9), multiply(divide(2, 9), divide(3, 4)))), divide(subtract(9, 2), 9))","linear_formula":"add(n0,n5)|divide(n0,n1)|divide(n2,n3)|subtract(n1,n0)|divide(n0,#0)|divide(#3,n1)|multiply(#1,#2)|subtract(#1,#6)|subtract(#4,#7)|divide(#8,#5)|","chain":"1 + 2<\/gadget>\n3<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2 \/ 9<\/gadget>\n2\/9 = around 0.222222<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(2\/9) * (3\/4)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(2\/9) - (1\/6)<\/gadget>\n1\/18 = around 0.055556<\/output>\n(2\/3) - (1\/18)<\/gadget>\n11\/18 = around 0.611111<\/output>\n9 - 2<\/gadget>\n7<\/output>\n7 \/ 9<\/gadget>\n7\/9 = around 0.777778<\/output>\n(11\/18) \/ (7\/9)<\/gadget>\n11\/14 = around 0.785714<\/output>\n11\/14 = around 0.785714<\/result>","index":3661} +{"problem":"tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 12 even numbers . his mother immediately worked out the answer . how much had he scored in math ?","rationale":"\"a 156 sum = ( n x n ) + n hence , 12 x 12 = 144 + 12 = 156\"","correct":"a","options":{"a":"156 ","b":"146 ","c":"158 ","d":"187","e":"160"},"options_float":{"a":156.0,"b":146.0,"c":158.0,"d":187.0,"e":160.0},"annotated_formula":"multiply(add(12, const_1), 12)","linear_formula":"add(n0,const_1)|multiply(n0,#0)|","chain":"12 + 1<\/gadget>\n13<\/output>\n13 * 12<\/gadget>\n156<\/output>\n156<\/result>","index":3663} +{"problem":"during a certain season , a team won 70 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ?","rationale":"\"we are first given that a team won 70 percent of its first 100 games . this means the team won 0.7 x 100 = 70 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 equals the number of remaining games in the season . thus we can say : 0.5 ( t – 100 ) = number of wins for remaining games 0.5 t – 50 = number of wins for remaining games lastly , we are given that team won 70 percent of all games played in the season . that is , they won 0.7 t games in the entire season . with this we can set up the equation : number of first 100 games won + number of games won for remaining games = total number of games won in the entire season 70 + 0.5 t – 50 = 0.7 t 20 = 0.2 t 200 = 2 t 100 = t answer is e .\"","correct":"e","options":{"a":"180 ","b":"170 ","c":"156 ","d":"150","e":"100"},"options_float":{"a":180.0,"b":170.0,"c":156.0,"d":150.0,"e":100.0},"annotated_formula":"add(100, divide(subtract(70, 70), subtract(divide(70, 100), divide(50, 100))))","linear_formula":"divide(n3,n1)|divide(n2,n1)|subtract(n0,n3)|subtract(#0,#1)|divide(#2,#3)|add(n1,#4)|","chain":"70 - 70<\/gadget>\n0<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(7\/10) - (1\/2)<\/gadget>\n1\/5 = around 0.2<\/output>\n0 \/ (1\/5)<\/gadget>\n0<\/output>\n100 + 0<\/gadget>\n100<\/output>\n100<\/result>","index":3664} +{"problem":"if x 2 + kx - 3 , is divisible by ( x - 1 ) , what is the value of k","rationale":"explanation : if x 2 + kx - 3 is divisible by ( x - 1 ) then x = 1 ( 1 ) 2 + 1 ( k ) - 3 = 0 1 + k - 3 = 0 k = 2 answer : option c","correct":"c","options":{"a":"- 1 ","b":"1 ","c":"2 ","d":"- 1","e":"0"},"options_float":{"a":-1.0,"b":1.0,"c":2.0,"d":-1.0,"e":0.0},"annotated_formula":"subtract(3, power(1, 2))","linear_formula":"power(n2,n0)|subtract(n1,#0)","chain":"1 ** 2<\/gadget>\n1<\/output>\n3 - 1<\/gadget>\n2<\/output>\n2<\/result>","index":3666} +{"problem":"if ( x - 10 ) ( y - 5 ) ( z - 2 ) = 1000 then what is the least value of x + y + z ? if x , y , z are all integer no .","rationale":"( x - 10 ) ( y - 5 ) ( z - 2 ) = 1000 put x = 0 y = - 5 z = 12 then ( 0 - 10 ) ( - 5 - 5 ) ( 12 - 2 ) = 1000 ( - 10 ) ( - 10 ) ( 10 ) = 1000 1000 = 1000 answer : a","correct":"a","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"subtract(divide(multiply(10, 5), 5), const_3)","linear_formula":"multiply(n0,n1)|divide(#0,n1)|subtract(#1,const_3)","chain":"10 * 5<\/gadget>\n50<\/output>\n50 \/ 5<\/gadget>\n10<\/output>\n10 - 3<\/gadget>\n7<\/output>\n7<\/result>","index":3667} +{"problem":"a can run 4 times as fast as b and gives b a start of 60 m . how long should the race course be so that a and b might reach in the same time ?","rationale":"\"speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 60 m he gains 60 * ( 4 \/ 3 ) i . e 80 m answer : c\"","correct":"c","options":{"a":"70 m ","b":"60 m ","c":"80 m ","d":"65 m","e":"75 m"},"options_float":{"a":70.0,"b":60.0,"c":80.0,"d":65.0,"e":75.0},"annotated_formula":"add(multiply(4, divide(divide(60, 4), subtract(4, const_1))), 60)","linear_formula":"divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)|","chain":"60 \/ 4<\/gadget>\n15<\/output>\n4 - 1<\/gadget>\n3<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n4 * 5<\/gadget>\n20<\/output>\n20 + 60<\/gadget>\n80<\/output>\n80<\/result>","index":3670} +{"problem":"a train 175 m long , running with a speed of 63 km \/ hr will pass a tree in ?","rationale":"\"speed = 63 * 5 \/ 18 = 35 \/ 2 m \/ sec time taken = 175 * 2 \/ 35 = 10 sec answer : e\"","correct":"e","options":{"a":"22 sec ","b":"16 sec ","c":"17 sec ","d":"88 sec","e":"10 sec"},"options_float":{"a":22.0,"b":16.0,"c":17.0,"d":88.0,"e":10.0},"annotated_formula":"multiply(divide(175, multiply(63, const_1000)), const_3600)","linear_formula":"multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|","chain":"63 * 1_000<\/gadget>\n63_000<\/output>\n175 \/ 63_000<\/gadget>\n1\/360 = around 0.002778<\/output>\n(1\/360) * 3_600<\/gadget>\n10<\/output>\n10<\/result>","index":3671} +{"problem":"an 80 - liter solution of cool - drink is made from 10 % jasmine water . if 8 liters of jasmine and 12 liters of water were added to the solution , what percent of the solution is jasmine ?","rationale":"\"the percent of jasmine in the resulting solution is : ( amount of jasmine ) \/ ( total volume ) ( 0.1 ( 80 ) + 8 ) \/ 100 = 16 \/ 100 = 16 % the answer is c .\"","correct":"c","options":{"a":"12 % ","b":"14 % ","c":"16 % ","d":"18 %","e":"20 %"},"options_float":{"a":12.0,"b":14.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"multiply(divide(add(multiply(80, divide(10, const_100)), 8), const_100), const_100)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n2,#1)|divide(#2,const_100)|multiply(#3,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n80 * (1\/10)<\/gadget>\n8<\/output>\n8 + 8<\/gadget>\n16<\/output>\n16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) * 100<\/gadget>\n16<\/output>\n16<\/result>","index":3672} +{"problem":"the length of rectangle is thrice its breadth and its perimeter is 120 m , find the area of the rectangle ?","rationale":"\"2 ( 3 x + x ) = 120 l = 45 b = 15 lb = 45 * 15 = 675 answer : e\"","correct":"e","options":{"a":"432 sq m ","b":"356 sq m ","c":"452 sq m ","d":"428 sq m","e":"675 sq m"},"options_float":{"a":432.0,"b":356.0,"c":452.0,"d":428.0,"e":675.0},"annotated_formula":"multiply(multiply(divide(120, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(120, add(multiply(const_3, const_2), multiply(const_1, const_2))))","linear_formula":"multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)|","chain":"3 * 2<\/gadget>\n6<\/output>\n1 * 2<\/gadget>\n2<\/output>\n6 + 2<\/gadget>\n8<\/output>\n120 \/ 8<\/gadget>\n15<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 * 15<\/gadget>\n675<\/output>\n675<\/result>","index":3674} +{"problem":"at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 164.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ?","rationale":"\"let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 164.5 subtracting the first equation from the second gives us b + 3 s = 44.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 31 . in any case , there is no necessity to know each item ' s price , just the sum . answer : c\"","correct":"c","options":{"a":"$ 21 ","b":"$ 27 ","c":"$ 31 ","d":"$ 41","e":"it can not be determined"},"options_float":{"a":21.0,"b":27.0,"c":31.0,"d":41.0,"e":null},"annotated_formula":"subtract(add(120, subtract(164.50, 120)), multiply(subtract(164.50, 120), 3))","linear_formula":"subtract(n3,n2)|add(n2,#0)|multiply(#0,n0)|subtract(#1,#2)|","chain":"164.5 - 120<\/gadget>\n44.5<\/output>\n120 + 44.5<\/gadget>\n164.5<\/output>\n44.5 * 3<\/gadget>\n133.5<\/output>\n164.5 - 133.5<\/gadget>\n31<\/output>\n31<\/result>","index":3676} +{"problem":"65 % of x = 20 % of 682.50 . find the value of x ?","rationale":"\"65 % of x = 20 % of 682.50 then , 65 \/ 100 * x = 20 \/ 100 * 6825 \/ 10 x = 210 answer is b\"","correct":"b","options":{"a":"100 ","b":"210 ","c":"150 ","d":"180","e":"199"},"options_float":{"a":100.0,"b":210.0,"c":150.0,"d":180.0,"e":199.0},"annotated_formula":"divide(multiply(multiply(divide(682.50, const_100), 20), const_100), 65)","linear_formula":"divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0)|","chain":"682.5 \/ 100<\/gadget>\n6.825<\/output>\n6.825 * 20<\/gadget>\n136.5<\/output>\n136.5 * 100<\/gadget>\n13_650<\/output>\n13_650 \/ 65<\/gadget>\n210<\/output>\n210<\/result>","index":3677} +{"problem":"a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 120 kilometers ?","rationale":"\"each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 120 kilometers ? 120 ÷ 40 = 3 × 1 gallon = 3 gallons correct answer d\"","correct":"d","options":{"a":"4.5 ","b":"5.5 ","c":"6.5 ","d":"3","e":"8.5"},"options_float":{"a":4.5,"b":5.5,"c":6.5,"d":3.0,"e":8.5},"annotated_formula":"divide(120, 40)","linear_formula":"divide(n1,n0)|","chain":"120 \/ 40<\/gadget>\n3<\/output>\n3<\/result>","index":3679} +{"problem":"a ranch has both horses and ponies . exactly 5 \/ 7 of the ponies have horseshoes , and exactly 2 \/ 3 of the ponies with horseshoes are from iceland . if there are 4 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ?","rationale":"\"5 \/ 7 * p are ponies with horseshoes , so p is a multiple of 7 . 2 \/ 3 * 5 \/ 7 * p = 10 \/ 21 * p are icelandic ponies with horseshoes , so p is a multiple of 21 . the minimum value of p is 21 . then h = p + 4 = 25 . the minimum number of horses and ponies is 46 . the answer is d .\"","correct":"d","options":{"a":"34 ","b":"38 ","c":"42 ","d":"46","e":"50"},"options_float":{"a":34.0,"b":38.0,"c":42.0,"d":46.0,"e":50.0},"annotated_formula":"add(add(lcm(7, multiply(divide(7, 2), 3)), 4), lcm(7, multiply(divide(7, 2), 3)))","linear_formula":"divide(n1,n2)|multiply(n3,#0)|lcm(n1,#1)|add(n4,#2)|add(#3,#2)|","chain":"7 \/ 2<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) * 3<\/gadget>\n21\/2 = around 10.5<\/output>\nlcm(7, 21\/2)<\/gadget>\n21<\/output>\n21 + 4<\/gadget>\n25<\/output>\n25 + 21<\/gadget>\n46<\/output>\n46<\/result>","index":3680} +{"problem":"on the first day of her vacation , louisa traveled 160 miles . on the second day , traveling at the same average speed , she traveled 280 miles . if the 160 - mile trip took 3 hours less than the 280 - mile trip , what was the average speed , in miles per hour ?","rationale":"( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) \/ ( time ) - - > given : ( rate ) = 160 \/ t = 280 \/ ( t + 3 ) - - > 4 \/ t = 7 \/ ( t + 3 ) - - > 4 t + 12 = 7 t - - - - > 3 t = 12 . t = 4 - - - - > ( rate ) = 160 \/ 4 = 40 answer : a","correct":"a","options":{"a":"40 ","b":"45 ","c":"46 ","d":"48","e":"50"},"options_float":{"a":40.0,"b":45.0,"c":46.0,"d":48.0,"e":50.0},"annotated_formula":"divide(subtract(280, 160), 3)","linear_formula":"subtract(n1,n0)|divide(#0,n3)","chain":"280 - 160<\/gadget>\n120<\/output>\n120 \/ 3<\/gadget>\n40<\/output>\n40<\/result>","index":3683} +{"problem":"evaluate : | 5 - 8 ( 3 - 12 ) | - | 5 - 11 | = ?","rationale":"\"according to order of operations , inner brackets first . hence | 5 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 5 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 5 + 72 | - | 5 - 11 | = | 77 | - | - 6 | = 77 - 6 = 71 correct answer d ) 71\"","correct":"d","options":{"a":"40 ","b":"50 ","c":"60 ","d":"71","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":71.0,"e":80.0},"annotated_formula":"subtract(subtract(5, multiply(8, subtract(3, 12))), negate(subtract(5, 11)))","linear_formula":"subtract(n2,n3)|subtract(n4,n5)|multiply(n1,#0)|negate(#1)|subtract(n0,#2)|subtract(#4,#3)|","chain":"3 - 12<\/gadget>\n-9<\/output>\n8 * (-9)<\/gadget>\n-72<\/output>\n5 - (-72)<\/gadget>\n77<\/output>\n5 - 11<\/gadget>\n-6<\/output>\n-(-6)<\/gadget>\n6<\/output>\n77 - 6<\/gadget>\n71<\/output>\n71<\/result>","index":3684} +{"problem":"a 10 % stock yielding 8 % is quoted at ?","rationale":"\"assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 10 , money invested = 100 × 10 \/ 8 = rs . 125 ie , market value of the stock = rs . 125 answer is a .\"","correct":"a","options":{"a":"125 ","b":"170 ","c":"175 ","d":"180","e":"185"},"options_float":{"a":125.0,"b":170.0,"c":175.0,"d":180.0,"e":185.0},"annotated_formula":"multiply(divide(const_100, 8), 10)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 8<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 10<\/gadget>\n125<\/output>\n125<\/result>","index":3685} +{"problem":"if a trader sold two cars each at rs . 325475 and gains 12 % on the first and loses 12 % on the second , then his profit or loss percent on the whole is ?","rationale":"sp of each car is rs . 325475 , he gains 12 % on first car and losses 12 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] \/ 100 = ( 12 ) ( 12 ) \/ 100 % = 1.44 % answer : a","correct":"a","options":{"a":"1.44 % ","b":"1.43 % ","c":"1.34 % ","d":"4.44 %","e":"1.74 %"},"options_float":{"a":1.44,"b":1.43,"c":1.34,"d":4.44,"e":1.74},"annotated_formula":"divide(multiply(12, 12), const_100)","linear_formula":"multiply(n1,n1)|divide(#0,const_100)","chain":"12 * 12<\/gadget>\n144<\/output>\n144 \/ 100<\/gadget>\n36\/25 = around 1.44<\/output>\n36\/25 = around 1.44<\/result>","index":3688} +{"problem":"i flew my tiny seaplane to visit my mother . on the flight up , i flew at 96 mph . on the way home , i flew 88 mph . what was my average speed for the trip ?","rationale":"\"( 96 mph + 88 mph ) \/ 2 = 92 mph correct answer is : c\"","correct":"c","options":{"a":"198 mph ","b":"110 mph ","c":"92 mph ","d":"100 mph","e":"99 mph"},"options_float":{"a":198.0,"b":110.0,"c":92.0,"d":100.0,"e":99.0},"annotated_formula":"divide(add(96, 88), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"96 + 88<\/gadget>\n184<\/output>\n184 \/ 2<\/gadget>\n92<\/output>\n92<\/result>","index":3689} +{"problem":"find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 20 cm ?","rationale":"\"area of a trapezium = 1 \/ 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 \/ 2 ( 20 + 18 ) * ( 20 ) = 380 cm 2 answer : b\"","correct":"b","options":{"a":"827 cm 2 ","b":"380 cm 2 ","c":"285 cm 2 ","d":"178 cm 2","e":"176 cm 2"},"options_float":{"a":827.0,"b":380.0,"c":285.0,"d":178.0,"e":176.0},"annotated_formula":"quadrilateral_area(20, 18, 20)","linear_formula":"quadrilateral_area(n2,n1,n0)|","chain":"(20 + 18) * 20 \/ 2<\/gadget>\n380<\/output>\n380<\/result>","index":3690} +{"problem":"if x < y < z and y - x > 11 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ?","rationale":"\"we have : 1 ) x < y < z 2 ) y - x > 11 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 11 , the least value for y - x must be 13 , the least value for x must be 2 , and , thus , the least possible value for y must be 15 ( y - 2 = 13 , y = 15 ) 2 < 15 < z , since z is odd , the least possible value for z is 17 z - x = 17 - 2 = 15 answer e\"","correct":"e","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"15"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":15.0},"annotated_formula":"add(add(11, const_2), const_2)","linear_formula":"add(n0,const_2)|add(#0,const_2)|","chain":"11 + 2<\/gadget>\n13<\/output>\n13 + 2<\/gadget>\n15<\/output>\n15<\/result>","index":3691} +{"problem":"find k if 24 \/ k = 4 .","rationale":"\"since 24 \/ k = 4 and 24 \/ 6 = 4 , then k = 6 correct answer a\"","correct":"a","options":{"a":"6 ","b":"7 ","c":"8 ","d":"4","e":"3"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":4.0,"e":3.0},"annotated_formula":"divide(24, 4)","linear_formula":"divide(n0,n1)|","chain":"24 \/ 4<\/gadget>\n6<\/output>\n6<\/result>","index":3692} +{"problem":"a merchandise feedback online portal accords ‘ ratings ’ to a product by recording the percentage of customers who have reported their views as either ‘ satisfied ’ or ‘ unsatisfied ’ only . the portal rounds off the exact percentage value calculated to the closest integral value . if the percentage of unsatisfied customers as recorded by the portal is 57 % , what is the least possible number of people that could have reported their views as ‘ satisfied ’ ?","rationale":"57 % unsatisfied . so ~ 43 % satisfied but 100 % should be an integer only and since the above percentages are approximations 100 % should be as close as possible to an integer . now substitute value [ 1 ] if 43 % is 2 what is 100 % - > 200 \/ 43 is not giving a number which is closest to an integer . [ 2 ] if 43 % is 3 what is 100 % - > 300 \/ 43 [ 43 * 7 = 301 ] . . . therefore closest possible to an integer . [ it also satisfies the least possible and hence we need not check for other choices which are greater than 3 ] therefore ans is b .","correct":"b","options":{"a":"2 ","b":"3 ","c":"12 ","d":"43","e":"57"},"options_float":{"a":2.0,"b":3.0,"c":12.0,"d":43.0,"e":57.0},"annotated_formula":"subtract(add(const_4, const_3), const_4)","linear_formula":"add(const_3,const_4)|subtract(#0,const_4)","chain":"4 + 3<\/gadget>\n7<\/output>\n7 - 4<\/gadget>\n3<\/output>\n3<\/result>","index":3693} +{"problem":"a bag consists of 30 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ?","rationale":"\"bag consists of 20 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 30 - 5 - 9 = 16 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 \/ 30 + 16 \/ 30 = 25 \/ 30 = 5 \/ 6 hence , answer will be a .\"","correct":"a","options":{"a":"5 \/ 6 ","b":"2 \/ 4 ","c":"1 \/ 4 ","d":"1 \/ 8","e":"1 \/ 16"},"options_float":{"a":0.8333333333,"b":0.5,"c":0.25,"d":0.125,"e":0.0625},"annotated_formula":"divide(add(subtract(30, add(5, 9)), 9), 30)","linear_formula":"add(n1,n2)|subtract(n0,#0)|add(n2,#1)|divide(#2,n0)|","chain":"5 + 9<\/gadget>\n14<\/output>\n30 - 14<\/gadget>\n16<\/output>\n16 + 9<\/gadget>\n25<\/output>\n25 \/ 30<\/gadget>\n5\/6 = around 0.833333<\/output>\n5\/6 = around 0.833333<\/result>","index":3694} +{"problem":"a present value of a machine is $ 1000 . its value depletiation rate is 10 % per annum then find the machine value after 2 years ?","rationale":"\"p = $ 1000 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r \/ 100 ) ^ t ] = 1000 * 9 \/ 10 * 9 \/ 10 = $ 810 answer is b\"","correct":"b","options":{"a":"$ 900 ","b":"$ 810 ","c":"$ 915 ","d":"$ 715","e":"$ 795"},"options_float":{"a":900.0,"b":810.0,"c":915.0,"d":715.0,"e":795.0},"annotated_formula":"multiply(1000, power(subtract(const_1, divide(10, const_100)), 2))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) ** 2<\/gadget>\n81\/100 = around 0.81<\/output>\n1_000 * (81\/100)<\/gadget>\n810<\/output>\n810<\/result>","index":3697} +{"problem":"in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 \/ 3 of the number of students of 8 years of age which is 48 . what is the total number of students in the school ?","rationale":"\"let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 48 + 2 \/ 3 of 48 ( 80 \/ 100 ) x = 80 x = 100 . answer d\"","correct":"d","options":{"a":"72 ","b":"80 ","c":"120 ","d":"100","e":"150"},"options_float":{"a":72.0,"b":80.0,"c":120.0,"d":100.0,"e":150.0},"annotated_formula":"divide(add(48, multiply(48, divide(2, 3))), subtract(const_1, divide(20, const_100)))","linear_formula":"divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n48 * (2\/3)<\/gadget>\n32<\/output>\n48 + 32<\/gadget>\n80<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n80 \/ (4\/5)<\/gadget>\n100<\/output>\n100<\/result>","index":3698} +{"problem":"a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 756 find the share of a .","rationale":"\"explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 \/ 21 * 756 = 288 answer : c\"","correct":"c","options":{"a":"240 ","b":"388 ","c":"288 ","d":"277","e":"122"},"options_float":{"a":240.0,"b":388.0,"c":288.0,"d":277.0,"e":122.0},"annotated_formula":"multiply(divide(756, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))))","linear_formula":"add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)|","chain":"3_000 * 8<\/gadget>\n24_000<\/output>\n3_000 - 1_000<\/gadget>\n2_000<\/output>\n12 - 8<\/gadget>\n4<\/output>\n2_000 * 4<\/gadget>\n8_000<\/output>\n24_000 + 8_000<\/gadget>\n32_000<\/output>\n4_000 * 8<\/gadget>\n32_000<\/output>\n4_000 + 1_000<\/gadget>\n5_000<\/output>\n5_000 * 4<\/gadget>\n20_000<\/output>\n32_000 + 20_000<\/gadget>\n52_000<\/output>\n32_000 + 52_000<\/gadget>\n84_000<\/output>\n756 \/ 84_000<\/gadget>\n9\/1_000 = around 0.009<\/output>\n(9\/1_000) * 32_000<\/gadget>\n288<\/output>\n288<\/result>","index":3700} +{"problem":"100 people are attending a newspaper conference . 35 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ?","rationale":"\"{ total } = { writers } + { editors } - { both } + { neither } . { total } = 100 ; { writers } = 35 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 100 = 35 + { editors } - x + 2 x - - > x = 65 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 39 , thus x = { both } = 65 - 39 = 26 . answer : b .\"","correct":"b","options":{"a":"21 ","b":"26 ","c":"28 ","d":"30","e":"32"},"options_float":{"a":21.0,"b":26.0,"c":28.0,"d":30.0,"e":32.0},"annotated_formula":"subtract(100, add(add(38, const_1), 35))","linear_formula":"add(n2,const_1)|add(n1,#0)|subtract(n0,#1)|","chain":"38 + 1<\/gadget>\n39<\/output>\n39 + 35<\/gadget>\n74<\/output>\n100 - 74<\/gadget>\n26<\/output>\n26<\/result>","index":3702} +{"problem":"there are 4 people of different heights standing in order of increasing height . the difference is 2 inches between the first person and the second person , and also between the second person and the third person . the difference between the third person and the fourth person is 6 inches and the average height is 78 . how tall is the fourth person ?","rationale":"\"let x be the height of the first person . then the heights are x , x + 2 , x + 4 , and x + 10 . 4 x + 16 = 4 ( 78 ) = 312 x = 74 and the fourth person has a height of 74 + 10 = 84 inches the answer is b .\"","correct":"b","options":{"a":"82 ","b":"84 ","c":"86 ","d":"88","e":"90"},"options_float":{"a":82.0,"b":84.0,"c":86.0,"d":88.0,"e":90.0},"annotated_formula":"add(divide(subtract(multiply(78, 4), add(6, add(4, 6))), 4), add(4, 6))","linear_formula":"add(n0,n2)|multiply(n0,n3)|add(n2,#0)|subtract(#1,#2)|divide(#3,n0)|add(#0,#4)|","chain":"78 * 4<\/gadget>\n312<\/output>\n4 + 6<\/gadget>\n10<\/output>\n6 + 10<\/gadget>\n16<\/output>\n312 - 16<\/gadget>\n296<\/output>\n296 \/ 4<\/gadget>\n74<\/output>\n74 + 10<\/gadget>\n84<\/output>\n84<\/result>","index":3703} +{"problem":"if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability w that this product will be a multiple of 45 ?","rationale":"\"there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability w = 1 \/ 24 . answer c .\"","correct":"c","options":{"a":"1 \/ 32 ","b":"1 \/ 28 ","c":"1 \/ 24 ","d":"1 \/ 16","e":"1 \/ 14"},"options_float":{"a":0.03125,"b":0.0357142857,"c":0.0416666667,"d":0.0625,"e":0.0714285714},"annotated_formula":"divide(const_1, add(20, const_4))","linear_formula":"add(n1,const_4)|divide(const_1,#0)|","chain":"20 + 4<\/gadget>\n24<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n1\/24 = around 0.041667<\/result>","index":3704} +{"problem":"during a pizza buffet where a eats more times 1.8 than b , and b eats 8 times less than c . find the leat number of times all the 3 has to eat","rationale":"a eats more than b if b eats 1 times than the ratio of a and b is a : b is 1.8 : 1 or 9 : 5 and as b eat 8 times less the c the the ratio of b : c is 5 : 40 the the least number of times all three has eat is the lcm of a , b , c that is 360 . . answer : c","correct":"c","options":{"a":"350 ","b":"260 ","c":"360 ","d":"380","e":"285"},"options_float":{"a":350.0,"b":260.0,"c":360.0,"d":380.0,"e":285.0},"annotated_formula":"multiply(multiply(divide(multiply(1.8, const_10), 3), divide(multiply(1.8, const_10), 3)), const_10)","linear_formula":"multiply(n0,const_10)|divide(#0,n2)|multiply(#1,#1)|multiply(#2,const_10)","chain":"1.8 * 10<\/gadget>\n18<\/output>\n18 \/ 3<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n36 * 10<\/gadget>\n360<\/output>\n360<\/result>","index":3705} +{"problem":"ashok secured average of 72 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ?","rationale":"explanation : number of subjects = 6 average of marks in 6 subjects = 72 therefore total marks in 6 subjects = 72 * 6 = 432 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 432 – 370 = 62 answer a","correct":"a","options":{"a":"62 ","b":"74 ","c":"78 ","d":"80","e":"none of these"},"options_float":{"a":62.0,"b":74.0,"c":78.0,"d":80.0,"e":null},"annotated_formula":"subtract(multiply(72, 6), multiply(74, 5))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)","chain":"72 * 6<\/gadget>\n432<\/output>\n74 * 5<\/gadget>\n370<\/output>\n432 - 370<\/gadget>\n62<\/output>\n62<\/result>","index":3706} +{"problem":"at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 85 . if the restaurant serves 120 customers today , raising the average to 90 customers per day , what is the value of x ?","rationale":"\"withoutusing the formula , we can see that today the restaurant served 30 customers above the average . the total amount above the average must equal total amount below the average . this additional 30 customers must offset the “ deficit ” below the average of 90 created on the x days the restaurant served only 85 customers per day . 30 \/ 5 = 6 days . choice ( a ) . withthe formula , we can set up the following : 90 = ( 85 x + 120 ) \/ ( x + 1 ) 90 x + 90 = 85 x + 120 5 x = 30 x = 6 answer choice ( c )\"","correct":"c","options":{"a":"2 ","b":"5 ","c":"6 ","d":"15","e":"30"},"options_float":{"a":2.0,"b":5.0,"c":6.0,"d":15.0,"e":30.0},"annotated_formula":"subtract(divide(subtract(120, 90), subtract(90, 85)), divide(subtract(120, const_100), const_100))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)|","chain":"120 - 90<\/gadget>\n30<\/output>\n90 - 85<\/gadget>\n5<\/output>\n30 \/ 5<\/gadget>\n6<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n6 - (1\/5)<\/gadget>\n29\/5 = around 5.8<\/output>\n29\/5 = around 5.8<\/result>","index":3707} +{"problem":"two boys starts from the same place walking at the rate of 5.5 kmph and 7.5 kmph respectively in the same direction . what time will they take to be 20 km apart ?","rationale":"\"explanation : relative speed = 7.5 - 5.5 = 2 kmph ( because they walk in the same direction ) distance = 20 km time = distance \/ speed = 20 \/ 2 = 10 hr answer : a\"","correct":"a","options":{"a":"10 ","b":"11 ","c":"14 ","d":"15","e":"20"},"options_float":{"a":10.0,"b":11.0,"c":14.0,"d":15.0,"e":20.0},"annotated_formula":"divide(20, subtract(7.5, 5.5))","linear_formula":"subtract(n1,n0)|divide(n2,#0)|","chain":"7.5 - 5.5<\/gadget>\n2<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":3708} +{"problem":"a man can row 6 kmph in still water . when the river is running at 2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?","rationale":"\"m = 6 s = 2 ds = 8 us = 4 x \/ 8 + x \/ 4 = 1 x = 2.66 d = 2.66 * 2 = 5.32 answer : a\"","correct":"a","options":{"a":"5.32 ","b":"5.7 ","c":"5.76 ","d":"5.74","e":"5.71"},"options_float":{"a":5.32,"b":5.7,"c":5.76,"d":5.74,"e":5.71},"annotated_formula":"multiply(divide(multiply(add(6, 2), subtract(6, 2)), add(add(6, 2), subtract(6, 2))), const_2)","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|","chain":"6 + 2<\/gadget>\n8<\/output>\n6 - 2<\/gadget>\n4<\/output>\n8 * 4<\/gadget>\n32<\/output>\n8 + 4<\/gadget>\n12<\/output>\n32 \/ 12<\/gadget>\n8\/3 = around 2.666667<\/output>\n(8\/3) * 2<\/gadget>\n16\/3 = around 5.333333<\/output>\n16\/3 = around 5.333333<\/result>","index":3709} +{"problem":"if 8 workers can build 8 cars in 8 days , then how many days would it take 6 workers to build 6 cars ?","rationale":"\"8 workers can build 1 car per day on average . 1 worker can build 1 \/ 8 of a car per day . 6 workers can build 6 \/ 8 car per day . the time required to build 6 cars is 6 \/ ( 6 \/ 8 ) = 8 days the answer is d .\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"6 ","d":"8","e":"12"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":8.0,"e":12.0},"annotated_formula":"multiply(divide(multiply(8, 8), 8), divide(6, 6))","linear_formula":"divide(n3,n3)|multiply(n0,n0)|divide(#1,n0)|multiply(#2,#0)|","chain":"8 * 8<\/gadget>\n64<\/output>\n64 \/ 8<\/gadget>\n8<\/output>\n6 \/ 6<\/gadget>\n1<\/output>\n8 * 1<\/gadget>\n8<\/output>\n8<\/result>","index":3710} +{"problem":"a person can swim in still water at 12 km \/ h . if the speed of water 10 km \/ h , how many hours will the man take to swim back against the current for 8 km ?","rationale":"\"m = 12 s = 10 us = 12 - 10 = 2 d = 8 t = 8 \/ 2 = 4 answer : a\"","correct":"a","options":{"a":"4 ","b":"7 ","c":"5 ","d":"9","e":"6"},"options_float":{"a":4.0,"b":7.0,"c":5.0,"d":9.0,"e":6.0},"annotated_formula":"divide(8, subtract(12, 10))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"12 - 10<\/gadget>\n2<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":3712} +{"problem":"the moon revolves around the earth at a speed of approximately 0.2 kilometers per second . this approximate speed is how many kilometers per hour ?","rationale":"\"moon revolves around the earth at a speed of 1.02 kilometers per second . one hour equal to 60 minutes . one minute equals to 60 seconds . so one hour equals to 3600 seconds . so one hour , speed = 0.2 * 3600 = 720 kilometers per hour . option b is correct\"","correct":"b","options":{"a":"60 ","b":"720 ","c":"62.5 ","d":"3,600","e":"3,672"},"options_float":{"a":60.0,"b":720.0,"c":62.5,"d":3600.0,"e":3672.0},"annotated_formula":"multiply(0.2, const_3600)","linear_formula":"multiply(n0,const_3600)|","chain":"0.2 * 3_600<\/gadget>\n720<\/output>\n720<\/result>","index":3713} +{"problem":"if 1.5 cm of a pencil is purple , 0.5 cm of the remaining is black and the remaining 2 cm is blue , what is the total length of the pencil ?","rationale":"purple is 1.5 cm black is 0.5 cm blue is 2 cm 1.5 + 0.5 + 2 = 4 cm answer is d ) 4 cm","correct":"d","options":{"a":"5 cm ","b":"3 cm ","c":"1 cm ","d":"4 cm","e":"2 cm"},"options_float":{"a":5.0,"b":3.0,"c":1.0,"d":4.0,"e":2.0},"annotated_formula":"add(add(1.5, 0.5), 2)","linear_formula":"add(n0,n1)|add(n2,#0)|","chain":"1.5 + 0.5<\/gadget>\n2<\/output>\n2 + 2<\/gadget>\n4<\/output>\n4<\/result>","index":3714} +{"problem":"of the 130 employees at company x , 80 are full - time , and 100 have worked at company x for at least a year . there are 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ?","rationale":"full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = 150 a + b = 80 i . e . c + d = 70 b + c = 100 i . e . a + d = 50 d = 20 i . e . c = 70 - 20 = 50 i . e . b = 100 - 50 = 50 i . e . a = 80 - 50 = 30 b = 30 answer : option b","correct":"b","options":{"a":"20 ","b":"30 ","c":"50 ","d":"80","e":"100"},"options_float":{"a":20.0,"b":30.0,"c":50.0,"d":80.0,"e":100.0},"annotated_formula":"subtract(subtract(130, 80), 20)","linear_formula":"subtract(n0,n1)|subtract(#0,n3)","chain":"130 - 80<\/gadget>\n50<\/output>\n50 - 20<\/gadget>\n30<\/output>\n30<\/result>","index":3715} +{"problem":"what is the rate percent when the simple interest on rs . 810 amount to rs . 155 in 4 years ?","rationale":"\"155 = ( 810 * 4 * r ) \/ 100 r = 4.78 % answer : a\"","correct":"a","options":{"a":"4.78 ","b":"4.68 ","c":"4.58 ","d":"4.48","e":"4.38"},"options_float":{"a":4.78,"b":4.68,"c":4.58,"d":4.48,"e":4.38},"annotated_formula":"divide(multiply(const_100, 155), multiply(810, 4))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 155<\/gadget>\n15_500<\/output>\n810 * 4<\/gadget>\n3_240<\/output>\n15_500 \/ 3_240<\/gadget>\n775\/162 = around 4.783951<\/output>\n775\/162 = around 4.783951<\/result>","index":3716} +{"problem":"a pharmaceutical company received $ 4 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ?","rationale":"\"solution : this is a percent decrease problem . we will use the formula : percent change = ( new – old ) \/ old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 20 million royalties \/ sales = 4 \/ 20 = 1 \/ 5 next 108 million royalties \/ sales = 9 \/ 108 = 1 \/ 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties \/ sales = ( 4 \/ 20 ) x 60 = 12 next 108 million royalties \/ sales = 9 \/ 108 = ( 1 \/ 12 ) x 60 = 5 we can plug 12 and 5 into our percent change formula : ( new – old ) \/ old x 100 [ ( 5 – 12 ) \/ 12 ] x 100 - 175 \/ 3 x 100 at this point we can stop and consider the answer choices . since we know that 175 \/ 3 is just a bit less than ½ , we know that - 175 \/ 3 x 100 is about a 58 % decrease . answer e .\"","correct":"e","options":{"a":"8 % ","b":"15 % ","c":"45 % ","d":"52 %","e":"58 %"},"options_float":{"a":8.0,"b":15.0,"c":45.0,"d":52.0,"e":58.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(4, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(4, 20), const_100)), const_100)","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,const_100)|multiply(#1,const_100)|subtract(#2,#3)|divide(#4,#2)|multiply(#5,const_100)|","chain":"4 \/ 20<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n9 \/ 108<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 100<\/gadget>\n25\/3 = around 8.333333<\/output>\n20 - (25\/3)<\/gadget>\n35\/3 = around 11.666667<\/output>\n(35\/3) \/ 20<\/gadget>\n7\/12 = around 0.583333<\/output>\n(7\/12) * 100<\/gadget>\n175\/3 = around 58.333333<\/output>\n175\/3 = around 58.333333<\/result>","index":3717} +{"problem":"a father said to his son , ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is 52 years now , the son ' s age 5 years back was :","rationale":"let the son ' s present age be x years . then , ( 52 - x ) = x 2 x = 52 . x = 26 . son ' s age 5 years back ( 26 - 5 ) = 21 years . answer : d","correct":"d","options":{"a":"14 ","b":"67 ","c":"56 ","d":"21","e":"34"},"options_float":{"a":14.0,"b":67.0,"c":56.0,"d":21.0,"e":34.0},"annotated_formula":"subtract(divide(52, const_2), 5)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)","chain":"52 \/ 2<\/gadget>\n26<\/output>\n26 - 5<\/gadget>\n21<\/output>\n21<\/result>","index":3718} +{"problem":"at a certain fruit stand , the price of each apple is 40 cents and the price of each orange is 60 cents . mary selects a total of 20 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 20 pieces of fruit is 56 cents . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 52 cents ?","rationale":"\"let number of apples = a number of oranges = b a + b = 20 - - - 1 . 56 = ( . 4 a + . 6 b ) \/ 20 = > 56 = 4 a + 6 b - - - - 2 solving 1 and 2 , we get a = 4 b = 16 let the number of oranges put back = c 52 * ( 20 - c ) = 40 * 4 + 60 ( 16 - c ) = > c = 10 answer e\"","correct":"e","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"divide(subtract(multiply(56, 20), multiply(52, 20)), subtract(60, 52))","linear_formula":"multiply(n2,n4)|multiply(n2,n5)|subtract(n1,n5)|subtract(#0,#1)|divide(#3,#2)|","chain":"56 * 20<\/gadget>\n1_120<\/output>\n52 * 20<\/gadget>\n1_040<\/output>\n1_120 - 1_040<\/gadget>\n80<\/output>\n60 - 52<\/gadget>\n8<\/output>\n80 \/ 8<\/gadget>\n10<\/output>\n10<\/result>","index":3719} +{"problem":"there are 20 cm , 21 cm , 22 cm , 23 cm , 24 cm , 25 cm rods of quantities 23 , 64 , 32 respectively . find average of this data .","rationale":"average = ( 20 * 2 + 21 * 3 + 22 * 6 + 23 * 4 + 24 * 3 + 25 * 2 ) \/ 20 = 22.45 answer : d","correct":"d","options":{"a":"32.25 ","b":"21.45 ","c":"45 ","d":"22.45","e":"22"},"options_float":{"a":32.25,"b":21.45,"c":45.0,"d":22.45,"e":22.0},"annotated_formula":"divide(add(add(add(add(add(multiply(20, 23), multiply(21, 23)), multiply(22, 64)), multiply(23, 64)), multiply(24, 32)), multiply(25, 32)), multiply(add(add(23, 64), 32), const_2))","linear_formula":"add(n3,n7)|multiply(n0,n3)|multiply(n1,n3)|multiply(n2,n7)|multiply(n3,n7)|multiply(n4,n8)|multiply(n5,n8)|add(#1,#2)|add(n8,#0)|add(#7,#3)|multiply(#8,const_2)|add(#9,#4)|add(#11,#5)|add(#12,#6)|divide(#13,#10)","chain":"20 * 23<\/gadget>\n460<\/output>\n21 * 23<\/gadget>\n483<\/output>\n460 + 483<\/gadget>\n943<\/output>\n22 * 64<\/gadget>\n1_408<\/output>\n943 + 1_408<\/gadget>\n2_351<\/output>\n23 * 64<\/gadget>\n1_472<\/output>\n2_351 + 1_472<\/gadget>\n3_823<\/output>\n24 * 32<\/gadget>\n768<\/output>\n3_823 + 768<\/gadget>\n4_591<\/output>\n25 * 32<\/gadget>\n800<\/output>\n4_591 + 800<\/gadget>\n5_391<\/output>\n23 + 64<\/gadget>\n87<\/output>\n87 + 32<\/gadget>\n119<\/output>\n119 * 2<\/gadget>\n238<\/output>\n5_391 \/ 238<\/gadget>\n5_391\/238 = around 22.651261<\/output>\n5_391\/238 = around 22.651261<\/result>","index":3720} +{"problem":"a can finish a work in 18 days and b can do the same work in half the time taken by a . then , working together what part of the same work they can finish in a day ?","rationale":"1 days work of a + b = ( 18 + 9 ) \/ 18 * 9 = 1 \/ 6 answer is c","correct":"c","options":{"a":"1 \/ 2 ","b":"1 \/ 3 ","c":"1 \/ 6 ","d":"2 \/ 5","e":"2 \/ 9"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.1666666667,"d":0.4,"e":0.2222222222},"annotated_formula":"add(divide(const_1, 18), divide(const_1, divide(18, const_2)))","linear_formula":"divide(const_1,n0)|divide(n0,const_2)|divide(const_1,#1)|add(#0,#2)","chain":"1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/18) + (1\/9)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":3721} +{"problem":"if n is a positive integer and n ^ 2 is divisible by 18 , then what is the largest positive integer that must divide n ?","rationale":"\"18 = 2 * 3 ^ 2 if 18 divides n ^ 2 , then n must be divisible by 2 * 3 = 6 the answer is c .\"","correct":"c","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"multiply(sqrt(divide(18, 2)), 2)","linear_formula":"divide(n1,n0)|sqrt(#0)|multiply(n0,#1)|","chain":"18 \/ 2<\/gadget>\n9<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6<\/result>","index":3722} +{"problem":"an employee ’ s annual salary was increased $ 25000 . if her new annual salary now equals $ 90000 , what was the percent increase ?","rationale":"new annual salary = $ 90,000 salary increase = $ 25,000 . original salary = $ 90,000 - $ 25,000 . = $ 75,000 % increase = ( $ 25,000 \/ $ 65,000 ) * 100 = 38.46 % hence d .","correct":"d","options":{"a":"15 % ","b":"162 ⁄ 3 % ","c":"20 % ","d":"38.46 %","e":"24 %"},"options_float":{"a":15.0,"b":162.0,"c":20.0,"d":38.46,"e":24.0},"annotated_formula":"multiply(divide(25000, subtract(90000, 25000)), const_100)","linear_formula":"subtract(n1,n0)|divide(n0,#0)|multiply(#1,const_100)","chain":"90_000 - 25_000<\/gadget>\n65_000<\/output>\n25_000 \/ 65_000<\/gadget>\n5\/13 = around 0.384615<\/output>\n(5\/13) * 100<\/gadget>\n500\/13 = around 38.461538<\/output>\n500\/13 = around 38.461538<\/result>","index":3723} +{"problem":"in a class of 45 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ?","rationale":"\"total = english + german - both + neither - - > 45 = english + 22 - 12 + 0 - - > english = 35 - - > only english = english - both = 35 - 12 = 23 . answer : d .\"","correct":"d","options":{"a":"30 ","b":"10 ","c":"18 ","d":"23","e":"32"},"options_float":{"a":30.0,"b":10.0,"c":18.0,"d":23.0,"e":32.0},"annotated_formula":"subtract(subtract(add(45, 12), 22), 12)","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)|","chain":"45 + 12<\/gadget>\n57<\/output>\n57 - 22<\/gadget>\n35<\/output>\n35 - 12<\/gadget>\n23<\/output>\n23<\/result>","index":3724} +{"problem":"mr . parkin invested in fund z and fund b . the total amount she invested , in both funds combined , was $ 100000 . in one year , fund z paid 23 % and fund b paid 17 % . the interest earned in fund b was exactly $ 200 greater than the interest earned in fund z . how much did ms . morris invest in fund z ?","rationale":"0.17 b - 0.23 z = 200 | * 100 17 b - 23 z = 20000 z + b = 100000 b = 100000 - z 17 ( 100000 - z ) - 23 z = 20000 1 , 700000 - 17 z - 23 z = 20000 1 , 680000 = 40 z divide by 40 first two digits : 42 . . . so answer is d .","correct":"d","options":{"a":"$ 32000 ","b":"$ 36000 ","c":"$ 40000 ","d":"$ 42000","e":"$ 45000"},"options_float":{"a":32000.0,"b":36000.0,"c":40000.0,"d":42000.0,"e":45000.0},"annotated_formula":"divide(multiply(add(23, 17), 100000), const_100)","linear_formula":"add(n1,n2)|multiply(n0,#0)|divide(#1,const_100)","chain":"23 + 17<\/gadget>\n40<\/output>\n40 * 100_000<\/gadget>\n4_000_000<\/output>\n4_000_000 \/ 100<\/gadget>\n40_000<\/output>\n40_000<\/result>","index":3725} +{"problem":"a train 100 meters long completely crosses a 300 meters long bridge in 36 seconds . what is the speed of the train is ?","rationale":"\"s = ( 100 + 300 ) \/ 36 = 400 \/ 36 * 18 \/ 5 = 40 answer : e\"","correct":"e","options":{"a":"32 kmph ","b":"76 kmph ","c":"34 kmph ","d":"43 kmph","e":"40 kmph"},"options_float":{"a":32.0,"b":76.0,"c":34.0,"d":43.0,"e":40.0},"annotated_formula":"divide(divide(add(100, 300), const_1000), divide(36, const_3600))","linear_formula":"add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)|","chain":"100 + 300<\/gadget>\n400<\/output>\n400 \/ 1_000<\/gadget>\n2\/5 = around 0.4<\/output>\n36 \/ 3_600<\/gadget>\n1\/100 = around 0.01<\/output>\n(2\/5) \/ (1\/100)<\/gadget>\n40<\/output>\n40<\/result>","index":3726} +{"problem":"a school has 11 maths 8 physics and 5 chemistry teachers each teacher can teach 3 subjects max what is he minimum number of teachers required","rationale":"\"total subjects = 11 + 8 + 5 = 24 max subjects by 1 teacher = 3 so , min of teachers required = 24 \/ 3 = 8 answer : d\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"divide(add(add(11, 8), 5), 3)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(#1,n3)|","chain":"11 + 8<\/gadget>\n19<\/output>\n19 + 5<\/gadget>\n24<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n8<\/result>","index":3727} +{"problem":"trapezoid jklm in the x - y plane has coordinates j = ( – 3 , – 4 ) , k = ( – 3 , 1 ) , l = ( 5 , 7 ) , and m = ( 5 , – 4 ) . what is its perimeter ?","rationale":"jk = 5 lm = 11 kl = using distance formula 10 jm = using distance formula 8 sum of all is 34 a","correct":"a","options":{"a":"34 ","b":"36 ","c":"38 ","d":"40","e":"( f ) 42"},"options_float":{"a":34.0,"b":36.0,"c":38.0,"d":40.0,"e":42.0},"annotated_formula":"add(add(add(const_10, add(1, 4)), add(3, 5)), add(7, 4))","linear_formula":"add(n1,n3)|add(n0,n4)|add(n1,n5)|add(#0,const_10)|add(#3,#1)|add(#4,#2)","chain":"1 + 4<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n3 + 5<\/gadget>\n8<\/output>\n15 + 8<\/gadget>\n23<\/output>\n7 + 4<\/gadget>\n11<\/output>\n23 + 11<\/gadget>\n34<\/output>\n34<\/result>","index":3728} +{"problem":"| - 2 | ( | - 25 | - | 5 | ) = ? ? source : preparation material mba center","rationale":"absolute value will turn negatives into their positive ' equivalents ' , and will leave positives unchanged , so | - 2 | = 2 , | - 25 | = 25 , and | 5 | = 5 . getting rid of our absolute values we have : | - 2 | ( | - 25 | - | 5 | ) = ( 2 ) ( 25 - 5 ) = 2 * 20 = 40","correct":"a","options":{"a":"40 ","b":"– 60 ","c":"60 ","d":"75","e":"100"},"options_float":{"a":40.0,"b":60.0,"c":60.0,"d":75.0,"e":100.0},"annotated_formula":"multiply(subtract(25, 5), 2)","linear_formula":"subtract(n1,n2)|multiply(n0,#0)","chain":"25 - 5<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n40<\/result>","index":3729} +{"problem":"at a certain fruit stand , the price of each apple is 40 cents and the price of each orange is 60 cents . mary selects a total of 30 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 30 pieces of fruit is 56 cents . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 52 cents ?","rationale":"\"let number of apples = a number of oranges = b a + b = 30 - - - 1 . 56 = ( . 4 a + . 6 b ) \/ 30 = > 56 = 4 a + 6 b - - - - 2 solving 1 and 2 , we get a = 6 b = 24 let the number of oranges put back = c 52 * ( 30 - c ) = 40 * 6 + 60 ( 24 - c ) = > c = 15 answer e\"","correct":"e","options":{"a":"10 ","b":"11 ","c":"13 ","d":"14","e":"15"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":14.0,"e":15.0},"annotated_formula":"divide(subtract(multiply(56, 30), multiply(52, 30)), subtract(60, 52))","linear_formula":"multiply(n2,n4)|multiply(n2,n5)|subtract(n1,n5)|subtract(#0,#1)|divide(#3,#2)|","chain":"56 * 30<\/gadget>\n1_680<\/output>\n52 * 30<\/gadget>\n1_560<\/output>\n1_680 - 1_560<\/gadget>\n120<\/output>\n60 - 52<\/gadget>\n8<\/output>\n120 \/ 8<\/gadget>\n15<\/output>\n15<\/result>","index":3730} +{"problem":"a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 30 % while selling . what is his percentage profit ?","rationale":"\"the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 70 kg ; profit : ( 112 - 70 ) \/ 70 * 100 = ~ 60 % answer : d .\"","correct":"d","options":{"a":"10.22 % ","b":"20.22 % ","c":"21.22 % ","d":"60 %","e":"ca n ' t be calculated"},"options_float":{"a":10.22,"b":20.22,"c":21.22,"d":60.0,"e":null},"annotated_formula":"divide(multiply(subtract(add(const_100, 12), subtract(const_100, 30)), const_100), subtract(const_100, 30))","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#1)|","chain":"100 + 12<\/gadget>\n112<\/output>\n100 - 30<\/gadget>\n70<\/output>\n112 - 70<\/gadget>\n42<\/output>\n42 * 100<\/gadget>\n4_200<\/output>\n4_200 \/ 70<\/gadget>\n60<\/output>\n60<\/result>","index":3731} +{"problem":"a statue is being carved by a sculptor . the original piece of marble weighed 180 kg . in the first week 28 percent is cut away . in the second week 18 percent of the remainder is cut away . in the third week the statue is completed when 20 percent of the remainder is cut away . what is the weight of the final statue ?","rationale":"\"b 85 kg 180 ã — 0.72 ã — 0.82 ã — 0.8 = 85 kg .\"","correct":"b","options":{"a":"105 kg ","b":"85 kg ","c":"108 kg ","d":"125 kg","e":"117 kg"},"options_float":{"a":105.0,"b":85.0,"c":108.0,"d":125.0,"e":117.0},"annotated_formula":"multiply(subtract(const_1, divide(20, const_100)), multiply(subtract(const_1, divide(18, const_100)), multiply(180, subtract(const_1, divide(28, const_100)))))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(n0,#3)|multiply(#6,#4)|multiply(#7,#5)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n1 - (9\/50)<\/gadget>\n41\/50 = around 0.82<\/output>\n28 \/ 100<\/gadget>\n7\/25 = around 0.28<\/output>\n1 - (7\/25)<\/gadget>\n18\/25 = around 0.72<\/output>\n180 * (18\/25)<\/gadget>\n648\/5 = around 129.6<\/output>\n(41\/50) * (648\/5)<\/gadget>\n13_284\/125 = around 106.272<\/output>\n(4\/5) * (13_284\/125)<\/gadget>\n53_136\/625 = around 85.0176<\/output>\n53_136\/625 = around 85.0176<\/result>","index":3733} +{"problem":"there are 22 kids in a classroom . 12 kids are wearing socks and 8 are wearing shoes . 6 kids are wearing both . how many are bare feet ?","rationale":"12 kids wear socks 8 kids wear shoes 6 wear both . 12 â € “ 6 = 6 ( wear only socks ) 8 â € “ 6 = 2 ( wear only shoes ) 6 ( wear both ) hence , 22 - 14 = 8 8 kids are with bare feet . e","correct":"e","options":{"a":"10 ","b":"2 ","c":"4 ","d":"5","e":"8"},"options_float":{"a":10.0,"b":2.0,"c":4.0,"d":5.0,"e":8.0},"annotated_formula":"subtract(22, add(add(subtract(12, 6), subtract(8, 6)), 6))","linear_formula":"subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)","chain":"12 - 6<\/gadget>\n6<\/output>\n8 - 6<\/gadget>\n2<\/output>\n6 + 2<\/gadget>\n8<\/output>\n8 + 6<\/gadget>\n14<\/output>\n22 - 14<\/gadget>\n8<\/output>\n8<\/result>","index":3734} +{"problem":"a and b together can do a piece of work in 6 days and a alone can do it in 13 days . in how many days can b alone can do it ?","rationale":"\"explanation : a and b can do work 1 \/ 6 in 1 day a alone can do 1 \/ 13 work in 1 day b alone can do ( 1 \/ 6 - 1 \/ 13 ) = 7 \/ 78 work in 1 day = > complete work can be done in 78 \/ 7 days by b answer : option c\"","correct":"c","options":{"a":"12 days ","b":"15 days ","c":"78 \/ 7 days ","d":"21 days","e":"22 \/ 7 days"},"options_float":{"a":12.0,"b":15.0,"c":11.1428571429,"d":21.0,"e":3.1428571429},"annotated_formula":"inverse(subtract(inverse(6), inverse(13)))","linear_formula":"inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 13<\/gadget>\n1\/13 = around 0.076923<\/output>\n(1\/6) - (1\/13)<\/gadget>\n7\/78 = around 0.089744<\/output>\n1 \/ (7\/78)<\/gadget>\n78\/7 = around 11.142857<\/output>\n78\/7 = around 11.142857<\/result>","index":3735} +{"problem":"the length of a side of a hexagon is 4 inches . what is the perimeter ?","rationale":"hexagon . it means 6 equal sides . p = 6 ( 4 ) = 24 inches answer c","correct":"c","options":{"a":"22 ","b":"23 ","c":"24 ","d":"25","e":"26"},"options_float":{"a":22.0,"b":23.0,"c":24.0,"d":25.0,"e":26.0},"annotated_formula":"multiply(4, multiply(const_3, const_2))","linear_formula":"multiply(const_2,const_3)|multiply(n0,#0)","chain":"3 * 2<\/gadget>\n6<\/output>\n4 * 6<\/gadget>\n24<\/output>\n24<\/result>","index":3739} +{"problem":"how long does a train 110 m long running at the speed of 72 km \/ hr takes to cross a bridge 112 m length ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec total distance covered = 110 + 112 = 222 m . required time = 222 \/ 20 = 11.1 sec . answer : d\"","correct":"d","options":{"a":"82.1 sec ","b":"12.1 sec ","c":"16.1 sec ","d":"11.1 sec","e":"12.15 sec"},"options_float":{"a":82.1,"b":12.1,"c":16.1,"d":11.1,"e":12.15},"annotated_formula":"divide(add(110, 112), multiply(72, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"110 + 112<\/gadget>\n222<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n222 \/ 20<\/gadget>\n111\/10 = around 11.1<\/output>\n111\/10 = around 11.1<\/result>","index":3741} +{"problem":"ketchup , mustard and mayo bottles at a hot dog stand are in the ratio of 3 : 3 : 2 . if there are 6 ketchup bottles , the number of mayo bottles at the hot dog stand is :","rationale":"explanation : let ketchup = 3 x , mustard = 3 x & mayo = 2 x . now , 3 x = 6 hence x = 2 . number of mayo bottles = 2 x which is 4 . answer : e","correct":"e","options":{"a":"12 ","b":"6 ","c":"3 ","d":"1","e":"4"},"options_float":{"a":12.0,"b":6.0,"c":3.0,"d":1.0,"e":4.0},"annotated_formula":"multiply(divide(6, 3), 2)","linear_formula":"divide(n3,n0)|multiply(n2,#0)","chain":"6 \/ 3<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4<\/result>","index":3742} +{"problem":"at a certain university , 65 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ?","rationale":"\"total women = 65 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 25 % but question wants to know the percent of men that are tenured 25 % \/ 40 % = 62.5 % d\"","correct":"d","options":{"a":"25 ","b":"37.5 ","c":"50 ","d":"62.5","e":"75"},"options_float":{"a":25.0,"b":37.5,"c":50.0,"d":62.5,"e":75.0},"annotated_formula":"add(subtract(const_100, 65), subtract(90, 65))","linear_formula":"subtract(const_100,n0)|subtract(n2,n0)|add(#0,#1)|","chain":"100 - 65<\/gadget>\n35<\/output>\n90 - 65<\/gadget>\n25<\/output>\n35 + 25<\/gadget>\n60<\/output>\n60<\/result>","index":3744} +{"problem":"a can do a work in 15 days and b in 20 days . if they work on it together for 5 days , then the fraction of the work that is left is","rationale":"\"person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( - ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 \/ 35 ) days to complete entire work for 1 st 5 days they work 35 * 5 = 175 remaining work is 300 - 175 = 125 remaining fraction of work is = 125 \/ 300 = 5 \/ 12 answer d\"","correct":"d","options":{"a":"8 \/ 17 ","b":"7 \/ 15 ","c":"3 \/ 15 ","d":"5 \/ 12","e":"1 \/ 4"},"options_float":{"a":0.4705882353,"b":0.4666666667,"c":0.2,"d":0.4166666667,"e":0.25},"annotated_formula":"subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 5))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/15) + (1\/20)<\/gadget>\n7\/60 = around 0.116667<\/output>\n(7\/60) * 5<\/gadget>\n7\/12 = around 0.583333<\/output>\n1 - (7\/12)<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":3745} +{"problem":"of the 200 stamps in a collection , 90 are foreign and 80 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ?","rationale":"\"20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 60 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 60 ) = 50 the answer is a .\"","correct":"a","options":{"a":"50 ","b":"80 ","c":"100 ","d":"130","e":"150"},"options_float":{"a":50.0,"b":80.0,"c":100.0,"d":130.0,"e":150.0},"annotated_formula":"subtract(200, subtract(add(90, 80), 20))","linear_formula":"add(n1,n2)|subtract(#0,n4)|subtract(n0,#1)|","chain":"90 + 80<\/gadget>\n170<\/output>\n170 - 20<\/gadget>\n150<\/output>\n200 - 150<\/gadget>\n50<\/output>\n50<\/result>","index":3747} +{"problem":"find the volume & curved surface area of a cylinder with diameter of base 14 cm and height 60 cm .","rationale":"volume = pier 2 h = pie x 72 x 60 = 9240 cm 3 curved surface area = 2 pierh = ( 2 x 3.14 x 7 x 60 ) cm 2 = 2640 cm 2 answer : d","correct":"d","options":{"a":"260 cm 2 ","b":"2610 cm 2 ","c":"2600 cm 2 ","d":"2640 cm 2","e":"264 cm 2"},"options_float":{"a":260.0,"b":2610.0,"c":2600.0,"d":2640.0,"e":264.0},"annotated_formula":"multiply(circumface(divide(14, const_2)), 60)","linear_formula":"divide(n0,const_2)|circumface(#0)|multiply(n1,#1)","chain":"14 \/ 2<\/gadget>\n7<\/output>\n2 * pi * 7<\/gadget>\n14*pi = around 43.982297<\/output>\n(14*pi) * 60<\/gadget>\n840*pi = around 2_638.937829<\/output>\n840*pi = around 2_638.937829<\/result>","index":3748} +{"problem":"a car traveled 560 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ?","rationale":"\"let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 \/ h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 \/ c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 \/ c = 560 \/ h = > 336 \/ c = 560 \/ ( c + 6 ) = > 336 c + 336 * 6 = 560 c = > c = 336 * 6 \/ 224 = 9 answer a .\"","correct":"a","options":{"a":"9 ","b":"16 ","c":"21 ","d":"22","e":"27"},"options_float":{"a":9.0,"b":16.0,"c":21.0,"d":22.0,"e":27.0},"annotated_formula":"divide(336, divide(subtract(560, 336), 6))","linear_formula":"subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)|","chain":"560 - 336<\/gadget>\n224<\/output>\n224 \/ 6<\/gadget>\n112\/3 = around 37.333333<\/output>\n336 \/ (112\/3)<\/gadget>\n9<\/output>\n9<\/result>","index":3749} +{"problem":"a hall 36 m long and 15 m broad is to be paved with stones , each measuring 2 dm by 5 dm . the number of stones required is :","rationale":"area of the hall = 3600 * 1500 area of each stone = ( 20 * 50 ) therefore , number of stones = ( 3600 * 1500 \/ 20 * 50 ) = 5400 answer : c","correct":"c","options":{"a":"180 ","b":"1800 ","c":"5400 ","d":"18000","e":"1.8"},"options_float":{"a":180.0,"b":1800.0,"c":5400.0,"d":18000.0,"e":1.8},"annotated_formula":"divide(multiply(36, 15), divide(multiply(2, 5), const_100))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#0,#2)","chain":"36 * 15<\/gadget>\n540<\/output>\n2 * 5<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n540 \/ (1\/10)<\/gadget>\n5_400<\/output>\n5_400<\/result>","index":3750} +{"problem":"by selling an article at rs . 900 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ?","rationale":"\"sp = 900 profit = 25 % cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 900 * [ 100 \/ 125 ] = 720 loss = 25 % = 25 % of 720 = rs . 180 sp = cp - loss = 720 - 180 = rs . 540 answer : d\"","correct":"d","options":{"a":"s . 600 ","b":"s . 480 ","c":"s . 500 ","d":"s . 540","e":"s . 550"},"options_float":{"a":600.0,"b":480.0,"c":500.0,"d":540.0,"e":550.0},"annotated_formula":"subtract(divide(multiply(900, const_100), add(25, const_100)), divide(multiply(divide(multiply(900, const_100), add(25, const_100)), 25), const_100))","linear_formula":"add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|","chain":"900 * 100<\/gadget>\n90_000<\/output>\n25 + 100<\/gadget>\n125<\/output>\n90_000 \/ 125<\/gadget>\n720<\/output>\n720 * 25<\/gadget>\n18_000<\/output>\n18_000 \/ 100<\/gadget>\n180<\/output>\n720 - 180<\/gadget>\n540<\/output>\n540<\/result>","index":3751} +{"problem":"a train 220 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ?","rationale":"\"speed of train relative to man = ( 60 + 6 ) km \/ hr = 66 km \/ hr [ 66 * 5 \/ 18 ] m \/ sec = [ 55 \/ 3 ] m \/ sec . time taken to pass the man = [ 220 * 3 \/ 55 ] sec = 12 sec answer : option d\"","correct":"d","options":{"a":"5 ","b":"6 ","c":"7 ","d":"12","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":12.0,"e":9.0},"annotated_formula":"multiply(const_3600, divide(divide(220, const_1000), add(60, 6)))","linear_formula":"add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)|","chain":"220 \/ 1_000<\/gadget>\n11\/50 = around 0.22<\/output>\n60 + 6<\/gadget>\n66<\/output>\n(11\/50) \/ 66<\/gadget>\n1\/300 = around 0.003333<\/output>\n3_600 * (1\/300)<\/gadget>\n12<\/output>\n12<\/result>","index":3752} +{"problem":"assume that f ( 1 ) = 0 and f ( m + n ) = f ( m ) + f ( n ) + 4 ( 9 mn - 1 ) . for all natural numbers ( integers > 0 ) m and n . what is the value of f ( 17 ) ?","rationale":"explanation : f ( 1 ) = 0 f ( 2 ) = f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) + 4 ( 9 × 1 × 1 – 1 ) = 0 + 0 + 4 × 8 = 32 f ( 4 ) = f ( 2 + 2 ) = f ( 2 ) + f ( 2 ) + 4 ( 9 × 2 × 2 – 1 ) = 32 + 32 + 4 × 35 = 204 f ( 8 ) = f ( 4 + 4 ) = f ( 4 ) + f ( 4 ) + 4 ( 9 × 4 × 4 – 1 ) = 204 + 204 + 4 × 143 = 980 f ( 16 ) = f ( 8 + 8 ) = f ( 8 ) + f ( 8 ) + 4 ( 9 × 8 × 8 – 1 ) = 980 + 980 + 4 × 575 = 4260 f ( 17 ) = f ( 1 + 16 ) = f ( 16 ) + f ( 1 ) + 4 ( 9 × 16 × 1 – 1 ) = 4260 + 0 + 4 × 143 = 4832 answer : a","correct":"a","options":{"a":"4832 ","b":"4855 ","c":"4825 ","d":"4895","e":"4862"},"options_float":{"a":4832.0,"b":4855.0,"c":4825.0,"d":4895.0,"e":4862.0},"annotated_formula":"add(add(add(add(add(add(add(multiply(4, multiply(const_4, const_2)), multiply(4, multiply(const_4, const_2))), multiply(add(multiply(4, multiply(const_4, const_2)), const_3), 4)), add(add(multiply(4, multiply(const_4, const_2)), multiply(4, multiply(const_4, const_2))), multiply(add(multiply(4, multiply(const_4, const_2)), const_3), 4))), multiply(add(const_100, add(multiply(4, const_10), const_3)), 4)), add(add(add(add(multiply(4, multiply(const_4, const_2)), multiply(4, multiply(const_4, const_2))), multiply(add(multiply(4, multiply(const_4, const_2)), const_3), 4)), add(add(multiply(4, multiply(const_4, const_2)), multiply(4, multiply(const_4, const_2))), multiply(add(multiply(4, multiply(const_4, const_2)), const_3), 4))), multiply(add(const_100, add(multiply(4, const_10), const_3)), 4))), multiply(add(multiply(add(const_4, const_1), const_100), subtract(add(multiply(17, const_4), 9), const_2)), const_4)), multiply(add(const_100, add(multiply(4, const_10), const_3)), 4))","linear_formula":"add(const_1,const_4)|multiply(const_2,const_4)|multiply(n2,const_10)|multiply(n6,const_4)|add(#2,const_3)|add(n3,#3)|multiply(n2,#1)|multiply(#0,const_100)|add(#6,#6)|add(#6,const_3)|add(#4,const_100)|subtract(#5,const_2)|add(#7,#11)|multiply(n2,#9)|multiply(n2,#10)|add(#8,#13)|multiply(#12,const_4)|add(#15,#15)|add(#17,#14)|add(#18,#18)|add(#19,#16)|add(#20,#14)","chain":"4 * 2<\/gadget>\n8<\/output>\n4 * 8<\/gadget>\n32<\/output>\n32 + 32<\/gadget>\n64<\/output>\n32 + 3<\/gadget>\n35<\/output>\n35 * 4<\/gadget>\n140<\/output>\n64 + 140<\/gadget>\n204<\/output>\n204 + 204<\/gadget>\n408<\/output>\n4 * 10<\/gadget>\n40<\/output>\n40 + 3<\/gadget>\n43<\/output>\n100 + 43<\/gadget>\n143<\/output>\n143 * 4<\/gadget>\n572<\/output>\n408 + 572<\/gadget>\n980<\/output>\n980 + 980<\/gadget>\n1_960<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n17 * 4<\/gadget>\n68<\/output>\n68 + 9<\/gadget>\n77<\/output>\n77 - 2<\/gadget>\n75<\/output>\n500 + 75<\/gadget>\n575<\/output>\n575 * 4<\/gadget>\n2_300<\/output>\n1_960 + 2_300<\/gadget>\n4_260<\/output>\n4_260 + 572<\/gadget>\n4_832<\/output>\n4_832<\/result>","index":3753} +{"problem":"allay ' s internet provider offers unlimited traffic which costs $ 0.5 per day charged off at 12 a . m . the service is discontinued when client ' s debt exceeds $ 5 . if today in the morning allay makes a payment of $ 7 , for how many days , including today , will she be connected to the internet without any additional payments , given her initial balance was $ 0 ?","rationale":"day 0 - payment of $ 7 is made in the morning ( after 12 am midnight ) day 1 onwards ( after 12 am midnight ) internet service will begin and will run for 7 * 2 = 14 days i . e . at the end of day 15 balance will be zero . now allay can still use the internet till her debt reaches $ 5 which means total of 5 * 2 = 10 days . now the question states for how many days , including today , will she be connected to the internet without any additional paymentsi . e . we will have to include day 0 also in our final answer . therefore total no . of days = 1 + 14 + 10 = 25 = c","correct":"c","options":{"a":"15 ","b":"24 ","c":"25 ","d":"26","e":"13"},"options_float":{"a":15.0,"b":24.0,"c":25.0,"d":26.0,"e":13.0},"annotated_formula":"add(divide(add(5, 7), 0.5), const_1)","linear_formula":"add(n2,n3)|divide(#0,n0)|add(#1,const_1)","chain":"5 + 7<\/gadget>\n12<\/output>\n12 \/ 0.5<\/gadget>\n24<\/output>\n24 + 1<\/gadget>\n25<\/output>\n25<\/result>","index":3754} +{"problem":"if ( 2 ^ 16 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ?","rationale":"given 2 ^ 16 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 16 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans d on comparing the power of 2 = > m = 16","correct":"d","options":{"a":"7 ","b":"8 ","c":"15 ","d":"16","e":"23"},"options_float":{"a":7.0,"b":8.0,"c":15.0,"d":16.0,"e":23.0},"annotated_formula":"power(divide(10, 5), const_4)","linear_formula":"divide(n4,n3)|power(#0,const_4)","chain":"10 \/ 5<\/gadget>\n2<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n16<\/result>","index":3757} +{"problem":"90 + 5 * 12 \/ ( 180 \/ 3 ) = ?","rationale":"\"explanation : 90 + 5 * 12 \/ ( 180 \/ 3 ) = 90 + 5 * 12 \/ ( 60 ) = 90 + ( 5 * 12 ) \/ 60 = 90 + 1 = 91 . answer : d\"","correct":"d","options":{"a":"23 ","b":"78 ","c":"27 ","d":"91","e":"81"},"options_float":{"a":23.0,"b":78.0,"c":27.0,"d":91.0,"e":81.0},"annotated_formula":"add(90, divide(multiply(5, 12), divide(180, 3)))","linear_formula":"divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)|","chain":"5 * 12<\/gadget>\n60<\/output>\n180 \/ 3<\/gadget>\n60<\/output>\n60 \/ 60<\/gadget>\n1<\/output>\n90 + 1<\/gadget>\n91<\/output>\n91<\/result>","index":3758} +{"problem":"if 5 % more is gained by selling an article for rs . 350 than by selling it for rs . 340 , the cost of the article is","rationale":"\"explanation : let c . p . be rs . x . then , 5 % of x = 350 - 340 = 10 x \/ 20 = 10 = > x = 200 answer : c\"","correct":"c","options":{"a":"289 ","b":"231 ","c":"200 ","d":"288","e":"111"},"options_float":{"a":289.0,"b":231.0,"c":200.0,"d":288.0,"e":111.0},"annotated_formula":"divide(subtract(350, 340), divide(5, const_100))","linear_formula":"divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)|","chain":"350 - 340<\/gadget>\n10<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n10 \/ (1\/20)<\/gadget>\n200<\/output>\n200<\/result>","index":3759} +{"problem":"a welder received an order to make a 1 million liter cube - shaped tank . if he has only 2 x 3 meter sheets of metal that can be cut , how many metal sheets will be required for this order ? ( 1 cubic meter = 1000 liters )","rationale":"the question say : a welder received an order to make a 1 million liter cube - shaped tank . ( 1 cubic meter = 1,000 liters ) in other words , the tank is going to have a volume of 1000 cubic meters . that would equal 1 million liters . a cube with a volume of 1000 cubic meters must be ( 10 meters ) x ( 10 meters ) x ( 10 meters ) . the question does n ' t specify whether this tank should have a closed top or an open top . ( the real gmat is good about specifying things like that . ) here , we are going to assume a closed top , only because if the top is open , we do n ' t use enough metal - - - we get an answer smaller than any of the answer choices . a closed - top cube , a full cube , has 6 sides , each of which is ( 10 meters ) x ( 10 meters ) . that ' s a total surface area of 600 sq m . the question tells us : he has only 2 x 3 meter sheets of metal that can be cut . each sheet has an area of 6 sq m . how many of these need to fill up 600 sq m ? 600 \/ 6 = 100 . he needs 100 sheets . answer = e","correct":"e","options":{"a":"92 ","b":"90 ","c":"82 ","d":"78","e":"100"},"options_float":{"a":92.0,"b":90.0,"c":82.0,"d":78.0,"e":100.0},"annotated_formula":"divide(multiply(multiply(2, 3), const_100), multiply(2, 3))","linear_formula":"multiply(n1,n2)|multiply(#0,const_100)|divide(#1,#0)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600 \/ 6<\/gadget>\n100<\/output>\n100<\/result>","index":3762} +{"problem":"if 18 persons can build a wall 140 m long in 42 days , the number of days that 30 persons will take to complete a similar wall 100 m long , is .","rationale":"explanation : ( length 140 : 100 ) : ( persons 30 : 18 ) : : 42 : x = > 140 * 30 * x = 100 * 18 * 42 = > or x = 18 answer : a","correct":"a","options":{"a":"18 ","b":"21 ","c":"24 ","d":"28","e":"30"},"options_float":{"a":18.0,"b":21.0,"c":24.0,"d":28.0,"e":30.0},"annotated_formula":"divide(multiply(multiply(18, 100), 42), multiply(140, 30))","linear_formula":"multiply(n0,n4)|multiply(n1,n3)|multiply(n2,#0)|divide(#2,#1)","chain":"18 * 100<\/gadget>\n1_800<\/output>\n1_800 * 42<\/gadget>\n75_600<\/output>\n140 * 30<\/gadget>\n4_200<\/output>\n75_600 \/ 4_200<\/gadget>\n18<\/output>\n18<\/result>","index":3763} +{"problem":"when x is multiplied by 3 , the result is 10 more than the result of subtracting x from 26 . what is the value of x ?","rationale":"\"the equation that can be formed is : 3 x - 10 = 26 - x or , 4 x = 36 or , x = 9 . b answer .\"","correct":"b","options":{"a":"- 4 ","b":"9 ","c":"11 ","d":"13","e":"22"},"options_float":{"a":-4.0,"b":9.0,"c":11.0,"d":13.0,"e":22.0},"annotated_formula":"divide(add(26, 10), add(3, const_1))","linear_formula":"add(n1,n2)|add(const_1,n0)|divide(#0,#1)|","chain":"26 + 10<\/gadget>\n36<\/output>\n3 + 1<\/gadget>\n4<\/output>\n36 \/ 4<\/gadget>\n9<\/output>\n9<\/result>","index":3764} +{"problem":"if the l . c . m of two numbers is 560 and their product is 42000 , find the h . c . f of the numbers .","rationale":"\"h . c . f = ( product of the numbers ) \/ ( their l . c . m ) = 42000 \/ 560 = 75 . answer : c\"","correct":"c","options":{"a":"50 ","b":"30 ","c":"75 ","d":"25","e":"none of these"},"options_float":{"a":50.0,"b":30.0,"c":75.0,"d":25.0,"e":null},"annotated_formula":"divide(42000, 560)","linear_formula":"divide(n1,n0)|","chain":"42_000 \/ 560<\/gadget>\n75<\/output>\n75<\/result>","index":3765} +{"problem":"1,000 ^ 2 + 1,001 ^ 2 + 1,002 ^ 2 + 1,003 ^ 2 + 1,004 ^ 2 =","rationale":"interesting problem . i think key is to notice that all the given answer choices differs in last two digits . therefore , our entire focus should be to figure out how the given terms contribute to last two digits of total . 1000 ^ 2 - > 00 1001 ^ 1 - > 01 . . . 1004 ^ 2 - > 16 total - > * 30 answer d .","correct":"d","options":{"a":"6 , 030,053 ","b":"6 , 030,054 ","c":"6 , 030,055 ","d":"6 , 020,030","e":"6 , 030,057"},"options_float":{"a":6.0,"b":6.0,"c":6.0,"d":6.0,"e":6.0},"annotated_formula":"multiply(2, const_3)","linear_formula":"multiply(n1,const_3)","chain":"2 * 3<\/gadget>\n6<\/output>\n6<\/result>","index":3766} +{"problem":"in a room filled with 6 people , all 6 people have exactly 1 sibling in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ?","rationale":"the 6 people consist of three sets of 2 siblings . the number of ways to choose 2 people is 6 c 2 = 15 . the number of ways to choose 2 from the first group of siblings is 2 c 2 = 1 . the number of ways to choose 2 from the second group of siblings is 2 c 2 = 1 . the number of ways to choose 2 from the third group of siblings is 2 c 2 = 1 . the number of ways to choose 2 people who are siblings is 1 + 1 + 1 = 3 . the number of ways to choose 2 people who are not siblings is 15 - 3 = 12 . p ( the two people are not siblings ) = 12 \/ 15 = 4 \/ 5 the answer is b .","correct":"b","options":{"a":"3 \/ 5 ","b":"4 \/ 5 ","c":"5 \/ 6 ","d":"5 \/ 12","e":"7 \/ 12"},"options_float":{"a":0.6,"b":0.8,"c":0.8333333333,"d":0.4166666667,"e":0.5833333333},"annotated_formula":"divide(multiply(6, subtract(subtract(6, const_1), const_1)), multiply(6, subtract(6, const_1)))","linear_formula":"subtract(n0,const_1)|multiply(n0,#0)|subtract(#0,const_1)|multiply(n0,#2)|divide(#3,#1)","chain":"6 - 1<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n6 * 4<\/gadget>\n24<\/output>\n6 * 5<\/gadget>\n30<\/output>\n24 \/ 30<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":3767} +{"problem":"find the least number of complete years in which a sum of money put out at 50 % compound interest will be more than double of itself ?","rationale":"2 years answer : a","correct":"a","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"floor(add(divide(log(const_2), log(add(const_1, divide(50, const_100)))), const_1))","linear_formula":"divide(n0,const_100)|log(const_2)|add(#0,const_1)|log(#2)|divide(#1,#3)|add(#4,const_1)|floor(#5)","chain":"log(2)<\/gadget>\nlog(2) = around 0.693147<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\nlog(3\/2)<\/gadget>\nlog(3\/2) = around 0.405465<\/output>\nlog(2) \/ log(3\/2)<\/gadget>\nlog(2)\/log(3\/2) = around 1.709511<\/output>\n(log(2)\/log(3\/2)) + 1<\/gadget>\n1 + log(2)\/log(3\/2) = around 2.709511<\/output>\nfloor(1 + log(2)\/log(3\/2))<\/gadget>\n2<\/output>\n2<\/result>","index":3768} +{"problem":"excluding stoppages , the average speed of a bus is 60 km \/ hr and including stoppages , the average speed of the bus is 40 km \/ hr . for how many minutes does the bus stop per hour ?","rationale":"\"in 1 hr , the bus covers 60 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 60 - 40 ) km i . e 20 km at 60 km \/ hr . stoppage time = 20 \/ 60 hrs = 20 min . answer : d\"","correct":"d","options":{"a":"18 min ","b":"17 min ","c":"19 min ","d":"20 min","e":"21 min"},"options_float":{"a":18.0,"b":17.0,"c":19.0,"d":20.0,"e":21.0},"annotated_formula":"subtract(multiply(const_1, const_60), multiply(divide(40, 60), const_60))","linear_formula":"divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)|","chain":"1 * 60<\/gadget>\n60<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 60<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20<\/result>","index":3769} +{"problem":"in an office , 50 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ?","rationale":"( 10 \/ 100 ) workers = 16 = > number of workers = 160 ( 50 \/ 100 ) * workers = x + 16 = > x = 8 answer a","correct":"a","options":{"a":"8 ","b":"64 ","c":"50 ","d":"144","e":"160"},"options_float":{"a":8.0,"b":64.0,"c":50.0,"d":144.0,"e":160.0},"annotated_formula":"divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4))","linear_formula":"divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n16 \/ (1\/10)<\/gadget>\n160<\/output>\n160 * 90<\/gadget>\n14_400<\/output>\n14_400 \/ 100<\/gadget>\n144<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n160 * (1\/2)<\/gadget>\n80<\/output>\n144 - 80<\/gadget>\n64<\/output>\n2 * 4<\/gadget>\n8<\/output>\n64 \/ 8<\/gadget>\n8<\/output>\n8<\/result>","index":3770} +{"problem":"bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne ’ s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many d hours does it currently take anne to clean the house on her own ?","rationale":"\"lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 \/ a + 1 \/ b portion of the work which equals 1 \/ 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 \/ a + 2 \/ b which is equal to 1 \/ 3 ( as the work is completed in d = 3 hours ) solving these 2 equations we can find b as 12 so , d\"","correct":"d","options":{"a":"6 ","b":"7 ","c":"8 ","d":"12","e":"14"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":12.0,"e":14.0},"annotated_formula":"inverse(subtract(divide(const_1, 3), divide(const_1, 4)))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/3) - (1\/4)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":3772} +{"problem":"the average of runs of a cricket player of 20 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 8 ?","rationale":"\"average = total runs \/ no . of innings = 32 so , total = average x no . of innings = 32 * 20 = 640 now increase in avg = 4 runs . so , new avg = 32 + 8 = 40 runs total runs = new avg x new no . of innings = 40 * 21 = 840 runs made in the 11 th inning = 840 - 640 = 200 answer : b\"","correct":"b","options":{"a":"96 ","b":"200 ","c":"128 ","d":"116","e":"122"},"options_float":{"a":96.0,"b":200.0,"c":128.0,"d":116.0,"e":122.0},"annotated_formula":"subtract(multiply(add(20, const_1), add(8, 32)), multiply(20, 32))","linear_formula":"add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"20 + 1<\/gadget>\n21<\/output>\n8 + 32<\/gadget>\n40<\/output>\n21 * 40<\/gadget>\n840<\/output>\n20 * 32<\/gadget>\n640<\/output>\n840 - 640<\/gadget>\n200<\/output>\n200<\/result>","index":3773} +{"problem":"a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 17 , the how old is b ? a . 7 b . 8 c . 9 d . 10 explanation :","rationale":"let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 17 5 x = 15 x = 3 . hence , b ' s age = 2 x = 6 years . answer : option a","correct":"a","options":{"a":"6 ","b":"7 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":6.0,"b":7.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"multiply(divide(subtract(17, const_2), add(const_4, const_1)), const_2)","linear_formula":"add(const_1,const_4)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)","chain":"17 - 2<\/gadget>\n15<\/output>\n4 + 1<\/gadget>\n5<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6<\/result>","index":3775} +{"problem":"two pipes can fill a tank in 12 minutes and 20 minutes respectively . both pipes are opened together and after some time the first pipe is closed and the tank is full in totally 10 minutes . for how many minutes was first pipe open ?","rationale":"second pipe is opened for 10 minutes . so , part of tank filled by the second pipe = 10 \/ 20 = 1 \/ 2 . so , 1 - 1 \/ 2 = 1 \/ 2 tank is filled by first pipe . first pipe can fill 1 \/ 2 of tank in 1 * 12 \/ 2 = 6 minutes . so , the first pipe is opened for 6 minutes . answer : b","correct":"b","options":{"a":"8 minutes ","b":"6 minutes ","c":"7 minutes ","d":"10 minutes","e":"9 minutes"},"options_float":{"a":8.0,"b":6.0,"c":7.0,"d":10.0,"e":9.0},"annotated_formula":"subtract(10, divide(subtract(multiply(add(inverse(12), inverse(20)), 10), const_1), inverse(12)))","linear_formula":"inverse(n0)|inverse(n1)|add(#0,#1)|multiply(n2,#2)|subtract(#3,const_1)|divide(#4,#0)|subtract(n2,#5)","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/12) + (1\/20)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(2\/15) * 10<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) - 1<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) \/ (1\/12)<\/gadget>\n4<\/output>\n10 - 4<\/gadget>\n6<\/output>\n6<\/result>","index":3776} +{"problem":"the average of 50 numbers id 20 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is :","rationale":"\"explanation : total of 50 numbers = ( 50 × 20 ) = 1000 total of 48 numbers = ( 1000 - ( 45 + 55 ) ] = 900 required average = 900 \/ 48 = 18.75 answer : e\"","correct":"e","options":{"a":"22.75 ","b":"33.25 ","c":"22.25 ","d":"11.75","e":"18.75"},"options_float":{"a":22.75,"b":33.25,"c":22.25,"d":11.75,"e":18.75},"annotated_formula":"divide(subtract(multiply(50, 20), add(45, 55)), subtract(50, const_2))","linear_formula":"add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2)|","chain":"50 * 20<\/gadget>\n1_000<\/output>\n45 + 55<\/gadget>\n100<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n50 - 2<\/gadget>\n48<\/output>\n900 \/ 48<\/gadget>\n75\/4 = around 18.75<\/output>\n75\/4 = around 18.75<\/result>","index":3779} +{"problem":"the area of a triangle is with base 10 m and height 10 m ?","rationale":"\"1 \/ 2 * 10 * 10 = 50 m 2 answer : c\"","correct":"c","options":{"a":"11 ","b":"10 ","c":"50 ","d":"122","e":"12"},"options_float":{"a":11.0,"b":10.0,"c":50.0,"d":122.0,"e":12.0},"annotated_formula":"triangle_area(10, 10)","linear_formula":"triangle_area(n0,n1)|","chain":"(10 * 10) \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":3780} +{"problem":"in what time will a railway train 40 m long moving at the rate of 36 kmph pass a telegraph post on its way ?","rationale":"\"t = 40 \/ 36 * 18 \/ 5 = 4 sec answer : b\"","correct":"b","options":{"a":"3 sec ","b":"4 sec ","c":"5 sec ","d":"6 sec","e":"7 sec"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(40, multiply(36, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n4<\/result>","index":3781} +{"problem":"find remainder if ( 67 ^ 67 + 67 ) is divided by 68","rationale":"= > ( 6767 + 1 ) is divisible by ( 67 + 1 ) = > ( 6767 + 1 ) is divisible by 68 = > ( 6767 + 1 ) ÷ 68 gives a remainder of 0 = > [ ( 6767 + 1 ) + 66 ] ÷ 68 gives a remainder of 66 = > ( 6767 + 67 ) ÷ 68 gives a remainder of 66 c","correct":"c","options":{"a":"60 ","b":"62 ","c":"66 ","d":"72","e":"78"},"options_float":{"a":60.0,"b":62.0,"c":66.0,"d":72.0,"e":78.0},"annotated_formula":"subtract(67, const_1)","linear_formula":"subtract(n0,const_1)","chain":"67 - 1<\/gadget>\n66<\/output>\n66<\/result>","index":3782} +{"problem":"last year a worker saved 6 % of her annual salary . this year , she made 20 % more money than last year and she saved 5 % of her salary . the amount she saved this year was what percent of the amount she saved last year ?","rationale":"let x be the worker ' s salary last year . last year , she saved 0.06 x . this year , she saved 0.05 ( 1.2 x ) = 0.06 x 0.06 x \/ 0.06 x = 100 % the answer is b .","correct":"b","options":{"a":"90 % ","b":"100 % ","c":"110 % ","d":"120 %","e":"130 %"},"options_float":{"a":90.0,"b":100.0,"c":110.0,"d":120.0,"e":130.0},"annotated_formula":"multiply(const_100, divide(6, multiply(add(20, const_100), divide(5, const_100))))","linear_formula":"add(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|divide(n0,#2)|multiply(#3,const_100)","chain":"20 + 100<\/gadget>\n120<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n120 * (1\/20)<\/gadget>\n6<\/output>\n6 \/ 6<\/gadget>\n1<\/output>\n100 * 1<\/gadget>\n100<\/output>\n100<\/result>","index":3783} +{"problem":"a walks at 10 kmph and 4 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ?","rationale":"\"suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 4 hours . x \/ 10 - x \/ 20 = 4 x = 80 km answer is c\"","correct":"c","options":{"a":"100 km ","b":"150 km ","c":"80 km ","d":"120 km","e":"200 km"},"options_float":{"a":100.0,"b":150.0,"c":80.0,"d":120.0,"e":200.0},"annotated_formula":"multiply(4, 20)","linear_formula":"multiply(n1,n2)|","chain":"4 * 20<\/gadget>\n80<\/output>\n80<\/result>","index":3784} +{"problem":"a man can swim in still water at 12 km \/ h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?","rationale":"\"m = 12 s = x ds = 12 + x us = 12 + x 12 + x = ( 12 - x ) 2 12 + x = 24 - 2 x 3 x = 12 x = 4 answer : e\"","correct":"e","options":{"a":"1.7 ","b":"2.67 ","c":"1.2 ","d":"1.5","e":"4"},"options_float":{"a":1.7,"b":2.67,"c":1.2,"d":1.5,"e":4.0},"annotated_formula":"divide(12, const_3)","linear_formula":"divide(n0,const_3)|","chain":"12 \/ 3<\/gadget>\n4<\/output>\n4<\/result>","index":3789} +{"problem":"nitin ranks 24 th in a class of 58 students . what is rank from the last ?","rationale":"\"explanation : number students behind the nitin in rank = ( 58 - 24 ) = 34 nitin is 34 nd from the last answer : b ) 34\"","correct":"b","options":{"a":"33 ","b":"34 ","c":"32 ","d":"28","e":"19"},"options_float":{"a":33.0,"b":34.0,"c":32.0,"d":28.0,"e":19.0},"annotated_formula":"subtract(58, 24)","linear_formula":"subtract(n1,n0)|","chain":"58 - 24<\/gadget>\n34<\/output>\n34<\/result>","index":3790} +{"problem":"a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 4 volts , what was the smallest possible recording in volts ?","rationale":"if average of 3 is 4 so sum of 3 should be 12 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 5 and 5 so a has to be 2 a","correct":"a","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"subtract(multiply(3, 4), 10)","linear_formula":"multiply(n2,n3)|subtract(#0,n1)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 - 10<\/gadget>\n2<\/output>\n2<\/result>","index":3791} +{"problem":"5 drainage pipes , each draining water from a pool at the same constant rate , together can drain a certain pool in 12 days . how many additional pipes q , each draining water at the same constant rate , will be needed to drain the pool in 4 days ?","rationale":"this is an inverse proportional problem . . . . . . 5 pipes in 12 days ; so for 4 days , it will be = 12 x 5 \/ 4 = 15 so , q = 15 - 5 = 10 . c","correct":"c","options":{"a":"6 ","b":"9 ","c":"10 ","d":"12","e":"15"},"options_float":{"a":6.0,"b":9.0,"c":10.0,"d":12.0,"e":15.0},"annotated_formula":"subtract(divide(multiply(12, 5), 4), 5)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|subtract(#1,n0)","chain":"12 * 5<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15 - 5<\/gadget>\n10<\/output>\n10<\/result>","index":3792} +{"problem":"set a { 3 , 3,3 , 4,5 , 5,5 } has a standard deviation of 1 . what will the standard deviation be if every number in the set is multiplied by 3 ?","rationale":"\"points to remember - 1 . if oneadd \/ subtractthe same amont from every term in a set , sd does n ' t change . 2 . if onemultiply \/ divideevery term by the same number in a set , sd changes by same number . hence the answer to the above question is d\"","correct":"d","options":{"a":"1 ","b":"2 ","c":"4 ","d":"3","e":"16"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":3.0,"e":16.0},"annotated_formula":"multiply(3, 1)","linear_formula":"multiply(n4,n5)|","chain":"3 * 1<\/gadget>\n3<\/output>\n3<\/result>","index":3793} +{"problem":"a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 45 % . staff are allowed a further 40 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ?","rationale":"effective discount = a + b + ab \/ 100 = - 45 - 40 + ( - 45 ) ( - 40 ) \/ 100 = - 67 sale price = d * ( 1 - 67 \/ 100 ) sale price = . 33 * d answer ( e )","correct":"e","options":{"a":"0.35 d ","b":"0.76 d ","c":"0.65 d ","d":"0.77 d","e":"0.33 d"},"options_float":{"a":0.35,"b":0.76,"c":0.65,"d":0.77,"e":0.33},"annotated_formula":"subtract(divide(subtract(const_100, 45), const_100), multiply(divide(subtract(const_100, 45), const_100), divide(40, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)","chain":"100 - 45<\/gadget>\n55<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(11\/20) * (2\/5)<\/gadget>\n11\/50 = around 0.22<\/output>\n(11\/20) - (11\/50)<\/gadget>\n33\/100 = around 0.33<\/output>\n33\/100 = around 0.33<\/result>","index":3794} +{"problem":"the average weight of 4 person ' s increases by 8.5 kg when a new person comes in place of one of them weighing 95 kg . what might be the weight of the new person ?","rationale":"\"total weight increased = ( 4 x 8.5 ) kg = 34 kg . weight of new person = ( 95 + 34 ) kg = 129 kg option a\"","correct":"a","options":{"a":"129 kg ","b":"134 kg ","c":"180 kg ","d":"185 kg","e":"190 kg"},"options_float":{"a":129.0,"b":134.0,"c":180.0,"d":185.0,"e":190.0},"annotated_formula":"add(multiply(4, 8.5), 95)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"4 * 8.5<\/gadget>\n34<\/output>\n34 + 95<\/gadget>\n129<\/output>\n129<\/result>","index":3798} +{"problem":"the s . i . on a certain sum of money for 2 years at 8 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ?","rationale":"\"explanation : c . i . = [ 4000 * ( 1 + 10 \/ 100 ) 2 - 4000 ] = ( 4000 * 11 \/ 10 * 11 \/ 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) \/ ( 2 * 8 ) = rs . 2625 answer : e\"","correct":"e","options":{"a":"2197 ","b":"1267 ","c":"1750 ","d":"2267","e":"2625"},"options_float":{"a":2197.0,"b":1267.0,"c":1750.0,"d":2267.0,"e":2625.0},"annotated_formula":"divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(2, divide(8, const_100)))","linear_formula":"divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 2<\/gadget>\n121\/100 = around 1.21<\/output>\n4_000 * (121\/100)<\/gadget>\n4_840<\/output>\n4_840 - 4_000<\/gadget>\n840<\/output>\n840 \/ 2<\/gadget>\n420<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n2 * (2\/25)<\/gadget>\n4\/25 = around 0.16<\/output>\n420 \/ (4\/25)<\/gadget>\n2_625<\/output>\n2_625<\/result>","index":3799} +{"problem":"two trains 120 m and 160 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ?","rationale":"relative speed = 60 + 40 = 100 km \/ hr . = 100 * 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = 120 + 160 = 280 m . required time = 280 * 9 \/ 250 = 252 \/ 25 = 10.08 sec . ' answer : d","correct":"d","options":{"a":"10.9 ","b":"10.7 ","c":"10.3 ","d":"10.08","e":"10.2"},"options_float":{"a":10.9,"b":10.7,"c":10.3,"d":10.08,"e":10.2},"annotated_formula":"divide(add(120, 160), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)","chain":"120 + 160<\/gadget>\n280<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n280 \/ (250\/9)<\/gadget>\n252\/25 = around 10.08<\/output>\n252\/25 = around 10.08<\/result>","index":3800} +{"problem":"in the first 20 overs of a cricket game , the run rate was only 4.6 . what should be the run rate in the remaining 30 overs to reach the target of 396 runs ?","rationale":"required run rate = 396 - ( 4.6 x 20 ) \/ 30 = 304 \/ 30 = 10.13 option a","correct":"a","options":{"a":"10.13 ","b":"9.13 ","c":"8.13 ","d":"11.13","e":"13.13"},"options_float":{"a":10.13,"b":9.13,"c":8.13,"d":11.13,"e":13.13},"annotated_formula":"divide(subtract(396, multiply(20, 4.6)), 30)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"20 * 4.6<\/gadget>\n92<\/output>\n396 - 92<\/gadget>\n304<\/output>\n304 \/ 30<\/gadget>\n152\/15 = around 10.133333<\/output>\n152\/15 = around 10.133333<\/result>","index":3802} +{"problem":"a club wants to mix 10 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ?","rationale":"\"let number of pounds of 5 $ candy to be used be w 6 = ( 10 * 8 + 5 * w ) \/ ( 10 + w ) = > 60 + 6 w = 80 + 5 w = > w = 20 answer a\"","correct":"a","options":{"a":"20 ","b":"30 ","c":"40 ","d":"50","e":"60"},"options_float":{"a":20.0,"b":30.0,"c":40.0,"d":50.0,"e":60.0},"annotated_formula":"subtract(multiply(10, 8.00), multiply(6.00, 10))","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|subtract(#0,#1)|","chain":"10 * 8<\/gadget>\n80<\/output>\n6 * 10<\/gadget>\n60<\/output>\n80 - 60<\/gadget>\n20<\/output>\n20<\/result>","index":3803} +{"problem":"if the annual increase in the population of a town is 10 % and the present number of people is 15000 , what will the population be in 2 years ?","rationale":"\"the required population is = 15000 ( 1 + 10 \/ 100 ) ^ 2 = 15000 * 11 \/ 10 * 11 \/ 10 = 18150 answer is c\"","correct":"c","options":{"a":"12100 ","b":"15240 ","c":"18150 ","d":"18452","e":"19002"},"options_float":{"a":12100.0,"b":15240.0,"c":18150.0,"d":18452.0,"e":19002.0},"annotated_formula":"multiply(multiply(divide(add(10, const_100), const_100), 15000), divide(add(10, const_100), const_100))","linear_formula":"add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|multiply(#1,#2)|","chain":"10 + 100<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) * 15_000<\/gadget>\n16_500<\/output>\n16_500 * (11\/10)<\/gadget>\n18_150<\/output>\n18_150<\/result>","index":3805} +{"problem":"a plant manager must assign 8 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ?","rationale":"\"whatever : my take selecting team of 2 out of 8 to assign to the shifts = 8 c 2 = 28 ways . now 2 out of 8 means total of 4 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 4 * 3 * 2 * 1 = 24 here alt and alt are the same : so 24 \/ 2 = 12 ways . total ways of selecting = ( selecting 2 out of 8 ) * arranging those teams in shifts = 28 * 12 = 336 ans : b\"","correct":"b","options":{"a":"243 ","b":"336 ","c":"350 ","d":"486","e":"540"},"options_float":{"a":243.0,"b":336.0,"c":350.0,"d":486.0,"e":540.0},"annotated_formula":"multiply(divide(factorial(divide(8, 2)), const_2), divide(factorial(8), multiply(factorial(subtract(8, 2)), factorial(2))))","linear_formula":"divide(n0,n1)|factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#0)|factorial(#3)|divide(#4,const_2)|multiply(#5,#2)|divide(#1,#7)|multiply(#6,#8)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\nfactorial(4)<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n8 - 2<\/gadget>\n6<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n720 * 2<\/gadget>\n1_440<\/output>\n40_320 \/ 1_440<\/gadget>\n28<\/output>\n12 * 28<\/gadget>\n336<\/output>\n336<\/result>","index":3806} +{"problem":"1 = 62 = 363 = 3634 = 3645 = 365 , then 36 = ?","rationale":"1 = 6,2 = 36,3 = 363,4 = 364,5 = 365 , then 36 = ? 36 = 2 check the second eqn . answer : b","correct":"b","options":{"a":"1 ","b":"2 ","c":"345 ","d":"445","e":"235"},"options_float":{"a":1.0,"b":2.0,"c":345.0,"d":445.0,"e":235.0},"annotated_formula":"subtract(const_3, 1)","linear_formula":"subtract(const_3,n0)","chain":"3 - 1<\/gadget>\n2<\/output>\n2<\/result>","index":3807} +{"problem":"find large number from below question the difference of two numbers is 1000 . on dividing the larger number by the smaller , we get 10 as quotient and the 10 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1000 = 10 x + 10 9 x = 990 x = 110 large number = 110 + 1000 = 1110 b\"","correct":"b","options":{"a":"1234 ","b":"1110 ","c":"1456 ","d":"1200","e":"1600"},"options_float":{"a":1234.0,"b":1110.0,"c":1456.0,"d":1200.0,"e":1600.0},"annotated_formula":"multiply(divide(subtract(1000, 10), subtract(10, const_1)), 10)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_000 - 10<\/gadget>\n990<\/output>\n10 - 1<\/gadget>\n9<\/output>\n990 \/ 9<\/gadget>\n110<\/output>\n110 * 10<\/gadget>\n1_100<\/output>\n1_100<\/result>","index":3811} +{"problem":"the average height of 30 girls out of a class of 40 is 160 cm . and that of the remaining girls is 156 cm . the average height of the whole class is :","rationale":"explanation : average height of the whole class = ( 36 × 160 + 10 × 156 \/ 40 ) = 159 cms answer c","correct":"c","options":{"a":"158 cms ","b":"158.5 cms ","c":"159 cms ","d":"159.5 cms","e":"none"},"options_float":{"a":158.0,"b":158.5,"c":159.0,"d":159.5,"e":null},"annotated_formula":"divide(add(multiply(160, 30), multiply(156, const_10)), 40)","linear_formula":"multiply(n0,n2)|multiply(n3,const_10)|add(#0,#1)|divide(#2,n1)","chain":"160 * 30<\/gadget>\n4_800<\/output>\n156 * 10<\/gadget>\n1_560<\/output>\n4_800 + 1_560<\/gadget>\n6_360<\/output>\n6_360 \/ 40<\/gadget>\n159<\/output>\n159<\/result>","index":3812} +{"problem":"a man owns 1 \/ 3 of market reserch beauro buzness , and sells 3 \/ 5 of his shares for 15000 rs , what is the value of buzness ?","rationale":"\"if value of business = x total sell ( 2 x \/ 3 ) ( 3 \/ 4 ) = 15000 - > x = 75000 answer : b\"","correct":"b","options":{"a":"55000 ","b":"75000 ","c":"65000 ","d":"190000","e":"250000"},"options_float":{"a":55000.0,"b":75000.0,"c":65000.0,"d":190000.0,"e":250000.0},"annotated_formula":"divide(15000, multiply(divide(1, 3), divide(3, 5)))","linear_formula":"divide(n0,n1)|divide(n1,n3)|multiply(#0,#1)|divide(n4,#2)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(1\/3) * (3\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n15_000 \/ (1\/5)<\/gadget>\n75_000<\/output>\n75_000<\/result>","index":3814} +{"problem":"jaylen was planning to save all summer for a pair of jordan 12 s . they cost $ 250 . jaylen figured that he could cut grass to help with the cost of the sneakers . how many yards would jaylen have to cut if he charges $ 2.15 per yard ?","rationale":"d","correct":"d","options":{"a":"116 yards ","b":"226 yards ","c":"97 yards ","d":"115.74 yards","e":"1 yard"},"options_float":{"a":116.0,"b":226.0,"c":97.0,"d":115.74,"e":1.0},"annotated_formula":"divide(250, 2.15)","linear_formula":"divide(n1,n2)","chain":"250 \/ 2.15<\/gadget>\n116.27907<\/output>\n116.27907<\/result>","index":3815} +{"problem":"at the end of the month , a certain ocean desalination plant ’ s reservoir contained 30 million gallons of water . this amount is twice the normal level . if this amount represents 75 % of the reservoir ’ s total capacity , how many million gallons short of total capacity is the normal level ?","rationale":"\"the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 30 mg = 75 % of total . . total = 30 \/ . 75 = 40 mg . . normal level = 1 \/ 2 of 40 = 20 mg . . shortage of normal level = 40 - 20 = 20 mg . . c\"","correct":"c","options":{"a":"10 ","b":"15 ","c":"20 ","d":"25","e":"30"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"divide(divide(30, const_2), divide(75, const_100))","linear_formula":"divide(n0,const_2)|divide(n1,const_100)|divide(#0,#1)|","chain":"30 \/ 2<\/gadget>\n15<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n15 \/ (3\/4)<\/gadget>\n20<\/output>\n20<\/result>","index":3817} +{"problem":"matt gets a $ 1000 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 250 , how many sales has matt made ?","rationale":"let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1000 new average = ( xy + 1000 ) \/ ( y + 1 ) = 150 + x i . e . ( xy + 1000 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1000 ) = ( xy + x + 150 y + 150 ) i . e . ( 850 ) = ( x + 150 y ) new commission = 250 = 150 + x i . e . x = 100 i . e . y = 5 new sales = y + 1 = 6 answer : option d","correct":"d","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"divide(subtract(1000, subtract(250, 150)), 150)","linear_formula":"subtract(n2,n1)|subtract(n0,#0)|divide(#1,n1)","chain":"250 - 150<\/gadget>\n100<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n900 \/ 150<\/gadget>\n6<\/output>\n6<\/result>","index":3818} +{"problem":"the present worth of $ 169 due in 2 years at 4 % per annum compound interest is ?","rationale":"\"present worth = 169 \/ ( 1 + 4 \/ 100 ) ^ 2 = 169 * 25 \/ 26 * 25 \/ 26 = $ 156.25 answer is c\"","correct":"c","options":{"a":"$ 145.89 ","b":"$ 138.62 ","c":"$ 156.25 ","d":"$ 160.23","e":"$ 158.95"},"options_float":{"a":145.89,"b":138.62,"c":156.25,"d":160.23,"e":158.95},"annotated_formula":"divide(169, power(add(divide(4, const_100), const_1), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) + 1<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 2<\/gadget>\n676\/625 = around 1.0816<\/output>\n169 \/ (676\/625)<\/gadget>\n625\/4 = around 156.25<\/output>\n625\/4 = around 156.25<\/result>","index":3819} +{"problem":"an urn contains 10 black and 5 white balls . two balls are drawn from the urn one after the other without replacement . what is the probability that both drawn balls are black ?","rationale":"let e and f denote respectively the events that first and second ball drawn are black . we have to find p ( e n f ) or p ( ef ) . now p ( e ) = p ( black ball in first draw ) = 10 \/ 15 also given that the first ball drawn is black , i . e . , event e has occurred , now there are 9 black balls and five white balls left in the urn . therefore , the probability that the second ball drawn is black , given that the ball in the first draw is black , is nothing but the conditional probability of f given that e has occurred . that is p ( f | e ) = 9 \/ 14 by multiplication rule of probability , we have p ( e n f ) = p ( e ) p ( f | e ) = 10 \/ 15 × 9 \/ 14 = 3 \/ 7 b","correct":"b","options":{"a":"1 \/ 5 ","b":"3 \/ 7 ","c":"2 \/ 5 ","d":"3 \/ 5","e":"2 \/ 7"},"options_float":{"a":0.2,"b":0.4285714286,"c":0.4,"d":0.6,"e":0.2857142857},"annotated_formula":"multiply(divide(10, add(10, 5)), divide(subtract(10, const_1), subtract(add(10, 5), const_1)))","linear_formula":"add(n0,n1)|subtract(n0,const_1)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#3)|multiply(#2,#4)","chain":"10 + 5<\/gadget>\n15<\/output>\n10 \/ 15<\/gadget>\n2\/3 = around 0.666667<\/output>\n10 - 1<\/gadget>\n9<\/output>\n15 - 1<\/gadget>\n14<\/output>\n9 \/ 14<\/gadget>\n9\/14 = around 0.642857<\/output>\n(2\/3) * (9\/14)<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":3820} +{"problem":"by selling 11 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 20 % ?","rationale":"90 % - - - 11 120 % - - - ? 90 \/ 120 * 11 = 8.25 answer : a","correct":"a","options":{"a":"8.25 ","b":"9.25 ","c":"7.25 ","d":"10.25","e":"3.25"},"options_float":{"a":8.25,"b":9.25,"c":7.25,"d":10.25,"e":3.25},"annotated_formula":"multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 10)))), 11)","linear_formula":"add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)|","chain":"100 + 20<\/gadget>\n120<\/output>\n100 - 10<\/gadget>\n90<\/output>\n1 \/ 90<\/gadget>\n1\/90 = around 0.011111<\/output>\n120 * (1\/90)<\/gadget>\n4\/3 = around 1.333333<\/output>\n1 \/ (4\/3)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 11<\/gadget>\n33\/4 = around 8.25<\/output>\n33\/4 = around 8.25<\/result>","index":3821} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 4.2 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?","rationale":"\"10 overs - run rate = 4.2 runs scored in first 10 overs = 42 remaining overs 40 total runs to be scored = 282 42 runs already scored 282 - 42 = 240 240 runs to be scored in 40 overs let required runrate be x 40 * x = 240 x = 240 \/ 40 x = 6.0 the required runrate is 6.0 answer : e\"","correct":"e","options":{"a":"6.25 ","b":"6.5 ","c":"6.75 ","d":"7","e":"6"},"options_float":{"a":6.25,"b":6.5,"c":6.75,"d":7.0,"e":6.0},"annotated_formula":"divide(subtract(282, multiply(10, 4.2)), 40)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 4.2<\/gadget>\n42<\/output>\n282 - 42<\/gadget>\n240<\/output>\n240 \/ 40<\/gadget>\n6<\/output>\n6<\/result>","index":3822} +{"problem":"average between two sets of numbers is closer to the set withmore numbers ?","rationale":"if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 × 90 + 2 × 805 = 4305 = 86.3 × 90 + 2 × 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 times . so the average is closer to 90 than to 80 as we have just calculated . b","correct":"b","options":{"a":"60 ","b":"80 ","c":"98 ","d":"89","e":"91"},"options_float":{"a":60.0,"b":80.0,"c":98.0,"d":89.0,"e":91.0},"annotated_formula":"multiply(multiply(const_2, const_4), const_10)","linear_formula":"multiply(const_2,const_4)|multiply(#0,const_10)","chain":"2 * 4<\/gadget>\n8<\/output>\n8 * 10<\/gadget>\n80<\/output>\n80<\/result>","index":3823} +{"problem":"crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 50 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it before the decrease ?","rationale":"\"deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $ 100 . they had this same selling price , $ 100 , before they made the change , and had a profit of 50 % , so the manufacturing must have been $ 50 . answer = ( c ) .\"","correct":"c","options":{"a":"$ 20 ","b":"$ 40 ","c":"$ 50 ","d":"$ 80","e":"$ 100"},"options_float":{"a":20.0,"b":40.0,"c":50.0,"d":80.0,"e":100.0},"annotated_formula":"subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(50, const_100)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n1,#0)|multiply(#2,#1)|subtract(#2,#3)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n50 \/ (1\/2)<\/gadget>\n100<\/output>\n100 * (1\/2)<\/gadget>\n50<\/output>\n100 - 50<\/gadget>\n50<\/output>\n50<\/result>","index":3824} +{"problem":"if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2240 less than the principal . what was the principal ?","rationale":"\"p - 2240 = ( p * 5 * 4 ) \/ 100 p = 2800 answer : d\"","correct":"d","options":{"a":"1500 ","b":"2500 ","c":"2507 ","d":"2800","e":"11500"},"options_float":{"a":1500.0,"b":2500.0,"c":2507.0,"d":2800.0,"e":11500.0},"annotated_formula":"divide(2240, subtract(const_1, divide(multiply(4, 5), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|","chain":"4 * 5<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n2_240 \/ (4\/5)<\/gadget>\n2_800<\/output>\n2_800<\/result>","index":3825} +{"problem":"a jeep takes 9 hours to cover a distance of 480 km . how much should the speed in kmph be maintained to cover the same direction in 3 \/ 2 th of the previous time ?","rationale":"time = 9 distance = 480 3 \/ 2 of 9 hours = 9 * 3 \/ 2 = 13.5 hours required speed = 480 \/ 13.5 = 35 kmph c )","correct":"c","options":{"a":"48 kmph ","b":"52 kmph ","c":"35 kmph ","d":"63 kmph","e":"65 kmph"},"options_float":{"a":48.0,"b":52.0,"c":35.0,"d":63.0,"e":65.0},"annotated_formula":"divide(480, multiply(divide(3, 2), 9))","linear_formula":"divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 9<\/gadget>\n27\/2 = around 13.5<\/output>\n480 \/ (27\/2)<\/gadget>\n320\/9 = around 35.555556<\/output>\n320\/9 = around 35.555556<\/result>","index":3826} +{"problem":"what is the tens digit of 6 ^ 22 ?","rationale":"\"the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . . . 56 is 5 . the tens digit of 6 ^ 7 = . . . 36 is 3 again . etc . . . 22 has the form 5 n + 2 , so the tens digit of 6 ^ 22 is 3 . the answer is b .\"","correct":"b","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"9"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":9.0},"annotated_formula":"floor(divide(reminder(power(6, reminder(22, add(const_4, const_1))), const_100), const_10))","linear_formula":"add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)|","chain":"4 + 1<\/gadget>\n5<\/output>\n22 % 5<\/gadget>\n2<\/output>\n6 ** 2<\/gadget>\n36<\/output>\n36 % 100<\/gadget>\n36<\/output>\n36 \/ 10<\/gadget>\n18\/5 = around 3.6<\/output>\nfloor(18\/5)<\/gadget>\n3<\/output>\n3<\/result>","index":3827} +{"problem":"in one hour , a boat goes 7 km along the steram and 5 km against the stream . the speed of the boat in still waer ( in km \/ hr ) is :","rationale":"\"sol . speed in still water = 1 \/ 2 ( 7 + 5 ) kmph = 6 kmph . answer c\"","correct":"c","options":{"a":"2 ","b":"4 ","c":"6 ","d":"12","e":"15"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":12.0,"e":15.0},"annotated_formula":"divide(add(7, 5), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"7 + 5<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6<\/result>","index":3828} +{"problem":"a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 36 sec . what is the length of the platform ?","rationale":"\"speed = 300 \/ 36 = 25 \/ 3 m \/ sec . let the length of the platform be x meters . then , ( x + 300 ) \/ 39 = 25 \/ 3 = > x = 25 m . answer : e\"","correct":"e","options":{"a":"389 m ","b":"350 m ","c":"289 m ","d":"299 m","e":"25 m"},"options_float":{"a":389.0,"b":350.0,"c":289.0,"d":299.0,"e":25.0},"annotated_formula":"subtract(multiply(speed(300, 36), 39), 300)","linear_formula":"speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|","chain":"300 \/ 36<\/gadget>\n25\/3 = around 8.333333<\/output>\n(25\/3) * 39<\/gadget>\n325<\/output>\n325 - 300<\/gadget>\n25<\/output>\n25<\/result>","index":3829} +{"problem":"a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 15 more than that received by b , find the total amount shared by a , b and c .","rationale":"\"a = 1 \/ 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 \/ 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 15 a = 15 + b ( 1 ) = = = > c = 45 + 3 b - b = 2 b + 45 = = > 2 b - c = - 45 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 15 = 2.5 b - 15 = = > 2.5 b - c = 15 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 60 b = $ 120 a = $ 135 c = 405 - 120 = $ 285 total amount = 135 + 120 + 285 = $ 540 answer : e\"","correct":"e","options":{"a":"$ 250 ","b":"$ 340 ","c":"$ 640 ","d":"$ 440","e":"$ 540"},"options_float":{"a":250.0,"b":340.0,"c":640.0,"d":440.0,"e":540.0},"annotated_formula":"add(add(add(multiply(multiply(const_3, 15), const_2), multiply(const_2, 15)), add(add(multiply(multiply(const_3, 15), const_2), multiply(const_2, 15)), 15)), subtract(multiply(add(add(multiply(multiply(const_3, 15), const_2), multiply(const_2, 15)), 15), const_3), add(multiply(multiply(const_3, 15), const_2), multiply(const_2, 15))))","linear_formula":"multiply(n0,const_3)|multiply(n0,const_2)|multiply(#0,const_2)|add(#2,#1)|add(n0,#3)|add(#3,#4)|multiply(#4,const_3)|subtract(#6,#3)|add(#5,#7)|","chain":"3 * 15<\/gadget>\n45<\/output>\n45 * 2<\/gadget>\n90<\/output>\n2 * 15<\/gadget>\n30<\/output>\n90 + 30<\/gadget>\n120<\/output>\n120 + 15<\/gadget>\n135<\/output>\n120 + 135<\/gadget>\n255<\/output>\n135 * 3<\/gadget>\n405<\/output>\n405 - 120<\/gadget>\n285<\/output>\n255 + 285<\/gadget>\n540<\/output>\n540<\/result>","index":3830} +{"problem":"williams works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 8 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ?","rationale":"first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 8 x total tips for all months = 6 * x + 8 x = 14 x tips for august \/ total tips for all months = 8 x \/ 14 x = 4 \/ 7 answer : e","correct":"e","options":{"a":"3 \/ 7 ","b":"5 \/ 7 ","c":"6 \/ 7 ","d":"1 \/ 7","e":"4 \/ 7"},"options_float":{"a":0.4285714286,"b":0.7142857143,"c":0.8571428571,"d":0.1428571429,"e":0.5714285714},"annotated_formula":"divide(8, add(8, subtract(add(const_3, const_4), const_1)))","linear_formula":"add(const_3,const_4)|subtract(#0,const_1)|add(n0,#1)|divide(n0,#2)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 - 1<\/gadget>\n6<\/output>\n8 + 6<\/gadget>\n14<\/output>\n8 \/ 14<\/gadget>\n4\/7 = around 0.571429<\/output>\n4\/7 = around 0.571429<\/result>","index":3832} +{"problem":"ajay can ride 50 km in 1 hour . in how many hours he can ride 1000 km ?","rationale":"\"1 hour he ride 50 km he ride 1000 km in = 1000 \/ 50 * 1 = 20 hours answer is c\"","correct":"c","options":{"a":"10 hrs ","b":"15 hrs ","c":"20 hrs ","d":"25 hrs","e":"18 hrs"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":25.0,"e":18.0},"annotated_formula":"divide(1000, 50)","linear_formula":"divide(n2,n0)|","chain":"1_000 \/ 50<\/gadget>\n20<\/output>\n20<\/result>","index":3833} +{"problem":"a ' s speed is 20 \/ 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 80 % of the length of the race ?","rationale":"\"first calculate the distance , b has covered with his speed in the time , in which a reached 20 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 20 \/ 13 of b , therefore , b ' s speed is 13 \/ 20 of a distance covered by b = speed x time = ( 13 \/ 20 ) x ( 0.2 ) = 0.13 % ( which means b would have covered 0.13 of the race length during the time in which a has covered 0.2 of the race length . therefore to win , b needs a headstart of ( 1 - 0.13 = ) 0.87 of the race length . so the answer is e\"","correct":"e","options":{"a":"44 % ","b":"48 % ","c":"52 % ","d":"42 %","e":"87 %"},"options_float":{"a":44.0,"b":48.0,"c":52.0,"d":42.0,"e":87.0},"annotated_formula":"multiply(subtract(const_1, multiply(divide(subtract(const_100, 80), const_100), divide(13, 20))), const_100)","linear_formula":"divide(n1,n0)|subtract(const_100,n2)|divide(#1,const_100)|multiply(#2,#0)|subtract(const_1,#3)|multiply(#4,const_100)|","chain":"100 - 80<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n13 \/ 20<\/gadget>\n13\/20 = around 0.65<\/output>\n(1\/5) * (13\/20)<\/gadget>\n13\/100 = around 0.13<\/output>\n1 - (13\/100)<\/gadget>\n87\/100 = around 0.87<\/output>\n(87\/100) * 100<\/gadget>\n87<\/output>\n87<\/result>","index":3837} +{"problem":"marginal cost is the cost of increasing the quantity produced ( or purchased ) by one unit . if the fixed cost for n products is $ 12,000 and the marginal cost is $ 200 , and the total cost is $ 16,000 , what is the value of n ?","rationale":"total cost for n products = fixed cost for n products + n * marginal cost - - > $ 16,000 = $ 12,000 + n * $ 200 - - > n = 20 . answer : a .","correct":"a","options":{"a":"20 ","b":"50 ","c":"60 ","d":"80","e":"100"},"options_float":{"a":20.0,"b":50.0,"c":60.0,"d":80.0,"e":100.0},"annotated_formula":"divide(200, const_10)","linear_formula":"divide(n1,const_10)","chain":"200 \/ 10<\/gadget>\n20<\/output>\n20<\/result>","index":3838} +{"problem":"90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 50 % of the total school population of x students , what is x ?","rationale":"\"90 = x \/ 100 * 50 \/ 100 * x = > x ^ 2 = 9 * 10000 \/ 5 = > x = 134 a\"","correct":"a","options":{"a":"134 ","b":"150 ","c":"225 ","d":"250","e":"500"},"options_float":{"a":134.0,"b":150.0,"c":225.0,"d":250.0,"e":500.0},"annotated_formula":"sqrt(divide(multiply(90, const_100), divide(50, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)|","chain":"90 * 100<\/gadget>\n9_000<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n9_000 \/ (1\/2)<\/gadget>\n18_000<\/output>\n18_000 ** (1\/2)<\/gadget>\n60*sqrt(5) = around 134.164079<\/output>\n60*sqrt(5) = around 134.164079<\/result>","index":3840} +{"problem":"the average salary of all the workers in a workshop is rs . 8000 . the average salary of 8 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is :","rationale":"\"explanation : lot the total number of workers be v then , 8 ooov = ( 12000 * 8 ) + 6000 ( v - 8 ) < = > 2000 v = 48000 < = > v = 24 answer : e ) 24\"","correct":"e","options":{"a":"22 ","b":"21 ","c":"88 ","d":"37","e":"24"},"options_float":{"a":22.0,"b":21.0,"c":88.0,"d":37.0,"e":24.0},"annotated_formula":"add(8, divide(multiply(8, subtract(12000, 8000)), subtract(8000, 6000)))","linear_formula":"subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)|","chain":"12_000 - 8_000<\/gadget>\n4_000<\/output>\n8 * 4_000<\/gadget>\n32_000<\/output>\n8_000 - 6_000<\/gadget>\n2_000<\/output>\n32_000 \/ 2_000<\/gadget>\n16<\/output>\n8 + 16<\/gadget>\n24<\/output>\n24<\/result>","index":3842} +{"problem":"what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 7 ?","rationale":"\"find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 7 seeing the options after dividing an finding the reminder of 2 my answer was a\"","correct":"a","options":{"a":"70,592 ","b":"64,850 ","c":"64,749 ","d":"149,700","e":"156,720"},"options_float":{"a":70592.0,"b":64850.0,"c":64749.0,"d":149700.0,"e":156720.0},"annotated_formula":"multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 7), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)))","linear_formula":"multiply(n0,n0)|subtract(const_1000,n0)|multiply(#0,const_10)|add(#2,#0)|add(n0,#3)|add(#4,#1)|subtract(#1,#4)|divide(#6,n2)|add(#7,const_1)|divide(#8,n1)|multiply(#5,#9)|","chain":"1_000 - 3<\/gadget>\n997<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n90 + 9<\/gadget>\n99<\/output>\n99 + 3<\/gadget>\n102<\/output>\n997 - 102<\/gadget>\n895<\/output>\n895 \/ 7<\/gadget>\n895\/7 = around 127.857143<\/output>\n(895\/7) + 1<\/gadget>\n902\/7 = around 128.857143<\/output>\n(902\/7) \/ 2<\/gadget>\n451\/7 = around 64.428571<\/output>\n997 + 102<\/gadget>\n1_099<\/output>\n(451\/7) * 1_099<\/gadget>\n70_807<\/output>\n70_807<\/result>","index":3843} +{"problem":"a tyre has two punctures . the first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes . if air leaks out at a constant rate , how long does it take both the punctures together to make it flat ?","rationale":"explanation : do not be confused , take this question same as that of work done question ' s . like work done by 1 st puncture in 1 minute and by second in 1 minute . lets solve it : 1 minute work done by both the punctures = ( 19 + 16 ) = ( 518 ) so both punctures will make the type flat in ( 185 ) mins = 335 mins answer : c","correct":"c","options":{"a":"315 min ","b":"325 min ","c":"335 min ","d":"345 min","e":"none of these"},"options_float":{"a":315.0,"b":325.0,"c":335.0,"d":345.0,"e":null},"annotated_formula":"add(multiply(const_100, const_3), multiply(multiply(divide(const_1, subtract(6, const_1)), add(9, 9)), const_10))","linear_formula":"add(n0,n0)|multiply(const_100,const_3)|subtract(n1,const_1)|divide(const_1,#2)|multiply(#0,#3)|multiply(#4,const_10)|add(#1,#5)","chain":"100 * 3<\/gadget>\n300<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n9 + 9<\/gadget>\n18<\/output>\n(1\/5) * 18<\/gadget>\n18\/5 = around 3.6<\/output>\n(18\/5) * 10<\/gadget>\n36<\/output>\n300 + 36<\/gadget>\n336<\/output>\n336<\/result>","index":3844} +{"problem":"if 15 % of a is the same as 40 % of b , then a : b is :","rationale":"\"expl : 15 % of a i = 40 % of b = 15 a \/ 100 = 40 b \/ 100 = 8 \/ 3 = 8 : 3 answer : b\"","correct":"b","options":{"a":"5 : 4 ","b":"8 : 3 ","c":"7 : 16 ","d":"2 : 17","e":"3 : 5"},"options_float":{"a":1.25,"b":2.6666666667,"c":0.4375,"d":0.1176470588,"e":0.6},"annotated_formula":"divide(divide(40, const_100), divide(15, const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(2\/5) \/ (3\/20)<\/gadget>\n8\/3 = around 2.666667<\/output>\n8\/3 = around 2.666667<\/result>","index":3848} +{"problem":"after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6500 . find the actual price of the good .","rationale":"\"let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6500 1 = 6500 \/ 68.4 100 = ( 6500 * 100 ) \/ 68.4 = rs . 9503 . answer : option d\"","correct":"d","options":{"a":"s . 6000 ","b":"s . 9000 ","c":"s . 10800 ","d":"s . 9503","e":"s . 9980"},"options_float":{"a":6000.0,"b":9000.0,"c":10800.0,"d":9503.0,"e":9980.0},"annotated_formula":"divide(multiply(6500, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100)))","linear_formula":"multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)|","chain":"6_500 * 100<\/gadget>\n650_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 * 10<\/gadget>\n800<\/output>\n800 \/ 100<\/gadget>\n8<\/output>\n80 - 8<\/gadget>\n72<\/output>\n72 * 5<\/gadget>\n360<\/output>\n360 \/ 100<\/gadget>\n18\/5 = around 3.6<\/output>\n72 - (18\/5)<\/gadget>\n342\/5 = around 68.4<\/output>\n650_000 \/ (342\/5)<\/gadget>\n1_625_000\/171 = around 9_502.923977<\/output>\n1_625_000\/171 = around 9_502.923977<\/result>","index":3849} +{"problem":"a box contains 10 tablets of medicine a and 10 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted .","rationale":"\"the worst case scenario will be if we remove all 15 tablets of medicine b first . the next 2 tablets we remove have to be of medicine a , so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 10 + 2 = 12 tablets . answer : a .\"","correct":"a","options":{"a":"12 ","b":"15 ","c":"17 ","d":"19","e":"21"},"options_float":{"a":12.0,"b":15.0,"c":17.0,"d":19.0,"e":21.0},"annotated_formula":"add(10, const_2)","linear_formula":"add(n1,const_2)|","chain":"10 + 2<\/gadget>\n12<\/output>\n12<\/result>","index":3852} +{"problem":"express a speed of 63 kmph in meters per second ?","rationale":"\"63 * 5 \/ 18 = 17.5 mps answer : c\"","correct":"c","options":{"a":"10 mps ","b":"14 mps ","c":"17.5 mps ","d":"19.4 mps","e":"10.8 mps"},"options_float":{"a":10.0,"b":14.0,"c":17.5,"d":19.4,"e":10.8},"annotated_formula":"multiply(const_0_2778, 63)","linear_formula":"multiply(n0,const_0_2778)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 63<\/gadget>\n35\/2 = around 17.5<\/output>\n35\/2 = around 17.5<\/result>","index":3853} +{"problem":"what least number must be subtracted from 42739 so that remaining no . is divisible by 15","rationale":"\"explanation : on dividing 42739 by 15 we get the remainder 4 , so 4 should be subtracted option b\"","correct":"b","options":{"a":"3 ","b":"4 ","c":"6 ","d":"8","e":"9"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":8.0,"e":9.0},"annotated_formula":"subtract(42739, multiply(floor(divide(42739, 15)), 15))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"42_739 \/ 15<\/gadget>\n42_739\/15 = around 2_849.266667<\/output>\nfloor(42_739\/15)<\/gadget>\n2_849<\/output>\n2_849 * 15<\/gadget>\n42_735<\/output>\n42_739 - 42_735<\/gadget>\n4<\/output>\n4<\/result>","index":3855} +{"problem":"18 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ?","rationale":"\"there are 18 business exec and in each handshake 2 business execs are involved . hence 18 c 2 = 153 also , each of 18 exec will shake hand with every 7 other chairmen for total of 126 handshake . total = 153 + 126 = 279 ans : c\"","correct":"c","options":{"a":"144 ","b":"131 ","c":"279 ","d":"90","e":"45"},"options_float":{"a":144.0,"b":131.0,"c":279.0,"d":90.0,"e":45.0},"annotated_formula":"add(divide(multiply(18, subtract(18, const_1)), const_2), multiply(18, 7))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)|","chain":"18 - 1<\/gadget>\n17<\/output>\n18 * 17<\/gadget>\n306<\/output>\n306 \/ 2<\/gadget>\n153<\/output>\n18 * 7<\/gadget>\n126<\/output>\n153 + 126<\/gadget>\n279<\/output>\n279<\/result>","index":3856} +{"problem":"for each color copy , print shop x charges $ 1.25 and print shop y charges $ 2.75 . how much greater is the charge for 40 color copies at print shop y than at print shop x ?","rationale":"\"the difference in the two prices is $ 2.75 - $ 1.25 = $ 1.50 for each color copy . each color copy will cost an extra $ 1.50 at print shop y . 40 * $ 1.50 = $ 60 the answer is c .\"","correct":"c","options":{"a":"$ 48 ","b":"$ 54 ","c":"$ 60 ","d":"$ 66","e":"$ 72"},"options_float":{"a":48.0,"b":54.0,"c":60.0,"d":66.0,"e":72.0},"annotated_formula":"subtract(multiply(2.75, 40), multiply(1.25, 40))","linear_formula":"multiply(n1,n2)|multiply(n0,n2)|subtract(#0,#1)|","chain":"2.75 * 40<\/gadget>\n110<\/output>\n1.25 * 40<\/gadget>\n50<\/output>\n110 - 50<\/gadget>\n60<\/output>\n60<\/result>","index":3857} +{"problem":"the probability of two events a and b are 0.20 and 0.40 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________","rationale":"\"we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 20 + . 40 - . 15 = . 45 but the probability of neither a nor b = 1 - . 45 = 0.55 answer : d\"","correct":"d","options":{"a":"0.45 ","b":"0.4 ","c":"0.5 ","d":"0.55","e":"0.6"},"options_float":{"a":0.45,"b":0.4,"c":0.5,"d":0.55,"e":0.6},"annotated_formula":"subtract(const_1, subtract(add(0.20, 0.40), 0.15))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)|","chain":"0.2 + 0.4<\/gadget>\n0.6<\/output>\n0.6 - 0.15<\/gadget>\n0.45<\/output>\n1 - 0.45<\/gadget>\n0.55<\/output>\n0.55<\/result>","index":3858} +{"problem":"of the diplomats who attended a summit conference : 20 spoke japanese , 32 did not speak russian and 20 % of the diplomats spoke neither japanese nor russian . if 10 % of the diplomats spoke both japanese and russian , then how many diplomats attended the conference ?","rationale":"2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer","correct":"d","options":{"a":"70 ","b":"96 ","c":"108 ","d":"120","e":"150"},"options_float":{"a":70.0,"b":96.0,"c":108.0,"d":120.0,"e":150.0},"annotated_formula":"divide(subtract(32, 20), subtract(const_1, divide(add(subtract(const_100, 20), 10), const_100)))","linear_formula":"subtract(n1,n0)|subtract(const_100,n0)|add(n3,#1)|divide(#2,const_100)|subtract(const_1,#3)|divide(#0,#4)","chain":"32 - 20<\/gadget>\n12<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 + 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n12 \/ (1\/10)<\/gadget>\n120<\/output>\n120<\/result>","index":3859} +{"problem":"if n = 2.0423 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * – n ?","rationale":"n * = 2.04 n * - n = 2.04 - 2.0423 - 0.0023 answer : b","correct":"b","options":{"a":"- 0.0053 ","b":"- 0.0023 ","c":"0.0007 ","d":"0.0047","e":"0.0153"},"options_float":{"a":-0.0053,"b":-0.0023,"c":0.0007,"d":0.0047,"e":0.0153},"annotated_formula":"subtract(subtract(2.0423, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0423)","linear_formula":"multiply(const_10,const_2)|add(#0,const_3)|divide(#1,const_100)|divide(#2,const_100)|subtract(n0,#3)|subtract(#4,n0)","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 3<\/gadget>\n23<\/output>\n23 \/ 100<\/gadget>\n23\/100 = around 0.23<\/output>\n(23\/100) \/ 100<\/gadget>\n23\/10_000 = around 0.0023<\/output>\n2.0423 - (23\/10_000)<\/gadget>\n2.04<\/output>\n2.04 - 2.0423<\/gadget>\n-0.0023<\/output>\n-0.0023<\/result>","index":3860} +{"problem":"car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 49 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?","rationale":"\"car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 49 t or 5 t + 6 = 7 t or t = 3 distance travelled by car x from the time car y began traveling until both cars stopped is 35 x 3 = 105 miles answer : - a\"","correct":"a","options":{"a":"105 ","b":"120 ","c":"140 ","d":"147","e":"168"},"options_float":{"a":105.0,"b":120.0,"c":140.0,"d":147.0,"e":168.0},"annotated_formula":"multiply(35, divide(multiply(divide(72, const_60), 35), subtract(49, 35)))","linear_formula":"divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)|","chain":"72 \/ 60<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 35<\/gadget>\n42<\/output>\n49 - 35<\/gadget>\n14<\/output>\n42 \/ 14<\/gadget>\n3<\/output>\n35 * 3<\/gadget>\n105<\/output>\n105<\/result>","index":3862} +{"problem":"in a restaurant , the profit is 160 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ?","rationale":"\"explanation : let c . p . = rs . 100 . then , profit = rs . 160 , s . p . = rs . 260 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 260 . profit = rs . ( 260 - 112 ) = rs . 148 . required percentage = ( 148 \/ 260 * 100 ) % = 57 % appox answer : b\"","correct":"b","options":{"a":"30 % ","b":"57 % ","c":"90 % ","d":"100 %","e":"none of these"},"options_float":{"a":30.0,"b":57.0,"c":90.0,"d":100.0,"e":null},"annotated_formula":"multiply(const_100, divide(subtract(add(multiply(const_100, divide(160, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(160, const_100)), const_100)))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#1,const_1)|multiply(#0,const_100)|add(#3,const_100)|multiply(#2,const_100)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100)|","chain":"160 \/ 100<\/gadget>\n8\/5 = around 1.6<\/output>\n100 * (8\/5)<\/gadget>\n160<\/output>\n160 + 100<\/gadget>\n260<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n1 + (3\/25)<\/gadget>\n28\/25 = around 1.12<\/output>\n100 * (28\/25)<\/gadget>\n112<\/output>\n260 - 112<\/gadget>\n148<\/output>\n148 \/ 260<\/gadget>\n37\/65 = around 0.569231<\/output>\n100 * (37\/65)<\/gadget>\n740\/13 = around 56.923077<\/output>\n740\/13 = around 56.923077<\/result>","index":3863} +{"problem":"25 * 25 * 25 * 25 * 25 = 5 ^ ?","rationale":"\"solution : 5 ^ 2 * 5 ^ 2 * 5 ^ 2 * 5 ^ 2 * 5 ^ 2 = 5 ^ ( 2 + 2 + 2 + 2 + 2 ) = 5 ^ 10 answer : e\"","correct":"e","options":{"a":"4 ","b":"6 ","c":"5 ","d":"3","e":"10"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":3.0,"e":10.0},"annotated_formula":"max(5, const_10)","linear_formula":"max(n5,const_10)|","chain":"max(5, 10)<\/gadget>\n10<\/output>\n10<\/result>","index":3867} +{"problem":"what quantity of water should taken out to concentrate 21 liters of 40 % acidic liquid to 60 % acidic liquid ?","rationale":"\"required answer is = 21 ( 60 - 40 ) \/ 60 = 7 liters answer is c\"","correct":"c","options":{"a":"5 liters ","b":"10 liters ","c":"7 liters ","d":"8 liters","e":"6 liters"},"options_float":{"a":5.0,"b":10.0,"c":7.0,"d":8.0,"e":6.0},"annotated_formula":"subtract(21, divide(multiply(21, 40), 60))","linear_formula":"multiply(n0,n1)|divide(#0,n2)|subtract(n0,#1)|","chain":"21 * 40<\/gadget>\n840<\/output>\n840 \/ 60<\/gadget>\n14<\/output>\n21 - 14<\/gadget>\n7<\/output>\n7<\/result>","index":3868} +{"problem":"a and b together can do a work in 6 days . if a alone can do it in 10 days . in how many days can b alone do it ?","rationale":"\"1 \/ 6 – 1 \/ 10 = 1 \/ 15 = > 15 answer : b\"","correct":"b","options":{"a":"10 ","b":"15 ","c":"77 ","d":"55","e":"21"},"options_float":{"a":10.0,"b":15.0,"c":77.0,"d":55.0,"e":21.0},"annotated_formula":"add(inverse(subtract(divide(const_1, 6), divide(const_1, 10))), divide(const_2, add(const_2, const_3)))","linear_formula":"add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/6) - (1\/10)<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ (1\/15)<\/gadget>\n15<\/output>\n2 + 3<\/gadget>\n5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n15 + (2\/5)<\/gadget>\n77\/5 = around 15.4<\/output>\n77\/5 = around 15.4<\/result>","index":3869} +{"problem":"sam and nik start from a and b respectively at the same time and travel towards each other at constant speeds along the same route . sam meets nik at point c on the road which is 600 m away from starting point a and 400 m away from point b . the speed of sam is 50 m \/ m ( meter \/ minute ) . how much time did sam take to cover the distance between a and b ?","rationale":"total distance from a to b - a - c = 600 m , b - c = 400 m ; therefore a - b = 1000 m ; vs = 50 m \/ m ; therefore , t = 1000 \/ 50 = 20 minutes . option c","correct":"c","options":{"a":"10 minutes ","b":"15 minutes ","c":"20 minutes ","d":"22 minutes","e":"17 minutes"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":22.0,"e":17.0},"annotated_formula":"divide(add(600, 400), 50)","linear_formula":"add(n0,n1)|divide(#0,n2)","chain":"600 + 400<\/gadget>\n1_000<\/output>\n1_000 \/ 50<\/gadget>\n20<\/output>\n20<\/result>","index":3870} +{"problem":"subtract . simplify your answer and write it as a proper fraction or as a whole or mixed number . 4 \/ 5 - 1 \/ 5 =","rationale":"here ' s how to subtract 1 \/ 5 from 4 \/ 5 : 4 \/ 5 − 1 \/ 5 since our denominators match , we can subtract the numerators . 4 − 1 = 3 so the answer is : 3 \/ 5 answer : c","correct":"c","options":{"a":"1 \/ 5 ","b":"2 \/ 5 ","c":"3 \/ 5 ","d":"4 \/ 5","e":"5 \/ 5"},"options_float":{"a":0.2,"b":0.4,"c":0.6,"d":0.8,"e":1.0},"annotated_formula":"divide(subtract(4, 1), 5)","linear_formula":"subtract(n0,n2)|divide(#0,n1)","chain":"4 - 1<\/gadget>\n3<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":3871} +{"problem":"ratio between two numbers is 3 : 4 and their sum is 420 . find the smaller number ?","rationale":"explanation : 3 x + 4 x = 420 x = 60 = > 3 x = 180 answer : option b","correct":"b","options":{"a":"378 ","b":"180 ","c":"820 ","d":"734","e":"840"},"options_float":{"a":378.0,"b":180.0,"c":820.0,"d":734.0,"e":840.0},"annotated_formula":"multiply(3, divide(420, add(3, 4)))","linear_formula":"add(n0,n1)|divide(n2,#0)|multiply(n0,#1)","chain":"3 + 4<\/gadget>\n7<\/output>\n420 \/ 7<\/gadget>\n60<\/output>\n3 * 60<\/gadget>\n180<\/output>\n180<\/result>","index":3872} +{"problem":"q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 30 , what is the greatest possible value of q minus r ?","rationale":"a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 30 . the greatest multiple of 9 which is less than 30 is 27 . the answer is e .","correct":"e","options":{"a":"23 ","b":"24 ","c":"25 ","d":"26","e":"27"},"options_float":{"a":23.0,"b":24.0,"c":25.0,"d":26.0,"e":27.0},"annotated_formula":"multiply(reminder(30, subtract(const_10, const_1)), subtract(const_10, const_1))","linear_formula":"subtract(const_10,const_1)|reminder(n0,#0)|multiply(#1,#0)","chain":"10 - 1<\/gadget>\n9<\/output>\n30 % 9<\/gadget>\n3<\/output>\n3 * 9<\/gadget>\n27<\/output>\n27<\/result>","index":3873} +{"problem":"in a vehicle ' s repair shed , 3 men and 8 boys can do a piece of work in 15 days ; 4 men and 9 boys can do the same work in 12 days . then , 5 men and 5 boys can do double the amount of this work in :","rationale":"explanation : solution : ( 3 * 15 ) men + ( 8 * 15 ) boys = ( 4 * 12 ) men + ( 9 * 12 ) boys 45 m + 120 b = 48 m + 108 b < = > 3 m = 12 b = > 1 m = 4 b . . ' . 3 m + 8 b = ( 3 * 4 ) b + 8 b = 20 b . 5 m + 5 b = ( 5 * 4 ) b + 5 b = 25 b . let the required number of days be x . now , more boys , less days ( ip ) more work , more days ( dp ) boys 25 : 20 } : : 15 : x work 1 : 2 . ' . ( 25 * 1 * x ) = ( 20 * 2 * 15 ) = > x = 40 * 15 \/ 25 = 24 . answer : c","correct":"c","options":{"a":"20 days ","b":"33 days ","c":"24 days ","d":"28 days","e":"none of these"},"options_float":{"a":20.0,"b":33.0,"c":24.0,"d":28.0,"e":null},"annotated_formula":"floor(multiply(divide(multiply(add(multiply(3, 3), 8), 15), add(15, 5)), const_2))","linear_formula":"add(n2,n6)|multiply(n0,n0)|add(n1,#1)|multiply(n2,#2)|divide(#3,#0)|multiply(#4,const_2)|floor(#5)","chain":"3 * 3<\/gadget>\n9<\/output>\n9 + 8<\/gadget>\n17<\/output>\n17 * 15<\/gadget>\n255<\/output>\n15 + 5<\/gadget>\n20<\/output>\n255 \/ 20<\/gadget>\n51\/4 = around 12.75<\/output>\n(51\/4) * 2<\/gadget>\n51\/2 = around 25.5<\/output>\nfloor(51\/2)<\/gadget>\n25<\/output>\n25<\/result>","index":3876} +{"problem":"increasing the original price of an article by 8 percent and then increasing the new price by 8 percent is equivalent to increasing the original price by","rationale":"\"1.08 * 1.08 * x = 1.1664 * x the answer is c .\"","correct":"c","options":{"a":"16.16 % ","b":"16.32 % ","c":"16.64 % ","d":"17.26 %","e":"18.56 %"},"options_float":{"a":16.16,"b":16.32,"c":16.64,"d":17.26,"e":18.56},"annotated_formula":"multiply(const_100, subtract(multiply(add(const_1, divide(8, const_100)), add(const_1, divide(8, const_100))), const_1))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n1 + (2\/25)<\/gadget>\n27\/25 = around 1.08<\/output>\n(27\/25) * (27\/25)<\/gadget>\n729\/625 = around 1.1664<\/output>\n(729\/625) - 1<\/gadget>\n104\/625 = around 0.1664<\/output>\n100 * (104\/625)<\/gadget>\n416\/25 = around 16.64<\/output>\n416\/25 = around 16.64<\/result>","index":3878} +{"problem":"if 6 women can color 180 m long cloth in 3 days , then 5 women can color 200 m long cloth in ?","rationale":"the length of cloth painted by one woman in one day = 180 \/ 6 × 3 = 10 m no . of days required to paint 200 m cloth by 5 women = 200 \/ 5 × 10 = 4 days answer : c","correct":"c","options":{"a":"6 days ","b":"5 days ","c":"4 days ","d":"2 days","e":"3 days"},"options_float":{"a":6.0,"b":5.0,"c":4.0,"d":2.0,"e":3.0},"annotated_formula":"divide(200, multiply(5, divide(180, multiply(6, 3))))","linear_formula":"multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)","chain":"6 * 3<\/gadget>\n18<\/output>\n180 \/ 18<\/gadget>\n10<\/output>\n5 * 10<\/gadget>\n50<\/output>\n200 \/ 50<\/gadget>\n4<\/output>\n4<\/result>","index":3879} +{"problem":"sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 210 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 3 of the days were hot , then what was the price of 1 cup on a hot day ?","rationale":"\"7 regular days - - > sales = 7 * 32 * x = 224 x ; 3 hot days - - > sales = 3 * 32 * ( 1.25 x ) = 120 x ; total sales = 224 x + 120 x = 344 x . total cost = 10 * 32 * 0.75 = 240 . profit = 344 x - 240 = 210 - - > x = 1.308 . 1.25 x = ~ 1.64 . answer : b .\"","correct":"b","options":{"a":"$ 1.50 ","b":"$ 1.64 ","c":"$ 2.25 ","d":"$ 2.50","e":"$ 3.25"},"options_float":{"a":1.5,"b":1.64,"c":2.25,"d":2.5,"e":3.25},"annotated_formula":"multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, 1), const_2)), 210), add(multiply(subtract(multiply(add(const_4, 1), const_2), 3), 32), multiply(multiply(divide(add(const_100, 25), const_100), 3), 32))), divide(add(const_100, 25), const_100))","linear_formula":"add(n5,const_4)|add(n1,const_100)|divide(n2,const_100)|divide(#1,const_100)|multiply(n3,#2)|multiply(#0,const_2)|multiply(#4,#5)|multiply(n4,#3)|subtract(#5,n4)|add(n0,#6)|multiply(n3,#8)|multiply(n3,#7)|add(#10,#11)|divide(#9,#12)|multiply(#13,#3)|","chain":"75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n32 * (3\/4)<\/gadget>\n24<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n24 * 10<\/gadget>\n240<\/output>\n240 + 210<\/gadget>\n450<\/output>\n10 - 3<\/gadget>\n7<\/output>\n7 * 32<\/gadget>\n224<\/output>\n100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 3<\/gadget>\n15\/4 = around 3.75<\/output>\n(15\/4) * 32<\/gadget>\n120<\/output>\n224 + 120<\/gadget>\n344<\/output>\n450 \/ 344<\/gadget>\n225\/172 = around 1.30814<\/output>\n(225\/172) * (5\/4)<\/gadget>\n1_125\/688 = around 1.635174<\/output>\n1_125\/688 = around 1.635174<\/result>","index":3880} +{"problem":"the price of a certain painting increased by 30 % during the first year and decreased by 15 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ?","rationale":"easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 30 = 130 price at the end of year 2 = 130 - 130 * 0.15 = 130 * 0.85 = 110.5 hence required answer = ( 110.5 \/ 100 ) * 100 % = 110.5 % answer is c .","correct":"c","options":{"a":"102 % ","b":"105 % ","c":"110.5 % ","d":"112.5 %","e":"100 %"},"options_float":{"a":102.0,"b":105.0,"c":110.5,"d":112.5,"e":100.0},"annotated_formula":"add(subtract(subtract(30, 15), divide(multiply(15, 30), const_100)), const_100)","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|add(#3,const_100)","chain":"30 - 15<\/gadget>\n15<\/output>\n15 * 30<\/gadget>\n450<\/output>\n450 \/ 100<\/gadget>\n9\/2 = around 4.5<\/output>\n15 - (9\/2)<\/gadget>\n21\/2 = around 10.5<\/output>\n(21\/2) + 100<\/gadget>\n221\/2 = around 110.5<\/output>\n221\/2 = around 110.5<\/result>","index":3883} +{"problem":"a student was asked to find 4 \/ 5 of a number . but the student divided the number by 4 \/ 5 , thus the student got 27 more than the correct answer . find the number .","rationale":"let the number be x . ( 5 \/ 4 ) * x = ( 4 \/ 5 ) * x + 27 25 x = 16 x + 540 9 x = 540 x = 60 the answer is c .","correct":"c","options":{"a":"40 ","b":"50 ","c":"60 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"divide(divide(multiply(multiply(27, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5))","linear_formula":"divide(n0,n1)|multiply(n4,#0)|multiply(#0,#0)|multiply(#0,#1)|subtract(const_1,#2)|divide(#3,#4)|divide(#5,#0)","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n27 * (4\/5)<\/gadget>\n108\/5 = around 21.6<\/output>\n(108\/5) * (4\/5)<\/gadget>\n432\/25 = around 17.28<\/output>\n(4\/5) * (4\/5)<\/gadget>\n16\/25 = around 0.64<\/output>\n1 - (16\/25)<\/gadget>\n9\/25 = around 0.36<\/output>\n(432\/25) \/ (9\/25)<\/gadget>\n48<\/output>\n48 \/ (4\/5)<\/gadget>\n60<\/output>\n60<\/result>","index":3884} +{"problem":"in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 \/ 10 . for a group of 150 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ?","rationale":"\"the probability of survival for each of the first 3 months of life is 1 - 1 \/ 10 = 9 \/ 10 , so of 150 newborn 150 * 9 \/ 10 * 9 \/ 10 * 9 \/ 10 = ~ 110 is expected to survive . answer : a .\"","correct":"a","options":{"a":"110 ","b":"146 ","c":"152 ","d":"162","e":"170"},"options_float":{"a":110.0,"b":146.0,"c":152.0,"d":162.0,"e":170.0},"annotated_formula":"multiply(multiply(multiply(150, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10)))","linear_formula":"divide(n1,n2)|subtract(n1,#0)|multiply(n3,#1)|multiply(#2,#1)|multiply(#3,#1)|","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n150 * (9\/10)<\/gadget>\n135<\/output>\n135 * (9\/10)<\/gadget>\n243\/2 = around 121.5<\/output>\n(243\/2) * (9\/10)<\/gadget>\n2_187\/20 = around 109.35<\/output>\n2_187\/20 = around 109.35<\/result>","index":3888} +{"problem":"the units digit of ( 35 ) ^ ( 7 ) + ( 93 ) ^ ( 45 ) is :","rationale":"\"any power of anything ending in 5 always has a units digit of 5 . so the first term has a units digit of 5 . done . the second term takes a little more work . we can ignore the tens digit , and just treat this base as 3 . here is the units digit patter for the powers of 3 . 3 ^ 45 has a units digit of 3 of course 5 + 3 = 8 , d\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"6 ","d":"8","e":"0"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":8.0,"e":0.0},"annotated_formula":"add(reminder(multiply(reminder(45, const_4), 93), const_10), reminder(35, const_10))","linear_formula":"reminder(n3,const_4)|reminder(n0,const_10)|multiply(n2,#0)|reminder(#2,const_10)|add(#3,#1)|","chain":"45 % 4<\/gadget>\n1<\/output>\n1 * 93<\/gadget>\n93<\/output>\n93 % 10<\/gadget>\n3<\/output>\n35 % 10<\/gadget>\n5<\/output>\n3 + 5<\/gadget>\n8<\/output>\n8<\/result>","index":3889} +{"problem":"uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought less of toyota than honda at the ratio of 7 : 3 . if 70 % of the toyota bought and 30 % of the honda bought were suv ã ¢ â ‚ ¬ â „ ¢ s . how many suv ã ¢ â ‚ ¬ â „ ¢ s did uba capital buy in the aforementioned purchase ?","rationale":"let total no of vehicles bought be 100 , toyota 70 and honda 30 , so total number of suv ' s bought for toyota and honda respectively 70 * 70 \/ 100 = 49 and 30 * 30 \/ 100 = 9 so total 58 suv ' s were bought out of 100 vehicles bought . . so required % is 58 % answer : b","correct":"b","options":{"a":"76 % ","b":"58 % ","c":"68 % ","d":"89 %","e":"49 %"},"options_float":{"a":76.0,"b":58.0,"c":68.0,"d":89.0,"e":49.0},"annotated_formula":"subtract(const_100, add(multiply(7, 3), multiply(3, 7)))","linear_formula":"multiply(n0,n1)|add(#0,#0)|subtract(const_100,#1)","chain":"7 * 3<\/gadget>\n21<\/output>\n3 * 7<\/gadget>\n21<\/output>\n21 + 21<\/gadget>\n42<\/output>\n100 - 42<\/gadget>\n58<\/output>\n58<\/result>","index":3890} +{"problem":"a can do a piece of work in 10 days , b in 15 days . they work for 5 days . the rest of work finished by c in 2 days . if they get rs 1500 for the whole work , the daily wages of b and c are ?","rationale":"explanation : let the total work = 30 units . then capacity of a = 3 units , b = 2 units . now they worked for 5 days . so they must have completed 25 units . rest of the work 5 units done by c in 2 days . so c capacity = 5 \/ 2 = 2.5 units . given that toatal wages are rs . 1500 for 30 units . so for 1 unit of work they get rs . 50 . now b and c per day work = ( 2 + 2.5 ) = 4.5 units . so their daily wages = 4.5 × 50 = rs . 225 answer : a","correct":"a","options":{"a":"225 ","b":"276 ","c":"126 ","d":"187","e":"1267"},"options_float":{"a":225.0,"b":276.0,"c":126.0,"d":187.0,"e":1267.0},"annotated_formula":"add(divide(multiply(1500, multiply(divide(subtract(const_1, multiply(5, add(divide(const_1, 10), divide(const_1, 15)))), 2), 2)), 2), divide(multiply(multiply(divide(const_1, 15), 5), 1500), 5))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#1)|multiply(n2,#2)|multiply(n4,#3)|divide(#5,n2)|subtract(const_1,#4)|divide(#7,n3)|multiply(n3,#8)|multiply(n4,#9)|divide(#10,n3)|add(#11,#6)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/10) + (1\/15)<\/gadget>\n1\/6 = around 0.166667<\/output>\n5 * (1\/6)<\/gadget>\n5\/6 = around 0.833333<\/output>\n1 - (5\/6)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) \/ 2<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 2<\/gadget>\n1\/6 = around 0.166667<\/output>\n1_500 * (1\/6)<\/gadget>\n250<\/output>\n250 \/ 2<\/gadget>\n125<\/output>\n(1\/15) * 5<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 1_500<\/gadget>\n500<\/output>\n500 \/ 5<\/gadget>\n100<\/output>\n125 + 100<\/gadget>\n225<\/output>\n225<\/result>","index":3893} +{"problem":"crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 20 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it before the decrease ?","rationale":"cost + profit = income ( selling price ) profit now = 0.5 x or x - 50 0.5 x = x - 50 50 = 0.5 x x = 100 $ is selling price so , profit before was 0.2 * 100 = 20 $ cost = 100 - 20 = 80 $ answer : d","correct":"d","options":{"a":"$ 20 ","b":"$ 40 ","c":"$ 50 ","d":"$ 80","e":"$ 100"},"options_float":{"a":20.0,"b":40.0,"c":50.0,"d":80.0,"e":100.0},"annotated_formula":"subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(20, const_100)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n1,#0)|multiply(#2,#1)|subtract(#2,#3)","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n50 \/ (1\/2)<\/gadget>\n100<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n100 * (1\/5)<\/gadget>\n20<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80<\/result>","index":3894} +{"problem":"if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 900 responses are needed , what is the minimum number of questionnaires that should be mailed ?","rationale":"\"let x be the minimum number of questionnaires to be mailed . 0.6 x = 900 x = 1500 the answer is e .\"","correct":"e","options":{"a":"1300 ","b":"1350 ","c":"1400 ","d":"1450","e":"1500"},"options_float":{"a":1300.0,"b":1350.0,"c":1400.0,"d":1450.0,"e":1500.0},"annotated_formula":"divide(900, divide(60, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,#0)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n900 \/ (3\/5)<\/gadget>\n1_500<\/output>\n1_500<\/result>","index":3895} +{"problem":"find the value o f 72519 x 9999 = m ?","rationale":"\"72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d\"","correct":"d","options":{"a":"434539873 ","b":"355797990 ","c":"435453490 ","d":"m = 725117481","e":"873430134"},"options_float":{"a":434539873.0,"b":355797990.0,"c":435453490.0,"d":725117481.0,"e":873430134.0},"annotated_formula":"multiply(subtract(9999, const_4), 72519)","linear_formula":"subtract(n1,const_4)|multiply(#0,n0)|","chain":"9_999 - 4<\/gadget>\n9_995<\/output>\n9_995 * 72_519<\/gadget>\n724_827_405<\/output>\n724_827_405<\/result>","index":3897} +{"problem":"( 4 ! ) ^ n is a factor of 12 ! , but ( 4 ! ) ^ ( n + 1 ) is not factor of 12 ! . what is the value of n ?","rationale":"12 ! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 4 ! = 24 question can be rephrased as how many 24 ' s are there in 12 ! 12 * 2 = 24 4 * 6 = 24 3 * 8 = 24 value of n = 3 answer : c","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"subtract(add(add(1, 4), divide(12, 12)), const_3)","linear_formula":"add(n0,n3)|divide(n1,n1)|add(#0,#1)|subtract(#2,const_3)","chain":"1 + 4<\/gadget>\n5<\/output>\n12 \/ 12<\/gadget>\n1<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6 - 3<\/gadget>\n3<\/output>\n3<\/result>","index":3898} +{"problem":"a motorcyclist goes from bombay to pune , a distance of 256 kms at an average of 32 kmph speed . another man starts from bombay by car 2 ½ hours after the first , and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ?","rationale":"\"t = 256 \/ 32 = 8 h t = 8 - 3 = 5 time ratio = 8 : 5 speed ratio = 5 : 8 answer : e\"","correct":"e","options":{"a":"1 : 2 ","b":"1 : 5 ","c":"1 : 4 ","d":"1 : 1","e":"5 : 8"},"options_float":{"a":0.5,"b":0.2,"c":0.25,"d":1.0,"e":0.625},"annotated_formula":"divide(divide(256, divide(256, 32)), divide(256, subtract(divide(256, 32), const_3)))","linear_formula":"divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3)|","chain":"256 \/ 32<\/gadget>\n8<\/output>\n256 \/ 8<\/gadget>\n32<\/output>\n8 - 3<\/gadget>\n5<\/output>\n256 \/ 5<\/gadget>\n256\/5 = around 51.2<\/output>\n32 \/ (256\/5)<\/gadget>\n5\/8 = around 0.625<\/output>\n5\/8 = around 0.625<\/result>","index":3899} +{"problem":"the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years x in which the population would be doubled assuming there is no emigration or immigration is","rationale":"\"suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 \/ 100 ) ) ^ n n = 35 answer is d .\"","correct":"d","options":{"a":"20 ","b":"x = 25 ","c":"x = 30 ","d":"x = 35","e":"40"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100))","linear_formula":"multiply(const_10,const_3)|subtract(n1,n2)|divide(#1,n0)|subtract(const_100,#0)|multiply(#2,const_100)|divide(#3,#4)|","chain":"10 * 3<\/gadget>\n30<\/output>\n100 - 30<\/gadget>\n70<\/output>\n39.4 - 19.4<\/gadget>\n20<\/output>\n20 \/ 1_000<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 100<\/gadget>\n2<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n35<\/result>","index":3900} +{"problem":"when the positive integer y is divided by 9 , the remainder is 5 . what is the remainder when 3 y is divided by 9 ?","rationale":"lets take the number as y when y is divided by 9 the remainder is 5 hence y can be written as y = 9 k + 5 multiplying by 3 will give 3 y = 27 k + 15 we can also write 3 y = 27 k + 9 + 6 now 27 k and 9 are divisible by 9 leaving the remainder as 6 hence e is the answer .","correct":"e","options":{"a":"0 ","b":"1 ","c":"3 ","d":"4","e":"6"},"options_float":{"a":0.0,"b":1.0,"c":3.0,"d":4.0,"e":6.0},"annotated_formula":"reminder(multiply(add(9, 5), 3), 9)","linear_formula":"add(n0,n1)|multiply(n2,#0)|reminder(#1,n0)","chain":"9 + 5<\/gadget>\n14<\/output>\n14 * 3<\/gadget>\n42<\/output>\n42 % 9<\/gadget>\n6<\/output>\n6<\/result>","index":3901} +{"problem":"let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 70 th number in this set ?","rationale":"\"the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 70 th number = 8 * ( 70 - 1 ) + 5 = 557 answer = a\"","correct":"a","options":{"a":"557 ","b":"608 ","c":"613 ","d":"616","e":"621"},"options_float":{"a":557.0,"b":608.0,"c":613.0,"d":616.0,"e":621.0},"annotated_formula":"add(multiply(subtract(70, const_1), 8), 5)","linear_formula":"subtract(n2,const_1)|multiply(n0,#0)|add(n1,#1)|","chain":"70 - 1<\/gadget>\n69<\/output>\n69 * 8<\/gadget>\n552<\/output>\n552 + 5<\/gadget>\n557<\/output>\n557<\/result>","index":3902} +{"problem":"julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 120 and $ 122 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same , what was the amount of julie ' s initial savings ?","rationale":"\"$ 120 for 2 years = $ 60 per year . extra $ 2 yearned with the compound interest is the percent yearned on percent . so , $ 2 is yearned on $ 60 , which means that the interest = 30 % . this on the other hand means that half of the savings = 60 * 30 = $ 1800 . twice of that = $ 3,600 . answer : e .\"","correct":"e","options":{"a":"600 ","b":"720 ","c":"1080 ","d":"2200","e":"3600"},"options_float":{"a":600.0,"b":720.0,"c":1080.0,"d":2200.0,"e":3600.0},"annotated_formula":"divide(120, divide(multiply(const_2, subtract(122, 120)), 120))","linear_formula":"subtract(n1,n0)|multiply(#0,const_2)|divide(#1,n0)|divide(n0,#2)|","chain":"122 - 120<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 \/ 120<\/gadget>\n1\/30 = around 0.033333<\/output>\n120 \/ (1\/30)<\/gadget>\n3_600<\/output>\n3_600<\/result>","index":3904} +{"problem":"a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 29 m * 2 m * 0.75 m ?","rationale":"\"29 * 2 * 0.75 = 20 \/ 100 * 10 \/ 100 * 7.5 \/ 100 * x 29 = 1 \/ 100 * x = > x = 29000 answer : a\"","correct":"a","options":{"a":"29000 ","b":"27908 ","c":"78902 ","d":"25000","e":"27991"},"options_float":{"a":29000.0,"b":27908.0,"c":78902.0,"d":25000.0,"e":27991.0},"annotated_formula":"divide(divide(divide(multiply(multiply(multiply(29, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5)","linear_formula":"multiply(n3,const_100)|multiply(n4,const_100)|multiply(n5,const_100)|multiply(#0,#1)|multiply(#3,#2)|divide(#4,n0)|divide(#5,n1)|divide(#6,n2)|","chain":"29 * 100<\/gadget>\n2_900<\/output>\n2 * 100<\/gadget>\n200<\/output>\n2_900 * 200<\/gadget>\n580_000<\/output>\n0.75 * 100<\/gadget>\n75<\/output>\n580_000 * 75<\/gadget>\n43_500_000<\/output>\n43_500_000 \/ 20<\/gadget>\n2_175_000<\/output>\n2_175_000 \/ 10<\/gadget>\n217_500<\/output>\n217_500 \/ 7.5<\/gadget>\n29_000<\/output>\n29_000<\/result>","index":3906} +{"problem":"on a map , 1 inch represents 28 miles . how many b inches would be necessary to represent a distance of 383.6 miles ?","rationale":"b inches necessary to represent a distance of 383.6 miles = 383.6 \/ 28 = 13.7 answer c","correct":"c","options":{"a":"5.2 ","b":"7.4 ","c":"13.7 ","d":"21.2","e":"28.7"},"options_float":{"a":5.2,"b":7.4,"c":13.7,"d":21.2,"e":28.7},"annotated_formula":"divide(383.6, 28)","linear_formula":"divide(n2,n1)","chain":"383.6 \/ 28<\/gadget>\n13.7<\/output>\n13.7<\/result>","index":3907} +{"problem":"a worker is paid a regular rate of rs . 30 for completing a survey . the worker completes 100 surveys per week . for any survey involving the use of her cellphone , she is paid at a rate of that is 20 % higher than her regular rate . if she completed 50 surveys involving the use of her cellphone , how much did she get that week ?","rationale":"\"amount earned using her cell phone = 50 * 36 = 1800 earned for remaining surveys = 50 * 30 = 1500 total earning = 3300 answer : e\"","correct":"e","options":{"a":"2200 ","b":"1100 ","c":"2500 ","d":"2800","e":"3300"},"options_float":{"a":2200.0,"b":1100.0,"c":2500.0,"d":2800.0,"e":3300.0},"annotated_formula":"add(multiply(30, 100), multiply(50, multiply(30, divide(20, 100))))","linear_formula":"divide(n2,n1)|multiply(n0,n1)|multiply(n0,#0)|multiply(n3,#2)|add(#1,#3)|","chain":"30 * 100<\/gadget>\n3_000<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n30 * (1\/5)<\/gadget>\n6<\/output>\n50 * 6<\/gadget>\n300<\/output>\n3_000 + 300<\/gadget>\n3_300<\/output>\n3_300<\/result>","index":3908} +{"problem":"a man goes from a to b at a speed of 20 kmph and comes back to a at a speed of 30 kmph . find his average speed for the entire journey ?","rationale":"\"distance from a and b be ' d ' average speed = total distance \/ total time average speed = ( 2 d ) \/ [ ( d \/ 20 ) + ( d \/ 30 ) ] = ( 2 d ) \/ [ 5 d \/ 60 ) = > 24 kmph . answer : b\"","correct":"b","options":{"a":"11 ","b":"24 ","c":"55 ","d":"66","e":"71"},"options_float":{"a":11.0,"b":24.0,"c":55.0,"d":66.0,"e":71.0},"annotated_formula":"divide(add(20, 30), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"20 + 30<\/gadget>\n50<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n25<\/result>","index":3909} +{"problem":"find the maximum value of the function f ( x ) = x − 5 if x is a number between - 5 and 13 .","rationale":"solution : since f ( x ) is a linear function whose slope is 1 , a positive number , it is strictly increasing for all x . therefore its maximal value is reached for the largest value of x , x = 13 and f ( x ) = 13 - 5 = 8 . answer c","correct":"c","options":{"a":"6 ","b":"7 ","c":"8 ","d":"9","e":"none"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":null},"annotated_formula":"subtract(13, 5)","linear_formula":"subtract(n2,n0)","chain":"13 - 5<\/gadget>\n8<\/output>\n8<\/result>","index":3910} +{"problem":"a shopkeeper sold 30 articles at the cost price of 35 articles . then find the profit % or lost %","rationale":"\"here 30 articles selling price = 35 articles cost price so the difference = 35 - 30 = 5 % of profit = 5 * 100 \/ 30 = 16.67 % correct option is c\"","correct":"c","options":{"a":"12 % ","b":"15 % ","c":"16.67 % ","d":"18 %","e":"22 %"},"options_float":{"a":12.0,"b":15.0,"c":16.67,"d":18.0,"e":22.0},"annotated_formula":"multiply(divide(subtract(35, 30), 30), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"35 - 30<\/gadget>\n5<\/output>\n5 \/ 30<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 100<\/gadget>\n50\/3 = around 16.666667<\/output>\n50\/3 = around 16.666667<\/result>","index":3911} +{"problem":"a cycle is bought for rs . 900 and sold for rs . 1140 , find the gain percent ?","rationale":"\"900 - - - - 240 100 - - - - ? = > 27 % answer : b\"","correct":"b","options":{"a":"11 ","b":"27 ","c":"99 ","d":"77","e":"18"},"options_float":{"a":11.0,"b":27.0,"c":99.0,"d":77.0,"e":18.0},"annotated_formula":"multiply(divide(subtract(1140, 900), 900), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_140 - 900<\/gadget>\n240<\/output>\n240 \/ 900<\/gadget>\n4\/15 = around 0.266667<\/output>\n(4\/15) * 100<\/gadget>\n80\/3 = around 26.666667<\/output>\n80\/3 = around 26.666667<\/result>","index":3912} +{"problem":"the average of 5 quantities is 12 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ?","rationale":"\"answer : a ( 5 x 12 - 3 x 4 ) \/ 2 = 24\"","correct":"a","options":{"a":"24 ","b":"10 ","c":"8 ","d":"9.5","e":"none of these"},"options_float":{"a":24.0,"b":10.0,"c":8.0,"d":9.5,"e":null},"annotated_formula":"divide(subtract(multiply(5, 12), multiply(3, 4)), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)|","chain":"5 * 12<\/gadget>\n60<\/output>\n3 * 4<\/gadget>\n12<\/output>\n60 - 12<\/gadget>\n48<\/output>\n48 \/ 2<\/gadget>\n24<\/output>\n24<\/result>","index":3913} +{"problem":"a trained covered x km at 30 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km .","rationale":"\"total time taken = x \/ 30 + 2 x \/ 20 hours = 2 x \/ 15 hours average speed = 3 x \/ ( 2 x \/ 15 ) = 22.5 kmph answer : a\"","correct":"a","options":{"a":"22.5 ","b":"99 ","c":"24 ","d":"66","e":"887"},"options_float":{"a":22.5,"b":99.0,"c":24.0,"d":66.0,"e":887.0},"annotated_formula":"divide(multiply(30, 3), add(divide(30, 30), divide(multiply(2, 30), 20)))","linear_formula":"divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)|","chain":"30 * 3<\/gadget>\n90<\/output>\n30 \/ 30<\/gadget>\n1<\/output>\n2 * 30<\/gadget>\n60<\/output>\n60 \/ 20<\/gadget>\n3<\/output>\n1 + 3<\/gadget>\n4<\/output>\n90 \/ 4<\/gadget>\n45\/2 = around 22.5<\/output>\n45\/2 = around 22.5<\/result>","index":3914} +{"problem":"the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 28,000 is made ?","rationale":"\"the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 28,000 is made = 96,000 + 28,000 = rs . 1 , 24,000 answer : d\"","correct":"d","options":{"a":"328897 ","b":"120000 ","c":"877888 ","d":"124000","e":"188871"},"options_float":{"a":328897.0,"b":120000.0,"c":877888.0,"d":124000.0,"e":188871.0},"annotated_formula":"add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100)))","linear_formula":"add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * (4\/5)<\/gadget>\n16\/25 = around 0.64<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n100 ** (1\/2)<\/gadget>\n10<\/output>\n10_000 * 10<\/gadget>\n100_000<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n500 * 100<\/gadget>\n50_000<\/output>\n100_000 + 50_000<\/gadget>\n150_000<\/output>\n(16\/25) * 150_000<\/gadget>\n96_000<\/output>\n20 + 2<\/gadget>\n22<\/output>\n22 * 100<\/gadget>\n2_200<\/output>\n2_200 * 10<\/gadget>\n22_000<\/output>\n96_000 + 22_000<\/gadget>\n118_000<\/output>\n118_000<\/result>","index":3915} +{"problem":"gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 60 and the turban . find the price of the turban .","rationale":"\"let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 60 . thus , 3434 ( 90 + x ) = 60 + x or 270 + 3 x = 240 + 4 x or x = 30 answer : b\"","correct":"b","options":{"a":"27 ","b":"30 ","c":"29 ","d":"10","e":"11"},"options_float":{"a":27.0,"b":30.0,"c":29.0,"d":10.0,"e":11.0},"annotated_formula":"subtract(multiply(divide(subtract(90, 60), subtract(const_12, 9)), const_12), 90)","linear_formula":"subtract(n0,n2)|subtract(const_12,n1)|divide(#0,#1)|multiply(#2,const_12)|subtract(#3,n0)|","chain":"90 - 60<\/gadget>\n30<\/output>\n12 - 9<\/gadget>\n3<\/output>\n30 \/ 3<\/gadget>\n10<\/output>\n10 * 12<\/gadget>\n120<\/output>\n120 - 90<\/gadget>\n30<\/output>\n30<\/result>","index":3917} +{"problem":"find the principle on a certain sum of money at 5 % per annum for 3 1 \/ 5 years if the amount being rs . 1450 ?","rationale":"\"explanation : 1450 = p [ 1 + ( 5 * 16 \/ 5 ) \/ 100 ] p = 1250 answer : option b\"","correct":"b","options":{"a":"rs . 1000 ","b":"rs . 1250 ","c":"rs . 1010 ","d":"rs . 10000","e":"none of these"},"options_float":{"a":1000.0,"b":1250.0,"c":1010.0,"d":10000.0,"e":null},"annotated_formula":"divide(1450, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1))","linear_formula":"multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|","chain":"3 * 5<\/gadget>\n15<\/output>\n15 + 3<\/gadget>\n18<\/output>\n18 \/ 5<\/gadget>\n18\/5 = around 3.6<\/output>\n(18\/5) * 5<\/gadget>\n18<\/output>\n18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n(9\/50) + 1<\/gadget>\n59\/50 = around 1.18<\/output>\n1_450 \/ (59\/50)<\/gadget>\n72_500\/59 = around 1_228.813559<\/output>\n72_500\/59 = around 1_228.813559<\/result>","index":3918} +{"problem":"how many seconds will a 600 metre long train take to cross a man walking with a speed of 3 km \/ hr in the direction of the moving train if the speed of the train is 63 km \/ hr ?","rationale":"\"relative speed of the train = 63 - 3 = 60 kmph = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec t = 600 * 3 \/ 50 = 36 sec answer : b\"","correct":"b","options":{"a":"25 ","b":"36 ","c":"40 ","d":"45","e":"50"},"options_float":{"a":25.0,"b":36.0,"c":40.0,"d":45.0,"e":50.0},"annotated_formula":"divide(600, multiply(const_0_2778, subtract(63, 3)))","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n63 - 3<\/gadget>\n60<\/output>\n(5\/18) * 60<\/gadget>\n50\/3 = around 16.666667<\/output>\n600 \/ (50\/3)<\/gadget>\n36<\/output>\n36<\/result>","index":3919} +{"problem":"a ship 450 m long is running at a speed of 24 km \/ hr . in what time will it pass a bridge 900 m long ?","rationale":"total length = 450 + 900 = 1350 speed = 24 km \/ h = ( 24 * 5 ) \/ 18 = 20 \/ 3 m \/ s time = ( 1350 * 3 ) \/ 20 = 202.5 sec . answer : a","correct":"a","options":{"a":"202.5 sec ","b":"205.5 sec ","c":"200 sec ","d":"1350 sec","e":"900 sec"},"options_float":{"a":202.5,"b":205.5,"c":200.0,"d":1350.0,"e":900.0},"annotated_formula":"divide(add(450, 900), multiply(24, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)","chain":"450 + 900<\/gadget>\n1_350<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n24 * (5\/18)<\/gadget>\n20\/3 = around 6.666667<\/output>\n1_350 \/ (20\/3)<\/gadget>\n405\/2 = around 202.5<\/output>\n405\/2 = around 202.5<\/result>","index":3920} +{"problem":"p has $ 32 more than what q and r together would have had if both b and c had 1 \/ 6 of what p has . how much does p have ?","rationale":"\"p = ( 2 \/ 6 ) * p + 32 ( 4 \/ 6 ) * p = 32 p = 48 the answer is b .\"","correct":"b","options":{"a":"$ 44 ","b":"$ 48 ","c":"$ 52 ","d":"$ 56","e":"$ 60"},"options_float":{"a":44.0,"b":48.0,"c":52.0,"d":56.0,"e":60.0},"annotated_formula":"divide(32, subtract(1, multiply(divide(1, 6), const_2)))","linear_formula":"divide(n1,n2)|multiply(#0,const_2)|subtract(n1,#1)|divide(n0,#2)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 2<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n32 \/ (2\/3)<\/gadget>\n48<\/output>\n48<\/result>","index":3925} +{"problem":"a salt manufacturing company produced a total of 2000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ?","rationale":"\"total production of salt by the company in that year = 2000 + 2100 + 2200 + . . . . + 3100 = 30600 . average monthly production of salt for that year = 30600 \/ 365 â ‰ ˆ 84 answer : c\"","correct":"c","options":{"a":"80 ","b":"85 ","c":"83 ","d":"95","e":"00"},"options_float":{"a":80.0,"b":85.0,"c":83.0,"d":95.0,"e":0.0},"annotated_formula":"divide(divide(multiply(add(multiply(2000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3)))","linear_formula":"add(const_2,const_3)|multiply(n0,const_2)|multiply(n1,const_3)|multiply(const_2,const_3)|subtract(const_12,const_1)|multiply(n1,#4)|multiply(#3,const_10)|add(#1,#5)|add(#2,#6)|add(#8,#0)|multiply(#7,const_12)|divide(#10,const_2)|divide(#11,#9)|","chain":"2_000 * 2<\/gadget>\n4_000<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11 * 100<\/gadget>\n1_100<\/output>\n4_000 + 1_100<\/gadget>\n5_100<\/output>\n5_100 * 12<\/gadget>\n61_200<\/output>\n61_200 \/ 2<\/gadget>\n30_600<\/output>\n3 * 100<\/gadget>\n300<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n300 + 60<\/gadget>\n360<\/output>\n2 + 3<\/gadget>\n5<\/output>\n360 + 5<\/gadget>\n365<\/output>\n30_600 \/ 365<\/gadget>\n6_120\/73 = around 83.835616<\/output>\n6_120\/73 = around 83.835616<\/result>","index":3926} +{"problem":"how many terminating zeroes e does 200 ! have ?","rationale":"you have 40 multiples of 5 , 8 of 25 and 1 of 125 . this will give 49 zeros . c","correct":"c","options":{"a":"40 ","b":"48 ","c":"49 ","d":"55","e":"64"},"options_float":{"a":40.0,"b":48.0,"c":49.0,"d":55.0,"e":64.0},"annotated_formula":"add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1))))","linear_formula":"add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3)|","chain":"4 + 1<\/gadget>\n5<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n5 * 5<\/gadget>\n25<\/output>\n200 \/ 25<\/gadget>\n8<\/output>\n40 + 8<\/gadget>\n48<\/output>\n48<\/result>","index":3927} +{"problem":"the average age of a group of n people is 14 years old . one more person aged 32 joins the group and the new average is 16 years old . what is the value of n ?","rationale":"\"14 n + 32 = 16 ( n + 1 ) 2 n = 16 n = 8 the answer is d .\"","correct":"d","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"divide(subtract(32, 16), subtract(16, 14))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"32 - 16<\/gadget>\n16<\/output>\n16 - 14<\/gadget>\n2<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":3928} +{"problem":"the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 216 . find the principal ?","rationale":"\"p = 216 ( 100 \/ 5 ) ^ 2 = > p = 5400 answer : c\"","correct":"c","options":{"a":"2277 ","b":"2667 ","c":"5400 ","d":"2766","e":"1811"},"options_float":{"a":2277.0,"b":2667.0,"c":5400.0,"d":2766.0,"e":1811.0},"annotated_formula":"divide(216, subtract(power(add(divide(20, const_100), const_1), 2), add(multiply(divide(20, const_100), 2), const_1)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#0)|add(#2,const_1)|power(#1,n0)|subtract(#4,#3)|divide(n2,#5)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 2<\/gadget>\n36\/25 = around 1.44<\/output>\n(1\/5) * 2<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) + 1<\/gadget>\n7\/5 = around 1.4<\/output>\n(36\/25) - (7\/5)<\/gadget>\n1\/25 = around 0.04<\/output>\n216 \/ (1\/25)<\/gadget>\n5_400<\/output>\n5_400<\/result>","index":3930} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 15 sec . what is the length of the train ?","rationale":"\"speed = 60 * 5 \/ 18 = 50 \/ 3 m \/ sec length of the train = speed * time = 50 \/ 3 * 12 = 250 m answer : c\"","correct":"c","options":{"a":"287 ","b":"699 ","c":"250 ","d":"168","e":"150"},"options_float":{"a":287.0,"b":699.0,"c":250.0,"d":168.0,"e":150.0},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 15)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"60 * 1_000<\/gadget>\n60_000<\/output>\n60_000 \/ 3_600<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) * 15<\/gadget>\n250<\/output>\n250<\/result>","index":3931} +{"problem":"rewrite ( 10 ^ 38 ) – 85 as a base 10 integer , what is the sum of the digits in that integer ?","rationale":"we know that ( 10 ^ 38 ) is ending 00 , so ( 10 ^ 38 ) – 85 = 9 . . . . 9915 total number of digits in ( 10 ^ 38 ) – 85 is 38 , or 36 digits of 9 and two digits 1 and 5 . answer choice is 36 * 9 + 6 = 330 answer is c 330","correct":"c","options":{"a":"354 ","b":"357 ","c":"330 ","d":"370","e":"360"},"options_float":{"a":354.0,"b":357.0,"c":330.0,"d":370.0,"e":360.0},"annotated_formula":"add(add(multiply(subtract(38, const_2), subtract(const_10, const_1)), const_1), add(const_2, const_3))","linear_formula":"add(const_2,const_3)|subtract(n1,const_2)|subtract(const_10,const_1)|multiply(#1,#2)|add(#3,const_1)|add(#4,#0)","chain":"38 - 2<\/gadget>\n36<\/output>\n10 - 1<\/gadget>\n9<\/output>\n36 * 9<\/gadget>\n324<\/output>\n324 + 1<\/gadget>\n325<\/output>\n2 + 3<\/gadget>\n5<\/output>\n325 + 5<\/gadget>\n330<\/output>\n330<\/result>","index":3932} +{"problem":"he average of 10 numbers is 23 . if each number is increased by 4 , what will the new average be ?","rationale":"sum of the 10 numbers = 230 if each number is increased by 4 , the total increase = 4 * 10 = 40 the new sum = 230 + 40 = 270 the new average = 270 \/ 10 = 27 . answer : b","correct":"b","options":{"a":"26 ","b":"27 ","c":"88 ","d":"99","e":"12"},"options_float":{"a":26.0,"b":27.0,"c":88.0,"d":99.0,"e":12.0},"annotated_formula":"add(23, 4)","linear_formula":"add(n1,n2)","chain":"23 + 4<\/gadget>\n27<\/output>\n27<\/result>","index":3933} +{"problem":"a man buy a book in rs 50 & sale it rs 70 . what is the rate of profit ? ? ?","rationale":"\"cp = 50 sp = 70 profit = 70 - 50 = 20 % = 20 \/ 50 * 100 = 40 % answer : b\"","correct":"b","options":{"a":"10 % ","b":"40 % ","c":"30 % ","d":"25 %","e":"28 %"},"options_float":{"a":10.0,"b":40.0,"c":30.0,"d":25.0,"e":28.0},"annotated_formula":"multiply(divide(subtract(70, 50), 50), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"70 - 50<\/gadget>\n20<\/output>\n20 \/ 50<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":3934} +{"problem":"a tank is filled by 3 pipes a , b , c in 6 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ?","rationale":"suppose pipe a alone take x hours to fill the tank then pipe b and c will take x \/ 2 and x \/ 4 hours respectively to fill the tank . 1 \/ x + 2 \/ x + 4 \/ x = 1 \/ 6 7 \/ x = 1 \/ 6 x = 42 hours answer is b","correct":"b","options":{"a":"25 hr ","b":"42 hr ","c":"40 hr ","d":"20 hr","e":"50 hr"},"options_float":{"a":25.0,"b":42.0,"c":40.0,"d":20.0,"e":50.0},"annotated_formula":"multiply(add(add(multiply(const_2, const_2), const_2), const_1), 6)","linear_formula":"multiply(const_2,const_2)|add(#0,const_2)|add(#1,const_1)|multiply(n1,#2)","chain":"2 * 2<\/gadget>\n4<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7 * 6<\/gadget>\n42<\/output>\n42<\/result>","index":3935} +{"problem":"first column specifies the no of days prior to departure information . second column specifies the percentage discount offered . 0 - 6 days = = > 0 % 7 - 13 days = = > 10 % 14 - 29 days = = > 25 % 30 days or more = = > 40 % the table above shows the discount structure for advanced purchase of tickets at a particular airline . a passenger bought a ticket at this airline for $ 1050 . the ticket agent informed her that , had she purchased the ticket one day later , she would have paid $ 210 more . how many days before her departure did she purchase her ticket .","rationale":"first i figured out the percentage change from the immediate previous amount for each of the period . thus it comes - 0 - 6 days = = > 0 % 7 - 13 days = = > 10 % ( 90 % of the price ) 14 - 29 days = = > 17 % ( approx ) ( went from 90 % to 75 % ) 30 days or more = = > 20 % ( went from 75 % to 60 % ) if the passenger had to pay the extra amount , she had to pay = ( 210 + 1050 ) = 1260 now 210 is somehow between 10 % and 20 % of 1260 . so 17 % is the choice here . correct ans : 14 days answer : d","correct":"d","options":{"a":"6 ","b":"7 ","c":"13 ","d":"14","e":"29"},"options_float":{"a":6.0,"b":7.0,"c":13.0,"d":14.0,"e":29.0},"annotated_formula":"divide(multiply(7, const_4), const_2)","linear_formula":"multiply(n3,const_4)|divide(#0,const_2)","chain":"7 * 4<\/gadget>\n28<\/output>\n28 \/ 2<\/gadget>\n14<\/output>\n14<\/result>","index":3936} +{"problem":"a no . when divided by 158 gives a remainder 50 , what remainder will be obtainedby dividingthe same no . 16 ?","rationale":"\"158 + 50 = 208 \/ 16 = 13 ( remainder ) d\"","correct":"d","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"divide(add(158, 50), 16)","linear_formula":"add(n0,n1)|divide(#0,n2)|","chain":"158 + 50<\/gadget>\n208<\/output>\n208 \/ 16<\/gadget>\n13<\/output>\n13<\/result>","index":3937} +{"problem":"if a = 2 , what is the value of – ( a ^ 2 + a ^ 3 + a ^ 4 + a ^ 5 ) ?","rationale":"\"if a = - 1 , then putting values in equation = - [ ( 2 ) ^ 2 + ( 2 ) ^ 3 + ( 2 ^ 4 ) + ( 2 ^ 5 ) ] = - [ 4 + 8 + 16 + 32 ] = - 60 answer = b = - 60\"","correct":"b","options":{"a":"- 14 ","b":"- 60 ","c":"0 ","d":"4","e":"14"},"options_float":{"a":-14.0,"b":-60.0,"c":0.0,"d":4.0,"e":14.0},"annotated_formula":"negate(add(add(add(power(2, 2), power(2, 3)), power(2, 4)), power(2, 5)))","linear_formula":"power(n0,n1)|power(n0,n2)|power(n0,n3)|power(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|negate(#6)|","chain":"2 ** 2<\/gadget>\n4<\/output>\n2 ** 3<\/gadget>\n8<\/output>\n4 + 8<\/gadget>\n12<\/output>\n2 ** 4<\/gadget>\n16<\/output>\n12 + 16<\/gadget>\n28<\/output>\n2 ** 5<\/gadget>\n32<\/output>\n28 + 32<\/gadget>\n60<\/output>\n-60<\/gadget>\n-60<\/output>\n-60<\/result>","index":3942} +{"problem":"a clock shows the time as 9 a . m . if the minute hand gains 7 minutes every hour , how many minutes will the clock gain by 6 p . m . ?","rationale":"\"there are 9 hours in between 9 a . m . to 6 p . m . 9 * 7 = 63 minutes . answer : c\"","correct":"c","options":{"a":"30 min ","b":"35 min ","c":"63 min ","d":"50 min","e":"55 min"},"options_float":{"a":30.0,"b":35.0,"c":63.0,"d":50.0,"e":55.0},"annotated_formula":"multiply(add(const_3, 6), 7)","linear_formula":"add(const_3,n2)|multiply(n1,#0)|","chain":"3 + 6<\/gadget>\n9<\/output>\n9 * 7<\/gadget>\n63<\/output>\n63<\/result>","index":3944} +{"problem":"how much interest will $ 10,000 earn in 9 months at an annual rate of 4 % ?","rationale":"soln : - 9 months = 3 \/ 4 of year ; 4 % = 4 \/ 100 = 1 \/ 25 ; $ 10,000 ( principal ) * 1 \/ 25 ( interest rate ) * 3 \/ 4 ( time ) = $ 300 . answer : a","correct":"a","options":{"a":"$ 300 ","b":"$ 350 ","c":"$ 450 ","d":"$ 550","e":"$ 650"},"options_float":{"a":300.0,"b":350.0,"c":450.0,"d":550.0,"e":650.0},"annotated_formula":"multiply(multiply(power(const_100, const_2), divide(4, const_100)), divide(const_3, const_4))","linear_formula":"divide(const_3,const_4)|divide(n2,const_100)|power(const_100,const_2)|multiply(#1,#2)|multiply(#0,#3)","chain":"100 ** 2<\/gadget>\n10_000<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n10_000 * (1\/25)<\/gadget>\n400<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n400 * (3\/4)<\/gadget>\n300<\/output>\n300<\/result>","index":3945} +{"problem":"if the perimeter of a rectangular garden is 700 m , its length when its breadth is 100 m is ?","rationale":"\"2 ( l + 100 ) = 700 = > l = 250 m answer : d\"","correct":"d","options":{"a":"299 m ","b":"777 m ","c":"200 m ","d":"250 m","e":"128 m"},"options_float":{"a":299.0,"b":777.0,"c":200.0,"d":250.0,"e":128.0},"annotated_formula":"subtract(divide(700, const_2), 100)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"700 \/ 2<\/gadget>\n350<\/output>\n350 - 100<\/gadget>\n250<\/output>\n250<\/result>","index":3946} +{"problem":"you have been given a physical balance and 7 weights of 40 , 42 , 36 , 33 , 47 , 52 and 67 kgs . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 169 kgs .","rationale":"\"67 + 52 + 47 = 166 answer : e\"","correct":"e","options":{"a":"163 ","b":"166 ","c":"162 ","d":"164","e":"165"},"options_float":{"a":163.0,"b":166.0,"c":162.0,"d":164.0,"e":165.0},"annotated_formula":"add(add(add(42, 36), 33), 52)","linear_formula":"add(n2,n3)|add(n4,#0)|add(n6,#1)|","chain":"42 + 36<\/gadget>\n78<\/output>\n78 + 33<\/gadget>\n111<\/output>\n111 + 52<\/gadget>\n163<\/output>\n163<\/result>","index":3947} +{"problem":"a shopkeeper loses 15 % , if an article is sold for rs . 102 . what should be the selling price of the article to gain 20 % ?","rationale":"\"given that sp = rs . 102 and loss = 15 % cp = [ 100 ( sp ) ] \/ ( 100 - l % ) = ( 100 * 102 ) \/ 85 = 20 * 6 = rs . 120 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] \/ 100 = ( 120 * 120 ) \/ 100 = rs . 144 answer : b\"","correct":"b","options":{"a":"rs . 146 ","b":"rs . 144 ","c":"rs . 129 ","d":"rs . 123","e":"rs . 128"},"options_float":{"a":146.0,"b":144.0,"c":129.0,"d":123.0,"e":128.0},"annotated_formula":"add(divide(102, subtract(const_1, divide(15, const_100))), multiply(divide(102, subtract(const_1, divide(15, const_100))), divide(20, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)|add(#3,#4)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 - (3\/20)<\/gadget>\n17\/20 = around 0.85<\/output>\n102 \/ (17\/20)<\/gadget>\n120<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n120 * (1\/5)<\/gadget>\n24<\/output>\n120 + 24<\/gadget>\n144<\/output>\n144<\/result>","index":3949} +{"problem":"sum of the squares of 3 no . ' s is 62 and the sum of their products taken two at a time is 131 . find the sum ?","rationale":"\"( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 62 + 2 * 131 a + b + c = √ 324 = 18 d\"","correct":"d","options":{"a":"20 ","b":"24 ","c":"26 ","d":"18","e":"30"},"options_float":{"a":20.0,"b":24.0,"c":26.0,"d":18.0,"e":30.0},"annotated_formula":"sqrt(add(multiply(131, const_2), 62))","linear_formula":"multiply(n2,const_2)|add(n1,#0)|sqrt(#1)|","chain":"131 * 2<\/gadget>\n262<\/output>\n262 + 62<\/gadget>\n324<\/output>\n324 ** (1\/2)<\/gadget>\n18<\/output>\n18<\/result>","index":3950} +{"problem":"some persons can do a piece of work in 12 days . two times the number of these people will do half of that work in ?","rationale":"\"12 \/ ( 2 * 2 ) = 3 days answer : a\"","correct":"a","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"multiply(multiply(12, divide(const_1, const_2)), divide(const_1, const_2))","linear_formula":"divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n12 * (1\/2)<\/gadget>\n6<\/output>\n6 * (1\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":3951} +{"problem":"how many multiples of 4 are there between 12 and 92 , inclusive ?","rationale":"\"the multiples of 4 are from 4 * 3 up to 4 * 23 . 23 - 3 + 1 = 21 . the answer is a .\"","correct":"a","options":{"a":"21 ","b":"22 ","c":"23 ","d":"24","e":"25"},"options_float":{"a":21.0,"b":22.0,"c":23.0,"d":24.0,"e":25.0},"annotated_formula":"add(divide(subtract(92, 12), 4), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"92 - 12<\/gadget>\n80<\/output>\n80 \/ 4<\/gadget>\n20<\/output>\n20 + 1<\/gadget>\n21<\/output>\n21<\/result>","index":3952} +{"problem":"a metered rikshaw charges rs 13.50 for the first 1 km and 2.50 per 1 \/ 3 km there after . what would the charge if this rikshaw is hired for 13 kms ? in approx ? ?","rationale":"for 1 st km : rs . 13.50 ramaining distance : 13 - 1 = 12 km 1 \/ 3 km costs 2.5 rs 1 km costs ( 2.5 * 3 ) remaining 12 km costs ( 2.5 * 3 ) * 12 = 90 rs total cost = 90 + 13.5 = 103.5 rs answer : a","correct":"a","options":{"a":"103.5 rs ","b":"104.5 rs ","c":"105.5 rs ","d":"106.5 rs","e":"107.5 rs"},"options_float":{"a":103.5,"b":104.5,"c":105.5,"d":106.5,"e":107.5},"annotated_formula":"add(13.5, multiply(divide(subtract(13, 1), divide(1, 3)), 2.5))","linear_formula":"divide(n1,n4)|subtract(n5,n1)|divide(#1,#0)|multiply(n2,#2)|add(n0,#3)","chain":"13 - 1<\/gadget>\n12<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n12 \/ (1\/3)<\/gadget>\n36<\/output>\n36 * 2.5<\/gadget>\n90<\/output>\n13.5 + 90<\/gadget>\n103.5<\/output>\n103.5<\/result>","index":3953} +{"problem":"two pipes a and b separately fill in 15 \/ 2 min and 5 min respactively and third pipe c can empty of 14 l \/ m if all the pipes are open when the cistern is full , it is emptied in 1 hr . how many litres does the cistern hold ?","rationale":"let total water in cistern = x liter speed of a = 2 x \/ 15 l \/ m ( speed = liter \/ time ) speed of b = x \/ 5 l \/ m speed of c = 14 l \/ m according to question x + ( x \/ 5 + 2 x \/ 15 ) * 60 = 60 * 14 x + = 60 * 14 x = 40 liter answer : a","correct":"a","options":{"a":"40 ","b":"30 ","c":"50 ","d":"80","e":"54"},"options_float":{"a":40.0,"b":30.0,"c":50.0,"d":80.0,"e":54.0},"annotated_formula":"divide(multiply(14, const_60), add(multiply(add(divide(2, 15), divide(1, 5)), const_60), 1))","linear_formula":"divide(n1,n0)|divide(n4,n2)|multiply(n3,const_60)|add(#0,#1)|multiply(#3,const_60)|add(n4,#4)|divide(#2,#5)","chain":"14 * 60<\/gadget>\n840<\/output>\n2 \/ 15<\/gadget>\n2\/15 = around 0.133333<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(2\/15) + (1\/5)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 60<\/gadget>\n20<\/output>\n20 + 1<\/gadget>\n21<\/output>\n840 \/ 21<\/gadget>\n40<\/output>\n40<\/result>","index":3954} +{"problem":"a fruit seller sells mangoes at the rate of rs . 12 per kg and thereby loses 10 % . at what price per kg , he should have sold them to make a profit of 15 % ?","rationale":"\"solution 90 : 12 = 115 : x x = ( 12 ã — 115 \/ 90 ) = rs . 15.33 hence , s . p per kg = rs . 15.33 answer c\"","correct":"c","options":{"a":"rs . 11.81 ","b":"rs . 12 ","c":"rs . 15.33 ","d":"rs . 12.31","e":"none"},"options_float":{"a":11.81,"b":12.0,"c":15.33,"d":12.31,"e":null},"annotated_formula":"divide(multiply(12, add(const_100, 15)), subtract(const_100, 10))","linear_formula":"add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)|","chain":"100 + 15<\/gadget>\n115<\/output>\n12 * 115<\/gadget>\n1_380<\/output>\n100 - 10<\/gadget>\n90<\/output>\n1_380 \/ 90<\/gadget>\n46\/3 = around 15.333333<\/output>\n46\/3 = around 15.333333<\/result>","index":3957} +{"problem":"two numbers have a h . c . f of 20 and a product of two numbers is 2560 . find the l . c . m of the two numbers ?","rationale":"l . c . m of two numbers is given by ( product of the two numbers ) \/ ( h . c . f of the two numbers ) = 2560 \/ 20 = 128 . answer : d","correct":"d","options":{"a":"140 ","b":"150 ","c":"160 ","d":"128","e":"180"},"options_float":{"a":140.0,"b":150.0,"c":160.0,"d":128.0,"e":180.0},"annotated_formula":"divide(2560, 20)","linear_formula":"divide(n1,n0)","chain":"2_560 \/ 20<\/gadget>\n128<\/output>\n128<\/result>","index":3959} +{"problem":"a cylindrical bucket of height 36 cm and radius 21 cm is filled with sand . the bucket is emptied on the ground and a conical heap of sand is formed , the height of the heap being 12 cm . the radius of the heap at the base is :","rationale":"volume of the bucket = volume of the sand emptied volume of sand = π ( 21 ) 2 × 36 let r be the radius of the conical heap . then , 1 ⁄ 3 π r 2 × 12 = π ( 21 ) 2 × 36 or r 2 = ( 21 ) 2 × 9 or r = 21 × 3 = 63 answer a","correct":"a","options":{"a":"63 cm ","b":"53 cm ","c":"56 cm ","d":"66 cm","e":"none of these"},"options_float":{"a":63.0,"b":53.0,"c":56.0,"d":66.0,"e":null},"annotated_formula":"sqrt(divide(multiply(multiply(21, 21), 36), const_4))","linear_formula":"multiply(n1,n1)|multiply(n0,#0)|divide(#1,const_4)|sqrt(#2)","chain":"21 * 21<\/gadget>\n441<\/output>\n441 * 36<\/gadget>\n15_876<\/output>\n15_876 \/ 4<\/gadget>\n3_969<\/output>\n3_969 ** (1\/2)<\/gadget>\n63<\/output>\n63<\/result>","index":3960} +{"problem":"a man opens a shop . he buys water bottles for 50 cents each , fruit for 25 cents each , and snacks for a dollar each . he wants to start a bundle of selling a bottle a water , 3 snacks and 2 fruits for d dollars . every 5 th bundle sold will be sold for 2 dollars and a complimentary snack at the owner ' s expense . what should he charge at least to not lose any money ?","rationale":"w = 0.50 f = 0.25 s = 1.00 b = w + 2 f + 3 s = 4.00 they pay , he pays 4 , net loss 0 fifth bundle = 2 they pay , he pays 5 , net loss 3 dollars 3 \/ 5 = 60 cents . charging 60 more cents on each bundle will be safe . answer : a","correct":"a","options":{"a":"4.6 ","b":"3.0 ","c":"4.1 ","d":"3.2","e":"2.9"},"options_float":{"a":4.6,"b":3.0,"c":4.1,"d":3.2,"e":2.9},"annotated_formula":"add(add(add(multiply(3, const_1), multiply(2, divide(5, const_10))), divide(5, const_10)), divide(const_1, const_10))","linear_formula":"divide(n4,const_10)|divide(const_1,const_10)|multiply(n2,const_1)|multiply(n3,#0)|add(#2,#3)|add(#4,#0)|add(#5,#1)","chain":"3 * 1<\/gadget>\n3<\/output>\n5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n2 * (1\/2)<\/gadget>\n1<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4 + (1\/2)<\/gadget>\n9\/2 = around 4.5<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(9\/2) + (1\/10)<\/gadget>\n23\/5 = around 4.6<\/output>\n23\/5 = around 4.6<\/result>","index":3961} +{"problem":"a person is traveling at 75 km \/ hr and reached his destiny in 4 hr then find the distance ?","rationale":"\"t = 4 hrs d = t * s = 75 * 4 = 300 km answer is a\"","correct":"a","options":{"a":"300 km ","b":"200 km ","c":"250 km ","d":"400 km","e":"220 km"},"options_float":{"a":300.0,"b":200.0,"c":250.0,"d":400.0,"e":220.0},"annotated_formula":"multiply(75, 4)","linear_formula":"multiply(n0,n1)|","chain":"75 * 4<\/gadget>\n300<\/output>\n300<\/result>","index":3963} +{"problem":"a box contains 11 apples , 10 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ?","rationale":"p ( choosing a red apple 6 times in a row ) = 10 \/ 11 * 9 \/ 10 * 8 \/ 9 * 7 \/ 8 * 6 \/ 7 * 5 \/ 6 = 5 \/ 11 < 0.5 the answer is d .","correct":"d","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"add(multiply(0.5, const_10), const_1)","linear_formula":"multiply(n2,const_10)|add(#0,const_1)","chain":"0.5 * 10<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6<\/result>","index":3965} +{"problem":"if the price of a certain computer increased 30 percent from x dollars to 351 dollars , then 2 x =","rationale":"\"before price increase price = x after 30 % price increase price = x + ( 30 \/ 100 ) * x = 1.3 x = 351 ( given ) i . e . x = 351 \/ 1.3 = $ 270 i . e . 2 x = 2 * 270 = 540 answer : option a\"","correct":"a","options":{"a":"540 ","b":"570 ","c":"619 ","d":"649","e":"700"},"options_float":{"a":540.0,"b":570.0,"c":619.0,"d":649.0,"e":700.0},"annotated_formula":"multiply(divide(351, divide(add(const_100, 30), const_100)), 2)","linear_formula":"add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)|","chain":"100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n351 \/ (13\/10)<\/gadget>\n270<\/output>\n270 * 2<\/gadget>\n540<\/output>\n540<\/result>","index":3967} +{"problem":"in the mundane goblet competition , 6 teams compete in a “ round robin ” format : that is , each team plays every other team exactly once . a team gets 3 points for a win , 1 point for a tie ( a draw ) , and 0 points for a loss . what is the difference between the maximum total points and the minimum total points that can be gained by all teams ( added together ) in the mundane goblet competition ?","rationale":"no oa yet : my method : max points : a b c d e f 15 12 9 6 3 0 = 45 points min points a b c d e f 5 5 5 5 5 5 = 30 45 - 30 = 15 points = = = = > ( a ) ans","correct":"a","options":{"a":"15 ","b":"30 ","c":"45 ","d":"60","e":"75"},"options_float":{"a":15.0,"b":30.0,"c":45.0,"d":60.0,"e":75.0},"annotated_formula":"add(multiply(6, const_2), const_3)","linear_formula":"multiply(n0,const_2)|add(#0,const_3)","chain":"6 * 2<\/gadget>\n12<\/output>\n12 + 3<\/gadget>\n15<\/output>\n15<\/result>","index":3968} +{"problem":"a can do a job in 15 days and b in 20 days . if they work on it together for 4 days , then the fraction of the work that is left is ?","rationale":"\"a ' s 1 day work = 1 \/ 15 b ' s 1 day work = 1 \/ 20 a + b 1 day work = 1 \/ 15 + 1 \/ 20 = 7 \/ 60 a + b 4 days work = 7 \/ 60 * 4 = 7 \/ 15 remaining work = 1 - 7 \/ 15 = 8 \/ 15 answer is b\"","correct":"b","options":{"a":"2 \/ 15 ","b":"8 \/ 15 ","c":"3 \/ 11 ","d":"1 \/ 12","e":"6 \/ 13"},"options_float":{"a":0.1333333333,"b":0.5333333333,"c":0.2727272727,"d":0.0833333333,"e":0.4615384615},"annotated_formula":"subtract(const_1, multiply(4, add(divide(const_1, 15), divide(const_1, 20))))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/15) + (1\/20)<\/gadget>\n7\/60 = around 0.116667<\/output>\n4 * (7\/60)<\/gadget>\n7\/15 = around 0.466667<\/output>\n1 - (7\/15)<\/gadget>\n8\/15 = around 0.533333<\/output>\n8\/15 = around 0.533333<\/result>","index":3970} +{"problem":"diana is painting statues . she has 3 \/ 6 of a gallon of paint remaining . each statue requires 1 \/ 6 gallon of paint . how many statues can she paint ?","rationale":"\"number of statues = all the paint ÷ amount used per statue = 3 \/ 6 ÷ 1 \/ 6 = 3 \/ 6 * 6 \/ 1 = 3 \/ 1 = 3 answer is a .\"","correct":"a","options":{"a":"3 ","b":"20 ","c":"28 ","d":"14","e":"19"},"options_float":{"a":3.0,"b":20.0,"c":28.0,"d":14.0,"e":19.0},"annotated_formula":"divide(divide(3, 6), divide(1, 6))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|","chain":"3 \/ 6<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/2) \/ (1\/6)<\/gadget>\n3<\/output>\n3<\/result>","index":3971} +{"problem":"jack and christina are standing 240 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 5 feet per second and christina walks in a straight line toward jack at a constant speed of 3 feet per second . lindy runs at a constant speed of 9 feet per second from christina to jack , back to christina , back to jack , and so forth . what is the total distance , in feet , that lindy has traveled when the three meet at one place ?","rationale":"the relative speed of jack and christina is 5 + 3 = 8 feet per second . the distance between them is 240 feet , hence they will meet in ( time ) = ( distance ) \/ ( relative speed ) = 240 \/ 8 = 30 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 9 * 30 = 270 feet . answer : d .","correct":"d","options":{"a":"180 ","b":"210 ","c":"240 ","d":"270","e":"300"},"options_float":{"a":180.0,"b":210.0,"c":240.0,"d":270.0,"e":300.0},"annotated_formula":"multiply(divide(240, add(5, 3)), 9)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n3,#1)","chain":"5 + 3<\/gadget>\n8<\/output>\n240 \/ 8<\/gadget>\n30<\/output>\n30 * 9<\/gadget>\n270<\/output>\n270<\/result>","index":3973} +{"problem":"the average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class .","rationale":"\"16 * 50.25 + 8 * 45.15 \/ 24 = 48.55 answer : c\"","correct":"c","options":{"a":"47.55 kg ","b":"48 kg ","c":"48.55 kg ","d":"49.25 kg","e":"59.25 kg"},"options_float":{"a":47.55,"b":48.0,"c":48.55,"d":49.25,"e":59.25},"annotated_formula":"divide(add(multiply(16, 50.25), multiply(8, 45.15)), add(16, 8))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"16 * 50.25<\/gadget>\n804<\/output>\n8 * 45.15<\/gadget>\n361.2<\/output>\n804 + 361.2<\/gadget>\n1_165.2<\/output>\n16 + 8<\/gadget>\n24<\/output>\n1_165.2 \/ 24<\/gadget>\n48.55<\/output>\n48.55<\/result>","index":3976} +{"problem":"how much water should be added to 13 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ?","rationale":"\"let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 \/ 4 ( new soln . = 13 + x ) ) ) 2 = 5 % * ( 13 + x ) - - - - - - - - > x = 27 ans b\"","correct":"b","options":{"a":"25 liters ","b":"27 liters ","c":"30 liters ","d":"32 liters","e":"35 liters"},"options_float":{"a":25.0,"b":27.0,"c":30.0,"d":32.0,"e":35.0},"annotated_formula":"subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 13)","linear_formula":"divide(n2,const_100)|multiply(const_100,const_2)|multiply(n1,#0)|subtract(n1,#2)|divide(#1,#3)|subtract(#4,n0)|","chain":"2 * 100<\/gadget>\n200<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 20<\/gadget>\n15<\/output>\n20 - 15<\/gadget>\n5<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n40 - 13<\/gadget>\n27<\/output>\n27<\/result>","index":3977} +{"problem":"the average age of students of a class is 15.8 years . the average age of boys in the class is 16.7 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is :","rationale":"\"let the ratio be k : 1 . then , k * 16.7 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.7 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 \/ 0.9 = 4 \/ 9 required ratio = 4 \/ 9 : 1 = 4 : 9 . answer : e\"","correct":"e","options":{"a":"2 : 5 ","b":"2 : 3 ","c":"2 : 4 ","d":"2 : 1","e":"4 : 9"},"options_float":{"a":0.4,"b":0.6666666667,"c":0.5,"d":2.0,"e":0.4444444444},"annotated_formula":"divide(subtract(15.8, 15.4), subtract(16.7, 15.8))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|","chain":"15.8 - 15.4<\/gadget>\n0.4<\/output>\n16.7 - 15.8<\/gadget>\n0.9<\/output>\n0.4 \/ 0.9<\/gadget>\n0.444444<\/output>\n0.444444<\/result>","index":3979} +{"problem":"a number when divided by a divisor leaves a remainder of 30 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ?","rationale":"let the number is n , the divisor = d , i will make the two equations - n = xd + 30 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 43 as d is also integer and 43 is a prime number , the d should be 43 to satisfy the above equation . hence answer is ' d '","correct":"d","options":{"a":"12 ","b":"13 ","c":"35 ","d":"43","e":"59"},"options_float":{"a":12.0,"b":13.0,"c":35.0,"d":43.0,"e":59.0},"annotated_formula":"add(30, 11)","linear_formula":"add(n0,n1)","chain":"30 + 11<\/gadget>\n41<\/output>\n41<\/result>","index":3982} +{"problem":"a student got twice as many sums wrong as he got right . if he attempted 36 sums in all , how many did he solve correctly ?","rationale":"\"explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 36 3 x = 36 x = 12 . answer : a\"","correct":"a","options":{"a":"12 ","b":"16 ","c":"18 ","d":"24","e":"26"},"options_float":{"a":12.0,"b":16.0,"c":18.0,"d":24.0,"e":26.0},"annotated_formula":"divide(36, add(const_1, const_2))","linear_formula":"add(const_1,const_2)|divide(n0,#0)|","chain":"1 + 2<\/gadget>\n3<\/output>\n36 \/ 3<\/gadget>\n12<\/output>\n12<\/result>","index":3983} +{"problem":"in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 8 ounces of corn syrup , how many ounces of water does it contain ?","rationale":"\"f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c \/ w = 12 \/ 180 = 3 ounces \/ x ounces x = 8 * 180 \/ 12 = 120 ounces of water e\"","correct":"e","options":{"a":"45 ","b":"50 ","c":"55 ","d":"60","e":"120"},"options_float":{"a":45.0,"b":50.0,"c":55.0,"d":60.0,"e":120.0},"annotated_formula":"multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 8)","linear_formula":"divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 3<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/30) \/ 2<\/gadget>\n1\/60 = around 0.016667<\/output>\n(1\/4) \/ (1\/60)<\/gadget>\n15<\/output>\n15 * 8<\/gadget>\n120<\/output>\n120<\/result>","index":3985} +{"problem":"in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability q this college graduate has a graduate degree ?","rationale":"\"in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability of cgd \/ ( cg ) q - > 3 \/ 19 answer d\"","correct":"d","options":{"a":"1 \/ 11 ","b":"1 \/ 12 ","c":"1 \/ 13 ","d":"3 \/ 19","e":"3 \/ 43"},"options_float":{"a":0.0909090909,"b":0.0833333333,"c":0.0769230769,"d":0.1578947368,"e":0.0697674419},"annotated_formula":"divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|add(#2,n0)|divide(#2,#3)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/8) \/ (2\/3)<\/gadget>\n3\/16 = around 0.1875<\/output>\n(3\/16) + 1<\/gadget>\n19\/16 = around 1.1875<\/output>\n(3\/16) \/ (19\/16)<\/gadget>\n3\/19 = around 0.157895<\/output>\n3\/19 = around 0.157895<\/result>","index":3986} +{"problem":"solution y is 20 percent liquid x and 80 percent water . if 2 kilograms of water evaporate from 8 kilograms of solution y and 2 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ?","rationale":"\"in 8 kilograms of solution y there are 0.2 * 8 = 1.6 kilograms of solution x ; after 2 kilograms of water are replaced by 2 kilograms of solution y , to the existing 2.4 kilograms of solution x , 0.2 * 2 = 0.4 kilograms of solution x are added , so in the new solution of 8 kilograms there are 1.6 + 0.4 = 2 kilograms of solution x , which is 2 \/ 8 * 100 = 25 % of this new solution . answer : e .\"","correct":"e","options":{"a":"20 % ","b":"21 1 \/ 2 % ","c":"22 % ","d":"24 %","e":"25 %"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":24.0,"e":25.0},"annotated_formula":"multiply(const_100, divide(add(multiply(8, divide(20, const_100)), multiply(2, divide(20, const_100))), add(add(multiply(8, divide(20, const_100)), multiply(2, divide(20, const_100))), add(subtract(multiply(8, divide(80, const_100)), 2), multiply(2, divide(80, const_100))))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(n3,#0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n3,#1)|add(#2,#3)|subtract(#5,n2)|add(#4,#7)|add(#6,#8)|divide(#6,#9)|multiply(#10,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n8 * (1\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n2 * (1\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(8\/5) + (2\/5)<\/gadget>\n2<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n8 * (4\/5)<\/gadget>\n32\/5 = around 6.4<\/output>\n(32\/5) - 2<\/gadget>\n22\/5 = around 4.4<\/output>\n2 * (4\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n(22\/5) + (8\/5)<\/gadget>\n6<\/output>\n2 + 6<\/gadget>\n8<\/output>\n2 \/ 8<\/gadget>\n1\/4 = around 0.25<\/output>\n100 * (1\/4)<\/gadget>\n25<\/output>\n25<\/result>","index":3988} +{"problem":"20.036 divided by 0.04 gives :","rationale":"\"= 20.036 \/ 0.04 = 2003.6 \/ 4 = 500.9 answer is e .\"","correct":"e","options":{"a":"50.09 ","b":"5.06 ","c":"50.06 ","d":"100.9","e":"500.9"},"options_float":{"a":50.09,"b":5.06,"c":50.06,"d":100.9,"e":500.9},"annotated_formula":"divide(20.036, 0.04)","linear_formula":"divide(n0,n1)|","chain":"20.036 \/ 0.04<\/gadget>\n500.9<\/output>\n500.9<\/result>","index":3989} +{"problem":"the ratio between the sale price and the cost price of an article is 12 : 7 . what is the ratio between the profit and the cost price of that article ?","rationale":"\"let c . p . = rs . 7 x and s . p . = rs . 12 x . then , gain = rs . 5 x required ratio = 5 x : 7 x = 5 : 7 answer : d\"","correct":"d","options":{"a":"2 : 9 ","b":"2 : 5 ","c":"3 : 6 ","d":"5 : 7","e":"2 : 1"},"options_float":{"a":0.2222222222,"b":0.4,"c":0.5,"d":0.7142857143,"e":2.0},"annotated_formula":"divide(subtract(12, 7), 7)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|","chain":"12 - 7<\/gadget>\n5<\/output>\n5 \/ 7<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":3990} +{"problem":"a grocer has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6800 ?","rationale":"\"total fr 5 mnths = ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs 34009 . reqd . sale = rs . [ ( 6800 * 6 ) - 34009 ] = rs . ( 40800 - 34009 ) = rs . 6791 . answer : d\"","correct":"d","options":{"a":"s . 4991 ","b":"s . 5991 ","c":"s . 6001 ","d":"s . 6791","e":"s . 7991"},"options_float":{"a":4991.0,"b":5991.0,"c":6001.0,"d":6791.0,"e":7991.0},"annotated_formula":"subtract(multiply(add(5, const_1), 6800), add(add(add(add(6435, 6927), 6855), 7230), 6562))","linear_formula":"add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|","chain":"5 + 1<\/gadget>\n6<\/output>\n6 * 6_800<\/gadget>\n40_800<\/output>\n6_435 + 6_927<\/gadget>\n13_362<\/output>\n13_362 + 6_855<\/gadget>\n20_217<\/output>\n20_217 + 7_230<\/gadget>\n27_447<\/output>\n27_447 + 6_562<\/gadget>\n34_009<\/output>\n40_800 - 34_009<\/gadget>\n6_791<\/output>\n6_791<\/result>","index":3991} +{"problem":"sakshi can do a piece of work in 12 days . tanya is 20 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work is ?","rationale":"\"ratio of times taken by sakshi and tanys = 120 : 100 = 6 : 5 suppose tanya takes x days to do the work . 6 : 5 : : 12 : x = > x = 10 days . hence , tanya takes 10 days to complete the work . answer : a\"","correct":"a","options":{"a":"10 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"divide(12, add(const_1, divide(20, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n12 \/ (6\/5)<\/gadget>\n10<\/output>\n10<\/result>","index":3993} +{"problem":"a river 4 m deep and 40 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is ?","rationale":"\"( 4000 * 4 * 40 ) \/ 60 = 10666 m 3 answer : a\"","correct":"a","options":{"a":"10666 ","b":"11678 ","c":"12345 ","d":"10678","e":"19099"},"options_float":{"a":10666.0,"b":11678.0,"c":12345.0,"d":10678.0,"e":19099.0},"annotated_formula":"divide(multiply(multiply(4, 40), multiply(4, const_1000)), multiply(const_1, const_60))","linear_formula":"multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)|","chain":"4 * 40<\/gadget>\n160<\/output>\n4 * 1_000<\/gadget>\n4_000<\/output>\n160 * 4_000<\/gadget>\n640_000<\/output>\n1 * 60<\/gadget>\n60<\/output>\n640_000 \/ 60<\/gadget>\n32_000\/3 = around 10_666.666667<\/output>\n32_000\/3 = around 10_666.666667<\/result>","index":3994} +{"problem":"in one hour , a boat goes 11 km along the stream and 5 km against the stream . the speed of the boat in still water is :","rationale":"\"solution speed in still water = 1 \/ 2 ( 11 + 5 ) kmph = 8 kmph . answer c\"","correct":"c","options":{"a":"3 ","b":"5 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":3.0,"b":5.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"divide(add(11, 5), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"11 + 5<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":3995} +{"problem":"what least number should be subtracted from 1398 so that the remainder when divided by 7 , 9 , and 11 will leave in each case the same remainder 5 ?","rationale":"\"the lcm of 7 , 9 , and 11 is 693 . the next multiple is 2 * 693 = 1386 . 1386 + { remainder } = 1386 + 5 = 1391 , which is 7 less than 1398 . answer : b .\"","correct":"b","options":{"a":"6 ","b":"7 ","c":"8 ","d":"10","e":"11"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":10.0,"e":11.0},"annotated_formula":"subtract(1398, add(multiply(multiply(multiply(7, 9), 11), const_2), 5))","linear_formula":"multiply(n1,n2)|multiply(n3,#0)|multiply(#1,const_2)|add(n4,#2)|subtract(n0,#3)|","chain":"7 * 9<\/gadget>\n63<\/output>\n63 * 11<\/gadget>\n693<\/output>\n693 * 2<\/gadget>\n1_386<\/output>\n1_386 + 5<\/gadget>\n1_391<\/output>\n1_398 - 1_391<\/gadget>\n7<\/output>\n7<\/result>","index":3996} +{"problem":"if the sides of a triangle are 32 cm , 27 cm and 12 cm , what is its area ?","rationale":"\"the triangle with sides 32 cm , 27 cm and 12 cm is right angled , where the hypotenuse is 32 cm . area of the triangle = 1 \/ 2 * 27 * 12 = 162 cm 2 answer : c\"","correct":"c","options":{"a":"120 cm 2 ","b":"112 cm 2 ","c":"162 cm 2 ","d":"666 cm 2","e":"886 cm 2"},"options_float":{"a":120.0,"b":112.0,"c":162.0,"d":666.0,"e":886.0},"annotated_formula":"divide(multiply(27, 12), const_2)","linear_formula":"multiply(n1,n2)|divide(#0,const_2)|","chain":"27 * 12<\/gadget>\n324<\/output>\n324 \/ 2<\/gadget>\n162<\/output>\n162<\/result>","index":3997} +{"problem":"a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1233 . find the number to be multiplied .","rationale":"\"let the required number be x . then , 43 x – 34 x = 1233 or 9 x = 1233 or x = 137 . required number = 137 . answer : d\"","correct":"d","options":{"a":"130 ","b":"132 ","c":"134 ","d":"137","e":"138"},"options_float":{"a":130.0,"b":132.0,"c":134.0,"d":137.0,"e":138.0},"annotated_formula":"divide(1233, subtract(43, 34))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"43 - 34<\/gadget>\n9<\/output>\n1_233 \/ 9<\/gadget>\n137<\/output>\n137<\/result>","index":3999} +{"problem":"two letters from the word barkhint are selected at random . what is the probability that at least one of the selected letters is a consonant ?","rationale":"there are 6 consonants and 2 vowels in barkhint . probability that at least one of the selected letters is a consonant = 1 - ( probability of both vowels ) probability of both vowels = 2 c 2 \/ 8 c 2 = 1 \/ 28 so , answer is 1 - 1 \/ 28 = 27 \/ 28 answer : d","correct":"d","options":{"a":"30 \/ 28 ","b":"40 \/ 28 ","c":"25 \/ 28 ","d":"27 \/ 28","e":"32 \/ 28"},"options_float":{"a":1.0714285714,"b":1.4285714286,"c":0.8928571429,"d":0.9642857143,"e":1.1428571429},"annotated_formula":"divide(subtract(choose(multiply(const_2, const_4), const_2), choose(const_2, const_2)), choose(multiply(const_2, const_4), const_2))","linear_formula":"choose(const_2,const_2)|multiply(const_2,const_4)|choose(#1,const_2)|subtract(#2,#0)|divide(#3,#2)","chain":"2 * 4<\/gadget>\n8<\/output>\nbinomial(8, 2)<\/gadget>\n28<\/output>\nbinomial(2, 2)<\/gadget>\n1<\/output>\n28 - 1<\/gadget>\n27<\/output>\n27 \/ 28<\/gadget>\n27\/28 = around 0.964286<\/output>\n27\/28 = around 0.964286<\/result>","index":4000} +{"problem":"if ( a + b ) = 4 , ( b + c ) = 8 and ( c + d ) = 3 , what is the value of ( a + d ) ?","rationale":"\"given a + b = 4 b + c = 8 c + d = 3 then ( a + b ) - ( b + c ) + ( c + d ) = 4 - 8 + 3 = - 1 option e . . .\"","correct":"e","options":{"a":"16 . ","b":"8 . ","c":"7 . ","d":"2 .","e":"- 1 ."},"options_float":{"a":16.0,"b":8.0,"c":7.0,"d":2.0,"e":-1.0},"annotated_formula":"subtract(add(4, 3), 8)","linear_formula":"add(n0,n2)|subtract(#0,n1)|","chain":"4 + 3<\/gadget>\n7<\/output>\n7 - 8<\/gadget>\n-1<\/output>\n-1<\/result>","index":4001} +{"problem":"( 69842 x 69842 - 30158 x 30158 ) \/ ( 69842 - 30158 ) = ?","rationale":"answer given expression = [ ( 69842 ) 2 - ( 30158 ) 2 ] \/ [ ( 69842 - 30158 ) ] = [ ( 69842 - 30158 ) ( 69842 + 30158 ) ] \/ ( 60842 - 30158 ) = 100000 correct option : a","correct":"a","options":{"a":"100000 ","b":"69842 ","c":"39684 ","d":"30158","e":"none"},"options_float":{"a":100000.0,"b":69842.0,"c":39684.0,"d":30158.0,"e":null},"annotated_formula":"divide(subtract(multiply(69842, 69842), multiply(30158, 30158)), subtract(69842, 30158))","linear_formula":"multiply(n0,n0)|multiply(n2,n2)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)","chain":"69_842 * 69_842<\/gadget>\n4_877_904_964<\/output>\n30_158 * 30_158<\/gadget>\n909_504_964<\/output>\n4_877_904_964 - 909_504_964<\/gadget>\n3_968_400_000<\/output>\n69_842 - 30_158<\/gadget>\n39_684<\/output>\n3_968_400_000 \/ 39_684<\/gadget>\n100_000<\/output>\n100_000<\/result>","index":4002} +{"problem":"two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 6 seconds . the length of the faster train is","rationale":"\"solution relative speed = ( 36 + 45 ) km \/ hr = ( 81 x 5 \/ 18 ) m \/ sec = ( 45 \/ 2 ) m \/ sec length of the train = ( 45 \/ 2 x 6 ) m = 135 m . answer d\"","correct":"d","options":{"a":"80 m ","b":"100 m ","c":"120 m ","d":"135 m","e":"none"},"options_float":{"a":80.0,"b":100.0,"c":120.0,"d":135.0,"e":null},"annotated_formula":"multiply(multiply(add(36, 45), const_0_2778), 6)","linear_formula":"add(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)|","chain":"36 + 45<\/gadget>\n81<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n81 * (5\/18)<\/gadget>\n45\/2 = around 22.5<\/output>\n(45\/2) * 6<\/gadget>\n135<\/output>\n135<\/result>","index":4003} +{"problem":"8 percent of the programmers in a startup company weigh 200 pounds or more . 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less . what percent of the programmers in the startup company weigh between 100 and 200 pounds ?","rationale":"initially 92 % and 8 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 % let total be ' 100 ' then that 75 % is ( 3 \/ 4 ) ∗ 92 so , the required % is [ ( 3 \/ 4 ) ∗ 92 \/ 100 ] ∗ 100 = 69 % answer : d","correct":"d","options":{"a":"15 % ","b":"20 % ","c":"25 % ","d":"69 %","e":"75 %"},"options_float":{"a":15.0,"b":20.0,"c":25.0,"d":69.0,"e":75.0},"annotated_formula":"subtract(subtract(100, 20), 8)","linear_formula":"subtract(n5,n2)|subtract(#0,n0)","chain":"100 - 20<\/gadget>\n80<\/output>\n80 - 8<\/gadget>\n72<\/output>\n72<\/result>","index":4005} +{"problem":"in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 20 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?","rationale":"\"0,16 r = x \/ 100 * 0.1 r answer b\"","correct":"b","options":{"a":"80 % ","b":"160 % ","c":"120 % ","d":"124.2 %","e":"138 %"},"options_float":{"a":80.0,"b":160.0,"c":120.0,"d":124.2,"e":138.0},"annotated_formula":"multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(20, const_100)), divide(10, const_100)), const_100)","linear_formula":"divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * (1\/5)<\/gadget>\n4\/25 = around 0.16<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(4\/25) \/ (1\/10)<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) * 100<\/gadget>\n160<\/output>\n160<\/result>","index":4006} +{"problem":"suppose you work for a manufacturing plant that pays you $ 12.50 an hour plus $ 0.16 for each widget you make . how many widgets must you produce in a 40 hour week to earn $ 750 ( before payroll deductions ) ?","rationale":"\"total pay = 40 * $ 12.50 + $ 0.16 * x = 760 x = 260 \/ 0.16 = 1625 the answer is b .\"","correct":"b","options":{"a":"1600 ","b":"1625 ","c":"1650 ","d":"1655","e":"1675"},"options_float":{"a":1600.0,"b":1625.0,"c":1650.0,"d":1655.0,"e":1675.0},"annotated_formula":"divide(subtract(750, multiply(12.50, 40)), 0.16)","linear_formula":"multiply(n0,n2)|subtract(n3,#0)|divide(#1,n1)|","chain":"12.5 * 40<\/gadget>\n500<\/output>\n750 - 500<\/gadget>\n250<\/output>\n250 \/ 0.16<\/gadget>\n1_562.5<\/output>\n1_562.5<\/result>","index":4007} +{"problem":"2 pipes can separately fill a tank in 20 hrs and 30 hrs respectively . both the pipes are opened to fillthe tank , a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ?","rationale":"1 \/ 20 + 1 \/ 30 = 1 \/ 12 1 + 1 \/ 3 = 4 \/ 3 1 - - - 12 4 \/ 3 - - - ? 4 \/ 3 * 12 = 16 hrs c","correct":"c","options":{"a":"10 hrs ","b":"12 hrs ","c":"16 hrs ","d":"18 hrs","e":"20 hrs"},"options_float":{"a":10.0,"b":12.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"divide(add(const_1, divide(const_1, const_3)), add(inverse(20), inverse(30)))","linear_formula":"divide(const_1,const_3)|inverse(n1)|inverse(n2)|add(#0,const_1)|add(#1,#2)|divide(#3,#4)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/20) + (1\/30)<\/gadget>\n1\/12 = around 0.083333<\/output>\n(4\/3) \/ (1\/12)<\/gadget>\n16<\/output>\n16<\/result>","index":4008} +{"problem":"the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 15 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 23 x 13 ) and ( 23 x 15 ) . larger number = ( 23 x 15 ) = 345 . answer : option d\"","correct":"d","options":{"a":"276 ","b":"299 ","c":"322 ","d":"345","e":"354"},"options_float":{"a":276.0,"b":299.0,"c":322.0,"d":345.0,"e":354.0},"annotated_formula":"multiply(23, 15)","linear_formula":"multiply(n0,n2)|","chain":"23 * 15<\/gadget>\n345<\/output>\n345<\/result>","index":4009} +{"problem":"an assembly line produces 15 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this whole time ?","rationale":"\"the time to produce the first 60 cogs is 60 \/ 15 = 4 hours . the time to produce the next 60 cogs is 60 \/ 60 = 1 hour . the average output is 120 cogs \/ 5 hours = 24 cogs per hour . the answer is c .\"","correct":"c","options":{"a":"20 ","b":"22 ","c":"24 ","d":"26","e":"28"},"options_float":{"a":20.0,"b":22.0,"c":24.0,"d":26.0,"e":28.0},"annotated_formula":"divide(multiply(60, const_2), add(divide(60, 15), const_1))","linear_formula":"divide(n1,n0)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#2)|","chain":"60 * 2<\/gadget>\n120<\/output>\n60 \/ 15<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n120 \/ 5<\/gadget>\n24<\/output>\n24<\/result>","index":4010} +{"problem":"210 students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ?","rationale":"we are told that 60 students claimed that they preferred mac to windows , which means that 60 preferred mac but not windows , so # of students who preferred mac ( p ( a ) as you wrote ) , does not equal to 60 , it equals to 60 + 20 ( 20 is # of students who equally preferred both brands ) . also we are asked to find # of the students who preferred windows to mac , so if you denote x as those who prefer windows then you should calculate x - 20 . so , if we use your formula it should be : 210 = { mac } + { windows } - { both } + { neither } = ( 60 + 20 ) + x - 20 + 90 - - > x = 60 ( # of student who prefer windows ) - - > # of the students who preferred windows to mac is x - 20 = 40 .","correct":"b","options":{"a":"25 ","b":"40 ","c":"50 ","d":"60","e":"75"},"options_float":{"a":25.0,"b":40.0,"c":50.0,"d":60.0,"e":75.0},"annotated_formula":"subtract(210, add(add(60, divide(60, const_3)), 90))","linear_formula":"divide(n1,const_3)|add(n1,#0)|add(n2,#1)|subtract(n0,#2)","chain":"60 \/ 3<\/gadget>\n20<\/output>\n60 + 20<\/gadget>\n80<\/output>\n80 + 90<\/gadget>\n170<\/output>\n210 - 170<\/gadget>\n40<\/output>\n40<\/result>","index":4011} +{"problem":"the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 19 - inch screen would be how many square inches greater than the area of a square 17 - inch screen ?","rationale":"\"pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 19 ^ 2 \/ 2 - 17 ^ 2 \/ 2 = 72 \/ 2 = 36 ans a .\"","correct":"a","options":{"a":"36 ","b":"4 ","c":"16 ","d":"38","e":"40"},"options_float":{"a":36.0,"b":4.0,"c":16.0,"d":38.0,"e":40.0},"annotated_formula":"divide(subtract(power(19, const_2), power(17, const_2)), const_2)","linear_formula":"power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|divide(#2,const_2)|","chain":"19 ** 2<\/gadget>\n361<\/output>\n17 ** 2<\/gadget>\n289<\/output>\n361 - 289<\/gadget>\n72<\/output>\n72 \/ 2<\/gadget>\n36<\/output>\n36<\/result>","index":4012} +{"problem":"increasing the original price of a certain item by 20 percent and then increasing the new price by 20 percent is equivalent to increasing the original price by what percent ?","rationale":"\"we ' re told that the original price of an item is increased by 20 % and then that price is increased by 20 % . . . . if . . . . starting value = $ 100 + 20 % = 100 + . 20 ( 100 ) = 120 + 20 % = 120 + . 20 ( 120 ) = 120 + 24 = 144 the question asks how the final price relates to the original price . this is essentially about percentage change , which means we should use the percentage change formula : percentage change = ( new - old ) \/ old = difference \/ original doing either calculation will yield the same result : 44 \/ 100 = 44 % final answer : e\"","correct":"e","options":{"a":"31.25 ","b":"37.5 ","c":"50.0 ","d":"52.5","e":"44.0"},"options_float":{"a":31.25,"b":37.5,"c":50.0,"d":52.5,"e":44.0},"annotated_formula":"multiply(subtract(multiply(add(divide(20, const_100), const_1), add(divide(20, const_100), const_1)), const_1), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * (6\/5)<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) - 1<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 100<\/gadget>\n44<\/output>\n44<\/result>","index":4013} +{"problem":"the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 63 ounces of this substance ?","rationale":"\"112 ounces of a substance has a volume of 48 cubic inches 63 ounces of a substance has a volume of ( 48 \/ 112 ) * 63 = 27 cubic inches answer a alternatively , we can use estimation 112 ounces of a substance has a volume of 48 cubic inches 56 ounces of a substance has a volume of 24 cubic inches therefore , 63 will have a volume a little more than 24 , that is 27 answer : a\"","correct":"a","options":{"a":"27 ","b":"36 ","c":"42 ","d":"64","e":"147"},"options_float":{"a":27.0,"b":36.0,"c":42.0,"d":64.0,"e":147.0},"annotated_formula":"multiply(divide(48, 112), 63)","linear_formula":"divide(n0,n1)|multiply(n2,#0)|","chain":"48 \/ 112<\/gadget>\n3\/7 = around 0.428571<\/output>\n(3\/7) * 63<\/gadget>\n27<\/output>\n27<\/result>","index":4014} +{"problem":"if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 266 m , what is its area ?","rationale":"\"length = breadth + 23 . therefore , 4 × breadth + 2 × 23 = 266 m ⇒ breadth = 55 m length = 55 + 23 = 78 m area = 78 × 55 = 4290 m 2 answer is a .\"","correct":"a","options":{"a":"4290 ","b":"2535 ","c":"2530 ","d":"2515","e":"2520"},"options_float":{"a":4290.0,"b":2535.0,"c":2530.0,"d":2515.0,"e":2520.0},"annotated_formula":"rectangle_area(add(divide(subtract(266, multiply(const_2, 23)), const_4), 23), divide(subtract(266, multiply(const_2, 23)), const_4))","linear_formula":"multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)|","chain":"2 * 23<\/gadget>\n46<\/output>\n266 - 46<\/gadget>\n220<\/output>\n220 \/ 4<\/gadget>\n55<\/output>\n55 + 23<\/gadget>\n78<\/output>\n78 * 55<\/gadget>\n4_290<\/output>\n4_290<\/result>","index":4015} +{"problem":"the chance of rain on any given day in tel - aviv is 50 % . what is the probability z that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?","rationale":"\"chances of rain on exactly 4 days and not rain on 2 days = ( 1 \/ 2 ) ^ 4 * ( 1 \/ 2 ) ^ 2 = 1 \/ 64 chosing 4 days out of 6 = 6 ! \/ ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days z = 15 \/ 64 ans a it is .\"","correct":"a","options":{"a":"15 \/ 64 ","b":"30 \/ 64 ","c":"1 \/ 2 ","d":"1 \/ 4","e":"52 \/ 64"},"options_float":{"a":0.234375,"b":0.46875,"c":0.5,"d":0.25,"e":0.8125},"annotated_formula":"divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), const_2))))","linear_formula":"choose(n2,n1)|divide(const_1,const_2)|power(#1,n1)|power(#1,const_2)|multiply(#2,#3)|inverse(#4)|divide(#0,#5)|","chain":"binomial(6, 4)<\/gadget>\n15<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 4<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/2) ** 2<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/16) * (1\/4)<\/gadget>\n1\/64 = around 0.015625<\/output>\n1 \/ (1\/64)<\/gadget>\n64<\/output>\n15 \/ 64<\/gadget>\n15\/64 = around 0.234375<\/output>\n15\/64 = around 0.234375<\/result>","index":4016} +{"problem":"little john had $ 7.10 . he spent $ 1.05 on sweets and gave to his two friends $ 1.00 each . how much money was left ?","rationale":"\"john spent and gave to his two friends a total of 1.05 + 1.00 + 1.00 = $ 3.05 money left 7.10 - 3.05 = $ 4.05 answer : e\"","correct":"e","options":{"a":"$ 4.85 ","b":"$ 4.15 ","c":"$ 4.65 ","d":"$ 4.55","e":"$ 4.05"},"options_float":{"a":4.85,"b":4.15,"c":4.65,"d":4.55,"e":4.05},"annotated_formula":"subtract(7.10, add(1.05, add(1.00, 1.00)))","linear_formula":"add(n2,n2)|add(n1,#0)|subtract(n0,#1)|","chain":"1 + 1<\/gadget>\n2<\/output>\n1.05 + 2<\/gadget>\n3.05<\/output>\n7.1 - 3.05<\/gadget>\n4.05<\/output>\n4.05<\/result>","index":4018} +{"problem":"in a class of 32 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 10 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ?","rationale":"\"total # of students = 32 avg # of books per student = 2 total # of books = 32 * 2 = 64 # of student borrowed at least 3 books = 32 - 2 - 12 - 10 = 8 # of books borrowed by above 8 students = 60 - ( 12 * 1 ) + ( 10 * 2 ) = 32 considering that 7 out of above 8 students borrowed only 3 books , # of books borrowed = 7 * 3 = 21 so maximum # of books borrowed by any single student = 32 - 21 = 11 option d\"","correct":"d","options":{"a":"3 ","b":"5 ","c":"8 ","d":"11","e":"15"},"options_float":{"a":3.0,"b":5.0,"c":8.0,"d":11.0,"e":15.0},"annotated_formula":"subtract(multiply(32, 2), add(multiply(subtract(subtract(32, add(add(multiply(12, 1), 10), 2)), 1), 3), add(multiply(12, 1), multiply(10, 2))))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)|","chain":"32 * 2<\/gadget>\n64<\/output>\n12 * 1<\/gadget>\n12<\/output>\n12 + 10<\/gadget>\n22<\/output>\n22 + 2<\/gadget>\n24<\/output>\n32 - 24<\/gadget>\n8<\/output>\n8 - 1<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n10 * 2<\/gadget>\n20<\/output>\n12 + 20<\/gadget>\n32<\/output>\n21 + 32<\/gadget>\n53<\/output>\n64 - 53<\/gadget>\n11<\/output>\n11<\/result>","index":4019} +{"problem":"the average of 10 numbers is calculated as 21 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ?","rationale":"\"explanation : 10 * 21 + 36 – 26 = 220 = > 220 \/ 10 = 22 d )\"","correct":"d","options":{"a":"16 ","b":"18 ","c":"19 ","d":"22","e":"24"},"options_float":{"a":16.0,"b":18.0,"c":19.0,"d":22.0,"e":24.0},"annotated_formula":"add(21, divide(subtract(36, 26), 10))","linear_formula":"subtract(n2,n3)|divide(#0,n0)|add(n1,#1)|","chain":"36 - 26<\/gadget>\n10<\/output>\n10 \/ 10<\/gadget>\n1<\/output>\n21 + 1<\/gadget>\n22<\/output>\n22<\/result>","index":4020} +{"problem":"- 45 * 29 + 108 = ?","rationale":"\"= > - 45 * ( 30 - 1 ) + 108 ; = > - ( 45 * 30 ) + 45 + 108 ; = > - 1350 + 153 = - 1197 . correct option : c\"","correct":"c","options":{"a":"1197 ","b":"1897 ","c":"- 1197 ","d":"- 1900","e":"none of these"},"options_float":{"a":1197.0,"b":1897.0,"c":-1197.0,"d":-1900.0,"e":null},"annotated_formula":"add(multiply(negate(45), 29), 108)","linear_formula":"negate(n0)|multiply(n1,#0)|add(n2,#1)|","chain":"-45<\/gadget>\n-45<\/output>\n(-45) * 29<\/gadget>\n-1_305<\/output>\n(-1_305) + 108<\/gadget>\n-1_197<\/output>\n-1_197<\/result>","index":4021} +{"problem":"in a family gathering , there is a basket in which there are oranges , bananas and apples . half of the people in the family eat oranges , half of the other half eat bananas and the rest eat apples . if the number of people who eat oranges are 10 less than the total number of people , find the number of people in the gathering .","rationale":"let the number of people who eat oranges be x . then , total number of people = x + ( x \/ 2 ) + ( x \/ 2 ) = 2 x = > 2 x - 10 = x = > x = 10 = > total number of people were 2 x = 20 option d","correct":"d","options":{"a":"5 ","b":"10 ","c":"15 ","d":"20","e":"25"},"options_float":{"a":5.0,"b":10.0,"c":15.0,"d":20.0,"e":25.0},"annotated_formula":"multiply(10, const_2)","linear_formula":"multiply(n0,const_2)","chain":"10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":4022} +{"problem":"excluding stoppages , the average speed of a bus is 60 km \/ hr and including stoppages , the average speed of the bus is 15 km \/ hr . for how many minutes does the bus stop per hour ?","rationale":"\"in 1 hr , the bus covers 60 km without stoppages and 15 km with stoppages . stoppage time = time take to travel ( 60 - 15 ) km i . e 45 km at 60 km \/ hr . stoppage time = 45 \/ 60 hrs = 45 min . answer : e\"","correct":"e","options":{"a":"22 ","b":"88 ","c":"77 ","d":"20","e":"45"},"options_float":{"a":22.0,"b":88.0,"c":77.0,"d":20.0,"e":45.0},"annotated_formula":"subtract(multiply(const_1, const_60), multiply(divide(15, 60), const_60))","linear_formula":"divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)|","chain":"1 * 60<\/gadget>\n60<\/output>\n15 \/ 60<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 60<\/gadget>\n15<\/output>\n60 - 15<\/gadget>\n45<\/output>\n45<\/result>","index":4023} +{"problem":"laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 20 kmph and prasanna moves southward at a speed of 30 kmph . how far will be prasanna from laxmi after 60 minutes ?","rationale":"explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 20 kmph x 1 hr = 20 km total southward prasanna ' s distance = 30 kmph x 1 hr = 30 km total distance between prasanna and laxmi is = 20 + 30 = 50 km . answer : b","correct":"b","options":{"a":"11 ","b":"50 ","c":"28 ","d":"27","e":"18"},"options_float":{"a":11.0,"b":50.0,"c":28.0,"d":27.0,"e":18.0},"annotated_formula":"add(20, 30)","linear_formula":"add(n0,n1)","chain":"20 + 30<\/gadget>\n50<\/output>\n50<\/result>","index":4025} +{"problem":"there is a train and car . the ratio between the speed of a train & a car is 16 : 15 respectively . also , a bus covered a distance of 320 km in 5 hours . the speed of the bus is 4 \/ 5 th of the speed of the train . how many kilometers will the car cover in 7 hours ?","rationale":"the speed of the bus is 320 \/ 5 = 64 km \/ hr the speed of the train is ( 64 * 5 ) \/ 4 = 80 km \/ hr the speed of the car is 80 \/ 16 * 15 = 75 km \/ hr the distance covered by the car in 7 hours is 75 × 7 = 525 km the answer is d .","correct":"d","options":{"a":"375 ","b":"425 ","c":"475 ","d":"525","e":"575"},"options_float":{"a":375.0,"b":425.0,"c":475.0,"d":525.0,"e":575.0},"annotated_formula":"multiply(divide(multiply(divide(multiply(divide(320, 5), 5), 4), 15), 16), 7)","linear_formula":"divide(n2,n3)|multiply(n3,#0)|divide(#1,n4)|multiply(n1,#2)|divide(#3,n0)|multiply(n6,#4)","chain":"320 \/ 5<\/gadget>\n64<\/output>\n64 * 5<\/gadget>\n320<\/output>\n320 \/ 4<\/gadget>\n80<\/output>\n80 * 15<\/gadget>\n1_200<\/output>\n1_200 \/ 16<\/gadget>\n75<\/output>\n75 * 7<\/gadget>\n525<\/output>\n525<\/result>","index":4026} +{"problem":"( ( ( 13 ! ) ^ 16 ) - ( ( 13 ! ) ^ 8 ) ) \/ ( ( ( 13 ! ) ^ 8 ) + ( ( 13 ! ) ^ 4 ) ) = a then what is the units digit for a \/ ( ( 13 ! ) ^ 4 ) =","rationale":"( ( ( 13 ! ) ^ 16 ) - ( ( 13 ! ) ^ 8 ) ) \/ ( ( ( 13 ! ) ^ 8 ) + ( ( 13 ! ) ^ 4 ) ) = a solving it , we get a \/ ( ( 13 ! ) ^ 4 ) = ( ( 13 ! ) ^ 4 - 1 ) last digit of ( 13 ! ) ^ 4 will be 0 . so last digit of ( ( 13 ! ) ^ 4 - 1 ) will be 9 . answer : c","correct":"c","options":{"a":"7 ","b":"8 ","c":"9 ","d":"6","e":"5"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":6.0,"e":5.0},"annotated_formula":"subtract(subtract(13, const_3), const_1)","linear_formula":"subtract(n0,const_3)|subtract(#0,const_1)","chain":"13 - 3<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9<\/result>","index":4027} +{"problem":"vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 6358 , how much amount did raghu invest ?","rationale":"\"let money invested by raghu = rs . x money invested by trishul = 9 \/ 10 x = 0.9 x money invested by vishal = 9 \/ 10 x * 110 \/ 100 = 0.99 x also , x + 0.9 x + 0.99 x = 6358 = x = 6358 \/ 2.89 = 2200 therefore , amount invested by raghu is rs . 2200 . answer : b\"","correct":"b","options":{"a":"1287 ","b":"2200 ","c":"2000 ","d":"1129","e":"1192"},"options_float":{"a":1287.0,"b":2200.0,"c":2000.0,"d":1129.0,"e":1192.0},"annotated_formula":"divide(6358, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|add(#3,#2)|add(#4,const_1)|divide(n2,#5)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(9\/10) * (11\/10)<\/gadget>\n99\/100 = around 0.99<\/output>\n(99\/100) + (9\/10)<\/gadget>\n189\/100 = around 1.89<\/output>\n(189\/100) + 1<\/gadget>\n289\/100 = around 2.89<\/output>\n6_358 \/ (289\/100)<\/gadget>\n2_200<\/output>\n2_200<\/result>","index":4028} +{"problem":"in n is a positive integer less than 200 , and 14 n \/ 60 is an integer , then n has how many different positive prime factors a ?","rationale":"\"i like to put the numbers in prime factors so it is easier and faster to visualize . 14 * n \/ 60 if we write the factors of 14 - - > 2 , 7 , and the factors of 60 - - > 2 , 2 , 3 , 5 , we have ( 2 * 7 * n ) \/ ( 2 ^ 2 * 3 * 5 ) simplifying 7 * n \/ ( 2 * 3 * 5 ) the only way the equation above has an integer value is if n has at least the factors 2 , 3 and 5 , so we can simplify again and we have the number 7 . the number could be 2 * 3 * 5 , or 2 * 3 * 5 * 2 , or 2 * 3 * 5 * . . . . . however to be less than 200 we can not add any prime number . 2 * 3 * 5 = 120 if we added the next prime factor 7 , we would have a = 2 * 3 * 5 * 7 = 840 thus , answer b\"","correct":"b","options":{"a":"2 ","b":"3 ","c":"5 ","d":"6","e":"8"},"options_float":{"a":2.0,"b":3.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2)","linear_formula":"add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)|","chain":"1 + 4<\/gadget>\n5<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n5 \/ 5<\/gadget>\n1<\/output>\n1 + 2<\/gadget>\n3<\/output>\n3<\/result>","index":4031} +{"problem":"the payment per hour for a daily - wages employee is increased by 33.33 and the working time is reduced by 33.33 % . what is the percentage change in the employee ' s income ?","rationale":"if payment per hr is increased by 33.33 % , then 11.11 % decrease in income if earlier he is getting rs 1 hr per hr and working for 1 hr , income = rs 1 now he is getting rs 1.3333 per hr and working for 0.66667 hrs , income = rs . 0.8889 answer : b","correct":"b","options":{"a":"rs . 0.7889 ","b":"rs . 0.8889 ","c":"rs . 0.9889 ","d":"rs . 1.8889","e":"rs . 2.8889"},"options_float":{"a":0.7889,"b":0.8889,"c":0.9889,"d":1.8889,"e":2.8889},"annotated_formula":"multiply(add(const_1, divide(33.33, const_100)), subtract(const_1, divide(33.33, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)","chain":"33.33 \/ 100<\/gadget>\n0.3333<\/output>\n1 + 0.3333<\/gadget>\n1.3333<\/output>\n1 - 0.3333<\/gadget>\n0.6667<\/output>\n1.3333 * 0.6667<\/gadget>\n0.888911<\/output>\n0.888911<\/result>","index":4032} +{"problem":"the sub - duplicate ratio of 16 : 64 is","rationale":"\"root ( 16 ) : root ( 64 ) = 4 : 8 = 2 : 4 answer : e\"","correct":"e","options":{"a":"4 : 3 ","b":"1 : 2 ","c":"1 : 3 ","d":"1 : 4","e":"2 : 4"},"options_float":{"a":1.3333333333,"b":0.5,"c":0.3333333333,"d":0.25,"e":0.5},"annotated_formula":"divide(sqrt(16), sqrt(64))","linear_formula":"sqrt(n0)|sqrt(n1)|divide(#0,#1)|","chain":"16 ** (1\/2)<\/gadget>\n4<\/output>\n64 ** (1\/2)<\/gadget>\n8<\/output>\n4 \/ 8<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":4033} +{"problem":"there are 38 students in a certain geometry class . if two thirds of the students are boys and 3 fourths of the boys are under 6 feet tall , how many boys in the class are under 6 feet tall ?","rationale":"38 * 2 \/ 3 * 3 \/ 4 = 19 answer : c","correct":"c","options":{"a":"6 ","b":"12 ","c":"19 ","d":"24","e":"27"},"options_float":{"a":6.0,"b":12.0,"c":19.0,"d":24.0,"e":27.0},"annotated_formula":"divide(multiply(multiply(38, const_2), 3), multiply(3, const_4))","linear_formula":"multiply(n0,const_2)|multiply(n1,const_4)|multiply(n1,#0)|divide(#2,#1)","chain":"38 * 2<\/gadget>\n76<\/output>\n76 * 3<\/gadget>\n228<\/output>\n3 * 4<\/gadget>\n12<\/output>\n228 \/ 12<\/gadget>\n19<\/output>\n19<\/result>","index":4034} +{"problem":"a boat crossed a lake from north to east at the speed of 7 km \/ h , entered a river and covered twice as much distance going upstream at 6 km \/ h . it then turned around and stopped at the south shore of the lake . if it averaged 3.8 km \/ h that day , what was its approximate downstream speed ?","rationale":"\"one way of solving this is : speed of boat on still water ( lake ) = 7 kmph speed upstream = 6 kmph = speed in still water - speed of river = > speed of river = 1 kmph = > speed downstream = speed in still water + speed of river = 7 + 1 = 8 kmph ans is e\"","correct":"e","options":{"a":"4 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"add(7, subtract(7, 6))","linear_formula":"subtract(n0,n1)|add(n0,#0)|","chain":"7 - 6<\/gadget>\n1<\/output>\n7 + 1<\/gadget>\n8<\/output>\n8<\/result>","index":4035} +{"problem":"a bullet train 120 m long is running with a speed of 50 kmph . in what time will it pass a man who is running at 4 kmph in the direction opposite to that in which the bullet train is going ?","rationale":"\"d 8 sec speed of the bullet train relative to man = ( 50 + 4 ) kmph = 54 * 5 \/ 18 m \/ sec = 45 \/ 3 m \/ sec . time taken by the bullet train to cross the man = time taken by it to cover 120 m at ( 45 \/ 3 ) m \/ sec = ( 120 * 3 \/ 45 ) sec = 8 sec\"","correct":"d","options":{"a":"23 sec ","b":"15 sec ","c":"12 sec ","d":"8 sec","e":"16 sec"},"options_float":{"a":23.0,"b":15.0,"c":12.0,"d":8.0,"e":16.0},"annotated_formula":"divide(120, divide(multiply(add(50, 4), const_1000), const_3600))","linear_formula":"add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|","chain":"50 + 4<\/gadget>\n54<\/output>\n54 * 1_000<\/gadget>\n54_000<\/output>\n54_000 \/ 3_600<\/gadget>\n15<\/output>\n120 \/ 15<\/gadget>\n8<\/output>\n8<\/result>","index":4036} +{"problem":"70,76 , 74,85 , 90,105 , 105,130 , 130,130 the list consists of the times in seconds that it took each of the 10 school children to run a distance of 400 mts . if the standard deviation of the 10 running times is 22.4 , rounded to the nearest tenth of a second , how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times .","rationale":"how many of the 10 running times are more than one sd below the meanmeans how many data points from given 10 are less thanmean - 1 sd . we are given that sd = 22.4 , so we should find mean - - > mean = 100 - - > there are only 3 data points below 100 - 22.4 = 77.6 , namely 70 75 , and 74 . answer : c .","correct":"c","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"add(1, const_2)","linear_formula":"add(n10,const_2)","chain":"1 + 2<\/gadget>\n3<\/output>\n3<\/result>","index":4039} +{"problem":"if the cost price of 60 articles is equal to the selling price of 40 articles , then what is the percent profit ?","rationale":"\"let x be the cost price of one article . let y be the selling price of one article . 40 y = 60 x y = 1.5 x the answer is c .\"","correct":"c","options":{"a":"30 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"70 %"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":70.0},"annotated_formula":"multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 40), 60)), divide(multiply(const_100, 40), 60)))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 * 40<\/gadget>\n4_000<\/output>\n4_000 \/ 60<\/gadget>\n200\/3 = around 66.666667<\/output>\n100 - (200\/3)<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) \/ (200\/3)<\/gadget>\n1\/2 = around 0.5<\/output>\n100 * (1\/2)<\/gadget>\n50<\/output>\n50<\/result>","index":4040} +{"problem":"john is the owner of a company and for thanksgiving he wants to give a turkey to each employee to celebrate the holiday . each turkey cost him $ 25 . if he has 85 employees , how much would john need to spend ?","rationale":"answer is ( a ) . if john wants to give each of his employees a turkey to help celebrate the thanksgiving holiday and each one costs him $ 25 multiplied by the number of employees ( 85 ) , he would have to pay a total of $ 2125 .","correct":"a","options":{"a":"$ 2125 ","b":"$ 1320 ","c":"$ 2000 ","d":"$ 1580","e":"$ 2000.75"},"options_float":{"a":2125.0,"b":1320.0,"c":2000.0,"d":1580.0,"e":2000.75},"annotated_formula":"multiply(25, 85)","linear_formula":"multiply(n0,n1)","chain":"25 * 85<\/gadget>\n2_125<\/output>\n2_125<\/result>","index":4041} +{"problem":"set a { 3 , 3,3 , 4,5 , 5,5 } has a standard deviation of 1 . what will the standard deviation be if every number in the set is multiplied by 5 ?","rationale":"\"points to remember - 1 . if oneadd \/ subtractthe same amont from every term in a set , sd does n ' t change . 2 . if onemultiply \/ divideevery term by the same number in a set , sd changes by same number . hence the answer to the above question is e\"","correct":"e","options":{"a":"1 ","b":"2 ","c":"4 ","d":"8","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":8.0,"e":5.0},"annotated_formula":"multiply(5, 1)","linear_formula":"multiply(n4,n5)|","chain":"5 * 1<\/gadget>\n5<\/output>\n5<\/result>","index":4042} +{"problem":"there are 28 balls which are red , blue or green . if 12 balls are green and the sum of red balls and green balls is less than 24 , at most how many red balls are there ?","rationale":"r + g + b = 28 g = 12 r + g < 24 = > r + 12 < 24 = > r < 12 = > at most 11 red balls answer : c","correct":"c","options":{"a":"10 ","b":"12 ","c":"11 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":12.0,"c":11.0,"d":13.0,"e":14.0},"annotated_formula":"subtract(12, const_1)","linear_formula":"subtract(n1,const_1)","chain":"12 - 1<\/gadget>\n11<\/output>\n11<\/result>","index":4043} +{"problem":"a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 480 . find the sum ?","rationale":"\"let the sum be rs . x . ( x * 18 * 2 ) \/ 100 - ( x * 12 * 2 ) \/ 100 = 480 = > 36 x \/ 100 - 24 x \/ 100 = 480 = > 12 x \/ 100 = 480 = > x = 4000 . answer : c\"","correct":"c","options":{"a":"7000 ","b":"7029 ","c":"4000 ","d":"2800","e":"2791"},"options_float":{"a":7000.0,"b":7029.0,"c":4000.0,"d":2800.0,"e":2791.0},"annotated_formula":"divide(480, divide(multiply(subtract(18, 12), const_2), const_100))","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)|","chain":"18 - 12<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n480 \/ (3\/25)<\/gadget>\n4_000<\/output>\n4_000<\/result>","index":4044} +{"problem":"a train 200 m long can cross an electric pole in 20 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 200 \/ 20 s = 10 m \/ sec speed = 10 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 36 kmph answer : e\"","correct":"e","options":{"a":"987 ","b":"271 ","c":"72 ","d":"27","e":"36"},"options_float":{"a":987.0,"b":271.0,"c":72.0,"d":27.0,"e":36.0},"annotated_formula":"divide(divide(200, const_1000), divide(20, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"200 \/ 1_000<\/gadget>\n1\/5 = around 0.2<\/output>\n20 \/ 3_600<\/gadget>\n1\/180 = around 0.005556<\/output>\n(1\/5) \/ (1\/180)<\/gadget>\n36<\/output>\n36<\/result>","index":4047} +{"problem":"a fair coin is tossed 5 times . what is the probability that it lands heads up at least twice ?","rationale":"fair coin is tossed 5 times . hence total number of outcomes = 2 ^ 5 = 32 . problem asks for probability of getting atleast heads twice . hence if we calculate probability of getting heads exactly once and probability of not getting heads at all and subract it from the total probability of the event which is 1 ( as total probability of certain event will be always 1 ) we can get the probability of getting atleast heads twice . probability of getting exactly one head and no heads = ( number of possible outcomes [ htttt , thttt , tthtt , tttht , tttth , ttttt ] = 6 ) \/ ( total possible outcomes = 32 ) = > 6 \/ 32 = 3 \/ 16 hence probability of getting atleast heads twice = 1 - ( 3 \/ 16 ) = 13 \/ 16 = > choice [ d ]","correct":"d","options":{"a":"1 \/ 16 ","b":"5 \/ 16 ","c":"2 \/ 5 ","d":"13 \/ 16","e":"27 \/ 32"},"options_float":{"a":0.0625,"b":0.3125,"c":0.4,"d":0.8125,"e":0.84375},"annotated_formula":"divide(add(add(add(choose(5, const_2), choose(5, const_3)), choose(5, const_4)), choose(5, 5)), power(const_2, 5))","linear_formula":"choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)","chain":"binomial(5, 2)<\/gadget>\n10<\/output>\nbinomial(5, 3)<\/gadget>\n10<\/output>\n10 + 10<\/gadget>\n20<\/output>\nbinomial(5, 4)<\/gadget>\n5<\/output>\n20 + 5<\/gadget>\n25<\/output>\nbinomial(5, 5)<\/gadget>\n1<\/output>\n25 + 1<\/gadget>\n26<\/output>\n2 ** 5<\/gadget>\n32<\/output>\n26 \/ 32<\/gadget>\n13\/16 = around 0.8125<\/output>\n13\/16 = around 0.8125<\/result>","index":4048} +{"problem":"what is 3 \/ 4 of 4 \/ 5 of 2 \/ 3 ?","rationale":"\"3 \/ 4 * 4 \/ 5 * 2 \/ 3 = 2 \/ 5 answer : b\"","correct":"b","options":{"a":"1 \/ 4 ","b":"2 \/ 5 ","c":"9 \/ 16 ","d":"5 \/ 8","e":"16 \/ 9"},"options_float":{"a":0.25,"b":0.4,"c":0.5625,"d":0.625,"e":1.7777777778},"annotated_formula":"multiply(divide(3, 4), multiply(divide(4, 5), divide(2, 3)))","linear_formula":"divide(n0,n1)|divide(n1,n3)|divide(n4,n0)|multiply(#1,#2)|multiply(#0,#3)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(4\/5) * (2\/3)<\/gadget>\n8\/15 = around 0.533333<\/output>\n(3\/4) * (8\/15)<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":4050} +{"problem":"if the numerator of a fraction is increased by 20 % and its denominator is diminished by 25 % value of the fraction is 2 \/ 15 . find the original fraction ?","rationale":"\"x * ( 120 \/ 100 ) - - - - - - - - - - - - - - - - = 2 \/ 15 y * ( 75 \/ 100 ) x \/ y = 1 \/ 12 answer : a\"","correct":"a","options":{"a":"1 \/ 12 ","b":"1 \/ 16 ","c":"1 \/ 19 ","d":"1 \/ 14","e":"1 \/ 13"},"options_float":{"a":0.0833333333,"b":0.0625,"c":0.0526315789,"d":0.0714285714,"e":0.0769230769},"annotated_formula":"divide(divide(2, 15), divide(divide(add(20, const_100), const_100), divide(subtract(const_100, 25), const_100)))","linear_formula":"add(n0,const_100)|divide(n2,n3)|subtract(const_100,n1)|divide(#0,const_100)|divide(#2,const_100)|divide(#3,#4)|divide(#1,#5)|","chain":"2 \/ 15<\/gadget>\n2\/15 = around 0.133333<\/output>\n20 + 100<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n(6\/5) \/ (3\/4)<\/gadget>\n8\/5 = around 1.6<\/output>\n(2\/15) \/ (8\/5)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1\/12 = around 0.083333<\/result>","index":4051} +{"problem":"when w is divided by 13 , the reminder is 0 . if w is 3 more than it value and when divided by 11 its remainder is 0 . what is the value of w ?","rationale":"w is divided by 13 so that is multiple of 14 as 13 , 26,39 . . . w + 3 is divided by 11 the remainder is 0 so it is divisible by 11 . consider from option let us take the number is 13 it is divisible by 14 but 13 + 3 is not divisible by 11 so it is not answers so let us take 2 nd option 52 which is divisible by 13 and 52 + 3 = 55 is divisible by 11 so ans is b","correct":"b","options":{"a":"13 ","b":"52 ","c":"39 ","d":"26","e":"35"},"options_float":{"a":13.0,"b":52.0,"c":39.0,"d":26.0,"e":35.0},"annotated_formula":"multiply(13, const_4)","linear_formula":"multiply(n0,const_4)","chain":"13 * 4<\/gadget>\n52<\/output>\n52<\/result>","index":4052} +{"problem":"a high school has 400 students 1 \/ 2 attend the airthmetic club , 5 \/ 8 attend the biology club and 3 \/ 4 attend the chemistry club . 1 \/ 8 attend all 3 clubs . if every student attends at least one club how many students attend exactly 2 clubs .","rationale":"basically , this question is asking you to figure out how many students are being double - counted . a - club has 200 members ( 1 \/ 2 of 400 ) b - club has 250 members ( 5 \/ 8 of 400 ) c - club has 300 members ( 3 \/ 4 of 400 ) we can create an equation to solve this : 200 + 250 + 300 = n + x + 2 y where n is the number of students , x is the number of students in two clubs , and y is the number of students in three clubs . the question provides y for us ( 50 ) . 750 = 400 + x + 100 x = 250 c","correct":"c","options":{"a":"300 ","b":"200 ","c":"250 ","d":"150","e":"160"},"options_float":{"a":300.0,"b":200.0,"c":250.0,"d":150.0,"e":160.0},"annotated_formula":"subtract(subtract(add(multiply(400, divide(3, 4)), add(multiply(400, divide(1, 2)), multiply(400, divide(5, 8)))), multiply(2, multiply(400, divide(1, 8)))), 400)","linear_formula":"divide(n1,n2)|divide(n3,n4)|divide(n5,n6)|divide(n1,n4)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|add(#4,#5)|multiply(n2,#7)|add(#8,#6)|subtract(#10,#9)|subtract(#11,n0)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n400 * (3\/4)<\/gadget>\n300<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n400 * (1\/2)<\/gadget>\n200<\/output>\n5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n400 * (5\/8)<\/gadget>\n250<\/output>\n200 + 250<\/gadget>\n450<\/output>\n300 + 450<\/gadget>\n750<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n400 * (1\/8)<\/gadget>\n50<\/output>\n2 * 50<\/gadget>\n100<\/output>\n750 - 100<\/gadget>\n650<\/output>\n650 - 400<\/gadget>\n250<\/output>\n250<\/result>","index":4054} +{"problem":"a is two years older than b who is twice as old as c . if the total of the ages of a , b nd c be 27 , then how old is b ?","rationale":"\"solution let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ∴ ( 2 x + 2 ) + 2 x + x = 27 ⇔ 5 x = 25 ⇔ x = 5 . hence , b ' s age = 2 x = 10 years . answer d\"","correct":"d","options":{"a":"7 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)|","chain":"27 - 2<\/gadget>\n25<\/output>\n25 * 2<\/gadget>\n50<\/output>\n4 + 1<\/gadget>\n5<\/output>\n50 \/ 5<\/gadget>\n10<\/output>\n10<\/result>","index":4056} +{"problem":"last year elaine spent 20 % of her annual earnings on rent . this year she earned 35 % more than last year and she spent 30 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ?","rationale":"\"for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 20 of this on rent . this year she earned 15 % more , or $ 135 . she would ' ve spent 30 % of this on rent , or $ 40.5 do $ 40.5 \/ $ 20 this will give you 202.5 % a is the correct answer .\"","correct":"a","options":{"a":"202.5 ","b":"194.5 ","c":"197.5 ","d":"182.5","e":"177.5"},"options_float":{"a":202.5,"b":194.5,"c":197.5,"d":182.5,"e":177.5},"annotated_formula":"multiply(divide(multiply(add(divide(35, const_100), const_1), 30), 20), const_100)","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n2,#1)|divide(#2,n0)|multiply(#3,const_100)|","chain":"35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n(7\/20) + 1<\/gadget>\n27\/20 = around 1.35<\/output>\n(27\/20) * 30<\/gadget>\n81\/2 = around 40.5<\/output>\n(81\/2) \/ 20<\/gadget>\n81\/40 = around 2.025<\/output>\n(81\/40) * 100<\/gadget>\n405\/2 = around 202.5<\/output>\n405\/2 = around 202.5<\/result>","index":4057} +{"problem":"an agent , gets a commission of 2.5 % on the sales of cloth . if on a certain day , he gets rs . 21 as commission , the cloth sold through him on that day is worth","rationale":"\"explanation : let the total sale be rs . x . then , 2.5 % . of x = 21 < = > ( 25 \/ 10 * 1 \/ 100 * x ) = 21 < = > x = 840 . answer : c\"","correct":"c","options":{"a":"333 ","b":"500 ","c":"840 ","d":"299","e":"132"},"options_float":{"a":333.0,"b":500.0,"c":840.0,"d":299.0,"e":132.0},"annotated_formula":"divide(21, divide(2.5, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,#0)|","chain":"2.5 \/ 100<\/gadget>\n0.025<\/output>\n21 \/ 0.025<\/gadget>\n840<\/output>\n840<\/result>","index":4059} +{"problem":"the average temperature for tuesday , wednesday and thursday was 32 ° c . the average temperature for wednesday , thursday and friday was 34 ° c . if the temperature on friday be 44 ° c , what was the temperature on tuesday ?","rationale":"\"explanation : t + w + t = 32 × 3 = 96 ° c w + t + f = 34 × 3 = 102 ° c also , temperature on friday = 44 ° c temperature on tuesday = 96 + 44 - 102 = 38 ° c answer : option c\"","correct":"c","options":{"a":"39 ° c ","b":"44 ° c ","c":"38 ° c ","d":"42 ° c","e":"none of these"},"options_float":{"a":39.0,"b":44.0,"c":38.0,"d":42.0,"e":null},"annotated_formula":"subtract(multiply(32, const_3), subtract(multiply(34, const_3), 44))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_3)|subtract(#1,n2)|subtract(#0,#2)|","chain":"32 * 3<\/gadget>\n96<\/output>\n34 * 3<\/gadget>\n102<\/output>\n102 - 44<\/gadget>\n58<\/output>\n96 - 58<\/gadget>\n38<\/output>\n38<\/result>","index":4060} +{"problem":"a cyclist bikes x distance at 10 miles per hour and returns over the same path at 9 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ?","rationale":"\"distance = d 1 = x miles speed = s 1 = 10 miles per hour time = t 1 = distance \/ speed = x \/ 10 2 . going from b to a distance = d 2 = x miles speed = s 2 = 9 miles per hour time = t 2 = distance \/ speed = x \/ 9 3 . average speed = total distance \/ total time total distance = x + x = 2 x total time = x \/ 10 + x \/ 9 = x ( 1 \/ 10 + 1 \/ 9 ) = x ( 4 + 5 ) \/ 40 = 19 x \/ 90 speed = 2 x \/ ( 19 x \/ 90 ) = ( 2 x * 90 ) \/ 9 x = 180 x \/ 19 x = 180 \/ 19 = 9.5 answer : e\"","correct":"e","options":{"a":"8.1 ","b":"8.3 ","c":"8.6 ","d":"8.9","e":"9.5"},"options_float":{"a":8.1,"b":8.3,"c":8.6,"d":8.9,"e":9.5},"annotated_formula":"divide(add(10, 9), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"10 + 9<\/gadget>\n19<\/output>\n19 \/ 2<\/gadget>\n19\/2 = around 9.5<\/output>\n19\/2 = around 9.5<\/result>","index":4061} +{"problem":"the diagonal of a cube is 8 √ 3 . find its volume and surface area .","rationale":"√ 3 . a = 8 √ 3 = > a = 8 volume = a ( power ) 3 = > ( 8 × 8 × 8 ) cm ( power ) 3 = > 512 cm ( power ) 3 surface area = > 6 a ( power ) 2 = > ( 6 × 8 × 8 ) cm ( power ) 2 = > 384 cm ( power ) 2 answer is b .","correct":"b","options":{"a":"354 ","b":"384 ","c":"324 ","d":"344","e":"364"},"options_float":{"a":354.0,"b":384.0,"c":324.0,"d":344.0,"e":364.0},"annotated_formula":"surface_cube(divide(multiply(8, sqrt(3)), sqrt(3)))","linear_formula":"sqrt(n1)|multiply(n0,#0)|divide(#1,#0)|surface_cube(#2)","chain":"3 ** (1\/2)<\/gadget>\nsqrt(3) = around 1.732051<\/output>\n8 * (sqrt(3))<\/gadget>\n8*sqrt(3) = around 13.856406<\/output>\n(8*sqrt(3)) \/ (sqrt(3))<\/gadget>\n8<\/output>\n6 * (8 ** 2)<\/gadget>\n384<\/output>\n384<\/result>","index":4062} +{"problem":"a group of boy scouts and girls scouts is going on a rafting trip . 70 % of the scouts arrived with signed permission slips . if 60 % of the scouts were boy scouts and 75 % of the boy scouts arrived with signed permission slips , then what percentage of the girl scouts arrived with signed permission slips ? round to the nearest percent .","rationale":"60 % were boy scouts so 40 % ( 100 - 60 = 40 ) were girl scouts . # of boy scouts with permission slips signed + # of girl scouts with permission slips signed = total # with permission slip signed ( 75 % of 60 % of the total going ) + ( ? % of 40 % of the total going ) = 70 % of the total going we can let the ` ` total going ' ' = 1,000 arbitrarily since we only care about the percent ( of girl scouts w \/ permission signed ) . 75 * . 60 * 1,000 + x * . 40 * 1,000 = . 70 * 1,000 450 + x * 400 = 700 400 x = 250 x = 250 \/ 400 x = . 625 converting this to a percent , we have 62.5 % . rounding , we have 63 % . c","correct":"c","options":{"a":"50 % ","b":"60 % ","c":"63 % ","d":"66 %","e":"70 %"},"options_float":{"a":50.0,"b":60.0,"c":63.0,"d":66.0,"e":70.0},"annotated_formula":"multiply(divide(subtract(70, divide(multiply(60, 75), const_100)), subtract(const_100, 60)), const_100)","linear_formula":"multiply(n1,n2)|subtract(const_100,n1)|divide(#0,const_100)|subtract(n0,#2)|divide(#3,#1)|multiply(#4,const_100)","chain":"60 * 75<\/gadget>\n4_500<\/output>\n4_500 \/ 100<\/gadget>\n45<\/output>\n70 - 45<\/gadget>\n25<\/output>\n100 - 60<\/gadget>\n40<\/output>\n25 \/ 40<\/gadget>\n5\/8 = around 0.625<\/output>\n(5\/8) * 100<\/gadget>\n125\/2 = around 62.5<\/output>\n125\/2 = around 62.5<\/result>","index":4064} +{"problem":"the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 216 ° of the circle , what percentage of megatek employees are in manufacturing ?","rationale":"\"answer : e 216 ° divided by 360 ° equals 0.6 , therefore the sector is equal to 60 % of the total\"","correct":"e","options":{"a":"20 % ","b":"25 % ","c":"30 % ","d":"35 %","e":"60 %"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":35.0,"e":60.0},"annotated_formula":"multiply(divide(216, divide(const_3600, const_10)), const_100)","linear_formula":"divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)|","chain":"3_600 \/ 10<\/gadget>\n360<\/output>\n216 \/ 360<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n60<\/result>","index":4065} +{"problem":"in a school of 850 boys , 34 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?","rationale":"\"34 + 28 + 10 = 72 % 100 – 72 = 28 % 850 * 28 \/ 100 = 238 answer : d\"","correct":"d","options":{"a":"125 ","b":"627 ","c":"153 ","d":"238","e":"159"},"options_float":{"a":125.0,"b":627.0,"c":153.0,"d":238.0,"e":159.0},"annotated_formula":"divide(multiply(850, subtract(const_100, add(add(34, 28), 10))), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)|","chain":"34 + 28<\/gadget>\n62<\/output>\n62 + 10<\/gadget>\n72<\/output>\n100 - 72<\/gadget>\n28<\/output>\n850 * 28<\/gadget>\n23_800<\/output>\n23_800 \/ 100<\/gadget>\n238<\/output>\n238<\/result>","index":4066} +{"problem":"2 ^ 46655 mod 9 = ?","rationale":"by remainder theorem remainder = { [ ( 2 ^ 3 ) ^ 1551 * 2 ^ 2 ] \/ 9 } = [ { ( - 1 ) ^ 1551 * 4 } \/ 9 ] = 5 answer : b","correct":"b","options":{"a":"4 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"reminder(power(2, add(const_2, const_3)), 9)","linear_formula":"add(const_2,const_3)|power(n0,#0)|reminder(#1,n2)","chain":"2 + 3<\/gadget>\n5<\/output>\n2 ** 5<\/gadget>\n32<\/output>\n32 % 9<\/gadget>\n5<\/output>\n5<\/result>","index":4068} +{"problem":"a cube has five of its faces painted half red and half white . the other face is completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ?","rationale":"let x be the area of each face of the cube . the area painted red is 5 ( x \/ 2 ) = 5 x \/ 2 the area painted white is 5 ( x \/ 2 ) + x = 7 x \/ 2 x the ratio of red to white is 5 x \/ 2 : 7 x \/ 2 which is 5 : 7 . the answer is d .","correct":"d","options":{"a":"6 : 11 ","b":"2 : 5 ","c":"3 : 4 ","d":"5 : 7","e":"2 : 3"},"options_float":{"a":0.5454545455,"b":0.4,"c":0.75,"d":0.7142857143,"e":0.6666666667},"annotated_formula":"divide(multiply(multiply(add(const_1, const_4), divide(const_1, const_2)), const_2), multiply(add(multiply(add(const_1, const_4), divide(const_1, const_2)), const_1), const_2))","linear_formula":"add(const_1,const_4)|divide(const_1,const_2)|multiply(#0,#1)|add(#2,const_1)|multiply(#2,const_2)|multiply(#3,const_2)|divide(#4,#5)","chain":"1 + 4<\/gadget>\n5<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n5 * (1\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) * 2<\/gadget>\n5<\/output>\n(5\/2) + 1<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) * 2<\/gadget>\n7<\/output>\n5 \/ 7<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":4069} +{"problem":"the h . c . f . of two numbers is 25 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 25 x 13 ) and ( 25 x 14 ) . larger number = ( 25 x 14 ) = 350 . answer : option e\"","correct":"e","options":{"a":"276 ","b":"299 ","c":"322 ","d":"345","e":"350"},"options_float":{"a":276.0,"b":299.0,"c":322.0,"d":345.0,"e":350.0},"annotated_formula":"multiply(25, 14)","linear_formula":"multiply(n0,n2)|","chain":"25 * 14<\/gadget>\n350<\/output>\n350<\/result>","index":4070} +{"problem":"increasing the original price of a certain item by 25 percent and then increasing the new price by 30 percent is equivalent to increasing the original price by what percent ?","rationale":"we ' re told that the original price of an item is increased by 25 % and then that price is increased by 30 % . . . . if . . . . starting value = $ 100 + 25 % = 100 + . 25 ( 100 ) = 125 + 30 % = 125 + . 30 ( 125 ) = 125 + 37.5 = 162.5 the question asks how the final price relates to the original price . this is essentially about percentage change , which means we should use the percentage change formula : percentage change = ( new - old ) \/ old = difference \/ original doing either calculation will yield the same result : 62.5 \/ 100 = 62.50 % final answer : e","correct":"e","options":{"a":"31.25 ","b":"37.5 ","c":"50.0 ","d":"52.5","e":"62.5"},"options_float":{"a":31.25,"b":37.5,"c":50.0,"d":52.5,"e":62.5},"annotated_formula":"multiply(subtract(multiply(add(divide(30, const_100), const_1), add(divide(25, const_100), const_1)), const_1), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) + 1<\/gadget>\n13\/10 = around 1.3<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(13\/10) * (5\/4)<\/gadget>\n13\/8 = around 1.625<\/output>\n(13\/8) - 1<\/gadget>\n5\/8 = around 0.625<\/output>\n(5\/8) * 100<\/gadget>\n125\/2 = around 62.5<\/output>\n125\/2 = around 62.5<\/result>","index":4071} +{"problem":"the population of a bacteria colony doubles every day . if it was started 10 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ?","rationale":"\"2 ^ 10 ( 2 ) = 2 ^ 11 = 2048 the answer is d .\"","correct":"d","options":{"a":"512 ","b":"768 ","c":"1024 ","d":"2048","e":"4096"},"options_float":{"a":512.0,"b":768.0,"c":1024.0,"d":2048.0,"e":4096.0},"annotated_formula":"subtract(power(2, add(10, const_1)), const_1)","linear_formula":"add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|","chain":"10 + 1<\/gadget>\n11<\/output>\n2 ** 11<\/gadget>\n2_048<\/output>\n2_048 - 1<\/gadget>\n2_047<\/output>\n2_047<\/result>","index":4072} +{"problem":"a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 2 per unit . further , the manufacturer has costs of $ 16,200 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ?","rationale":"\"this question can be solved easily using a equation model . 150 * p = 16200 ( fixed cost ) + 150 * 80 + 150 * 2 . = 16200 + 12000 + 300 = 28500 p = $ 190 . d\"","correct":"d","options":{"a":"$ 28 ","b":"$ 82 ","c":"$ 110 ","d":"$ 190","e":"$ 192"},"options_float":{"a":28.0,"b":82.0,"c":110.0,"d":190.0,"e":192.0},"annotated_formula":"divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(2, const_100)), multiply(add(80, 2), 150)), 150)","linear_formula":"add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16 * 1_000<\/gadget>\n16_000<\/output>\n2 * 100<\/gadget>\n200<\/output>\n16_000 + 200<\/gadget>\n16_200<\/output>\n80 + 2<\/gadget>\n82<\/output>\n82 * 150<\/gadget>\n12_300<\/output>\n16_200 + 12_300<\/gadget>\n28_500<\/output>\n28_500 \/ 150<\/gadget>\n190<\/output>\n190<\/result>","index":4073} +{"problem":"in a sports club with 42 members , 20 play badminton and 23 play tennis and 6 do not play either . how many members play both badminton and tennis ?","rationale":"\"20 + 23 = 43 but where as total number is 42 - 6 = 36 therefore answer is 43 - 36 = 7 hence answer is d\"","correct":"d","options":{"a":"3 ","b":"5 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"subtract(add(add(20, 23), 6), 42)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)|","chain":"20 + 23<\/gadget>\n43<\/output>\n43 + 6<\/gadget>\n49<\/output>\n49 - 42<\/gadget>\n7<\/output>\n7<\/result>","index":4075} +{"problem":"a tempo , while travelling at 70 kmph , uses 30 % more petrol than it uses while travelling at 50 kmph for travelling a certain distance . using one litre of petrol , it can travel 19.5 km at 50 kmph . how far can the tempo travel on 10 litres of petrol at a speed of 70 kmph ?","rationale":"@ 50 kmph , one km distance uses 1 \/ 19.5 ltr petrol . @ 70 kmph , one km dist . uses 1.3 \/ 19.5 ltr petrol distance travelled with 10 ltrs @ 70 kmph = 10 * 19.5 \/ 1.3 = 150 kms . answer : a","correct":"a","options":{"a":"150 kms ","b":"160 kms ","c":"170 kms ","d":"180 kms","e":"190 kms"},"options_float":{"a":150.0,"b":160.0,"c":170.0,"d":180.0,"e":190.0},"annotated_formula":"divide(multiply(10, 19.5), add(const_1, divide(30, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n3,n5)|add(#0,const_1)|divide(#1,#2)","chain":"10 * 19.5<\/gadget>\n195<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n195 \/ (13\/10)<\/gadget>\n150<\/output>\n150<\/result>","index":4076} +{"problem":"in an examination it is necessary for a candidate to get 45 % of the maximum marks to pass . a candidate who gets 180 marks , fails by 45 marks . find the maximum marks .","rationale":"let max . mark = x pass mark = 45 x \/ 100 180 = 45 x \/ 100 - 45 x = 180 * 100 - 4500 \/ 45 = 500 answer : c","correct":"c","options":{"a":"450 ","b":"600 ","c":"500 ","d":"550","e":"650"},"options_float":{"a":450.0,"b":600.0,"c":500.0,"d":550.0,"e":650.0},"annotated_formula":"divide(add(180, 45), divide(45, const_100))","linear_formula":"add(n0,n1)|divide(n0,const_100)|divide(#0,#1)","chain":"180 + 45<\/gadget>\n225<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n225 \/ (9\/20)<\/gadget>\n500<\/output>\n500<\/result>","index":4077} +{"problem":"in a box of 9 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?","rationale":"p ( neither pen is defective ) = 6 \/ 9 * 5 \/ 8 = 5 \/ 12 the answer is c .","correct":"c","options":{"a":"3 \/ 10 ","b":"4 \/ 11 ","c":"5 \/ 12 ","d":"6 \/ 13","e":"7 \/ 15"},"options_float":{"a":0.3,"b":0.3636363636,"c":0.4166666667,"d":0.4615384615,"e":0.4666666667},"annotated_formula":"multiply(divide(subtract(9, 3), 9), divide(subtract(subtract(9, 3), const_1), subtract(9, const_1)))","linear_formula":"subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)","chain":"9 - 3<\/gadget>\n6<\/output>\n6 \/ 9<\/gadget>\n2\/3 = around 0.666667<\/output>\n6 - 1<\/gadget>\n5<\/output>\n9 - 1<\/gadget>\n8<\/output>\n5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n(2\/3) * (5\/8)<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":4078} +{"problem":"find a two digit number , given that the sum of the digits is 9 and the difference of the digits is 3 . ?","rationale":"using elimination method find which of the options fit the description of the number . . . from the option only 63 meets this description sum of digits - - - 6 + 3 = 9 difference of digits - - - 6 - 3 = 3 answer b .","correct":"b","options":{"a":"75 ","b":"63 ","c":"83 ","d":"94","e":"39"},"options_float":{"a":75.0,"b":63.0,"c":83.0,"d":94.0,"e":39.0},"annotated_formula":"multiply(multiply(const_3, const_3), add(const_3, const_4))","linear_formula":"add(const_3,const_4)|multiply(const_3,const_3)|multiply(#0,#1)","chain":"3 * 3<\/gadget>\n9<\/output>\n3 + 4<\/gadget>\n7<\/output>\n9 * 7<\/gadget>\n63<\/output>\n63<\/result>","index":4080} +{"problem":"fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 25 percent , what is the discount rate on pony jeans ?","rationale":"\"x discount on pony jeans , ( 0.25 - x ) discount on fox jeans . set the equation : 3 * 15 ( 0.25 - x ) + 2 * 18 x = 9 - - > x = 0.25 = 25 % answer : e .\"","correct":"e","options":{"a":"9 % ","b":"10 % ","c":"11 % ","d":"12 %","e":"25 %"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":12.0,"e":25.0},"annotated_formula":"multiply(subtract(divide(25, const_100), divide(subtract(9, multiply(divide(25, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100)","linear_formula":"divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n18 * 2<\/gadget>\n36<\/output>\n(1\/4) * 36<\/gadget>\n9<\/output>\n9 - 9<\/gadget>\n0<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 - 36<\/gadget>\n9<\/output>\n0 \/ 9<\/gadget>\n0<\/output>\n(1\/4) - 0<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":4081} +{"problem":"if x is an integer and 2.13 × 10 ^ x is less than 2100 , what is the greatest possible value for x ?","rationale":"\"if x = 3 2.13 × 10 ^ 3 = 2130 > 2,100 so , x = 2 answer : a\"","correct":"a","options":{"a":"2 ","b":"6 ","c":"5 ","d":"4","e":"3"},"options_float":{"a":2.0,"b":6.0,"c":5.0,"d":4.0,"e":3.0},"annotated_formula":"floor(divide(log(divide(2100, 2.13)), log(10)))","linear_formula":"divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3)|","chain":"2_100 \/ 2.13<\/gadget>\n985.915493<\/output>\nlog(985.915493)<\/gadget>\n6.893571<\/output>\nlog(10)<\/gadget>\nlog(10) = around 2.302585<\/output>\n6.893571 \/ log(10)<\/gadget>\n6.893571\/log(10) = around 2.99384<\/output>\nfloor(6.893571\/log(10))<\/gadget>\n2<\/output>\n2<\/result>","index":4083} +{"problem":"if 5 % more is gained by selling an article for rs . 360 than by selling it for rs . 340 , the cost of the article is ?","rationale":"\"let c . p . be rs . x . then , 5 % of x = 360 - 340 = 20 x \/ 20 = 20 = > x = 400 answer : c\"","correct":"c","options":{"a":"127 ","b":"688 ","c":"400 ","d":"121","e":"123"},"options_float":{"a":127.0,"b":688.0,"c":400.0,"d":121.0,"e":123.0},"annotated_formula":"divide(subtract(360, 340), divide(5, const_100))","linear_formula":"divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)|","chain":"360 - 340<\/gadget>\n20<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n20 \/ (1\/20)<\/gadget>\n400<\/output>\n400<\/result>","index":4084} +{"problem":"3 candidates in an election and received 6136 , 7636 and 11628 votes respectively . what % of the total votes did the winning candidate got in that election ?","rationale":"\"total number of votes polled = ( 6136 + 7636 + 11628 ) = 25400 so , required percentage = 11628 \/ 25400 * 100 = 46 % a\"","correct":"a","options":{"a":"46 % ","b":"49 % ","c":"57 % ","d":"59 %","e":"61 %"},"options_float":{"a":46.0,"b":49.0,"c":57.0,"d":59.0,"e":61.0},"annotated_formula":"multiply(divide(11628, add(add(6136, 7636), 11628)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)|","chain":"6_136 + 7_636<\/gadget>\n13_772<\/output>\n13_772 + 11_628<\/gadget>\n25_400<\/output>\n11_628 \/ 25_400<\/gadget>\n2_907\/6_350 = around 0.457795<\/output>\n(2_907\/6_350) * 100<\/gadget>\n5_814\/127 = around 45.779528<\/output>\n5_814\/127 = around 45.779528<\/result>","index":4088} +{"problem":"the ratio of males to females in a class is 2 : 3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one - fourth of the males and one - half of the females in the class ?","rationale":"1 \/ 4 * 2 \/ 5 + 1 \/ 2 * 3 \/ 5 = 2 \/ 20 + 6 \/ 20 = 2 \/ 5 the number of degrees is 2 \/ 5 * 360 = 144 degrees the answer is c .","correct":"c","options":{"a":"128 ","b":"136 ","c":"144 ","d":"152","e":"160"},"options_float":{"a":128.0,"b":136.0,"c":144.0,"d":152.0,"e":160.0},"annotated_formula":"divide(multiply(2, const_360), add(2, 3))","linear_formula":"add(n0,n1)|multiply(n0,const_360)|divide(#1,#0)","chain":"2 * 360<\/gadget>\n720<\/output>\n2 + 3<\/gadget>\n5<\/output>\n720 \/ 5<\/gadget>\n144<\/output>\n144<\/result>","index":4089} +{"problem":"300 first - time customers of a fashion store were surveyed for their shopping experience right after leaving the store . 60 % of the customers in the survey had purchased clothes for less than $ 100 . 40 % of the customers in the survey reported they were overall satisfied with their purchase . 45 % of the customers that had purchased clothes for less than $ 100 reported they were overall satisfied with their purchase . what percent of the customers surveyed purchased clothes for at least $ 100 and reported that they were not overall satisfied with their purchase ?","rationale":"\"out of 300 - 180 purchased for less than 100 $ 120 for more out of 300 - 120 responded as satisfied and 180 responded disatisfied out of 180 ( purchased less than 100 $ ) - 45 % = 81 responded as satisfied , so remaining satisfied are 120 - 81 = 39 so 39 is what percentage of 300 - 13 % so the answer should be c\"","correct":"c","options":{"a":"15 ","b":"25 ","c":"13 ","d":"45","e":"75"},"options_float":{"a":15.0,"b":25.0,"c":13.0,"d":45.0,"e":75.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(40, 100), 300), multiply(divide(45, 100), multiply(300, divide(60, 100)))), 300), 100)","linear_formula":"divide(n3,n2)|divide(n4,n2)|divide(n1,n2)|multiply(n0,#0)|multiply(n0,#2)|multiply(#1,#4)|subtract(#3,#5)|divide(#6,n0)|multiply(#7,n2)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 300<\/gadget>\n120<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n300 * (3\/5)<\/gadget>\n180<\/output>\n(9\/20) * 180<\/gadget>\n81<\/output>\n120 - 81<\/gadget>\n39<\/output>\n39 \/ 300<\/gadget>\n13\/100 = around 0.13<\/output>\n(13\/100) * 100<\/gadget>\n13<\/output>\n13<\/result>","index":4091} +{"problem":"solution a is 30 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ?","rationale":"\"forget the volumes for the time being . you have to mix 30 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 \/ w 2 = ( a 2 - aavg ) \/ ( aavg - a 1 ) w 1 \/ w 2 = ( 80 - 50 ) \/ ( 50 - 30 ) = 3 \/ 2 so the volume of the two solutions will be equal . answer has to be 3 : 2 e\"","correct":"e","options":{"a":"6 : 4 ","b":"6 : 14 ","c":"4 : 4 ","d":"4 : 6","e":"3 : 2"},"options_float":{"a":1.5,"b":0.4285714286,"c":1.0,"d":0.6666666667,"e":1.5},"annotated_formula":"divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(30, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(30, const_100)))))","linear_formula":"divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n50 * (4\/5)<\/gadget>\n40<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n50 * (1\/2)<\/gadget>\n25<\/output>\n40 - 25<\/gadget>\n15<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(4\/5) - (3\/10)<\/gadget>\n1\/2 = around 0.5<\/output>\n15 \/ (1\/2)<\/gadget>\n30<\/output>\n50 - 30<\/gadget>\n20<\/output>\n30 \/ 20<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":4093} +{"problem":"the monthly rent of a shop of dimension 18 feet × 22 feet is rs . 2244 . what is the annual rent per square foot of the shop ? ( a ) rs . 5 ( 2 \/ 3 )","rationale":"\"sol . monthly rent per square feet = 2244 \/ ( 18 * 22 ) & annual rent per square feet = 12 * ( 2244 \/ ( 18 * 22 ) ) = 68 c\"","correct":"c","options":{"a":"43 ","b":"56 ","c":"68 ","d":"87","e":"92"},"options_float":{"a":43.0,"b":56.0,"c":68.0,"d":87.0,"e":92.0},"annotated_formula":"multiply(add(const_10, const_2), divide(2244, rectangle_area(18, 22)))","linear_formula":"add(const_10,const_2)|rectangle_area(n0,n1)|divide(n2,#1)|multiply(#0,#2)|","chain":"10 + 2<\/gadget>\n12<\/output>\n18 * 22<\/gadget>\n396<\/output>\n2_244 \/ 396<\/gadget>\n17\/3 = around 5.666667<\/output>\n12 * (17\/3)<\/gadget>\n68<\/output>\n68<\/result>","index":4097} +{"problem":"mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 135 mangoes .","rationale":"mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 \/ 108 = 3.33 original price of 135 mangoes = 135 * 3.33 = 450 . answer d","correct":"d","options":{"a":"360 ","b":"380 ","c":"400 ","d":"450","e":"412"},"options_float":{"a":360.0,"b":380.0,"c":400.0,"d":450.0,"e":412.0},"annotated_formula":"multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, const_10))), 360), 12), 135)","linear_formula":"divide(const_1,const_10)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|divide(#3,n2)|multiply(n3,#4)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n360 \/ (9\/10)<\/gadget>\n400<\/output>\n400 - 360<\/gadget>\n40<\/output>\n40 \/ 12<\/gadget>\n10\/3 = around 3.333333<\/output>\n(10\/3) * 135<\/gadget>\n450<\/output>\n450<\/result>","index":4099} +{"problem":"a gambler has won 40 % of his 20 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 80 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ?","rationale":"\"let x be the number of additional games the gambler needs to play . 0.4 ( 20 ) + 0.8 x = 0.6 ( x + 20 ) 0.2 x = 4 x = 20 the answer is a .\"","correct":"a","options":{"a":"20 ","b":"25 ","c":"30 ","d":"35","e":"40"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"divide(subtract(multiply(20, divide(60, const_100)), multiply(20, divide(40, const_100))), subtract(divide(80, const_100), divide(60, const_100)))","linear_formula":"divide(n3,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n20 * (3\/5)<\/gadget>\n12<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n20 * (2\/5)<\/gadget>\n8<\/output>\n12 - 8<\/gadget>\n4<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) - (3\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n4 \/ (1\/5)<\/gadget>\n20<\/output>\n20<\/result>","index":4100} +{"problem":"the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 60 feet wide by 100 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ?","rationale":"\"volume of pool = 60 * 100 * 10 cu . ft , 80 % full = 60 * 100 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft \/ min therefore time taken = 60 * 100 * 10 * 0.8 \/ 60 min = 800 min d\"","correct":"d","options":{"a":"1000 ","b":"1400 ","c":"1200 ","d":"800","e":"1000"},"options_float":{"a":1000.0,"b":1400.0,"c":1200.0,"d":800.0,"e":1000.0},"annotated_formula":"divide(multiply(divide(80, const_100), multiply(multiply(60, 100), 10)), 60)","linear_formula":"divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n60 * 100<\/gadget>\n6_000<\/output>\n6_000 * 10<\/gadget>\n60_000<\/output>\n(4\/5) * 60_000<\/gadget>\n48_000<\/output>\n48_000 \/ 60<\/gadget>\n800<\/output>\n800<\/result>","index":4101} +{"problem":"walking at 4 \/ 5 of her normal speed , a worker is 10 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is","rationale":"\"let v be her normal speed and let t be her normal time . d = ( 4 \/ 5 ) v * ( t + 10 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 4 \/ 5 ) v * ( t + 10 ) t \/ 5 = 8 t = 40 the answer is c .\"","correct":"c","options":{"a":"30 ","b":"35 ","c":"40 ","d":"45","e":"50"},"options_float":{"a":30.0,"b":35.0,"c":40.0,"d":45.0,"e":50.0},"annotated_formula":"multiply(4, 10)","linear_formula":"multiply(n0,n2)|","chain":"4 * 10<\/gadget>\n40<\/output>\n40<\/result>","index":4102} +{"problem":"find the length of the longest pole that can be placed in a room 12 m long , 8 m broad and 9 m high .","rationale":"solution length of longest pole = length of the diagonal of the room = √ ( 12 ) 2 + 82 + 92 ‹ = › √ 289 = 17 m . answer c","correct":"c","options":{"a":"10 m ","b":"12 m ","c":"17 m ","d":"19 m","e":"none"},"options_float":{"a":10.0,"b":12.0,"c":17.0,"d":19.0,"e":null},"annotated_formula":"sqrt(add(power(9, const_2), add(power(12, const_2), power(8, const_2))))","linear_formula":"power(n0,const_2)|power(n1,const_2)|power(n2,const_2)|add(#0,#1)|add(#3,#2)|sqrt(#4)","chain":"9 ** 2<\/gadget>\n81<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n144 + 64<\/gadget>\n208<\/output>\n81 + 208<\/gadget>\n289<\/output>\n289 ** (1\/2)<\/gadget>\n17<\/output>\n17<\/result>","index":4103} +{"problem":"a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 4 m , the volume of the box ( in m 3 ) is :","rationale":"\"clearly , l = ( 48 - 8 ) m = 40 m , b = ( 36 - 8 ) m = 28 m , h = 8 m . volume of the box = ( 40 x 28 x 4 ) m 3 = 4480 m 3 . answer : option a\"","correct":"a","options":{"a":"4480 ","b":"5120 ","c":"6420 ","d":"8960","e":"7960"},"options_float":{"a":4480.0,"b":5120.0,"c":6420.0,"d":8960.0,"e":7960.0},"annotated_formula":"volume_rectangular_prism(subtract(48, multiply(4, const_2)), subtract(36, multiply(4, const_2)), 4)","linear_formula":"multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|","chain":"4 * 2<\/gadget>\n8<\/output>\n48 - 8<\/gadget>\n40<\/output>\n36 - 8<\/gadget>\n28<\/output>\n40 * 28 * 4<\/gadget>\n4_480<\/output>\n4_480<\/result>","index":4104} +{"problem":"a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 15 days , a finished the remaining work . in how many days a alone can finish the job ?","rationale":"\"a + b 10 days work = 10 * 1 \/ 40 = 1 \/ 4 remaining work = 1 - 1 \/ 4 = 3 \/ 4 3 \/ 4 work is done by a in 15 days whole work will be done by a in 15 * 4 \/ 3 = 20 days answer is e\"","correct":"e","options":{"a":"10 ","b":"25 ","c":"60 ","d":"30","e":"20"},"options_float":{"a":10.0,"b":25.0,"c":60.0,"d":30.0,"e":20.0},"annotated_formula":"divide(multiply(15, 40), subtract(40, 10))","linear_formula":"multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)|","chain":"15 * 40<\/gadget>\n600<\/output>\n40 - 10<\/gadget>\n30<\/output>\n600 \/ 30<\/gadget>\n20<\/output>\n20<\/result>","index":4106} +{"problem":"if the radius of a cylinder is made 3 times and height is doubled , what is the new volume of the cylinder divided by the old one ?","rationale":"let v and v ' be the original and the changed volume now v = pir ^ 2 h v ' = pi ( 3 r ) ^ 2 ( 2 h ) v ' = 18 v a ) 18","correct":"a","options":{"a":"18 . ","b":"2 . ","c":"6 . ","d":"4 .","e":"10 ."},"options_float":{"a":18.0,"b":2.0,"c":6.0,"d":4.0,"e":10.0},"annotated_formula":"add(multiply(3, const_4), multiply(const_3, const_2))","linear_formula":"multiply(n0,const_4)|multiply(const_2,const_3)|add(#0,#1)","chain":"3 * 4<\/gadget>\n12<\/output>\n3 * 2<\/gadget>\n6<\/output>\n12 + 6<\/gadget>\n18<\/output>\n18<\/result>","index":4107} +{"problem":"a train 110 m long is running with a speed of 82 km \/ hr . in what time will it pass a man who is running at 6 km \/ hr in the direction opposite to that in which the train is going ?","rationale":"\"speed of train relative to man = 82 + 6 = 88 km \/ hr . = 88 * 5 \/ 18 = 24.44 m \/ sec . time taken to pass the men = 110 \/ 24.44 = 4.5 sec . answer : d\"","correct":"d","options":{"a":"7 sec ","b":"6 sec ","c":"8 sec ","d":"4.5 sec","e":"2 sec"},"options_float":{"a":7.0,"b":6.0,"c":8.0,"d":4.5,"e":2.0},"annotated_formula":"divide(110, multiply(add(82, 6), const_0_2778))","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"82 + 6<\/gadget>\n88<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n88 * (5\/18)<\/gadget>\n220\/9 = around 24.444444<\/output>\n110 \/ (220\/9)<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":4108} +{"problem":"the average marks of a class of 35 students is 40 and that of another class of 45 students is 60 . find the average marks of all the students ?","rationale":"\"sum of the marks for the class of 35 students = 35 * 40 = 1400 sum of the marks for the class of 45 students = 45 * 60 = 2700 sum of the marks for the class of 80 students = 1400 + 2700 = 4100 average marks of all the students = 4100 \/ 80 = 51.2 . answer : d\"","correct":"d","options":{"a":"52.6 ","b":"52.4 ","c":"52.1 ","d":"51.2","e":"52.9"},"options_float":{"a":52.6,"b":52.4,"c":52.1,"d":51.2,"e":52.9},"annotated_formula":"divide(add(multiply(35, 40), multiply(45, 60)), add(35, 45))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"35 * 40<\/gadget>\n1_400<\/output>\n45 * 60<\/gadget>\n2_700<\/output>\n1_400 + 2_700<\/gadget>\n4_100<\/output>\n35 + 45<\/gadget>\n80<\/output>\n4_100 \/ 80<\/gadget>\n205\/4 = around 51.25<\/output>\n205\/4 = around 51.25<\/result>","index":4109} +{"problem":"abcd is a square where ab = √ 2016 . let x be a point on ab and y be a point on cd such that ax = cy . compute the area of trapezoid axyd .","rationale":"note that trapezoids axy d and bxy c are congruent , so the area of axy d is always 2016 \/ 2 = 1008 . correct answer c","correct":"c","options":{"a":"3008 ","b":"2002 ","c":"1008 ","d":"2016","e":"3000"},"options_float":{"a":3008.0,"b":2002.0,"c":1008.0,"d":2016.0,"e":3000.0},"annotated_formula":"divide(2016, const_2)","linear_formula":"divide(n0,const_2)","chain":"2_016 \/ 2<\/gadget>\n1_008<\/output>\n1_008<\/result>","index":4110} +{"problem":"the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 8600 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8600 ( 100 \/ 120 ) = rs . 7166 . answer : d\"","correct":"d","options":{"a":"rs . 6289 ","b":"rs . 6298 ","c":"rs . 6290 ","d":"rs . 7166","e":"rs . 7708"},"options_float":{"a":6289.0,"b":6298.0,"c":6290.0,"d":7166.0,"e":7708.0},"annotated_formula":"divide(8600, add(const_1, divide(20, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n8_600 \/ (6\/5)<\/gadget>\n21_500\/3 = around 7_166.666667<\/output>\n21_500\/3 = around 7_166.666667<\/result>","index":4112} +{"problem":"what is the perimeter , in meters , of a rectangular garden 16 meters wide that has the same area as a rectangular playground 16 meters long and 12 meters wide ?","rationale":"\"say length of garden is l meters . then ( l * 16 ) = 16 * 12 [ given area is same . area = l * b ] therefore l works out to 12 meters . perimeter of garden = 2 * ( 12 + 16 ) = 56 meter . answer is d\"","correct":"d","options":{"a":"48 ","b":"56 ","c":"60 ","d":"56","e":"192"},"options_float":{"a":48.0,"b":56.0,"c":60.0,"d":56.0,"e":192.0},"annotated_formula":"rectangle_perimeter(divide(rectangle_area(16, 12), 16), 16)","linear_formula":"rectangle_area(n1,n2)|divide(#0,n0)|rectangle_perimeter(n0,#1)|","chain":"16 * 12<\/gadget>\n192<\/output>\n192 \/ 16<\/gadget>\n12<\/output>\n2 * (12 + 16)<\/gadget>\n56<\/output>\n56<\/result>","index":4113} +{"problem":"mrs . evans gave a test to her freshmen economics class , which has 29 students enrolled and 19 of them answered question 1 correctly . if 24 answered question 2 correctly and 5 did not take the test then how many answered both questions correctly ?","rationale":"\"total number of enrolled students = 29 number of students who did not take test = 5 hence , number of students who took test = 29 - 5 = 24 number of students who answered q 2 correctly = 24 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 19 . number of students who answered both q 1 & q 2 correctly = 19 . answer : e\"","correct":"e","options":{"a":"3 ","b":"9 ","c":"11 ","d":"17","e":"19"},"options_float":{"a":3.0,"b":9.0,"c":11.0,"d":17.0,"e":19.0},"annotated_formula":"subtract(add(add(19, 24), 5), 29)","linear_formula":"add(n1,n3)|add(n5,#0)|subtract(#1,n0)|","chain":"19 + 24<\/gadget>\n43<\/output>\n43 + 5<\/gadget>\n48<\/output>\n48 - 29<\/gadget>\n19<\/output>\n19<\/result>","index":4114} +{"problem":"if m is a positive integer and m ^ 2 is divisible by 33 , then the largest positive integer that must divide m is ?","rationale":"\"m ^ 2 is divisible by 33 so m ^ 2 must be multiple of 48 . if the value of m is multiples of 11 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is d but if m = 33 then answer should be 16 . is the question right ? or am i missing some thing ? d\"","correct":"d","options":{"a":"3 ","b":"6 ","c":"8 ","d":"11","e":"16"},"options_float":{"a":3.0,"b":6.0,"c":8.0,"d":11.0,"e":16.0},"annotated_formula":"multiply(const_3, divide(divide(33, const_3), const_3))","linear_formula":"divide(n1,const_3)|divide(#0,const_3)|multiply(#1,const_3)|","chain":"33 \/ 3<\/gadget>\n11<\/output>\n11 \/ 3<\/gadget>\n11\/3 = around 3.666667<\/output>\n3 * (11\/3)<\/gadget>\n11<\/output>\n11<\/result>","index":4116} +{"problem":"from a total of 5 boys and 6 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ?","rationale":"answer = e = 150 no of 4 person committees that can be formed = 5 c 2 * 6 c 2 = 150 answer e","correct":"e","options":{"a":"16 ","b":"24 ","c":"60 ","d":"120","e":"150"},"options_float":{"a":16.0,"b":24.0,"c":60.0,"d":120.0,"e":150.0},"annotated_formula":"multiply(multiply(const_3, 5), multiply(5, 2))","linear_formula":"multiply(n0,const_3)|multiply(n0,n3)|multiply(#0,#1)","chain":"3 * 5<\/gadget>\n15<\/output>\n5 * 2<\/gadget>\n10<\/output>\n15 * 10<\/gadget>\n150<\/output>\n150<\/result>","index":4120} +{"problem":"the simple interest on a sum of money will be rs . 1400 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ?","rationale":"\"p - - - 10 - - - - 1400 p - - - 5 - - - - - 700 3 p - - - 5 - - - - - 2100 - - - - - - = > 2800 answer : a\"","correct":"a","options":{"a":"2800 ","b":"3799 ","c":"1200 ","d":"2693","e":"1732"},"options_float":{"a":2800.0,"b":3799.0,"c":1200.0,"d":2693.0,"e":1732.0},"annotated_formula":"add(multiply(multiply(divide(1400, 10), 5), const_3), multiply(divide(1400, 10), 5))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)|","chain":"1_400 \/ 10<\/gadget>\n140<\/output>\n140 * 5<\/gadget>\n700<\/output>\n700 * 3<\/gadget>\n2_100<\/output>\n2_100 + 700<\/gadget>\n2_800<\/output>\n2_800<\/result>","index":4121} +{"problem":"two cars start at the same time from opposite ends of a highway that is 175 miles long . one car is riding at 25 mph and the second car is riding at 45 mph . how long after they begin will they meet ?","rationale":"\"as cars are moving in opposite directions their speeds will be added . so their relative speeds : 45 + 25 = 70 mph total distance to be covered = 175 miles . time taken would be : 175 miles \/ 70 mph = 2.5 hours b is the answer .\"","correct":"b","options":{"a":"2 ","b":"2.5 ","c":"3 ","d":"3.5","e":"4"},"options_float":{"a":2.0,"b":2.5,"c":3.0,"d":3.5,"e":4.0},"annotated_formula":"divide(175, add(25, 45))","linear_formula":"add(n1,n2)|divide(n0,#0)|","chain":"25 + 45<\/gadget>\n70<\/output>\n175 \/ 70<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":4123} +{"problem":"in a school of 850 boys , 46 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?","rationale":"\"46 + 28 + 10 = 84 % 100 – 84 = 16 % 850 * 16 \/ 100 = 136 answer : a\"","correct":"a","options":{"a":"136 ","b":"627 ","c":"153 ","d":"721","e":"159"},"options_float":{"a":136.0,"b":627.0,"c":153.0,"d":721.0,"e":159.0},"annotated_formula":"divide(multiply(850, subtract(const_100, add(add(46, 28), 10))), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)|","chain":"46 + 28<\/gadget>\n74<\/output>\n74 + 10<\/gadget>\n84<\/output>\n100 - 84<\/gadget>\n16<\/output>\n850 * 16<\/gadget>\n13_600<\/output>\n13_600 \/ 100<\/gadget>\n136<\/output>\n136<\/result>","index":4127} +{"problem":"45 % of x is greater than 1 \/ 3 rd of x by 110 . what is x ?","rationale":"\"45 x \/ 100 - x \/ 3 = 110 7 x \/ 60 = 110 x = 943 answer : e\"","correct":"e","options":{"a":"100 ","b":"200 ","c":"300 ","d":"350","e":"943"},"options_float":{"a":100.0,"b":200.0,"c":300.0,"d":350.0,"e":943.0},"annotated_formula":"divide(110, subtract(divide(45, const_100), divide(1, 3)))","linear_formula":"divide(n0,const_100)|divide(n1,n2)|subtract(#0,#1)|divide(n3,#2)|","chain":"45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(9\/20) - (1\/3)<\/gadget>\n7\/60 = around 0.116667<\/output>\n110 \/ (7\/60)<\/gadget>\n6_600\/7 = around 942.857143<\/output>\n6_600\/7 = around 942.857143<\/result>","index":4128} +{"problem":"a \/ ( b \/ c ) in the expression above , a , b , and c are different numbers and each is one of the numbers 2 , 3 , or 6 . what is the largest possible value of the expression ?","rationale":"a \/ ( b \/ c ) = ( a * c ) \/ b the expression will have the largest value when numerator ( a * c ) is the largest . = ( 6 * 3 ) \/ 2 = 18 \/ 2 answer d","correct":"d","options":{"a":"2 \/ 18 ","b":"6 \/ 3 ","c":"30 \/ 2 ","d":"18 \/ 2","e":"6 \/ 18"},"options_float":{"a":0.1111111111,"b":2.0,"c":15.0,"d":9.0,"e":0.3333333333},"annotated_formula":"divide(3, divide(2, 6))","linear_formula":"divide(n0,n2)|divide(n1,#0)","chain":"2 \/ 6<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 \/ (1\/3)<\/gadget>\n9<\/output>\n9<\/result>","index":4130} +{"problem":"ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 5 percent and the mobile phone at a profit of 10 percent . overall he make a .","rationale":"\"let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 5 \/ 100 ) = 15000 - 750 m = 8000 ( 1 + 10 \/ 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 750 + 800 = rs . 50 as this is positive , an overall profit of rs . 50 was made . answer : a\"","correct":"a","options":{"a":"50 ","b":"288 ","c":"27 ","d":"200","e":"881"},"options_float":{"a":50.0,"b":288.0,"c":27.0,"d":200.0,"e":881.0},"annotated_formula":"subtract(add(multiply(15000, subtract(const_1, divide(5, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (1\/20)<\/gadget>\n19\/20 = around 0.95<\/output>\n15_000 * (19\/20)<\/gadget>\n14_250<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n8_000 * (11\/10)<\/gadget>\n8_800<\/output>\n14_250 + 8_800<\/gadget>\n23_050<\/output>\n15_000 + 8_000<\/gadget>\n23_000<\/output>\n23_050 - 23_000<\/gadget>\n50<\/output>\n50<\/result>","index":4132} +{"problem":"a batsman makes a score of 87 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning .","rationale":"\"explanation : consider the avg for first 16 innings is x . then total runs scored till 16 innings is 16 x . total runs after 17 innings = 16 x + 87 . thus , 16 x + 8717 = x + 316 x + 8717 = x + 3 ⇒ x = 36 so his average after 17 innings = 39 . answer : a\"","correct":"a","options":{"a":"39 ","b":"88 ","c":"26 ","d":"87","e":"11"},"options_float":{"a":39.0,"b":88.0,"c":26.0,"d":87.0,"e":11.0},"annotated_formula":"add(subtract(87, multiply(17, 3)), 3)","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|","chain":"17 * 3<\/gadget>\n51<\/output>\n87 - 51<\/gadget>\n36<\/output>\n36 + 3<\/gadget>\n39<\/output>\n39<\/result>","index":4133} +{"problem":"the area of a square is equal to five times the area of a rectangle of dimensions 32 cm * 10 cm . what is the perimeter of the square ?","rationale":"area of the square = s * s = 5 ( 32 * 10 ) = 1600 = > s = 40 = 40 cm perimeter of the square = 4 * 40 = 160 cm . answer : b","correct":"b","options":{"a":"289 cm ","b":"160 cm ","c":"829 cm ","d":"288 cm","e":"289 cm"},"options_float":{"a":289.0,"b":160.0,"c":829.0,"d":288.0,"e":289.0},"annotated_formula":"multiply(sqrt(multiply(rectangle_area(32, 10), divide(const_10, const_2))), const_4)","linear_formula":"divide(const_10,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)","chain":"32 * 10<\/gadget>\n320<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n320 * 5<\/gadget>\n1_600<\/output>\n1_600 ** (1\/2)<\/gadget>\n40<\/output>\n40 * 4<\/gadget>\n160<\/output>\n160<\/result>","index":4134} +{"problem":"working alone , mary can pave a driveway in 4 hours and hillary can pave the same driveway in 3 hours . when they work together , mary thrives on teamwork so her rate increases by 33.33 % , but hillary becomes distracted and her rate decreases by 50 % . if they both work together , how many hours will it take to pave the driveway ?","rationale":"initial working rates : mary = 1 \/ 4 per hour hillary = 1 \/ 3 per hour rate when working together : mary = 1 \/ 4 + ( 1 \/ 3 * 1 \/ 4 ) = 1 \/ 3 per hour hillary = 1 \/ 3 - ( 1 \/ 2 * 1 \/ 3 ) = 1 \/ 6 per hour together they work 1 \/ 3 + 1 \/ 6 = 1 \/ 2 per hour so they will need 2 hours to complete the driveway . the correct answer is a .","correct":"a","options":{"a":"2 hours ","b":"4 hours ","c":"5 hours ","d":"6 hours","e":"7 hours"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":7.0},"annotated_formula":"inverse(add(multiply(divide(const_1, 4), add(divide(33.33, const_100), const_1)), multiply(divide(const_1, 3), divide(50, const_100))))","linear_formula":"divide(n2,const_100)|divide(const_1,n0)|divide(const_1,n1)|divide(n3,const_100)|add(#0,const_1)|multiply(#2,#3)|multiply(#4,#1)|add(#6,#5)|inverse(#7)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n33.33 \/ 100<\/gadget>\n0.3333<\/output>\n0.3333 + 1<\/gadget>\n1.3333<\/output>\n(1\/4) * 1.3333<\/gadget>\n0.333325<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/3) * (1\/2)<\/gadget>\n1\/6 = around 0.166667<\/output>\n0.333325 + (1\/6)<\/gadget>\n0.499992<\/output>\n1 \/ 0.499992<\/gadget>\n2.000032<\/output>\n2.000032<\/result>","index":4135} +{"problem":"if y is a positive number and 1 \/ 2 the square root of x is the cube root of y , then y =","rationale":"1 \/ 2 the square root of x is cube root of y . if y = 64 . . then 1 \/ 2 the square root of y = 4 and cube of y is 64 . option a .","correct":"a","options":{"a":"64 ","b":"32 ","c":"16 ","d":"4","e":"1"},"options_float":{"a":64.0,"b":32.0,"c":16.0,"d":4.0,"e":1.0},"annotated_formula":"multiply(power(2, const_3), power(2, const_3))","linear_formula":"power(n1,const_3)|multiply(#0,#0)","chain":"2 ** 3<\/gadget>\n8<\/output>\n8 * 8<\/gadget>\n64<\/output>\n64<\/result>","index":4136} +{"problem":"the perimeter of a semi circle is 180 cm then the radius is ?","rationale":"\"36 \/ 7 r = 180 = > r = 35 answer : c\"","correct":"c","options":{"a":"22 ","b":"28 ","c":"35 ","d":"37","e":"13"},"options_float":{"a":22.0,"b":28.0,"c":35.0,"d":37.0,"e":13.0},"annotated_formula":"divide(180, add(const_2, const_pi))","linear_formula":"add(const_2,const_pi)|divide(n0,#0)|","chain":"2 + pi<\/gadget>\n2 + pi = around 5.141593<\/output>\n180 \/ (2 + pi)<\/gadget>\n180\/(2 + pi) = around 35.008608<\/output>\n180\/(2 + pi) = around 35.008608<\/result>","index":4137} +{"problem":"the sum of the ages of 4 children born at the intervals of 1 years each is 12 years . what is the age of the youngest child ?","rationale":"\"let x = the youngest child . each of the other four children will then be x + 1 , x + 2 , x + 3 we know that the sum of their ages is 12 so , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 12 therefore the youngest child is 1.5 years old answer : b\"","correct":"b","options":{"a":"2.5 ","b":"1.5 ","c":"1 ","d":"3.5","e":"4.5"},"options_float":{"a":2.5,"b":1.5,"c":1.0,"d":3.5,"e":4.5},"annotated_formula":"divide(subtract(divide(12, divide(4, const_2)), multiply(subtract(4, const_1), 1)), const_2)","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,const_2)|","chain":"4 \/ 2<\/gadget>\n2<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n4 - 1<\/gadget>\n3<\/output>\n3 * 1<\/gadget>\n3<\/output>\n6 - 3<\/gadget>\n3<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":4138} +{"problem":"what is the decimal equivalent of ( 1 \/ 2 ) ^ 2 ?","rationale":"\"( 1 \/ 2 ) ² = ( 1 \/ 2 ) ( 1 \/ 2 ) = 1 \/ 4 approach # 1 : use long division to divide 4 into 1 to get 1 \/ 4 = 0.25 d\"","correct":"d","options":{"a":"0.0016 ","b":"0.0625 ","c":"0.16 ","d":"0.25","e":"0.5"},"options_float":{"a":0.0016,"b":0.0625,"c":0.16,"d":0.25,"e":0.5},"annotated_formula":"power(divide(1, 2), 2)","linear_formula":"divide(n0,n1)|power(#0,n2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 2<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":4139} +{"problem":"when x is even , [ x ] = x \/ 2 + 1 , when x is odd [ x ] = 2 x + 1 then [ 6 ] * [ 3 ] = ?","rationale":"[ 6 ] * [ 3 ] = ( 6 \/ 2 + 1 ) ( 3 * 2 + 1 ) = [ 28 ] . ans - d","correct":"d","options":{"a":"[ 22 ] ","b":"[ 44 ] ","c":"[ 45 ] ","d":"[ 28 ]","e":"[ 10 ]"},"options_float":{"a":22.0,"b":44.0,"c":45.0,"d":28.0,"e":10.0},"annotated_formula":"multiply(add(divide(6, 2), const_1), add(multiply(2, 3), const_1))","linear_formula":"divide(n4,n0)|multiply(n0,n5)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)","chain":"6 \/ 2<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n4 * 7<\/gadget>\n28<\/output>\n28<\/result>","index":4140} +{"problem":"the sum of all the integers a such that - 26 < a < 24 is","rationale":"easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 10 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 a = - 49 . d is the answer .","correct":"d","options":{"a":"0 ","b":"- 2 ","c":"- 25 ","d":"- 49","e":"- 51"},"options_float":{"a":0.0,"b":-2.0,"c":-25.0,"d":-49.0,"e":-51.0},"annotated_formula":"add(add(negate(26), const_1), add(add(negate(26), const_1), const_1))","linear_formula":"negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)","chain":"-26<\/gadget>\n-26<\/output>\n(-26) + 1<\/gadget>\n-25<\/output>\n(-25) + 1<\/gadget>\n-24<\/output>\n(-25) + (-24)<\/gadget>\n-49<\/output>\n-49<\/result>","index":4141} +{"problem":"in what time will a railway train 52 m long moving at the rate of 24 kmph pass a telegraph post on its way ?","rationale":"t = 52 \/ 24 * 18 \/ 5 = 8 sec answer : c","correct":"c","options":{"a":"3 sec ","b":"4 sec ","c":"8 sec ","d":"6 sec","e":"7 sec"},"options_float":{"a":3.0,"b":4.0,"c":8.0,"d":6.0,"e":7.0},"annotated_formula":"divide(52, multiply(24, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n24 * (5\/18)<\/gadget>\n20\/3 = around 6.666667<\/output>\n52 \/ (20\/3)<\/gadget>\n39\/5 = around 7.8<\/output>\n39\/5 = around 7.8<\/result>","index":4142} +{"problem":"6 \/ 9 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded if necessary to the nearest percent , the probability that a juliet supporter chosen at random resides in capulet is","rationale":"total population = 90 ( assume ) . 6 \/ 9 * 90 = 60 people from montague . 3 \/ 9 * 90 = 30 people from capulet . 0.2 * 60 = 12 people from montague support juliet . 0.7 * 30 = 21 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 21 \/ ( 12 + 21 ) = ~ 64 . answer : a","correct":"a","options":{"a":"64 % ","b":"41 % ","c":"45 % ","d":"72 %","e":"78 %"},"options_float":{"a":64.0,"b":41.0,"c":45.0,"d":72.0,"e":78.0},"annotated_formula":"multiply(divide(multiply(divide(70, const_100), divide(subtract(9, 6), 9)), add(multiply(divide(subtract(const_100, 80), const_100), divide(6, 9)), multiply(divide(70, const_100), divide(subtract(9, 6), 9)))), const_100)","linear_formula":"divide(n3,const_100)|divide(n0,n1)|subtract(n1,n0)|subtract(const_100,n2)|divide(#2,n1)|divide(#3,const_100)|multiply(#0,#4)|multiply(#5,#1)|add(#7,#6)|divide(#6,#8)|multiply(#9,const_100)","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n9 - 6<\/gadget>\n3<\/output>\n3 \/ 9<\/gadget>\n1\/3 = around 0.333333<\/output>\n(7\/10) * (1\/3)<\/gadget>\n7\/30 = around 0.233333<\/output>\n100 - 80<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n6 \/ 9<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/5) * (2\/3)<\/gadget>\n2\/15 = around 0.133333<\/output>\n(2\/15) + (7\/30)<\/gadget>\n11\/30 = around 0.366667<\/output>\n(7\/30) \/ (11\/30)<\/gadget>\n7\/11 = around 0.636364<\/output>\n(7\/11) * 100<\/gadget>\n700\/11 = around 63.636364<\/output>\n700\/11 = around 63.636364<\/result>","index":4145} +{"problem":"if a 2 - b 2 = 10 and a * b = 5 , find a 4 + b 4 .","rationale":"a 2 - b 2 = 10 : given a 4 + b 4 - 2 a 2 b 2 = 102 : square both sides and expand . a * b = 5 : given a 2 b 2 = 52 : square both sides . a 4 + b 4 - 2 ( 25 ) = 100 : substitute a 4 + b 4 = 150 correct answer c","correct":"c","options":{"a":"50 ","b":"100 ","c":"150 ","d":"92","e":"25"},"options_float":{"a":50.0,"b":100.0,"c":150.0,"d":92.0,"e":25.0},"annotated_formula":"add(power(10, const_2), multiply(power(5, const_2), const_2))","linear_formula":"power(n3,const_2)|power(n2,const_2)|multiply(#0,const_2)|add(#2,#1)","chain":"10 ** 2<\/gadget>\n100<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n25 * 2<\/gadget>\n50<\/output>\n100 + 50<\/gadget>\n150<\/output>\n150<\/result>","index":4146} +{"problem":"in an office , totally there are 2800 employees and 25 % of the total employees are males . 30 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ?","rationale":"number of male employees = 2800 * 25 \/ 100 = 700 required number of male employees who are less than 50 years old = 700 * ( 100 - 30 ) % = 700 * 70 \/ 100 = 490 . answer : b","correct":"b","options":{"a":"390 ","b":"490 ","c":"400 ","d":"460","e":"none of these"},"options_float":{"a":390.0,"b":490.0,"c":400.0,"d":460.0,"e":null},"annotated_formula":"multiply(divide(multiply(2800, 25), const_100), divide(subtract(const_100, 30), const_100))","linear_formula":"multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)","chain":"2_800 * 25<\/gadget>\n70_000<\/output>\n70_000 \/ 100<\/gadget>\n700<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n700 * (7\/10)<\/gadget>\n490<\/output>\n490<\/result>","index":4147} +{"problem":"when 3 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ?","rationale":"explanation : let the number be 3 + 1 \/ 2 [ 1 \/ 3 ( a \/ 5 ) ] = a \/ 15 = > 3 = a \/ 30 = > a = 90 answer : e","correct":"e","options":{"a":"32 ","b":"81 ","c":"60 ","d":"27","e":"90"},"options_float":{"a":32.0,"b":81.0,"c":60.0,"d":27.0,"e":90.0},"annotated_formula":"divide(3, divide(divide(const_1, multiply(const_3, add(const_2, const_3))), const_2))","linear_formula":"add(const_2,const_3)|multiply(#0,const_3)|divide(const_1,#1)|divide(#2,const_2)|divide(n0,#3)","chain":"2 + 3<\/gadget>\n5<\/output>\n3 * 5<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/15) \/ 2<\/gadget>\n1\/30 = around 0.033333<\/output>\n3 \/ (1\/30)<\/gadget>\n90<\/output>\n90<\/result>","index":4148} +{"problem":"a certain psychologist charges $ 20 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 300 , what is the total charge to a patient who receives only 3 hours of therapy ?","rationale":"\"let the charge for first hour = x + 20 then charge for each additional hour = x x + 20 + 4 x = 300 = > 5 x = 280 = > x = 56 total charge for patient for 3 hours of therapy = x + 20 + 2 x = 3 x + 20 = 188 $ answer e\"","correct":"e","options":{"a":"$ 120 ","b":"$ 135 ","c":"$ 150 ","d":"$ 165","e":"$ 188"},"options_float":{"a":120.0,"b":135.0,"c":150.0,"d":165.0,"e":188.0},"annotated_formula":"add(multiply(divide(subtract(300, 20), 5), 3), 20)","linear_formula":"subtract(n2,n0)|divide(#0,n1)|multiply(n3,#1)|add(n0,#2)|","chain":"300 - 20<\/gadget>\n280<\/output>\n280 \/ 5<\/gadget>\n56<\/output>\n56 * 3<\/gadget>\n168<\/output>\n168 + 20<\/gadget>\n188<\/output>\n188<\/result>","index":4150} +{"problem":"a bus trip of 550 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ?","rationale":"the time is the distance \/ speed . the time difference is 1 hour . 550 \/ v - 550 \/ ( v + 5 ) = 1 550 ( v + 5 ) - 550 v = ( v ) ( v + 5 ) 2750 = ( v ) ( v + 5 ) 50 * 55 = ( v ) ( v + 5 ) v = 50 mph the answer is c .","correct":"c","options":{"a":"40 ","b":"45 ","c":"50 ","d":"55","e":"60"},"options_float":{"a":40.0,"b":45.0,"c":50.0,"d":55.0,"e":60.0},"annotated_formula":"divide(subtract(sqrt(add(multiply(multiply(550, 5), const_4), power(5, const_2))), 5), const_2)","linear_formula":"multiply(n0,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)","chain":"550 * 5<\/gadget>\n2_750<\/output>\n2_750 * 4<\/gadget>\n11_000<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n11_000 + 25<\/gadget>\n11_025<\/output>\n11_025 ** (1\/2)<\/gadget>\n105<\/output>\n105 - 5<\/gadget>\n100<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":4152} +{"problem":"the sum of the squares of 3 consecutive natural number is 2030 . what is the middle number ?","rationale":"let the numbers be x , x + 1 and x + 2 x 2 + ( x + 1 ) 2 + ( x + 2 ) 2 = 2030 3 x 2 + 6 x - 2025 = 0 ( x + 27 ) ( x - 25 ) = 0 x = 25 the middle number is 26 answer b 26","correct":"b","options":{"a":"25 ","b":"26 ","c":"27 ","d":"28","e":"29"},"options_float":{"a":25.0,"b":26.0,"c":27.0,"d":28.0,"e":29.0},"annotated_formula":"add(multiply(multiply(const_4, 3), const_2), const_2)","linear_formula":"multiply(n0,const_4)|multiply(#0,const_2)|add(#1,const_2)","chain":"4 * 3<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 + 2<\/gadget>\n26<\/output>\n26<\/result>","index":4153} +{"problem":"( √ 112 + √ 567 ) \/ √ 175 = ?","rationale":"( √ 112 + √ 567 ) \/ √ 175 = ( 4 √ 7 + 9 √ 7 ) \/ 5 √ 7 = 13 \/ 5 hence , the correct answer is b .","correct":"b","options":{"a":"2 √ 2 ","b":"13 \/ 5 ","c":"3 √ 2 ","d":"3 √ 3","e":"√ 2"},"options_float":{"a":2.0,"b":2.6,"c":3.0,"d":3.0,"e":2.0},"annotated_formula":"divide(add(sqrt(112), sqrt(567)), sqrt(175))","linear_formula":"sqrt(n0)|sqrt(n1)|sqrt(n2)|add(#0,#1)|divide(#3,#2)","chain":"112 ** (1\/2)<\/gadget>\n4*sqrt(7) = around 10.583005<\/output>\n567 ** (1\/2)<\/gadget>\n9*sqrt(7) = around 23.811762<\/output>\n(4*sqrt(7)) + (9*sqrt(7))<\/gadget>\n13*sqrt(7) = around 34.394767<\/output>\n175 ** (1\/2)<\/gadget>\n5*sqrt(7) = around 13.228757<\/output>\n(13*sqrt(7)) \/ (5*sqrt(7))<\/gadget>\n13\/5 = around 2.6<\/output>\n13\/5 = around 2.6<\/result>","index":4154} +{"problem":"a statue is being carved by a sculptor . the original piece of marble weighed 250 kg . in the first week 30 percent is cut away . in the second week 20 percent of the remainder is cut away . in the third week the statue is completed when 25 percent of the remainder is cut away . what is the weight of the final statue ?","rationale":"\"a 105 kg 250 × 0.7 × 0.8 × 0.75 = 105 kg .\"","correct":"a","options":{"a":"105 kg ","b":"103 kg ","c":"108 kg ","d":"125 kg","e":"117 kg"},"options_float":{"a":105.0,"b":103.0,"c":108.0,"d":125.0,"e":117.0},"annotated_formula":"multiply(subtract(const_1, divide(25, const_100)), multiply(subtract(const_1, divide(20, const_100)), multiply(250, subtract(const_1, divide(30, const_100)))))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(n0,#3)|multiply(#6,#4)|multiply(#7,#5)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n250 * (7\/10)<\/gadget>\n175<\/output>\n(4\/5) * 175<\/gadget>\n140<\/output>\n(3\/4) * 140<\/gadget>\n105<\/output>\n105<\/result>","index":4155} +{"problem":"dan can do a job alone in 12 hours . annie , working alone , can do the same job in just 9 hours . if dan works alone for 8 hours and then stops , how many hours will it take annie , working alone , to complete the job ?","rationale":"dan can complete 1 \/ 12 of the job per hour . in 8 hours , dan completes 8 ( 1 \/ 12 ) = 2 \/ 3 of the job . annie can complete 1 \/ 9 of the job per hour . to complete the job , annie will take 1 \/ 3 \/ 1 \/ 9 = 3 hours . the answer is b .","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"5","e":"6"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":5.0,"e":6.0},"annotated_formula":"multiply(subtract(const_1, divide(8, 12)), 9)","linear_formula":"divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)","chain":"8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 9<\/gadget>\n3<\/output>\n3<\/result>","index":4156} +{"problem":"two cars start from the opposite places of a main road , 105 km apart . first car runs for 25 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run only 35 km along the main road . what would be the distance between two cars at this point ?","rationale":"answer : c ) 20 km","correct":"c","options":{"a":"65 ","b":"38 ","c":"20 ","d":"28","e":"21"},"options_float":{"a":65.0,"b":38.0,"c":20.0,"d":28.0,"e":21.0},"annotated_formula":"subtract(subtract(105, 35), add(25, 25))","linear_formula":"add(n1,n1)|subtract(n0,n4)|subtract(#1,#0)|","chain":"105 - 35<\/gadget>\n70<\/output>\n25 + 25<\/gadget>\n50<\/output>\n70 - 50<\/gadget>\n20<\/output>\n20<\/result>","index":4157} +{"problem":"every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 126 , how many more green disks than blue disks are in the bag ?","rationale":"let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 126 - - > x = 7 . g - b = 8 x - 3 x = 5 x = 35 . the answer is d .","correct":"d","options":{"a":"25 ","b":"28 ","c":"30 ","d":"35","e":"40"},"options_float":{"a":25.0,"b":28.0,"c":30.0,"d":35.0,"e":40.0},"annotated_formula":"multiply(divide(126, add(add(3, 7), 8)), subtract(8, 3))","linear_formula":"add(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(n3,#2)|multiply(#3,#1)","chain":"3 + 7<\/gadget>\n10<\/output>\n10 + 8<\/gadget>\n18<\/output>\n126 \/ 18<\/gadget>\n7<\/output>\n8 - 3<\/gadget>\n5<\/output>\n7 * 5<\/gadget>\n35<\/output>\n35<\/result>","index":4159} +{"problem":"ben left a job paying $ 75000 per year to accept a sales job paying $ 45000 per year plus 15 percent commission . if each of his sales is for $ 750 , what is the least number of sales he must make per year if he is not to lose money because of the change ?","rationale":"in order not to lose money because of the change ben ' s total commission must beat least $ 75,000 - $ 45,000 = $ 30,000 , so total sales must beat least $ 30,000 \/ 0.15 = $ 200,000 . which means that he must makeat least $ 200,000 \/ 750 = 800 \/ 3 = 266.6 sales , so 267 sales . answer : d .","correct":"d","options":{"a":"40 ","b":"200 ","c":"266 ","d":"267","e":"600"},"options_float":{"a":40.0,"b":200.0,"c":266.0,"d":267.0,"e":600.0},"annotated_formula":"divide(divide(subtract(75000, 45000), divide(15, const_100)), 750)","linear_formula":"divide(n2,const_100)|subtract(n0,n1)|divide(#1,#0)|divide(#2,n3)","chain":"75_000 - 45_000<\/gadget>\n30_000<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n30_000 \/ (3\/20)<\/gadget>\n200_000<\/output>\n200_000 \/ 750<\/gadget>\n800\/3 = around 266.666667<\/output>\n800\/3 = around 266.666667<\/result>","index":4160} +{"problem":"if there are 86 workers in a factory , and on a certain day , 72 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) .","rationale":"\"72 \/ 86 * 100 = 83.72 83.7 % correct answer c\"","correct":"c","options":{"a":"82.7 % ","b":"81.7 % ","c":"83.7 % ","d":"89.7 %","e":"88.7 %"},"options_float":{"a":82.7,"b":81.7,"c":83.7,"d":89.7,"e":88.7},"annotated_formula":"multiply(divide(72, 86), const_100)","linear_formula":"divide(n1,n0)|multiply(#0,const_100)|","chain":"72 \/ 86<\/gadget>\n36\/43 = around 0.837209<\/output>\n(36\/43) * 100<\/gadget>\n3_600\/43 = around 83.72093<\/output>\n3_600\/43 = around 83.72093<\/result>","index":4161} +{"problem":"x does a work in 20 days . y does the same work in 40 days . in how many days they together will do the same work ?","rationale":"x ' s 1 day ' s work = 1 \/ 20 y ' s 1 day ' s work = 1 \/ 40 ( x + y ) ' s 1 day ' s work = ( 1 \/ 20 + 1 \/ 40 ) = 1 \/ 13.33 both together will finish the work in 13.33 days . correct option is e","correct":"e","options":{"a":"10 ","b":"12 ","c":"20 ","d":"30","e":"13.33"},"options_float":{"a":10.0,"b":12.0,"c":20.0,"d":30.0,"e":13.33},"annotated_formula":"inverse(add(divide(const_1, 20), divide(const_1, 40)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)","chain":"1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/20) + (1\/40)<\/gadget>\n3\/40 = around 0.075<\/output>\n1 \/ (3\/40)<\/gadget>\n40\/3 = around 13.333333<\/output>\n40\/3 = around 13.333333<\/result>","index":4162} +{"problem":"in a school the ratio of boys and girls are in the ration 5 : 6 . 25 % of boys will get scholar ship and 20 % of girls will get scholar ship what % of students wont get scholarship ?","rationale":"boys : girls = > 5 x : 6 x total no . of students including boys nd girls = 5 x + 6 x = 11 x 25 % of boys = > 5 x * 25 \/ 100 = 5 x \/ 4 20 % of girls = > 6 x * 20 \/ 100 = 6 x \/ 5 no . of students who are getting scholarship = > 5 x \/ 4 + 6 x \/ 5 = 49 x \/ 20 no . of students who are not getting scholarship = 11 x - 49 x \/ 20 = 171 x \/ 20 % of students not getting scolorship = > 171 x \/ 20 \/ 11 x * 100 = > 100 * 171 x \/ 220 x = 77.7 % answer : b","correct":"b","options":{"a":"60 % ","b":"78 % ","c":"88 % ","d":"98 %","e":"99 %"},"options_float":{"a":60.0,"b":78.0,"c":88.0,"d":98.0,"e":99.0},"annotated_formula":"multiply(add(multiply(divide(subtract(const_100, 25), const_100), divide(5, add(5, 6))), multiply(divide(subtract(const_100, 20), const_100), divide(6, add(5, 6)))), const_100)","linear_formula":"add(n0,n1)|subtract(const_100,n2)|subtract(const_100,n3)|divide(#1,const_100)|divide(n0,#0)|divide(#2,const_100)|divide(n1,#0)|multiply(#3,#4)|multiply(#5,#6)|add(#7,#8)|multiply(#9,const_100)","chain":"100 - 25<\/gadget>\n75<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n5 + 6<\/gadget>\n11<\/output>\n5 \/ 11<\/gadget>\n5\/11 = around 0.454545<\/output>\n(3\/4) * (5\/11)<\/gadget>\n15\/44 = around 0.340909<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n6 \/ 11<\/gadget>\n6\/11 = around 0.545455<\/output>\n(4\/5) * (6\/11)<\/gadget>\n24\/55 = around 0.436364<\/output>\n(15\/44) + (24\/55)<\/gadget>\n171\/220 = around 0.777273<\/output>\n(171\/220) * 100<\/gadget>\n855\/11 = around 77.727273<\/output>\n855\/11 = around 77.727273<\/result>","index":4163} +{"problem":"a start walking from a place at a uniform speed of 7 kmph in a particular direction . after half an hour , b starts from the same place and walks in the same direction as a at a uniform speed and overtakes a after 1 hour 48 minutes . find the speed of b .","rationale":"distance covered by a in 30 min = 1 km b covers extra 1 km in 1 hour 48 minutes ( 9 \/ 5 hr ) i . e . relative speed of b over a = 1 \/ ( 9 \/ 5 ) = 5 \/ 9 so the speed of b = speed of a + 5 \/ 9 = 7 + 5 \/ 9 = 7.55 answer b","correct":"b","options":{"a":"4.7 kmph ","b":"7.6 kmph ","c":"4 kmph ","d":"7 kmph","e":"5.3 kmph"},"options_float":{"a":4.7,"b":7.6,"c":4.0,"d":7.0,"e":5.3},"annotated_formula":"add(divide(1, divide(add(const_60, 48), const_60)), 7)","linear_formula":"add(n2,const_60)|divide(#0,const_60)|divide(n1,#1)|add(n0,#2)","chain":"60 + 48<\/gadget>\n108<\/output>\n108 \/ 60<\/gadget>\n9\/5 = around 1.8<\/output>\n1 \/ (9\/5)<\/gadget>\n5\/9 = around 0.555556<\/output>\n(5\/9) + 7<\/gadget>\n68\/9 = around 7.555556<\/output>\n68\/9 = around 7.555556<\/result>","index":4164} +{"problem":"find large number from below question the difference of two numbers is 1355 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1355 ) . x + 1355 = 6 x + 15 5 x = 1340 x = 268 large number = 268 + 1355 = 1623 e\"","correct":"e","options":{"a":"1235 ","b":"1345 ","c":"1678 ","d":"1767","e":"1623"},"options_float":{"a":1235.0,"b":1345.0,"c":1678.0,"d":1767.0,"e":1623.0},"annotated_formula":"multiply(divide(subtract(1355, 15), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_355 - 15<\/gadget>\n1_340<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_340 \/ 5<\/gadget>\n268<\/output>\n268 * 6<\/gadget>\n1_608<\/output>\n1_608<\/result>","index":4165} +{"problem":"solve below question 2 x + 1 = - 13","rationale":"\"2 x + 1 = - 13 x = - 7 e\"","correct":"e","options":{"a":"- 8 ","b":"- 9 ","c":"9 ","d":"8","e":"- 7"},"options_float":{"a":-8.0,"b":-9.0,"c":9.0,"d":8.0,"e":-7.0},"annotated_formula":"divide(negate(add(13, 1)), 2)","linear_formula":"add(n1,n2)|negate(#0)|divide(#1,n0)|","chain":"13 + 1<\/gadget>\n14<\/output>\n-14<\/gadget>\n-14<\/output>\n(-14) \/ 2<\/gadget>\n-7<\/output>\n-7<\/result>","index":4167} +{"problem":"car a is 16 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ?","rationale":"relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 16 miles and drive 8 miles ahead so to drive 24 miles it ' ll need 24 \/ 8 = 3 hours . answer : d","correct":"d","options":{"a":"1.5 ","b":"2.0 ","c":"2.5 ","d":"3.0","e":"3.5"},"options_float":{"a":1.5,"b":2.0,"c":2.5,"d":3.0,"e":3.5},"annotated_formula":"divide(add(16, 8), subtract(58, 50))","linear_formula":"add(n0,n3)|subtract(n1,n2)|divide(#0,#1)","chain":"16 + 8<\/gadget>\n24<\/output>\n58 - 50<\/gadget>\n8<\/output>\n24 \/ 8<\/gadget>\n3<\/output>\n3<\/result>","index":4168} +{"problem":"if x + y = 290 , x - y = 200 , for integers of x and y , y = ?","rationale":"\"x + y = 290 x - y = 200 2 x = 90 x = 45 y = 245 answer is b\"","correct":"b","options":{"a":"200 ","b":"245 ","c":"50 ","d":"115","e":"150"},"options_float":{"a":200.0,"b":245.0,"c":50.0,"d":115.0,"e":150.0},"annotated_formula":"divide(add(290, 200), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"290 + 200<\/gadget>\n490<\/output>\n490 \/ 2<\/gadget>\n245<\/output>\n245<\/result>","index":4169} +{"problem":"the bankers discount of a certain sum of money is rs . 18 and the true discount on the same sum for the same time is rs . 15 . the sum due is :","rationale":"\"sum = ( b . d * t . d ) \/ ( b . d - t . d ) ( 18 * 15 ) \/ 18 - 15 ; 90 answer : b\"","correct":"b","options":{"a":"89 ","b":"90 ","c":"91 ","d":"92","e":"93"},"options_float":{"a":89.0,"b":90.0,"c":91.0,"d":92.0,"e":93.0},"annotated_formula":"divide(multiply(15, 18), subtract(18, 15))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"15 * 18<\/gadget>\n270<\/output>\n18 - 15<\/gadget>\n3<\/output>\n270 \/ 3<\/gadget>\n90<\/output>\n90<\/result>","index":4172} +{"problem":"the mean of 50 observations is 200 . but later he found that there is decrements of 34 from each observations . what is the the updated mean is ?","rationale":"\"166 answer is b\"","correct":"b","options":{"a":"165 ","b":"166 ","c":"190 ","d":"198","e":"199"},"options_float":{"a":165.0,"b":166.0,"c":190.0,"d":198.0,"e":199.0},"annotated_formula":"subtract(200, 34)","linear_formula":"subtract(n1,n2)|","chain":"200 - 34<\/gadget>\n166<\/output>\n166<\/result>","index":4173} +{"problem":"the sum of first 45 natural numbers is","rationale":"solution we know that ( 1 + 2 + 3 + . . . . . + 45 ) = n ( n + 1 ) \/ 2 therefore ( 1 + 2 + 3 + . . . . + 45 ) = ( 45 × 46 \/ 2 ) = 1035 . answer a","correct":"a","options":{"a":"1035 ","b":"1280 ","c":"2070 ","d":"2140","e":"none"},"options_float":{"a":1035.0,"b":1280.0,"c":2070.0,"d":2140.0,"e":null},"annotated_formula":"divide(multiply(45, add(45, const_1)), const_2)","linear_formula":"add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)","chain":"45 + 1<\/gadget>\n46<\/output>\n45 * 46<\/gadget>\n2_070<\/output>\n2_070 \/ 2<\/gadget>\n1_035<\/output>\n1_035<\/result>","index":4174} +{"problem":"what is the dividend . divisor 13 , the quotient is 17 and the remainder is 1","rationale":"\"a = d * q + r a = 13 * 17 + 1 a = 221 + 1 a = 222\"","correct":"a","options":{"a":"222 ","b":"224 ","c":"226 ","d":"228","e":"230"},"options_float":{"a":222.0,"b":224.0,"c":226.0,"d":228.0,"e":230.0},"annotated_formula":"add(multiply(13, 17), 1)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"13 * 17<\/gadget>\n221<\/output>\n221 + 1<\/gadget>\n222<\/output>\n222<\/result>","index":4175} +{"problem":"a man performs 3 \/ 5 of the total journey by rail , 17 \/ 20 by bus and the remaining 6.5 km on foot . his total journey is :","rationale":"\"sol . let the total journey be x km . then 3 x \/ 5 + 7 x \/ 20 + 6.5 = x ⇔ 12 x + 7 x + 20 * 6.5 = 20 x ⇔ x = 130 km . answer b\"","correct":"b","options":{"a":"125 km ","b":"130 km ","c":"135 km ","d":"140 km","e":"none"},"options_float":{"a":125.0,"b":130.0,"c":135.0,"d":140.0,"e":null},"annotated_formula":"multiply(20, 6.5)","linear_formula":"multiply(n3,n4)|","chain":"20 * 6.5<\/gadget>\n130<\/output>\n130<\/result>","index":4176} +{"problem":"there are 300 female managers in a certain company . find the total number of female employees in the company , if 2 \/ 5 of all the employees are managers and 2 \/ 5 of all male employees are managers .","rationale":"as per question stem 2 \/ 5 m ( portion of men employees who are managers ) + 200 ( portion of female employees who are managers ) = 2 \/ 5 t ( portion of total number of employees who are managers ) , thus we get that 2 \/ 5 m + 300 = 2 \/ 5 t , or 2 \/ 5 ( t - m ) = 300 , from here we get that t - m = 750 , that would be total number of female employees and the answer ( c )","correct":"c","options":{"a":"650 ","b":"700 ","c":"750 ","d":"800","e":"none of these"},"options_float":{"a":650.0,"b":700.0,"c":750.0,"d":800.0,"e":null},"annotated_formula":"divide(300, divide(2, 5))","linear_formula":"divide(n1,n2)|divide(n0,#0)|","chain":"2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n300 \/ (2\/5)<\/gadget>\n750<\/output>\n750<\/result>","index":4177} +{"problem":"the true discount on a bill of rs . 2260 is rs . 360 . what is the banker ' s discount ?","rationale":"\"explanation : f = rs . 2260 td = rs . 360 pw = f - td = 2260 - 360 = rs . 1900 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 1900 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2260 for unexpired time = 360 \/ 1900 × 2260 = 0.19 × 2260 = rs . 428 answer : option a\"","correct":"a","options":{"a":"rs . 428 ","b":"rs . 422 ","c":"rs . 412 ","d":"rs . 442","e":"none of these"},"options_float":{"a":428.0,"b":422.0,"c":412.0,"d":442.0,"e":null},"annotated_formula":"multiply(divide(360, subtract(2260, 360)), 2260)","linear_formula":"subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)|","chain":"2_260 - 360<\/gadget>\n1_900<\/output>\n360 \/ 1_900<\/gadget>\n18\/95 = around 0.189474<\/output>\n(18\/95) * 2_260<\/gadget>\n8_136\/19 = around 428.210526<\/output>\n8_136\/19 = around 428.210526<\/result>","index":4179} +{"problem":"in the first 20 overs of a cricket game , the run rate was only 4.8 . what should be the run rate in the remaining 30 overs to reach the target of 302 runs ?","rationale":"required run rate = 302 - ( 4.8 x 20 ) \/ 30 = 206 \/ 30 = 6.87 option d","correct":"d","options":{"a":"5 ","b":"6.25 ","c":"6.75 ","d":"6.87","e":"7.25"},"options_float":{"a":5.0,"b":6.25,"c":6.75,"d":6.87,"e":7.25},"annotated_formula":"divide(subtract(302, multiply(20, 4.8)), 30)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)","chain":"20 * 4.8<\/gadget>\n96<\/output>\n302 - 96<\/gadget>\n206<\/output>\n206 \/ 30<\/gadget>\n103\/15 = around 6.866667<\/output>\n103\/15 = around 6.866667<\/result>","index":4180} +{"problem":"a trader sells 85 meters of cloth for rs . 8500 at the profit of rs . 15 per metre of cloth . what is the cost price of one metre of cloth ?","rationale":"\"sp of 1 m of cloth = 8500 \/ 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 100 - rs . 15 = rs . 85 . answer : b\"","correct":"b","options":{"a":"rs . 80 ","b":"rs . 85 ","c":"rs . 90 ","d":"rs . 95","e":"none of these"},"options_float":{"a":80.0,"b":85.0,"c":90.0,"d":95.0,"e":null},"annotated_formula":"subtract(divide(8500, 85), 15)","linear_formula":"divide(n1,n0)|subtract(#0,n2)|","chain":"8_500 \/ 85<\/gadget>\n100<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85<\/result>","index":4181} +{"problem":"anne bought doughnuts for a class breakfast party . she bought 9 chocolate doughnuts , 7 coconut doughnuts , and 9 jam - filled doughnuts . how many doughnuts did anne buy in all ?","rationale":"\"add the numbers of doughnuts . 9 + 7 + 9 = 25 . answer is a .\"","correct":"a","options":{"a":"25 ","b":"36 ","c":"39 ","d":"21","e":"11"},"options_float":{"a":25.0,"b":36.0,"c":39.0,"d":21.0,"e":11.0},"annotated_formula":"add(add(9, 7), 9)","linear_formula":"add(n0,n1)|add(n2,#0)|","chain":"9 + 7<\/gadget>\n16<\/output>\n16 + 9<\/gadget>\n25<\/output>\n25<\/result>","index":4182} +{"problem":"the length of a rectangle i s two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ?","rationale":"given that the area of the square = 1225 sq . units = > side of square = √ 1225 = 35 units the radius of the circle = side of the square = 35 units length of the rectangle = 2 \/ 5 * 35 = 14 units given that breadth = 10 units area of the rectangle = lb = 14 * 10 = 140 sq . units answer : option a","correct":"a","options":{"a":"140 sq . units ","b":"149 ","c":"148 ","d":"17","e":"143"},"options_float":{"a":140.0,"b":149.0,"c":148.0,"d":17.0,"e":143.0},"annotated_formula":"multiply(multiply(divide(const_2, add(const_2, const_3)), sqrt(1225)), 10)","linear_formula":"add(const_2,const_3)|sqrt(n0)|divide(const_2,#0)|multiply(#2,#1)|multiply(n1,#3)","chain":"2 + 3<\/gadget>\n5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n1_225 ** (1\/2)<\/gadget>\n35<\/output>\n(2\/5) * 35<\/gadget>\n14<\/output>\n14 * 10<\/gadget>\n140<\/output>\n140<\/result>","index":4183} +{"problem":"a cyclist bikes x distance at 8 miles per hour and returns over the same path at 10 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ?","rationale":"distance = d 1 = x miles speed = s 1 = 8 miles per hour time = t 1 = distance \/ speed = x \/ 8 2 . going from b to a distance = d 2 = x miles speed = s 2 = 10 miles per hour time = t 2 = distance \/ speed = x \/ 10 3 . average speed = total distance \/ total time total distance = x + x = 2 x total time = x \/ 10 + x \/ 8 = x ( 1 \/ 10 + 1 \/ 8 ) = = 9 x \/ 40 speed = 2 x \/ ( 9 x \/ 40 ) = 80 \/ 9 = 8.9 answer : b","correct":"b","options":{"a":"8.1 ","b":"8.9 ","c":"8.6 ","d":"8.3","e":"9.0"},"options_float":{"a":8.1,"b":8.9,"c":8.6,"d":8.3,"e":9.0},"annotated_formula":"divide(add(8, 10), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"8 + 10<\/gadget>\n18<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n9<\/result>","index":4184} +{"problem":"right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 38 years . what is molly ' s age right now ?","rationale":"now , sandy is 38 - 6 = 32 molly ' s age is ( 3 \/ 4 ) * 32 = 24 the answer is d .","correct":"d","options":{"a":"15 ","b":"18 ","c":"21 ","d":"24","e":"27"},"options_float":{"a":15.0,"b":18.0,"c":21.0,"d":24.0,"e":27.0},"annotated_formula":"multiply(divide(subtract(38, 6), 4), 3)","linear_formula":"subtract(n3,n2)|divide(#0,n0)|multiply(n1,#1)","chain":"38 - 6<\/gadget>\n32<\/output>\n32 \/ 4<\/gadget>\n8<\/output>\n8 * 3<\/gadget>\n24<\/output>\n24<\/result>","index":4187} +{"problem":"in a survey of parents , exactly 5 \/ 6 of the mothers and 3 \/ 4 of the fathers held full - time jobs . if 40 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ?","rationale":"\"fathers without full - time jobs are 1 \/ 4 * 3 \/ 5 = 3 \/ 20 of all the parents surveyed . mothers without full - time jobs are 1 \/ 6 * 2 \/ 5 = 1 \/ 15 of all the parents surveyed . the percent of parents without full - time jobs is 3 \/ 20 + 1 \/ 15 = 13 \/ 60 which is about 22 % the answer is e .\"","correct":"e","options":{"a":"10 % ","b":"12 % ","c":"15 % ","d":"18 %","e":"22 %"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":18.0,"e":22.0},"annotated_formula":"add(subtract(subtract(const_100, 40), multiply(divide(3, 4), subtract(const_100, 40))), subtract(40, multiply(divide(5, 6), 40)))","linear_formula":"divide(n2,n3)|divide(n0,n1)|subtract(const_100,n4)|multiply(#0,#2)|multiply(n4,#1)|subtract(#2,#3)|subtract(n4,#4)|add(#5,#6)|","chain":"100 - 40<\/gadget>\n60<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 60<\/gadget>\n45<\/output>\n60 - 45<\/gadget>\n15<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(5\/6) * 40<\/gadget>\n100\/3 = around 33.333333<\/output>\n40 - (100\/3)<\/gadget>\n20\/3 = around 6.666667<\/output>\n15 + (20\/3)<\/gadget>\n65\/3 = around 21.666667<\/output>\n65\/3 = around 21.666667<\/result>","index":4191} +{"problem":"circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 45 more revolutions than gear p ?","rationale":"\"gear q makes 30 more revolutions each 60 seconds . the time to make 45 = 30 * 1.5 more revolutions is 60 * 1.5 = 90 seconds . the answer is c .\"","correct":"c","options":{"a":"45 ","b":"80 ","c":"90 ","d":"100","e":"120"},"options_float":{"a":45.0,"b":80.0,"c":90.0,"d":100.0,"e":120.0},"annotated_formula":"divide(45, subtract(divide(40, const_60), divide(10, const_60)))","linear_formula":"divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|","chain":"40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n10 \/ 60<\/gadget>\n1\/6 = around 0.166667<\/output>\n(2\/3) - (1\/6)<\/gadget>\n1\/2 = around 0.5<\/output>\n45 \/ (1\/2)<\/gadget>\n90<\/output>\n90<\/result>","index":4194} +{"problem":"for the symbol , m ” n = n ^ 2 − m for all values of m and n . what is the value of 6 ” 3 ?","rationale":"6 ” 3 = 9 - 6 = 3 answer : b","correct":"b","options":{"a":"5 ","b":"3 ","c":"2 ","d":"1","e":"0"},"options_float":{"a":5.0,"b":3.0,"c":2.0,"d":1.0,"e":0.0},"annotated_formula":"subtract(power(3, 2), 6)","linear_formula":"power(n2,n0)|subtract(#0,n1)","chain":"3 ** 2<\/gadget>\n9<\/output>\n9 - 6<\/gadget>\n3<\/output>\n3<\/result>","index":4195} +{"problem":"the sides of a square region , measured to the nearest centimeter , are 4 centimeters long . the least possible value of the actual area of the square region is","rationale":"\"though there might be some technicalities concerning the termnearest ( as 3.5 is equidistant from both 3 and 4 ) the answer still should be : 3.5 ^ 2 = 12.25 . answer : c\"","correct":"c","options":{"a":"15.25 sq cm ","b":"18.25 sq cm ","c":"12.25 sq cm ","d":"21.25 sq cm","e":"25.25 sq cm"},"options_float":{"a":15.25,"b":18.25,"c":12.25,"d":21.25,"e":25.25},"annotated_formula":"power(subtract(subtract(4, const_0_25), const_0_25), const_2)","linear_formula":"subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n4 - (1\/4)<\/gadget>\n15\/4 = around 3.75<\/output>\n(15\/4) - (1\/4)<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) ** 2<\/gadget>\n49\/4 = around 12.25<\/output>\n49\/4 = around 12.25<\/result>","index":4196} +{"problem":"what sum of money will produce rs . 80 as simple interest in 4 years at 3 1 \/ 2 percent ?","rationale":"\"80 = ( p * 4 * 7 \/ 2 ) \/ 100 p = 571.4 answer : b\"","correct":"b","options":{"a":"263 ","b":"571.4 ","c":"367 ","d":"368","e":"339"},"options_float":{"a":263.0,"b":571.4,"c":367.0,"d":368.0,"e":339.0},"annotated_formula":"divide(80, divide(multiply(4, add(3, divide(1, 2))), const_100))","linear_formula":"divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n3 + (1\/2)<\/gadget>\n7\/2 = around 3.5<\/output>\n4 * (7\/2)<\/gadget>\n14<\/output>\n14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n80 \/ (7\/50)<\/gadget>\n4_000\/7 = around 571.428571<\/output>\n4_000\/7 = around 571.428571<\/result>","index":4197} +{"problem":"two - third of a positive number and 144 \/ 216 of its reciprocal are equal . the number is :","rationale":"\"let the number be x . then , 2 \/ 3 x = 144 \/ 216 * 1 \/ x x 2 = 144 \/ 216 * 3 \/ 2 = 144 \/ 144 x = 1 answer : e\"","correct":"e","options":{"a":"5 \/ 12 ","b":"12 \/ 5 ","c":"25 \/ 144 ","d":"144 \/ 25","e":"1"},"options_float":{"a":0.4166666667,"b":2.4,"c":0.1736111111,"d":5.76,"e":1.0},"annotated_formula":"sqrt(divide(multiply(144, const_3), multiply(216, const_2)))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_2)|divide(#0,#1)|sqrt(#2)|","chain":"144 * 3<\/gadget>\n432<\/output>\n216 * 2<\/gadget>\n432<\/output>\n432 \/ 432<\/gadget>\n1<\/output>\n1 ** (1\/2)<\/gadget>\n1<\/output>\n1<\/result>","index":4198} +{"problem":"from a container having pure milk , 20 % is replaced by water and the process is repeated thrice . at the end of the third operation , the milk is ?","rationale":"let the total quantity of original milk = 1000 gm milk after 1 st operation = 80 % of 1000 = 800 gm milk after second operation = 80 % of 800 = 640 gm milk after third operation = 80 % of 640 = 512 gm strength of final mixture = 51.2 % answer is c","correct":"c","options":{"a":"45 % ","b":"49.6 % ","c":"51.2 % ","d":"55 %","e":"60.3 %"},"options_float":{"a":45.0,"b":49.6,"c":51.2,"d":55.0,"e":60.3},"annotated_formula":"multiply(power(divide(subtract(const_100, 20), const_100), const_3), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|power(#1,const_3)|multiply(#2,const_100)","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) ** 3<\/gadget>\n64\/125 = around 0.512<\/output>\n(64\/125) * 100<\/gadget>\n256\/5 = around 51.2<\/output>\n256\/5 = around 51.2<\/result>","index":4199} +{"problem":"after 6 games , team b had an average of 70 points per game . if it got only 47 points in game 7 , how many more points does it need to score to get its total above 500 ?","rationale":"\"( 6 * 70 ) + 47 + x > 500 420 + 47 + x > 500 467 + x > 500 = > x > 33 option d\"","correct":"d","options":{"a":"85 ","b":"74 ","c":"67 ","d":"33","e":"28"},"options_float":{"a":85.0,"b":74.0,"c":67.0,"d":33.0,"e":28.0},"annotated_formula":"subtract(500, add(multiply(6, 70), 47))","linear_formula":"multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)|","chain":"6 * 70<\/gadget>\n420<\/output>\n420 + 47<\/gadget>\n467<\/output>\n500 - 467<\/gadget>\n33<\/output>\n33<\/result>","index":4200} +{"problem":"if a ' s height is 40 % less than that of b , how much percent b ' s height is more than that of a ?","rationale":"\"excess of b ' s height over a ' s = [ ( 40 \/ ( 100 - 40 ) ] x 100 % = 66.66 % answer : a ) 66.66 %\"","correct":"a","options":{"a":"66.66 % ","b":"66.68 % ","c":"66.766 % ","d":"86.66 %","e":"66.65 %"},"options_float":{"a":66.66,"b":66.68,"c":66.766,"d":86.66,"e":66.65},"annotated_formula":"multiply(divide(40, subtract(const_100, 40)), const_100)","linear_formula":"subtract(const_100,n0)|divide(n0,#0)|multiply(#1,const_100)|","chain":"100 - 40<\/gadget>\n60<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 100<\/gadget>\n200\/3 = around 66.666667<\/output>\n200\/3 = around 66.666667<\/result>","index":4201} +{"problem":"how long will a boy take to run round a square field of side 35 meters , if he runs at the rate of 9 km \/ hr ?","rationale":"answer : d ) 56 sec","correct":"d","options":{"a":"22 ","b":"38 ","c":"37 ","d":"56","e":"28"},"options_float":{"a":22.0,"b":38.0,"c":37.0,"d":56.0,"e":28.0},"annotated_formula":"divide(multiply(35, const_4), multiply(9, divide(const_1000, const_3600)))","linear_formula":"divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)|","chain":"35 * 4<\/gadget>\n140<\/output>\n1_000 \/ 3_600<\/gadget>\n5\/18 = around 0.277778<\/output>\n9 * (5\/18)<\/gadget>\n5\/2 = around 2.5<\/output>\n140 \/ (5\/2)<\/gadget>\n56<\/output>\n56<\/result>","index":4202} +{"problem":"how many prime numbers are between 25 \/ 3 and 84 \/ 6 ?","rationale":"\"25 \/ 3 = 8 . xxx 84 \/ 6 = 14 . xxx so we need to find prime numbers between 8 ( exclusive ) - 13 ( inclusive ) there are 2 prime numbers 1113 hence answer will be ( b ) 2 b\"","correct":"b","options":{"a":"1 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"floor(const_2)","linear_formula":"floor(const_2)|","chain":"floor(2)<\/gadget>\n2<\/output>\n2<\/result>","index":4207} +{"problem":"a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 400 and $ 350 per unit , respectively . what is the fewest number of computer printers that the company can purchase ?","rationale":"\"the smallest amount that the company can spend is the lcm of 400 and 350 , which is 2800 for each , which is total 5600 . the number of 1 st type of computers which costing $ 400 = 2800 \/ 400 = 7 . the number of 2 nd type of computers which costing $ 350 = 2800 \/ 350 = 8 . total = 7 + 8 = 15 answer is c .\"","correct":"c","options":{"a":"11 ","b":"13 ","c":"15 ","d":"17","e":"19"},"options_float":{"a":11.0,"b":13.0,"c":15.0,"d":17.0,"e":19.0},"annotated_formula":"add(divide(lcm(400, 350), 400), divide(lcm(400, 350), 350))","linear_formula":"lcm(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|","chain":"lcm(400, 350)<\/gadget>\n2_800<\/output>\n2_800 \/ 400<\/gadget>\n7<\/output>\n2_800 \/ 350<\/gadget>\n8<\/output>\n7 + 8<\/gadget>\n15<\/output>\n15<\/result>","index":4208} +{"problem":"a man can row upstream at 25 kmph and downstream at 55 kmph , and then find the speed of the man in still water ?","rationale":"\"us = 25 ds = 55 m = ( 55 + 25 ) \/ 2 = 40 answer : b\"","correct":"b","options":{"a":"65 ","b":"40 ","c":"30 ","d":"78","e":"38"},"options_float":{"a":65.0,"b":40.0,"c":30.0,"d":78.0,"e":38.0},"annotated_formula":"divide(add(25, 55), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"25 + 55<\/gadget>\n80<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40<\/result>","index":4210} +{"problem":"what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 60 ?","rationale":"\"let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 60 = rs . 4080 . answer : c\"","correct":"c","options":{"a":"s . 3944 ","b":"s . 3948 ","c":"s . 4080 ","d":"s . 3965","e":"s . 3929"},"options_float":{"a":3944.0,"b":3948.0,"c":4080.0,"d":3965.0,"e":3929.0},"annotated_formula":"multiply(square_perimeter(sqrt(289)), 60)","linear_formula":"sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|","chain":"289 ** (1\/2)<\/gadget>\n17<\/output>\n4 * 17<\/gadget>\n68<\/output>\n68 * 60<\/gadget>\n4_080<\/output>\n4_080<\/result>","index":4212} +{"problem":"an electrical appliances store sold this month 400 % more than the average of all the other months in the year . the sales total for this month was approximately what percent of total sales for this year ?","rationale":"the average sales for all the other months ( 11 months ) of the year is x . so , the total sales of those 11 months will be = 11 x now , this month ' s sales , which is 400 % more than the average of other months ' sales , is 5 x , ( x + 400 % of x ) . thus , this month ' s ' sales are ( 5 x \/ 16 x ) * 100 = 31.25 % of the total yearly sales . [ total sales = 11 x + 5 x = 16 x ] answer is b","correct":"b","options":{"a":"14 % ","b":"21 % . ","c":"31 % . ","d":"37 % .","e":"43 % ."},"options_float":{"a":14.0,"b":21.0,"c":31.0,"d":37.0,"e":43.0},"annotated_formula":"subtract(multiply(divide(add(const_100, 400), add(multiply(const_100, const_12), 400)), const_100), const_10)","linear_formula":"add(n0,const_100)|multiply(const_100,const_12)|add(n0,#1)|divide(#0,#2)|multiply(#3,const_100)|subtract(#4,const_10)","chain":"100 + 400<\/gadget>\n500<\/output>\n100 * 12<\/gadget>\n1_200<\/output>\n1_200 + 400<\/gadget>\n1_600<\/output>\n500 \/ 1_600<\/gadget>\n5\/16 = around 0.3125<\/output>\n(5\/16) * 100<\/gadget>\n125\/4 = around 31.25<\/output>\n(125\/4) - 10<\/gadget>\n85\/4 = around 21.25<\/output>\n85\/4 = around 21.25<\/result>","index":4214} +{"problem":"a bag contains 10 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is","rationale":"drawing two balls of same color from ten green balls can be done in 10 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 10 c ₂ \/ ¹ ⁵ c ₂ + ⁸ c ₂ \/ ¹ ⁵ c ₂ = 21 \/ 44 answer : b","correct":"b","options":{"a":"8 \/ 15 ","b":"21 \/ 44 ","c":"3 \/ 5 ","d":"11 \/ 15","e":"7 \/ 15"},"options_float":{"a":0.5333333333,"b":0.4772727273,"c":0.6,"d":0.7333333333,"e":0.4666666667},"annotated_formula":"add(multiply(divide(8, add(10, 8)), divide(subtract(8, const_1), subtract(add(10, 8), const_1))), multiply(divide(10, add(10, 8)), divide(subtract(10, const_1), subtract(add(10, 8), const_1))))","linear_formula":"add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)","chain":"10 + 8<\/gadget>\n18<\/output>\n8 \/ 18<\/gadget>\n4\/9 = around 0.444444<\/output>\n8 - 1<\/gadget>\n7<\/output>\n18 - 1<\/gadget>\n17<\/output>\n7 \/ 17<\/gadget>\n7\/17 = around 0.411765<\/output>\n(4\/9) * (7\/17)<\/gadget>\n28\/153 = around 0.183007<\/output>\n10 \/ 18<\/gadget>\n5\/9 = around 0.555556<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 \/ 17<\/gadget>\n9\/17 = around 0.529412<\/output>\n(5\/9) * (9\/17)<\/gadget>\n5\/17 = around 0.294118<\/output>\n(28\/153) + (5\/17)<\/gadget>\n73\/153 = around 0.477124<\/output>\n73\/153 = around 0.477124<\/result>","index":4215} +{"problem":"two bike riders ride in opposite directions around a circular track , starting at the same time from the same point . biker a rides at a speed of 16 kmph and biker b rides at a speed of 14 kmph . if the track has a diameter of 40 km , after how much time ( in hours ) will the two bikers meet ?","rationale":"explanation : distance to be covered = = 40 km relative speed of bikers = 16 + 14 = 30 kmph . now , = = 4.18 hrs . answer : c","correct":"c","options":{"a":"8.18 hrs . ","b":"6.18 hrs . ","c":"4.18 hrs . ","d":"1.18 hrs .","e":"7.18 hrs ."},"options_float":{"a":8.18,"b":6.18,"c":4.18,"d":1.18,"e":7.18},"annotated_formula":"multiply(divide(subtract(multiply(multiply(const_3, const_4), const_2), const_2), add(const_3, const_4)), divide(40, add(16, 14)))","linear_formula":"add(const_3,const_4)|add(n0,n1)|multiply(const_3,const_4)|divide(n2,#1)|multiply(#2,const_2)|subtract(#4,const_2)|divide(#5,#0)|multiply(#6,#3)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 - 2<\/gadget>\n22<\/output>\n3 + 4<\/gadget>\n7<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n16 + 14<\/gadget>\n30<\/output>\n40 \/ 30<\/gadget>\n4\/3 = around 1.333333<\/output>\n(22\/7) * (4\/3)<\/gadget>\n88\/21 = around 4.190476<\/output>\n88\/21 = around 4.190476<\/result>","index":4216} +{"problem":"a cistern 12 m long and 14 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is :","rationale":"\"area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 14 x 1.25 + 12 x 1.25 ) + 12 x 14 ] m 2 = 233 m 2 . answer : e\"","correct":"e","options":{"a":"224 m 2 ","b":"250 m 2 ","c":"220 m 2 ","d":"230 m 2","e":"233 m 2"},"options_float":{"a":224.0,"b":250.0,"c":220.0,"d":230.0,"e":233.0},"annotated_formula":"add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 14), multiply(add(divide(25, const_100), 1), 12))), multiply(14, 12))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 14<\/gadget>\n35\/2 = around 17.5<\/output>\n(5\/4) * 12<\/gadget>\n15<\/output>\n(35\/2) + 15<\/gadget>\n65\/2 = around 32.5<\/output>\n2 * (65\/2)<\/gadget>\n65<\/output>\n14 * 12<\/gadget>\n168<\/output>\n65 + 168<\/gadget>\n233<\/output>\n233<\/result>","index":4217} +{"problem":"if a \/ b = 1 \/ 3 , b \/ c = 2 , c \/ d = 1 \/ 2 , d \/ e = 3 and e \/ f = 1 \/ 10 , then what is the value of abc \/ def ?","rationale":"say a = 2 . then : a \/ b = 1 \/ 3 - - > b = 6 ; b \/ c = 2 - - > c = 3 ; c \/ d = 1 \/ 2 - - > d = 6 ; d \/ e = 3 - - > e = 2 ; e \/ f = 1 \/ 10 - - > f = 20 . abc \/ def = ( 2 * 6 * 3 ) \/ ( 6 * 2 * 20 ) = 3 \/ 20 . answer : d .","correct":"d","options":{"a":"27 \/ 4 ","b":"27 \/ 8 ","c":"3 \/ 4 ","d":"3 \/ 20","e":"1 \/ 4"},"options_float":{"a":6.75,"b":3.375,"c":0.75,"d":0.15,"e":0.25},"annotated_formula":"divide(multiply(3, 3), multiply(2, multiply(3, 10)))","linear_formula":"multiply(n1,n1)|multiply(n1,n7)|multiply(n2,#1)|divide(#0,#2)","chain":"3 * 3<\/gadget>\n9<\/output>\n3 * 10<\/gadget>\n30<\/output>\n2 * 30<\/gadget>\n60<\/output>\n9 \/ 60<\/gadget>\n3\/20 = around 0.15<\/output>\n3\/20 = around 0.15<\/result>","index":4218} +{"problem":"find the compound interest on $ 10000 in 2 years at 4 % per annum , the interest being compounded half - yearly ?","rationale":"\"principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 10000 * ( 1 + 2 \/ 100 ) ^ 4 = 10000 * 51 \/ 50 * 51 \/ 50 * 51 \/ 50 * 51 \/ 50 = $ 10824.32 c . i . = 10824.32 - 10000 = $ 824.32 answer is b\"","correct":"b","options":{"a":"$ 645.56 ","b":"$ 824.32 ","c":"$ 954.26 ","d":"$ 745.69","e":"$ 1020.45"},"options_float":{"a":645.56,"b":824.32,"c":954.26,"d":745.69,"e":1020.45},"annotated_formula":"subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 10000), 10000)","linear_formula":"divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) \/ 2<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) + 1<\/gadget>\n51\/50 = around 1.02<\/output>\n(51\/50) ** 4<\/gadget>\n6_765_201\/6_250_000 = around 1.082432<\/output>\n(6_765_201\/6_250_000) * 10_000<\/gadget>\n6_765_201\/625 = around 10_824.3216<\/output>\n(6_765_201\/625) - 10_000<\/gadget>\n515_201\/625 = around 824.3216<\/output>\n515_201\/625 = around 824.3216<\/result>","index":4220} +{"problem":"if a = 105 and a ^ 3 = 21 × 25 × 45 × z , what is the value of z ?","rationale":"\"a = 105 = 3 * 5 * 7 a ^ 3 = 21 × 25 × 45 × z = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x z = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x z = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x z z = 7 ^ 2 = 49 answer d\"","correct":"d","options":{"a":"35 ","b":"42 ","c":"45 ","d":"49","e":"54"},"options_float":{"a":35.0,"b":42.0,"c":45.0,"d":49.0,"e":54.0},"annotated_formula":"divide(power(105, 3), multiply(multiply(21, 25), 45))","linear_formula":"multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)|","chain":"105 ** 3<\/gadget>\n1_157_625<\/output>\n21 * 25<\/gadget>\n525<\/output>\n525 * 45<\/gadget>\n23_625<\/output>\n1_157_625 \/ 23_625<\/gadget>\n49<\/output>\n49<\/result>","index":4221} +{"problem":"if the cost price of 50 articles is equal to the selling price of 45 articles , then the gain or loss percent is ?","rationale":"\"given that , cost price of 50 article is equal to selling price of 45 articles . let cost price of one article = rs . 1 selling price of 45 articles = rs . 50 but cost price of 45 articles = rs . 45 therefore , the trader made profit . \\ percentage of profit = 5 \/ 45 * 100 = 11.11 % answer : a\"","correct":"a","options":{"a":"11.11 ","b":"65 ","c":"78 ","d":"33","e":"25"},"options_float":{"a":11.11,"b":65.0,"c":78.0,"d":33.0,"e":25.0},"annotated_formula":"multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 45), 50)), divide(multiply(const_100, 45), 50)))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 * 45<\/gadget>\n4_500<\/output>\n4_500 \/ 50<\/gadget>\n90<\/output>\n100 - 90<\/gadget>\n10<\/output>\n10 \/ 90<\/gadget>\n1\/9 = around 0.111111<\/output>\n100 * (1\/9)<\/gadget>\n100\/9 = around 11.111111<\/output>\n100\/9 = around 11.111111<\/result>","index":4222} +{"problem":"a train moves with a speed of 162 kmph . its speed in metres per second is","rationale":"\"solution speed = 108 kmph = ( 162 x 5 \/ 18 ) m \/ sec = 45 m \/ sec . answer e\"","correct":"e","options":{"a":"10.8 ","b":"18 ","c":"30 ","d":"38.8","e":"45"},"options_float":{"a":10.8,"b":18.0,"c":30.0,"d":38.8,"e":45.0},"annotated_formula":"multiply(162, const_0_2778)","linear_formula":"multiply(n0,const_0_2778)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n162 * (5\/18)<\/gadget>\n45<\/output>\n45<\/result>","index":4224} +{"problem":"which number should replace both the asterisks in ( * \/ 20 ) x ( * \/ 80 ) = 1 ?","rationale":"\"let ( y \/ 20 ) x ( y \/ 80 ) = 1 y ^ 2 = 20 x 80 = 20 x 20 x 4 y = ( 20 x 2 ) = 40 the answer is b .\"","correct":"b","options":{"a":"20 ","b":"40 ","c":"60 ","d":"80","e":"120"},"options_float":{"a":20.0,"b":40.0,"c":60.0,"d":80.0,"e":120.0},"annotated_formula":"sqrt(multiply(80, 20))","linear_formula":"multiply(n0,n1)|sqrt(#0)|","chain":"80 * 20<\/gadget>\n1_600<\/output>\n1_600 ** (1\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":4225} +{"problem":"a couple spent $ 211.20 in total while dining out and paid this amount using a credit card . the $ 211.20 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ?","rationale":"\"let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = $ 211.20 x = $ 160 the correct answer is d .\"","correct":"d","options":{"a":"$ 120 ","b":"$ 140 ","c":"$ 150 ","d":"$ 160","e":"$ 180"},"options_float":{"a":120.0,"b":140.0,"c":150.0,"d":160.0,"e":180.0},"annotated_formula":"divide(multiply(divide(multiply(211.20, const_100), add(const_100, 20)), const_100), add(const_100, 10))","linear_formula":"add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1)|","chain":"211.2 * 100<\/gadget>\n21_120<\/output>\n100 + 20<\/gadget>\n120<\/output>\n21_120 \/ 120<\/gadget>\n176<\/output>\n176 * 100<\/gadget>\n17_600<\/output>\n100 + 10<\/gadget>\n110<\/output>\n17_600 \/ 110<\/gadget>\n160<\/output>\n160<\/result>","index":4227} +{"problem":"at a school , 3 - fifths of the male students and two - thirds of the female students speak a foreign language . if the number of males is equal to the number of females , what fraction of the students speak a foreign language ?","rationale":"let x be the number of students in the school . the number of males who speak a foreign language is ( 3 \/ 5 ) ( 1 \/ 2 ) x = ( 3 \/ 10 ) x the number of females who speak a foreign language is ( 2 \/ 3 ) ( 1 \/ 2 ) x = ( 1 \/ 3 ) x the total number of students who speak a foreign language is ( 3 \/ 10 ) x + ( 1 \/ 3 ) x = ( 19 \/ 30 ) x the answer is b .","correct":"b","options":{"a":"17 \/ 30 ","b":"19 \/ 30 ","c":"11 \/ 15 ","d":"7 \/ 15","e":"2 \/ 5"},"options_float":{"a":0.5666666667,"b":0.6333333333,"c":0.7333333333,"d":0.4666666667,"e":0.4},"annotated_formula":"divide(add(multiply(divide(3, add(const_2, const_3)), divide(const_100, const_2)), multiply(divide(const_2, const_3), divide(const_100, const_2))), const_100)","linear_formula":"add(const_2,const_3)|divide(const_100,const_2)|divide(const_2,const_3)|divide(n0,#0)|multiply(#2,#1)|multiply(#3,#1)|add(#5,#4)|divide(#6,const_100)","chain":"2 + 3<\/gadget>\n5<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n(3\/5) * 50<\/gadget>\n30<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 50<\/gadget>\n100\/3 = around 33.333333<\/output>\n30 + (100\/3)<\/gadget>\n190\/3 = around 63.333333<\/output>\n(190\/3) \/ 100<\/gadget>\n19\/30 = around 0.633333<\/output>\n19\/30 = around 0.633333<\/result>","index":4228} +{"problem":"the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 60 kmph , find the speed of the stream ?","rationale":"\"the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) \/ ( 2 - 1 ) = 3 \/ 1 = 3 : 1 speed of the stream = 60 \/ 3 = 20 kmph . answer : e\"","correct":"e","options":{"a":"18 kmph ","b":"19 kmph ","c":"21 kmph ","d":"22 kmph","e":"20 kmph"},"options_float":{"a":18.0,"b":19.0,"c":21.0,"d":22.0,"e":20.0},"annotated_formula":"subtract(60, divide(multiply(60, const_2), const_3))","linear_formula":"multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)|","chain":"60 * 2<\/gadget>\n120<\/output>\n120 \/ 3<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20<\/result>","index":4231} +{"problem":"the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 432 sq meters , then what is the width ( in meters ) of the rectangular plot ?","rationale":"area = l * w = 3 w ^ 2 = 432 w ^ 2 = 144 w = 12 the answer is c .","correct":"c","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"divide(divide(divide(432, const_3), const_3), const_4)","linear_formula":"divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)","chain":"432 \/ 3<\/gadget>\n144<\/output>\n144 \/ 3<\/gadget>\n48<\/output>\n48 \/ 4<\/gadget>\n12<\/output>\n12<\/result>","index":4233} +{"problem":"a certain car uses 12 gallons of gasoline in traveling 120 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car ’ s gas mileage be increased ?","rationale":"120 \/ 10 = 12 . the difference is 12 - 10 = 2 . answer a","correct":"a","options":{"a":"2 ","b":"4 ","c":"6 ","d":"8","e":"10"},"options_float":{"a":2.0,"b":4.0,"c":6.0,"d":8.0,"e":10.0},"annotated_formula":"subtract(divide(120, 10), divide(120, 12))","linear_formula":"divide(n1,n2)|divide(n1,n0)|subtract(#0,#1)","chain":"120 \/ 10<\/gadget>\n12<\/output>\n120 \/ 12<\/gadget>\n10<\/output>\n12 - 10<\/gadget>\n2<\/output>\n2<\/result>","index":4234} +{"problem":"the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . find out the average age of the team .","rationale":"\"number of members in the team = 11 let the average age of of the team = x = > ( sum of ages of all 11 members ) \/ 11 = x = > sum of the ages of all 11 members = 11 x age of the captain = 26 age of the wicket keeper = 26 + 3 = 29 sum of the ages of 9 members of the team excluding captain and wicket keeper = 11 x − 26 − 29 = 11 x − 55 average age of 9 members of the team excluding captain and wicket keeper = ( 11 x − 55 ) \/ 9 given that ( 11 x − 55 ) \/ 9 = ( x − 1 ) ⇒ 11 x − 55 = 9 ( x − 1 ) ⇒ 11 x − 55 = 9 x − 9 ⇒ 2 x = 46 ⇒ x = 46 \/ 2 = 23 years answer is d .\"","correct":"d","options":{"a":"20 ","b":"21 ","c":"22 ","d":"23","e":"25"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":23.0,"e":25.0},"annotated_formula":"divide(subtract(add(26, add(26, 3)), multiply(3, 3)), const_2)","linear_formula":"add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)|","chain":"26 + 3<\/gadget>\n29<\/output>\n26 + 29<\/gadget>\n55<\/output>\n3 * 3<\/gadget>\n9<\/output>\n55 - 9<\/gadget>\n46<\/output>\n46 \/ 2<\/gadget>\n23<\/output>\n23<\/result>","index":4235} +{"problem":"a box contains 7 pairs of shoes ( 14 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ?","rationale":"\"the problem with your solution is that we do n ' t choose 1 shoe from 14 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 \/ 1 * 1 \/ 13 ( as after taking one at random there are 13 shoes left and only one is the pair of the first one ) = 1 \/ 13 answer : d .\"","correct":"d","options":{"a":"1 \/ 190 ","b":"1 \/ 20 ","c":"1 \/ 19 ","d":"1 \/ 13","e":"1 \/ 9"},"options_float":{"a":0.0052631579,"b":0.05,"c":0.0526315789,"d":0.0769230769,"e":0.1111111111},"annotated_formula":"divide(const_1, subtract(14, const_1))","linear_formula":"subtract(n1,const_1)|divide(const_1,#0)|","chain":"14 - 1<\/gadget>\n13<\/output>\n1 \/ 13<\/gadget>\n1\/13 = around 0.076923<\/output>\n1\/13 = around 0.076923<\/result>","index":4236} +{"problem":"when positive integer x is divided by positive integer y , the remainder is 3 . if x \/ y = 96.12 , what is the value of y ?","rationale":"\"when positive integer x is divided by positive integer y , the remainder is 3 - - > x = qy + 3 ; x \/ y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 3 - - > y = 25 . answer : d .\"","correct":"d","options":{"a":"96 ","b":"75 ","c":"48 ","d":"25","e":"12"},"options_float":{"a":96.0,"b":75.0,"c":48.0,"d":25.0,"e":12.0},"annotated_formula":"divide(3, subtract(96.12, floor(96.12)))","linear_formula":"floor(n1)|subtract(n1,#0)|divide(n0,#1)|","chain":"floor(96.12)<\/gadget>\n96<\/output>\n96.12 - 96<\/gadget>\n0.12<\/output>\n3 \/ 0.12<\/gadget>\n25<\/output>\n25<\/result>","index":4238} +{"problem":"if john makes a contribution to a charity fund at school , the average contribution size will increase by 50 % reaching $ 75 per person . if there were 1 other contributions made before john ' s , what is the size of his donation ?","rationale":"\"cavg = average contribution before john cavg * 1.5 = 75 , therefore the average cont is $ 50 before john . if he needs to increase the average contribution by $ 25 , he must put in $ 25 for each of the 1 people . so $ 25 . but , he also has to put in the average for himself ( the twoth person ) , so add $ 75 . so $ 100 is your answer . answer a\"","correct":"a","options":{"a":"$ 100 ","b":"$ 150 ","c":"$ 200 ","d":"$ 250","e":"$ 450"},"options_float":{"a":100.0,"b":150.0,"c":200.0,"d":250.0,"e":450.0},"annotated_formula":"add(subtract(multiply(add(1, const_1), 75), multiply(add(1, const_1), 50)), 50)","linear_formula":"add(n2,const_1)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|add(n0,#3)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 * 75<\/gadget>\n150<\/output>\n2 * 50<\/gadget>\n100<\/output>\n150 - 100<\/gadget>\n50<\/output>\n50 + 50<\/gadget>\n100<\/output>\n100<\/result>","index":4241} +{"problem":"15 men can complete a work in 10 days while 20 boys can complete the same work in 15 days . how many days will 10 men and 10 boys together take to complete the same work","rationale":"solution : work done by one man in one day = 1 \/ ( 15 * 10 ) work done by one boy in one day = 1 \/ ( 20 * 15 ) work done by 10 men and 10 boys in one day = 10 [ 1 \/ ( 15 * 10 ) + 1 \/ ( 20 * 15 ) ] days taken = 1 \/ work done by 10 men and 10 boys in one day answer b","correct":"b","options":{"a":"9 days ","b":"10 days ","c":"11 days ","d":"12 days","e":"13 days"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":12.0,"e":13.0},"annotated_formula":"inverse(add(divide(const_1, divide(multiply(15, 10), 10)), divide(const_1, multiply(15, const_2))))","linear_formula":"multiply(n0,n1)|multiply(n0,const_2)|divide(#0,n1)|divide(const_1,#1)|divide(const_1,#2)|add(#4,#3)|inverse(#5)","chain":"15 * 10<\/gadget>\n150<\/output>\n150 \/ 10<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n15 * 2<\/gadget>\n30<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n(1\/15) + (1\/30)<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ (1\/10)<\/gadget>\n10<\/output>\n10<\/result>","index":4242} +{"problem":"a rectangular tiled patio is composed of 96 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 96 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ?","rationale":"\"suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 96 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 96 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 96 ( 2 r - 4 ) * r = 96 2 r ^ 2 - 4 r - 96 = 0 r can not be negative so r = 8 and c = 12 so , answer will be c\"","correct":"c","options":{"a":"5 ","b":"6 ","c":"8 ","d":"13","e":"28"},"options_float":{"a":5.0,"b":6.0,"c":8.0,"d":13.0,"e":28.0},"annotated_formula":"divide(96, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(96, 2))))), 2))","linear_formula":"multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)|","chain":"-4<\/gadget>\n-4<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n96 * 2<\/gadget>\n192<\/output>\n4 * 192<\/gadget>\n768<\/output>\n16 + 768<\/gadget>\n784<\/output>\n784 ** (1\/2)<\/gadget>\n28<\/output>\n(-4) + 28<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n96 \/ 12<\/gadget>\n8<\/output>\n8<\/result>","index":4243} +{"problem":"find the area of a parallelogram with base 12 cm and height 48 cm ?","rationale":"\"area of a parallelogram = base * height = 12 * 48 = 576 cm 2 answer : b\"","correct":"b","options":{"a":"122 cm 2 ","b":"576 cm 2 ","c":"246 cm 2 ","d":"42 cm 2","e":"39 cm 2"},"options_float":{"a":122.0,"b":576.0,"c":246.0,"d":42.0,"e":39.0},"annotated_formula":"multiply(12, 48)","linear_formula":"multiply(n0,n1)|","chain":"12 * 48<\/gadget>\n576<\/output>\n576<\/result>","index":4245} +{"problem":"of the diplomats attending a summit conference , 25 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ?","rationale":"\"{ total } = { french } + { russian } - { both } + { neither } { total } = 25 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 70 . answer : b .\"","correct":"b","options":{"a":"72 ","b":"70 ","c":"108 ","d":"120","e":"150"},"options_float":{"a":72.0,"b":70.0,"c":108.0,"d":120.0,"e":150.0},"annotated_formula":"divide(subtract(32, 25), subtract(divide(20, const_100), divide(10, const_100)))","linear_formula":"divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)|","chain":"32 - 25<\/gadget>\n7<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/5) - (1\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n7 \/ (1\/10)<\/gadget>\n70<\/output>\n70<\/result>","index":4246} +{"problem":"a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price q per t - shirt that the vendor paid for these purchases ?","rationale":"\"correct answer : a explanation : the relevant formula for this problem is average q = ( sum ) \/ ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ' s cost was $ 165 , since 11 x 15 = 165 . the grand sum is then $ 140 + $ 165 , which equals $ 305 . the total number of shirts purchased was 25 , so to get the average price per shirt , we divide 305 by 25 , which equals $ 12.20 . as a result , the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of 11 and 14 is 12.5 , but the vendor sells more shirts at the lower price than at the higher price , the weighted average must be less than $ 12.50 ; only answer choice a is a possibility .\"","correct":"a","options":{"a":"$ 12.20 ","b":"$ 12.50 ","c":"$ 12.55 ","d":"$ 12.70","e":"$ 13.00"},"options_float":{"a":12.2,"b":12.5,"c":12.55,"d":12.7,"e":13.0},"annotated_formula":"add(divide(add(10, 14), const_2), add(const_0_25, const_0_25))","linear_formula":"add(const_0_25,const_0_25)|add(n0,n1)|divide(#1,const_2)|add(#0,#2)|","chain":"10 + 14<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n12 + (1\/2)<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":4247} +{"problem":"excluding the stoppages , the speed of a bus is 48 km \/ hr and including the stoppages the speed of the bus is 12 km \/ hr . for how many minutes does the bus stop per hour ?","rationale":"\"speed of the bus without stoppage = 48 km \/ hr speed of the bus with stoppage = 12 km \/ hr difference in speed = 36 km \/ hr so , the time taken in the stoppages = time taken to cover 36 km = ( 36 \/ 48 ) hr = 3 \/ 4 hr = 45 min answer : a\"","correct":"a","options":{"a":"45 min ","b":"10 min ","c":"12 min ","d":"20 min","e":"18 min"},"options_float":{"a":45.0,"b":10.0,"c":12.0,"d":20.0,"e":18.0},"annotated_formula":"subtract(const_60, multiply(divide(12, 48), const_60))","linear_formula":"divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|","chain":"12 \/ 48<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 60<\/gadget>\n15<\/output>\n60 - 15<\/gadget>\n45<\/output>\n45<\/result>","index":4248} +{"problem":"the area of the square formed on the diagonal of a rectangle as its side is 108 1 \/ 3 % more than the area of the rectangle . if the perimeter of the rectangle is 28 units , find the difference between the sides of the rectangle ?","rationale":"let the sides of the rectangle be l and b respectively . from the given data , ( √ l 2 + b 2 ) = ( 1 + 108 1 \/ 3 % ) lb = > l 2 + b 2 = ( 1 + 325 \/ 3 * 1 \/ 100 ) lb = ( 1 + 13 \/ 12 ) lb = 25 \/ 12 lb = > ( l 2 + b 2 ) \/ lb = 25 \/ 12 12 ( l 2 + b 2 ) = 25 lb adding 24 lb on both sides 12 l 2 + 12 b 2 + 24 lb = 49 lb 12 ( l 2 + b 2 + 2 lb ) = 49 lb but 2 ( l + b ) = 28 = > l + b = 14 12 ( l + b ) 2 = 49 lb = > 12 ( 14 ) 2 = 49 lb = > lb = 48 since l + b = 14 , l = 8 and b = 6 l - b = 8 - 6 = 2 m . answer : d","correct":"d","options":{"a":"5 ","b":"4 ","c":"7 ","d":"2","e":"1"},"options_float":{"a":5.0,"b":4.0,"c":7.0,"d":2.0,"e":1.0},"annotated_formula":"divide(subtract(divide(108, 3), 28), const_4)","linear_formula":"divide(n0,n2)|subtract(#0,n3)|divide(#1,const_4)","chain":"108 \/ 3<\/gadget>\n36<\/output>\n36 - 28<\/gadget>\n8<\/output>\n8 \/ 4<\/gadget>\n2<\/output>\n2<\/result>","index":4251} +{"problem":"the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 8800 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8800 ( 100 \/ 110 ) = rs . 8000 . answer : c\"","correct":"c","options":{"a":"rs . 6725 ","b":"rs . 6727 ","c":"rs . 8000 ","d":"rs . 6725","e":"rs . 6728"},"options_float":{"a":6725.0,"b":6727.0,"c":8000.0,"d":6725.0,"e":6728.0},"annotated_formula":"divide(8800, add(const_1, divide(10, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n8_800 \/ (11\/10)<\/gadget>\n8_000<\/output>\n8_000<\/result>","index":4253} +{"problem":"if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 15 percent , how much would she earn per week ?","rationale":"\"( 348 \/ 116 ) 115 = 345 in this case long division does not take much time . ( 348 \/ 116 ) = 3 3 * 115 = 345 ( 300 + 45 ) answer c\"","correct":"c","options":{"a":"$ 374 ","b":"$ 382 ","c":"$ 345 ","d":"$ 392","e":"$ 399"},"options_float":{"a":374.0,"b":382.0,"c":345.0,"d":392.0,"e":399.0},"annotated_formula":"add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(15, const_100), divide(348, add(const_1, divide(16, const_100)))))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4)|","chain":"16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n1 + (4\/25)<\/gadget>\n29\/25 = around 1.16<\/output>\n348 \/ (29\/25)<\/gadget>\n300<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 300<\/gadget>\n45<\/output>\n300 + 45<\/gadget>\n345<\/output>\n345<\/result>","index":4254} +{"problem":"the average monthly salary of 8 workers and one supervisor in a factory was 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . whenthesupervisor , whosesalarywas 430 . when the supervisor , whose salary was 870 per month , retired , a new person was appointed and then the average salary of 9 people was $ $ 440 per month . the salary of the new supervisor is :","rationale":"explanation : total salary of 8 workers and supervisor together = 9 ã — 430 = 3870 now total salary of 8 workers = 3870 â ˆ ’ 870 = 3000 total salary of 9 workers including the new supervisor = 9 ã — 440 = 3960 salary of the new supervisor = 3960 â ˆ ’ 3000 = 960 answer : b","correct":"b","options":{"a":"233 ","b":"960 ","c":"287 ","d":"771","e":"191"},"options_float":{"a":233.0,"b":960.0,"c":287.0,"d":771.0,"e":191.0},"annotated_formula":"subtract(multiply(440, 9), subtract(multiply(430, 9), 870))","linear_formula":"multiply(n6,n7)|multiply(n1,n6)|subtract(#1,n5)|subtract(#0,#2)","chain":"440 * 9<\/gadget>\n3_960<\/output>\n430 * 9<\/gadget>\n3_870<\/output>\n3_870 - 870<\/gadget>\n3_000<\/output>\n3_960 - 3_000<\/gadget>\n960<\/output>\n960<\/result>","index":4255} +{"problem":"if ( m - 8 ) is a factor of m ^ 2 - hm - 24 , then h =","rationale":"( m - 8 ) ( m - a ) = m ^ 2 - hm - 24 a = - 3 h = 8 + a = 5 = b","correct":"b","options":{"a":"3 ","b":"5 ","c":"6 ","d":"11","e":"16"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":11.0,"e":16.0},"annotated_formula":"subtract(8, divide(24, 8))","linear_formula":"divide(n2,n0)|subtract(n0,#0)","chain":"24 \/ 8<\/gadget>\n3<\/output>\n8 - 3<\/gadget>\n5<\/output>\n5<\/result>","index":4256} +{"problem":"the length of the bridge , which a train 500 metres long and travelling at 42 km \/ hr can cross in 60 seconds , is ?","rationale":"\"speed = [ 42 x 5 \/ 18 ] m \/ sec = [ 35 \/ 3 ] m \/ sec time = 30 sec let the length of bridge be x metres . then , ( 130 + x ) \/ 60 35 \/ 3 = ( 130 + x ) \/ 60 500 + x = 700 = > x = 200 m answer : b\"","correct":"b","options":{"a":"100 m ","b":"200 m ","c":"300 m ","d":"250 m","e":"350 m"},"options_float":{"a":100.0,"b":200.0,"c":300.0,"d":250.0,"e":350.0},"annotated_formula":"subtract(multiply(divide(multiply(42, speed(const_1000, const_1)), speed(const_3600, const_1)), 60), 500)","linear_formula":"speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|","chain":"1_000 \/ 1<\/gadget>\n1_000<\/output>\n42 * 1_000<\/gadget>\n42_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n42_000 \/ 3_600<\/gadget>\n35\/3 = around 11.666667<\/output>\n(35\/3) * 60<\/gadget>\n700<\/output>\n700 - 500<\/gadget>\n200<\/output>\n200<\/result>","index":4259} +{"problem":"a bowl contains equal numbers of red , orange , green , blue , and yellow candies . kaz eats all of the green candies and half of the orange ones . next , he eats half of the remaining pieces of each color . finally , he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 20 % of the original number . what percent of the red candies remain ?","rationale":"\"let x be the original number of each color . kaz eats all of the green candies and half of the orange ones . there are 0 green candies and 0.5 x orange candies remaining . he eats half of the remaining pieces of each color . there are 0.25 x orange candies , and 0.5 x each of red , yellow , and blue candies . he eats red and yellow candies in equal proportions . orange + blue + red + yellow = 0.75 x + red + yellow = x red + yellow = 0.25 x red = 0.125 x , since red = yellow . the answer is d .\"","correct":"d","options":{"a":"5 % ","b":"7.5 % ","c":"10 % ","d":"12.5 %","e":"15 %"},"options_float":{"a":5.0,"b":7.5,"c":10.0,"d":12.5,"e":15.0},"annotated_formula":"multiply(divide(divide(subtract(20, add(divide(divide(const_100, add(const_2, const_3)), const_2), divide(divide(divide(const_100, add(const_2, const_3)), const_2), const_2))), const_2), divide(const_100, add(const_2, const_3))), const_100)","linear_formula":"add(const_2,const_3)|divide(const_100,#0)|divide(#1,const_2)|divide(#2,const_2)|add(#2,#3)|subtract(n0,#4)|divide(#5,const_2)|divide(#6,#1)|multiply(#7,const_100)|","chain":"2 + 3<\/gadget>\n5<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n20 - 15<\/gadget>\n5<\/output>\n5 \/ 2<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) \/ 20<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":4260} +{"problem":"the average of 2 , 76 and x is 5 . find the value of x ?","rationale":"average = ( 2 + 7 + 6 + x ) \/ 4 = 5 x = 5 answer is e","correct":"e","options":{"a":"10 ","b":"8 ","c":"12 ","d":"15","e":"5"},"options_float":{"a":10.0,"b":8.0,"c":12.0,"d":15.0,"e":5.0},"annotated_formula":"subtract(multiply(5, const_4), add(add(floor(divide(76, const_10)), reminder(76, const_10)), 2))","linear_formula":"divide(n1,const_10)|multiply(n2,const_4)|reminder(n1,const_10)|floor(#0)|add(#3,#2)|add(n0,#4)|subtract(#1,#5)","chain":"5 * 4<\/gadget>\n20<\/output>\n76 \/ 10<\/gadget>\n38\/5 = around 7.6<\/output>\nfloor(38\/5)<\/gadget>\n7<\/output>\n76 % 10<\/gadget>\n6<\/output>\n7 + 6<\/gadget>\n13<\/output>\n13 + 2<\/gadget>\n15<\/output>\n20 - 15<\/gadget>\n5<\/output>\n5<\/result>","index":4263} +{"problem":"if y > 0 , ( 6 y ) \/ 20 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"can be reduced to 3 y \/ 10 + 3 y \/ 10 = 3 y \/ 5 = 60 % c\"","correct":"c","options":{"a":"40 % ","b":"50 % ","c":"60 % ","d":"70 %","e":"80 %"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(const_100, add(divide(6, 20), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"6 \/ 20<\/gadget>\n3\/10 = around 0.3<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) + (3\/10)<\/gadget>\n3\/5 = around 0.6<\/output>\n100 * (3\/5)<\/gadget>\n60<\/output>\n60<\/result>","index":4265} +{"problem":"if the number is divided by 3 , it reduced by 36 . the number is","rationale":"\"explanation : let the number be x . then , x - ( x \/ 3 ) = 36 = > 2 x \/ 3 = 36 = > x = 54 answer : option d\"","correct":"d","options":{"a":"51 ","b":"50 ","c":"45 ","d":"54","e":"36"},"options_float":{"a":51.0,"b":50.0,"c":45.0,"d":54.0,"e":36.0},"annotated_formula":"divide(multiply(36, 3), subtract(3, const_1))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)|","chain":"36 * 3<\/gadget>\n108<\/output>\n3 - 1<\/gadget>\n2<\/output>\n108 \/ 2<\/gadget>\n54<\/output>\n54<\/result>","index":4266} +{"problem":"in a group of 150 readers who read science fiction or literacy works or both , 120 read science fiction and 90 read literacy works . how many read both science fiction and literacy works ?","rationale":"consider total number of reader n ( s u l ) = 150 people who read science fiction n ( s ) = 120 people who read literacy works n ( l ) = 90 both science fiction and literacy n ( s â ˆ © l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s â ˆ © l ) 150 = 120 + 90 - n ( s â ˆ © l ) n ( s â ˆ © l ) = 210 - 150 n ( s â ˆ © l ) = 60 so people read both science fiction and literacy works are 160 answer : b","correct":"b","options":{"a":"70 ","b":"60 ","c":"90 ","d":"100","e":"110"},"options_float":{"a":70.0,"b":60.0,"c":90.0,"d":100.0,"e":110.0},"annotated_formula":"subtract(add(120, 90), 150)","linear_formula":"add(n1,n2)|subtract(#0,n0)","chain":"120 + 90<\/gadget>\n210<\/output>\n210 - 150<\/gadget>\n60<\/output>\n60<\/result>","index":4268} +{"problem":"the speed of a boat in still water in 30 km \/ hr and the rate of current is 7 km \/ hr . the distance travelled downstream in 36 minutes is :","rationale":"\"explanation : speed downstream = ( 30 + 7 ) = 37 kmph time = 36 minutes = 36 \/ 60 hour = 3 \/ 5 hour distance travelled = time × speed = ( 3 \/ 5 ) × 37 = 22.2 km answer : option e\"","correct":"e","options":{"a":"18.4 km ","b":"20.6 km ","c":"21.4 km ","d":"1111 km","e":"22.2 km"},"options_float":{"a":18.4,"b":20.6,"c":21.4,"d":1111.0,"e":22.2},"annotated_formula":"multiply(add(30, 7), divide(36, const_60))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"30 + 7<\/gadget>\n37<\/output>\n36 \/ 60<\/gadget>\n3\/5 = around 0.6<\/output>\n37 * (3\/5)<\/gadget>\n111\/5 = around 22.2<\/output>\n111\/5 = around 22.2<\/result>","index":4269} +{"problem":"39 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 3 hours a day , complete the work ?","rationale":"\"let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 39 : : 12 : x working hours \/ day 3 : 5 30 x 3 x x = 39 x 5 x 12 x = ( 39 x 5 x 12 ) \/ ( 30 x 3 ) x = 26 . answer : b\"","correct":"b","options":{"a":"10 ","b":"26 ","c":"14 ","d":"15","e":"16"},"options_float":{"a":10.0,"b":26.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"divide(multiply(multiply(39, 12), 5), multiply(30, 3))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|","chain":"39 * 12<\/gadget>\n468<\/output>\n468 * 5<\/gadget>\n2_340<\/output>\n30 * 3<\/gadget>\n90<\/output>\n2_340 \/ 90<\/gadget>\n26<\/output>\n26<\/result>","index":4270} +{"problem":"cole drove from home to work at an average speed of 80 kmh . he then returned home at an average speed of 120 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ?","rationale":"\"first round distance travelled ( say ) = d speed = 80 k \/ h time taken , t 2 = d \/ 80 hr second round distance traveled = d ( same distance ) speed = 120 k \/ h time taken , t 2 = d \/ 120 hr total time taken = 2 hrs therefore , 2 = d \/ 80 + d \/ 120 lcm of 80 and 120 = 240 2 = d \/ 80 + d \/ 120 = > 2 = 3 d \/ 240 + 2 d \/ 240 = > d = 240 * 2 \/ 5 km therefore , t 1 = d \/ 80 = > t 1 = ( 240 * 2 ) \/ ( 5 x 80 ) = > t 1 = ( 6 x 60 ) \/ 5 - - in minutes = > t 1 = 72 minutes . c\"","correct":"c","options":{"a":"66 ","b":"70 ","c":"72 ","d":"75","e":"78"},"options_float":{"a":66.0,"b":70.0,"c":72.0,"d":75.0,"e":78.0},"annotated_formula":"multiply(divide(multiply(120, 2), add(80, 120)), const_60)","linear_formula":"add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)|","chain":"120 * 2<\/gadget>\n240<\/output>\n80 + 120<\/gadget>\n200<\/output>\n240 \/ 200<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 60<\/gadget>\n72<\/output>\n72<\/result>","index":4272} +{"problem":"a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 25 % , what should be the sale price of the mixed tea ?","rationale":"\"c . p . of mixture = 80 × 15 + 20 × 20 \/ 80 + 20 = 16 ∴ s . p . = ( 100 + 25 ) \/ 100 × 16 = 20 answer c\"","correct":"c","options":{"a":"23.75 ","b":"22 ","c":"20 ","d":"19.20","e":"none of these"},"options_float":{"a":23.75,"b":22.0,"c":20.0,"d":19.2,"e":null},"annotated_formula":"add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 25), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n2)|add(#1,#2)|divide(#3,#0)|multiply(n4,#4)|divide(#5,const_100)|add(#6,#4)|","chain":"80 * 15<\/gadget>\n1_200<\/output>\n20 * 20<\/gadget>\n400<\/output>\n1_200 + 400<\/gadget>\n1_600<\/output>\n80 + 20<\/gadget>\n100<\/output>\n1_600 \/ 100<\/gadget>\n16<\/output>\n16 * 25<\/gadget>\n400<\/output>\n400 \/ 100<\/gadget>\n4<\/output>\n4 + 16<\/gadget>\n20<\/output>\n20<\/result>","index":4274} +{"problem":"pipe a fills a tank of capacity 800 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ?","rationale":"\"in one cycle they fill 40 + 30 - 20 = 50 liters 800 = 50 * n = > n = 16 here n = number of cycles . total time = 16 * 3 = 48 as in one cycle there are 3 minutes . thus 48 minutes answer : a\"","correct":"a","options":{"a":"48 minutes ","b":"14 minutes ","c":"39 minutes ","d":"40 minutes 20 seconds","e":"none of these"},"options_float":{"a":48.0,"b":14.0,"c":39.0,"d":40.0,"e":null},"annotated_formula":"multiply(divide(800, subtract(add(40, 30), 20)), const_3)","linear_formula":"add(n1,n2)|subtract(#0,n3)|divide(n0,#1)|multiply(#2,const_3)|","chain":"40 + 30<\/gadget>\n70<\/output>\n70 - 20<\/gadget>\n50<\/output>\n800 \/ 50<\/gadget>\n16<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48<\/result>","index":4275} +{"problem":"if w \/ x = 1 \/ 3 and w \/ y = 3 \/ 4 , then ( x + y ) \/ y =","rationale":"\"w \/ x = 1 \/ 3 = > x = 3 w and w \/ y = 3 \/ 4 = > y = ( 4 \/ 3 ) w ( x + y ) \/ y = ( 3 w + 4 \/ 3 w ) \/ ( 4 \/ 3 w ) = ( 13 \/ 3 w ) \/ ( 4 \/ 3 w ) = 13 \/ 4 correct option : a\"","correct":"a","options":{"a":"13 \/ 4 ","b":"6 \/ 5 ","c":"7 \/ 5 ","d":"8 \/ 5","e":"9 \/ 5"},"options_float":{"a":3.25,"b":1.2,"c":1.4,"d":1.6,"e":1.8},"annotated_formula":"add(divide(divide(3, 1), divide(4, 3)), const_1)","linear_formula":"divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)|","chain":"3 \/ 1<\/gadget>\n3<\/output>\n4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 \/ (4\/3)<\/gadget>\n9\/4 = around 2.25<\/output>\n(9\/4) + 1<\/gadget>\n13\/4 = around 3.25<\/output>\n13\/4 = around 3.25<\/result>","index":4277} +{"problem":"the speed of a boat in still water is 15 km \/ hr and the rate of current is 3 km \/ hr . the distance travelled downstream in 24 minutes is","rationale":"\"speed of a boat in still water = 15 km \/ hr speed of the current = 3 km \/ hr speed downstream = ( 15 + 3 ) = 18 km \/ hr distance travelled downstream in 24 minutes = ( 24 \/ 60 ) × 18 = ( 2 × 18 ) \/ 5 = 7.2 km answer is b\"","correct":"b","options":{"a":"17.2 km ","b":"7.2 km ","c":"7.6 km ","d":"17.6 km","e":"27.2 km"},"options_float":{"a":17.2,"b":7.2,"c":7.6,"d":17.6,"e":27.2},"annotated_formula":"multiply(divide(24, const_60), add(15, 3))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"24 \/ 60<\/gadget>\n2\/5 = around 0.4<\/output>\n15 + 3<\/gadget>\n18<\/output>\n(2\/5) * 18<\/gadget>\n36\/5 = around 7.2<\/output>\n36\/5 = around 7.2<\/result>","index":4278} +{"problem":"a man is 20 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ?","rationale":"\"let the son ' s present age be x years . then , man ' s present age = ( x + 20 ) years . ( x + 20 ) + 2 = 2 ( x + 2 ) x + 22 = 2 x + 4 = > x = 18 . answer : d\"","correct":"d","options":{"a":"11 ","b":"25 ","c":"27 ","d":"18","e":"91"},"options_float":{"a":11.0,"b":25.0,"c":27.0,"d":18.0,"e":91.0},"annotated_formula":"divide(subtract(20, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))","linear_formula":"multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|","chain":"2 * 2<\/gadget>\n4<\/output>\n4 - 2<\/gadget>\n2<\/output>\n20 - 2<\/gadget>\n18<\/output>\n2 - 1<\/gadget>\n1<\/output>\n18 \/ 1<\/gadget>\n18<\/output>\n18<\/result>","index":4279} +{"problem":"a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes 1 square feet of area in her garden . this year , she has increased her output by 193 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ?","rationale":"\"let the side for growing cabbages this year be x ft . thus the area is x ^ 2 . let the side for growing cabbages last year be y ft . thus , the area was y ^ 2 . the area would have increased by 193 sq ft as each cabbage takes 1 sq ft space . x ^ 2 - y ^ 2 = 193 ( x + y ) ( x - y ) = 193 193 is a prime number and thus it will be ( 97 + 96 ) * ( 97 - 96 ) . thus x = 97 and y = 96 x ^ 2 = 97 ^ 2 = 9409 the answer is b .\"","correct":"b","options":{"a":"8,208 ","b":"9,409 ","c":"11,424 ","d":"12,586","e":"can not be determined"},"options_float":{"a":8208.0,"b":9409.0,"c":11424.0,"d":12586.0,"e":null},"annotated_formula":"power(add(divide(193, const_2), add(const_0_25, const_0_25)), const_2)","linear_formula":"add(const_0_25,const_0_25)|divide(n1,const_2)|add(#0,#1)|power(#2,const_2)|","chain":"193 \/ 2<\/gadget>\n193\/2 = around 96.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n(193\/2) + (1\/2)<\/gadget>\n97<\/output>\n97 ** 2<\/gadget>\n9_409<\/output>\n9_409<\/result>","index":4280} +{"problem":"if d is the smallest positive integer such that 3150 multiplied by d is the square of an integer , then d must be","rationale":"solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3150 . 3150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that the exponents of both 2 and 7 are not even numbers . this tells us that 3150 itself is not a perfect square . ) we also are given that 3150 multiplied by d is the square of an integer . we can write this as : 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 x d = square of an integer according to our rule , we need all unique prime factors ' exponents to be even numbers . thus , we need one more 2 and one more 7 . therefore , d = 7 x 2 = 14 answer is e .","correct":"e","options":{"a":"2 ","b":"5 ","c":"6 ","d":"7","e":"14"},"options_float":{"a":2.0,"b":5.0,"c":6.0,"d":7.0,"e":14.0},"annotated_formula":"divide(3150, multiply(power(const_3, const_2), power(add(const_4, const_1), const_2)))","linear_formula":"add(const_1,const_4)|power(const_3,const_2)|power(#0,const_2)|multiply(#1,#2)|divide(n0,#3)","chain":"3 ** 2<\/gadget>\n9<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n9 * 25<\/gadget>\n225<\/output>\n3_150 \/ 225<\/gadget>\n14<\/output>\n14<\/result>","index":4282} +{"problem":"at garage sale , all of the prices of the items sold were different . if the price of a radio sold at the garage sale was both the 9 th highest price and the 35 th lowest price among the prices of the items sold , how many items were sold at the garage sale ?","rationale":"8 + 34 + 1 = 43 answer : a","correct":"a","options":{"a":"43 ","b":"28 ","c":"44 ","d":"49","e":"46"},"options_float":{"a":43.0,"b":28.0,"c":44.0,"d":49.0,"e":46.0},"annotated_formula":"subtract(add(9, 35), const_1)","linear_formula":"add(n0,n1)|subtract(#0,const_1)","chain":"9 + 35<\/gadget>\n44<\/output>\n44 - 1<\/gadget>\n43<\/output>\n43<\/result>","index":4283} +{"problem":"excluding the stoppages , the speed of a bus is 12 km \/ hr and including the stoppages the speed of the bus is 6 km \/ hr . for how many minutes does the bus stop per hour ?","rationale":"\"speed of the bus without stoppage = 12 km \/ hr speed of the bus with stoppage = 6 km \/ hr difference in speed = 6 km \/ hr so , the time taken in the stoppages = time taken to cover 6 km = ( 6 \/ 12 ) hr = 1 \/ 2 hr = 30 min answer : b\"","correct":"b","options":{"a":"15 min ","b":"30 min ","c":"12 min ","d":"20 min","e":"18 min"},"options_float":{"a":15.0,"b":30.0,"c":12.0,"d":20.0,"e":18.0},"annotated_formula":"subtract(const_60, multiply(divide(6, 12), const_60))","linear_formula":"divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|","chain":"6 \/ 12<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 60<\/gadget>\n30<\/output>\n60 - 30<\/gadget>\n30<\/output>\n30<\/result>","index":4284} +{"problem":"in a park there are two ponds with both brown ducks and green ducks . in the smaller pond there are 45 ducks and in the larger pond there are 55 ducks . if 20 % of the ducks in the smaller pond are green and 40 % of the ducks in the larger pond are green , then what percentage of ducks are green ?","rationale":"\"number of ducks in small pond = 45 green ducks in small pond = 20 % of 45 = 9 ducks number of ducks in large pond = 55 green ducks in large pond = 40 % of 55 = 22 ducks total number of ducks = 45 + 55 = 100 total number of green ducks = 9 + 22 = 31 ducks percentage of green ducks = 31 \/ 100 * 100 = 31 % answer : d\"","correct":"d","options":{"a":"25 % ","b":"27 % ","c":"29 % ","d":"31 %","e":"33 %"},"options_float":{"a":25.0,"b":27.0,"c":29.0,"d":31.0,"e":33.0},"annotated_formula":"multiply(divide(add(multiply(45, divide(20, const_100)), multiply(55, divide(40, const_100))), add(45, 55)), const_100)","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n45 * (1\/5)<\/gadget>\n9<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n55 * (2\/5)<\/gadget>\n22<\/output>\n9 + 22<\/gadget>\n31<\/output>\n45 + 55<\/gadget>\n100<\/output>\n31 \/ 100<\/gadget>\n31\/100 = around 0.31<\/output>\n(31\/100) * 100<\/gadget>\n31<\/output>\n31<\/result>","index":4286} +{"problem":"rice weighing 35 \/ 2 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces )","rationale":"35 \/ 2 ÷ 4 = 35 \/ 8 pounds in each container 35 \/ 8 pounds * 16 ounces \/ pound = 70 ounces in each container the answer is d .","correct":"d","options":{"a":"40 ","b":"50 ","c":"60 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"divide(multiply(divide(35, 2), 16), 4)","linear_formula":"divide(n0,n1)|multiply(n4,#0)|divide(#1,n2)","chain":"35 \/ 2<\/gadget>\n35\/2 = around 17.5<\/output>\n(35\/2) * 16<\/gadget>\n280<\/output>\n280 \/ 4<\/gadget>\n70<\/output>\n70<\/result>","index":4287} +{"problem":"ashok secured average of 74 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ?","rationale":"\"explanation : number of subjects = 6 average of marks in 6 subjects = 74 therefore total marks in 6 subjects = 74 * 6 = 444 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 444 – 370 = 74 answer b\"","correct":"b","options":{"a":"66 ","b":"74 ","c":"78 ","d":"80","e":"none of these"},"options_float":{"a":66.0,"b":74.0,"c":78.0,"d":80.0,"e":null},"annotated_formula":"subtract(multiply(74, 6), multiply(74, 5))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|","chain":"74 * 6<\/gadget>\n444<\/output>\n74 * 5<\/gadget>\n370<\/output>\n444 - 370<\/gadget>\n74<\/output>\n74<\/result>","index":4290} +{"problem":"susan made a block with small cubes of 5 cubic cm volume to make a block 7 small cubes long , 7 small cubes wide and 6 small cubes deep . she realise that she has used more small cubes than she really needed . she realised that she could have glued a fewer number of cubes together ot look like a block with same dimensions , if it were made","rationale":"volume of cube of dimensions 7 7 6 is 7 * 7 * 6 = 294 if it were to look hollow the imaginary inside cube volume is to be deducted which is ( 7 - 2 ) * ( 7 - 2 ) * ( 6 - 2 ) = 100 ( here 2 is deducted because the inside cube which has to be hollow is 2 cubes more on every dimension ) so minimum number of cubes required to make hollow cube is 294 - 100 = 194 cubes answer : a","correct":"a","options":{"a":"194 ","b":"195 ","c":"196 ","d":"197","e":"198"},"options_float":{"a":194.0,"b":195.0,"c":196.0,"d":197.0,"e":198.0},"annotated_formula":"subtract(multiply(multiply(7, 7), 6), multiply(multiply(subtract(7, const_2), subtract(7, const_2)), subtract(6, const_2)))","linear_formula":"multiply(n1,n1)|subtract(n1,const_2)|subtract(n3,const_2)|multiply(n3,#0)|multiply(#1,#1)|multiply(#4,#2)|subtract(#3,#5)","chain":"7 * 7<\/gadget>\n49<\/output>\n49 * 6<\/gadget>\n294<\/output>\n7 - 2<\/gadget>\n5<\/output>\n5 * 5<\/gadget>\n25<\/output>\n6 - 2<\/gadget>\n4<\/output>\n25 * 4<\/gadget>\n100<\/output>\n294 - 100<\/gadget>\n194<\/output>\n194<\/result>","index":4291} +{"problem":"in a dairy farm , 55 cows eat 55 bags of husk in 55 days . in how many days one cow will eat one bag of husk ?","rationale":"\"explanation : less cows , more days ( indirect proportion ) less bags , less days ( direct proportion ) [ cows 1 55 bags 55 1 ] : : 55 : x = > x ∗ 55 ∗ 1 = 55 ∗ 1 ∗ 55 = > x = 55 option e\"","correct":"e","options":{"a":"34 days ","b":"40 days ","c":"46 days ","d":"50 days","e":"55 days"},"options_float":{"a":34.0,"b":40.0,"c":46.0,"d":50.0,"e":55.0},"annotated_formula":"multiply(divide(55, 55), 55)","linear_formula":"divide(n0,n0)|multiply(n0,#0)|","chain":"55 \/ 55<\/gadget>\n1<\/output>\n1 * 55<\/gadget>\n55<\/output>\n55<\/result>","index":4292} +{"problem":"a tank is filled in 10 hours by 3 pipes a , b and c . the pipe c is twice as fast as b and b is twice as fast as a . how much time will pipe a alone take to fill the tank ?","rationale":"1 \/ a + 1 \/ b + 1 \/ c = 1 \/ 10 ( a , b and c 1 hour work ) given . . 1 \/ b = 2 \/ a and 1 \/ c = 2 \/ b = 4 \/ a hence 1 \/ a + 2 \/ a + 4 \/ a = 1 \/ 10 on solving . . . . we get 1 \/ a = 1 \/ 70 ( work done by a in 1 hour ) therefore time taken by a is 70 hours to finish the whole work answer : a","correct":"a","options":{"a":"70 hours ","b":"30 hours ","c":"35 hours ","d":"50 hours","e":"60 hours"},"options_float":{"a":70.0,"b":30.0,"c":35.0,"d":50.0,"e":60.0},"annotated_formula":"multiply(add(const_4, add(const_1, const_2)), 10)","linear_formula":"add(const_1,const_2)|add(#0,const_4)|multiply(n0,#1)","chain":"1 + 2<\/gadget>\n3<\/output>\n4 + 3<\/gadget>\n7<\/output>\n7 * 10<\/gadget>\n70<\/output>\n70<\/result>","index":4297} +{"problem":"in a certain production lot , 40 percent of the toys are red and the remaining toys are green . half of toys are small and half are large . if 10 percent of the toys are red and small , and 40 toys are green and large . how many of the toys are red and large ?","rationale":"40 % toys are red . 50 % toys are small - 50 % are large red and small - 10 % red and large must be 30 % ( since there are total 40 % red toys ) but 50 % toys are large . if 30 % are red and large , the rest 20 % large toys must be green . so large and green toys are 20 % . these 20 % large and green toys are actually 40 so total number of toys must be 200 . since red and large toys are 30 % , they must be 60 . answer : b","correct":"b","options":{"a":"40 ","b":"60 ","c":"70 ","d":"80","e":"90"},"options_float":{"a":40.0,"b":60.0,"c":70.0,"d":80.0,"e":90.0},"annotated_formula":"multiply(subtract(divide(40, const_100), divide(10, const_100)), divide(40, subtract(const_1, subtract(add(divide(40, const_100), divide(const_1, const_2)), divide(10, const_100)))))","linear_formula":"divide(n0,const_100)|divide(const_1,const_2)|divide(n1,const_100)|add(#0,#1)|subtract(#0,#2)|subtract(#3,#2)|subtract(const_1,#5)|divide(n0,#6)|multiply(#7,#4)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(2\/5) - (1\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(2\/5) + (1\/2)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) - (1\/10)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n40 \/ (1\/5)<\/gadget>\n200<\/output>\n(3\/10) * 200<\/gadget>\n60<\/output>\n60<\/result>","index":4298} +{"problem":"if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4030 , r will receive","rationale":"\"explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p \/ 3 r = 2 p \/ 5 p : q : r = p : 2 p \/ 3 : 2 p \/ 5 = 15 : 10 : 6 r ' s share = 4030 * ( 6 \/ 31 ) = 130 * 6 = 780 . answer : option b\"","correct":"b","options":{"a":"600 ","b":"780 ","c":"800 ","d":"900","e":"none of these"},"options_float":{"a":600.0,"b":780.0,"c":800.0,"d":900.0,"e":null},"annotated_formula":"multiply(4030, divide(6, add(add(add(10, add(4, const_1)), 10), 6)))","linear_formula":"add(n0,const_1)|add(n2,#0)|add(n2,#1)|add(n1,#2)|divide(n1,#3)|multiply(n3,#4)|","chain":"4 + 1<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15 + 10<\/gadget>\n25<\/output>\n25 + 6<\/gadget>\n31<\/output>\n6 \/ 31<\/gadget>\n6\/31 = around 0.193548<\/output>\n4_030 * (6\/31)<\/gadget>\n780<\/output>\n780<\/result>","index":4299} +{"problem":"mark and ann together were allocated n boxes of cookies to sell for a club project . mark sold 11 boxes less than n and ann sold 2 boxes less than n . if mark and ann have each sold at least one box of cookies , but together they have sold less than n boxes , what is the value of n ?","rationale":"if n = 12 mark sold 1 box and ann sold 10 boxes total 11 < 12 answer : c","correct":"c","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"add(11, const_1)","linear_formula":"add(n0,const_1)","chain":"11 + 1<\/gadget>\n12<\/output>\n12<\/result>","index":4301} +{"problem":"in a forest 10 rabbits were caught , tagged with electronic markers , then released . a month later , 10 rabbits were captured in the same forest . of these 10 rabbits , it was found that 2 had been tagged with the electronic markers . if the percentage of tagged rabbits in the second sample approximates the percentage of tagged rabbits in the forest , and if no rabbits had either left or entered the forest over the preceding month , what is the approximate number of rabbits in the forest ?","rationale":"given that 10 rabbits were caught and tagged in the first sample , and 10 were caught in the second sample of which 2 had tags , then 2 \/ 10 approximates the fraction of the population of tagged rabbits in the forest . proportion : a \/ x = c \/ d a = rabbits tagged first sample c \/ d = the fraction of the population of tagged rabbits in the forest x = the population of rabbits in the forest proportion : 10 \/ x = 1 \/ 5 x = ( 1 ) ( 10 ) ( 5 ) = 50 . there are approximately 50 rabbits in the forest . answer is c","correct":"c","options":{"a":"150 ","b":"70 ","c":"50 ","d":"1500","e":"500"},"options_float":{"a":150.0,"b":70.0,"c":50.0,"d":1500.0,"e":500.0},"annotated_formula":"add(add(add(10, 10), 10), multiply(divide(2, 10), const_100))","linear_formula":"add(n0,n0)|divide(n3,n0)|add(n0,#0)|multiply(#1,const_100)|add(#2,#3)","chain":"10 + 10<\/gadget>\n20<\/output>\n20 + 10<\/gadget>\n30<\/output>\n2 \/ 10<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n30 + 20<\/gadget>\n50<\/output>\n50<\/result>","index":4302} +{"problem":"train a leaves the station traveling at 30 miles per hour . two hours later train в leaves the same station traveling in the same direction at 45 miles per hour . how many miles from the station was train a overtaken by train b ?","rationale":"\"after two hours , train a is ahead by 60 miles . train b can catch up at a rate of 15 miles per hour . the time to catch up is 60 \/ 15 = 4 hours . in 4 hours , train a travels another 30 * 4 = 120 miles for a total of 180 miles . the answer is e .\"","correct":"e","options":{"a":"100 ","b":"120 ","c":"140 ","d":"160","e":"180"},"options_float":{"a":100.0,"b":120.0,"c":140.0,"d":160.0,"e":180.0},"annotated_formula":"multiply(divide(multiply(30, const_2), subtract(45, 30)), 45)","linear_formula":"multiply(n0,const_2)|subtract(n1,n0)|divide(#0,#1)|multiply(n1,#2)|","chain":"30 * 2<\/gadget>\n60<\/output>\n45 - 30<\/gadget>\n15<\/output>\n60 \/ 15<\/gadget>\n4<\/output>\n4 * 45<\/gadget>\n180<\/output>\n180<\/result>","index":4304} +{"problem":"micheal and adam can do together a piece of work in 20 days . after they have worked together for 15 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately .","rationale":"\"rate of both = 1 \/ 20 together they do = 1 \/ 20 * 15 = 3 \/ 4 left work = 1 - 3 \/ 4 = 1 \/ 4 adam completes 1 \/ 4 work in 10 day so he took 10 * 4 = 40 days to complete the left work alone . thus the rate of adam is 40 \/ 1 rate of micheal = 1 \/ 20 - 1 \/ 40 = 1 \/ 40 thus micheal takes 40 days to complete the whole work . ans . a .\"","correct":"a","options":{"a":"40 days ","b":"100 days ","c":"120 days ","d":"110 days","e":"90 days"},"options_float":{"a":40.0,"b":100.0,"c":120.0,"d":110.0,"e":90.0},"annotated_formula":"inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 15))), 10))))","linear_formula":"inverse(n0)|multiply(n1,#0)|subtract(const_1,#1)|inverse(#2)|multiply(n2,#3)|inverse(#4)|subtract(#0,#5)|inverse(#6)|","chain":"1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 15<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ (1\/4)<\/gadget>\n4<\/output>\n4 * 10<\/gadget>\n40<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/20) - (1\/40)<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":4307} +{"problem":"pipe a can fill a tank in 6 hours . due to a leak at the bottom , it takes 10 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ?","rationale":"\"let the leak can empty the full tank in x hours 1 \/ 6 - 1 \/ x = 1 \/ 10 = > 1 \/ x = 1 \/ 6 - 1 \/ 10 = ( 5 - 3 ) \/ 30 = 2 \/ 30 = 1 \/ 15 = > x = 15 . answer : c\"","correct":"c","options":{"a":"13 ","b":"17 ","c":"15 ","d":"19","e":"12"},"options_float":{"a":13.0,"b":17.0,"c":15.0,"d":19.0,"e":12.0},"annotated_formula":"divide(multiply(10, 6), subtract(10, 6))","linear_formula":"multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|","chain":"10 * 6<\/gadget>\n60<\/output>\n10 - 6<\/gadget>\n4<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15<\/result>","index":4308} +{"problem":"how many terminating zeroes w does 200 ! have ?","rationale":"you have 40 multiples of 5 , 8 of 25 and 1 of 125 . this will give 49 zeros . c","correct":"c","options":{"a":"40 ","b":"48 ","c":"49 ","d":"55","e":"64"},"options_float":{"a":40.0,"b":48.0,"c":49.0,"d":55.0,"e":64.0},"annotated_formula":"add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1))))","linear_formula":"add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3)|","chain":"4 + 1<\/gadget>\n5<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n5 * 5<\/gadget>\n25<\/output>\n200 \/ 25<\/gadget>\n8<\/output>\n40 + 8<\/gadget>\n48<\/output>\n48<\/result>","index":4311} +{"problem":"two trains 500 m and 750 m long run at the speed of 60 km \/ hr and 40 km \/ hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ?","rationale":"\"relative speed = 60 + 40 = 100 km \/ hr . = 100 * 5 \/ 18 = 250 \/ 9 m \/ sec . distance covered in crossing each other = 500 + 750 = 1250 m . required time = 1250 * 9 \/ 250 = 45 sec . answer : b\"","correct":"b","options":{"a":"10.6 ","b":"45 ","c":"10.4 ","d":"10.8","e":"10.1"},"options_float":{"a":10.6,"b":45.0,"c":10.4,"d":10.8,"e":10.1},"annotated_formula":"divide(add(500, 750), multiply(add(60, 40), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"500 + 750<\/gadget>\n1_250<\/output>\n60 + 40<\/gadget>\n100<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n100 * (5\/18)<\/gadget>\n250\/9 = around 27.777778<\/output>\n1_250 \/ (250\/9)<\/gadget>\n45<\/output>\n45<\/result>","index":4312} +{"problem":"a train 110 m long is running with a speed of 36 km \/ hr . in what time will it pass a man who is running at 6 km \/ hr in the direction opposite to that in which the train is going ?","rationale":"\"speed of train relative to man = 36 + 6 = 42 km \/ hr . = 42 * 5 \/ 18 = 35 \/ 3 m \/ sec . time taken to pass the men = 110 * 3 \/ 35 = 9 sec . answer : d\"","correct":"d","options":{"a":"5 ","b":"6 ","c":"7 ","d":"9","e":"5"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":9.0,"e":5.0},"annotated_formula":"divide(110, multiply(add(36, 6), const_0_2778))","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"36 + 6<\/gadget>\n42<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n42 * (5\/18)<\/gadget>\n35\/3 = around 11.666667<\/output>\n110 \/ (35\/3)<\/gadget>\n66\/7 = around 9.428571<\/output>\n66\/7 = around 9.428571<\/result>","index":4313} +{"problem":"tough and tricky questions : remainders . 1 ^ 1 + 2 ^ 2 + 3 ^ 3 + . . . + 7 ^ 7 is divided by 7 . what is the remainder ?","rationale":"a number is divisible by 5 , if its last digit is divisible by 5 let ' s look into the sum of last digits of each term of the given expression 1 ^ 1 = 1 2 ^ 2 = 4 3 ^ 3 = 7 4 ^ 4 = 6 5 ^ 5 = 5 6 ^ 6 = 6 7 ^ 7 = 3 adding all these numbers we get 32 which gives a remainder of 4 when divided by 7 . so answer must be 4 . bunuel , can you please confirm the answer of this question . yes , the oa is c . clicked the wrong button when posting . edited . thank you . e","correct":"e","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"subtract(add(add(add(add(add(add(1, const_4), 7), add(const_4, const_2)), add(const_4, const_1)), add(const_4, const_2)), const_3), multiply(7, floor(divide(add(add(add(add(add(add(1, const_4), 7), add(const_4, const_2)), add(const_4, const_1)), add(const_4, const_2)), const_3), 7))))","linear_formula":"add(n0,const_4)|add(const_2,const_4)|add(const_1,const_4)|add(n6,#0)|add(#3,#1)|add(#4,#2)|add(#5,#1)|add(#6,const_3)|divide(#7,n6)|floor(#8)|multiply(n6,#9)|subtract(#7,#10)","chain":"1 + 4<\/gadget>\n5<\/output>\n5 + 7<\/gadget>\n12<\/output>\n4 + 2<\/gadget>\n6<\/output>\n12 + 6<\/gadget>\n18<\/output>\n4 + 1<\/gadget>\n5<\/output>\n18 + 5<\/gadget>\n23<\/output>\n23 + 6<\/gadget>\n29<\/output>\n29 + 3<\/gadget>\n32<\/output>\n32 \/ 7<\/gadget>\n32\/7 = around 4.571429<\/output>\nfloor(32\/7)<\/gadget>\n4<\/output>\n7 * 4<\/gadget>\n28<\/output>\n32 - 28<\/gadget>\n4<\/output>\n4<\/result>","index":4315} +{"problem":"a sells a bicycle to b and makes a profit of 50 % . b sells the same bicycle to c at a profit of 25 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a .","rationale":"\"explanation : let cp be 100 a sells at 50 % profit so sp = 150 b sells at 25 % profit = 150 x ( 1 + 25 \/ 100 ) = 187.50 cp - - - sp 100 - - - 187.50 x - - - 225 cp = 225 x 100 \/ 187.50 = 120 answer : e\"","correct":"e","options":{"a":"237 ","b":"126 ","c":"971 ","d":"611","e":"120"},"options_float":{"a":237.0,"b":126.0,"c":971.0,"d":611.0,"e":120.0},"annotated_formula":"divide(divide(225, divide(add(const_100, 25), const_100)), divide(add(50, const_100), const_100))","linear_formula":"add(n1,const_100)|add(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#2)|divide(#4,#3)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n225 \/ (5\/4)<\/gadget>\n180<\/output>\n50 + 100<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n180 \/ (3\/2)<\/gadget>\n120<\/output>\n120<\/result>","index":4316} +{"problem":"average marks of a , b , c is 48 . when d joins average becomes 47 . e has 3 more marks than d . average marks of b , c , d , e is 48 . what is the marks of a ?","rationale":"a + b + c = 144 a + b + c + d = 188 d = 44 e = 47 b + c + d + e = 192 b + c = 101 a = 43 answer : a","correct":"a","options":{"a":"43 ","b":"44 ","c":"45 ","d":"46","e":"47"},"options_float":{"a":43.0,"b":44.0,"c":45.0,"d":46.0,"e":47.0},"annotated_formula":"subtract(multiply(const_3, 48), subtract(multiply(48, const_4), add(add(subtract(multiply(47, const_4), multiply(const_3, 48)), 3), subtract(multiply(47, const_4), multiply(const_3, 48)))))","linear_formula":"multiply(n0,const_3)|multiply(n0,const_4)|multiply(n1,const_4)|subtract(#2,#0)|add(n2,#3)|add(#4,#3)|subtract(#1,#5)|subtract(#0,#6)","chain":"3 * 48<\/gadget>\n144<\/output>\n48 * 4<\/gadget>\n192<\/output>\n47 * 4<\/gadget>\n188<\/output>\n188 - 144<\/gadget>\n44<\/output>\n44 + 3<\/gadget>\n47<\/output>\n47 + 44<\/gadget>\n91<\/output>\n192 - 91<\/gadget>\n101<\/output>\n144 - 101<\/gadget>\n43<\/output>\n43<\/result>","index":4320} +{"problem":"3 candidates in an election and received 3136 , 7636 and 11628 votes respectively . what % of the total votes did the winning candidate got in that election ?","rationale":"\"total number of votes polled = ( 3136 + 7636 + 11628 ) = 22400 so , required percentage = 11628 \/ 22400 * 100 = 52 % c\"","correct":"c","options":{"a":"45 % ","b":"49 % ","c":"52 % ","d":"59 %","e":"61 %"},"options_float":{"a":45.0,"b":49.0,"c":52.0,"d":59.0,"e":61.0},"annotated_formula":"multiply(divide(11628, add(add(3136, 7636), 11628)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)|","chain":"3_136 + 7_636<\/gadget>\n10_772<\/output>\n10_772 + 11_628<\/gadget>\n22_400<\/output>\n11_628 \/ 22_400<\/gadget>\n2_907\/5_600 = around 0.519107<\/output>\n(2_907\/5_600) * 100<\/gadget>\n2_907\/56 = around 51.910714<\/output>\n2_907\/56 = around 51.910714<\/result>","index":4322} +{"problem":"two pipes p and q can fill a cistern in 12 and 15 minutes respectively . both are opened together , but at the end of 2 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ?","rationale":"\"let x be the total time it takes for the cistern to fill . 2 \/ 12 + x \/ 15 = 1 x \/ 15 = 5 \/ 6 x = 12.5 after the first pipe is turned off , it takes 10.5 more minutes to fill the cistern . the answer is c .\"","correct":"c","options":{"a":"8.5 ","b":"9.5 ","c":"10.5 ","d":"11.5","e":"12.5"},"options_float":{"a":8.5,"b":9.5,"c":10.5,"d":11.5,"e":12.5},"annotated_formula":"multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 15)), 2)), 15)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/12) + (1\/15)<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 2<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) * 15<\/gadget>\n21\/2 = around 10.5<\/output>\n21\/2 = around 10.5<\/result>","index":4323} +{"problem":"a company conducted a survey about its two brands , a and b . x percent of respondents liked product a , ( x – 20 ) percent liked product b , 23 percent liked both products , and 23 percent liked neither product . what is the minimum number g of people surveyed by the company ?","rationale":"\"100 = x + x - 20 + 23 - 23 x = 60 , so , product a = 60 % , product b = 40 % , both = 23 % , neither = 23 % 23 % of the total no . of people should be an integer . so , a , bc are out . 60 % of d and 40 % of d are both integers . so , d satisfies all conditions . so , answer is d .\"","correct":"d","options":{"a":"46 ","b":"g = 80 ","c":"g = 90 ","d":"g = 100","e":"200"},"options_float":{"a":46.0,"b":80.0,"c":90.0,"d":100.0,"e":200.0},"annotated_formula":"add(subtract(divide(add(add(subtract(const_100, 23), 23), 20), const_2), 20), divide(add(add(subtract(const_100, 23), 23), 20), const_2))","linear_formula":"subtract(const_100,n1)|add(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#3,n0)|add(#3,#4)|","chain":"100 - 23<\/gadget>\n77<\/output>\n77 + 23<\/gadget>\n100<\/output>\n100 + 20<\/gadget>\n120<\/output>\n120 \/ 2<\/gadget>\n60<\/output>\n60 - 20<\/gadget>\n40<\/output>\n40 + 60<\/gadget>\n100<\/output>\n100<\/result>","index":4326} +{"problem":"a train traveling at 72 kmph crosses a platform in 33 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ?","rationale":"\"speed of train = 72 * ( 5 \/ 18 ) = 20 m \/ s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 33 = 660 m . so , length of platform = 660 - 360 = 300 m imo , answer d\"","correct":"d","options":{"a":"240 meters ","b":"360 meters ","c":"420 meters ","d":"300 meters","e":"can not be determined"},"options_float":{"a":240.0,"b":360.0,"c":420.0,"d":300.0,"e":null},"annotated_formula":"subtract(multiply(divide(multiply(72, const_1000), const_3600), 33), multiply(divide(multiply(72, const_1000), const_3600), 18))","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)|","chain":"72 * 1_000<\/gadget>\n72_000<\/output>\n72_000 \/ 3_600<\/gadget>\n20<\/output>\n20 * 33<\/gadget>\n660<\/output>\n20 * 18<\/gadget>\n360<\/output>\n660 - 360<\/gadget>\n300<\/output>\n300<\/result>","index":4328} +{"problem":"the weights of one liter vegetable ghee packet of two brands ‘ a ’ and ‘ b ’ are 800 gm and 850 gm respectively . if they are mixed in the ratio of 3 : 2 by volumes to form a mixture of 3 liters , what is the weight ( in kg ) of the mixture ?","rationale":"\"here ' s how i did it . my notes from reading the problem were : 1 l a = 800 gm 1 l b = 850 gm we are mixing five parts ( 3 parts a plus 2 parts b , 5 parts total ) to get 3 l , so 5 x = 3 - - - > x = 3 \/ 5 . each part is 3 \/ 5 of a liter . so if we have 3 parts a , we have 800 * 3 * ( 3 \/ 5 ) = 1440 if we have 2 parts b , we have 850 * 2 * ( 3 \/ 5 ) = 1020 1440 + 1020 = 2460 solving for units gives us 2.46 so the answer is d\"","correct":"d","options":{"a":"3.84 ","b":"1.75 ","c":"3.52 ","d":"2.46","e":"none of these"},"options_float":{"a":3.84,"b":1.75,"c":3.52,"d":2.46,"e":null},"annotated_formula":"divide(divide(multiply(3, add(multiply(3, 800), multiply(2, 850))), add(3, 2)), const_1000)","linear_formula":"add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#3)|divide(#4,#0)|divide(#5,const_1000)|","chain":"3 * 800<\/gadget>\n2_400<\/output>\n2 * 850<\/gadget>\n1_700<\/output>\n2_400 + 1_700<\/gadget>\n4_100<\/output>\n3 * 4_100<\/gadget>\n12_300<\/output>\n3 + 2<\/gadget>\n5<\/output>\n12_300 \/ 5<\/gadget>\n2_460<\/output>\n2_460 \/ 1_000<\/gadget>\n123\/50 = around 2.46<\/output>\n123\/50 = around 2.46<\/result>","index":4330} +{"problem":"$ 9,000 is deposited in a savings account that pays 14 % annual interest compounded semiannually . to the nearest dollar , how much is in the account at the end of the year ?","rationale":"this is the case of semi - annual compoundingso , multiply time period by 2 and divide rate by 2 so , new time = 2 periods and new rate = 14 \/ 2 = 7 % now , ca = 9000 ( 1 + 7 \/ 100 ) ^ 2 = $ 10304.10 hence answer is d","correct":"d","options":{"a":"$ 11304.12 ","b":"$ 12304.54 ","c":"$ 11604.14 ","d":"$ 10304.10","e":"$ 9304.41"},"options_float":{"a":11304.12,"b":12304.54,"c":11604.14,"d":10304.1,"e":9304.41},"annotated_formula":"multiply(const_100, const_100)","linear_formula":"multiply(const_100,const_100)","chain":"100 * 100<\/gadget>\n10_000<\/output>\n10_000<\/result>","index":4332} +{"problem":"a certain bacteria colony doubles in size every day for 25 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ?","rationale":"\"if there is one bacteria colony , then it will reach the limit of its habitat in 25 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 24 times . answer : e . similar questions to practice : hope it helps .\"","correct":"e","options":{"a":"6.33 ","b":"7.5 ","c":"10 ","d":"15","e":"24"},"options_float":{"a":6.33,"b":7.5,"c":10.0,"d":15.0,"e":24.0},"annotated_formula":"subtract(25, divide(25, 25))","linear_formula":"divide(n0,n0)|subtract(n0,#0)|","chain":"25 \/ 25<\/gadget>\n1<\/output>\n25 - 1<\/gadget>\n24<\/output>\n24<\/result>","index":4333} +{"problem":"if daily wages of a man is 4 times to that of a woman , how many men should work for 25 days to earn rs . 14400 ? given that wages for 40 women for 30 days are rs . 21600 .","rationale":"explanation : wages of 1 woman for 1 day = 21600 \/ 40 ã — 30 wages of 1 man for 1 day = 21600 ã — 4 \/ 40 ã — 30 wages of 1 man for 25 days = 21600 ã — 4 ã — 25 \/ 40 ã — 30 number of men = 14400 \/ ( 21600 ã — 4 ã — 25 \/ 40 ã — 30 ) = 144 \/ ( 216 ã — 100 \/ 40 ã — 30 ) = 8 answer : option d","correct":"d","options":{"a":"12 ","b":"4 ","c":"6 ","d":"8","e":"20"},"options_float":{"a":12.0,"b":4.0,"c":6.0,"d":8.0,"e":20.0},"annotated_formula":"divide(14400, multiply(multiply(divide(21600, multiply(40, 30)), 4), 25))","linear_formula":"multiply(n3,n4)|divide(n5,#0)|multiply(n0,#1)|multiply(n1,#2)|divide(n2,#3)","chain":"40 * 30<\/gadget>\n1_200<\/output>\n21_600 \/ 1_200<\/gadget>\n18<\/output>\n18 * 4<\/gadget>\n72<\/output>\n72 * 25<\/gadget>\n1_800<\/output>\n14_400 \/ 1_800<\/gadget>\n8<\/output>\n8<\/result>","index":4335} +{"problem":"jerry went to a shop and bought things worth rs . 20 , out of which 30 % went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ?","rationale":"total cost of the items he purchased = rs . 20 given that out of this rs . 20 , 30 % is given as tax = > total tax incurred = 30 % = rs . 30 \/ 100 let the cost of the tax free items = x given that tax rate = 6 % ∴ ( 20 − 30 \/ 100 − x ) 6 \/ 100 = 30 \/ 100 ⇒ 6 ( 20 − 0.3 − x ) = 30 ⇒ ( 20 − 0.3 − x ) = 5 ⇒ x = 20 − 0.3 − 5 = 14.7 b )","correct":"b","options":{"a":"14 ","b":"14.7 ","c":"21.3 ","d":"21.5","e":"21.7"},"options_float":{"a":14.0,"b":14.7,"c":21.3,"d":21.5,"e":21.7},"annotated_formula":"subtract(20, add(multiply(divide(30, const_100), divide(const_100, 6)), divide(30, const_100)))","linear_formula":"divide(n1,const_100)|divide(const_100,n2)|multiply(#0,#1)|add(#0,#2)|subtract(n0,#3)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n100 \/ 6<\/gadget>\n50\/3 = around 16.666667<\/output>\n(3\/10) * (50\/3)<\/gadget>\n5<\/output>\n5 + (3\/10)<\/gadget>\n53\/10 = around 5.3<\/output>\n20 - (53\/10)<\/gadget>\n147\/10 = around 14.7<\/output>\n147\/10 = around 14.7<\/result>","index":4336} +{"problem":"a random 10 - letter code is to be formed using the letters a , b , c , d , e , f , g , h , i and i ( only the “ i ” will be used twice ) . what is the probability that a code that has the two i ’ s adjacent to one another will be formed ?","rationale":"total number of combinations possible = 10 ! \/ 2 ! 10 ! \/ 2 ! total favorable combinations = 9 ! 9 ! - we have considered the two is as one element . since we have two is , they can interchange their positions , without affecting the code . probability = 9 ! \/ ( 10 ! \/ 2 ! ) = 9 ! ∗ 2 ! \/ 10 ! = 1 \/ 59 ! \/ ( 10 ! \/ 2 ! ) = 9 ! ∗ 2 ! \/ 10 ! = 1 \/ 5 ans : c","correct":"c","options":{"a":"1 \/ 10 ","b":"1 \/ 8 ","c":"1 \/ 5 ","d":"1 \/ 4","e":"1 \/ 2"},"options_float":{"a":0.1,"b":0.125,"c":0.2,"d":0.25,"e":0.5},"annotated_formula":"divide(multiply(subtract(10, const_1), factorial(subtract(10, const_2))), multiply(divide(divide(factorial(10), factorial(subtract(10, const_2))), const_2), factorial(subtract(10, const_2))))","linear_formula":"factorial(n0)|subtract(n0,const_2)|subtract(n0,const_1)|factorial(#1)|divide(#0,#3)|multiply(#3,#2)|divide(#4,const_2)|multiply(#6,#3)|divide(#5,#7)","chain":"10 - 1<\/gadget>\n9<\/output>\n10 - 2<\/gadget>\n8<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n9 * 40_320<\/gadget>\n362_880<\/output>\nfactorial(10)<\/gadget>\n3_628_800<\/output>\n3_628_800 \/ 40_320<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n45 * 40_320<\/gadget>\n1_814_400<\/output>\n362_880 \/ 1_814_400<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":4339} +{"problem":"at the end of the day , february 14 th , a florist had 90 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 15 of which were white and 10 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ?","rationale":"r + w + p = 90 s + l = 90 1 \/ 3 * 90 = 30 short - stemmed white = 15 short - stemmed pink = 10 = > short - stemmed red = 5 10 \/ p = 5 \/ r = > r = p \/ 2 , p = 2 r so total long stemmed = 60 and long stemmed red + long stemmed pink = 60 so long stemmed red \/ long stemmed = ? total white = 15 ( as no long stemmed white ) = > r + 2 r + 15 = 90 = > 3 r = 75 and r = 25 long stemmed r = 25 - 5 = 20 so long stemmed red \/ r = 20 \/ 60 = 33.3 % answer - d","correct":"d","options":{"a":"20 % ","b":"75 % ","c":"50 % ","d":"33.3 %","e":"80 %"},"options_float":{"a":20.0,"b":75.0,"c":50.0,"d":33.3,"e":80.0},"annotated_formula":"multiply(divide(divide(multiply(subtract(90, divide(90, const_3)), subtract(subtract(divide(90, const_3), 15), 10)), add(10, subtract(subtract(divide(90, const_3), 15), 10))), subtract(90, divide(90, const_3))), const_100)","linear_formula":"divide(n1,const_3)|subtract(n1,#0)|subtract(#0,n2)|subtract(#2,n3)|add(n3,#3)|multiply(#1,#3)|divide(#5,#4)|divide(#6,#1)|multiply(#7,const_100)","chain":"90 \/ 3<\/gadget>\n30<\/output>\n90 - 30<\/gadget>\n60<\/output>\n30 - 15<\/gadget>\n15<\/output>\n15 - 10<\/gadget>\n5<\/output>\n60 * 5<\/gadget>\n300<\/output>\n10 + 5<\/gadget>\n15<\/output>\n300 \/ 15<\/gadget>\n20<\/output>\n20 \/ 60<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 100<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":4340} +{"problem":"the average salary \/ head of allthe workers in a workshop is rs . 1200 , if the average salary \/ head of 10 technician is rs . 1400 and the average salary \/ head of the rest is rs . 900 , the total no . of workers in the work - shop is ?","rationale":"let the total number of workers be y . so sum of salary for all workers = sum of salary of 10 technician + sum of salary for other y - 10 workers . 10 x 1400 + 900 ( y - 10 ) = 1200 y ⇒ 14000 + 900 y - 9000 = 1200 y ⇒ 300 y = 5000 ∴ y = 17 so total number of workers = 17 b","correct":"b","options":{"a":"18 ","b":"17 ","c":"15 ","d":"19","e":"21"},"options_float":{"a":18.0,"b":17.0,"c":15.0,"d":19.0,"e":21.0},"annotated_formula":"divide(subtract(multiply(10, 1400), multiply(10, 900)), subtract(1200, 900))","linear_formula":"multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)","chain":"10 * 1_400<\/gadget>\n14_000<\/output>\n10 * 900<\/gadget>\n9_000<\/output>\n14_000 - 9_000<\/gadget>\n5_000<\/output>\n1_200 - 900<\/gadget>\n300<\/output>\n5_000 \/ 300<\/gadget>\n50\/3 = around 16.666667<\/output>\n50\/3 = around 16.666667<\/result>","index":4341} +{"problem":"two cubes have their volumes are in the ratio 1 : 27 . find the ratio of their surface area ?","rationale":"let their edges be a and b then , a ^ 3 \/ b ^ 3 = 1 \/ 27 ( a \/ b ) ^ 3 = ( 1 \/ 3 ) ^ 3 a \/ b = 1 \/ 3 ratio of their surface areas = 6 a ^ 2 \/ 6 b ^ 2 = ( a \/ b ) ^ 2 = 1 \/ 9 = 1 : 9 answer is c","correct":"c","options":{"a":"1 : 2 ","b":"2 : 3 ","c":"1 : 9 ","d":"2 : 7","e":"3 : 11"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.1111111111,"d":0.2857142857,"e":0.2727272727},"annotated_formula":"multiply(divide(1, 27), const_3)","linear_formula":"divide(n0,n1)|multiply(#0,const_3)","chain":"1 \/ 27<\/gadget>\n1\/27 = around 0.037037<\/output>\n(1\/27) * 3<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":4342} +{"problem":"if the integer n has exactly 4 positive divisors , including 1 and n , how many positive divisors does n ^ 3 have ?","rationale":"take the example of 6 . . . it has 4 positive divisors ( 1,2 , 3,4 ) now , take the example of 216 . . . it has 16 divisors . . so a is the ans","correct":"a","options":{"a":"16 ","b":"11 ","c":"10 ","d":"20","e":"9"},"options_float":{"a":16.0,"b":11.0,"c":10.0,"d":20.0,"e":9.0},"annotated_formula":"power(4, subtract(4, const_2))","linear_formula":"subtract(n0,const_2)|power(n0,#0)","chain":"4 - 2<\/gadget>\n2<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n16<\/result>","index":4343} +{"problem":"a father and his son are waiting at a bus stop in the evening . there is a lamp post behind them . the lamp post , the father and his son stand on the same straight line . the father observes that the shadows of his head and his son ’ s head are incident at the same point on the ground . if the heights of the lamp post , the father and his son are 6 metres , 1.8 metres and 0.9 metres respectively , and the father is standing 2.1 metres away from the post , then how far ( in metres ) is the son standing from his father ?","rationale":"explanation : l is the lamp post position , f is father and s is son ’ s position . x is the point where the shadow falls . ld = 0.9 = son ’ s height , lb = 1.8 = father ’ s height . so ab = 6 – 1.8 = 4.2 also bc = lf = 2.1 we observe that ∆ abc ~ ∆ ade ( two triangles are similar ) . hence , the corresponding sides are proportional . so , ab \/ ad = dc \/ de . 4.2 \/ 5.1 = 2.1 \/ de . de = 5.1 * 2.1 \/ 4.2 = 5.1 \/ 2 = 2.55 . ls = de = 2.55 fs = ls – lf = 2.55 – 2.1 = 0.45 . answer : d","correct":"d","options":{"a":"0.9 ","b":"0.75 ","c":"0.6 ","d":"0.45","e":"none of these"},"options_float":{"a":0.9,"b":0.75,"c":0.6,"d":0.45,"e":null},"annotated_formula":"divide(multiply(0.9, 2.1), subtract(6, 1.8))","linear_formula":"multiply(n2,n3)|subtract(n0,n1)|divide(#0,#1)","chain":"0.9 * 2.1<\/gadget>\n1.89<\/output>\n6 - 1.8<\/gadget>\n4.2<\/output>\n1.89 \/ 4.2<\/gadget>\n0.45<\/output>\n0.45<\/result>","index":4346} +{"problem":"a rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 5 per sq m ?","rationale":"\"area = ( l + b â € “ d ) d ( 80 + 60 â € “ 10 ) 10 = > 1300 m 2 1300 * 5 = rs . 6500 answer : a\"","correct":"a","options":{"a":"6500 ","b":"2779 ","c":"2779 ","d":"3900","e":"2781"},"options_float":{"a":6500.0,"b":2779.0,"c":2779.0,"d":3900.0,"e":2781.0},"annotated_formula":"multiply(multiply(subtract(add(80, 60), 10), 10), 5)","linear_formula":"add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)|","chain":"80 + 60<\/gadget>\n140<\/output>\n140 - 10<\/gadget>\n130<\/output>\n130 * 10<\/gadget>\n1_300<\/output>\n1_300 * 5<\/gadget>\n6_500<\/output>\n6_500<\/result>","index":4348} +{"problem":"at their respective rates , pump a , b , and c can fulfill an empty tank , or pump - out the full tank in 2 , 3 , and 6 hours . if a and b are used to pump - out water from the half - full tank , while c is used to fill water into the tank , in how many hours , the tank will be empty ?","rationale":"rate of a = 1 \/ 2 tank \/ hour ; rate of b = 1 \/ 3 tank \/ hour ; rate of c = 1 \/ 6 tank \/ hour . combined rate when a and b are used to pump - out water , while c is used to fill water into the tank is 1 \/ 2 + 1 \/ 3 - 1 \/ 6 = 2 \/ 3 tank hour . so , to empty the full tank 3 \/ 2 hours ( reciprocal of rate ) are needed . to empty the half - full tank half of that time would be needed : 1 \/ 2 * 3 \/ 2 = 3 \/ 4 hours . answer : c .","correct":"c","options":{"a":"2 \/ 3 ","b":"1 ","c":"3 \/ 4 ","d":"3 \/ 2","e":"2"},"options_float":{"a":0.6666666667,"b":1.0,"c":0.75,"d":1.5,"e":2.0},"annotated_formula":"divide(divide(6, subtract(add(divide(6, 2), divide(6, 3)), const_1)), const_2)","linear_formula":"divide(n2,n0)|divide(n2,n1)|add(#0,#1)|subtract(#2,const_1)|divide(n2,#3)|divide(#4,const_2)","chain":"6 \/ 2<\/gadget>\n3<\/output>\n6 \/ 3<\/gadget>\n2<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n6 \/ 4<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) \/ 2<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":4349} +{"problem":"a boat can travel with a speed of 22 km \/ hr in still water . if the speed of the stream is 5 km \/ hr , find the time taken by the boat to go 108 km downstream","rationale":"\"explanation : speed of the boat in still water = 22 km \/ hr speed of the stream = 5 km \/ hr speed downstream = ( 22 + 5 ) = 27 km \/ hr distance travelled downstream = 108 km time taken = distance \/ speed = 108 \/ 27 = 4 hours . answer : option b\"","correct":"b","options":{"a":"5 hours ","b":"4 hours ","c":"3 hours ","d":"2 hours","e":"none of these"},"options_float":{"a":5.0,"b":4.0,"c":3.0,"d":2.0,"e":null},"annotated_formula":"divide(108, add(22, 5))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"22 + 5<\/gadget>\n27<\/output>\n108 \/ 27<\/gadget>\n4<\/output>\n4<\/result>","index":4350} +{"problem":"company c sells a line of 25 products with an average retail price of $ 1,200 . if none of these products sells for less than $ 400 , and exactly 10 of the products sell for less than $ 1,000 , what is the greatest possible selling price of the most expensive product ?","rationale":"\"the average price of 25 products is $ 1,200 means that the total price of 25 products is 25 * 1,200 = $ 30,000 . next , since exactly 10 of the products sell for less than $ 1,000 , then let ' s make these 10 items to be at $ 400 each ( min possible ) . now , the remaining 14 items can not be priced less than $ 1,000 , thus the minimum possible price of each of these 14 items is $ 1,000 . thus the minimum possible value of 24 products is 10 * 400 + 14 * 1,000 = $ 18,000 . therefore , the greatest possible selling price of the most expensive product is $ 30,000 - $ 18,000 = $ 12,000 . answer : a .\"","correct":"a","options":{"a":"12000 ","b":"13000 ","c":"14000 ","d":"15000","e":"16000"},"options_float":{"a":12000.0,"b":13000.0,"c":14000.0,"d":15000.0,"e":16000.0},"annotated_formula":"subtract(multiply(multiply(const_12, const_100), 25), add(multiply(400, 10), multiply(subtract(subtract(25, 10), const_1), 1,000)))","linear_formula":"multiply(const_100,const_12)|multiply(n2,n3)|subtract(n0,n3)|multiply(n0,#0)|subtract(#2,const_1)|multiply(#4,n4)|add(#1,#5)|subtract(#3,#6)|","chain":"12 * 100<\/gadget>\n1_200<\/output>\n1_200 * 25<\/gadget>\n30_000<\/output>\n400 * 10<\/gadget>\n4_000<\/output>\n25 - 10<\/gadget>\n15<\/output>\n15 - 1<\/gadget>\n14<\/output>\n14 * 1_000<\/gadget>\n14_000<\/output>\n4_000 + 14_000<\/gadget>\n18_000<\/output>\n30_000 - 18_000<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":4351} +{"problem":"a man purchased earphone on friday . but he forgot how much it cost . the data he have are mean expenditure of all 7 days is 500 . expenditure on mon , tue , wed , thurs , sat , sun are 450600 , 400500 , 550300 . on friday he purchased 3 things one earphone , one pen and one notebook . pen cost 30 and notebook cost 50 . then what is cost of earphone ?","rationale":"solution let expenditure on friday be x ( 450 + 600 + 400 + 500 + x + 550 + 300 ) \/ 7 = 500 2800 + x = 3500 x = 700 pen + notebook + earphone = 700 earphone = 700 - 80 = 620 answer a","correct":"a","options":{"a":"620 ","b":"530 ","c":"450 ","d":"470","e":"500"},"options_float":{"a":620.0,"b":530.0,"c":450.0,"d":470.0,"e":500.0},"annotated_formula":"subtract(subtract(subtract(multiply(7, 500), add(add(add(add(add(subtract(500, 50), add(500, const_100)), subtract(500, const_100)), 500), add(500, 50)), subtract(subtract(500, const_100), const_100))), 30), 50)","linear_formula":"add(n1,const_100)|add(n1,n7)|multiply(n0,n1)|subtract(n1,n7)|subtract(n1,const_100)|add(#0,#3)|subtract(#4,const_100)|add(#5,#4)|add(n1,#7)|add(#8,#1)|add(#9,#6)|subtract(#2,#10)|subtract(#11,n6)|subtract(#12,n7)","chain":"7 * 500<\/gadget>\n3_500<\/output>\n500 - 50<\/gadget>\n450<\/output>\n500 + 100<\/gadget>\n600<\/output>\n450 + 600<\/gadget>\n1_050<\/output>\n500 - 100<\/gadget>\n400<\/output>\n1_050 + 400<\/gadget>\n1_450<\/output>\n1_450 + 500<\/gadget>\n1_950<\/output>\n500 + 50<\/gadget>\n550<\/output>\n1_950 + 550<\/gadget>\n2_500<\/output>\n400 - 100<\/gadget>\n300<\/output>\n2_500 + 300<\/gadget>\n2_800<\/output>\n3_500 - 2_800<\/gadget>\n700<\/output>\n700 - 30<\/gadget>\n670<\/output>\n670 - 50<\/gadget>\n620<\/output>\n620<\/result>","index":4352} +{"problem":"how many prime numbers are between 51 \/ 13 and 89 \/ 9 ?","rationale":"\"51 \/ 13 = 4 - 89 \/ 9 = 10 - prime numbers between 4 and 10 are 5 and 7 - sign signifies that the number is marginally less . answer c\"","correct":"c","options":{"a":"0 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"floor(const_2)","linear_formula":"floor(const_2)|","chain":"floor(2)<\/gadget>\n2<\/output>\n2<\/result>","index":4355} +{"problem":"a soccer team played 130 games and won 60 percent of them . how many games did it win ?","rationale":"\"60 % of 130 = x 0.60 * 130 = x 78 = x answer : e\"","correct":"e","options":{"a":"84 ","b":"94 ","c":"100 ","d":"125","e":"78"},"options_float":{"a":84.0,"b":94.0,"c":100.0,"d":125.0,"e":78.0},"annotated_formula":"divide(multiply(60, 130), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|","chain":"60 * 130<\/gadget>\n7_800<\/output>\n7_800 \/ 100<\/gadget>\n78<\/output>\n78<\/result>","index":4357} +{"problem":"what least no . must be subtracted from 5026 so that remaining no . is divisible by 5 ?","rationale":"explanation : on dividing 5026 by 5 we get the remainder 1 , so 1 should be subtracted option e","correct":"e","options":{"a":"3 ","b":"4 ","c":"5 ","d":"6","e":"1"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":1.0},"annotated_formula":"subtract(5026, multiply(floor(divide(5026, 5)), 5))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)","chain":"5_026 \/ 5<\/gadget>\n5_026\/5 = around 1_005.2<\/output>\nfloor(5_026\/5)<\/gadget>\n1_005<\/output>\n1_005 * 5<\/gadget>\n5_025<\/output>\n5_026 - 5_025<\/gadget>\n1<\/output>\n1<\/result>","index":4358} +{"problem":"a gambler has won 40 % of his 40 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 70 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ?","rationale":"\"let x be the number of additional games the gambler needs to play . 0.4 ( 40 ) + 0.7 x = 0.6 ( x + 40 ) 0.1 x = 8 x = 80 the answer is d .\"","correct":"d","options":{"a":"50 ","b":"60 ","c":"70 ","d":"80","e":"90"},"options_float":{"a":50.0,"b":60.0,"c":70.0,"d":80.0,"e":90.0},"annotated_formula":"divide(subtract(multiply(40, divide(60, const_100)), multiply(40, divide(40, const_100))), subtract(divide(70, const_100), divide(60, const_100)))","linear_formula":"divide(n3,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n40 * (3\/5)<\/gadget>\n24<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n40 * (2\/5)<\/gadget>\n16<\/output>\n24 - 16<\/gadget>\n8<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) - (3\/5)<\/gadget>\n1\/10 = around 0.1<\/output>\n8 \/ (1\/10)<\/gadget>\n80<\/output>\n80<\/result>","index":4359} +{"problem":"when 2 \/ 9 of the votes on a certain resolution have been counted , 3 \/ 4 of those counted are in favor of the resolution . what fraction x of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution ?","rationale":"\"if we use variable for total votes there will be too many fractions to manipulate with , so pick some smart # : let set total # of votes is 18 . 2 \/ 9 of the votes on a certain resolution have been counted - - > 4 counted and 18 - 4 = 14 votes left to be counted ; 3 \/ 4 of those counted are in favor of the resolution - - > 3 in favor and 1 against ; ratio of those who voted against to those who voted for to be 2 to 1 there should be total of 18 * 2 \/ 3 = 12 people who voted against , so in the remaining 14 votes there should be 12 - 1 = 11 people who voted against . thus x = 11 \/ 14 of the remaining votes must be against . answer : a .\"","correct":"a","options":{"a":"11 \/ 14 ","b":"13 \/ 18 ","c":"4 \/ 7 ","d":"3 \/ 7","e":"3 \/ 14"},"options_float":{"a":0.7857142857,"b":0.7222222222,"c":0.5714285714,"d":0.4285714286,"e":0.2142857143},"annotated_formula":"divide(subtract(divide(2, add(1, 2)), subtract(divide(2, 9), multiply(divide(2, 9), divide(3, 4)))), divide(subtract(9, 2), 9))","linear_formula":"add(n0,n5)|divide(n0,n1)|divide(n2,n3)|subtract(n1,n0)|divide(n0,#0)|divide(#3,n1)|multiply(#1,#2)|subtract(#1,#6)|subtract(#4,#7)|divide(#8,#5)|","chain":"1 + 2<\/gadget>\n3<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2 \/ 9<\/gadget>\n2\/9 = around 0.222222<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(2\/9) * (3\/4)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(2\/9) - (1\/6)<\/gadget>\n1\/18 = around 0.055556<\/output>\n(2\/3) - (1\/18)<\/gadget>\n11\/18 = around 0.611111<\/output>\n9 - 2<\/gadget>\n7<\/output>\n7 \/ 9<\/gadget>\n7\/9 = around 0.777778<\/output>\n(11\/18) \/ (7\/9)<\/gadget>\n11\/14 = around 0.785714<\/output>\n11\/14 = around 0.785714<\/result>","index":4362} +{"problem":"the speed of the boat in still water in 12 kmph . it can travel downstream through 54 kms in 3 hrs . in what time would it cover the same distance upstream ?","rationale":"\"still water = 12 km \/ hr downstream = 54 \/ 3 = 18 km \/ hr upstream = > > still water = ( u + v \/ 2 ) = > > 12 = u + 18 \/ 2 = 6 km \/ hr so time taken in upstream = 545 \/ 6 = 9 hrs answer : e\"","correct":"e","options":{"a":"8 hours ","b":"6 hours ","c":"4 hours ","d":"5 hours","e":"9 hours"},"options_float":{"a":8.0,"b":6.0,"c":4.0,"d":5.0,"e":9.0},"annotated_formula":"divide(54, subtract(12, subtract(divide(54, 3), 12)))","linear_formula":"divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)|","chain":"54 \/ 3<\/gadget>\n18<\/output>\n18 - 12<\/gadget>\n6<\/output>\n12 - 6<\/gadget>\n6<\/output>\n54 \/ 6<\/gadget>\n9<\/output>\n9<\/result>","index":4363} +{"problem":"if shares of two persons in profits are rs . 400 and rs . 600 then ratio of their capitals is","rationale":"\"profit = time * capital so 400 : 600 = 2 : 3 answer : b\"","correct":"b","options":{"a":"3 : 4 ","b":"2 : 3 ","c":"4 : 3 ","d":"1 : 3","e":"1 : 2"},"options_float":{"a":0.75,"b":0.6666666667,"c":1.3333333333,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(400, 600)","linear_formula":"divide(n0,n1)|","chain":"400 \/ 600<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":4364} +{"problem":"f ( x ) is a function such that f ( 2 x + 1 ) = 2 f ( x ) + 1 for all real numbers x and f ( 0 ) = 2 . find the value of f ( 3 ) .","rationale":"f ( 2 x + 1 ) = 2 f ( x ) + 1 : given f ( 3 ) = 2 f ( 1 ) + 1 : x = 1 in a f ( 1 ) = 2 f ( 0 ) + 1 : x = 0 in a f ( 3 ) = 11 : substitute correct answer b","correct":"b","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"14"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":14.0},"annotated_formula":"subtract(multiply(multiply(2, 3), 2), 1)","linear_formula":"multiply(n0,n6)|multiply(n0,#0)|subtract(#1,n1)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11<\/result>","index":4367} +{"problem":"a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 2662 in 3 years . find the rate percent ?","rationale":"\"2420 - - - 242 100 - - - ? = > 10 % answer : b\"","correct":"b","options":{"a":"23 % ","b":"10 % ","c":"32 % ","d":"98 %","e":"31 %"},"options_float":{"a":23.0,"b":10.0,"c":32.0,"d":98.0,"e":31.0},"annotated_formula":"multiply(divide(subtract(2662, 2420), 2420), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"2_662 - 2_420<\/gadget>\n242<\/output>\n242 \/ 2_420<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":4371} +{"problem":"a car started running at a speed of 36 km \/ hr and the speed of the car was increased by 2 km \/ hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey .","rationale":"\"a 45 km the total distance covered by the car in the first 10 hours = 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 = sum of 10 terms in ap whose first term is 36 and last term is 54 = 10 \/ 2 [ 36 + 54 ] = 450 km .\"","correct":"a","options":{"a":"45 km ","b":"76 km ","c":"25 km ","d":"15 km","e":"30 km"},"options_float":{"a":45.0,"b":76.0,"c":25.0,"d":15.0,"e":30.0},"annotated_formula":"divide(add(add(36, multiply(2, 10)), 36), 2)","linear_formula":"multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n36 + 20<\/gadget>\n56<\/output>\n56 + 36<\/gadget>\n92<\/output>\n92 \/ 2<\/gadget>\n46<\/output>\n46<\/result>","index":4372} +{"problem":"the profits of qrs company rose 30 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ?","rationale":"\"assume 100 in march , then 130 in april as 30 % increase , then 104 in may as 20 % decrease from april , and then 156 in june which is 150 % of 104 . so overall increase is from 100 to 156 is 56 % answer c\"","correct":"c","options":{"a":"15 % ","b":"32 % ","c":"56 % ","d":"62 %","e":"80 %"},"options_float":{"a":15.0,"b":32.0,"c":56.0,"d":62.0,"e":80.0},"annotated_formula":"multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(30, const_100)), subtract(const_1, divide(20, const_100)))), const_1))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(13\/10) * (4\/5)<\/gadget>\n26\/25 = around 1.04<\/output>\n(3\/2) * (26\/25)<\/gadget>\n39\/25 = around 1.56<\/output>\n(39\/25) - 1<\/gadget>\n14\/25 = around 0.56<\/output>\n100 * (14\/25)<\/gadget>\n56<\/output>\n56<\/result>","index":4374} +{"problem":"a pupil ' s marks were wrongly entered as 83 instead of 63 . due to that the average marks for the class got increased by half . the number of pupils in the class is :","rationale":"\"explanation : let there be x pupils in the class . total increase in marks = ( x * 1 \/ 2 ) = x \/ 2 . x \/ 2 = ( 83 - 63 ) = > x \/ 2 = 20 = > x = 40 . answer : b\"","correct":"b","options":{"a":"45 ","b":"40 ","c":"39 ","d":"37","e":"38"},"options_float":{"a":45.0,"b":40.0,"c":39.0,"d":37.0,"e":38.0},"annotated_formula":"multiply(subtract(83, 63), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|","chain":"83 - 63<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n40<\/result>","index":4378} +{"problem":"if 40 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ?","rationale":"40 % of ( x - y ) = 30 % of ( x + y ) ( 40 \/ 100 ) ( x - y ) = ( 30 \/ 100 ) ( x + y ) 4 ( x - y ) = 3 ( x + y ) x = 7 y x = 7 y therefore required percentage = ( ( y \/ x ) x 100 ) % = ( ( y \/ 7 y ) x 100 ) = 14 % answer is d .","correct":"d","options":{"a":"2.5 % ","b":"10 % ","c":"5 % ","d":"14 %","e":"25 %"},"options_float":{"a":2.5,"b":10.0,"c":5.0,"d":14.0,"e":25.0},"annotated_formula":"multiply(divide(subtract(40, 30), add(40, 30)), const_100)","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)","chain":"40 - 30<\/gadget>\n10<\/output>\n40 + 30<\/gadget>\n70<\/output>\n10 \/ 70<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/7) * 100<\/gadget>\n100\/7 = around 14.285714<\/output>\n100\/7 = around 14.285714<\/result>","index":4379} +{"problem":"there are 20 players on a rugby team . if the average weight of the players is 180 pounds , what will be the new average height if a 210 pound player will join the team ?","rationale":"( sum of the 20 weights ) \/ 20 = 180 sum of the 20 weights = 3600 new sum = 3600 + 210 = 3810 new average = 3810 \/ 21 = 181.43 ans : d","correct":"d","options":{"a":"185 ","b":"183.22 ","c":"184.96 ","d":"181.43","e":"182.78"},"options_float":{"a":185.0,"b":183.22,"c":184.96,"d":181.43,"e":182.78},"annotated_formula":"divide(add(multiply(20, 180), 210), add(20, const_1))","linear_formula":"add(n0,const_1)|multiply(n0,n1)|add(n2,#1)|divide(#2,#0)","chain":"20 * 180<\/gadget>\n3_600<\/output>\n3_600 + 210<\/gadget>\n3_810<\/output>\n20 + 1<\/gadget>\n21<\/output>\n3_810 \/ 21<\/gadget>\n1_270\/7 = around 181.428571<\/output>\n1_270\/7 = around 181.428571<\/result>","index":4381} +{"problem":"if in a 20 over cricket match , there were no wide balls , no wides , no extras and no overthrows . what is the maximum number of runs that a batsman can score in an ideal case scenario ?","rationale":"solution : 663 explanation : for an ideal case , the batsman will hit a six on each ball . but if he hits six on the last ball of the over , the strike will change in the next over . thus , the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over . thus the total runs that he can score in each over : 6 * 5 + 3 = 33 but this will have to go like it only till the 19 th over . in the last over , he can hit a six in the last ball as well as that will be the last ball of the match . thus runs for the last over will be 6 * 6 = 36 . hence the maximum runs = 33 * 19 + 36 = 663 answer a","correct":"a","options":{"a":"663 ","b":"720 ","c":"625 ","d":"640","e":"none"},"options_float":{"a":663.0,"b":720.0,"c":625.0,"d":640.0,"e":null},"annotated_formula":"multiply(const_3, add(add(multiply(20, const_10), 20), const_1))","linear_formula":"multiply(n0,const_10)|add(n0,#0)|add(#1,const_1)|multiply(#2,const_3)","chain":"20 * 10<\/gadget>\n200<\/output>\n200 + 20<\/gadget>\n220<\/output>\n220 + 1<\/gadget>\n221<\/output>\n3 * 221<\/gadget>\n663<\/output>\n663<\/result>","index":4382} +{"problem":"what annual payment will discharge a debt of rs . 1025 due in 2 years at the rate of 5 % compound interest ?","rationale":"\"let each installment be rs . x . then , x \/ ( 1 + 5 \/ 100 ) + x \/ ( 1 + 5 \/ 100 ) 2 = 1025 820 x + 1025 * 441 x = 551.25 so , value of each installment = rs . 551.25 answer : b\"","correct":"b","options":{"a":"rs . 550 ","b":"rs . 551.25 ","c":"rs . 560 ","d":"rs . 560.75","e":"rs . 660.75"},"options_float":{"a":550.0,"b":551.25,"c":560.0,"d":560.75,"e":660.75},"annotated_formula":"divide(multiply(power(add(divide(5, const_100), const_1), 2), 1025), 2)","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|divide(#3,n1)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) + 1<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) ** 2<\/gadget>\n441\/400 = around 1.1025<\/output>\n(441\/400) * 1_025<\/gadget>\n18_081\/16 = around 1_130.0625<\/output>\n(18_081\/16) \/ 2<\/gadget>\n18_081\/32 = around 565.03125<\/output>\n18_081\/32 = around 565.03125<\/result>","index":4383} +{"problem":"after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 5130 . what was the number of original inhabitants ?","rationale":"\"let the total number of original inhabitants be x . ( 75 \/ 100 ) * ( 90 \/ 100 ) * x = 5130 ( 27 \/ 40 ) * x = 5130 x = 5130 * 40 \/ 27 = 7600 the answer is b .\"","correct":"b","options":{"a":"7300 ","b":"7600 ","c":"7900 ","d":"8200","e":"8500"},"options_float":{"a":7300.0,"b":7600.0,"c":7900.0,"d":8200.0,"e":8500.0},"annotated_formula":"divide(5130, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#0)|multiply(#1,#2)|subtract(#2,#3)|divide(n2,#4)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(9\/10) * (1\/4)<\/gadget>\n9\/40 = around 0.225<\/output>\n(9\/10) - (9\/40)<\/gadget>\n27\/40 = around 0.675<\/output>\n5_130 \/ (27\/40)<\/gadget>\n7_600<\/output>\n7_600<\/result>","index":4384} +{"problem":"40 is subtracted from 70 % of a number , the result is 30 . find the number ?","rationale":"\"( 70 \/ 100 ) * x – 40 = 30 7 x = 700 x = 100 answer : e\"","correct":"e","options":{"a":"150 ","b":"997 ","c":"266 ","d":"288","e":"100"},"options_float":{"a":150.0,"b":997.0,"c":266.0,"d":288.0,"e":100.0},"annotated_formula":"divide(add(40, 30), divide(70, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"40 + 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n70 \/ (7\/10)<\/gadget>\n100<\/output>\n100<\/result>","index":4385} +{"problem":"if a train , travelling at a speed of 120 kmph , crosses a pole in 6 sec , then the length of train is ?","rationale":"\"d = 120 * 5 \/ 18 * 6 = 200 m answer : c\"","correct":"c","options":{"a":"281 ","b":"125 ","c":"200 ","d":"266","e":"121"},"options_float":{"a":281.0,"b":125.0,"c":200.0,"d":266.0,"e":121.0},"annotated_formula":"multiply(multiply(120, const_0_2778), 6)","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n120 * (5\/18)<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) * 6<\/gadget>\n200<\/output>\n200<\/result>","index":4387} +{"problem":"sakshi can do a piece of work in 5 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work :","rationale":"\"solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 5 : x ⇒ x = ( 5 x 4 \/ 5 ) ⇒ x = 4 days . hence , tanya takes 16 days is complete the work . answer a\"","correct":"a","options":{"a":"4 ","b":"16 ","c":"18 ","d":"25","e":"10"},"options_float":{"a":4.0,"b":16.0,"c":18.0,"d":25.0,"e":10.0},"annotated_formula":"divide(5, add(const_1, divide(25, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n5 \/ (5\/4)<\/gadget>\n4<\/output>\n4<\/result>","index":4388} +{"problem":"two pipes a and b can separately fill a cistern in 60 minutes and 80 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 40 minutes . in how much time , the third pipe alone can empty the cistern ?","rationale":"1 \/ 40 - ( 1 \/ 60 + 1 \/ 80 ) = - 1 \/ 240 third pipe can empty in 240 minutes answer : c","correct":"c","options":{"a":"90 min ","b":"100 min ","c":"240 min ","d":"120 min","e":"130 min"},"options_float":{"a":90.0,"b":100.0,"c":240.0,"d":120.0,"e":130.0},"annotated_formula":"inverse(subtract(add(inverse(60), inverse(80)), inverse(40)))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)","chain":"1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n1 \/ 80<\/gadget>\n1\/80 = around 0.0125<\/output>\n(1\/60) + (1\/80)<\/gadget>\n7\/240 = around 0.029167<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n(7\/240) - (1\/40)<\/gadget>\n1\/240 = around 0.004167<\/output>\n1 \/ (1\/240)<\/gadget>\n240<\/output>\n240<\/result>","index":4389} +{"problem":"13 buckets of water fill a tank when the capacity of each bucket is 51 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 17 litres ?","rationale":"\"capacity of the tank = ( 13 × 51 ) litre number of buckets required of capacity of each bucket is 17 litre = 13 × 51 \/ 17 = 13 × 3 = 39 answer is a\"","correct":"a","options":{"a":"39 ","b":"28 ","c":"23 ","d":"12","e":"24"},"options_float":{"a":39.0,"b":28.0,"c":23.0,"d":12.0,"e":24.0},"annotated_formula":"divide(multiply(51, 13), 17)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"51 * 13<\/gadget>\n663<\/output>\n663 \/ 17<\/gadget>\n39<\/output>\n39<\/result>","index":4390} +{"problem":"a candidate appearing for an examination has to secure 55 % marks to pass paper i . but he secured only 45 marks and failed by 25 marks . what is the maximum mark for paper i ?","rationale":"he secured 45 marks nd fail by 25 marks so total marks for pass the examinatn = 70 let toal marks x x * 55 \/ 100 = 70 x = 127 answer : a","correct":"a","options":{"a":"127 ","b":"120 ","c":"130 ","d":"140","e":"150"},"options_float":{"a":127.0,"b":120.0,"c":130.0,"d":140.0,"e":150.0},"annotated_formula":"divide(add(45, 25), divide(55, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)","chain":"45 + 25<\/gadget>\n70<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n70 \/ (11\/20)<\/gadget>\n1_400\/11 = around 127.272727<\/output>\n1_400\/11 = around 127.272727<\/result>","index":4391} +{"problem":"by selling 15 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ?","rationale":"\"80 % - - - 15 120 % - - - ? 80 \/ 120 * 15 = 10 answer : e\"","correct":"e","options":{"a":"8 ","b":"7 ","c":"6 ","d":"4","e":"10"},"options_float":{"a":8.0,"b":7.0,"c":6.0,"d":4.0,"e":10.0},"annotated_formula":"multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 15)","linear_formula":"add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)|","chain":"100 + 20<\/gadget>\n120<\/output>\n100 - 20<\/gadget>\n80<\/output>\n1 \/ 80<\/gadget>\n1\/80 = around 0.0125<\/output>\n120 * (1\/80)<\/gadget>\n3\/2 = around 1.5<\/output>\n1 \/ (3\/2)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 15<\/gadget>\n10<\/output>\n10<\/result>","index":4392} +{"problem":"a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money n did bob receive ?","rationale":"\"notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 \/ 3 of that as well . 1 \/ 3 of k - 4 is ( k - 4 ) \/ 3 so altogether , bob receives 4 + ( k - 4 ) \/ 3 so , our equation becomes : k - [ 4 + ( k - 4 ) \/ 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) \/ 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ n = 20 answer : b\"","correct":"b","options":{"a":"20 ","b":"22 ","c":"24 ","d":"26","e":"52"},"options_float":{"a":20.0,"b":22.0,"c":24.0,"d":26.0,"e":52.0},"annotated_formula":"divide(multiply(32, const_2), const_3)","linear_formula":"multiply(n2,const_2)|divide(#0,const_3)|","chain":"32 * 2<\/gadget>\n64<\/output>\n64 \/ 3<\/gadget>\n64\/3 = around 21.333333<\/output>\n64\/3 = around 21.333333<\/result>","index":4394} +{"problem":"two trains leave a station traveling the same direction . train a leaves the station traveling at a constant speed of 60 mph , and train b leaves the station traveling at a constant speed of 80 mph . if both trains maintain constant speeds , in how many minutes will train b overtake train a if train a left 40 minutes prior to train b ?","rationale":"we can use d = rt [ distance = rate * time ] train a will have traveled for 40 minutes longer than train b when b overtakes a , so time of train a : t + 40 minutes = t + ( 2 \/ 3 ) hours ( use hours since the rate is in hours ) time of train b : t rate of train a : 60 mph rate of train b : 80 mph the distances traveled by each will be equal when train b overtakes train a , so we can set the right sides of d = rt equal for the two trains ( rate of train a ) * ( time of train a ) = ( rate of train b ) * ( time of train b ) 60 * ( t + 2 \/ 3 ) = 80 * t 60 t + 40 = 80 t 40 = 20 t 2 = t 2 hours = 120 minutes d","correct":"d","options":{"a":"90 ","b":"100 ","c":"110 ","d":"120","e":"130"},"options_float":{"a":90.0,"b":100.0,"c":110.0,"d":120.0,"e":130.0},"annotated_formula":"multiply(divide(multiply(divide(40, const_60), 60), subtract(80, 60)), const_60)","linear_formula":"divide(n2,const_60)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(#3,const_60)","chain":"40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 60<\/gadget>\n40<\/output>\n80 - 60<\/gadget>\n20<\/output>\n40 \/ 20<\/gadget>\n2<\/output>\n2 * 60<\/gadget>\n120<\/output>\n120<\/result>","index":4395} +{"problem":"if the average of 5 positive integers is 50 and the difference between the largest and the smallest of these 5 numbers is 10 , what is the maximum value possible for the largest of these 5 integers ?","rationale":"sum of 5 integer ( a , b , c , d , e ) = 5 * 50 = 250 e - a = 10 i . e . e = a + 10 for e to be maximum remaining 4 must be as small as possible since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers i . e . a + a + a + a + ( a + 10 ) = 250 i . e . 5 a = 240 i . e . a = 48 i . e . largest e = 48 + 10 = 58 answer : option b","correct":"b","options":{"a":"50 ","b":"58 ","c":"49 ","d":"48","e":"44"},"options_float":{"a":50.0,"b":58.0,"c":49.0,"d":48.0,"e":44.0},"annotated_formula":"add(divide(subtract(multiply(50, 5), 10), 5), 10)","linear_formula":"multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)","chain":"50 * 5<\/gadget>\n250<\/output>\n250 - 10<\/gadget>\n240<\/output>\n240 \/ 5<\/gadget>\n48<\/output>\n48 + 10<\/gadget>\n58<\/output>\n58<\/result>","index":4396} +{"problem":"what is the smallest number which , when increased by 7 , is divisible by 8 , 11 , and 24 ?","rationale":"\"lcm ( 8 , 11,24 ) = 24 x 11 = 264 so the least divisible number is 264 , and the number we are looking for is 264 - 7 = 257 . the answer is b .\"","correct":"b","options":{"a":"264 ","b":"257 ","c":"269 ","d":"270","e":"282"},"options_float":{"a":264.0,"b":257.0,"c":269.0,"d":270.0,"e":282.0},"annotated_formula":"subtract(lcm(24, 11), 7)","linear_formula":"lcm(n2,n3)|subtract(#0,n0)|","chain":"lcm(24, 11)<\/gadget>\n264<\/output>\n264 - 7<\/gadget>\n257<\/output>\n257<\/result>","index":4397} +{"problem":"in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 210 girls , the total number of students in the college is ?","rationale":"\"let the number of boys and girls be 8 x and 5 x then , 5 x = 210 x = 42 total number of students = 13 x = 13 * 42 = 546 answer is e\"","correct":"e","options":{"a":"562 ","b":"356 ","c":"452 ","d":"416","e":"546"},"options_float":{"a":562.0,"b":356.0,"c":452.0,"d":416.0,"e":546.0},"annotated_formula":"add(multiply(divide(8, 5), 210), 210)","linear_formula":"divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|","chain":"8 \/ 5<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) * 210<\/gadget>\n336<\/output>\n336 + 210<\/gadget>\n546<\/output>\n546<\/result>","index":4398} +{"problem":"the grade point average of one third of the classroom is 15 ; the grade point average of the rest is 18 . what is the grade point average of the whole class ?","rationale":"\"let n = total students in class total points for 1 \/ 3 class = 15 n \/ 3 = 5 n total points for 2 \/ 3 class = 18 * 2 n \/ 3 = 12 n total points for whole class = 5 n + 12 n = 17 n 17 n total class points \/ n total students = 17 grade point average for total class answer : a\"","correct":"a","options":{"a":"17 ","b":"19 ","c":"21 ","d":"23","e":"25"},"options_float":{"a":17.0,"b":19.0,"c":21.0,"d":23.0,"e":25.0},"annotated_formula":"add(multiply(divide(18, const_3), const_2), divide(15, const_3))","linear_formula":"divide(n0,const_3)|divide(n1,const_3)|multiply(#1,const_2)|add(#0,#2)|","chain":"18 \/ 3<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n12 + 5<\/gadget>\n17<\/output>\n17<\/result>","index":4399} +{"problem":"if the operation @ is defined for all a and b by the equation a @ b = ( a - b ) \/ 5 , then ( 3 @ - 2 ) @ 6 = ?","rationale":"work within the parenthesis first so solve ( 3 @ - 2 ) first ( 3 @ - 2 ) = ( 3 - ( - 2 ) ) \/ 5 = ( 3 + 2 ) \/ 5 = 1 now take 1 plug back into equation and solve the rest ( 1 @ 6 ) = ( 1 - 6 ) \/ 5 = - 5 \/ 5 = - 1 so - 1 is the answer . . . . this question is merely testing order of operations remember pemdas answer is option : e","correct":"e","options":{"a":"0 ","b":"1 ","c":"2 ","d":"- 2","e":"- 1"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":-2.0,"e":-1.0},"annotated_formula":"subtract(const_2, const_3)","linear_formula":"subtract(const_2,const_3)","chain":"2 - 3<\/gadget>\n-1<\/output>\n-1<\/result>","index":4401} +{"problem":"a goods train runs at the speed of 72 km \/ hr and crosses a 250 m long platform in 15 sec . what is the length of the goods train ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec . time = 15 sec . let the length of the train be x meters . then , ( x + 250 ) \/ 15 = 20 x = 50 m . answer : e\"","correct":"e","options":{"a":"299 ","b":"277 ","c":"276 ","d":"270","e":"50"},"options_float":{"a":299.0,"b":277.0,"c":276.0,"d":270.0,"e":50.0},"annotated_formula":"subtract(multiply(multiply(divide(72, const_3600), const_1000), 15), 250)","linear_formula":"divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)|","chain":"72 \/ 3_600<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 1_000<\/gadget>\n20<\/output>\n20 * 15<\/gadget>\n300<\/output>\n300 - 250<\/gadget>\n50<\/output>\n50<\/result>","index":4403} +{"problem":"a goods train runs at the speed of 72 km \/ hr and crosses a 290 m long platform in 26 sec . what is the length of the goods train ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec . time = 26 sec . let the length of the train be x meters . then , ( x + 290 ) \/ 26 = 20 x = 230 m . answer : d\"","correct":"d","options":{"a":"278 ","b":"166 ","c":"151 ","d":"230","e":"109"},"options_float":{"a":278.0,"b":166.0,"c":151.0,"d":230.0,"e":109.0},"annotated_formula":"subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 290)","linear_formula":"divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)|","chain":"72 \/ 3_600<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 1_000<\/gadget>\n20<\/output>\n20 * 26<\/gadget>\n520<\/output>\n520 - 290<\/gadget>\n230<\/output>\n230<\/result>","index":4404} +{"problem":"the contents of a certain box consist of 24 oranges and 30 kiwis . how many kiwis must be added to the box so that exactly 30 % of the pieces of fruit in the box will be oranges ?","rationale":"oranges = ( oranges + kiwi + x ) * 0.3 24 = ( 30 + 24 + x ) * 0.3 x = 26 . answer : b .","correct":"b","options":{"a":"24 ","b":"26 ","c":"30 ","d":"46","e":"50"},"options_float":{"a":24.0,"b":26.0,"c":30.0,"d":46.0,"e":50.0},"annotated_formula":"divide(subtract(24, multiply(divide(30, const_100), add(24, 30))), divide(30, const_100))","linear_formula":"add(n0,n1)|divide(n1,const_100)|multiply(#0,#1)|subtract(n0,#2)|divide(#3,#1)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n24 + 30<\/gadget>\n54<\/output>\n(3\/10) * 54<\/gadget>\n81\/5 = around 16.2<\/output>\n24 - (81\/5)<\/gadget>\n39\/5 = around 7.8<\/output>\n(39\/5) \/ (3\/10)<\/gadget>\n26<\/output>\n26<\/result>","index":4405} +{"problem":"if s is the sum of reciprocals of a list of consecutive integers from 55 to 64 , inclusive , s is approximately equal to","rationale":"we need to find the approximate value of 1 \/ 55 + 1 \/ 56 + 1 \/ 57 + 1 \/ 58 + 1 \/ 59 + 1 \/ 60 + 1 \/ 61 + 1 \/ 62 + 1 \/ 63 + 1 \/ 64 . now , the sum of these 10 terms will be very close to 10 times 1 \/ 60 , which is 0.016 * 10 = 0.16 . answer : a .","correct":"a","options":{"a":"0.16 ","b":"0.2 ","c":"0.3 ","d":"0.4","e":"0.5"},"options_float":{"a":0.16,"b":0.2,"c":0.3,"d":0.4,"e":0.5},"annotated_formula":"add(add(add(add(add(add(add(add(divide(const_1, 55), divide(const_1, add(55, const_1))), divide(const_1, add(add(55, const_1), const_1))), divide(const_1, add(add(add(55, const_1), const_1), const_1))), divide(const_1, add(add(add(add(55, const_1), const_1), const_1), const_1))), divide(const_1, add(add(add(add(add(55, const_1), const_1), const_1), const_1), const_1))), divide(const_1, add(add(add(add(add(add(add(55, const_1), const_1), const_1), const_1), const_1), const_1), const_1))), divide(const_1, add(add(add(add(add(add(add(add(55, const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1))), divide(const_1, 64))","linear_formula":"add(n0,const_1)|divide(const_1,n0)|divide(const_1,n1)|add(#0,const_1)|divide(const_1,#0)|add(#1,#4)|add(#3,const_1)|divide(const_1,#3)|add(#5,#7)|add(#6,const_1)|divide(const_1,#6)|add(#8,#10)|add(#9,const_1)|divide(const_1,#9)|add(#11,#13)|add(#12,const_1)|divide(const_1,#12)|add(#14,#16)|add(#15,const_1)|add(#18,const_1)|divide(const_1,#18)|add(#17,#20)|divide(const_1,#19)|add(#21,#22)|add(#23,#2)","chain":"1 \/ 55<\/gadget>\n1\/55 = around 0.018182<\/output>\n55 + 1<\/gadget>\n56<\/output>\n1 \/ 56<\/gadget>\n1\/56 = around 0.017857<\/output>\n(1\/55) + (1\/56)<\/gadget>\n111\/3_080 = around 0.036039<\/output>\n56 + 1<\/gadget>\n57<\/output>\n1 \/ 57<\/gadget>\n1\/57 = around 0.017544<\/output>\n(111\/3_080) + (1\/57)<\/gadget>\n9_407\/175_560 = around 0.053583<\/output>\n57 + 1<\/gadget>\n58<\/output>\n1 \/ 58<\/gadget>\n1\/58 = around 0.017241<\/output>\n(9_407\/175_560) + (1\/58)<\/gadget>\n360_583\/5_091_240 = around 0.070824<\/output>\n58 + 1<\/gadget>\n59<\/output>\n1 \/ 59<\/gadget>\n1\/59 = around 0.016949<\/output>\n(360_583\/5_091_240) + (1\/59)<\/gadget>\n26_365_637\/300_383_160 = around 0.087773<\/output>\n59 + 1<\/gadget>\n60<\/output>\n1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n(26_365_637\/300_383_160) + (1\/60)<\/gadget>\n10_457_341\/100_127_720 = around 0.10444<\/output>\n60 + 1<\/gadget>\n61<\/output>\n61 + 1<\/gadget>\n62<\/output>\n1 \/ 62<\/gadget>\n1\/62 = around 0.016129<\/output>\n(10_457_341\/100_127_720) + (1\/62)<\/gadget>\n374_241_431\/3_103_959_320 = around 0.120569<\/output>\n62 + 1<\/gadget>\n63<\/output>\n1 \/ 63<\/gadget>\n1\/63 = around 0.015873<\/output>\n(374_241_431\/3_103_959_320) + (1\/63)<\/gadget>\n3_811_595_639\/27_935_633_880 = around 0.136442<\/output>\n1 \/ 64<\/gadget>\n1\/64 = around 0.015625<\/output>\n(3_811_595_639\/27_935_633_880) + (1\/64)<\/gadget>\n33_984_719_347\/223_485_071_040 = around 0.152067<\/output>\n33_984_719_347\/223_485_071_040 = around 0.152067<\/result>","index":4406} +{"problem":"a swimmer swims downstream 55 km and upstream 10 km taking 5 hours each time ; what is the speed of the current ?","rationale":"55 - - - 5 ds = 11 ? - - - - 1 10 - - - - 5 us = 2 ? - - - - 1 s = ? s = ( 11 - 2 ) \/ 2 = 4.5 answer : b","correct":"b","options":{"a":"4.6 ","b":"4.5 ","c":"4.4 ","d":"4.3","e":"4.2"},"options_float":{"a":4.6,"b":4.5,"c":4.4,"d":4.3,"e":4.2},"annotated_formula":"divide(subtract(divide(55, 5), divide(10, 5)), const_2)","linear_formula":"divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)","chain":"55 \/ 5<\/gadget>\n11<\/output>\n10 \/ 5<\/gadget>\n2<\/output>\n11 - 2<\/gadget>\n9<\/output>\n9 \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":4408} +{"problem":"a is half good a work man as b and together they finish a job in 13 days . in how many days working alone b finish the job ?","rationale":"c 39 \/ 2 wc = 1 : 2 2 x + x = 1 \/ 13 = > x = 1 \/ 39 2 x = 2 \/ 39 = > 39 \/ 2 days","correct":"c","options":{"a":"23 ","b":"22 ","c":"39 \/ 2 ","d":"36","e":"48"},"options_float":{"a":23.0,"b":22.0,"c":19.5,"d":36.0,"e":48.0},"annotated_formula":"multiply(13, divide(const_3, const_2))","linear_formula":"divide(const_3,const_2)|multiply(n0,#0)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n13 * (3\/2)<\/gadget>\n39\/2 = around 19.5<\/output>\n39\/2 = around 19.5<\/result>","index":4410} +{"problem":"33 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ?","rationale":"let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 33 : : 12 : x working hours \/ day 6 : 5 30 x 6 x x = 33 x 5 x 12 x = ( 33 x 5 x 12 ) \/ ( 30 x 6 ) x = 11 answer b","correct":"b","options":{"a":"10 ","b":"11 ","c":"13 ","d":"18","e":"19"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":18.0,"e":19.0},"annotated_formula":"divide(multiply(multiply(33, 12), 5), multiply(30, 6))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)","chain":"33 * 12<\/gadget>\n396<\/output>\n396 * 5<\/gadget>\n1_980<\/output>\n30 * 6<\/gadget>\n180<\/output>\n1_980 \/ 180<\/gadget>\n11<\/output>\n11<\/result>","index":4411} +{"problem":"the diagonal of a rectangle is √ 41 cm and its area is 20 sq . cm . the perimeter of the rectangle must be :","rationale":"explanation : l 2 + b 2 = 41 also , lb = 20 ( l + b ) 2 = l 2 + b 2 + 2 lb = 41 + 40 = 81 ( l + b ) = 9 perimeter = 2 ( l + b ) = 18 cm . answer : a","correct":"a","options":{"a":"18 ","b":"28 ","c":"38 ","d":"48","e":"58"},"options_float":{"a":18.0,"b":28.0,"c":38.0,"d":48.0,"e":58.0},"annotated_formula":"multiply(sqrt(add(power(sqrt(41), const_2), multiply(20, const_2))), const_2)","linear_formula":"multiply(n1,const_2)|sqrt(n0)|power(#1,const_2)|add(#0,#2)|sqrt(#3)|multiply(#4,const_2)","chain":"41 ** (1\/2)<\/gadget>\nsqrt(41) = around 6.403124<\/output>\n(sqrt(41)) ** 2<\/gadget>\n41<\/output>\n20 * 2<\/gadget>\n40<\/output>\n41 + 40<\/gadget>\n81<\/output>\n81 ** (1\/2)<\/gadget>\n9<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18<\/result>","index":4412} +{"problem":"there is a square with sides of 12 . what is the circumference of the biggest circle that can be cut out of this square ?","rationale":"circumference of a circle = c = 2 ï € r square is 12 wide , so circle ' s diameter would be 12 , and radius would be 6 c = 2 ï € 6 which is approximately 37.7 answer is e","correct":"e","options":{"a":"75.4 ","b":"31.42 ","c":"36 ","d":"18.85","e":"37.7"},"options_float":{"a":75.4,"b":31.42,"c":36.0,"d":18.85,"e":37.7},"annotated_formula":"circumface(divide(12, const_2))","linear_formula":"divide(n0,const_2)|circumface(#0)","chain":"12 \/ 2<\/gadget>\n6<\/output>\n2 * pi * 6<\/gadget>\n12*pi = around 37.699112<\/output>\n12*pi = around 37.699112<\/result>","index":4414} +{"problem":"convert 0.36 in to a vulgar fraction ?","rationale":"\"answer 0.36 = 36 \/ 100 = 9 \/ 25 correct option : d\"","correct":"d","options":{"a":"18 \/ 50 ","b":"16 \/ 50 ","c":"17 \/ 50 ","d":"9 \/ 25","e":"none"},"options_float":{"a":0.36,"b":0.32,"c":0.34,"d":0.36,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.36), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))","linear_formula":"add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|","chain":"3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n100 * 0.36<\/gadget>\n36<\/output>\n36 \/ 100<\/gadget>\n9\/25 = around 0.36<\/output>\n9\/25 = around 0.36<\/result>","index":4415} +{"problem":"a box contain the number of balls which is as much greater than 44 is less than 70 . the number of balls is ?","rationale":"answer let the number be x . then , x - 44 = 70 - x â ‡ ’ 2 x = 70 + 44 = 114 â ‡ ’ 2 x = 114 â ˆ ´ x = 57 correct option : e","correct":"e","options":{"a":"47 ","b":"114 ","c":"74 ","d":"67","e":"57"},"options_float":{"a":47.0,"b":114.0,"c":74.0,"d":67.0,"e":57.0},"annotated_formula":"divide(add(70, 44), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"70 + 44<\/gadget>\n114<\/output>\n114 \/ 2<\/gadget>\n57<\/output>\n57<\/result>","index":4416} +{"problem":"robert ' s salary was decreased by 20 % and subsequently increased by 20 % . how much percentage does he lose ?","rationale":"\"let original salary be $ 100 salary after decreasing 20 % = 100 - 100 x 20 \/ 100 = $ 80 salary after increasing 20 % on $ 80 = 80 + 80 x 20 \/ 100 = $ 96 percentage of loss = 100 - 96 = 4 % answer : a\"","correct":"a","options":{"a":"4 % ","b":"20 % ","c":"25 % ","d":"30 %","e":"50 %"},"options_float":{"a":4.0,"b":20.0,"c":25.0,"d":30.0,"e":50.0},"annotated_formula":"multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 20), add(const_100, 20))), multiply(const_100, const_100)), const_100)","linear_formula":"add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#1)|multiply(#5,const_100)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n100 + 20<\/gadget>\n120<\/output>\n80 * 120<\/gadget>\n9_600<\/output>\n10_000 - 9_600<\/gadget>\n400<\/output>\n400 \/ 10_000<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) * 100<\/gadget>\n4<\/output>\n4<\/result>","index":4417} +{"problem":"12 . what is the dividend . divisor 17 , the quotient is 9 and the remainder is 6 .","rationale":"\"d = d * q + r d = 17 * 9 + 6 d = 153 + 6 d = 159 answer d\"","correct":"d","options":{"a":"150 ","b":"152 ","c":"154 ","d":"159","e":"144"},"options_float":{"a":150.0,"b":152.0,"c":154.0,"d":159.0,"e":144.0},"annotated_formula":"add(multiply(17, 9), 6)","linear_formula":"multiply(n1,n2)|add(n3,#0)|","chain":"17 * 9<\/gadget>\n153<\/output>\n153 + 6<\/gadget>\n159<\/output>\n159<\/result>","index":4419} +{"problem":"what is the probability that the sum of two dice will yield a 5 , and then when both are thrown again , their sum will again yield a 5 ? assume that each die has 4 sides with faces numbered 1 to 4 .","rationale":"\"solution - rolling dices is an independent event . the combinations to get 5 are ( 1,4 ) , ( 4,1 ) , ( 2,3 ) , ( 3,2 ) , and total combinations of both dices is 16 . the probability of getting 5 in first attempt is 4 \/ 16 = 1 \/ 4 . probability of getting 5 again in second attempt = ( 1 \/ 4 ) * ( 1 \/ 4 ) = 1 \/ 16 . ans d\"","correct":"d","options":{"a":"1 \/ 144 ","b":"1 \/ 36 ","c":"1 \/ 12 ","d":"1 \/ 16","e":"1 \/ 3"},"options_float":{"a":0.0069444444,"b":0.0277777778,"c":0.0833333333,"d":0.0625,"e":0.3333333333},"annotated_formula":"multiply(divide(4, power(4, const_2)), divide(4, power(4, const_2)))","linear_formula":"power(n2,const_2)|divide(n2,#0)|multiply(#1,#1)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n4 \/ 16<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * (1\/4)<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":4420} +{"problem":"a , b , c and d enter into partnership . a subscribes 1 \/ 3 of the capital b 1 \/ 4 , c 1 \/ 5 and d the rest . how much share did a get in a profit of rs . 2400 ?","rationale":"\"2400 * 1 \/ 3 = 800 option a\"","correct":"a","options":{"a":"s . 800 ","b":"s . 810 ","c":"s . 820 ","d":"s . 900","e":"s . 920"},"options_float":{"a":800.0,"b":810.0,"c":820.0,"d":900.0,"e":920.0},"annotated_formula":"multiply(2400, divide(1, 3))","linear_formula":"divide(n0,n1)|multiply(n6,#0)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n2_400 * (1\/3)<\/gadget>\n800<\/output>\n800<\/result>","index":4421} +{"problem":"a shopkeeper sells 200 metres of cloth for rs . 12000 at a loss of rs . 12 per metre . find his cost price for one metre of cloth ?","rationale":"\"sp per metre = 12000 \/ 200 = rs . 60 loss per metre = rs . 12 cp per metre = 60 + 12 = rs . 72 answer : b\"","correct":"b","options":{"a":"s . 59 ","b":"s . 72 ","c":"s . 62 ","d":"s . 50","e":"s . 13"},"options_float":{"a":59.0,"b":72.0,"c":62.0,"d":50.0,"e":13.0},"annotated_formula":"add(divide(12000, 200), 12)","linear_formula":"divide(n1,n0)|add(n2,#0)|","chain":"12_000 \/ 200<\/gadget>\n60<\/output>\n60 + 12<\/gadget>\n72<\/output>\n72<\/result>","index":4423} +{"problem":"a pipe can empty 2 \/ 3 rd of a cistern in 12 mins . in 6 mins , what part of the cistern will be empty ?","rationale":"\"2 \/ 3 - - - - 12 ? - - - - - 6 = = > 1 \/ 3 e\"","correct":"e","options":{"a":"2 \/ 3 ","b":"3 \/ 7 ","c":"4 \/ 9 ","d":"3 \/ 8","e":"1 \/ 3"},"options_float":{"a":0.6666666667,"b":0.4285714286,"c":0.4444444444,"d":0.375,"e":0.3333333333},"annotated_formula":"divide(multiply(divide(2, 3), 6), 12)","linear_formula":"divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 6<\/gadget>\n4<\/output>\n4 \/ 12<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":4425} +{"problem":"a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6400 ?","rationale":"\"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6400 * 6 ) – 30,130 ] = rs . ( 38400 – 30,130 ) = rs . 8270 answer b\"","correct":"b","options":{"a":"rs . 5870 ","b":"rs . 8270 ","c":"rs . 6020 ","d":"rs . 6850","e":"none of these"},"options_float":{"a":5870.0,"b":8270.0,"c":6020.0,"d":6850.0,"e":null},"annotated_formula":"subtract(multiply(add(5, const_1), 6400), add(add(add(add(5420, 5660), 6200), 6350), 6500))","linear_formula":"add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|","chain":"5 + 1<\/gadget>\n6<\/output>\n6 * 6_400<\/gadget>\n38_400<\/output>\n5_420 + 5_660<\/gadget>\n11_080<\/output>\n11_080 + 6_200<\/gadget>\n17_280<\/output>\n17_280 + 6_350<\/gadget>\n23_630<\/output>\n23_630 + 6_500<\/gadget>\n30_130<\/output>\n38_400 - 30_130<\/gadget>\n8_270<\/output>\n8_270<\/result>","index":4426} +{"problem":"the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 9 and 39 inclusive .","rationale":"\"we ' re dealing with a sequence of consecutive odd integers : 9 to 39 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 9 + 39 = 48 2 ) now look at the ' next smallest ' and the ' next biggest ' : 11 + 37 = 48 now we have proof that there is no middle term . we have 8 bunches of 48 8 ( 48 ) = 384 b\"","correct":"b","options":{"a":"351 ","b":"384 ","c":"410 ","d":"424","e":"450"},"options_float":{"a":351.0,"b":384.0,"c":410.0,"d":424.0,"e":450.0},"annotated_formula":"subtract(power(divide(add(39, const_1), 2), 2), power(divide(add(subtract(9, 2), const_1), 2), 2))","linear_formula":"add(n2,const_1)|subtract(n1,n0)|add(#1,const_1)|divide(#0,n0)|divide(#2,n0)|power(#3,n0)|power(#4,n0)|subtract(#5,#6)|","chain":"39 + 1<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20 ** 2<\/gadget>\n400<\/output>\n9 - 2<\/gadget>\n7<\/output>\n7 + 1<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n400 - 16<\/gadget>\n384<\/output>\n384<\/result>","index":4428} +{"problem":"huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 12 inches on a side and costs $ 10 , and a large pizza that measures 18 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ?","rationale":"\"in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 12 * 12 = 144 ) for the small pizza and ( 18 * 18 = 324 ) for the large pizza . in total sq inches that would be ( 144 + 324 ) * 2 = 936 sq inches . in the second case if they pool their money together they can buy 3 large pizzas . in terms of square inches that would be 3 * 324 = 972 sq inches . hence , the difference is 36 square inches more ( 972 - 936 ) . the correct answer is b\"","correct":"b","options":{"a":"5 square inches ","b":"36 square inches ","c":"15 square inches ","d":"25 square inches","e":"350 square inches"},"options_float":{"a":5.0,"b":36.0,"c":15.0,"d":25.0,"e":350.0},"annotated_formula":"subtract(multiply(power(18, const_2), const_3), add(add(power(12, const_2), power(18, const_2)), add(power(12, const_2), power(18, const_2))))","linear_formula":"power(n2,const_2)|power(n0,const_2)|add(#1,#0)|multiply(#0,const_3)|add(#2,#2)|subtract(#3,#4)|","chain":"18 ** 2<\/gadget>\n324<\/output>\n324 * 3<\/gadget>\n972<\/output>\n12 ** 2<\/gadget>\n144<\/output>\n144 + 324<\/gadget>\n468<\/output>\n468 + 468<\/gadget>\n936<\/output>\n972 - 936<\/gadget>\n36<\/output>\n36<\/result>","index":4429} +{"problem":"if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 153 games will be played during the tournament . find the number of participants .","rationale":"\"if the number of participants is 3 ( say a , b , c ) the number of games played will be 2 ( a plays against b and c ) + 1 ( b plays against c ) = 3 using the same logic , if the number of participants is n , the number of games played will be ( n - 1 ) + ( n - 2 ) + ( n - 3 ) + . . . 3 + 2 + 1 given that this sum = 153 = 1 + 2 + 3 + . . . ( n - 1 ) sum of first m positive integers is given by m ( m + 1 ) \/ 2 . so sum of first ( n - 1 ) positive integers is ( n - 1 ) * n \/ 2 153 = ( n - 1 ) * n \/ 2 ( n - 1 ) * n = 306 17 * 18 = 306 ( we know that 15 ^ 2 = 225 so the two consecutive numbers must be greater than 15 . also , 20 ^ 2 = 400 so the two numbers must be less than 20 . the pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18 ) so n = 18 answer ( d )\"","correct":"d","options":{"a":"15 ","b":"16 ","c":"17 ","d":"18","e":"19"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":19.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(153, const_2), const_4), const_1)), const_1), const_2)","linear_formula":"multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|","chain":"153 * 2<\/gadget>\n306<\/output>\n306 * 4<\/gadget>\n1_224<\/output>\n1_224 + 1<\/gadget>\n1_225<\/output>\n1_225 ** (1\/2)<\/gadget>\n35<\/output>\n35 + 1<\/gadget>\n36<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n18<\/result>","index":4431} +{"problem":"if a lends rs . 2000 to b at 15 % per annum and b lends the same sum to c at 17 % per annum then the gain of b in a period of 4 years is ?","rationale":"\"( 2000 * 2 * 4 ) \/ 100 = > 160 answer : d\"","correct":"d","options":{"a":"155 ","b":"162 ","c":"157.5 ","d":"160","e":"157.12"},"options_float":{"a":155.0,"b":162.0,"c":157.5,"d":160.0,"e":157.12},"annotated_formula":"subtract(divide(multiply(multiply(2000, 17), 4), const_100), divide(multiply(multiply(2000, 15), 4), const_100))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|","chain":"2_000 * 17<\/gadget>\n34_000<\/output>\n34_000 * 4<\/gadget>\n136_000<\/output>\n136_000 \/ 100<\/gadget>\n1_360<\/output>\n2_000 * 15<\/gadget>\n30_000<\/output>\n30_000 * 4<\/gadget>\n120_000<\/output>\n120_000 \/ 100<\/gadget>\n1_200<\/output>\n1_360 - 1_200<\/gadget>\n160<\/output>\n160<\/result>","index":4433} +{"problem":"how much time will a train of length 400 m moving at a speed of 72 kmph take to cross another train of length 300 m , moving at 36 kmph in the same direction ?","rationale":"\"the distance to be covered = sum of their lengths = 400 + 300 = 700 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 \/ 18 = 10 mps . time required = d \/ s = 700 \/ 10 = 70 sec . answer : d\"","correct":"d","options":{"a":"40 sec ","b":"50 sec ","c":"60 sec ","d":"70 sec","e":"80 sec"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"divide(add(400, 300), multiply(subtract(72, 36), const_0_2778))","linear_formula":"add(n0,n2)|subtract(n1,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"400 + 300<\/gadget>\n700<\/output>\n72 - 36<\/gadget>\n36<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n700 \/ 10<\/gadget>\n70<\/output>\n70<\/result>","index":4434} +{"problem":"length of a rectangular plot is 26 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ?","rationale":"\"let breadth = x metres . then , length = ( x + 26 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 26 ) + x ] = 200 2 x + 26 = 100 2 x = 74 x = 37 hence , length = x + 26 = 63 m b\"","correct":"b","options":{"a":"53 m ","b":"63 m ","c":"73 m ","d":"83 m","e":"93 m"},"options_float":{"a":53.0,"b":63.0,"c":73.0,"d":83.0,"e":93.0},"annotated_formula":"divide(add(divide(5300, 26.50), multiply(const_2, 26)), const_4)","linear_formula":"divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)|","chain":"5_300 \/ 26.5<\/gadget>\n200<\/output>\n2 * 26<\/gadget>\n52<\/output>\n200 + 52<\/gadget>\n252<\/output>\n252 \/ 4<\/gadget>\n63<\/output>\n63<\/result>","index":4435} +{"problem":"on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 34 inch from the base of the mountain . how many km is he from the base of the mountain ?","rationale":"explanation : since 312 inch = 136 km so 1 inch = 136 \/ 312 km so 34 inch = ( 136 × 34 ) \/ 312 = 14.82 km answer : d","correct":"d","options":{"a":"14.83 ","b":"14.81 ","c":"14.8 ","d":"14.82","e":"14.12"},"options_float":{"a":14.83,"b":14.81,"c":14.8,"d":14.82,"e":14.12},"annotated_formula":"divide(multiply(34, 136), 312)","linear_formula":"multiply(n1,n2)|divide(#0,n0)","chain":"34 * 136<\/gadget>\n4_624<\/output>\n4_624 \/ 312<\/gadget>\n578\/39 = around 14.820513<\/output>\n578\/39 = around 14.820513<\/result>","index":4436} +{"problem":"the sum of the present ages of a father and his son is 42 years . 6 years ago , father ' s age was 4 times the age of the son . after 6 years , son ' s age will be :","rationale":"let the present ages of son and father be x and ( 42 - x ) years respectively . then , ( 42 - x ) - 6 = 4 ( x - 6 ) 36 - x = 4 x - 24 x = 12 the son ' s age after 6 years will be 18 . the answer is c .","correct":"c","options":{"a":"14 ","b":"16 ","c":"18 ","d":"20","e":"22"},"options_float":{"a":14.0,"b":16.0,"c":18.0,"d":20.0,"e":22.0},"annotated_formula":"add(divide(add(42, subtract(multiply(4, 6), 6)), add(4, const_1)), 6)","linear_formula":"add(n2,const_1)|multiply(n1,n2)|subtract(#1,n1)|add(n0,#2)|divide(#3,#0)|add(n1,#4)","chain":"4 * 6<\/gadget>\n24<\/output>\n24 - 6<\/gadget>\n18<\/output>\n42 + 18<\/gadget>\n60<\/output>\n4 + 1<\/gadget>\n5<\/output>\n60 \/ 5<\/gadget>\n12<\/output>\n12 + 6<\/gadget>\n18<\/output>\n18<\/result>","index":4437} +{"problem":"a recipe requires 2 1 \/ 2 ( mixed number ) cups of flour 2 3 \/ 4 ( mixed number ) cups of sugar and 1 1 \/ 3 ( mixed number ) cups of milk to make one cake . victor has 15 cups if flour , 16 cups of sugar and 8 cups of milk . what is the greatest number of cakes tom can make using this recipe ?","rationale":"less work up front : go through each item and see what the greatest number of cakes you can make with each . the lowest of these will be the right answer . flour : 15 cups , we need 2.5 cups each . just keep going up the line to see how many cakes we can make : that means i can make 2 cakes with 5 cups , so 6 cakes overall with 15 cups . i ' ve already got the answer narrowed to either a or b . sugar : 16 cups , we need 2.75 cups each . same principle . i can make 2 cups with 5.5 cups , so to make 6 cakes i ' d need 16.5 cups . i do n ' t have that much sugar , so we ' re limited to 5 cakes . no need to even do milk because we ' re already at 5 . sugar will be the limiting factor . answer is a","correct":"a","options":{"a":"5 ","b":"6 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":5.0,"b":6.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"min(min(divide(15, add(2, divide(const_1, const_2))), floor(divide(16, add(divide(const_3, const_4), 2)))), divide(8, add(divide(1, 3), 1)))","linear_formula":"divide(n1,n4)|divide(const_1,const_2)|divide(const_3,const_4)|add(n1,#0)|add(n0,#1)|add(n0,#2)|divide(n11,#3)|divide(n9,#4)|divide(n10,#5)|floor(#8)|min(#7,#9)|min(#6,#10)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n2 + (1\/2)<\/gadget>\n5\/2 = around 2.5<\/output>\n15 \/ (5\/2)<\/gadget>\n6<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) + 2<\/gadget>\n11\/4 = around 2.75<\/output>\n16 \/ (11\/4)<\/gadget>\n64\/11 = around 5.818182<\/output>\nfloor(64\/11)<\/gadget>\n5<\/output>\nmin(6, 5)<\/gadget>\n5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) + 1<\/gadget>\n4\/3 = around 1.333333<\/output>\n8 \/ (4\/3)<\/gadget>\n6<\/output>\nmin(5, 6)<\/gadget>\n5<\/output>\n5<\/result>","index":4439} +{"problem":"the c . p of 10 pens is equal to the s . p of 5 pens . find his gain % or loss % ?","rationale":"10 cp = 5 sp 10 - - - 5 cp gain 100 - - - ? = > 50 % ( gain ) answer : a","correct":"a","options":{"a":"50 % ","b":"40 % ","c":"30 % ","d":"60 %","e":"20 %"},"options_float":{"a":50.0,"b":40.0,"c":30.0,"d":60.0,"e":20.0},"annotated_formula":"multiply(divide(subtract(10, 5), 10), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)","chain":"10 - 5<\/gadget>\n5<\/output>\n5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":4440} +{"problem":"the average of 6 no . ' s is 3.95 . the average of 2 of them is 4.2 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ?","rationale":"\"sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 4.2 * 2 ) + ( 3.85 * 2 ) ] = 7.60 . required average = ( 7.6 \/ 2 ) = 3.8 . d\"","correct":"d","options":{"a":"4.2 ","b":"4.4 ","c":"4.6 ","d":"3.8","e":"5.7"},"options_float":{"a":4.2,"b":4.4,"c":4.6,"d":3.8,"e":5.7},"annotated_formula":"divide(subtract(multiply(6, 3.95), add(multiply(2, 4.2), multiply(2, 3.85))), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2)|","chain":"6 * 3.95<\/gadget>\n23.7<\/output>\n2 * 4.2<\/gadget>\n8.4<\/output>\n2 * 3.85<\/gadget>\n7.7<\/output>\n8.4 + 7.7<\/gadget>\n16.1<\/output>\n23.7 - 16.1<\/gadget>\n7.6<\/output>\n7.6 \/ 2<\/gadget>\n3.8<\/output>\n3.8<\/result>","index":4441} +{"problem":"20 one identical machines , running continuously at the same constant rate , take 9 days to complete a shipment . how many additional machines , each running at the same constant rate , would be needed to reduce the time required to complete a shipment by two days ?","rationale":"from the question you know that 21 r = 1 \/ 9 . the question asks you ( partially ) to make the rate from 1 \/ 9 to 1 \/ 7 ( drop from 9 day to 7 ) . so the only thing that you need to do is to find the magic number than can convert 1 \/ 9 to 1 \/ 7 . so 1 \/ 9 * x = 1 \/ 7 ( 1 equation with one unknown ) . so by solving this you get x = 9 \/ 7 . thats it then ! take the magic number 9 \/ 7 and multiply both sides of the original equation and you have : 21 * ( 9 \/ 7 ) * r = ( 9 \/ 7 ) * 1 \/ 9 27 * r = 1 \/ 7 , hence 27 r = 1 \/ 7 , therefore 6 more machines ! d","correct":"d","options":{"a":"2 ","b":"3 ","c":"4 ","d":"6","e":"9"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":6.0,"e":9.0},"annotated_formula":"subtract(divide(multiply(20, 9), subtract(9, const_2)), 20)","linear_formula":"multiply(n0,n1)|subtract(n1,const_2)|divide(#0,#1)|subtract(#2,n0)","chain":"20 * 9<\/gadget>\n180<\/output>\n9 - 2<\/gadget>\n7<\/output>\n180 \/ 7<\/gadget>\n180\/7 = around 25.714286<\/output>\n(180\/7) - 20<\/gadget>\n40\/7 = around 5.714286<\/output>\n40\/7 = around 5.714286<\/result>","index":4442} +{"problem":"the number of sailors on a ship is 78 % more than the number of officers . the ratio of sailors to officers would be","rationale":"sailor = 1.78 * officer sailor \/ officer = 1.78 \/ 1 = 178 \/ 100 = 89 \/ 50 answer will be d","correct":"d","options":{"a":"33 : 25 ","b":"47 : 20 ","c":"47 : 25 ","d":"89 : 50","e":"97 : 50"},"options_float":{"a":1.32,"b":2.35,"c":1.88,"d":1.78,"e":1.94},"annotated_formula":"divide(add(78, const_100), const_100)","linear_formula":"add(n0,const_100)|divide(#0,const_100)","chain":"78 + 100<\/gadget>\n178<\/output>\n178 \/ 100<\/gadget>\n89\/50 = around 1.78<\/output>\n89\/50 = around 1.78<\/result>","index":4444} +{"problem":"how much more would rs . 10000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ?","rationale":"\"10000 ( 11 \/ 10 ) 4 - 10000 ( 6 \/ 5 ) 2 = 241 answer : a\"","correct":"a","options":{"a":"241 ","b":"725 ","c":"992 ","d":"837","e":"923"},"options_float":{"a":241.0,"b":725.0,"c":992.0,"d":837.0,"e":923.0},"annotated_formula":"subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 10000), multiply(power(add(divide(20, const_100), const_1), const_2), 10000))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(#0,const_2)|add(#2,const_1)|power(#1,const_2)|multiply(n0,#4)|power(#3,const_4)|multiply(n0,#6)|subtract(#7,#5)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ 2<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) ** 4<\/gadget>\n14_641\/10_000 = around 1.4641<\/output>\n(14_641\/10_000) * 10_000<\/gadget>\n14_641<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 2<\/gadget>\n36\/25 = around 1.44<\/output>\n(36\/25) * 10_000<\/gadget>\n14_400<\/output>\n14_641 - 14_400<\/gadget>\n241<\/output>\n241<\/result>","index":4446} +{"problem":"what is the decimal equivalent of ( 1 \/ 4 ) ^ 2 ?","rationale":"\"1 \/ 4 = 25 \/ 100 = > ( 25 \/ 100 ) ^ 2 = 625 \/ 10000 = 0.0625 answer : b\"","correct":"b","options":{"a":"0.0016 ","b":"0.0625 ","c":"0.16 ","d":"0.25","e":"0.5"},"options_float":{"a":0.0016,"b":0.0625,"c":0.16,"d":0.25,"e":0.5},"annotated_formula":"power(divide(1, 4), 2)","linear_formula":"divide(n0,n1)|power(#0,n2)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) ** 2<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":4447} +{"problem":"a paint store mixes 3 \/ 4 pint of red paint and 2 \/ 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 33 pints of perfect pink paint ?","rationale":"\"3 \/ 4 pint is required to make 3 \/ 4 + 2 \/ 3 = 17 \/ 12 pint of perfect pink so 17 \/ 12 pint requires 3 \/ 4 pint of red . . 1 pint will require 3 \/ 4 * 12 \/ 17 = 9 \/ 17 . . 33 pints will require 9 \/ 17 * 33 = 17 pints . . c\"","correct":"c","options":{"a":"9 ","b":"16 ","c":"17 ","d":"25 1 \/ 3","e":"28 1 \/ 2"},"options_float":{"a":9.0,"b":16.0,"c":17.0,"d":25.0,"e":28.0},"annotated_formula":"multiply(33, divide(multiply(3, 2), multiply(4, 3)))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|divide(#0,#1)|multiply(n4,#2)|","chain":"3 * 2<\/gadget>\n6<\/output>\n4 * 3<\/gadget>\n12<\/output>\n6 \/ 12<\/gadget>\n1\/2 = around 0.5<\/output>\n33 * (1\/2)<\/gadget>\n33\/2 = around 16.5<\/output>\n33\/2 = around 16.5<\/result>","index":4448} +{"problem":"if 5 + 7 \/ x = 6 - 5 \/ x , then x =","rationale":"we ' re given the equation 5 + 7 \/ x = 6 - 5 \/ x . we ' re asked for the value of x . the common - denominator of these 4 numbers is x , so we need to multiply both sides of the equation by x , giving us . . . 5 x + 7 x \/ x = 6 x - 5 x \/ x we can then eliminate that denominator , which gives us . . . . 5 x + 7 = 6 x - 5 12 = x c","correct":"c","options":{"a":"7 ","b":"14 ","c":"12 ","d":"6 \/ 7","e":"5"},"options_float":{"a":7.0,"b":14.0,"c":12.0,"d":0.8571428571,"e":5.0},"annotated_formula":"divide(add(7, 5), subtract(6, 5))","linear_formula":"add(n0,n1)|subtract(n2,n0)|divide(#0,#1)","chain":"7 + 5<\/gadget>\n12<\/output>\n6 - 5<\/gadget>\n1<\/output>\n12 \/ 1<\/gadget>\n12<\/output>\n12<\/result>","index":4450} +{"problem":"kelly has had 3 pay cuts in her salary in the past 6 months . if the first pay cut was 8 % , the second pay cut was 14 % and the third was 18 % . what will be the percentage decrease , if the salary is decreased in a single shot ?","rationale":"let rs . 100 be initial salary . salary after 1 st decrease , 8 % = 92 salary after 2 nd decrease , 14 % = 79.12 i . e . reduced by 14 percent of 92 salary after 3 rd decrease , 18 % = 64.88 i . e . reduced by 18 percent of 79.12 so if its decreased in single shot = i = ( ( b - a ) \/ b ) * 100 = 35.12 % answer : d","correct":"d","options":{"a":"35.9 % ","b":"34.12 % ","c":"32.12 % ","d":"35.12 %","e":"31.12 %"},"options_float":{"a":35.9,"b":34.12,"c":32.12,"d":35.12,"e":31.12},"annotated_formula":"multiply(subtract(const_1, multiply(multiply(subtract(const_1, divide(8, const_100)), subtract(const_1, divide(14, const_100))), subtract(const_1, divide(18, const_100)))), const_100)","linear_formula":"divide(n2,const_100)|divide(n3,const_100)|divide(n4,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#6,#5)|subtract(const_1,#7)|multiply(#8,const_100)","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n1 - (2\/25)<\/gadget>\n23\/25 = around 0.92<\/output>\n14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n1 - (7\/50)<\/gadget>\n43\/50 = around 0.86<\/output>\n(23\/25) * (43\/50)<\/gadget>\n989\/1_250 = around 0.7912<\/output>\n18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n1 - (9\/50)<\/gadget>\n41\/50 = around 0.82<\/output>\n(989\/1_250) * (41\/50)<\/gadget>\n40_549\/62_500 = around 0.648784<\/output>\n1 - (40_549\/62_500)<\/gadget>\n21_951\/62_500 = around 0.351216<\/output>\n(21_951\/62_500) * 100<\/gadget>\n21_951\/625 = around 35.1216<\/output>\n21_951\/625 = around 35.1216<\/result>","index":4452} +{"problem":"a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of a ?","rationale":"\"( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 \/ 21 * 630 = 240 answer : a\"","correct":"a","options":{"a":"240 ","b":"278 ","c":"209 ","d":"267","e":"262"},"options_float":{"a":240.0,"b":278.0,"c":209.0,"d":267.0,"e":262.0},"annotated_formula":"multiply(divide(630, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))))","linear_formula":"add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)|","chain":"3_000 * 8<\/gadget>\n24_000<\/output>\n3_000 - 1_000<\/gadget>\n2_000<\/output>\n12 - 8<\/gadget>\n4<\/output>\n2_000 * 4<\/gadget>\n8_000<\/output>\n24_000 + 8_000<\/gadget>\n32_000<\/output>\n4_000 * 8<\/gadget>\n32_000<\/output>\n4_000 + 1_000<\/gadget>\n5_000<\/output>\n5_000 * 4<\/gadget>\n20_000<\/output>\n32_000 + 20_000<\/gadget>\n52_000<\/output>\n32_000 + 52_000<\/gadget>\n84_000<\/output>\n630 \/ 84_000<\/gadget>\n3\/400 = around 0.0075<\/output>\n(3\/400) * 32_000<\/gadget>\n240<\/output>\n240<\/result>","index":4456} +{"problem":"the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1000 to 1000000 bacteria","rationale":"the question basically asks how many minutes it takes for a population to increase by factor 1000 ( 1000000 \/ 1000 = 1000 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 1000 , where x represents the number of times the population doubles . a lot of people remember that 2 ^ 10 = 1024 . i . e . the population has to double 10 times . since it takes the population 2 minutes to double once it takes 10 * 2 minutes = 20 minutes to double nine times . thus , solution e = 20 is correct .","correct":"e","options":{"a":"10 ","b":"12 ","c":"14 ","d":"16","e":"20"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":16.0,"e":20.0},"annotated_formula":"multiply(divide(divide(1000000, 1000), const_100), const_2)","linear_formula":"divide(n2,n1)|divide(#0,const_100)|multiply(#1,const_2)","chain":"1_000_000 \/ 1_000<\/gadget>\n1_000<\/output>\n1_000 \/ 100<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":4458} +{"problem":"a woman bought a pack of 36 eggs , but the seller told her that there were 3 rotten eggs in the pack . she forgot it and begun to prepare a recipe that needs 2 eggs . what is the probability that both choosen eggs will be rotten ?","rationale":"there are 3 rotten eggs in a pack of 36 . thus the probability of picking 2 rotten eggs is : ( 3 \/ 36 ) ( 2 \/ 35 ) = 1 \/ 210 . the correct answer is a .","correct":"a","options":{"a":"1 \/ 210 ","b":"2 \/ 210 ","c":"3 \/ 210 ","d":"1 \/ 119","e":"2 \/ 119"},"options_float":{"a":0.0047619048,"b":0.0095238095,"c":0.0142857143,"d":0.0084033613,"e":0.0168067227},"annotated_formula":"multiply(divide(2, subtract(36, const_1)), divide(3, 36))","linear_formula":"divide(n1,n0)|subtract(n0,const_1)|divide(n2,#1)|multiply(#2,#0)","chain":"36 - 1<\/gadget>\n35<\/output>\n2 \/ 35<\/gadget>\n2\/35 = around 0.057143<\/output>\n3 \/ 36<\/gadget>\n1\/12 = around 0.083333<\/output>\n(2\/35) * (1\/12)<\/gadget>\n1\/210 = around 0.004762<\/output>\n1\/210 = around 0.004762<\/result>","index":4459} +{"problem":"if a person walks at 20 km \/ hr instead of 10 km \/ hr , he would have walked 20 km more . the actual distance traveled by him is :","rationale":"\"let the actual distance travelled be x km . x \/ 10 = ( x + 20 ) \/ 14 20 x = 10 x + 200 10 x = 200 x = 20 km . answer : a\"","correct":"a","options":{"a":"20 km ","b":"56 km ","c":"60 km ","d":"70 km","e":"80 km"},"options_float":{"a":20.0,"b":56.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"multiply(10, divide(20, subtract(20, 10)))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|","chain":"20 - 10<\/gadget>\n10<\/output>\n20 \/ 10<\/gadget>\n2<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":4460} +{"problem":"20 % of a number is more than 30 % of 120 by 80 . find the number ?","rationale":"( 20 \/ 100 ) * x – ( 30 \/ 100 ) * 120 = 80 1 \/ 5 x = 116 x = 580 answer : c","correct":"c","options":{"a":"520 ","b":"560 ","c":"580 ","d":"620","e":"610"},"options_float":{"a":520.0,"b":560.0,"c":580.0,"d":620.0,"e":610.0},"annotated_formula":"divide(multiply(add(divide(multiply(120, 30), const_100), 80), const_100), 20)","linear_formula":"multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)","chain":"120 * 30<\/gadget>\n3_600<\/output>\n3_600 \/ 100<\/gadget>\n36<\/output>\n36 + 80<\/gadget>\n116<\/output>\n116 * 100<\/gadget>\n11_600<\/output>\n11_600 \/ 20<\/gadget>\n580<\/output>\n580<\/result>","index":4461} +{"problem":"the population of a bacteria colony doubles every day . if it was started 9 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ?","rationale":"\"it started 9 days ago . so bacterias keep growing for 9 days + today = 10 days 2 ^ 10 = 1024 c is the answer\"","correct":"c","options":{"a":"512 ","b":"768 ","c":"1024 ","d":"2048","e":"4096"},"options_float":{"a":512.0,"b":768.0,"c":1024.0,"d":2048.0,"e":4096.0},"annotated_formula":"subtract(power(2, add(9, const_1)), const_1)","linear_formula":"add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|","chain":"9 + 1<\/gadget>\n10<\/output>\n2 ** 10<\/gadget>\n1_024<\/output>\n1_024 - 1<\/gadget>\n1_023<\/output>\n1_023<\/result>","index":4463} +{"problem":"on dividing 190 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?","rationale":"\"d = ( d - r ) \/ q = ( 190 - 1 ) \/ 9 = 189 \/ 9 = 21 e )\"","correct":"e","options":{"a":"12 ","b":"15 ","c":"16 ","d":"17","e":"21"},"options_float":{"a":12.0,"b":15.0,"c":16.0,"d":17.0,"e":21.0},"annotated_formula":"floor(divide(190, 9))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"190 \/ 9<\/gadget>\n190\/9 = around 21.111111<\/output>\nfloor(190\/9)<\/gadget>\n21<\/output>\n21<\/result>","index":4465} +{"problem":"exactly 28 % of the reporters for a certain wire service cover local politics in country x . if 30 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ?","rationale":"\"let ' s assume there are 100 reporters - - > 28 reporters cover local politics . now , as 30 % of the reporters who cover all politics do not cover local politics then the rest 70 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 70 % of them equal to 28 ( # of reporters who cover local politics ) : 0.7 x = 28 - - > x = 40 , hence 40 reporters cover politics and the rest 100 - 40 = 60 reporters do not cover politics at all . answer : d .\"","correct":"d","options":{"a":"20 % ","b":"42 % ","c":"44 % ","d":"60 %","e":"84 %"},"options_float":{"a":20.0,"b":42.0,"c":44.0,"d":60.0,"e":84.0},"annotated_formula":"multiply(subtract(const_1, divide(28, subtract(const_100, 30))), const_100)","linear_formula":"subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"100 - 30<\/gadget>\n70<\/output>\n28 \/ 70<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n60<\/result>","index":4467} +{"problem":"a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 98.01 , the number of the member is the group is :","rationale":"\"money collected = ( 98.01 x 100 ) paise = 9801 paise . number of members = square root of 9801 = 99 . answer : option d\"","correct":"d","options":{"a":"57 ","b":"67 ","c":"77 ","d":"99","e":"97"},"options_float":{"a":57.0,"b":67.0,"c":77.0,"d":99.0,"e":97.0},"annotated_formula":"sqrt(multiply(98.01, const_100))","linear_formula":"multiply(n0,const_100)|sqrt(#0)|","chain":"98.01 * 100<\/gadget>\n9_801<\/output>\n9_801 ** (1\/2)<\/gadget>\n99<\/output>\n99<\/result>","index":4468} +{"problem":"the speed of a car is 60 km in the first hour and 30 km in the second hour . what is the average speed of the car ?","rationale":"\"explanation : s = ( 60 + 30 ) \/ 2 = 45 kmph a )\"","correct":"a","options":{"a":"45 kmph ","b":"65 kmph ","c":"75 kmph ","d":"85 kmph","e":"90 kmph"},"options_float":{"a":45.0,"b":65.0,"c":75.0,"d":85.0,"e":90.0},"annotated_formula":"divide(add(60, 30), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"60 + 30<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n45<\/result>","index":4469} +{"problem":"how many figures are required for numbering the pages of a book containing 1000 pages ?","rationale":"\"explanation : 1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 999 = 900 * 3 = 2700 1000 = 4 - - - - - - - - - - - 2893 answer : c\"","correct":"c","options":{"a":"2726 ","b":"2878 ","c":"2893 ","d":"2765","e":"1811"},"options_float":{"a":2726.0,"b":2878.0,"c":2893.0,"d":2765.0,"e":1811.0},"annotated_formula":"add(add(subtract(divide(divide(1000, const_10), const_10), const_1), subtract(subtract(divide(1000, const_10), const_1), subtract(divide(divide(1000, const_10), const_10), const_1))), multiply(subtract(subtract(1000, const_1), subtract(divide(1000, const_10), const_1)), const_3))","linear_formula":"divide(n0,const_10)|subtract(n0,const_1)|divide(#0,const_10)|subtract(#0,const_1)|subtract(#2,const_1)|subtract(#1,#3)|multiply(#5,const_3)|subtract(#3,#4)|add(#4,#7)|add(#8,#6)|","chain":"1_000 \/ 10<\/gadget>\n100<\/output>\n100 \/ 10<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n100 - 1<\/gadget>\n99<\/output>\n99 - 9<\/gadget>\n90<\/output>\n9 + 90<\/gadget>\n99<\/output>\n1_000 - 1<\/gadget>\n999<\/output>\n999 - 99<\/gadget>\n900<\/output>\n900 * 3<\/gadget>\n2_700<\/output>\n99 + 2_700<\/gadget>\n2_799<\/output>\n2_799<\/result>","index":4470} +{"problem":"if the length of the longest chord of a certain circle is 10 , what is the radius of that certain circle ?","rationale":"longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 10 = 2 * 5 so radius of the circle = 5 correct answer - b","correct":"b","options":{"a":"2.5 ","b":"5 ","c":"10 ","d":"15","e":"20"},"options_float":{"a":2.5,"b":5.0,"c":10.0,"d":15.0,"e":20.0},"annotated_formula":"divide(10, const_2)","linear_formula":"divide(n0,const_2)","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":4471}