diff --git "a/test.jsonl" "b/test.jsonl" new file mode 100644--- /dev/null +++ "b/test.jsonl" @@ -0,0 +1,2029 @@ +{"problem":"a shopkeeper sold an article offering a discount of 5 % and earned a profit of 31.1 % . what would have been the percentage of profit earned if no discount had been offered ?","rationale":"\"giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 ã — ( 100 + 31.1 ) \/ ( 100 â ˆ ’ 5 ) = 138 hence , required % profit = 138 â € “ 100 = 38 % answer a\"","correct":"a","options":{"a":"38 ","b":"27.675 ","c":"30 ","d":"data inadequate","e":"none of these"},"options_float":{"a":38.0,"b":27.675,"c":30.0,"d":null,"e":null},"annotated_formula":"subtract(divide(multiply(add(const_100, 31.1), 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 + 31.1<\/gadget>\n131.1<\/output>\n131.1 * 100<\/gadget>\n13_110<\/output>\n100 - 5<\/gadget>\n95<\/output>\n13_110 \/ 95<\/gadget>\n138<\/output>\n138 - 100<\/gadget>\n38<\/output>\n38<\/result>","index":0} +{"problem":"what will be the difference between simple and compound interest at 14 % per annum on a sum of rs . 1000 after 4 years ?","rationale":"\"s . i . = ( 1000 * 14 * 4 ) \/ 100 = rs . 560 c . i . = [ 1000 * ( 1 + 14 \/ 100 ) 4 - 1000 ] = rs . 689 difference = ( 689 - 560 ) = rs . 129 answer : a\"","correct":"a","options":{"a":"129 ","b":"130 ","c":"124 ","d":"133","e":"145"},"options_float":{"a":129.0,"b":130.0,"c":124.0,"d":133.0,"e":145.0},"annotated_formula":"subtract(subtract(multiply(1000, power(add(divide(14, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(14, const_100)), 4))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)|","chain":"14 \/ 100<\/gadget>\n7\/50 = around 0.14<\/output>\n(7\/50) + 1<\/gadget>\n57\/50 = around 1.14<\/output>\n(57\/50) ** 4<\/gadget>\n10_556_001\/6_250_000 = around 1.68896<\/output>\n1_000 * (10_556_001\/6_250_000)<\/gadget>\n10_556_001\/6_250 = around 1_688.96016<\/output>\n(10_556_001\/6_250) - 1_000<\/gadget>\n4_306_001\/6_250 = around 688.96016<\/output>\n1_000 * (7\/50)<\/gadget>\n140<\/output>\n140 * 4<\/gadget>\n560<\/output>\n(4_306_001\/6_250) - 560<\/gadget>\n806_001\/6_250 = around 128.96016<\/output>\n806_001\/6_250 = around 128.96016<\/result>","index":1} +{"problem":"there are 28 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 = 30 from 30 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 3 ⁰ p ₂ ways . 30 p ₂ = 30 * 29 = 870 . answer : c\"","correct":"c","options":{"a":"156 ","b":"167 ","c":"870 ","d":"352","e":"380"},"options_float":{"a":156.0,"b":167.0,"c":870.0,"d":352.0,"e":380.0},"annotated_formula":"multiply(add(28, const_1), add(add(28, const_1), const_1))","linear_formula":"add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)|","chain":"28 + 1<\/gadget>\n29<\/output>\n29 + 1<\/gadget>\n30<\/output>\n29 * 30<\/gadget>\n870<\/output>\n870<\/result>","index":2} +{"problem":"the sum of all the integers s such that - 26 < s < 24 is","rationale":"\"easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 s = - 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":5} +{"problem":"a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 6 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ?","rationale":"\"the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 36 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 360 \/ 10,000 = 36 \/ 1000 = 0.036 ft the answer is a .\"","correct":"a","options":{"a":"0.036 ","b":"0.36 ","c":"0.6 ","d":"6","e":"3.6"},"options_float":{"a":0.036,"b":0.36,"c":0.6,"d":6.0,"e":3.6},"annotated_formula":"divide(volume_cylinder(6, 10), circle_area(100))","linear_formula":"circle_area(n0)|volume_cylinder(n2,n3)|divide(#1,#0)|","chain":"pi * (6 ** 2) * 10<\/gadget>\n360*pi = around 1_130.973355<\/output>\npi * (100 ** 2)<\/gadget>\n10000*pi = around 31_415.926536<\/output>\n(360*pi) \/ (10000*pi)<\/gadget>\n9\/250 = around 0.036<\/output>\n9\/250 = around 0.036<\/result>","index":6} +{"problem":"if n is an integer and 101 n ^ 2 is less than or equal to 10000 , what is the greatest possible value of n ?","rationale":"\"101 * n ^ 2 < = 10000 n ^ 2 < = 10000 \/ 101 which will be less than 100 since 10000 \/ 100 = 100 which is the square of 9 next closest value of n where n ^ 2 < = 100 is 9 ans 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":"floor(sqrt(divide(10000, 101)))","linear_formula":"divide(n2,n0)|sqrt(#0)|floor(#1)|","chain":"10_000 \/ 101<\/gadget>\n10_000\/101 = around 99.009901<\/output>\n(10_000\/101) ** (1\/2)<\/gadget>\n100*sqrt(101)\/101 = around 9.950372<\/output>\nfloor(100*sqrt(101)\/101)<\/gadget>\n9<\/output>\n9<\/result>","index":8} +{"problem":"the percentage profit earned by selling an article for rs . 1920 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 40 % profit ?","rationale":"\"let c . p . be rs . x . then , ( 1920 - x ) \/ x * 100 = ( x - 1280 ) \/ x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 140 % of rs . 1600 = 140 \/ 100 * 1600 = rs . 2240 . answer : e\"","correct":"e","options":{"a":"2000 ","b":"2778 ","c":"2299 ","d":"2778","e":"2240"},"options_float":{"a":2000.0,"b":2778.0,"c":2299.0,"d":2778.0,"e":2240.0},"annotated_formula":"multiply(divide(add(const_100, 40), const_100), divide(add(1920, 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 + 40<\/gadget>\n140<\/output>\n140 \/ 100<\/gadget>\n7\/5 = around 1.4<\/output>\n1_920 + 1_280<\/gadget>\n3_200<\/output>\n3_200 \/ 2<\/gadget>\n1_600<\/output>\n(7\/5) * 1_600<\/gadget>\n2_240<\/output>\n2_240<\/result>","index":11} +{"problem":"running at the same constant rate , 6 identical machines can produce a total of 360 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 : : 360 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 360 x = ( 10 x 4 x 360 ) \/ ( 6 ) x = 2400 . answer : c\"","correct":"c","options":{"a":"648 ","b":"1800 ","c":"2400 ","d":"10800","e":"10900"},"options_float":{"a":648.0,"b":1800.0,"c":2400.0,"d":10800.0,"e":10900.0},"annotated_formula":"multiply(multiply(divide(360, 6), 4), 10)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"360 \/ 6<\/gadget>\n60<\/output>\n60 * 4<\/gadget>\n240<\/output>\n240 * 10<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":12} +{"problem":"there are 1000 buildings in a street . a sign - maker is contracted to number the houses from 1 to 1000 . how many zeroes will he need ?","rationale":"divide as ( 1 - 100 ) ( 100 - 200 ) . . . . ( 900 - 1000 ) total 192 answer : c","correct":"c","options":{"a":"190 ","b":"191 ","c":"192 ","d":"193","e":"194"},"options_float":{"a":190.0,"b":191.0,"c":192.0,"d":193.0,"e":194.0},"annotated_formula":"add(add(divide(1000, const_10), multiply(subtract(const_10, 1), const_10)), const_2)","linear_formula":"divide(n0,const_10)|subtract(const_10,n1)|multiply(#1,const_10)|add(#0,#2)|add(#3,const_2)","chain":"1_000 \/ 10<\/gadget>\n100<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n100 + 90<\/gadget>\n190<\/output>\n190 + 2<\/gadget>\n192<\/output>\n192<\/result>","index":13} +{"problem":"? % of 360 = 108","rationale":"\"? % of 360 = 108 or , ? = 108 × 100 \/ 360 = 30 answer a\"","correct":"a","options":{"a":"30 ","b":"36 ","c":"64 ","d":"72","e":"none of these"},"options_float":{"a":30.0,"b":36.0,"c":64.0,"d":72.0,"e":null},"annotated_formula":"divide(multiply(108, const_100), 360)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"108 * 100<\/gadget>\n10_800<\/output>\n10_800 \/ 360<\/gadget>\n30<\/output>\n30<\/result>","index":15} +{"problem":"a train running at the speed of 120 km \/ hr crosses a pole in 18 seconds . what is the length of the train ?","rationale":"\"speed = ( 120 x ( 5 \/ 18 ) m \/ sec = ( 100 \/ 3 ) m \/ sec . length of the train = ( speed x time ) . length of the train = ( ( 100 \/ 3 ) x 18 ) m = 600 m e\"","correct":"e","options":{"a":"560 ","b":"570 ","c":"580 ","d":"590","e":"600"},"options_float":{"a":560.0,"b":570.0,"c":580.0,"d":590.0,"e":600.0},"annotated_formula":"multiply(divide(multiply(120, const_1000), const_3600), 18)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"120 * 1_000<\/gadget>\n120_000<\/output>\n120_000 \/ 3_600<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) * 18<\/gadget>\n600<\/output>\n600<\/result>","index":18} +{"problem":"a train 100 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ?","rationale":"\"s = ( 100 + 300 ) \/ 45 = 400 \/ 45 * 18 \/ 5 = 32 answer : a\"","correct":"a","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(45, 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>\n45 \/ 3_600<\/gadget>\n1\/80 = around 0.0125<\/output>\n(2\/5) \/ (1\/80)<\/gadget>\n32<\/output>\n32<\/result>","index":19} +{"problem":"each month a retailer sells 100 identical items . on each item he makes a profit of $ 40 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 + $ 40 sell price = $ 400 + $ 40 thus , the sell price is $ 440 for each item . selling all 100 items gives the retailer . . . 100 ( $ 40 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 440 ) = $ 22 discount $ 418 = $ 400 + $ 18 now , the retailer makes a profit of just $ 18 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 18 ( x ) = $ 2,000 x = 2,000 \/ 18 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, 40), subtract(40, divide(multiply(add(divide(multiply(100, 40), 10), 40), 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 * 40<\/gadget>\n4_000<\/output>\n4_000 \/ 10<\/gadget>\n400<\/output>\n400 + 40<\/gadget>\n440<\/output>\n440 * 5<\/gadget>\n2_200<\/output>\n2_200 \/ 100<\/gadget>\n22<\/output>\n40 - 22<\/gadget>\n18<\/output>\n4_000 \/ 18<\/gadget>\n2_000\/9 = around 222.222222<\/output>\n2_000\/9 = around 222.222222<\/result>","index":20} +{"problem":"bookman purchased 55 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 460 , how many paperback copies were sold ?","rationale":"\"the bookman had 10 hardback ad 55 - 10 = 45 paperback copies ; 14 copies were sold , hence 55 - 14 = 41 copies were left . let # of paperback copies left be p then 10 p + 20 ( 41 - p ) = 460 - - > 10 p = 360 - - > p = 36 # of paperback copies sold is 45 - 36 = 9 answer : e\"","correct":"e","options":{"a":"10 ","b":"11 ","c":"12 ","d":"13","e":"9"},"options_float":{"a":10.0,"b":11.0,"c":12.0,"d":13.0,"e":9.0},"annotated_formula":"divide(subtract(subtract(add(multiply(subtract(55, 10), 10), multiply(10, 20)), 460), multiply(gcd(55, 10), 20)), 10)","linear_formula":"gcd(n0,n1)|multiply(n1,n2)|subtract(n0,n1)|multiply(n1,#2)|multiply(n2,#0)|add(#3,#1)|subtract(#5,n5)|subtract(#6,#4)|divide(#7,n1)|","chain":"55 - 10<\/gadget>\n45<\/output>\n45 * 10<\/gadget>\n450<\/output>\n10 * 20<\/gadget>\n200<\/output>\n450 + 200<\/gadget>\n650<\/output>\n650 - 460<\/gadget>\n190<\/output>\ngcd(55, 10)<\/gadget>\n5<\/output>\n5 * 20<\/gadget>\n100<\/output>\n190 - 100<\/gadget>\n90<\/output>\n90 \/ 10<\/gadget>\n9<\/output>\n9<\/result>","index":22} +{"problem":"diana is painting statues . she has 1 \/ 2 of a gallon of paint remaining . each statue requires 1 \/ 16 gallon of paint . how many statues can she paint ?","rationale":"\"number of statues = all the paint ÷ amount used per statue = 1 \/ 2 ÷ 1 \/ 16 = 8 \/ 16 * 16 \/ 1 = 8 \/ 1 = 8 answer is a .\"","correct":"a","options":{"a":"8 ","b":"20 ","c":"28 ","d":"14","e":"19"},"options_float":{"a":8.0,"b":20.0,"c":28.0,"d":14.0,"e":19.0},"annotated_formula":"divide(divide(1, 2), divide(1, 16))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/2) \/ (1\/16)<\/gadget>\n8<\/output>\n8<\/result>","index":23} +{"problem":"if the price of gasoline increases by 25 % and a driver intends to spend only 20 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ?","rationale":"\"let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.25 * p ) y = 1.2 ( p * x ) y = ( 1.2 \/ 1.25 ) x = 0.96 x which is a reduction of 4 % . the answer is c .\"","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":"multiply(divide(subtract(add(25, const_100), add(20, const_100)), add(25, const_100)), const_100)","linear_formula":"add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)|","chain":"25 + 100<\/gadget>\n125<\/output>\n20 + 100<\/gadget>\n120<\/output>\n125 - 120<\/gadget>\n5<\/output>\n5 \/ 125<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) * 100<\/gadget>\n4<\/output>\n4<\/result>","index":24} +{"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 1000 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 . 1000 sculptures are not on display , so according to ( iii ) 2 \/ 3 * { not on display } = 1000 - - > { not on display } = 1500 . according to ( i ) 2 \/ 3 * { total } = 1500 - - > { total } = 2250 . answer : b .","correct":"b","options":{"a":"360 ","b":"2250 ","c":"540 ","d":"640","e":"720"},"options_float":{"a":360.0,"b":2250.0,"c":540.0,"d":640.0,"e":720.0},"annotated_formula":"divide(divide(1000, subtract(const_1, divide(1, 3))), subtract(const_1, divide(1, 3)))","linear_formula":"divide(n0,n1)|subtract(const_1,#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>\n1_000 \/ (2\/3)<\/gadget>\n1_500<\/output>\n1_500 \/ (2\/3)<\/gadget>\n2_250<\/output>\n2_250<\/result>","index":25} +{"problem":"john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 60000 and ingrid makes $ 72000 , what is their combined tax rate ?","rationale":"\"( 1 ) when 30 and 40 has equal weight or weight = 1 \/ 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 72000 \/ 132000 = 6 \/ 11 . 6 \/ 11 is a little above 6 \/ 12 = 1 \/ 2 . thus , our answer is just a little above 35 . answer : d\"","correct":"d","options":{"a":"32 % ","b":"34.4 % ","c":"35 % ","d":"35.6 %","e":"36.4 %"},"options_float":{"a":32.0,"b":34.4,"c":35.0,"d":35.6,"e":36.4},"annotated_formula":"multiply(divide(add(multiply(divide(30, const_100), 60000), multiply(divide(40, const_100), 72000)), add(72000, 60000)), const_100)","linear_formula":"add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 60_000<\/gadget>\n18_000<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 72_000<\/gadget>\n28_800<\/output>\n18_000 + 28_800<\/gadget>\n46_800<\/output>\n72_000 + 60_000<\/gadget>\n132_000<\/output>\n46_800 \/ 132_000<\/gadget>\n39\/110 = around 0.354545<\/output>\n(39\/110) * 100<\/gadget>\n390\/11 = around 35.454545<\/output>\n390\/11 = around 35.454545<\/result>","index":26} +{"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 . 900 per sq . metre .","rationale":"\"area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 900 ã — 22 ) = rs . 19800 answer : option a\"","correct":"a","options":{"a":"s . 19,800 ","b":"s . 15,600 ","c":"s . 16,500 ","d":"s . 17,600","e":"s . 17,900"},"options_float":{"a":19800.0,"b":15600.0,"c":16500.0,"d":17600.0,"e":17900.0},"annotated_formula":"multiply(900, multiply(5.5, 4))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 4<\/gadget>\n22<\/output>\n900 * 22<\/gadget>\n19_800<\/output>\n19_800<\/result>","index":27} +{"problem":"a corporation double its annual bonus to 100 of its employees . what percent of the employees ’ new bonus is the increase ?","rationale":"let the annual bonus be x . a corporation double its annual bonus . so new bonus = 2 x . increase = 2 x - x = x the increase is what percent of the employees ’ new bonus = ( x \/ 2 x ) * 100 = 50 % hence a .","correct":"a","options":{"a":"50 % ","b":"12 % ","c":"8 % ","d":"6 %","e":"5 %"},"options_float":{"a":50.0,"b":12.0,"c":8.0,"d":6.0,"e":5.0},"annotated_formula":"multiply(divide(subtract(const_2, const_1), const_2), 100)","linear_formula":"subtract(const_2,const_1)|divide(#0,const_2)|multiply(n0,#1)","chain":"2 - 1<\/gadget>\n1<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":29} +{"problem":"calculate the ratio between x and y if 30 % of x equal to 50 % of y ?","rationale":"\"explanation : 30 x = 50 y x : y = 30 : 50 = 3 : 5 answer : b\"","correct":"b","options":{"a":"4 : 5 ","b":"3 : 5 ","c":"3 : 7 ","d":"3 : 2","e":"4 : 5"},"options_float":{"a":0.8,"b":0.6,"c":0.4285714286,"d":1.5,"e":0.8},"annotated_formula":"divide(30, 50)","linear_formula":"divide(n0,n1)|","chain":"30 \/ 50<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":32} +{"problem":"three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 34 square meters . what is the area that is covered with three layers of wallpaper ?","rationale":"\"300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 34 = 86 sq m of the wallpaper overlaps in three layers . 86 sq m makes two extra layers hence the area over which it makes two extra layers is 43 sq m . answer ( a ) .\"","correct":"a","options":{"a":"43 square meters ","b":"36 square meters ","c":"42 square meters ","d":"83.3 square meters","e":"120 square meters"},"options_float":{"a":43.0,"b":36.0,"c":42.0,"d":83.3,"e":120.0},"annotated_formula":"divide(subtract(subtract(300, 180), 34), const_2)","linear_formula":"subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|","chain":"300 - 180<\/gadget>\n120<\/output>\n120 - 34<\/gadget>\n86<\/output>\n86 \/ 2<\/gadget>\n43<\/output>\n43<\/result>","index":33} +{"problem":"a trader bought a car at 20 % discount on its original price . he sold it at a 40 % 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 * ( 140 \/ 100 ) = 112 100 - 112 = 12 % answer : c\"","correct":"c","options":{"a":"82 % ","b":"52 % ","c":"12 % ","d":"19 %","e":"22 %"},"options_float":{"a":82.0,"b":52.0,"c":12.0,"d":19.0,"e":22.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 40)), 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 + 40<\/gadget>\n140<\/output>\n80 * 140<\/gadget>\n11_200<\/output>\n11_200 \/ 100<\/gadget>\n112<\/output>\n112 \/ 100<\/gadget>\n28\/25 = around 1.12<\/output>\n(28\/25) - 1<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) * 100<\/gadget>\n12<\/output>\n12<\/result>","index":35} +{"problem":"what is the unit digit in the product ( 3 ^ 65 x 6 ^ 59 x 7 ^ 71 ) ?","rationale":"explanation : unit digit in 3 ^ 4 = 1 unit digit in ( 3 ^ 4 ) 16 = 1 unit digit in 3 ^ 65 = unit digit in [ ( 3 ^ 4 ) 16 x 3 ] = ( 1 x 3 ) = 3 unit digit in 6 ^ 59 = 6 unit digit in 7 ^ 4 unit digit in ( 7 ^ 4 ) 17 is 1 . unit digit in 7 ^ 71 = unit digit in [ ( 7 ^ 4 ) 17 x 73 ] = ( 1 x 3 ) = 3 required digit = unit digit in ( 3 x 6 x 3 ) = 4 e","correct":"e","options":{"a":"18 ","b":"12 ","c":"69 ","d":"32","e":"4"},"options_float":{"a":18.0,"b":12.0,"c":69.0,"d":32.0,"e":4.0},"annotated_formula":"subtract(multiply(multiply(3, 6), 3), subtract(multiply(multiply(3, 6), 3), const_4))","linear_formula":"multiply(n0,n2)|multiply(n0,#0)|subtract(#1,const_4)|subtract(#1,#2)","chain":"3 * 6<\/gadget>\n18<\/output>\n18 * 3<\/gadget>\n54<\/output>\n54 - 4<\/gadget>\n50<\/output>\n54 - 50<\/gadget>\n4<\/output>\n4<\/result>","index":36} +{"problem":"a start walking from a place at a uniform speed of 6 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 = 6 + 5 \/ 9 = 6.55 answer b\"","correct":"b","options":{"a":"4.7 kmph ","b":"6.6 kmph ","c":"4 kmph ","d":"7 kmph","e":"5.3 kmph"},"options_float":{"a":4.7,"b":6.6,"c":4.0,"d":7.0,"e":5.3},"annotated_formula":"add(divide(1, divide(add(const_60, 48), const_60)), 6)","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) + 6<\/gadget>\n59\/9 = around 6.555556<\/output>\n59\/9 = around 6.555556<\/result>","index":37} +{"problem":"oak street begins at pine street and runs directly east for 2 kilometers until it ends when it meets maple street . oak street is intersected every 400 meters by a perpendicular street , and each of those streets other than pine street and maple street is given a number beginning at 1 st street ( one block east of pine street ) and continuing consecutively ( 2 nd street , 3 rd street , etc . . . ) until the highest - numbered street one block west of maple street . what is the highest - numbered street that intersects oak street ?","rationale":"2 km \/ 400 m = 5 . however , the street at the 2 - km mark is not 5 th street ; it is maple street . therefore , the highest numbered street is 4 th street . the answer is a .","correct":"a","options":{"a":"4 th ","b":"5 th ","c":"6 th ","d":"7 th","e":"8 th"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"subtract(divide(multiply(2, const_1000), 400), const_1)","linear_formula":"multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1)","chain":"2 * 1_000<\/gadget>\n2_000<\/output>\n2_000 \/ 400<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n4<\/result>","index":38} +{"problem":"50 percent of the members of a study group are women , and 30 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ?","rationale":"\"say there are 100 people in that group , then there would be 0.5 * 0.30 * 100 = 15 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable \/ total = 15 \/ 100 . answer : e\"","correct":"e","options":{"a":"0.16 ","b":"0.25 ","c":"0.45 ","d":"0.35","e":"0.15"},"options_float":{"a":0.16,"b":0.25,"c":0.45,"d":0.35,"e":0.15},"annotated_formula":"multiply(divide(50, multiply(multiply(const_5, const_5), const_4)), divide(30, multiply(multiply(const_5, const_5), const_4)))","linear_formula":"multiply(const_5,const_5)|multiply(#0,const_4)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)|","chain":"5 * 5<\/gadget>\n25<\/output>\n25 * 4<\/gadget>\n100<\/output>\n50 \/ 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>\n3\/20 = around 0.15<\/result>","index":39} +{"problem":"the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if arthur and david have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ?","rationale":"\"if arthur and david submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combined no discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 \/ 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) .\"","correct":"b","options":{"a":"$ 0.32 ","b":"$ 0.40 ","c":"$ 0.45 ","d":"$ 0.48","e":"$ 0.54"},"options_float":{"a":0.32,"b":0.4,"c":0.45,"d":0.48,"e":0.54},"annotated_formula":"divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2)","linear_formula":"multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2)|","chain":"80 * 0.02<\/gadget>\n1.6<\/output>\n2 * 1.6<\/gadget>\n3.2<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n160 * (3\/4)<\/gadget>\n120<\/output>\n120 * 0.02<\/gadget>\n2.4<\/output>\n3.2 - 2.4<\/gadget>\n0.8<\/output>\n0.8 \/ 2<\/gadget>\n0.4<\/output>\n0.4<\/result>","index":44} +{"problem":"if 6 men and 8 women can do a piece of work in 10 days while 26 men and 48 women can do the same in 2 days , the time taken by 15 men and 20 women in doing the same type of work will be ?","rationale":"let 1 man ' s 1 day ' s work = x and 1 women ' s 1 day ' s work = y . then , 6 x + 8 y = 1 and 26 x + 48 y = 1 . 10 2 solving these two equations , we get : x = 1 and y = 1 . 100 200 ( 15 men + 20 women ) ' s 1 day ' s work = 15 + 20 = 1 . 100 200 4 15 men and 20 women can do the work in 4 days . hence answer will be b","correct":"b","options":{"a":"5 ","b":"4 ","c":"6 ","d":"7","e":"8"},"options_float":{"a":5.0,"b":4.0,"c":6.0,"d":7.0,"e":8.0},"annotated_formula":"divide(multiply(add(divide(8, divide(subtract(multiply(48, 2), multiply(8, 10)), subtract(multiply(6, 10), multiply(26, 2)))), 6), 10), add(divide(20, divide(subtract(multiply(48, 2), multiply(8, 10)), subtract(multiply(6, 10), multiply(26, 2)))), 15))","linear_formula":"multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|divide(n1,#6)|divide(n7,#6)|add(n0,#7)|add(n6,#8)|multiply(n2,#9)|divide(#11,#10)","chain":"48 * 2<\/gadget>\n96<\/output>\n8 * 10<\/gadget>\n80<\/output>\n96 - 80<\/gadget>\n16<\/output>\n6 * 10<\/gadget>\n60<\/output>\n26 * 2<\/gadget>\n52<\/output>\n60 - 52<\/gadget>\n8<\/output>\n16 \/ 8<\/gadget>\n2<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4 + 6<\/gadget>\n10<\/output>\n10 * 10<\/gadget>\n100<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10 + 15<\/gadget>\n25<\/output>\n100 \/ 25<\/gadget>\n4<\/output>\n4<\/result>","index":46} +{"problem":"the maximum number of students among them 1345 pens and 775 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :","rationale":"\"explanation : required number of students = h . c . f of 1345 and 775 = 5 . answer : d\"","correct":"d","options":{"a":"91 ","b":"10 ","c":"6 ","d":"5","e":"none of these"},"options_float":{"a":91.0,"b":10.0,"c":6.0,"d":5.0,"e":null},"annotated_formula":"gcd(1345, 775)","linear_formula":"gcd(n0,n1)|","chain":"gcd(1_345, 775)<\/gadget>\n5<\/output>\n5<\/result>","index":47} +{"problem":"a sum of rs . 1360 has been divided among a , b and c such that a gets 2 \/ 3 of what b gets and b gets 1 \/ 4 of what c gets . b ' s share is :","rationale":"\"let c ' s share = rs . x then , b ' s share = rs . x \/ 4 ; a ' s share = rs . 2 \/ 3 * x \/ 4 = rs . x \/ 6 therefore x \/ 6 + x \/ 4 + x = 1360 17 x \/ 12 = 1360 x = 1360 * 12 \/ 17 = rs . 960 hence , b ' s share = rs . 960 \/ 4 = rs . 240 answer : c\"","correct":"c","options":{"a":"rs . 120 ","b":"rs . 160 ","c":"rs . 240 ","d":"rs . 300","e":"rs . 500"},"options_float":{"a":120.0,"b":160.0,"c":240.0,"d":300.0,"e":500.0},"annotated_formula":"subtract(subtract(multiply(divide(1360, const_10), const_2), const_12), const_12)","linear_formula":"divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)|","chain":"1_360 \/ 10<\/gadget>\n136<\/output>\n136 * 2<\/gadget>\n272<\/output>\n272 - 12<\/gadget>\n260<\/output>\n260 - 12<\/gadget>\n248<\/output>\n248<\/result>","index":48} +{"problem":"two - third of a positive number and 16 \/ 216 of its reciprocal are equal . find the positive number .","rationale":"\"explanation : let the positive number be x . then , 2 \/ 3 x = 16 \/ 216 * 1 \/ x x 2 = 16 \/ 216 * 3 \/ 2 = 16 \/ 144 x = √ 16 \/ 144 = 4 \/ 12 . answer : a\"","correct":"a","options":{"a":"4 \/ 12 ","b":"4 \/ 17 ","c":"4 \/ 15 ","d":"4 \/ 11","e":"4 \/ 03"},"options_float":{"a":0.3333333333,"b":0.2352941176,"c":0.2666666667,"d":0.3636363636,"e":null},"annotated_formula":"sqrt(divide(multiply(16, const_3), multiply(216, const_2)))","linear_formula":"multiply(n0,const_3)|multiply(n1,const_2)|divide(#0,#1)|sqrt(#2)|","chain":"16 * 3<\/gadget>\n48<\/output>\n216 * 2<\/gadget>\n432<\/output>\n48 \/ 432<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/9) ** (1\/2)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":50} +{"problem":"find large number 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 d\"","correct":"d","options":{"a":"1235 ","b":"1346 ","c":"1378 ","d":"1635","e":"1489"},"options_float":{"a":1235.0,"b":1346.0,"c":1378.0,"d":1635.0,"e":1489.0},"annotated_formula":"multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","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<\/result>","index":51} +{"problem":"in a recent election , james received 0.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 50 percent of the vote . how many additional votes would james have needed to win the election ?","rationale":"james = ( 0.5 \/ 100 ) * 2000 = 10 votes to win = ( 50 \/ 100 ) * total votes + 1 = ( 50 \/ 100 ) * 2000 + 1 = 1001 remaining voted needed to win election = 1001 - 10 = 991 answer : option d","correct":"d","options":{"a":"901 ","b":"989 ","c":"990 ","d":"991","e":"1,001"},"options_float":{"a":901.0,"b":989.0,"c":990.0,"d":991.0,"e":1001.0},"annotated_formula":"subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 0.5))","linear_formula":"add(const_1000,const_1000)|multiply(n0,#0)|subtract(#0,#1)","chain":"1_000 + 1_000<\/gadget>\n2_000<\/output>\n2_000 * 0.5<\/gadget>\n1_000<\/output>\n2_000 - 1_000<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":53} +{"problem":"what least number must be subtracted from 3832 so that the remaining number is divisible by 5 ?","rationale":"\"on dividing 3832 by 5 , we get remainder = 2 . required number be subtracted = 2 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(3832, multiply(floor(divide(3832, 5)), 5))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"3_832 \/ 5<\/gadget>\n3_832\/5 = around 766.4<\/output>\nfloor(3_832\/5)<\/gadget>\n766<\/output>\n766 * 5<\/gadget>\n3_830<\/output>\n3_832 - 3_830<\/gadget>\n2<\/output>\n2<\/result>","index":55} +{"problem":"find the length of the wire required to go 14 times round a square field containing 5625 m 2 .","rationale":"\"a 2 = 5625 = > a = 75 4 a = 300 300 * 14 = 4200 answer : c\"","correct":"c","options":{"a":"15840 ","b":"3388 ","c":"4200 ","d":"8766","e":"66711"},"options_float":{"a":15840.0,"b":3388.0,"c":4200.0,"d":8766.0,"e":66711.0},"annotated_formula":"multiply(square_perimeter(square_edge_by_area(5625)), 14)","linear_formula":"square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)|","chain":"5_625 ** (1\/2)<\/gadget>\n75<\/output>\n4 * 75<\/gadget>\n300<\/output>\n300 * 14<\/gadget>\n4_200<\/output>\n4_200<\/result>","index":56} +{"problem":"how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 1 m x 2 m x 20 cm ?","rationale":"\"number of bricks = volume of the wall \/ volume of 1 brick = ( 100 x 200 x 20 ) \/ ( 25 x 11.25 x 6 ) = 237 . answer : option c\"","correct":"c","options":{"a":"5600 ","b":"6000 ","c":"237 ","d":"7200","e":"8600"},"options_float":{"a":5600.0,"b":6000.0,"c":237.0,"d":7200.0,"e":8600.0},"annotated_formula":"divide(multiply(multiply(multiply(1, const_100), multiply(2, const_100)), 20), 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":"1 * 100<\/gadget>\n100<\/output>\n2 * 100<\/gadget>\n200<\/output>\n100 * 200<\/gadget>\n20_000<\/output>\n20_000 * 20<\/gadget>\n400_000<\/output>\n25 * 11.25<\/gadget>\n281.25<\/output>\n281.25 * 6<\/gadget>\n1_687.5<\/output>\n400_000 \/ 1_687.5<\/gadget>\n237.037037<\/output>\n237.037037<\/result>","index":57} +{"problem":"spanish language broadcast records last 90 min on each of two sides . if it takes 3 hours to translate one hour of broadcast , how long will it take to translate 16 full records ?","rationale":"records last 90 min on each of 2 sides , = = > record last 90 * 2 = 180 min = 3 hours 16 full records - - > 16 * 3 = 48 hour broadcast given , 3 hours to translate 1 hour of broadcast let x be the time required to translate 48 hour broadcast ( 16 full records ) x = 48 * 3 = 144 hours answer : a","correct":"a","options":{"a":"144 hours ","b":"124 hours ","c":"134 hours ","d":"154 hours","e":"164 hours"},"options_float":{"a":144.0,"b":124.0,"c":134.0,"d":154.0,"e":164.0},"annotated_formula":"multiply(multiply(divide(multiply(90, const_2), const_60), 16), 3)","linear_formula":"multiply(n0,const_2)|divide(#0,const_60)|multiply(n2,#1)|multiply(n1,#2)","chain":"90 * 2<\/gadget>\n180<\/output>\n180 \/ 60<\/gadget>\n3<\/output>\n3 * 16<\/gadget>\n48<\/output>\n48 * 3<\/gadget>\n144<\/output>\n144<\/result>","index":58} +{"problem":"a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and c together can do it in 2 hours . how long will b alone take to do it ?","rationale":"\"a ' s 1 hour work = 1 \/ 4 ; ( b + c ) ' s 1 hour work = 1 \/ 3 ; ( a + c ) ' s 1 hour work = 1 \/ 2 ( a + b + c ) ' s 1 hour work = ( 1 \/ 4 + 1 \/ 3 ) = 7 \/ 12 b ' s 1 hour work = ( 7 \/ 12 + 1 \/ 2 ) = 1 \/ 12 b alone will take 12 hours to do the work . answer : c\"","correct":"c","options":{"a":"15 hours ","b":"14 hours ","c":"12 hours ","d":"74 hours","e":"79 hours"},"options_float":{"a":15.0,"b":14.0,"c":12.0,"d":74.0,"e":79.0},"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":60} +{"problem":"how many seconds will a 650 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 = 650 mts speed s = 63 - 3 = 60 kmph = 60 x 5 \/ 18 m \/ s time t = = 39 sec . answer : d\"","correct":"d","options":{"a":"48 ","b":"36 ","c":"26 ","d":"39","e":"18"},"options_float":{"a":48.0,"b":36.0,"c":26.0,"d":39.0,"e":18.0},"annotated_formula":"divide(650, 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>\n650 \/ (50\/3)<\/gadget>\n39<\/output>\n39<\/result>","index":61} +{"problem":"today is thursday . i came home from a trip 3 days before the day after last monday . how many days have i been home ?","rationale":"d 6 days the day after last monday was tuesday . if i came home 3 days before that , i came home on saturday , sunday , monday , tuesday , wednesday , and thursday = 6 days .","correct":"d","options":{"a":"1 day ","b":"2 days ","c":"7 days ","d":"6 days","e":"10 days"},"options_float":{"a":1.0,"b":2.0,"c":7.0,"d":6.0,"e":10.0},"annotated_formula":"add(add(3, const_1), const_2)","linear_formula":"add(n0,const_1)|add(#0,const_2)","chain":"3 + 1<\/gadget>\n4<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6<\/result>","index":62} +{"problem":"in one hour , a boat goes 19 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km \/ hr ) is :","rationale":"\"sol . speed in still water = 1 \/ 2 ( 19 + 5 ) kmph = 12 kmph . answer d\"","correct":"d","options":{"a":"2 ","b":"4 ","c":"7 ","d":"12","e":"15"},"options_float":{"a":2.0,"b":4.0,"c":7.0,"d":12.0,"e":15.0},"annotated_formula":"divide(add(19, 5), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"19 + 5<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12<\/result>","index":64} +{"problem":"a profit of rs . 900 is divided between x and y in the ratio of 1 \/ 2 : 1 \/ 3 . what is the difference between their profit shares ?","rationale":"a profit of rs . 900 is divided between x and y in the ratio of 1 \/ 2 : 1 \/ 3 or 3 : 2 . so profits are 540 and 360 . difference in profit share = 540 - 360 = 180 answer : b","correct":"b","options":{"a":"s . 280 ","b":"s . 180 ","c":"s . 380 ","d":"s . 50","e":"s . 90"},"options_float":{"a":280.0,"b":180.0,"c":380.0,"d":50.0,"e":90.0},"annotated_formula":"subtract(divide(divide(900, add(divide(1, 2), divide(1, 3))), 2), divide(divide(900, add(divide(1, 2), divide(1, 3))), 3))","linear_formula":"divide(n1,n2)|divide(n1,n4)|add(#0,#1)|divide(n0,#2)|divide(#3,n2)|divide(#3,n4)|subtract(#4,#5)","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>\n5\/6 = around 0.833333<\/output>\n900 \/ (5\/6)<\/gadget>\n1_080<\/output>\n1_080 \/ 2<\/gadget>\n540<\/output>\n1_080 \/ 3<\/gadget>\n360<\/output>\n540 - 360<\/gadget>\n180<\/output>\n180<\/result>","index":67} +{"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 27 , the how old is b ?","rationale":"\"explanation : 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 : e\"","correct":"e","options":{"a":"7 ","b":"9 ","c":"8 ","d":"11","e":"10"},"options_float":{"a":7.0,"b":9.0,"c":8.0,"d":11.0,"e":10.0},"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":68} +{"problem":"calculate the share of y , if rs . 2880 is divided among x , y and z in the ratio 3 : 5 : 8 ?","rationale":"3 + 5 + 8 = 16 2880 \/ 16 = 180 so y ' s share = 3 * 180 = 540 answer : a","correct":"a","options":{"a":"540 ","b":"520 ","c":"140 ","d":"560","e":"240"},"options_float":{"a":540.0,"b":520.0,"c":140.0,"d":560.0,"e":240.0},"annotated_formula":"multiply(divide(2880, add(add(3, 5), 8)), 3)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n1,#2)","chain":"3 + 5<\/gadget>\n8<\/output>\n8 + 8<\/gadget>\n16<\/output>\n2_880 \/ 16<\/gadget>\n180<\/output>\n180 * 3<\/gadget>\n540<\/output>\n540<\/result>","index":69} +{"problem":"3 years ago , paula was 3 times as old as karl . in 9 years , paula will be twice as old as karl . what is the sum of their ages now ?","rationale":"\"p - 3 = 3 ( k - 3 ) and so p = 3 k - 6 p + 9 = 2 ( k + 9 ) ( 3 k - 6 ) + 9 = 2 k + 18 k = 15 p = 39 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(9, multiply(9, const_2))), subtract(multiply(3, 3), 3)), 3), subtract(multiply(3, 3), 3)), add(negate(subtract(9, multiply(9, const_2))), subtract(multiply(3, 3), 3)))","linear_formula":"multiply(n2,const_2)|multiply(n0,n1)|subtract(n2,#0)|subtract(#1,n0)|negate(#2)|add(#4,#3)|multiply(#5,n1)|subtract(#6,#3)|add(#5,#7)|","chain":"9 * 2<\/gadget>\n18<\/output>\n9 - 18<\/gadget>\n-9<\/output>\n-(-9)<\/gadget>\n9<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 - 3<\/gadget>\n6<\/output>\n9 + 6<\/gadget>\n15<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 - 6<\/gadget>\n39<\/output>\n39 + 15<\/gadget>\n54<\/output>\n54<\/result>","index":70} +{"problem":"if soundharya rows 49 km upstream and 77 km down steam taking 7 hours each , then the speed of the stream","rationale":"speed upstream = 49 \/ 7 = 7 kmph speed down stream = 77 \/ 7 = 11 kmph speed of stream = ½ ( 11 - 7 ) = 2 kmph answer : c","correct":"c","options":{"a":"6 kmph ","b":"5 kmph ","c":"2 kmph ","d":"3 kmph","e":"4 kmph"},"options_float":{"a":6.0,"b":5.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"divide(subtract(77, 49), multiply(7, const_2))","linear_formula":"multiply(n2,const_2)|subtract(n1,n0)|divide(#1,#0)","chain":"77 - 49<\/gadget>\n28<\/output>\n7 * 2<\/gadget>\n14<\/output>\n28 \/ 14<\/gadget>\n2<\/output>\n2<\/result>","index":72} +{"problem":"the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 140 inclusive is the digital sum a multiple of 7 ?","rationale":"is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 95 106 115 124 133 15 ways . . d","correct":"d","options":{"a":"7 ","b":"8 ","c":"14 ","d":"15","e":"20"},"options_float":{"a":7.0,"b":8.0,"c":14.0,"d":15.0,"e":20.0},"annotated_formula":"subtract(subtract(24, 7), const_2)","linear_formula":"subtract(n0,n2)|subtract(#0,const_2)","chain":"24 - 7<\/gadget>\n17<\/output>\n17 - 2<\/gadget>\n15<\/output>\n15<\/result>","index":73} +{"problem":"20 beavers , working together in a constant pace , can build a dam in 6 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ?","rationale":"\"total work = 20 * 6 = 120 beaver hours 12 beaver * x = 120 beaver hours x = 120 \/ 12 = 10 answer : a\"","correct":"a","options":{"a":"10 . ","b":"4 . ","c":"5 . ","d":"6","e":"8 ."},"options_float":{"a":10.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(multiply(6, 20), 12)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"6 * 20<\/gadget>\n120<\/output>\n120 \/ 12<\/gadget>\n10<\/output>\n10<\/result>","index":75} +{"problem":"if a student loses 6 kilograms , he will weigh twice as much as his sister . together they now weigh 126 kilograms . what is the student ' s present weight in kilograms ?","rationale":"\"let x be the weight of the sister . then the student ' s weight is 2 x + 6 . x + ( 2 x + 6 ) = 126 3 x = 120 x = 40 kg then the student ' s weight is 86 kg . the answer is c .\"","correct":"c","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":"subtract(126, divide(subtract(126, 6), const_3))","linear_formula":"subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)|","chain":"126 - 6<\/gadget>\n120<\/output>\n120 \/ 3<\/gadget>\n40<\/output>\n126 - 40<\/gadget>\n86<\/output>\n86<\/result>","index":76} +{"problem":"a person buys an article at rs . 575 . at what price should he sell the article so as to make a profit of 15 % ?","rationale":"\"cost price = rs . 575 profit = 15 % of 575 = rs . 86.25 selling price = cost price + profit = 575 + 86.25 = 661.25 answer : d\"","correct":"d","options":{"a":"600 ","b":"277 ","c":"269 ","d":"661.25","e":"281"},"options_float":{"a":600.0,"b":277.0,"c":269.0,"d":661.25,"e":281.0},"annotated_formula":"add(575, multiply(575, divide(15, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n575 * (3\/20)<\/gadget>\n345\/4 = around 86.25<\/output>\n575 + (345\/4)<\/gadget>\n2_645\/4 = around 661.25<\/output>\n2_645\/4 = around 661.25<\/result>","index":77} +{"problem":"two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many t hours will it take if jim types the report and mary edits it immediately after he is done ?","rationale":"\"break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 \/ 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 \/ 30 + 1 \/ x = 1 \/ 12.5 = 2 \/ 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 \/ 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 \/ 12 ; mary and jim can edit a report in 7.5 and - - > 1 \/ y + 1 \/ 12 = 1 \/ 7.5 = 2 \/ 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many t hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 \/ 7 + 20 = ~ 41.4 answer : a .\"","correct":"a","options":{"a":"41.4 ","b":"34.1 ","c":"13.4 ","d":"12.4","e":"10.8"},"options_float":{"a":41.4,"b":34.1,"c":13.4,"d":12.4,"e":10.8},"annotated_formula":"add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12))))","linear_formula":"divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|","chain":"1 \/ 12.5<\/gadget>\n0.08<\/output>\n1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n0.08 - (1\/30)<\/gadget>\n0.046667<\/output>\n1 \/ 0.046667<\/gadget>\n21.428418<\/output>\n1 \/ 7.5<\/gadget>\n0.133333<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n0.133333 - (1\/12)<\/gadget>\n0.05<\/output>\n1 \/ 0.05<\/gadget>\n20<\/output>\n21.428418 + 20<\/gadget>\n41.428418<\/output>\n41.428418<\/result>","index":78} +{"problem":"an amount of rs . 1638 was divided among a , b and c , in the ratio 1 \/ 2 : 1 \/ 3 : 1 \/ 4 . find the share of a ?","rationale":"let the shares of a , b and c be a , b and c respectively . a : b : c = 1 \/ 2 : 1 \/ 3 : 1 \/ 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 \/ 12 : 4 \/ 12 : 3 \/ 12 = 6 : 4 : 3 . share of a = 6 \/ 13 * 1638 = rs . 756 answer : c","correct":"c","options":{"a":"656 ","b":"456 ","c":"756 ","d":"745","e":"720"},"options_float":{"a":656.0,"b":456.0,"c":756.0,"d":745.0,"e":720.0},"annotated_formula":"multiply(divide(1638, add(add(2, 3), 4)), 4)","linear_formula":"add(n2,n4)|add(n6,#0)|divide(n0,#1)|multiply(n6,#2)","chain":"2 + 3<\/gadget>\n5<\/output>\n5 + 4<\/gadget>\n9<\/output>\n1_638 \/ 9<\/gadget>\n182<\/output>\n182 * 4<\/gadget>\n728<\/output>\n728<\/result>","index":79} +{"problem":"in two triangles , the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4 . find the ratio of their bases .","rationale":"sol . let the bases of the two triangles be x and y and their heights be 3 h and 4 h respectively . then , ( ( 1 \/ 2 ) x xx 3 h ) \/ ( 1 \/ 2 ) x y x 4 h ) = 4 \/ 3  x \/ y = ( 4 \/ 3 x 4 \/ 3 ) = 16 \/ 9 required ratio = 16 : 9 . ans : c","correct":"c","options":{"a":"2 : 3 ","b":"4 : 5 ","c":"16 : 9 ","d":"7 : 9","e":"8 : 5"},"options_float":{"a":0.6666666667,"b":0.8,"c":1.7777777778,"d":0.7777777778,"e":1.6},"annotated_formula":"multiply(divide(4, 3), inverse(divide(3, 4)))","linear_formula":"divide(n0,n1)|divide(n1,n0)|inverse(#1)|multiply(#0,#2)","chain":"4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ (3\/4)<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) * (4\/3)<\/gadget>\n16\/9 = around 1.777778<\/output>\n16\/9 = around 1.777778<\/result>","index":82} +{"problem":"a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 15 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ?","rationale":"\"explanation : sp of 1 metre cloth = 8200 \/ 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 15 = rs . 190 cp on 40 metres = 190 x 40 = rs . 7600 profit earned on 40 metres cloth = rs . 8200 – rs . 7600 = rs . 600 . answer : option c\"","correct":"c","options":{"a":"rs . 950 ","b":"rs . 1500 ","c":"rs . 600 ","d":"rs . 1200","e":"none of these"},"options_float":{"a":950.0,"b":1500.0,"c":600.0,"d":1200.0,"e":null},"annotated_formula":"multiply(15, 40)","linear_formula":"multiply(n0,n2)|","chain":"15 * 40<\/gadget>\n600<\/output>\n600<\/result>","index":83} +{"problem":"a thief goes away with a santro car at a speed of 50 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 60 kmph when will the owner over take the thief from the start ?","rationale":"\"explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 60 50 d = 20 rs = 60 – 50 = 10 t = 20 \/ 10 = 2 hours answer : option a\"","correct":"a","options":{"a":"2 ","b":"5 ","c":"7 ","d":"5","e":"8"},"options_float":{"a":2.0,"b":5.0,"c":7.0,"d":5.0,"e":8.0},"annotated_formula":"subtract(divide(multiply(divide(const_1, const_2), 50), subtract(60, 50)), divide(const_1, const_2))","linear_formula":"divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 50<\/gadget>\n25<\/output>\n60 - 50<\/gadget>\n10<\/output>\n25 \/ 10<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) - (1\/2)<\/gadget>\n2<\/output>\n2<\/result>","index":85} +{"problem":"the average weight of 18 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 18 boys = 50.25 total weight of 18 boys = 50.25 × 18 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 × 18 ) + ( 45.15 × 8 ) total boys = 18 + 8 = 26 average weight of all the boys = ( 50.25 × 18 ) + ( 45.15 × 8 ) \/ 26 = 48.68077 answer : option a\"","correct":"a","options":{"a":"48.68077 ","b":"42.25983 ","c":"50 ","d":"51.25388","e":"52.25"},"options_float":{"a":48.68077,"b":42.25983,"c":50.0,"d":51.25388,"e":52.25},"annotated_formula":"divide(add(multiply(18, 50.25), multiply(8, 45.15)), add(18, 8))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"18 * 50.25<\/gadget>\n904.5<\/output>\n8 * 45.15<\/gadget>\n361.2<\/output>\n904.5 + 361.2<\/gadget>\n1_265.7<\/output>\n18 + 8<\/gadget>\n26<\/output>\n1_265.7 \/ 26<\/gadget>\n48.680769<\/output>\n48.680769<\/result>","index":86} +{"problem":"the ratio of the area of a square to that of the square drawn on its diagonal is","rationale":"answer : a ) 1 : 2","correct":"a","options":{"a":"1 : 2 ","b":"1 : 0 ","c":"1 : 7 ","d":"1 : 5","e":"1 : 6"},"options_float":{"a":0.5,"b":null,"c":0.1428571429,"d":0.2,"e":0.1666666667},"annotated_formula":"power(divide(const_1, sqrt(const_2)), const_2)","linear_formula":"sqrt(const_2)|divide(const_1,#0)|power(#1,const_2)|","chain":"2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n1 \/ (sqrt(2))<\/gadget>\nsqrt(2)\/2 = around 0.707107<\/output>\n(sqrt(2)\/2) ** 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":89} +{"problem":"a 4 digit number divisible by 7 becomes divisible by 3 when 19 is added to it . the largest such number is :","rationale":"out of all the 5 options , only 4487 is not divisible by 3 . all others are divisible so answer = d ( no further calculation required ) addition of any two non - divisible numbers by 3 gives the resultant divisible by 3 19 is non - divisible by 3 ; we are adding a number to that so that the resultant becomes divisible by 3 applying the above rule , it means that the number which we are going to add should be non - divisible by 3 so comes the answer = 4487 answer : d","correct":"d","options":{"a":"4461 ","b":"4473 ","c":"4479 ","d":"4487","e":"4491"},"options_float":{"a":4461.0,"b":4473.0,"c":4479.0,"d":4487.0,"e":4491.0},"annotated_formula":"add(multiply(multiply(multiply(4, 7), multiply(3, 19)), const_2), subtract(multiply(multiply(4, 7), multiply(3, 19)), multiply(const_2, const_100)))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(const_100,const_2)|multiply(#0,#1)|multiply(#3,const_2)|subtract(#3,#2)|add(#4,#5)","chain":"4 * 7<\/gadget>\n28<\/output>\n3 * 19<\/gadget>\n57<\/output>\n28 * 57<\/gadget>\n1_596<\/output>\n1_596 * 2<\/gadget>\n3_192<\/output>\n2 * 100<\/gadget>\n200<\/output>\n1_596 - 200<\/gadget>\n1_396<\/output>\n3_192 + 1_396<\/gadget>\n4_588<\/output>\n4_588<\/result>","index":90} +{"problem":"what is the probability for a family with 3 children to have a girl and two boys ( assuming the probability of having a boy or a girl is equal ) ?","rationale":"one possible case is : girl - boy - boy the probability of this is 1 \/ 2 * 1 \/ 2 * 1 \/ 2 = 1 \/ 8 there are 3 c 2 = 3 such cases so we should multiply by 3 . p ( one girl and two boys ) = 3 \/ 8 the answer is d .","correct":"d","options":{"a":"1 \/ 8 ","b":"1 \/ 4 ","c":"1 \/ 2 ","d":"3 \/ 8","e":"5 \/ 8"},"options_float":{"a":0.125,"b":0.25,"c":0.5,"d":0.375,"e":0.625},"annotated_formula":"divide(subtract(const_1, multiply(power(divide(const_1, const_2), 3), const_2)), const_2)","linear_formula":"divide(const_1,const_2)|power(#0,n0)|multiply(#1,const_2)|subtract(const_1,#2)|divide(#3,const_2)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 3<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/8) * 2<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) \/ 2<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":91} +{"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 \/ 5 - inch increase in the length of the shoe . if the largest size of this shoe is 30 % 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 0.3 x = 9 \/ 5 x = 90 \/ 15 = 6 inches the size 15 shoe has a length of 6 + 7 \/ 5 = 7.4 inches the answer is b .\"","correct":"b","options":{"a":"6.8 ","b":"7.4 ","c":"7.7 ","d":"8.2","e":"8.6"},"options_float":{"a":6.8,"b":7.4,"c":7.7,"d":8.2,"e":8.6},"annotated_formula":"add(divide(multiply(divide(1, 5), subtract(17, 8)), divide(30, const_100)), multiply(subtract(15, 8), divide(1, 5)))","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 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n17 - 8<\/gadget>\n9<\/output>\n(1\/5) * 9<\/gadget>\n9\/5 = around 1.8<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(9\/5) \/ (3\/10)<\/gadget>\n6<\/output>\n15 - 8<\/gadget>\n7<\/output>\n7 * (1\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n6 + (7\/5)<\/gadget>\n37\/5 = around 7.4<\/output>\n37\/5 = around 7.4<\/result>","index":92} +{"problem":"a 240 m long train running at the speed of 120 km \/ hr crosses another train running in opposite direction at the speed of 80 km \/ hr in 9 sec . what is the length of the other train ?","rationale":"\"relative speed = 120 + 80 = 200 km \/ hr . = 200 * 5 \/ 18 = 500 \/ 9 m \/ sec . let the length of the other train be x m . then , ( x + 240 ) \/ 9 = 500 \/ 9 = > x = 260 . answer : option a\"","correct":"a","options":{"a":"260 ","b":"250 ","c":"240 ","d":"230","e":"220"},"options_float":{"a":260.0,"b":250.0,"c":240.0,"d":230.0,"e":220.0},"annotated_formula":"subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 240)","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 - 240<\/gadget>\n260<\/output>\n260<\/result>","index":94} +{"problem":"34 . the side surface of a cylinder is rolled with a rectangular plate . if the perimeter of the circular base is 9 feet , and the diagonal of the rectangular plate was 15 ft . what is height of the of the cylinder ?","rationale":"think of a pringles can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both , one side and thehypothenuse use pythagoreans theorem or properties of 3 - 4 - 5 triangles to solve for the other side , 12 . correct answer a","correct":"a","options":{"a":"12 ","b":"15 ","c":"10 ","d":"8","e":"14"},"options_float":{"a":12.0,"b":15.0,"c":10.0,"d":8.0,"e":14.0},"annotated_formula":"sqrt(subtract(power(15, const_2), power(9, const_2)))","linear_formula":"power(n2,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)","chain":"15 ** 2<\/gadget>\n225<\/output>\n9 ** 2<\/gadget>\n81<\/output>\n225 - 81<\/gadget>\n144<\/output>\n144 ** (1\/2)<\/gadget>\n12<\/output>\n12<\/result>","index":95} +{"problem":"what quantity of water should be added to reduce 9 liters of 50 % acidic liquid to 30 % acidic liquid ?","rationale":"acid in 9 liters = 50 % of 9 = 4.5 liters suppose x liters of water be added . then 4.5 liters of acid in 9 + x liters of diluted solution 30 % of 9 + x = 4.5 27 + 3 x = 45 x = 6 liters answer is a","correct":"a","options":{"a":"6 liters ","b":"8 liters ","c":"10 liters ","d":"12 liters","e":"15 liters"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":12.0,"e":15.0},"annotated_formula":"subtract(divide(multiply(multiply(9, divide(50, const_100)), const_100), 30), 9)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_100)|divide(#2,n2)|subtract(#3,n0)","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n9 * (1\/2)<\/gadget>\n9\/2 = around 4.5<\/output>\n(9\/2) * 100<\/gadget>\n450<\/output>\n450 \/ 30<\/gadget>\n15<\/output>\n15 - 9<\/gadget>\n6<\/output>\n6<\/result>","index":96} +{"problem":"the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 46 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ?","rationale":"\"20 x + 46 = 21 ( x – 5 ) x = 59 answer : e\"","correct":"e","options":{"a":"55 ","b":"56 ","c":"57 ","d":"58","e":"59"},"options_float":{"a":55.0,"b":56.0,"c":57.0,"d":58.0,"e":59.0},"annotated_formula":"subtract(multiply(add(20, const_1), 5), 46)","linear_formula":"add(n0,const_1)|multiply(n2,#0)|subtract(#1,n1)|","chain":"20 + 1<\/gadget>\n21<\/output>\n21 * 5<\/gadget>\n105<\/output>\n105 - 46<\/gadget>\n59<\/output>\n59<\/result>","index":98} +{"problem":"a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 18 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 18 days whole work will be done by a in 18 * 4 \/ 3 = 24 days answer is b","correct":"b","options":{"a":"10 ","b":"24 ","c":"60 ","d":"30","e":"20"},"options_float":{"a":10.0,"b":24.0,"c":60.0,"d":30.0,"e":20.0},"annotated_formula":"divide(multiply(18, 40), subtract(40, 10))","linear_formula":"multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)","chain":"18 * 40<\/gadget>\n720<\/output>\n40 - 10<\/gadget>\n30<\/output>\n720 \/ 30<\/gadget>\n24<\/output>\n24<\/result>","index":99} +{"problem":"a student gets 55 % in one subject , 65 % in the other . to get an overall of 55 % how much should get in third subject .","rationale":"\"let the 3 rd subject % = x 55 + 65 + x = 3 * 55 120 + x = 165 x = 165 - 120 = 45 answer : c\"","correct":"c","options":{"a":"75 % ","b":"25 % ","c":"45 % ","d":"55 %","e":"65 %"},"options_float":{"a":75.0,"b":25.0,"c":45.0,"d":55.0,"e":65.0},"annotated_formula":"subtract(multiply(55, const_3), add(55, 65))","linear_formula":"add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)|","chain":"55 * 3<\/gadget>\n165<\/output>\n55 + 65<\/gadget>\n120<\/output>\n165 - 120<\/gadget>\n45<\/output>\n45<\/result>","index":100} +{"problem":"in an election between two candidates , the winner has a margin of 10 % of the votes polled . if 4000 people change their mind and vote for the loser , the loser would have won by a margin of 10 % of the votes polled . find the total number of votes polled in the election ?","rationale":"\"winner - looser 55 % - 45 % if 4000 people change their mind and vote for the loser : winner - looser 45 % - 55 % thus 4,000 people compose 25 % of all voters , which means that the total number of votes is 40,000 answer : c\"","correct":"c","options":{"a":"16000 ","b":"10000 ","c":"40000 ","d":"12000","e":"5000"},"options_float":{"a":16000.0,"b":10000.0,"c":40000.0,"d":12000.0,"e":5000.0},"annotated_formula":"divide(4000, divide(10, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,#0)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n4_000 \/ (1\/10)<\/gadget>\n40_000<\/output>\n40_000<\/result>","index":101} +{"problem":"there are 28 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ?","rationale":"\"the total number of stations = 30 from 30 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 30 p 2 ways . 30 p 2 = 30 * 29 = 870 answer : d\"","correct":"d","options":{"a":"800 ","b":"820 ","c":"850 ","d":"870","e":"900"},"options_float":{"a":800.0,"b":820.0,"c":850.0,"d":870.0,"e":900.0},"annotated_formula":"multiply(add(28, const_2), subtract(add(28, const_2), const_1))","linear_formula":"add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)|","chain":"28 + 2<\/gadget>\n30<\/output>\n30 - 1<\/gadget>\n29<\/output>\n30 * 29<\/gadget>\n870<\/output>\n870<\/result>","index":102} +{"problem":"the wages earned by robin is 40 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ?","rationale":"\"let wage of erica = 10 wage of robin = 1.4 * 10 = 14 wage of charles = 1.6 * 10 = 16 percentage by which wage earned by charles is more than that earned by robin = ( 16 - 14 ) \/ 14 * 100 % = 2 \/ 14 * 100 % = 14 % answer a\"","correct":"a","options":{"a":"14 % ","b":"23 % ","c":"30 % ","d":"50 %","e":"100 %"},"options_float":{"a":14.0,"b":23.0,"c":30.0,"d":50.0,"e":100.0},"annotated_formula":"multiply(divide(subtract(add(const_100, 60), add(const_100, 40)), add(const_100, 40)), const_100)","linear_formula":"add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 + 60<\/gadget>\n160<\/output>\n100 + 40<\/gadget>\n140<\/output>\n160 - 140<\/gadget>\n20<\/output>\n20 \/ 140<\/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":104} +{"problem":"25 is subtracted from 75.00001 % of a number , the result is 50 . find the number ?","rationale":"\"( 75 \/ 100 ) * x â € “ 25 = 50 7.5 x = 750 x = 100 answer : e\"","correct":"e","options":{"a":"150 ","b":"75 ","c":"125 ","d":"95","e":"100"},"options_float":{"a":150.0,"b":75.0,"c":125.0,"d":95.0,"e":100.0},"annotated_formula":"divide(add(25, 50), divide(75.00001, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"25 + 50<\/gadget>\n75<\/output>\n75.00001 \/ 100<\/gadget>\n0.75<\/output>\n75 \/ 0.75<\/gadget>\n100<\/output>\n100<\/result>","index":105} +{"problem":"half a number plus 7 is 11 . what is the number ?","rationale":"\"let x be the number . always replace ` ` is ' ' with an equal sign ( 1 \/ 2 ) x + 7 = 11 ( 1 \/ 2 ) x = 11 - 7 ( 1 \/ 2 ) x = 4 x = 8 correct answer is a\"","correct":"a","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(subtract(11, 7), const_2)","linear_formula":"subtract(n1,n0)|multiply(#0,const_2)|","chain":"11 - 7<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":107} +{"problem":"of the total amount that jill spent on a shopping trip , excluding taxes , she spent 25 percent on clothing , 25 percent on food , and 50 percent on other items . if jill paid a 10 percent tax on the clothing , no tax on the food , and an 2 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ?","rationale":"\"assume she has $ 200 to spend . tax clothing = 25 % = $ 50 = $ 5 food = 25 % = $ 50 = $ 0.00 items = 50 % = $ 100 = $ 2.00 total tax = $ 20.00 % of total amount = 7 \/ 200 * 100 = 3.5 % answer d\"","correct":"d","options":{"a":"3 ","b":"4.5 ","c":"4 ","d":"3.5","e":"5"},"options_float":{"a":3.0,"b":4.5,"c":4.0,"d":3.5,"e":5.0},"annotated_formula":"multiply(divide(add(multiply(25, divide(10, const_100)), multiply(50, divide(2, const_100))), const_100), const_100)","linear_formula":"divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n25 * (1\/10)<\/gadget>\n5\/2 = around 2.5<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n50 * (1\/50)<\/gadget>\n1<\/output>\n(5\/2) + 1<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) \/ 100<\/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":109} +{"problem":"in a certain pond , 55 fish were caught , tagged , and returned to the pond . a few days later , 55 fish were caught again , of which 2 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 ` s the approximate number of fish in the pond ?","rationale":"\"the percent of tagged fish in the second catch is 2 \/ 55 * 100 = 3.64 % . we are told that 3.64 % approximates the percent of tagged fish in the pond . since there are 55 tagged fish , then we have 0.036 x = 55 - - > x = 1,528 . answer : d .\"","correct":"d","options":{"a":"400 ","b":"625 ","c":"1250 ","d":"1528","e":"10 000"},"options_float":{"a":400.0,"b":625.0,"c":1250.0,"d":1528.0,"e":10.0},"annotated_formula":"divide(55, divide(2, 55))","linear_formula":"divide(n2,n1)|divide(n0,#0)|","chain":"2 \/ 55<\/gadget>\n2\/55 = around 0.036364<\/output>\n55 \/ (2\/55)<\/gadget>\n3_025\/2 = around 1_512.5<\/output>\n3_025\/2 = around 1_512.5<\/result>","index":110} +{"problem":"how many 1 \/ 6 s are there in 37 1 \/ 2 ?","rationale":"required number = ( 75 \/ 2 ) \/ ( 1 \/ 6 ) = ( 75 \/ 2 x 6 \/ 1 ) = 225 . answer : a","correct":"a","options":{"a":"225 ","b":"425 ","c":"520 ","d":"600","e":"700"},"options_float":{"a":225.0,"b":425.0,"c":520.0,"d":600.0,"e":700.0},"annotated_formula":"divide(add(37, divide(1, 2)), divide(1, 6))","linear_formula":"divide(n0,n4)|divide(n0,n1)|add(n2,#0)|divide(#2,#1)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n37 + (1\/2)<\/gadget>\n75\/2 = around 37.5<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(75\/2) \/ (1\/6)<\/gadget>\n225<\/output>\n225<\/result>","index":111} +{"problem":"the ratio between the present ages of a and b is 7 : 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 7 x and 3 x years respectively . then , ( 7 x - 4 ) \/ ( 3 x + 4 ) = 1 \/ 1 4 x = 8 = > x = 2 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 18 : 2 = 9 : 1 . answer : c\"","correct":"c","options":{"a":"3 : 4 ","b":"3 : 0 ","c":"9 : 1 ","d":"9 : 2","e":"3 : 9"},"options_float":{"a":0.75,"b":null,"c":9.0,"d":4.5,"e":0.3333333333},"annotated_formula":"divide(add(multiply(7, divide(add(7, 7), subtract(7, 3))), 7), subtract(multiply(3, divide(add(7, 7), subtract(7, 3))), 7))","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":"7 + 7<\/gadget>\n14<\/output>\n7 - 3<\/gadget>\n4<\/output>\n14 \/ 4<\/gadget>\n7\/2 = around 3.5<\/output>\n7 * (7\/2)<\/gadget>\n49\/2 = around 24.5<\/output>\n(49\/2) + 7<\/gadget>\n63\/2 = around 31.5<\/output>\n3 * (7\/2)<\/gadget>\n21\/2 = around 10.5<\/output>\n(21\/2) - 7<\/gadget>\n7\/2 = around 3.5<\/output>\n(63\/2) \/ (7\/2)<\/gadget>\n9<\/output>\n9<\/result>","index":113} +{"problem":"two vessels contains equal number of mixtures milk and water in the ratio 3 : 2 and 4 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ?","rationale":"\"the ratio of milk and water in the new vessel is = ( 3 \/ 5 + 4 \/ 5 ) : ( 2 \/ 5 + 1 \/ 5 ) = 7 \/ 5 : 3 \/ 5 = 7 : 3 answer is d\"","correct":"d","options":{"a":"1 : 3 ","b":"9 : 13 ","c":"5 : 11 ","d":"7 : 3","e":"15 : 4"},"options_float":{"a":0.3333333333,"b":0.6923076923,"c":0.4545454545,"d":2.3333333333,"e":3.75},"annotated_formula":"divide(add(multiply(3, divide(add(4, 1), add(3, 2))), 4), add(multiply(2, divide(add(4, 1), add(3, 2))), 1))","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)|","chain":"4 + 1<\/gadget>\n5<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 \/ 5<\/gadget>\n1<\/output>\n3 * 1<\/gadget>\n3<\/output>\n3 + 4<\/gadget>\n7<\/output>\n2 * 1<\/gadget>\n2<\/output>\n2 + 1<\/gadget>\n3<\/output>\n7 \/ 3<\/gadget>\n7\/3 = around 2.333333<\/output>\n7\/3 = around 2.333333<\/result>","index":114} +{"problem":"if sharon ' s weekly salary increased by 15 percent , she would earn $ 460 per week . if instead , her weekly salary were to increase by 10 percent , how much would she earn per week ?","rationale":"soln : - 460 \/ 115 ) 110 = 385 in this case long division does not take much time . ( 4 \/ 1 ) 110 = rs . 440 answer : b","correct":"b","options":{"a":"rs . 400 ","b":"rs . 440 ","c":"rs . 150 ","d":"rs . 460","e":"rs . 520"},"options_float":{"a":400.0,"b":440.0,"c":150.0,"d":460.0,"e":520.0},"annotated_formula":"add(divide(460, add(const_1, divide(15, const_100))), multiply(divide(10, const_100), divide(460, add(const_1, divide(15, 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":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n460 \/ (23\/20)<\/gadget>\n400<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 400<\/gadget>\n40<\/output>\n400 + 40<\/gadget>\n440<\/output>\n440<\/result>","index":116} +{"problem":"a fruit seller had some oranges . he sells 10 % oranges and still has 360 oranges . how many oranges he had originally ?","rationale":"\"explanation : he sells 10 % of oranges and still there are 360 oranges remaining = > 90 % of oranges = 360 ⇒ ( 90 × total oranges ) \/ 100 = 360 ⇒ total oranges \/ 100 = 4 ⇒ total oranges = 4 × 100 = 400 answer : option d\"","correct":"d","options":{"a":"420 ","b":"700 ","c":"220 ","d":"400","e":"none of these"},"options_float":{"a":420.0,"b":700.0,"c":220.0,"d":400.0,"e":null},"annotated_formula":"add(360, multiply(360, divide(10, const_100)))","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n360 * (1\/10)<\/gadget>\n36<\/output>\n360 + 36<\/gadget>\n396<\/output>\n396<\/result>","index":117} +{"problem":"because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 20 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ?","rationale":"\"say morks income is - 100 so tax paid will be 40 say mindys income is 4 * 100 = 400 so tax paid is 20 % * 400 = 80 total tax paid = 40 + 80 = 120 . combined tax % will be 120 \/ 100 + 400 = 24 %\"","correct":"b","options":{"a":"22.5 % ","b":"24 % ","c":"30 % ","d":"33 %","e":"20 %"},"options_float":{"a":22.5,"b":24.0,"c":30.0,"d":33.0,"e":20.0},"annotated_formula":"multiply(const_100, divide(add(divide(40, const_100), multiply(4, divide(20, const_100))), add(const_1, 4)))","linear_formula":"add(n2,const_1)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|multiply(#5,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n4 * (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(2\/5) + (4\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n1 + 4<\/gadget>\n5<\/output>\n(6\/5) \/ 5<\/gadget>\n6\/25 = around 0.24<\/output>\n100 * (6\/25)<\/gadget>\n24<\/output>\n24<\/result>","index":118} +{"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 10 minutes , how much time will she spend doing yoga ? ( rounded to minutes )","rationale":"\"the ratio is 2 : 3 = gym : ride , so ( 10 ) ( 3 \/ 2 ) = 15 minutes at the gym , and 15 + 10 = 25 minutes exercise , so ( 2 \/ 3 ) ( 25 ) = 17 minutes yoga . answer : c\"","correct":"c","options":{"a":"10 min . ","b":"41 min . ","c":"17 min . ","d":"23 min .","e":"25 min ."},"options_float":{"a":10.0,"b":41.0,"c":17.0,"d":23.0,"e":25.0},"annotated_formula":"divide(multiply(10, 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>\n10 * (3\/5)<\/gadget>\n6<\/output>\n(3\/5) * (3\/5)<\/gadget>\n9\/25 = around 0.36<\/output>\n6 \/ (9\/25)<\/gadget>\n50\/3 = around 16.666667<\/output>\n50\/3 = around 16.666667<\/result>","index":119} +{"problem":"the length of the bridge , which a train 180 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 , ( 180 + x ) \/ 30 = 25 \/ 2 = > 2 ( 180 + x ) = 750 = > x = 195 m . answer : c\"","correct":"c","options":{"a":"876 m ","b":"157 m ","c":"195 m ","d":"156 m","e":"167 m"},"options_float":{"a":876.0,"b":157.0,"c":195.0,"d":156.0,"e":167.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 180)","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 - 180<\/gadget>\n195<\/output>\n195<\/result>","index":120} +{"problem":"the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the difference the sum and the number if the ratio between the digits of the number is 1 : 2 ?","rationale":"let the number be xy . given xy – yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten ’ s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) – ( 10 y + x ) = 36 = > x – y = 4 = > 2 y – y = 4 . hence , ( x + y ) – ( x – y ) = 3 y – y = 2 y = 8 b","correct":"b","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":"multiply(divide(36, subtract(multiply(subtract(const_10, const_1), multiply(2, const_1)), subtract(const_10, const_1))), const_2)","linear_formula":"multiply(n0,const_1)|subtract(const_10,const_1)|multiply(#0,#1)|subtract(#2,#1)|divide(n1,#3)|multiply(#4,const_2)","chain":"10 - 1<\/gadget>\n9<\/output>\n2 * 1<\/gadget>\n2<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18 - 9<\/gadget>\n9<\/output>\n36 \/ 9<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":121} +{"problem":"at the faculty of aerospace engineering , 312 students study random - processing methods , 232 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ?","rationale":"\"students studying random - processing methods = 312 students studying scramjet rocket engines = 232 students studying them both = 112 therefore ; students studying only random processing methods = 312 - 112 = 200 students studying only scramjet rocket engines = 232 - 112 = 120 students studying both = 112 students studying none = 0 ( as mentioned in question that every student in the faculty has to study one of the two subjects ) total students in faculty of aerospace engineering = students of only random processing methods + students of only scramjet rocket engines + both + none total number of students = 200 + 120 + 112 + 0 = 432 . . . . answer d\"","correct":"d","options":{"a":"424 . ","b":"428 . ","c":"430 . ","d":"432 .","e":"436"},"options_float":{"a":424.0,"b":428.0,"c":430.0,"d":432.0,"e":436.0},"annotated_formula":"add(subtract(312, divide(112, const_2)), subtract(232, divide(112, const_2)))","linear_formula":"divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|add(#1,#2)|","chain":"112 \/ 2<\/gadget>\n56<\/output>\n312 - 56<\/gadget>\n256<\/output>\n232 - 56<\/gadget>\n176<\/output>\n256 + 176<\/gadget>\n432<\/output>\n432<\/result>","index":122} +{"problem":"if n = 2 ^ 0.20 and n ^ b = 16 , b must equal","rationale":"\"20 \/ 100 = 1 \/ 5 n = 2 ^ 1 \/ 5 n ^ b = 2 ^ 4 ( 2 ^ 1 \/ 5 ) ^ b = 2 ^ 4 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(16), log(power(2, 0.20)))","linear_formula":"log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)|","chain":"log(16)<\/gadget>\nlog(16) = around 2.772589<\/output>\n2 ** 0.2<\/gadget>\n1.148698<\/output>\nlog(1.148698)<\/gadget>\n0.138629<\/output>\nlog(16) \/ 0.138629<\/gadget>\n7.21349789726536*log(16) = around 20.000063<\/output>\n7.21349789726536*log(16) = around 20.000063<\/result>","index":125} +{"problem":"a man buys an item at rs . 800 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 800 = 80 \/ 100 * 800 = 640 option a\"","correct":"a","options":{"a":"rs . 640 ","b":"rs . 760 ","c":"rs . 860 ","d":"rs . 960","e":"none of these"},"options_float":{"a":640.0,"b":760.0,"c":860.0,"d":960.0,"e":null},"annotated_formula":"multiply(800, 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>\n800 * (4\/5)<\/gadget>\n640<\/output>\n640<\/result>","index":126} +{"problem":"find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 14 per cubic meter ?","rationale":"\"22 \/ 7 * 14 * 3 \/ 2 * 3 \/ 2 = 99 m 2 99 * 14 = 1386 answer : a\"","correct":"a","options":{"a":"1386 ","b":"2799 ","c":"2890 ","d":"1485","e":"2780"},"options_float":{"a":1386.0,"b":2799.0,"c":2890.0,"d":1485.0,"e":2780.0},"annotated_formula":"multiply(volume_cylinder(divide(3, const_2), 14), 14)","linear_formula":"divide(n1,const_2)|volume_cylinder(#0,n0)|multiply(n2,#1)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\npi * ((3\/2) ** 2) * 14<\/gadget>\n63*pi\/2 = around 98.960169<\/output>\n(63*pi\/2) * 14<\/gadget>\n441*pi = around 1_385.44236<\/output>\n441*pi = around 1_385.44236<\/result>","index":127} +{"problem":"rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage z .","rationale":"\"given : selling price of an apple = 23 - - > cost price = 23 \/ 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 \/ 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % z = ( 97 \/ 556 ) * 100 = 17.4 % answer : b\"","correct":"b","options":{"a":"16.8 % ","b":"17.4 % ","c":"17.9 % ","d":"18.5 %","e":"19.1 %"},"options_float":{"a":16.8,"b":17.4,"c":17.9,"d":18.5,"e":19.1},"annotated_formula":"multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100)","linear_formula":"add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100)|","chain":"2 * 10<\/gadget>\n20<\/output>\n20 + 1<\/gadget>\n21<\/output>\n20 * 21<\/gadget>\n420<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n10 \/ (5\/4)<\/gadget>\n8<\/output>\n15 + 2<\/gadget>\n17<\/output>\n8 * 17<\/gadget>\n136<\/output>\n420 + 136<\/gadget>\n556<\/output>\n653 - 556<\/gadget>\n97<\/output>\n97 \/ 556<\/gadget>\n97\/556 = around 0.17446<\/output>\n(97\/556) * 100<\/gadget>\n2_425\/139 = around 17.446043<\/output>\n2_425\/139 = around 17.446043<\/result>","index":128} +{"problem":"if 8 men or 12 women can do a piece of work in 25 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 \/ 25 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 25 = 1 \/ 300 so , 20 women work = 20 \/ 300 = 1 \/ 15 ans : 15 days answer : d\"","correct":"d","options":{"a":"10 days ","b":"11 days ","c":"13 days ","d":"15 days","e":"17 days"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":15.0,"e":17.0},"annotated_formula":"inverse(add(divide(6, multiply(8, 25)), divide(11, multiply(12, 25))))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)|","chain":"8 * 25<\/gadget>\n200<\/output>\n6 \/ 200<\/gadget>\n3\/100 = around 0.03<\/output>\n12 * 25<\/gadget>\n300<\/output>\n11 \/ 300<\/gadget>\n11\/300 = around 0.036667<\/output>\n(3\/100) + (11\/300)<\/gadget>\n1\/15 = around 0.066667<\/output>\n1 \/ (1\/15)<\/gadget>\n15<\/output>\n15<\/result>","index":130} +{"problem":"a train 140 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train .","rationale":"\"explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 \/ 18 m \/ sec . therefore 140 \/ ( ( x + 5 ) × 5 \/ 18 ) = 6 < = > 30 ( x + 5 ) = 2520 < = > x = 79 speed of the train is 79 kmph . answer : option e\"","correct":"e","options":{"a":"45 kmph ","b":"50 kmph ","c":"55 kmph ","d":"60 kmph","e":"79 kmph"},"options_float":{"a":45.0,"b":50.0,"c":55.0,"d":60.0,"e":79.0},"annotated_formula":"subtract(divide(140, multiply(6, const_0_2778)), 5)","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|subtract(#1,n2)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n6 * (5\/18)<\/gadget>\n5\/3 = around 1.666667<\/output>\n140 \/ (5\/3)<\/gadget>\n84<\/output>\n84 - 5<\/gadget>\n79<\/output>\n79<\/result>","index":131} +{"problem":"find the average of all the number between 6 and 34 which are divisible by 5 .","rationale":"\"solution average = ( 10 + 15 + 20 + 25 + 30 \/ 5 ) = 100 \/ 2 = 20 answer b\"","correct":"b","options":{"a":"18 ","b":"20 ","c":"24 ","d":"30","e":"32"},"options_float":{"a":18.0,"b":20.0,"c":24.0,"d":30.0,"e":32.0},"annotated_formula":"divide(add(add(6, const_4), subtract(34, const_4)), const_2)","linear_formula":"add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|","chain":"6 + 4<\/gadget>\n10<\/output>\n34 - 4<\/gadget>\n30<\/output>\n10 + 30<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":134} +{"problem":"if the wheel is 14 cm then the number of revolutions to cover a distance of 880 cm is ?","rationale":"\"2 * 22 \/ 7 * 14 * x = 880 = > x = 10 answer : b\"","correct":"b","options":{"a":"15 ","b":"10 ","c":"14 ","d":"12","e":"11"},"options_float":{"a":15.0,"b":10.0,"c":14.0,"d":12.0,"e":11.0},"annotated_formula":"divide(880, 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>\n880 \/ (2_198\/25)<\/gadget>\n11_000\/1_099 = around 10.009099<\/output>\n11_000\/1_099 = around 10.009099<\/result>","index":135} +{"problem":"a and b can do a piece of work in 7 days , b and c in 8 days , c and a in 9 days . how long will c take to do it ?","rationale":"\"2 c = 1 \/ 8 + 1 \/ 9 – 1 \/ 7 = 47 \/ 504 c = 47 \/ 1008 = > 1008 \/ 47 = 21.4 days the answer is c .\"","correct":"c","options":{"a":"14.1 days ","b":"18.8 days ","c":"21.4 days ","d":"24.3 days","e":"26.5 days"},"options_float":{"a":14.1,"b":18.8,"c":21.4,"d":24.3,"e":26.5},"annotated_formula":"divide(multiply(7, const_3), subtract(divide(add(divide(multiply(7, const_3), 9), add(divide(multiply(7, const_3), 7), divide(multiply(7, const_3), 8))), const_2), divide(multiply(7, const_3), 7)))","linear_formula":"multiply(n0,const_3)|divide(#0,n0)|divide(#0,n1)|divide(#0,n2)|add(#1,#2)|add(#4,#3)|divide(#5,const_2)|subtract(#6,#1)|divide(#0,#7)|","chain":"7 * 3<\/gadget>\n21<\/output>\n21 \/ 9<\/gadget>\n7\/3 = around 2.333333<\/output>\n21 \/ 7<\/gadget>\n3<\/output>\n21 \/ 8<\/gadget>\n21\/8 = around 2.625<\/output>\n3 + (21\/8)<\/gadget>\n45\/8 = around 5.625<\/output>\n(7\/3) + (45\/8)<\/gadget>\n191\/24 = around 7.958333<\/output>\n(191\/24) \/ 2<\/gadget>\n191\/48 = around 3.979167<\/output>\n(191\/48) - 3<\/gadget>\n47\/48 = around 0.979167<\/output>\n21 \/ (47\/48)<\/gadget>\n1_008\/47 = around 21.446809<\/output>\n1_008\/47 = around 21.446809<\/result>","index":136} +{"problem":"oil cans x and y are right circular cylinders and the height and radius of y are each 5 times those of x . if the oil in can x filled to capacity sells for $ 1 , how much does the oil in y sell for if y is only 1 \/ 5 th filled ?","rationale":"\"formula for vol of a cyl is pi * r ^ 2 * h so vy = 125 * vy y when half filled will cost 25 times x so ans is a\"","correct":"a","options":{"a":"$ 25 ","b":"$ 26 ","c":"$ 27 ","d":"$ 28","e":"$ 30"},"options_float":{"a":25.0,"b":26.0,"c":27.0,"d":28.0,"e":30.0},"annotated_formula":"multiply(power(5, 1), 5)","linear_formula":"power(n0,n1)|multiply(n0,#0)|","chain":"5 ** 1<\/gadget>\n5<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25<\/result>","index":137} +{"problem":"car x began traveling at an average speed of 35 miles per hour . after 36 minutes , car y began traveling at an average speed of 38 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":"\"in 36 minutes , car x travels 21 miles . car y gains 3 miles each hour , so it takes 7 hours to catch car x . in 7 hours , car x travels 245 miles . the answer is e .\"","correct":"e","options":{"a":"105 ","b":"140 ","c":"175 ","d":"210","e":"245"},"options_float":{"a":105.0,"b":140.0,"c":175.0,"d":210.0,"e":245.0},"annotated_formula":"multiply(35, divide(multiply(divide(36, const_60), 35), subtract(38, 35)))","linear_formula":"divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)|","chain":"36 \/ 60<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 35<\/gadget>\n21<\/output>\n38 - 35<\/gadget>\n3<\/output>\n21 \/ 3<\/gadget>\n7<\/output>\n35 * 7<\/gadget>\n245<\/output>\n245<\/result>","index":139} +{"problem":"cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 51 after it went on sale for 15 % off , what is the difference between the original price and the final price ?","rationale":"\"0.85 * { original price } = $ 51 - - > { original price } = $ 60 . { final price } = $ 51 * 1.25 = $ 63.75 . the difference = $ 63.75 - 60 $ = $ 3.75 . answer : c .\"","correct":"c","options":{"a":"$ 0.00 ","b":"$ 1.00 ","c":"$ 3.75 ","d":"$ 5.00","e":"$ 6.80"},"options_float":{"a":0.0,"b":1.0,"c":3.75,"d":5.0,"e":6.8},"annotated_formula":"subtract(multiply(51, divide(add(const_100, 25), const_100)), divide(51, divide(subtract(const_100, 15), const_100)))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#3)|multiply(n2,#2)|subtract(#5,#4)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n51 * (5\/4)<\/gadget>\n255\/4 = around 63.75<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n51 \/ (17\/20)<\/gadget>\n60<\/output>\n(255\/4) - 60<\/gadget>\n15\/4 = around 3.75<\/output>\n15\/4 = around 3.75<\/result>","index":140} +{"problem":"2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations . what is the ratio of their speeds ?","rationale":"in same time , they cover 110 km & 90 km respectively so ratio of their speed = 110 : 90 = 11 : 9 answer : a","correct":"a","options":{"a":"11 : 9 ","b":"11 : 2 ","c":"91 : 9 ","d":"11 : 1","e":"11 : 5"},"options_float":{"a":1.2222222222,"b":5.5,"c":10.1111111111,"d":11.0,"e":2.2},"annotated_formula":"inverse(divide(subtract(200, 110), 110))","linear_formula":"subtract(n2,n3)|divide(#0,n3)|inverse(#1)","chain":"200 - 110<\/gadget>\n90<\/output>\n90 \/ 110<\/gadget>\n9\/11 = around 0.818182<\/output>\n1 \/ (9\/11)<\/gadget>\n11\/9 = around 1.222222<\/output>\n11\/9 = around 1.222222<\/result>","index":141} +{"problem":"a grocer has a sale of rs . 4435 , rs . 4927 , rs . 4855 , rs . 5230 and rs . 4562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 4500 ?","rationale":"\"total sale for 5 months = rs . ( 4435 + 4927 + 4855 + 5230 + 4562 ) = rs . 24009 . required sale = rs . [ ( 4500 x 6 ) - 24009 ] = rs . ( 27000 - 24009 ) = rs . 2991 answer : option b\"","correct":"b","options":{"a":"1991 ","b":"2991 ","c":"3991 ","d":"4521","e":"5991"},"options_float":{"a":1991.0,"b":2991.0,"c":3991.0,"d":4521.0,"e":5991.0},"annotated_formula":"subtract(multiply(add(5, const_1), 4500), add(add(add(add(4435, 4927), 4855), 5230), 4562))","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 * 4_500<\/gadget>\n27_000<\/output>\n4_435 + 4_927<\/gadget>\n9_362<\/output>\n9_362 + 4_855<\/gadget>\n14_217<\/output>\n14_217 + 5_230<\/gadget>\n19_447<\/output>\n19_447 + 4_562<\/gadget>\n24_009<\/output>\n27_000 - 24_009<\/gadget>\n2_991<\/output>\n2_991<\/result>","index":142} +{"problem":"what is the sum of all 3 digit integers formed using the digits 34 and 5 ( repetition is allowed )","rationale":"n = 3 * 3 * 3 = 27 = ( 555 + 333 ) \/ 2 = 444 sum = number of integers x average value n * = 27 * 444 = 11988 answer = d","correct":"d","options":{"a":"11982 ","b":"11984 ","c":"11985 ","d":"11988","e":"11986"},"options_float":{"a":11982.0,"b":11984.0,"c":11985.0,"d":11988.0,"e":11986.0},"annotated_formula":"multiply(multiply(add(add(const_100, const_10), const_1), add(add(3, 5), const_4)), power(const_3, const_2))","linear_formula":"add(const_10,const_100)|add(n0,n2)|power(const_3,const_2)|add(#0,const_1)|add(#1,const_4)|multiply(#3,#4)|multiply(#5,#2)","chain":"100 + 10<\/gadget>\n110<\/output>\n110 + 1<\/gadget>\n111<\/output>\n3 + 5<\/gadget>\n8<\/output>\n8 + 4<\/gadget>\n12<\/output>\n111 * 12<\/gadget>\n1_332<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n1_332 * 9<\/gadget>\n11_988<\/output>\n11_988<\/result>","index":145} +{"problem":"two men are going along a track rail in the opposite direction . one goods train crossed the first person in 20 sec . after 10 min the train crossed the other person who is coming in opposite direction in 18 sec . after the train has passed , when the two persons will meet ?","rationale":"explanation : let us consider that speed of train , first man and second man are respectively t , f and s . according to first given condition goods train crossed the first person moving in same direction in 20 sec . so length of the will be 20 ( t - f ) similarly train crossed the second man in 18 sec . so length of the train will be 18 ( t + s ) on comparing these two equations , we get 20 ( t - f ) = 18 ( t + s ) = > 2 t = 20 f + 18 s = > t = 10 f + 9 s now it is given that after 10 min the train crossed the other person who is coming in opposite direction . so , if we consider this way of train as l then the next equation will be l = 600 t ( here 600 is used for 10 minutes ) finally as asked in the question the time required to meet the two man after the train has passed will be given by time = ( l - 600 f ) \/ ( f + s ) { here 600 f is used for the distance traveled by first man in 10 minutes } = > = ( 600 t - 600 f ) \/ ( f + s ) = > = [ 600 ( 10 f + 9 s ) - 600 f ] \/ ( f + s ) = > = 600 ( 10 f + 9 s - f ) \/ ( f + s ) = 600 * 9 ( f + s ) \/ ( f + s ) = > = 600 * 9 seconds = > = 600 * 9 \/ 60 min = > = 90 minutes hence ( b ) is the correct answer . answer : b","correct":"b","options":{"a":"95 minutes ","b":"90 minutes ","c":"88 minutes ","d":"95 minutes","e":"none of these"},"options_float":{"a":95.0,"b":90.0,"c":88.0,"d":95.0,"e":null},"annotated_formula":"divide(multiply(multiply(const_60, 10), divide(18, const_2)), const_60)","linear_formula":"divide(n2,const_2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,const_60)","chain":"60 * 10<\/gadget>\n600<\/output>\n18 \/ 2<\/gadget>\n9<\/output>\n600 * 9<\/gadget>\n5_400<\/output>\n5_400 \/ 60<\/gadget>\n90<\/output>\n90<\/result>","index":146} +{"problem":"there are 6 people in the elevator . their average weight is 150 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 150 ) = 1057 - 900 = 157 . answer a\"","correct":"a","options":{"a":"157 ","b":"168 ","c":"189 ","d":"190","e":"200"},"options_float":{"a":157.0,"b":168.0,"c":189.0,"d":190.0,"e":200.0},"annotated_formula":"subtract(multiply(151, 7), multiply(6, 150))","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|","chain":"151 * 7<\/gadget>\n1_057<\/output>\n6 * 150<\/gadget>\n900<\/output>\n1_057 - 900<\/gadget>\n157<\/output>\n157<\/result>","index":150} +{"problem":"average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 860 . what was original rent of friend whose rent is increased ?","rationale":"\"0.2 x = 4 ( 860 - 800 ) 0.2 x = 240 x = 1200 answer e\"","correct":"e","options":{"a":"800 ","b":"900 ","c":"1000 ","d":"1100","e":"1200"},"options_float":{"a":800.0,"b":900.0,"c":1000.0,"d":1100.0,"e":1200.0},"annotated_formula":"divide(multiply(subtract(860, 800), 4), divide(20, const_100))","linear_formula":"divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0)|","chain":"860 - 800<\/gadget>\n60<\/output>\n60 * 4<\/gadget>\n240<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n240 \/ (1\/5)<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":151} +{"problem":"if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 10 m of the same rod ?","rationale":"\"explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 10 : : 42.75 : x = > 11.25 x x = 10 x 42.75 = > x = ( 10 x 42.75 ) \/ 11.25 = > x = 38 answer : a\"","correct":"a","options":{"a":"38 kg ","b":"25.6 kg ","c":"28 kg ","d":"26.5 kg","e":"none of these"},"options_float":{"a":38.0,"b":25.6,"c":28.0,"d":26.5,"e":null},"annotated_formula":"divide(multiply(10, 42.75), 11.25)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"10 * 42.75<\/gadget>\n427.5<\/output>\n427.5 \/ 11.25<\/gadget>\n38<\/output>\n38<\/result>","index":152} +{"problem":"8 men can dig a pit in 20 days . if a man works half as much again a s a boy , then 4 men and 9 boys can dig a similar pit in :","rationale":"explanation : 1 work done = 8 × 20 1 man = 3 \/ 2 boys 1 boy = 2 \/ 3 men then , 9 boys = 9 × 2 \/ 3 men = 6 men then , 4 men + 9 boys = 10 men then , 8 × 20 = 10 × ? days ? days = 8 × 20 \/ 10 = 16 days . answer : option d","correct":"d","options":{"a":"10 days ","b":"12 days ","c":"15 days ","d":"16 days","e":"20 days"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":16.0,"e":20.0},"annotated_formula":"divide(multiply(multiply(8, divide(const_3, const_2)), 20), add(multiply(4, divide(const_3, const_2)), 9))","linear_formula":"divide(const_3,const_2)|multiply(n0,#0)|multiply(n2,#0)|add(n3,#2)|multiply(n1,#1)|divide(#4,#3)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n8 * (3\/2)<\/gadget>\n12<\/output>\n12 * 20<\/gadget>\n240<\/output>\n4 * (3\/2)<\/gadget>\n6<\/output>\n6 + 9<\/gadget>\n15<\/output>\n240 \/ 15<\/gadget>\n16<\/output>\n16<\/result>","index":153} +{"problem":"john makes $ 40 a week from his job . he earns a raise and now makes $ 70 a week . what is the % increase ?","rationale":"\"increase = ( 30 \/ 40 ) * 100 = ( 3 \/ 4 ) * 100 = 75 % . e\"","correct":"e","options":{"a":"16 % ","b":"16.66 % ","c":"17.9 % ","d":"18.12 %","e":"75 %"},"options_float":{"a":16.0,"b":16.66,"c":17.9,"d":18.12,"e":75.0},"annotated_formula":"multiply(divide(subtract(70, 40), 40), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"70 - 40<\/gadget>\n30<\/output>\n30 \/ 40<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n75<\/result>","index":154} +{"problem":"a train is 410 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 = 410 + 140 = 550 meter time = distance \/ speed = 550 ∗ 2 \/ 25 = 44 seconds option b\"","correct":"b","options":{"a":"20 seconds ","b":"44 seconds ","c":"40 seconds ","d":"50 seconds","e":"none of these"},"options_float":{"a":20.0,"b":44.0,"c":40.0,"d":50.0,"e":null},"annotated_formula":"divide(add(410, 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":"410 + 140<\/gadget>\n550<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n550 \/ (25\/2)<\/gadget>\n44<\/output>\n44<\/result>","index":155} +{"problem":"the average ( arithmetic mean ) of 4 positive integers is 50 . if the average of 2 of these integers is 40 , what is the greatest possible value that one of the other 2 integers can have ?","rationale":"\"a + b + c + d = 200 a + b = 80 c + d = 120 greatest possible = 119 ( just less than 1 ) answer = c\"","correct":"c","options":{"a":"55 ","b":"65 ","c":"119 ","d":"109","e":"115"},"options_float":{"a":55.0,"b":65.0,"c":119.0,"d":109.0,"e":115.0},"annotated_formula":"subtract(multiply(50, 4), multiply(40, 2))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|","chain":"50 * 4<\/gadget>\n200<\/output>\n40 * 2<\/gadget>\n80<\/output>\n200 - 80<\/gadget>\n120<\/output>\n120<\/result>","index":157} +{"problem":"an uneducated retailer marks all his goods at 50 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ?","rationale":"\"sol . let c . p . = rs . 100 . then , marked price = rs . 150 . s . p . = 75 % of rs . 150 = rs . 112.50 . ∴ gain % = 12.50 % . answer a\"","correct":"a","options":{"a":"12.50 % ","b":"13.50 % ","c":"14 % ","d":"14.50 %","e":"none"},"options_float":{"a":12.5,"b":13.5,"c":14.0,"d":14.5,"e":null},"annotated_formula":"multiply(subtract(subtract(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(50, const_100)), divide(25, const_100))), const_1), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(#2,#3)|subtract(#4,const_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>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(3\/2) * (1\/4)<\/gadget>\n3\/8 = around 0.375<\/output>\n(3\/2) - (3\/8)<\/gadget>\n9\/8 = around 1.125<\/output>\n(9\/8) - 1<\/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":159} +{"problem":"the mass of 1 cubic meter of a substance is 300 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters )","rationale":"\"300 kg - 1 cubic meter ; 300,000 g - 1 cubic meter ; 300,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 \/ 300,000 = 10 \/ 3 = 3.33 cubic centimeters . answer : a .\"","correct":"a","options":{"a":"3.33 ","b":"3.34 ","c":"3.53 ","d":"5.32","e":"3.92"},"options_float":{"a":3.33,"b":3.34,"c":3.53,"d":5.32,"e":3.92},"annotated_formula":"divide(multiply(1,000, 1,000), multiply(300, 1,000))","linear_formula":"multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)|","chain":"1_000 * 1_000<\/gadget>\n1_000_000<\/output>\n300 * 1_000<\/gadget>\n300_000<\/output>\n1_000_000 \/ 300_000<\/gadget>\n10\/3 = around 3.333333<\/output>\n10\/3 = around 3.333333<\/result>","index":160} +{"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 50 % 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 50 % increase over cost . here , we need to think about percentage increases as multipliers . using multipliers ( cost ) * 1.50 = $ 36 cost = 36 \/ 1.5 = 360 \/ 12 = $ 24 if he wants a 20 % increase over cost on the sale of one liter of his milk - water , the cost has to be $ 24 . well , a liter of milk costs $ 33 , so if he is going to use just $ 30 of milk in his mixture , that ' s 24 \/ 33 = 8 \/ 11 of a liter . if milk is 8 \/ 11 of the liter , then water is 3 \/ 11 of the liter , and the ratio of water to milk is 3 : 8 . answer choice ( c )\"","correct":"c","options":{"a":"1 : 20 ","b":"1 : 10 ","c":"3 : 8 ","d":"3 : 4","e":"3 : 2"},"options_float":{"a":0.05,"b":0.1,"c":0.375,"d":0.75,"e":1.5},"annotated_formula":"divide(subtract(33, divide(36, divide(add(const_100, 50), const_100))), divide(36, divide(add(const_100, 50), const_100)))","linear_formula":"add(n1,const_100)|divide(#0,const_100)|divide(n2,#1)|subtract(n0,#2)|divide(#3,#2)|","chain":"100 + 50<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n36 \/ (3\/2)<\/gadget>\n24<\/output>\n33 - 24<\/gadget>\n9<\/output>\n9 \/ 24<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":161} +{"problem":"a train 520 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 = 520 \/ 20 s = 26 m \/ sec speed = 26 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 94 kmph answer : b\"","correct":"b","options":{"a":"88 kmph ","b":"94 kmph ","c":"72 kmph ","d":"16 kmph","e":"18 kmph"},"options_float":{"a":88.0,"b":94.0,"c":72.0,"d":16.0,"e":18.0},"annotated_formula":"divide(divide(520, const_1000), divide(20, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"520 \/ 1_000<\/gadget>\n13\/25 = around 0.52<\/output>\n20 \/ 3_600<\/gadget>\n1\/180 = around 0.005556<\/output>\n(13\/25) \/ (1\/180)<\/gadget>\n468\/5 = around 93.6<\/output>\n468\/5 = around 93.6<\/result>","index":162} +{"problem":"in a throw of dice what is the probability of ge æ « ng number greater than 4","rationale":"\"explanation : number greater than 4 is 5 & 6 , so only 2 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 2 \/ 6 = 1 \/ 3 answer : b\"","correct":"b","options":{"a":"1 \/ 2 ","b":"1 \/ 3 ","c":"1 \/ 5 ","d":"1 \/ 6","e":"none of these"},"options_float":{"a":0.5,"b":0.3333333333,"c":0.2,"d":0.1666666667,"e":null},"annotated_formula":"divide(subtract(const_6, 4), const_6)","linear_formula":"subtract(const_6,n0)|divide(#0,const_6)|","chain":"6 - 4<\/gadget>\n2<\/output>\n2 \/ 6<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":163} +{"problem":"in a particular state , 60 % of the counties received some rain on monday , and 65 % of the counties received some rain on tuesday . no rain fell either day in 25 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ?","rationale":"\"60 + 65 + 25 = 150 % the number is 50 % above 100 % because 50 % of the counties were counted twice . the answer is c .\"","correct":"c","options":{"a":"12.5 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"67.5 %"},"options_float":{"a":12.5,"b":40.0,"c":50.0,"d":60.0,"e":67.5},"annotated_formula":"subtract(add(60, 65), subtract(const_100, 25))","linear_formula":"add(n0,n1)|subtract(const_100,n2)|subtract(#0,#1)|","chain":"60 + 65<\/gadget>\n125<\/output>\n100 - 25<\/gadget>\n75<\/output>\n125 - 75<\/gadget>\n50<\/output>\n50<\/result>","index":164} +{"problem":"two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction . what time will they take to be 8.5 km apart ?","rationale":"\"relative speed = 5.5 - 5 = 0.5 kmph ( because they walk in the same direction ) distance = 8.5 km time = distance \/ speed = 8.5 \/ 0.5 = 17 hr answer is a\"","correct":"a","options":{"a":"17 hr ","b":"14 hr ","c":"12 hr ","d":"19 hr","e":"23 hr"},"options_float":{"a":17.0,"b":14.0,"c":12.0,"d":19.0,"e":23.0},"annotated_formula":"divide(8.5, subtract(5.5, 5))","linear_formula":"subtract(n1,n0)|divide(n2,#0)|","chain":"5.5 - 5<\/gadget>\n0.5<\/output>\n8.5 \/ 0.5<\/gadget>\n17<\/output>\n17<\/result>","index":165} +{"problem":"two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team q . if team a ' s average speed was 5 mph greater than team q ' s , what was team q ' s average mph ?","rationale":"\"this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team q ' s speed was s . then team q ' s time is 300 \/ s . team a ' s speed was then s + 5 , and team a ' s time was then 300 \/ ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team q . that is , we need an answer choice so that 300 \/ ( s + 5 ) = ( 300 \/ s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team q is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct .\"","correct":"d","options":{"a":"12 ","b":"15 ","c":"18 ","d":"20","e":"25"},"options_float":{"a":12.0,"b":15.0,"c":18.0,"d":20.0,"e":25.0},"annotated_formula":"divide(divide(300, 5), 3)","linear_formula":"divide(n0,n2)|divide(#0,n1)|","chain":"300 \/ 5<\/gadget>\n60<\/output>\n60 \/ 3<\/gadget>\n20<\/output>\n20<\/result>","index":166} +{"problem":"the average salary of the employees in a office is rs . 120 \/ month . the avg salary of officers is rs . 460 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 * 460 + x * 110 = ( x + 15 ) 120 x = 510 e\"","correct":"e","options":{"a":"400 ","b":"420 ","c":"430 ","d":"450","e":"510"},"options_float":{"a":400.0,"b":420.0,"c":430.0,"d":450.0,"e":510.0},"annotated_formula":"divide(subtract(multiply(15, 460), 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 * 460<\/gadget>\n6_900<\/output>\n15 * 120<\/gadget>\n1_800<\/output>\n6_900 - 1_800<\/gadget>\n5_100<\/output>\n120 - 110<\/gadget>\n10<\/output>\n5_100 \/ 10<\/gadget>\n510<\/output>\n510<\/result>","index":167} +{"problem":"the h . c . f . of two numbers is 10 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 ( 10 x 13 ) and ( 10 x 14 ) . larger number = ( 10 x 14 ) = 140 . answer : option d\"","correct":"d","options":{"a":"100 ","b":"120 ","c":"180 ","d":"140","e":"60"},"options_float":{"a":100.0,"b":120.0,"c":180.0,"d":140.0,"e":60.0},"annotated_formula":"multiply(10, 14)","linear_formula":"multiply(n0,n2)|","chain":"10 * 14<\/gadget>\n140<\/output>\n140<\/result>","index":168} +{"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 . 672 find the share of a .","rationale":"\"explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 \/ 21 * 672 = 256 answer : a\"","correct":"a","options":{"a":"256 ","b":"388 ","c":"379 ","d":"277","e":"122"},"options_float":{"a":256.0,"b":388.0,"c":379.0,"d":277.0,"e":122.0},"annotated_formula":"multiply(divide(672, 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>\n672 \/ 84_000<\/gadget>\n1\/125 = around 0.008<\/output>\n(1\/125) * 32_000<\/gadget>\n256<\/output>\n256<\/result>","index":169} +{"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":"17 ","b":"98 ","c":"88 ","d":"85","e":"83"},"options_float":{"a":17.0,"b":98.0,"c":88.0,"d":85.0,"e":83.0},"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":170} +{"problem":"a began business with rs . 27000 and was joined afterwards by b with rs . 54000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ?","rationale":"27 * 12 : 54 * x = 2 : 1 x = 3 12 - 3 = 9 answer : a","correct":"a","options":{"a":"9 ","b":"6 ","c":"7 ","d":"8","e":"2"},"options_float":{"a":9.0,"b":6.0,"c":7.0,"d":8.0,"e":2.0},"annotated_formula":"subtract(multiply(const_4, const_3), divide(divide(multiply(27000, multiply(const_4, const_3)), 54000), 2))","linear_formula":"multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3)","chain":"4 * 3<\/gadget>\n12<\/output>\n27_000 * 12<\/gadget>\n324_000<\/output>\n324_000 \/ 54_000<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n12 - 3<\/gadget>\n9<\/output>\n9<\/result>","index":171} +{"problem":"a , band c can do a piece of work in 11 days , 20 days and 20 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ?","rationale":"\"( a + b ) ' s 1 day ' s work = 1 \/ 11 + 1 \/ 20 = 31 \/ 220 ( a + c ) ' s 1 day ' s work = 1 \/ 11 + 1 \/ 20 = 31 \/ 220 work done in 2 day ' s = 31 \/ 220 + 31 \/ 220 = 31 \/ 110 31 \/ 110 th work done in 2 days work done = 110 \/ 31 * 2 = 7 days ( approx ) answer : a\"","correct":"a","options":{"a":"7 days ","b":"8 days ","c":"9 days ","d":"10 days","e":"11 days"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"divide(20, divide(add(add(divide(20, 11), divide(20, 20)), add(divide(20, 11), divide(20, 20))), const_2))","linear_formula":"divide(n2,n0)|divide(n2,n1)|divide(n2,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(#5,const_2)|divide(n2,#6)|","chain":"20 \/ 11<\/gadget>\n20\/11 = around 1.818182<\/output>\n20 \/ 20<\/gadget>\n1<\/output>\n(20\/11) + 1<\/gadget>\n31\/11 = around 2.818182<\/output>\n(31\/11) + (31\/11)<\/gadget>\n62\/11 = around 5.636364<\/output>\n(62\/11) \/ 2<\/gadget>\n31\/11 = around 2.818182<\/output>\n20 \/ (31\/11)<\/gadget>\n220\/31 = around 7.096774<\/output>\n220\/31 = around 7.096774<\/result>","index":173} +{"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 t 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 ) t = 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":175} +{"problem":"two numbers a and b are such that the sum of 5 % of a and 2 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b .","rationale":"explanation : 5 % of a + 2 % of b = 2 \/ 3 ( 6 % of a + 8 % of b ) 5 a \/ 100 + 2 b \/ 100 = 2 \/ 3 ( 6 a \/ 100 + 8 b \/ 100 ) ⇒ 5 a + 2 b = 2 \/ 3 ( 6 a + 8 b ) ⇒ 15 a + 6 b = 12 a + 16 b ⇒ 3 a = 10 b ⇒ ab = 10 \/ 3 ⇒ a : b = 10 : 3 answer : option d","correct":"d","options":{"a":"2 : 1 ","b":"1 : 2 ","c":"4 : 3 ","d":"10 : 3","e":"3 : 2"},"options_float":{"a":2.0,"b":0.5,"c":1.3333333333,"d":3.3333333333,"e":1.5},"annotated_formula":"divide(subtract(divide(multiply(divide(8, const_100), const_2), const_3), divide(2, const_100)), subtract(divide(5, const_100), divide(multiply(divide(6, const_100), const_2), const_3)))","linear_formula":"divide(n3,const_100)|divide(n1,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(#0,const_2)|multiply(#3,const_2)|divide(#4,const_3)|divide(#5,const_3)|subtract(#6,#1)|subtract(#2,#7)|divide(#8,#9)","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) * 2<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) \/ 3<\/gadget>\n4\/75 = around 0.053333<\/output>\n2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n(4\/75) - (1\/50)<\/gadget>\n1\/30 = around 0.033333<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n(3\/50) * 2<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) \/ 3<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/20) - (1\/25)<\/gadget>\n1\/100 = around 0.01<\/output>\n(1\/30) \/ (1\/100)<\/gadget>\n10\/3 = around 3.333333<\/output>\n10\/3 = around 3.333333<\/result>","index":176} +{"problem":"20 men do a work in 20 days . how many men are needed to finish the work in 10 days ?","rationale":"\"men required to finish the work in 10 days = 20 * 20 \/ 10 = 40 answer is e\"","correct":"e","options":{"a":"50 ","b":"20 ","c":"30 ","d":"10","e":"40"},"options_float":{"a":50.0,"b":20.0,"c":30.0,"d":10.0,"e":40.0},"annotated_formula":"divide(multiply(20, 20), 10)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"20 * 20<\/gadget>\n400<\/output>\n400 \/ 10<\/gadget>\n40<\/output>\n40<\/result>","index":178} +{"problem":"if 5 a = 6 b and ab ≠ 0 , what is the ratio of a \/ 6 to b \/ 5 ?","rationale":"\"a nice fast approach is the first find a pair of numbers that satisfy the given equation : 5 a = 6 b here ' s one pair : a = 6 and b = 5 what is the ratio of a \/ 6 to b \/ 5 ? in other words , what is the value of ( a \/ 6 ) \/ ( b \/ 5 ) ? plug in values to get : ( a \/ 6 ) \/ ( b \/ 5 ) = ( 6 \/ 6 ) \/ ( 5 \/ 5 ) = 1 \/ 1 = 1 c\"","correct":"c","options":{"a":"36 \/ 25 ","b":"10 \/ 12 ","c":"1 ","d":"1 \/ 5","e":"25 \/ 36"},"options_float":{"a":1.44,"b":0.8333333333,"c":1.0,"d":0.2,"e":0.6944444444},"annotated_formula":"divide(multiply(5, 6), multiply(6, 5))","linear_formula":"multiply(n0,n1)|divide(#0,#0)|","chain":"5 * 6<\/gadget>\n30<\/output>\n6 * 5<\/gadget>\n30<\/output>\n30 \/ 30<\/gadget>\n1<\/output>\n1<\/result>","index":179} +{"problem":"x does a work in 20 days . y does the same work in 30 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 \/ 30 ( x + y ) ' s 1 day ' s work = ( 1 \/ 20 + 1 \/ 30 ) = 1 \/ 12 both together will finish the work in 12 days . correct option is b\"","correct":"b","options":{"a":"10 ","b":"12 ","c":"20 ","d":"30","e":"15"},"options_float":{"a":10.0,"b":12.0,"c":20.0,"d":30.0,"e":15.0},"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":182} +{"problem":"jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 3.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill )","rationale":"average speed of jack = distance \/ time = 42 \/ ( 7 \/ 2 ) = 84 \/ 7 average speed of jill = 42 \/ ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 \/ 7 ) \/ 10 = 84 \/ 70 = 6 \/ 5 answer d","correct":"d","options":{"a":"14 \/ 15 ","b":"15 \/ 14 ","c":"4 \/ 5 ","d":"6 \/ 5","e":"can not be determined"},"options_float":{"a":0.9333333333,"b":1.0714285714,"c":0.8,"d":1.2,"e":null},"annotated_formula":"divide(divide(42, 3.5), divide(42, 4.2))","linear_formula":"divide(n0,n1)|divide(n0,n2)|divide(#0,#1)","chain":"42 \/ 3.5<\/gadget>\n12<\/output>\n42 \/ 4.2<\/gadget>\n10<\/output>\n12 \/ 10<\/gadget>\n6\/5 = around 1.2<\/output>\n6\/5 = around 1.2<\/result>","index":183} +{"problem":"a container contains 1000 liters of milk , from this container 10 liters of milk was taken out and replaced by water . this process was repeated further 2 times . how much milk is now contained by the container ?","rationale":"amount of milk left after 3 operations = 1000 ( 1 - 10 \/ 1000 ) ^ 3 = 1000 * 99 \/ 100 * 99 \/ 100 * 99 \/ 100 = 970.3 liters answer is a","correct":"a","options":{"a":"970.3 liters ","b":"1000.45 liters ","c":"879.65 liters ","d":"1020.56 liters","e":"910.95 liters"},"options_float":{"a":970.3,"b":1000.45,"c":879.65,"d":1020.56,"e":910.95},"annotated_formula":"subtract(subtract(subtract(1000, 10), 10), 10)","linear_formula":"subtract(n0,n1)|subtract(#0,n1)|subtract(#1,n1)","chain":"1_000 - 10<\/gadget>\n990<\/output>\n990 - 10<\/gadget>\n980<\/output>\n980 - 10<\/gadget>\n970<\/output>\n970<\/result>","index":184} +{"problem":"a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 3 feet . how many feet does the squirrel travels if the post is 18 feet tall and 3 feet in circumference ?","rationale":"\"total circuit = 18 \/ 3 = 6 total feet squirrel travels = 6 * 3 = 18 feet answer : e\"","correct":"e","options":{"a":"10 feet ","b":"12 feet ","c":"13 feet ","d":"15 feet","e":"18 feet"},"options_float":{"a":10.0,"b":12.0,"c":13.0,"d":15.0,"e":18.0},"annotated_formula":"multiply(divide(18, 3), 3)","linear_formula":"divide(n1,n0)|multiply(n2,#0)|","chain":"18 \/ 3<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":187} +{"problem":"in an electric circuit , two resistors with resistances 3 ohm and 5 ohm are connected in parallel . in this case , if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of two resistors . what is the value ?","rationale":"the wording is a bit confusing , though basically we are told that 1 \/ r = 1 \/ 3 + 1 \/ 5 , from which it follows that r = 15 \/ 8 ohms . answer : b .","correct":"b","options":{"a":"15 ohms ","b":"15 \/ 8 ohms ","c":"1 \/ 8 ohms ","d":"8 \/ 15 ohms","e":"8 ohms"},"options_float":{"a":15.0,"b":1.875,"c":0.125,"d":0.5333333333,"e":8.0},"annotated_formula":"divide(multiply(3, 5), add(3, 5))","linear_formula":"add(n0,n1)|multiply(n0,n1)|divide(#1,#0)","chain":"3 * 5<\/gadget>\n15<\/output>\n3 + 5<\/gadget>\n8<\/output>\n15 \/ 8<\/gadget>\n15\/8 = around 1.875<\/output>\n15\/8 = around 1.875<\/result>","index":188} +{"problem":"a and b enterd into a partnership investing rs . 16000 and rs . 12000 respectively . after 3 months , a withdrew rs . 5000 while b invested rs . 5000 more . after 3 more months . c joins the business with a capital of rs . 21000 . the share of b exceeds that of c , out of a total profit of rs . 26400 after one year by","rationale":"solution a : b : c = ( 16000 x 3 + 11000 x 9 ) : ( 12000 x 3 + 17000 x 9 ) : ( 21000 x 6 ) = 147 : 180 : 126 = 7 : 9 : 6 . ∴ difference of b and c ’ s shares = rs . ( 26400 x 9 \/ 22 - 26400 x 6 \/ 22 ) = rs . 3600 . answer c","correct":"c","options":{"a":"rs . 2400 ","b":"rs . 3000 ","c":"rs . 3600 ","d":"rs . 4800","e":"none of these"},"options_float":{"a":2400.0,"b":3000.0,"c":3600.0,"d":4800.0,"e":null},"annotated_formula":"subtract(multiply(26400, divide(add(multiply(12000, 3), multiply(add(12000, 5000), subtract(const_12, 3))), add(add(add(multiply(16000, 3), multiply(subtract(16000, 5000), subtract(const_12, 3))), add(multiply(12000, 3), multiply(add(12000, 5000), subtract(const_12, 3)))), multiply(21000, subtract(subtract(const_12, 3), 3))))), multiply(26400, divide(multiply(21000, subtract(subtract(const_12, 3), 3)), add(add(add(multiply(16000, 3), multiply(subtract(16000, 5000), subtract(const_12, 3))), add(multiply(12000, 3), multiply(add(12000, 5000), subtract(const_12, 3)))), multiply(21000, subtract(subtract(const_12, 3), 3))))))","linear_formula":"add(n1,n3)|multiply(n1,n2)|multiply(n0,n2)|subtract(const_12,n2)|subtract(n0,n3)|multiply(#0,#3)|multiply(#4,#3)|subtract(#3,n2)|add(#1,#5)|add(#2,#6)|multiply(n6,#7)|add(#9,#8)|add(#11,#10)|divide(#8,#12)|divide(#10,#12)|multiply(n7,#13)|multiply(n7,#14)|subtract(#15,#16)","chain":"12_000 * 3<\/gadget>\n36_000<\/output>\n12_000 + 5_000<\/gadget>\n17_000<\/output>\n12 - 3<\/gadget>\n9<\/output>\n17_000 * 9<\/gadget>\n153_000<\/output>\n36_000 + 153_000<\/gadget>\n189_000<\/output>\n16_000 * 3<\/gadget>\n48_000<\/output>\n16_000 - 5_000<\/gadget>\n11_000<\/output>\n11_000 * 9<\/gadget>\n99_000<\/output>\n48_000 + 99_000<\/gadget>\n147_000<\/output>\n147_000 + 189_000<\/gadget>\n336_000<\/output>\n9 - 3<\/gadget>\n6<\/output>\n21_000 * 6<\/gadget>\n126_000<\/output>\n336_000 + 126_000<\/gadget>\n462_000<\/output>\n189_000 \/ 462_000<\/gadget>\n9\/22 = around 0.409091<\/output>\n26_400 * (9\/22)<\/gadget>\n10_800<\/output>\n126_000 \/ 462_000<\/gadget>\n3\/11 = around 0.272727<\/output>\n26_400 * (3\/11)<\/gadget>\n7_200<\/output>\n10_800 - 7_200<\/gadget>\n3_600<\/output>\n3_600<\/result>","index":189} +{"problem":"in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 155 m . then , a wins by :","rationale":"\"to reach the winning post a will have to cover a distance of ( 500 - 155 ) m , i . e . , 345 m . while a covers 3 m , b covers 4 m . while a covers 345 m , b covers 4 x 345 \/ 3 m = 460 m . thus , when a reaches the winning post , b covers 460 m and therefore remains 40 m behind . a wins by 40 m . answer : c\"","correct":"c","options":{"a":"60 m ","b":"20 m ","c":"40 m ","d":"20 m","e":"23 m"},"options_float":{"a":60.0,"b":20.0,"c":40.0,"d":20.0,"e":23.0},"annotated_formula":"subtract(500, divide(multiply(subtract(500, 155), 4), 3))","linear_formula":"subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)|","chain":"500 - 155<\/gadget>\n345<\/output>\n345 * 4<\/gadget>\n1_380<\/output>\n1_380 \/ 3<\/gadget>\n460<\/output>\n500 - 460<\/gadget>\n40<\/output>\n40<\/result>","index":192} +{"problem":"dan ' s age after 16 years will be 6 times his age 4 years ago . what is the present age of dan ?","rationale":"\"let dan ' s present age be x . x + 16 = 6 ( x - 4 ) 5 x = 40 x = 8 the answer is a .\"","correct":"a","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":"divide(add(16, multiply(4, 6)), subtract(6, const_1))","linear_formula":"multiply(n1,n2)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)|","chain":"4 * 6<\/gadget>\n24<\/output>\n16 + 24<\/gadget>\n40<\/output>\n6 - 1<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8<\/result>","index":193} +{"problem":"by selling 20 pencils for a rupee a man loses 60 % . how many for a rupee should he sell in order to gain 60 % ?","rationale":"\"40 % - - - 20 160 % - - - ? 40 \/ 160 * 20 = 5 answer : e\"","correct":"e","options":{"a":"8 ","b":"9 ","c":"7 ","d":"6","e":"5"},"options_float":{"a":8.0,"b":9.0,"c":7.0,"d":6.0,"e":5.0},"annotated_formula":"multiply(divide(const_1, multiply(add(const_100, 60), divide(const_1, subtract(const_100, 60)))), 20)","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 + 60<\/gadget>\n160<\/output>\n100 - 60<\/gadget>\n40<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n160 * (1\/40)<\/gadget>\n4<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 20<\/gadget>\n5<\/output>\n5<\/result>","index":195} +{"problem":"after decreasing 24 % in the price of an article costs rs . 1216 . find the actual cost of an article ?","rationale":"\"cp * ( 76 \/ 100 ) = 1216 cp = 16 * 100 = > cp = 1600 answer : d\"","correct":"d","options":{"a":"1667 ","b":"6789 ","c":"1200 ","d":"1600","e":"1421"},"options_float":{"a":1667.0,"b":6789.0,"c":1200.0,"d":1600.0,"e":1421.0},"annotated_formula":"divide(1216, 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>\n1_216 \/ (19\/25)<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":196} +{"problem":"if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 3 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ?","rationale":"18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 3 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 3 ) ( 3 ^ b ) just compare powers of 2 from both sides ( no need to calculate powers of 3 , 9 as value of b is not asked ) answer = 3 answer : e","correct":"e","options":{"a":"22 ","b":"11 ","c":"9 ","d":"6","e":"3"},"options_float":{"a":22.0,"b":11.0,"c":9.0,"d":6.0,"e":3.0},"annotated_formula":"multiply(3, 1)","linear_formula":"multiply(n2,n3)","chain":"3 * 1<\/gadget>\n3<\/output>\n3<\/result>","index":197} +{"problem":"a student traveled 10 percent of the distance of the trip alone , continued another 30 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ?","rationale":"let x be the total length of the trip . 0.1 x + 30 miles + 0.5 x = x 30 miles = 0.4 x x = 75 miles the answer is a .","correct":"a","options":{"a":"75 ","b":"100 ","c":"125 ","d":"150","e":"175"},"options_float":{"a":75.0,"b":100.0,"c":125.0,"d":150.0,"e":175.0},"annotated_formula":"divide(30, subtract(subtract(const_1, inverse(10)), divide(const_1, const_2)))","linear_formula":"divide(const_1,const_2)|inverse(n0)|subtract(const_1,#1)|subtract(#2,#0)|divide(n1,#3)","chain":"1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(9\/10) - (1\/2)<\/gadget>\n2\/5 = around 0.4<\/output>\n30 \/ (2\/5)<\/gadget>\n75<\/output>\n75<\/result>","index":198} +{"problem":"a bag marked at $ 250 is sold for $ 120 . the rate of discount is ?","rationale":"\"rate of discount = 130 \/ 250 * 100 = 52 % answer is d\"","correct":"d","options":{"a":"10 % ","b":"25 % ","c":"20 % ","d":"52 %","e":"45 %"},"options_float":{"a":10.0,"b":25.0,"c":20.0,"d":52.0,"e":45.0},"annotated_formula":"multiply(divide(subtract(250, 120), 250), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"250 - 120<\/gadget>\n130<\/output>\n130 \/ 250<\/gadget>\n13\/25 = around 0.52<\/output>\n(13\/25) * 100<\/gadget>\n52<\/output>\n52<\/result>","index":199} +{"problem":"a can do a piece of work 30 days . b can do work in 45 days . in how many days they will complete the work together ?","rationale":"\"lcm = 90 , ratio = 30 : 45 = 2 : 3 no of days = 90 \/ ( 2 + 3 ) = 90 \/ 5 = 18 days answer : e\"","correct":"e","options":{"a":"15 days ","b":"16 days ","c":"19 days ","d":"17 days","e":"18 days"},"options_float":{"a":15.0,"b":16.0,"c":19.0,"d":17.0,"e":18.0},"annotated_formula":"divide(const_1, add(divide(const_1, 30), divide(const_1, 45)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ 45<\/gadget>\n1\/45 = around 0.022222<\/output>\n(1\/30) + (1\/45)<\/gadget>\n1\/18 = around 0.055556<\/output>\n1 \/ (1\/18)<\/gadget>\n18<\/output>\n18<\/result>","index":200} +{"problem":"average between two sets of numbers is closer to the set with morenumbers ?","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":"85 ","d":"90","e":"95"},"options_float":{"a":70.0,"b":80.0,"c":85.0,"d":90.0,"e":95.0},"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":202} +{"problem":"what is the probability that the sum of two dice will yield a 6 , and then when both are thrown again , their sum will again yield a 6 ? assume that each die has 5 sides with faces numbered 1 to 5 .","rationale":"solution - rolling dices is an independent event . the combinations to get 6 are ( 1,5 ) , ( 5,1 ) , ( 2,4 ) , ( 4,2 ) , ( 3,3 ) and total combinations of both dices is 25 . the probability of getting 6 in first attempt is 5 \/ 25 = 1 \/ 5 . probability of getting 6 again in second attempt = ( 1 \/ 5 ) * ( 1 \/ 5 ) = 1 \/ 25 . ans b","correct":"b","options":{"a":"1 \/ 144 ","b":"1 \/ 25 ","c":"1 \/ 12 ","d":"1 \/ 6","e":"1 \/ 3"},"options_float":{"a":0.0069444444,"b":0.04,"c":0.0833333333,"d":0.1666666667,"e":0.3333333333},"annotated_formula":"multiply(divide(5, power(5, const_2)), divide(5, power(5, const_2)))","linear_formula":"power(n2,const_2)|divide(n2,#0)|multiply(#1,#1)","chain":"5 ** 2<\/gadget>\n25<\/output>\n5 \/ 25<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * (1\/5)<\/gadget>\n1\/25 = around 0.04<\/output>\n1\/25 = around 0.04<\/result>","index":204} +{"problem":"the average of runs of a cricket player of 10 innings was 34 . 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 = 34 so , total = average x no . of innings = 34 x 10 = 340 . now increase in avg = 4 runs . so , new avg = 34 + 4 = 38 runs total runs = new avg x new no . of innings = 38 x 11 = 418 runs made in the 11 th inning = 418 - 340 = 78 answer : a\"","correct":"a","options":{"a":"78 ","b":"79 ","c":"85 ","d":"87","e":"89"},"options_float":{"a":78.0,"b":79.0,"c":85.0,"d":87.0,"e":89.0},"annotated_formula":"subtract(multiply(add(10, const_1), add(4, 34)), multiply(10, 34))","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 + 34<\/gadget>\n38<\/output>\n11 * 38<\/gadget>\n418<\/output>\n10 * 34<\/gadget>\n340<\/output>\n418 - 340<\/gadget>\n78<\/output>\n78<\/result>","index":205} +{"problem":"a truck covers a distance of 288 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 90 km more than that travelled by the truck ?","rationale":"\"explanation : speed of the truck = distance \/ time = 288 \/ 8 = 36 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 36 + 18 ) = 54 kmph distance travelled by car = 288 + 90 = 378 km time taken by car = distance \/ speed = 378 \/ 54 = 7 hours . answer – c\"","correct":"c","options":{"a":"6 hours ","b":"5 hours ","c":"7 hours ","d":"8 hours","e":"none"},"options_float":{"a":6.0,"b":5.0,"c":7.0,"d":8.0,"e":null},"annotated_formula":"divide(add(288, 90), add(divide(288, 8), 18))","linear_formula":"add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2)|","chain":"288 + 90<\/gadget>\n378<\/output>\n288 \/ 8<\/gadget>\n36<\/output>\n36 + 18<\/gadget>\n54<\/output>\n378 \/ 54<\/gadget>\n7<\/output>\n7<\/result>","index":206} +{"problem":"a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? a . 15 b . 18","rationale":"a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 6 \/ 25 * 75 = 18 answer : b","correct":"b","options":{"a":"22 ","b":"18 ","c":"99 ","d":"77","e":"24"},"options_float":{"a":22.0,"b":18.0,"c":99.0,"d":77.0,"e":24.0},"annotated_formula":"multiply(divide(75, add(add(multiply(divide(2, 3), divide(2, 5)), divide(2, 5)), const_1)), divide(2, 5))","linear_formula":"divide(n0,n3)|divide(n0,n1)|multiply(#1,#0)|add(#0,#2)|add(#3,const_1)|divide(n4,#4)|multiply(#5,#0)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/3) * (2\/5)<\/gadget>\n4\/15 = around 0.266667<\/output>\n(4\/15) + (2\/5)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) + 1<\/gadget>\n5\/3 = around 1.666667<\/output>\n75 \/ (5\/3)<\/gadget>\n45<\/output>\n45 * (2\/5)<\/gadget>\n18<\/output>\n18<\/result>","index":207} +{"problem":"if 65 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":"\"65 % answered the first question correctly and 20 % answered neither correctly . then 15 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 50 % - 15 % = 35 % . the answer is c .\"","correct":"c","options":{"a":"25 % ","b":"30 % ","c":"35 % ","d":"40 %","e":"45 %"},"options_float":{"a":25.0,"b":30.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"subtract(add(add(65, 50), 20), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,const_100)|","chain":"65 + 50<\/gadget>\n115<\/output>\n115 + 20<\/gadget>\n135<\/output>\n135 - 100<\/gadget>\n35<\/output>\n35<\/result>","index":208} +{"problem":"john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 56000 and ingrid makes $ 73000 , what is their combined tax rate ?","rationale":"\"( 1 ) when 30 and 40 has equal weight or weight = 1 \/ 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 73000 \/ 129000 = 73 \/ 129 is a little above 1 \/ 2 . thus , our answer is just a little above 35 . answer : d\"","correct":"d","options":{"a":"32 % ","b":"34.4 % ","c":"35 % ","d":"35.6 %","e":"36.4 %"},"options_float":{"a":32.0,"b":34.4,"c":35.0,"d":35.6,"e":36.4},"annotated_formula":"multiply(divide(add(multiply(divide(30, const_100), 56000), multiply(divide(40, const_100), 73000)), add(73000, 56000)), const_100)","linear_formula":"add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 56_000<\/gadget>\n16_800<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 73_000<\/gadget>\n29_200<\/output>\n16_800 + 29_200<\/gadget>\n46_000<\/output>\n73_000 + 56_000<\/gadget>\n129_000<\/output>\n46_000 \/ 129_000<\/gadget>\n46\/129 = around 0.356589<\/output>\n(46\/129) * 100<\/gadget>\n4_600\/129 = around 35.658915<\/output>\n4_600\/129 = around 35.658915<\/result>","index":209} +{"problem":"the average ( arithmetic mean ) of 20 , 40 , and 60 is 6 more than the average of 10 , 70 , and what number ?","rationale":"\"a 1 = 120 \/ 3 = 40 a 2 = a 1 - 6 = 34 sum of second list = 34 * 3 = 102 therefore the number = 102 - 80 = 22 answer : d\"","correct":"d","options":{"a":"15 ","b":"25 ","c":"35 ","d":"22","e":"55"},"options_float":{"a":15.0,"b":25.0,"c":35.0,"d":22.0,"e":55.0},"annotated_formula":"subtract(add(add(20, 40), 60), add(add(multiply(6, const_3), 10), 70))","linear_formula":"add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)|","chain":"20 + 40<\/gadget>\n60<\/output>\n60 + 60<\/gadget>\n120<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18 + 10<\/gadget>\n28<\/output>\n28 + 70<\/gadget>\n98<\/output>\n120 - 98<\/gadget>\n22<\/output>\n22<\/result>","index":210} +{"problem":"if 150 ! \/ 10 ^ n is an integer , what is the largest possible value of n ?","rationale":"the question actually asks the highest power of 10 which divides 150 ! ( for a number to be an integer - without any remainder all the trailing zeroe ' s must be divided by the denominator ) 10 = 2 x 5 150 factorial will have 37 as - 150 \/ 5 = 30 30 \/ 5 = 6 6 \/ 5 = 1 so answer will be ( e ) 37","correct":"e","options":{"a":"38 ","b":"47 ","c":"32 ","d":"36","e":"37"},"options_float":{"a":38.0,"b":47.0,"c":32.0,"d":36.0,"e":37.0},"annotated_formula":"add(divide(divide(150, add(const_4, const_1)), add(const_4, const_1)), divide(150, add(const_4, const_1)))","linear_formula":"add(const_1,const_4)|divide(n0,#0)|divide(#1,#0)|add(#2,#1)","chain":"4 + 1<\/gadget>\n5<\/output>\n150 \/ 5<\/gadget>\n30<\/output>\n30 \/ 5<\/gadget>\n6<\/output>\n6 + 30<\/gadget>\n36<\/output>\n36<\/result>","index":211} +{"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 131 cm . what was the actual average height of the students in the class ?","rationale":"\"the total height was 20 cm too much . the average height should be reduced by 20 cm \/ 20 = 1 cm the answer is a .\"","correct":"a","options":{"a":"174 cm ","b":"173 cm ","c":"172 cm ","d":"171 cm","e":"170 cm"},"options_float":{"a":174.0,"b":173.0,"c":172.0,"d":171.0,"e":170.0},"annotated_formula":"divide(subtract(multiply(20, 175), subtract(151, 131)), 20)","linear_formula":"multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)|","chain":"20 * 175<\/gadget>\n3_500<\/output>\n151 - 131<\/gadget>\n20<\/output>\n3_500 - 20<\/gadget>\n3_480<\/output>\n3_480 \/ 20<\/gadget>\n174<\/output>\n174<\/result>","index":212} +{"problem":"a certain number of workers can do a work in 85 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ?","rationale":"\"number of workers = 10 * ( 85 - 10 ) \/ 10 = 75 answer is a\"","correct":"a","options":{"a":"75 ","b":"30 ","c":"28 ","d":"24","e":"32"},"options_float":{"a":75.0,"b":30.0,"c":28.0,"d":24.0,"e":32.0},"annotated_formula":"divide(multiply(subtract(85, 10), 10), subtract(85, subtract(85, 10)))","linear_formula":"subtract(n0,n1)|multiply(n1,#0)|subtract(n0,#0)|divide(#1,#2)|","chain":"85 - 10<\/gadget>\n75<\/output>\n75 * 10<\/gadget>\n750<\/output>\n85 - 75<\/gadget>\n10<\/output>\n750 \/ 10<\/gadget>\n75<\/output>\n75<\/result>","index":213} +{"problem":"car z travels 48 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 45 miles \/ hour , the car will run 48 miles \/ gallon and at 60 miles \/ hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 38.4 miles \/ gallon at a constant speed of 60 miles \/ gallon . with 10 gallons of fuel , the car can therefore travel 38.4 miles \/ gallon * 10 gallons = 384 miles . answer b .\"","correct":"b","options":{"a":"320 ","b":"384 ","c":"400 ","d":"408.3","e":"440"},"options_float":{"a":320.0,"b":384.0,"c":400.0,"d":408.3,"e":440.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), 48), 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) * 48<\/gadget>\n192\/5 = around 38.4<\/output>\n(192\/5) * 10<\/gadget>\n384<\/output>\n384<\/result>","index":215} +{"problem":"mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 24000 , how much amount per month would he have spent on clothes and transport ?","rationale":"\"∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 24000 ⁄ 12 = 2000 answer a\"","correct":"a","options":{"a":"2000 ","b":"8076 ","c":"9691.2 ","d":"4845.6","e":"none of these"},"options_float":{"a":2000.0,"b":8076.0,"c":9691.2,"d":4845.6,"e":null},"annotated_formula":"multiply(divide(divide(24000, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100))","linear_formula":"multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)|","chain":"100 - 60<\/gadget>\n40<\/output>\n40 * 50<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n24_000 \/ (1\/5)<\/gadget>\n120_000<\/output>\n3 * 4<\/gadget>\n12<\/output>\n120_000 \/ 12<\/gadget>\n10_000<\/output>\n10_000 * (1\/5)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":216} +{"problem":"in the class of 50 students , 30 speak tamil and 40 speak telugu . what is the lowest possible number of students who speak both the languages ?","rationale":"let the student who speaks tamil - x let the student who speaks telugu - y as ( xuy ) - ( xny ) = total 30 + 40 - ( xny ) = 50 = 20 c )","correct":"c","options":{"a":"8 ","b":"10 ","c":"20 ","d":"30","e":"32"},"options_float":{"a":8.0,"b":10.0,"c":20.0,"d":30.0,"e":32.0},"annotated_formula":"subtract(add(40, 30), 50)","linear_formula":"add(n1,n2)|subtract(#0,n0)","chain":"40 + 30<\/gadget>\n70<\/output>\n70 - 50<\/gadget>\n20<\/output>\n20<\/result>","index":217} +{"problem":"two trains start simultaneously from opposite ends of a 175 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 175 - km trip in 4 hours . train y , travelling at a constant rate , completes the 175 - km trip in 3 hours . how many kilometers had train x traveled when it met train y ?","rationale":"if the two trains cover a total distance d , then train x travels ( 3 \/ 7 ) * d while train y travels ( 4 \/ 7 ) * d . if the trains travel 175 km to the meeting point , then train x travels ( 3 \/ 7 ) * 175 = 75 km . the answer is d .","correct":"d","options":{"a":"66 ","b":"69 ","c":"72 ","d":"75","e":"78"},"options_float":{"a":66.0,"b":69.0,"c":72.0,"d":75.0,"e":78.0},"annotated_formula":"multiply(divide(175, 4), divide(175, add(divide(175, 4), divide(175, 3))))","linear_formula":"divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(n0,#2)|multiply(#0,#3)|","chain":"175 \/ 4<\/gadget>\n175\/4 = around 43.75<\/output>\n175 \/ 3<\/gadget>\n175\/3 = around 58.333333<\/output>\n(175\/4) + (175\/3)<\/gadget>\n1_225\/12 = around 102.083333<\/output>\n175 \/ (1_225\/12)<\/gadget>\n12\/7 = around 1.714286<\/output>\n(175\/4) * (12\/7)<\/gadget>\n75<\/output>\n75<\/result>","index":218} +{"problem":"what is the rate percent when the simple interest on rs . 2000 amount to rs . 320 in 4 years ?","rationale":"\"interest for 1 year = 320 \/ 4 = 80 interest on rs 2000 p \/ a = 80 interest rate = 80 \/ 2000 * 100 = 4 % answer : c\"","correct":"c","options":{"a":"4.5 % ","b":"4.25 % ","c":"4 % ","d":"4.3 %","e":"4.1 %"},"options_float":{"a":4.5,"b":4.25,"c":4.0,"d":4.3,"e":4.1},"annotated_formula":"divide(multiply(const_100, 320), multiply(2000, 4))","linear_formula":"multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|","chain":"100 * 320<\/gadget>\n32_000<\/output>\n2_000 * 4<\/gadget>\n8_000<\/output>\n32_000 \/ 8_000<\/gadget>\n4<\/output>\n4<\/result>","index":221} +{"problem":"a ratio between two numbers is 4 : 5 and their l . c . m . is 200 . the first number is","rationale":"\"sol . let the required numbers be 4 x and 5 x . then , their l . c . m . is 20 x . ∴ 20 x = 200 ⇔ x = 10 . hence , the first number is 40 . answer b\"","correct":"b","options":{"a":"60 ","b":"40 ","c":"20 ","d":"15","e":"none"},"options_float":{"a":60.0,"b":40.0,"c":20.0,"d":15.0,"e":null},"annotated_formula":"multiply(divide(200, multiply(4, 5)), 4)","linear_formula":"multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|","chain":"4 * 5<\/gadget>\n20<\/output>\n200 \/ 20<\/gadget>\n10<\/output>\n10 * 4<\/gadget>\n40<\/output>\n40<\/result>","index":222} +{"problem":"10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 170000 , find the cost of a camel ?","rationale":"\"cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 170000 p = rs . [ ( 24 * 4 * 4 * 170000 ) \/ ( 10 * 16 * 6 * 10 ) ] p = rs . ( 65280000 \/ 9600 ) = > p = rs . 6800 answer : d\"","correct":"d","options":{"a":"rs . 9800 ","b":"rs . 3800 ","c":"rs . 9800 ","d":"rs . 6800","e":"rs . 6880"},"options_float":{"a":9800.0,"b":3800.0,"c":9800.0,"d":6800.0,"e":6880.0},"annotated_formula":"divide(multiply(multiply(multiply(24, 4), 4), 170000), multiply(multiply(multiply(10, 16), 6), 10))","linear_formula":"multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)|","chain":"24 * 4<\/gadget>\n96<\/output>\n96 * 4<\/gadget>\n384<\/output>\n384 * 170_000<\/gadget>\n65_280_000<\/output>\n10 * 16<\/gadget>\n160<\/output>\n160 * 6<\/gadget>\n960<\/output>\n960 * 10<\/gadget>\n9_600<\/output>\n65_280_000 \/ 9_600<\/gadget>\n6_800<\/output>\n6_800<\/result>","index":223} +{"problem":"when the positive integer k is divided by the positive integer n , the remainder is 11 . if k \/ n = 81.1 , what is the value of n ?","rationale":"\"here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k \/ n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . eliminate a and e . with the remaining answers , we can test the answers and find the one that fits the rest of the info ( k \/ n = 81.2 and k \/ n has a remainder of 11 ) answer b : if n = 20 , then k = 1624 ; 1624 \/ 20 has a remainder of 4 not a match answer c : if n = 55 , then k = 4466 ; 4466 \/ 55 has a remainder of 11 match . final answer : e\"","correct":"e","options":{"a":"9 ","b":"20 ","c":"55 ","d":"70","e":"81"},"options_float":{"a":9.0,"b":20.0,"c":55.0,"d":70.0,"e":81.0},"annotated_formula":"divide(add(multiply(divide(subtract(multiply(81.1, const_10), const_2), const_10), divide(11, divide(const_2, const_10))), 11), divide(11, divide(const_2, const_10)))","linear_formula":"divide(const_2,const_10)|multiply(n1,const_10)|divide(n0,#0)|subtract(#1,const_2)|divide(#3,const_10)|multiply(#4,#2)|add(n0,#5)|divide(#6,#2)|","chain":"81.1 * 10<\/gadget>\n811<\/output>\n811 - 2<\/gadget>\n809<\/output>\n809 \/ 10<\/gadget>\n809\/10 = around 80.9<\/output>\n2 \/ 10<\/gadget>\n1\/5 = around 0.2<\/output>\n11 \/ (1\/5)<\/gadget>\n55<\/output>\n(809\/10) * 55<\/gadget>\n8_899\/2 = around 4_449.5<\/output>\n(8_899\/2) + 11<\/gadget>\n8_921\/2 = around 4_460.5<\/output>\n(8_921\/2) \/ 55<\/gadget>\n811\/10 = around 81.1<\/output>\n811\/10 = around 81.1<\/result>","index":225} +{"problem":"if a person walks at 10 km \/ hr instead of 8 km \/ hr , he would have walked 14 km more . the actual distance traveled by him is ?","rationale":"\"let the actual distance traveled be x km . then , x \/ 8 = ( x + 14 ) \/ 10 2 x - 112 = > x = 56 km . answer : c\"","correct":"c","options":{"a":"50 ","b":"40 ","c":"56 ","d":"16","e":"20"},"options_float":{"a":50.0,"b":40.0,"c":56.0,"d":16.0,"e":20.0},"annotated_formula":"multiply(8, divide(14, subtract(10, 8)))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|","chain":"10 - 8<\/gadget>\n2<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n8 * 7<\/gadget>\n56<\/output>\n56<\/result>","index":226} +{"problem":"if twice of a number divided by 3 d gives 20 as the remainder , and 5 times of the same number gives 32 as the remainder . what will be the value of d ?","rationale":"remainder in second case is , 32 . so , 3 d > = 33 . so , minimum value of d should 11 . if number = 28 . double of number = 56 and take d = 12 so 56 \/ 36 remainder = 20 . now 5 times of number = 140 . so 140 \/ 36 remainder = 32 . that is , 3 d = 36 satisfy the conditions . so d = 12 . answer : b","correct":"b","options":{"a":"11 ","b":"12 ","c":"15 ","d":"14","e":"18"},"options_float":{"a":11.0,"b":12.0,"c":15.0,"d":14.0,"e":18.0},"annotated_formula":"subtract(multiply(divide(32, 5), const_2), divide(const_1, const_4))","linear_formula":"divide(n3,n2)|divide(const_1,const_4)|multiply(#0,const_2)|subtract(#2,#1)","chain":"32 \/ 5<\/gadget>\n32\/5 = around 6.4<\/output>\n(32\/5) * 2<\/gadget>\n64\/5 = around 12.8<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(64\/5) - (1\/4)<\/gadget>\n251\/20 = around 12.55<\/output>\n251\/20 = around 12.55<\/result>","index":227} +{"problem":"3 people candidates contested an election and they received 1136 , 7636 and 11628 votes respectively . what is the percentage of the total votes did the winning candidate get ?","rationale":"tot no of votes = ( 1136 + 7636 + 11628 ) = 20400 req = > ( 11628 \/ 20400 * 100 ) = > 57 % answer c","correct":"c","options":{"a":"40 % ","b":"45 % ","c":"57 % ","d":"58 %","e":"60 %"},"options_float":{"a":40.0,"b":45.0,"c":57.0,"d":58.0,"e":60.0},"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":229} +{"problem":"a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 40 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ?","rationale":"\"yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 40 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 20 cents . fraction of spirit is 1 \/ 2 . 20 = k * ( 1 \/ 2 ) k = 40 cost per liter = 40 * ( 1 \/ 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 40 * ( 1 \/ 3 ) * 3 = 40 cents a . 40 cents\"","correct":"a","options":{"a":"40 ","b":"33 ","c":"50 ","d":"51","e":"52"},"options_float":{"a":40.0,"b":33.0,"c":50.0,"d":51.0,"e":52.0},"annotated_formula":"multiply(multiply(40, divide(1, add(1, 2))), add(1, 2))","linear_formula":"add(n0,n4)|divide(n0,#0)|multiply(n2,#1)|multiply(#0,#2)|","chain":"1 + 2<\/gadget>\n3<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n40 * (1\/3)<\/gadget>\n40\/3 = around 13.333333<\/output>\n(40\/3) * 3<\/gadget>\n40<\/output>\n40<\/result>","index":233} +{"problem":"if 12 men or 20 women can do a piece of work in 54 days , then in how many days can 9 men and 12 women together do the work ?","rationale":"\"e 40 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 54 days . so , 27 women can do it in ( 20 * 54 ) \/ 27 = 40 days .\"","correct":"e","options":{"a":"10 days ","b":"30 days ","c":"20 days ","d":"80 days","e":"40 days"},"options_float":{"a":10.0,"b":30.0,"c":20.0,"d":80.0,"e":40.0},"annotated_formula":"inverse(add(divide(9, multiply(12, 54)), divide(12, multiply(20, 54))))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)|","chain":"12 * 54<\/gadget>\n648<\/output>\n9 \/ 648<\/gadget>\n1\/72 = around 0.013889<\/output>\n20 * 54<\/gadget>\n1_080<\/output>\n12 \/ 1_080<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/72) + (1\/90)<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":234} +{"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 13 times as heavy as water ?","rationale":"\"g = 19 w c = 9 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 19 w + 9 wx = x + 1 * 13 w 19 + 9 x = 13 ( x + 1 ) x = 3 \/ 2 ratio of gold with copper = 1 : 3 \/ 2 = 2 : 3 answer is b\"","correct":"b","options":{"a":"1 : 2 ","b":"2 : 3 ","c":"4 : 1 ","d":"5 : 2","e":"6 : 5"},"options_float":{"a":0.5,"b":0.6666666667,"c":4.0,"d":2.5,"e":1.2},"annotated_formula":"divide(subtract(13, 9), subtract(19, 13))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"13 - 9<\/gadget>\n4<\/output>\n19 - 13<\/gadget>\n6<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":236} +{"problem":"a and b started a business jointly a ' s investment was thrice the investment of b and the period of his investment was two times the period of investment of b . if b received rs . 4000 as profit , then their total profit is","rationale":"\"explanation : suppose b invested rs . x for y months . then , a invested rs . 3 x for 2 y months . so , a : b = ( 3 x * 2 y ) : ( x * y ) = 6 xy : xy = 6 : 1 . b ' s profit : total profit = 1 : 7 . let the total profit be rs . x then , 1 \/ 7 = 4000 \/ x or x = 28000 . answer : b ) 28000\"","correct":"b","options":{"a":"23477 ","b":"28000 ","c":"28877 ","d":"1987","e":"1771"},"options_float":{"a":23477.0,"b":28000.0,"c":28877.0,"d":1987.0,"e":1771.0},"annotated_formula":"multiply(add(multiply(const_2, const_3), const_1), 4000)","linear_formula":"multiply(const_2,const_3)|add(#0,const_1)|multiply(n0,#1)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7 * 4_000<\/gadget>\n28_000<\/output>\n28_000<\/result>","index":238} +{"problem":"excluding stoppages , the average speed of a bus is 100 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 100 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 100 - 40 ) km i . e 60 km at 100 km \/ hr . stoppage time = 60 \/ 100 hrs = 36 min answer : e","correct":"e","options":{"a":"15 min ","b":"18 min ","c":"16 min ","d":"20 min","e":"36 min"},"options_float":{"a":15.0,"b":18.0,"c":16.0,"d":20.0,"e":36.0},"annotated_formula":"subtract(multiply(const_1, const_60), multiply(divide(40, 100), 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 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 60<\/gadget>\n24<\/output>\n60 - 24<\/gadget>\n36<\/output>\n36<\/result>","index":239} +{"problem":"tom traveled the entire 60 miles trip . if he did the first 15 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 45 miles per hour , what is the his average speed , in miles per hour ?","rationale":"avg speed = total distance \/ total time = ( d 1 + d 2 ) \/ ( t 1 + t 2 ) = ( 15 + 45 ) \/ ( ( 15 \/ 24 ) + ( 45 \/ 45 ) ) = 60 * 2 \/ 3 = 36.92 mph d","correct":"d","options":{"a":"35.62 mph ","b":"25.45 mph ","c":"34.52 mph ","d":"36.92 mph","e":"36.29 mph"},"options_float":{"a":35.62,"b":25.45,"c":34.52,"d":36.92,"e":36.29},"annotated_formula":"divide(60, add(divide(45, subtract(60, 15)), divide(15, 24)))","linear_formula":"divide(n1,n2)|subtract(n0,n1)|divide(n3,#1)|add(#2,#0)|divide(n0,#3)","chain":"60 - 15<\/gadget>\n45<\/output>\n45 \/ 45<\/gadget>\n1<\/output>\n15 \/ 24<\/gadget>\n5\/8 = around 0.625<\/output>\n1 + (5\/8)<\/gadget>\n13\/8 = around 1.625<\/output>\n60 \/ (13\/8)<\/gadget>\n480\/13 = around 36.923077<\/output>\n480\/13 = around 36.923077<\/result>","index":240} +{"problem":"if w is the set of all the integers between 49 and 114 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ?","rationale":"\"official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 114 , the smallest is 51 ) : 114 - 51 = 63 step 2 . divide by 3 : 63 \/ 3 = 21 step 3 . add 1 : 21 + 1 = 22 . so there are 22 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 114 , the smallest is 50 ) : 114 - 50 = 64 step 2 . divide by 2 : 64 \/ 2 = 32 step 3 . add 1 : 32 + 1 = 33 . so there are 33 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 22 multiples of 3 and the 33 multiples of 2 : 22 + 33 = 55 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 22 + 33 = 55 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 55 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 114 - 54 = 60 step 2 . divide by 6 : 60 \/ 6 = 10 step 3 . add 1 : 10 + 1 = 11 . so there are 11 multiples of 6 within the range : we counted 11 numbers twice . subtract the 11 multiples of 6 from the sum of the multiples of 2 and 3 : = 22 + 33 - 11 = 55 - 11 = 44 therefore , the final number of multiples of 2 , 3 or 6 is 44 . hence , this is the correct answer . ( b )\"","correct":"b","options":{"a":"34 ","b":"44 ","c":"45 ","d":"55","e":"72"},"options_float":{"a":34.0,"b":44.0,"c":45.0,"d":55.0,"e":72.0},"annotated_formula":"subtract(add(floor(divide(subtract(114, 49), 3)), divide(subtract(114, 49), 2)), floor(divide(subtract(114, 49), multiply(2, 3))))","linear_formula":"multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)|","chain":"114 - 49<\/gadget>\n65<\/output>\n65 \/ 3<\/gadget>\n65\/3 = around 21.666667<\/output>\nfloor(65\/3)<\/gadget>\n21<\/output>\n65 \/ 2<\/gadget>\n65\/2 = around 32.5<\/output>\n21 + (65\/2)<\/gadget>\n107\/2 = around 53.5<\/output>\n2 * 3<\/gadget>\n6<\/output>\n65 \/ 6<\/gadget>\n65\/6 = around 10.833333<\/output>\nfloor(65\/6)<\/gadget>\n10<\/output>\n(107\/2) - 10<\/gadget>\n87\/2 = around 43.5<\/output>\n87\/2 = around 43.5<\/result>","index":243} +{"problem":"a shopkeeper sold 8 articles at the cost price of 10 articles . then find the profit % or lost %","rationale":"\"here 8 articles selling price = 10 articles cost price so the difference = 10 - 8 = 2 % of profit = 2 * 100 \/ 8 = 25 % correct option is b\"","correct":"b","options":{"a":"10 % ","b":"25 % ","c":"20 % ","d":"30 %","e":"50 %"},"options_float":{"a":10.0,"b":25.0,"c":20.0,"d":30.0,"e":50.0},"annotated_formula":"multiply(divide(subtract(10, 8), 8), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"10 - 8<\/gadget>\n2<\/output>\n2 \/ 8<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":245} +{"problem":"the general hospital is comprised of , 3 \/ 5 pediatricians , 1 \/ 4 surgeons , and the rest are gp doctors . if 1 \/ 4 of the surgeons are heart surgeons , and the hospital doubles the number of gp doctors , what proportion of the hospital are now heart surgeons ?","rationale":"ped = 3 \/ 5 = 12 \/ 20 sur = 1 \/ 4 = 4 \/ 20 gp = 1 - ( 12 \/ 20 + 4 \/ 20 ) = 1 - 16 \/ 20 = 4 \/ 20 hsur = ( 1 \/ 4 ) ( 4 \/ 20 ) = 1 \/ 20 if gp doubled = > ( 2 ) ( 4 ) = 8 ; then , total = 12 + 4 + 8 = 24 , and 1 is hsur = > proportion = 1 \/ 24 . answer : d","correct":"d","options":{"a":"2 \/ 5 ","b":"1 \/ 4 ","c":"1 \/ 2 ","d":"1 \/ 24","e":"1 \/ 25"},"options_float":{"a":0.4,"b":0.25,"c":0.5,"d":0.0416666667,"e":0.04},"annotated_formula":"divide(subtract(3, add(1, 1)), add(multiply(4, 5), 4))","linear_formula":"add(n2,n2)|multiply(n1,n3)|add(n3,#1)|subtract(n0,#0)|divide(#3,#2)","chain":"1 + 1<\/gadget>\n2<\/output>\n3 - 2<\/gadget>\n1<\/output>\n4 * 5<\/gadget>\n20<\/output>\n20 + 4<\/gadget>\n24<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n1\/24 = around 0.041667<\/result>","index":247} +{"problem":"45 pupil , out of them 12 in debate only and 22 in singing only . then how many in both ?","rationale":"the intersection for two = 45 - 12 - 22 = 11 play both games . answer : c","correct":"c","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(45, add(12, 22))","linear_formula":"add(n1,n2)|subtract(n0,#0)","chain":"12 + 22<\/gadget>\n34<\/output>\n45 - 34<\/gadget>\n11<\/output>\n11<\/result>","index":248} +{"problem":"of the 75 cars on a car lot , 45 have air - conditioning , 35 have power steering , and 12 have both air - conditioning and power steering . how many of the cars on the lot have neither air - conditioning nor power steering ?","rationale":"\"total - neither = all air conditioning + all power steering - both or 75 - neither = 45 + 35 - 12 = 68 . = > neither = 7 , hence a . answer : a\"","correct":"a","options":{"a":"7 ","b":"8 ","c":"10 ","d":"15","e":"18"},"options_float":{"a":7.0,"b":8.0,"c":10.0,"d":15.0,"e":18.0},"annotated_formula":"subtract(75, subtract(add(45, 35), 12))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|","chain":"45 + 35<\/gadget>\n80<\/output>\n80 - 12<\/gadget>\n68<\/output>\n75 - 68<\/gadget>\n7<\/output>\n7<\/result>","index":249} +{"problem":"a corporation 5 times its annual bonus to 10 of its employees . what percent of the employees ’ new bonus is the increase ?","rationale":"let the annual bonus be x . a corporation triples its annual bonus . so new bonus = 5 x . increase = 5 x - x = 4 x the increase is what percent of the employees ’ new bonus = ( 4 x \/ 5 x ) * 100 = 80 % hence c .","correct":"c","options":{"a":"12 % ","b":"18 % ","c":"80 % ","d":"20 %","e":"15 %"},"options_float":{"a":12.0,"b":18.0,"c":80.0,"d":20.0,"e":15.0},"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":250} +{"problem":"the sum of ages of 5 children born at the intervals of 3 years each is 50 years . what is the age of the youngest child ?","rationale":"\"explanation 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 ) = 50 5 x = 20 x = 4 . age of the youngest child = x = 4 years . answer a\"","correct":"a","options":{"a":"4 years ","b":"8 years ","c":"10 years ","d":"none of these","e":"can not be determined"},"options_float":{"a":4.0,"b":8.0,"c":10.0,"d":null,"e":null},"annotated_formula":"subtract(subtract(divide(50, 5), 3), 3)","linear_formula":"divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|","chain":"50 \/ 5<\/gadget>\n10<\/output>\n10 - 3<\/gadget>\n7<\/output>\n7 - 3<\/gadget>\n4<\/output>\n4<\/result>","index":251} +{"problem":"yearly subscription to professional magazines cost a company $ 940.00 . to make a 25 % cut in the magazine budget , how much less must be spent ?","rationale":"total cost 940 940 * 25 \/ 100 = 235 so the cut in amount is 235 the less amount to be spend is 940 - 235 = 705 answer : a","correct":"a","options":{"a":"705 ","b":"655 ","c":"656 ","d":"657","e":"658"},"options_float":{"a":705.0,"b":655.0,"c":656.0,"d":657.0,"e":658.0},"annotated_formula":"multiply(divide(subtract(const_100, 25), const_100), 940)","linear_formula":"subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1)","chain":"100 - 25<\/gadget>\n75<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 940<\/gadget>\n705<\/output>\n705<\/result>","index":252} +{"problem":"the mean of 50 observations was 40 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is","rationale":"\"sol . therefore correct sum = ( 40 × 50 + 48 – 23 ) = 2023 . therefore correct mean = 2023 \/ 50 = 40.46 . answer a\"","correct":"a","options":{"a":"40.46 ","b":"36.1 ","c":"36.5 ","d":"39.1","e":"none"},"options_float":{"a":40.46,"b":36.1,"c":36.5,"d":39.1,"e":null},"annotated_formula":"divide(add(multiply(40, 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":"40 * 50<\/gadget>\n2_000<\/output>\n50 - 2<\/gadget>\n48<\/output>\n48 - 23<\/gadget>\n25<\/output>\n2_000 + 25<\/gadget>\n2_025<\/output>\n2_025 \/ 50<\/gadget>\n81\/2 = around 40.5<\/output>\n81\/2 = around 40.5<\/result>","index":255} +{"problem":"a certain tests consists 8 sections with 25 questions , numbered from 1 to 25 , in each section . if a student answered all of the even - numbered questions correctly and 3 \/ 4 of the odd - numbered questions correctly , what was the total number of questions he answered correctly ? a . 150 b . 172 c . 174 d . 175 e . 176","rationale":"each set has 12 even and 13 odd numbered questions leading to total 96 even and 104 odd questions . 96 + 3 \/ 4 â ˆ — 104 = 96 + 78 = 17496 + 3 \/ 4 â ˆ — 104 = 96 + 78 = 174 answer : a","correct":"a","options":{"a":"174 ","b":"150 ","c":"180 ","d":"175","e":"190"},"options_float":{"a":174.0,"b":150.0,"c":180.0,"d":175.0,"e":190.0},"annotated_formula":"add(divide(multiply(8, 25), const_2), multiply(divide(multiply(8, 25), const_2), divide(3, 4)))","linear_formula":"divide(n4,n5)|multiply(n0,n1)|divide(#1,const_2)|multiply(#2,#0)|add(#2,#3)","chain":"8 * 25<\/gadget>\n200<\/output>\n200 \/ 2<\/gadget>\n100<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n100 * (3\/4)<\/gadget>\n75<\/output>\n100 + 75<\/gadget>\n175<\/output>\n175<\/result>","index":256} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 30 seconds . what is the length of the train ?","rationale":"\"speed = ( 60 * 5 \/ 18 ) m \/ sec = ( 50 \/ 3 ) m \/ sec length of the train = ( speed x time ) = ( 50 \/ 3 * 30 ) m = 500 m . answer : a\"","correct":"a","options":{"a":"500 m ","b":"620 m ","c":"872 m ","d":"150 m","e":"765 m"},"options_float":{"a":500.0,"b":620.0,"c":872.0,"d":150.0,"e":765.0},"annotated_formula":"multiply(divide(multiply(60, const_1000), const_3600), 30)","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) * 30<\/gadget>\n500<\/output>\n500<\/result>","index":257} +{"problem":"when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.60 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ?","rationale":"\"total no . of boxes = 3060000 \/ ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 0.6 = $ 306 answer a\"","correct":"a","options":{"a":"$ 306 ","b":"$ 275 ","c":"$ 510 ","d":"$ 1,250","e":"$ 2,550"},"options_float":{"a":306.0,"b":275.0,"c":510.0,"d":1250.0,"e":2550.0},"annotated_formula":"multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 0.60)","linear_formula":"multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)|","chain":"1_000 * 1_000<\/gadget>\n1_000_000<\/output>\n3.06 * 1_000_000<\/gadget>\n3_060_000<\/output>\n20 * 20<\/gadget>\n400<\/output>\n400 * 15<\/gadget>\n6_000<\/output>\n3_060_000 \/ 6_000<\/gadget>\n510<\/output>\n510 * 0.6<\/gadget>\n306<\/output>\n306<\/result>","index":258} +{"problem":"the sum of ages of 5 children born at the intervals of 3 years each is 80 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 ) = 80 5 x = 50 x = 10 . age of the youngest child = x = 10 years . e )\"","correct":"e","options":{"a":"3 years ","b":"4 years ","c":"6 years ","d":"7 years","e":"10 years"},"options_float":{"a":3.0,"b":4.0,"c":6.0,"d":7.0,"e":10.0},"annotated_formula":"subtract(subtract(divide(80, 5), 3), 3)","linear_formula":"divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|","chain":"80 \/ 5<\/gadget>\n16<\/output>\n16 - 3<\/gadget>\n13<\/output>\n13 - 3<\/gadget>\n10<\/output>\n10<\/result>","index":260} +{"problem":"2 is what percent of 40 ?","rationale":"\"2 = x * 40 \/ 100 x = 5 % ans ; c\"","correct":"c","options":{"a":"0.2 % ","b":"2 % ","c":"5 % ","d":"20 %","e":"500 %"},"options_float":{"a":0.2,"b":2.0,"c":5.0,"d":20.0,"e":500.0},"annotated_formula":"multiply(divide(2, 40), const_100)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|","chain":"2 \/ 40<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n5<\/result>","index":261} +{"problem":"tea worth rs . 126 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","rationale":"\"explanation : since first 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 . cost of 1 kg tea of 1 st kind cost of 1 kg tea of 2 nd kind x - 153 \/ 22.50 = 1 = > x - 153 = 22.50 = > x = 175.50 . hence , price of the third variety = rs . 175.50 per kg . answer : c ) rs . 175.50\"","correct":"c","options":{"a":"175.59 ","b":"175.5 ","c":"175.57 ","d":"175.52","e":"175.11"},"options_float":{"a":175.59,"b":175.5,"c":175.57,"d":175.52,"e":175.11},"annotated_formula":"divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2)","linear_formula":"add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)|","chain":"1 + 1<\/gadget>\n2<\/output>\n2 + 2<\/gadget>\n4<\/output>\n153 * 4<\/gadget>\n612<\/output>\n126 + 126<\/gadget>\n252<\/output>\n612 - 252<\/gadget>\n360<\/output>\n360 \/ 2<\/gadget>\n180<\/output>\n180<\/result>","index":263} +{"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 120 marks , the number of questions he attempts correctly , is :","rationale":"\"let the number of correct answers be x . number of incorrect answers = ( 80 â € “ x ) . 4 x â € “ 1 ( 80 â € “ x ) = 120 or 5 x = 200 or x = 40 . answer : e\"","correct":"e","options":{"a":"35 ","b":"46 ","c":"42 ","d":"30","e":"40"},"options_float":{"a":35.0,"b":46.0,"c":42.0,"d":30.0,"e":40.0},"annotated_formula":"divide(add(120, 80), add(4, 1))","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|","chain":"120 + 80<\/gadget>\n200<\/output>\n4 + 1<\/gadget>\n5<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n40<\/result>","index":264} +{"problem":"a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 10 as remainder . find the no . is ?","rationale":"\"( 555 + 445 ) * 2 * 110 + 10 = 220000 + 10 = 220010 a\"","correct":"a","options":{"a":"220010 ","b":"145778 ","c":"220110 ","d":"235467","e":"220001"},"options_float":{"a":220010.0,"b":145778.0,"c":220110.0,"d":235467.0,"e":220001.0},"annotated_formula":"add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 10)","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 + 10<\/gadget>\n220_010<\/output>\n220_010<\/result>","index":266} +{"problem":"in a sports club with 55 members , 23 play badminton and 29 play tennis and 7 do not play either . how many members play both badminton and tennis ?","rationale":"\"23 + 29 = 52 but where as total number is 55 - 7 = 48 therefore answer is 52 - 48 = 4 hence answer is c\"","correct":"c","options":{"a":"3 ","b":"5 ","c":"4 ","d":"7","e":"8"},"options_float":{"a":3.0,"b":5.0,"c":4.0,"d":7.0,"e":8.0},"annotated_formula":"subtract(add(add(23, 29), 7), 55)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)|","chain":"23 + 29<\/gadget>\n52<\/output>\n52 + 7<\/gadget>\n59<\/output>\n59 - 55<\/gadget>\n4<\/output>\n4<\/result>","index":267} +{"problem":"a garrison of 500 men has provisions for 20 days . at the end of 14 days , a reinforcement arrives , and it is now found that the provisions will last only for 3 days more . what is the reinforcement ?","rationale":"\"500 - - - - 20 500 - - - - 6 x - - - - - 3 x * 3 = 500 * 6 x = 1000 500 - - - - - - - 500 answer : a\"","correct":"a","options":{"a":"500 ","b":"1000 ","c":"1500 ","d":"3000","e":"2000"},"options_float":{"a":500.0,"b":1000.0,"c":1500.0,"d":3000.0,"e":2000.0},"annotated_formula":"subtract(divide(subtract(multiply(500, 20), multiply(500, 14)), 3), 500)","linear_formula":"multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)|","chain":"500 * 20<\/gadget>\n10_000<\/output>\n500 * 14<\/gadget>\n7_000<\/output>\n10_000 - 7_000<\/gadget>\n3_000<\/output>\n3_000 \/ 3<\/gadget>\n1_000<\/output>\n1_000 - 500<\/gadget>\n500<\/output>\n500<\/result>","index":268} +{"problem":"how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ?","rationale":"\"t = ( 660 + 165 ) \/ 54 * 18 \/ 5 t = 55 answer a\"","correct":"a","options":{"a":"55 sec ","b":"40 sec ","c":"45 sec ","d":"30 sec","e":"35 sec"},"options_float":{"a":55.0,"b":40.0,"c":45.0,"d":30.0,"e":35.0},"annotated_formula":"divide(add(165, 660), multiply(54, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"165 + 660<\/gadget>\n825<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n54 * (5\/18)<\/gadget>\n15<\/output>\n825 \/ 15<\/gadget>\n55<\/output>\n55<\/result>","index":269} +{"problem":"in covering a distance of 54 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is :","rationale":"\"let abhay ' s speed be x km \/ hr . then , 54 \/ x - 54 \/ 2 x = 3 6 x = 54 x = 9 km \/ hr . answer : option e\"","correct":"e","options":{"a":"5 kmph ","b":"6 kmph ","c":"6.25 kmph ","d":"7.5 kmph","e":"9 kmph"},"options_float":{"a":5.0,"b":6.0,"c":6.25,"d":7.5,"e":9.0},"annotated_formula":"divide(subtract(54, divide(54, 2)), add(1, 2))","linear_formula":"add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0)|","chain":"54 \/ 2<\/gadget>\n27<\/output>\n54 - 27<\/gadget>\n27<\/output>\n1 + 2<\/gadget>\n3<\/output>\n27 \/ 3<\/gadget>\n9<\/output>\n9<\/result>","index":270} +{"problem":"john has $ 1600 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":"divide(subtract(1600, 600), const_2)","linear_formula":"subtract(n0,n1)|divide(#0,const_2)","chain":"1_600 - 600<\/gadget>\n1_000<\/output>\n1_000 \/ 2<\/gadget>\n500<\/output>\n500<\/result>","index":272} +{"problem":"students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 181 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 181 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 + 181 m ) \/ ( w + m ) = 180 170 w + 181 m = 180 w + 180 m 1 m = 10 w m \/ w = 10 \/ 1 the ratio of men to women is 10 to 1 . e","correct":"e","options":{"a":"5 : 2 ","b":"5 : 1 ","c":"4 : 3 ","d":"4 : 1","e":"10 : 1"},"options_float":{"a":2.5,"b":5.0,"c":1.3333333333,"d":4.0,"e":10.0},"annotated_formula":"divide(subtract(180, 170), subtract(181, 180))","linear_formula":"subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)","chain":"180 - 170<\/gadget>\n10<\/output>\n181 - 180<\/gadget>\n1<\/output>\n10 \/ 1<\/gadget>\n10<\/output>\n10<\/result>","index":274} +{"problem":"if the length of the longest chord of a certain circle is 18 , 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 18 = 2 * 9 so radius of the circle = 9 correct answer - b\"","correct":"b","options":{"a":"2.5 ","b":"9 ","c":"10 ","d":"15","e":"20"},"options_float":{"a":2.5,"b":9.0,"c":10.0,"d":15.0,"e":20.0},"annotated_formula":"divide(18, const_2)","linear_formula":"divide(n0,const_2)|","chain":"18 \/ 2<\/gadget>\n9<\/output>\n9<\/result>","index":275} +{"problem":"a cistern 10 meters long and 4 meters wide contains water up to a depth of 1 meter 25 cm . what is the total area of the wet surface ?","rationale":"area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 10 x 1.25 ) + 10 x 4 ] = 75 the answer is d .","correct":"d","options":{"a":"45 ","b":"55 ","c":"65 ","d":"75","e":"85"},"options_float":{"a":45.0,"b":55.0,"c":65.0,"d":75.0,"e":85.0},"annotated_formula":"add(add(multiply(10, 4), multiply(const_2, multiply(4, add(1, divide(25, const_100))))), multiply(const_2, multiply(10, add(1, divide(25, const_100)))))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|multiply(#3,const_2)|multiply(#4,const_2)|add(#1,#5)|add(#7,#6)","chain":"10 * 4<\/gadget>\n40<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n4 * (5\/4)<\/gadget>\n5<\/output>\n2 * 5<\/gadget>\n10<\/output>\n40 + 10<\/gadget>\n50<\/output>\n10 * (5\/4)<\/gadget>\n25\/2 = around 12.5<\/output>\n2 * (25\/2)<\/gadget>\n25<\/output>\n50 + 25<\/gadget>\n75<\/output>\n75<\/result>","index":276} +{"problem":"if 50 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ?","rationale":"\"50 % of ( x - y ) = 30 % of ( x + y ) ( 50 \/ 100 ) ( x - y ) = ( 30 \/ 100 ) ( x + y ) 5 ( x - y ) = 3 ( x + y ) 2 x = 8 y x = 4 y therefore required percentage = ( ( y \/ x ) x 100 ) % = ( ( y \/ 4 y ) x 100 ) = 25 % answer is e .\"","correct":"e","options":{"a":"2.5 % ","b":"10 % ","c":"5 % ","d":"15 %","e":"25 %"},"options_float":{"a":2.5,"b":10.0,"c":5.0,"d":15.0,"e":25.0},"annotated_formula":"multiply(divide(subtract(50, 30), add(50, 30)), const_100)","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|","chain":"50 - 30<\/gadget>\n20<\/output>\n50 + 30<\/gadget>\n80<\/output>\n20 \/ 80<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":278} +{"problem":"how many seconds does sandy take to cover a distance of 600 meters , if sandy runs at a speed of 18 km \/ hr ?","rationale":"18 km \/ hr = 18000 m \/ 3600 s = 5 m \/ s time = 600 \/ 5 = 120 seconds the answer is b .","correct":"b","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":"divide(600, multiply(18, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n18 * (5\/18)<\/gadget>\n5<\/output>\n600 \/ 5<\/gadget>\n120<\/output>\n120<\/result>","index":279} +{"problem":"calculate how many days it will take for 10 boys to paint a 80 m long wall if 6 boys can paint a 70 m long wall in 8 days ,","rationale":"\"the length of wall painted by one boy in one day = 70 \/ 6 * 1 \/ 8 = 1.46 m no . of days required to paint 50 m cloth by 8 boys = 80 \/ 10 * 1 \/ 1.46 = 5.48 days . d\"","correct":"d","options":{"a":"9.48 days ","b":"3.48 days ","c":"7.48 days ","d":"5.48 days","e":"6.48 days"},"options_float":{"a":9.48,"b":3.48,"c":7.48,"d":5.48,"e":6.48},"annotated_formula":"divide(multiply(multiply(6, 8), 80), multiply(70, 10))","linear_formula":"multiply(n2,n4)|multiply(n0,n3)|multiply(n1,#0)|divide(#2,#1)|","chain":"6 * 8<\/gadget>\n48<\/output>\n48 * 80<\/gadget>\n3_840<\/output>\n70 * 10<\/gadget>\n700<\/output>\n3_840 \/ 700<\/gadget>\n192\/35 = around 5.485714<\/output>\n192\/35 = around 5.485714<\/result>","index":280} +{"problem":"a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 60 as remainder . find the no . is ?","rationale":"\"( 555 + 445 ) * 2 * 110 + 60 = 220000 + 60 = 220060 e\"","correct":"e","options":{"a":"145646 ","b":"236578 ","c":"645353 ","d":"456546","e":"220060"},"options_float":{"a":145646.0,"b":236578.0,"c":645353.0,"d":456546.0,"e":220060.0},"annotated_formula":"add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 60)","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 + 60<\/gadget>\n220_060<\/output>\n220_060<\/result>","index":281} +{"problem":"a monkey start climbing up a tree 18 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top .","rationale":"\"if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft \/ hr . similarly in 15 hrs it wil be 15 ft . bt since the height of the tree is 18 ft , so if the monkey hops up the tree in the next hr i . e 16 th hr then it reaches at the top of the tree . hence it takes 16 hrs for monkey to reach at the top answer : e\"","correct":"e","options":{"a":"15 hrs ","b":"18 hrs ","c":"19 hrs ","d":"17 hrs","e":"16 hrs"},"options_float":{"a":15.0,"b":18.0,"c":19.0,"d":17.0,"e":16.0},"annotated_formula":"subtract(divide(18, subtract(3, 2)), 2)","linear_formula":"subtract(n1,n2)|divide(n0,#0)|subtract(#1,n2)|","chain":"3 - 2<\/gadget>\n1<\/output>\n18 \/ 1<\/gadget>\n18<\/output>\n18 - 2<\/gadget>\n16<\/output>\n16<\/result>","index":282} +{"problem":"jane and ashley take 8 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an eight day leave from work . this led to jane ' s working for four extra days on her own to complete the project . how long did it take to finish the project ?","rationale":"\"let us assume that the work is laying 40 bricks . jane = 5 bricks per day ashley = 1 brick per day together = 6 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 20 remaining bricks = 40 - 28 = 12 so together , they would take 12 \/ 6 = 2 total no . of days = 8 + 4 + 2 = 14 answer is a\"","correct":"a","options":{"a":"14 days ","b":"15 days ","c":"16 days ","d":"18 days","e":"20 days"},"options_float":{"a":14.0,"b":15.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"add(add(divide(subtract(subtract(const_1, multiply(const_4, divide(const_1, 8))), multiply(add(const_4, const_4), divide(const_1, 40))), add(divide(const_1, 8), divide(const_1, 40))), add(const_4, const_4)), const_4)","linear_formula":"add(const_4,const_4)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|multiply(#1,const_4)|multiply(#0,#2)|subtract(const_1,#4)|subtract(#6,#5)|divide(#7,#3)|add(#0,#8)|add(#9,const_4)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n4 * (1\/8)<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n4 + 4<\/gadget>\n8<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n8 * (1\/40)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/2) - (1\/5)<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/8) + (1\/40)<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/10) \/ (3\/20)<\/gadget>\n2<\/output>\n2 + 8<\/gadget>\n10<\/output>\n10 + 4<\/gadget>\n14<\/output>\n14<\/result>","index":283} +{"problem":"if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 32 ?","rationale":"\"because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 \/ 2 that it will be divisible by 4 and , thus , a probability of 1 \/ 4 that it will be divisible by 8 and , thus , a probability of 1 \/ 8 that it will be divisible by 16 and , thus , a probability of 1 \/ 16 that it will be divisible by 32 1 * 1 \/ 16 = 1 \/ 16 answer : c\"","correct":"c","options":{"a":"2 \/ 7 ","b":"3 \/ 7 ","c":"1 \/ 16 ","d":"1 \/ 14","e":"1 \/ 12"},"options_float":{"a":0.2857142857,"b":0.4285714286,"c":0.0625,"d":0.0714285714,"e":0.0833333333},"annotated_formula":"divide(const_2, 32)","linear_formula":"divide(const_2,n3)|","chain":"2 \/ 32<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":284} +{"problem":"today jim is twice as old as fred , and sam is 4 years younger than fred . 4 years ago jim was 8 times as old as sam . how old is jim now ?","rationale":"we ' re asked how old jim is now . we ' re given three facts to work with : 1 ) today , jim is twice as old as fred 2 ) today , sam is 4 years younger than fred 3 ) four years ago , jim was 8 times as old as sam . let ' s test answer d : 20 if . . . . jim is currently 20 years old . . . . fred is 10 years old sam is 6 years old 4 years ago , jim was 16 and sam was 2 , so jim was 8 times sam ' s age . this is an exact match for what we were told , so this must be the answer . d","correct":"d","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":"multiply(divide(subtract(multiply(8, 8), 4), subtract(8, const_2)), const_2)","linear_formula":"multiply(n2,n2)|subtract(n2,const_2)|subtract(#0,n0)|divide(#2,#1)|multiply(#3,const_2)","chain":"8 * 8<\/gadget>\n64<\/output>\n64 - 4<\/gadget>\n60<\/output>\n8 - 2<\/gadget>\n6<\/output>\n60 \/ 6<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":285} +{"problem":"a bullet train 150 m long is running with a speed of 30 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the bullet train is going ?","rationale":"e 15 sec speed of the bullet train relative to man = ( 30 + 6 ) kmph = 36 * 5 \/ 18 m \/ sec = 30 \/ 3 m \/ sec . time taken by the bullet train to cross the man = time taken by it to cover 150 m at ( 30 \/ 3 ) m \/ sec = ( 150 * 3 \/ 30 ) sec = 15 sec","correct":"e","options":{"a":"23 sec ","b":"15 sec ","c":"12 sec ","d":"11 sec","e":"15 sec"},"options_float":{"a":23.0,"b":15.0,"c":12.0,"d":11.0,"e":15.0},"annotated_formula":"divide(150, divide(multiply(add(30, 6), const_1000), const_3600))","linear_formula":"add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)","chain":"30 + 6<\/gadget>\n36<\/output>\n36 * 1_000<\/gadget>\n36_000<\/output>\n36_000 \/ 3_600<\/gadget>\n10<\/output>\n150 \/ 10<\/gadget>\n15<\/output>\n15<\/result>","index":286} +{"problem":"the sum of all the integers k such that – 21 < k < 24 is","rationale":"\"- 20 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 20 - 19 sum of which is - 39 . hence option d . d\"","correct":"d","options":{"a":"0 ","b":"- 2 ","c":"- 25 ","d":"- 39","e":"- 51"},"options_float":{"a":0.0,"b":-2.0,"c":-25.0,"d":-39.0,"e":-51.0},"annotated_formula":"add(add(negate(21), const_1), add(add(negate(21), const_1), const_1))","linear_formula":"negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)|","chain":"-21<\/gadget>\n-21<\/output>\n(-21) + 1<\/gadget>\n-20<\/output>\n(-20) + 1<\/gadget>\n-19<\/output>\n(-20) + (-19)<\/gadget>\n-39<\/output>\n-39<\/result>","index":287} +{"problem":"a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 30 . 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 . 150 ( 30 * 5 ) profit percentage = ( 150 - 100 ) \/ 100 * 100 = 50 % answer : b\"","correct":"b","options":{"a":"25 % ","b":"50 % ","c":"20 % ","d":"5 %","e":"none of these"},"options_float":{"a":25.0,"b":50.0,"c":20.0,"d":5.0,"e":null},"annotated_formula":"subtract(multiply(30, 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>\n30 * 5<\/gadget>\n150<\/output>\n25 * 4<\/gadget>\n100<\/output>\n150 - 100<\/gadget>\n50<\/output>\n50<\/result>","index":288} +{"problem":"a worker ' s daily wage is increased by 50 % and the new wage is $ 30 per day . what was the worker ' s daily wage before the increase ?","rationale":"let x be the daily wage before the increase . 1.5 x = $ 30 x = $ 20 the answer is c .","correct":"c","options":{"a":"$ 15 ","b":"$ 18 ","c":"$ 20 ","d":"$ 22","e":"$ 25"},"options_float":{"a":15.0,"b":18.0,"c":20.0,"d":22.0,"e":25.0},"annotated_formula":"divide(30, add(const_1, divide(50, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n30 \/ (3\/2)<\/gadget>\n20<\/output>\n20<\/result>","index":290} +{"problem":"if the weight of 12 meters long rod is 13.4 kg . what is the weight of 6 meters long rod ?","rationale":"\"answer ∵ weight of 12 m long rod = 13.4 kg ∴ weight of 1 m long rod = 13.4 \/ 12 kg ∴ weight of 6 m long rod = 13.4 x 6 \/ 12 = 6.7 kg option : a\"","correct":"a","options":{"a":"6.7 kg . ","b":"10.8 kg . ","c":"12.4 kg . ","d":"18.0 kg","e":"none"},"options_float":{"a":6.7,"b":10.8,"c":12.4,"d":18.0,"e":null},"annotated_formula":"divide(multiply(6, 13.4), 12)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"6 * 13.4<\/gadget>\n80.4<\/output>\n80.4 \/ 12<\/gadget>\n6.7<\/output>\n6.7<\/result>","index":292} +{"problem":"a store sells 2 kinds of jelly beans mixes ( a and b ) both made up of red and yellow beans . if b contains 20 % more red beans than a but 10 % fewer yellow beans . and jar a contains twice as many red beans as yellow by what percent is the number of beans in jar b larger than the number in jar a","rationale":"a has 10 yellows 20 reds total = 30 so b has 1.2 x 20 = 24 reds 0.9 x 10 = 9 yellows total = 33 difference = 3 \/ 30 = 10 % answer : e","correct":"e","options":{"a":"5 ","b":"6 ","c":"8 ","d":"9","e":"10"},"options_float":{"a":5.0,"b":6.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"multiply(subtract(divide(add(add(const_100, 20), subtract(const_100, 10)), const_100), const_2), const_100)","linear_formula":"add(n1,const_100)|subtract(const_100,n2)|add(#0,#1)|divide(#2,const_100)|subtract(#3,const_2)|multiply(#4,const_100)","chain":"100 + 20<\/gadget>\n120<\/output>\n100 - 10<\/gadget>\n90<\/output>\n120 + 90<\/gadget>\n210<\/output>\n210 \/ 100<\/gadget>\n21\/10 = around 2.1<\/output>\n(21\/10) - 2<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":295} +{"problem":"find the greatest common factor ( gfc ) of 24 , 40 and 60 .","rationale":"we first write the prime factorization of each given number 24 = 2 × 2 × 2 × 3 = 23 * cubic * × 3 40 = 2 × 2 × 2 × 5 = 23 * cubic * × 5 60 = 2 × 2 × 3 × 5 = 22 * square * × 3 × 5 gfc = 22 * square * = 4 corect answer is d ) 4","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":"gcd(gcd(24, 40), 60)","linear_formula":"gcd(n0,n1)|gcd(n2,#0)","chain":"gcd(24, 40)<\/gadget>\n8<\/output>\ngcd(8, 60)<\/gadget>\n4<\/output>\n4<\/result>","index":296} +{"problem":"when 242 is divided by a certain divisor the remainder obtained is 12 . when 698 is divided by the same divisor the remainder obtained is 16 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 10 . what is the value of the divisor ?","rationale":"\"let that divisor be x since remainder is 12 or 16 it means divisor is greater than 16 . now 242 - 12 = 230 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 16 = 682 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 230 + 682 ) + 12 + 16 = x ( k + l ) + 28 when we divide this number by x then remainder will be equal to remainder of ( 28 divided by x ) = 10 hence x = 28 - 10 = 18 hence b\"","correct":"b","options":{"a":"11 ","b":"18 ","c":"13 ","d":"23","e":"none of these"},"options_float":{"a":11.0,"b":18.0,"c":13.0,"d":23.0,"e":null},"annotated_formula":"subtract(add(12, 16), 10)","linear_formula":"add(n1,n3)|subtract(#0,n6)|","chain":"12 + 16<\/gadget>\n28<\/output>\n28 - 10<\/gadget>\n18<\/output>\n18<\/result>","index":297} +{"problem":"the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 20 % ?","rationale":"\"explanation : let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 80 gain % = 16 \/ 64 * 100 = 25 % . answer : option e\"","correct":"e","options":{"a":"37.5 % ","b":"48 % ","c":"50.5 % ","d":"52 %","e":"25 %"},"options_float":{"a":37.5,"b":48.0,"c":50.5,"d":52.0,"e":25.0},"annotated_formula":"multiply(subtract(divide(subtract(const_100, 20), 64), const_1), const_100)","linear_formula":"subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 64<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) - 1<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":298} +{"problem":"by selling 22 pens for a rupee a woman loses 10 % . how many for a rupee should he sell in order to gain 50 % ?","rationale":"\"d 90 % - - - 22 150 % - - - ? 90 \/ 150 * 22 = 13\"","correct":"d","options":{"a":"12 ","b":"14 ","c":"45 ","d":"13","e":"65"},"options_float":{"a":12.0,"b":14.0,"c":45.0,"d":13.0,"e":65.0},"annotated_formula":"inverse(add(divide(divide(const_100, subtract(const_100, 10)), 22), divide(multiply(divide(divide(const_100, subtract(const_100, 10)), 22), 50), 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 - 10<\/gadget>\n90<\/output>\n100 \/ 90<\/gadget>\n10\/9 = around 1.111111<\/output>\n(10\/9) \/ 22<\/gadget>\n5\/99 = around 0.050505<\/output>\n(5\/99) * 50<\/gadget>\n250\/99 = around 2.525253<\/output>\n(250\/99) \/ 100<\/gadget>\n5\/198 = around 0.025253<\/output>\n(5\/99) + (5\/198)<\/gadget>\n5\/66 = around 0.075758<\/output>\n1 \/ (5\/66)<\/gadget>\n66\/5 = around 13.2<\/output>\n66\/5 = around 13.2<\/result>","index":299} +{"problem":"a and b start a business jointly . a invests rs 16000 for 8 month and b remains in the business for 4 months . out of total profit , b claims 2 \/ 7 of the profit . how much money was contributed by b ?","rationale":"b claims 2 \/ 7 of the profit a claims remaining 5 \/ 7 of the profit = > a : b = 5 \/ 7 : 2 \/ 7 = 5 : 2 let the money contributed by b = b then a : b = 16000 × 8 : b × 4 therefore , 16000 × 8 : b × 4 = 5 : 2 16000 × 8 × 2 = b × 4 × 5 16000 × 2 × 2 = b × 5 3200 × 2 × 2 = b b = 12800 answer is a .","correct":"a","options":{"a":"12800 ","b":"13000 ","c":"11500 ","d":"12500","e":"12000"},"options_float":{"a":12800.0,"b":13000.0,"c":11500.0,"d":12500.0,"e":12000.0},"annotated_formula":"divide(multiply(multiply(16000, 8), divide(8, 4)), multiply(4, add(4, const_1)))","linear_formula":"add(n2,const_1)|divide(n1,n2)|multiply(n0,n1)|multiply(#1,#2)|multiply(n2,#0)|divide(#3,#4)","chain":"16_000 * 8<\/gadget>\n128_000<\/output>\n8 \/ 4<\/gadget>\n2<\/output>\n128_000 * 2<\/gadget>\n256_000<\/output>\n4 + 1<\/gadget>\n5<\/output>\n4 * 5<\/gadget>\n20<\/output>\n256_000 \/ 20<\/gadget>\n12_800<\/output>\n12_800<\/result>","index":301} +{"problem":"john and steve are speed walkers in a race . john is 10 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 + 10 + 2 - - - > 0.5 t = 12 - - - > t = 24 seconds . c is the correct answer .\"","correct":"c","options":{"a":"13 seconds ","b":"17 seconds ","c":"24 seconds ","d":"34 seconds","e":"51 seconds"},"options_float":{"a":13.0,"b":17.0,"c":24.0,"d":34.0,"e":51.0},"annotated_formula":"divide(add(divide(multiply(3.7, add(10, 2)), subtract(4.2, 3.7)), add(10, 2)), 4.2)","linear_formula":"add(n0,n3)|subtract(n1,n2)|multiply(n2,#0)|divide(#2,#1)|add(#0,#3)|divide(#4,n1)|","chain":"10 + 2<\/gadget>\n12<\/output>\n3.7 * 12<\/gadget>\n44.4<\/output>\n4.2 - 3.7<\/gadget>\n0.5<\/output>\n44.4 \/ 0.5<\/gadget>\n88.8<\/output>\n88.8 + 12<\/gadget>\n100.8<\/output>\n100.8 \/ 4.2<\/gadget>\n24<\/output>\n24<\/result>","index":303} +{"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 3 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 = 3 * 180 \/ 12 = 45 ounces of water answer : a","correct":"a","options":{"a":"45 ","b":"50 ","c":"55 ","d":"60","e":"63"},"options_float":{"a":45.0,"b":50.0,"c":55.0,"d":60.0,"e":63.0},"annotated_formula":"multiply(multiply(30, const_2), divide(3, const_4))","linear_formula":"divide(n3,const_4)|multiply(n2,const_2)|multiply(#0,#1)","chain":"30 * 2<\/gadget>\n60<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n60 * (3\/4)<\/gadget>\n45<\/output>\n45<\/result>","index":306} +{"problem":"two employees x and y are paid a total of rs . 440 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 = 440 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 â ˆ ´ 12 y \/ 10 + y = 440 â ‡ ’ y [ 12 \/ 10 + 1 ] = 440 â ‡ ’ 22 y \/ 10 = 440 â ‡ ’ 22 y = 4400 â ‡ ’ y = 4400 \/ 22 = 400 \/ 2 = rs . 200 b )\"","correct":"b","options":{"a":"s . 150 ","b":"s . 200 ","c":"s . 250 ","d":"s . 350","e":"s . 400"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":350.0,"e":400.0},"annotated_formula":"divide(multiply(440, 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>\n440 * 10<\/gadget>\n4_400<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n4_400 \/ 22<\/gadget>\n200<\/output>\n200<\/result>","index":308} +{"problem":"how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 25 percent butterfat to obtain milk that is 20 percent butterfat ?","rationale":"\"equate the fat : 0.1 x + 0.25 * 8 = 0.2 ( x + 8 ) - - > x = 4 . answer : a .\"","correct":"a","options":{"a":"4 ","b":"12 ","c":"14 ","d":"16","e":"28"},"options_float":{"a":4.0,"b":12.0,"c":14.0,"d":16.0,"e":28.0},"annotated_formula":"divide(multiply(subtract(25, 20), 8), 10)","linear_formula":"subtract(n2,n3)|multiply(n1,#0)|divide(#1,n0)|","chain":"25 - 20<\/gadget>\n5<\/output>\n5 * 8<\/gadget>\n40<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n4<\/result>","index":309} +{"problem":"a envelop weight 8.5 gm , if 820 of these envelop are sent with an advertisement mail . how much wieght ?","rationale":"\"820 * 8.5 6970.0 gm 6.97 kg answer : d\"","correct":"d","options":{"a":"6.6 kg ","b":"6.8 kg ","c":"6.7 kg ","d":"6.97 kg","e":"7.8 kg"},"options_float":{"a":6.6,"b":6.8,"c":6.7,"d":6.97,"e":7.8},"annotated_formula":"divide(multiply(8.5, 820), const_1000)","linear_formula":"multiply(n0,n1)|divide(#0,const_1000)|","chain":"8.5 * 820<\/gadget>\n6_970<\/output>\n6_970 \/ 1_000<\/gadget>\n697\/100 = around 6.97<\/output>\n697\/100 = around 6.97<\/result>","index":310} +{"problem":"the area of a rectangular field is equal to 500 square meters . its perimeter is equal to 90 meters . find the width of this rectangle .","rationale":"\"l * w = 500 : area , l is the length and w is the width . 2 l + 2 w = 90 : perimeter l = 45 - w : solve for l ( 45 - w ) * w = 500 : substitute in the area equation w = 20 and l = 25 correct answer 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":"divide(subtract(divide(90, const_2), sqrt(subtract(multiply(divide(90, const_2), divide(90, const_2)), multiply(const_4, 500)))), const_2)","linear_formula":"divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)|","chain":"90 \/ 2<\/gadget>\n45<\/output>\n45 * 45<\/gadget>\n2_025<\/output>\n4 * 500<\/gadget>\n2_000<\/output>\n2_025 - 2_000<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n45 - 5<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":311} +{"problem":"the angle between the minute hand and the hour hand of a clock when the time is 11.30 , is","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 = 11 , m = 30 and minute hand is behind the hour hand . hence the angle = 30 ( h − m \/ 5 ) + m \/ 2 = 30 ( 11 − 30 \/ 5 ) + 30 \/ 2 = 30 ( 11 − 6 ) + 15 = 30 × 5 + 15 = 165 ° answer is d .","correct":"d","options":{"a":"35 ° ","b":"65 ° ","c":"45 ° ","d":"165 °","e":"95 °"},"options_float":{"a":35.0,"b":65.0,"c":45.0,"d":165.0,"e":95.0},"annotated_formula":"divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2)","linear_formula":"add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 - 1<\/gadget>\n11<\/output>\n12 \/ 11<\/gadget>\n12\/11 = around 1.090909<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 11<\/gadget>\n55<\/output>\n(12\/11) * 55<\/gadget>\n60<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n60 - 30<\/gadget>\n30<\/output>\n30 * 11<\/gadget>\n330<\/output>\n330 \/ 2<\/gadget>\n165<\/output>\n165<\/result>","index":312} +{"problem":"there are 16 bees in the hive , then 6 more fly . how many bees are there in all ?","rationale":"16 + 6 = 22 . answer is c .","correct":"c","options":{"a":"7 ","b":"33 ","c":"22 ","d":"17","e":"25"},"options_float":{"a":7.0,"b":33.0,"c":22.0,"d":17.0,"e":25.0},"annotated_formula":"add(16, 6)","linear_formula":"add(n0,n1)|","chain":"16 + 6<\/gadget>\n22<\/output>\n22<\/result>","index":313} +{"problem":"a train moves with a speed of 108 kmph . its speed in metres per second is :","rationale":"\"explanation : 108 kmph = ( 108 x 5 \/ 18 ) m \/ sec = 30 m \/ s . answer : c\"","correct":"c","options":{"a":"10.8 ","b":"18 ","c":"30 ","d":"38.8","e":"none of these"},"options_float":{"a":10.8,"b":18.0,"c":30.0,"d":38.8,"e":null},"annotated_formula":"multiply(108, const_0_2778)","linear_formula":"multiply(n0,const_0_2778)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n108 * (5\/18)<\/gadget>\n30<\/output>\n30<\/result>","index":314} +{"problem":"20 is subtracted from 60 % of a number , the result is 88 . find the number ?","rationale":"\"( 60 \/ 100 ) * x â € “ 20 = 88 6 x = 1080 x = 180 answer : c\"","correct":"c","options":{"a":"120 ","b":"300 ","c":"180 ","d":"170","e":"148"},"options_float":{"a":120.0,"b":300.0,"c":180.0,"d":170.0,"e":148.0},"annotated_formula":"divide(add(20, 88), divide(60, const_100))","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|","chain":"20 + 88<\/gadget>\n108<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n108 \/ (3\/5)<\/gadget>\n180<\/output>\n180<\/result>","index":315} +{"problem":"by how much is 50 % of 250 greater than 25 % of 400 .","rationale":"\"( 50 \/ 100 ) * 250 â € “ ( 25 \/ 100 ) * 400 125 - 100 = 25 answer : b\"","correct":"b","options":{"a":"25 ","b":"26 ","c":"29 ","d":"39","e":"26"},"options_float":{"a":25.0,"b":26.0,"c":29.0,"d":39.0,"e":26.0},"annotated_formula":"subtract(multiply(250, divide(50, const_100)), multiply(divide(25, const_100), 400))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n250 * (1\/2)<\/gadget>\n125<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 400<\/gadget>\n100<\/output>\n125 - 100<\/gadget>\n25<\/output>\n25<\/result>","index":316} +{"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":"solution required ratio = 60 : 90 = 2 : 3 answer b","correct":"b","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(6.3, 5.7), subtract(7.2, 6.3))","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":317} +{"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 35 % lemonade syrup ?","rationale":"\"let the total solution is 150 l with 80 l water 70 l syrup . to make 35 % syrup solution , the result solution must have 97.5 l syrup and 52.5 l syrup . therefore we are taking 17.5 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 17.5 l syrup in 37.5 l solution we started by multiplying 10 now to get to the result we need to divide by 17.5 = > amount of solution to be replaced with water ( 37.5 \/ 17.5 ) = 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":320} +{"problem":"running at the same constant rate , 100 identical machines can produce a total of 500 coffee bar per minute . at this rate , how many bottles could 20 such machines produce in 2 minutes ?","rationale":"let ' s take the approach that uses the answer choices to eliminate wasted time . 500 \/ 100 = 5 coffee bar per minute per machine . 20 machines = 100 per minute . 2 minutes worth = 200 coffe bar . looking at the answers it is clear . . . we can only choose ( d ) the correct answer is d .","correct":"d","options":{"a":"110 ","b":"220 ","c":"330 ","d":"200","e":"789"},"options_float":{"a":110.0,"b":220.0,"c":330.0,"d":200.0,"e":789.0},"annotated_formula":"multiply(multiply(divide(500, 100), 2), 20)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)","chain":"500 \/ 100<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 20<\/gadget>\n200<\/output>\n200<\/result>","index":321} +{"problem":"find the constant k so that : - x 2 - ( k + 8 ) x - 8 = - ( x - 2 ) ( x - 4 )","rationale":"\"- x 2 - ( k + 8 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 8 ) x - 8 = - x 2 + 6 x - 8 - ( k + 8 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 14 : solve the above for k correct answer c\"","correct":"c","options":{"a":"11 ","b":"12 ","c":"14 ","d":"19","e":"15"},"options_float":{"a":11.0,"b":12.0,"c":14.0,"d":19.0,"e":15.0},"annotated_formula":"add(8, add(4, 2))","linear_formula":"add(n0,n4)|add(n1,#0)|","chain":"4 + 2<\/gadget>\n6<\/output>\n8 + 6<\/gadget>\n14<\/output>\n14<\/result>","index":324} +{"problem":"a , b and c invested rs . 6000 , rs . 4000 and rs . 10000 respectively , in a partnership business . find the share of a in profit of rs . 11000 after a year ?","rationale":"\"explanation : 6000 : 4000 : 10000 3 : 2 : 5 3 \/ 10 * 11000 = 3300 answer : a\"","correct":"a","options":{"a":"3300 ","b":"1100 ","c":"2667 ","d":"600","e":"4000"},"options_float":{"a":3300.0,"b":1100.0,"c":2667.0,"d":600.0,"e":4000.0},"annotated_formula":"multiply(divide(6000, add(add(6000, 4000), 10000)), 11000)","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|","chain":"6_000 + 4_000<\/gadget>\n10_000<\/output>\n10_000 + 10_000<\/gadget>\n20_000<\/output>\n6_000 \/ 20_000<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 11_000<\/gadget>\n3_300<\/output>\n3_300<\/result>","index":325} +{"problem":"a group of people participate in some curriculum , 30 of them practice yoga , 20 study cooking , 15 study weaving , 5 of them study cooking only , 8 of them study both the cooking and yoga , 5 of them participate all curriculums . how many people study both cooking and weaving ?","rationale":"\"both cooking and weaving = 20 - ( 5 + 8 + 5 ) = 2 so , the correct answer is 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(subtract(subtract(20, 8), 5), 5)","linear_formula":"subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)|","chain":"20 - 8<\/gadget>\n12<\/output>\n12 - 5<\/gadget>\n7<\/output>\n7 - 5<\/gadget>\n2<\/output>\n2<\/result>","index":326} +{"problem":"at what rate percent on simple interest will rs . 750 amount to rs . 825 in 5 years ?","rationale":"\"75 = ( 750 * 5 * r ) \/ 100 r = 2 % . 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":"multiply(divide(divide(subtract(825, 750), 750), 5), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|","chain":"825 - 750<\/gadget>\n75<\/output>\n75 \/ 750<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) \/ 5<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/50) * 100<\/gadget>\n2<\/output>\n2<\/result>","index":327} +{"problem":"two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of b crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of b ?","rationale":"lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( b ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e","correct":"e","options":{"a":"b \/ 2 ","b":"b ","c":"3 b \/ 2 ","d":"2 b","e":"3 b"},"options_float":{"a":2.0,"b":null,"c":3.0,"d":2.0,"e":3.0},"annotated_formula":"multiply(const_2, add(const_1, divide(const_1, const_2)))","linear_formula":"divide(const_1,const_2)|add(#0,const_1)|multiply(#1,const_2)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n2 * (3\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":329} +{"problem":"a certain number of badges were distributed among a class of students . the student who got 1 \/ 6 th of the total number of badges actually got 5 times the average number of badges the others got ! how many students were there in the class ?","rationale":"detailed solution let the total students be ( n + 1 ) let total badges be x let the average of ‘ n ’ students be y the student who got 1 \/ 6 th of x = 5 y or y = x \/ 30 therefore ‘ n ’ students got 1 \/ 30 th of total share each or n * x \/ 30 + 1 * x \/ 6 = x nx + 5 x = 30 x n + 5 = 30 or n = 25 total = n + 1 = 26 correct answer : b","correct":"b","options":{"a":"30 ","b":"26 ","c":"11 ","d":"31","e":"32"},"options_float":{"a":30.0,"b":26.0,"c":11.0,"d":31.0,"e":32.0},"annotated_formula":"add(subtract(multiply(6, 5), 5), 1)","linear_formula":"multiply(n1,n2)|subtract(#0,n2)|add(n0,#1)","chain":"6 * 5<\/gadget>\n30<\/output>\n30 - 5<\/gadget>\n25<\/output>\n25 + 1<\/gadget>\n26<\/output>\n26<\/result>","index":330} +{"problem":"tough and tricky questions : work \/ rate problems . a group of 4 junior lawyers require 7 hours to complete a legal research assignment . how many hours would it take a group of 3 legal assistants to complete the same research assignment assuming that a legal assistant works at two - thirds the rate of a junior lawyer ? source : chili hot gmat","rationale":"# of people times the # of hours : 4 * 7 = 28 - - > 4 lawyers do 28 worksin 7 hours . 3 * 14 \/ 3 = 14 - - > 3 assistants do 14 worksin 4 hours so , since the amount of work the assistants do is half the work the lawyers do , the time will be double , soans a","correct":"a","options":{"a":"14 ","b":"10 ","c":"9 ","d":"6","e":"5"},"options_float":{"a":14.0,"b":10.0,"c":9.0,"d":6.0,"e":5.0},"annotated_formula":"multiply(multiply(divide(const_2, const_3), 3), 7)","linear_formula":"divide(const_2,const_3)|multiply(n2,#0)|multiply(n1,#1)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 3<\/gadget>\n2<\/output>\n2 * 7<\/gadget>\n14<\/output>\n14<\/result>","index":332} +{"problem":"mahesh can do a piece of work in 30 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ?","rationale":"\"work done by mahesh in 30 days = 20 * 1 \/ 30 = 2 \/ 3 remaining work = 1 - 2 \/ 3 = 1 \/ 3 1 \/ 3 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 3 = 90 days answer is a\"","correct":"a","options":{"a":"90 ","b":"25 ","c":"37 ","d":"41","e":"30"},"options_float":{"a":90.0,"b":25.0,"c":37.0,"d":41.0,"e":30.0},"annotated_formula":"divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 30))), 30))","linear_formula":"divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)|","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n20 * (1\/30)<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) \/ 30<\/gadget>\n1\/90 = around 0.011111<\/output>\n1 \/ (1\/90)<\/gadget>\n90<\/output>\n90<\/result>","index":333} +{"problem":"if x and y are numbers such that ( x + 6 ) ( y - 6 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2","rationale":"from ( x + 6 ) ( y - 6 ) = 0 it follows that either x = - 6 or y = 6 . thus either x ^ 2 = 36 or y ^ 2 = 36 . now , if x ^ 2 = 36 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 36 + 0 = 36 . similarly if y ^ 2 = 36 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 + 36 = 36 . answer : d .","correct":"d","options":{"a":"0 ","b":"16 ","c":"25 ","d":"36","e":"49"},"options_float":{"a":0.0,"b":16.0,"c":25.0,"d":36.0,"e":49.0},"annotated_formula":"power(6, 2)","linear_formula":"power(n0,n3)","chain":"6 ** 2<\/gadget>\n36<\/output>\n36<\/result>","index":334} +{"problem":"the difference of a larger number and a smaller number is 6 . the sum of the larger number and twice the smaller is 15 . what is the larger number ?","rationale":"let x be the larger number and y be the smaller number . x - y = 6 x + 2 ( y ) = 15 solve by substitution : y = x - 6 x + 2 ( x - 6 ) = 15 x + 2 x - 12 = 15 3 x = 27 x = 9 the larger number is 9 , so answer c is correct .","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":"divide(add(15, multiply(const_2, 6)), add(const_1, const_2))","linear_formula":"add(const_1,const_2)|multiply(n0,const_2)|add(n1,#1)|divide(#2,#0)","chain":"2 * 6<\/gadget>\n12<\/output>\n15 + 12<\/gadget>\n27<\/output>\n1 + 2<\/gadget>\n3<\/output>\n27 \/ 3<\/gadget>\n9<\/output>\n9<\/result>","index":335} +{"problem":"a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ?","rationale":"\"explanation : length of the platform = speed of train * extra time taken to cross the platform . length of platform = 72 kmph * 12 seconds convert 72 kmph into m \/ sec 1 kmph = 518518 m \/ s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . ∴ 72 kmph = 518 ∗ 72518 ∗ 72 = 20 m \/ sec therefore , length of the platform = 20 m \/ s * 12 sec = 240 meters . correct answer : a\"","correct":"a","options":{"a":"240 meters ","b":"360 meters ","c":"420 meters ","d":"600 meters","e":"can not be determined"},"options_float":{"a":240.0,"b":360.0,"c":420.0,"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), 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 * 30<\/gadget>\n600<\/output>\n20 * 18<\/gadget>\n360<\/output>\n600 - 360<\/gadget>\n240<\/output>\n240<\/result>","index":336} +{"problem":"a vessel of capacity 2 litre has 16 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ?","rationale":"\"16 % of 2 litres = 0.32 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.72 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 27.2 % answer : d\"","correct":"d","options":{"a":"31 % . ","b":"71 % . ","c":"49 % . ","d":"27.2 % .","e":"51 % ."},"options_float":{"a":31.0,"b":71.0,"c":49.0,"d":27.2,"e":51.0},"annotated_formula":"multiply(divide(add(multiply(divide(16, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100)","linear_formula":"divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)|","chain":"16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) * 2<\/gadget>\n8\/25 = around 0.32<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 6<\/gadget>\n12\/5 = around 2.4<\/output>\n(8\/25) + (12\/5)<\/gadget>\n68\/25 = around 2.72<\/output>\n(68\/25) \/ 10<\/gadget>\n34\/125 = around 0.272<\/output>\n(34\/125) * 100<\/gadget>\n136\/5 = around 27.2<\/output>\n136\/5 = around 27.2<\/result>","index":337} +{"problem":"the diameters of two spheres are in the ratio 1 : 2 what is the ratio of their surface area ?","rationale":"1 : 4 answer : b","correct":"b","options":{"a":"1 : 0 ","b":"1 : 4 ","c":"1 : 6 ","d":"1 : 2","e":"1 : 1"},"options_float":{"a":null,"b":0.25,"c":0.1666666667,"d":0.5,"e":1.0},"annotated_formula":"divide(1, const_4)","linear_formula":"divide(n0,const_4)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1\/4 = around 0.25<\/result>","index":338} +{"problem":"there are 3 prizes to be distributed among 10 students . if no students gets more than one prize , then this can be done in ?","rationale":"explanation : 3 prize among 10 students can be distributed in 10 c 3 ways = 120 ways . answer : d","correct":"d","options":{"a":"10 ","b":"45 ","c":"95 ","d":"120","e":"none of these"},"options_float":{"a":10.0,"b":45.0,"c":95.0,"d":120.0,"e":null},"annotated_formula":"add(multiply(10, 3), multiply(subtract(10, const_1), 10))","linear_formula":"multiply(n0,n1)|subtract(n1,const_1)|multiply(n1,#1)|add(#0,#2)","chain":"10 * 3<\/gadget>\n30<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 * 10<\/gadget>\n90<\/output>\n30 + 90<\/gadget>\n120<\/output>\n120<\/result>","index":339} +{"problem":"a student got 72 % in math and 82 % in history . to get an overall average of 75 % , how much should the student get in the third subject ?","rationale":"\"72 + 82 + x = 3 * 75 x = 71 the answer is b .\"","correct":"b","options":{"a":"69 % ","b":"71 % ","c":"73 % ","d":"75 %","e":"77 %"},"options_float":{"a":69.0,"b":71.0,"c":73.0,"d":75.0,"e":77.0},"annotated_formula":"subtract(multiply(const_3, 75), add(72, 82))","linear_formula":"add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)|","chain":"3 * 75<\/gadget>\n225<\/output>\n72 + 82<\/gadget>\n154<\/output>\n225 - 154<\/gadget>\n71<\/output>\n71<\/result>","index":340} +{"problem":"a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ' s contribution in the capital ?","rationale":"let b ' s capital be rs . x . { 3500 \\ 12 } \/ { 7 x } = { 2 } \/ { 3 } = > x = 9000 . answer : d","correct":"d","options":{"a":"rs . 9228 ","b":"rs . 9129 ","c":"rs . 9120 ","d":"rs . 9000","e":"rs . 1922"},"options_float":{"a":9228.0,"b":9129.0,"c":9120.0,"d":9000.0,"e":1922.0},"annotated_formula":"divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 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 - 5<\/gadget>\n7<\/output>\n7 * 2<\/gadget>\n14<\/output>\n126_000 \/ 14<\/gadget>\n9_000<\/output>\n9_000<\/result>","index":341} +{"problem":"a car dealership has 40 cars on the lot , 15 % of which are silver . if the dealership receives a new shipment of 80 cars , 45 % of which are not silver , what percentage of total number of cars are silver ?","rationale":"\"the number of silver cars is 0.15 * 40 + 0.45 * 80 = 42 the percentage of cars which are silver is 42 \/ 120 = 35 % the answer is b .\"","correct":"b","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(divide(add(multiply(40, divide(15, const_100)), multiply(80, divide(45, const_100))), add(40, 80)), const_100)","linear_formula":"add(n0,n2)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n40 * (3\/20)<\/gadget>\n6<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n80 * (9\/20)<\/gadget>\n36<\/output>\n6 + 36<\/gadget>\n42<\/output>\n40 + 80<\/gadget>\n120<\/output>\n42 \/ 120<\/gadget>\n7\/20 = around 0.35<\/output>\n(7\/20) * 100<\/gadget>\n35<\/output>\n35<\/result>","index":343} +{"problem":"a person travels from p to q a speed of 60 km \/ hr and returns by increasing his speed by 20 % . what is his average speed for both the trips ?","rationale":"\"speed on return trip = 120 % of 60 = 72 km \/ hr . average speed of trip = 60 + 72 \/ 2 = 132 \/ 2 = 66 km \/ hr answer : d\"","correct":"d","options":{"a":"33 ","b":"77 ","c":"48 ","d":"66","e":"21"},"options_float":{"a":33.0,"b":77.0,"c":48.0,"d":66.0,"e":21.0},"annotated_formula":"divide(add(multiply(60, add(const_1, divide(20, const_100))), 60), const_2)","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n60 * (6\/5)<\/gadget>\n72<\/output>\n72 + 60<\/gadget>\n132<\/output>\n132 \/ 2<\/gadget>\n66<\/output>\n66<\/result>","index":345} +{"problem":"a and b can together finish a work in 10 days . they worked together for 5 days and then b left . after another 5 days , a finished the remaining work . in how many days a alone can finish the job ?","rationale":"\"a + b 5 days work = 5 * 1 \/ 10 = 1 \/ 2 remaining work = 1 - 1 \/ 2 = 1 \/ 2 1 \/ 2 work is done by a in 5 days whole work will be done by a in 5 * 2 = 10 days answer is a\"","correct":"a","options":{"a":"10 ","b":"15 ","c":"20 ","d":"5","e":"30"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":5.0,"e":30.0},"annotated_formula":"divide(multiply(5, 10), subtract(10, 5))","linear_formula":"multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)|","chain":"5 * 10<\/gadget>\n50<\/output>\n10 - 5<\/gadget>\n5<\/output>\n50 \/ 5<\/gadget>\n10<\/output>\n10<\/result>","index":346} +{"problem":"a rectangular field has a length 10 meters more than it is width . if the area of the field is 200 , what is the length ( in meters ) of the rectangular field ?","rationale":"\"area = l * w = ( l ) * ( l - 10 ) = 200 trial and error : 19 * 9 = 171 ( too low ) 20 * 10 = 200 the length is 20 meters . the answer is a .\"","correct":"a","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":"add(10, add(const_0_25, add(const_0_33, divide(divide(200, 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>\n200 \/ 10<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n(1\/3) + 10<\/gadget>\n31\/3 = around 10.333333<\/output>\n(1\/4) + (31\/3)<\/gadget>\n127\/12 = around 10.583333<\/output>\n10 + (127\/12)<\/gadget>\n247\/12 = around 20.583333<\/output>\n247\/12 = around 20.583333<\/result>","index":347} +{"problem":"a bag contains 7 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 seven green balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ways . 7 \/ 15 answer : e\"","correct":"e","options":{"a":"7 \/ 18 ","b":"7 \/ 19 ","c":"7 \/ 11 ","d":"7 \/ 12","e":"7 \/ 15"},"options_float":{"a":0.3888888889,"b":0.3684210526,"c":0.6363636364,"d":0.5833333333,"e":0.4666666667},"annotated_formula":"add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 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":"7 + 8<\/gadget>\n15<\/output>\n8 \/ 15<\/gadget>\n8\/15 = around 0.533333<\/output>\n8 - 1<\/gadget>\n7<\/output>\n15 - 1<\/gadget>\n14<\/output>\n7 \/ 14<\/gadget>\n1\/2 = around 0.5<\/output>\n(8\/15) * (1\/2)<\/gadget>\n4\/15 = around 0.266667<\/output>\n7 \/ 15<\/gadget>\n7\/15 = around 0.466667<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6 \/ 14<\/gadget>\n3\/7 = around 0.428571<\/output>\n(7\/15) * (3\/7)<\/gadget>\n1\/5 = around 0.2<\/output>\n(4\/15) + (1\/5)<\/gadget>\n7\/15 = around 0.466667<\/output>\n7\/15 = around 0.466667<\/result>","index":348} +{"problem":"if 4 men working 10 hours a day earn rs . 1600 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 ) : : 1600 : x hence 4 * 10 * x = 9 * 6 * 1600 or x = 9 * 6 * 1600 \/ 4 * 10 = 2160 answer : d\"","correct":"d","options":{"a":"rs 840 ","b":"rs 1320 ","c":"rs 1620 ","d":"rs 2160","e":"none of these"},"options_float":{"a":840.0,"b":1320.0,"c":1620.0,"d":2160.0,"e":null},"annotated_formula":"multiply(divide(multiply(9, 6), multiply(4, 10)), 1600)","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) * 1_600<\/gadget>\n2_160<\/output>\n2_160<\/result>","index":349} +{"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 9 more than the correct answer . find the number .","rationale":"\"let the number be x . ( 5 \/ 4 ) * x = ( 4 \/ 5 ) * x + 9 25 x = 16 x + 180 9 x = 180 x = 20 the answer is c .\"","correct":"c","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(divide(multiply(multiply(9, 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>\n9 * (4\/5)<\/gadget>\n36\/5 = around 7.2<\/output>\n(36\/5) * (4\/5)<\/gadget>\n144\/25 = around 5.76<\/output>\n(4\/5) * (4\/5)<\/gadget>\n16\/25 = around 0.64<\/output>\n1 - (16\/25)<\/gadget>\n9\/25 = around 0.36<\/output>\n(144\/25) \/ (9\/25)<\/gadget>\n16<\/output>\n16 \/ (4\/5)<\/gadget>\n20<\/output>\n20<\/result>","index":350} +{"problem":"raman mixed 24 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":"cp per kg of mixture = [ 24 ( 150 ) + 36 ( 125 ) ] \/ ( 24 + 36 ) = rs . 135 sp = cp [ ( 100 + profit % ) \/ 100 ] = 135 * [ ( 100 + 40 ) \/ 100 ] = rs . 189 . answer : c","correct":"c","options":{"a":"rs . 135 ","b":"rs . 162 ","c":"rs . 189 ","d":"rs . 198","e":"none of these"},"options_float":{"a":135.0,"b":162.0,"c":189.0,"d":198.0,"e":null},"annotated_formula":"add(divide(add(multiply(24, 150), multiply(36, 125)), add(36, 24)), multiply(divide(add(multiply(24, 150), multiply(36, 125)), add(36, 24)), 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":"24 * 150<\/gadget>\n3_600<\/output>\n36 * 125<\/gadget>\n4_500<\/output>\n3_600 + 4_500<\/gadget>\n8_100<\/output>\n36 + 24<\/gadget>\n60<\/output>\n8_100 \/ 60<\/gadget>\n135<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n135 * (2\/5)<\/gadget>\n54<\/output>\n135 + 54<\/gadget>\n189<\/output>\n189<\/result>","index":351} +{"problem":"if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 284 pounds . what is jake ’ s present weight , in pounds ?","rationale":"\"j + k = 284 and so k = 284 - j j - 8 = 2 k j - 8 = 2 ( 284 - j ) 3 j = 576 j = 192 the answer is e .\"","correct":"e","options":{"a":"176 ","b":"180 ","c":"184 ","d":"188","e":"192"},"options_float":{"a":176.0,"b":180.0,"c":184.0,"d":188.0,"e":192.0},"annotated_formula":"add(multiply(divide(subtract(284, 8), const_3), const_2), 8)","linear_formula":"subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)|","chain":"284 - 8<\/gadget>\n276<\/output>\n276 \/ 3<\/gadget>\n92<\/output>\n92 * 2<\/gadget>\n184<\/output>\n184 + 8<\/gadget>\n192<\/output>\n192<\/result>","index":352} +{"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 40 days .","rationale":"\"they together completed 4 \/ 10 work in 4 days . balance 6 \/ 10 work will be completed by arun alone in 40 * 6 \/ 10 = 24 days . answer : e\"","correct":"e","options":{"a":"16 days . ","b":"17 days . ","c":"18 days . ","d":"19 days .","e":"24 days ."},"options_float":{"a":16.0,"b":17.0,"c":18.0,"d":19.0,"e":24.0},"annotated_formula":"subtract(40, multiply(divide(40, 10), 4))","linear_formula":"divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)|","chain":"40 \/ 10<\/gadget>\n4<\/output>\n4 * 4<\/gadget>\n16<\/output>\n40 - 16<\/gadget>\n24<\/output>\n24<\/result>","index":353} +{"problem":"a towel , when bleached , lost 30 % of its length and 20 % of its breadth . what is the percentage decrease in area ?","rationale":"percentage change in area = ( − 30 − 20 + ( 30 × 20 ) \/ 100 ) % = − 44 % i . e . , area is decreased by 44 % answer : c","correct":"c","options":{"a":"24 % ","b":"30 % ","c":"44 % ","d":"54 %","e":"64 %"},"options_float":{"a":24.0,"b":30.0,"c":44.0,"d":54.0,"e":64.0},"annotated_formula":"divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), subtract(const_100, 20))), 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 - 20<\/gadget>\n80<\/output>\n70 * 80<\/gadget>\n5_600<\/output>\n10_000 - 5_600<\/gadget>\n4_400<\/output>\n4_400 \/ 100<\/gadget>\n44<\/output>\n44<\/result>","index":355} +{"problem":"a man is 30 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":"\"solution let the son ' s present age be x years . then , man ' s present age = ( x + 30 ) years . then â € ¹ = â € º ( x + 30 ) + 2 = 2 ( x + 2 ) â € ¹ = â € º x + 32 = 2 x + 4 x = 28 . answer b\"","correct":"b","options":{"a":"14 years ","b":"28 years ","c":"20 years ","d":"22 years","e":"none"},"options_float":{"a":14.0,"b":28.0,"c":20.0,"d":22.0,"e":null},"annotated_formula":"divide(subtract(30, 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>\n30 - 2<\/gadget>\n28<\/output>\n2 - 1<\/gadget>\n1<\/output>\n28 \/ 1<\/gadget>\n28<\/output>\n28<\/result>","index":356} +{"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 $ 8.73 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 ) = 8.73 9.9 - 45 x + 36 x = 8.73 9 x = 1.17 x = 0.13 the answer is e .\"","correct":"e","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(8.73, 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>\n8.73 - (198\/25)<\/gadget>\n0.81<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45 - 36<\/gadget>\n9<\/output>\n0.81 \/ 9<\/gadget>\n0.09<\/output>\n(11\/50) - 0.09<\/gadget>\n0.13<\/output>\n0.13 * 100<\/gadget>\n13<\/output>\n13<\/result>","index":357} +{"problem":"a trader bought a car at 20 % 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 = 80 s = 80 * ( 180 \/ 100 ) = 112 100 - 144 = 44 % answer : e\"","correct":"e","options":{"a":"18 % ","b":"13 % ","c":"12 % ","d":"32 %","e":"44 %"},"options_float":{"a":18.0,"b":13.0,"c":12.0,"d":32.0,"e":44.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 20), 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 - 20<\/gadget>\n80<\/output>\n100 + 80<\/gadget>\n180<\/output>\n80 * 180<\/gadget>\n14_400<\/output>\n14_400 \/ 100<\/gadget>\n144<\/output>\n144 \/ 100<\/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":358} +{"problem":"800 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ?","rationale":"\"800 * 15 = 1000 * x x = 12 answer : e\"","correct":"e","options":{"a":"11.5 ","b":"12.5 ","c":"10.5 ","d":"11","e":"12"},"options_float":{"a":11.5,"b":12.5,"c":10.5,"d":11.0,"e":12.0},"annotated_formula":"divide(multiply(15, 800), add(800, 200))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"15 * 800<\/gadget>\n12_000<\/output>\n800 + 200<\/gadget>\n1_000<\/output>\n12_000 \/ 1_000<\/gadget>\n12<\/output>\n12<\/result>","index":359} +{"problem":"n = 1 ! + 2 ! + 3 ! . . . . . + 10 ! . what is the last digit of n ^ n ?","rationale":"as you see 5 ! till 10 ! each unit digit is zero . so 1 ! + 2 ! + 3 ! + 4 ! = 33 so unit digit 3 + 0 = 3 n = 3 n ^ n = 3 ^ 3 = 27 so last digit is 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":"power(subtract(10, 3), const_1)","linear_formula":"subtract(n3,n2)|power(#0,const_1)","chain":"10 - 3<\/gadget>\n7<\/output>\n7 ** 1<\/gadget>\n7<\/output>\n7<\/result>","index":360} +{"problem":"by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 250 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ?","rationale":"i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 250 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 250 ) \/ 100 + ( 1.8 * 700 ) \/ 100 = 14.6 gms % of liquid x in resultant mixture = ( 14.6 \/ 1000 ) * 100 = 1.46 % a","correct":"a","options":{"a":"1.46 % ","b":"1.93 % ","c":"10 % ","d":"15 %","e":"19 %"},"options_float":{"a":1.46,"b":1.93,"c":10.0,"d":15.0,"e":19.0},"annotated_formula":"divide(add(multiply(250, 0.8), multiply(700, 1.8)), const_1000)","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)","chain":"250 * 0.8<\/gadget>\n200<\/output>\n700 * 1.8<\/gadget>\n1_260<\/output>\n200 + 1_260<\/gadget>\n1_460<\/output>\n1_460 \/ 1_000<\/gadget>\n73\/50 = around 1.46<\/output>\n73\/50 = around 1.46<\/result>","index":361} +{"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 3 , 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 3 then total increased to 62 + 3 = 65 new avg - 65 \/ 10 = 6.5 . hence answer is a .\"","correct":"a","options":{"a":"6.5 ","b":"6.7 ","c":"6.8 ","d":"6.85","e":"6.9"},"options_float":{"a":6.5,"b":6.7,"c":6.8,"d":6.85,"e":6.9},"annotated_formula":"divide(add(multiply(10, 6.2), 3), 10)","linear_formula":"multiply(n0,n1)|add(n2,#0)|divide(#1,n0)|","chain":"10 * 6.2<\/gadget>\n62<\/output>\n62 + 3<\/gadget>\n65<\/output>\n65 \/ 10<\/gadget>\n13\/2 = around 6.5<\/output>\n13\/2 = around 6.5<\/result>","index":362} +{"problem":"a group of people participate in some curriculum , 30 of them practice yoga , 25 study cooking , 15 study weaving , 6 of them study cooking only , 8 of them study both the cooking and yoga , 7 of them participate all curriculums . how many people study both cooking and weaving ?","rationale":"\"both cooking and weaving = 25 - ( 6 + 8 + 7 ) = 4 so , the correct 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(subtract(subtract(25, 8), 7), 6)","linear_formula":"subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)|","chain":"25 - 8<\/gadget>\n17<\/output>\n17 - 7<\/gadget>\n10<\/output>\n10 - 6<\/gadget>\n4<\/output>\n4<\/result>","index":364} +{"problem":"a carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet . if the carpenter were to make a similar sandbox twice as long , twice as wide , and twice as high as the first sandbox , what would be the capacity , in cubic feet , of the second sandbox ?","rationale":"\"a quick note on doubling . when you double a length you have 2 * l 1 . when you double all lengths of a rectangle you have ( 2 * l 1 ) ( 2 * l 2 ) = a . an increase of 2 ^ 2 or 4 . when you double all lengths of a rectangular prism you have ( 2 * l 1 ) ( 2 * l 2 ) ( 2 * l 3 ) = v . an increase of 2 ^ 3 or 8 . this leads to the basic relationship : line : 2 * original size rectangle : 4 * original size rectangular prism : 8 * original size 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(power(const_2, const_3), 10)","linear_formula":"power(const_2,const_3)|multiply(n0,#0)|","chain":"2 ** 3<\/gadget>\n8<\/output>\n8 * 10<\/gadget>\n80<\/output>\n80<\/result>","index":365} +{"problem":"find the remainder of the division ( 3 ^ 50 ) \/ 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 ^ 50 ) \/ 7 gives us a remainder of 2 final answer : c ) 2\"","correct":"c","options":{"a":"5 ","b":"3 ","c":"2 ","d":"1","e":"7"},"options_float":{"a":5.0,"b":3.0,"c":2.0,"d":1.0,"e":7.0},"annotated_formula":"reminder(power(3, 50), 7)","linear_formula":"power(n0,n1)|reminder(#0,n2)|","chain":"3 ** 50<\/gadget>\n717_897_987_691_852_588_770_249<\/output>\n717_897_987_691_852_588_770_249 % 7<\/gadget>\n2<\/output>\n2<\/result>","index":371} +{"problem":"what is the area inscribed by the lines y = 2 , x = 2 , y = 10 - x on an xy - coordinate plane ?","rationale":"first , let ' s graph the lines y = 2 and x = 2 at this point , we need to find the points where the line y = 10 - x intersects the other two lines . for the vertical line , we know that x = 2 , so we ' ll plug x = 2 into the equation y = 10 - x to get y = 10 - 2 = 8 perfect , when x = 2 , y = 8 , so one point of intersection is ( 28 ) for the horizontal line , we know that y = 2 , so we ' ll plug y = 2 into the equation y = 10 - x to get 2 = 10 - x . solve to get : x = 8 so , when y = 2 , x = 8 , so one point of intersection is ( 82 ) now add these points to our graph and sketch the line y = 10 - x at this point , we can see that we have the following triangle . the base has length 6 and the height is 6 area = ( 1 \/ 2 ) ( base ) ( height ) = ( 1 \/ 2 ) ( 6 ) ( 6 ) = 18 answer : e","correct":"e","options":{"a":"8 ","b":"10 ","c":"12 ","d":"14","e":"18"},"options_float":{"a":8.0,"b":10.0,"c":12.0,"d":14.0,"e":18.0},"annotated_formula":"multiply(subtract(subtract(10, 2), 2), multiply(multiply(const_2, const_0_25), subtract(subtract(10, 2), 2)))","linear_formula":"multiply(const_0_25,const_2)|subtract(n2,n0)|subtract(#1,n0)|multiply(#0,#2)|multiply(#3,#2)","chain":"10 - 2<\/gadget>\n8<\/output>\n8 - 2<\/gadget>\n6<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n2 * (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 6<\/gadget>\n3<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":372} +{"problem":"if n is a positive integer , what is the remainder when ( 4 ^ ( 4 n + 3 ) ) ( 6 ^ n ) is divided by 10 ?","rationale":"\"this one took me bout 3 1 \/ 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * 7 = 9 7 : 7 * 9 = 3 pattern repeats every @ 5 . notice every ^ 4 or multiple of 4 is always going to be 1 . this is just for future notice for similar problems . so 7 ^ 4 n + 3 - - - > if n = 1 then its ( ( 7 ^ 7 ) * 6 ) ) \/ 10 which can say is going to be 3 * 8 - - > 18 \/ 10 - - > r = 8 now from here if id double check just to make sure . 7 ^ 4 ( 2 ) + 3 * 6 ^ 2 - - - > 7 ^ 11 * 36 or we can just say again 7 ^ 11 * 6 ( b \/ c we are only interested in the units digit ) . since ^ 12 is going to be 1 that means ^ 11 is going to be 3 ( as taken from our pattern ) so again 3 * 6 = 18 \/ 10 - - - > r = 8 . c or j in this problem .\"","correct":"c","options":{"a":"1 ","b":"2 ","c":"4 ","d":"6","e":"8"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":6.0,"e":8.0},"annotated_formula":"reminder(multiply(multiply(const_3, const_3), 6), 10)","linear_formula":"multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4)|","chain":"3 * 3<\/gadget>\n9<\/output>\n9 * 6<\/gadget>\n54<\/output>\n54 % 10<\/gadget>\n4<\/output>\n4<\/result>","index":374} +{"problem":"the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 20 percent , and on day 3 , it is discounted an additional 40 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?","rationale":"let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 20 % discount = 720 price in day 3 after 40 % discount = 432 so , price in day 3 as percentage of the sale price on day 1 will be = 432 \/ 900 * 100 = > 48 % answer will definitely be ( b )","correct":"b","options":{"a":"28 % ","b":"48 % ","c":"64.8 % ","d":"70 %","e":"72 %"},"options_float":{"a":28.0,"b":48.0,"c":64.8,"d":70.0,"e":72.0},"annotated_formula":"add(multiply(divide(divide(40, const_100), subtract(1, divide(1, 10))), const_100), 2)","linear_formula":"divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(2\/5) \/ (9\/10)<\/gadget>\n4\/9 = around 0.444444<\/output>\n(4\/9) * 100<\/gadget>\n400\/9 = around 44.444444<\/output>\n(400\/9) + 2<\/gadget>\n418\/9 = around 46.444444<\/output>\n418\/9 = around 46.444444<\/result>","index":375} +{"problem":"each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 28 have long fur , 17 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ?","rationale":"\"no of dogs = 45 long fur = 28 brown = 17 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 28 + 17 - both both = 8 answer d\"","correct":"d","options":{"a":"26 ","b":"19 ","c":"11 ","d":"8","e":"6"},"options_float":{"a":26.0,"b":19.0,"c":11.0,"d":8.0,"e":6.0},"annotated_formula":"subtract(add(28, 17), subtract(45, 8))","linear_formula":"add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|","chain":"28 + 17<\/gadget>\n45<\/output>\n45 - 8<\/gadget>\n37<\/output>\n45 - 37<\/gadget>\n8<\/output>\n8<\/result>","index":378} +{"problem":"when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 4 times , what is the probability that it will land twice on heads and twice tails ?","rationale":"must be twice on heads and twice on tails 1 \/ 2 * 1 \/ 2 * 1 \/ 2 * 1 \/ 2 = 1 \/ 16 answer : c","correct":"c","options":{"a":"1 \/ 8 ","b":"1 \/ 4 ","c":"1 \/ 16 ","d":"1 \/ 32","e":"1 \/ 2"},"options_float":{"a":0.125,"b":0.25,"c":0.0625,"d":0.03125,"e":0.5},"annotated_formula":"divide(const_1, power(const_2, 4))","linear_formula":"power(const_2,n0)|divide(const_1,#0)","chain":"2 ** 4<\/gadget>\n16<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n1\/16 = around 0.0625<\/result>","index":379} +{"problem":"the effective annual rate of interest corresponding to a nominal rate of 14 % per annum payable half - yearly is ?","rationale":"\"amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 7 \/ 100 ) 2 ] = rs . 14.49 effective rate = ( 114.49 - 100 ) = 14.49 % answer : d\"","correct":"d","options":{"a":"16.06 % ","b":"16.07 % ","c":"16.08 % ","d":"14.49 %","e":"16.19 %"},"options_float":{"a":16.06,"b":16.07,"c":16.08,"d":14.49,"e":16.19},"annotated_formula":"add(add(divide(14, const_2), divide(14, const_2)), divide(multiply(divide(14, const_2), divide(14, const_2)), const_100))","linear_formula":"divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)|","chain":"14 \/ 2<\/gadget>\n7<\/output>\n7 + 7<\/gadget>\n14<\/output>\n7 * 7<\/gadget>\n49<\/output>\n49 \/ 100<\/gadget>\n49\/100 = around 0.49<\/output>\n14 + (49\/100)<\/gadget>\n1_449\/100 = around 14.49<\/output>\n1_449\/100 = around 14.49<\/result>","index":380} +{"problem":"mary ' s income is 60 percent more than tim ' s income , and tim ' s income is 50 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 = 50 ( 50 percent less than juan ' s income ) ; mary ' s income = 80 ( 60 percent more than tim ' s income ) . thus , mary ' s income ( 80 ) is 80 % of juan ' s income ( 100 ) . answer : d .\"","correct":"d","options":{"a":"124 % ","b":"120 % ","c":"96 % ","d":"80 %","e":"64 %"},"options_float":{"a":124.0,"b":120.0,"c":96.0,"d":80.0,"e":64.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(50, 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":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n(1\/2) * (8\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":381} +{"problem":"the circumferences of the fore and hind - wheels of a carriage are 2 2 \/ 5 and 3 3 \/ 7 meters respectively . a chalk mark is put on the point of contact of each wheel with the ground at any given moment . how far will the carriage have travelled so that their chalk marks may be again on the ground at the same time ?","rationale":"a little reflection will show that chalk marks will touch the ground together for the first time after the wheels have passed over a distance which is the lcm of 2 2 \/ 5 metres and 3 3 \/ 7 metres . lcm of 12 \/ 5 metres and 24 \/ 7 metres = 24 metres . answer is e","correct":"e","options":{"a":"18 metres ","b":"16 metres ","c":"38 metres ","d":"42 metres","e":"24 metres"},"options_float":{"a":18.0,"b":16.0,"c":38.0,"d":42.0,"e":24.0},"annotated_formula":"add(multiply(7, 3), 3)","linear_formula":"multiply(n3,n5)|add(n3,#0)","chain":"7 * 3<\/gadget>\n21<\/output>\n21 + 3<\/gadget>\n24<\/output>\n24<\/result>","index":382} +{"problem":"a 100 - litre mixture of milk and water contains 30.25 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 25 % . what is the value of x ?","rationale":"working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 30.25 % ( 1 - x \/ 100 ) ^ 2 = 25 % * 100 % ( 1 - x \/ 100 ) ^ 2 = 25 \/ 30.25 - - - - - - > ( 1 - x \/ 100 ) ^ 2 = ( 5 \/ 5.5 ) ^ 2 100 - x = 500 \/ 5.5 x = 9.1 . . . ans a","correct":"a","options":{"a":"9.1 litres ","b":"10 litres ","c":"8 litres ","d":"12 litres","e":"6 litres"},"options_float":{"a":9.1,"b":10.0,"c":8.0,"d":12.0,"e":6.0},"annotated_formula":"multiply(100, subtract(const_1, sqrt(divide(25, 30.25))))","linear_formula":"divide(n2,n1)|sqrt(#0)|subtract(const_1,#1)|multiply(n0,#2)","chain":"25 \/ 30.25<\/gadget>\n0.826446<\/output>\n0.826446 ** (1\/2)<\/gadget>\n0.909091<\/output>\n1 - 0.909091<\/gadget>\n0.090909<\/output>\n100 * 0.090909<\/gadget>\n9.0909<\/output>\n9.0909<\/result>","index":385} +{"problem":"m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 12 ) both lie on the line defined by the equation x = ( y \/ 4 ) - ( 2 \/ 5 ) , what is the value of p ?","rationale":"\"x = ( y \/ 4 ) - ( 2 \/ 5 ) , and so y = 4 x + 8 \/ 5 . the slope is 4 . ( n + 12 - n ) \/ ( m + p - m ) = 3 p = 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":"divide(12, 4)","linear_formula":"divide(n0,n1)|","chain":"12 \/ 4<\/gadget>\n3<\/output>\n3<\/result>","index":386} +{"problem":"1000 men have provisions for 15 days . if 300 more men join them , for how many days will the provisions last now ?","rationale":"\"1000 * 15 = 1300 * x x = 11.5 answer : a\"","correct":"a","options":{"a":"11.5 ","b":"12.5 ","c":"12.6 ","d":"12.2","e":"12.1"},"options_float":{"a":11.5,"b":12.5,"c":12.6,"d":12.2,"e":12.1},"annotated_formula":"divide(multiply(15, 1000), add(1000, 300))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"15 * 1_000<\/gadget>\n15_000<\/output>\n1_000 + 300<\/gadget>\n1_300<\/output>\n15_000 \/ 1_300<\/gadget>\n150\/13 = around 11.538462<\/output>\n150\/13 = around 11.538462<\/result>","index":387} +{"problem":"the equation of line a is y = 4 \/ 3 * x - 100 . what is the smallest possible distance in the xy - plane from the point with coordinates ( 0 , 0 ) to any point on line a ?","rationale":"this can be solve in two steps and without any complex calculation . given : equation of line a as y = ( 4 \/ 3 ) x - 100 . so the line intercept the axes at ( 0 , - 100 ) and ( 750 ) . this can be considered a right angle triangle with right angle at ( 00 ) . so base = 100 , height = 75 and hypotenuse = 125 ( by pythagoras triplet ) so a perpendicular from the ( 00 ) to hypotenuse will be the answer . area of triangle = 0.5 * 100 * 75 = 0.5 * 125 * x = > x = 60 ; so answer is 60 = c","correct":"c","options":{"a":"48 ","b":"50 ","c":"60 ","d":"75","e":"100"},"options_float":{"a":48.0,"b":50.0,"c":60.0,"d":75.0,"e":100.0},"annotated_formula":"divide(multiply(100, 3), sqrt(add(power(4, const_2), power(3, const_2))))","linear_formula":"multiply(n1,n2)|power(n0,const_2)|power(n1,const_2)|add(#1,#2)|sqrt(#3)|divide(#0,#4)","chain":"100 * 3<\/gadget>\n300<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n3 ** 2<\/gadget>\n9<\/output>\n16 + 9<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n300 \/ 5<\/gadget>\n60<\/output>\n60<\/result>","index":388} +{"problem":"the average age of applicants for a new job is 31 , with a standard deviation of 6 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ?","rationale":"\"within one standard deviation of the average age means 31 + \/ - 7 25 - - 31 - - 37 number of dif . ages - 25 26 27 28 29 30 31 32 33 34 35 36 37 total = 13 b\"","correct":"b","options":{"a":"8 ","b":"13 ","c":"15 ","d":"18","e":"30"},"options_float":{"a":8.0,"b":13.0,"c":15.0,"d":18.0,"e":30.0},"annotated_formula":"add(multiply(6, const_2), const_1)","linear_formula":"multiply(n1,const_2)|add(#0,const_1)|","chain":"6 * 2<\/gadget>\n12<\/output>\n12 + 1<\/gadget>\n13<\/output>\n13<\/result>","index":389} +{"problem":"20 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ?","rationale":"\"3 w = 2 m 20 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 20 * 21 * 8 = 14 * x * 6 x = 40 answer : c\"","correct":"c","options":{"a":"32 ","b":"87 ","c":"40 ","d":"99","e":"77"},"options_float":{"a":32.0,"b":87.0,"c":40.0,"d":99.0,"e":77.0},"annotated_formula":"add(floor(divide(multiply(multiply(21, 8), multiply(20, 3)), multiply(multiply(21, 2), 6))), const_1)","linear_formula":"multiply(n1,n2)|multiply(n0,n5)|multiply(n4,n6)|multiply(#0,#1)|multiply(n3,#2)|divide(#3,#4)|floor(#5)|add(#6,const_1)|","chain":"21 * 8<\/gadget>\n168<\/output>\n20 * 3<\/gadget>\n60<\/output>\n168 * 60<\/gadget>\n10_080<\/output>\n21 * 2<\/gadget>\n42<\/output>\n42 * 6<\/gadget>\n252<\/output>\n10_080 \/ 252<\/gadget>\n40<\/output>\nfloor(40)<\/gadget>\n40<\/output>\n40 + 1<\/gadget>\n41<\/output>\n41<\/result>","index":390} +{"problem":"machine a produces 100 parts thrice as fast as machine b does . machine b produces 100 parts in 30 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 6 minutes ?","rationale":"machine b produces 100 part in 30 minutes . machine a produces 100 parts thrice as fast as b , so machine a produces 100 parts in 30 \/ 3 = 10 minutes . now , machine a produces 100 parts in 10 minutes which is 100 \/ 10 = 10 parts \/ minute . 10 parts x a total of 6 minutes = 60 d","correct":"d","options":{"a":"20 ","b":"80 ","c":"40 ","d":"60","e":"50"},"options_float":{"a":20.0,"b":80.0,"c":40.0,"d":60.0,"e":50.0},"annotated_formula":"multiply(multiply(divide(100, 30), const_3), 6)","linear_formula":"divide(n0,n2)|multiply(#0,const_3)|multiply(n3,#1)","chain":"100 \/ 30<\/gadget>\n10\/3 = around 3.333333<\/output>\n(10\/3) * 3<\/gadget>\n10<\/output>\n10 * 6<\/gadget>\n60<\/output>\n60<\/result>","index":392} +{"problem":"in the faculty of reverse - engineering , 226 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":"\"solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 515 , so 100 % is approx . 644 . answer is d : 644\"","correct":"d","options":{"a":"515 ","b":"545 ","c":"618 . ","d":"644 .","e":"666"},"options_float":{"a":515.0,"b":545.0,"c":618.0,"d":644.0,"e":666.0},"annotated_formula":"add(226, 423)","linear_formula":"add(n0,n1)|","chain":"226 + 423<\/gadget>\n649<\/output>\n649<\/result>","index":394} +{"problem":"the average of 5 quantities is 9 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ?","rationale":"\"answer : a ( 5 x 9 - 3 x 4 ) \/ 2 = 16.5\"","correct":"a","options":{"a":"16.5 ","b":"10 ","c":"8 ","d":"9.5","e":"none of these"},"options_float":{"a":16.5,"b":10.0,"c":8.0,"d":9.5,"e":null},"annotated_formula":"divide(subtract(multiply(5, 9), multiply(3, 4)), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)|","chain":"5 * 9<\/gadget>\n45<\/output>\n3 * 4<\/gadget>\n12<\/output>\n45 - 12<\/gadget>\n33<\/output>\n33 \/ 2<\/gadget>\n33\/2 = around 16.5<\/output>\n33\/2 = around 16.5<\/result>","index":395} +{"problem":"the average height of 30 students in a class was calculated as 177 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 106 cm . what was the actual average height of the students in the class ?","rationale":"\"the total height was 45 cm too much . the average height should be reduced by 45 cm \/ 30 = 1.5 cm the answer is b .\"","correct":"b","options":{"a":"176.5 cm ","b":"175.5 cm ","c":"174.5 cm ","d":"173.5 cm","e":"172.5 cm"},"options_float":{"a":176.5,"b":175.5,"c":174.5,"d":173.5,"e":172.5},"annotated_formula":"divide(subtract(multiply(30, 177), subtract(151, 106)), 30)","linear_formula":"multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)|","chain":"30 * 177<\/gadget>\n5_310<\/output>\n151 - 106<\/gadget>\n45<\/output>\n5_310 - 45<\/gadget>\n5_265<\/output>\n5_265 \/ 30<\/gadget>\n351\/2 = around 175.5<\/output>\n351\/2 = around 175.5<\/result>","index":397} +{"problem":"what is the max number of rectangular boxes , each measuring 5 inches by 2 inches by 7 inches , that can be packed into a rectangular packing box measuring 15 inches by 20 inches by 35 inches , if all boxes are aligned in the same direction ?","rationale":"\"the 5 inch side should be aligned to the 15 inch side ( 3 layer ) 2 inch side should be aligned to the 20 inch side . ( 10 layer ) 7 inch side should be aligned to the 35 inch side . ( 5 layer ) maximum number of rectangles = 3 * 10 * 5 = 150 answer is d\"","correct":"d","options":{"a":"200 ","b":"350 ","c":"100 ","d":"150","e":"120"},"options_float":{"a":200.0,"b":350.0,"c":100.0,"d":150.0,"e":120.0},"annotated_formula":"divide(multiply(multiply(15, 20), 35), multiply(multiply(5, 2), 7))","linear_formula":"multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)|","chain":"15 * 20<\/gadget>\n300<\/output>\n300 * 35<\/gadget>\n10_500<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 7<\/gadget>\n70<\/output>\n10_500 \/ 70<\/gadget>\n150<\/output>\n150<\/result>","index":400} +{"problem":"the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 288 . find the principal ?","rationale":"\"p = 288 ( 100 \/ 5 ) ^ 2 = > p = 7200 answer : d\"","correct":"d","options":{"a":"2277 ","b":"2667 ","c":"3600 ","d":"7200","e":"1811"},"options_float":{"a":2277.0,"b":2667.0,"c":3600.0,"d":7200.0,"e":1811.0},"annotated_formula":"divide(288, 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>\n288 \/ (1\/25)<\/gadget>\n7_200<\/output>\n7_200<\/result>","index":401} +{"problem":"63.2 is what percent of 867 ?","rationale":"\"we assume that 867 is 100 % assume ' x ' is value we looking for here , 867 = 100 % and x % = 63.2 therefore , 100 \/ x = 867 \/ 63.2 100 \/ x = 13.71 x = 7.29 c\"","correct":"c","options":{"a":"6.9 ","b":"8.99 ","c":"7.29 ","d":"7.98","e":"9.21"},"options_float":{"a":6.9,"b":8.99,"c":7.29,"d":7.98,"e":9.21},"annotated_formula":"multiply(divide(63.2, 867), const_100)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|","chain":"63.2 \/ 867<\/gadget>\n0.072895<\/output>\n0.072895 * 100<\/gadget>\n7.2895<\/output>\n7.2895<\/result>","index":402} +{"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 15 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 \/ 15 work done by p , q and r in 1 day = 1 \/ 10 + 1 \/ 15 = 1 \/ 6 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 = 1 \/ 6 = > work done by p in 1 day = 1 \/ 12 = > work done by q and r in 1 day = 1 \/ 12 hence work done by q in 1 day = 1 \/ 12 â € “ 1 \/ 15 = 1 \/ 60 so q alone can do the work in 60 days answer is e .\"","correct":"e","options":{"a":"20 ","b":"22 ","c":"25 ","d":"27","e":"60"},"options_float":{"a":20.0,"b":22.0,"c":25.0,"d":27.0,"e":60.0},"annotated_formula":"divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 15)), const_2), divide(const_1, 15)))","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 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/10) + (1\/15)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) \/ 2<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) - (1\/15)<\/gadget>\n1\/60 = around 0.016667<\/output>\n1 \/ (1\/60)<\/gadget>\n60<\/output>\n60<\/result>","index":403} +{"problem":"a customer purchased a package of ground beef at a cost of $ 1.80 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 20 percent less than the package of ground beef . what was the cost per pound of the steak ?","rationale":"for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.80 . let x be the price per pound for the steak . then 0.8 x = 180 x = 180 \/ 0.8 = $ 2.25 the answer is c .","correct":"c","options":{"a":"$ 2.05 ","b":"$ 2.15 ","c":"$ 2.25 ","d":"$ 2.35","e":"$ 2.45"},"options_float":{"a":2.05,"b":2.15,"c":2.25,"d":2.35,"e":2.45},"annotated_formula":"divide(1.8, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1)))","linear_formula":"multiply(const_0_25,const_2)|multiply(const_0_33,const_1)|add(#0,#1)|divide(n0,#2)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 1<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/2) + (1\/3)<\/gadget>\n5\/6 = around 0.833333<\/output>\n1.8 \/ (5\/6)<\/gadget>\n2.16<\/output>\n2.16<\/result>","index":406} +{"problem":"at a wedding reception , 125 guests ate chicken and 75 guests ate beef . if exactly 100 guests ate only one of the two types of meat , how many guests ate both types of meat ?","rationale":"say x guests ate both types of meat . ( 125 - x ) + ( 75 - x ) = 100 - - > x = 50 . answer : e .","correct":"e","options":{"a":"5 ","b":"100 ","c":"7 ","d":"4","e":"50"},"options_float":{"a":5.0,"b":100.0,"c":7.0,"d":4.0,"e":50.0},"annotated_formula":"add(subtract(125, 100), subtract(100, 75))","linear_formula":"subtract(n0,n2)|subtract(n2,n1)|add(#0,#1)","chain":"125 - 100<\/gadget>\n25<\/output>\n100 - 75<\/gadget>\n25<\/output>\n25 + 25<\/gadget>\n50<\/output>\n50<\/result>","index":408} +{"problem":"the average temperature for monday , tuesday and wednsday is 36.3 degrees c . the average temperature for tuesday , wednesday and thursday is 36.7 degrees c . if monday ’ s temperature recorded as 39 degrees c , find the thursday ’ s temperature ?","rationale":"explanation : mon + tue + wed temperature = 3 x 36.3 = 108.9 tue + wed temperature = 108.9 – 39 = 69.9 tue + wed + thu temperature = 3 x 36.7 = 110.1 so , thursday ’ s temperature = 110.1 – 69.9 = 40.2 degrees c answer : c","correct":"c","options":{"a":"60.2 degrees c ","b":"50.2 degrees c ","c":"40.2 degrees c ","d":"70.2 degrees c","e":"none of these"},"options_float":{"a":60.2,"b":50.2,"c":40.2,"d":70.2,"e":null},"annotated_formula":"subtract(multiply(36.7, const_3), subtract(multiply(36.3, const_3), 39))","linear_formula":"multiply(n1,const_3)|multiply(n0,const_3)|subtract(#1,n2)|subtract(#0,#2)","chain":"36.7 * 3<\/gadget>\n110.1<\/output>\n36.3 * 3<\/gadget>\n108.9<\/output>\n108.9 - 39<\/gadget>\n69.9<\/output>\n110.1 - 69.9<\/gadget>\n40.2<\/output>\n40.2<\/result>","index":409} +{"problem":"a palindrome is a number that reads the same forward and backward , such as 343 . how many odd , 6 - digit numbers are palindromes ?","rationale":"\"first recognize you only need to consider the first three digits ( because the second three are just the first three flipped ) there are 900 possibilities for the first three digits of a 6 digit number , 100 - 999 inclusive . everything starting with a 1 , 3,5 , 7,9 will be odd , which is 5 \/ 9 ths of the combinations . 5 \/ 9 * 900 = 500 answer : c\"","correct":"c","options":{"a":"400 ","b":"450 ","c":"500 ","d":"900","e":"2500"},"options_float":{"a":400.0,"b":450.0,"c":500.0,"d":900.0,"e":2500.0},"annotated_formula":"divide(power(const_10, divide(6, const_2)), const_2)","linear_formula":"divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)|","chain":"6 \/ 2<\/gadget>\n3<\/output>\n10 ** 3<\/gadget>\n1_000<\/output>\n1_000 \/ 2<\/gadget>\n500<\/output>\n500<\/result>","index":411} +{"problem":"p , q and r have $ 6000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ?","rationale":"\"a 2400 let the amount with r be $ r r = 2 \/ 3 ( total amount with p and q ) r = 2 \/ 3 ( 6000 - r ) = > 3 r = 12000 - 2 r = > 5 r = 12000 = > r = 2400 .\"","correct":"a","options":{"a":"2400 ","b":"2403 ","c":"3998 ","d":"2539","e":"1930"},"options_float":{"a":2400.0,"b":2403.0,"c":3998.0,"d":2539.0,"e":1930.0},"annotated_formula":"divide(multiply(6000, 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>\n6_000 * 4<\/gadget>\n24_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>\n24_000 \/ 10<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":412} +{"problem":"light glows for every 15 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am .","rationale":"\"time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 \/ 15 ) = 331 times . answer : e\"","correct":"e","options":{"a":"380 times ","b":"381 times ","c":"382 times ","d":"392 times","e":"331 times"},"options_float":{"a":380.0,"b":381.0,"c":382.0,"d":392.0,"e":331.0},"annotated_formula":"divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 15)","linear_formula":"add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)|","chain":"2 + 47<\/gadget>\n49<\/output>\n2 + 60<\/gadget>\n62<\/output>\n20 + 62<\/gadget>\n82<\/output>\n82 * 60<\/gadget>\n4_920<\/output>\n49 + 4_920<\/gadget>\n4_969<\/output>\n4_969 \/ 15<\/gadget>\n4_969\/15 = around 331.266667<\/output>\n4_969\/15 = around 331.266667<\/result>","index":413} +{"problem":"a boy goes to his school from his house at a speed of 3 km \/ hr and returns at a speed of 2 km \/ hr . if he takes 5 hours in going and coming . the distance between his house and school is :","rationale":"\"sol . average speed = [ 2 * 3 * 2 \/ 3 + 2 ] km \/ hr = 12 \/ 5 km \/ hr . distance travelled = [ 12 \/ 5 * 5 ] km = 12 km . ∴ distance between house and school = [ 12 \/ 2 ] km = 6 km . answer c\"","correct":"c","options":{"a":"4.5 km ","b":"5.5 km ","c":"6 km ","d":"7 km","e":"none"},"options_float":{"a":4.5,"b":5.5,"c":6.0,"d":7.0,"e":null},"annotated_formula":"multiply(divide(5, add(divide(3, 2), const_1)), 3)","linear_formula":"divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) + 1<\/gadget>\n5\/2 = around 2.5<\/output>\n5 \/ (5\/2)<\/gadget>\n2<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6<\/result>","index":414} +{"problem":"of the 120 passengers on flight 750 , 60 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 \/ 3 of the passengers in first class are male , how many females are there in coach class ?","rationale":"\"number of passengers on flight = 120 number of female passengers = . 6 * 120 = 72 number of passengers in first class = ( 10 \/ 100 ) * 120 = 12 number of passengers in coach class = ( 90 \/ 100 ) * 120 = 108 number of male passengers in first class = 1 \/ 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 72 - 8 = 64 answer d\"","correct":"d","options":{"a":"44 ","b":"48 ","c":"50 ","d":"64","e":"56"},"options_float":{"a":44.0,"b":48.0,"c":50.0,"d":64.0,"e":56.0},"annotated_formula":"subtract(multiply(120, divide(60, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3)))","linear_formula":"divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|divide(#3,n5)|subtract(#3,#4)|subtract(#2,#5)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n120 * (3\/5)<\/gadget>\n72<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n120 * (1\/10)<\/gadget>\n12<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n12 - 4<\/gadget>\n8<\/output>\n72 - 8<\/gadget>\n64<\/output>\n64<\/result>","index":415} +{"problem":"the ratio of radius of a circle and the side of a square is 2 : 7 . find the ratio of their areas :","rationale":"\"radius \/ side = 2 \/ 7 â ‡ ’ area of circle \/ area of square = 4 \/ 49 answer : d\"","correct":"d","options":{"a":"2 : 1 ","b":"4 : 7 ","c":"8 : 77 ","d":"4 : 49","e":"none"},"options_float":{"a":2.0,"b":0.5714285714,"c":0.1038961039,"d":0.0816326531,"e":null},"annotated_formula":"power(divide(2, 7), 2)","linear_formula":"divide(n0,n1)|power(#0,n0)|","chain":"2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n(2\/7) ** 2<\/gadget>\n4\/49 = around 0.081633<\/output>\n4\/49 = around 0.081633<\/result>","index":416} +{"problem":"how many kilograms of sugar costing rs . 9 per kg must be mixed with 27 kg of sugar costing rs . 7 per kg so that there may be a gain of 10 % by selling the mixture at rs . 9.24 per kg ?","rationale":"by the rule of alligation : c . p . of 1 kg sugar of 1 st kind c . p . of 1 kg sugar of 2 nd kind { \\ color { blue } \\ therefore } ratio of quantities of 1 st and 2 nd kind = 14 : 6 = 7 : 3 . let x kg of sugar of 1 st kind be mixed with 27 kg of 2 nd kind . then , 7 : 3 = x : 27 or x = ( 7 x 27 \/ 3 ) = 63 kg . answer : d ) 63 kg","correct":"d","options":{"a":"33 ","b":"39 ","c":"38 ","d":"63","e":"01"},"options_float":{"a":33.0,"b":39.0,"c":38.0,"d":63.0,"e":1.0},"annotated_formula":"divide(subtract(multiply(27, divide(9.24, add(divide(10, const_100), const_1))), multiply(27, 7)), subtract(9, divide(9.24, add(divide(10, const_100), const_1))))","linear_formula":"divide(n3,const_100)|multiply(n1,n2)|add(#0,const_1)|divide(n4,#2)|multiply(n1,#3)|subtract(n0,#3)|subtract(#4,#1)|divide(#6,#5)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n9.24 \/ (11\/10)<\/gadget>\n8.4<\/output>\n27 * 8.4<\/gadget>\n226.8<\/output>\n27 * 7<\/gadget>\n189<\/output>\n226.8 - 189<\/gadget>\n37.8<\/output>\n9 - 8.4<\/gadget>\n0.6<\/output>\n37.8 \/ 0.6<\/gadget>\n63<\/output>\n63<\/result>","index":417} +{"problem":"a sum of money lent out at s . i . amounts to rs . 820 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ?","rationale":"\"s . i for 5 years = ( 1020 - 820 ) = rs . 200 . s . i . for 2 years = 200 \/ 5 * 2 = rs . 80 . principal = ( 820 - 80 ) = rs . 740 . answer : d\"","correct":"d","options":{"a":"rs . 440 ","b":"rs . 500 ","c":"rs . 540 ","d":"rs . 740","e":"rs . 840"},"options_float":{"a":440.0,"b":500.0,"c":540.0,"d":740.0,"e":840.0},"annotated_formula":"subtract(820, multiply(divide(subtract(1020, 820), 5), 2))","linear_formula":"subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|","chain":"1_020 - 820<\/gadget>\n200<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n40 * 2<\/gadget>\n80<\/output>\n820 - 80<\/gadget>\n740<\/output>\n740<\/result>","index":418} +{"problem":"a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 60 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ?","rationale":"\"let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.15 * 1.07 x = 60 x = 48.7 hence , the correct option is e\"","correct":"e","options":{"a":"39.55 $ ","b":"40.63 $ ","c":"41.63 $ ","d":"42.15 $","e":"48.7 $"},"options_float":{"a":39.55,"b":40.63,"c":41.63,"d":42.15,"e":48.7},"annotated_formula":"divide(60, add(divide(add(7, 15), const_100), const_1))","linear_formula":"add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2)|","chain":"7 + 15<\/gadget>\n22<\/output>\n22 \/ 100<\/gadget>\n11\/50 = around 0.22<\/output>\n(11\/50) + 1<\/gadget>\n61\/50 = around 1.22<\/output>\n60 \/ (61\/50)<\/gadget>\n3_000\/61 = around 49.180328<\/output>\n3_000\/61 = around 49.180328<\/result>","index":419} +{"problem":"for 2 consecutive yrs , my incomes are in the ratio of 4 : 7 and expenses in the ratio of 3 : 5 . if my income in the 2 nd yr is rs . 42000 & my expenses in the first yr in rs . 21000 , my total savings for the two - year is","rationale":"sol . income in first year = * x 42000 = rs . 24000 expenses in second year = \\ x 21000 = rs . 35000 total savings = total income - total expenses = ( 42000 + 24000 ) - ( 21000 + 35000 ) = 66000 - s 6000 = rs . 10000 e","correct":"e","options":{"a":"rs . 8000 ","b":"rs . 9000 ","c":"rs . 9800 ","d":"rs . 9900","e":"rs . 10000"},"options_float":{"a":8000.0,"b":9000.0,"c":9800.0,"d":9900.0,"e":10000.0},"annotated_formula":"add(subtract(42000, divide(multiply(21000, 5), 3)), subtract(divide(multiply(42000, 4), 7), 21000))","linear_formula":"multiply(n4,n7)|multiply(n1,n6)|divide(#0,n3)|divide(#1,n2)|subtract(n6,#2)|subtract(#3,n7)|add(#4,#5)","chain":"21_000 * 5<\/gadget>\n105_000<\/output>\n105_000 \/ 3<\/gadget>\n35_000<\/output>\n42_000 - 35_000<\/gadget>\n7_000<\/output>\n42_000 * 4<\/gadget>\n168_000<\/output>\n168_000 \/ 7<\/gadget>\n24_000<\/output>\n24_000 - 21_000<\/gadget>\n3_000<\/output>\n7_000 + 3_000<\/gadget>\n10_000<\/output>\n10_000<\/result>","index":420} +{"problem":"a boat can travel with a speed of 12 km \/ hr in still water . if the speed of the stream is 4 km \/ hr , find the time taken by the boat to go 68 km downstream .","rationale":"\"speed of boat in still water = 12 km \/ hr speed of the stream = 4 km \/ hr speed downstream = ( 12 + 4 ) = 16 km \/ hr time taken to travel 68 km downstream = 68 ⁄ 16 = 17 ⁄ 4 = 4.25 hours answer is a\"","correct":"a","options":{"a":"4.25 hr ","b":"5.25 hr ","c":"8.25 hr ","d":"2.25 hr","e":"2.50 hr"},"options_float":{"a":4.25,"b":5.25,"c":8.25,"d":2.25,"e":2.5},"annotated_formula":"divide(68, add(12, 4))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"12 + 4<\/gadget>\n16<\/output>\n68 \/ 16<\/gadget>\n17\/4 = around 4.25<\/output>\n17\/4 = around 4.25<\/result>","index":422} +{"problem":"the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if saran and david have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ?","rationale":"if saran and david submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combinedno discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 \/ 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) .","correct":"b","options":{"a":"$ 0.32 ","b":"$ 0.40 ","c":"$ 0.45 ","d":"$ 0.48","e":"$ 0.54"},"options_float":{"a":0.32,"b":0.4,"c":0.45,"d":0.48,"e":0.54},"annotated_formula":"divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2)","linear_formula":"multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2)","chain":"80 * 0.02<\/gadget>\n1.6<\/output>\n2 * 1.6<\/gadget>\n3.2<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n160 * (3\/4)<\/gadget>\n120<\/output>\n120 * 0.02<\/gadget>\n2.4<\/output>\n3.2 - 2.4<\/gadget>\n0.8<\/output>\n0.8 \/ 2<\/gadget>\n0.4<\/output>\n0.4<\/result>","index":423} +{"problem":"p is able to do a piece of work in 10 days and q can do the same work in 12 days . if they can work together for 5 days , what is the fraction of work left ?","rationale":"\"explanation : amount of work p can do in 1 day = 1 \/ 10 amount of work q can do in 1 day = 1 \/ 12 amount of work p and q can do in 1 day = 1 \/ 10 + 1 \/ 12 = 11 \/ 60 amount of work p and q can together do in 5 days = 5 × ( 11 \/ 60 ) = 11 \/ 12 fraction of work left = 1 – 11 \/ 12 = 1 \/ 12 answer : option c\"","correct":"c","options":{"a":"7 \/ 12 ","b":"5 \/ 12 ","c":"1 \/ 12 ","d":"3 \/ 12","e":"1 \/ 2"},"options_float":{"a":0.5833333333,"b":0.4166666667,"c":0.0833333333,"d":0.25,"e":0.5},"annotated_formula":"subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 10)), 5))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/12) + (1\/10)<\/gadget>\n11\/60 = around 0.183333<\/output>\n(11\/60) * 5<\/gadget>\n11\/12 = around 0.916667<\/output>\n1 - (11\/12)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1\/12 = around 0.083333<\/result>","index":425} +{"problem":"find the area of a parallelogram with base 24 cm and height 12 cm ?","rationale":"\"area of a parallelogram = base * height = 24 * 12 = 288 cm 2 answer : e\"","correct":"e","options":{"a":"297 cm 2 ","b":"384 cm 2 ","c":"672 cm 2 ","d":"267 cm 2","e":"288 cm 2"},"options_float":{"a":297.0,"b":384.0,"c":672.0,"d":267.0,"e":288.0},"annotated_formula":"multiply(24, 12)","linear_formula":"multiply(n0,n1)|","chain":"24 * 12<\/gadget>\n288<\/output>\n288<\/result>","index":427} +{"problem":"for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 2 \/ 3 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ?","rationale":"p ( of not burning out in a six mnth period ) = 1 \/ 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 2 \/ 3 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 \/ 3 = 1 \/ 3 - - - - > p ( of not burning out in a six mnth period ) = 1 \/ 2 * 1 \/ 3 = 1 \/ 6 - - - > p ( of burning out in a six mnth period ) = 1 - 1 \/ 3 = 2 \/ 3 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 2 \/ 3 * 1 \/ 6 = 2 \/ 7 ans e","correct":"e","options":{"a":"5 \/ 27 ","b":"2 \/ 9 ","c":"1 \/ 3 ","d":"4 \/ 9","e":"2 \/ 7"},"options_float":{"a":0.1851851852,"b":0.2222222222,"c":0.3333333333,"d":0.4444444444,"e":0.2857142857},"annotated_formula":"multiply(subtract(1, divide(2, 3)), subtract(1, divide(subtract(1, divide(2, 3)), 2)))","linear_formula":"divide(n3,n4)|subtract(n6,#0)|divide(#1,n3)|subtract(n6,#2)|multiply(#1,#3)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) \/ 2<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 - (1\/6)<\/gadget>\n5\/6 = around 0.833333<\/output>\n(1\/3) * (5\/6)<\/gadget>\n5\/18 = around 0.277778<\/output>\n5\/18 = around 0.277778<\/result>","index":428} +{"problem":"a glass was filled with 10 ounces of water , and 0.05 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 0.05 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 0.05 by 20 . this gives us : 0.05 x 20 = 1 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 % ( 1 \/ 10 ) x 100 % ( 10 \/ 100 ) x 100 % = 10 % answer e\"","correct":"e","options":{"a":"0.002 % ","b":"0.02 % ","c":"0.2 % ","d":"2 %","e":"10 %"},"options_float":{"a":0.002,"b":0.02,"c":0.2,"d":2.0,"e":10.0},"annotated_formula":"multiply(divide(multiply(0.05, 20), 10), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"0.05 * 20<\/gadget>\n1<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":429} +{"problem":"a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ' s contribution in the capital","rationale":"\"explanation : let b contribution is x . 3500 * 12 \/ 7 x = 2 \/ 3 = > 14 x = 126000 = > x = rs 9000 option a\"","correct":"a","options":{"a":"rs 9000 ","b":"rs 7000 ","c":"rs 5000 ","d":"rs 4000","e":"none of these"},"options_float":{"a":9000.0,"b":7000.0,"c":5000.0,"d":4000.0,"e":null},"annotated_formula":"divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 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 - 5<\/gadget>\n7<\/output>\n7 * 2<\/gadget>\n14<\/output>\n126_000 \/ 14<\/gadget>\n9_000<\/output>\n9_000<\/result>","index":430} +{"problem":"a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 10 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 \/ 5 - 1 \/ 10 = 1 \/ 10 the cistern will be filled in 10 hr answer is b\"","correct":"b","options":{"a":"20 hr ","b":"10 hr ","c":"5 hr ","d":"4 hr","e":"15 hr"},"options_float":{"a":20.0,"b":10.0,"c":5.0,"d":4.0,"e":15.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 5), divide(const_1, 10)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|","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>\n1\/10 = around 0.1<\/output>\n1 \/ (1\/10)<\/gadget>\n10<\/output>\n10<\/result>","index":433} +{"problem":"the mall charges 50 cents for the first hour of parking and $ 3 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 1 per hour . if a certain customer parked his in the mall for 7 hours and 30 minutes , how much is he going to pay ?","rationale":"charges for 7 hours = ( first hour @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 ) charges for 7 hours = ( 1 @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 ) charges for 7 hours = ( $ 0.50 ) + ( $ 9 ) + ( $ 3.5 ) charges for 7 hours = ( $ 0.50 ) + ( $ 9 ) + ( $ 3.50 ) charges for 7 hours = $ 13 hence correct answer must be ( c )","correct":"c","options":{"a":"$ 11.5 . ","b":"$ 12 . ","c":"$ 13 . ","d":"$ 14.5","e":"$ 15 ."},"options_float":{"a":11.5,"b":12.0,"c":13.0,"d":14.5,"e":15.0},"annotated_formula":"add(add(multiply(3, 3), multiply(add(subtract(7, 4), divide(50, const_100)), 1)), divide(50, const_100))","linear_formula":"divide(n0,const_100)|multiply(n1,n1)|subtract(n4,n2)|add(#0,#2)|multiply(n3,#3)|add(#1,#4)|add(#5,#0)","chain":"3 * 3<\/gadget>\n9<\/output>\n7 - 4<\/gadget>\n3<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n3 + (1\/2)<\/gadget>\n7\/2 = around 3.5<\/output>\n(7\/2) * 1<\/gadget>\n7\/2 = around 3.5<\/output>\n9 + (7\/2)<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) + (1\/2)<\/gadget>\n13<\/output>\n13<\/result>","index":434} +{"problem":"x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40000 , the amount invested by y is","rationale":"explanation : suppose y invested rs . y . then 40000 \/ y = 2 \/ 3 or y = 60000 . answer : c ) 60000","correct":"c","options":{"a":"33488 ","b":"63809 ","c":"60000 ","d":"37887","e":"77824"},"options_float":{"a":33488.0,"b":63809.0,"c":60000.0,"d":37887.0,"e":77824.0},"annotated_formula":"multiply(divide(multiply(40000, add(2, 3)), 2), divide(3, add(2, 3)))","linear_formula":"add(n0,n1)|divide(n1,#0)|multiply(n2,#0)|divide(#2,n0)|multiply(#3,#1)","chain":"2 + 3<\/gadget>\n5<\/output>\n40_000 * 5<\/gadget>\n200_000<\/output>\n200_000 \/ 2<\/gadget>\n100_000<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n100_000 * (3\/5)<\/gadget>\n60_000<\/output>\n60_000<\/result>","index":435} +{"problem":"an article with cost price of 245 is sold at 34 % profit . what is the selling price ?","rationale":"sp = 1.34 * 245 = 328 answer : a","correct":"a","options":{"a":"328 ","b":"320 ","c":"300 ","d":"207","e":"310"},"options_float":{"a":328.0,"b":320.0,"c":300.0,"d":207.0,"e":310.0},"annotated_formula":"add(245, multiply(245, divide(34, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)","chain":"34 \/ 100<\/gadget>\n17\/50 = around 0.34<\/output>\n245 * (17\/50)<\/gadget>\n833\/10 = around 83.3<\/output>\n245 + (833\/10)<\/gadget>\n3_283\/10 = around 328.3<\/output>\n3_283\/10 = around 328.3<\/result>","index":437} +{"problem":"two stations p and q are 155 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ?","rationale":"\"assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 155 = > 20 x + 25 ( x - 1 ) = 155 = > 45 x = 180 = > x = 4 means , they meet after 3 hours after 7 am , ie , they meet at 11 am answer is e .\"","correct":"e","options":{"a":"10 am ","b":"12 am ","c":"10.30 am ","d":"12.30 am","e":"11 am"},"options_float":{"a":10.0,"b":12.0,"c":10.3,"d":12.3,"e":11.0},"annotated_formula":"add(divide(add(155, 25), add(20, 25)), 7)","linear_formula":"add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)|","chain":"155 + 25<\/gadget>\n180<\/output>\n20 + 25<\/gadget>\n45<\/output>\n180 \/ 45<\/gadget>\n4<\/output>\n4 + 7<\/gadget>\n11<\/output>\n11<\/result>","index":438} +{"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 age of the team ?","rationale":"\"a 23 years let the average age of the whole team by 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\"","correct":"a","options":{"a":"23 years ","b":"20 years ","c":"25 years ","d":"22 years","e":"28 years"},"options_float":{"a":23.0,"b":20.0,"c":25.0,"d":22.0,"e":28.0},"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":439} +{"problem":"the number of students in each section of a school is 24 . after admitting new students , 3 new sections were started . now , the total number of sections is 16 and there are 21 students in each section . the number of new students admitted is :","rationale":"original number of sections = 16 - 3 = 13 original number of students = 24 x 13 = 312 present number of students = 21 x 16 = 336 number of new students admitted = 336 - 312 = 24 so the answer is option c ) 24 .","correct":"c","options":{"a":"12 ","b":"42 ","c":"24 ","d":"28","e":"26"},"options_float":{"a":12.0,"b":42.0,"c":24.0,"d":28.0,"e":26.0},"annotated_formula":"subtract(multiply(21, 16), multiply(24, subtract(16, 3)))","linear_formula":"multiply(n2,n3)|subtract(n2,n1)|multiply(n0,#1)|subtract(#0,#2)","chain":"21 * 16<\/gadget>\n336<\/output>\n16 - 3<\/gadget>\n13<\/output>\n24 * 13<\/gadget>\n312<\/output>\n336 - 312<\/gadget>\n24<\/output>\n24<\/result>","index":440} +{"problem":"simplify : 81 x 81 + 68 x 68 - 2 x 81 x 68 .","rationale":"= ( 81 ) ^ 2 + ( 68 ) ^ 2 – 2 x 81 x 68 = a ^ 2 + b ^ 2 – 2 ab , where a = 81 , b = 68 = ( a - b ) ^ 2 = ( 81 – 68 ) ^ 2 = ( 13 ) ^ 2 = 169 . answer is a .","correct":"a","options":{"a":"169 ","b":"159 ","c":"189 ","d":"179","e":"219"},"options_float":{"a":169.0,"b":159.0,"c":189.0,"d":179.0,"e":219.0},"annotated_formula":"add(81, 81)","linear_formula":"add(n0,n0)","chain":"81 + 81<\/gadget>\n162<\/output>\n162<\/result>","index":441} +{"problem":"- 24 * 29 + 1240 = ?","rationale":"\"= > - 24 * ( 30 - 1 ) + 1240 ; = > - ( 24 * 30 ) + 24 + 1240 ; = > - 720 + 1264 = 544 . correct option : c\"","correct":"c","options":{"a":"- 544 ","b":"584 ","c":"544 ","d":"345","e":"none of these"},"options_float":{"a":-544.0,"b":584.0,"c":544.0,"d":345.0,"e":null},"annotated_formula":"add(multiply(negate(24), 29), 1240)","linear_formula":"negate(n0)|multiply(n1,#0)|add(n2,#1)|","chain":"-24<\/gadget>\n-24<\/output>\n(-24) * 29<\/gadget>\n-696<\/output>\n(-696) + 1_240<\/gadget>\n544<\/output>\n544<\/result>","index":442} +{"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 . 13750 ?","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 = 13750 × 16 \/ 55 = rs . 4000 . share of z in the profit is rs . 4000 . option b","correct":"b","options":{"a":"3200 ","b":"4000 ","c":"3250 ","d":"3825","e":"3985"},"options_float":{"a":3200.0,"b":4000.0,"c":3250.0,"d":3825.0,"e":3985.0},"annotated_formula":"multiply(multiply(48000, subtract(multiply(const_3, const_4), const_4)), divide(13750, add(add(multiply(36000, multiply(const_3, const_4)), multiply(42000, multiply(const_3, const_4))), multiply(48000, subtract(multiply(const_3, const_4), const_4)))))","linear_formula":"multiply(const_3,const_4)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,const_4)|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>\n13_750 \/ 1_320_000<\/gadget>\n1\/96 = around 0.010417<\/output>\n384_000 * (1\/96)<\/gadget>\n4_000<\/output>\n4_000<\/result>","index":444} +{"problem":"machine a and machine b are each used to manufacture 550 sprockets . it takes machine a 10 hours longer to produce 550 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ?","rationale":"\"machine b : takes x hours to produce 550 sprockets machine a : takes ( x + 10 ) hours to produce 550 sprockets machine b : in 1 hour , b makes 550 \/ x sprockets machine a : in 1 hour , a makes 550 \/ ( x + 10 ) sprockets equating : 1.1 ( 550 \/ ( x + 10 ) ) = 550 \/ x 605 \/ ( x + 10 ) = 550 \/ x 605 x = 550 x + 5500 55 x = 5500 x = 100 a makes 550 \/ ( 110 ) = 5 sprockets per hour answer : b\"","correct":"b","options":{"a":"6 ","b":"5 ","c":"7 ","d":"8","e":"9"},"options_float":{"a":6.0,"b":5.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"divide(550, divide(multiply(multiply(10, 550), divide(add(const_100, 10), const_100)), subtract(multiply(550, divide(add(const_100, 10), const_100)), 550)))","linear_formula":"add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)|","chain":"10 * 550<\/gadget>\n5_500<\/output>\n100 + 10<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n5_500 * (11\/10)<\/gadget>\n6_050<\/output>\n550 * (11\/10)<\/gadget>\n605<\/output>\n605 - 550<\/gadget>\n55<\/output>\n6_050 \/ 55<\/gadget>\n110<\/output>\n550 \/ 110<\/gadget>\n5<\/output>\n5<\/result>","index":445} +{"problem":"how long does a train 100 m long travelling at 60 kmph takes to cross a bridge of 120 m in length ?","rationale":"\"b 13.2 sec d = 100 + 120 = 220 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 220 * 3 \/ 50 = 13.2 sec answer is b\"","correct":"b","options":{"a":"15.8 sec ","b":"13.2 sec ","c":"12.4 sec ","d":"16.8 sec","e":"11.8 sec"},"options_float":{"a":15.8,"b":13.2,"c":12.4,"d":16.8,"e":11.8},"annotated_formula":"divide(add(100, 120), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"100 + 120<\/gadget>\n220<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n220 \/ (50\/3)<\/gadget>\n66\/5 = around 13.2<\/output>\n66\/5 = around 13.2<\/result>","index":446} +{"problem":"a rainstorm increased the amount of water stored in state j reservoirs from 5 billion gallons to 7.5 billion gallons . if the storm increased the amount of water in the reservoirs to 55 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ?","rationale":"after the reservoir is filled to 7.5 gallons the amount of water is at 55 % - which means that 45 % of the reservoir is empty . to figure out what that 45 % is approximate : 7.5 gallons \/ 55 percent = x gallons \/ 45 percent , therefore , x = 6.136 gallons , answer choices e , b , c , d are below 6.136 . we know that the reservoir must be short more than 6.136 gallons , therefore , the only possible choice is a .","correct":"a","options":{"a":"6.9 ","b":"1.4 ","c":"2.5 ","d":"3.0","e":"4.4"},"options_float":{"a":6.9,"b":1.4,"c":2.5,"d":3.0,"e":4.4},"annotated_formula":"divide(divide(multiply(7.5, const_100), 55), const_2)","linear_formula":"multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2)","chain":"7.5 * 100<\/gadget>\n750<\/output>\n750 \/ 55<\/gadget>\n150\/11 = around 13.636364<\/output>\n(150\/11) \/ 2<\/gadget>\n75\/11 = around 6.818182<\/output>\n75\/11 = around 6.818182<\/result>","index":447} +{"problem":"kathleen can paint a room in 2 hours , and anthony can paint an identical room in 3 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ?","rationale":"\"( 1 \/ 2 + 1 \/ 3 ) t = 2 t = 12 \/ 5 answer : c\"","correct":"c","options":{"a":"8 \/ 15 ","b":"4 \/ 3 ","c":"12 \/ 5 ","d":"9 \/ 4","e":"15 \/ 4"},"options_float":{"a":0.5333333333,"b":1.3333333333,"c":2.4,"d":2.25,"e":3.75},"annotated_formula":"multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 3))), 2)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,n0)|","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>\n5\/6 = around 0.833333<\/output>\n1 \/ (5\/6)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 2<\/gadget>\n12\/5 = around 2.4<\/output>\n12\/5 = around 2.4<\/result>","index":449} +{"problem":"a parallelogram has a base that is four time the size of it ' s height . the total area of this parallelogram is 2,304 sq ft . what is the height of the parallelogram ?","rationale":"4 x * x = 2304 = > x = 24 answer : c","correct":"c","options":{"a":"19 ","b":"23 ","c":"24 ","d":"16","e":"17"},"options_float":{"a":19.0,"b":23.0,"c":24.0,"d":16.0,"e":17.0},"annotated_formula":"sqrt(divide(add(add(multiply(const_1000, const_2), multiply(const_100, const_3)), const_4), const_4))","linear_formula":"multiply(const_1000,const_2)|multiply(const_100,const_3)|add(#0,#1)|add(#2,const_4)|divide(#3,const_4)|sqrt(#4)","chain":"1_000 * 2<\/gadget>\n2_000<\/output>\n100 * 3<\/gadget>\n300<\/output>\n2_000 + 300<\/gadget>\n2_300<\/output>\n2_300 + 4<\/gadget>\n2_304<\/output>\n2_304 \/ 4<\/gadget>\n576<\/output>\n576 ** (1\/2)<\/gadget>\n24<\/output>\n24<\/result>","index":452} +{"problem":"how long does a train 150 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ?","rationale":"\"d = 150 + 170 = 320 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 320 * 3 \/ 50 = 19.2 sec answer : b\"","correct":"b","options":{"a":"16.5 ","b":"19.2 ","c":"16.4 ","d":"16.8","e":"16.1"},"options_float":{"a":16.5,"b":19.2,"c":16.4,"d":16.8,"e":16.1},"annotated_formula":"divide(add(150, 170), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"150 + 170<\/gadget>\n320<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n320 \/ (50\/3)<\/gadget>\n96\/5 = around 19.2<\/output>\n96\/5 = around 19.2<\/result>","index":453} +{"problem":"if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 7 } . what is the probability that the median of the set will increase but the range still remains the same ?","rationale":"\"we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the same . the median of the set is ( 3 + 4 ) \/ 2 = 3.5 , thus the number selected must be 4 or greater . for the range to remain the same , the number must be between 2 and 7 inclusive . to satisfy both conditions , the number selected must be 4 , 5 , 6 , or 7 . the probability is 4 \/ 10 = 0.4 the answer is c .\"","correct":"c","options":{"a":"0.2 ","b":"0.3 ","c":"0.4 ","d":"0.5","e":"0.6"},"options_float":{"a":0.2,"b":0.3,"c":0.4,"d":0.5,"e":0.6},"annotated_formula":"divide(const_4, const_10)","linear_formula":"divide(const_4,const_10)|","chain":"4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":454} +{"problem":"a student chose a number , multiplied it by 6 , then subtracted 250 from the result and got 122 . what was the number he chose ?","rationale":"\"let x be the number he chose , then 6 ⋅ x − 250 = 122 6 x = 372 x = 62 correct answer d\"","correct":"d","options":{"a":"59 ","b":"60 ","c":"61 ","d":"62","e":"63"},"options_float":{"a":59.0,"b":60.0,"c":61.0,"d":62.0,"e":63.0},"annotated_formula":"divide(add(122, 250), 6)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"122 + 250<\/gadget>\n372<\/output>\n372 \/ 6<\/gadget>\n62<\/output>\n62<\/result>","index":457} +{"problem":"if the average marks of 3 batches of 55 , 60 and 45 students respectively is 40 , 62 , 58 , then the average marks of all the students is","rationale":"explanation : ( 55 ã — 40 ) + ( 60 ã — 62 ) + ( 45 ã — 58 ) \/ 55 + 60 + 45 8530 \/ 160 = 53.3 option b","correct":"b","options":{"a":"54.48 ","b":"53.31 ","c":"54.6 ","d":"54.58","e":"none of these"},"options_float":{"a":54.48,"b":53.31,"c":54.6,"d":54.58,"e":null},"annotated_formula":"divide(add(add(multiply(55, 40), multiply(60, 62)), multiply(40, 58)), add(add(55, 60), 45))","linear_formula":"add(n1,n2)|multiply(n1,n4)|multiply(n2,n5)|multiply(n4,n6)|add(#1,#2)|add(n3,#0)|add(#4,#3)|divide(#6,#5)","chain":"55 * 40<\/gadget>\n2_200<\/output>\n60 * 62<\/gadget>\n3_720<\/output>\n2_200 + 3_720<\/gadget>\n5_920<\/output>\n40 * 58<\/gadget>\n2_320<\/output>\n5_920 + 2_320<\/gadget>\n8_240<\/output>\n55 + 60<\/gadget>\n115<\/output>\n115 + 45<\/gadget>\n160<\/output>\n8_240 \/ 160<\/gadget>\n103\/2 = around 51.5<\/output>\n103\/2 = around 51.5<\/result>","index":459} +{"problem":"in an election contested by two parties , party d secured 12 % of the total votes more than party r . if party r got 132000 votes , by how many votes did it lose the election ?","rationale":"explanatory answer let the percentage of the total votes secured by party d be x % then the percentage of total votes secured by party r = ( x - 12 ) % as there are only two parties contesting in the election , the sum total of the votes secured by the two parties should total up to 100 % i . e . , x + x - 12 = 100 2 x - 12 = 100 or 2 x = 112 or x = 56 % . if party d got 56 % of the votes , then party got ( 56 - 12 ) = 44 % of the total votes . 44 % of the total votes = 132,000 i . e . , 44 \/ 100 * t = 132,000 = > t = 132000 * 100 \/ 44 = 300,000 votes . the margin by which party r lost the election = 12 % of the total votes = 12 % of 300,000 = 36,000 . the correct choice is ( d )","correct":"d","options":{"a":"240000 ","b":"300000 ","c":"168000 ","d":"36000","e":"24,000"},"options_float":{"a":240000.0,"b":300000.0,"c":168000.0,"d":36000.0,"e":24000.0},"annotated_formula":"multiply(divide(132000, divide(subtract(const_100, 12), const_2)), 12)","linear_formula":"subtract(const_100,n0)|divide(#0,const_2)|divide(n1,#1)|multiply(n0,#2)","chain":"100 - 12<\/gadget>\n88<\/output>\n88 \/ 2<\/gadget>\n44<\/output>\n132_000 \/ 44<\/gadget>\n3_000<\/output>\n3_000 * 12<\/gadget>\n36_000<\/output>\n36_000<\/result>","index":462} +{"problem":"in measuring the sides of a rectangle , one side is taken 4 % in excess , and the other 3 % in deficit . find the error percent in the area calculated from these measurements .","rationale":"\"let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 26 \/ 25 ) x ( 32 \/ 33 ) y = ( 344 \/ 341 ) ( xy ) error in measurement = ( 344 \/ 341 ) xy - xy = ( 3 \/ 341 ) xy error percentage = [ ( 3 \/ 341 ) xy ( 1 \/ xy ) 100 ] % = ( 22 \/ 25 ) % = 0.88 % . answer is e .\"","correct":"e","options":{"a":"0.11 % ","b":"0.7 % ","c":"0.4 % ","d":"0.6 %","e":"0.88 %"},"options_float":{"a":0.11,"b":0.7,"c":0.4,"d":0.6,"e":0.88},"annotated_formula":"subtract(subtract(4, 3), divide(multiply(4, 3), const_100))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|","chain":"4 - 3<\/gadget>\n1<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n1 - (3\/25)<\/gadget>\n22\/25 = around 0.88<\/output>\n22\/25 = around 0.88<\/result>","index":463} +{"problem":"a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many e ways can the test be completed if every question is unanswered ?","rationale":"\"5 choices for each of the 4 questions , thus total e 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":464} +{"problem":"a and b undertake to do a piece of work for $ 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of a ?","rationale":"c ' s 1 day work = ( 1 \/ 3 ) - ( 1 \/ 6 + 1 \/ 8 ) = 1 \/ 24 a : b : c = 1 \/ 6 : 1 \/ 8 : 1 \/ 24 = 4 : 3 : 1 a ' s share = 600 * 4 \/ 8 = $ 300 answer is c","correct":"c","options":{"a":"$ 100 ","b":"$ 150 ","c":"$ 300 ","d":"$ 250","e":"$ 350"},"options_float":{"a":100.0,"b":150.0,"c":300.0,"d":250.0,"e":350.0},"annotated_formula":"multiply(divide(multiply(multiply(3, 8), inverse(6)), add(add(multiply(multiply(3, 8), subtract(inverse(3), add(inverse(6), inverse(8)))), multiply(multiply(3, 8), inverse(6))), multiply(multiply(3, 8), inverse(8)))), 600)","linear_formula":"inverse(n1)|inverse(n3)|inverse(n2)|multiply(n2,n3)|add(#0,#2)|multiply(#0,#3)|multiply(#2,#3)|subtract(#1,#4)|multiply(#3,#7)|add(#8,#5)|add(#9,#6)|divide(#5,#10)|multiply(n0,#11)","chain":"3 * 8<\/gadget>\n24<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n24 * (1\/6)<\/gadget>\n4<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/6) + (1\/8)<\/gadget>\n7\/24 = around 0.291667<\/output>\n(1\/3) - (7\/24)<\/gadget>\n1\/24 = around 0.041667<\/output>\n24 * (1\/24)<\/gadget>\n1<\/output>\n1 + 4<\/gadget>\n5<\/output>\n24 * (1\/8)<\/gadget>\n3<\/output>\n5 + 3<\/gadget>\n8<\/output>\n4 \/ 8<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 600<\/gadget>\n300<\/output>\n300<\/result>","index":465} +{"problem":"two employees x and y are paid a total of rs . 660 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 = 660 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 â ˆ ´ 12 y \/ 10 + y = 660 â ‡ ’ y [ 12 \/ 10 + 1 ] = 660 â ‡ ’ 22 y \/ 10 = 660 â ‡ ’ 22 y = 6600 â ‡ ’ y = 6600 \/ 22 = 600 \/ 2 = rs . 300 e\"","correct":"e","options":{"a":"s . 150 ","b":"s . 200 ","c":"s . 250 ","d":"s . 350","e":"s . 300"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":350.0,"e":300.0},"annotated_formula":"divide(multiply(660, 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>\n660 * 10<\/gadget>\n6_600<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n6_600 \/ 22<\/gadget>\n300<\/output>\n300<\/result>","index":466} +{"problem":"a 250 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":"\"speed = ( 120 + 80 ) km \/ h ( because direction is opposite hence relative velocity is added ) = 500 \/ 9 m \/ s time = 9 sec let the lenght of second train is x total distance covered = 250 + x therefore , d = speed * time thus 250 + x = 500 \/ 9 * 9 x = 500 - 250 = 250 m answer : c\"","correct":"c","options":{"a":"230 m ","b":"240 m ","c":"250 m ","d":"260 m","e":"270 m"},"options_float":{"a":230.0,"b":240.0,"c":250.0,"d":260.0,"e":270.0},"annotated_formula":"subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 250)","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 - 250<\/gadget>\n250<\/output>\n250<\/result>","index":468} +{"problem":"the value of x + x ( xx ) when x = 7","rationale":"x + x ( xx ) put the value of x = 7 in the above expression we get , 7 + 7 ( 77 ) = 7 + 7 ( 7 ã — 7 ) = 7 + 7 ( 49 ) = 7 + 343 = 350 the answer is ( a )","correct":"a","options":{"a":"350 ","b":"346 ","c":"358 ","d":"336","e":"364"},"options_float":{"a":350.0,"b":346.0,"c":358.0,"d":336.0,"e":364.0},"annotated_formula":"add(multiply(7, multiply(7, 7)), 7)","linear_formula":"multiply(n0,n0)|multiply(n0,#0)|add(n0,#1)","chain":"7 * 7<\/gadget>\n49<\/output>\n7 * 49<\/gadget>\n343<\/output>\n343 + 7<\/gadget>\n350<\/output>\n350<\/result>","index":469} +{"problem":"in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 11 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":"\"the class borrowed a total of 40 * 2 = 80 books . the 25 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 11 * 2 = 34 . to maximize the number of books borrowed by 1 student , let ' s assume that 14 students borrowed 3 books and 1 student borrowed the rest . 80 - 34 - 3 * 14 = 4 the maximum number of books borrowed by any student is 4 . the answer is 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(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 11), 2)), 1), 3), add(multiply(12, 1), multiply(11, 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":"40 * 2<\/gadget>\n80<\/output>\n12 * 1<\/gadget>\n12<\/output>\n12 + 11<\/gadget>\n23<\/output>\n23 + 2<\/gadget>\n25<\/output>\n40 - 25<\/gadget>\n15<\/output>\n15 - 1<\/gadget>\n14<\/output>\n14 * 3<\/gadget>\n42<\/output>\n11 * 2<\/gadget>\n22<\/output>\n12 + 22<\/gadget>\n34<\/output>\n42 + 34<\/gadget>\n76<\/output>\n80 - 76<\/gadget>\n4<\/output>\n4<\/result>","index":470} +{"problem":"if ( t - 8 ) is a factor of t ^ 2 - kt - 45 , then k =","rationale":"t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 \/ 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 12 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ) ( t + m ) b","correct":"b","options":{"a":"16 ","b":"12 ","c":"2 ","d":"6","e":"14"},"options_float":{"a":16.0,"b":12.0,"c":2.0,"d":6.0,"e":14.0},"annotated_formula":"add(const_10, 2)","linear_formula":"add(n1,const_10)","chain":"10 + 2<\/gadget>\n12<\/output>\n12<\/result>","index":471} +{"problem":"set a contains all the even numbers between 12 and 50 inclusive . set b contains all the even numbers between 112 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ?","rationale":"set a contains 12,14 , 16 . . . 50 set b contains 112 , 114 , 116 . . . 150 number of terms in each set = 20 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 20 = 2000 answer a","correct":"a","options":{"a":"2000 ","b":"2550 ","c":"5050 ","d":"6275","e":"11325"},"options_float":{"a":2000.0,"b":2550.0,"c":5050.0,"d":6275.0,"e":11325.0},"annotated_formula":"multiply(subtract(112, 12), add(divide(subtract(50, 12), const_2), const_1))","linear_formula":"subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)","chain":"112 - 12<\/gadget>\n100<\/output>\n50 - 12<\/gadget>\n38<\/output>\n38 \/ 2<\/gadget>\n19<\/output>\n19 + 1<\/gadget>\n20<\/output>\n100 * 20<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":472} +{"problem":"how much greater is the combined area in square inches of the front and back of a rectangular sheet of paper measuring 11 inches by 15 inches than that of a rectangular sheet of paper measuring 7.5 inches by 11 inches ?","rationale":"\"let ' s just look at the dimensions ( no calculation needed ) . with dimension 11 the same , the other dimension 15 is twice 7.5 then the area will be double which means 100 % greater . the answer is c .\"","correct":"c","options":{"a":"50 % ","b":"87 % ","c":"100 % ","d":"187 %","e":"200 %"},"options_float":{"a":50.0,"b":87.0,"c":100.0,"d":187.0,"e":200.0},"annotated_formula":"multiply(divide(subtract(multiply(rectangle_area(11, 15), const_2), multiply(rectangle_area(7.5, 11), const_2)), rectangle_area(11, 15)), const_100)","linear_formula":"rectangle_area(n0,n1)|rectangle_area(n0,n2)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#5,const_100)|","chain":"11 * 15<\/gadget>\n165<\/output>\n165 * 2<\/gadget>\n330<\/output>\n7.5 * 11<\/gadget>\n82.5<\/output>\n82.5 * 2<\/gadget>\n165<\/output>\n330 - 165<\/gadget>\n165<\/output>\n165 \/ 165<\/gadget>\n1<\/output>\n1 * 100<\/gadget>\n100<\/output>\n100<\/result>","index":473} +{"problem":"a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 30 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ?","rationale":"\"yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 30 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 15 cents . fraction of spirit is 1 \/ 2 . 15 = k * ( 1 \/ 2 ) k = 30 cost per liter = 30 * ( 1 \/ 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 30 * ( 1 \/ 3 ) * 3 = 50 cents b . 30 cents\"","correct":"b","options":{"a":"13 ","b":"30 ","c":"50 ","d":"51","e":"52"},"options_float":{"a":13.0,"b":30.0,"c":50.0,"d":51.0,"e":52.0},"annotated_formula":"multiply(multiply(30, divide(1, add(1, 2))), add(1, 2))","linear_formula":"add(n0,n4)|divide(n0,#0)|multiply(n2,#1)|multiply(#0,#2)|","chain":"1 + 2<\/gadget>\n3<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n30 * (1\/3)<\/gadget>\n10<\/output>\n10 * 3<\/gadget>\n30<\/output>\n30<\/result>","index":480} +{"problem":"you need to print a document of the area 216 sq cm . condition is 3 cm margin is to be left at both top & bottom and 2 cm at the sides . what is the optimized size of your paper ?","rationale":"let us consider it is a rectangle . so area = 24 * 9 = 216 sq . cm now ( 24 - 3 * 2 ) * ( 9 - 2 * 2 ) = 18 * 5 = 90 sq . cm answer : e","correct":"e","options":{"a":"60 sq . cm ","b":"70 sq . cm ","c":"95 sq . cm ","d":"80 sq . cm","e":"90 sq . cm"},"options_float":{"a":60.0,"b":70.0,"c":95.0,"d":80.0,"e":90.0},"annotated_formula":"multiply(subtract(divide(216, power(const_3, const_2)), multiply(3, const_2)), subtract(power(const_3, const_2), multiply(2, const_2)))","linear_formula":"multiply(n1,const_2)|multiply(n2,const_2)|power(const_3,const_2)|divide(n0,#2)|subtract(#2,#1)|subtract(#3,#0)|multiply(#5,#4)","chain":"3 ** 2<\/gadget>\n9<\/output>\n216 \/ 9<\/gadget>\n24<\/output>\n3 * 2<\/gadget>\n6<\/output>\n24 - 6<\/gadget>\n18<\/output>\n2 * 2<\/gadget>\n4<\/output>\n9 - 4<\/gadget>\n5<\/output>\n18 * 5<\/gadget>\n90<\/output>\n90<\/result>","index":481} +{"problem":"if x \/ y = 5 \/ 3 , then ( x + y ) \/ ( x - y ) = ?","rationale":"\"any x and y satisfying x \/ y = 5 \/ 3 should give the same value for ( x + y ) \/ ( x - y ) . say x = 5 and y = 3 , then ( x + y ) \/ ( x - y ) = ( 5 + 3 ) \/ ( 5 - 3 ) = 4 . answer : a .\"","correct":"a","options":{"a":"4 ","b":"1 \/ 5 ","c":"- 1 \/ 6 ","d":"- 1 \/ 5","e":"- 5"},"options_float":{"a":4.0,"b":0.2,"c":-0.1666666667,"d":-0.2,"e":-5.0},"annotated_formula":"divide(add(5, 3), subtract(5, 3))","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"5 + 3<\/gadget>\n8<\/output>\n5 - 3<\/gadget>\n2<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":483} +{"problem":"a number , when 35 is subtracted from it , reduces to its 80 percent . what is 4 - fifth of that number ?","rationale":"explanation : x - 35 = 80 x \/ 100 = > x = 175 = > 4 x \/ 5 = 4 x 175 \/ 5 = 140 . answer d","correct":"d","options":{"a":"130 ","b":"155 ","c":"490 ","d":"140","e":"160"},"options_float":{"a":130.0,"b":155.0,"c":490.0,"d":140.0,"e":160.0},"annotated_formula":"multiply(divide(4, add(const_4, const_1)), multiply(35, add(const_4, const_1)))","linear_formula":"add(const_1,const_4)|divide(n2,#0)|multiply(n0,#0)|multiply(#1,#2)","chain":"4 + 1<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n35 * 5<\/gadget>\n175<\/output>\n(4\/5) * 175<\/gadget>\n140<\/output>\n140<\/result>","index":484} +{"problem":"a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 40 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ?","rationale":"\"the actual answer is obtained by multiplying 140 % by 70 % and subtracting 100 % from this total . that is : 140 % × 70 % = 98 % ; 98 % − 100 % = - 2 % . answer : b\"","correct":"b","options":{"a":"− 5 % ","b":"− 2 % ","c":"15 % ","d":"20 %","e":"80 %"},"options_float":{"a":-5.0,"b":-2.0,"c":15.0,"d":20.0,"e":80.0},"annotated_formula":"multiply(subtract(multiply(add(const_1, divide(40, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/5) * (7\/10)<\/gadget>\n49\/50 = around 0.98<\/output>\n(49\/50) - 1<\/gadget>\n-1\/50 = around -0.02<\/output>\n(-1\/50) * 100<\/gadget>\n-2<\/output>\n-2<\/result>","index":485} +{"problem":"a library has an average of 425 visitors on sundays and 325 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is :","rationale":"explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 415 * 5 + 325 * 25 ) \/ 30 = 10200 \/ 30 = 340 answer : e ) 340","correct":"e","options":{"a":"140 ","b":"240 ","c":"260 ","d":"280","e":"340"},"options_float":{"a":140.0,"b":240.0,"c":260.0,"d":280.0,"e":340.0},"annotated_formula":"divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 425), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 325)), 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 * 425<\/gadget>\n2_125<\/output>\n30 - 5<\/gadget>\n25<\/output>\n25 * 325<\/gadget>\n8_125<\/output>\n2_125 + 8_125<\/gadget>\n10_250<\/output>\n10_250 \/ 30<\/gadget>\n1_025\/3 = around 341.666667<\/output>\n1_025\/3 = around 341.666667<\/result>","index":486} +{"problem":"the sector of a circle has radius of 14 cm and its perimeter 50 cm . find its central angel ?","rationale":"lte central angle = x perimeter of the sector = length of the arc + 2 ( radius ) 50 = ( x \/ 360 * 2 * 22 \/ 7 * 14 ) + 2 ( 14 ) 50 = 88 x \/ 360 + 28 88 x \/ 360 = 22 88 x = 7920 x = 90 answer : e","correct":"e","options":{"a":"180 o ","b":"225 o ","c":"270 o ","d":"150 o","e":"90 o"},"options_float":{"a":180.0,"b":225.0,"c":270.0,"d":150.0,"e":90.0},"annotated_formula":"multiply(multiply(const_2, divide(multiply(subtract(14, const_3), const_2), add(const_4, const_3))), 14)","linear_formula":"add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)","chain":"14 - 3<\/gadget>\n11<\/output>\n11 * 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) * 14<\/gadget>\n88<\/output>\n88<\/result>","index":488} +{"problem":"a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction q of the sum of the 21 numbers in the list ?","rationale":"this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction q of n to the total would be 4 a \/ 24 a or 1 \/ 6 answer b","correct":"b","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, 4), 21), 4), const_2), 4), const_3))","linear_formula":"divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|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 - 4<\/gadget>\n9<\/output>\n9 - 3<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":489} +{"problem":"if 12 men can reap 120 acres of land in 16 days , how many acres of land can 36 men reap in 32 days ?","rationale":"12 men 120 acres 16 days 36 men ? 32 days 120 * 36 \/ 12 * 32 \/ 16 120 * 3 * 2 120 * 6 = 720 answer : d","correct":"d","options":{"a":"269 ","b":"512 ","c":"369 ","d":"720","e":"450"},"options_float":{"a":269.0,"b":512.0,"c":369.0,"d":720.0,"e":450.0},"annotated_formula":"multiply(120, multiply(divide(36, 12), divide(32, 16)))","linear_formula":"divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2)","chain":"36 \/ 12<\/gadget>\n3<\/output>\n32 \/ 16<\/gadget>\n2<\/output>\n3 * 2<\/gadget>\n6<\/output>\n120 * 6<\/gadget>\n720<\/output>\n720<\/result>","index":490} +{"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 4 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 = ( 4 \/ 50 ) * 100 = 8 % so , x * 8 % = 50 x = 625 ans d .\"","correct":"d","options":{"a":"200 ","b":"325 ","c":"565 ","d":"625","e":"700"},"options_float":{"a":200.0,"b":325.0,"c":565.0,"d":625.0,"e":700.0},"annotated_formula":"divide(50, divide(4, 50))","linear_formula":"divide(n2,n1)|divide(n0,#0)|","chain":"4 \/ 50<\/gadget>\n2\/25 = around 0.08<\/output>\n50 \/ (2\/25)<\/gadget>\n625<\/output>\n625<\/result>","index":491} +{"problem":"an inspector rejects 0.04 % of the meters as defective . how many will he examine to reject 2 ?","rationale":"\"let the number of meters to be examined be x then , 0.04 % of x = 2 ( 4 \/ 100 ) * ( ( 1 \/ 100 ) * x = 2 x = 5000 answer is d\"","correct":"d","options":{"a":"1500 ","b":"2000 ","c":"2500 ","d":"5000","e":"3100"},"options_float":{"a":1500.0,"b":2000.0,"c":2500.0,"d":5000.0,"e":3100.0},"annotated_formula":"divide(multiply(2, const_100), 0.04)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"2 * 100<\/gadget>\n200<\/output>\n200 \/ 0.04<\/gadget>\n5_000<\/output>\n5_000<\/result>","index":494} +{"problem":"john makes $ 50 a week from his job . he earns a raise andnow makes $ 90 a week . what is the % increase ?","rationale":"\"increase = ( 40 \/ 50 ) * 100 = ( 4 \/ 5 ) * 100 = 80 % . d\"","correct":"d","options":{"a":"15 % ","b":"16.66 % ","c":"17.8 % ","d":"80 %","e":"21 %"},"options_float":{"a":15.0,"b":16.66,"c":17.8,"d":80.0,"e":21.0},"annotated_formula":"multiply(divide(subtract(90, 50), 50), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"90 - 50<\/gadget>\n40<\/output>\n40 \/ 50<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":495} +{"problem":"a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 48 now , the son age 5 years back was","rationale":"\"let the son ' s present age be x years . then , ( 48 - x ) = x x = 24 . son ' s age 5 years back = ( 24 - 5 ) = 19 years answer : d\"","correct":"d","options":{"a":"14 ","b":"17 ","c":"11 ","d":"19","e":"99"},"options_float":{"a":14.0,"b":17.0,"c":11.0,"d":19.0,"e":99.0},"annotated_formula":"subtract(divide(48, const_2), 5)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"48 \/ 2<\/gadget>\n24<\/output>\n24 - 5<\/gadget>\n19<\/output>\n19<\/result>","index":496} +{"problem":"a man can row upstream at 25 kmph and downstream at 43 kmph , and then find the speed of the man in still water ?","rationale":"\"us = 25 ds = 43 m = ( 43 + 25 ) \/ 2 = 34 answer : b\"","correct":"b","options":{"a":"86 ","b":"34 ","c":"30 ","d":"15","e":"17"},"options_float":{"a":86.0,"b":34.0,"c":30.0,"d":15.0,"e":17.0},"annotated_formula":"divide(add(25, 43), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"25 + 43<\/gadget>\n68<\/output>\n68 \/ 2<\/gadget>\n34<\/output>\n34<\/result>","index":497} +{"problem":"if | x - 20 | = 40 what is the sum of all the values of x .","rationale":"\"there will be two cases x - 20 = 40 and x - 20 = - 40 solve for x = > x = 40 + 20 = > x = 60 or x = - 40 + 20 = > x = - 20 the sum of both values will be 60 + - 20 = 40 answer is d\"","correct":"d","options":{"a":"0 ","b":"60 ","c":"- 80 ","d":"40","e":"80"},"options_float":{"a":0.0,"b":60.0,"c":-80.0,"d":40.0,"e":80.0},"annotated_formula":"subtract(add(40, 20), subtract(40, 20))","linear_formula":"add(n0,n1)|subtract(n1,n0)|subtract(#0,#1)|","chain":"40 + 20<\/gadget>\n60<\/output>\n40 - 20<\/gadget>\n20<\/output>\n60 - 20<\/gadget>\n40<\/output>\n40<\/result>","index":498} +{"problem":"the length and breadth of a rectangular courtyard is 75 m and 32 m . find the cost of leveling it at the rate of $ 3 per m 2 . also , find the distance covered by a boy to take 4 rounds of the courtyard .","rationale":"length of the courtyard = 75 m breadth of the courtyard = 32 m perimeter of the courtyard = 2 ( 75 + 32 ) m = 2 × 107 m = 214 m distance covered by the boy in taking 4 rounds = 4 × perimeter of courtyard = 4 × 214 = 856 m we know that area of the courtyard = length × breadth = 75 × 32 m 2 = 2400 m 2 for 1 m 2 , the cost of levelling = $ 3 for 2400 m 2 , the cost of levelling = $ 3 × 2400 = $ 7200 answer : e","correct":"e","options":{"a":"3573 ","b":"3455 ","c":"8600 ","d":"7000","e":"7200"},"options_float":{"a":3573.0,"b":3455.0,"c":8600.0,"d":7000.0,"e":7200.0},"annotated_formula":"multiply(3, multiply(75, 32))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)","chain":"75 * 32<\/gadget>\n2_400<\/output>\n3 * 2_400<\/gadget>\n7_200<\/output>\n7_200<\/result>","index":501} +{"problem":"city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 4 : 40 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ?","rationale":"\"train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi \/ 60 mph = 2 hrs 4 : 40 pm + 2 hrs = 6 : 40 pm answer : c . 6 : 40\"","correct":"c","options":{"a":"5 : 00 ","b":"5 : 30 ","c":"6 : 40 ","d":"6 : 30","e":"7 : 00"},"options_float":{"a":null,"b":0.1666666667,"c":0.15,"d":0.2,"e":null},"annotated_formula":"divide(add(4, const_2), 40)","linear_formula":"add(n4,const_2)|divide(#0,n5)|","chain":"4 + 2<\/gadget>\n6<\/output>\n6 \/ 40<\/gadget>\n3\/20 = around 0.15<\/output>\n3\/20 = around 0.15<\/result>","index":502} +{"problem":"a corporation that had $ 2 billion in profits for the year paid out $ 100 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 = [ ( 100 million ) \/ ( 2 billion ) ] * 100 = [ ( 100 * 10 ^ 6 ) \/ ( 2 * 10 ^ 9 ) ] * 100 = ( 50 \/ 1000 ) * 100 = 5 % so answer is ( c )\"","correct":"c","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(100, power(10, add(const_3, const_3))), multiply(2, 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>\n100 * 1_000_000<\/gadget>\n100_000_000<\/output>\n10 ** 9<\/gadget>\n1_000_000_000<\/output>\n2 * 1_000_000_000<\/gadget>\n2_000_000_000<\/output>\n100_000_000 \/ 2_000_000_000<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n5<\/result>","index":503} +{"problem":"a man and a boy complete a work together in 24 days . if for the last 6 days man alone does the work then it is completed in 26 days . how long the boy will take to complete the work alone ?","rationale":"explanation : ( man + boy ) ’ s 1 day ’ s work = 1 \/ 24 their 20 day ’ s work = 1 \/ 24 × 20 = 5 \/ 6 the remaining 1 \/ 6 work is done by the man in 6 days therefore , the man alone will finish the work in 6 × 6 days = 36 days man ’ s 1 day ’ s work = 1 \/ 36 therefore , boy ’ s 1 day ’ s work = 1 \/ 24 – 1 \/ 36 = 3 – 2 \/ 72 = 1 \/ 72 therefore , the boy alone will finish the work in 72 days . answer : option a","correct":"a","options":{"a":"72 days ","b":"20 days ","c":"24 days ","d":"36 days","e":"34 days"},"options_float":{"a":72.0,"b":20.0,"c":24.0,"d":36.0,"e":34.0},"annotated_formula":"add(subtract(26, 6), multiply(26, const_2))","linear_formula":"multiply(n2,const_2)|subtract(n2,n1)|add(#0,#1)","chain":"26 - 6<\/gadget>\n20<\/output>\n26 * 2<\/gadget>\n52<\/output>\n20 + 52<\/gadget>\n72<\/output>\n72<\/result>","index":504} +{"problem":"mathew is planning a vacation trip to london next year from today for 5 days , he has calculated he would need about $ 3000 for expenses , including a round - trip plane ticket from l . a to london . he nets around $ 1500 monthly in gross income , after all bills are paid , he is left with about $ 350 each month free for whatever he desires . how much money would mathew need to evenly save from his $ 350 to have $ 3000 in his bank within 12 months ?","rationale":"answer is ( d ) . if mathew is left with about $ 350 after all expenses each month , he would need to divide the total expense budget to london ( $ 3000 ) by 12 months to determine how much he would need to put away every single month to hit his target . $ 3000 \/ 12 = $ 250 .","correct":"d","options":{"a":"$ 240 ","b":"$ 350 ","c":"$ 217 ","d":"$ 250","e":"$ 340"},"options_float":{"a":240.0,"b":350.0,"c":217.0,"d":250.0,"e":340.0},"annotated_formula":"divide(3000, 12)","linear_formula":"divide(n1,n6)","chain":"3_000 \/ 12<\/gadget>\n250<\/output>\n250<\/result>","index":505} +{"problem":"the present worth of rs . 1014 due in 2 years at 4 % per annum compound interest is","rationale":"\"solution present worth = rs . [ 1014 \/ ( 1 + 4 \/ 100 ) ² ] = rs . ( 1014 x 25 \/ 26 x 25 \/ 26 ) = rs . 937.5 answer b\"","correct":"b","options":{"a":"rs . 150.50 ","b":"rs . 937.5 ","c":"rs . 156.25 ","d":"rs . 158","e":"none"},"options_float":{"a":150.5,"b":937.5,"c":156.25,"d":158.0,"e":null},"annotated_formula":"divide(1014, 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>\n1_014 \/ (676\/625)<\/gadget>\n1_875\/2 = around 937.5<\/output>\n1_875\/2 = around 937.5<\/result>","index":507} +{"problem":"if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 281 pounds . what is jake ’ s present weight , in pounds ?","rationale":"\"j + k = 281 and so k = 281 - j j - 8 = 2 k j - 8 = 2 ( 281 - j ) 3 j = 570 j = 190 the answer is e .\"","correct":"e","options":{"a":"125 ","b":"135 ","c":"140 ","d":"165","e":"190"},"options_float":{"a":125.0,"b":135.0,"c":140.0,"d":165.0,"e":190.0},"annotated_formula":"add(multiply(divide(subtract(281, 8), const_3), const_2), 8)","linear_formula":"subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)|","chain":"281 - 8<\/gadget>\n273<\/output>\n273 \/ 3<\/gadget>\n91<\/output>\n91 * 2<\/gadget>\n182<\/output>\n182 + 8<\/gadget>\n190<\/output>\n190<\/result>","index":509} +{"problem":"the total cost of 100 paper plates and 200 paper cups is $ 8.00 at the same rates what is the total cost of 20 of the plates and 40 of the cups ?","rationale":"\"u dont need to go through all this what u have with u is 100 p + 200 c = $ 8.00 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 1.60 therefore oa is e\"","correct":"e","options":{"a":"$ . 90 ","b":"$ 1.00 ","c":"$ 1.20 ","d":"$ 1.50","e":"$ 1.60"},"options_float":{"a":90.0,"b":1.0,"c":1.2,"d":1.5,"e":1.6},"annotated_formula":"multiply(divide(20, 100), 8.00)","linear_formula":"divide(n3,n0)|multiply(n2,#0)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 8<\/gadget>\n8\/5 = around 1.6<\/output>\n8\/5 = around 1.6<\/result>","index":510} +{"problem":"2 cow ’ s and 5 goats are brought for rs . 1050 . if the average price of a goat be rs . 90 . what is the average price of a cow .","rationale":"\"explanation : average price of a goat = rs . 90 total price of 5 goats = 5 * 90 = rs . 450 but total price of 2 cows and 5 goats = rs . 1050 total price of 2 cows is = 1050 - 450 = 600 average price of a cow = 600 \/ 2 = rs . 300 answer : a\"","correct":"a","options":{"a":"300 ","b":"320 ","c":"330 ","d":"350","e":"375"},"options_float":{"a":300.0,"b":320.0,"c":330.0,"d":350.0,"e":375.0},"annotated_formula":"divide(subtract(1050, multiply(5, 90)), 2)","linear_formula":"multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0)|","chain":"5 * 90<\/gadget>\n450<\/output>\n1_050 - 450<\/gadget>\n600<\/output>\n600 \/ 2<\/gadget>\n300<\/output>\n300<\/result>","index":512} +{"problem":"24 oz of juice p and 25 oz of juice v are mixed to make smothies x and y . the ratio of p to v in smothie x is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie x ?","rationale":"let us now solve for x : ( 4 \/ 5 ) x + ( 1 \/ 6 ) ( 49 - x ) = 24 24 x + 5 ( 49 - x ) = ( 24 ) ( 30 ) 24 x + 245 - 5 x = ( 24 ) ( 30 ) 19 x = 720 - 245 19 x = 475 x = 25 answer : e","correct":"e","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(25, 1)","linear_formula":"multiply(n1,n3)","chain":"25 * 1<\/gadget>\n25<\/output>\n25<\/result>","index":513} +{"problem":"a certain industrial loom weaves 0.13 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 ) 0.13 : 15 : : 1 : x = > 0.13 * x = 15 = > x = 15 \/ 0.13 = > x = 115 answer : b\"","correct":"b","options":{"a":"114 ","b":"115 ","c":"116 ","d":"117","e":"118"},"options_float":{"a":114.0,"b":115.0,"c":116.0,"d":117.0,"e":118.0},"annotated_formula":"divide(15, 0.13)","linear_formula":"divide(n1,n0)|","chain":"15 \/ 0.13<\/gadget>\n115.384615<\/output>\n115.384615<\/result>","index":514} +{"problem":"in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 \/ 8 and picking a straight flower is 1 \/ 2 , then what is the probability of picking a flower which is yellow and straight","rationale":"\"good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 \/ 8 of the garden is green , so , since all the flowers must be either green or yellow , we know that 7 \/ 8 are yellow . we ' re also told there is an equal probability of straight or curved , 1 \/ 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 7 \/ 8 * 1 \/ 2 = 4 \/ 9 , or e .\"","correct":"e","options":{"a":"1 \/ 7 ","b":"1 \/ 8 ","c":"1 \/ 4 ","d":"3 \/ 4","e":"4 \/ 9"},"options_float":{"a":0.1428571429,"b":0.125,"c":0.25,"d":0.75,"e":0.4444444444},"annotated_formula":"multiply(subtract(1, divide(1, 8)), divide(1, 2))","linear_formula":"divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 - (1\/8)<\/gadget>\n7\/8 = around 0.875<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(7\/8) * (1\/2)<\/gadget>\n7\/16 = around 0.4375<\/output>\n7\/16 = around 0.4375<\/result>","index":515} +{"problem":"how many 3 digit positive integers with distinct digits are there , which are not multiples of 10 ?","rationale":"a number not to be a multiple of 10 should not have the units digit of 0 . xxx 9 options for the first digit ( from 1 to 9 inclusive ) . 8 options for the third digit ( from 1 to 9 inclusive minus the one we used for the first digit ) . 8 options for the second digit ( from 0 to 9 inclusive minus 2 digits we used for the first and the third digits ) 9 * 8 * 8 = 576 . answer : a .","correct":"a","options":{"a":"576 ","b":"520 ","c":"504 ","d":"432","e":"348"},"options_float":{"a":576.0,"b":520.0,"c":504.0,"d":432.0,"e":348.0},"annotated_formula":"multiply(multiply(multiply(3, 3), subtract(multiply(3, 3), const_1)), subtract(multiply(3, 3), const_1))","linear_formula":"multiply(n0,n0)|subtract(#0,const_1)|multiply(#0,#1)|multiply(#2,#1)","chain":"3 * 3<\/gadget>\n9<\/output>\n9 - 1<\/gadget>\n8<\/output>\n9 * 8<\/gadget>\n72<\/output>\n72 * 8<\/gadget>\n576<\/output>\n576<\/result>","index":516} +{"problem":"3 pounds of 05 grass seed contain 1 percent herbicide . a different type of grass seed , 20 , which contains 20 percent herbicide , will be mixed with 3 pounds of 05 grass seed . how much grass seed of type 20 should be added to the 3 pounds of 05 grass seed so that the mixture contains 15 percent herbicide ?","rationale":"05 grass seed contains 5 % herbicide and its amount is 3 pound 20 grass seed contains 20 % herbicide and its amount is x when these two types of grass seeds are mixed , their average becomes 15 % thus we have 3 ( 1 ) + x ( 20 ) \/ ( x + 3 ) = 15 3 + 20 x = 15 x + 45 5 x = 42 or x = 8.4 d","correct":"d","options":{"a":"3 ","b":"3.75 ","c":"4.5 ","d":"8.4","e":"9"},"options_float":{"a":3.0,"b":3.75,"c":4.5,"d":8.4,"e":9.0},"annotated_formula":"divide(subtract(multiply(15, 3), 3), subtract(20, 15))","linear_formula":"multiply(n0,n10)|subtract(n3,n10)|subtract(#0,n0)|divide(#2,#1)","chain":"15 * 3<\/gadget>\n45<\/output>\n45 - 3<\/gadget>\n42<\/output>\n20 - 15<\/gadget>\n5<\/output>\n42 \/ 5<\/gadget>\n42\/5 = around 8.4<\/output>\n42\/5 = around 8.4<\/result>","index":517} +{"problem":"if f ( f ( n ) ) + f ( n ) = 2 n + 3 and f ( 0 ) = 1 , what is the value of f ( 2012 ) ?","rationale":"\"put n = 0 then f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 ⇒ ⇒ f ( 1 ) + 1 = 3 ⇒ ⇒ f ( 1 ) = 2 put n = 1 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 ⇒ ⇒ f ( 2 ) + 2 = 5 ⇒ ⇒ f ( 2 ) = 3 put n = 2 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 ⇒ ⇒ f ( 3 ) + 3 = 7 ⇒ ⇒ f ( 3 ) = 4 . . . . . . f ( 2012 ) = 2013 answer : c\"","correct":"c","options":{"a":"222 ","b":"2787 ","c":"2013 ","d":"2778","e":"10222"},"options_float":{"a":222.0,"b":2787.0,"c":2013.0,"d":2778.0,"e":10222.0},"annotated_formula":"add(1, 2012)","linear_formula":"add(n3,n4)|","chain":"1 + 2_012<\/gadget>\n2_013<\/output>\n2_013<\/result>","index":518} +{"problem":"a certain barrel , which is a right circular cylinder , is filled to capacity with 60 gallons of oil . the first barrel is poured into a second barrel , also a right circular cylinder , which is empty . the second barrel is twice as tall as the first barrel and has twice the diameter of the first barrel . if all of the oil in the first barrel is poured into the second barrel , how much empty capacity , in gallons , is left in the second barrel ?","rationale":"radius of first cylinder = r , diameter = 2 r , height = h radius of second cylinder = 2 r , diamter = 2 d and height = 2 h volume of first cylinder = pie ( r ^ 2 ) * h = 60 volume of second cylinder = pie ( 2 r ^ 2 ) 2 h put the value of pie ( r ^ 2 ) * h = 60 in the second cylinder , volume = pie ( r ^ 2 ) * 4 * 2 = 60 * 8 = 480 gallons empty capacity = 420 gallons answer d","correct":"d","options":{"a":"there is no empty capacity ","b":"100 gallons ","c":"300 gallons ","d":"420 gallons","e":"840 gallons"},"options_float":{"a":null,"b":100.0,"c":300.0,"d":420.0,"e":840.0},"annotated_formula":"subtract(multiply(60, power(const_2, const_3)), 60)","linear_formula":"power(const_2,const_3)|multiply(n0,#0)|subtract(#1,n0)","chain":"2 ** 3<\/gadget>\n8<\/output>\n60 * 8<\/gadget>\n480<\/output>\n480 - 60<\/gadget>\n420<\/output>\n420<\/result>","index":519} +{"problem":"concentrated apples juice comes inside a cylinder tube with a radius of 2.5 inches and a height of 15 inches . the tubes are packed into wooden boxes , each with dimensions of 11 inches by 10 inches by 31 inches . how many tubes of concentrated apples juice , at the most , can fit into 3 wooden boxes ?","rationale":"concentrated apples juice comes inside a cylinder tube since height of a tube is 15 inches , the tubes can fit only in one way now , diameter of each tube = 5 inches therefore , 4 * 2 can be put in each wooden box in 3 boxes 3 * 4 * 2 can be accommodated = 24 = a","correct":"a","options":{"a":"24 . ","b":"28 . ","c":"36 . ","d":"42 .","e":"48 ."},"options_float":{"a":24.0,"b":28.0,"c":36.0,"d":42.0,"e":48.0},"annotated_formula":"subtract(divide(multiply(multiply(multiply(11, 10), 31), 3), multiply(multiply(divide(multiply(add(const_10, const_1), const_2), add(const_3, const_4)), power(2.5, const_2)), 15)), 10)","linear_formula":"add(const_1,const_10)|add(const_3,const_4)|multiply(n2,n3)|power(n0,const_2)|multiply(n4,#2)|multiply(#0,const_2)|divide(#5,#1)|multiply(n5,#4)|multiply(#6,#3)|multiply(n1,#8)|divide(#7,#9)|subtract(#10,n3)","chain":"11 * 10<\/gadget>\n110<\/output>\n110 * 31<\/gadget>\n3_410<\/output>\n3_410 * 3<\/gadget>\n10_230<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n3 + 4<\/gadget>\n7<\/output>\n22 \/ 7<\/gadget>\n22\/7 = around 3.142857<\/output>\n2.5 ** 2<\/gadget>\n6.25<\/output>\n(22\/7) * 6.25<\/gadget>\n19.642857<\/output>\n19.642857 * 15<\/gadget>\n294.642855<\/output>\n10_230 \/ 294.642855<\/gadget>\n34.72<\/output>\n34.72 - 10<\/gadget>\n24.72<\/output>\n24.72<\/result>","index":520} +{"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 . 51.84 , the number of the member is the group is :","rationale":"\"explanation : money collected = ( 51.84 x 100 ) paise = 5184 paise . ∴ number of members = √ ( 5184 ) = 72 . answer : c\"","correct":"c","options":{"a":"57 ","b":"67 ","c":"72 ","d":"87","e":"97"},"options_float":{"a":57.0,"b":67.0,"c":72.0,"d":87.0,"e":97.0},"annotated_formula":"sqrt(multiply(51.84, const_100))","linear_formula":"multiply(n0,const_100)|sqrt(#0)|","chain":"51.84 * 100<\/gadget>\n5_184<\/output>\n5_184 ** (1\/2)<\/gadget>\n72<\/output>\n72<\/result>","index":521} +{"problem":"a cube has a volume of 125 cubic feet . if a similar cube is twice as long , twice as wide , and twice as high , then the volume , in cubic feet of such cube is ?","rationale":"\"volume = 125 = side ^ 3 i . e . side of cube = 5 new cube has dimensions 10 , 10 , and 10 as all sides are twice of teh side of first cube volume = 10 * 10 * 10 = 1000 square feet answer : option e\"","correct":"e","options":{"a":"24 ","b":"48 ","c":"64 ","d":"80","e":"1000"},"options_float":{"a":24.0,"b":48.0,"c":64.0,"d":80.0,"e":1000.0},"annotated_formula":"volume_cube(multiply(const_2, cube_edge_by_volume(125)))","linear_formula":"cube_edge_by_volume(n0)|multiply(#0,const_2)|volume_cube(#1)|","chain":"125 ** (1\/3)<\/gadget>\n5<\/output>\n2 * 5<\/gadget>\n10<\/output>\n10 ** 3<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":523} +{"problem":"in an election , candidate a got 65 % 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 = 65 % therefore , the number of valid votes polled in favour of candidate a = 65 % of 476000 = 65 \/ 100 × 476000 = 30940000 \/ 100 = 309400 d )\"","correct":"d","options":{"a":"355600 ","b":"355800 ","c":"356500 ","d":"309400","e":"357000"},"options_float":{"a":355600.0,"b":355800.0,"c":356500.0,"d":309400.0,"e":357000.0},"annotated_formula":"multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(65, 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>\n65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n476_000 * (13\/20)<\/gadget>\n309_400<\/output>\n309_400<\/result>","index":525} +{"problem":"in a manufacturing plant , it takes 36 machines 4 hours of continuous work to fill 6 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ?","rationale":"the choices give away the answer . . 36 machines take 4 hours to fill 8 standard orders . . in next eq we aredoubling the machines from 36 to 72 , but thework is not doubling ( only 1 1 \/ 2 times ) , = 4 * 48 \/ 72 * 12 \/ 6 = 4 ans a","correct":"a","options":{"a":"4 ","b":"6 ","c":"8 ","d":"9","e":"12"},"options_float":{"a":4.0,"b":6.0,"c":8.0,"d":9.0,"e":12.0},"annotated_formula":"divide(divide(multiply(multiply(36, 12), const_4), 72), 6)","linear_formula":"multiply(n0,n4)|multiply(#0,const_4)|divide(#1,n3)|divide(#2,n2)","chain":"36 * 12<\/gadget>\n432<\/output>\n432 * 4<\/gadget>\n1_728<\/output>\n1_728 \/ 72<\/gadget>\n24<\/output>\n24 \/ 6<\/gadget>\n4<\/output>\n4<\/result>","index":527} +{"problem":"how many 3 digit number formed by using 23 , 45 , 67 once such that number is divisible by 15 .","rationale":"4 * 2 * 1 = 8 at one ' s place only 5 will come and at ten ' s place 4 and 7 can be placed , and at 100 th place rest of the 4 digits can come . . . so the answer is 8 answer : a","correct":"a","options":{"a":"8 ","b":"13 ","c":"12 ","d":"20","e":"22"},"options_float":{"a":8.0,"b":13.0,"c":12.0,"d":20.0,"e":22.0},"annotated_formula":"add(divide(divide(45, 3), 3), const_3)","linear_formula":"divide(n2,n0)|divide(#0,n0)|add(#1,const_3)","chain":"45 \/ 3<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n5 + 3<\/gadget>\n8<\/output>\n8<\/result>","index":529} +{"problem":"how many 3 digit positive integers t exist that when divided by 7 leave a remainder of 5 ?","rationale":"\"minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e .\"","correct":"e","options":{"a":"128 ","b":"142 ","c":"143 ","d":"141","e":"129"},"options_float":{"a":128.0,"b":142.0,"c":143.0,"d":141.0,"e":129.0},"annotated_formula":"divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7)","linear_formula":"add(const_10,const_4)|multiply(const_10,const_100)|multiply(n1,#0)|subtract(#1,const_1)|add(n2,#2)|subtract(#3,#4)|divide(#5,n1)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 - 1<\/gadget>\n999<\/output>\n10 + 4<\/gadget>\n14<\/output>\n14 * 7<\/gadget>\n98<\/output>\n98 + 5<\/gadget>\n103<\/output>\n999 - 103<\/gadget>\n896<\/output>\n896 \/ 7<\/gadget>\n128<\/output>\n128<\/result>","index":530} +{"problem":"the average of 10 numbers is calculated as 16 . 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 * 16 + 36 – 26 = 170 = > 170 \/ 10 = 17 a )\"","correct":"a","options":{"a":"17 ","b":"18 ","c":"19 ","d":"22","e":"24"},"options_float":{"a":17.0,"b":18.0,"c":19.0,"d":22.0,"e":24.0},"annotated_formula":"add(16, 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>\n16 + 1<\/gadget>\n17<\/output>\n17<\/result>","index":533} +{"problem":"48 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 : 48 : : 12 : x working hours \/ day 6 : 5 30 x 6 x x = 48 x 5 x 12 x = ( 48 x 5 x 12 ) \/ ( 30 x 6 ) x = 16 answer b\"","correct":"b","options":{"a":"10 ","b":"16 ","c":"13 ","d":"18","e":"19"},"options_float":{"a":10.0,"b":16.0,"c":13.0,"d":18.0,"e":19.0},"annotated_formula":"divide(multiply(multiply(48, 12), 5), multiply(30, 6))","linear_formula":"multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|","chain":"48 * 12<\/gadget>\n576<\/output>\n576 * 5<\/gadget>\n2_880<\/output>\n30 * 6<\/gadget>\n180<\/output>\n2_880 \/ 180<\/gadget>\n16<\/output>\n16<\/result>","index":534} +{"problem":"the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if x 20 = 190 , what is the value of x ?","rationale":"\"ab = ( a - 1 ) ( b - 1 ) x 20 = ( x - 1 ) ( 20 - 1 ) = 190 - - > x - 1 = 10 - - > x = 11 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":"add(divide(190, subtract(20, 1)), 1)","linear_formula":"subtract(n2,n0)|divide(n3,#0)|add(n0,#1)|","chain":"20 - 1<\/gadget>\n19<\/output>\n190 \/ 19<\/gadget>\n10<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11<\/result>","index":541} +{"problem":"how many 5 digit nos are there if the 2 leftmost digits are odd and the digit 4 ca n ' t appear more than once in the number ? could someone please provide a solution using a approach other than ( 1 - x ( none ) ) approach ?","rationale":"a . 4 is used once : oo * * 4 - - > ( 5 * 5 * 9 * 9 ) * 3 : 5 choices for the first digit as there are 5 odd numbers , 5 choices for the second digit for the same reason , 9 choices for one of the two * ( not - 4 digit ) , 9 choices for another * ( not - 4 digit ) , multiplied by 3 as 4 can take the place of any of the three last digits ( oo * * 4 , oo * 4 * , oo 4 * * ) ; b . 4 is not used : oo * * * - - > 5 * 5 * 9 * 9 * 9 : the same logic as above . 5 * 5 * 9 * 9 * 3 + 5 * 5 * 9 * 9 * 9 = 24300 . answer : a","correct":"a","options":{"a":"24300 ","b":"25700 ","c":"26500 ","d":"24400","e":"26300"},"options_float":{"a":24300.0,"b":25700.0,"c":26500.0,"d":24400.0,"e":26300.0},"annotated_formula":"multiply(multiply(multiply(multiply(subtract(multiply(4, 5), 1), 5), 5), 5), const_10)","linear_formula":"multiply(n0,n2)|subtract(#0,n3)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|multiply(#4,const_10)","chain":"4 * 5<\/gadget>\n20<\/output>\n20 - 1<\/gadget>\n19<\/output>\n19 * 5<\/gadget>\n95<\/output>\n95 * 5<\/gadget>\n475<\/output>\n475 * 5<\/gadget>\n2_375<\/output>\n2_375 * 10<\/gadget>\n23_750<\/output>\n23_750<\/result>","index":542} +{"problem":"a cat leaps 6 leaps for every 5 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ?","rationale":"\"given ; 2 dog = 3 cat ; or , dog \/ cat = 3 \/ 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 6 \/ 3 * 5 = 4 : 5 ' ' answer : 4 : 5 ;\"","correct":"a","options":{"a":"4 : 5 ","b":"2 : 3 ","c":"4 : 1 ","d":"1 : 9","e":"3 : 2"},"options_float":{"a":0.8,"b":0.6666666667,"c":4.0,"d":0.1111111111,"e":1.5},"annotated_formula":"divide(multiply(divide(2, 3), 6), 5)","linear_formula":"divide(n2,n3)|multiply(n0,#0)|divide(#1,n1)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 6<\/gadget>\n4<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":544} +{"problem":"a horse is tethered to one corner of a rectangular grassy field 36 m by 20 m with a rope 18 m long . over how much area of the field can it graze ?","rationale":"\"area of the shaded portion = 1 ⁄ 4 × π × ( 18 ) 2 = 254 m 2 answer c\"","correct":"c","options":{"a":"154 cm 2 ","b":"308 m 2 ","c":"254 m 2 ","d":"260 m 2","e":"none of these"},"options_float":{"a":154.0,"b":308.0,"c":254.0,"d":260.0,"e":null},"annotated_formula":"divide(multiply(power(18, const_2), const_pi), const_4)","linear_formula":"power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)|","chain":"18 ** 2<\/gadget>\n324<\/output>\n324 * pi<\/gadget>\n324*pi = around 1_017.87602<\/output>\n(324*pi) \/ 4<\/gadget>\n81*pi = around 254.469005<\/output>\n81*pi = around 254.469005<\/result>","index":545} +{"problem":"if the arithmetic mean of seventy 5 numbers is calculated , it is 35 . if each number is increased by 5 , then mean of new number is ?","rationale":"a . m . of 75 numbers = 35 sum of 75 numbers = 75 * 35 = 2625 total increase = 75 * 5 = 375 increased sum = 2625 + 375 = 3000 increased average = 3000 \/ 75 = 40 . answer : b","correct":"b","options":{"a":"87 ","b":"40 ","c":"37 ","d":"28","e":"26"},"options_float":{"a":87.0,"b":40.0,"c":37.0,"d":28.0,"e":26.0},"annotated_formula":"add(35, 5)","linear_formula":"add(n0,n1)","chain":"35 + 5<\/gadget>\n40<\/output>\n40<\/result>","index":546} +{"problem":"8873 + x = 13200 , then x is ?","rationale":"answer x = 13200 - 8873 = 4327 option : b","correct":"b","options":{"a":"3327 ","b":"4327 ","c":"3337 ","d":"2337","e":"none of these"},"options_float":{"a":3327.0,"b":4327.0,"c":3337.0,"d":2337.0,"e":null},"annotated_formula":"subtract(13200, 8873)","linear_formula":"subtract(n1,n0)","chain":"13_200 - 8_873<\/gadget>\n4_327<\/output>\n4_327<\/result>","index":547} +{"problem":"a box contains 8 pairs of shoes ( 16 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 16 , 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 \/ 15 ( as after taking one at random there are 15 shoes left and only one is the pair of the first one ) = 1 \/ 15 answer : e .\"","correct":"e","options":{"a":"1 \/ 190 ","b":"1 \/ 20 ","c":"1 \/ 19 ","d":"1 \/ 10","e":"1 \/ 15"},"options_float":{"a":0.0052631579,"b":0.05,"c":0.0526315789,"d":0.1,"e":0.0666666667},"annotated_formula":"divide(const_1, subtract(16, const_1))","linear_formula":"subtract(n1,const_1)|divide(const_1,#0)|","chain":"16 - 1<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1\/15 = around 0.066667<\/result>","index":548} +{"problem":"sales price is $ 60 , gross profit is 140 % of cost , what is the value of gross profit ?","rationale":"\"cost + profit = sales cost + ( 140 \/ 100 ) cost = 60 cost = 25 profit = 60 - 25 = 35 answer ( d )\"","correct":"d","options":{"a":"32 ","b":"33 ","c":"39 ","d":"35","e":"42"},"options_float":{"a":32.0,"b":33.0,"c":39.0,"d":35.0,"e":42.0},"annotated_formula":"subtract(60, divide(60, add(const_1, divide(140, const_100))))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|","chain":"140 \/ 100<\/gadget>\n7\/5 = around 1.4<\/output>\n1 + (7\/5)<\/gadget>\n12\/5 = around 2.4<\/output>\n60 \/ (12\/5)<\/gadget>\n25<\/output>\n60 - 25<\/gadget>\n35<\/output>\n35<\/result>","index":549} +{"problem":"a , b , k start from the same place and travel in the same direction at speeds of 30 km \/ hr , 40 km \/ hr , 60 km \/ hr respectively . b starts 6 hours after a . if b and k overtake a at the same instant , how many hours after a did k start ?","rationale":"the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 6 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 6 + t ) = 40 * t t = 18 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 18 \/ 60 = 12 hrs time taken by a = 6 + t = 6 + 18 = 24 hrs time taken by k = 12 hrs so k starts 24 - 12 = 12 hrs after a . ( answer d )","correct":"d","options":{"a":"3 ","b":"4.5 ","c":"4 ","d":"12","e":"5"},"options_float":{"a":3.0,"b":4.5,"c":4.0,"d":12.0,"e":5.0},"annotated_formula":"divide(multiply(30, add(6, divide(multiply(30, 6), subtract(40, 30)))), 60)","linear_formula":"multiply(n0,n3)|subtract(n1,n0)|divide(#0,#1)|add(n3,#2)|multiply(n0,#3)|divide(#4,n2)","chain":"30 * 6<\/gadget>\n180<\/output>\n40 - 30<\/gadget>\n10<\/output>\n180 \/ 10<\/gadget>\n18<\/output>\n6 + 18<\/gadget>\n24<\/output>\n30 * 24<\/gadget>\n720<\/output>\n720 \/ 60<\/gadget>\n12<\/output>\n12<\/result>","index":550} +{"problem":"in an exam 80 % of the boys and 40 % of the girls passed . the number of girls who passed is 120 , which is 2 \/ 3 rd of the number of boys who failed . what is the total number of students who appeared for the exam ?","rationale":"let the number of boys = x , number of girls = y 40 y \/ 100 = 120 y = 300 120 = 2 \/ 3 * 20 x \/ 100 = 2 x \/ 15 x = 900 total = x + y = 300 + 900 = 1200 answer : a","correct":"a","options":{"a":"1200 ","b":"380 ","c":"3800 ","d":"2180","e":"3180"},"options_float":{"a":1200.0,"b":380.0,"c":3800.0,"d":2180.0,"e":3180.0},"annotated_formula":"add(divide(120, multiply(divide(subtract(const_100, 80), const_100), divide(2, 3))), divide(120, divide(40, const_100)))","linear_formula":"divide(n3,n4)|divide(n1,const_100)|subtract(const_100,n0)|divide(#2,const_100)|divide(n2,#1)|multiply(#3,#0)|divide(n2,#5)|add(#6,#4)","chain":"100 - 80<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/5) * (2\/3)<\/gadget>\n2\/15 = around 0.133333<\/output>\n120 \/ (2\/15)<\/gadget>\n900<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n120 \/ (2\/5)<\/gadget>\n300<\/output>\n900 + 300<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":551} +{"problem":"pipe a fills a tank in 42 minutes . pipe b can fill the same tank 6 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 \/ 42 and b ' s rate is 1 \/ 7 . the combined rate is 1 \/ 42 + 1 \/ 7 = 1 \/ 6 the pipes will fill the tank in 6 minutes . 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":"inverse(add(divide(const_1, 42), divide(6, 42)))","linear_formula":"divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 42<\/gadget>\n1\/42 = around 0.02381<\/output>\n6 \/ 42<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/42) + (1\/7)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ (1\/6)<\/gadget>\n6<\/output>\n6<\/result>","index":555} +{"problem":"by weight , liquid x makes up 1.5 percent of solution p and 6.5 percent of solution q . if 200 grams of solution p are mixed with 800 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ?","rationale":"\"the number of grams of liquid x is 1.5 ( 200 ) \/ 100 + 6.5 ( 800 ) \/ 100 = 3 + 52 = 55 grams . 55 \/ 1000 = 5.5 % the answer is c .\"","correct":"c","options":{"a":"4.5 % ","b":"5.0 % ","c":"5.5 % ","d":"5.8 %","e":"6.0 %"},"options_float":{"a":4.5,"b":5.0,"c":5.5,"d":5.8,"e":6.0},"annotated_formula":"multiply(divide(add(const_1, divide(multiply(6.5, 800), const_100)), const_1000), const_100)","linear_formula":"multiply(n1,n3)|divide(#0,const_100)|add(#1,const_1)|divide(#2,const_1000)|multiply(#3,const_100)|","chain":"6.5 * 800<\/gadget>\n5_200<\/output>\n5_200 \/ 100<\/gadget>\n52<\/output>\n1 + 52<\/gadget>\n53<\/output>\n53 \/ 1_000<\/gadget>\n53\/1_000 = around 0.053<\/output>\n(53\/1_000) * 100<\/gadget>\n53\/10 = around 5.3<\/output>\n53\/10 = around 5.3<\/result>","index":556} +{"problem":"900 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ?","rationale":"\"900 * 15 = 1100 * x x = 12.27 . answer : e\"","correct":"e","options":{"a":"12.88 ","b":"12.6 ","c":"12.55 ","d":"12.21","e":"12.27"},"options_float":{"a":12.88,"b":12.6,"c":12.55,"d":12.21,"e":12.27},"annotated_formula":"divide(multiply(15, 900), add(900, 200))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"15 * 900<\/gadget>\n13_500<\/output>\n900 + 200<\/gadget>\n1_100<\/output>\n13_500 \/ 1_100<\/gadget>\n135\/11 = around 12.272727<\/output>\n135\/11 = around 12.272727<\/result>","index":557} +{"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 2 ?","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 b","correct":"b","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(2, 1)","linear_formula":"multiply(n4,n5)","chain":"2 * 1<\/gadget>\n2<\/output>\n2<\/result>","index":558} +{"problem":"a snail , climbing a 24 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?","rationale":"total transaction in two days = 4 - 2 = 2 feet in 20 days it will climb 20 feet on the 21 st day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 21 e","correct":"e","options":{"a":"12 ","b":"16 ","c":"17 ","d":"20","e":"21"},"options_float":{"a":12.0,"b":16.0,"c":17.0,"d":20.0,"e":21.0},"annotated_formula":"subtract(24, 4)","linear_formula":"subtract(n0,n1)","chain":"24 - 4<\/gadget>\n20<\/output>\n20<\/result>","index":559} +{"problem":"a ’ s speed is 25 \/ 18 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":"\"we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 25 meters , b covers 18 meters in that time . so if the race is 25 meters long , when a reaches the finish line , b would be 7 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 7 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 25 meters ) would be the same as the time required by b ( to cover 18 meters ) to reach the finish line . so b should get a head start of 7 \/ 25 th of the race . answer ( c )\"","correct":"c","options":{"a":"1 \/ 18 ","b":"7 \/ 18 ","c":"7 \/ 25 ","d":"3 \/ 25","e":"1 \/ 25"},"options_float":{"a":0.0555555556,"b":0.3888888889,"c":0.28,"d":0.12,"e":0.04},"annotated_formula":"divide(subtract(25, 18), 25)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|","chain":"25 - 18<\/gadget>\n7<\/output>\n7 \/ 25<\/gadget>\n7\/25 = around 0.28<\/output>\n7\/25 = around 0.28<\/result>","index":560} +{"problem":"the marked price of a book is 20 % more than the cost price . after the book is sold , the vendor realizes that he had wrongly raised the cost price by a margin of 25 % . if the marked price of the book is rs . 30 , what is the original cost price of the book ?","rationale":"let the incorrect cost price be c 1 and let the original cost price be c 2 . marked price of book is rs . 30 . it is 20 % more than c 1 . therefore , ( 120 \/ 100 ) x c 1 = 30 or c 1 = 25 . c 1 is more than c 2 margin of 25 % . or c 1 = ( 125 \/ 100 ) c 2 therefore , c 2 = ( 100 \/ 125 ) x 25 = rs 20 answer : d","correct":"d","options":{"a":"rs . 30 ","b":"rs . 25 ","c":"rs . 45 ","d":"rs . 20","e":"rs . 10"},"options_float":{"a":30.0,"b":25.0,"c":45.0,"d":20.0,"e":10.0},"annotated_formula":"divide(divide(30, add(const_1, divide(20, const_100))), add(const_1, divide(25, const_100)))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n2,#2)|divide(#4,#3)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n30 \/ (6\/5)<\/gadget>\n25<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n25 \/ (5\/4)<\/gadget>\n20<\/output>\n20<\/result>","index":561} +{"problem":"the h . c . f . of two numbers is 18 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 ( 18 x 11 ) and ( 18 x 15 ) . larger number = ( 18 x 15 ) = 270 . answer : a\"","correct":"a","options":{"a":"270 ","b":"300 ","c":"299 ","d":"322","e":"345"},"options_float":{"a":270.0,"b":300.0,"c":299.0,"d":322.0,"e":345.0},"annotated_formula":"multiply(18, 15)","linear_formula":"multiply(n0,n2)|","chain":"18 * 15<\/gadget>\n270<\/output>\n270<\/result>","index":562} +{"problem":"the ratio of buses to cars on river road is 1 to 3 . if there are 20 fewer buses than cars on river road , how many cars are on river road ?","rationale":"\"b \/ c = 1 \/ 3 c - b = 20 . . . . . . . . . > b = c - 20 ( c - 20 ) \/ c = 1 \/ 3 testing answers . clearly eliminate abce put c = 30 . . . . . . . . . > ( 30 - 20 ) \/ 30 = 10 \/ 30 = 1 \/ 3 answer : d\"","correct":"d","options":{"a":"100 ","b":"120 ","c":"140 ","d":"30","e":"150"},"options_float":{"a":100.0,"b":120.0,"c":140.0,"d":30.0,"e":150.0},"annotated_formula":"multiply(divide(20, subtract(3, 1)), 3)","linear_formula":"subtract(n1,n0)|divide(n2,#0)|multiply(n1,#1)|","chain":"3 - 1<\/gadget>\n2<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10 * 3<\/gadget>\n30<\/output>\n30<\/result>","index":563} +{"problem":"the total of 324 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 = ( 324 - x ) . 0.20 * ( x ) + 0.25 ( 324 - x ) = 71 = > x = 200 . . answer : d ) 200\"","correct":"d","options":{"a":"238 ","b":"277 ","c":"278 ","d":"200","e":"288"},"options_float":{"a":238.0,"b":277.0,"c":278.0,"d":200.0,"e":288.0},"annotated_formula":"divide(subtract(multiply(324, 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":"324 * 25<\/gadget>\n8_100<\/output>\n71 * 100<\/gadget>\n7_100<\/output>\n8_100 - 7_100<\/gadget>\n1_000<\/output>\n25 - 20<\/gadget>\n5<\/output>\n1_000 \/ 5<\/gadget>\n200<\/output>\n200<\/result>","index":565} +{"problem":"if annual decrease in the population of a town is 5 % and the present number of people is 40000 what will the population be in 2 years ?","rationale":"population in 2 years = 40000 ( 1 - 5 \/ 100 ) ^ 2 = 40000 * 19 * 19 \/ 20 * 20 = 36100 answer is c","correct":"c","options":{"a":"24560 ","b":"26450 ","c":"36100 ","d":"38920","e":"45200"},"options_float":{"a":24560.0,"b":26450.0,"c":36100.0,"d":38920.0,"e":45200.0},"annotated_formula":"multiply(power(divide(subtract(const_100, 5), const_100), 2), 40000)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|power(#1,n2)|multiply(n1,#2)","chain":"100 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n(19\/20) ** 2<\/gadget>\n361\/400 = around 0.9025<\/output>\n(361\/400) * 40_000<\/gadget>\n36_100<\/output>\n36_100<\/result>","index":566} +{"problem":"the price of lunch for 15 people was $ 206.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ?","rationale":"take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 206 is for 115 then we have to calculate for 100 for 115 206 for 100 x so by cross multiplication we get 115 x = 100 * 206 = > x = 100 * 206 \/ 110 by simplifying we get x as 187.27 which is the price of lunch before gratuity so the gratuity is 18.73 so as the question ask the average price person excluding gratuity is 187.27 \/ 15 = 12.48 so our answer is b )","correct":"b","options":{"a":"$ 11.73 ","b":"$ 12.48 ","c":"$ 13.80 ","d":"$ 14.00","e":"$ 15.87"},"options_float":{"a":11.73,"b":12.48,"c":13.8,"d":14.0,"e":15.87},"annotated_formula":"multiply(multiply(divide(206, add(const_100, 15)), const_100), divide(const_1, 15))","linear_formula":"add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3)","chain":"100 + 15<\/gadget>\n115<\/output>\n206 \/ 115<\/gadget>\n206\/115 = around 1.791304<\/output>\n(206\/115) * 100<\/gadget>\n4_120\/23 = around 179.130435<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(4_120\/23) * (1\/15)<\/gadget>\n824\/69 = around 11.942029<\/output>\n824\/69 = around 11.942029<\/result>","index":567} +{"problem":"joe drives 600 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ?","rationale":"\"t 1 = 600 \/ 60 = 10 hours t 2 = 120 \/ 40 = 3 hours t = t 1 + t 2 = 13 hours avg speed = total distance \/ t = 720 \/ 13 = 55 mph = b\"","correct":"b","options":{"a":"42 ","b":"55 ","c":"50 ","d":"54","e":"56"},"options_float":{"a":42.0,"b":55.0,"c":50.0,"d":54.0,"e":56.0},"annotated_formula":"divide(add(600, 120), add(divide(600, 60), divide(120, 40)))","linear_formula":"add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)|","chain":"600 + 120<\/gadget>\n720<\/output>\n600 \/ 60<\/gadget>\n10<\/output>\n120 \/ 40<\/gadget>\n3<\/output>\n10 + 3<\/gadget>\n13<\/output>\n720 \/ 13<\/gadget>\n720\/13 = around 55.384615<\/output>\n720\/13 = around 55.384615<\/result>","index":569} +{"problem":"a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer ’ s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer ’ s suggested retail price o f $ 20.00 ?","rationale":"\"for retail price = $ 20 first maximum discounted price = 20 - 30 % of 20 = 20 - 6 = 14 price after additional discount of 20 % = 14 - 20 % of 14 = 14 - 2.8 = 11.2 answer : option b\"","correct":"b","options":{"a":"$ 10.00 ","b":"$ 11.20 ","c":"$ 14.40 ","d":"$ 16.00","e":"$ 18.00"},"options_float":{"a":10.0,"b":11.2,"c":14.4,"d":16.0,"e":18.0},"annotated_formula":"multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 20.00))","linear_formula":"subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(n3,#3)|multiply(#2,#4)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) * 20<\/gadget>\n14<\/output>\n(4\/5) * 14<\/gadget>\n56\/5 = around 11.2<\/output>\n56\/5 = around 11.2<\/result>","index":570} +{"problem":"if 20 liters of chemical x are added to 80 liters of a mixture that is 15 % chemical x and 85 % chemical y , then what percentage of the resulting mixture is chemical x ?","rationale":"\"the amount of chemical x in the solution is 20 + 0.15 ( 80 ) = 32 liters . 32 liters \/ 100 liters = 32 % the answer is b .\"","correct":"b","options":{"a":"30 % ","b":"32 % ","c":"35 % ","d":"38 %","e":"40 %"},"options_float":{"a":30.0,"b":32.0,"c":35.0,"d":38.0,"e":40.0},"annotated_formula":"add(20, multiply(divide(15, const_100), 80))","linear_formula":"divide(n2,const_100)|multiply(n1,#0)|add(n0,#1)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 80<\/gadget>\n12<\/output>\n20 + 12<\/gadget>\n32<\/output>\n32<\/result>","index":571} +{"problem":"a set consists of 12 numbers , all are even or multiple of 5 . if 4 numbers are even and 10 numbers are multiple of 5 , how many numbers is multiple of 10 ?","rationale":"\"{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 12 = 4 + 10 - { both } + 0 ; { both } = 2 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : c .\"","correct":"c","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":"subtract(12, 10)","linear_formula":"subtract(n0,n3)|","chain":"12 - 10<\/gadget>\n2<\/output>\n2<\/result>","index":572} +{"problem":"two employees x and y are paid a total of rs . 506 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 = 506 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 ∴ 12 y \/ 10 + y = 506 ⇒ y [ 12 \/ 10 + 1 ] = 506 ⇒ 22 y \/ 10 = 506 ⇒ 22 y = 5060 ⇒ y = 5060 \/ 22 = 460 \/ 2 = rs . 230 e\"","correct":"e","options":{"a":"s . 250 ","b":"s . 280 ","c":"s . 290 ","d":"s . 299","e":"s . 230"},"options_float":{"a":250.0,"b":280.0,"c":290.0,"d":299.0,"e":230.0},"annotated_formula":"divide(multiply(506, 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>\n506 * 10<\/gadget>\n5_060<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n5_060 \/ 22<\/gadget>\n230<\/output>\n230<\/result>","index":573} +{"problem":"the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 10 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 4 + 16 x 10 ) ] = ( 225 - 216 ) = 9 years . answer e\"","correct":"e","options":{"a":"9 years ","b":"11 years ","c":"14 years ","d":"21 years","e":"9 years"},"options_float":{"a":9.0,"b":11.0,"c":14.0,"d":21.0,"e":9.0},"annotated_formula":"subtract(multiply(15, 15), add(multiply(4, 14), multiply(10, 16)))","linear_formula":"multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|","chain":"15 * 15<\/gadget>\n225<\/output>\n4 * 14<\/gadget>\n56<\/output>\n10 * 16<\/gadget>\n160<\/output>\n56 + 160<\/gadget>\n216<\/output>\n225 - 216<\/gadget>\n9<\/output>\n9<\/result>","index":576} +{"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 . 40 , find the sum of money ?","rationale":"\"a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 40 41 - - - - ? = > rs . 205 answer : d\"","correct":"d","options":{"a":"288 ","b":"262 ","c":"72 ","d":"205","e":"267"},"options_float":{"a":288.0,"b":262.0,"c":72.0,"d":205.0,"e":267.0},"annotated_formula":"multiply(divide(40, 40), add(add(const_100, 65), 40))","linear_formula":"add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)|","chain":"40 \/ 40<\/gadget>\n1<\/output>\n100 + 65<\/gadget>\n165<\/output>\n165 + 40<\/gadget>\n205<\/output>\n1 * 205<\/gadget>\n205<\/output>\n205<\/result>","index":577} +{"problem":"two trains leave the train station at the same time . one train , on the blue line , heads east - while the other , on the red line , heads west . if the train on the blue line averages 40 km \/ hr and the other train averages 40 km \/ hr - how long will it take for the trains to be 100 km apart ?","rationale":"each train is averaging 40 km \/ hour in an opposite direction . after 1 hour , they will be 80 km apart , and after 1.25 hours , they will be 100 km apart . ( 80 * 1.25 = 100 ) answer is d","correct":"d","options":{"a":"2 hours ","b":"2.25 hours ","c":"1 hour ","d":"1.25 hours","e":"not enough information"},"options_float":{"a":2.0,"b":2.25,"c":1.0,"d":1.25,"e":null},"annotated_formula":"divide(divide(100, const_2), 40)","linear_formula":"divide(n2,const_2)|divide(#0,n0)","chain":"100 \/ 2<\/gadget>\n50<\/output>\n50 \/ 40<\/gadget>\n5\/4 = around 1.25<\/output>\n5\/4 = around 1.25<\/result>","index":578} +{"problem":"a certain college ' s enrollment at the beginning of 1992 was 30 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 10 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ?","rationale":"\"suppose enrollment in 1991 was 100 then enrollment in 1992 will be 130 and enrollment in 1993 will be 130 * 1.1 = 143 increase in 1993 from 1991 = 143 - 100 = 43 answer : a\"","correct":"a","options":{"a":"43 % ","b":"45 % ","c":"50 % ","d":"35 %","e":"38 %"},"options_float":{"a":43.0,"b":45.0,"c":50.0,"d":35.0,"e":38.0},"annotated_formula":"subtract(multiply(add(const_100, 30), divide(add(const_100, 10), const_100)), const_100)","linear_formula":"add(n1,const_100)|add(n4,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)|","chain":"100 + 30<\/gadget>\n130<\/output>\n100 + 10<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n130 * (11\/10)<\/gadget>\n143<\/output>\n143 - 100<\/gadget>\n43<\/output>\n43<\/result>","index":579} +{"problem":"a can give b 120 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 meters while b runs 880 meters and c runs 800 meters . therefore , b runs 880 meters while c runs 800 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 800 ) \/ 880 = 909.09 meters thus , b can give c ( 1000 - 909.09 ) = 90.09 meters start answer : a\"","correct":"a","options":{"a":"90.09 meters ","b":"111.12 meters ","c":"112.12 meters ","d":"113.12 meters","e":"none of these"},"options_float":{"a":90.09,"b":111.12,"c":112.12,"d":113.12,"e":null},"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), 120)))","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 - 120<\/gadget>\n880<\/output>\n800_000 \/ 880<\/gadget>\n10_000\/11 = around 909.090909<\/output>\n1_000 - (10_000\/11)<\/gadget>\n1_000\/11 = around 90.909091<\/output>\n1_000\/11 = around 90.909091<\/result>","index":582} +{"problem":"a merchant marks his goods up by 30 % and then offers a discount of 20 % 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 20 % on 130 . profit = 104 - 100 4 % answer e\"","correct":"e","options":{"a":"8 % ","b":"10 % ","c":"21 % ","d":"15 %","e":"4 %"},"options_float":{"a":8.0,"b":10.0,"c":21.0,"d":15.0,"e":4.0},"annotated_formula":"subtract(subtract(add(30, const_100), divide(multiply(add(30, const_100), 20), 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 * 20<\/gadget>\n2_600<\/output>\n2_600 \/ 100<\/gadget>\n26<\/output>\n130 - 26<\/gadget>\n104<\/output>\n104 - 100<\/gadget>\n4<\/output>\n4<\/result>","index":584} +{"problem":"the owner of a furniture shop charges his customer 42 % more than the cost price . if a customer paid rs . 8300 for a computer table , then what was the cost price of the computer table ?","rationale":"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8300 ( 100 \/ 142 ) = rs . 5845 . answer : b","correct":"b","options":{"a":"rs . 5725 ","b":"rs . 5845 ","c":"rs . 6275 ","d":"rs . 6725","e":"none of these"},"options_float":{"a":5725.0,"b":5845.0,"c":6275.0,"d":6725.0,"e":null},"annotated_formula":"divide(8300, add(const_1, divide(42, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)","chain":"42 \/ 100<\/gadget>\n21\/50 = around 0.42<\/output>\n1 + (21\/50)<\/gadget>\n71\/50 = around 1.42<\/output>\n8_300 \/ (71\/50)<\/gadget>\n415_000\/71 = around 5_845.070423<\/output>\n415_000\/71 = around 5_845.070423<\/result>","index":585} +{"problem":"an article with cost price of 320 is sold at 18 % profit . what is the selling price ?","rationale":"\"sp = 1.18 * 320 = 378 answer : d\"","correct":"d","options":{"a":"198 ","b":"200 ","c":"204 ","d":"378","e":"347"},"options_float":{"a":198.0,"b":200.0,"c":204.0,"d":378.0,"e":347.0},"annotated_formula":"add(320, multiply(320, divide(18, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n320 * (9\/50)<\/gadget>\n288\/5 = around 57.6<\/output>\n320 + (288\/5)<\/gadget>\n1_888\/5 = around 377.6<\/output>\n1_888\/5 = around 377.6<\/result>","index":586} +{"problem":"given that 100.48 = x , 100.70 = y and xz = y 2 , then the value of z is close to :","rationale":"\"xz = y 2 10 ( 0.48 z ) = 10 ( 2 x 0.70 ) = 101.40 0.48 z = 1.40 z = 140 = 35 = 2.9 ( approx . ) 48 12 answer : c\"","correct":"c","options":{"a":"2.2 ","b":"8.2 ","c":"2.9 ","d":"2.1","e":"2.6"},"options_float":{"a":2.2,"b":8.2,"c":2.9,"d":2.1,"e":2.6},"annotated_formula":"divide(multiply(subtract(100.70, const_100), const_2), subtract(100.48, const_100))","linear_formula":"subtract(n1,const_100)|subtract(n0,const_100)|multiply(#0,const_2)|divide(#2,#1)|","chain":"100.7 - 100<\/gadget>\n0.7<\/output>\n0.7 * 2<\/gadget>\n1.4<\/output>\n100.48 - 100<\/gadget>\n0.48<\/output>\n1.4 \/ 0.48<\/gadget>\n2.916667<\/output>\n2.916667<\/result>","index":591} +{"problem":"how many digits are required to number a book containing 250 pages ?","rationale":"\"9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 250 - ( 90 + 9 ) = 151 pages will require 151 * 3 = 453 digits . the total number of digits is 9 + 180 + 453 = 642 . the answer is b .\"","correct":"b","options":{"a":"756 ","b":"642 ","c":"492 ","d":"372","e":"250"},"options_float":{"a":756.0,"b":642.0,"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(250, 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>\n250 - 100<\/gadget>\n150<\/output>\n150 + 1<\/gadget>\n151<\/output>\n151 * 3<\/gadget>\n453<\/output>\n189 + 453<\/gadget>\n642<\/output>\n642<\/result>","index":592} +{"problem":"at scratch and dents rent - a - car , it costs $ 34.95 a day plus $ 0.23 per mile to rent a car . at rent - a - lemon , the charge is $ 25.00 a day plus $ 1.31 per mile . if you need to rent a car for 3 days , how many miles ( to nearest tenth ) must you drive for a car from both agencies to cost the same amount ?","rationale":"for sad : saddaily = $ 34.95 \/ day sadmile = $ 0.23 \/ mile for ral : raldaily = $ 25.00 \/ day ralmile = $ 1.31 \/ mile we want the raltotal = sadtotal , so we get ( raldaily * days ) + ( ralmile * miles ) = ( saddaily * days ) + ( sadmile * miles ) = > miles = ( ( saddaily * days ) - ( raldaily * days ) ) \/ ( ralmiles - sadmiles ) = ( ( saddaily - raldaily ) * days ) \/ ( ralmiles - sadmiles ) miles = ( ( $ 34.95 * 3 ) - ( $ 25.00 * 3 ) ) \/ ( $ 1.31 - $ 0.23 ) = 27.6 miles c . 27.6 miles","correct":"c","options":{"a":"25.7 miles ","b":"26.2 miles ","c":"27.6 miles ","d":"27.9 miles","e":"29.9 miles"},"options_float":{"a":25.7,"b":26.2,"c":27.6,"d":27.9,"e":29.9},"annotated_formula":"divide(subtract(multiply(34.95, 3), multiply(25, 3)), subtract(1.31, 0.23))","linear_formula":"multiply(n0,n4)|multiply(n2,n4)|subtract(n3,n1)|subtract(#0,#1)|divide(#3,#2)","chain":"34.95 * 3<\/gadget>\n104.85<\/output>\n25 * 3<\/gadget>\n75<\/output>\n104.85 - 75<\/gadget>\n29.85<\/output>\n1.31 - 0.23<\/gadget>\n1.08<\/output>\n29.85 \/ 1.08<\/gadget>\n27.638889<\/output>\n27.638889<\/result>","index":593} +{"problem":"pipe a can fill a tank in 9 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 \/ 9 - 1 \/ x = 1 \/ 12 = > 1 \/ x = 1 \/ 9 - 1 \/ 12 = 1 \/ 12 = > x = 36 . answer : a\"","correct":"a","options":{"a":"36 ","b":"88 ","c":"18 ","d":"26","e":"12"},"options_float":{"a":36.0,"b":88.0,"c":18.0,"d":26.0,"e":12.0},"annotated_formula":"divide(multiply(12, 9), subtract(12, 9))","linear_formula":"multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|","chain":"12 * 9<\/gadget>\n108<\/output>\n12 - 9<\/gadget>\n3<\/output>\n108 \/ 3<\/gadget>\n36<\/output>\n36<\/result>","index":594} +{"problem":"if p is a prime number greater than 3 , find the remainder when p ^ 2 + 16 is divided by 12 .","rationale":"every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 16 = 36 n ^ 2 + 12 n + 1 + 16 = 36 n ^ 2 + 12 n + 12 + 5 if p = 6 n - 1 , then p ^ 2 + 16 = 36 n ^ 2 - 12 n + 1 + 16 = 36 n ^ 2 - 12 n + 12 + 5 when divided by 12 , it must leave a remainder of 5 . the answer is a .","correct":"a","options":{"a":"5 ","b":"1 ","c":"0 ","d":"8","e":"7"},"options_float":{"a":5.0,"b":1.0,"c":0.0,"d":8.0,"e":7.0},"annotated_formula":"subtract(add(16, power(add(const_1, const_4), const_2)), multiply(12, 3))","linear_formula":"add(const_1,const_4)|multiply(n0,n3)|power(#0,const_2)|add(n2,#2)|subtract(#3,#1)","chain":"1 + 4<\/gadget>\n5<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n16 + 25<\/gadget>\n41<\/output>\n12 * 3<\/gadget>\n36<\/output>\n41 - 36<\/gadget>\n5<\/output>\n5<\/result>","index":595} +{"problem":"if x is 20 percent greater than 55 , then x =","rationale":"x is 20 % greater than 55 means x is 1.2 times 55 ( in other words 55 + 20 \/ 100 * 55 = 1.2 * 55 ) therefore , x = 1.2 * 55 = 66 answer : d","correct":"d","options":{"a":"68 ","b":"70.4 ","c":"86 ","d":"66","e":"108"},"options_float":{"a":68.0,"b":70.4,"c":86.0,"d":66.0,"e":108.0},"annotated_formula":"add(55, multiply(divide(20, const_100), 55))","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 55<\/gadget>\n11<\/output>\n55 + 11<\/gadget>\n66<\/output>\n66<\/result>","index":599} +{"problem":"find the value of 72519 x 9999 = y ?","rationale":"\"72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d\"","correct":"d","options":{"a":"354517481 ","b":"457687783 ","c":"246567689 ","d":"725117481","e":"565776879"},"options_float":{"a":354517481.0,"b":457687783.0,"c":246567689.0,"d":725117481.0,"e":565776879.0},"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":600} +{"problem":"a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 49 \/ 64 . how many marbles are in the bag ?","rationale":"\"ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) \/ x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probability of pulling a non - red marble out is ( x - 12 ) \/ x probability theorm states that if the probability of event a occuring is m and the probability of event b occuring is n then the probability of both a and b occuring is m * n so therefore the probability of 2 non - red marbles getting pulled out is [ ( x - 12 ) \/ x ] * [ ( x - 12 ) \/ x ] this is given as 49 \/ 64 - - > ( x - 12 ) ^ 2 = 49 \/ 64 x ^ 2 square rooting u have x - 12 \/ x = 7 \/ 8 or x = 96 d\"","correct":"d","options":{"a":"24 ","b":"48 ","c":"60 ","d":"96","e":"84"},"options_float":{"a":24.0,"b":48.0,"c":60.0,"d":96.0,"e":84.0},"annotated_formula":"divide(12, subtract(const_1, sqrt(divide(49, 64))))","linear_formula":"divide(n2,n3)|sqrt(#0)|subtract(const_1,#1)|divide(n0,#2)|","chain":"49 \/ 64<\/gadget>\n49\/64 = around 0.765625<\/output>\n(49\/64) ** (1\/2)<\/gadget>\n7\/8 = around 0.875<\/output>\n1 - (7\/8)<\/gadget>\n1\/8 = around 0.125<\/output>\n12 \/ (1\/8)<\/gadget>\n96<\/output>\n96<\/result>","index":601} +{"problem":"average of 15 results is 43 . if the average of first 7 results is 41 and average of last 7 results is 45 then find the eighth result ?","rationale":"option ' c '","correct":"c","options":{"a":"41 ","b":"39 ","c":"43 ","d":"45","e":"47"},"options_float":{"a":41.0,"b":39.0,"c":43.0,"d":45.0,"e":47.0},"annotated_formula":"subtract(multiply(15, 43), add(multiply(7, 41), multiply(7, 45)))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)","chain":"15 * 43<\/gadget>\n645<\/output>\n7 * 41<\/gadget>\n287<\/output>\n7 * 45<\/gadget>\n315<\/output>\n287 + 315<\/gadget>\n602<\/output>\n645 - 602<\/gadget>\n43<\/output>\n43<\/result>","index":602} +{"problem":"the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 432 . find the principal ?","rationale":"p = 432 ( 100 \/ 5 ) ^ 2 = > p = 10800 answer : e","correct":"e","options":{"a":"2277 ","b":"2667 ","c":"3600 ","d":"9766","e":"10800"},"options_float":{"a":2277.0,"b":2667.0,"c":3600.0,"d":9766.0,"e":10800.0},"annotated_formula":"divide(432, 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>\n432 \/ (1\/25)<\/gadget>\n10_800<\/output>\n10_800<\/result>","index":603} +{"problem":"a sprinter starts running on a circular path of radius r metres . her average speed ( in metres \/ minute ) is π r during the first 30 seconds , π r \/ 2 during next one minute , π r \/ 4 during next 2 minutes , π r \/ 8 during next 4 minutes , and so on . what is the ratio of the time taken for the nth round to that for the previous round ?","rationale":"explanation : there is more than 1 way to approach the solution ; however , i will detail the easiest way to go about it here . we want to find the ratio of time taken for nth round : time taken for ( n - 1 ) th round it will be same as finding the ratio of time taken for 2 nd round : time taken for 1 st round . 1 round = circumference of the circle = 2 π r 1 st round : speed = π r for 30 seconds . so , total distance travelled = π r \/ 2 . speed = π r \/ 2 for 1 minute . so , total distance travelled = π r \/ 2 . speed = π r \/ 4 for 2 minutes . so , total distance travelled = π r \/ 2 . speed = π r \/ 8 for 4 minutes . so , total distance travelled = π r \/ 2 . so , for a distance of 2 π r , time taken is 7.5 minutes . 2 nd round : speed = π r \/ 16 for 8 minutes . so , total distance travelled = π r \/ 2 . speed = π r \/ 32 for 16 minutes . so , total distance travelled = π r \/ 2 . speed = π r \/ 64 for 32 minutes . so , total distance travelled = π r \/ 2 . speed = π r \/ 128 for 64 minutes . so , total distance travelled = π r \/ 2 . so , for a distance of 2 π r , time taken is 120 minutes . ratio is 120 : 7.5 = 16 : 1 . answer : c","correct":"c","options":{"a":"4 ","b":"8 ","c":"16 ","d":"32","e":"36"},"options_float":{"a":4.0,"b":8.0,"c":16.0,"d":32.0,"e":36.0},"annotated_formula":"power(2, 4)","linear_formula":"power(n1,n2)","chain":"2 ** 4<\/gadget>\n16<\/output>\n16<\/result>","index":604} +{"problem":". a car covers a distance of 1028 km in 4 hours . find its speed ?","rationale":"\"1028 \/ 4 = 257 kmph answer : d\"","correct":"d","options":{"a":"104 ","b":"255 ","c":"266 ","d":"257","e":"276"},"options_float":{"a":104.0,"b":255.0,"c":266.0,"d":257.0,"e":276.0},"annotated_formula":"divide(1028, 4)","linear_formula":"divide(n0,n1)|","chain":"1_028 \/ 4<\/gadget>\n257<\/output>\n257<\/result>","index":605} +{"problem":"the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 π · r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams t ?","rationale":"\"weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )\"","correct":"b","options":{"a":"t = 16 ","b":"t = 32 ","c":"t = 64 ","d":"128","e":"512"},"options_float":{"a":16.0,"b":32.0,"c":64.0,"d":128.0,"e":512.0},"annotated_formula":"multiply(8, 4)","linear_formula":"multiply(n0,n3)|","chain":"8 * 4<\/gadget>\n32<\/output>\n32<\/result>","index":608} +{"problem":"an enterprising businessman earns an income of re 5 on the first day of his business . on every subsequent day , he earns an income which is just thrice of that made on the previous day . on the 10 th day of business , he earns an income of :","rationale":"2 nd day he earns = 3 ( 2 – 5 ) 3 rd day he earns = 3 ( 3 – 5 ) on 20 th day he earns 3 ( 20 - 5 ) = 45 rupees answer : d","correct":"d","options":{"a":"21 ","b":"22 ","c":"20 ","d":"45","e":"30"},"options_float":{"a":21.0,"b":22.0,"c":20.0,"d":45.0,"e":30.0},"annotated_formula":"subtract(multiply(5, 10), 5)","linear_formula":"multiply(n0,n1)|subtract(#0,n0)","chain":"5 * 10<\/gadget>\n50<\/output>\n50 - 5<\/gadget>\n45<\/output>\n45<\/result>","index":609} +{"problem":"a student chose a number , multiplied it by 3 , then subtracted 138 from the result and got 102 . what was the number he chose ?","rationale":"\"solution : let xx be the number he chose , then 3 ⋅ x − 138 = 102 3 x = 240 x = 80 answer a\"","correct":"a","options":{"a":"80 ","b":"120 ","c":"130 ","d":"140","e":"150"},"options_float":{"a":80.0,"b":120.0,"c":130.0,"d":140.0,"e":150.0},"annotated_formula":"divide(add(102, 138), 3)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"102 + 138<\/gadget>\n240<\/output>\n240 \/ 3<\/gadget>\n80<\/output>\n80<\/result>","index":610} +{"problem":"two goods train each 500 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":"solution relative speed = ( 45 + 30 ) km \/ hr = ( 75 x 5 \/ 18 ) m \/ sec = ( 125 \/ 6 ) m \/ sec total distance covered = ( 500 + 500 ) m = 1000 m required time = ( 1000 x 6 \/ 125 ) sec = 48 sec answer c","correct":"c","options":{"a":"12 sec ","b":"24 sec ","c":"48 sec ","d":"60 sec","e":"none"},"options_float":{"a":12.0,"b":24.0,"c":48.0,"d":60.0,"e":null},"annotated_formula":"multiply(divide(500, divide(multiply(const_1000, add(45, 30)), const_3600)), const_2)","linear_formula":"add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|multiply(#3,const_2)","chain":"45 + 30<\/gadget>\n75<\/output>\n1_000 * 75<\/gadget>\n75_000<\/output>\n75_000 \/ 3_600<\/gadget>\n125\/6 = around 20.833333<\/output>\n500 \/ (125\/6)<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\n48<\/result>","index":611} +{"problem":"the population of a village is 14300 . it increases annually at the rate of 15 % p . a . what will be its population after 2 years ?","rationale":"formula : ( after = 100 denominator ago = 100 numerator ) 14300 × 115 \/ 100 × 115 \/ 100 = 18911 a","correct":"a","options":{"a":"18911 ","b":"18788 ","c":"19898 ","d":"14000","e":"14400"},"options_float":{"a":18911.0,"b":18788.0,"c":19898.0,"d":14000.0,"e":14400.0},"annotated_formula":"multiply(14300, power(add(const_1, divide(15, const_100)), 2))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n(23\/20) ** 2<\/gadget>\n529\/400 = around 1.3225<\/output>\n14_300 * (529\/400)<\/gadget>\n75_647\/4 = around 18_911.75<\/output>\n75_647\/4 = around 18_911.75<\/result>","index":612} +{"problem":"what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 20 ?","rationale":"\"first of all | x - 3 | ^ 2 = ( x - 3 ) ^ 2 , so we have : ( x - 3 ) ^ 2 + | x - 3 | = 20 . when x < 3 , x - 3 is negative , thus | x - 3 | = - ( x - 3 ) . in this case we ' ll have ( x - 3 ) ^ 2 - ( x - 3 ) = 20 - - > x = - 1 or x = 8 . discard x = 8 because it ' s not in the range we consider ( < 3 ) . when x > = 3 , x - 3 is non - negative , thus | x - 3 | = x - 3 . in this case we ' ll have ( x - 3 ) ^ 2 + ( x - 3 ) = 20 - - > x = - 2 or x = 7 . discard x = - 2 because it ' s not in the range we consider ( > = 3 ) . thus there are two solutions : x = - 1 and x = 7 - - > the sum = 6 . answer : b .\"","correct":"b","options":{"a":"- 1 ","b":"6 ","c":"7 ","d":"12","e":"14"},"options_float":{"a":-1.0,"b":6.0,"c":7.0,"d":12.0,"e":14.0},"annotated_formula":"add(add(const_4, 3), subtract(3, const_4))","linear_formula":"add(n0,const_4)|subtract(n0,const_4)|add(#0,#1)|","chain":"4 + 3<\/gadget>\n7<\/output>\n3 - 4<\/gadget>\n-1<\/output>\n7 + (-1)<\/gadget>\n6<\/output>\n6<\/result>","index":613} +{"problem":"a closed cylindrical tank contains 36 pie cubic feet of water and its filled to half its capacity . when the tank is placed upright on its circular base on level ground , the height of water in the tank is 4 feet . when the tank is placed on its side on level ground , what is the height , in feet , of the surface of the water above the ground ?","rationale":"36 pie cubic feet of water and its filled to half tank ' s capacity . . . volume of tank = 72 pie cubic feet height of tank = 4 * 2 = 8 feet ( since tank is placed upright on its circular base on level ground , the height of water in the tank is 4 feet . ) 72 pie = pie * r 2 * 8 r 2 = 9 r = 3 feet answer : d","correct":"d","options":{"a":"0 feet ","b":"1 feet ","c":"2 feet ","d":"3 feet","e":"4 feet"},"options_float":{"a":0.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"sqrt(divide(divide(multiply(36, const_pi), 4), const_pi))","linear_formula":"multiply(n0,const_pi)|divide(#0,n1)|divide(#1,const_pi)|sqrt(#2)","chain":"36 * pi<\/gadget>\n36*pi = around 113.097336<\/output>\n(36*pi) \/ 4<\/gadget>\n9*pi = around 28.274334<\/output>\n(9*pi) \/ pi<\/gadget>\n9<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n3<\/result>","index":616} +{"problem":"shopkeeper rise price by 33 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ?","rationale":"\"let d initial price be 100 33 % rise now price = 133 \/ 100 * 100 = 133 10 % discount then price = 133 * 90 \/ 100 = 119.7 15 % discount then price = 119.7 * 85 \/ 100 = 101.745 so gain = 101.745 - 100 = 1.745 gain % = gain * 100 \/ cp = = > 1.745 * 100 \/ 100 = 1.745 % answer : a\"","correct":"a","options":{"a":"1.745 % ","b":"4.745 % ","c":"3.745 % ","d":"6.745 %","e":"7.745 %"},"options_float":{"a":1.745,"b":4.745,"c":3.745,"d":6.745,"e":7.745},"annotated_formula":"subtract(multiply(multiply(add(const_100, 33), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100)","linear_formula":"add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)|","chain":"100 + 33<\/gadget>\n133<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n133 * (9\/10)<\/gadget>\n1_197\/10 = around 119.7<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n(1_197\/10) * (17\/20)<\/gadget>\n20_349\/200 = around 101.745<\/output>\n(20_349\/200) - 100<\/gadget>\n349\/200 = around 1.745<\/output>\n349\/200 = around 1.745<\/result>","index":618} +{"problem":"in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 3 : 7 : 13 . if the shop carries 352 dogs and bunnies total in stock , how many dogs are there ?","rationale":"\"let us assume the number of dogs , cats and bunnies to be 3 x , 7 x and 13 x total dogs and bunnies = 16 x . and we are given that 16 x = 352 . hence x = 22 . dogs = 3 x = 3 * 22 = 66 ( option b )\"","correct":"b","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(352, add(3, 13)), 3)","linear_formula":"add(n0,n2)|divide(n3,#0)|multiply(n0,#1)|","chain":"3 + 13<\/gadget>\n16<\/output>\n352 \/ 16<\/gadget>\n22<\/output>\n22 * 3<\/gadget>\n66<\/output>\n66<\/result>","index":619} +{"problem":"when the price of an article was reduced by 25 % its sale increased by 80 % . what was the net effect on the sale ?","rationale":"\"if n items are sold for $ p each , revenue is $ np . if we reduce the price by 25 % , the new price is 0.75 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.75 p ) ( 1.8 n ) = 1.35 np , which is 1.35 times the old revenue , so is 35 % greater . answer : a\"","correct":"a","options":{"a":"35 % increase ","b":"44 % decrease ","c":"60 % increase ","d":"66 % increase","e":"66 % decrease"},"options_float":{"a":35.0,"b":44.0,"c":60.0,"d":66.0,"e":66.0},"annotated_formula":"subtract(divide(multiply(add(80, const_100), subtract(const_100, 25)), 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":"80 + 100<\/gadget>\n180<\/output>\n100 - 25<\/gadget>\n75<\/output>\n180 * 75<\/gadget>\n13_500<\/output>\n13_500 \/ 100<\/gadget>\n135<\/output>\n135 - 100<\/gadget>\n35<\/output>\n35<\/result>","index":621} +{"problem":"5 years ago , the average age of a and b was 15 years . average age of a , b and c today is 20 years . how old will c be after 14 years ?","rationale":"explanation : ( a + b ) , five years ago = ( 15 * 2 ) = 30 years . ( a + b ) , now = ( 30 + 5 * 2 ) years = 40 years . ( a + b + c ) , now = ( 20 x 3 ) years = 60 years . c , now = ( 60 - 40 ) years = 20 years . c , after 14 years = ( 20 + 14 ) years = 34 years . answer : b","correct":"b","options":{"a":"30 ","b":"34 ","c":"40 ","d":"50","e":"60"},"options_float":{"a":30.0,"b":34.0,"c":40.0,"d":50.0,"e":60.0},"annotated_formula":"add(subtract(multiply(20, const_3), add(add(multiply(15, const_2), 5), 5)), 14)","linear_formula":"multiply(n2,const_3)|multiply(n1,const_2)|add(n0,#1)|add(n0,#2)|subtract(#0,#3)|add(n3,#4)","chain":"20 * 3<\/gadget>\n60<\/output>\n15 * 2<\/gadget>\n30<\/output>\n30 + 5<\/gadget>\n35<\/output>\n35 + 5<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20 + 14<\/gadget>\n34<\/output>\n34<\/result>","index":622} +{"problem":"one fourth of a solution that was 10 % sugar by weight was replaced with 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 * 0.1 + 1 \/ 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : a .\"","correct":"a","options":{"a":"34 % ","b":"24 % ","c":"22 % ","d":"18 %","e":"8.5 %"},"options_float":{"a":34.0,"b":24.0,"c":22.0,"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(10, 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>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n75 * (1\/10)<\/gadget>\n15\/2 = around 7.5<\/output>\n16 - (15\/2)<\/gadget>\n17\/2 = around 8.5<\/output>\n(17\/2) \/ 25<\/gadget>\n17\/50 = around 0.34<\/output>\n(17\/50) * 100<\/gadget>\n34<\/output>\n34<\/result>","index":623} +{"problem":"a can give b 100 meters start and c 120 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"explanation : a runs 1000 meters while b runs 900 meters and c runs 880 meters . therefore , b runs 900 meters while c runs 880 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 880 ) \/ 900 = 977.778 meters thus , b can give c ( 1000 - 977.77 ) = 22.22 meters start answer : c\"","correct":"c","options":{"a":"10.22 meters ","b":"11.22 meters ","c":"22.22 meters ","d":"33.22 meters","e":"none of these"},"options_float":{"a":10.22,"b":11.22,"c":22.22,"d":33.22,"e":null},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 120)), 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 - 120<\/gadget>\n880<\/output>\n1_000 * 880<\/gadget>\n880_000<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n880_000 \/ 900<\/gadget>\n8_800\/9 = around 977.777778<\/output>\n1_000 - (8_800\/9)<\/gadget>\n200\/9 = around 22.222222<\/output>\n200\/9 = around 22.222222<\/result>","index":626} +{"problem":"a train 450 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 = ( 450 × 2 \/ 15 ) sec = 60 sec answer : e\"","correct":"e","options":{"a":"30 sec ","b":"32 sec ","c":"36 sec ","d":"38 sec","e":"60 sec"},"options_float":{"a":30.0,"b":32.0,"c":36.0,"d":38.0,"e":60.0},"annotated_formula":"multiply(const_3600, divide(divide(450, const_1000), add(25, 2)))","linear_formula":"add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)|","chain":"450 \/ 1_000<\/gadget>\n9\/20 = around 0.45<\/output>\n25 + 2<\/gadget>\n27<\/output>\n(9\/20) \/ 27<\/gadget>\n1\/60 = around 0.016667<\/output>\n3_600 * (1\/60)<\/gadget>\n60<\/output>\n60<\/result>","index":627} +{"problem":"a rectangular with dimensions 35 inches by 45 inches is to be divided into squares of equal size . which of the following could be a length of a side of the squares ?","rationale":"you need to divide the width and length of the rectangular to equal pieces where l = w you can solve this using gcf 35 = 5 * 7 45 = 3 * 3 * 5 gcf = 5 p . s you can make squares with side of 5 answer : c","correct":"c","options":{"a":"4 inches ","b":"6 inches ","c":"5 inches ","d":"8 inches","e":"10 inches"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":8.0,"e":10.0},"annotated_formula":"divide(subtract(45, 35), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,const_2)","chain":"45 - 35<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":628} +{"problem":"the function f ( p ) represents the number of ways that prime numbers can be uniquely summed to form a certain number p such that p = a + b + c + d … where those summed variables are each prime and a ≤ b ≤ c ≤ d . . . for instance f ( 8 ) = 3 and the unique ways are 2 + 2 + 2 + 2 and 2 + 3 + 3 and 3 + 5 . what is f ( 12 ) ?","rationale":"so we can start with 2 and check whether sum of two primes is primes is even . 1 ) 2 ( 6 times ) 2 ) 2 ( 3 times ) + 3 ( 2 times ) 3 ) 2 ( 2 times ) + 3 + 5 4 ) 2 + 3 + 7 5 ) 2 + 5 + 5 6 ) 3 ( 4 times ) 7 ) 5 + 7 answer : 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(divide(12, const_4), const_4)","linear_formula":"divide(n11,const_4)|add(#0,const_4)","chain":"12 \/ 4<\/gadget>\n3<\/output>\n3 + 4<\/gadget>\n7<\/output>\n7<\/result>","index":632} +{"problem":"in what time will a train 175 m long cross an electric pole , it its speed be 144 km \/ hr ?","rationale":"\"speed = 144 * 5 \/ 18 = 40 m \/ sec time taken = 175 \/ 40 = 4.37 sec . answer : c\"","correct":"c","options":{"a":"2.58 sec ","b":"2.91 sec ","c":"4.37 sec ","d":"2.9 sec","e":"1.8 sec"},"options_float":{"a":2.58,"b":2.91,"c":4.37,"d":2.9,"e":1.8},"annotated_formula":"divide(175, 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>\n175 \/ 40<\/gadget>\n35\/8 = around 4.375<\/output>\n35\/8 = around 4.375<\/result>","index":633} +{"problem":"a student chose a number , multiplied it by 5 , then subtracted 275 from the result and got 135 . what was the number he chose ?","rationale":"\"let x be the number he chose , then 5 ⋅ x − 275 = 135 5 x = 410 x = 82 correct answer b\"","correct":"b","options":{"a":"80 ","b":"82 ","c":"84 ","d":"86","e":"88"},"options_float":{"a":80.0,"b":82.0,"c":84.0,"d":86.0,"e":88.0},"annotated_formula":"divide(add(135, 275), 5)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"135 + 275<\/gadget>\n410<\/output>\n410 \/ 5<\/gadget>\n82<\/output>\n82<\/result>","index":634} +{"problem":"in a can , there is a mixture of milk and water in the ratio 3 : 2 . if the can is filled with an additional 6 liters of milk , the can would be full and the ratio of milk and water would become 2 : 1 . find the capacity of the can ?","rationale":"\"let c be the capacity of the can . ( 3 \/ 5 ) * ( c - 6 ) + 6 = ( 2 \/ 3 ) * c 9 c - 54 + 90 = 10 c c = 36 the answer is a .\"","correct":"a","options":{"a":"36 ","b":"34 ","c":"32 ","d":"30","e":"28"},"options_float":{"a":36.0,"b":34.0,"c":32.0,"d":30.0,"e":28.0},"annotated_formula":"add(add(multiply(divide(multiply(1, 6), subtract(multiply(2, 2), multiply(1, 3))), 3), 6), multiply(divide(multiply(1, 6), subtract(multiply(2, 2), multiply(1, 3))), 2))","linear_formula":"multiply(n2,n4)|multiply(n1,n3)|multiply(n0,n4)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)|multiply(n1,#4)|add(n2,#5)|add(#7,#6)|","chain":"1 * 6<\/gadget>\n6<\/output>\n2 * 2<\/gadget>\n4<\/output>\n1 * 3<\/gadget>\n3<\/output>\n4 - 3<\/gadget>\n1<\/output>\n6 \/ 1<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18 + 6<\/gadget>\n24<\/output>\n6 * 2<\/gadget>\n12<\/output>\n24 + 12<\/gadget>\n36<\/output>\n36<\/result>","index":635} +{"problem":"susan drives from city a to city b . after two hours of driving she noticed that she covered 80 km and calculated that , if she continued driving at the same speed , she would end up been 15 minutes late . so she increased her speed by 10 km \/ hr and she arrived at city b 36 minutes earlier than she planned . find the distance between cities a and b .","rationale":"let xx be the distance between a and b . since susan covered 80 km in 2 hours , her speed was v = 802 = 40 v = 802 = 40 km \/ hr . if she continued at the same speed she would be 1515 minutes late , i . e . the planned time on the road is x 40 − 1560 x 40 − 1560 hr . the rest of the distance is ( x − 80 ) ( x − 80 ) km . v = 40 + 10 = 50 v = 40 + 10 = 50 km \/ hr . so , she covered the distance between a and b in 2 + x − 80502 + x − 8050 hr , and it was 36 min less than planned . therefore , the planned time was 2 + x − 8050 + 36602 + x − 8050 + 3660 . when we equalize the expressions for the scheduled time , we get the equation : x 40 − 1560 = 2 + x − 8050 + 3660 x 40 − 1560 = 2 + x − 8050 + 3660 x − 1040 = 100 + x − 80 + 3050 x − 1040 = 100 + x − 80 + 3050 x − 104 = x + 505 x − 104 = x + 505 5 x − 50 = 4 x + 2005 x − 50 = 4 x + 200 x = 250 x = 250 so , the distance between cities a and b is 250 km . answer : c","correct":"c","options":{"a":"223 ","b":"376 ","c":"250 ","d":"378","e":"271"},"options_float":{"a":223.0,"b":376.0,"c":250.0,"d":378.0,"e":271.0},"annotated_formula":"add(divide(subtract(add(subtract(divide(36, const_60), divide(80, add(divide(80, const_2), 10))), const_2), divide(15, const_60)), subtract(divide(const_1, divide(80, const_2)), divide(const_1, add(divide(80, const_2), 10)))), const_100)","linear_formula":"divide(n3,const_60)|divide(n0,const_2)|divide(n1,const_60)|add(n2,#1)|divide(const_1,#1)|divide(n0,#3)|divide(const_1,#3)|subtract(#0,#5)|subtract(#4,#6)|add(#7,const_2)|subtract(#9,#2)|divide(#10,#8)|add(#11,const_100)","chain":"36 \/ 60<\/gadget>\n3\/5 = around 0.6<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40 + 10<\/gadget>\n50<\/output>\n80 \/ 50<\/gadget>\n8\/5 = around 1.6<\/output>\n(3\/5) - (8\/5)<\/gadget>\n-1<\/output>\n(-1) + 2<\/gadget>\n1<\/output>\n15 \/ 60<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ 50<\/gadget>\n1\/50 = around 0.02<\/output>\n(1\/40) - (1\/50)<\/gadget>\n1\/200 = around 0.005<\/output>\n(3\/4) \/ (1\/200)<\/gadget>\n150<\/output>\n150 + 100<\/gadget>\n250<\/output>\n250<\/result>","index":636} +{"problem":"a train running at the speed of 40 km \/ hr crosses a pole in 9 sec . what is the length of the train ?","rationale":"\"speed = 40 * 5 \/ 18 = 100 \/ 9 m \/ sec length of the train = speed * time = 100 \/ 9 * 9 = 100 m answer : a\"","correct":"a","options":{"a":"100 m ","b":"150 m ","c":"187 m ","d":"167 m","e":"197 m"},"options_float":{"a":100.0,"b":150.0,"c":187.0,"d":167.0,"e":197.0},"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":637} +{"problem":"what is the remainder if 7 ^ 16 is divided by 100 ?","rationale":"\"7 ^ 16 can be written as ( 7 ^ 4 ) ^ 4 if we divide 7 ^ 4 by 100 the reminder is 1 so , ( 7 ^ 4 ) ^ 4 by 100 , the reminder is 1 ^ 4 = 1 answer : d\"","correct":"d","options":{"a":"3 ","b":"4 ","c":"2 ","d":"1","e":"5"},"options_float":{"a":3.0,"b":4.0,"c":2.0,"d":1.0,"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":638} +{"problem":"a large box contains 17 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ?","rationale":"\"the number of chocolate bars is equal to 17 * 25 = 425 correct answer c\"","correct":"c","options":{"a":"250 ","b":"350 ","c":"425 ","d":"550","e":"650"},"options_float":{"a":250.0,"b":350.0,"c":425.0,"d":550.0,"e":650.0},"annotated_formula":"multiply(17, 25)","linear_formula":"multiply(n0,n1)|","chain":"17 * 25<\/gadget>\n425<\/output>\n425<\/result>","index":639} +{"problem":"what is the product of all the possible values of x if x ^ 2 + 5 x + 6 ?","rationale":"explanation : = > y = x ^ 2 + 5 x + 6 = > y = ( x + 2 ) ( x + 3 ) = > x = - 2 , x = - 3 product x = ( - 2 ) ( - 3 ) = 6 answer option 6 answer : d","correct":"d","options":{"a":"12 ","b":"18 ","c":"15 ","d":"6","e":"9"},"options_float":{"a":12.0,"b":18.0,"c":15.0,"d":6.0,"e":9.0},"annotated_formula":"divide(6, const_1)","linear_formula":"divide(n2,const_1)","chain":"6 \/ 1<\/gadget>\n6<\/output>\n6<\/result>","index":640} +{"problem":"jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 6 to 2 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ?","rationale":"\"u \/ i = 6 \/ 2 i \/ b = 5 \/ 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 30 : 18 : 2 i . e . u : b = 30 : 2 answer : option b\"","correct":"b","options":{"a":"5 : 1 ","b":"30 : 2 ","c":"15 : 2 ","d":"20 : 2","e":"25 : 2"},"options_float":{"a":5.0,"b":15.0,"c":7.5,"d":10.0,"e":12.5},"annotated_formula":"divide(multiply(6, 5), multiply(1, 2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)|","chain":"6 * 5<\/gadget>\n30<\/output>\n1 * 2<\/gadget>\n2<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15<\/result>","index":641} +{"problem":"a man walking at a constant rate of 9 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 15 miles per hour . the woman stops to wait for the man 3 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ?","rationale":"when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 3 * 15 \/ 60 = 15 \/ 20 miles and the man walks the distance mm , which is 3 * 9 \/ 60 = 9 \/ 20 mile . the difference of 15 \/ 20 - 9 \/ 20 = 3 \/ 10 miles ( mw ) will be covered by the man in ( 3 \/ 10 ) \/ 9 = 1 \/ 30 of an hour , which is 2 minutes . 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":"multiply(const_60, divide(multiply(divide(3, const_60), subtract(15, 9)), 9))","linear_formula":"divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)","chain":"3 \/ 60<\/gadget>\n1\/20 = around 0.05<\/output>\n15 - 9<\/gadget>\n6<\/output>\n(1\/20) * 6<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) \/ 9<\/gadget>\n1\/30 = around 0.033333<\/output>\n60 * (1\/30)<\/gadget>\n2<\/output>\n2<\/result>","index":642} +{"problem":"in a certain state , the ratio of registered republicans to registered democrats is 3 to 2 , and every registered voter is either a republican or a democrat . if 80 percent of the republicans and 20 percent of the democrats are expected to vote for candidate x , and everyone else is expected to vote for candidate y , by what percent is candidate x expected to win the election ?","rationale":"\"since we were expected to find a percentage figure - it thought that it might be easier to pick a ' smart number ' to represent the total number of voters ( republicans and democrats ) . therefore , i picked 100 ( as the total number of voters ) and thus 30 : 20 represents the number ratio of republicans : democrats . if 80 % of republicans ( which is ( 60 * 0.8 ) = 48 ) and 20 % of democrats ( 40 * 0.2 = 8 ) voted for candidate x , means that out of total of 100 voters ; 56 ( 48 + 8 ) voters voted for candidate x and 44 voted for candidate y . thus we can infer that candidate x is expected to win the election by 12 ( 56 - 44 ) votes . therefore candidate x is expected to win the election by ( 12 \/ 100 ) votes which is equivalent to 12 % . i think the answer is e .\"","correct":"e","options":{"a":"2 % ","b":"5 % ","c":"8 % ","d":"10 %","e":"12 %"},"options_float":{"a":2.0,"b":5.0,"c":8.0,"d":10.0,"e":12.0},"annotated_formula":"multiply(divide(subtract(add(multiply(divide(20, const_100), 2), multiply(divide(80, const_100), 3)), add(subtract(3, multiply(divide(80, const_100), 3)), subtract(2, multiply(divide(20, const_100), 2)))), add(3, 2)), const_100)","linear_formula":"add(n0,n1)|divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#1)|multiply(n0,#2)|add(#3,#4)|subtract(n0,#4)|subtract(n1,#3)|add(#6,#7)|subtract(#5,#8)|divide(#9,#0)|multiply(#10,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 2<\/gadget>\n2\/5 = around 0.4<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 3<\/gadget>\n12\/5 = around 2.4<\/output>\n(2\/5) + (12\/5)<\/gadget>\n14\/5 = around 2.8<\/output>\n3 - (12\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n2 - (2\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n(3\/5) + (8\/5)<\/gadget>\n11\/5 = around 2.2<\/output>\n(14\/5) - (11\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n3 + 2<\/gadget>\n5<\/output>\n(3\/5) \/ 5<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) * 100<\/gadget>\n12<\/output>\n12<\/result>","index":643} +{"problem":"the sides of a square region , measured to the nearest centimeter , are 10 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 9.5 is equidistant from both 9 and 10 ) the answer still should be : 9.5 ^ 2 = 90.25 . answer : e\"","correct":"e","options":{"a":"96.25 sq cm ","b":"98.25 sq cm ","c":"92.25 sq cm ","d":"100.25 sq cm","e":"90.25 sq cm"},"options_float":{"a":96.25,"b":98.25,"c":92.25,"d":100.25,"e":90.25},"annotated_formula":"power(subtract(subtract(10, 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>\n10 - (1\/4)<\/gadget>\n39\/4 = around 9.75<\/output>\n(39\/4) - (1\/4)<\/gadget>\n19\/2 = around 9.5<\/output>\n(19\/2) ** 2<\/gadget>\n361\/4 = around 90.25<\/output>\n361\/4 = around 90.25<\/result>","index":645} +{"problem":"if tim had lunch at $ 50.50 and he gave 10 % tip , how much did he spend ?","rationale":"\"the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.50 = ( 10 \/ 100 ) * 50.50 = $ 5.05 total spent 50.50 + 5.05 = $ 55.55 correct answer d\"","correct":"d","options":{"a":"$ 30.60 ","b":"$ 60.60 ","c":"$ 70.60 ","d":"$ 55.55","e":"$ 50.60"},"options_float":{"a":30.6,"b":60.6,"c":70.6,"d":55.55,"e":50.6},"annotated_formula":"add(50.50, divide(multiply(50.50, 10), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|","chain":"50.5 * 10<\/gadget>\n505<\/output>\n505 \/ 100<\/gadget>\n101\/20 = around 5.05<\/output>\n50.5 + (101\/20)<\/gadget>\n55.55<\/output>\n55.55<\/result>","index":648} +{"problem":"a 240 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train .","rationale":"\"relative speeds = ( 120 + 80 ) km \/ hr = 200 km \/ hr = ( 200 * 5 \/ 18 ) m \/ s = ( 500 \/ 9 ) m \/ s let length of train be xm x + 240 \/ 9 = 500 \/ 9 x = 260 ans is 260 m answer : a\"","correct":"a","options":{"a":"260 m ","b":"220 m ","c":"230 m ","d":"240 m","e":"250 m"},"options_float":{"a":260.0,"b":220.0,"c":230.0,"d":240.0,"e":250.0},"annotated_formula":"subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 240)","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>\n9 * (500\/9)<\/gadget>\n500<\/output>\n500 - 240<\/gadget>\n260<\/output>\n260<\/result>","index":649} +{"problem":"company a imported 12,000 widgets made of either brass or aluminum . the widgets are painted blue , red or green . if 10 percent of the widgets are made of brass and of those 20 percent are painted green and 40 percent are painted red how many brass widgets painted blue were imported ?","rationale":"answer a . we are told that 10 % of all imported widgets are made of brass and of those , 20 % are green and 40 % are red . since we know that there are only three colors , the remaining 40 % must be blue . 40 % blue of 10 % brass widgets leads to 4 % blue brass widgets out of the total 10,550 widgets . 12,000 \/ 100 * 4 = 480 . answer b .","correct":"b","options":{"a":"420 ","b":"480 ","c":"1050 ","d":"1680","e":"2100"},"options_float":{"a":420.0,"b":480.0,"c":1050.0,"d":1680.0,"e":2100.0},"annotated_formula":"multiply(multiply(multiply(multiply(divide(10, const_100), divide(40, const_100)), divide(add(10, const_2), 10)), const_100), const_100)","linear_formula":"add(n1,const_2)|divide(n1,const_100)|divide(n3,const_100)|divide(#0,n1)|multiply(#1,#2)|multiply(#3,#4)|multiply(#5,const_100)|multiply(#6,const_100)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(1\/10) * (2\/5)<\/gadget>\n1\/25 = around 0.04<\/output>\n10 + 2<\/gadget>\n12<\/output>\n12 \/ 10<\/gadget>\n6\/5 = around 1.2<\/output>\n(1\/25) * (6\/5)<\/gadget>\n6\/125 = around 0.048<\/output>\n(6\/125) * 100<\/gadget>\n24\/5 = around 4.8<\/output>\n(24\/5) * 100<\/gadget>\n480<\/output>\n480<\/result>","index":650} +{"problem":"mr . das decided to walk down the escalator of a mall . he found that if he walks down 26 steps , he requires 30 seconds to reach the bottom . however , if he steps down 34 stair she would only require 18 seconds to get to the bottom . if the time is measured from the moment the top step begins to descend to the time he steps off the last step at the bottom , find out the height of the stair way insteps ?","rationale":"here when he step down 26 steps he has 30 seconds for remaining steps . if he step down 34 stairs he has only 18 sec . 30 - 18 = 12 12 secs for 8 steps . . 18 secs for 12 steps . 12 + 34 = 46 so ans is 46 . . answer : b","correct":"b","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(add(multiply(divide(subtract(34, 26), subtract(30, 18)), 30), 26), multiply(const_4, const_10))","linear_formula":"multiply(const_10,const_4)|subtract(n2,n0)|subtract(n1,n3)|divide(#1,#2)|multiply(n1,#3)|add(n0,#4)|subtract(#5,#0)","chain":"34 - 26<\/gadget>\n8<\/output>\n30 - 18<\/gadget>\n12<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 30<\/gadget>\n20<\/output>\n20 + 26<\/gadget>\n46<\/output>\n4 * 10<\/gadget>\n40<\/output>\n46 - 40<\/gadget>\n6<\/output>\n6<\/result>","index":651} +{"problem":"solution x is 30 % chemical a and 70 % chemical b by volume . solution y is 40 % chemical a and 60 % chemical b by volume . if a mixture of x and y is 36 % chemical a , what percent of the mixture is solution x ?","rationale":"\"the volume of the mixture be x + y . 0.3 x + 0.4 y = 0.36 ( x + y ) x = 2 y \/ 3 x \/ ( x + y ) = ( 2 y \/ 3 ) \/ ( 5 y \/ 3 ) = 2 \/ 5 = 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(divide(divide(subtract(40, 36), subtract(36, 30)), add(divide(subtract(40, 36), subtract(36, 30)), const_1)), const_100)","linear_formula":"subtract(n2,n4)|subtract(n4,n0)|divide(#0,#1)|add(#2,const_1)|divide(#2,#3)|multiply(#4,const_100)|","chain":"40 - 36<\/gadget>\n4<\/output>\n36 - 30<\/gadget>\n6<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) + 1<\/gadget>\n5\/3 = around 1.666667<\/output>\n(2\/3) \/ (5\/3)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":652} +{"problem":"jim ’ s taxi service charges an initial fee of $ 2.45 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 = 2.45 $ and charge per 2 \/ 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.45 + ( 3.6 \/ . 4 ) * . 35 = 2.45 + 9 * . 35 = 5.6 $ answer e\"","correct":"e","options":{"a":"$ 3.15 ","b":"$ 4.45 ","c":"$ 4.80 ","d":"$ 5.05","e":"$ 5.6"},"options_float":{"a":3.15,"b":4.45,"c":4.8,"d":5.05,"e":5.6},"annotated_formula":"add(2.45, 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>\n2.45 + 3.15<\/gadget>\n5.6<\/output>\n5.6<\/result>","index":653} +{"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 24 % of the original number . what percent of the red candies remain ?","rationale":"let x be the original number of each color so there are a total of 5 x candies . 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 = 1.2 x red + yellow = 0.45 x red = 0.225 x , since red = yellow . the answer is c .","correct":"c","options":{"a":"12.5 % ","b":"16.7 % ","c":"22.5 % ","d":"27.5 %","e":"33.3 %"},"options_float":{"a":12.5,"b":16.7,"c":22.5,"d":27.5,"e":33.3},"annotated_formula":"multiply(divide(divide(subtract(24, 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>\n24 - 15<\/gadget>\n9<\/output>\n9 \/ 2<\/gadget>\n9\/2 = around 4.5<\/output>\n(9\/2) \/ 20<\/gadget>\n9\/40 = around 0.225<\/output>\n(9\/40) * 100<\/gadget>\n45\/2 = around 22.5<\/output>\n45\/2 = around 22.5<\/result>","index":654} +{"problem":"on selling 9 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is :","rationale":"\"( c . p . of 9 balls ) - ( s . p . of 9 balls ) = ( c . p . of 5 balls ) c . p . of 4 balls = s . p . of 9 balls = rs . 720 . c . p . of 1 ball = rs . 720 \/ 4 = rs . 180 . answer : option e\"","correct":"e","options":{"a":"s . 145 ","b":"s . 150 ","c":"s . 155 ","d":"s . 160","e":"s . 180"},"options_float":{"a":145.0,"b":150.0,"c":155.0,"d":160.0,"e":180.0},"annotated_formula":"divide(720, subtract(9, 5))","linear_formula":"subtract(n0,n2)|divide(n1,#0)|","chain":"9 - 5<\/gadget>\n4<\/output>\n720 \/ 4<\/gadget>\n180<\/output>\n180<\/result>","index":655} +{"problem":"boy sells a book for rs . 630 he gets a loss of 10 % , to gain 10 % , what should be the sp ?","rationale":"\"cost price = 630 \/ 90 x 100 = 700 to gain 10 % = 700 x 10 \/ 100 = 70 sp = cp + gain = 700 + 70 = 770 answer : d\"","correct":"d","options":{"a":"430 ","b":"450 ","c":"550 ","d":"770","e":"660"},"options_float":{"a":430.0,"b":450.0,"c":550.0,"d":770.0,"e":660.0},"annotated_formula":"add(divide(630, subtract(const_1, divide(10, const_100))), multiply(divide(630, 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>\n630 \/ (9\/10)<\/gadget>\n700<\/output>\n700 * (1\/10)<\/gadget>\n70<\/output>\n700 + 70<\/gadget>\n770<\/output>\n770<\/result>","index":656} +{"problem":"a train 310 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 = [ 310 * 3 \/ 55 ] sec = 17 sec answer : d\"","correct":"d","options":{"a":"4 ","b":"6 ","c":"5 ","d":"17","e":"13"},"options_float":{"a":4.0,"b":6.0,"c":5.0,"d":17.0,"e":13.0},"annotated_formula":"multiply(const_3600, divide(divide(310, const_1000), add(60, 6)))","linear_formula":"add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)|","chain":"310 \/ 1_000<\/gadget>\n31\/100 = around 0.31<\/output>\n60 + 6<\/gadget>\n66<\/output>\n(31\/100) \/ 66<\/gadget>\n31\/6_600 = around 0.004697<\/output>\n3_600 * (31\/6_600)<\/gadget>\n186\/11 = around 16.909091<\/output>\n186\/11 = around 16.909091<\/result>","index":658} +{"problem":"how many odd factors does 210 have ?","rationale":"start with the prime factorization : 210 = 2 * 3 * 5 * 7 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 , 1 } plus 1 to each = { 2 , 2 , 2 } product = 2 * 2 * 2 = 8 therefore , there are 8 odd factors of 210 . in case you are curious , they are { 1 , 3 , 5 , 7 , 15 , 21 , 35 , and 105 } answer : e .","correct":"e","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":"add(add(add(const_4, const_2), const_1), const_1)","linear_formula":"add(const_2,const_4)|add(#0,const_1)|add(#1,const_1)","chain":"4 + 2<\/gadget>\n6<\/output>\n6 + 1<\/gadget>\n7<\/output>\n7 + 1<\/gadget>\n8<\/output>\n8<\/result>","index":659} +{"problem":"sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 350 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 = 350 - - > x = 1.715 . 1.25 x = ~ 2.14 . answer : c .\"","correct":"c","options":{"a":"$ 1.50 ","b":"$ 1.88 ","c":"$ 2.14 ","d":"$ 2.50","e":"$ 3.25"},"options_float":{"a":1.5,"b":1.88,"c":2.14,"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)), 350), 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 + 350<\/gadget>\n590<\/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>\n590 \/ 344<\/gadget>\n295\/172 = around 1.715116<\/output>\n(295\/172) * (5\/4)<\/gadget>\n1_475\/688 = around 2.143895<\/output>\n1_475\/688 = around 2.143895<\/result>","index":660} +{"problem":"a man has $ 480 in the denominations of one - dollar , 5 - dollar notes and 10 - dollar . the number of dollars of each denomination is equal . what is the total number of dollar that he has ?","rationale":"c $ 90 let number of notes of each denomination be x . then x + 5 x + 10 x = 480 16 x = 480 x = 30 . hence , total number of notes = 3 x = 90 .","correct":"c","options":{"a":"50 ","b":"60 ","c":"90 ","d":"48","e":"67"},"options_float":{"a":50.0,"b":60.0,"c":90.0,"d":48.0,"e":67.0},"annotated_formula":"add(divide(multiply(480, 10), const_60), 10)","linear_formula":"multiply(n0,n2)|divide(#0,const_60)|add(n2,#1)","chain":"480 * 10<\/gadget>\n4_800<\/output>\n4_800 \/ 60<\/gadget>\n80<\/output>\n80 + 10<\/gadget>\n90<\/output>\n90<\/result>","index":664} +{"problem":"find how many positive integers less than 10000 are there such thatthe sum of the digits of the no . is divisible by 3 ?","rationale":"if sum of the digits is divisible by 3 , the number is divisible by 3 . therefore , required number of non - negative integers is equal to count of numbers less than 10000 which are divisible by 3 . such numbers are ( 3 , 6 , 9 , . . . , 9999 ) ( arithmetic progression with first term = 3 , last term = 9999 , common difference = 3 ) . count of such numbers = 9999 3 = 3333 99993 = 3333 but zero is also divisible by 3 . this makes our total count 3334 d","correct":"d","options":{"a":"2468 ","b":"2789 ","c":"2987 ","d":"3334","e":"3568"},"options_float":{"a":2468.0,"b":2789.0,"c":2987.0,"d":3334.0,"e":3568.0},"annotated_formula":"add(floor(divide(10000, 3)), const_1)","linear_formula":"divide(n0,n1)|floor(#0)|add(#1,const_1)","chain":"10_000 \/ 3<\/gadget>\n10_000\/3 = around 3_333.333333<\/output>\nfloor(10_000\/3)<\/gadget>\n3_333<\/output>\n3_333 + 1<\/gadget>\n3_334<\/output>\n3_334<\/result>","index":666} +{"problem":"two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 6 days ?","rationale":"\"a ' s one day ' s work = 1 \/ 30 b ' s one day ' s work = 1 \/ 45 ( a + b ) ' s one day ' s work = 1 \/ 30 + 1 \/ 45 = 1 \/ 18 the part of the work completed in 6 days = 6 ( 1 \/ 18 ) = 1 \/ 3 . answer c\"","correct":"c","options":{"a":"2 \/ 5 ","b":"1 \/ 6 ","c":"1 \/ 3 ","d":"1 \/ 9","e":"2 \/ 6"},"options_float":{"a":0.4,"b":0.1666666667,"c":0.3333333333,"d":0.1111111111,"e":0.3333333333},"annotated_formula":"multiply(6, add(divide(const_1, 30), divide(const_1, 45)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ 45<\/gadget>\n1\/45 = around 0.022222<\/output>\n(1\/30) + (1\/45)<\/gadget>\n1\/18 = around 0.055556<\/output>\n6 * (1\/18)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":670} +{"problem":"in how many ways 4 boys and 4 girls can be seated in a row so that they are alternate .","rationale":"\"solution : let the arrangement be , b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 4 ! ways . required number of ways , = 4 ! * 4 ! = 576 . answer : option d\"","correct":"d","options":{"a":"144 ","b":"288 ","c":"12 ","d":"576","e":"none"},"options_float":{"a":144.0,"b":288.0,"c":12.0,"d":576.0,"e":null},"annotated_formula":"multiply(factorial(4), factorial(4))","linear_formula":"factorial(n0)|factorial(n1)|multiply(#0,#1)|","chain":"factorial(4)<\/gadget>\n24<\/output>\n24 * 24<\/gadget>\n576<\/output>\n576<\/result>","index":672} +{"problem":"kavi spends 50 % of his monthly salary on food and saves 80 % of the remaining amount . if his monthly salary is rs . 19000 , how much money does he save every month ?","rationale":"explanation : kavi ' s monthly income = rs . 19,000 he spends 50 % on food . the total money spent on food = 50 \/ 100 * 19000 = rs . 9500 now , his monthly remaining income = rs . 19000 – rs . 9500 = rs . 9500 out of rs . 9500 , he saves 40 % . amount saved = 40 \/ 100 * 9500 = rs . 3800 answer : d","correct":"d","options":{"a":"rs . 2000 ","b":"rs . 600 ","c":"rs . 8000 ","d":"rs . 3800","e":"rs . 1200"},"options_float":{"a":2000.0,"b":600.0,"c":8000.0,"d":3800.0,"e":1200.0},"annotated_formula":"divide(divide(multiply(divide(multiply(19000, 50), const_100), 80), const_100), const_2)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_2)","chain":"19_000 * 50<\/gadget>\n950_000<\/output>\n950_000 \/ 100<\/gadget>\n9_500<\/output>\n9_500 * 80<\/gadget>\n760_000<\/output>\n760_000 \/ 100<\/gadget>\n7_600<\/output>\n7_600 \/ 2<\/gadget>\n3_800<\/output>\n3_800<\/result>","index":674} +{"problem":"a man can row upstream at 40 kmph and downstream at 52 kmph , and then find the speed of the man in still water ?","rationale":"\"us = 40 ds = 52 m = ( 40 + 52 ) \/ 2 = 46 answer : d\"","correct":"d","options":{"a":"27 ","b":"40 ","c":"42 ","d":"46","e":"24"},"options_float":{"a":27.0,"b":40.0,"c":42.0,"d":46.0,"e":24.0},"annotated_formula":"divide(add(40, 52), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"40 + 52<\/gadget>\n92<\/output>\n92 \/ 2<\/gadget>\n46<\/output>\n46<\/result>","index":675} +{"problem":"a can do a piece of work in 10 days and b alone can do it in 20 days . how much time will both take to finish the work ?","rationale":"\"this question can be solved by different methods . we need to conserve time in exams so solving this problem using equations is the good idea . time taken to finish the job = xy \/ ( x + y ) = 10 x 20 \/ ( 10 + 20 ) = 200 \/ 30 = 6.666 days answer : c\"","correct":"c","options":{"a":"5.333 ","b":"6 ","c":"6.666 ","d":"8.333","e":"9"},"options_float":{"a":5.333,"b":6.0,"c":6.666,"d":8.333,"e":9.0},"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":676} +{"problem":"a ferry can transport 100 tons of vehicles . automobiles range in weight from 1,600 to 3,000 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ?","rationale":"\"to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 100 tons will be 200,000 pounds from the answer choices : let max number of vehicles be 120 total weight will be = 120 * 1600 = 192000 pounds , which is lesser than the maximum weight allowed . ans : e\"","correct":"e","options":{"a":"110 ","b":"128 ","c":"115 ","d":"130","e":"120"},"options_float":{"a":110.0,"b":128.0,"c":115.0,"d":130.0,"e":120.0},"annotated_formula":"divide(multiply(multiply(100, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100))","linear_formula":"add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)|","chain":"100 * 2<\/gadget>\n200<\/output>\n200 * 1_000<\/gadget>\n200_000<\/output>\n1_000 + 100<\/gadget>\n1_100<\/output>\n1_100 + 100<\/gadget>\n1_200<\/output>\n1_200 + 100<\/gadget>\n1_300<\/output>\n1_300 + 100<\/gadget>\n1_400<\/output>\n1_400 + 100<\/gadget>\n1_500<\/output>\n1_500 + 100<\/gadget>\n1_600<\/output>\n200_000 \/ 1_600<\/gadget>\n125<\/output>\n125<\/result>","index":677} +{"problem":"the simple interest on rs . 10 for 4 months at the rate of 3 paise per rupeeper month is","rationale":"\"sol . s . i . = rs . [ 10 * 3 \/ 100 * 4 ] = rs . 1.20 answer a\"","correct":"a","options":{"a":"1.2 ","b":"1.4 ","c":"2.25 ","d":"3.21","e":"none"},"options_float":{"a":1.2,"b":1.4,"c":2.25,"d":3.21,"e":null},"annotated_formula":"divide(multiply(multiply(10, 4), 3), const_100)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|","chain":"10 * 4<\/gadget>\n40<\/output>\n40 * 3<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n6\/5 = around 1.2<\/result>","index":678} +{"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 25 % 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 = 1.25 x red + yellow = 0.5 x red = 0.25 x , since red = yellow . the answer is c .\"","correct":"c","options":{"a":"10 % ","b":"15 % ","c":"25 % ","d":"35 %","e":"40 %"},"options_float":{"a":10.0,"b":15.0,"c":25.0,"d":35.0,"e":40.0},"annotated_formula":"multiply(divide(divide(subtract(25, 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>\n25 - 15<\/gadget>\n10<\/output>\n5 \/ 20<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":680} +{"problem":"if - 4 and - 8 are negative integers , then - 4 * - 8 + 2 is","rationale":"answer : c","correct":"c","options":{"a":"- 32 ","b":"- 30 ","c":"34 ","d":"- 24","e":"- 10"},"options_float":{"a":-32.0,"b":-30.0,"c":34.0,"d":-24.0,"e":-10.0},"annotated_formula":"add(multiply(negate(4), negate(8)), 2)","linear_formula":"negate(n0)|negate(n1)|multiply(#0,#1)|add(n4,#2)","chain":"-4<\/gadget>\n-4<\/output>\n-8<\/gadget>\n-8<\/output>\n(-4) * (-8)<\/gadget>\n32<\/output>\n32 + 2<\/gadget>\n34<\/output>\n34<\/result>","index":682} +{"problem":"the maximum number of students among them 1234 pens and 874 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :","rationale":"\"explanation : required number of students = h . c . f of 1234 and 874 = 2 . answer : b\"","correct":"b","options":{"a":"91 ","b":"2 ","c":"1001 ","d":"1911","e":"none of these"},"options_float":{"a":91.0,"b":2.0,"c":1001.0,"d":1911.0,"e":null},"annotated_formula":"gcd(1234, 874)","linear_formula":"gcd(n0,n1)|","chain":"gcd(1_234, 874)<\/gadget>\n2<\/output>\n2<\/result>","index":683} +{"problem":"a boatman selling a boat along river flow . if he sell boat in steal water at 3 m \/ sec and flow of river is 2 m \/ sec . how much time he will take to sell 100 m .","rationale":"net speed = 3 + 2 = 5 m \/ sec distance = 100 m time = 100 \/ 5 = 20 sec answer d","correct":"d","options":{"a":"30 ","b":"10 ","c":"15 ","d":"20","e":"25"},"options_float":{"a":30.0,"b":10.0,"c":15.0,"d":20.0,"e":25.0},"annotated_formula":"divide(100, add(3, 2))","linear_formula":"add(n0,n1)|divide(n2,#0)","chain":"3 + 2<\/gadget>\n5<\/output>\n100 \/ 5<\/gadget>\n20<\/output>\n20<\/result>","index":684} +{"problem":"a man can row his boat with the stream at 12 km \/ h and against the stream in 8 km \/ h . the man ' s rate is ?","rationale":"explanation : ds = 12 us = 8 s = ? s = ( 12 - 8 ) \/ 2 = 2 kmph answer : a","correct":"a","options":{"a":"2 kmph ","b":"6 kmph ","c":"7 kmph ","d":"4 kmph","e":"9 kmph"},"options_float":{"a":2.0,"b":6.0,"c":7.0,"d":4.0,"e":9.0},"annotated_formula":"divide(subtract(12, 8), const_2)","linear_formula":"subtract(n0,n1)|divide(#0,const_2)","chain":"12 - 8<\/gadget>\n4<\/output>\n4 \/ 2<\/gadget>\n2<\/output>\n2<\/result>","index":685} +{"problem":"a vessel of capacity 2 litre has 25 % of alcohol and another vessel of capacity 6 litre had 30 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ?","rationale":"\"25 % of 2 litres = 0.5 litres 30 % of 6 litres = 1.8 litres therefore , total quantity of alcohol is 2.3 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 23 % answer : a\"","correct":"a","options":{"a":"23 % . ","b":"71 % . ","c":"49 % . ","d":"29 % .","e":"51 % ."},"options_float":{"a":23.0,"b":71.0,"c":49.0,"d":29.0,"e":51.0},"annotated_formula":"multiply(divide(add(multiply(divide(25, const_100), 2), multiply(divide(30, const_100), 6)), 10), const_100)","linear_formula":"divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 2<\/gadget>\n1\/2 = around 0.5<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 6<\/gadget>\n9\/5 = around 1.8<\/output>\n(1\/2) + (9\/5)<\/gadget>\n23\/10 = around 2.3<\/output>\n(23\/10) \/ 10<\/gadget>\n23\/100 = around 0.23<\/output>\n(23\/100) * 100<\/gadget>\n23<\/output>\n23<\/result>","index":688} +{"problem":"jack and jill work at a hospital with 4 other workers . for an internal review , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ?","rationale":"\"total number of ways to choose 2 out of 6 workers = 6 ! \/ 2 ! 4 ! = 15 number of ways to choose both jack and jill = 1 probability = 1 \/ 15 c should be the answer\"","correct":"c","options":{"a":"1 \/ 3 ","b":"1 \/ 4 ","c":"1 \/ 15 ","d":"3 \/ 8","e":"2 \/ 3"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.0666666667,"d":0.375,"e":0.6666666667},"annotated_formula":"inverse(divide(factorial(6), multiply(factorial(2), factorial(4))))","linear_formula":"factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)|","chain":"factorial(6)<\/gadget>\n720<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\nfactorial(4)<\/gadget>\n24<\/output>\n2 * 24<\/gadget>\n48<\/output>\n720 \/ 48<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n1\/15 = around 0.066667<\/result>","index":690} +{"problem":"a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 35 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ?","rationale":"\"find the product of the two increases : ( 1.4 ) * ( 1.35 ) which is 1.89 and a 89 % increase . d\"","correct":"d","options":{"a":"65 % ","b":"70 % ","c":"75 % ","d":"89 %","e":"95 %"},"options_float":{"a":65.0,"b":70.0,"c":75.0,"d":89.0,"e":95.0},"annotated_formula":"multiply(subtract(divide(multiply(multiply(const_100, add(const_1, divide(40, const_100))), add(const_1, divide(35, const_100))), const_100), const_1), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,const_100)|multiply(#2,#4)|divide(#5,const_100)|subtract(#6,const_1)|multiply(#7,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n100 * (7\/5)<\/gadget>\n140<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n1 + (7\/20)<\/gadget>\n27\/20 = around 1.35<\/output>\n140 * (27\/20)<\/gadget>\n189<\/output>\n189 \/ 100<\/gadget>\n189\/100 = around 1.89<\/output>\n(189\/100) - 1<\/gadget>\n89\/100 = around 0.89<\/output>\n(89\/100) * 100<\/gadget>\n89<\/output>\n89<\/result>","index":691} +{"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 150 milliliters of solution x to create a solution that is 25 percent alcohol by volume ?","rationale":"\"we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 150 then y = 450 answer : d\"","correct":"d","options":{"a":"250 \/ 3 ","b":"500 \/ 3 ","c":"400 ","d":"450","e":"600"},"options_float":{"a":83.3333333333,"b":166.6666666667,"c":400.0,"d":450.0,"e":600.0},"annotated_formula":"multiply(divide(subtract(25, 10), subtract(30, 25)), 150)","linear_formula":"subtract(n3,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n2,#2)|","chain":"25 - 10<\/gadget>\n15<\/output>\n30 - 25<\/gadget>\n5<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\n3 * 150<\/gadget>\n450<\/output>\n450<\/result>","index":693} +{"problem":"a cycle is bought for rs . 750 and sold for rs . 1080 , find the gain percent ?","rationale":"\"750 - - - - 180 100 - - - - ? = > 44 % answer : b\"","correct":"b","options":{"a":"22 ","b":"44 ","c":"99 ","d":"88","e":"11"},"options_float":{"a":22.0,"b":44.0,"c":99.0,"d":88.0,"e":11.0},"annotated_formula":"multiply(divide(subtract(1080, 750), 750), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_080 - 750<\/gadget>\n330<\/output>\n330 \/ 750<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 100<\/gadget>\n44<\/output>\n44<\/result>","index":694} +{"problem":"the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 62.5 . when one student dropped the class , the average of the remaining scores increased to 65.0 . what is the quiz score of the student who dropped the class ?","rationale":"\"total score of 16 students is 16 * 62.50 = 1000 total score of 15 students is 15 * 65 = 975 so , the score of the person who left is 25 ( 1000 - 975 ) answer will be ( b )\"","correct":"b","options":{"a":"10 ","b":"25 ","c":"40 ","d":"55","e":"70"},"options_float":{"a":10.0,"b":25.0,"c":40.0,"d":55.0,"e":70.0},"annotated_formula":"subtract(multiply(16, 62.5), multiply(subtract(16, const_1), 65.0))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)|","chain":"16 * 62.5<\/gadget>\n1_000<\/output>\n16 - 1<\/gadget>\n15<\/output>\n15 * 65<\/gadget>\n975<\/output>\n1_000 - 975<\/gadget>\n25<\/output>\n25<\/result>","index":696} +{"problem":"when positive integer x is divided by positive integer y , the remainder is 8 . 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 8 - - > x = qy + 8 ; x \/ y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 8 - - > y = 66.7 . answer : c .\"","correct":"c","options":{"a":"96 ","b":"75 ","c":"66.7 ","d":"25","e":"12"},"options_float":{"a":96.0,"b":75.0,"c":66.7,"d":25.0,"e":12.0},"annotated_formula":"divide(8, 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>\n8 \/ 0.12<\/gadget>\n66.666667<\/output>\n66.666667<\/result>","index":698} +{"problem":"a bullock cart has to cover a distance of 80 km in 10 hrs . if it covers half of the journey in 3 \/ 5 th time . what should be its speed to cover the remaining distance in the time left .","rationale":"a 10 kmph time left = 10 - 3 \/ 5 * 10 = 4 hr 10 km \/ h speed = 40 km \/ 4 hr = 10 kmph","correct":"a","options":{"a":"10 kmph ","b":"20 kmph ","c":"30 kmph ","d":"40 kmph","e":"50 kmph"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(divide(80, const_2), subtract(10, multiply(divide(10, 5), 3)))","linear_formula":"divide(n0,const_2)|divide(n1,n3)|multiply(n2,#1)|subtract(n1,#2)|divide(#0,#3)","chain":"80 \/ 2<\/gadget>\n40<\/output>\n10 \/ 5<\/gadget>\n2<\/output>\n2 * 3<\/gadget>\n6<\/output>\n10 - 6<\/gadget>\n4<\/output>\n40 \/ 4<\/gadget>\n10<\/output>\n10<\/result>","index":699} +{"problem":"reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 300 as interest at the end of the loan period , what was the rate of interest ?","rationale":"\"let rate = r % and time = r years . then , 1200 x r x r \/ 100 = 300 12 r 2 = 300 r 2 = 25 r = 5 . answer : a\"","correct":"a","options":{"a":"5 ","b":"6 ","c":"18 ","d":"can not be determined","e":"none of these"},"options_float":{"a":5.0,"b":6.0,"c":18.0,"d":null,"e":null},"annotated_formula":"sqrt(divide(multiply(300, const_100), 1200))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)|","chain":"300 * 100<\/gadget>\n30_000<\/output>\n30_000 \/ 1_200<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n5<\/result>","index":700} +{"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 20 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 + 20 x = 5.5 minutes answer : a\"","correct":"a","options":{"a":"5.5 ","b":"15 ","c":"25.5 ","d":"30","e":"60"},"options_float":{"a":5.5,"b":15.0,"c":25.5,"d":30.0,"e":60.0},"annotated_formula":"subtract(subtract(divide(const_60, const_2), 20), 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 - 20<\/gadget>\n10<\/output>\n20 \/ 1.5<\/gadget>\n13.333333<\/output>\n60 \/ 13.333333<\/gadget>\n4.5<\/output>\n10 - 4.5<\/gadget>\n5.5<\/output>\n5.5<\/result>","index":702} +{"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 150 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 ) = 150 2.5 h = 150 h = 60 . a is the answer .\"","correct":"a","options":{"a":"60 cm ","b":"90 cm ","c":"100 cm ","d":"120 cm","e":"130 cm"},"options_float":{"a":60.0,"b":90.0,"c":100.0,"d":120.0,"e":130.0},"annotated_formula":"divide(150, 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>\n150 \/ (5\/2)<\/gadget>\n60<\/output>\n60<\/result>","index":703} +{"problem":"a furniture manufacturer has two machines , but only one can be used at a time . machine w is utilized during the first shift and machine b during the second shift , while both work half of the third shift . if machine w can do the job in 12 days working two shifts and machine b can do the job in 15 days working two shifts , how many days will it take to do the job with the current work schedule ?","rationale":"' approximately ' could actually make such a question ambiguous . not this one though but a similar question with the answer as 9.2 days . you round off 8.89 days as 9 days and everything is fine in this question . what do you do when you get 9.2 days ? do you need 9 days or 10 days ? can you round off 9.2 as 9 even though that is what you do with numbers ? no , because in 9 days your work is not over . you do need 10 days . to finish a work machine w say you need to work full 9 days and a part of the 10 th day . if i ask you how many days do you need to complete the work , will you say 9 or 10 ? you will say 10 even if you do n ' t use the 10 th day fully = d","correct":"d","options":{"a":"14 ","b":"13 ","c":"11 ","d":"9","e":"7"},"options_float":{"a":14.0,"b":13.0,"c":11.0,"d":9.0,"e":7.0},"annotated_formula":"inverse(add(inverse(divide(multiply(12, const_2), add(const_1, divide(const_1, const_2)))), inverse(divide(multiply(15, const_2), add(const_1, divide(const_1, const_2))))))","linear_formula":"divide(const_1,const_2)|multiply(n0,const_2)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#3)|divide(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|inverse(#8)","chain":"12 * 2<\/gadget>\n24<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n24 \/ (3\/2)<\/gadget>\n16<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n15 * 2<\/gadget>\n30<\/output>\n30 \/ (3\/2)<\/gadget>\n20<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/16) + (1\/20)<\/gadget>\n9\/80 = around 0.1125<\/output>\n1 \/ (9\/80)<\/gadget>\n80\/9 = around 8.888889<\/output>\n80\/9 = around 8.888889<\/result>","index":704} +{"problem":"calculate 469200 x 9999 = ?","rationale":"\"answer 469200 x 9999 = 469200 x ( 10000 - 1 ) = 4692000000 - 469200 = 4691100843 . option : e\"","correct":"e","options":{"a":"4586970843 ","b":"4686970743 ","c":"4691100843 ","d":"4586870843","e":"4691530800"},"options_float":{"a":4586970843.0,"b":4686970743.0,"c":4691100843.0,"d":4586870843.0,"e":4691530800.0},"annotated_formula":"multiply(469200, 9999)","linear_formula":"multiply(n0,n1)|","chain":"469_200 * 9_999<\/gadget>\n4_691_530_800<\/output>\n4_691_530_800<\/result>","index":706} +{"problem":"how many seconds will a train 100 meters long take to cross a bridge 130 meters long if the speed of the train is 36 kmph ?","rationale":"\"explanation : d = 100 + 130 = 230 s = 36 * 5 \/ 18 = 10 mps t = 230 \/ 10 = 23 sec answer : option d\"","correct":"d","options":{"a":"26 ","b":"72 ","c":"25 ","d":"23","e":"27"},"options_float":{"a":26.0,"b":72.0,"c":25.0,"d":23.0,"e":27.0},"annotated_formula":"divide(add(130, 100), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"130 + 100<\/gadget>\n230<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n230 \/ 10<\/gadget>\n23<\/output>\n23<\/result>","index":707} +{"problem":"a is two years older than b who is twice as old as c . if the total ages of a , b and c be 27 . what is the age of b ?","rationale":"\"c age x , then b age is 2 x so a age is 2 x + 2 . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 x = 5 so b is 2 x = 2 ( 5 ) 2 x 5 = 10 answer : b\"","correct":"b","options":{"a":"12 years ","b":"10 years ","c":"8 years ","d":"14 years","e":"16 years"},"options_float":{"a":12.0,"b":10.0,"c":8.0,"d":14.0,"e":16.0},"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":709} +{"problem":"a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 22 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 22 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ?","rationale":"total number of installments = 52 payment per installment for the first 22 installments = 410 payment per installment for the remaining 30 installments = 410 + 65 = 475 average = ( 22 * 410 + 30 * 475 ) \/ 52 = 447.50 answer a","correct":"a","options":{"a":"447.5 ","b":"450 ","c":"465 ","d":"468","e":"475"},"options_float":{"a":447.5,"b":450.0,"c":465.0,"d":468.0,"e":475.0},"annotated_formula":"divide(add(multiply(22, 410), multiply(add(410, 65), subtract(52, 22))), 52)","linear_formula":"add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0)","chain":"22 * 410<\/gadget>\n9_020<\/output>\n410 + 65<\/gadget>\n475<\/output>\n52 - 22<\/gadget>\n30<\/output>\n475 * 30<\/gadget>\n14_250<\/output>\n9_020 + 14_250<\/gadget>\n23_270<\/output>\n23_270 \/ 52<\/gadget>\n895\/2 = around 447.5<\/output>\n895\/2 = around 447.5<\/result>","index":710} +{"problem":"the ratio of numbers is 5 : 6 and their h . c . f is 4 . their l . c . m is :","rationale":"\"let the numbers be 5 x and 6 x . then their h . c . f = x . so , x = 4 . so , the numbers are 20 and 24 . l . c . m of 20 and 24 = 120 . answer : e\"","correct":"e","options":{"a":"12 ","b":"16 ","c":"24 ","d":"48","e":"120"},"options_float":{"a":12.0,"b":16.0,"c":24.0,"d":48.0,"e":120.0},"annotated_formula":"lcm(multiply(5, 4), multiply(6, 4))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|lcm(#0,#1)|","chain":"5 * 4<\/gadget>\n20<\/output>\n6 * 4<\/gadget>\n24<\/output>\nlcm(20, 24)<\/gadget>\n120<\/output>\n120<\/result>","index":711} +{"problem":"a number when divided by a certain divisor left remainder 241 , when twice the number was divided by the same divisor , the remainder was 102 . find the divisor ?","rationale":"\"easy solution : n = dq 1 + 241 2 n = 2 dq 1 + 482 - ( 1 ) 2 n = dq 2 + 102 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 380 d * some integer = 380 checking all options only ( c ) syncs with it . answer c\"","correct":"c","options":{"a":"370 ","b":"365 ","c":"380 ","d":"456","e":"460"},"options_float":{"a":370.0,"b":365.0,"c":380.0,"d":456.0,"e":460.0},"annotated_formula":"subtract(multiply(241, const_2), 102)","linear_formula":"multiply(n0,const_2)|subtract(#0,n1)|","chain":"241 * 2<\/gadget>\n482<\/output>\n482 - 102<\/gadget>\n380<\/output>\n380<\/result>","index":712} +{"problem":"the banker â € ™ s discount of a certain sum of money is rs . 90 and the true discount on the same sum for the same time is rs . 60 . the sum due is","rationale":"\"sol . sum = b . d . * t . d . \/ b . d . - t . d . = rs . [ 90 * 60 \/ 90 - 60 ] = rs . [ 90 * 60 \/ 30 ] = rs . 180 answer b\"","correct":"b","options":{"a":"210 ","b":"180 ","c":"360 ","d":"450","e":"none"},"options_float":{"a":210.0,"b":180.0,"c":360.0,"d":450.0,"e":null},"annotated_formula":"divide(multiply(90, 60), subtract(90, 60))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"90 * 60<\/gadget>\n5_400<\/output>\n90 - 60<\/gadget>\n30<\/output>\n5_400 \/ 30<\/gadget>\n180<\/output>\n180<\/result>","index":713} +{"problem":"s is a positive integer and multiple of 2 ; p = 4 ^ s , what is the remainder when p is divided by 10 ?","rationale":"it is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer . to help see this , consider the following examples : 4 \/ 10 is 0 with a remainder of 4 14 \/ 10 is 1 with a remainder of 4 5 \/ 10 is 0 with a remainder of 5 105 \/ 10 is 10 with a remainder of 5 it is also essential to remember that the s is a positive integer and multiple of 2 . any integer that is a multiple of 2 is an even number . so , s must be a positive even integer . with these two observations , the question can be simplified to : what is the units digit of 4 raised to an even positive integer ? the units digit of 4 raised to an integer follows a specific repeating pattern : 4 ^ 1 = 4 4 ^ 2 = 16 4 ^ 3 = 64 4 ^ 4 = 256 4 ^ ( odd number ) - - > units digit of 4 4 ^ ( even number ) - - > units digit of 6 there is a clear pattern regarding the units digit . 4 raised to any odd integer has a units digit of 4 while 4 raised to any even integer has a units digit of 6 . since s must be an even integer , the units digit of p = 4 ^ s will always be 6 . consequently , the remainder when p = 4 ^ s is divided by 10 will always be 6 . in case this is too theoretical , consider the following examples : s = 2 - - > p = 4 ^ z = 16 - - > s \/ 10 = 1 with a remainder of 6 s = 4 - - > p = 4 ^ z = 256 - - > s \/ 10 = 25 with a remainder of 6 s = 6 - - > p = 4 ^ z = 4096 - - > s \/ 10 = 409 with a remainder of 6 s = 8 - - > p = 4 ^ z = 65536 - - > s \/ 10 = 6553 with a remainder of 6 answer : b .","correct":"b","options":{"a":"10 ","b":"6 ","c":"4 ","d":"0","e":"it can not be determined"},"options_float":{"a":10.0,"b":6.0,"c":4.0,"d":0.0,"e":null},"annotated_formula":"reminder(power(4, 2), 10)","linear_formula":"power(n1,n0)|reminder(#0,n2)","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 % 10<\/gadget>\n6<\/output>\n6<\/result>","index":714} +{"problem":"if the sum of two numbers is 22 and the sum of their squares is 386 , then the product of the numbers is","rationale":"\"sol . let the numbers be x and y . then , ( x + y ) = 22 and x 2 + y 2 = 386 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 22 ) 2 - 386 = 484 - 386 = 98 xy = 49 . answer b\"","correct":"b","options":{"a":"40 ","b":"49 ","c":"80 ","d":"88","e":"90"},"options_float":{"a":40.0,"b":49.0,"c":80.0,"d":88.0,"e":90.0},"annotated_formula":"divide(subtract(power(22, const_2), 386), const_2)","linear_formula":"power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|","chain":"22 ** 2<\/gadget>\n484<\/output>\n484 - 386<\/gadget>\n98<\/output>\n98 \/ 2<\/gadget>\n49<\/output>\n49<\/result>","index":715} +{"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 80 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 80 % of 50 % i . e 40 % . so he completed 50 + 40 = 90 % of total trip . e\"","correct":"e","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":"add(divide(const_100, const_2), divide(multiply(80, 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>\n80 * 50<\/gadget>\n4_000<\/output>\n4_000 \/ 100<\/gadget>\n40<\/output>\n50 + 40<\/gadget>\n90<\/output>\n90<\/result>","index":717} +{"problem":"if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 15 - m ) then m = ?","rationale":"\"( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 15 - m ) = 4 ^ ( ( 15 - m ) \/ 2 ) therefore , m = ( 15 - m ) \/ 2 2 m = 15 - m m = 5 answer d\"","correct":"d","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(15, add(2, const_1))","linear_formula":"add(n0,const_1)|divide(n3,#0)|","chain":"2 + 1<\/gadget>\n3<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n5<\/result>","index":718} +{"problem":"$ 378 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ?","rationale":"\"let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 378 x = 54 4 x = 216 the answer is c .\"","correct":"c","options":{"a":"$ 200 ","b":"$ 208 ","c":"$ 216 ","d":"$ 224","e":"$ 232"},"options_float":{"a":200.0,"b":208.0,"c":216.0,"d":224.0,"e":232.0},"annotated_formula":"multiply(divide(378, add(add(divide(const_1, const_2), const_1), const_2)), const_2)","linear_formula":"divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) + 1<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) + 2<\/gadget>\n7\/2 = around 3.5<\/output>\n378 \/ (7\/2)<\/gadget>\n108<\/output>\n108 * 2<\/gadget>\n216<\/output>\n216<\/result>","index":720} +{"problem":"harkamal purchased 8 kg of grapes at the rate of 75 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 = 75 × 8 = 600 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 600 + 495 = 1095 . e )\"","correct":"e","options":{"a":"1000 ","b":"1055 ","c":"1065 ","d":"1075","e":"1095"},"options_float":{"a":1000.0,"b":1055.0,"c":1065.0,"d":1075.0,"e":1095.0},"annotated_formula":"add(multiply(8, 75), multiply(9, 55))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"8 * 75<\/gadget>\n600<\/output>\n9 * 55<\/gadget>\n495<\/output>\n600 + 495<\/gadget>\n1_095<\/output>\n1_095<\/result>","index":721} +{"problem":"if the side length of square b is sqrt ( 5 ) times that of square a , the area of square b is how many times the area of square a ?","rationale":"\"let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( sqrt ( 5 ) x ) ^ 2 = 5 x ^ 2 . the answer is e .\"","correct":"e","options":{"a":"1.5 ","b":"2 ","c":"3 ","d":"4","e":"5"},"options_float":{"a":1.5,"b":2.0,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"power(sqrt(5), const_2)","linear_formula":"sqrt(n0)|power(#0,const_2)|","chain":"5 ** (1\/2)<\/gadget>\nsqrt(5) = around 2.236068<\/output>\n(sqrt(5)) ** 2<\/gadget>\n5<\/output>\n5<\/result>","index":722} +{"problem":"a certain quantity of 50 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ?","rationale":"\"let ' s say that the total original mixture a is 100 ml the original mixture a thus has 50 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 50 ml and 25 ml is 25 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 25 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 15 % . therefore , 15 out of 25 is 3 \/ 5 and e is the answer .\"","correct":"e","options":{"a":"1 \/ 4 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"2 \/ 3","e":"3 \/ 5"},"options_float":{"a":0.25,"b":0.3333333333,"c":0.5,"d":0.6666666667,"e":0.6},"annotated_formula":"inverse(add(divide(subtract(35, 25), subtract(50, 35)), const_1))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|add(#2,const_1)|inverse(#3)|","chain":"35 - 25<\/gadget>\n10<\/output>\n50 - 35<\/gadget>\n15<\/output>\n10 \/ 15<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) + 1<\/gadget>\n5\/3 = around 1.666667<\/output>\n1 \/ (5\/3)<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":723} +{"problem":"a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 20 % so that the customer could not buy the required amount of the product . the customer managed to buy only 80 % of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ?","rationale":"\"let x be the amount of money paid for the first purchase . the second time , the customer paid 0.8 ( 1.2 x ) = 0.96 x . the difference is 4 % . the answer is d .\"","correct":"d","options":{"a":"10 % ","b":"8 % ","c":"6 % ","d":"4 %","e":"2 %"},"options_float":{"a":10.0,"b":8.0,"c":6.0,"d":4.0,"e":2.0},"annotated_formula":"multiply(subtract(const_1, multiply(add(divide(20, const_100), const_1), divide(80, const_100))), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(6\/5) * (4\/5)<\/gadget>\n24\/25 = around 0.96<\/output>\n1 - (24\/25)<\/gadget>\n1\/25 = around 0.04<\/output>\n(1\/25) * 100<\/gadget>\n4<\/output>\n4<\/result>","index":724} +{"problem":"there is a lot of speculation that the economy of a country depends on how fast people spend their money in addition to how much they save . auggie was very curious to test this theory . auggie spent all of his money in 5 stores . in each store , he spent rs . 4 more than one - half of what he had when he went in . how many rupees did auggie have when he entered the first store ?","rationale":"amount left = 0.5 x − 4 for fifth store this is zero . so x = 8 . that means he entered fifth store with 8 . now for fourth store , amount left = 8 so 0.5 x − 4 = 8 ⇒ x = 24 for third store , amount left = 24 so 12 x − 4 = 24 ⇒ x = 56 for second store , amount left = 56 so 0.5 x − 4 = 56 ⇒ x = 120 for first store , amount left = 120 so 0.5 x − 4 = 120 ⇒ x = 248 so he entered first store with 248 . answer : a","correct":"a","options":{"a":"248 ","b":"120 ","c":"252 ","d":"250","e":"350"},"options_float":{"a":248.0,"b":120.0,"c":252.0,"d":250.0,"e":350.0},"annotated_formula":"multiply(add(multiply(add(multiply(add(multiply(add(multiply(4, const_2), 4), const_2), 4), const_2), 4), const_2), 4), const_2)","linear_formula":"multiply(n1,const_2)|add(n1,#0)|multiply(#1,const_2)|add(n1,#2)|multiply(#3,const_2)|add(n1,#4)|multiply(#5,const_2)|add(n1,#6)|multiply(#7,const_2)","chain":"4 * 2<\/gadget>\n8<\/output>\n8 + 4<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 + 4<\/gadget>\n28<\/output>\n28 * 2<\/gadget>\n56<\/output>\n56 + 4<\/gadget>\n60<\/output>\n60 * 2<\/gadget>\n120<\/output>\n120 + 4<\/gadget>\n124<\/output>\n124 * 2<\/gadget>\n248<\/output>\n248<\/result>","index":725} +{"problem":"3 people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car , then the share of each of the remaining persons increased by :","rationale":"original share of 1 person = 1 \/ 3 new share of 1 person = 1 \/ 2 increase = ( 1 \/ 2 - 1 \/ 3 = 1 \/ 6 ) therefore , required fraction = ( 1 \/ 6 ) \/ ( 1 \/ 3 ) = ( 1 \/ 6 ) x ( 3 \/ 1 ) = 1 \/ 2 answer is a .","correct":"a","options":{"a":"1 \/ 2 ","b":"2 \/ 7 ","c":"3 \/ 2 ","d":"4 \/ 7","e":"none of them"},"options_float":{"a":0.5,"b":0.2857142857,"c":1.5,"d":0.5714285714,"e":null},"annotated_formula":"divide(subtract(divide(const_1, const_2), divide(const_1, 3)), divide(const_1, 3))","linear_formula":"divide(const_1,const_2)|divide(const_1,n0)|subtract(#0,#1)|divide(#2,#1)","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>\n(1\/6) \/ (1\/3)<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":728} +{"problem":"two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively . in how much time will they cross each other , if they are running in the same direction ?","rationale":"solution relative speed = ( 45 - 40 ) kmph = 5 kmph = ( 5 x 5 \/ 18 ) m \/ sec = ( 25 \/ 18 ) m \/ sec time taken = ( 350 x 18 \/ 25 ) sec = 252 sec . answer d","correct":"d","options":{"a":"72 sec ","b":"132 sec ","c":"192 sec ","d":"252 sec","e":"none"},"options_float":{"a":72.0,"b":132.0,"c":192.0,"d":252.0,"e":null},"annotated_formula":"multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(45, 40)))","linear_formula":"add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)","chain":"200 + 150<\/gadget>\n350<\/output>\n350 \/ 1_000<\/gadget>\n7\/20 = around 0.35<\/output>\n45 - 40<\/gadget>\n5<\/output>\n(7\/20) \/ 5<\/gadget>\n7\/100 = around 0.07<\/output>\n3_600 * (7\/100)<\/gadget>\n252<\/output>\n252<\/result>","index":729} +{"problem":"a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 35 . 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 . 175 ( 35 * 5 ) profit percentage = ( 175 - 100 ) \/ 100 * 100 = 75 % answer : c\"","correct":"c","options":{"a":"80 % ","b":"50 % ","c":"75 % ","d":"40 %","e":"53 %"},"options_float":{"a":80.0,"b":50.0,"c":75.0,"d":40.0,"e":53.0},"annotated_formula":"subtract(multiply(35, 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>\n35 * 5<\/gadget>\n175<\/output>\n25 * 4<\/gadget>\n100<\/output>\n175 - 100<\/gadget>\n75<\/output>\n75<\/result>","index":730} +{"problem":"how many 1 \/ 8 s are there in 37 1 \/ 2 ?","rationale":"\"required number = ( 75 \/ 2 ) \/ ( 1 \/ 8 ) = ( 75 \/ 2 x 8 \/ 1 ) = 300 . answer : a\"","correct":"a","options":{"a":"300 ","b":"400 ","c":"500 ","d":"600","e":"700"},"options_float":{"a":300.0,"b":400.0,"c":500.0,"d":600.0,"e":700.0},"annotated_formula":"divide(add(37, divide(1, 2)), divide(1, 8))","linear_formula":"divide(n0,n4)|divide(n0,n1)|add(n2,#0)|divide(#2,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n37 + (1\/2)<\/gadget>\n75\/2 = around 37.5<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(75\/2) \/ (1\/8)<\/gadget>\n300<\/output>\n300<\/result>","index":731} +{"problem":"in triangle pqr , the angle q = 90 degree , pq = 5 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 : 5 = 8 : 10 - - > qy = 4.0 . answer : c .\"","correct":"c","options":{"a":"3.6 cm ","b":"2.4 cm ","c":"4.0 cm ","d":"2.16 cm","e":"3.2 cm"},"options_float":{"a":3.6,"b":2.4,"c":4.0,"d":2.16,"e":3.2},"annotated_formula":"divide(multiply(5, 8), const_10)","linear_formula":"multiply(n1,n2)|divide(#0,const_10)|","chain":"5 * 8<\/gadget>\n40<\/output>\n40 \/ 10<\/gadget>\n4<\/output>\n4<\/result>","index":733} +{"problem":"the average runs of a cricket player of 10 innings was 20 . how many runs must he make in his next innings so as to increase his average of runs by 4 ?","rationale":"\"explanation : average after 11 innings = 24 required number of runs = ( 24 * 11 ) – ( 20 * 10 ) = 264 – 200 = 64 answer : c\"","correct":"c","options":{"a":"22 ","b":"77 ","c":"64 ","d":"19","e":"17"},"options_float":{"a":22.0,"b":77.0,"c":64.0,"d":19.0,"e":17.0},"annotated_formula":"subtract(multiply(add(10, const_1), add(20, 4)), multiply(20, 10))","linear_formula":"add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"10 + 1<\/gadget>\n11<\/output>\n20 + 4<\/gadget>\n24<\/output>\n11 * 24<\/gadget>\n264<\/output>\n20 * 10<\/gadget>\n200<\/output>\n264 - 200<\/gadget>\n64<\/output>\n64<\/result>","index":734} +{"problem":"the sum of the even numbers between 1 and k is 79 * 80 , where k is an odd number , then k =","rationale":"\"the number of terms in this set would be : n = ( k - 1 ) \/ 2 ( as k is odd ) last term : k - 1 average would be first term + last term \/ 2 = ( 2 + k - 1 ) \/ 2 = ( k + 1 ) \/ 2 also average : sum \/ number of terms = 79 * 80 \/ ( ( k - 1 ) \/ 2 ) = 158 * 80 \/ ( k - 1 ) ( k + 1 ) \/ 2 = 158 * 80 \/ ( k - 1 ) - - > ( k - 1 ) ( k + 1 ) = 158 * 160 - - > k = 159 answer e .\"","correct":"e","options":{"a":"79 ","b":"80 ","c":"81 ","d":"157","e":"159"},"options_float":{"a":79.0,"b":80.0,"c":81.0,"d":157.0,"e":159.0},"annotated_formula":"add(multiply(79, const_2), 1)","linear_formula":"multiply(n1,const_2)|add(#0,n0)|","chain":"79 * 2<\/gadget>\n158<\/output>\n158 + 1<\/gadget>\n159<\/output>\n159<\/result>","index":736} +{"problem":"how long does a train 125 m long running at the speed of 78 km \/ hr takes to cross a bridge 125 m length ?","rationale":"\"speed = 78 * 5 \/ 18 = 21.7 m \/ sec total distance covered = 125 + 125 = 250 m . required time = 250 \/ 21.7 ' = 11.5 sec . answer : c\"","correct":"c","options":{"a":"12.7 sec ","b":"10.1 sec ","c":"11.5 sec ","d":"12.1 sec","e":"11.7 sec"},"options_float":{"a":12.7,"b":10.1,"c":11.5,"d":12.1,"e":11.7},"annotated_formula":"divide(add(125, 125), multiply(78, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"125 + 125<\/gadget>\n250<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n78 * (5\/18)<\/gadget>\n65\/3 = around 21.666667<\/output>\n250 \/ (65\/3)<\/gadget>\n150\/13 = around 11.538462<\/output>\n150\/13 = around 11.538462<\/result>","index":737} +{"problem":"ele , the circus elephant , is currently 3 times older than lyn , the circus lion . in 15 years from now , lyn the circus lion will be exactly half as old as ele , the circus elephant . how old is ele today ?","rationale":"ele , the circus elephant , is currently three times older than lyn , the circus lion . ele = 3 * lyn usually , ages are integers so there is a good possibility that the age of ele is 45 ( the only option that is a multiple of 3 ) . then age of lyn would be 15 . in 15 yrs , ele would be 60 and lyn would be 30 - so lyn would be half as old as ele . answer ( d )","correct":"d","options":{"a":"40 ","b":"48 ","c":"43 ","d":"45","e":"41"},"options_float":{"a":40.0,"b":48.0,"c":43.0,"d":45.0,"e":41.0},"annotated_formula":"multiply(subtract(multiply(const_2, 15), 15), 3)","linear_formula":"multiply(n1,const_2)|subtract(#0,n1)|multiply(n0,#1)","chain":"2 * 15<\/gadget>\n30<\/output>\n30 - 15<\/gadget>\n15<\/output>\n15 * 3<\/gadget>\n45<\/output>\n45<\/result>","index":738} +{"problem":"a man is 30 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is","rationale":"\"explanation : let ' s son age is x , then father age is x + 30 . = > 2 ( x + 2 ) = ( x + 30 + 2 ) = > 2 x + 4 = x + 32 = > x = 28 years option a\"","correct":"a","options":{"a":"28 years ","b":"22 years ","c":"23 years ","d":"24 years","e":"26 years"},"options_float":{"a":28.0,"b":22.0,"c":23.0,"d":24.0,"e":26.0},"annotated_formula":"divide(subtract(30, 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>\n30 - 2<\/gadget>\n28<\/output>\n2 - 1<\/gadget>\n1<\/output>\n28 \/ 1<\/gadget>\n28<\/output>\n28<\/result>","index":739} +{"problem":"the length of the bridge , which a train 110 meters long and travelling at 45 km \/ hr can cross in 30 seconds , is :","rationale":"\"speed = ( 45 * 5 \/ 18 ) m \/ sec = ( 25 \/ 2 ) m \/ sec . time = 30 sec . let the length of bridge be x meters . then , ( 110 + x ) \/ 30 = 25 \/ 2 = = > 2 ( 110 + x ) = 750 = = > x = 265 m . answer : option a\"","correct":"a","options":{"a":"265 ","b":"244 ","c":"245 ","d":"238","e":"236"},"options_float":{"a":265.0,"b":244.0,"c":245.0,"d":238.0,"e":236.0},"annotated_formula":"subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 110)","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 - 110<\/gadget>\n265<\/output>\n265<\/result>","index":741} +{"problem":"cole drove from home to work at an average speed of 60 kmh . he then returned home at an average speed of 100 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ?","rationale":"\"let the distance one way be x time from home to work = x \/ 60 time from work to home = x \/ 100 total time = 2 hrs ( x \/ 60 ) + ( x \/ 100 ) = 2 solving for x , we get x = 75 time from home to work in minutes = ( 75 ) * 60 \/ 60 = 75 minutes ans = d\"","correct":"d","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(100, 2), add(60, 100)), const_60)","linear_formula":"add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)|","chain":"100 * 2<\/gadget>\n200<\/output>\n60 + 100<\/gadget>\n160<\/output>\n200 \/ 160<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 60<\/gadget>\n75<\/output>\n75<\/result>","index":742} +{"problem":"if the radius of a circle is decreased 50 % , what happens to the area ?","rationale":"\"area of square = pi * radius ^ 2 new radius = 0.5 * old radius so new area = ( 0.5 ) ^ 2 old area = > 0.25 of old area = > 25 % old area ans : c\"","correct":"c","options":{"a":"10 % decrease ","b":"20 % decrease ","c":"75 % decrease ","d":"40 % decrease","e":"50 % decrease"},"options_float":{"a":10.0,"b":20.0,"c":75.0,"d":40.0,"e":50.0},"annotated_formula":"subtract(const_100, multiply(power(divide(50, const_100), const_2), const_100))","linear_formula":"divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 2<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n100 - 25<\/gadget>\n75<\/output>\n75<\/result>","index":743} +{"problem":"calculate the ratio between x and y if 90 % of x equal to 60 % of y ?","rationale":"\"explanation : 90 x = 60 y x : y = 90 : 60 = 3 : 2 answer : c\"","correct":"c","options":{"a":"1 : 2 ","b":"3 : 5 ","c":"3 : 2 ","d":"3 : 4","e":"5 : 2"},"options_float":{"a":0.5,"b":0.6,"c":1.5,"d":0.75,"e":2.5},"annotated_formula":"divide(90, 60)","linear_formula":"divide(n0,n1)|","chain":"90 \/ 60<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":746} +{"problem":"find the area of the quadrilateral of one of its diagonals is 50 cm and its off sets 15 cm and 5 cm ?","rationale":"\"1 \/ 2 * 50 ( 15 + 5 ) = 500 cm 2 answer : e\"","correct":"e","options":{"a":"100 cm 2 ","b":"150 cm 2 ","c":"200 cm 2 ","d":"250 cm 2","e":"500 cm 2"},"options_float":{"a":100.0,"b":150.0,"c":200.0,"d":250.0,"e":500.0},"annotated_formula":"multiply(multiply(divide(const_1, const_2), add(5, 15)), 50)","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>\n5 + 15<\/gadget>\n20<\/output>\n(1\/2) * 20<\/gadget>\n10<\/output>\n10 * 50<\/gadget>\n500<\/output>\n500<\/result>","index":747} +{"problem":"a right circular cone is exactly fitted inside a cube in such away that the edges of the base of the cone are touching the edges of one of the faces of the cube and the vertex is on the opposite face of the cube . if the volume of the cube is 343 cc , what approximately is the volume of the cone ?","rationale":"edge of the cube = 3 √ 334 = 7 cm ∴ radius of cone = 3.5 cm height = 7 cm volume of cone = 1 ⁄ 3 π r 2 h 1 ⁄ 3 π r 2 h = 1 ⁄ 3 × 22 ⁄ 7 × ( 3.5 ) 2 × 7 = 1 ⁄ 3 × 22 × 12.25 ≈ 90 sec answer b","correct":"b","options":{"a":"80 cc ","b":"90 cc ","c":"110 cc ","d":"105 cc","e":"100 cc"},"options_float":{"a":80.0,"b":90.0,"c":110.0,"d":105.0,"e":100.0},"annotated_formula":"volume_cone(divide(cube_edge_by_volume(343), const_2), cube_edge_by_volume(343))","linear_formula":"cube_edge_by_volume(n0)|divide(#0,const_2)|volume_cone(#1,#0)","chain":"343 ** (1\/3)<\/gadget>\n7<\/output>\n7 \/ 2<\/gadget>\n7\/2 = around 3.5<\/output>\npi * ((7\/2) ** 2) * 7 \/ 3<\/gadget>\n343*pi\/12 = around 89.79719<\/output>\n343*pi\/12 = around 89.79719<\/result>","index":750} +{"problem":"if the given two numbers are respectively 7 % and 28 % of a third number , then what percentage is the first of the second ?","rationale":"\"here , l = 7 and m = 28 therefore , first number = l \/ m x 100 % of second number = 7 \/ 28 x 100 % of second number = 25 % of second number answer : b\"","correct":"b","options":{"a":"20 % ","b":"25 % ","c":"18 % ","d":"30 %","e":"none of these"},"options_float":{"a":20.0,"b":25.0,"c":18.0,"d":30.0,"e":null},"annotated_formula":"multiply(divide(divide(7, const_100), divide(28, const_100)), const_100)","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|","chain":"7 \/ 100<\/gadget>\n7\/100 = around 0.07<\/output>\n28 \/ 100<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/100) \/ (7\/25)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":751} +{"problem":"a pupil ' s marks were wrongly entered as 83 instead of 53 . due to 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 = ( 83 - 53 ) = > x \/ 2 = 30 = > x = 60 . answer : c\"","correct":"c","options":{"a":"18 ","b":"82 ","c":"60 ","d":"27","e":"29"},"options_float":{"a":18.0,"b":82.0,"c":60.0,"d":27.0,"e":29.0},"annotated_formula":"multiply(subtract(83, 53), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|","chain":"83 - 53<\/gadget>\n30<\/output>\n30 * 2<\/gadget>\n60<\/output>\n60<\/result>","index":752} +{"problem":"the area of a rhombus is equal to the area of a rectangle whose length is 20 cm and the breadth is 10 cm . if one of the diagonals is 32 cm what is the length of other diagonal ?","rationale":"area of rectangle = 20 x 10 = 200 cm â ² let ' l ' the length of other diagonal = 0.5 x 32 xl = 200 which gives x = 12.5 cm answer : b","correct":"b","options":{"a":"10 ","b":"12.5 ","c":"15 ","d":"16","e":"17.5"},"options_float":{"a":10.0,"b":12.5,"c":15.0,"d":16.0,"e":17.5},"annotated_formula":"divide(multiply(rectangle_area(20, 10), const_2), 32)","linear_formula":"rectangle_area(n0,n1)|multiply(#0,const_2)|divide(#1,n2)","chain":"20 * 10<\/gadget>\n200<\/output>\n200 * 2<\/gadget>\n400<\/output>\n400 \/ 32<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":753} +{"problem":"if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2160 less than the principal . what was the principal ?","rationale":"p - 2160 = ( p * 5 * 4 ) \/ 100 p = 2700 answer : c","correct":"c","options":{"a":"1500 ","b":"2500 ","c":"2700 ","d":"3200","e":"11500"},"options_float":{"a":1500.0,"b":2500.0,"c":2700.0,"d":3200.0,"e":11500.0},"annotated_formula":"divide(2160, 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_160 \/ (4\/5)<\/gadget>\n2_700<\/output>\n2_700<\/result>","index":754} +{"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 $ 948 in total compensation , how many total hours did he work that week ?","rationale":"\"for 40 hrs = 40 * 16 = 640 excess = 948 - 640 = 308 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 308 \/ 28 = 77 \/ 7 = 11 total hrs = 40 + 11 = 51 answer e 51\"","correct":"e","options":{"a":"36 ","b":"40 ","c":"44 ","d":"48","e":"51"},"options_float":{"a":36.0,"b":40.0,"c":44.0,"d":48.0,"e":51.0},"annotated_formula":"add(40, divide(subtract(948, 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>\n948 - 640<\/gadget>\n308<\/output>\n100 + 75<\/gadget>\n175<\/output>\n16 * 175<\/gadget>\n2_800<\/output>\n2_800 \/ 100<\/gadget>\n28<\/output>\n308 \/ 28<\/gadget>\n11<\/output>\n40 + 11<\/gadget>\n51<\/output>\n51<\/result>","index":755} +{"problem":"a line has a slope of 3 \/ 4 and intersects the point w ( - 12 , - 39 ) . at which point does this line intersect the x - axis ?","rationale":"\"assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 \/ 4 as the slope is given to be 3 \/ 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point w of intersection is ( 40,0 ) . a is the correct answer .\"","correct":"a","options":{"a":"( 40,0 ) ","b":"( 30,0 ) ","c":"( 0,40 ) ","d":"( 40,30 )","e":"( 0,30 )"},"options_float":{"a":400.0,"b":300.0,"c":40.0,"d":4030.0,"e":30.0},"annotated_formula":"multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10)","linear_formula":"divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10)|","chain":"-39<\/gadget>\n-39<\/output>\n-12<\/gadget>\n-12<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(-12) * (3\/4)<\/gadget>\n-9<\/output>\n(-39) - (-9)<\/gadget>\n-30<\/output>\n(-30) \/ (3\/4)<\/gadget>\n-40<\/output>\n-(-40)<\/gadget>\n40<\/output>\n40 * 10<\/gadget>\n400<\/output>\n400<\/result>","index":756} +{"problem":"a salesman commission is 5 % on all sales upto $ 10000 and 4 % on all sales exceeding this . he remits $ 31100 to his parent company after deducting his commission . find the total sales ?","rationale":"let his total sales be x total sales - commission = $ 31100 x - [ ( 5 % of 10000 ) + 4 % of ( x - 10000 ) ] = 31100 x - 500 - ( x - 10000 ) \/ 25 = 31100 x = 32500 answer is b","correct":"b","options":{"a":"$ 30000 ","b":"$ 32500 ","c":"$ 35120 ","d":"$ 41520","e":"$ 25460"},"options_float":{"a":30000.0,"b":32500.0,"c":35120.0,"d":41520.0,"e":25460.0},"annotated_formula":"divide(add(31100, subtract(divide(multiply(5, 10000), const_100), divide(multiply(4, 10000), const_100))), subtract(const_1, divide(4, const_100)))","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,const_100)|subtract(const_1,#0)|subtract(#3,#4)|add(n3,#6)|divide(#7,#5)","chain":"5 * 10_000<\/gadget>\n50_000<\/output>\n50_000 \/ 100<\/gadget>\n500<\/output>\n4 * 10_000<\/gadget>\n40_000<\/output>\n40_000 \/ 100<\/gadget>\n400<\/output>\n500 - 400<\/gadget>\n100<\/output>\n31_100 + 100<\/gadget>\n31_200<\/output>\n4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 - (1\/25)<\/gadget>\n24\/25 = around 0.96<\/output>\n31_200 \/ (24\/25)<\/gadget>\n32_500<\/output>\n32_500<\/result>","index":757} +{"problem":"how much 60 % of 50 is greater than 34 % of 30 ?","rationale":"( 60 \/ 100 ) * 50 – ( 34 \/ 100 ) * 30 30 - 10.2 = 19.8 answer : b","correct":"b","options":{"a":"18 ","b":"19.8 ","c":"11 ","d":"17","e":"12"},"options_float":{"a":18.0,"b":19.8,"c":11.0,"d":17.0,"e":12.0},"annotated_formula":"subtract(divide(multiply(60, 50), const_100), divide(multiply(34, 30), const_100))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)","chain":"60 * 50<\/gadget>\n3_000<\/output>\n3_000 \/ 100<\/gadget>\n30<\/output>\n34 * 30<\/gadget>\n1_020<\/output>\n1_020 \/ 100<\/gadget>\n51\/5 = around 10.2<\/output>\n30 - (51\/5)<\/gadget>\n99\/5 = around 19.8<\/output>\n99\/5 = around 19.8<\/result>","index":758} +{"problem":"a man can row 5 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 = 5 s = 2 ds = 7 us = 3 x \/ 7 + x \/ 3 = 1 x = 2.1 d = 2.1 * 2 = 4.2 answer : d\"","correct":"d","options":{"a":"5.75 ","b":"5.7 ","c":"5.76 ","d":"4.2","e":"5.71"},"options_float":{"a":5.75,"b":5.7,"c":5.76,"d":4.2,"e":5.71},"annotated_formula":"multiply(divide(multiply(add(5, 2), subtract(5, 2)), add(add(5, 2), subtract(5, 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":"5 + 2<\/gadget>\n7<\/output>\n5 - 2<\/gadget>\n3<\/output>\n7 * 3<\/gadget>\n21<\/output>\n7 + 3<\/gadget>\n10<\/output>\n21 \/ 10<\/gadget>\n21\/10 = around 2.1<\/output>\n(21\/10) * 2<\/gadget>\n21\/5 = around 4.2<\/output>\n21\/5 = around 4.2<\/result>","index":759} +{"problem":"how long does a truck of 200 m long traveling at 60 kmph takes to cross a bridge of 180 m in length ?","rationale":"d = 200 + 180 = 380 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 380 * 3 \/ 50 = 22.8 sec answer : c","correct":"c","options":{"a":"36.7 ","b":"26.8 ","c":"22.8 ","d":"21.1","e":"16.2"},"options_float":{"a":36.7,"b":26.8,"c":22.8,"d":21.1,"e":16.2},"annotated_formula":"divide(add(200, 180), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)","chain":"200 + 180<\/gadget>\n380<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n380 \/ (50\/3)<\/gadget>\n114\/5 = around 22.8<\/output>\n114\/5 = around 22.8<\/result>","index":760} +{"problem":"difference between the length & breadth of a rectangle is 10 m . if its perimeter is 206 m , then its area is ?","rationale":"solving the two equations , we get : l = 30 and b = 40 . area = ( l x b ) = ( 30 x 40 ) m 2 = 1200 m ^ 2 d","correct":"d","options":{"a":"2400 m ^ 2 ","b":"1500 m ^ 2 ","c":"2520 m ^ 2 ","d":"1200 m ^ 2","e":"2580 m ^ 2"},"options_float":{"a":2400.0,"b":1500.0,"c":2520.0,"d":1200.0,"e":2580.0},"annotated_formula":"multiply(multiply(10, const_4), multiply(10, const_3))","linear_formula":"multiply(n0,const_4)|multiply(n0,const_3)|multiply(#0,#1)","chain":"10 * 4<\/gadget>\n40<\/output>\n10 * 3<\/gadget>\n30<\/output>\n40 * 30<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":761} +{"problem":"for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.367 ] = 1 and [ 1.999 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 \/ 4000 ?","rationale":"\"factorization of 45 = 3 * 3 * 5 factorization of 80 = 5 * 2 ^ 4 means the smallest value of k is \\ sqrt { 3 ^ 2 * 5 * 2 ^ 4 } = 3 * 5 * 2 ^ 2 k ^ 3 = 3 ^ 3 * 5 ^ 3 * 2 ^ 6 * x = 3 ^ 3 * 2 * 4000 * x where x can be any integer k ^ 3 \/ 4000 = 3 ^ 3 * 2 * x = 54 x which obviously has different units digit depending on x answer : d\"","correct":"d","options":{"a":"0 ","b":"1 ","c":"27 ","d":"54","e":"can not be determined"},"options_float":{"a":0.0,"b":1.0,"c":27.0,"d":54.0,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4)","linear_formula":"multiply(n6,n9)|multiply(#0,#0)|multiply(#1,#0)|divide(#2,const_4)|","chain":"3 * 2<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n36 * 6<\/gadget>\n216<\/output>\n216 \/ 4<\/gadget>\n54<\/output>\n54<\/result>","index":764} +{"problem":"paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 40 % of his income to sam and 60 % of his income to quentin , quentin ' s new income would be what fraction of sam ' s new income ?","rationale":"make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get s = 7.6 and q 10.8 so 10.8 \/ 7.6 = 2.7 \/ 1.9 ans : e","correct":"e","options":{"a":"11 \/ 12 ","b":"13 \/ 17 ","c":"13 \/ 19 ","d":"12 \/ 19","e":"27 \/ 19"},"options_float":{"a":0.9166666667,"b":0.7647058824,"c":0.6842105263,"d":0.6315789474,"e":1.4210526316},"annotated_formula":"divide(add(multiply(60, const_100), multiply(60, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 60)))","linear_formula":"add(n0,n1)|multiply(n4,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|multiply(n4,#3)|multiply(n4,#0)|add(#1,#4)|add(#2,#5)|divide(#6,#7)","chain":"60 * 100<\/gadget>\n6_000<\/output>\n100 - 20<\/gadget>\n80<\/output>\n60 * 80<\/gadget>\n4_800<\/output>\n6_000 + 4_800<\/gadget>\n10_800<\/output>\n40 * 100<\/gadget>\n4_000<\/output>\n40 + 20<\/gadget>\n60<\/output>\n60 * 60<\/gadget>\n3_600<\/output>\n4_000 + 3_600<\/gadget>\n7_600<\/output>\n10_800 \/ 7_600<\/gadget>\n27\/19 = around 1.421053<\/output>\n27\/19 = around 1.421053<\/result>","index":765} +{"problem":"if p \/ q = 3 \/ 7 , then 2 p + q = ?","rationale":"\"let p = 3 , q = 7 then 2 * 3 + 7 = 13 so 2 p + q = 13 . answer : c\"","correct":"c","options":{"a":"12 ","b":"14 ","c":"13 ","d":"15","e":"16"},"options_float":{"a":12.0,"b":14.0,"c":13.0,"d":15.0,"e":16.0},"annotated_formula":"add(multiply(3, 2), 7)","linear_formula":"multiply(n0,n2)|add(n1,#0)|","chain":"3 * 2<\/gadget>\n6<\/output>\n6 + 7<\/gadget>\n13<\/output>\n13<\/result>","index":766} +{"problem":"in a box of 9 pens , a total of 2 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":"\"# defective pens = 2 # good pens = 7 probability of the 1 st pen being good = 7 \/ 9 probability of the 2 nd pen being good = 6 \/ 8 total probability = 7 \/ 9 * 6 \/ 8 = 7 \/ 12 the answer is d .\"","correct":"d","options":{"a":"4 \/ 7 ","b":"5 \/ 9 ","c":"6 \/ 11 ","d":"7 \/ 12","e":"8 \/ 15"},"options_float":{"a":0.5714285714,"b":0.5555555556,"c":0.5454545455,"d":0.5833333333,"e":0.5333333333},"annotated_formula":"multiply(divide(subtract(9, 2), 9), divide(subtract(subtract(9, 2), 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 - 2<\/gadget>\n7<\/output>\n7 \/ 9<\/gadget>\n7\/9 = around 0.777778<\/output>\n7 - 1<\/gadget>\n6<\/output>\n9 - 1<\/gadget>\n8<\/output>\n6 \/ 8<\/gadget>\n3\/4 = around 0.75<\/output>\n(7\/9) * (3\/4)<\/gadget>\n7\/12 = around 0.583333<\/output>\n7\/12 = around 0.583333<\/result>","index":767} +{"problem":"if 36 men can do a piece of work in 25 hours , in how mwny hours will 15 men do it ?","rationale":"\"explanation : let the required no of hours be x . then less men , more hours ( indirct proportion ) \\ inline \\ fn _ jvn \\ therefore 15 : 36 : : 25 : x \\ inline \\ fn _ jvn \\ leftrightarrow ( 15 x x ) = ( 36 x 25 ) \\ inline \\ fn _ jvn \\ leftrightarrow \\ inline \\ fn _ jvn x = \\ frac { 36 \\ times 25 } { 15 } = 60 hence , 15 men can do it in 60 hours . answer : c ) 60\"","correct":"c","options":{"a":"22 ","b":"38 ","c":"60 ","d":"88","e":"72"},"options_float":{"a":22.0,"b":38.0,"c":60.0,"d":88.0,"e":72.0},"annotated_formula":"divide(multiply(36, 25), 15)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"36 * 25<\/gadget>\n900<\/output>\n900 \/ 15<\/gadget>\n60<\/output>\n60<\/result>","index":769} +{"problem":"last year the price range ( per kg ) for 100 varieties of apples in wholesale market was $ 100 . if the prices of each of the 100 varieties increased by 10 percent this year over what it was last year , what is the range of the wholesale prices of the 1000 varieties of apples this year ?","rationale":"let the lowest price be x . therefore , highest price is x + 100 . now price of each variety is increased by 10 % . therefore the price will remain arranged in the same order as before . or lowest price = 1.1 x and highest = 1.1 * ( x + 100 ) or range = highest - lowest = 1.1 * ( x + 100 ) - 1.1 x = 110 , hence , c","correct":"c","options":{"a":"$ 50 ","b":"$ 100 ","c":"$ 110 ","d":"$ 600","e":"$ 300"},"options_float":{"a":50.0,"b":100.0,"c":110.0,"d":600.0,"e":300.0},"annotated_formula":"multiply(100, divide(add(10, const_100), const_100))","linear_formula":"add(n3,const_100)|divide(#0,const_100)|multiply(n0,#1)","chain":"10 + 100<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n100 * (11\/10)<\/gadget>\n110<\/output>\n110<\/result>","index":771} +{"problem":"the percentage profit earned by selling an article for rs . 1920 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":"\"explanation : let the c . p . of the article be rs . x given that % profit earned by selling article at rs . 1920 = % loss incurred by selling article at rs . 1280 ( 1920 − x \/ x ) ∗ 100 = ( x − 1280 \/ x ) ∗ 100 = > 1920 - x = x - 1280 = > 2 x = 3200 = > x = 1600 s . p . for 25 % profit = rs . 1600 + 25 % of rs . 1600 = rs . 1600 * ( 125 \/ 100 ) = rs . 2000 answer : a\"","correct":"a","options":{"a":"rs . 2000 ","b":"rs . 2200 ","c":"rs . 2400 ","d":"data inadequate","e":"can not be determined"},"options_float":{"a":2000.0,"b":2200.0,"c":2400.0,"d":null,"e":null},"annotated_formula":"multiply(divide(add(const_100, 25), const_100), divide(add(1920, 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_920 + 1_280<\/gadget>\n3_200<\/output>\n3_200 \/ 2<\/gadget>\n1_600<\/output>\n(5\/4) * 1_600<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":772} +{"problem":"the tailor has a 10 meter long piece of fabric for which to sew a ball room dress . she has to cuts this fabric into strips of 200 centimeters each . how long will it take the tailor to complete this tasks if each 200 centimeter took 5 minutes to cut ?","rationale":"the tailors would need to cut the fabric 49 times thus the total amount spent would be 245 minutes . the answer is d","correct":"d","options":{"a":"150 ","b":"200 ","c":"188 ","d":"245","e":"123"},"options_float":{"a":150.0,"b":200.0,"c":188.0,"d":245.0,"e":123.0},"annotated_formula":"multiply(subtract(divide(multiply(10, const_1000), 200), const_1), 5)","linear_formula":"multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1)|multiply(n3,#2)","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 200<\/gadget>\n50<\/output>\n50 - 1<\/gadget>\n49<\/output>\n49 * 5<\/gadget>\n245<\/output>\n245<\/result>","index":773} +{"problem":"in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 4720 votes were polled , then how many valid votes did b get ?","rationale":"\"let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 4720 32.5 k = 32.5 k \/ 100 k * 4720 = 1534 the number of valid votes polled in favour of b is 1534 . answer : a\"","correct":"a","options":{"a":"1534 ","b":"2999 ","c":"2834 ","d":"2777","e":"2991"},"options_float":{"a":1534.0,"b":2999.0,"c":2834.0,"d":2777.0,"e":2991.0},"annotated_formula":"divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 4720), divide(multiply(15, 4720), const_100)), const_2)","linear_formula":"divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(const_1,#0)|multiply(n2,#3)|subtract(#4,#2)|divide(#5,const_2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 4_720<\/gadget>\n3_776<\/output>\n15 * 4_720<\/gadget>\n70_800<\/output>\n70_800 \/ 100<\/gadget>\n708<\/output>\n3_776 - 708<\/gadget>\n3_068<\/output>\n3_068 \/ 2<\/gadget>\n1_534<\/output>\n1_534<\/result>","index":776} +{"problem":"if $ 1088 are divided between worker a and worker b in the ratio 5 : 11 , what is the share that worker b will get ?","rationale":"\"worker b will get 11 \/ 16 = 68.75 % the answer is d .\"","correct":"d","options":{"a":"62.45 % ","b":"64.55 % ","c":"66.65 % ","d":"68.75 %","e":"70.85 %"},"options_float":{"a":62.45,"b":64.55,"c":66.65,"d":68.75,"e":70.85},"annotated_formula":"divide(1088, add(5, 11))","linear_formula":"add(n1,n2)|divide(n0,#0)|","chain":"5 + 11<\/gadget>\n16<\/output>\n1_088 \/ 16<\/gadget>\n68<\/output>\n68<\/result>","index":777} +{"problem":"the average age of 6 men increases by 2 years when two women are included in place of two men of ages 20 and 24 years . find the average age of the women ?","rationale":"\"20 + 24 + 6 * 2 = 56 \/ 2 = 28 answer : e\"","correct":"e","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":"divide(add(add(20, 24), multiply(6, 2)), const_2)","linear_formula":"add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)|","chain":"20 + 24<\/gadget>\n44<\/output>\n6 * 2<\/gadget>\n12<\/output>\n44 + 12<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":778} +{"problem":"operation # is defined as : a # b = 4 a ^ 2 + 4 b ^ 2 + 8 ab for all non - negative integers . what is the value of ( a + b ) + 4 , when a # b = 100 ?","rationale":"official solution : ( b ) we know that a # b = 100 and a # b = 4 a ² + 4 b ² + 8 ab . so 4 a ² + 4 b ² + 8 ab = 100 we can see that 4 a ² + 4 b ² + 8 ab is a well - known formula for ( 2 a + 2 b ) ² . therefore ( 2 a + 2 b ) ² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbers . so we can conclude that 2 ( a + b ) = 10 . ( a + b ) + 4 = 10 \/ 2 + 4 = 9 . the correct answer is b .","correct":"b","options":{"a":"5 ","b":"9 ","c":"10 ","d":"13","e":"17"},"options_float":{"a":5.0,"b":9.0,"c":10.0,"d":13.0,"e":17.0},"annotated_formula":"add(sqrt(divide(100, 4)), 4)","linear_formula":"divide(n6,n0)|sqrt(#0)|add(n0,#1)","chain":"100 \/ 4<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n5 + 4<\/gadget>\n9<\/output>\n9<\/result>","index":779} +{"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 $ 500 per sq . metre .","rationale":"\"area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 500 hence , total cost = 5.5 × 3.75 × 500 = $ 10312.50 a\"","correct":"a","options":{"a":"$ 10312.50 ","b":"$ 13512.50 ","c":"$ 16512.50 ","d":"$ 14512.50","e":"$ 18512.50"},"options_float":{"a":10312.5,"b":13512.5,"c":16512.5,"d":14512.5,"e":18512.5},"annotated_formula":"multiply(500, multiply(5.5, 3.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 3.75<\/gadget>\n20.625<\/output>\n500 * 20.625<\/gadget>\n10_312.5<\/output>\n10_312.5<\/result>","index":780} +{"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 10 sq . feet , how many feet of fencing will be required ?","rationale":"\"we have : l = 20 ft and lb = 10 sq . ft . so , b = 0.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 1 ) ft = 21 ft . answer : d\"","correct":"d","options":{"a":"34 ","b":"40 ","c":"68 ","d":"21","e":"78"},"options_float":{"a":34.0,"b":40.0,"c":68.0,"d":21.0,"e":78.0},"annotated_formula":"add(multiply(divide(10, 20), const_2), 20)","linear_formula":"divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|","chain":"10 \/ 20<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 2<\/gadget>\n1<\/output>\n1 + 20<\/gadget>\n21<\/output>\n21<\/result>","index":781} +{"problem":"each of 3 investments has a 20 % of becoming worthless within a year of purchase , independently of what happens to the other two investments . if simone invests an equal sum k in each of these 3 investments on january 1 , the approximate chance that by the end of the year , she loses no more than 1 \/ 3 of her original investment is","rationale":"the problem asks for the approximate chance that no more than 1 \/ 3 of the original investment is lost . we can apply the “ 1 – x ” technique : what ’ s the chance that more than 1 \/ 3 of the original investment is lost ? there are two outcomes we have to separately measure : ( a ) all 3 investments become worthless . ( b ) 2 of the 3 investments become worthless , while 1 doesn ’ t . outcome ( a ) : the probability is ( 0.2 ) ( 0.2 ) ( 0.2 ) = 0.008 , or a little less than 1 % . outcome ( b ) : call the investments x , y , and z . the probability that x retains value , while y and z become worthless , is ( 0.8 ) ( 0.2 ) ( 0.2 ) = 0.032 . now , we have to do the same thing for the specific scenarios in which y retains value ( while x and z don ’ t ) and in which z retains value ( while x and y don ’ t ) . each of those scenarios results in the same math : 0.032 . thus , we can simply multiply 0.032 by 3 to get 0.096 , or a little less than 10 % . the sum of these two probabilities is 0.008 + 0.096 = 0.104 , or a little more than 10 % . finally , subtracting from 100 % and rounding , we find that the probability we were looking for is approximately 90 % . the correct answer is a . this problem illustrates the power of diversification in financial investments . all else being equal , it ’ s less risky to hold a third of your money in three uncorrelated ( independent ) but otherwise equivalent investments than to put all your eggs in one of the baskets . that said , be wary of historical correlations ! housing price changes in different us cities were not so correlated — and then they became highly correlated during the recent housing crisis ( they all fell together ) , fatally undermining spreadsheet models that assumed that these price changes were independent .","correct":"a","options":{"a":"90 % ","b":"80 % ","c":"70 % ","d":"60 %","e":"40 %"},"options_float":{"a":90.0,"b":80.0,"c":70.0,"d":60.0,"e":40.0},"annotated_formula":"subtract(add(multiply(20, const_2), multiply(20, 3)), const_10)","linear_formula":"multiply(n1,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#2,const_10)","chain":"20 * 2<\/gadget>\n40<\/output>\n20 * 3<\/gadget>\n60<\/output>\n40 + 60<\/gadget>\n100<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90<\/result>","index":782} +{"problem":"a ferry can transport 50 tons of vehicles . automobiles range in weight from 1,600 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ?","rationale":"to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 50 tons will be 100,000 pounds from the answer choices : let max number of vehicles be 62 total weight will be = 62 * 1600 = 99200 pounds , which is lesser than the maximum weight allowed . ans : d","correct":"d","options":{"a":"23 ","b":"41 ","c":"48 ","d":"62","e":"86"},"options_float":{"a":23.0,"b":41.0,"c":48.0,"d":62.0,"e":86.0},"annotated_formula":"divide(multiply(multiply(50, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100))","linear_formula":"add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)","chain":"50 * 2<\/gadget>\n100<\/output>\n100 * 1_000<\/gadget>\n100_000<\/output>\n1_000 + 100<\/gadget>\n1_100<\/output>\n1_100 + 100<\/gadget>\n1_200<\/output>\n1_200 + 100<\/gadget>\n1_300<\/output>\n1_300 + 100<\/gadget>\n1_400<\/output>\n1_400 + 100<\/gadget>\n1_500<\/output>\n1_500 + 100<\/gadget>\n1_600<\/output>\n100_000 \/ 1_600<\/gadget>\n125\/2 = around 62.5<\/output>\n125\/2 = around 62.5<\/result>","index":784} +{"problem":"what is the ratio between perimeters of two squares one having 8 times the diagonal then the other ?","rationale":"\"d = 8 d d = d a √ 2 = 8 d a √ 2 = d a = 8 d \/ √ 2 a = d \/ √ 2 = > 8 : 1 answer : d\"","correct":"d","options":{"a":"3 : 6 ","b":"3 : 3 ","c":"3 : 8 ","d":"8 : 1","e":"3 : 2"},"options_float":{"a":0.5,"b":1.0,"c":0.375,"d":8.0,"e":1.5},"annotated_formula":"divide(8, divide(8, 8))","linear_formula":"divide(n0,n0)|divide(n0,#0)|","chain":"8 \/ 8<\/gadget>\n1<\/output>\n8 \/ 1<\/gadget>\n8<\/output>\n8<\/result>","index":785} +{"problem":"a bank pays interest to its customers on the last day of the year . the interest paid to a customer is calculated as 10 % of the average monthly balance maintained by the customer . john is a customer at the bank . on the last day , when the interest was accumulated into his account , his bank balance doubled to $ 5080 . what is the average monthly balance maintained by john in his account during the year ?","rationale":"bank balance is doubled with accumulation of interest tp 5080 . . this means interest is 5080 \/ 2 = 2540 for entire year . . although since interest is 10 % of avg monthly balance , it becomes 25400 . . d","correct":"d","options":{"a":"2840 ","b":"5680 ","c":"6840 ","d":"25400","e":"28400"},"options_float":{"a":2840.0,"b":5680.0,"c":6840.0,"d":25400.0,"e":28400.0},"annotated_formula":"multiply(5080, divide(10, const_2))","linear_formula":"divide(n0,const_2)|multiply(n1,#0)","chain":"10 \/ 2<\/gadget>\n5<\/output>\n5_080 * 5<\/gadget>\n25_400<\/output>\n25_400<\/result>","index":786} +{"problem":"a bus 75 m long is running with a speed of 21 km \/ hr . in what time will it pass a woman who is walking at 3 km \/ hr in the direction opposite to that in which the bus is going ?","rationale":"\"speed of bus relative to woman = 21 + 3 = 24 km \/ hr . = 24 * 5 \/ 18 = 20 \/ 3 m \/ sec . time taken to pass the woman = 75 * 3 \/ 20 = 11.25 sec . answer : c\"","correct":"c","options":{"a":"5.75 ","b":"7.62 ","c":"11.25 ","d":"4.25","e":"3.25"},"options_float":{"a":5.75,"b":7.62,"c":11.25,"d":4.25,"e":3.25},"annotated_formula":"divide(divide(multiply(75, const_3600), add(21, 3)), const_1000)","linear_formula":"add(n1,n2)|multiply(n0,const_3600)|divide(#1,#0)|divide(#2,const_1000)|","chain":"75 * 3_600<\/gadget>\n270_000<\/output>\n21 + 3<\/gadget>\n24<\/output>\n270_000 \/ 24<\/gadget>\n11_250<\/output>\n11_250 \/ 1_000<\/gadget>\n45\/4 = around 11.25<\/output>\n45\/4 = around 11.25<\/result>","index":787} +{"problem":"reeya obtained 40 , 60 , 70 , 80 and 80 out of 100 in different subjects , what will be the average","rationale":"\"explanation : ( 40 + 60 + 70 + 80 + 80 \/ 5 ) = 66 option a\"","correct":"a","options":{"a":"66 ","b":"75 ","c":"80 ","d":"85","e":"90"},"options_float":{"a":66.0,"b":75.0,"c":80.0,"d":85.0,"e":90.0},"annotated_formula":"divide(add(add(add(add(40, 60), 70), 80), 80), 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":"40 + 60<\/gadget>\n100<\/output>\n100 + 70<\/gadget>\n170<\/output>\n170 + 80<\/gadget>\n250<\/output>\n250 + 80<\/gadget>\n330<\/output>\n4 + 1<\/gadget>\n5<\/output>\n330 \/ 5<\/gadget>\n66<\/output>\n66<\/result>","index":789} +{"problem":"the distance from city a to city b is 100 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ?","rationale":"\"the time it takes bob to drive to city b is 100 \/ 40 = 2.5 hours . alice needs to take less than 2 hours for the trip . alice needs to exceed a constant speed of 100 \/ 2 = 50 miles per hour . the answer is c .\"","correct":"c","options":{"a":"45 ","b":"48 ","c":"50 ","d":"52","e":"54"},"options_float":{"a":45.0,"b":48.0,"c":50.0,"d":52.0,"e":54.0},"annotated_formula":"divide(100, subtract(divide(100, 40), divide(30, const_60)))","linear_formula":"divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2)|","chain":"100 \/ 40<\/gadget>\n5\/2 = around 2.5<\/output>\n30 \/ 60<\/gadget>\n1\/2 = around 0.5<\/output>\n(5\/2) - (1\/2)<\/gadget>\n2<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":790} +{"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 24 , 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 } ( less than 24 ) 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 } ( less than 24 ) 17 is the only common number in both the sets . hence k = 17 answer : d\"","correct":"d","options":{"a":"2 ","b":"5 ","c":"6 ","d":"3","e":"8"},"options_float":{"a":2.0,"b":5.0,"c":6.0,"d":3.0,"e":8.0},"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":793} +{"problem":"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 32 , then how old is b ?","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 = 32 5 x = 30 = > x = 6 hence , b ' s age = 2 x = 12 years . 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":"multiply(divide(subtract(32, const_2), add(const_3, const_2)), const_2)","linear_formula":"add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|","chain":"32 - 2<\/gadget>\n30<\/output>\n3 + 2<\/gadget>\n5<\/output>\n30 \/ 5<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n12<\/result>","index":794} +{"problem":"how long does a train 110 m long running at the speed of 72 km \/ hr takes to cross a bridge 132 m length ?","rationale":"\"speed = 72 * 5 \/ 18 = 20 m \/ sec total distance covered = 110 + 132 = 242 m . required time = 242 \/ 20 = 12.1 sec . answer : b\"","correct":"b","options":{"a":"12.6 ","b":"12.0 ","c":"12.1 ","d":"12.3","e":"12.2"},"options_float":{"a":12.6,"b":12.0,"c":12.1,"d":12.3,"e":12.2},"annotated_formula":"divide(add(110, 132), multiply(72, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"110 + 132<\/gadget>\n242<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n242 \/ 20<\/gadget>\n121\/10 = around 12.1<\/output>\n121\/10 = around 12.1<\/result>","index":795} +{"problem":"mary ' s income is 60 % more than tim ' s income and tim ' s income is 60 % less than juan ' s income . what % of juan ' s income is mary ' s income .","rationale":"\"even i got 96 % j = 100 t = 100 * 0.4 = 40 m = 40 * 1.6 = 64 if mary ' s income is x percent of j m = j * x \/ 100 x = m * 100 \/ j = 64 * 100 \/ 100 = 64 ans : c\"","correct":"c","options":{"a":"124 % ","b":"b . 120 % ","c":"64 % ","d":"80 %","e":"64 %"},"options_float":{"a":124.0,"b":120.0,"c":64.0,"d":80.0,"e":64.0},"annotated_formula":"add(subtract(const_100, 60), multiply(subtract(const_100, 60), divide(60, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)|","chain":"100 - 60<\/gadget>\n40<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n40 * (3\/5)<\/gadget>\n24<\/output>\n40 + 24<\/gadget>\n64<\/output>\n64<\/result>","index":797} +{"problem":"11 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ?","rationale":"\"no . of ways of picking 2 biology books ( from 11 books ) = 11 c 2 = ( 11 * 10 ) \/ 2 = 55 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) \/ 2 = 28 total ways of picking 2 books of each type = 55 * 28 = 1540 ( option e )\"","correct":"e","options":{"a":"80 ","b":"160 ","c":"720 ","d":"1100","e":"1540"},"options_float":{"a":80.0,"b":160.0,"c":720.0,"d":1100.0,"e":1540.0},"annotated_formula":"multiply(divide(divide(factorial(11), factorial(subtract(11, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2))","linear_formula":"factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9)|","chain":"factorial(11)<\/gadget>\n39_916_800<\/output>\n11 - 2<\/gadget>\n9<\/output>\nfactorial(9)<\/gadget>\n362_880<\/output>\n39_916_800 \/ 362_880<\/gadget>\n110<\/output>\n110 \/ 2<\/gadget>\n55<\/output>\nfactorial(8)<\/gadget>\n40_320<\/output>\n8 - 2<\/gadget>\n6<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\n40_320 \/ 720<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n55 * 28<\/gadget>\n1_540<\/output>\n1_540<\/result>","index":798} +{"problem":"pipe a can fill the tank in 30 minutes and pipe b can empty the tank in 90 minutes . how long it will take to fill the tank if both pipes are operating together ?","rationale":"pipe a fills 1 \/ 30 th of the tank in a minute and pipe b empties 1 \/ 90 th of the tank ( 1 \/ 30 ) - ( 1 \/ 90 ) = ( 1 \/ x ) 2 \/ 90 = 1 \/ x = > x = 45 answer : d","correct":"d","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":"divide(const_1, subtract(divide(const_1, 30), divide(const_1, 90)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n1 \/ 90<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/30) - (1\/90)<\/gadget>\n1\/45 = around 0.022222<\/output>\n1 \/ (1\/45)<\/gadget>\n45<\/output>\n45<\/result>","index":799} +{"problem":"an electric pump can fill a tank in 4 hours . because of a leak in the tank , it took 8 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ?","rationale":"work done by the leak in 1 hour = 1 \/ 4 - 1 \/ 8 = 1 \/ 8 the leak will empty the tank in 8 hours answer is c","correct":"c","options":{"a":"10 hours ","b":"12 hours ","c":"8 hours ","d":"5 hours","e":"15 hours"},"options_float":{"a":10.0,"b":12.0,"c":8.0,"d":5.0,"e":15.0},"annotated_formula":"divide(8, const_1)","linear_formula":"divide(n1,const_1)","chain":"8 \/ 1<\/gadget>\n8<\/output>\n8<\/result>","index":800} +{"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 $ 203 ?","rationale":"\"i think it should be d . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 6 20 - pack for 3.05 * 5 = $ 18.30 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 5 = 2120\"","correct":"d","options":{"a":"1,108 ","b":"2,100 ","c":"2,108 ","d":"2,120","e":"2,256"},"options_float":{"a":1108.0,"b":2100.0,"c":2108.0,"d":2120.0,"e":2256.0},"annotated_formula":"multiply(divide(203, 22.95), 250)","linear_formula":"divide(n6,n5)|multiply(n4,#0)|","chain":"203 \/ 22.95<\/gadget>\n8.845316<\/output>\n8.845316 * 250<\/gadget>\n2_211.329<\/output>\n2_211.329<\/result>","index":801} +{"problem":"if 3 girls can do 3 times of a particular work in 3 days , then , 7 girls can do 7 times of that work in","rationale":"answer : option ' d ' that is , 1 girl can do one time of the work in 3 days . therefore , 7 girls can do 7 times work in the same 3 days itself .","correct":"d","options":{"a":"1 1 \/ 5 days ","b":"2 days ","c":"2 1 \/ 5 days ","d":"3 days","e":"4 days"},"options_float":{"a":1.0,"b":2.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"multiply(divide(3, 3), 3)","linear_formula":"divide(n0,n0)|multiply(n0,#0)","chain":"3 \/ 3<\/gadget>\n1<\/output>\n1 * 3<\/gadget>\n3<\/output>\n3<\/result>","index":803} +{"problem":"a sum fetched a total simple interest of 4043.25 at the rate of 9 % . p . a . in 5 years . what is the sum ?","rationale":"\"principal = ( 100 x 4043.25 ) \/ ( 9 x 5 ) = 404325 \/ 45 = 8985 . answer b\"","correct":"b","options":{"a":"5768 ","b":"8985 ","c":"2345 ","d":"6474","e":"8723"},"options_float":{"a":5768.0,"b":8985.0,"c":2345.0,"d":6474.0,"e":8723.0},"annotated_formula":"divide(divide(multiply(4043.25, const_100), 9), 5)","linear_formula":"multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|","chain":"4_043.25 * 100<\/gadget>\n404_325<\/output>\n404_325 \/ 9<\/gadget>\n44_925<\/output>\n44_925 \/ 5<\/gadget>\n8_985<\/output>\n8_985<\/result>","index":804} +{"problem":"arun is travelling on his cycle and has calculated to reach point a at 2 pm if he travels at 10 kmph . he will reach there at 12 noon if he travels at 15 kmph . at what speed must he travel to reach a at 1 pm ?","rationale":"let distance be x km travelling at 10 kmph reach at 2 pm travelling at 15 kmph reach at 12 noon = > time taken when travelling at 10 km - time taken when travelling at 15 km = 2 hrs x \/ 10 - x \/ 15 = 2 3 x - 2 x * 30 x = 60 time needed if travelled at 10 kmph = 60 \/ 10 = 6 hrs = > reach at 1 pm = > ( 6 - 1 ) = 5 hrs req speed = 60 \/ 5 = 12 kmph answer b","correct":"b","options":{"a":"8 kmph ","b":"12 kmph ","c":"10 kmph ","d":"14 kmph","e":"15 kmph"},"options_float":{"a":8.0,"b":12.0,"c":10.0,"d":14.0,"e":15.0},"annotated_formula":"divide(add(15, 10), 2)","linear_formula":"add(n1,n3)|divide(#0,n0)","chain":"15 + 10<\/gadget>\n25<\/output>\n25 \/ 2<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":806} +{"problem":"rs . 600 amounts to rs . 900 in 3 years at simple interest . if the interest is increased by 4 % , it would amount to how much ?","rationale":"\"( 600 * 3 * 4 ) \/ 100 = 72 600 + 72 = 672 answer : a\"","correct":"a","options":{"a":"672 ","b":"246 ","c":"258 ","d":"856","e":"653"},"options_float":{"a":672.0,"b":246.0,"c":258.0,"d":856.0,"e":653.0},"annotated_formula":"multiply(power(add(const_1, divide(4, const_100)), 3), 600)","linear_formula":"divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n1 + (1\/25)<\/gadget>\n26\/25 = around 1.04<\/output>\n(26\/25) ** 3<\/gadget>\n17_576\/15_625 = around 1.124864<\/output>\n(17_576\/15_625) * 600<\/gadget>\n421_824\/625 = around 674.9184<\/output>\n421_824\/625 = around 674.9184<\/result>","index":807} +{"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 = 84403 cm = 844.03 m answer : b\"","correct":"b","options":{"a":"843.03 m ","b":"844.03 m ","c":"845.03 m ","d":"846.03 m","e":"847.03 m"},"options_float":{"a":843.03,"b":844.03,"c":845.03,"d":846.03,"e":847.03},"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":808} +{"problem":"a certain bag contains 100 balls â € ” 50 white , 25 green , 10 yellow , 7 red , and 8 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ?","rationale":"according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) \/ ( total ) = ( 50 + 25 + 10 ) \/ 100 = 85 \/ 100 = 0.85 . answer is b","correct":"b","options":{"a":"0.9 ","b":"0.85 ","c":"0.6 ","d":"0.8","e":"0.5"},"options_float":{"a":0.9,"b":0.85,"c":0.6,"d":0.8,"e":0.5},"annotated_formula":"divide(subtract(100, add(7, 8)), 100)","linear_formula":"add(n4,n5)|subtract(n0,#0)|divide(#1,n0)","chain":"7 + 8<\/gadget>\n15<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n17\/20 = around 0.85<\/result>","index":811} +{"problem":"a 1 k . m . long wire is held by n poles . if one pole is removed , the length of the gap becomes 12 \/ 3 m . what is the number of poles initially ?","rationale":"length after removing pole is 12 \/ 3 = 4 then before removing pole is 2 ( ' coz | 2 | 2 | is | 4 | ) i . e . gap between two poles is 2 m 1 km = 1000 m then split 1000 m by 2 m = > we have 500 sections or gaps then no . of poles is 500 + 1 st pole = 501 poles therefore n = 501 . answer : b","correct":"b","options":{"a":"500 ","b":"501 ","c":"502 ","d":"503","e":"504"},"options_float":{"a":500.0,"b":501.0,"c":502.0,"d":503.0,"e":504.0},"annotated_formula":"subtract(add(add(add(add(multiply(multiply(12, 3), const_12), const_10), multiply(const_10, const_4)), const_10), const_10), 1)","linear_formula":"multiply(n1,n2)|multiply(const_10,const_4)|multiply(#0,const_12)|add(#2,const_10)|add(#3,#1)|add(#4,const_10)|add(#5,const_10)|subtract(#6,n0)","chain":"12 * 3<\/gadget>\n36<\/output>\n36 * 12<\/gadget>\n432<\/output>\n432 + 10<\/gadget>\n442<\/output>\n10 * 4<\/gadget>\n40<\/output>\n442 + 40<\/gadget>\n482<\/output>\n482 + 10<\/gadget>\n492<\/output>\n492 + 10<\/gadget>\n502<\/output>\n502 - 1<\/gadget>\n501<\/output>\n501<\/result>","index":812} +{"problem":"a certain deep blue paint contains 45 percent blue pigment and 55 percent red pigment by weight . a certain green paint contains 35 percent blue pigment and 65 percent yellow pigment . when these paints are mixed to produce a brown paint , the brown paint contains 40 percent blue pigment . if the brown paint weighs 10 grams , then the red pigment contributes how many grams of that weight ?","rationale":"10 grams of combined mixture and 40 % blue pigment means that the mixtures were mixed 50 % each . thus 5 grams a piece . out of the 5 grams of the dark blue paint , 60 % is red . therefore , 5 * . 55 = 2.75 grams of red pigment","correct":"d","options":{"a":"1.5 ","b":"2.5 ","c":"3.5 ","d":"2.75","e":"4.5"},"options_float":{"a":1.5,"b":2.5,"c":3.5,"d":2.75,"e":4.5},"annotated_formula":"multiply(divide(55, multiply(const_100, const_2)), 10)","linear_formula":"multiply(const_100,const_2)|divide(n1,#0)|multiply(n5,#1)","chain":"100 * 2<\/gadget>\n200<\/output>\n55 \/ 200<\/gadget>\n11\/40 = around 0.275<\/output>\n(11\/40) * 10<\/gadget>\n11\/4 = around 2.75<\/output>\n11\/4 = around 2.75<\/result>","index":813} +{"problem":"the average expenditure of a labourer for 6 months was 95 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i","rationale":"\"income of 6 months = ( 6 × 95 ) – debt = 570 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 840 average monthly income = 840 ÷ 10 = 84 answer e\"","correct":"e","options":{"a":"70 ","b":"72 ","c":"75 ","d":"78","e":"84"},"options_float":{"a":70.0,"b":72.0,"c":75.0,"d":78.0,"e":84.0},"annotated_formula":"divide(add(add(multiply(95, 6), multiply(60, 4)), 30), add(6, 4))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)|","chain":"95 * 6<\/gadget>\n570<\/output>\n60 * 4<\/gadget>\n240<\/output>\n570 + 240<\/gadget>\n810<\/output>\n810 + 30<\/gadget>\n840<\/output>\n6 + 4<\/gadget>\n10<\/output>\n840 \/ 10<\/gadget>\n84<\/output>\n84<\/result>","index":816} +{"problem":"if 60 % of 3 \/ 5 of a number is 18 , then the number is ?","rationale":"\"let the number be x . then 60 % of 3 \/ 5 of x = 18 60 \/ 100 * 3 \/ 5 * x = 18 x = ( 18 * 25 \/ 9 ) = 50 required number = 50 . correct option : d\"","correct":"d","options":{"a":"80 ","b":"100 ","c":"75 ","d":"50","e":"none of these"},"options_float":{"a":80.0,"b":100.0,"c":75.0,"d":50.0,"e":null},"annotated_formula":"divide(18, multiply(divide(60, const_100), divide(3, 5)))","linear_formula":"divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * (3\/5)<\/gadget>\n9\/25 = around 0.36<\/output>\n18 \/ (9\/25)<\/gadget>\n50<\/output>\n50<\/result>","index":819} +{"problem":"a driver just filled the car ' s gas tank with 18 liters of gasohol , a mixture consisting of 5 % ethanol and 95 % gasoline . if the car runs best on a mixture consisting of 10 % ethanol and 90 % gasoline , how many liters of ethanol must be added into the gas tank for the car to achieve optimum performance ?","rationale":"\"let x be the number of liters of ethanol added to the gas tank . 0.05 ( 18 ) + x = 0.1 ( 18 + x ) 0.9 x = 1.8 - 0.9 x = 1 liter the answer is a .\"","correct":"a","options":{"a":"1 ","b":"1.5 ","c":"1.8 ","d":"2.4","e":"3"},"options_float":{"a":1.0,"b":1.5,"c":1.8,"d":2.4,"e":3.0},"annotated_formula":"divide(multiply(18, 5), subtract(const_100, 10))","linear_formula":"multiply(n0,n1)|subtract(const_100,n3)|divide(#0,#1)|","chain":"18 * 5<\/gadget>\n90<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 90<\/gadget>\n1<\/output>\n1<\/result>","index":820} +{"problem":"in covering a distance of 30 km , arun takes 22 hours more than anil . if arun doubles his speed , then he would take 1 hour less than anil . what is arun ' s speed ?","rationale":"explanation : if arun doubles his speed , he needs 3 hour less . double speed means half time . hence , half of the time required by arun to cover 30 km = 3 hour i . e . , time required by arun to cover 30 km = 6 hour arun ' s speed = 30 \/ 6 = 5 kmph answer is b","correct":"b","options":{"a":"8 kmph ","b":"5 kmph ","c":"4 kmph ","d":"7 kmph","e":"9 kmph"},"options_float":{"a":8.0,"b":5.0,"c":4.0,"d":7.0,"e":9.0},"annotated_formula":"divide(30, multiply(add(1, const_2), const_2))","linear_formula":"add(n2,const_2)|multiply(#0,const_2)|divide(n0,#1)","chain":"1 + 2<\/gadget>\n3<\/output>\n3 * 2<\/gadget>\n6<\/output>\n30 \/ 6<\/gadget>\n5<\/output>\n5<\/result>","index":821} +{"problem":"in the new budget the price of wheat rose by 8 % . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ?","rationale":"reduce in consumption = r \/ ( 100 + r ) * 100 % = 8 \/ 108 * 100 = 7.41 % answer is b","correct":"b","options":{"a":"7.5 % ","b":"7.41 % ","c":"10.9 % ","d":"12.6 %","e":"15 %"},"options_float":{"a":7.5,"b":7.41,"c":10.9,"d":12.6,"e":15.0},"annotated_formula":"multiply(divide(divide(8, const_100), add(divide(8, const_100), const_1)), const_100)","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(#0,#1)|multiply(#2,const_100)","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) + 1<\/gadget>\n27\/25 = around 1.08<\/output>\n(2\/25) \/ (27\/25)<\/gadget>\n2\/27 = around 0.074074<\/output>\n(2\/27) * 100<\/gadget>\n200\/27 = around 7.407407<\/output>\n200\/27 = around 7.407407<\/result>","index":822} +{"problem":"find the length of the wire required to go 12 times round a square field containing 104976 m 2 .","rationale":"\"a 2 = 104976 = > a = 324 4 a = 1296 1296 * 12 = 15552 answer : e\"","correct":"e","options":{"a":"15840 ","b":"3388 ","c":"2667 ","d":"8766","e":"15552"},"options_float":{"a":15840.0,"b":3388.0,"c":2667.0,"d":8766.0,"e":15552.0},"annotated_formula":"multiply(square_perimeter(square_edge_by_area(104976)), 12)","linear_formula":"square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)|","chain":"104_976 ** (1\/2)<\/gadget>\n324<\/output>\n4 * 324<\/gadget>\n1_296<\/output>\n1_296 * 12<\/gadget>\n15_552<\/output>\n15_552<\/result>","index":823} +{"problem":"a , b and c started a business with a total investment of rs . 72000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find a ' s share .","rationale":"explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 72000 = > x + ( x - 3000 ) + ( x + 3000 ) = 72000 = > 3 x = 72000 = > x = 24000 hence , a ' s investment = rs . 27000 b ' s investment = rs . 21000 c ' s investment = rs . 24000 ratio of the capitals of a , b and c = 27000 : 21000 : 24000 = 9 : 7 : 8 a ' s share = rs . [ ( 9 \/ 24 ) × 8640 ] = rs . 3240 answer : option a","correct":"a","options":{"a":"rs . 3240 ","b":"rs . 2520 ","c":"rs . 2880 ","d":"rs . 3360","e":"none of these"},"options_float":{"a":3240.0,"b":2520.0,"c":2880.0,"d":3360.0,"e":null},"annotated_formula":"multiply(8640, divide(add(divide(subtract(72000, add(6000, 3000)), const_3), 6000), 72000))","linear_formula":"add(n1,n2)|subtract(n0,#0)|divide(#1,const_3)|add(n1,#2)|divide(#3,n0)|multiply(n3,#4)","chain":"6_000 + 3_000<\/gadget>\n9_000<\/output>\n72_000 - 9_000<\/gadget>\n63_000<\/output>\n63_000 \/ 3<\/gadget>\n21_000<\/output>\n21_000 + 6_000<\/gadget>\n27_000<\/output>\n27_000 \/ 72_000<\/gadget>\n3\/8 = around 0.375<\/output>\n8_640 * (3\/8)<\/gadget>\n3_240<\/output>\n3_240<\/result>","index":825} +{"problem":"a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction e of the sum of the 21 numbers in the list ?","rationale":"\"this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction e of n to the total would be 4 a \/ 24 a or 1 \/ 6 answer b\"","correct":"b","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, 4), 21), 4), const_2), 4), const_3))","linear_formula":"divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|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 - 4<\/gadget>\n9<\/output>\n9 - 3<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":826} +{"problem":"the s . i . on a certain sum of money for 5 years at 6 % per annum is half the c . i . on rs . 3000 for 2 years at 15 % per annum . the sum placed on s . i . is ?","rationale":"c . i . = [ 3000 * ( 1 + 15 \/ 100 ) 2 - 3000 ] = ( 3000 * 23 \/ 20 * 23 \/ 20 - 3000 ) = rs . 967.5 . sum = ( 483.75 * 100 ) \/ ( 5 * 6 ) = rs . 1612.5 answer : c","correct":"c","options":{"a":"1525.2 ","b":"1256.3 ","c":"1612.5 ","d":"1548.5","e":"1254.5"},"options_float":{"a":1525.2,"b":1256.3,"c":1612.5,"d":1548.5,"e":1254.5},"annotated_formula":"divide(divide(subtract(multiply(3000, power(add(const_1, divide(15, const_100)), 2)), 3000), const_2), multiply(5, divide(6, 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,const_2)|divide(#7,#3)","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n(23\/20) ** 2<\/gadget>\n529\/400 = around 1.3225<\/output>\n3_000 * (529\/400)<\/gadget>\n7_935\/2 = around 3_967.5<\/output>\n(7_935\/2) - 3_000<\/gadget>\n1_935\/2 = around 967.5<\/output>\n(1_935\/2) \/ 2<\/gadget>\n1_935\/4 = around 483.75<\/output>\n6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n5 * (3\/50)<\/gadget>\n3\/10 = around 0.3<\/output>\n(1_935\/4) \/ (3\/10)<\/gadget>\n3_225\/2 = around 1_612.5<\/output>\n3_225\/2 = around 1_612.5<\/result>","index":830} +{"problem":"what is rate of interest if principal . amount be 400 , simple interest 80 and time 2 year .","rationale":"\"s . i = ( p * r * t ) \/ 100 80 = 800 r \/ 100 r = 80 \/ 8 = 10 % answer a\"","correct":"a","options":{"a":"10 ","b":"12.5 ","c":"25 ","d":"12","e":"14.5"},"options_float":{"a":10.0,"b":12.5,"c":25.0,"d":12.0,"e":14.5},"annotated_formula":"multiply(divide(80, multiply(400, 2)), const_100)","linear_formula":"multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)|","chain":"400 * 2<\/gadget>\n800<\/output>\n80 \/ 800<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":834} +{"problem":"if 0.20 : x : : 4 : 2 , then x is equal to","rationale":"\"sol . ( x × 4 ) = ( 0.20 × 2 ) ⇒ x = 0.4 \/ 4 = 0.1 . answer c\"","correct":"c","options":{"a":"0.2 ","b":"0.3 ","c":"0.1 ","d":"0.5","e":"none"},"options_float":{"a":0.2,"b":0.3,"c":0.1,"d":0.5,"e":null},"annotated_formula":"divide(multiply(0.20, 2), 4)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"0.2 * 2<\/gadget>\n0.4<\/output>\n0.4 \/ 4<\/gadget>\n0.1<\/output>\n0.1<\/result>","index":835} +{"problem":"along a yard 225 metres long , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees","rationale":"\"explanation : 26 trees have 25 gaps between them , required distance ( 225 \/ 25 ) = 10 option 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(225, subtract(26, const_1))","linear_formula":"subtract(n1,const_1)|divide(n0,#0)|","chain":"26 - 1<\/gadget>\n25<\/output>\n225 \/ 25<\/gadget>\n9<\/output>\n9<\/result>","index":836} +{"problem":"a retailer marks her goods in such a way that the profit made by selling 50 articles is equal to the selling price of 20 articles . what is the percentage of profit made by the retailer ?","rationale":"let cost price = x profit = y selling price = x + y 50 y = 20 ( x + y ) 30 y = 20 x percentage profit = y \/ x ∗ 100 = 20 \/ 30 ∗ 100 = 66.667 answer = a","correct":"a","options":{"a":"66.67 % ","b":"33.33 % ","c":"40 % ","d":"25 %","e":"20 %"},"options_float":{"a":66.67,"b":33.33,"c":40.0,"d":25.0,"e":20.0},"annotated_formula":"multiply(subtract(divide(50, subtract(50, 20)), const_1), const_100)","linear_formula":"subtract(n0,n1)|divide(n0,#0)|subtract(#1,const_1)|multiply(#2,const_100)","chain":"50 - 20<\/gadget>\n30<\/output>\n50 \/ 30<\/gadget>\n5\/3 = around 1.666667<\/output>\n(5\/3) - 1<\/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":837} +{"problem":"a basket contains 10 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?","rationale":"\"the total number of ways to choose 2 apples is 10 c 2 = 45 the number of ways that include the spoiled apple is 9 c 1 = 9 p ( the spoiled apple is included ) = 9 \/ 45 = 1 \/ 5 the answer is e .\"","correct":"e","options":{"a":"2 \/ 9 ","b":"3 \/ 8 ","c":"2 \/ 7 ","d":"1 \/ 6","e":"1 \/ 5"},"options_float":{"a":0.2222222222,"b":0.375,"c":0.2857142857,"d":0.1666666667,"e":0.2},"annotated_formula":"divide(choose(subtract(10, 1), 1), choose(10, 2))","linear_formula":"choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)|","chain":"10 - 1<\/gadget>\n9<\/output>\nbinomial(9, 1)<\/gadget>\n9<\/output>\nbinomial(10, 2)<\/gadget>\n45<\/output>\n9 \/ 45<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":838} +{"problem":"s is a set of 85 consecutive multiples of 5 . if the smallest number in s is 90 , then the greatest number in s is","rationale":"\"last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 85 consecutive multiples of 5 . if the smallest number in s is 90 , then the greatest number in s is first term = 90 ; total terms = 85 ; difference = 5 90 + ( 84 ) 5 = 510 ans e\"","correct":"e","options":{"a":"158 ","b":"597 ","c":"599 ","d":"402","e":"510"},"options_float":{"a":158.0,"b":597.0,"c":599.0,"d":402.0,"e":510.0},"annotated_formula":"add(90, multiply(subtract(85, const_1), 5))","linear_formula":"subtract(n0,const_1)|multiply(n1,#0)|add(n2,#1)|","chain":"85 - 1<\/gadget>\n84<\/output>\n84 * 5<\/gadget>\n420<\/output>\n90 + 420<\/gadget>\n510<\/output>\n510<\/result>","index":839} +{"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 191 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 191 sq ft as each cabbage takes 1 sq ft space . x ^ 2 - y ^ 2 = 191 ( x + y ) ( x - y ) = 191 191 is a prime number and thus it will be ( 96 + 95 ) * ( 96 - 95 ) . thus x = 96 and y = 95 x ^ 2 = 96 ^ 2 = 9216 the answer is c .\"","correct":"c","options":{"a":"7,251 ","b":"8406 ","c":"9216 ","d":"10,348","e":"can not be determined"},"options_float":{"a":7251.0,"b":8406.0,"c":9216.0,"d":10348.0,"e":null},"annotated_formula":"power(add(divide(191, 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":"191 \/ 2<\/gadget>\n191\/2 = around 95.5<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n(191\/2) + (1\/2)<\/gadget>\n96<\/output>\n96 ** 2<\/gadget>\n9_216<\/output>\n9_216<\/result>","index":841} +{"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":"288 ","c":"277 ","d":"877","e":"361"},"options_float":{"a":240.0,"b":288.0,"c":277.0,"d":877.0,"e":361.0},"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":842} +{"problem":"on a certain day , tim invested $ 1,000 at 10 percent annual interest , compounded annually , and lana invested 2,000 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 1,000 ( 1.10 ) ^ 2 = 1,000 ( 1.21 ) = 1,210 and lana would have 2,000 ( 1.05 ) ^ 2 = 2,000 ( 1.1025 ) = 2,205 thus , tim earned 210 dollars , while lana earned 205 dollars the difference is $ 5 and the answer is a .\"","correct":"a","options":{"a":"$ 5 ","b":"$ 15 ","c":"$ 50 ","d":"$ 100","e":"$ 105"},"options_float":{"a":5.0,"b":15.0,"c":50.0,"d":100.0,"e":105.0},"annotated_formula":"subtract(subtract(multiply(1,000, power(add(const_1, divide(10, const_100)), 2)), 1,000), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 2,000), 2,000))","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>\n1_000 * (121\/100)<\/gadget>\n1_210<\/output>\n1_210 - 1_000<\/gadget>\n210<\/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) * 2_000<\/gadget>\n2_205<\/output>\n2_205 - 2_000<\/gadget>\n205<\/output>\n210 - 205<\/gadget>\n5<\/output>\n5<\/result>","index":843} +{"problem":"what is the angle between the minute and the hour hand of the clock which shows 12 : 24 ?","rationale":"at 12 : 24 - minute hand will be at 24 * 6 = 144 degrees from position of 12 . - hour hand will move by 2 * 6 = 12 degree during the same time so the difference between the two hands will be 144 - 12 = 132 degrees . answer : e","correct":"e","options":{"a":"115 ","b":"120 ","c":"124 ","d":"130","e":"132"},"options_float":{"a":115.0,"b":120.0,"c":124.0,"d":130.0,"e":132.0},"annotated_formula":"subtract(multiply(24, multiply(const_3, const_2)), 12)","linear_formula":"multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)","chain":"3 * 2<\/gadget>\n6<\/output>\n24 * 6<\/gadget>\n144<\/output>\n144 - 12<\/gadget>\n132<\/output>\n132<\/result>","index":844} +{"problem":"find the value of ( 70 + 28 \/ 100 ) × 100","rationale":"( 7000 + 28 ) \/ 100 * 100 = 7028 answer : a","correct":"a","options":{"a":"7028 ","b":"4028 ","c":"3128 ","d":"3256","e":"5264"},"options_float":{"a":7028.0,"b":4028.0,"c":3128.0,"d":3256.0,"e":5264.0},"annotated_formula":"multiply(add(divide(28, 100), 70), const_100)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(#1,const_100)","chain":"28 \/ 100<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) + 70<\/gadget>\n1_757\/25 = around 70.28<\/output>\n(1_757\/25) * 100<\/gadget>\n7_028<\/output>\n7_028<\/result>","index":846} +{"problem":"a train with 120 wagons crosses john who is going in the same direction , in 36 seconds . it travels for half an hour from the time it starts ove ( who is also riding on his horse ) coming from the opposite direction in 24 seconds . in how much time after the train has crossed the mike do the john meets to mike ? rtaking the john ( he is riding on the horse ) before it starts overtaking the mike","rationale":"let the length of the train be l metres and speeds of the train arun and sriram be r , a and s respectively , then - - - - - - - - - - ( i ) and - - - - - - - - - ( ii ) from eq . ( i ) and ( ii ) 3 ( r - a ) = 2 ( r + k ) r = 3 a + 2 k in 30 minutes ( i . e 1800 seconds ) , the train covers 1800 r ( distance ) but the arun also covers 1800 a ( distance ) in the same time . therefore distance between arun and sriram , when the train has just crossed sriram = 1800 ( r - a ) - 24 ( a + k ) time required = = ( 3600 - 24 ) = 3576 s e","correct":"e","options":{"a":"2534 ","b":"3545 ","c":"3521 ","d":"4564","e":"3576"},"options_float":{"a":2534.0,"b":3545.0,"c":3521.0,"d":4564.0,"e":3576.0},"annotated_formula":"subtract(divide(multiply(subtract(divide(add(36, 24), subtract(36, 24)), const_1), multiply(multiply(const_10, const_3), const_60)), const_2), 24)","linear_formula":"add(n1,n2)|multiply(const_10,const_3)|subtract(n1,n2)|divide(#0,#2)|multiply(#1,const_60)|subtract(#3,const_1)|multiply(#4,#5)|divide(#6,const_2)|subtract(#7,n2)","chain":"36 + 24<\/gadget>\n60<\/output>\n36 - 24<\/gadget>\n12<\/output>\n60 \/ 12<\/gadget>\n5<\/output>\n5 - 1<\/gadget>\n4<\/output>\n10 * 3<\/gadget>\n30<\/output>\n30 * 60<\/gadget>\n1_800<\/output>\n4 * 1_800<\/gadget>\n7_200<\/output>\n7_200 \/ 2<\/gadget>\n3_600<\/output>\n3_600 - 24<\/gadget>\n3_576<\/output>\n3_576<\/result>","index":847} +{"problem":"the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 5808 sq cm . what is the ratio of the width and the area of the rectangle ?","rationale":"\"let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 5808 12 x ^ 2 = 5808 x ^ 2 = 484 x = 22 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 88 the answer is d .\"","correct":"d","options":{"a":"1 : 76 ","b":"1 : 80 ","c":"1 : 84 ","d":"1 : 88","e":"1 : 92"},"options_float":{"a":0.0131578947,"b":0.0125,"c":0.0119047619,"d":0.0113636364,"e":0.0108695652},"annotated_formula":"divide(divide(sqrt(multiply(3, 5808)), const_2), 5808)","linear_formula":"multiply(n2,n1)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2)|","chain":"3 * 5_808<\/gadget>\n17_424<\/output>\n17_424 ** (1\/2)<\/gadget>\n132<\/output>\n132 \/ 2<\/gadget>\n66<\/output>\n66 \/ 5_808<\/gadget>\n1\/88 = around 0.011364<\/output>\n1\/88 = around 0.011364<\/result>","index":852} +{"problem":"a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 990 grams . find the profit percent .","rationale":"\"explanation : ( 100 + g ) \/ ( 100 + x ) = true measure \/ faulty measure x = 0 true measure = 1000 faulty measure = 990 100 + g \/ 100 + 0 = 1000 \/ 990 100 + g = 100 \/ 99 * 100 g = 1.01 answer : c\"","correct":"c","options":{"a":"1.05 ","b":"1.06 ","c":"1.01 ","d":"1.08","e":"1.09"},"options_float":{"a":1.05,"b":1.06,"c":1.01,"d":1.08,"e":1.09},"annotated_formula":"multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 990), 990), const_100)","linear_formula":"add(const_1,const_4)|add(#0,#0)|multiply(#1,const_100)|subtract(#2,n0)|divide(#3,n0)|multiply(#4,const_100)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 + 5<\/gadget>\n10<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n1_000 - 990<\/gadget>\n10<\/output>\n10 \/ 990<\/gadget>\n1\/99 = around 0.010101<\/output>\n(1\/99) * 100<\/gadget>\n100\/99 = around 1.010101<\/output>\n100\/99 = around 1.010101<\/result>","index":853} +{"problem":"a basket contains 9 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?","rationale":"the total number of ways to choose 2 apples is 9 c 2 = 36 the number of ways that include the spoiled apple is 8 c 1 = 8 p ( the spoiled apple is included ) = 8 \/ 36 = 2 \/ 9 the answer is d .","correct":"d","options":{"a":"2 \/ 3 ","b":"2 \/ 5 ","c":"2 \/ 7 ","d":"2 \/ 9","e":"2 \/ 11"},"options_float":{"a":0.6666666667,"b":0.4,"c":0.2857142857,"d":0.2222222222,"e":0.1818181818},"annotated_formula":"divide(choose(subtract(9, 1), 1), choose(9, 2))","linear_formula":"choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)","chain":"9 - 1<\/gadget>\n8<\/output>\nbinomial(8, 1)<\/gadget>\n8<\/output>\nbinomial(9, 2)<\/gadget>\n36<\/output>\n8 \/ 36<\/gadget>\n2\/9 = around 0.222222<\/output>\n2\/9 = around 0.222222<\/result>","index":854} +{"problem":"26 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ?","rationale":"\"think of 100 people total : from the first fact , 26 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 ) * 26 equals the number of fair haired men , which is 39 men with fair hair . add this 39 to the 26 women and get 65 fair haired men and women out of 100 total men and women . 65 % e\"","correct":"e","options":{"a":"25 ","b":"30 ","c":"50 ","d":"55","e":"65"},"options_float":{"a":25.0,"b":30.0,"c":50.0,"d":55.0,"e":65.0},"annotated_formula":"multiply(divide(26, 40), const_100)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|","chain":"26 \/ 40<\/gadget>\n13\/20 = around 0.65<\/output>\n(13\/20) * 100<\/gadget>\n65<\/output>\n65<\/result>","index":855} +{"problem":"a work as fast as b . if b can complete a work in 24 days independently , the number of days in which a and b can together finish the work in ?","rationale":"\"ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 \/ 12 b ' s 1 day work = 1 \/ 24 a + b 1 day work = 1 \/ 12 + 1 \/ 24 = 3 \/ 24 = 1 \/ 8 a and b can finish the work in 8 days answer is c\"","correct":"c","options":{"a":"2 days ","b":"3 days ","c":"8 days ","d":"5 days","e":"6 days"},"options_float":{"a":2.0,"b":3.0,"c":8.0,"d":5.0,"e":6.0},"annotated_formula":"inverse(add(inverse(24), multiply(const_2, inverse(24))))","linear_formula":"inverse(n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n2 * (1\/24)<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/24) + (1\/12)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n8<\/result>","index":856} +{"problem":"if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 20 , then n =","rationale":"\"2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 20 = > 4 x 2 ^ 2 n = 4 ^ 20 = 2 ^ 40 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 40 = > 2 ^ ( 2 n + 2 ) = 2 ^ 40 = > 2 n + 2 = 40 = > n = 19 so . answer will be c .\"","correct":"c","options":{"a":"3 ","b":"6 ","c":"19 ","d":"23","e":"24"},"options_float":{"a":3.0,"b":6.0,"c":19.0,"d":23.0,"e":24.0},"annotated_formula":"divide(subtract(multiply(20, 2), 2), 2)","linear_formula":"multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)|","chain":"20 * 2<\/gadget>\n40<\/output>\n40 - 2<\/gadget>\n38<\/output>\n38 \/ 2<\/gadget>\n19<\/output>\n19<\/result>","index":857} +{"problem":"in a division sum , the remainder is 8 and the divisor is 6 times the quotient and is obtained by adding 3 to the thrice of the remainder . the dividend is :","rationale":"\"diver = ( 8 * 3 ) + 3 = 27 6 * quotient = 27 quotient = 4.5 dividend = ( divisor * quotient ) + remainder dividend = ( 27 * 4.5 ) + 8 = 129.5 b\"","correct":"b","options":{"a":"110.6 ","b":"129.5 ","c":"130.5 ","d":"86","e":"88"},"options_float":{"a":110.6,"b":129.5,"c":130.5,"d":86.0,"e":88.0},"annotated_formula":"add(multiply(add(multiply(8, const_3), 3), divide(add(multiply(8, const_3), 3), 6)), 8)","linear_formula":"multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3)|","chain":"8 * 3<\/gadget>\n24<\/output>\n24 + 3<\/gadget>\n27<\/output>\n27 \/ 6<\/gadget>\n9\/2 = around 4.5<\/output>\n27 * (9\/2)<\/gadget>\n243\/2 = around 121.5<\/output>\n(243\/2) + 8<\/gadget>\n259\/2 = around 129.5<\/output>\n259\/2 = around 129.5<\/result>","index":858} +{"problem":"a rectangular swimming pool is 20 feet by 20 feet . a deck that has uniform width surrounds the pool . the total area of the pool and deck is 576 square feet . what is the width of the deck ?","rationale":"\"let the width = w total area of the pool and deck = ( 2 w + 20 ) ( 2 w + 20 ) we can test the answer choices along with unit digit method a ) 2 feet . . . . . . . . . . . 24 * 24 has unit digit 6 . . . . . . . . . . hold b ) 2.5 feet . . . . . . . . . 25 * 25 has unit digit 5 . . . . . . . . . . eliminate c ) 3 feet . . . . . . . . . . . . 26 * 26 has unit digit 6 . . . . . . . . . . . elimate ( area is more than stipulated ) d ) 4 feet . . . . . . . . . . . . 28 * 28 has unit digit 4 . . . . . . . . . . . eliminate e ) 5 feet . . . . . . . . . . . . 30 * 30 has unit digit 0 . . . . . . . . . . . eliminate 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":"divide(subtract(sqrt(add(power(subtract(20, const_1), const_2), subtract(576, rectangle_area(20, 20)))), subtract(20, const_1)), const_2)","linear_formula":"rectangle_area(n0,n1)|subtract(n1,const_1)|power(#1,const_2)|subtract(n2,#0)|add(#2,#3)|sqrt(#4)|subtract(#5,#1)|divide(#6,const_2)|","chain":"20 - 1<\/gadget>\n19<\/output>\n19 ** 2<\/gadget>\n361<\/output>\n20 * 20<\/gadget>\n400<\/output>\n576 - 400<\/gadget>\n176<\/output>\n361 + 176<\/gadget>\n537<\/output>\n537 ** (1\/2)<\/gadget>\nsqrt(537) = around 23.17326<\/output>\n(sqrt(537)) - 19<\/gadget>\n-19 + sqrt(537) = around 4.17326<\/output>\n(-19 + sqrt(537)) \/ 2<\/gadget>\n-19\/2 + sqrt(537)\/2 = around 2.08663<\/output>\n-19\/2 + sqrt(537)\/2 = around 2.08663<\/result>","index":860} +{"problem":"a circle graph shows how the budget of a certain company was spent : 61 percent for salaries , 10 percent for research and development , 6 percent for utilities , 5 percent for equipment , 3 percent for supplies , and the remainder for transportation . if the area of each sector of the graph is proportional to the percent of the budget it represents , how many degrees of the circle are used to represent transportation ?","rationale":"the percent of the budget for transportation is 100 - ( 61 + 10 + 6 + 5 + 3 ) = 15 % 100 % of the circle is 360 degrees . then ( 15 % \/ 100 % ) * 360 = 54 degrees the answer is c .","correct":"c","options":{"a":"18 ° ","b":"36 ° ","c":"54 ° ","d":"72 °","e":"90 °"},"options_float":{"a":18.0,"b":36.0,"c":54.0,"d":72.0,"e":90.0},"annotated_formula":"divide(multiply(const_360, subtract(const_100, add(add(add(add(61, 10), 6), 5), 3))), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|subtract(const_100,#3)|multiply(#4,const_360)|divide(#5,const_100)","chain":"61 + 10<\/gadget>\n71<\/output>\n71 + 6<\/gadget>\n77<\/output>\n77 + 5<\/gadget>\n82<\/output>\n82 + 3<\/gadget>\n85<\/output>\n100 - 85<\/gadget>\n15<\/output>\n360 * 15<\/gadget>\n5_400<\/output>\n5_400 \/ 100<\/gadget>\n54<\/output>\n54<\/result>","index":861} +{"problem":"10 stickers numbered 1 to 10 are placed in a bowl , mixed up thoroughly and then one sticker is drawn randomly . if it is known that the number on the drawn sticker is more than 3 , what is the probability that it is an even number ?","rationale":"let a be the event ‘ the number on the card drawn is even ’ and b be the event ‘ the number on the card drawn is greater than 3 ’ . we have to find p ( a | b ) . now , the sample space of the experiment is s = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } then a = { 2 , 4 , 6 , 8 , 10 } , b = { 4 , 5 , 6 , 7 , 8 , 9 , 10 } and a n b = { 4 , 6 , 8 , 10 } also p ( a ) = 5 \/ 2 , p ( b ) = 7 \/ 10 and p ( a n b ) = 4 \/ 10 then p ( a | b ) = p ( a n b ) \/ p ( b ) = ( 4 \/ 10 ) \/ ( 7 \/ 10 ) = 4 \/ 7 b )","correct":"b","options":{"a":"3 \/ 7 ","b":"4 \/ 7 ","c":"5 \/ 7 ","d":"7 \/ 11","e":"9 \/ 11"},"options_float":{"a":0.4285714286,"b":0.5714285714,"c":0.7142857143,"d":0.6363636364,"e":0.8181818182},"annotated_formula":"multiply(divide(const_4, 10), divide(10, subtract(10, 3)))","linear_formula":"divide(const_4,n0)|subtract(n0,n3)|divide(n0,#1)|multiply(#0,#2)","chain":"4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n10 - 3<\/gadget>\n7<\/output>\n10 \/ 7<\/gadget>\n10\/7 = around 1.428571<\/output>\n(2\/5) * (10\/7)<\/gadget>\n4\/7 = around 0.571429<\/output>\n4\/7 = around 0.571429<\/result>","index":862} +{"problem":"for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score w for the exam ?","rationale":"\"a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean w = 74 . answer : a .\"","correct":"a","options":{"a":"74 ","b":"76 ","c":"78 ","d":"80","e":"82"},"options_float":{"a":74.0,"b":76.0,"c":78.0,"d":80.0,"e":82.0},"annotated_formula":"divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3))","linear_formula":"add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)|","chain":"58 * 3<\/gadget>\n174<\/output>\n98 * 2<\/gadget>\n196<\/output>\n174 + 196<\/gadget>\n370<\/output>\n2 + 3<\/gadget>\n5<\/output>\n370 \/ 5<\/gadget>\n74<\/output>\n74<\/result>","index":863} +{"problem":"tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 9 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":"\"for a , r = 8 \/ 2 pi . its capacity = ( 4 pi ) ^ 2 * 9 = 144 pi for b , r = 10 \/ pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a \/ b = 144 pi \/ 200 pi = 0.72 a\"","correct":"a","options":{"a":"72 % ","b":"80 % ","c":"100 % ","d":"120 %","e":"125 %"},"options_float":{"a":72.0,"b":80.0,"c":100.0,"d":120.0,"e":125.0},"annotated_formula":"multiply(multiply(power(divide(8, 10), const_2), divide(9, 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>\n9 \/ 8<\/gadget>\n9\/8 = around 1.125<\/output>\n(16\/25) * (9\/8)<\/gadget>\n18\/25 = around 0.72<\/output>\n(18\/25) * 100<\/gadget>\n72<\/output>\n72<\/result>","index":865} +{"problem":"220 metres long yard , 21 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees","rationale":"\"21 trees have 20 gaps between them , required distance ( 220 \/ 20 ) = 11 d\"","correct":"d","options":{"a":"10 ","b":"12 ","c":"14 ","d":"11","e":"17"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":11.0,"e":17.0},"annotated_formula":"divide(220, add(subtract(21, 2), const_1))","linear_formula":"subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)|","chain":"21 - 2<\/gadget>\n19<\/output>\n19 + 1<\/gadget>\n20<\/output>\n220 \/ 20<\/gadget>\n11<\/output>\n11<\/result>","index":867} +{"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 10 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 10 = 50 km imo , correct answer is ` ` d . ' '\"","correct":"d","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, 10), 60))","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|","chain":"5 * 10<\/gadget>\n50<\/output>\n50 \/ 60<\/gadget>\n5\/6 = around 0.833333<\/output>\n60 * (5\/6)<\/gadget>\n50<\/output>\n50<\/result>","index":868} +{"problem":"a student travels from his house to school at 10 km \/ hr and reaches school 1 hour late . the next day he travels 12 km \/ hr and reaches school 1 hour early . what is the distance between his house and the school ?","rationale":"let x be the distance from his house to the school . x \/ 10 = x \/ 12 + 2 6 x = 5 x + 120 x = 120 km the answer is e .","correct":"e","options":{"a":"100 ","b":"105 ","c":"110 ","d":"115","e":"120"},"options_float":{"a":100.0,"b":105.0,"c":110.0,"d":115.0,"e":120.0},"annotated_formula":"multiply(multiply(10, 12), divide(subtract(12, 10), add(1, 1)))","linear_formula":"add(n1,n1)|multiply(n0,n2)|subtract(n2,n0)|divide(#2,#0)|multiply(#3,#1)","chain":"10 * 12<\/gadget>\n120<\/output>\n12 - 10<\/gadget>\n2<\/output>\n1 + 1<\/gadget>\n2<\/output>\n2 \/ 2<\/gadget>\n1<\/output>\n120 * 1<\/gadget>\n120<\/output>\n120<\/result>","index":869} +{"problem":"two bullet trains of equal lengths take 10 seconds and 30 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 \/ 30 m \/ sec = 4 m \/ sec . relative speed = ( 12 + 4 ) = 16 m \/ sec . required time = ( 120 + 120 ) \/ 16 sec = 15 sec . d\"","correct":"d","options":{"a":"13 sec . ","b":"14 sec . ","c":"12 sec . ","d":"15 sec .","e":"19 sec ."},"options_float":{"a":13.0,"b":14.0,"c":12.0,"d":15.0,"e":19.0},"annotated_formula":"divide(add(120, 120), add(speed(120, 10), speed(120, 30)))","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 \/ 30<\/gadget>\n4<\/output>\n12 + 4<\/gadget>\n16<\/output>\n240 \/ 16<\/gadget>\n15<\/output>\n15<\/result>","index":871} +{"problem":"kathleen can paint a room in 2 hours , and anthony can paint an identical room in 7 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ?","rationale":"( 1 \/ 2 + 1 \/ 7 ) t = 2 t = 28 \/ 9 answer : a","correct":"a","options":{"a":"28 \/ 9 ","b":"4 \/ 3 ","c":"15 \/ 8 ","d":"9 \/ 4","e":"15 \/ 4"},"options_float":{"a":3.1111111111,"b":1.3333333333,"c":1.875,"d":2.25,"e":3.75},"annotated_formula":"multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 7))), const_2)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_2)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/2) + (1\/7)<\/gadget>\n9\/14 = around 0.642857<\/output>\n1 \/ (9\/14)<\/gadget>\n14\/9 = around 1.555556<\/output>\n(14\/9) * 2<\/gadget>\n28\/9 = around 3.111111<\/output>\n28\/9 = around 3.111111<\/result>","index":872} +{"problem":"according to a recent student poll , 4 \/ 6 out of 24 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ?","rationale":"\"16 students are interested , 8 are not interested prob = 1 - 8 c 2 \/ 24 c 2 = 1 - ( 8 * 7 \/ ( 24 * 23 ) ) = 1 - 7 \/ 69 = 62 \/ 69 answer : d\"","correct":"d","options":{"a":"1 \/ 14 ","b":"4 \/ 49 ","c":"2 \/ 7 ","d":"62 \/ 69","e":"13 \/ 14"},"options_float":{"a":0.0714285714,"b":0.0816326531,"c":0.2857142857,"d":0.8985507246,"e":0.9285714286},"annotated_formula":"divide(subtract(choose(24, const_2), choose(subtract(24, multiply(24, divide(4, 6))), const_2)), choose(24, const_2))","linear_formula":"choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0)|","chain":"binomial(24, 2)<\/gadget>\n276<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n24 * (2\/3)<\/gadget>\n16<\/output>\n24 - 16<\/gadget>\n8<\/output>\nbinomial(8, 2)<\/gadget>\n28<\/output>\n276 - 28<\/gadget>\n248<\/output>\n248 \/ 276<\/gadget>\n62\/69 = around 0.898551<\/output>\n62\/69 = around 0.898551<\/result>","index":873} +{"problem":"in school there are some bicycles and 4 wheeler wagons . one tuesday there are 190 wheels in the campus . how many bicycles are there ?","rationale":"let no . of bicycles be x & no . of wagons be y so , 2 x + 4 y = 190 by solving , we get no . of bicycles = 39 ( wheels = > 2 * 39 = 78 ) no . of wagons = 28 ( wheels = > 4 * 28 = 112 ) answer : e","correct":"e","options":{"a":"35 ","b":"36 ","c":"37 ","d":"38","e":"39"},"options_float":{"a":35.0,"b":36.0,"c":37.0,"d":38.0,"e":39.0},"annotated_formula":"multiply(divide(190, add(multiply(4, const_2), const_2)), const_2)","linear_formula":"multiply(n0,const_2)|add(#0,const_2)|divide(n1,#1)|multiply(#2,const_2)","chain":"4 * 2<\/gadget>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n190 \/ 10<\/gadget>\n19<\/output>\n19 * 2<\/gadget>\n38<\/output>\n38<\/result>","index":876} +{"problem":"in a certain pond , 90.00001 fish were caught , tagged , and returned to the pond . a few days later , 90 fish were caught again , of which 2 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 ` s the approximate number of fish in the pond ?","rationale":"\"the percent of tagged fish in the second catch is 2 \/ 90 * 100 = 2.22 % . we are told that 2.22 % approximates the percent of tagged fish in the pond . since there are 90 tagged fish , then we have 0.022 x = 90 - - > x = 4,091 . answer : e .\"","correct":"e","options":{"a":"400 ","b":"625 ","c":"1250 ","d":"2500","e":"4091"},"options_float":{"a":400.0,"b":625.0,"c":1250.0,"d":2500.0,"e":4091.0},"annotated_formula":"divide(90.00001, divide(2, 90))","linear_formula":"divide(n2,n1)|divide(n0,#0)|","chain":"2 \/ 90<\/gadget>\n1\/45 = around 0.022222<\/output>\n90.00001 \/ (1\/45)<\/gadget>\n4_050.00045<\/output>\n4_050.00045<\/result>","index":878} +{"problem":"in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 \/ 5 and picking a straight flower is 1 \/ 2 , then what is the probability of picking a flower which is yellow and straight","rationale":"\"good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 \/ 5 of the garden is green , so , since all the flowers must be either green or yellow , we know that 4 \/ 5 are yellow . we ' re also told there is an equal probability of straight or curved , 1 \/ 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 4 \/ 5 * 1 \/ 2 = 2 \/ 5 , or c .\"","correct":"c","options":{"a":"1 \/ 7 ","b":"1 \/ 8 ","c":"2 \/ 5 ","d":"3 \/ 4","e":"7 \/ 8"},"options_float":{"a":0.1428571429,"b":0.125,"c":0.4,"d":0.75,"e":0.875},"annotated_formula":"multiply(subtract(1, divide(1, 5)), divide(1, 2))","linear_formula":"divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)|","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(4\/5) * (1\/2)<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":879} +{"problem":"in what time will a railway train 110 m long moving at the rate of 36 kmph pass a telegraph post on its way ?","rationale":"\"t = 110 \/ 36 * 18 \/ 5 = 11 sec answer : d\"","correct":"d","options":{"a":"6 sec ","b":"7 sec ","c":"8 sec ","d":"11 sec","e":"2 sec"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":11.0,"e":2.0},"annotated_formula":"divide(110, 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>\n110 \/ 10<\/gadget>\n11<\/output>\n11<\/result>","index":881} +{"problem":"there are two groups of students in the sixth grade . there are 45 students in group a , and 55 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 40 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their homework ?","rationale":"\"number of students in group a = 45 students who forget homework in group a = 20 % of 45 = 9 students number of students in group b = 55 students who forget homework in group b = 40 % of 55 = 22 students total number of students = 45 + 55 = 100 students who forgot homework = 9 + 22 = 31 students percentage of students who forget homework = 31 \/ 100 * 100 = 31 % answer : d\"","correct":"d","options":{"a":"23 % ","b":"25 % ","c":"29 % ","d":"31 %","e":"36 %"},"options_float":{"a":23.0,"b":25.0,"c":29.0,"d":31.0,"e":36.0},"annotated_formula":"multiply(divide(add(divide(multiply(45, 20), const_100), divide(multiply(55, 40), const_100)), add(45, 55)), const_100)","linear_formula":"add(n0,n1)|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":"45 * 20<\/gadget>\n900<\/output>\n900 \/ 100<\/gadget>\n9<\/output>\n55 * 40<\/gadget>\n2_200<\/output>\n2_200 \/ 100<\/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":883} +{"problem":"in a factory , an average of 50 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 43 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 43 - 25 x 50 = 40 ∴ average for last 5 days = 40 \/ 5 = 8 c\"","correct":"c","options":{"a":"20 ","b":"36 ","c":"8 ","d":"50","e":"59"},"options_float":{"a":20.0,"b":36.0,"c":8.0,"d":50.0,"e":59.0},"annotated_formula":"divide(subtract(multiply(add(25, 5), 43), multiply(25, 50)), 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 * 43<\/gadget>\n1_290<\/output>\n25 * 50<\/gadget>\n1_250<\/output>\n1_290 - 1_250<\/gadget>\n40<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8<\/result>","index":884} +{"problem":"find the principle on a certain sum of money at 5 % per annum for 2 2 \/ 5 years if the amount being rs . 1904 ?","rationale":"\"1904 = p [ 1 + ( 5 * 12 \/ 5 ) \/ 100 ] p = 1700 . answer : a\"","correct":"a","options":{"a":"1700 ","b":"2777 ","c":"2889 ","d":"27670","e":"2771"},"options_float":{"a":1700.0,"b":2777.0,"c":2889.0,"d":27670.0,"e":2771.0},"annotated_formula":"divide(1904, 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_904 \/ (28\/25)<\/gadget>\n1_700<\/output>\n1_700<\/result>","index":888} +{"problem":"the overall age of x and y is 11 year greater than the overall age of y and z . z is how many decades younger that x ?","rationale":"\"a 11 ( x + y ) â € “ ( y + z ) = 11 x â € “ z = 11\"","correct":"a","options":{"a":"11 ","b":"15 ","c":"12 ","d":"17","e":"19"},"options_float":{"a":11.0,"b":15.0,"c":12.0,"d":17.0,"e":19.0},"annotated_formula":"divide(11, const_1)","linear_formula":"divide(n0,const_1)|","chain":"11 \/ 1<\/gadget>\n11<\/output>\n11<\/result>","index":889} +{"problem":"the mean of 50 observations was 36 . it was found later that an observation 47 was wrongly taken as 23 . the corrected new mean is","rationale":"\"solution correct sum = ( 36 x 50 + 47 - 23 ) = 1824 . â ˆ ´ correct mean = 1824 \/ 50 = 36.48 . answer d\"","correct":"d","options":{"a":"35.24 ","b":"36.16 ","c":"36.22 ","d":"36.48","e":"none"},"options_float":{"a":35.24,"b":36.16,"c":36.22,"d":36.48,"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":892} +{"problem":"the length of a rectangular floor is more than its breadth by 200 % . if rs . 324 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 = 324 \/ 3 = 108 sq m l b = 108 i . e . , l * l \/ 3 = 108 l 2 = 324 = > l = 18 . answer : c\"","correct":"c","options":{"a":"27 m ","b":"24 m ","c":"18 m ","d":"21 m","e":"none of these"},"options_float":{"a":27.0,"b":24.0,"c":18.0,"d":21.0,"e":null},"annotated_formula":"multiply(sqrt(divide(divide(324, 3), const_3)), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)|","chain":"324 \/ 3<\/gadget>\n108<\/output>\n108 \/ 3<\/gadget>\n36<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":893} +{"problem":"a watch was sold at a loss of 10 % . if it was sold for rs . 182 more , there would have been a gain of 4 % . what is the cost price ?","rationale":"\"90 % 104 % - - - - - - - - 14 % - - - - 182 100 % - - - - ? = > rs : 1300 answer : d\"","correct":"d","options":{"a":"s : 1000 ","b":"s : 1067 ","c":"s : 1278 ","d":"s : 1300","e":"s : 1027"},"options_float":{"a":1000.0,"b":1067.0,"c":1278.0,"d":1300.0,"e":1027.0},"annotated_formula":"divide(multiply(182, const_100), subtract(add(const_100, 4), subtract(const_100, 10)))","linear_formula":"add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)|","chain":"182 * 100<\/gadget>\n18_200<\/output>\n100 + 4<\/gadget>\n104<\/output>\n100 - 10<\/gadget>\n90<\/output>\n104 - 90<\/gadget>\n14<\/output>\n18_200 \/ 14<\/gadget>\n1_300<\/output>\n1_300<\/result>","index":894} +{"problem":"a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 10 - 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 = 10 . plugging it in , we get 35 diagonals ! answer ( b ) .\"","correct":"b","options":{"a":"875 ","b":"35 ","c":"1425 ","d":"2025","e":"2500"},"options_float":{"a":875.0,"b":35.0,"c":1425.0,"d":2025.0,"e":2500.0},"annotated_formula":"divide(multiply(subtract(10, const_3), 10), const_2)","linear_formula":"subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)|","chain":"10 - 3<\/gadget>\n7<\/output>\n7 * 10<\/gadget>\n70<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n35<\/result>","index":896} +{"problem":"a and b start walking towards each other at 5 am at speed of 4 kmph and 8 kmph . they were initially 36 km apart . at what time do they meet ?","rationale":"time of meeting = distance \/ relative speed = 36 \/ 8 + 4 = 36 \/ 12 = 3 hrs after 5 am = 8 am answer is a","correct":"a","options":{"a":"8 am ","b":"6 am ","c":"7 am ","d":"10 am","e":"8 pm"},"options_float":{"a":8.0,"b":6.0,"c":7.0,"d":10.0,"e":8.0},"annotated_formula":"add(5, divide(36, add(4, 8)))","linear_formula":"add(n1,n2)|divide(n3,#0)|add(n0,#1)","chain":"4 + 8<\/gadget>\n12<\/output>\n36 \/ 12<\/gadget>\n3<\/output>\n5 + 3<\/gadget>\n8<\/output>\n8<\/result>","index":898} +{"problem":"a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 7 with 3 students left over . what is the sum of the two smallest possible values of n ?","rationale":"n = 4 k + 1 = 7 j + 3 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 7 j + 3 . 1 , 5 , 9 , 13 , 17 = 7 ( 2 ) + 3 the next such number is 17 + 4 * 7 = 45 . 17 + 45 = 62 the answer is c .","correct":"c","options":{"a":"54 ","b":"58 ","c":"62 ","d":"66","e":"70"},"options_float":{"a":54.0,"b":58.0,"c":62.0,"d":66.0,"e":70.0},"annotated_formula":"add(add(multiply(7, const_2), 3), add(multiply(7, multiply(const_2, const_3)), 3))","linear_formula":"multiply(n2,const_2)|multiply(const_2,const_3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)","chain":"7 * 2<\/gadget>\n14<\/output>\n14 + 3<\/gadget>\n17<\/output>\n2 * 3<\/gadget>\n6<\/output>\n7 * 6<\/gadget>\n42<\/output>\n42 + 3<\/gadget>\n45<\/output>\n17 + 45<\/gadget>\n62<\/output>\n62<\/result>","index":899} +{"problem":"a train 150 m long running at 72 kmph crosses a platform in 28 sec . what is the length of the platform ?","rationale":"\"a 410 a = ( 72 * 5 \/ 18 ) * 28 - 150 = 410\"","correct":"a","options":{"a":"410 m ","b":"354 m ","c":"450 m ","d":"350 m","e":"250 m"},"options_float":{"a":410.0,"b":354.0,"c":450.0,"d":350.0,"e":250.0},"annotated_formula":"subtract(multiply(28, multiply(72, const_0_2778)), 150)","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>\n28 * 20<\/gadget>\n560<\/output>\n560 - 150<\/gadget>\n410<\/output>\n410<\/result>","index":900} +{"problem":"the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 170 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 € “ 170 x + ( x - 170 ) = 208 2 x = 378 x = 189 runs answer : c\"","correct":"c","options":{"a":"179 ","b":"367 ","c":"189 ","d":"177","e":"191"},"options_float":{"a":179.0,"b":367.0,"c":189.0,"d":177.0,"e":191.0},"annotated_formula":"divide(add(170, 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>\n170 + 208<\/gadget>\n378<\/output>\n378 \/ 2<\/gadget>\n189<\/output>\n189<\/result>","index":901} +{"problem":"the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 1000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ?","rationale":"\"x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 1000 mi \/ hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 1000 ) * t = = = > 3 d = ts + 1000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 1000 t - 2 d = - 2 st + 1000 t 2 d = 2 st - 1000 t d = st - 500 t x 13 : d = s * t st - 500 t = s * t s - 500 = s - 250 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 \/ 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d \/ t 3 d \/ t - d \/ t = 1000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 1000 2 = 500 d = s * t d = 500 * ( 1 \/ 2 ) d = 250 answer : a\"","correct":"a","options":{"a":"250 ","b":"1,600 \/ 3 ","c":"1,000 ","d":"1,500","e":"2,500"},"options_float":{"a":250.0,"b":533.3333333333,"c":1000.0,"d":1500.0,"e":2500.0},"annotated_formula":"multiply(divide(1000, 2), divide(const_1, 2))","linear_formula":"divide(n8,n3)|divide(const_1,n3)|multiply(#0,#1)|","chain":"1_000 \/ 2<\/gadget>\n500<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n500 * (1\/2)<\/gadget>\n250<\/output>\n250<\/result>","index":902} +{"problem":"in town x , 64 percent of the population are employed , and 35 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 35 are employed males , hence 29 % are employed females . ( employed females ) \/ ( total employed people ) = 29 \/ 64 = 45 % answer : d .\"","correct":"d","options":{"a":"16 % ","b":"25 % ","c":"32 % ","d":"45 %","e":"52 %"},"options_float":{"a":16.0,"b":25.0,"c":32.0,"d":45.0,"e":52.0},"annotated_formula":"multiply(divide(subtract(64, 35), 64), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"64 - 35<\/gadget>\n29<\/output>\n29 \/ 64<\/gadget>\n29\/64 = around 0.453125<\/output>\n(29\/64) * 100<\/gadget>\n725\/16 = around 45.3125<\/output>\n725\/16 = around 45.3125<\/result>","index":903} +{"problem":"find the value of ( 950 + 220 \/ 900 ) × 900","rationale":"\"( 855000 + 220 ) \/ 900 * 900 = 855000 + 220 = 855220 answer : d\"","correct":"d","options":{"a":"854542 ","b":"856945 ","c":"758965 ","d":"855220","e":"826450"},"options_float":{"a":854542.0,"b":856945.0,"c":758965.0,"d":855220.0,"e":826450.0},"annotated_formula":"multiply(add(divide(220, 900), 950), 900)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|","chain":"220 \/ 900<\/gadget>\n11\/45 = around 0.244444<\/output>\n(11\/45) + 950<\/gadget>\n42_761\/45 = around 950.244444<\/output>\n(42_761\/45) * 900<\/gadget>\n855_220<\/output>\n855_220<\/result>","index":907} +{"problem":"__ 2 a x __ b ____ cc in the multiplication problem above , a , b , and c represent distinct digits . if the sum of a and b is equal to 4.6 , what is the value of c ?","rationale":"if a + b = 4.6 assuming a and b are positive then a * b < 10 ( they could be either 1,2 , 3,4 ) therefore a * b = c 2 * b = c a + b = 4.6 three simple equations - divide the 1 st \/ 2 nd - - > a = 2 plug it the 3 rd - - > b = 2.6 - - > c = 5.2 ( answer b )","correct":"b","options":{"a":"6 ","b":"5.2 ","c":"4 ","d":"3","e":"2"},"options_float":{"a":6.0,"b":5.2,"c":4.0,"d":3.0,"e":2.0},"annotated_formula":"divide(multiply(subtract(4.6, 2), const_10), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|subtract(n1,n0)|multiply(#1,const_10)|divide(#2,#0)","chain":"4.6 - 2<\/gadget>\n2.6<\/output>\n2.6 * 10<\/gadget>\n26<\/output>\n4 + 1<\/gadget>\n5<\/output>\n26 \/ 5<\/gadget>\n26\/5 = around 5.2<\/output>\n26\/5 = around 5.2<\/result>","index":908} +{"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 $ 18,000 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(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(9, const_100))), subtract(divide(12, const_100), divide(9, const_100)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)","chain":"2 * 3<\/gadget>\n6<\/output>\n3 * 6<\/gadget>\n18<\/output>\n18 * 1_000<\/gadget>\n18_000<\/output>\n9 \/ 100<\/gadget>\n9\/100 = around 0.09<\/output>\n18_000 * (9\/100)<\/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":909} +{"problem":"a scuba diver descends at a rate of 40 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 3000 feet below the sea level . . how long will he take to reach the ship ?","rationale":"\"time taken to reach = 3000 \/ 40 = 75 minutes answer : c\"","correct":"c","options":{"a":"70 minutes ","b":"72 minutes ","c":"75 minutes ","d":"76 minutes","e":"77 minutes"},"options_float":{"a":70.0,"b":72.0,"c":75.0,"d":76.0,"e":77.0},"annotated_formula":"divide(3000, 40)","linear_formula":"divide(n1,n0)|","chain":"3_000 \/ 40<\/gadget>\n75<\/output>\n75<\/result>","index":912} +{"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 boy alone in 18 days , how long will a woman 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 boy ’ s 1 day ’ s work = 1 \/ 18 ( 1 man + 1 boy ) ‘ s 1 day ’ s work = 1 \/ 6 + 1 \/ 18 = 2 \/ 9 therefore , 1 woman ’ s 1 day ’ s work = 1 \/ 3 – 2 \/ 9 = 3 - 2 \/ 9 = 1 \/ 9 therefore , the woman alone can finish the work in 9 days . answer : option a\"","correct":"a","options":{"a":"9 days ","b":"21 days ","c":"24 days ","d":"27 days","e":"28 days"},"options_float":{"a":9.0,"b":21.0,"c":24.0,"d":27.0,"e":28.0},"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":913} +{"problem":"running at the same constant rate , 6 identical machines can produce a total of 270 pens per minute . at this rate , how many pens could 10 such machines produce in 4 minutes ?","rationale":"\"explanation : let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 | | : : 270 : x time 1 : 4 | = > 6 * 1 * x = 10 * 4 * 270 = > x = ( 10 * 4 * 270 ) \/ 6 = > x = 1800 answer : d\"","correct":"d","options":{"a":"1500 ","b":"1545.6 ","c":"1640.33 ","d":"1800","e":"none of these"},"options_float":{"a":1500.0,"b":1545.6,"c":1640.33,"d":1800.0,"e":null},"annotated_formula":"multiply(multiply(divide(270, 6), 4), 10)","linear_formula":"divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|","chain":"270 \/ 6<\/gadget>\n45<\/output>\n45 * 4<\/gadget>\n180<\/output>\n180 * 10<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":914} +{"problem":"from the sale of sleeping bags , a retailer made a gross profit of 13 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ?","rationale":"\"cost price * 1.13 = selling price - - > cost price * 1.13 = $ 28 - - > cost price = $ 24.78 . answer : c .\"","correct":"c","options":{"a":"3.0 ","b":"3.36 ","c":"24.78 ","d":"25.0","e":"31.36"},"options_float":{"a":3.0,"b":3.36,"c":24.78,"d":25.0,"e":31.36},"annotated_formula":"divide(multiply(28, const_100), add(const_100, 13))","linear_formula":"add(n0,const_100)|multiply(n1,const_100)|divide(#1,#0)|","chain":"28 * 100<\/gadget>\n2_800<\/output>\n100 + 13<\/gadget>\n113<\/output>\n2_800 \/ 113<\/gadget>\n2_800\/113 = around 24.778761<\/output>\n2_800\/113 = around 24.778761<\/result>","index":916} +{"problem":"a light has a rating of 86 watts , it is replaced with a new light that has 18 % higher wattage . how many watts does the new light have ?","rationale":"final number = initial number + 18 % ( original number ) = 86 + 18 % ( 86 ) = 86 + 15 = 101 . answer e","correct":"e","options":{"a":"105 ","b":"95 ","c":"80 ","d":"60","e":"101"},"options_float":{"a":105.0,"b":95.0,"c":80.0,"d":60.0,"e":101.0},"annotated_formula":"multiply(86, add(const_1, divide(18, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)","chain":"18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n1 + (9\/50)<\/gadget>\n59\/50 = around 1.18<\/output>\n86 * (59\/50)<\/gadget>\n2_537\/25 = around 101.48<\/output>\n2_537\/25 = around 101.48<\/result>","index":917} +{"problem":"a boat running upstream takes 8 hours 48 minutes to cover a certain distance , while it takes 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":"let the man ' s rate upstream be x kmph and that downstream be y kmph . then , distance covered upstream in 8 hrs 48 min = distance covered downstream in 4 hrs . 44 * x \/ 5 = 4 * y y = 11 \/ 5 * x required ratio = ( y + x ) \/ 2 : ( y - x ) \/ 2 = 8 \/ 5 : 3 \/ 5 = 8 \/ 3 ans - b","correct":"b","options":{"a":"8 \/ 5 ","b":"8 \/ 3 ","c":"3 \/ 5 ","d":"5 \/ 8","e":"5 \/ 3"},"options_float":{"a":1.6,"b":2.6666666667,"c":0.6,"d":0.625,"e":1.6666666667},"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":918} +{"problem":"a worker makes a toy in every 1 h . if he works for 100 h , then how many toys will he make ?","rationale":"\"no . of toys = 100 \/ 1 = 100 answer : e\"","correct":"e","options":{"a":"40 ","b":"54 ","c":"45 ","d":"39","e":"100"},"options_float":{"a":40.0,"b":54.0,"c":45.0,"d":39.0,"e":100.0},"annotated_formula":"divide(100, 1)","linear_formula":"divide(n1,n0)|","chain":"100 \/ 1<\/gadget>\n100<\/output>\n100<\/result>","index":919} +{"problem":"there are 3000 students in a school and among them 20 % of them attends chess class . 40 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ?","rationale":"\"20 % of 3000 gives 600 . so 600 attends chess and 40 % of 600 gives 240 so 240 enrolled for swimming answer : a\"","correct":"a","options":{"a":"240 ","b":"10 ","c":"100 ","d":"50","e":"20"},"options_float":{"a":240.0,"b":10.0,"c":100.0,"d":50.0,"e":20.0},"annotated_formula":"divide(multiply(divide(multiply(20, 3000), const_100), 40), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)|","chain":"20 * 3_000<\/gadget>\n60_000<\/output>\n60_000 \/ 100<\/gadget>\n600<\/output>\n600 * 40<\/gadget>\n24_000<\/output>\n24_000 \/ 100<\/gadget>\n240<\/output>\n240<\/result>","index":920} +{"problem":"the present population of a town is 280 . population increase rate is 10 % p . a . find the population of town after 1 years ?","rationale":"\"p = 280 r = 10 % required population of town = p * ( 1 + r \/ 100 ) ^ t = 280 * ( 1 + 10 \/ 100 ) = 280 * ( 11 \/ 10 ) = 308 answer is e\"","correct":"e","options":{"a":"100 ","b":"120 ","c":"200 ","d":"220","e":"308"},"options_float":{"a":100.0,"b":120.0,"c":200.0,"d":220.0,"e":308.0},"annotated_formula":"add(280, divide(multiply(280, 10), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|","chain":"280 * 10<\/gadget>\n2_800<\/output>\n2_800 \/ 100<\/gadget>\n28<\/output>\n280 + 28<\/gadget>\n308<\/output>\n308<\/result>","index":921} +{"problem":"mary sold boxes of butter cookies . ann sold 5 times as much as she did . 18 boxes of cookies were sold that day , how many boxes did mary sell ?","rationale":"# of boxes of cookies mary sold = x ann sold 5 times more = 5 x x + 5 x = 18 6 x = 18 x = 18 \/ 6 = 3 answer : a","correct":"a","options":{"a":"3 ","b":"5 ","c":"6 ","d":"10","e":"18"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":10.0,"e":18.0},"annotated_formula":"divide(18, add(5, const_1))","linear_formula":"add(n0,const_1)|divide(n1,#0)","chain":"5 + 1<\/gadget>\n6<\/output>\n18 \/ 6<\/gadget>\n3<\/output>\n3<\/result>","index":922} +{"problem":"a batsman in his 9 th inning makes a score of 75 and their by increasing his average by 7 . what is his average after the 9 th inning ?","rationale":"8 x + 75 = 9 ( x + 7 ) x = 12 + 7 = 19 answer : b","correct":"b","options":{"a":"12 ","b":"19 ","c":"26 ","d":"33","e":"40"},"options_float":{"a":12.0,"b":19.0,"c":26.0,"d":33.0,"e":40.0},"annotated_formula":"add(subtract(75, multiply(7, 9)), 7)","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)","chain":"7 * 9<\/gadget>\n63<\/output>\n75 - 63<\/gadget>\n12<\/output>\n12 + 7<\/gadget>\n19<\/output>\n19<\/result>","index":923} +{"problem":"a man can row at 5 kmph in still water . if the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back , how far is the place ?","rationale":"\"explanation : speed in still water = 5 kmph speed of the current = 1 kmph speed downstream = ( 5 + 1 ) = 6 kmph speed upstream = ( 5 - 1 ) = 4 kmph let the requited distance be x km total time taken = 1 hour = > x \/ 6 + x \/ 4 = 1 = > 2 x + 3 x = 12 = > 5 x = 12 = > x = 2.4 km . answer : option c\"","correct":"c","options":{"a":"3.2 km ","b":"3 km ","c":"2.4 km ","d":"3.6 km","e":"none of these"},"options_float":{"a":3.2,"b":3.0,"c":2.4,"d":3.6,"e":null},"annotated_formula":"divide(multiply(subtract(5, 1), const_3), 5)","linear_formula":"subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)|","chain":"5 - 1<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12 \/ 5<\/gadget>\n12\/5 = around 2.4<\/output>\n12\/5 = around 2.4<\/result>","index":924} +{"problem":"working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 2 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 4 x boxes ?","rationale":"rate = work \/ time given rate of machine a = x \/ 10 min machine b produces 2 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 4 x \/ 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x \/ 10 min + rate of machine b = 4 x \/ 10 = 5 x \/ 10 now combine time = combine work needs to be done \/ combine rate = 4 x \/ 5 x * 10 = 8 min ans : e","correct":"e","options":{"a":"3 minutes ","b":"4 minutes ","c":"5 minutes ","d":"6 minutes","e":"8 minutes"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":6.0,"e":8.0},"annotated_formula":"divide(multiply(4, 10), add(speed(10, 10), speed(multiply(2, 10), 5)))","linear_formula":"multiply(n0,n3)|multiply(n0,n1)|speed(n0,n0)|speed(#1,n2)|add(#2,#3)|divide(#0,#4)","chain":"4 * 10<\/gadget>\n40<\/output>\n10 \/ 10<\/gadget>\n1<\/output>\n2 * 10<\/gadget>\n20<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8<\/result>","index":925} +{"problem":"in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 4 rows in all and a total of 134 bricks in the wall , how many bricks does the bottom row contain ?","rationale":"\"the bottom row has x bricks x + x - 1 + x - 2 + x - 3 = 134 4 x - 6 = 134 4 x = 128 x = 32 answer : c\"","correct":"c","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(subtract(subtract(subtract(subtract(134, const_1), const_2), const_3), const_4), 4)","linear_formula":"subtract(n1,const_1)|subtract(#0,const_2)|subtract(#1,const_3)|subtract(#2,const_4)|divide(#3,n0)|","chain":"134 - 1<\/gadget>\n133<\/output>\n133 - 2<\/gadget>\n131<\/output>\n131 - 3<\/gadget>\n128<\/output>\n128 - 4<\/gadget>\n124<\/output>\n124 \/ 4<\/gadget>\n31<\/output>\n31<\/result>","index":926} +{"problem":"if paint costs $ 3.10 per quart , and a quart covers 20 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ?","rationale":"\"total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 \/ 20 = 30 cost will be 30 * 3.1 = $ 93 ans : c\"","correct":"c","options":{"a":"$ 1.60 ","b":"$ 16.00 ","c":"$ 93.00 ","d":"$ 108.00","e":"$ 196.00"},"options_float":{"a":1.6,"b":16.0,"c":93.0,"d":108.0,"e":196.0},"annotated_formula":"multiply(divide(3.10, 20), surface_cube(10))","linear_formula":"divide(n0,n1)|surface_cube(n2)|multiply(#0,#1)|","chain":"3.1 \/ 20<\/gadget>\n0.155<\/output>\n6 * (10 ** 2)<\/gadget>\n600<\/output>\n0.155 * 600<\/gadget>\n93<\/output>\n93<\/result>","index":927} +{"problem":"a man engaged a servant on the condition that he would pay him rs . 1000 and a uniform after one year service . he served only for 9 months and received uniform and rs . 550 , find the price of the uniform ?","rationale":"\"9 \/ 12 = 3 \/ 4 * 1000 = 750 550 - - - - - - - - - - - - - 200 1 \/ 4 - - - - - - - - 200 1 - - - - - - - - - ? = > rs . 800 answer : d\"","correct":"d","options":{"a":"s . 80 ","b":"s . 85 ","c":"s . 90 ","d":"s . 800","e":"s . 120"},"options_float":{"a":80.0,"b":85.0,"c":90.0,"d":800.0,"e":120.0},"annotated_formula":"multiply(divide(subtract(multiply(9, 1000), multiply(multiply(const_3, const_4), 550)), 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 * 1_000<\/gadget>\n9_000<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 * 550<\/gadget>\n6_600<\/output>\n9_000 - 6_600<\/gadget>\n2_400<\/output>\n12 * 1<\/gadget>\n12<\/output>\n2_400 \/ 12<\/gadget>\n200<\/output>\n200 * 4<\/gadget>\n800<\/output>\n800<\/result>","index":928} +{"problem":"a glass was filled with 24 ounces of water , and 0.06 ounce of the water evaporated each day during a 12 - day period . what percent of the original amount of water evaporated during this period ?","rationale":"\"in 12 days 12 * 0.06 = 0.72 ounces of water evaporated , which is 0.72 \/ 24 â ˆ — 100 = 3 of the original amount of water . answer : c .\"","correct":"c","options":{"a":"0.003 % ","b":"0.03 % ","c":"3 % ","d":"2 %","e":"30 %"},"options_float":{"a":0.003,"b":0.03,"c":3.0,"d":2.0,"e":30.0},"annotated_formula":"multiply(divide(multiply(0.06, 12), 24), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"0.06 * 12<\/gadget>\n0.72<\/output>\n0.72 \/ 24<\/gadget>\n0.03<\/output>\n0.03 * 100<\/gadget>\n3<\/output>\n3<\/result>","index":930} +{"problem":"i chose a number and divide it by 4 . then i subtracted 18 from the result and got 7 . what was the number i chose ?","rationale":"solution : let x be the number i chose , then x \/ 4 â ˆ ’ 18 = 7 x \/ 4 = 25 x = 100 answer b","correct":"b","options":{"a":"600 ","b":"100 ","c":"800 ","d":"900","e":"none"},"options_float":{"a":600.0,"b":100.0,"c":800.0,"d":900.0,"e":null},"annotated_formula":"multiply(add(18, 7), 4)","linear_formula":"add(n1,n2)|multiply(n0,#0)","chain":"18 + 7<\/gadget>\n25<\/output>\n25 * 4<\/gadget>\n100<\/output>\n100<\/result>","index":931} +{"problem":"there are 76 persons . 54 can read hindu , 43 can read times , 37 can read deccan and 15 can read all . if 24 can read hindu and deccan and 27 can read deccan and times then what is the number of persons who read only times and hindu .","rationale":"let ' a ' can be read hindu , let ' b ' can be read times , let ' c ' can be read deccan , from the given data : n ( aubuc ) = 76 , n ( a ) = 54 , n ( b ) = 43 , n ( c ) = 37 , n ( anbnc ) = 15 , n ( anc ) = 24 , n ( bnc ) = 27 , n ( anb ) = ? n ( aubuc ) = n ( a ) + n ( b ) + n ( c ) - n ( anb ) - n ( bnc ) - n ( anc ) + n ( anbnc ) = = > 76 = 54 + 43 + 37 - n ( anb ) - 24 - 27 + 15 = = > n ( anb ) = 54 + 43 + 37 + 15 - 24 - 27 - 76 = = > n ( anb ) = 149 - 127 = = > n ( anb ) = 22 answer : b","correct":"b","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(subtract(24, 15), subtract(27, 15))","linear_formula":"subtract(n5,n4)|subtract(n6,n4)|add(#0,#1)","chain":"24 - 15<\/gadget>\n9<\/output>\n27 - 15<\/gadget>\n12<\/output>\n9 + 12<\/gadget>\n21<\/output>\n21<\/result>","index":932} +{"problem":"the original price of a suit is $ 100 . the price increased 20 % , and after this increase , the store published a 20 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 20 % off the increased price , how much did these consumers pay for the suit ?","rationale":"\"0.8 * ( 1.2 * 100 ) = $ 96 the answer is b .\"","correct":"b","options":{"a":"$ 88 ","b":"$ 96 ","c":"$ 100 ","d":"$ 106","e":"$ 110"},"options_float":{"a":88.0,"b":96.0,"c":100.0,"d":106.0,"e":110.0},"annotated_formula":"subtract(add(100, divide(multiply(100, 20), const_100)), divide(multiply(add(100, divide(multiply(100, 20), const_100)), 20), 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":"100 * 20<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n100 + 20<\/gadget>\n120<\/output>\n120 * 20<\/gadget>\n2_400<\/output>\n2_400 \/ 100<\/gadget>\n24<\/output>\n120 - 24<\/gadget>\n96<\/output>\n96<\/result>","index":933} +{"problem":"the average runs scored by a batsman in 20 matches is 40 . in the next 30 matches the batsman scored an average of 20 runs . find his average in all the 50 matches ?","rationale":"\"total score of the batsman in 20 matches = 800 . total score of the batsman in the next 30 matches = 600 . total score of the batsman in the 50 matches = 1400 . average score of the batsman = 1400 \/ 50 = 28 . answer : c\"","correct":"c","options":{"a":"31 ","b":"46 ","c":"28 ","d":"13","e":"12"},"options_float":{"a":31.0,"b":46.0,"c":28.0,"d":13.0,"e":12.0},"annotated_formula":"divide(add(multiply(40, 20), multiply(20, 30)), add(20, 30))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"40 * 20<\/gadget>\n800<\/output>\n20 * 30<\/gadget>\n600<\/output>\n800 + 600<\/gadget>\n1_400<\/output>\n20 + 30<\/gadget>\n50<\/output>\n1_400 \/ 50<\/gadget>\n28<\/output>\n28<\/result>","index":934} +{"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 5000 , the number of valid votes that the other candidate got , was :","rationale":"\"a number of valid votes = 80 % of 5000 = 4000 . valid votes polled by other candidate = 45 % of 4000 = ( 45 \/ 100 x 4000 ) = 1800 .\"","correct":"a","options":{"a":"1800 ","b":"2700 ","c":"2900 ","d":"2200","e":"2300"},"options_float":{"a":1800.0,"b":2700.0,"c":2900.0,"d":2200.0,"e":2300.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 5000)","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) * 5_000<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":936} +{"problem":"for a group of n people , k of whom are of the same sex , the ( n - k ) \/ n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 8 of whom are females , by how much does the index for the females exceed the index for the males in the group ?","rationale":"index for females = ( 20 - 8 ) \/ 20 = 3 \/ 5 = 0.6 index for males = ( 20 - 12 \/ 20 = 2 \/ 5 = 0.4 index for females exceeds males by 0.6 - 0.4 = 0.2 answer : c","correct":"c","options":{"a":"0.05 ","b":"0.0625 ","c":"0.2 ","d":"0.25","e":"0.6"},"options_float":{"a":0.05,"b":0.0625,"c":0.2,"d":0.25,"e":0.6},"annotated_formula":"subtract(divide(subtract(20, 8), 20), divide(8, 20))","linear_formula":"divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)","chain":"20 - 8<\/gadget>\n12<\/output>\n12 \/ 20<\/gadget>\n3\/5 = around 0.6<\/output>\n8 \/ 20<\/gadget>\n2\/5 = around 0.4<\/output>\n(3\/5) - (2\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":938} +{"problem":"chocolate bars are sold in packages of 4 or 9 only . if mark bought 97 chocolate bars exactly , what could be the number of large packs mark bought ?","rationale":"let number of packs of four = f let number of packs of nine = n 4 f + 9 n = 97 now , we need to test for values of n . since sum 97 is odd and 4 f will always be even , n ca n ' t be even . now , we can test for values e = 2 , 4 and 6 4 * 4 + 9 * 9 = 16 + 81 = 97 answer d","correct":"d","options":{"a":"3 ","b":"4 ","c":"8 ","d":"9","e":"13"},"options_float":{"a":3.0,"b":4.0,"c":8.0,"d":9.0,"e":13.0},"annotated_formula":"divide(subtract(97, multiply(4, 4)), 9)","linear_formula":"multiply(n0,n0)|subtract(n2,#0)|divide(#1,n1)","chain":"4 * 4<\/gadget>\n16<\/output>\n97 - 16<\/gadget>\n81<\/output>\n81 \/ 9<\/gadget>\n9<\/output>\n9<\/result>","index":939} +{"problem":"the percentage profit earned by selling an article for rs . 1920 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 20 % profit ?","rationale":"\"let c . p . be rs . x . then , ( 1920 - x ) \/ x * 100 = ( x - 1280 ) \/ x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 120 % of rs . 1600 = 120 \/ 100 * 1600 = rs . 1920 . answer : d\"","correct":"d","options":{"a":"2000 ","b":"2778 ","c":"2299 ","d":"1920","e":"2771"},"options_float":{"a":2000.0,"b":2778.0,"c":2299.0,"d":1920.0,"e":2771.0},"annotated_formula":"multiply(divide(add(const_100, 20), const_100), divide(add(1920, 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 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n1_920 + 1_280<\/gadget>\n3_200<\/output>\n3_200 \/ 2<\/gadget>\n1_600<\/output>\n(6\/5) * 1_600<\/gadget>\n1_920<\/output>\n1_920<\/result>","index":940} +{"problem":"find the least number must be subtracted from 531742 so that remaining no . is divisible by 3 ?","rationale":"\"on dividing 531742 by 3 we get the remainder 1 , so 1 should be subtracted c\"","correct":"c","options":{"a":"4 ","b":"5 ","c":"1 ","d":"2","e":"3"},"options_float":{"a":4.0,"b":5.0,"c":1.0,"d":2.0,"e":3.0},"annotated_formula":"subtract(531742, multiply(floor(divide(531742, 3)), 3))","linear_formula":"divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|","chain":"531_742 \/ 3<\/gadget>\n531_742\/3 = around 177_247.333333<\/output>\nfloor(531_742\/3)<\/gadget>\n177_247<\/output>\n177_247 * 3<\/gadget>\n531_741<\/output>\n531_742 - 531_741<\/gadget>\n1<\/output>\n1<\/result>","index":942} +{"problem":"10 men can cut 10 trees in 2 hours . if 2 men leave the job , how many trees will be cut in 3 hours ?","rationale":"10 men - working 2 hrs - cut 10 trees 1 men - working 1 hr - cuts = 10 \/ 10 * 2 thus 8 men - working 3 hrs - cut = 10 * 8 * 3 \/ 10 * 2 = 12 trees answer is a","correct":"a","options":{"a":"12 ","b":"15 ","c":"16 ","d":"18","e":"20"},"options_float":{"a":12.0,"b":15.0,"c":16.0,"d":18.0,"e":20.0},"annotated_formula":"multiply(multiply(subtract(10, 2), divide(divide(10, 2), 10)), 3)","linear_formula":"divide(n0,n2)|subtract(n0,n2)|divide(#0,n0)|multiply(#2,#1)|multiply(n4,#3)","chain":"10 - 2<\/gadget>\n8<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n8 * (1\/2)<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12<\/result>","index":943} +{"problem":"if x and y are integers such that x ^ 2 = 2 y and xy = 32 , then x – y = ?","rationale":"\"here x and y are integers . x ^ 2 = 2 y , xy = 32 . substitute ( x ^ 2 ) \/ 2 = y in xy = > x ^ 3 = 32 * 2 = > x ^ 3 = 64 . here x 3 is positive , x is also positive . x = 4 then y = 8 . x - y = - 4 so option d is correct\"","correct":"d","options":{"a":"- 30 ","b":"- 20 ","c":"- 5 ","d":"- 4","e":"20"},"options_float":{"a":-30.0,"b":-20.0,"c":-5.0,"d":-4.0,"e":20.0},"annotated_formula":"subtract(power(multiply(32, 2), const_0_33), divide(32, power(multiply(32, 2), const_0_33)))","linear_formula":"multiply(n1,n2)|power(#0,const_0_33)|divide(n2,#1)|subtract(#1,#2)|","chain":"32 * 2<\/gadget>\n64<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n64 ** (1\/3)<\/gadget>\n4<\/output>\n32 \/ 4<\/gadget>\n8<\/output>\n4 - 8<\/gadget>\n-4<\/output>\n-4<\/result>","index":946} +{"problem":"a shopkeeper loses 15 % , if an article is sold for $ 102 . what should be the selling price of the article to gain 20 % ?","rationale":"\"c $ 144 given that sp = $ 102 and loss = 15 % cp = [ 100 ( sp ) ] \/ ( 100 - l % ) = ( 100 * 102 ) \/ 85 = 20 * 6 = $ 120 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] \/ 100 = ( 120 * 120 ) \/ 100 = $ 144\"","correct":"c","options":{"a":"$ 165 ","b":"$ 174 ","c":"$ 144 ","d":"$ 164","e":"$ 183"},"options_float":{"a":165.0,"b":174.0,"c":144.0,"d":164.0,"e":183.0},"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":947} +{"problem":"after decreasing 24 % in the price of an article costs rs . 912 . find the actual cost of an article ?","rationale":"\"cp * ( 76 \/ 100 ) = 912 cp = 12 * 100 = > cp = 1200 answer : c\"","correct":"c","options":{"a":"226 ","b":"255 ","c":"1200 ","d":"2771","e":"332"},"options_float":{"a":226.0,"b":255.0,"c":1200.0,"d":2771.0,"e":332.0},"annotated_formula":"divide(912, 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>\n912 \/ (19\/25)<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":948} +{"problem":"a man sells an article at 10 % gain . had be sold at for rs . 60 \/ - more he could have gained 20 % what is cost price of article","rationale":"first selling price = 110 % - - - - - > x rupees = sold at for rs . 60 \/ - = 120 % - - - - - > x + 60 rupees ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 10 % - - - - - - - - > 60 100 % - - - - - - - > rs . 600 \/ - option ' b '","correct":"b","options":{"a":"rs . 500 ","b":"rs . 600 ","c":"rs . 650 ","d":"rs . 760","e":"rs . 800"},"options_float":{"a":500.0,"b":600.0,"c":650.0,"d":760.0,"e":800.0},"annotated_formula":"multiply(divide(60, subtract(20, 10)), const_100)","linear_formula":"subtract(n2,n0)|divide(n1,#0)|multiply(#1,const_100)","chain":"20 - 10<\/gadget>\n10<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600<\/result>","index":949} +{"problem":"if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 7 ¤ √ 7 =","rationale":"\"x = √ 7 and y also = √ 7 applying the function ( √ 7 + √ 7 ) ^ 2 - ( √ 7 - √ 7 ) ^ 2 = ( 2 √ 7 ) ^ 2 - 0 = 4 x 7 = 28 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = √ 7 and y = √ 7 you get the answer 28 . answer e\"","correct":"e","options":{"a":"0 ","b":"5 ","c":"10 ","d":"15","e":"28"},"options_float":{"a":0.0,"b":5.0,"c":10.0,"d":15.0,"e":28.0},"annotated_formula":"power(add(sqrt(7), sqrt(7)), 2)","linear_formula":"sqrt(n2)|add(#0,#0)|power(#1,n0)|","chain":"7 ** (1\/2)<\/gadget>\nsqrt(7) = around 2.645751<\/output>\n(sqrt(7)) + (sqrt(7))<\/gadget>\n2*sqrt(7) = around 5.291503<\/output>\n(2*sqrt(7)) ** 2<\/gadget>\n28<\/output>\n28<\/result>","index":951} +{"problem":"6 people can do work in 80 days how much people they required to complete the work in 16 days ?","rationale":"man and days concept . . . 6 m * 80 d = m * 16 d solve it , total no of people required is 30 ; 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":"divide(multiply(6, 80), 16)","linear_formula":"multiply(n0,n1)|divide(#0,n2)","chain":"6 * 80<\/gadget>\n480<\/output>\n480 \/ 16<\/gadget>\n30<\/output>\n30<\/result>","index":952} +{"problem":"the average of 10 consecutive integers is 15 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ?","rationale":"\"the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) \/ 2 = 45 on average , each number will be reduced by 45 \/ 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is b .\"","correct":"b","options":{"a":"10 ","b":"10.5 ","c":"11 ","d":"11.5","e":"12"},"options_float":{"a":10.0,"b":10.5,"c":11.0,"d":11.5,"e":12.0},"annotated_formula":"divide(subtract(multiply(10, 15), multiply(add(const_4, const_1), 9)), 10)","linear_formula":"add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)|","chain":"10 * 15<\/gadget>\n150<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5 * 9<\/gadget>\n45<\/output>\n150 - 45<\/gadget>\n105<\/output>\n105 \/ 10<\/gadget>\n21\/2 = around 10.5<\/output>\n21\/2 = around 10.5<\/result>","index":953} +{"problem":"on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 140 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 = 140 km so 1 inch = 140 \/ 312 km so 34 inch = ( 140 ã — 34 ) \/ 312 = 15.25 km answer : d\"","correct":"d","options":{"a":"14.83 ","b":"14.81 ","c":"14.8 ","d":"15.25","e":"14.12"},"options_float":{"a":14.83,"b":14.81,"c":14.8,"d":15.25,"e":14.12},"annotated_formula":"divide(multiply(34, 140), 312)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|","chain":"34 * 140<\/gadget>\n4_760<\/output>\n4_760 \/ 312<\/gadget>\n595\/39 = around 15.25641<\/output>\n595\/39 = around 15.25641<\/result>","index":954} +{"problem":"a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 8 % dividend at the end of the year , then how much does he get ?","rationale":"\"solution number of shares = ( 14400 \/ 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 8 \/ 100 x 12000 ) = rs . 960 . answer c\"","correct":"c","options":{"a":"rs . 500 ","b":"rs . 600 ","c":"rs . 960 ","d":"rs . 720","e":"none"},"options_float":{"a":500.0,"b":600.0,"c":960.0,"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(20, 100))))), divide(8, 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>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n100 * (1\/5)<\/gadget>\n20<\/output>\n100 + 20<\/gadget>\n120<\/output>\n14_400 \/ 120<\/gadget>\n120<\/output>\n100 * 120<\/gadget>\n12_000<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n12_000 * (2\/25)<\/gadget>\n960<\/output>\n960<\/result>","index":956} +{"problem":"a man took loan from a bank at the rate of 12 % p . a . s . i . after 20 years he had to pay rs . 1500 interest only for the period . the principal amount borrowed by him was ?","rationale":"principal = ( 100 * 1500 ) \/ ( 12 * 20 ) = rs . 625 answer : a","correct":"a","options":{"a":"625 ","b":"700 ","c":"950 ","d":"825","e":"630"},"options_float":{"a":625.0,"b":700.0,"c":950.0,"d":825.0,"e":630.0},"annotated_formula":"divide(divide(multiply(1500, const_100), 20), 12)","linear_formula":"multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)","chain":"1_500 * 100<\/gadget>\n150_000<\/output>\n150_000 \/ 20<\/gadget>\n7_500<\/output>\n7_500 \/ 12<\/gadget>\n625<\/output>\n625<\/result>","index":957} +{"problem":"david and lewis leave chennai for tirupati simultaneously at 7 a . m in the morning driving in two cars at speeds of 60 mph and 80 mph respectively . as soon as lewis reaches tirupati he returns back to chennai along the same route and meets david on the way back . if the distance between the two cities is 160 miles , how far from chennai did david and lewis meet ?","rationale":"time taken by lewis to reach tirupati = 160 \/ 80 = 2 hours in 2 hours , david travels 60 * 2 = 120 miles so distance at which they meet should be greater than 120 miles . only b satisfies . answer is b .","correct":"b","options":{"a":"100 mlies ","b":"120 miles ","c":"90 miles ","d":"95 miles","e":"110 miles"},"options_float":{"a":100.0,"b":120.0,"c":90.0,"d":95.0,"e":110.0},"annotated_formula":"multiply(const_2, 60)","linear_formula":"multiply(n1,const_2)","chain":"2 * 60<\/gadget>\n120<\/output>\n120<\/result>","index":958} +{"problem":"the product of x and y is a constant . if the value of x is increased by 60 % , by what percentage must the value of y be decreased ?","rationale":"\"x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.6 x ( 120 ) i . e . 160 * new ' y ' = 10000 i . e . new ' y ' = 10000 \/ 160 = 62.5 i . e . y decreases from 100 to 62.5 i . e . decrease of 37.5 % answer : option d\"","correct":"d","options":{"a":"30 % ","b":"32 % ","c":"35 % ","d":"37.5 %","e":"40 %"},"options_float":{"a":30.0,"b":32.0,"c":35.0,"d":37.5,"e":40.0},"annotated_formula":"multiply(subtract(const_1, divide(const_100, add(const_100, 60))), const_100)","linear_formula":"add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"100 + 60<\/gadget>\n160<\/output>\n100 \/ 160<\/gadget>\n5\/8 = around 0.625<\/output>\n1 - (5\/8)<\/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":961} +{"problem":"linda spent 3 \/ 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 400 , what were her original savings ?","rationale":"if linda spent 3 \/ 4 of her savings on furnitute , the rest 4 \/ 4 - 3 \/ 4 = 1 \/ 4 on a tv but the tv cost her $ 400 . so 1 \/ 4 of her savings is $ 400 . so her original savings are 4 times $ 400 = $ 1600 correct answer b","correct":"b","options":{"a":"$ 1500 ","b":"$ 1600 ","c":"$ 1700 ","d":"$ 1800","e":"$ 1900"},"options_float":{"a":1500.0,"b":1600.0,"c":1700.0,"d":1800.0,"e":1900.0},"annotated_formula":"divide(400, subtract(const_1, divide(3, 4)))","linear_formula":"divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 - (3\/4)<\/gadget>\n1\/4 = around 0.25<\/output>\n400 \/ (1\/4)<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":962} +{"problem":"city a to city b , andrew drove for 1 hour at 50 mph and for 3 hours at 60 mph . what was the average speed for the whole trip ?","rationale":"the total distance is 1 × 50 + 3 × 60 = 2301 × 50 + 3 × 60 = 230 . and the total time is 4 hours . hence , average speed = ( total distancetotal time ) = 2304 = 57.5 b","correct":"b","options":{"a":"56 ","b":"57.5 ","c":"58.9 ","d":"61.4","e":"62"},"options_float":{"a":56.0,"b":57.5,"c":58.9,"d":61.4,"e":62.0},"annotated_formula":"divide(add(multiply(50, 1), multiply(60, 3)), add(3, 1))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)","chain":"50 * 1<\/gadget>\n50<\/output>\n60 * 3<\/gadget>\n180<\/output>\n50 + 180<\/gadget>\n230<\/output>\n3 + 1<\/gadget>\n4<\/output>\n230 \/ 4<\/gadget>\n115\/2 = around 57.5<\/output>\n115\/2 = around 57.5<\/result>","index":963} +{"problem":"laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 15 kmph and prasanna moves southward at a speed of 45 kmph . how far will be prasanna from laxmi after 60 minutes ?","rationale":"\"explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 15 kmph x 1 hr = 15 km total southward prasanna ' s distance = 45 kmph x 1 hr = 45 km total distance between prasanna and laxmi is = 15 + 45 = 60 km . answer : e\"","correct":"e","options":{"a":"11 ","b":"50 ","c":"28 ","d":"27","e":"60"},"options_float":{"a":11.0,"b":50.0,"c":28.0,"d":27.0,"e":60.0},"annotated_formula":"add(15, 45)","linear_formula":"add(n0,n1)|","chain":"15 + 45<\/gadget>\n60<\/output>\n60<\/result>","index":964} +{"problem":"car x began traveling at an average speed of 35 miles per hour . after 84 minutes , car y began traveling at an average speed of 42 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":"in 84 minutes , car x travels 49 miles . car y gains 7 miles each hour , so it takes 7 hours to catch car x . in 7 hours , car x travels 245 miles . the answer is d .","correct":"d","options":{"a":"140 ","b":"175 ","c":"210 ","d":"245","e":"270"},"options_float":{"a":140.0,"b":175.0,"c":210.0,"d":245.0,"e":270.0},"annotated_formula":"multiply(35, divide(multiply(divide(84, const_60), 35), subtract(42, 35)))","linear_formula":"divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)","chain":"84 \/ 60<\/gadget>\n7\/5 = around 1.4<\/output>\n(7\/5) * 35<\/gadget>\n49<\/output>\n42 - 35<\/gadget>\n7<\/output>\n49 \/ 7<\/gadget>\n7<\/output>\n35 * 7<\/gadget>\n245<\/output>\n245<\/result>","index":965} +{"problem":"the cost of producing x tools by a company is given by c ( x ) = 600 x + 5500 ( in $ ) a ) what is the cost of 100 tools ?","rationale":"solution c ( 100 ) = 600 * 100 + 5500 = 125500 $ answer a","correct":"a","options":{"a":"65500 $ ","b":"125800 $ ","c":"125900 $ ","d":"6500 $","e":"122500 $"},"options_float":{"a":65500.0,"b":125800.0,"c":125900.0,"d":6500.0,"e":122500.0},"annotated_formula":"add(multiply(100, 600), 5500)","linear_formula":"multiply(n0,n2)|add(n1,#0)","chain":"100 * 600<\/gadget>\n60_000<\/output>\n60_000 + 5_500<\/gadget>\n65_500<\/output>\n65_500<\/result>","index":966} +{"problem":"how many odd numbers between 10 and 1,000 are the squares of integers ?","rationale":"the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 14 numbers . answer : c .","correct":"c","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(10, const_4)","linear_formula":"add(n0,const_4)","chain":"10 + 4<\/gadget>\n14<\/output>\n14<\/result>","index":967} +{"problem":"a man can row 11 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?","rationale":"\"m = 11 s = 1.2 ds = 12.2 us = 9.8 x \/ 12.2 + x \/ 9.8 = 1 x = 5.43 d = 5.43 * 2 = 10.87 answer : e\"","correct":"e","options":{"a":"6.24 km ","b":"6 km ","c":"5.76 km ","d":"5.66 km","e":"10.87 km"},"options_float":{"a":6.24,"b":6.0,"c":5.76,"d":5.66,"e":10.87},"annotated_formula":"multiply(divide(multiply(add(11, 1.2), subtract(11, 1.2)), add(add(11, 1.2), subtract(11, 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":"11 + 1.2<\/gadget>\n12.2<\/output>\n11 - 1.2<\/gadget>\n9.8<\/output>\n12.2 * 9.8<\/gadget>\n119.56<\/output>\n12.2 + 9.8<\/gadget>\n22<\/output>\n119.56 \/ 22<\/gadget>\n5.434545<\/output>\n5.434545 * 2<\/gadget>\n10.86909<\/output>\n10.86909<\/result>","index":968} +{"problem":"if the l . c . m of two numbers is 720 and their product is 38880 , find the h . c . f of the numbers .","rationale":"\"h . c . f = ( product of the numbers ) \/ ( their l . c . m ) = 38880 \/ 720 = 54 . answer : d\"","correct":"d","options":{"a":"50 ","b":"30 ","c":"125 ","d":"54","e":"none of these"},"options_float":{"a":50.0,"b":30.0,"c":125.0,"d":54.0,"e":null},"annotated_formula":"divide(38880, 720)","linear_formula":"divide(n1,n0)|","chain":"38_880 \/ 720<\/gadget>\n54<\/output>\n54<\/result>","index":969} +{"problem":"a , b and c have rs . 400 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ?","rationale":"\"a + b + c = 400 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 400 - - - - - - - - - - - - - - - - c = 150 answer : a\"","correct":"a","options":{"a":"150 ","b":"140 ","c":"130 ","d":"120","e":"110"},"options_float":{"a":150.0,"b":140.0,"c":130.0,"d":120.0,"e":110.0},"annotated_formula":"subtract(add(200, 350), 400)","linear_formula":"add(n1,n2)|subtract(#0,n0)|","chain":"200 + 350<\/gadget>\n550<\/output>\n550 - 400<\/gadget>\n150<\/output>\n150<\/result>","index":970} +{"problem":"an agent , gets a commission of 2.5 % on the sales of cloth . if on a certain day , he gets rs . 12.50 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 = 12.50 < = > ( 25 \/ 100 * 1 \/ 100 * x ) = 125 \/ 10 < = > x = 500 . answer : b ) 500","correct":"b","options":{"a":"333 ","b":"500 ","c":"887 ","d":"299","e":"132"},"options_float":{"a":333.0,"b":500.0,"c":887.0,"d":299.0,"e":132.0},"annotated_formula":"divide(12.5, divide(2.5, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,#0)","chain":"2.5 \/ 100<\/gadget>\n0.025<\/output>\n12.5 \/ 0.025<\/gadget>\n500<\/output>\n500<\/result>","index":971} +{"problem":"an empty wooden vessel weighs 20 % 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 20 % of its total weight when filled with paint : vessel = 0.2 ( vessel + paint ) ; 20 v = v + p ( so the weight of completely filled vessel is 10 v ) p = 19 v ( so the weight of the paint when the vessels is completely filled is 19 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 \/ 2 * 20 v ; p ' = 9 v ( so the weight of the paint when the vessels is partially filled is 9 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 = 9 v \/ 19 v = 9 \/ 19 . answer : d .","correct":"d","options":{"a":"3 \/ 5 ","b":"5 \/ 9 ","c":"1 \/ 24 ","d":"9 \/ 19","e":"2 \/ 5"},"options_float":{"a":0.6,"b":0.5555555556,"c":0.0416666667,"d":0.4736842105,"e":0.4},"annotated_formula":"divide(subtract(divide(20, const_2), const_1), subtract(20, const_1))","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|subtract(#0,const_1)|divide(#2,#1)","chain":"20 \/ 2<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n20 - 1<\/gadget>\n19<\/output>\n9 \/ 19<\/gadget>\n9\/19 = around 0.473684<\/output>\n9\/19 = around 0.473684<\/result>","index":972} +{"problem":"a cycle is bought for rs . 900 and sold for rs . 990 , find the gain percent ?","rationale":"\"900 - - - - 90 100 - - - - ? = > 10 % answer : e\"","correct":"e","options":{"a":"39 % ","b":"20 % ","c":"23 % ","d":"74 %","e":"10 %"},"options_float":{"a":39.0,"b":20.0,"c":23.0,"d":74.0,"e":10.0},"annotated_formula":"multiply(divide(subtract(990, 900), 900), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"990 - 900<\/gadget>\n90<\/output>\n90 \/ 900<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":973} +{"problem":"( 0.0066 ) ( 3.6 ) \/ ( 0.04 ) ( 0.1 ) ( 0.006 ) =","rationale":"\"( 0.0066 ) ( 3.6 ) \/ ( 0.04 ) ( 0.1 ) ( 0.006 ) = 0.0060 * 360 \/ 4 * ( 0.1 ) ( 0.006 ) = 0.066 * 90 \/ 1 * 0.006 = 66 * 90 \/ 6 = 11 * 90 = 990 answer : a\"","correct":"a","options":{"a":"990 ","b":"99.0 ","c":"9.9 ","d":"0.99","e":"0.099"},"options_float":{"a":990.0,"b":99.0,"c":9.9,"d":0.99,"e":0.099},"annotated_formula":"divide(multiply(0.0066, 3.6), multiply(multiply(0.04, 0.1), 0.006))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2)|","chain":"0.0066 * 3.6<\/gadget>\n0.02376<\/output>\n0.04 * 0.1<\/gadget>\n0.004<\/output>\n0.004 * 0.006<\/gadget>\n0.000024<\/output>\n0.02376 \/ 0.000024<\/gadget>\n990<\/output>\n990<\/result>","index":974} +{"problem":"in the manufacture of a certain product , 9 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":"\"0.09 * 0.05 = 0.0045 = 0.45 % the answer is b .\"","correct":"b","options":{"a":"0.15 % ","b":"0.45 % ","c":"0.8 % ","d":"1.25 %","e":"2.0 %"},"options_float":{"a":0.15,"b":0.45,"c":0.8,"d":1.25,"e":2.0},"annotated_formula":"multiply(9, divide(5, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n9 * (1\/20)<\/gadget>\n9\/20 = around 0.45<\/output>\n9\/20 = around 0.45<\/result>","index":975} +{"problem":"if the sales tax reduced from 3 1 \/ 2 % to 3 1 \/ 3 % , then what difference does it make to a person who purchases an article with market price of rs . 9600 ?","rationale":"required difference = [ 3 ½ % of rs . 9600 ] – [ 3 1 \/ 3 % of rs . 9600 ] = [ ( 7 \/ 20 - ( 10 \/ 3 ) ] % of rs . 9600 = 1 \/ 6 % of rs . 9600 = rs . [ ( 1 \/ 6 ) 8 ( 1 \/ 100 ) * 9600 ] = rs . 16 . answer is e .","correct":"e","options":{"a":"11 ","b":"13 ","c":"14 ","d":"18","e":"16"},"options_float":{"a":11.0,"b":13.0,"c":14.0,"d":18.0,"e":16.0},"annotated_formula":"divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 9600), const_100)","linear_formula":"divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) + 3<\/gadget>\n7\/2 = around 3.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) + 3<\/gadget>\n10\/3 = around 3.333333<\/output>\n(7\/2) - (10\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 9_600<\/gadget>\n1_600<\/output>\n1_600 \/ 100<\/gadget>\n16<\/output>\n16<\/result>","index":977} +{"problem":"a 6 % stock yields 8 % . the market value of the stock is :","rationale":"for an income of rs . 8 , investment = rs . 100 . for an income of rs . 6 , investment = rs . 100 x 6 = rs . 75 . 8 market value of rs . 100 stock = rs . 75 . answer : b","correct":"b","options":{"a":"33 ","b":"75 ","c":"44 ","d":"27","e":"91"},"options_float":{"a":33.0,"b":75.0,"c":44.0,"d":27.0,"e":91.0},"annotated_formula":"multiply(divide(const_100, 8), 6)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)","chain":"100 \/ 8<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 6<\/gadget>\n75<\/output>\n75<\/result>","index":978} +{"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 $ 16 , 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 16 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 16 solve for x x = $ 28.44 correct answer b\"","correct":"b","options":{"a":"$ 18.44 ","b":"$ 28.44 ","c":"$ 48.44 ","d":"$ 58.44","e":"$ 38.44"},"options_float":{"a":18.44,"b":28.44,"c":48.44,"d":58.44,"e":38.44},"annotated_formula":"divide(multiply(multiply(const_100, const_100), 16), 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 * 16<\/gadget>\n160_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>\n160_000 \/ 5_625<\/gadget>\n256\/9 = around 28.444444<\/output>\n256\/9 = around 28.444444<\/result>","index":979} +{"problem":"in an election between the two candidates , the candidates who gets 60 % of votes polled is winned 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":"1600 ","c":"1800 ","d":"2000","e":"2100"},"options_float":{"a":1400.0,"b":1600.0,"c":1800.0,"d":2000.0,"e":2100.0},"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":980} +{"problem":"the height of a cylinder is 60 cm and the diameter of its base is 5 cm . the total surface area of the cylinder is","rationale":"given h = 60 cm and r = 5 \/ 2 cm total surface area = 2 π rh + 2 & pir ( power 2 ) = 2 π r ( h + r ) = [ 2 × 22 \/ 7 × 5 \/ 2 × ( 60 + 5 \/ 2 ) ] cm ( power 2 ) = [ 44 \/ 7 × 5 \/ 2 × ( ( 120 + 5 ) \/ 2 ) ] cm ( power 2 ) = 22 \/ 7 × 5 × 125 \/ 2 cm ( power 2 ) = ( 55 × 125 ) \/ 7 cm ( power 2 ) = 6875 \/ 7 cm ( power 2 ) = 982.14 cm ( power 2 ) answer is c .","correct":"c","options":{"a":"918.14 ","b":"981.41 ","c":"982.14 ","d":"928.41","e":"none of them"},"options_float":{"a":918.14,"b":981.41,"c":982.14,"d":928.41,"e":null},"annotated_formula":"surface_cylinder(divide(5, const_2), 60)","linear_formula":"divide(n1,const_2)|surface_cylinder(#0,n0)","chain":"5 \/ 2<\/gadget>\n5\/2 = around 2.5<\/output>\n2 * pi * (5\/2) * ((5\/2) + 60)<\/gadget>\n625*pi\/2 = around 981.747704<\/output>\n625*pi\/2 = around 981.747704<\/result>","index":982} +{"problem":"a tank is filled by 3 pipes a , b , c in 7 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 \/ 7 7 \/ x = 1 \/ 7 x = 49 hours answer is d\"","correct":"d","options":{"a":"25 hr ","b":"35 hr ","c":"40 hr ","d":"49 hr","e":"50 hr"},"options_float":{"a":25.0,"b":35.0,"c":40.0,"d":49.0,"e":50.0},"annotated_formula":"multiply(add(add(multiply(const_2, const_2), const_2), const_1), 7)","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 * 7<\/gadget>\n49<\/output>\n49<\/result>","index":983} +{"problem":"a worker is paid a regular rate of rs . 20 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 10 % higher than her regular rate . if she completed 70 surveys involving the use of her cellphone , how much did she get that week ?","rationale":"amount earned using her cell phone = 70 * 22 = 1540 earned for remaining surveys = 30 * 20 = 600 total earning = 2140 answer : a","correct":"a","options":{"a":"2140 ","b":"1140 ","c":"550 ","d":"650","e":"750"},"options_float":{"a":2140.0,"b":1140.0,"c":550.0,"d":650.0,"e":750.0},"annotated_formula":"add(multiply(20, 100), multiply(70, multiply(20, divide(10, const_100))))","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n3,#2)|add(#1,#3)","chain":"20 * 100<\/gadget>\n2_000<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n20 * (1\/10)<\/gadget>\n2<\/output>\n70 * 2<\/gadget>\n140<\/output>\n2_000 + 140<\/gadget>\n2_140<\/output>\n2_140<\/result>","index":985} +{"problem":"if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 38 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ?","rationale":"\"0.1 x = 0.08 ( 2000 - x ) + 38 0.18 x = 198 x = 1100 then the amount invested at 8 % is $ 2000 - $ 1100 = $ 900 the answer is c .\"","correct":"c","options":{"a":"$ 700 ","b":"$ 800 ","c":"$ 900 ","d":"$ 1100","e":"$ 1200"},"options_float":{"a":700.0,"b":800.0,"c":900.0,"d":1100.0,"e":1200.0},"annotated_formula":"subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 38), add(divide(10, const_100), divide(8, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n1_000 * 2<\/gadget>\n2_000<\/output>\n10 * 8<\/gadget>\n80<\/output>\n80 * 2<\/gadget>\n160<\/output>\n160 + 38<\/gadget>\n198<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(1\/10) + (2\/25)<\/gadget>\n9\/50 = around 0.18<\/output>\n198 \/ (9\/50)<\/gadget>\n1_100<\/output>\n2_000 - 1_100<\/gadget>\n900<\/output>\n900<\/result>","index":986} +{"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 42 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 = 42 \/ 3 = 14 kmph . answer : c\"","correct":"c","options":{"a":"16 kmph ","b":"18 kmph ","c":"14 kmph ","d":"79 kmph","e":"27 kmph"},"options_float":{"a":16.0,"b":18.0,"c":14.0,"d":79.0,"e":27.0},"annotated_formula":"subtract(42, divide(multiply(42, const_2), const_3))","linear_formula":"multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)|","chain":"42 * 2<\/gadget>\n84<\/output>\n84 \/ 3<\/gadget>\n28<\/output>\n42 - 28<\/gadget>\n14<\/output>\n14<\/result>","index":988} +{"problem":"by selling an article at rs . 300 , a profit of 25 % is made . find its cost price ?","rationale":"\"sp = 300 cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 300 * [ 100 \/ ( 100 + 25 ) ] = 300 * [ 100 \/ 125 ] = rs . 240 answer : c\"","correct":"c","options":{"a":"s . 486 ","b":"s . 455 ","c":"s . 240 ","d":"s . 480","e":"s . 489"},"options_float":{"a":486.0,"b":455.0,"c":240.0,"d":480.0,"e":489.0},"annotated_formula":"divide(multiply(300, const_100), add(const_100, 25))","linear_formula":"add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|","chain":"300 * 100<\/gadget>\n30_000<\/output>\n100 + 25<\/gadget>\n125<\/output>\n30_000 \/ 125<\/gadget>\n240<\/output>\n240<\/result>","index":989} +{"problem":"a batsman scores 26 runs and increases his average from 14 to 15 . find the runs to be made if he wants top increasing the average to 19 in the same match ?","rationale":"number of runs scored more to increase the ratio by 1 is 26 - 14 = 12 to raise the average by one ( from 14 to 15 ) , he scored 12 more than the existing average . therefore , to raise the average by five ( from 14 to 19 ) , he should score 12 x 5 = 60 more than the existing average . thus he should score 14 + 60 = 74 . answer d","correct":"d","options":{"a":"12 ","b":"18 ","c":"25 ","d":"74","e":"88"},"options_float":{"a":12.0,"b":18.0,"c":25.0,"d":74.0,"e":88.0},"annotated_formula":"subtract(multiply(19, add(subtract(26, 15), const_1)), multiply(14, subtract(26, 15)))","linear_formula":"subtract(n0,n2)|add(#0,const_1)|multiply(n1,#0)|multiply(n3,#1)|subtract(#3,#2)","chain":"26 - 15<\/gadget>\n11<\/output>\n11 + 1<\/gadget>\n12<\/output>\n19 * 12<\/gadget>\n228<\/output>\n14 * 11<\/gadget>\n154<\/output>\n228 - 154<\/gadget>\n74<\/output>\n74<\/result>","index":991} +{"problem":"a total of 520 players participated in a single tennis knock out tournament . what is the total number of matches played in the tournament ? ( knockout means if a player loses , he is out of the tournament ) . no match ends in a tie .","rationale":"there are 520 players , only 1 person wins , 519 players lose . in order to lose , you must have lost a game . 519 games . ans - b","correct":"b","options":{"a":"511 ","b":"519 ","c":"256 ","d":"255","e":"1023"},"options_float":{"a":511.0,"b":519.0,"c":256.0,"d":255.0,"e":1023.0},"annotated_formula":"add(add(add(add(add(add(add(divide(divide(divide(520, const_2), const_2), const_2), add(divide(520, const_2), divide(divide(520, const_2), const_2))), divide(divide(divide(divide(520, const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2))","linear_formula":"divide(n0,const_2)|divide(#0,const_2)|add(#0,#1)|divide(#1,const_2)|add(#2,#3)|divide(#3,const_2)|add(#4,#5)|divide(#5,const_2)|add(#6,#7)|divide(#7,const_2)|add(#8,#9)|divide(#9,const_2)|add(#10,#11)|divide(#11,const_2)|add(#12,#13)|divide(#13,const_2)|add(#14,#15)","chain":"520 \/ 2<\/gadget>\n260<\/output>\n260 \/ 2<\/gadget>\n130<\/output>\n130 \/ 2<\/gadget>\n65<\/output>\n260 + 130<\/gadget>\n390<\/output>\n65 + 390<\/gadget>\n455<\/output>\n65 \/ 2<\/gadget>\n65\/2 = around 32.5<\/output>\n455 + (65\/2)<\/gadget>\n975\/2 = around 487.5<\/output>\n(65\/2) \/ 2<\/gadget>\n65\/4 = around 16.25<\/output>\n(975\/2) + (65\/4)<\/gadget>\n2_015\/4 = around 503.75<\/output>\n(65\/4) \/ 2<\/gadget>\n65\/8 = around 8.125<\/output>\n(2_015\/4) + (65\/8)<\/gadget>\n4_095\/8 = around 511.875<\/output>\n(65\/8) \/ 2<\/gadget>\n65\/16 = around 4.0625<\/output>\n(4_095\/8) + (65\/16)<\/gadget>\n8_255\/16 = around 515.9375<\/output>\n(65\/16) \/ 2<\/gadget>\n65\/32 = around 2.03125<\/output>\n(8_255\/16) + (65\/32)<\/gadget>\n16_575\/32 = around 517.96875<\/output>\n(65\/32) \/ 2<\/gadget>\n65\/64 = around 1.015625<\/output>\n(16_575\/32) + (65\/64)<\/gadget>\n33_215\/64 = around 518.984375<\/output>\n33_215\/64 = around 518.984375<\/result>","index":993} +{"problem":"solving a linear equation with several occurrences of the variable , solve for w . simplify answer as much as possible . ( 7 w + 6 ) \/ 6 + ( 9 w + 8 ) \/ 2 = 22","rationale":"( 7 w + 6 ) \/ 6 + ( 9 w + 8 ) \/ 2 = 22 or , [ 7 w + 6 + 3 ( 9 w + 8 ) ] \/ 6 = 22 or , 7 w + 6 + 27 w + 24 = 132 or , 34 w + 30 = 132 or , 34 w = 132 - 30 or , 34 w = 102 or , w = 102 \/ 34 therefore , w = 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":"divide(subtract(multiply(multiply(2, 6), 22), add(multiply(8, 6), multiply(2, 6))), add(multiply(9, 6), multiply(7, const_2)))","linear_formula":"multiply(n1,n5)|multiply(n1,n4)|multiply(n1,n3)|multiply(n0,const_2)|add(#1,#0)|add(#2,#3)|multiply(n6,#0)|subtract(#6,#4)|divide(#7,#5)","chain":"2 * 6<\/gadget>\n12<\/output>\n12 * 22<\/gadget>\n264<\/output>\n8 * 6<\/gadget>\n48<\/output>\n48 + 12<\/gadget>\n60<\/output>\n264 - 60<\/gadget>\n204<\/output>\n9 * 6<\/gadget>\n54<\/output>\n7 * 2<\/gadget>\n14<\/output>\n54 + 14<\/gadget>\n68<\/output>\n204 \/ 68<\/gadget>\n3<\/output>\n3<\/result>","index":994} +{"problem":"two good train each 750 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":"\"sol . relative speed = ( 45 + 30 ) km \/ hr = ( 75 x 5 \/ 18 ) m \/ sec = ( 125 \/ 6 ) m \/ sec . distance covered = ( 750 + 750 ) m = 1500 m . required time = ( 1500 x 6 \/ 125 ) sec = 72 sec . answer d\"","correct":"d","options":{"a":"12 sec ","b":"24 sec ","c":"48 sec ","d":"72 sec","e":"none"},"options_float":{"a":12.0,"b":24.0,"c":48.0,"d":72.0,"e":null},"annotated_formula":"multiply(multiply(750, inverse(multiply(add(45, 30), const_0_2778))), const_2)","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2)|","chain":"45 + 30<\/gadget>\n75<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n75 * (5\/18)<\/gadget>\n125\/6 = around 20.833333<\/output>\n1 \/ (125\/6)<\/gadget>\n6\/125 = around 0.048<\/output>\n750 * (6\/125)<\/gadget>\n36<\/output>\n36 * 2<\/gadget>\n72<\/output>\n72<\/result>","index":996} +{"problem":"the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 4800 sq cm . what is the ratio of the width and the area of the rectangle ?","rationale":"let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 4800 12 x ^ 2 = 4800 x ^ 2 = 400 x = 20 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 80 the answer is c .","correct":"c","options":{"a":"1 : 72 ","b":"1 : 76 ","c":"1 : 80 ","d":"1 : 84","e":"1 : 88"},"options_float":{"a":0.0138888889,"b":0.0131578947,"c":0.0125,"d":0.0119047619,"e":0.0113636364},"annotated_formula":"divide(divide(sqrt(multiply(const_3, 4800)), const_2), 4800)","linear_formula":"multiply(n2,const_3)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2)","chain":"3 * 4_800<\/gadget>\n14_400<\/output>\n14_400 ** (1\/2)<\/gadget>\n120<\/output>\n120 \/ 2<\/gadget>\n60<\/output>\n60 \/ 4_800<\/gadget>\n1\/80 = around 0.0125<\/output>\n1\/80 = around 0.0125<\/result>","index":997} +{"problem":"a work which could be finished in 7 days was finished 3 days earlier after 10 more men joined . the number of men employed was ?","rationale":"\"x - - - - - - - 7 ( x + 10 ) - - - - 6 x * 7 = ( x + 10 ) 6 x = 60 \\ answer : c\"","correct":"c","options":{"a":"22 ","b":"20 ","c":"60 ","d":"71","e":"11"},"options_float":{"a":22.0,"b":20.0,"c":60.0,"d":71.0,"e":11.0},"annotated_formula":"divide(multiply(multiply(3, const_2), 10), subtract(7, multiply(3, const_2)))","linear_formula":"multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2)|","chain":"3 * 2<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n7 - 6<\/gadget>\n1<\/output>\n60 \/ 1<\/gadget>\n60<\/output>\n60<\/result>","index":998} +{"problem":"if ( 6 ) ( x ^ 2 ) has 3 different prime factors , at most how many different prime factors does x have ?","rationale":"x can have at most 3 prime factors , namely the prime factors 2 and 3 , plus one other . if x had more than this number of prime factors , then ( 6 ) ( x ^ 2 ) would have more than 3 prime factors . 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":"multiply(3, const_1)","linear_formula":"multiply(n2,const_1)","chain":"3 * 1<\/gadget>\n3<\/output>\n3<\/result>","index":999} +{"problem":"a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 190 kilometers ?","rationale":"each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 180 ã · 40 = 9.5 ã — 1 gallon = 9.5 gallons correct answer is b ) 9.5 gallons","correct":"b","options":{"a":"3.5 gallons ","b":"9.5 gallons ","c":"8.7 gallons ","d":"4.5 gallons","e":"9.2 gallons"},"options_float":{"a":3.5,"b":9.5,"c":8.7,"d":4.5,"e":9.2},"annotated_formula":"divide(190, 20)","linear_formula":"divide(n1,n0)","chain":"190 \/ 20<\/gadget>\n19\/2 = around 9.5<\/output>\n19\/2 = around 9.5<\/result>","index":1001} +{"problem":"the list price of an article is rs . 66 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?","rationale":"\"66 * ( 90 \/ 100 ) * ( ( 100 - x ) \/ 100 ) = 56.16 x = 5.45 % answer : c\"","correct":"c","options":{"a":"3.45 % ","b":"4.45 % ","c":"5.45 % ","d":"6.45 %","e":"7.45 %"},"options_float":{"a":3.45,"b":4.45,"c":5.45,"d":6.45,"e":7.45},"annotated_formula":"multiply(divide(subtract(subtract(66, multiply(66, divide(10, const_100))), 56.16), subtract(66, multiply(66, 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>\n66 * (1\/10)<\/gadget>\n33\/5 = around 6.6<\/output>\n66 - (33\/5)<\/gadget>\n297\/5 = around 59.4<\/output>\n(297\/5) - 56.16<\/gadget>\n3.24<\/output>\n3.24 \/ (297\/5)<\/gadget>\n0.054545<\/output>\n0.054545 * 100<\/gadget>\n5.4545<\/output>\n5.4545<\/result>","index":1004} +{"problem":"a trader marked the selling price of an article at 60 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of :","rationale":"\"sol . let c . p . = rs . 100 . then , marked price = rs . 160 , s . p . = rs . 99 . ∴ discount % = [ 11 \/ 160 * 100 ] % = 6.8 % answer c\"","correct":"c","options":{"a":"10 % ","b":"10.5 % ","c":"6.8 % ","d":"12.5 %","e":"none"},"options_float":{"a":10.0,"b":10.5,"c":6.8,"d":12.5,"e":null},"annotated_formula":"multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 60)))","linear_formula":"add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)|","chain":"2 + 3<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 + 1<\/gadget>\n11<\/output>\n100 + 60<\/gadget>\n160<\/output>\n11 \/ 160<\/gadget>\n11\/160 = around 0.06875<\/output>\n100 * (11\/160)<\/gadget>\n55\/8 = around 6.875<\/output>\n55\/8 = around 6.875<\/result>","index":1005} +{"problem":"a person saved $ 10 in buying an item on sale . if he spent $ 200 for the item , approximately how much percent he saved in the transaction ?","rationale":"\"actual price = 200 + 10 = $ 210 saving = 10 \/ 210 * 100 = 100 \/ 21 = 5 % approximately answer is d\"","correct":"d","options":{"a":"8 % ","b":"10 % ","c":"2 % ","d":"5 %","e":"6 %"},"options_float":{"a":8.0,"b":10.0,"c":2.0,"d":5.0,"e":6.0},"annotated_formula":"add(floor(multiply(divide(10, add(10, 200)), const_100)), const_1)","linear_formula":"add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|floor(#2)|add(#3,const_1)|","chain":"10 + 200<\/gadget>\n210<\/output>\n10 \/ 210<\/gadget>\n1\/21 = around 0.047619<\/output>\n(1\/21) * 100<\/gadget>\n100\/21 = around 4.761905<\/output>\nfloor(100\/21)<\/gadget>\n4<\/output>\n4 + 1<\/gadget>\n5<\/output>\n5<\/result>","index":1007} +{"problem":"a is 1.5 times as fast as b . a alone can do the work in 20 days . if a and b working together , in how many days will the work be completed ?","rationale":"a can finish 1 work in 20 days b can finish 1 \/ 1.5 work in 20 days - since a is 1.5 faster than b this means b can finish 1 work in 20 * 1.5 days = 30 days now using the awesome gmat formula when two machines work together they can finish the job in = ab \/ ( a + b ) = 20 * 30 \/ ( 20 + 30 ) = 20 * 30 \/ 50 = 12 days so answer is c","correct":"c","options":{"a":"23 ","b":"22 ","c":"12 ","d":"24","e":"25"},"options_float":{"a":23.0,"b":22.0,"c":12.0,"d":24.0,"e":25.0},"annotated_formula":"divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 1.5)))","linear_formula":"divide(const_1,n1)|divide(#0,n0)|add(#0,#1)|divide(const_1,#2)","chain":"1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) \/ 1.5<\/gadget>\n0.033333<\/output>\n(1\/20) + 0.033333<\/gadget>\n0.083333<\/output>\n1 \/ 0.083333<\/gadget>\n12.000048<\/output>\n12.000048<\/result>","index":1010} +{"problem":"if a speaks the truth 65 % 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 65 \/ 100 = 0.65 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 ) = 0.65 × 0.6 = 0.39 answer : a\"","correct":"a","options":{"a":"0.39 ","b":"0.48 ","c":"0.41 ","d":"0.482","e":"0.411"},"options_float":{"a":0.39,"b":0.48,"c":0.41,"d":0.482,"e":0.411},"annotated_formula":"multiply(divide(65, 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>\n65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(13\/20) * (3\/5)<\/gadget>\n39\/100 = around 0.39<\/output>\n39\/100 = around 0.39<\/result>","index":1011} +{"problem":"if a = 1 , what is the value of – ( a ^ 2 + a ^ 3 + a ^ 4 + a ^ 5 ) ?","rationale":"if a = 1 , then putting values in equation = - [ ( 1 ) ^ 2 + ( 1 ) ^ 3 + ( 1 ^ 4 ) + ( 1 ^ 5 ) ] = - [ 1 + 1 + 1 + 1 ] = - 4 answer = b = - 4","correct":"b","options":{"a":"- 14 ","b":"- 4 ","c":"0 ","d":"4","e":"14"},"options_float":{"a":-14.0,"b":-4.0,"c":0.0,"d":4.0,"e":14.0},"annotated_formula":"negate(add(add(add(power(1, 2), power(1, 3)), power(1, 4)), power(1, 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":"1 ** 2<\/gadget>\n1<\/output>\n1 ** 3<\/gadget>\n1<\/output>\n1 + 1<\/gadget>\n2<\/output>\n1 ** 4<\/gadget>\n1<\/output>\n2 + 1<\/gadget>\n3<\/output>\n1 ** 5<\/gadget>\n1<\/output>\n3 + 1<\/gadget>\n4<\/output>\n-4<\/gadget>\n-4<\/output>\n-4<\/result>","index":1013} +{"problem":"what profit percent is made by selling an article at a certain price , if by selling at 2 \/ 3 rd of that price , there would be a loss of 30 % ?","rationale":"\"sp 2 = 2 \/ 3 sp 1 cp = 100 sp 2 = 70 2 \/ 3 sp 1 = 70 sp 1 = 105 100 - - - 105 = > 5 % answer : e\"","correct":"e","options":{"a":"20 % ","b":"29 % ","c":"70 % ","d":"27 %","e":"5 %"},"options_float":{"a":20.0,"b":29.0,"c":70.0,"d":27.0,"e":5.0},"annotated_formula":"subtract(divide(subtract(const_100, 30), divide(2, 3)), const_100)","linear_formula":"divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|","chain":"100 - 30<\/gadget>\n70<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n70 \/ (2\/3)<\/gadget>\n105<\/output>\n105 - 100<\/gadget>\n5<\/output>\n5<\/result>","index":1014} +{"problem":"an astronomer noted that a grouping of red giant stars had an average solar mass of 8 m each , and a grouping of white dwarf stars had an average solar mass of 1.5 m each . if the astronomer calculated the total solar mass of both groupings to be 49 m , what total number of red giant stars and white dwarf stars did the astronomer note ?","rationale":"we can determine quickly that total number should range between 49 \/ 8 < = n < = 49 \/ 1.5 , so ans should be between 6 and 33 . now solving the expression 8 a + 1.5 b = 49 decreasing 49 in multiple of 8 and checking divisibility of that number by 1.5 . this way we get 2 red giants , 22 white dwarfs we get 49 , but 2 + 22 = 24 and 24 is not an option . next we get 5 red giants and 6 white dwarfs to get 49 , 5 * 8 + 6 * 1.5 = 49 hence total number is 5 + 6 = 11 ans 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":"add(divide(subtract(49, multiply(1.5, add(const_2, const_4))), 8), add(const_2, const_4))","linear_formula":"add(const_2,const_4)|multiply(n1,#0)|subtract(n2,#1)|divide(#2,n0)|add(#0,#3)","chain":"2 + 4<\/gadget>\n6<\/output>\n1.5 * 6<\/gadget>\n9<\/output>\n49 - 9<\/gadget>\n40<\/output>\n40 \/ 8<\/gadget>\n5<\/output>\n5 + 6<\/gadget>\n11<\/output>\n11<\/result>","index":1015} +{"problem":"village a ’ s population is 300 greater than village b ' s population . if village b ’ s population were reduced by 600 people , then village a ’ s population would be 4 times as large as village b ' s population . what is village b ' s current population ?","rationale":"a = b + 300 . a = 4 ( b - 600 ) . 4 ( b - 600 ) = b + 300 . 3 b = 2700 . b = 900 . the answer is a .","correct":"a","options":{"a":"900 ","b":"1000 ","c":"1100 ","d":"1200","e":"1300"},"options_float":{"a":900.0,"b":1000.0,"c":1100.0,"d":1200.0,"e":1300.0},"annotated_formula":"divide(add(multiply(600, 4), 300), subtract(4, const_1))","linear_formula":"multiply(n1,n2)|subtract(n2,const_1)|add(n0,#0)|divide(#2,#1)","chain":"600 * 4<\/gadget>\n2_400<\/output>\n2_400 + 300<\/gadget>\n2_700<\/output>\n4 - 1<\/gadget>\n3<\/output>\n2_700 \/ 3<\/gadget>\n900<\/output>\n900<\/result>","index":1016} +{"problem":"a sum amounts to rs . 4851 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ?","rationale":"\"ci = 4851 , r = 5 , n = 2 ci = p [ 1 + r \/ 100 ] ^ 2 = p [ 1 + 5 \/ 100 ] ^ 2 4851 = p [ 21 \/ 20 ] ^ 2 4851 [ 20 \/ 21 ] ^ 2 4400 answer : c\"","correct":"c","options":{"a":"s . 4000 ","b":"s . 5000 ","c":"s . 4400 ","d":"s . 4800","e":"s . 5800"},"options_float":{"a":4000.0,"b":5000.0,"c":4400.0,"d":4800.0,"e":5800.0},"annotated_formula":"divide(4851, power(add(divide(5, const_100), const_1), 2))","linear_formula":"divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|","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>\n4_851 \/ (441\/400)<\/gadget>\n4_400<\/output>\n4_400<\/result>","index":1018} +{"problem":"the average monthly salary of 10 employees in an organisation is rs . 1600 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?","rationale":"\"manager ' s monthly salary = rs . ( 1700 * 11 - 1600 * 10 ) = rs . 2700 answer : c\"","correct":"c","options":{"a":"rs . 3601 ","b":"rs . 3618 ","c":"rs . 2700 ","d":"rs . 3619","e":"rs . 3610"},"options_float":{"a":3601.0,"b":3618.0,"c":2700.0,"d":3619.0,"e":3610.0},"annotated_formula":"subtract(multiply(add(1600, 100), add(10, const_1)), multiply(1600, 10))","linear_formula":"add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"1_600 + 100<\/gadget>\n1_700<\/output>\n10 + 1<\/gadget>\n11<\/output>\n1_700 * 11<\/gadget>\n18_700<\/output>\n1_600 * 10<\/gadget>\n16_000<\/output>\n18_700 - 16_000<\/gadget>\n2_700<\/output>\n2_700<\/result>","index":1019} +{"problem":"if a lends rs . 1200 to b at 10 % per annum and b lends the same sum to c at 18 % per annum then the gain of b in a period of 2 years is ?","rationale":"\"( 1200 * 8 * 2 ) \/ 100 = > 192 answer : c\"","correct":"c","options":{"a":"190 ","b":"188 ","c":"192 ","d":"145","e":"188"},"options_float":{"a":190.0,"b":188.0,"c":192.0,"d":145.0,"e":188.0},"annotated_formula":"subtract(divide(multiply(multiply(1200, 18), 2), const_100), divide(multiply(multiply(1200, 10), 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":"1_200 * 18<\/gadget>\n21_600<\/output>\n21_600 * 2<\/gadget>\n43_200<\/output>\n43_200 \/ 100<\/gadget>\n432<\/output>\n1_200 * 10<\/gadget>\n12_000<\/output>\n12_000 * 2<\/gadget>\n24_000<\/output>\n24_000 \/ 100<\/gadget>\n240<\/output>\n432 - 240<\/gadget>\n192<\/output>\n192<\/result>","index":1020} +{"problem":"if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 6 ?","rationale":"i get 5 \/ 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e \/ 6 = integer , therefore we have 48 \/ 96 numbers divisible by 6 o e o \/ 6 = not integer we can not forget multiples of 6 from 1 to 96 we have 16 numbers that are multiple of 8 therefore , 48 \/ 96 + 16 \/ 96 = 64 \/ 96 = 2 \/ 3 answer : a","correct":"a","options":{"a":"2 \/ 3 ","b":"3 \/ 8 ","c":"1 \/ 2 ","d":"5 \/ 8","e":"3 \/ 4"},"options_float":{"a":0.6666666667,"b":0.375,"c":0.5,"d":0.625,"e":0.75},"annotated_formula":"divide(add(divide(96, 2), divide(96, 6)), 96)","linear_formula":"divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)","chain":"96 \/ 2<\/gadget>\n48<\/output>\n96 \/ 6<\/gadget>\n16<\/output>\n48 + 16<\/gadget>\n64<\/output>\n64 \/ 96<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":1021} +{"problem":"how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 5 ?","rationale":"\"the numbers should be of the form 5 c + 3 . the minimum is 3 when c = 0 . the maximum is 48 when c = 9 . there are 10 such numbers . the answer is 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(const_100.0, const_10)","linear_formula":"divide(const_100.0,const_10)|","chain":"100 \/ 10<\/gadget>\n10<\/output>\n10<\/result>","index":1022} +{"problem":"6 students wrote science exam . their average marks are 70 . 5 students got 65 , 75 , 55 , 72 and 69 marks respectively . therefore what is the marks of the sixth student ?","rationale":"explanation total marks of 5 students = ( 65 + 75 + 55 + 72 + 69 ) = 336 required marks = [ ( 70 x 6 ) – 336 ] = ( 420 – 336 ) = 84 answer a","correct":"a","options":{"a":"84 ","b":"68 ","c":"85 ","d":"75","e":"42"},"options_float":{"a":84.0,"b":68.0,"c":85.0,"d":75.0,"e":42.0},"annotated_formula":"subtract(multiply(70, 6), add(add(add(add(65, 75), 55), 72), 69))","linear_formula":"add(n3,n4)|multiply(n0,n1)|add(n5,#0)|add(n6,#2)|add(n7,#3)|subtract(#1,#4)","chain":"70 * 6<\/gadget>\n420<\/output>\n65 + 75<\/gadget>\n140<\/output>\n140 + 55<\/gadget>\n195<\/output>\n195 + 72<\/gadget>\n267<\/output>\n267 + 69<\/gadget>\n336<\/output>\n420 - 336<\/gadget>\n84<\/output>\n84<\/result>","index":1023} +{"problem":"a bowl was filled with 10 ounces of water , and 0.010 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 = . 008 * 50 = . 010 * 100 \/ 2 = 1.0 \/ 2 = . 5 percent of the original amount of water evaporated during this period = ( . 5 \/ 10 ) * 100 % = 5 % answer d\"","correct":"d","options":{"a":"0.004 % ","b":"0.04 % ","c":"0.40 % ","d":"5 %","e":"40 %"},"options_float":{"a":0.004,"b":0.04,"c":0.4,"d":5.0,"e":40.0},"annotated_formula":"multiply(divide(multiply(50, 0.010), 10), const_100)","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|","chain":"50 * 0.01<\/gadget>\n0.5<\/output>\n0.5 \/ 10<\/gadget>\n0.05<\/output>\n0.05 * 100<\/gadget>\n5<\/output>\n5<\/result>","index":1025} +{"problem":"the sum of 7 th and 23 rd term of a . p . is equal to the sum of 8 th , 15 th and 13 th term . find the term which is 0","rationale":"t 7 + t 23 = t 8 + t 15 + t 13 = > a + 6 d + a + 22 d = a + 7 d + a + 14 d + a + 12 d = > a + 5 d = 0 = > t 6 = 0 i . e . 6 th term is zero . answer : a","correct":"a","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(add(13, add(8, 15)), add(7, 23))","linear_formula":"add(n2,n3)|add(n0,n1)|add(n4,#0)|subtract(#2,#1)","chain":"8 + 15<\/gadget>\n23<\/output>\n13 + 23<\/gadget>\n36<\/output>\n7 + 23<\/gadget>\n30<\/output>\n36 - 30<\/gadget>\n6<\/output>\n6<\/result>","index":1027} +{"problem":"let c be defined as the sum of all prime numbers between 0 and 35 . what is c \/ 3","rationale":"prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 33 sum , c = 193 c \/ 3 = 64.3 answer c","correct":"c","options":{"a":"155 ","b":"129 ","c":"64.3 ","d":"47","e":"43"},"options_float":{"a":155.0,"b":129.0,"c":64.3,"d":47.0,"e":43.0},"annotated_formula":"add(divide(3, const_10), power(const_2, add(const_2, const_4)))","linear_formula":"add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)","chain":"3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n2 + 4<\/gadget>\n6<\/output>\n2 ** 6<\/gadget>\n64<\/output>\n(3\/10) + 64<\/gadget>\n643\/10 = around 64.3<\/output>\n643\/10 = around 64.3<\/result>","index":1031} +{"problem":"the speed of a boat in still water in 22 km \/ hr and the rate of current is 5 km \/ hr . the distance travelled downstream in 24 minutes is :","rationale":"speed downstream = ( 22 + 5 ) = 27 kmph time = 24 minutes = 24 \/ 60 hour = 2 \/ 5 hour distance travelled = time × speed = 2 \/ 5 × 27 = 10.8 km answer is c .","correct":"c","options":{"a":"10.6 ","b":"10.2 ","c":"10.8 ","d":"10.4","e":"10.0"},"options_float":{"a":10.6,"b":10.2,"c":10.8,"d":10.4,"e":10.0},"annotated_formula":"multiply(add(22, 5), divide(24, const_60))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)","chain":"22 + 5<\/gadget>\n27<\/output>\n24 \/ 60<\/gadget>\n2\/5 = around 0.4<\/output>\n27 * (2\/5)<\/gadget>\n54\/5 = around 10.8<\/output>\n54\/5 = around 10.8<\/result>","index":1032} +{"problem":"a boat having a length 3 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 = ( 3 x 3 x 0.01 ) m 3 = 0.09 m 3 . mass of man = volume of water displaced x density of water = ( 0.09 x 1000 ) kg = 90 kg . answer : d\"","correct":"d","options":{"a":"100 kg ","b":"120 kg ","c":"89 kg ","d":"90 kg","e":"110 kg"},"options_float":{"a":100.0,"b":120.0,"c":89.0,"d":90.0,"e":110.0},"annotated_formula":"multiply(multiply(multiply(3, 3), divide(1, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"3 * 3<\/gadget>\n9<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n9 * (1\/100)<\/gadget>\n9\/100 = around 0.09<\/output>\n(9\/100) * 1_000<\/gadget>\n90<\/output>\n90<\/result>","index":1033} +{"problem":"by selling 12 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ?","rationale":"\"80 % - - - 12 120 % - - - ? 80 \/ 120 * 12 = 8 answer : a\"","correct":"a","options":{"a":"8 ","b":"76 ","c":"17 ","d":"7","e":"77"},"options_float":{"a":8.0,"b":76.0,"c":17.0,"d":7.0,"e":77.0},"annotated_formula":"multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 12)","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) * 12<\/gadget>\n8<\/output>\n8<\/result>","index":1034} +{"problem":"walking with 4 \/ 5 of my usual speed , i miss the bus by 7 minutes . what is my usual time ?","rationale":"\"speed ratio = 1 : 4 \/ 5 = 5 : 4 time ratio = 4 : 51 - - - - - - - - 7 4 - - - - - - - - - ? è 28 answer : c\"","correct":"c","options":{"a":"16 min ","b":"26 min ","c":"28 min ","d":"20 min","e":"12 min"},"options_float":{"a":16.0,"b":26.0,"c":28.0,"d":20.0,"e":12.0},"annotated_formula":"multiply(divide(7, divide(5, 4)), 5)","linear_formula":"divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)|","chain":"5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n7 \/ (5\/4)<\/gadget>\n28\/5 = around 5.6<\/output>\n(28\/5) * 5<\/gadget>\n28<\/output>\n28<\/result>","index":1036} +{"problem":"if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 150 then sum is","rationale":"\"p ( r \/ 100 ) ^ 2 = c . i - s . i p ( 10 \/ 100 ) ^ 2 = 150 15000 answer : a\"","correct":"a","options":{"a":"s . 15000 ","b":"s . 15100 ","c":"s . 15800 ","d":"s . 16000","e":"s . 16200"},"options_float":{"a":15000.0,"b":15100.0,"c":15800.0,"d":16000.0,"e":16200.0},"annotated_formula":"divide(150, multiply(divide(10, const_100), divide(10, const_100)))","linear_formula":"divide(n0,const_100)|multiply(#0,#0)|divide(n2,#1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * (1\/10)<\/gadget>\n1\/100 = around 0.01<\/output>\n150 \/ (1\/100)<\/gadget>\n15_000<\/output>\n15_000<\/result>","index":1037} +{"problem":"in one hour , a boat goes 8 km along the stream and 2 km against the stream . the sped of the boat in still water ( in km \/ hr ) is :","rationale":"\"solution speed in still water = 1 \/ 2 ( 8 + 2 ) km \/ hr = 5 kmph . answer b\"","correct":"b","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(8, 2), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)|","chain":"8 + 2<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":1038} +{"problem":"coconuts were purchased at 150 per 100 and sold at 2 per coconut . if 2000 coconuts were sold , what was the total profit made ?","rationale":"c . p . for one coconut = 150 ⁄ 100 = 3 ⁄ 2 s . p . for one coconut = 2 profit on one coconut = 2 - 3 ⁄ 2 = 1 ⁄ 2 ∴ profit on 2000 coconut = 1 ⁄ 2 × 2000 = 1000 answer b","correct":"b","options":{"a":"500 ","b":"1000 ","c":"1500 ","d":"2000","e":"none of these"},"options_float":{"a":500.0,"b":1000.0,"c":1500.0,"d":2000.0,"e":null},"annotated_formula":"multiply(2000, subtract(2, divide(150, 100)))","linear_formula":"divide(n0,n1)|subtract(n2,#0)|multiply(n3,#1)","chain":"150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n2 - (3\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n2_000 * (1\/2)<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1039} +{"problem":"a train is 435 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 = 435 + 140 = 575 meter time = distance \/ speed = 575 ∗ 2 \/ 25 = 46 seconds option c\"","correct":"c","options":{"a":"20 seconds ","b":"30 seconds ","c":"46 seconds ","d":"50 seconds","e":"none of these"},"options_float":{"a":20.0,"b":30.0,"c":46.0,"d":50.0,"e":null},"annotated_formula":"divide(add(435, 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":"435 + 140<\/gadget>\n575<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n575 \/ (25\/2)<\/gadget>\n46<\/output>\n46<\/result>","index":1042} +{"problem":"if the l . c . m of two numbers is 450 and their product is 22500 , find the h . c . f of the numbers .","rationale":"\"h . c . f = ( product of the numbers ) \/ ( their l . c . m ) = 22500 \/ 450 = 50 . answer : a\"","correct":"a","options":{"a":"50 ","b":"30 ","c":"125 ","d":"25","e":"none of these"},"options_float":{"a":50.0,"b":30.0,"c":125.0,"d":25.0,"e":null},"annotated_formula":"divide(22500, 450)","linear_formula":"divide(n1,n0)|","chain":"22_500 \/ 450<\/gadget>\n50<\/output>\n50<\/result>","index":1043} +{"problem":"the area of a parallelogram is 72 cm ^ 2 and its altitude is twice the corresponding base . what is the length of the base ?","rationale":"let base = x cm height = 2 x cm area = x ã — 2 x = 2 x ^ 2 area = x ã — 2 x = 2 x ^ 2 area is given as 72 cm ^ 2 2 x ^ 2 = 72 x ^ 2 = 36 x = 6 cm answer : c","correct":"c","options":{"a":"1 cm ","b":"3 cm ","c":"6 cm ","d":"4 cm","e":"2 cm"},"options_float":{"a":1.0,"b":3.0,"c":6.0,"d":4.0,"e":2.0},"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":1044} +{"problem":"martin buys a pencil and a notebook for 80 cents . at the same store , gloria buys a notebook and an eraser for $ 1.15 cents , and zachary buys a pencil and an eraser for 75 cents . how many cents would it cost to buy 3 pencils , 3 notebooks , and 3 erasers ? ( assume that there is no volume discount . )","rationale":"pencil + notebook = 80 notebook + eraser = 115 pencil + eraser = 75 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 270 cents the cost to buy 3 of each would be ( 3 \/ 2 ) ( 270 ) = 405 the answer is e .","correct":"e","options":{"a":"325 ","b":"345 ","c":"365 ","d":"385","e":"405"},"options_float":{"a":325.0,"b":345.0,"c":365.0,"d":385.0,"e":405.0},"annotated_formula":"multiply(divide(add(add(multiply(1.15, const_100), 80), 75), const_2), 3)","linear_formula":"multiply(n1,const_100)|add(n0,#0)|add(n2,#1)|divide(#2,const_2)|multiply(n3,#3)","chain":"1.15 * 100<\/gadget>\n115<\/output>\n115 + 80<\/gadget>\n195<\/output>\n195 + 75<\/gadget>\n270<\/output>\n270 \/ 2<\/gadget>\n135<\/output>\n135 * 3<\/gadget>\n405<\/output>\n405<\/result>","index":1045} +{"problem":"when 200 is divided by positive integer x , the remainder is 3 . what is the remainder when 297 is divided by x ?","rationale":"\"f 200 \/ x leaves a reminder 3 then ( 200 - 3 ) i . e . 197 is divisible by x so ( 200 + 197 ) \/ x leaves a reminder rem ( 200 \/ x ) + rem ( 197 \/ x ) = > 3 + 0 = 3 answer : b\"","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"6","e":"8"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":6.0,"e":8.0},"annotated_formula":"subtract(const_100.0, subtract(297, 200))","linear_formula":"subtract(n2,const_100.0)|subtract(n0,#0)|","chain":"297 - 200<\/gadget>\n97<\/output>\n100 - 97<\/gadget>\n3<\/output>\n3<\/result>","index":1046} +{"problem":"if a fast song has 160 beats per minute , and a slow song has 90 beats per minute , how many minutes total would you play a fast and a slow song to have a stream of music that had a total of 1020 beats ?","rationale":"we can determine quickly that total number should range between 1020 \/ 160 < = n < = 1020 \/ 90 , so ans should be between 6 and 12 . now solving the expression 160 a + 90 b = 1020 decreasing 1020 by multiples of 160 and checking divisibility of that number by 9 , we get fast song plays for 3 minutes and slow somg plays for 6 minutes , 3 * 160 + 6 * 90 = 1020 hence total number of minutes stream of music plays is 3 + 6 = 9 minutes 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(floor(multiply(divide(1020, add(160, 90)), const_2)), const_1)","linear_formula":"add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)|floor(#2)|add(#3,const_1)","chain":"160 + 90<\/gadget>\n250<\/output>\n1_020 \/ 250<\/gadget>\n102\/25 = around 4.08<\/output>\n(102\/25) * 2<\/gadget>\n204\/25 = around 8.16<\/output>\nfloor(204\/25)<\/gadget>\n8<\/output>\n8 + 1<\/gadget>\n9<\/output>\n9<\/result>","index":1047} +{"problem":"in a party every person shakes hands with every other person . if there were a total of 190 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 = 190 n ( n - 1 ) \/ 2 = 190 â ˆ ´ n = 20 answer : option e\"","correct":"e","options":{"a":"15 ","b":"16 ","c":"17 ","d":"18","e":"20"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":20.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(190, 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":"190 * 2<\/gadget>\n380<\/output>\n380 * 4<\/gadget>\n1_520<\/output>\n1_520 + 1<\/gadget>\n1_521<\/output>\n1_521 ** (1\/2)<\/gadget>\n39<\/output>\n39 + 1<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":1048} +{"problem":"ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 30 years . what is deepak present age ?","rationale":"\"present age is 4 x and 3 x , = > 4 x + 6 = 30 = > x = 6 so deepak age is = 3 ( 6 ) = 18 answer : a\"","correct":"a","options":{"a":"18 ","b":"15 ","c":"77 ","d":"266","e":"182"},"options_float":{"a":18.0,"b":15.0,"c":77.0,"d":266.0,"e":182.0},"annotated_formula":"divide(multiply(subtract(30, 6), 3), 4)","linear_formula":"subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|","chain":"30 - 6<\/gadget>\n24<\/output>\n24 * 3<\/gadget>\n72<\/output>\n72 \/ 4<\/gadget>\n18<\/output>\n18<\/result>","index":1050} +{"problem":"what is remainder of the division ( 1525 * 1527 * 1529 ) \/ 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 : d\"","correct":"d","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(multiply(multiply(1525, 1527), 1529), subtract(multiply(multiply(1525, 1527), 1529), const_3))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)|","chain":"1_525 * 1_527<\/gadget>\n2_328_675<\/output>\n2_328_675 * 1_529<\/gadget>\n3_560_544_075<\/output>\n3_560_544_075 - 3<\/gadget>\n3_560_544_072<\/output>\n3_560_544_075 - 3_560_544_072<\/gadget>\n3<\/output>\n3<\/result>","index":1052} +{"problem":"a pharmaceutical company received $ 5 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 = 5 \/ 20 = 1 \/ 4 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 = ( 5 \/ 20 ) x 12 = 3 next 108 million royalties \/ sales = 9 \/ 108 = ( 1 \/ 12 ) x 12 = 1 we can plug 15 and 5 into our percent change formula : ( new – old ) \/ old x 100 [ ( 1 – 3 ) \/ 3 ] x 100 - 200 \/ 3 x 100 at this point we can stop and consider the answer choices . since we know that 200 \/ 3 is just a bit less than ½ , we know that - 200 \/ 3 x 100 is about a 67 % decrease . answer e .","correct":"e","options":{"a":"8 % ","b":"15 % ","c":"45 % ","d":"52 %","e":"67 %"},"options_float":{"a":8.0,"b":15.0,"c":45.0,"d":52.0,"e":67.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(5, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(5, 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":"5 \/ 20<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n9 \/ 108<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/12) * 100<\/gadget>\n25\/3 = around 8.333333<\/output>\n25 - (25\/3)<\/gadget>\n50\/3 = around 16.666667<\/output>\n(50\/3) \/ 25<\/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":1053} +{"problem":"in the first 10 overs of a cricket game , the run rate was only 5.2 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?","rationale":"\"10 overs - run rate = 5.2 runs scored in first 10 overs = 52 remaining overs 40 total runs to be scored = 282 52 runs already scored 282 - 52 = 230 230 runs to be scored in 40 overs let required runrate be x 40 * x = 230 x = 230 \/ 40 x = 5.75 the required runrate is 5.75 answer : d\"","correct":"d","options":{"a":"6.25 ","b":"6.5 ","c":"6.75 ","d":"5.75","e":"8"},"options_float":{"a":6.25,"b":6.5,"c":6.75,"d":5.75,"e":8.0},"annotated_formula":"divide(subtract(282, multiply(10, 5.2)), 40)","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|","chain":"10 * 5.2<\/gadget>\n52<\/output>\n282 - 52<\/gadget>\n230<\/output>\n230 \/ 40<\/gadget>\n23\/4 = around 5.75<\/output>\n23\/4 = around 5.75<\/result>","index":1054} +{"problem":"of the diplomats attending a summit conference , 30 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 } = 30 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 20 . answer : a .\"","correct":"a","options":{"a":"20 ","b":"96 ","c":"108 ","d":"120","e":"150"},"options_float":{"a":20.0,"b":96.0,"c":108.0,"d":120.0,"e":150.0},"annotated_formula":"divide(subtract(32, 30), 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 - 30<\/gadget>\n2<\/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>\n2 \/ (1\/10)<\/gadget>\n20<\/output>\n20<\/result>","index":1055} +{"problem":"if 0.2 of a number is equal to 0.08 of another number , the ratio of the numbers is :","rationale":"\"0.2 a = 0.08 b - > a \/ b = 0.08 \/ 0.20 = 8 \/ 20 = 2 \/ 5 : . a : b = 2 : 5 answer : c\"","correct":"c","options":{"a":"2 : 3 ","b":"3 : 4 ","c":"2 : 5 ","d":"20 : 3","e":"30 : 7"},"options_float":{"a":0.6666666667,"b":0.75,"c":0.4,"d":6.6666666667,"e":4.2857142857},"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":1057} +{"problem":"a man covered a certain distance at some speed . had he moved 3 kmph faster , he would have taken 40 minutes less . if he had moved 2 kmph slower , he would have taken 40 minutes more . the distance ( in km ) is","rationale":"explanation : let distance = x km and usual rate = y kmph . then , x \/ y - x \/ ( y + 3 ) = 40 \/ 60 - - > 2 y ( y + 3 ) = 9 x - - - - - ( i ) also , x \/ ( y - 2 ) - x \/ y = 40 \/ 60 - - > y ( y - 2 ) = 3 x - - - - - - - - ( ii ) on dividing ( i ) by ( ii ) , we get : x = 40 km . answer : c","correct":"c","options":{"a":"27 ","b":"87 ","c":"40 ","d":"18","e":"17"},"options_float":{"a":27.0,"b":87.0,"c":40.0,"d":18.0,"e":17.0},"annotated_formula":"multiply(multiply(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), divide(40, const_60)), add(const_1, divide(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), 3)))","linear_formula":"divide(n1,const_60)|multiply(n0,n2)|subtract(n0,n2)|multiply(n2,#1)|divide(#3,#2)|divide(#4,n0)|multiply(#4,#0)|add(#5,const_1)|multiply(#7,#6)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n3 - 2<\/gadget>\n1<\/output>\n12 \/ 1<\/gadget>\n12<\/output>\n40 \/ 60<\/gadget>\n2\/3 = around 0.666667<\/output>\n12 * (2\/3)<\/gadget>\n8<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n8 * 5<\/gadget>\n40<\/output>\n40<\/result>","index":1058} +{"problem":"if the average of w , b , c , 14 and 15 is 12 . what is the average value of w , b , c and 29","rationale":"w + b + c + 14 + 15 = 12 * 5 = 60 = > w + b + c = 60 - 29 = 31 w + b + c + 29 = 31 + 29 = 60 average = 60 \/ 4 = 15 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":"divide(add(subtract(subtract(multiply(add(const_1, const_4), 12), 15), 14), 29), const_4)","linear_formula":"add(const_1,const_4)|multiply(n2,#0)|subtract(#1,n1)|subtract(#2,n0)|add(n3,#3)|divide(#4,const_4)","chain":"1 + 4<\/gadget>\n5<\/output>\n5 * 12<\/gadget>\n60<\/output>\n60 - 15<\/gadget>\n45<\/output>\n45 - 14<\/gadget>\n31<\/output>\n31 + 29<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15<\/result>","index":1059} +{"problem":"set a of 8 positive integers may have the same element and have 40 . and set b of 8 positive integers must have different elements and have 40 . when m and n are the greatest possible differences between 40 and other elements ’ sums in set a and set b , respectively , m - n = ?","rationale":"this is maximum - minimum . hence , 40 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 32 and 40 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 11 . so , 32 - 11 = 21 . the correct answer is a .","correct":"a","options":{"a":"21 ","b":"29 ","c":"23 ","d":"25","e":"26"},"options_float":{"a":21.0,"b":29.0,"c":23.0,"d":25.0,"e":26.0},"annotated_formula":"subtract(40, add(add(8, const_2), 8))","linear_formula":"add(n0,const_2)|add(n0,#0)|subtract(n1,#1)|","chain":"8 + 2<\/gadget>\n10<\/output>\n10 + 8<\/gadget>\n18<\/output>\n40 - 18<\/gadget>\n22<\/output>\n22<\/result>","index":1060} +{"problem":"after 6 games , team b had an average of 65 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 * 65 ) + 47 + x > 500 390 + 47 + x > 500 437 + x > 500 = > x > 63 option d\"","correct":"d","options":{"a":"85 ","b":"74 ","c":"67 ","d":"63","e":"28"},"options_float":{"a":85.0,"b":74.0,"c":67.0,"d":63.0,"e":28.0},"annotated_formula":"subtract(500, add(multiply(6, 65), 47))","linear_formula":"multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)|","chain":"6 * 65<\/gadget>\n390<\/output>\n390 + 47<\/gadget>\n437<\/output>\n500 - 437<\/gadget>\n63<\/output>\n63<\/result>","index":1061} +{"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 16 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 16 more transactions than cal . jade = 66 + 16 = 82 jade handled = 82 transactions . answer : c\"","correct":"c","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), 3), 16)","linear_formula":"add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|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 + 16<\/gadget>\n82<\/output>\n82<\/result>","index":1062} +{"problem":"a man buys an article for $ 100 . and sells it for $ 125 . find the gain percent ?","rationale":"\"c . p . = $ 100 s . p . = $ 125 gain = $ 25 gain % = 25 \/ 100 * 100 = 25 % answer is c\"","correct":"c","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(125, 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>\n125 \/ 1<\/gadget>\n125<\/output>\n125 - 100<\/gadget>\n25<\/output>\n25<\/result>","index":1063} +{"problem":"list a consists of 12 consecutive integers . if - 4 is the least integer in list a , what is the range of positive integers in list a ?","rationale":"since - 4 is the least integer in list a , then 7 is the largest integer in that list . thus the range of the positive integers in the list is 7 - 1 = 6 . answer : b .","correct":"b","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(12, add(4, const_1)), const_1)","linear_formula":"add(n1,const_1)|subtract(n0,#0)|subtract(#1,const_1)","chain":"4 + 1<\/gadget>\n5<\/output>\n12 - 5<\/gadget>\n7<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6<\/result>","index":1067} +{"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 \/ 2 of all the paint . during the second week , he uses 1 \/ 5 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 \/ 2 ) * 360 = 180 gallons . remaning paint = 180 gallons paint used in the second week = ( 1 \/ 5 ) * 180 = 36 gallons total paint used = 216 gallons . option d\"","correct":"d","options":{"a":"18 ","b":"144 ","c":"175 ","d":"216","e":"250"},"options_float":{"a":18.0,"b":144.0,"c":175.0,"d":216.0,"e":250.0},"annotated_formula":"add(multiply(divide(360, 2), 1), divide(subtract(360, multiply(divide(360, 2), 1)), 5))","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)|","chain":"360 \/ 2<\/gadget>\n180<\/output>\n180 * 1<\/gadget>\n180<\/output>\n360 - 180<\/gadget>\n180<\/output>\n180 \/ 5<\/gadget>\n36<\/output>\n180 + 36<\/gadget>\n216<\/output>\n216<\/result>","index":1068} +{"problem":"if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 4 ?","rationale":"\"i get 5 \/ 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e \/ 4 = integer , therefore we have 48 \/ 96 numbers divisible by 8 o e o \/ 4 = not integer we can not forget multiples of 8 from 1 to 96 we have 24 numbers that are multiple of 4 therefore , 48 \/ 96 + 24 \/ 96 = 72 \/ 96 = 3 \/ 4 answer : e\"","correct":"e","options":{"a":"1 \/ 4 ","b":"3 \/ 8 ","c":"1 \/ 2 ","d":"5 \/ 8","e":"3 \/ 4"},"options_float":{"a":0.25,"b":0.375,"c":0.5,"d":0.625,"e":0.75},"annotated_formula":"divide(add(divide(96, 2), divide(96, 4)), 96)","linear_formula":"divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|","chain":"96 \/ 2<\/gadget>\n48<\/output>\n96 \/ 4<\/gadget>\n24<\/output>\n48 + 24<\/gadget>\n72<\/output>\n72 \/ 96<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":1071} +{"problem":"when y is divided by 288 , the remainder is 44 . what is the remainder when the same y is divided by 24 ?","rationale":"y = 288 * a + 44 = ( 24 * 12 ) * a + 24 + 20 the answer is a .","correct":"a","options":{"a":"20 ","b":"21 ","c":"23 ","d":"25","e":"26"},"options_float":{"a":20.0,"b":21.0,"c":23.0,"d":25.0,"e":26.0},"annotated_formula":"reminder(44, 24)","linear_formula":"reminder(n1,n2)","chain":"44 % 24<\/gadget>\n20<\/output>\n20<\/result>","index":1073} +{"problem":"after decreasing 24 % in the price of an article costs rs . 1140 . find the actual cost of an article ?","rationale":"\"cp * ( 76 \/ 100 ) = 1140 cp = 15 * 100 = > cp = 1500 answer : a\"","correct":"a","options":{"a":"1500 ","b":"6789 ","c":"1200 ","d":"6151","e":"1421"},"options_float":{"a":1500.0,"b":6789.0,"c":1200.0,"d":6151.0,"e":1421.0},"annotated_formula":"divide(1140, 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>\n1_140 \/ (19\/25)<\/gadget>\n1_500<\/output>\n1_500<\/result>","index":1076} +{"problem":"a room 5 m 44 cm long and 3 m 74 cm broad needs to be paved with square tiles . what will be the least number of square tiles required to cover the floor ?","rationale":"\"length = 5 m 44 cm = 544 cm breadth = 3 m 74 cm = 374 cm area = 544 * 374 hcf = 34 area of square = 34 * 34 cm 2 no of tiles req = 544 * 374 \/ 34 * 34 = 16 * 11 = 176 answer a\"","correct":"a","options":{"a":"176 ","b":"124 ","c":"224 ","d":"186","e":"190"},"options_float":{"a":176.0,"b":124.0,"c":224.0,"d":186.0,"e":190.0},"annotated_formula":"divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), multiply(subtract(44, add(multiply(const_2, const_4), const_2)), subtract(44, add(multiply(const_2, const_4), const_2))))","linear_formula":"multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)|","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>\n8<\/output>\n8 + 2<\/gadget>\n10<\/output>\n44 - 10<\/gadget>\n34<\/output>\n34 * 34<\/gadget>\n1_156<\/output>\n203_456 \/ 1_156<\/gadget>\n176<\/output>\n176<\/result>","index":1079} +{"problem":"the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 8 , is :","rationale":"\"explanation : required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 8 = 540 + 8 = 548 . answer : option d\"","correct":"d","options":{"a":"504 ","b":"536 ","c":"544 ","d":"548","e":"none of these"},"options_float":{"a":504.0,"b":536.0,"c":544.0,"d":548.0,"e":null},"annotated_formula":"multiply(54, const_10)","linear_formula":"multiply(n3,const_10)|","chain":"54 * 10<\/gadget>\n540<\/output>\n540<\/result>","index":1082} +{"problem":"jonathan can type a 30 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 : 30 \/ 40 + 30 \/ 30 + 30 \/ 24 = 30 \/ 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(30, add(divide(30, 24), add(divide(30, 40), divide(30, 30))))","linear_formula":"divide(n0,n1)|divide(n0,n2)|divide(n0,n3)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|","chain":"30 \/ 24<\/gadget>\n5\/4 = around 1.25<\/output>\n30 \/ 40<\/gadget>\n3\/4 = around 0.75<\/output>\n30 \/ 30<\/gadget>\n1<\/output>\n(3\/4) + 1<\/gadget>\n7\/4 = around 1.75<\/output>\n(5\/4) + (7\/4)<\/gadget>\n3<\/output>\n30 \/ 3<\/gadget>\n10<\/output>\n10<\/result>","index":1087} +{"problem":"a box contains 10 tablets of medicine a and 17 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 17 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 17 + 2 = 19 tablets . answer : d .\"","correct":"d","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(17, const_2)","linear_formula":"add(n1,const_2)|","chain":"17 + 2<\/gadget>\n19<\/output>\n19<\/result>","index":1089} +{"problem":"the milk level in a rectangular box measuring 60 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 60 * 25 * 1 \/ 2 = 750 feet ^ 3 of milk must be removed , which equals to 750 * 7.5 = 5625 gallons . answer : d .\"","correct":"d","options":{"a":"100 ","b":"250 ","c":"750 ","d":"5625","e":"5635"},"options_float":{"a":100.0,"b":250.0,"c":750.0,"d":5625.0,"e":5635.0},"annotated_formula":"multiply(multiply(multiply(60, 25), divide(1, const_2)), 7.5)","linear_formula":"divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)|","chain":"60 * 25<\/gadget>\n1_500<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1_500 * (1\/2)<\/gadget>\n750<\/output>\n750 * 7.5<\/gadget>\n5_625<\/output>\n5_625<\/result>","index":1090} +{"problem":"rhonda picked 2 pen from the table , if there were 7 pens on the table and 5 belongs to jill , what is the probability that the 2 pen she picked does not belong to jill ? .","rationale":"since jill owns 5 of the pen , the subset from which the 2 pens hould be chosen are the 2 pens not owned by jill fom the universe of 7 . the first pen can be one of the 2 from the 7 with probability 2 \/ 7 . the second pen can be one of the 1 from the 6 remaining with probability 1 \/ 6 , the total probability will be 2 \/ 7 × 1 \/ 6 . on cancellation , this comes to 2 \/ 42 . thus , the answer is b - 2 \/ 42 .","correct":"b","options":{"a":"5 \/ 42 ","b":"2 \/ 42 ","c":"7 \/ 42 ","d":"2 \/ 7","e":"5 \/ 7"},"options_float":{"a":0.119047619,"b":0.0476190476,"c":0.1666666667,"d":0.2857142857,"e":0.7142857143},"annotated_formula":"multiply(divide(subtract(7, 5), 7), divide(subtract(subtract(7, 5), const_1), subtract(7, const_1)))","linear_formula":"subtract(n1,n2)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)","chain":"7 - 5<\/gadget>\n2<\/output>\n2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n2 - 1<\/gadget>\n1<\/output>\n7 - 1<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(2\/7) * (1\/6)<\/gadget>\n1\/21 = around 0.047619<\/output>\n1\/21 = around 0.047619<\/result>","index":1092} +{"problem":"country x taxes each of its citizens an amount equal to 13 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ?","rationale":"\"equation is correct , so math must be a problem . 0.13 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 5,200 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 10,800 - - > x = 54,000 . answer : a .\"","correct":"a","options":{"a":"$ 54,000 ","b":"$ 56,000 ","c":"$ 64,000 ","d":"$ 66,667","e":"$ 80,000"},"options_float":{"a":54000.0,"b":56000.0,"c":64000.0,"d":66667.0,"e":80000.0},"annotated_formula":"add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(13, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|multiply(const_2,const_4)|multiply(const_10,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(#4,#6)|divide(#7,#1)|add(#8,#5)|","chain":"4 * 10<\/gadget>\n40<\/output>\n40 * 1_000<\/gadget>\n40_000<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8 * 1_000<\/gadget>\n8_000<\/output>\n13 \/ 100<\/gadget>\n13\/100 = around 0.13<\/output>\n(13\/100) * 40_000<\/gadget>\n5_200<\/output>\n8_000 - 5_200<\/gadget>\n2_800<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n2_800 \/ (1\/5)<\/gadget>\n14_000<\/output>\n40_000 + 14_000<\/gadget>\n54_000<\/output>\n54_000<\/result>","index":1094} +{"problem":"the surface of a cube is 294 sq cm . find its volume ?","rationale":"\"6 a 2 = 294 = 6 * 49 a = 7 = > a 3 = 343 cc answer : d\"","correct":"d","options":{"a":"8 cc ","b":"9 cc ","c":"2 cc ","d":"343 cc","e":"6 cc"},"options_float":{"a":8.0,"b":9.0,"c":2.0,"d":343.0,"e":6.0},"annotated_formula":"volume_cube(sqrt(divide(294, add(const_2, const_4))))","linear_formula":"add(const_2,const_4)|divide(n0,#0)|sqrt(#1)|volume_cube(#2)|","chain":"2 + 4<\/gadget>\n6<\/output>\n294 \/ 6<\/gadget>\n49<\/output>\n49 ** (1\/2)<\/gadget>\n7<\/output>\n7 ** 3<\/gadget>\n343<\/output>\n343<\/result>","index":1096} +{"problem":"the average marks of 20 students in a class is 100 . but a student mark is wrongly noted as 50 instead of 10 then find the correct average marks ?","rationale":"\"correct avg marks = 100 + ( 10 - 50 ) \/ 20 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, 20), 50), 10), 20)","linear_formula":"multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|","chain":"100 * 20<\/gadget>\n2_000<\/output>\n2_000 - 50<\/gadget>\n1_950<\/output>\n1_950 + 10<\/gadget>\n1_960<\/output>\n1_960 \/ 20<\/gadget>\n98<\/output>\n98<\/result>","index":1098} +{"problem":"a parking garage rents parking spaces for $ 10 per week or $ 20 per month . how much does a person save in a year by renting by the month rather than by the week ?","rationale":"\"10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 20 = 240 $ 520 - 240 = 280 ans e\"","correct":"e","options":{"a":"$ 140 ","b":"$ 160 ","c":"$ 220 ","d":"$ 240","e":"$ 280"},"options_float":{"a":140.0,"b":160.0,"c":220.0,"d":240.0,"e":280.0},"annotated_formula":"subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(20, const_12))","linear_formula":"add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n10 * 5<\/gadget>\n50<\/output>\n50 + 2<\/gadget>\n52<\/output>\n52 * 10<\/gadget>\n520<\/output>\n20 * 12<\/gadget>\n240<\/output>\n520 - 240<\/gadget>\n280<\/output>\n280<\/result>","index":1099} +{"problem":"jerry travels 8 miles at an average speed of 40 miles per hour , stops for 14 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ?","rationale":"\"total time taken by jerry = ( 8 \/ 40 ) * 60 minutes + 14 minutes + ( 20 \/ 60 ) * 60 minutes = 46 minutes average speed = total distance \/ total time = ( 8 + 20 ) miles \/ ( 46 \/ 60 ) hours = 28 * 60 \/ 46 = 36 miles per hour answer : option a\"","correct":"a","options":{"a":"36 ","b":"42.5 ","c":"44 ","d":"50","e":"52.5"},"options_float":{"a":36.0,"b":42.5,"c":44.0,"d":50.0,"e":52.5},"annotated_formula":"divide(add(8, 20), add(add(divide(8, 40), divide(14, 60)), divide(20, 60)))","linear_formula":"add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|","chain":"8 + 20<\/gadget>\n28<\/output>\n8 \/ 40<\/gadget>\n1\/5 = around 0.2<\/output>\n14 \/ 60<\/gadget>\n7\/30 = around 0.233333<\/output>\n(1\/5) + (7\/30)<\/gadget>\n13\/30 = around 0.433333<\/output>\n20 \/ 60<\/gadget>\n1\/3 = around 0.333333<\/output>\n(13\/30) + (1\/3)<\/gadget>\n23\/30 = around 0.766667<\/output>\n28 \/ (23\/30)<\/gadget>\n840\/23 = around 36.521739<\/output>\n840\/23 = around 36.521739<\/result>","index":1100} +{"problem":"the total surface area of a cuboid length 12 m , breadth 10 m and height 8 m .","rationale":"total surface area of cuboid = 2 ( lb + bh + lh ) = 2 ( 120 + 80 + 96 ) = 2 ( 296 ) = > 596 m ( power 2 ) answer is c .","correct":"c","options":{"a":"576 ","b":"566 ","c":"596 ","d":"556","e":"586"},"options_float":{"a":576.0,"b":566.0,"c":596.0,"d":556.0,"e":586.0},"annotated_formula":"surface_rectangular_prism(12, 10, 8)","linear_formula":"surface_rectangular_prism(n0,n1,n2)","chain":"2 * (12 * 10 + 10 * 8 + 12 * 8)<\/gadget>\n592<\/output>\n592<\/result>","index":1101} +{"problem":"60 boys can complete a work in 24 days . how many men need to complete twice the work in 20 days","rationale":"\"one man can complete the work in 24 * 60 = 1440 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 1440 m = 144 workers answer : a\"","correct":"a","options":{"a":"144 ","b":"170 ","c":"180 ","d":"190","e":"200"},"options_float":{"a":144.0,"b":170.0,"c":180.0,"d":190.0,"e":200.0},"annotated_formula":"divide(multiply(60, multiply(24, const_2)), 20)","linear_formula":"multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2)|","chain":"24 * 2<\/gadget>\n48<\/output>\n60 * 48<\/gadget>\n2_880<\/output>\n2_880 \/ 20<\/gadget>\n144<\/output>\n144<\/result>","index":1105} +{"problem":"jill has 21 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 = 21 1 3 \/ 4 w = 21 w = 12 the total number of jars = 3 w = 36 answer : d","correct":"d","options":{"a":"3 ","b":"6 ","c":"9 ","d":"36","e":"14"},"options_float":{"a":3.0,"b":6.0,"c":9.0,"d":36.0,"e":14.0},"annotated_formula":"multiply(divide(21, 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>\n21 \/ (7\/4)<\/gadget>\n12<\/output>\n12 * 3<\/gadget>\n36<\/output>\n36<\/result>","index":1106} +{"problem":"a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 20 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ?","rationale":"\"explanation : sp of 1 metre cloth = 8200 \/ 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 20 = rs . 185 cp on 40 metres = 185 x 40 = rs . 7400 profit earned on 40 metres cloth = rs . 8200 – rs . 7400 = rs . 800 . answer : option a\"","correct":"a","options":{"a":"rs . 800 ","b":"rs . 1500 ","c":"rs . 1000 ","d":"rs . 1200","e":"none of these"},"options_float":{"a":800.0,"b":1500.0,"c":1000.0,"d":1200.0,"e":null},"annotated_formula":"multiply(20, 40)","linear_formula":"multiply(n0,n2)|","chain":"20 * 40<\/gadget>\n800<\/output>\n800<\/result>","index":1109} +{"problem":"24 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ?","rationale":"\"required number of machines = 24 * 10 \/ 40 = 6 answer is b\"","correct":"b","options":{"a":"10 ","b":"6 ","c":"4 ","d":"7","e":"5"},"options_float":{"a":10.0,"b":6.0,"c":4.0,"d":7.0,"e":5.0},"annotated_formula":"divide(multiply(24, 10), 40)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"24 * 10<\/gadget>\n240<\/output>\n240 \/ 40<\/gadget>\n6<\/output>\n6<\/result>","index":1111} +{"problem":"in a company of 170 employees , 110 are females . a total of 80 employees have advanced degrees and the rest have a college degree only . if 25 employees are males with college degree only , how many employees are females with advanced degrees ?","rationale":"the number of males is 170 - 110 = 60 . the number of males with advanced degrees is 60 - 25 = 35 . the number of females with advanced degrees is 80 - 35 = 45 . the answer is b .","correct":"b","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":"subtract(80, subtract(subtract(170, 110), 25))","linear_formula":"subtract(n0,n1)|subtract(#0,n3)|subtract(n2,#1)","chain":"170 - 110<\/gadget>\n60<\/output>\n60 - 25<\/gadget>\n35<\/output>\n80 - 35<\/gadget>\n45<\/output>\n45<\/result>","index":1112} +{"problem":"at a particular graduation party with 300 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 36 males did not bring gifts to the party , how many females did bring gifts ?","rationale":"\"the correct method total = 300 . . 70 % of 300 = 210 got gifts . . 90 did not get gift , out of which 48 are males , so remaining 90 - 36 = 54 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 54 , 100 % = 54 * 100 \/ 60 = 90 . . ans 40 % of 90 = 36 b\"","correct":"b","options":{"a":"18 ","b":"36 ","c":"42 ","d":"68","e":"70"},"options_float":{"a":18.0,"b":36.0,"c":42.0,"d":68.0,"e":70.0},"annotated_formula":"divide(multiply(divide(multiply(subtract(subtract(300, divide(multiply(70, 300), const_100)), 36), const_100), subtract(const_100, 40)), 40), const_100)","linear_formula":"multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|subtract(n0,#2)|subtract(#3,n3)|multiply(#4,const_100)|divide(#5,#1)|multiply(n2,#6)|divide(#7,const_100)|","chain":"70 * 300<\/gadget>\n21_000<\/output>\n21_000 \/ 100<\/gadget>\n210<\/output>\n300 - 210<\/gadget>\n90<\/output>\n90 - 36<\/gadget>\n54<\/output>\n54 * 100<\/gadget>\n5_400<\/output>\n100 - 40<\/gadget>\n60<\/output>\n5_400 \/ 60<\/gadget>\n90<\/output>\n90 * 40<\/gadget>\n3_600<\/output>\n3_600 \/ 100<\/gadget>\n36<\/output>\n36<\/result>","index":1113} +{"problem":"if 8 a = 9 b and ab ≠ 0 , what is the ratio of a \/ 9 to b \/ 8 ?","rationale":"\"if ab ≠ 0 then a and b has two integer sets of pair if a = 9 then b = 8 and if a = - 9 then b = - 8 also in fraction if a = 1 \/ 8 then b = 1 \/ 9 any of the pair we check the ratio 8 a \/ 9 b = 1 answer : c\"","correct":"c","options":{"a":"64 \/ 81 ","b":"8 \/ 9 ","c":"1 ","d":"9 \/ 8","e":"81 \/ 64"},"options_float":{"a":0.7901234568,"b":0.8888888889,"c":1.0,"d":1.125,"e":1.265625},"annotated_formula":"divide(multiply(8, 9), multiply(9, 8))","linear_formula":"multiply(n0,n1)|divide(#0,#0)|","chain":"8 * 9<\/gadget>\n72<\/output>\n9 * 8<\/gadget>\n72<\/output>\n72 \/ 72<\/gadget>\n1<\/output>\n1<\/result>","index":1114} +{"problem":"in a theater , the first row has 17 seats and each row has 3 more seats than previous row . if the last row has 44 seats , what is the total number of seats in the theater ?","rationale":"\"the number of seats in the theater is 17 + ( 17 + 3 ) + . . . + ( 17 + 27 ) = 10 ( 17 ) + 3 ( 1 + 2 + . . . + 9 ) = 10 ( 17 ) + 3 ( 9 ) ( 10 ) \/ 2 = 170 + 135 = 305 the answer is e .\"","correct":"e","options":{"a":"265 ","b":"275 ","c":"285 ","d":"295","e":"305"},"options_float":{"a":265.0,"b":275.0,"c":285.0,"d":295.0,"e":305.0},"annotated_formula":"multiply(divide(add(17, 44), const_2), divide(add(subtract(44, 17), 3), 3))","linear_formula":"add(n0,n2)|subtract(n2,n0)|add(n1,#1)|divide(#0,const_2)|divide(#2,n1)|multiply(#3,#4)|","chain":"17 + 44<\/gadget>\n61<\/output>\n61 \/ 2<\/gadget>\n61\/2 = around 30.5<\/output>\n44 - 17<\/gadget>\n27<\/output>\n27 + 3<\/gadget>\n30<\/output>\n30 \/ 3<\/gadget>\n10<\/output>\n(61\/2) * 10<\/gadget>\n305<\/output>\n305<\/result>","index":1115} +{"problem":"the list price of an article is rs . 69 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?","rationale":"\"69 * ( 90 \/ 100 ) * ( ( 100 - x ) \/ 100 ) = 56.16 x = 9.56 % answer : d\"","correct":"d","options":{"a":"9.33 % ","b":"9.44 % ","c":"9.45 % ","d":"9.56 %","e":"9.67 %"},"options_float":{"a":9.33,"b":9.44,"c":9.45,"d":9.56,"e":9.67},"annotated_formula":"multiply(divide(subtract(subtract(69, multiply(69, divide(10, const_100))), 56.16), subtract(69, multiply(69, 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>\n69 * (1\/10)<\/gadget>\n69\/10 = around 6.9<\/output>\n69 - (69\/10)<\/gadget>\n621\/10 = around 62.1<\/output>\n(621\/10) - 56.16<\/gadget>\n5.94<\/output>\n5.94 \/ (621\/10)<\/gadget>\n0.095652<\/output>\n0.095652 * 100<\/gadget>\n9.5652<\/output>\n9.5652<\/result>","index":1116} +{"problem":"how many seconds will a 200 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 = 200 * 3 \/ 50 = 12 sec . answer : c\"","correct":"c","options":{"a":"26 sec ","b":"30 sec ","c":"12 sec ","d":"19 sec","e":"12 sec"},"options_float":{"a":26.0,"b":30.0,"c":12.0,"d":19.0,"e":12.0},"annotated_formula":"divide(200, 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>\n200 \/ (50\/3)<\/gadget>\n12<\/output>\n12<\/result>","index":1117} +{"problem":"the radius of a semi circle is 70 cm then its perimeter is ?","rationale":"\"diameter = 140 cm 1 \/ 2 * 22 \/ 7 * 140 + 140 = 360 answer : c\"","correct":"c","options":{"a":"32.8 ","b":"32.4 ","c":"360 ","d":"32.2","e":"32.9"},"options_float":{"a":32.8,"b":32.4,"c":360.0,"d":32.2,"e":32.9},"annotated_formula":"add(divide(circumface(70), const_2), multiply(70, const_2))","linear_formula":"circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)|","chain":"2 * pi * 70<\/gadget>\n140*pi = around 439.822972<\/output>\n(140*pi) \/ 2<\/gadget>\n70*pi = around 219.911486<\/output>\n70 * 2<\/gadget>\n140<\/output>\n(70*pi) + 140<\/gadget>\n140 + 70*pi = around 359.911486<\/output>\n140 + 70*pi = around 359.911486<\/result>","index":1120} +{"problem":"running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w \/ 4 widgets in 3 days , how many days would it take machine x alone to produce 6 w widgets .","rationale":"i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 \/ ( y + 2 ) + 1 \/ y = 5 \/ 12 ( 5 \/ 12 is because ( 5 \/ 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 \/ 5 days or 5 \/ 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . so , he will take 36 days to doing 6 w widgets . answer : b","correct":"b","options":{"a":"4 ","b":"36 ","c":"8 ","d":"10","e":"12"},"options_float":{"a":4.0,"b":36.0,"c":8.0,"d":10.0,"e":12.0},"annotated_formula":"multiply(multiply(6, 2), 3)","linear_formula":"multiply(n0,n4)|multiply(n3,#0)","chain":"6 * 2<\/gadget>\n12<\/output>\n12 * 3<\/gadget>\n36<\/output>\n36<\/result>","index":1123} +{"problem":"the cost price of a radio is rs . 1500 and it was sold for rs . 1230 , find the loss % ?","rationale":"\"1500 - - - - 270 100 - - - - ? = > 18 % answer : a\"","correct":"a","options":{"a":"18 % ","b":"19 % ","c":"68 % ","d":"19 %","e":"38 %"},"options_float":{"a":18.0,"b":19.0,"c":68.0,"d":19.0,"e":38.0},"annotated_formula":"multiply(divide(subtract(1500, 1230), 1500), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_500 - 1_230<\/gadget>\n270<\/output>\n270 \/ 1_500<\/gadget>\n9\/50 = around 0.18<\/output>\n(9\/50) * 100<\/gadget>\n18<\/output>\n18<\/result>","index":1124} +{"problem":"the sale price sarees listed for rs . 500 after successive discount is 10 % and 5 % is ?","rationale":"\"500 * ( 90 \/ 100 ) * ( 95 \/ 100 ) = 427.5 answer : a\"","correct":"a","options":{"a":"427.5 ","b":"277 ","c":"342 ","d":"882","e":"212"},"options_float":{"a":427.5,"b":277.0,"c":342.0,"d":882.0,"e":212.0},"annotated_formula":"subtract(subtract(500, divide(multiply(500, 10), const_100)), divide(multiply(subtract(500, divide(multiply(500, 10), const_100)), 5), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|","chain":"500 * 10<\/gadget>\n5_000<\/output>\n5_000 \/ 100<\/gadget>\n50<\/output>\n500 - 50<\/gadget>\n450<\/output>\n450 * 5<\/gadget>\n2_250<\/output>\n2_250 \/ 100<\/gadget>\n45\/2 = around 22.5<\/output>\n450 - (45\/2)<\/gadget>\n855\/2 = around 427.5<\/output>\n855\/2 = around 427.5<\/result>","index":1126} +{"problem":"the number of timeshare condos available at sunset beach is 2 \/ 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 350 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ?","rationale":"let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 \/ 5 x we know , x + 2 \/ 5 x = 350 hence , x = 250 . so , number of timeshare condos available at playa del mar = 250 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 2 \/ 5 x = 3 \/ 5 x = 3 \/ 5 ( 250 ) = 150 the correct answer is d","correct":"d","options":{"a":"60 ","b":"90 ","c":"120 ","d":"150","e":"240"},"options_float":{"a":60.0,"b":90.0,"c":120.0,"d":150.0,"e":240.0},"annotated_formula":"add(divide(multiply(350, 2), 5), multiply(2, 5))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1)","chain":"350 * 2<\/gadget>\n700<\/output>\n700 \/ 5<\/gadget>\n140<\/output>\n2 * 5<\/gadget>\n10<\/output>\n140 + 10<\/gadget>\n150<\/output>\n150<\/result>","index":1127} +{"problem":"a can do a piece of work in 8 days and b alone can do it in 12 days . how much time will both take to finish the work ?","rationale":"\"this question can be solved by different methods . we need to conserve time in exams so solving this problem using equations is the good idea . time taken to finish the job = xy \/ ( x + y ) = 8 x 12 \/ ( 8 + 12 ) = 96 \/ 20 = 4.8 days answer : a\"","correct":"a","options":{"a":"4.8 ","b":"6.333 ","c":"7.333 ","d":"8.5","e":"9"},"options_float":{"a":4.8,"b":6.333,"c":7.333,"d":8.5,"e":9.0},"annotated_formula":"divide(const_1, add(divide(const_1, 8), divide(const_1, 12)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/8) + (1\/12)<\/gadget>\n5\/24 = around 0.208333<\/output>\n1 \/ (5\/24)<\/gadget>\n24\/5 = around 4.8<\/output>\n24\/5 = around 4.8<\/result>","index":1128} +{"problem":"the measurement of a rectangular box with lid is 25 cmx 6 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere )","rationale":"d = 6 , r = 2 ; volume of the largest sphere = 4 \/ 3 π r 3 = 4 \/ 3 * π * 3 * 3 * 3 = 36 π cm 3 answer : c","correct":"c","options":{"a":"288 ","b":"48 ","c":"36 ","d":"864","e":"964"},"options_float":{"a":288.0,"b":48.0,"c":36.0,"d":864.0,"e":964.0},"annotated_formula":"multiply(divide(const_4, const_3), power(3, const_3))","linear_formula":"divide(const_4,const_3)|power(n3,const_3)|multiply(#0,#1)","chain":"4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 ** 3<\/gadget>\n27<\/output>\n(4\/3) * 27<\/gadget>\n36<\/output>\n36<\/result>","index":1129} +{"problem":"a and b go around a circular track of length 150 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 { 150 \/ ( 36 * 5 \/ 18 ) , 150 \/ ( 54 * 5 \/ 18 ) } = lcm ( 15 , 10 ) = 30 sec . answer : a\"","correct":"a","options":{"a":"30 sec ","b":"198 sec ","c":"178 sec ","d":"665 sec","e":"276 sec"},"options_float":{"a":30.0,"b":198.0,"c":178.0,"d":665.0,"e":276.0},"annotated_formula":"divide(150, 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>\n150 \/ 5<\/gadget>\n30<\/output>\n30<\/result>","index":1130} +{"problem":"two trains of length 150 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 = 150 + 280 = 430 m . the time required = d \/ s = 430 \/ 20 = 21.5 sec . answer : d\"","correct":"d","options":{"a":"22 sec ","b":"77 sec ","c":"76 sec ","d":"21.5 sec","e":"66 sec"},"options_float":{"a":22.0,"b":77.0,"c":76.0,"d":21.5,"e":66.0},"annotated_formula":"divide(add(150, 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":"150 + 280<\/gadget>\n430<\/output>\n42 + 30<\/gadget>\n72<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n430 \/ 20<\/gadget>\n43\/2 = around 21.5<\/output>\n43\/2 = around 21.5<\/result>","index":1132} +{"problem":"source : knewton a cyclist ' s speed varies , depending on the terrain , between 6.0 miles per hour and 14.0 miles per hour , inclusive . what is the maximum distance , in miles , that the cyclist could travel in 5 hours ?","rationale":"we are told that : generallya cyclist ' s speed varies , depending on the terrain , between 6.0 miles per hour and 14.0 miles per hour , inclusive . is it possible the cyclist to travel with maximum speed for some time ? why not , if there is right terrain for that . so , if there is long enough terrain for the maximum speed of 14 mph then the maximum distance , in miles , that the cyclist could travel in 5 hours would be 5 * 14 = 70 miles . answer : c .","correct":"c","options":{"a":"42 ","b":"56 ","c":"70 ","d":"98","e":"140"},"options_float":{"a":42.0,"b":56.0,"c":70.0,"d":98.0,"e":140.0},"annotated_formula":"multiply(14, 5)","linear_formula":"multiply(n1,n2)","chain":"14 * 5<\/gadget>\n70<\/output>\n70<\/result>","index":1133} +{"problem":"the length of a rectangle is two - seventh of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 5929 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 25 units ?","rationale":"given that the area of the square = 5929 sq . units = > side of square = â ˆ š 5929 = 77 units the radius of the circle = side of the square = 77 units length of the rectangle = 2 \/ 7 * 77 = 22 units given that breadth = 25 units area of the rectangle = lb = 22 * 25 = 550 sq . units answer : d","correct":"d","options":{"a":"660 sq . units ","b":"440 sq . units ","c":"770 sq . units ","d":"550 sq . units","e":"220 sq . units"},"options_float":{"a":660.0,"b":440.0,"c":770.0,"d":550.0,"e":220.0},"annotated_formula":"rectangle_area(25, multiply(sqrt(5929), divide(const_2, add(const_3, const_4))))","linear_formula":"add(const_3,const_4)|sqrt(n0)|divide(const_2,#0)|multiply(#2,#1)|rectangle_area(n1,#3)","chain":"5_929 ** (1\/2)<\/gadget>\n77<\/output>\n3 + 4<\/gadget>\n7<\/output>\n2 \/ 7<\/gadget>\n2\/7 = around 0.285714<\/output>\n77 * (2\/7)<\/gadget>\n22<\/output>\n25 * 22<\/gadget>\n550<\/output>\n550<\/result>","index":1135} +{"problem":"if 20 men take 15 days to to complete a job , in how many days can 25 men finish that work ?","rationale":"ans . 12 days","correct":"a","options":{"a":"12 ","b":"66 ","c":"77 ","d":"8","e":"6"},"options_float":{"a":12.0,"b":66.0,"c":77.0,"d":8.0,"e":6.0},"annotated_formula":"divide(multiply(20, 15), 25)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"20 * 15<\/gadget>\n300<\/output>\n300 \/ 25<\/gadget>\n12<\/output>\n12<\/result>","index":1136} +{"problem":"on a certain transatlantic crossing , 20 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 50 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":"\"let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.5 x had round trip tickets and took their cars . 0.5 x = 0.2 t x = 0.4 t the answer is b .\"","correct":"b","options":{"a":"30 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"65 %"},"options_float":{"a":30.0,"b":40.0,"c":50.0,"d":60.0,"e":65.0},"annotated_formula":"divide(20, subtract(const_1, divide(50, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n20 \/ (1\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":1138} +{"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 7 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 7 = 35 km imo , correct answer is ` ` b . ' '\"","correct":"b","options":{"a":"20 km ","b":"35 km ","c":"40 km ","d":"50 km","e":"60 km"},"options_float":{"a":20.0,"b":35.0,"c":40.0,"d":50.0,"e":60.0},"annotated_formula":"multiply(60, divide(multiply(5, 7), 60))","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|","chain":"5 * 7<\/gadget>\n35<\/output>\n35 \/ 60<\/gadget>\n7\/12 = around 0.583333<\/output>\n60 * (7\/12)<\/gadget>\n35<\/output>\n35<\/result>","index":1139} +{"problem":"the fuel indicator in a car shows 1 \/ 5 th of the fuel tank as full . when 22 more liters of fuel are poured in to the tank , the indicator rests at the 3 \/ 4 of the full mark . find the capacity of the tank .","rationale":"x \/ 5 + 22 = 3 x \/ 4 = > x = 40 litres answer : d","correct":"d","options":{"a":"25 litres ","b":"35 litres ","c":"30 litres ","d":"40 litres","e":"none of these"},"options_float":{"a":25.0,"b":35.0,"c":30.0,"d":40.0,"e":null},"annotated_formula":"divide(22, subtract(divide(3, 4), divide(1, 5)))","linear_formula":"divide(n3,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(3\/4) - (1\/5)<\/gadget>\n11\/20 = around 0.55<\/output>\n22 \/ (11\/20)<\/gadget>\n40<\/output>\n40<\/result>","index":1141} +{"problem":"a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 450 kilograms of pears that day , how many kilograms did he sell in the afternoon ?","rationale":"3 x = 450 x = 150 therefore , the salesman sold 150 kg in the morning and 2 ⋅ 150 = 300 kg in the afternoon . so answer is c .","correct":"c","options":{"a":"120 ","b":"180 ","c":"300 ","d":"280","e":"320"},"options_float":{"a":120.0,"b":180.0,"c":300.0,"d":280.0,"e":320.0},"annotated_formula":"multiply(divide(450, const_3), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_2)","chain":"450 \/ 3<\/gadget>\n150<\/output>\n150 * 2<\/gadget>\n300<\/output>\n300<\/result>","index":1142} +{"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 30 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 30 5 x = 1335 x = 267 large number = 267 + 1365 = 1632 d\"","correct":"d","options":{"a":"1235 ","b":"1456 ","c":"1567 ","d":"1632","e":"1635"},"options_float":{"a":1235.0,"b":1456.0,"c":1567.0,"d":1632.0,"e":1635.0},"annotated_formula":"add(multiply(divide(subtract(1365, 30), subtract(6, const_1)), 6), 30)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)|","chain":"1_365 - 30<\/gadget>\n1_335<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_335 \/ 5<\/gadget>\n267<\/output>\n267 * 6<\/gadget>\n1_602<\/output>\n1_602 + 30<\/gadget>\n1_632<\/output>\n1_632<\/result>","index":1145} +{"problem":"a rectangle with width 8 and diagonal 30 . find the area ?","rationale":"then the area is : 8 ' ' x 30 ' ' = 240 square inches , or 240 square units hence a","correct":"a","options":{"a":"240 square units ","b":"180 square units ","c":"100 square units ","d":"150 square units","e":"160 square units"},"options_float":{"a":240.0,"b":180.0,"c":100.0,"d":150.0,"e":160.0},"annotated_formula":"rectangle_area(sqrt(subtract(power(30, const_2), power(8, const_2))), 8)","linear_formula":"power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)","chain":"30 ** 2<\/gadget>\n900<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n900 - 64<\/gadget>\n836<\/output>\n836 ** (1\/2)<\/gadget>\n2*sqrt(209) = around 28.913665<\/output>\n(2*sqrt(209)) * 8<\/gadget>\n16*sqrt(209) = around 231.309317<\/output>\n16*sqrt(209) = around 231.309317<\/result>","index":1148} +{"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 35 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 5 miles per hour . the time to catch up is 60 \/ 5 = 12 hours . in 12 hours , train a travels another 30 * 12 = 360 miles for a total of 420 miles . the answer is a .\"","correct":"a","options":{"a":"420 ","b":"450 ","c":"480 ","d":"510","e":"540"},"options_float":{"a":420.0,"b":450.0,"c":480.0,"d":510.0,"e":540.0},"annotated_formula":"multiply(divide(multiply(30, const_2), subtract(35, 30)), 35)","linear_formula":"multiply(n0,const_2)|subtract(n1,n0)|divide(#0,#1)|multiply(n1,#2)|","chain":"30 * 2<\/gadget>\n60<\/output>\n35 - 30<\/gadget>\n5<\/output>\n60 \/ 5<\/gadget>\n12<\/output>\n12 * 35<\/gadget>\n420<\/output>\n420<\/result>","index":1149} +{"problem":"a box contains 25 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 ) = 21 c 2 \/ 25 c 2 = 7 \/ 10 p ( at least one is defective ) = ( 1 - 7 \/ 10 ) = 3 \/ 10 . answer a\"","correct":"a","options":{"a":"3 \/ 10 ","b":"7 \/ 19 ","c":"12 \/ 19 ","d":"21 \/ 95","e":"none"},"options_float":{"a":0.3,"b":0.3684210526,"c":0.6315789474,"d":0.2210526316,"e":null},"annotated_formula":"subtract(const_1, divide(choose(subtract(25, 4), const_2), choose(25, const_2)))","linear_formula":"choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)|","chain":"25 - 4<\/gadget>\n21<\/output>\nbinomial(21, 2)<\/gadget>\n210<\/output>\nbinomial(25, 2)<\/gadget>\n300<\/output>\n210 \/ 300<\/gadget>\n7\/10 = around 0.7<\/output>\n1 - (7\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n3\/10 = around 0.3<\/result>","index":1150} +{"problem":"how many prime numbers are between 28 \/ 3 and 86 \/ 6 ?","rationale":"\"28 \/ 3 = 9 . xxx 86 \/ 6 = 14 . xxx so we need to find prime numbers between 9 ( 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":1151} +{"problem":"in a school of 800 students , 44 % wear blue shirts , 28 % wear red shirts , 10 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ?","rationale":"\"44 + 28 + 10 = 82 % 100 – 82 = 18 % 800 * 18 \/ 100 = 144 the answer is c .\"","correct":"c","options":{"a":"120 ","b":"132 ","c":"144 ","d":"156","e":"168"},"options_float":{"a":120.0,"b":132.0,"c":144.0,"d":156.0,"e":168.0},"annotated_formula":"subtract(800, add(add(multiply(divide(44, const_100), 800), multiply(divide(28, const_100), 800)), multiply(divide(10, const_100), 800)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)|","chain":"44 \/ 100<\/gadget>\n11\/25 = around 0.44<\/output>\n(11\/25) * 800<\/gadget>\n352<\/output>\n28 \/ 100<\/gadget>\n7\/25 = around 0.28<\/output>\n(7\/25) * 800<\/gadget>\n224<\/output>\n352 + 224<\/gadget>\n576<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 800<\/gadget>\n80<\/output>\n576 + 80<\/gadget>\n656<\/output>\n800 - 656<\/gadget>\n144<\/output>\n144<\/result>","index":1152} +{"problem":"in what ratio must rice of rs . 25 per kg be mixed with rice of rs . 12 per kg so that cost of mixture is rs . 20 per kg ?","rationale":"\"( 20 - 12 ) \/ ( 25 - 20 = 8 \/ 5 = 8 : 5 answer : a\"","correct":"a","options":{"a":"8 : 5 ","b":"5 : 8 ","c":"20 : 25 ","d":"12 : 20","e":"25 : 12"},"options_float":{"a":1.6,"b":0.625,"c":0.8,"d":0.6,"e":2.0833333333},"annotated_formula":"divide(divide(subtract(20, 12), subtract(25, 12)), subtract(const_1, divide(subtract(20, 12), subtract(25, 12))))","linear_formula":"subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|","chain":"20 - 12<\/gadget>\n8<\/output>\n25 - 12<\/gadget>\n13<\/output>\n8 \/ 13<\/gadget>\n8\/13 = around 0.615385<\/output>\n1 - (8\/13)<\/gadget>\n5\/13 = around 0.384615<\/output>\n(8\/13) \/ (5\/13)<\/gadget>\n8\/5 = around 1.6<\/output>\n8\/5 = around 1.6<\/result>","index":1154} +{"problem":"10 men and 15 women together can complete a work in 4 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 \/ 4 15 women ' s 1 day work = ( 1 \/ 4 - 10 \/ 100 ) = 3 \/ 20 1 woman ' s 1 day work = 1 \/ 100 1 woman alone can complete the work in 100 days . answer : a\"","correct":"a","options":{"a":"100 days ","b":"667 days ","c":"177 days ","d":"187 days","e":"225 days"},"options_float":{"a":100.0,"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, 4))), multiply(15, 4))","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 * 4<\/gadget>\n40<\/output>\n100 - 40<\/gadget>\n60<\/output>\n100 \/ 60<\/gadget>\n5\/3 = around 1.666667<\/output>\n15 * 4<\/gadget>\n60<\/output>\n(5\/3) * 60<\/gadget>\n100<\/output>\n100<\/result>","index":1155} +{"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 4 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 4 people . so $ 100 . but , he also has to put in the average for himself ( the fiveth person ) , so add $ 75 . so $ 175 is your answer . answer b","correct":"b","options":{"a":"$ 100 ","b":"$ 175 ","c":"$ 200 ","d":"$ 250","e":"$ 450"},"options_float":{"a":100.0,"b":175.0,"c":200.0,"d":250.0,"e":450.0},"annotated_formula":"add(subtract(multiply(add(4, const_1), 75), multiply(add(4, const_1), 50)), 50)","linear_formula":"add(n2,const_1)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|add(n0,#3)","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 75<\/gadget>\n375<\/output>\n5 * 50<\/gadget>\n250<\/output>\n375 - 250<\/gadget>\n125<\/output>\n125 + 50<\/gadget>\n175<\/output>\n175<\/result>","index":1156} +{"problem":"if 0.5 : x : : 5 : 8 , then x is equal to :","rationale":"\"( x * 5 ) = ( 0.5 * 8 ) x = 4 \/ 5 x = 0.8 answer = e\"","correct":"e","options":{"a":"1.12 ","b":"1.2 ","c":"1.25 ","d":"0.9","e":"0.8"},"options_float":{"a":1.12,"b":1.2,"c":1.25,"d":0.9,"e":0.8},"annotated_formula":"divide(multiply(0.5, 8), 5)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"0.5 * 8<\/gadget>\n4<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n4\/5 = around 0.8<\/result>","index":1157} +{"problem":"the average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 75 kg . what is the weight of the new person ?","rationale":"\"the total weight increase = ( 8 x 2.5 ) kg = 20 kg weight of new person = ( 75 + 20 ) kg = 95 kg the answer is c .\"","correct":"c","options":{"a":"75 kg ","b":"85 kg ","c":"95 kg ","d":"65 kg","e":"55 kg"},"options_float":{"a":75.0,"b":85.0,"c":95.0,"d":65.0,"e":55.0},"annotated_formula":"add(multiply(2.5, 8), 75)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"2.5 * 8<\/gadget>\n20<\/output>\n20 + 75<\/gadget>\n95<\/output>\n95<\/result>","index":1158} +{"problem":"the radius of a wheel is 12.6 cm . what is the distance covered by the wheel in making 200 resolutions ?","rationale":"\"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 200 resolutions . = 200 * 2 * 22 \/ 7 * 12.6 = 15825.6 cm = 158.256 m answer : d\"","correct":"d","options":{"a":"158.210 m ","b":"704 m ","c":"153.256 m ","d":"158.256 m","e":"204 m"},"options_float":{"a":158.21,"b":704.0,"c":153.256,"d":158.256,"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)), 12.6), const_2), 200), 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) * 12.6<\/gadget>\n39.6<\/output>\n39.6 * 2<\/gadget>\n79.2<\/output>\n79.2 * 200<\/gadget>\n15_840<\/output>\n15_840 \/ 100<\/gadget>\n792\/5 = around 158.4<\/output>\n792\/5 = around 158.4<\/result>","index":1159} +{"problem":"on my sister ' s birthday , she was 143 cm in height , having grown 10 % since the year before . how tall was she the previous year ?","rationale":"\"let the previous year ' s height be x . 1.1 x = 143 x = 130 the answer is c .\"","correct":"c","options":{"a":"140 cm ","b":"136 cm ","c":"130 cm ","d":"127 cm","e":"125 cm"},"options_float":{"a":140.0,"b":136.0,"c":130.0,"d":127.0,"e":125.0},"annotated_formula":"subtract(143, divide(multiply(143, 10), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|","chain":"143 * 10<\/gadget>\n1_430<\/output>\n1_430 \/ 100<\/gadget>\n143\/10 = around 14.3<\/output>\n143 - (143\/10)<\/gadget>\n1_287\/10 = around 128.7<\/output>\n1_287\/10 = around 128.7<\/result>","index":1161} +{"problem":"the length of each side of an equilateral triangle having an area of 4 â ˆ š 3 cm 2 is ?","rationale":"explanation : â ˆ š 3 \/ 4 a 2 = 4 â ˆ š 3 - > a = 4 answer is d","correct":"d","options":{"a":"4 \/ 3 cm ","b":"3 \/ 4 cm ","c":"3 cm ","d":"4 cm","e":"5 cm"},"options_float":{"a":1.3333333333,"b":0.75,"c":3.0,"d":4.0,"e":5.0},"annotated_formula":"sqrt(divide(multiply(4, sqrt(3)), multiply(multiply(divide(const_1, 2), divide(const_1, 2)), sqrt(3))))","linear_formula":"divide(const_1,n2)|sqrt(n1)|multiply(n0,#1)|multiply(#0,#0)|multiply(#3,#1)|divide(#2,#4)|sqrt(#5)","chain":"3 ** (1\/2)<\/gadget>\nsqrt(3) = around 1.732051<\/output>\n4 * (sqrt(3))<\/gadget>\n4*sqrt(3) = around 6.928203<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * (1\/2)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * (sqrt(3))<\/gadget>\nsqrt(3)\/4 = around 0.433013<\/output>\n(4*sqrt(3)) \/ (sqrt(3)\/4)<\/gadget>\n16<\/output>\n16 ** (1\/2)<\/gadget>\n4<\/output>\n4<\/result>","index":1163} +{"problem":"if the sides of a rectangle are increased by 25 % , what is the percentage increase in the area ?","rationale":"\"if sides are a and b , after increase sides would be 1.25 a and 1.25 b . percentage increase in area = ( 1.25 a * 1.25 b - ab ) * 100 \/ ab = 56.25 % answer : b\"","correct":"b","options":{"a":"54.25 % ","b":"56.25 $ ","c":"53.25 % ","d":"55.25 %","e":"58.25 %"},"options_float":{"a":54.25,"b":56.25,"c":53.25,"d":55.25,"e":58.25},"annotated_formula":"multiply(subtract(power(add(divide(25, 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":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) ** 2<\/gadget>\n25\/16 = around 1.5625<\/output>\n(25\/16) - 1<\/gadget>\n9\/16 = around 0.5625<\/output>\n(9\/16) * 100<\/gadget>\n225\/4 = around 56.25<\/output>\n225\/4 = around 56.25<\/result>","index":1164} +{"problem":"the charge for a single room at hotel p is 30 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 = 70 x g = 100 y p = 90 y thus 70 x = 90 y or x = 1.28 y ans r = 128 y so increase = 28 % answer : a\"","correct":"a","options":{"a":"28 % ","b":"20 % ","c":"40 % ","d":"50 %","e":"150 %"},"options_float":{"a":28.0,"b":20.0,"c":40.0,"d":50.0,"e":150.0},"annotated_formula":"multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 30), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 30), 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 - 30<\/gadget>\n70<\/output>\n100 - 10<\/gadget>\n90<\/output>\n70 \/ 90<\/gadget>\n7\/9 = around 0.777778<\/output>\n(7\/9) * 100<\/gadget>\n700\/9 = around 77.777778<\/output>\n100 - (700\/9)<\/gadget>\n200\/9 = around 22.222222<\/output>\n(200\/9) \/ (700\/9)<\/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":1165} +{"problem":"a trader marked the selling price of an article at 11 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of :","rationale":"\"sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . ∴ discount % = [ 11 \/ 111 * 100 ] % = 9.9 % answer c\"","correct":"c","options":{"a":"10 % ","b":"10.5 % ","c":"9.9 % ","d":"12.5 %","e":"none"},"options_float":{"a":10.0,"b":10.5,"c":9.9,"d":12.5,"e":null},"annotated_formula":"multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 11)))","linear_formula":"add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)|","chain":"2 + 3<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 + 1<\/gadget>\n11<\/output>\n100 + 11<\/gadget>\n111<\/output>\n11 \/ 111<\/gadget>\n11\/111 = around 0.099099<\/output>\n100 * (11\/111)<\/gadget>\n1_100\/111 = around 9.90991<\/output>\n1_100\/111 = around 9.90991<\/result>","index":1166} +{"problem":"bottle r contains 250 capsules and costs $ 5.25 . 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 5.25 \/ 250 = 0.525 \/ 25 = 0.021 cost per capsule in t is 2.99 \/ 130 = 0.023 the difference is 0.002 the answer is e","correct":"e","options":{"a":"$ 0.25 ","b":"$ 0.12 ","c":"$ 0.05 ","d":"$ 0.03","e":"$ 0.002"},"options_float":{"a":0.25,"b":0.12,"c":0.05,"d":0.03,"e":0.002},"annotated_formula":"subtract(divide(2.99, 130), divide(5.25, 250))","linear_formula":"divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)","chain":"2.99 \/ 130<\/gadget>\n0.023<\/output>\n5.25 \/ 250<\/gadget>\n0.021<\/output>\n0.023 - 0.021<\/gadget>\n0.002<\/output>\n0.002<\/result>","index":1167} +{"problem":"area of four walls of a room is 99 m 2 . the length and breadth of the room are 7.5 m and 3.5 m respectively . the height of the room is :","rationale":"2 ( 7.5 + 3.5 ) × h = 99 2 ( 11 ) h = 99 22 h = 99 h = 99 \/ 22 = 9 \/ 2 = 4.5 m answer is d .","correct":"d","options":{"a":"2.5 m ","b":"3.5 m ","c":"1.5 m ","d":"4.5 m","e":"5.5 m"},"options_float":{"a":2.5,"b":3.5,"c":1.5,"d":4.5,"e":5.5},"annotated_formula":"divide(99, add(multiply(7.5, const_2), multiply(3.5, const_2)))","linear_formula":"multiply(n2,const_2)|multiply(n3,const_2)|add(#0,#1)|divide(n0,#2)","chain":"7.5 * 2<\/gadget>\n15<\/output>\n3.5 * 2<\/gadget>\n7<\/output>\n15 + 7<\/gadget>\n22<\/output>\n99 \/ 22<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":1170} +{"problem":"x is a positive integer less than 300 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ?","rationale":"\"the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 300 n = 22.13 or n = 22 answer : - b\"","correct":"b","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":"subtract(add(multiply(reminder(7, 300), 3), reminder(3, 300)), reminder(1, 300))","linear_formula":"reminder(n1,n0)|reminder(n3,n0)|reminder(n2,n0)|multiply(n3,#0)|add(#3,#1)|subtract(#4,#2)|","chain":"7 % 300<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n3 % 300<\/gadget>\n3<\/output>\n21 + 3<\/gadget>\n24<\/output>\n1 % 300<\/gadget>\n1<\/output>\n24 - 1<\/gadget>\n23<\/output>\n23<\/result>","index":1171} +{"problem":"how many different ways can 2 students be seated in a row of 4 desks , so that there is always at least one empty desk between the students ?","rationale":"\"total cases : 12 ( student one has 4 options and student two has three options , 4 x 3 = 12 ) non - favourable cases : 6 ( when two students sit together . students in desk 1 and desk 2 , in desk 2 and desk 3 , in desk 3 and desk 4 ) for each of these cases there are two possibilities because the positions can be interchanged . hence 2 x 3 = 6 . so favourable cases : 12 - 6 = 6 . answer : d\"","correct":"d","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(4, const_1), 2)","linear_formula":"subtract(n1,const_1)|permutation(#0,n0)|","chain":"4 - 1<\/gadget>\n3<\/output>\nfactorial(3) \/ factorial(3 - 2)<\/gadget>\n6<\/output>\n6<\/result>","index":1172} +{"problem":"a dishonest dealer professes to sell goods at the cost price but uses a weight of 720 grams per kg , what is his percent ?","rationale":"\"explanation : 720 - - - 280 100 - - - ? = > 38.9 % answer : c\"","correct":"c","options":{"a":"15 % ","b":"25 % ","c":"38.9 % ","d":"45 %","e":"35 %"},"options_float":{"a":15.0,"b":25.0,"c":38.9,"d":45.0,"e":35.0},"annotated_formula":"subtract(multiply(divide(const_100, 720), 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 \/ 720<\/gadget>\n5\/36 = around 0.138889<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n(5\/36) * 1_000<\/gadget>\n1_250\/9 = around 138.888889<\/output>\n(1_250\/9) - 100<\/gadget>\n350\/9 = around 38.888889<\/output>\n350\/9 = around 38.888889<\/result>","index":1174} +{"problem":"machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 80 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 20 minutes ?","rationale":"machine b produces 100 part in 80 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 80 \/ 2 = 40 minutes . now , machine a produces 100 parts in 40 minutes which is 100 \/ 40 = 10 \/ 4 parts \/ minute . 10 \/ 4 parts x a total of 20 minutes = 10 \/ 4 * 20 = 50 e","correct":"e","options":{"a":"60 ","b":"80 ","c":"70 ","d":"40","e":"50"},"options_float":{"a":60.0,"b":80.0,"c":70.0,"d":40.0,"e":50.0},"annotated_formula":"multiply(multiply(divide(100, 80), const_2), 20)","linear_formula":"divide(n0,n2)|multiply(#0,const_2)|multiply(n3,#1)","chain":"100 \/ 80<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) * 2<\/gadget>\n5\/2 = around 2.5<\/output>\n(5\/2) * 20<\/gadget>\n50<\/output>\n50<\/result>","index":1176} +{"problem":"a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 66 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ?","rationale":"\"this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 66 % is one st . dev , then on either side of m we have 66 \/ 2 = 33 % . so , 33 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 33 = 83 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 83 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon \/ theory to it c\"","correct":"c","options":{"a":"16 % ","b":"32 % ","c":"83 % ","d":"84 %","e":"92 %"},"options_float":{"a":16.0,"b":32.0,"c":83.0,"d":84.0,"e":92.0},"annotated_formula":"subtract(const_100, divide(subtract(const_100, 66), const_2))","linear_formula":"subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)|","chain":"100 - 66<\/gadget>\n34<\/output>\n34 \/ 2<\/gadget>\n17<\/output>\n100 - 17<\/gadget>\n83<\/output>\n83<\/result>","index":1178} +{"problem":"in a class of 54 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 - - > 54 = english + 22 - 12 + 0 - - > english = 44 - - > only english = english - both = 44 - 12 = 32 . answer : e .\"","correct":"e","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(54, 12), 22), 12)","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)|","chain":"54 + 12<\/gadget>\n66<\/output>\n66 - 22<\/gadget>\n44<\/output>\n44 - 12<\/gadget>\n32<\/output>\n32<\/result>","index":1179} +{"problem":"usc invited each south carolina high school to send up to 39 students to watch a football game . a section which has 199 seats in each row is reserved for those students . what is the least number of rows needed to guarantee that if 2006 students show up , then all students from the same high school can be seated in the same row ?","rationale":"the answer is 12 rows . if 59 schools send 34 students each , then we can sit at most 5 groups of students in the same row , so we will need 12 rows . next , 12 rows are sufficient . assume that this is not the case . suppose the groups of students are seated like this : first the largest group , then the second largest group , then the third largest group , etc . suppose we run out of space - there are not enough seats in any row to seat together the next group . suppose the first group that can not be seated together is the kth group and it consists of n students . then k 61 since any row fits at least 5 groups . also , n 2006 \/ k 2006 \/ 61 < 33 ( all groups already seated are no smaller than the kth group ) . so , n 32 . since there is not enough space in any of the 12 rows to seat the kth group , then there must be at least 168 students seated in each of the 12 rows . now , 12 × 168 = 2016 > 2006 a contradiction . so , 12 rows are sufficient . correct answer 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":"add(divide(2006, 199), const_2)","linear_formula":"divide(n2,n1)|add(#0,const_2)","chain":"2_006 \/ 199<\/gadget>\n2_006\/199 = around 10.080402<\/output>\n(2_006\/199) + 2<\/gadget>\n2_404\/199 = around 12.080402<\/output>\n2_404\/199 = around 12.080402<\/result>","index":1180} +{"problem":"a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 3 inches wide on each side , the total area would have been ( m + 64 ) square inches . what is the perimeter of the photograph , in inches ?","rationale":"let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 6 ) ( y + 6 ) = m and so ( 2 ) xy + 6 x + 6 y + 36 = m + 64 let ' s subtract equation ( 1 ) from equation ( 2 ) . 4 x + 4 y + 32 = 64 2 x + 2 y = 16 , which is the perimeter of the photograph . the answer is a .","correct":"a","options":{"a":"16 ","b":"24 ","c":"32 ","d":"40","e":"48"},"options_float":{"a":16.0,"b":24.0,"c":32.0,"d":40.0,"e":48.0},"annotated_formula":"divide(subtract(64, subtract(power(multiply(3, const_2), const_2), power(multiply(1, const_2), const_2))), const_2)","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)","chain":"3 * 2<\/gadget>\n6<\/output>\n6 ** 2<\/gadget>\n36<\/output>\n1 * 2<\/gadget>\n2<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n36 - 4<\/gadget>\n32<\/output>\n64 - 32<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16<\/result>","index":1181} +{"problem":"the unit digit in the product 4556 * 3432 * 4581 * 2784 is ?","rationale":"unit digit in the given product = unit digit in 6 * 2 * 1 * 4 = 8 answer is a","correct":"a","options":{"a":"8 ","b":"1 ","c":"2 ","d":"3","e":"4"},"options_float":{"a":8.0,"b":1.0,"c":2.0,"d":3.0,"e":4.0},"annotated_formula":"subtract(multiply(multiply(multiply(4556, 3432), 4581), 2784), subtract(multiply(multiply(multiply(4556, 3432), 4581), 2784), add(const_4, const_4)))","linear_formula":"add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)","chain":"4_556 * 3_432<\/gadget>\n15_636_192<\/output>\n15_636_192 * 4_581<\/gadget>\n71_629_395_552<\/output>\n71_629_395_552 * 2_784<\/gadget>\n199_416_237_216_768<\/output>\n4 + 4<\/gadget>\n8<\/output>\n199_416_237_216_768 - 8<\/gadget>\n199_416_237_216_760<\/output>\n199_416_237_216_768 - 199_416_237_216_760<\/gadget>\n8<\/output>\n8<\/result>","index":1183} +{"problem":"a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 276 metres long in 20 seconds . the length of the train ( in metres ) is :","rationale":"\"explanation : let the length of train be l m . acc . to question ( 276 + l ) \/ 20 = l \/ 8 2208 + 8 l = 20 l l = 2208 \/ 12 = 184 m answer a\"","correct":"a","options":{"a":"184 ","b":"176 ","c":"175 ","d":"96","e":"none of these"},"options_float":{"a":184.0,"b":176.0,"c":175.0,"d":96.0,"e":null},"annotated_formula":"multiply(divide(276, subtract(20, 8)), 8)","linear_formula":"subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)|","chain":"20 - 8<\/gadget>\n12<\/output>\n276 \/ 12<\/gadget>\n23<\/output>\n23 * 8<\/gadget>\n184<\/output>\n184<\/result>","index":1185} +{"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 760 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 760 . as per question we need to take 1 \/ 4 of 760 to be considered for top 10 movies = 190 approximate the value we 190 . imo option b is the correct answer . . .\"","correct":"b","options":{"a":"191 ","b":"190 ","c":"193 ","d":"212","e":"213"},"options_float":{"a":191.0,"b":190.0,"c":193.0,"d":212.0,"e":213.0},"annotated_formula":"divide(760, 4)","linear_formula":"divide(n3,n1)|","chain":"760 \/ 4<\/gadget>\n190<\/output>\n190<\/result>","index":1186} +{"problem":"each child has 4 crayons and 14 apples . if there are 9 children , how many crayons are there in total ?","rationale":"4 * 9 = 36 . answer is d .","correct":"d","options":{"a":"22 ","b":"65 ","c":"12 ","d":"36","e":"10"},"options_float":{"a":22.0,"b":65.0,"c":12.0,"d":36.0,"e":10.0},"annotated_formula":"multiply(9, 4)","linear_formula":"multiply(n0,n2)","chain":"9 * 4<\/gadget>\n36<\/output>\n36<\/result>","index":1187} +{"problem":"a man cycles round the boundary of a rectangular park at the rate of 12 kmph and completes one full round in 8 minutes . if the ratio between the length and breadth of the park be 3 : 2 , then its area is :","rationale":"perimeter = distance covered in 8 min = ( 12000 \/ 60 * 8 ) m = 1600 m let , length = 3 x meters and breadth = 2 x meters then , 2 ( 3 x + 2 x ) = 1600 or x = 160 therefore , length = 480 m and breadth = 320 m therefore , area = ( 480 * 320 ) m 2 = 153600 m 2 answer : c","correct":"c","options":{"a":"1536 m 2 ","b":"15360 m 2 ","c":"153600 m 2 ","d":"163600 m 2","e":"none of these"},"options_float":{"a":1536.0,"b":15360.0,"c":153600.0,"d":163600.0,"e":null},"annotated_formula":"multiply(multiply(divide(multiply(8, divide(multiply(12, const_1000), multiply(multiply(const_3, const_2), const_10))), const_10), 2), multiply(divide(multiply(8, divide(multiply(12, const_1000), multiply(multiply(const_3, const_2), const_10))), const_10), 3))","linear_formula":"multiply(n0,const_1000)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(#0,#2)|multiply(n1,#3)|divide(#4,const_10)|multiply(n3,#5)|multiply(n2,#5)|multiply(#6,#7)","chain":"12 * 1_000<\/gadget>\n12_000<\/output>\n3 * 2<\/gadget>\n6<\/output>\n6 * 10<\/gadget>\n60<\/output>\n12_000 \/ 60<\/gadget>\n200<\/output>\n8 * 200<\/gadget>\n1_600<\/output>\n1_600 \/ 10<\/gadget>\n160<\/output>\n160 * 2<\/gadget>\n320<\/output>\n160 * 3<\/gadget>\n480<\/output>\n320 * 480<\/gadget>\n153_600<\/output>\n153_600<\/result>","index":1188} +{"problem":"find the value of m 12519 x 9999 = m ?","rationale":"\"12519 x 9999 = 12519 x ( 10000 - 1 ) = 12519 x 10000 - 12519 x 1 = 125190000 - 12519 = 125177481 a\"","correct":"a","options":{"a":"125177481 ","b":"353654655 ","c":"545463251 ","d":"725117481","e":"477899932"},"options_float":{"a":125177481.0,"b":353654655.0,"c":545463251.0,"d":725117481.0,"e":477899932.0},"annotated_formula":"multiply(subtract(9999, const_4), 12519)","linear_formula":"subtract(n1,const_4)|multiply(#0,n0)|","chain":"9_999 - 4<\/gadget>\n9_995<\/output>\n9_995 * 12_519<\/gadget>\n125_127_405<\/output>\n125_127_405<\/result>","index":1189} +{"problem":"a is twice as good a workman as b and they took 9 days together to do the work b alone can do it in .","rationale":"\"wc = 2 : 1 2 x + x = 1 \/ 9 x = 1 \/ 27 = > 27 days answer : e\"","correct":"e","options":{"a":"17 days ","b":"12 days ","c":"29 days ","d":"25 days","e":"27 days"},"options_float":{"a":17.0,"b":12.0,"c":29.0,"d":25.0,"e":27.0},"annotated_formula":"multiply(divide(multiply(9, 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>\n9 * 3<\/gadget>\n27<\/output>\n27 \/ 2<\/gadget>\n27\/2 = around 13.5<\/output>\n(27\/2) * 2<\/gadget>\n27<\/output>\n27<\/result>","index":1190} +{"problem":"find the value of ( 875 233 \/ 899 ) × 899","rationale":"\"( 875 233 \/ 899 ) × 899 ( 786625 + 233 ) \/ 899 × 899 786858 \/ 899 × 899 786858 c\"","correct":"c","options":{"a":"786845 ","b":"786857 ","c":"786858 ","d":"786859","e":"786860"},"options_float":{"a":786845.0,"b":786857.0,"c":786858.0,"d":786859.0,"e":786860.0},"annotated_formula":"multiply(add(divide(233, 899), 875), 899)","linear_formula":"divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|","chain":"233 \/ 899<\/gadget>\n233\/899 = around 0.259177<\/output>\n(233\/899) + 875<\/gadget>\n786_858\/899 = around 875.259177<\/output>\n(786_858\/899) * 899<\/gadget>\n786_858<\/output>\n786_858<\/result>","index":1191} +{"problem":"the length of a rectangular floor is more than its breadth by 200 % . if rs . 150 is required to paint the floor at the rate of rs . 2 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 = 150 \/ 2 = 75 sq m l b = 75 i . e . , l * l \/ 3 = 75 l 2 = 225 = > l = 15 answer : b\"","correct":"b","options":{"a":"65 ","b":"15 ","c":"18 ","d":"16","e":"14"},"options_float":{"a":65.0,"b":15.0,"c":18.0,"d":16.0,"e":14.0},"annotated_formula":"multiply(sqrt(divide(divide(150, 2), const_3)), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)|","chain":"150 \/ 2<\/gadget>\n75<\/output>\n75 \/ 3<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n5 * 3<\/gadget>\n15<\/output>\n15<\/result>","index":1196} +{"problem":"a car traveled from san diego to san francisco at an average speed of 66 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 = 66 x total distance traveled = 2 * 66 x = 132 x total time taken = x + 2 x = 3 x average speed = 132 x \/ 3 x = 44 answer : e\"","correct":"e","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(66), divide(inverse(66), const_2)))","linear_formula":"inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)|","chain":"1 \/ 66<\/gadget>\n1\/66 = around 0.015152<\/output>\n(1\/66) \/ 2<\/gadget>\n1\/132 = around 0.007576<\/output>\n(1\/66) + (1\/132)<\/gadget>\n1\/44 = around 0.022727<\/output>\n1 \/ (1\/44)<\/gadget>\n44<\/output>\n44<\/result>","index":1197} +{"problem":"rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 2 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ?","rationale":"\"for 100 - 40 - 10 = 50 pages only cost is 5 $ per page for the first time page is typed - 50 * 5 = 250 $ ; for 40 pages the cost is : first time 5 $ + 2 $ of the first revision - 40 * ( 5 + 2 ) = 280 $ ; for 10 pages the cost is : first time 5 $ + 2 $ of the first revision + 2 $ of the second revision - 10 ( 5 + 2 + 2 ) = 90 $ ; total : 250 + 280 + 90 = 620 $ . answer : b .\"","correct":"b","options":{"a":"$ 430 ","b":"$ 620 ","c":"$ 650 ","d":"$ 680","e":"$ 770"},"options_float":{"a":430.0,"b":620.0,"c":650.0,"d":680.0,"e":770.0},"annotated_formula":"add(add(multiply(100, 5), multiply(40, 2)), multiply(multiply(10, 2), const_2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)|","chain":"100 * 5<\/gadget>\n500<\/output>\n40 * 2<\/gadget>\n80<\/output>\n500 + 80<\/gadget>\n580<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20 * 2<\/gadget>\n40<\/output>\n580 + 40<\/gadget>\n620<\/output>\n620<\/result>","index":1199} +{"problem":"some persons can do a piece of work in 32 days . two times the number of these people will do half of that work in ?","rationale":"32 \/ ( 2 * 2 ) = 8 days answer : e","correct":"e","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(32, 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>\n32 * (1\/2)<\/gadget>\n16<\/output>\n16 * (1\/2)<\/gadget>\n8<\/output>\n8<\/result>","index":1200} +{"problem":"a department of 10 people - 6 men and 4 women - needs to send a team of 5 to a conference . if they want to make sure that there are no more than 3 members of the team from any one gender , how many distinct groups are possible to send ?","rationale":"they can make a team of 3 men and 2 women . the number of ways to do this is 6 c 3 * 4 c 2 = 20 * 6 = 120 they can make a team of 2 men and 3 women . the number of ways to do this is 6 c 2 * 4 c 3 = 15 * 4 = 60 the total number of distinct groups is 180 . the answer is c .","correct":"c","options":{"a":"120 ","b":"150 ","c":"180 ","d":"210","e":"240"},"options_float":{"a":120.0,"b":150.0,"c":180.0,"d":210.0,"e":240.0},"annotated_formula":"add(add(multiply(multiply(6, 5), 4), multiply(6, 5)), multiply(6, 5))","linear_formula":"multiply(n1,n3)|multiply(n2,#0)|add(#1,#0)|add(#2,#0)","chain":"6 * 5<\/gadget>\n30<\/output>\n30 * 4<\/gadget>\n120<\/output>\n120 + 30<\/gadget>\n150<\/output>\n150 + 30<\/gadget>\n180<\/output>\n180<\/result>","index":1202} +{"problem":"forks , spoons , and knives in drawer are in the ratio of 4 : 4 : 3 . if there are 16 forks , the number of knives in the drawer is :","rationale":"explanation : let forks = 4 x , spoons = 4 x & knives = 3 x . now , 4 x = 16 hence x = 4 . number of knives = 3 x = 12 . answer : c","correct":"c","options":{"a":"8 ","b":"4 ","c":"12 ","d":"16","e":"14"},"options_float":{"a":8.0,"b":4.0,"c":12.0,"d":16.0,"e":14.0},"annotated_formula":"multiply(divide(16, 4), 3)","linear_formula":"divide(n3,n0)|multiply(n2,#0)","chain":"16 \/ 4<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12<\/result>","index":1209} +{"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 6000 , the number of valid votes that the other candidate got , was :","rationale":"\"c number of valid votes = 80 % of 6000 = 4800 . valid votes polled by other candidate = 45 % of 4800 = ( 45 \/ 100 x 4800 ) = 2160 .\"","correct":"c","options":{"a":"2800 ","b":"2700 ","c":"2160 ","d":"2200","e":"2300"},"options_float":{"a":2800.0,"b":2700.0,"c":2160.0,"d":2200.0,"e":2300.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6000)","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) * 6_000<\/gadget>\n2_160<\/output>\n2_160<\/result>","index":1211} +{"problem":"10 : 4 seconds : : ? : 6 minutes","rationale":"\"10 * 6 = 4 * x x = 15 answer : b\"","correct":"b","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(6, divide(10, 4))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|","chain":"10 \/ 4<\/gadget>\n5\/2 = around 2.5<\/output>\n6 * (5\/2)<\/gadget>\n15<\/output>\n15<\/result>","index":1212} +{"problem":"a train passes a platform in 36 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 × 36 = 540 = > l = 540 - 300 = 240 meter answer is a .\"","correct":"a","options":{"a":"240 ","b":"250 ","c":"260 ","d":"230","e":"220"},"options_float":{"a":240.0,"b":250.0,"c":260.0,"d":230.0,"e":220.0},"annotated_formula":"multiply(multiply(const_0_2778, 54), subtract(36, 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>\n36 - 20<\/gadget>\n16<\/output>\n15 * 16<\/gadget>\n240<\/output>\n240<\/result>","index":1213} +{"problem":"if two dice are thrown together , the probability of getting a doublet on the dice is","rationale":"\"the number of exhaustive outcomes is 36 . let e be the event of getting doublet on the dies is 6 \/ 36 = 1 \/ 6 p ( e ) = 1 \/ 6 . a )\"","correct":"a","options":{"a":"1 \/ 6 ","b":"1 \/ 5 ","c":"1 \/ 4 ","d":"1 \/ 3","e":"1 \/ 2"},"options_float":{"a":0.1666666667,"b":0.2,"c":0.25,"d":0.3333333333,"e":0.5},"annotated_formula":"divide(const_6, multiply(const_6, const_6))","linear_formula":"multiply(const_6,const_6)|divide(const_6,#0)|","chain":"6 * 6<\/gadget>\n36<\/output>\n6 \/ 36<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":1215} +{"problem":"a 6 % stock yields 9 % . the market value of the stock is :","rationale":"\"explanation : for an income of rs . 9 , investment = rs . 100 . for an income of rs 6 , investment = rs . 100 \/ 9 x 6 = rs 66.66 market value of rs . 100 stock = rs . 66.66 answer is e\"","correct":"e","options":{"a":"rs 66.55 ","b":"rs 68.55 ","c":"rs 69.55 ","d":"rs 65.55","e":"rs 66.66"},"options_float":{"a":66.55,"b":68.55,"c":69.55,"d":65.55,"e":66.66},"annotated_formula":"multiply(divide(const_100, 9), 6)","linear_formula":"divide(const_100,n1)|multiply(n0,#0)|","chain":"100 \/ 9<\/gadget>\n100\/9 = around 11.111111<\/output>\n(100\/9) * 6<\/gadget>\n200\/3 = around 66.666667<\/output>\n200\/3 = around 66.666667<\/result>","index":1218} +{"problem":"jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 25 \/ 36 , what is the probability that the next one she draws will be pink ?","rationale":"\"the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 \/ 36 = ( 5 \/ 6 ) * ( 5 \/ 6 ) therefore probability that the next one she draws is pink = 1 \/ 6 option ( a )\"","correct":"a","options":{"a":"1 \/ 6 ","b":"4 \/ 7 ","c":"3 \/ 7 ","d":"16 \/ 49","e":"40 \/ 49"},"options_float":{"a":0.1666666667,"b":0.5714285714,"c":0.4285714286,"d":0.3265306122,"e":0.8163265306},"annotated_formula":"subtract(const_1, sqrt(divide(25, 36)))","linear_formula":"divide(n0,n1)|sqrt(#0)|subtract(const_1,#1)|","chain":"25 \/ 36<\/gadget>\n25\/36 = around 0.694444<\/output>\n(25\/36) ** (1\/2)<\/gadget>\n5\/6 = around 0.833333<\/output>\n1 - (5\/6)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":1219} +{"problem":"if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 253 games will be played during the tournament . what is the number of participants ?","rationale":"\"let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) \/ 2 = 253 n * ( n - 1 ) = 506 = 23 * 22 ( trial and error ) the answer is c .\"","correct":"c","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(sqrt(add(multiply(multiply(253, 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":"253 * 2<\/gadget>\n506<\/output>\n506 * 4<\/gadget>\n2_024<\/output>\n2_024 + 1<\/gadget>\n2_025<\/output>\n2_025 ** (1\/2)<\/gadget>\n45<\/output>\n45 + 1<\/gadget>\n46<\/output>\n46 \/ 2<\/gadget>\n23<\/output>\n23<\/result>","index":1220} +{"problem":"it takes 10 days for digging a trench of 100 m long , 50 m broad and 10 m deep . what length of trench , 25 m broad and 15 m deep can be dug in 30 days ?","rationale":"more days , more length ( direct ) less breadth , more length ( indirect ) more depth , less length ( indirect days 10 : 30 ; breadth 25 : 50 ; : : 100 : x depth 15 : 10 ; : . 10 * 25 * 15 * x = 30 * 50 * 10 * 100 x = ( 30 * 50 * 10 * 100 ) \/ 10 * 25 * 15 = 400 so the required length = 400 m answer : a","correct":"a","options":{"a":"400 m ","b":"200 m ","c":"100 m ","d":"89 m","e":"79 m"},"options_float":{"a":400.0,"b":200.0,"c":100.0,"d":89.0,"e":79.0},"annotated_formula":"divide(multiply(multiply(multiply(30, 50), 10), 100), multiply(15, multiply(10, 25)))","linear_formula":"multiply(n2,n6)|multiply(n0,n4)|multiply(n0,#0)|multiply(n5,#1)|multiply(n1,#2)|divide(#4,#3)","chain":"30 * 50<\/gadget>\n1_500<\/output>\n1_500 * 10<\/gadget>\n15_000<\/output>\n15_000 * 100<\/gadget>\n1_500_000<\/output>\n10 * 25<\/gadget>\n250<\/output>\n15 * 250<\/gadget>\n3_750<\/output>\n1_500_000 \/ 3_750<\/gadget>\n400<\/output>\n400<\/result>","index":1221} +{"problem":"if 20 men can build a water fountain 56 metres long in 3 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 3 : 3 : : 56 : x therefore ( 20 x 3 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) \/ 60 = 98 hence , the required length is 98 m . answer : e\"","correct":"e","options":{"a":"40 m ","b":"64 m ","c":"77 m ","d":"89 m","e":"98 m"},"options_float":{"a":40.0,"b":64.0,"c":77.0,"d":89.0,"e":98.0},"annotated_formula":"multiply(divide(56, multiply(20, 3)), multiply(35, 3))","linear_formula":"multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)|","chain":"20 * 3<\/gadget>\n60<\/output>\n56 \/ 60<\/gadget>\n14\/15 = around 0.933333<\/output>\n35 * 3<\/gadget>\n105<\/output>\n(14\/15) * 105<\/gadget>\n98<\/output>\n98<\/result>","index":1222} +{"problem":"p has $ 63 more than what q and r together would have had if both b and c had 1 \/ 9 of what p has . how much does p have ?","rationale":"p = ( 2 \/ 9 ) * p + 63 ( 7 \/ 9 ) * p = 63 p = 81 the answer is e .","correct":"e","options":{"a":"$ 69 ","b":"$ 72 ","c":"$ 75 ","d":"$ 78","e":"$ 81"},"options_float":{"a":69.0,"b":72.0,"c":75.0,"d":78.0,"e":81.0},"annotated_formula":"divide(63, subtract(const_1, multiply(divide(1, 9), const_2)))","linear_formula":"divide(n1,n2)|multiply(#0,const_2)|subtract(const_1,#1)|divide(n0,#2)","chain":"1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/9) * 2<\/gadget>\n2\/9 = around 0.222222<\/output>\n1 - (2\/9)<\/gadget>\n7\/9 = around 0.777778<\/output>\n63 \/ (7\/9)<\/gadget>\n81<\/output>\n81<\/result>","index":1223} +{"problem":"a train running at the speed of 110 km \/ hr crosses a pole in 9 sec . what is the length of the train ?","rationale":"\"speed = 110 * 5 \/ 18 = 275 \/ 9 m \/ sec length of the train = speed * time = 275 \/ 9 * 9 = 275 m answer : b\"","correct":"b","options":{"a":"298 m ","b":"275 m ","c":"208 m ","d":"988 m","e":"299 m"},"options_float":{"a":298.0,"b":275.0,"c":208.0,"d":988.0,"e":299.0},"annotated_formula":"multiply(divide(multiply(110, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"110 * 1_000<\/gadget>\n110_000<\/output>\n110_000 \/ 3_600<\/gadget>\n275\/9 = around 30.555556<\/output>\n(275\/9) * 9<\/gadget>\n275<\/output>\n275<\/result>","index":1224} +{"problem":"if 2 ^ y = 4 ^ ( 5 x + 3 ) and 3 ^ ( x - 7 ) = 9 ^ y , what is the value of x + y ?","rationale":"2 ^ y = 4 ^ ( 5 x + 3 ) 2 ^ y = 2 ^ 2 ( 5 x + 3 ) y = 10 x + 6 . . . . . . . . . . . 1 3 ^ ( x - 7 ) = 9 ^ y 3 ^ ( x - 7 ) = 3 ^ 2 y x - 7 = 2 y . . . . . . . . . . . . . 2 put value of y = 10 x + 6 in eq 2 x - 7 = 2 ( 10 x + 6 ) x - 7 = 20 x + 12 19 x = - 19 x = - 1 therefore , y = - 10 + 6 y = - 4 x + y = - 1 - 4 = - 5 answer : b","correct":"b","options":{"a":"- 10 ","b":"- 5 ","c":"- 4 ","d":"3","e":"7"},"options_float":{"a":-10.0,"b":-5.0,"c":-4.0,"d":3.0,"e":7.0},"annotated_formula":"add(divide(add(divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1)), negate(7)), 2), divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1)))","linear_formula":"multiply(n0,n3)|multiply(n0,const_10)|negate(n5)|multiply(n0,#0)|subtract(#1,const_1)|negate(#3)|add(#2,#5)|divide(#6,#4)|add(#7,#2)|divide(#8,n0)|add(#9,#7)","chain":"-7<\/gadget>\n-7<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6 * 2<\/gadget>\n12<\/output>\n-12<\/gadget>\n-12<\/output>\n(-7) + (-12)<\/gadget>\n-19<\/output>\n2 * 10<\/gadget>\n20<\/output>\n20 - 1<\/gadget>\n19<\/output>\n(-19) \/ 19<\/gadget>\n-1<\/output>\n(-1) + (-7)<\/gadget>\n-8<\/output>\n(-8) \/ 2<\/gadget>\n-4<\/output>\n(-4) + (-1)<\/gadget>\n-5<\/output>\n-5<\/result>","index":1225} +{"problem":"a jogger running at 9 km \/ hr along side a railway track is 240 m ahead of the engine of a 140 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 = 240 + 140 = 380 m . time taken = 380 \/ 10 = 38 sec . answer : c\"","correct":"c","options":{"a":"28 sec ","b":"16 sec ","c":"38 sec ","d":"18 sec","e":"17 sec"},"options_float":{"a":28.0,"b":16.0,"c":38.0,"d":18.0,"e":17.0},"annotated_formula":"divide(add(240, 140), 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":"240 + 140<\/gadget>\n380<\/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>\n380 \/ 10<\/gadget>\n38<\/output>\n38<\/result>","index":1226} +{"problem":"of the 75 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ?","rationale":"\"neither car nor garage = total - garage - ( swim - common ) = 75 - 50 - ( 40 - 35 ) = 75 - 55 = 20 answer 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":"subtract(75, add(add(subtract(50, 35), subtract(40, 35)), 35))","linear_formula":"subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)|","chain":"50 - 35<\/gadget>\n15<\/output>\n40 - 35<\/gadget>\n5<\/output>\n15 + 5<\/gadget>\n20<\/output>\n20 + 35<\/gadget>\n55<\/output>\n75 - 55<\/gadget>\n20<\/output>\n20<\/result>","index":1227} +{"problem":"if the price of a certain computer increased 30 percent from a dollars to 351 dollars , then 2 a =","rationale":"\"before price increase price = a after 30 % price increase price = a + ( 30 \/ 100 ) * a = 1.3 a = 351 ( given ) i . e . a = 351 \/ 1.3 = $ 270 i . e . 2 a = 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":1228} +{"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 w 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 w of cgd \/ ( cg ) - > 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":1229} +{"problem":"what profit percent is made by selling an article at a certain price , if by selling at 2 \/ 3 rd of that price , there would be a loss of 25 % ?","rationale":"\"sp 2 = 2 \/ 3 sp 1 cp = 100 sp 2 = 75 2 \/ 3 sp 1 = 75 sp 1 = 112.50 100 - - - 12.5 = > 12.5 % answer : e\"","correct":"e","options":{"a":"20 % ","b":"26 % ","c":"42 % ","d":"27 %","e":"12.5 %"},"options_float":{"a":20.0,"b":26.0,"c":42.0,"d":27.0,"e":12.5},"annotated_formula":"subtract(divide(subtract(const_100, 25), divide(2, 3)), const_100)","linear_formula":"divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|","chain":"100 - 25<\/gadget>\n75<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n75 \/ (2\/3)<\/gadget>\n225\/2 = around 112.5<\/output>\n(225\/2) - 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":1230} +{"problem":"how many 5 - digit numbers are there , if the two leftmost digits are even , the other digits are odd and the digit 2 can not appear more than once in the number .","rationale":"n = ( 4 * 5 - 1 ) * 5 * 5 * 5 = 2375 where 4 cases of first digit { 2,4 , 6,8 } 5 cases of second digit { 0 , 2,4 , 6,8 } 1 case of 22 for two leftmost digit 5 cases of third digit { 1 , 3,5 , 7,9 } 5 cases of fourth digit { 1 , 3,5 , 7,9 } 5 cases of fifth digit { 1 , 3,5 , 7,9 } c","correct":"c","options":{"a":"2200 ","b":"2295 ","c":"2375 ","d":"2380","e":"2385"},"options_float":{"a":2200.0,"b":2295.0,"c":2375.0,"d":2380.0,"e":2385.0},"annotated_formula":"multiply(add(multiply(const_3, const_4), add(const_3, const_4)), power(5, const_3))","linear_formula":"add(const_3,const_4)|multiply(const_3,const_4)|power(n0,const_3)|add(#0,#1)|multiply(#3,#2)","chain":"3 * 4<\/gadget>\n12<\/output>\n3 + 4<\/gadget>\n7<\/output>\n12 + 7<\/gadget>\n19<\/output>\n5 ** 3<\/gadget>\n125<\/output>\n19 * 125<\/gadget>\n2_375<\/output>\n2_375<\/result>","index":1231} +{"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 . 1081 . 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 ) = 1081 96 + 225 + 490 + 6 x = 1081 6 x = 270 = > x = 45 . answer : e\"","correct":"e","options":{"a":"25 ","b":"66 ","c":"77 ","d":"99","e":"45"},"options_float":{"a":25.0,"b":66.0,"c":77.0,"d":99.0,"e":45.0},"annotated_formula":"divide(subtract(subtract(subtract(1081, 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>\n1_081 - 96<\/gadget>\n985<\/output>\n5 * 45<\/gadget>\n225<\/output>\n985 - 225<\/gadget>\n760<\/output>\n7 * 70<\/gadget>\n490<\/output>\n760 - 490<\/gadget>\n270<\/output>\n270 \/ 6<\/gadget>\n45<\/output>\n45<\/result>","index":1233} +{"problem":"selling an kite for rs . 30 , a shop keeper gains 40 % . 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 \/ 140 * 30 = rs . 21.42 sale price = 90 % of rs . 30 = rs . 27 required gain % = 5.57 \/ 21.42 * 100 = 26 % . answer : e\"","correct":"e","options":{"a":"8 % ","b":"10 % ","c":"11 % ","d":"15 %","e":"26 %"},"options_float":{"a":8.0,"b":10.0,"c":11.0,"d":15.0,"e":26.0},"annotated_formula":"multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(40, const_100))), divide(multiply(30, const_100), add(40, 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>\n40 + 100<\/gadget>\n140<\/output>\n3_000 \/ 140<\/gadget>\n150\/7 = around 21.428571<\/output>\n27 - (150\/7)<\/gadget>\n39\/7 = around 5.571429<\/output>\n(39\/7) \/ (150\/7)<\/gadget>\n13\/50 = around 0.26<\/output>\n(13\/50) * 100<\/gadget>\n26<\/output>\n26<\/result>","index":1234} +{"problem":"at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 75 . if the restaurant serves 120 customers today , raising the average to 80 customers per day , what is the value of x ?","rationale":"\"withoutusing the formula , we can see that today the restaurant served 40 customers above the average . the total amount above the average must equal total amount below the average . this additional 40 customers must offset the “ deficit ” below the average of 80 created on the x days the restaurant served only 75 customers per day . 40 \/ 5 = 8 days . choice ( a ) . withthe formula , we can set up the following : 80 = ( 75 x + 120 ) \/ ( x + 1 ) 80 x + 80 = 75 x + 120 5 x = 40 x = 8 answer choice ( d )\"","correct":"d","options":{"a":"2 ","b":"5 ","c":"7 ","d":"8","e":"20"},"options_float":{"a":2.0,"b":5.0,"c":7.0,"d":8.0,"e":20.0},"annotated_formula":"subtract(divide(subtract(120, 80), subtract(80, 75)), 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 - 80<\/gadget>\n40<\/output>\n80 - 75<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n120 - 100<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n8 - (1\/5)<\/gadget>\n39\/5 = around 7.8<\/output>\n39\/5 = around 7.8<\/result>","index":1236} +{"problem":"an inspector rejects 15 % of the meters as defective . how many will he examine to reject 15 ?","rationale":"\"then , 15 % of x = 15 ( 15 \/ 100 ) x = 15 x = ( 15 * 100 * ) \/ 15 = 100 answer is a\"","correct":"a","options":{"a":"100 ","b":"120 ","c":"250 ","d":"200","e":"160"},"options_float":{"a":100.0,"b":120.0,"c":250.0,"d":200.0,"e":160.0},"annotated_formula":"divide(multiply(15, const_100), 15)","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|","chain":"15 * 100<\/gadget>\n1_500<\/output>\n1_500 \/ 15<\/gadget>\n100<\/output>\n100<\/result>","index":1238} +{"problem":"pipe a fills a tank of capacity 700 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 700 = 50 * n = > n = 14 here n = number of cycles . total time = 14 * 3 = 42 as in one cycle there are 3 minutes . thus 42 minutes answer : a\"","correct":"a","options":{"a":"42 minutes ","b":"14 minutes ","c":"39 minutes ","d":"40 minutes 20 seconds","e":"none of these"},"options_float":{"a":42.0,"b":14.0,"c":39.0,"d":40.0,"e":null},"annotated_formula":"multiply(divide(700, 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>\n700 \/ 50<\/gadget>\n14<\/output>\n14 * 3<\/gadget>\n42<\/output>\n42<\/result>","index":1239} +{"problem":"the area of a parallelogram is 72 sq m and its altitude is twice the corresponding base . then the length of the base is ?","rationale":"\"2 x * x = 72 = > x = 6 answer : a\"","correct":"a","options":{"a":"6 ","b":"16 ","c":"8 ","d":"36","e":"none"},"options_float":{"a":6.0,"b":16.0,"c":8.0,"d":36.0,"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":1240} +{"problem":"a and b can finish a work together in 12 days , and b and c together in 16 days . if a alone works for 5 days and then b alone continues for 7 days , then remaining work is done by c in 13 days . in how many days can c alone finish the complete work ?","rationale":"here lcm of 12 and 16 is taken as total work . ( becomes easy to solve ) assume total work = 48 units then workdone by ( a + b ) in one day = 48 \/ 12 = 4 units similarly , by ( b + c ) in one day = 48 \/ 16 = 3 units now according to question , a works 5 days , b for 7 days and c for 13 days to complete total work so , 5 a + 7 b + 13 c = 48 units 5 ( a + b ) + 2 ( b + c ) + 11 c = 48 units 5 * 4 + 2 * 3 + 11 c = 48 units 11 c = 22 units c = 2 units ( c does 2 units of work daily ) therefore , 48 \/ 2 = 24 days c requires 24 days to complete the work alone . answer d","correct":"d","options":{"a":"22 days ","b":"21 days ","c":"25 days ","d":"24 days","e":"23 days"},"options_float":{"a":22.0,"b":21.0,"c":25.0,"d":24.0,"e":23.0},"annotated_formula":"divide(const_1, divide(subtract(const_1, add(divide(5, 12), divide(const_2, 16))), subtract(add(5, 13), 7)))","linear_formula":"add(n2,n4)|divide(n2,n0)|divide(const_2,n1)|add(#1,#2)|subtract(#0,n3)|subtract(const_1,#3)|divide(#5,#4)|divide(const_1,#6)","chain":"5 \/ 12<\/gadget>\n5\/12 = around 0.416667<\/output>\n2 \/ 16<\/gadget>\n1\/8 = around 0.125<\/output>\n(5\/12) + (1\/8)<\/gadget>\n13\/24 = around 0.541667<\/output>\n1 - (13\/24)<\/gadget>\n11\/24 = around 0.458333<\/output>\n5 + 13<\/gadget>\n18<\/output>\n18 - 7<\/gadget>\n11<\/output>\n(11\/24) \/ 11<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ (1\/24)<\/gadget>\n24<\/output>\n24<\/result>","index":1242} +{"problem":"the ages of two person , differ by 20 years . if 5 years ag , 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 + 20 ) years ( x - 5 ) * 5 = ( x + 20 - 5 ) after solving this we get x = 10 years the age of elder one = 10 + 20 = 30 years so the present ages are 30 and 10 years answer : a","correct":"a","options":{"a":"30 , 10 ","b":"2010 ","c":"3515 ","d":"5117","e":"20,17"},"options_float":{"a":30.0,"b":2010.0,"c":3515.0,"d":5117.0,"e":2017.0},"annotated_formula":"add(20, const_10)","linear_formula":"add(n0,const_10)","chain":"20 + 10<\/gadget>\n30<\/output>\n30<\/result>","index":1243} +{"problem":"the price of stock increased by 8 % last year and decreased by 6 % this year . what is the net percentage change in the price of the stock ?","rationale":"( 100 % + 8 % ) * ( 100 % - 6 % ) = 1.08 * 0.94 = 1.0152 = 101.52 % . the net percentage change in the price of the stock is ( + ) 1.52 % the answer is d","correct":"d","options":{"a":"0.2 % ","b":"0.8 % ","c":"1.2 % ","d":"1.52 %","e":"2 %"},"options_float":{"a":0.2,"b":0.8,"c":1.2,"d":1.52,"e":2.0},"annotated_formula":"subtract(multiply(multiply(divide(add(const_100, 8), const_100), divide(subtract(const_100, 6), const_100)), const_100), const_100)","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100)","chain":"100 + 8<\/gadget>\n108<\/output>\n108 \/ 100<\/gadget>\n27\/25 = around 1.08<\/output>\n100 - 6<\/gadget>\n94<\/output>\n94 \/ 100<\/gadget>\n47\/50 = around 0.94<\/output>\n(27\/25) * (47\/50)<\/gadget>\n1_269\/1_250 = around 1.0152<\/output>\n(1_269\/1_250) * 100<\/gadget>\n2_538\/25 = around 101.52<\/output>\n(2_538\/25) - 100<\/gadget>\n38\/25 = around 1.52<\/output>\n38\/25 = around 1.52<\/result>","index":1244} +{"problem":"the mass of 1 cubic meter of a substance is 400 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters )","rationale":"\"400 kg - 1 cubic meter ; 400,000 g - 1 cubic meter ; 400,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 \/ 400,000 = 10 \/ 4 = 2.5 cubic centimeters . answer : b .\"","correct":"b","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":"divide(multiply(1,000, 1,000), multiply(400, 1,000))","linear_formula":"multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)|","chain":"1_000 * 1_000<\/gadget>\n1_000_000<\/output>\n400 * 1_000<\/gadget>\n400_000<\/output>\n1_000_000 \/ 400_000<\/gadget>\n5\/2 = around 2.5<\/output>\n5\/2 = around 2.5<\/result>","index":1245} +{"problem":"a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 24 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 ( 24 ) ^ 2 * h = ( 4 \/ 3 ) π ( 12 ) ^ 3 = > h = 4 cm answer : b\"","correct":"b","options":{"a":"6 cm ","b":"4 cm ","c":"8 cm ","d":"3 cm","e":"9 cm"},"options_float":{"a":6.0,"b":4.0,"c":8.0,"d":3.0,"e":9.0},"annotated_formula":"divide(multiply(const_4, divide(power(12, const_3), power(24, 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":"12 ** 3<\/gadget>\n1_728<\/output>\n24 ** 2<\/gadget>\n576<\/output>\n1_728 \/ 576<\/gadget>\n3<\/output>\n4 * 3<\/gadget>\n12<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n4<\/result>","index":1248} +{"problem":"boys and girls in a class are writing letters . there are twice as many girls as boys in the class , and each girl writes 3 more letters than each boy . if boys write 24 of the 90 total letters written by the class , how many letters does each boy write ?","rationale":"there are twice as many girls as boys in the class - - > g = 2 b . each girl writes 3 more letters than each boy - - > boys write x letters , girls write x + 3 letters . boys write 24 letters - - > bx = 24 . girls write 90 - 24 = 66 letters - - > ( 2 b ) ( x + 3 ) = 66 - - > 2 bx + 6 b = 66 - - > 2 * 24 + 6 b = 66 - - > b = 3 . bx = 24 - - > 3 x = 24 - - > x = 8 . answer : d .","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":"divide(24, divide(subtract(90, multiply(3, 24)), multiply(3, const_2)))","linear_formula":"multiply(n0,n1)|multiply(n0,const_2)|subtract(n2,#0)|divide(#2,#1)|divide(n1,#3)","chain":"3 * 24<\/gadget>\n72<\/output>\n90 - 72<\/gadget>\n18<\/output>\n3 * 2<\/gadget>\n6<\/output>\n18 \/ 6<\/gadget>\n3<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n8<\/result>","index":1249} +{"problem":"a man walking at the rate of 5 km \/ hr crosses a bridge in 15 minutes . the length of the bridge ( in meters ) is :","rationale":"\"speed = ( 5 * 5 \/ 18 ) m \/ sec = 25 \/ 18 m \/ sec . distance covered in 15 minutes = ( 25 \/ 18 * 15 * 60 ) m = 1250 m . correct option : d\"","correct":"d","options":{"a":"600 ","b":"750 ","c":"1000 ","d":"1250","e":"none of these"},"options_float":{"a":600.0,"b":750.0,"c":1000.0,"d":1250.0,"e":null},"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":1250} +{"problem":"mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8010 \/ - . find out the amount that he borrowed .","rationale":"\"explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt \/ 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) \/ 100 = 54 p \/ 100 amount = principal + s . i . 8010 = p + ( 54 p \/ 100 ) 8010 = ( 100 p + 54 p ) \/ 100 8010 = 154 p \/ 100 p = ( 8010 * 100 ) \/ 154 = rs . 5201.298 answer : d\"","correct":"d","options":{"a":"s . 5266 ","b":"s . 5269 ","c":"s . 5228 ","d":"s . 5201","e":"s . 52192"},"options_float":{"a":5266.0,"b":5269.0,"c":5228.0,"d":5201.0,"e":52192.0},"annotated_formula":"divide(8010, add(const_1, divide(multiply(6, 9), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)|","chain":"6 * 9<\/gadget>\n54<\/output>\n54 \/ 100<\/gadget>\n27\/50 = around 0.54<\/output>\n1 + (27\/50)<\/gadget>\n77\/50 = around 1.54<\/output>\n8_010 \/ (77\/50)<\/gadget>\n400_500\/77 = around 5_201.298701<\/output>\n400_500\/77 = around 5_201.298701<\/result>","index":1252} +{"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 6500 , the number of valid votes that the other candidate got , was :","rationale":"\"d number of valid votes = 80 % of 6500 = 5200 . valid votes polled by other candidate = 45 % of 5200 = ( 45 \/ 100 x 5200 ) = 2340 .\"","correct":"d","options":{"a":"2800 ","b":"2700 ","c":"2900 ","d":"2340","e":"2300"},"options_float":{"a":2800.0,"b":2700.0,"c":2900.0,"d":2340.0,"e":2300.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6500)","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) * 6_500<\/gadget>\n2_340<\/output>\n2_340<\/result>","index":1253} +{"problem":"if 9 ! \/ 3 ^ x is an integer , what is the greatest possible value of x ?","rationale":"9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 4 is allowed . imo 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":"add(divide(9, 3), divide(3, divide(9, 3)))","linear_formula":"divide(n0,n1)|divide(n1,#0)|add(#0,#1)","chain":"9 \/ 3<\/gadget>\n3<\/output>\n3 \/ 3<\/gadget>\n1<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4<\/result>","index":1254} +{"problem":"the average of 13 result is 60 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th result ?","rationale":"sum of all the 13 results = 13 * 60 = 780 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 780 = 46 . c","correct":"c","options":{"a":"35 ","b":"37 ","c":"46 ","d":"48","e":"50"},"options_float":{"a":35.0,"b":37.0,"c":46.0,"d":48.0,"e":50.0},"annotated_formula":"subtract(add(multiply(7, 57), multiply(7, 61)), multiply(13, 60))","linear_formula":"multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)","chain":"7 * 57<\/gadget>\n399<\/output>\n7 * 61<\/gadget>\n427<\/output>\n399 + 427<\/gadget>\n826<\/output>\n13 * 60<\/gadget>\n780<\/output>\n826 - 780<\/gadget>\n46<\/output>\n46<\/result>","index":1256} +{"problem":"after decreasing 25 % in the price of an article costs rs . 1500 . find the actual cost of an article ?","rationale":"\"cp * ( 75 \/ 100 ) = 1500 cp = 20 * 100 = > cp = 2000 answer : d\"","correct":"d","options":{"a":"1400 ","b":"1300 ","c":"1200 ","d":"2000","e":"1500"},"options_float":{"a":1400.0,"b":1300.0,"c":1200.0,"d":2000.0,"e":1500.0},"annotated_formula":"divide(1500, subtract(const_1, divide(25, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n1_500 \/ (3\/4)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":1259} +{"problem":"if the sides of a triangle are 52 cm , 48 cm and 20 cm , what is its area ?","rationale":"\"the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 \/ 2 * 48 * 20 = 480 cm 2 answer : a\"","correct":"a","options":{"a":"480 cm 2 ","b":"765 cm 2 ","c":"216 cm 2 ","d":"197 cm 2","e":"275 cm 2"},"options_float":{"a":480.0,"b":765.0,"c":216.0,"d":197.0,"e":275.0},"annotated_formula":"divide(multiply(48, 20), const_2)","linear_formula":"multiply(n1,n2)|divide(#0,const_2)|","chain":"48 * 20<\/gadget>\n960<\/output>\n960 \/ 2<\/gadget>\n480<\/output>\n480<\/result>","index":1260} +{"problem":"5 % 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 3443 , how much was it in the beginning ?","rationale":"\"x * ( 95 \/ 100 ) * ( 85 \/ 100 ) = 3443 x = 4264 answer : a\"","correct":"a","options":{"a":"4264 ","b":"2776 ","c":"4400 ","d":"2871","e":"881"},"options_float":{"a":4264.0,"b":2776.0,"c":4400.0,"d":2871.0,"e":881.0},"annotated_formula":"floor(divide(3443, multiply(divide(subtract(const_100, 5), 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 - 5<\/gadget>\n95<\/output>\n95 \/ 100<\/gadget>\n19\/20 = around 0.95<\/output>\n100 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/output>\n(19\/20) * (17\/20)<\/gadget>\n323\/400 = around 0.8075<\/output>\n3_443 \/ (323\/400)<\/gadget>\n1_377_200\/323 = around 4_263.77709<\/output>\nfloor(1_377_200\/323)<\/gadget>\n4_263<\/output>\n4_263<\/result>","index":1261} +{"problem":"8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 6 men , 4 women and 10 children together complete the piece of work ?","rationale":"\"explanation : 1 man ’ s 1 day ’ s work = 1 \/ 8 × 12 = 1 \/ 96 6 men ’ s 1 day ’ s work = 1 × 6 \/ 96 = 1 \/ 16 1 woman ’ s 1 day ’ s work = 1 \/ 192 4 women ’ s 1 day ’ s work = 1 \/ 192 × 4 = 1 \/ 48 1 child ’ s 1 day ’ s work = 1 \/ 240 10 children ’ s 1 day ’ s work = 1 \/ 24 therefore , ( 6 men + 4 women + 10 children ) ’ s 1 day ’ s work = 1 \/ 16 + 1 \/ 48 + 1 \/ 24 = 1 \/ 8 the required no . of days = 8 days answer : option d\"","correct":"d","options":{"a":"5 days ","b":"15 days ","c":"28 days ","d":"8 days","e":"7 days"},"options_float":{"a":5.0,"b":15.0,"c":28.0,"d":8.0,"e":7.0},"annotated_formula":"inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 6), multiply(inverse(multiply(48, 4)), 4))))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n6,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10)|","chain":"24 * 10<\/gadget>\n240<\/output>\n1 \/ 240<\/gadget>\n1\/240 = around 0.004167<\/output>\n10 * (1\/240)<\/gadget>\n1\/24 = around 0.041667<\/output>\n12 * 8<\/gadget>\n96<\/output>\n1 \/ 96<\/gadget>\n1\/96 = around 0.010417<\/output>\n(1\/96) * 6<\/gadget>\n1\/16 = around 0.0625<\/output>\n48 * 4<\/gadget>\n192<\/output>\n1 \/ 192<\/gadget>\n1\/192 = around 0.005208<\/output>\n(1\/192) * 4<\/gadget>\n1\/48 = around 0.020833<\/output>\n(1\/16) + (1\/48)<\/gadget>\n1\/12 = around 0.083333<\/output>\n(1\/24) + (1\/12)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n8<\/result>","index":1263} +{"problem":"if henry were to add 6 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 = 6 galls 1 \/ 8 * x = 6 gallons x = 48 gallons answer b\"","correct":"b","options":{"a":"25 ","b":"48 ","c":"64 ","d":"80","e":"96"},"options_float":{"a":25.0,"b":48.0,"c":64.0,"d":80.0,"e":96.0},"annotated_formula":"multiply(6, 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>\n6 * 8<\/gadget>\n48<\/output>\n48<\/result>","index":1264} +{"problem":"in what time will a railway train 120 m long moving at the rate of 70 kmph pass a telegraph post on its way ?","rationale":"\"t = 120 \/ 70 * 18 \/ 5 = 6 sec answer : d\"","correct":"d","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(120, multiply(70, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n70 * (5\/18)<\/gadget>\n175\/9 = around 19.444444<\/output>\n120 \/ (175\/9)<\/gadget>\n216\/35 = around 6.171429<\/output>\n216\/35 = around 6.171429<\/result>","index":1265} +{"problem":"there are 400 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 396 managers and 4 others . the 4 others would compose 2 % of the total number of people if there were 200 people in the room . thus 200 managers must leave . the answer is d .\"","correct":"d","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(subtract(multiply(400, divide(99, const_100)), multiply(400, 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>\n400 * (99\/100)<\/gadget>\n396<\/output>\n98 \/ 100<\/gadget>\n49\/50 = around 0.98<\/output>\n400 * (49\/50)<\/gadget>\n392<\/output>\n396 - 392<\/gadget>\n4<\/output>\n1 - (49\/50)<\/gadget>\n1\/50 = around 0.02<\/output>\n4 \/ (1\/50)<\/gadget>\n200<\/output>\n200<\/result>","index":1266} +{"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 97 % 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 97 % water and 3 % of non - water , so now 1 pound of non - water composes 3 % of cucucmbers , which means that the new weight of cucumbers is 1 \/ 0.03 = 34 pounds . answer : b .\"","correct":"b","options":{"a":"2 ","b":"33 ","c":"92 ","d":"96","e":"98"},"options_float":{"a":2.0,"b":33.0,"c":92.0,"d":96.0,"e":98.0},"annotated_formula":"multiply(divide(subtract(100, 99), subtract(100, 97)), 100)","linear_formula":"subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|","chain":"100 - 99<\/gadget>\n1<\/output>\n100 - 97<\/gadget>\n3<\/output>\n1 \/ 3<\/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":1267} +{"problem":"a light flashes every 5 seconds , how many times will it flash in ? of an hour ?","rationale":"\"1 flash = 5 sec for 1 min = 12 flashes so for 1 hour = 12 * 60 = 720 flashes . answer : a\"","correct":"a","options":{"a":"720 ","b":"600 ","c":"650 ","d":"700","e":"750"},"options_float":{"a":720.0,"b":600.0,"c":650.0,"d":700.0,"e":750.0},"annotated_formula":"divide(const_3600, 5)","linear_formula":"divide(const_3600,n0)|","chain":"3_600 \/ 5<\/gadget>\n720<\/output>\n720<\/result>","index":1269} +{"problem":"a 6 litre sol is 40 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?","rationale":"\"40 % of 6 = 2.4 50 % of 6 = 3 shortage is 0.6 so we need to have 0.6 \/ 50 % to get 50 % alcohol content . = 1.2 b\"","correct":"b","options":{"a":"a . 0.6 ","b":"b . 1.2 ","c":"c . 2.1 ","d":"d . 3","e":"e . 5.4"},"options_float":{"a":0.6,"b":1.2,"c":2.1,"d":3.0,"e":5.4},"annotated_formula":"subtract(6, multiply(const_2, multiply(divide(40, const_100), 6)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 6<\/gadget>\n12\/5 = around 2.4<\/output>\n2 * (12\/5)<\/gadget>\n24\/5 = around 4.8<\/output>\n6 - (24\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n6\/5 = around 1.2<\/result>","index":1271} +{"problem":"in arun ' s opinion , his weight is greater than 63 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 66 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ?","rationale":"\"let arun ' s weight be x kg . according to arun , 63 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 66 . the values satisfying all the above conditions are 64 and 65 . required average = ( 64 + 65 ) \/ 2 = 64.5 kg answer : b\"","correct":"b","options":{"a":"86.5 kg ","b":"64.5 kg ","c":"46.5 kg ","d":"26.5 kg","e":"16.5 kg"},"options_float":{"a":86.5,"b":64.5,"c":46.5,"d":26.5,"e":16.5},"annotated_formula":"divide(add(66, add(63, const_1)), const_2)","linear_formula":"add(n0,const_1)|add(n4,#0)|divide(#1,const_2)|","chain":"63 + 1<\/gadget>\n64<\/output>\n66 + 64<\/gadget>\n130<\/output>\n130 \/ 2<\/gadget>\n65<\/output>\n65<\/result>","index":1274} +{"problem":"a cat leaps 5 leaps for every 4 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ?","rationale":"given ; 2 dog = 3 cat ; or , dog \/ cat = 3 \/ 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 5 \/ 3 * 4 = 5 : 6 . ' ' answer : 5 : 6 ;","correct":"a","options":{"a":"5 : 6 ","b":"3 : 2 ","c":"4 : 8 ","d":"1 : 2","e":"7 : 8"},"options_float":{"a":0.8333333333,"b":1.5,"c":0.5,"d":0.5,"e":0.875},"annotated_formula":"divide(multiply(divide(2, 3), 5), 4)","linear_formula":"divide(n2,n3)|multiply(n0,#0)|divide(#1,n1)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 5<\/gadget>\n10\/3 = around 3.333333<\/output>\n(10\/3) \/ 4<\/gadget>\n5\/6 = around 0.833333<\/output>\n5\/6 = around 0.833333<\/result>","index":1275} +{"problem":"the percentage profit earned by selling an article for rs . 1920 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 15 % profit ?","rationale":"\"let c . p . be rs . x . then , ( 1920 - x ) \/ x * 100 = ( x - 1280 ) \/ x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 115 % of rs . 1600 = 115 \/ 100 * 1600 = rs . 1840 . answer : c\"","correct":"c","options":{"a":"2000 ","b":"2778 ","c":"1840 ","d":"2778","e":"2771"},"options_float":{"a":2000.0,"b":2778.0,"c":1840.0,"d":2778.0,"e":2771.0},"annotated_formula":"multiply(divide(add(const_100, 15), const_100), divide(add(1920, 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 + 15<\/gadget>\n115<\/output>\n115 \/ 100<\/gadget>\n23\/20 = around 1.15<\/output>\n1_920 + 1_280<\/gadget>\n3_200<\/output>\n3_200 \/ 2<\/gadget>\n1_600<\/output>\n(23\/20) * 1_600<\/gadget>\n1_840<\/output>\n1_840<\/result>","index":1276} +{"problem":"rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 125 . how much was lent at 5 % ?","rationale":"\"( x * 5 * 1 ) \/ 100 + [ ( 2500 - x ) * 6 * 1 ] \/ 100 = 125 x = 2500 answer : e\"","correct":"e","options":{"a":"2333 ","b":"2777 ","c":"2688 ","d":"1000","e":"2500"},"options_float":{"a":2333.0,"b":2777.0,"c":2688.0,"d":1000.0,"e":2500.0},"annotated_formula":"divide(subtract(125, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|multiply(n0,n2)|divide(#2,const_100)|subtract(#0,#1)|subtract(n3,#3)|divide(#5,#4)|","chain":"6 * 2_500<\/gadget>\n15_000<\/output>\n15_000 \/ 100<\/gadget>\n150<\/output>\n125 - 150<\/gadget>\n-25<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n(1\/20) - (3\/50)<\/gadget>\n-1\/100 = around -0.01<\/output>\n(-25) \/ (-1\/100)<\/gadget>\n2_500<\/output>\n2_500<\/result>","index":1278} +{"problem":"surface area of two spheres are in the ratio 1 : 4 what is the ratio of their volumes ?","rationale":"1 : 8 answer : b","correct":"b","options":{"a":"1 : 9 ","b":"1 : 8 ","c":"1 : 3 ","d":"1 : 4","e":"1 : 5"},"options_float":{"a":0.1111111111,"b":0.125,"c":0.3333333333,"d":0.25,"e":0.2},"annotated_formula":"power(sqrt(divide(1, 4)), const_3)","linear_formula":"divide(n0,n1)|sqrt(#0)|power(#1,const_3)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) ** (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 3<\/gadget>\n1\/8 = around 0.125<\/output>\n1\/8 = around 0.125<\/result>","index":1281} +{"problem":"if the area of a circle decreases by 30 % , then the radius of a circle decreases by","rationale":"\"if area of a circle decreased by x % then the radius of a circle decreases by ( 100 − 10 √ 100 − x ) % = ( 100 − 10 √ 100 − 30 ) % = ( 100 − 10 √ 70 ) % = 100 - 84 = 16 % answer b\"","correct":"b","options":{"a":"20 % ","b":"16 % ","c":"36 % ","d":"64 %","e":"none of these"},"options_float":{"a":20.0,"b":16.0,"c":36.0,"d":64.0,"e":null},"annotated_formula":"multiply(subtract(const_1, sqrt(divide(subtract(const_100, 30), const_100))), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,const_100)|sqrt(#1)|subtract(const_1,#2)|multiply(#3,const_100)|","chain":"100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) ** (1\/2)<\/gadget>\nsqrt(70)\/10 = around 0.83666<\/output>\n1 - (sqrt(70)\/10)<\/gadget>\n1 - sqrt(70)\/10 = around 0.16334<\/output>\n(1 - sqrt(70)\/10) * 100<\/gadget>\n100 - 10*sqrt(70) = around 16.333997<\/output>\n100 - 10*sqrt(70) = around 16.333997<\/result>","index":1282} +{"problem":"the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 25 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ?","rationale":"\"let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 25 = 2 r a = 12 \/ 5 r and r = 75 a = 180 required total cost = 4 * 180 + 3 * 75 + 5 * 25 = 720 + 225 + 125 = $ 1070 c\"","correct":"c","options":{"a":"347 ","b":"987 ","c":"1070 ","d":"1371","e":"1667"},"options_float":{"a":347.0,"b":987.0,"c":1070.0,"d":1371.0,"e":1667.0},"annotated_formula":"add(add(multiply(4, multiply(divide(24, 10), divide(multiply(25, 6), 2))), multiply(3, divide(multiply(25, 6), 2))), multiply(5, 25))","linear_formula":"divide(n1,n0)|multiply(n2,n4)|multiply(n4,n7)|divide(#1,n3)|multiply(#0,#3)|multiply(n6,#3)|multiply(n5,#4)|add(#6,#5)|add(#7,#2)|","chain":"24 \/ 10<\/gadget>\n12\/5 = around 2.4<\/output>\n25 * 6<\/gadget>\n150<\/output>\n150 \/ 2<\/gadget>\n75<\/output>\n(12\/5) * 75<\/gadget>\n180<\/output>\n4 * 180<\/gadget>\n720<\/output>\n3 * 75<\/gadget>\n225<\/output>\n720 + 225<\/gadget>\n945<\/output>\n5 * 25<\/gadget>\n125<\/output>\n945 + 125<\/gadget>\n1_070<\/output>\n1_070<\/result>","index":1284} +{"problem":"in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 40 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ?","rationale":"\"percent of students who are 25 years old or older is 0.4 * 48 + 0.2 * 52 = ~ 30 , so percent of people who are less than 25 years old is 100 - 30 = 70 . answer : b .\"","correct":"b","options":{"a":"0.9 ","b":"0.7 ","c":"0.45 ","d":"0.3","e":"0.25"},"options_float":{"a":0.9,"b":0.7,"c":0.45,"d":0.3,"e":0.25},"annotated_formula":"subtract(const_1, multiply(divide(40, const_100), divide(subtract(const_100, 20), const_100)))","linear_formula":"divide(n2,const_100)|subtract(const_100,n3)|divide(#1,const_100)|multiply(#0,#2)|subtract(const_1,#3)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(2\/5) * (4\/5)<\/gadget>\n8\/25 = around 0.32<\/output>\n1 - (8\/25)<\/gadget>\n17\/25 = around 0.68<\/output>\n17\/25 = around 0.68<\/result>","index":1285} +{"problem":"a train 300 m long , running with a speed of 54 km \/ hr will pass a tree in ?","rationale":"\"speed = 54 * 5 \/ 18 = 15 m \/ sec time taken = 300 * 1 \/ 15 = 20 sec answer : c\"","correct":"c","options":{"a":"17 sec ","b":"16 sec ","c":"20 sec ","d":"14 sec","e":"12 sec"},"options_float":{"a":17.0,"b":16.0,"c":20.0,"d":14.0,"e":12.0},"annotated_formula":"multiply(divide(300, multiply(54, const_1000)), const_3600)","linear_formula":"multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|","chain":"54 * 1_000<\/gadget>\n54_000<\/output>\n300 \/ 54_000<\/gadget>\n1\/180 = around 0.005556<\/output>\n(1\/180) * 3_600<\/gadget>\n20<\/output>\n20<\/result>","index":1286} +{"problem":"a student got twice as many sums wrong as he got right . if he attempted 27 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 = 27 3 x = 27 x = 9 . answer : d\"","correct":"d","options":{"a":"12 ","b":"16 ","c":"18 ","d":"9","e":"12"},"options_float":{"a":12.0,"b":16.0,"c":18.0,"d":9.0,"e":12.0},"annotated_formula":"divide(27, add(const_1, const_2))","linear_formula":"add(const_1,const_2)|divide(n0,#0)|","chain":"1 + 2<\/gadget>\n3<\/output>\n27 \/ 3<\/gadget>\n9<\/output>\n9<\/result>","index":1288} +{"problem":"the compound interest on a sum for 2 years is rs . 832 and the simple interest on the same sum for the same period is rs . 800 . the difference between the compound and simple interest for 3 years will be","rationale":"explanation : given that simple interest for 2 years is rs . 800 i . e . , simple interest for 1 st year is rs . 400 and simple interest for 2 nd year is also rs . 400 compound interest for 1 st year will be 400 and compound interest for 2 nd year will be 832 - 400 = 432 you can see that compound interest for 2 nd year is more than simple interest for 2 nd year by 432 - 400 = rs . 32 i . e , rs . 32 is the interest obtained for rs . 400 for 1 year rate , r = 100 × si \/ pt = ( 100 × 32 ) \/ ( 400 × 1 ) = 8 % difference between compound and simple interest for the 3 rd year = simple interest obtained for rs . 832 = prt \/ 100 = ( 832 × 8 × 1 ) \/ 100 = rs . 66.56 total difference between the compound and simple interest for 3 years = 32 + 66.56 = rs . 98.56 answer : option b","correct":"b","options":{"a":"rs . 48 ","b":"rs . 98.56 ","c":"rs . 66.56 ","d":"rs . 66.58","e":"none of these"},"options_float":{"a":48.0,"b":98.56,"c":66.56,"d":66.58,"e":null},"annotated_formula":"add(subtract(832, 800), multiply(832, divide(subtract(832, 800), divide(800, 2))))","linear_formula":"divide(n2,n0)|subtract(n1,n2)|divide(#1,#0)|multiply(n1,#2)|add(#3,#1)","chain":"832 - 800<\/gadget>\n32<\/output>\n800 \/ 2<\/gadget>\n400<\/output>\n32 \/ 400<\/gadget>\n2\/25 = around 0.08<\/output>\n832 * (2\/25)<\/gadget>\n1_664\/25 = around 66.56<\/output>\n32 + (1_664\/25)<\/gadget>\n2_464\/25 = around 98.56<\/output>\n2_464\/25 = around 98.56<\/result>","index":1289} +{"problem":"the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8463 for a computer table , then what was the cost price of the computer table ?","rationale":"\": cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8463 ( 100 \/ 124 ) = rs . 6825 . answer : b\"","correct":"b","options":{"a":"6727 ","b":"6825 ","c":"6728 ","d":"6725","e":"2871"},"options_float":{"a":6727.0,"b":6825.0,"c":6728.0,"d":6725.0,"e":2871.0},"annotated_formula":"divide(8463, add(const_1, divide(24, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n1 + (6\/25)<\/gadget>\n31\/25 = around 1.24<\/output>\n8_463 \/ (31\/25)<\/gadget>\n6_825<\/output>\n6_825<\/result>","index":1291} +{"problem":"15 beavers , working together in a constant pace , can build a dam in 4 hours . how many hours will it take 20 beavers that work at the same pace , to build the same dam ?","rationale":"\"total work = 15 * 4 = 60 beaver hours 20 beaver * x = 60 beaver hours x = 60 \/ 20 = 3 answer : e\"","correct":"e","options":{"a":"2 . ","b":"4 . ","c":"5 . ","d":"6","e":"3 ."},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":3.0},"annotated_formula":"divide(multiply(4, 15), 20)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"4 * 15<\/gadget>\n60<\/output>\n60 \/ 20<\/gadget>\n3<\/output>\n3<\/result>","index":1292} +{"problem":"the sum q of prime numbers that are greater than 60 but less than 70 is","rationale":"a prime number is a number that has only two factors : 1 and itself . therefore , a prime number is divisible by two numbers only . let ' s list the numbers from 61 to 69 . 61 , 62 , 63 , 64 , 65 , 66 , 67 , 68 , 69 immediately we can eliminate the even numbers because they are divisible by 2 and thus are not prime . we are now left with : 61 , 63 , 65 , 67 , 69 we can next eliminate 65 because 65 is a multiple of 5 . we are now left with 61 , 63 , 67 , 69 . to eliminate any remaining values , we would look at those that are multiples of 3 . if you don ’ t know an easy way to do this , just start with a number that is an obvious multiple of 3 , such as 60 , and then keep adding 3 . we see that 60 , 63 , 66 , 69 are all multiples of 3 and therefore are not prime . thus , we can eliminate 63 and 69 from the list because they are not prime . finally , we are left with 61 and 67 , and we must determine whether they are divisible by 7 . they are not , and therefore they must be both prime . thus , the sum q of 61 and 67 is 128 . answer b .","correct":"b","options":{"a":"67 ","b":"128 ","c":"191 ","d":"197","e":"260"},"options_float":{"a":67.0,"b":128.0,"c":191.0,"d":197.0,"e":260.0},"annotated_formula":"add(add(60, const_1), subtract(70, const_3))","linear_formula":"add(n0,const_1)|subtract(n1,const_3)|add(#0,#1)","chain":"60 + 1<\/gadget>\n61<\/output>\n70 - 3<\/gadget>\n67<\/output>\n61 + 67<\/gadget>\n128<\/output>\n128<\/result>","index":1294} +{"problem":"the probability that a man will be alive for 10 more yrs is 1 \/ 2 & the probability that his wife will alive for 10 more yrs is 1 \/ 3 . the probability that none of them will be alive for 10 more yrs , is","rationale":"\"sol . required probability = pg . ) x p ( b ) = ( 1 — d x ( 1 — i ) = : x 1 = 1 \/ 3 ans . ( c )\"","correct":"c","options":{"a":"1 \/ 2 ","b":"1 ","c":"1 \/ 3 ","d":"3 \/ 4","e":"2"},"options_float":{"a":0.5,"b":1.0,"c":0.3333333333,"d":0.75,"e":2.0},"annotated_formula":"multiply(subtract(1, divide(1, 2)), subtract(1, divide(1, 3)))","linear_formula":"divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1\/2) * (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":1297} +{"problem":"if a * b = 2 a - 3 b + ab , then 3 * 5 + 5 * 3 is equal to","rationale":"3 * 5 + 5 * 3 = ( 2 x 3 - 3 x 5 + 3 x 5 ) + ( 2 x 5 - 3 x 3 + 5 x 3 ) = 22 answer a 22","correct":"a","options":{"a":"22 ","b":"25 ","c":"26 ","d":"28","e":"23"},"options_float":{"a":22.0,"b":25.0,"c":26.0,"d":28.0,"e":23.0},"annotated_formula":"add(multiply(2, 3), multiply(3, 5))","linear_formula":"multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)","chain":"2 * 3<\/gadget>\n6<\/output>\n3 * 5<\/gadget>\n15<\/output>\n6 + 15<\/gadget>\n21<\/output>\n21<\/result>","index":1298} +{"problem":"if 25 ^ 5 × 5 ^ ( - 1 ) = ( 125 ) ^ x , then what is the value of x ?","rationale":"25 ^ 5 × 5 ^ ( - 1 ) = ( 125 ) ^ x ( 5 ^ 2 ) ^ 5 × 5 ^ ( - 1 ) = 5 ^ 3 x 5 ^ 10 x 5 ^ ( - 1 ) = 5 ^ 3 x ; since all of the bases are the same now , we can equate the exponents in the next step 10 - 1 = 3 x 9 = 3 x x = 3 ans . b ) 3","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":"divide(subtract(multiply(const_2, 5), 1), const_3)","linear_formula":"multiply(n1,const_2)|subtract(#0,n3)|divide(#1,const_3)","chain":"2 * 5<\/gadget>\n10<\/output>\n10 - 1<\/gadget>\n9<\/output>\n9 \/ 3<\/gadget>\n3<\/output>\n3<\/result>","index":1300} +{"problem":"a lady starts from p towards q and realizes that at a point r , if he walks 50 km further he will be at a point s , which is as far away from r as it is from q . what is the distance between p and q if the distance between p and r is half the distance from r to q ? ( assume that p , q , r and s are all on the same straight line )","rationale":"p ___ 50 _____ r ___ 50 _____ s ____ 50 ___ q the above figure gives the locations of p , r , s & q in relation to each other . answer : a","correct":"a","options":{"a":"150 km ","b":"200 km ","c":"250 km ","d":"125 km","e":"155 km"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":125.0,"e":155.0},"annotated_formula":"add(multiply(50, const_2), divide(multiply(50, const_2), const_2))","linear_formula":"multiply(n0,const_2)|divide(#0,const_2)|add(#1,#0)","chain":"50 * 2<\/gadget>\n100<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n100 + 50<\/gadget>\n150<\/output>\n150<\/result>","index":1301} +{"problem":"in kaya ' s teacher ' s desk there are 24 pink highlighters , 28 yellow highlighters , and 25 blue highlighters . how many highlighters are there in all ?","rationale":"\"add the numbers of highlighters . 24 + 28 + 25 = 77 . answer is c .\"","correct":"c","options":{"a":"11 ","b":"22 ","c":"77 ","d":"33","e":"88"},"options_float":{"a":11.0,"b":22.0,"c":77.0,"d":33.0,"e":88.0},"annotated_formula":"add(add(24, 28), 25)","linear_formula":"add(n0,n1)|add(n2,#0)|","chain":"24 + 28<\/gadget>\n52<\/output>\n52 + 25<\/gadget>\n77<\/output>\n77<\/result>","index":1303} +{"problem":"a train speeds past a pole in 10 seconds and a platform 50 m long in 20 seconds . its length is :","rationale":"\"let the length of the train be x meters and its speed be y m \/ sec . they , x \/ y = 10 = > y = x \/ 10 x + 50 \/ 20 = x \/ 10 x = 50 m . answer : option d\"","correct":"d","options":{"a":"30 m . ","b":"40 m . ","c":"60 m . ","d":"50 m .","e":"70 m ."},"options_float":{"a":30.0,"b":40.0,"c":60.0,"d":50.0,"e":70.0},"annotated_formula":"multiply(50, subtract(const_2, const_1))","linear_formula":"subtract(const_2,const_1)|multiply(n1,#0)|","chain":"2 - 1<\/gadget>\n1<\/output>\n50 * 1<\/gadget>\n50<\/output>\n50<\/result>","index":1304} +{"problem":"the denominator of a fraction is 6 greater than the numerator . if the numerator and the denominator are increased by 1 , the resulting fraction is equal to 4 ⠁ „ 5 . what is the value of the original fraction ?","rationale":"\"let the numerator be x . then the denominator is x + 6 . x + 1 \/ x + 7 = 4 \/ 5 . 5 x + 5 = 4 x + 28 . x = 23 . the original fraction is 23 \/ 29 . the answer is b .\"","correct":"b","options":{"a":"17 \/ 23 ","b":"23 \/ 29 ","c":"29 \/ 35 ","d":"31 \/ 37","e":"33 \/ 39"},"options_float":{"a":0.7391304348,"b":0.7931034483,"c":0.8285714286,"d":0.8378378378,"e":0.8461538462},"annotated_formula":"divide(divide(subtract(multiply(4, add(1, 6)), 5), subtract(5, 4)), add(divide(subtract(multiply(4, add(1, 6)), 5), subtract(5, 4)), 6))","linear_formula":"add(n0,n1)|subtract(n3,n2)|multiply(n2,#0)|subtract(#2,n3)|divide(#3,#1)|add(n0,#4)|divide(#4,#5)|","chain":"1 + 6<\/gadget>\n7<\/output>\n4 * 7<\/gadget>\n28<\/output>\n28 - 5<\/gadget>\n23<\/output>\n5 - 4<\/gadget>\n1<\/output>\n23 \/ 1<\/gadget>\n23<\/output>\n23 + 6<\/gadget>\n29<\/output>\n23 \/ 29<\/gadget>\n23\/29 = around 0.793103<\/output>\n23\/29 = around 0.793103<\/result>","index":1307} +{"problem":"consider the sets tn = { n , n + 1 , n + 2 , n + 3 , n + 4 ) , where n = 1 , 2 , 3 , … , 96 . how many of these sets contain 6 or any integral multiple thereof ( i . e . , any one of the numbers 6 , 12 , 18 , … ) ?","rationale":"explanation : if n = 1 , then the set t 1 = { 1 , 23 , 45 } , and it does not have 6 or any multiples . n = 2 to n = 6 has 6 in the set . n = 7 , has the set t 7 = { 7 , 89 , 1011 } , and no 6 or multiples . so 1 in every 6 members do not have 6 or multiples of 6 . so , till n = 96 , there are 16 sets of “ 6 members ” ( 16 * 6 = 96 ) and 16 sets do not have 6 or its multiples , while the remaining 80 sets have . answer : a","correct":"a","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":"multiply(divide(add(const_2, const_3), 6), 96)","linear_formula":"add(const_2,const_3)|divide(#0,n8)|multiply(n7,#1)","chain":"2 + 3<\/gadget>\n5<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(5\/6) * 96<\/gadget>\n80<\/output>\n80<\/result>","index":1308} +{"problem":"car z travels 55 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 45 miles \/ hour , the car will run 55 miles \/ gallon and at 60 miles \/ hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 44 miles \/ gallon at a constant speed of 60 miles \/ gallon . with 10 gallons of fuel , the car can therefore travel 44 miles \/ gallon * 10 gallons = 440 miles . answer e .","correct":"e","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)), 55), 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) * 55<\/gadget>\n44<\/output>\n44 * 10<\/gadget>\n440<\/output>\n440<\/result>","index":1309} +{"problem":"a clock shows the time as 9 a . m . if the minute hand gains 6 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 * 6 = 54 minutes . answer : e\"","correct":"e","options":{"a":"30 min ","b":"35 min ","c":"45 min ","d":"50 min","e":"54 min"},"options_float":{"a":30.0,"b":35.0,"c":45.0,"d":50.0,"e":54.0},"annotated_formula":"multiply(add(const_3, 6), 6)","linear_formula":"add(const_3,n2)|multiply(n1,#0)|","chain":"3 + 6<\/gadget>\n9<\/output>\n9 * 6<\/gadget>\n54<\/output>\n54<\/result>","index":1312} +{"problem":"20 people went to a hotel for combine dinner party 12 of them spent rs . 70 each on their dinner and rest spent 4 more than the average expenditure of all the 20 . what was the total money spent by them .","rationale":"\"solution : let average expenditure of 20 people be x . then , 20 x = 12 * 70 + 8 * ( x + 4 ) ; or , 20 x = 12 * 70 + 8 x + 32 ; or , x = 72.667 ; so , total money spent = 72.67 * 20 = rs . 1453.4 . answer : option c\"","correct":"c","options":{"a":"1628.4 ","b":"1534 ","c":"1453 ","d":"1496","e":"none of these"},"options_float":{"a":1628.4,"b":1534.0,"c":1453.0,"d":1496.0,"e":null},"annotated_formula":"multiply(divide(add(multiply(12, 70), multiply(subtract(20, 12), 4)), subtract(20, subtract(20, 12))), 20)","linear_formula":"multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|subtract(n0,#1)|add(#0,#2)|divide(#4,#3)|multiply(n0,#5)|","chain":"12 * 70<\/gadget>\n840<\/output>\n20 - 12<\/gadget>\n8<\/output>\n8 * 4<\/gadget>\n32<\/output>\n840 + 32<\/gadget>\n872<\/output>\n20 - 8<\/gadget>\n12<\/output>\n872 \/ 12<\/gadget>\n218\/3 = around 72.666667<\/output>\n(218\/3) * 20<\/gadget>\n4_360\/3 = around 1_453.333333<\/output>\n4_360\/3 = around 1_453.333333<\/result>","index":1313} +{"problem":"what annual payment will discharge a debt of rs . 1060 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 = 1060 820 x + 1060 * 441 x = 570.07 so , value of each installment = rs . 570.07 answer : option c\"","correct":"c","options":{"a":"993.2 ","b":"551.25 ","c":"570.07 ","d":"543.33","e":"646.33"},"options_float":{"a":993.2,"b":551.25,"c":570.07,"d":543.33,"e":646.33},"annotated_formula":"divide(multiply(power(add(divide(5, const_100), const_1), 2), 1060), 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_060<\/gadget>\n23_373\/20 = around 1_168.65<\/output>\n(23_373\/20) \/ 2<\/gadget>\n23_373\/40 = around 584.325<\/output>\n23_373\/40 = around 584.325<\/result>","index":1316} +{"problem":"50 % of the population of a village is 23040 . the total population of the village is ?","rationale":"\"answer ∵ 50 % of p = 23040 ∴ p = ( 23040 x 100 ) \/ 50 = 46080 correct option : d\"","correct":"d","options":{"a":"32256 ","b":"24000 ","c":"44936 ","d":"46080","e":"none"},"options_float":{"a":32256.0,"b":24000.0,"c":44936.0,"d":46080.0,"e":null},"annotated_formula":"multiply(divide(const_100, 50), 23040)","linear_formula":"divide(const_100,n0)|multiply(n1,#0)|","chain":"100 \/ 50<\/gadget>\n2<\/output>\n2 * 23_040<\/gadget>\n46_080<\/output>\n46_080<\/result>","index":1317} +{"problem":"if the perimeter of a rectangular house is 1400 m , its length when its breadth is 300 m is ?","rationale":"2 ( l + 300 ) = 1400 = > l = 400 m answer : b","correct":"b","options":{"a":"300 ","b":"400 ","c":"500 ","d":"600","e":"700"},"options_float":{"a":300.0,"b":400.0,"c":500.0,"d":600.0,"e":700.0},"annotated_formula":"subtract(divide(1400, const_2), 300)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"1_400 \/ 2<\/gadget>\n700<\/output>\n700 - 300<\/gadget>\n400<\/output>\n400<\/result>","index":1318} +{"problem":"if a # b = ab – b + b ^ 2 , then 3 # 4 =","rationale":"\"solution - simply substitute 3 and 4 in equation in the place of a and b respectively . 3 # 4 = 3 * 4 - 4 + 4 ^ 2 = 12 - 4 + 16 = 24 . ans d\"","correct":"d","options":{"a":"2 ","b":"8 ","c":"15 ","d":"24","e":"35"},"options_float":{"a":2.0,"b":8.0,"c":15.0,"d":24.0,"e":35.0},"annotated_formula":"add(subtract(multiply(3, 4), 4), power(4, 2))","linear_formula":"multiply(n1,n2)|power(n2,n0)|subtract(#0,n2)|add(#1,#2)|","chain":"3 * 4<\/gadget>\n12<\/output>\n12 - 4<\/gadget>\n8<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n8 + 16<\/gadget>\n24<\/output>\n24<\/result>","index":1320} +{"problem":"what number has a 4 : 1 ratio to the number 100 ?","rationale":"\"4 : 1 = x : 100 x = 4 * 100 x = 400 answer : c\"","correct":"c","options":{"a":"40 ","b":"200 ","c":"400 ","d":"800","e":"4000"},"options_float":{"a":40.0,"b":200.0,"c":400.0,"d":800.0,"e":4000.0},"annotated_formula":"multiply(100, 4)","linear_formula":"multiply(n0,n2)|","chain":"100 * 4<\/gadget>\n400<\/output>\n400<\/result>","index":1321} +{"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 5535 . what was the number of original inhabitants ?","rationale":"\"let the total number of original inhabitants be x . ( 75 \/ 100 ) * ( 90 \/ 100 ) * x = 5535 ( 27 \/ 40 ) * x = 5535 x = 5535 * 40 \/ 27 = 8200 the answer is b .\"","correct":"b","options":{"a":"7900 ","b":"8200 ","c":"8500 ","d":"8800","e":"9100"},"options_float":{"a":7900.0,"b":8200.0,"c":8500.0,"d":8800.0,"e":9100.0},"annotated_formula":"divide(5535, 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_535 \/ (27\/40)<\/gadget>\n8_200<\/output>\n8_200<\/result>","index":1322} +{"problem":"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 42 , then how old is b ?","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 = 42 5 x = 40 = > x = 8 hence , b ' s age = 2 x = 16 years . answer : a\"","correct":"a","options":{"a":"16 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":16.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"multiply(divide(subtract(42, const_2), add(const_3, const_2)), const_2)","linear_formula":"add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|","chain":"42 - 2<\/gadget>\n40<\/output>\n3 + 2<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16<\/result>","index":1323} +{"problem":"last year the range of the annual bonus of the 100 employees at company x was $ 20000 . if the annual bonus of each of the 100 employees this year is 10 percent greater than it was last year , what is the range of the annual bonus of the 100 employees this year ?","rationale":"let the lowest bonus be x . therefore , highest bonus is x + 20000 . now bonus of each employee is increased by 10 % . therefore the bonus will remain arranged in the same order as before . or lowest bonus = 1.1 x and highest = 1.1 * ( x + 20000 ) or range = highest - lowest = 1.1 * ( x + 20000 ) - 1.1 x = 22000 , hence , b","correct":"b","options":{"a":"$ 27000 ","b":"$ 22000 ","c":"$ 33000 ","d":"$ 16000","e":"$ 43000"},"options_float":{"a":27000.0,"b":22000.0,"c":33000.0,"d":16000.0,"e":43000.0},"annotated_formula":"multiply(20000, add(const_1, divide(10, const_100)))","linear_formula":"divide(n3,const_100)|add(#0,const_1)|multiply(n1,#1)","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n20_000 * (11\/10)<\/gadget>\n22_000<\/output>\n22_000<\/result>","index":1325} +{"problem":"the list price of an article is rs . 65 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?","rationale":"\"explanation : 65 * ( 90 \/ 100 ) * ( ( 100 - x ) \/ 100 ) = 56.16 x = 4 % answer : option b\"","correct":"b","options":{"a":"3 % ","b":"4 % ","c":"7 % ","d":"8 %","e":"9 %"},"options_float":{"a":3.0,"b":4.0,"c":7.0,"d":8.0,"e":9.0},"annotated_formula":"multiply(divide(subtract(subtract(65, multiply(65, divide(10, const_100))), 56.16), subtract(65, multiply(65, 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>\n65 * (1\/10)<\/gadget>\n13\/2 = around 6.5<\/output>\n65 - (13\/2)<\/gadget>\n117\/2 = around 58.5<\/output>\n(117\/2) - 56.16<\/gadget>\n2.34<\/output>\n2.34 \/ (117\/2)<\/gadget>\n0.04<\/output>\n0.04 * 100<\/gadget>\n4<\/output>\n4<\/result>","index":1326} +{"problem":"a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 50 cents . how many cents does a solution of 1 liter of spirit and 3 liters of water cost ?","rationale":"c . 50 cents yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 50 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 25 cents . fraction of spirit is 1 \/ 2 . 25 = k * ( 1 \/ 2 ) k = 50 cost per liter = 50 * ( 1 \/ 4 ) ( 1 part spirit , 3 parts water ) cost for 4 liters = 50 * ( 1 \/ 4 ) * 4 = 50 cents d . 50 cents","correct":"d","options":{"a":"13 ","b":"33 ","c":"56 ","d":"50","e":"52"},"options_float":{"a":13.0,"b":33.0,"c":56.0,"d":50.0,"e":52.0},"annotated_formula":"multiply(multiply(50, divide(1, add(1, 3))), add(1, 3))","linear_formula":"add(n0,n4)|divide(n0,#0)|multiply(n2,#1)|multiply(#0,#2)","chain":"1 + 3<\/gadget>\n4<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n50 * (1\/4)<\/gadget>\n25\/2 = around 12.5<\/output>\n(25\/2) * 4<\/gadget>\n50<\/output>\n50<\/result>","index":1327} +{"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 $ 1004 in total compensation , how many total hours did he work that week ?","rationale":"\"for 40 hrs = 40 * 16 = 640 excess = 1004 - 640 = 364 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 364 \/ 28 = 13 total hrs = 40 + 13 = 53 answer e 53\"","correct":"e","options":{"a":"36 ","b":"40 ","c":"44 ","d":"48","e":"53"},"options_float":{"a":36.0,"b":40.0,"c":44.0,"d":48.0,"e":53.0},"annotated_formula":"add(40, divide(subtract(1004, 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_004 - 640<\/gadget>\n364<\/output>\n100 + 75<\/gadget>\n175<\/output>\n16 * 175<\/gadget>\n2_800<\/output>\n2_800 \/ 100<\/gadget>\n28<\/output>\n364 \/ 28<\/gadget>\n13<\/output>\n40 + 13<\/gadget>\n53<\/output>\n53<\/result>","index":1331} +{"problem":"you have to send 3000 grapes 1000 kilometers from grapecity to appleland . your truck can carry 1000 grapes at a time . every time you travel a kilometer towards appleland you must pay a tax of 1 grape but you pay nothing when going in the other direction ( towards grapecity ) . what is highest number of grapes you can get to appleland ?","rationale":"step one : first you want to make 3 trips of 1,000 grapes 333 kilometers . you will be left with 2,001 grapes and 667 kilometers to go . step two : next you want to take 2 trips of 1,000 grapes 500 kilometers . you will be left with 1,000 grapes and 167 kilometers to go ( you have to leave a grape behind ) . step three : finally , you travel the last 167 kilometers with one load of 1,000 grapes and are left with 833 grapes in appleland . correct answer is a ) 833","correct":"a","options":{"a":"833 ","b":"765 ","c":"665 ","d":"679","e":"874"},"options_float":{"a":833.0,"b":765.0,"c":665.0,"d":679.0,"e":874.0},"annotated_formula":"subtract(1000, subtract(subtract(1000, floor(divide(1000, const_3))), divide(1000, const_2)))","linear_formula":"divide(n1,const_3)|divide(n1,const_2)|floor(#0)|subtract(n1,#2)|subtract(#3,#1)|subtract(n1,#4)","chain":"1_000 \/ 3<\/gadget>\n1_000\/3 = around 333.333333<\/output>\nfloor(1_000\/3)<\/gadget>\n333<\/output>\n1_000 - 333<\/gadget>\n667<\/output>\n1_000 \/ 2<\/gadget>\n500<\/output>\n667 - 500<\/gadget>\n167<\/output>\n1_000 - 167<\/gadget>\n833<\/output>\n833<\/result>","index":1332} +{"problem":"the contents of a certain box consist of 14 apples and 25 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ?","rationale":"\"the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 \/ 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 25 oranges , so to get to 6 oranges , we should remove 25 - 6 = 19 oranges . answer d\"","correct":"d","options":{"a":"3 ","b":"6 ","c":"14 ","d":"19","e":"20"},"options_float":{"a":3.0,"b":6.0,"c":14.0,"d":19.0,"e":20.0},"annotated_formula":"subtract(add(14, 25), divide(14, divide(70, const_100)))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)|","chain":"14 + 25<\/gadget>\n39<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n14 \/ (7\/10)<\/gadget>\n20<\/output>\n39 - 20<\/gadget>\n19<\/output>\n19<\/result>","index":1334} +{"problem":"jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 7 to 2 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ?","rationale":"u \/ i = 7 \/ 2 i \/ b = 5 \/ 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 35 : 21 : 2 i . e . u : b = 35 : 2 answer : option e","correct":"e","options":{"a":"5 : 1 ","b":"10 : 5 ","c":"15 : 2 ","d":"20 : 2","e":"35 : 2"},"options_float":{"a":5.0,"b":2.0,"c":7.5,"d":10.0,"e":17.5},"annotated_formula":"divide(multiply(7, 5), multiply(1, 2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)","chain":"7 * 5<\/gadget>\n35<\/output>\n1 * 2<\/gadget>\n2<\/output>\n35 \/ 2<\/gadget>\n35\/2 = around 17.5<\/output>\n35\/2 = around 17.5<\/result>","index":1335} +{"problem":"how many prime numbers are between 13 \/ 3 and 83 \/ 6 ?","rationale":"\"13 \/ 3 = 4 . xxx 83 \/ 6 = 13 . xxx so we need to find prime numbers between 4 ( exclusive ) - 12 ( inclusive ) there are 2 prime numbers 7 11 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":1336} +{"problem":"john makes $ 60 a week from his job . he earns a raise andnow makes $ 80 a week . what is the % increase ?","rationale":"\"increase = ( 20 \/ 60 ) * 100 = ( 1 \/ 3 ) * 100 = 33.33 % . b\"","correct":"b","options":{"a":"15 % ","b":"33.33 % ","c":"17.8 % ","d":"19 %","e":"21 %"},"options_float":{"a":15.0,"b":33.33,"c":17.8,"d":19.0,"e":21.0},"annotated_formula":"multiply(divide(subtract(80, 60), 60), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"80 - 60<\/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":1337} +{"problem":"a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c which is at midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ?","rationale":"\"explanation : speed in downstream = ( 14 + 4 ) km \/ hr = 18 km \/ hr ; speed in upstream = ( 14 â € “ 4 ) km \/ hr = 10 km \/ hr . let the distance between a and b be x km . then , x \/ 18 + ( x \/ 2 ) \/ 10 = 19 â ‡ ” x \/ 18 + x \/ 20 = 19 â ‡ ’ x = 180 km . answer : a\"","correct":"a","options":{"a":"180 km ","b":"127 km ","c":"178 km ","d":"188 km","e":"111 km"},"options_float":{"a":180.0,"b":127.0,"c":178.0,"d":188.0,"e":111.0},"annotated_formula":"divide(19, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2))))","linear_formula":"add(n1,n2)|subtract(n2,n1)|divide(const_1,#0)|multiply(#1,const_2)|divide(const_1,#3)|add(#2,#4)|divide(n0,#5)|","chain":"14 + 4<\/gadget>\n18<\/output>\n1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n14 - 4<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/18) + (1\/20)<\/gadget>\n19\/180 = around 0.105556<\/output>\n19 \/ (19\/180)<\/gadget>\n180<\/output>\n180<\/result>","index":1339} +{"problem":"in a division sum , the quotient is 18 , the divisor 43 and the remainder 12 , find the dividend ?","rationale":"\"explanation : 18 * 43 + 12 = 786 answer : d\"","correct":"d","options":{"a":"586 ","b":"766 ","c":"796 ","d":"786","e":"686"},"options_float":{"a":586.0,"b":766.0,"c":796.0,"d":786.0,"e":686.0},"annotated_formula":"add(multiply(18, 43), 12)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"18 * 43<\/gadget>\n774<\/output>\n774 + 12<\/gadget>\n786<\/output>\n786<\/result>","index":1340} +{"problem":"niall ' s income is 60 % less than rex ' s income , and sam ' s income is 25 % less than niall ' s income . if rex gave 60 % of his income to sam and 40 % of his income to niall , niall ' s new income would be what fraction of sam ' s new income ?","rationale":"we can take some easy numbers and make calculations simpler . let r ( rex ' s income ) = 100 q ( niall ' s income ) = 40 % r = 40 s ( sam ' s income ) = 75 % q = ( 3 \/ 4 ) * 40 = 30 now , if rex gives 40 % to niall - - > q = 40 + 40 = 80 60 % given to sam - - > s = 30 + 60 = 90 the ratio is : q \/ s = 80 \/ 90 = 8 \/ 9 = a","correct":"a","options":{"a":"8 \/ 9 ","b":"11 \/ 12 ","c":"8 \/ 13 ","d":"11 \/ 13","e":"12 \/ 13"},"options_float":{"a":0.8888888889,"b":0.9166666667,"c":0.6153846154,"d":0.8461538462,"e":0.9230769231},"annotated_formula":"divide(add(40, 40), add(add(40, 40), const_10))","linear_formula":"add(n3,n3)|add(#0,const_10)|divide(#0,#1)","chain":"40 + 40<\/gadget>\n80<\/output>\n80 + 10<\/gadget>\n90<\/output>\n80 \/ 90<\/gadget>\n8\/9 = around 0.888889<\/output>\n8\/9 = around 0.888889<\/result>","index":1344} +{"problem":"if 2 a = 4 b = 10 , then 40 ab =","rationale":"2 a * 4 b = 10 * 10 = 100 8 ab = 100 i . e . 40 ab = 500 answer : option e","correct":"e","options":{"a":"50 ","b":"100 ","c":"250 ","d":"450","e":"500"},"options_float":{"a":50.0,"b":100.0,"c":250.0,"d":450.0,"e":500.0},"annotated_formula":"multiply(40, multiply(divide(10, 2), divide(10, 4)))","linear_formula":"divide(n2,n0)|divide(n2,n1)|multiply(#0,#1)|multiply(n3,#2)","chain":"10 \/ 2<\/gadget>\n5<\/output>\n10 \/ 4<\/gadget>\n5\/2 = around 2.5<\/output>\n5 * (5\/2)<\/gadget>\n25\/2 = around 12.5<\/output>\n40 * (25\/2)<\/gadget>\n500<\/output>\n500<\/result>","index":1345} +{"problem":"rahim bought 52 books for rs . 1200 from one shop and 32 books for rs . 480 from another . what is the average price he paid per book ?","rationale":"\"average price per book = ( 1200 + 480 ) \/ ( 52 + 32 ) = 1680 \/ 84 = rs . 20 answer : d\"","correct":"d","options":{"a":"s . 17 ","b":"s . 18 ","c":"s . 12 ","d":"s . 20","e":"s . 10"},"options_float":{"a":17.0,"b":18.0,"c":12.0,"d":20.0,"e":10.0},"annotated_formula":"divide(add(1200, 480), add(52, 32))","linear_formula":"add(n1,n3)|add(n0,n2)|divide(#0,#1)|","chain":"1_200 + 480<\/gadget>\n1_680<\/output>\n52 + 32<\/gadget>\n84<\/output>\n1_680 \/ 84<\/gadget>\n20<\/output>\n20<\/result>","index":1346} +{"problem":"if 2 ^ 5 , 4 ^ 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 * 4 * 4 = 208 smash d\"","correct":"d","options":{"a":"26 ","b":"39 ","c":"42 ","d":"208","e":"156"},"options_float":{"a":26.0,"b":39.0,"c":42.0,"d":208.0,"e":156.0},"annotated_formula":"multiply(multiply(multiply(power(2, 2), 4), 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 * 4<\/gadget>\n16<\/output>\n13 \/ 2<\/gadget>\n13\/2 = around 6.5<\/output>\n16 * (13\/2)<\/gadget>\n104<\/output>\n104 * 2<\/gadget>\n208<\/output>\n208<\/result>","index":1348} +{"problem":"if 3 < x < 6 < y < 11 , then what is the greatest possible positive integer difference of x and y ?","rationale":"\"3 < x < 6 < y < 11 ; 3 < x y < 11 3 + y < x + 11 y - x < 8 . positive integer difference is 7 ( for example y = 10.5 and x = 3.5 ) 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":"subtract(subtract(11, 3), const_1)","linear_formula":"subtract(n2,n0)|subtract(#0,const_1)|","chain":"11 - 3<\/gadget>\n8<\/output>\n8 - 1<\/gadget>\n7<\/output>\n7<\/result>","index":1349} +{"problem":"chris age after 20 years will be 5 times his age 5 years back . what is the present age of chris ?","rationale":"chris present age = x after 20 years = x + 20 5 years back = x - 5 x + 20 = 5 ( x - 5 ) x = 5 answer is e","correct":"e","options":{"a":"20 ","b":"25 ","c":"15 ","d":"22","e":"5"},"options_float":{"a":20.0,"b":25.0,"c":15.0,"d":22.0,"e":5.0},"annotated_formula":"subtract(divide(add(multiply(5, 5), 20), subtract(5, const_1)), subtract(divide(add(multiply(5, 5), 20), subtract(5, const_1)), 5))","linear_formula":"multiply(n1,n1)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)|subtract(#3,n1)|subtract(#3,#4)","chain":"5 * 5<\/gadget>\n25<\/output>\n25 + 20<\/gadget>\n45<\/output>\n5 - 1<\/gadget>\n4<\/output>\n45 \/ 4<\/gadget>\n45\/4 = around 11.25<\/output>\n(45\/4) - 5<\/gadget>\n25\/4 = around 6.25<\/output>\n(45\/4) - (25\/4)<\/gadget>\n5<\/output>\n5<\/result>","index":1350} +{"problem":"the weight of a glass of jar is 30 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ?","rationale":"let weight of jar filled with beans = 100 g weight of jar = 30 g weight of coffee beans = 70 g weight of jar and remaining beans = 60 g weight of remaining beans = 30 g fraction remaining = 30 \/ 70 = 3 \/ 7 answer is e .","correct":"e","options":{"a":"1 \/ 5 ","b":"1 \/ 3 ","c":"2 \/ 5 ","d":"1 \/ 2","e":"3 \/ 7"},"options_float":{"a":0.2,"b":0.3333333333,"c":0.4,"d":0.5,"e":0.4285714286},"annotated_formula":"divide(subtract(60, 30), subtract(const_100, 30))","linear_formula":"subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)","chain":"60 - 30<\/gadget>\n30<\/output>\n100 - 30<\/gadget>\n70<\/output>\n30 \/ 70<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":1354} +{"problem":"a number increased by 15 % gives 1150 . the number is","rationale":"\"formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 15 % = 115 % 115 % - - - - - - - > 1150 ( 115 × 100 = 1150 ) 100 % - - - - - - - > 1000 ( 100 × 100 = 1000 ) b )\"","correct":"b","options":{"a":"250 ","b":"1000 ","c":"450 ","d":"500","e":"520"},"options_float":{"a":250.0,"b":1000.0,"c":450.0,"d":500.0,"e":520.0},"annotated_formula":"divide(1150, add(const_1, divide(15, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 + (3\/20)<\/gadget>\n23\/20 = around 1.15<\/output>\n1_150 \/ (23\/20)<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1356} +{"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 390 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 + 390 ) \/ 36 * 5 \/ 18 = 51 min . answer : c\"","correct":"c","options":{"a":"19 ","b":"27 ","c":"51 ","d":"47","e":"28"},"options_float":{"a":19.0,"b":27.0,"c":51.0,"d":47.0,"e":28.0},"annotated_formula":"divide(add(390, 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>\n390 + 120<\/gadget>\n510<\/output>\n510 \/ 10<\/gadget>\n51<\/output>\n51<\/result>","index":1357} +{"problem":"the average of 7 numbers is 23 . if each number be multiplied by 5 . find the average of new set of numbers ?","rationale":"\"explanation : average of new numbers = 23 * 5 = 115 answer : option d\"","correct":"d","options":{"a":"110 ","b":"122 ","c":"120 ","d":"115","e":"145"},"options_float":{"a":110.0,"b":122.0,"c":120.0,"d":115.0,"e":145.0},"annotated_formula":"multiply(23, 5)","linear_formula":"multiply(n1,n2)|","chain":"23 * 5<\/gadget>\n115<\/output>\n115<\/result>","index":1358} +{"problem":"if 8 men or 12 women can do a piece of work in 35 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 \/ 35 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 35 = 1 \/ 420 so , 20 women work = 20 \/ 420 = 1 \/ 21 ans : 21 days answer : e\"","correct":"e","options":{"a":"10 days ","b":"11 days ","c":"13 days ","d":"15 days","e":"21 days"},"options_float":{"a":10.0,"b":11.0,"c":13.0,"d":15.0,"e":21.0},"annotated_formula":"inverse(add(divide(6, multiply(8, 35)), divide(11, multiply(12, 35))))","linear_formula":"multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)|","chain":"8 * 35<\/gadget>\n280<\/output>\n6 \/ 280<\/gadget>\n3\/140 = around 0.021429<\/output>\n12 * 35<\/gadget>\n420<\/output>\n11 \/ 420<\/gadget>\n11\/420 = around 0.02619<\/output>\n(3\/140) + (11\/420)<\/gadget>\n1\/21 = around 0.047619<\/output>\n1 \/ (1\/21)<\/gadget>\n21<\/output>\n21<\/result>","index":1360} +{"problem":"the total age of a and b is 18 years more than the total age of b and c . c is how many year younger than","rationale":"\"given that a + b = 182 + b + c = > a â € “ c = 18 + b â € “ b = 18 = > c is younger than a by 18 years answer : e\"","correct":"e","options":{"a":"14 years ","b":"12 years ","c":"56 years ","d":"66 years","e":"18 years"},"options_float":{"a":14.0,"b":12.0,"c":56.0,"d":66.0,"e":18.0},"annotated_formula":"multiply(18, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"18 * 1<\/gadget>\n18<\/output>\n18<\/result>","index":1361} +{"problem":"the greatest number which on dividing 1657 and 2037 leaves remainders 9 and 5 respectively , is :","rationale":"\"explanation : required number = h . c . f . of ( 1657 - 9 ) and ( 2037 - 5 ) = h . c . f . of 1648 and 2032 = 16 . answer : a\"","correct":"a","options":{"a":"16 ","b":"127 ","c":"235 ","d":"305","e":"505"},"options_float":{"a":16.0,"b":127.0,"c":235.0,"d":305.0,"e":505.0},"annotated_formula":"gcd(subtract(2037, 5), subtract(1657, 9))","linear_formula":"subtract(n1,n3)|subtract(n0,n2)|gcd(#0,#1)|","chain":"2_037 - 5<\/gadget>\n2_032<\/output>\n1_657 - 9<\/gadget>\n1_648<\/output>\ngcd(2_032, 1_648)<\/gadget>\n16<\/output>\n16<\/result>","index":1363} +{"problem":"a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 10 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":"\"l * 30 = 1200 l = 40 40 + 30 + 50 = 120 120 * 10 = 1200 e\"","correct":"e","options":{"a":"2387 ","b":"1298 ","c":"1128 ","d":"1237","e":"1200"},"options_float":{"a":2387.0,"b":1298.0,"c":1128.0,"d":1237.0,"e":1200.0},"annotated_formula":"multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 10)","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 * 10<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":1364} +{"problem":"the length of a rectangle is 2 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 2 times the width , then the length is 2 * 4 = 8 in the area is given by 4 * 8 = 32 square inches correct answer e\"","correct":"e","options":{"a":"30 square inches ","b":"75 square inches ","c":"68 square inches ","d":"89 square inches","e":"32 square inches"},"options_float":{"a":30.0,"b":75.0,"c":68.0,"d":89.0,"e":32.0},"annotated_formula":"rectangle_area(4, multiply(2, 4))","linear_formula":"multiply(n0,n1)|rectangle_area(n1,#0)|","chain":"2 * 4<\/gadget>\n8<\/output>\n4 * 8<\/gadget>\n32<\/output>\n32<\/result>","index":1365} +{"problem":"the ratio between the sale price and the cost price of an article is 7 : 4 . what is the ratio between the profit and the cost price of that article ?","rationale":"\"c . p . = rs . 4 x and s . p . = rs . 7 x . then , gain = rs . 3 x required ratio = 3 x : 4 x = 3 : 4 a\"","correct":"a","options":{"a":"3 : 4 ","b":"1 : 2 ","c":"2 : 5 ","d":"3 : 5","e":"25"},"options_float":{"a":0.75,"b":0.5,"c":0.4,"d":0.6,"e":25.0},"annotated_formula":"divide(subtract(7, 4), 4)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|","chain":"7 - 4<\/gadget>\n3<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":1366} +{"problem":"barbata invests $ 2400 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 ; 2400 + 0.05 * 2400 + x + 0.08 x = 2400 + x + 0.06 ( 2400 + x ) 0.05 * 2400 + 0.08 x = 0.06 x + 0.06 * 2400 0.02 x = 2400 ( 0.06 - 0.05 ) x = 2400 * 0.01 \/ 0.02 = 1200 ans : ` ` a ' '\"","correct":"a","options":{"a":"1200 ","b":"3000 ","c":"1000 ","d":"3600","e":"2400"},"options_float":{"a":1200.0,"b":3000.0,"c":1000.0,"d":3600.0,"e":2400.0},"annotated_formula":"divide(subtract(multiply(divide(6, const_100), 2400), multiply(2400, 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_400<\/gadget>\n144<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n2_400 * (1\/20)<\/gadget>\n120<\/output>\n144 - 120<\/gadget>\n24<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) - (3\/50)<\/gadget>\n1\/50 = around 0.02<\/output>\n24 \/ (1\/50)<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":1367} +{"problem":"difference between a two - digit number and the number obtained by interchanging the two digits is 36 , what is the difference between two numbers","rationale":"explanation : let the ten digit be x , unit digit is y . then ( 10 x + y ) - ( 10 y + x ) = 36 = > 9 x - 9 y = 36 = > x - y = 4 . option b","correct":"b","options":{"a":"2 ","b":"4 ","c":"8 ","d":"12","e":"14"},"options_float":{"a":2.0,"b":4.0,"c":8.0,"d":12.0,"e":14.0},"annotated_formula":"divide(36, subtract(const_10, const_1))","linear_formula":"subtract(const_10,const_1)|divide(n0,#0)","chain":"10 - 1<\/gadget>\n9<\/output>\n36 \/ 9<\/gadget>\n4<\/output>\n4<\/result>","index":1370} +{"problem":"the average score of a cricketer in 6 matches is 27 and in other 4 matches is 32 . then find the average score in all the 10 matches ?","rationale":"\"average in 10 matches = ( 6 * 27 + 4 * 32 ) \/ 6 + 4 = 162 + 128 \/ 10 = 290 \/ 10 = 29 answer is d\"","correct":"d","options":{"a":"25 ","b":"27 ","c":"30 ","d":"29","e":"42"},"options_float":{"a":25.0,"b":27.0,"c":30.0,"d":29.0,"e":42.0},"annotated_formula":"divide(add(multiply(6, 27), multiply(4, 32)), 10)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)|","chain":"6 * 27<\/gadget>\n162<\/output>\n4 * 32<\/gadget>\n128<\/output>\n162 + 128<\/gadget>\n290<\/output>\n290 \/ 10<\/gadget>\n29<\/output>\n29<\/result>","index":1371} +{"problem":"rs . 900 amounts to rs . 920 in 3 years at simple interest . if the interest is increased by 3 % , it would amount to how much ?","rationale":"\"( 900 * 3 * 3 ) \/ 100 = 81 920 + 81 = 1001 answer : b\"","correct":"b","options":{"a":"rs . 1056 ","b":"rs . 1001 ","c":"rs . 2056 ","d":"rs . 1026","e":"rs . 1856"},"options_float":{"a":1056.0,"b":1001.0,"c":2056.0,"d":1026.0,"e":1856.0},"annotated_formula":"multiply(power(add(const_1, divide(3, const_100)), 3), 900)","linear_formula":"divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|","chain":"3 \/ 100<\/gadget>\n3\/100 = around 0.03<\/output>\n1 + (3\/100)<\/gadget>\n103\/100 = around 1.03<\/output>\n(103\/100) ** 3<\/gadget>\n1_092_727\/1_000_000 = around 1.092727<\/output>\n(1_092_727\/1_000_000) * 900<\/gadget>\n9_834_543\/10_000 = around 983.4543<\/output>\n9_834_543\/10_000 = around 983.4543<\/result>","index":1372} +{"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 136 , 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 = 136 = > . 1 x + 3 \/ 4 x = 136 = > x = 160 160 - 136 = pure gala = 24 . a","correct":"a","options":{"a":"24 ","b":"33 ","c":"55 ","d":"77","e":"88"},"options_float":{"a":24.0,"b":33.0,"c":55.0,"d":77.0,"e":88.0},"annotated_formula":"subtract(divide(136, add(divide(10, const_100), divide(3, 4))), 136)","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>\n136 \/ (17\/20)<\/gadget>\n160<\/output>\n160 - 136<\/gadget>\n24<\/output>\n24<\/result>","index":1375} +{"problem":"at a monthly meeting , 1 \/ 3 of the attendees were males and 4 \/ 5 of the male attendees arrived on time . if 5 \/ 6 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 \/ 5 * 1 \/ 3 = 1 \/ 15 of the attendees . females who did not arrive on time are 1 \/ 6 * 2 \/ 3 = 1 \/ 9 of the attendees . the fraction of all attendees who did not arrive on time is 1 \/ 15 + 1 \/ 9 = 8 \/ 45 the answer is e .\"","correct":"e","options":{"a":"1 \/ 6 ","b":"2 \/ 15 ","c":"3 \/ 20 ","d":"7 \/ 30","e":"8 \/ 45"},"options_float":{"a":0.1666666667,"b":0.1333333333,"c":0.15,"d":0.2333333333,"e":0.1777777778},"annotated_formula":"add(multiply(subtract(const_1, divide(5, 6)), subtract(const_1, divide(1, 3))), multiply(subtract(const_1, divide(4, 5)), divide(1, 3)))","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":"5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n1 - (5\/6)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ 3<\/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>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * (1\/3)<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/9) + (1\/15)<\/gadget>\n8\/45 = around 0.177778<\/output>\n8\/45 = around 0.177778<\/result>","index":1376} +{"problem":"a shopkeeper sold an book offering a discount of 5 % and earned a profit of 30 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"\"let c . p . be $ 100 . then , s . p . = $ 130 let marked price be $ x . then , 95 \/ 100 x = 130 x = 13000 \/ 95 = $ 136.8 now , s . p . = $ 136.8 , c . p . = $ 100 profit % = 136.8 % . d\"","correct":"d","options":{"a":"140 ","b":"120 ","c":"130 ","d":"136.8","e":"150"},"options_float":{"a":140.0,"b":120.0,"c":130.0,"d":136.8,"e":150.0},"annotated_formula":"multiply(const_100, divide(add(const_100, 30), subtract(const_100, 5)))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|","chain":"100 + 30<\/gadget>\n130<\/output>\n100 - 5<\/gadget>\n95<\/output>\n130 \/ 95<\/gadget>\n26\/19 = around 1.368421<\/output>\n100 * (26\/19)<\/gadget>\n2_600\/19 = around 136.842105<\/output>\n2_600\/19 = around 136.842105<\/result>","index":1377} +{"problem":"a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 300 and $ 200 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 300 and 200 , which is 600 for each , which is total 1200 . the number of 1 st type of computers which costing $ 300 = 600 \/ 300 = 2 . the number of 2 nd type of computers which costing $ 200 = 600 \/ 200 = 3 . total = 2 + 3 = 5 answer is b .\"","correct":"b","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":"add(divide(lcm(300, 200), 300), divide(lcm(300, 200), 200))","linear_formula":"lcm(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|","chain":"lcm(300, 200)<\/gadget>\n600<\/output>\n600 \/ 300<\/gadget>\n2<\/output>\n600 \/ 200<\/gadget>\n3<\/output>\n2 + 3<\/gadget>\n5<\/output>\n5<\/result>","index":1378} +{"problem":"if p is a prime number greater than 3 , find the remainder when p ^ 2 + 14 is divided by 12 .","rationale":"\"every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 14 = 36 n ^ 2 + 12 n + 1 + 14 = 36 n ^ 2 + 12 n + 12 + 3 if p = 6 n - 1 , then p ^ 2 + 14 = 36 n ^ 2 - 12 n + 1 + 14 = 36 n ^ 2 - 12 n + 12 + 3 when divided by 12 , it must leave a remainder of 3 . the answer is c .\"","correct":"c","options":{"a":"6 ","b":"1 ","c":"3 ","d":"8","e":"7"},"options_float":{"a":6.0,"b":1.0,"c":3.0,"d":8.0,"e":7.0},"annotated_formula":"subtract(add(14, power(add(const_1, const_4), 2)), multiply(12, 3))","linear_formula":"add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)|","chain":"1 + 4<\/gadget>\n5<\/output>\n5 ** 2<\/gadget>\n25<\/output>\n14 + 25<\/gadget>\n39<\/output>\n12 * 3<\/gadget>\n36<\/output>\n39 - 36<\/gadget>\n3<\/output>\n3<\/result>","index":1379} +{"problem":"a man has some hens and cows . if the number of heads be 42 and the number of feet equals 124 , then the number of hens will be","rationale":"\"explanation : let number of hens = h and number of cows = c number of heads = 42 = > h + c = 42 - - - ( equation 1 ) number of feet = 124 = > 2 h + 4 c = 124 = > h + 2 c = 62 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 62 - 42 = > c = 20 substituting the value of c in equation 1 , we get h + 22 = 42 = > h = 42 - 20 = 22 i . e . , number of hens = 22 answer : a\"","correct":"a","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(42, const_4), 124), const_2)","linear_formula":"multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|","chain":"42 * 4<\/gadget>\n168<\/output>\n168 - 124<\/gadget>\n44<\/output>\n44 \/ 2<\/gadget>\n22<\/output>\n22<\/result>","index":1385} +{"problem":"find large number from below question the difference of two numbers is 1385 . 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 + 1385 ) . x + 1385 = 6 x + 15 5 x = 1370 x = 274 large number = 274 + 1385 = 1659 e\"","correct":"e","options":{"a":"1235 ","b":"1345 ","c":"1678 ","d":"1767","e":"1659"},"options_float":{"a":1235.0,"b":1345.0,"c":1678.0,"d":1767.0,"e":1659.0},"annotated_formula":"multiply(divide(subtract(1385, 15), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_385 - 15<\/gadget>\n1_370<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_370 \/ 5<\/gadget>\n274<\/output>\n274 * 6<\/gadget>\n1_644<\/output>\n1_644<\/result>","index":1386} +{"problem":"convert 0.30 in to a vulgar fraction ?","rationale":"answer 0.30 = 30 \/ 100 = 3 \/ 10 correct option : c","correct":"c","options":{"a":"18 \/ 50 ","b":"16 \/ 50 ","c":"3 \/ 10 ","d":"19 \/ 50","e":"none"},"options_float":{"a":0.36,"b":0.32,"c":0.3,"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)), 0.3), 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 * 0.3<\/gadget>\n30<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n3\/10 = around 0.3<\/result>","index":1388} +{"problem":"a room is 30 m long and 24 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , the volume of the hall is :","rationale":"let the height be h 2 ( 30 + 24 ) x h – 2 ( 30 - 24 ) h = ( 2 ( 30 x 24 ) ) \/ ( 2 ( 30 + 24 ) ) = ( 30 x 24 ) \/ 54 = 40 \/ 3 m volume = 30 x 24 x 40 \/ 3 = 9600 m 3 answer : d","correct":"d","options":{"a":"9.6 m 3 ","b":"96 m 3 ","c":"960 m 3 ","d":"9600 m 3","e":"96000 m 3"},"options_float":{"a":9.6,"b":96.0,"c":960.0,"d":9600.0,"e":96000.0},"annotated_formula":"volume_rectangular_prism(30, 24, divide(multiply(rectangle_area(30, 24), const_2), rectangle_perimeter(30, 24)))","linear_formula":"rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|multiply(#0,const_2)|divide(#2,#1)|volume_rectangular_prism(n0,n1,#3)","chain":"30 * 24<\/gadget>\n720<\/output>\n720 * 2<\/gadget>\n1_440<\/output>\n2 * (30 + 24)<\/gadget>\n108<\/output>\n1_440 \/ 108<\/gadget>\n40\/3 = around 13.333333<\/output>\n30 * 24 * (40\/3)<\/gadget>\n9_600<\/output>\n9_600<\/result>","index":1390} +{"problem":"if $ 120 invested at a certain rate of simple interest amounts to $ 180 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ?","rationale":"\"120 amounts to 180 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 180 120 + 120 * ( r \/ 100 ) * ( 3 ) = 140 = > r = 50 \/ 3 150 in 6 years = principal + interest = 300 answer is e .\"","correct":"e","options":{"a":"$ 190 ","b":"$ 180 ","c":"$ 200 ","d":"$ 240","e":"$ 300"},"options_float":{"a":190.0,"b":180.0,"c":200.0,"d":240.0,"e":300.0},"annotated_formula":"add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(180, 120), 120), 120), 3)), 120))","linear_formula":"multiply(n3,n4)|subtract(n1,n0)|multiply(#1,n0)|divide(#2,n0)|divide(#3,n2)|multiply(#4,#0)|divide(#5,n0)|add(n3,#6)|","chain":"150 * 6<\/gadget>\n900<\/output>\n180 - 120<\/gadget>\n60<\/output>\n60 * 120<\/gadget>\n7_200<\/output>\n7_200 \/ 120<\/gadget>\n60<\/output>\n60 \/ 3<\/gadget>\n20<\/output>\n900 * 20<\/gadget>\n18_000<\/output>\n18_000 \/ 120<\/gadget>\n150<\/output>\n150 + 150<\/gadget>\n300<\/output>\n300<\/result>","index":1391} +{"problem":"the price of a certain product increased by the same percent from 1960 to 1970 as from 1970 to 1980 . if its price of $ 1.20 in 1970 was 150 percent of its price in 1960 , what was its price in 1980 ?","rationale":"the price in 1970 was 150 percent of its price in 1960 , means that the percent increase was 50 % from 1960 to 1970 ( and from 1970 to 1980 ) . therefore the price in 1980 = $ 1.2 * 1.5 = $ 1.8 . answer : a .","correct":"a","options":{"a":"$ 1.80 ","b":"$ 2.00 ","c":"$ 2.40 ","d":"$ 2.70","e":"$ 3.00"},"options_float":{"a":1.8,"b":2.0,"c":2.4,"d":2.7,"e":3.0},"annotated_formula":"multiply(divide(150, const_100), 1.2)","linear_formula":"divide(n6,const_100)|multiply(n4,#0)","chain":"150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 1.2<\/gadget>\n1.8<\/output>\n1.8<\/result>","index":1393} +{"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 37 , the how old is b ?","rationale":"explanation : 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 = 37 ⇒ 5 x = 35 ⇒ x = 7 . hence , b ' s age = 2 x = 14 years . answer : d","correct":"d","options":{"a":"7 ","b":"9 ","c":"8 ","d":"14","e":"10"},"options_float":{"a":7.0,"b":9.0,"c":8.0,"d":14.0,"e":10.0},"annotated_formula":"divide(multiply(subtract(37, 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":"37 - 2<\/gadget>\n35<\/output>\n35 * 2<\/gadget>\n70<\/output>\n4 + 1<\/gadget>\n5<\/output>\n70 \/ 5<\/gadget>\n14<\/output>\n14<\/result>","index":1394} +{"problem":"if c and t are positive integers , ct + c + t can not be","rationale":"let ct + t + c = x add 1 on both sides : ct + t + c + 1 = x + 1 t ( c + 1 ) + c + 1 = x + 1 ( c + 1 ) ( t + 1 ) = x + 1 minimum value of ( c + 1 ) = 2 minimum value of ( t + 1 ) = 2 hence x + 1 can not be prime substitute x from the given options : 6 + 1 = 7 - - > prime - - > ct + t + s can not be 6 answer : b","correct":"b","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(const_2, const_3)","linear_formula":"multiply(const_2,const_3)","chain":"2 * 3<\/gadget>\n6<\/output>\n6<\/result>","index":1395} +{"problem":"the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 2 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 2 ) = rs . 2972 . answer : option a\"","correct":"a","options":{"a":"2972 ","b":"4567 ","c":"4235 ","d":"4547","e":"4675"},"options_float":{"a":2972.0,"b":4567.0,"c":4235.0,"d":4547.0,"e":4675.0},"annotated_formula":"multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 2)","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)) * 2<\/gadget>\n1676.18614718056*sqrt(pi) = around 2_970.962591<\/output>\n1676.18614718056*sqrt(pi) = around 2_970.962591<\/result>","index":1396} +{"problem":"a man swims downstream 30 km and upstream 12 km taking 3 hours each time , what is the speed of the man in still water ?","rationale":"\"30 - - - 3 ds = 10 ? - - - - 1 12 - - - - 3 us = 4 ? - - - - 1 m = ? m = ( 10 + 4 ) \/ 2 = 7 answer : b\"","correct":"b","options":{"a":"8 ","b":"7 ","c":"5 ","d":"2","e":"4"},"options_float":{"a":8.0,"b":7.0,"c":5.0,"d":2.0,"e":4.0},"annotated_formula":"divide(add(divide(12, 3), divide(30, 3)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"12 \/ 3<\/gadget>\n4<\/output>\n30 \/ 3<\/gadget>\n10<\/output>\n4 + 10<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":1397} +{"problem":"a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 7 x km .","rationale":"\"total time taken = x \/ 40 + 2 x \/ 20 hours = 5 x \/ 40 = x \/ 8 hours average speed = 7 x \/ ( x \/ 8 ) = 56 kmph answer : a\"","correct":"a","options":{"a":"56 ","b":"18 ","c":"24 ","d":"19","e":"12"},"options_float":{"a":56.0,"b":18.0,"c":24.0,"d":19.0,"e":12.0},"annotated_formula":"divide(multiply(40, 7), add(divide(40, 40), divide(multiply(2, 40), 20)))","linear_formula":"divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)|","chain":"40 * 7<\/gadget>\n280<\/output>\n40 \/ 40<\/gadget>\n1<\/output>\n2 * 40<\/gadget>\n80<\/output>\n80 \/ 20<\/gadget>\n4<\/output>\n1 + 4<\/gadget>\n5<\/output>\n280 \/ 5<\/gadget>\n56<\/output>\n56<\/result>","index":1398} +{"problem":"if the average of 6 digits is 16 and the average of 4 of them is 10 , calculate the average of the remaining 2 numbers ?","rationale":"\"explanation : total of the 6 digits - 6 * 16 = 96 total of the 4 digits - 4 * 10 = 40 total of the remaining 2 digits - 96 - 40 = 56 average of the remaining 2 numbers = 56 \/ 2 = 28 answer : c\"","correct":"c","options":{"a":"36 ","b":"35 ","c":"28 ","d":"33","e":"21"},"options_float":{"a":36.0,"b":35.0,"c":28.0,"d":33.0,"e":21.0},"annotated_formula":"divide(subtract(multiply(6, 16), multiply(4, 10)), 2)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)|","chain":"6 * 16<\/gadget>\n96<\/output>\n4 * 10<\/gadget>\n40<\/output>\n96 - 40<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n28<\/result>","index":1399} +{"problem":"a company pays 20.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ?","rationale":"\"explanation : dividend on 1 share = ( 20.5 * 50 ) \/ 100 = rs . 10.25 rs . 25 is income on an investment of rs . 100 rs . 10.25 is income on an investment of rs . ( 10.25 * 100 ) \/ 25 = rs . 41 answer : e\"","correct":"e","options":{"a":"25 ","b":"66 ","c":"18 ","d":"19","e":"41"},"options_float":{"a":25.0,"b":66.0,"c":18.0,"d":19.0,"e":41.0},"annotated_formula":"divide(multiply(divide(multiply(20.5, 50), const_100), const_100), 25)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)|","chain":"20.5 * 50<\/gadget>\n1_025<\/output>\n1_025 \/ 100<\/gadget>\n41\/4 = around 10.25<\/output>\n(41\/4) * 100<\/gadget>\n1_025<\/output>\n1_025 \/ 25<\/gadget>\n41<\/output>\n41<\/result>","index":1400} +{"problem":"how many diagonals does a 59 - sided convex polygon have ?","rationale":"a 59 - sided convex polygon has 59 vertices . if we examine a single vertex , we can see that we can connect it with 56 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 59 vertices can be connected with 56 vertices to create a diagonal then the total number of diagonals would be ( 59 ) ( 56 ) = 3304 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3304 by 2 to get 1652 . the answer is b .","correct":"b","options":{"a":"1168 ","b":"1652 ","c":"2452 ","d":"3304","e":"4256"},"options_float":{"a":1168.0,"b":1652.0,"c":2452.0,"d":3304.0,"e":4256.0},"annotated_formula":"divide(factorial(59), multiply(factorial(subtract(59, 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(59)<\/gadget>\n138_683_118_545_689_835_737_939_019_720_389_406_345_902_876_772_687_432_540_821_294_940_160_000_000_000_000<\/output>\n59 - 2<\/gadget>\n57<\/output>\nfactorial(57)<\/gadget>\n40_526_919_504_877_216_755_680_601_905_432_322_134_980_384_796_226_602_145_184_481_280_000_000_000_000<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n40_526_919_504_877_216_755_680_601_905_432_322_134_980_384_796_226_602_145_184_481_280_000_000_000_000 * 2<\/gadget>\n81_053_839_009_754_433_511_361_203_810_864_644_269_960_769_592_453_204_290_368_962_560_000_000_000_000<\/output>\n138_683_118_545_689_835_737_939_019_720_389_406_345_902_876_772_687_432_540_821_294_940_160_000_000_000_000 \/ 81_053_839_009_754_433_511_361_203_810_864_644_269_960_769_592_453_204_290_368_962_560_000_000_000_000<\/gadget>\n1_711<\/output>\n1_711<\/result>","index":1401} +{"problem":"a shopkeeper sold an book offering a discount of 5 % and earned a profit of 33 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"\"let c . p . be $ 100 . then , s . p . = $ 133 let marked price be $ x . then , 95 \/ 100 x = 133 x = 13300 \/ 95 = $ 140 now , s . p . = $ 140 , c . p . = $ 100 profit % = 40 % . a\"","correct":"a","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, 33), subtract(const_100, 5)))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|","chain":"100 + 33<\/gadget>\n133<\/output>\n100 - 5<\/gadget>\n95<\/output>\n133 \/ 95<\/gadget>\n7\/5 = around 1.4<\/output>\n100 * (7\/5)<\/gadget>\n140<\/output>\n140<\/result>","index":1402} +{"problem":"calculate how long it will take a swimmer to swim a distance of 10 km against the current of a river which flows at 3 km \/ hr , given that he can swim in still water at 5 km \/ h","rationale":"\"swim in still water at = 5 speed of river = 3 us = 5 - 3 = 2 distance = 10 t = 10 \/ 2 = 5 answer : 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":"divide(10, subtract(5, 3))","linear_formula":"subtract(n2,n1)|divide(n0,#0)|","chain":"5 - 3<\/gadget>\n2<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":1403} +{"problem":"a team of 8 persons joins in a shooting competition . the best marksman scored 85 points . if he had scored 92 points , the average score for the team would have been 84 . the number of points , the team scored was :","rationale":"explanation : let the total score be x . ( x + 92 - 85 ) \/ 8 = 84 . so , x + 7 = 672 = > x = 665 . answer : a ) 665","correct":"a","options":{"a":"665 ","b":"376 ","c":"998 ","d":"1277","e":"1991"},"options_float":{"a":665.0,"b":376.0,"c":998.0,"d":1277.0,"e":1991.0},"annotated_formula":"subtract(add(multiply(84, 8), 85), 92)","linear_formula":"multiply(n0,n3)|add(n1,#0)|subtract(#1,n2)","chain":"84 * 8<\/gadget>\n672<\/output>\n672 + 85<\/gadget>\n757<\/output>\n757 - 92<\/gadget>\n665<\/output>\n665<\/result>","index":1408} +{"problem":"at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 20 seconds . how long does it tick at 12 ′ o clock","rationale":"\"explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 20 \/ 5 = 4 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 4 = 44 secs . answer : c\"","correct":"c","options":{"a":"47 ","b":"76 ","c":"44 ","d":"66","e":"11"},"options_float":{"a":47.0,"b":76.0,"c":44.0,"d":66.0,"e":11.0},"annotated_formula":"multiply(divide(20, subtract(6, const_1)), subtract(12, const_1))","linear_formula":"subtract(n0,const_1)|subtract(n3,const_1)|divide(n2,#0)|multiply(#2,#1)|","chain":"6 - 1<\/gadget>\n5<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n12 - 1<\/gadget>\n11<\/output>\n4 * 11<\/gadget>\n44<\/output>\n44<\/result>","index":1410} +{"problem":"the price of a t . v . set worth rs . 30000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ?","rationale":"\"money paid in cash = rs . 1000 balance payment = ( 30000 - 1000 ) = rs . 29000 answer : c\"","correct":"c","options":{"a":"22678 ","b":"26699 ","c":"29000 ","d":"19000","e":"26711"},"options_float":{"a":22678.0,"b":26699.0,"c":29000.0,"d":19000.0,"e":26711.0},"annotated_formula":"subtract(30000, 1000)","linear_formula":"subtract(n0,n2)|","chain":"30_000 - 1_000<\/gadget>\n29_000<\/output>\n29_000<\/result>","index":1411} +{"problem":"55 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be :","rationale":"\"sol . let the length of the wire b h . radius = 1 \/ 2 mm = 1 \/ 20 cm . then , 22 \/ 7 * 1 \/ 20 * 1 \/ 20 * h = 55 ⇔ = [ 55 * 20 * 20 * 7 \/ 22 ] = 7000 cm = 70 m . answer d\"","correct":"d","options":{"a":"84 m ","b":"88 m ","c":"120 m ","d":"70 m","e":"none"},"options_float":{"a":84.0,"b":88.0,"c":120.0,"d":70.0,"e":null},"annotated_formula":"divide(55, multiply(power(divide(1, const_2), const_2), const_pi))","linear_formula":"divide(n1,const_2)|power(#0,const_2)|multiply(#1,const_pi)|divide(n0,#2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 2<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * pi<\/gadget>\npi\/4 = around 0.785398<\/output>\n55 \/ (pi\/4)<\/gadget>\n220\/pi = around 70.028175<\/output>\n220\/pi = around 70.028175<\/result>","index":1412} +{"problem":"a metallic sheet is of rectangular shape with dimensions 48 m x 34 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 = ( 48 - 16 ) m = 32 m , b = ( 34 - 16 ) m = 18 m , h = 8 m . volume of the box = ( 32 x 18 x 8 ) m 3 = 4608 m 3 . answer : option c\"","correct":"c","options":{"a":"4830 ","b":"5120 ","c":"4608 ","d":"7500","e":"8960"},"options_float":{"a":4830.0,"b":5120.0,"c":4608.0,"d":7500.0,"e":8960.0},"annotated_formula":"volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(34, 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>\n48 - 16<\/gadget>\n32<\/output>\n34 - 16<\/gadget>\n18<\/output>\n32 * 18 * 8<\/gadget>\n4_608<\/output>\n4_608<\/result>","index":1413} +{"problem":"a certain fruit stand sold apples for $ 0.70 each and guava for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ?","rationale":"some multiple of 7 + some multiple of 5 should yield 63 . to get to a some multiple of 5 , we should ensure that a 3 or 8 ( 5 + 3 ) should be a multiple of 7 . 63 is a direct multiple of 7 , however in this case there wo n ' t be any guava . hence the next option is to look for a multiple of 7 that has 8 as the unit digit . 28 satisfies this hence no . of apples is 4 and no of bananas is 7 . c","correct":"c","options":{"a":"12 ","b":"13 ","c":"11 ","d":"14","e":"5"},"options_float":{"a":12.0,"b":13.0,"c":11.0,"d":14.0,"e":5.0},"annotated_formula":"add(divide(subtract(6.3, multiply(0.7, const_4)), 0.5), const_4)","linear_formula":"multiply(n0,const_4)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_4)","chain":"0.7 * 4<\/gadget>\n2.8<\/output>\n6.3 - 2.8<\/gadget>\n3.5<\/output>\n3.5 \/ 0.5<\/gadget>\n7<\/output>\n7 + 4<\/gadget>\n11<\/output>\n11<\/result>","index":1415} +{"problem":"if 25 % 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":"\"we know 25 % people study biology , therefore the no of people not studying = 100 - 25 = 75 % > therefore the people not studying biology out of a total 880 people are = 75 % of 880 > ( 75 \/ 100 ) * 880 = 660 people e\"","correct":"e","options":{"a":"500 ","b":"600 ","c":"620 ","d":"640","e":"660"},"options_float":{"a":500.0,"b":600.0,"c":620.0,"d":640.0,"e":660.0},"annotated_formula":"multiply(divide(880, const_100), subtract(const_100, 25))","linear_formula":"divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|","chain":"880 \/ 100<\/gadget>\n44\/5 = around 8.8<\/output>\n100 - 25<\/gadget>\n75<\/output>\n(44\/5) * 75<\/gadget>\n660<\/output>\n660<\/result>","index":1416} +{"problem":"find large number from below question the difference of two numbers is 1380 . 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 + 1380 ) . x + 1380 = 6 x + 15 5 x = 1365 x = 273 large number = 273 + 1380 = 1653 e\"","correct":"e","options":{"a":"1245 ","b":"1345 ","c":"1455 ","d":"1577","e":"1653"},"options_float":{"a":1245.0,"b":1345.0,"c":1455.0,"d":1577.0,"e":1653.0},"annotated_formula":"multiply(divide(subtract(1380, 15), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_380 - 15<\/gadget>\n1_365<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_365 \/ 5<\/gadget>\n273<\/output>\n273 * 6<\/gadget>\n1_638<\/output>\n1_638<\/result>","index":1417} +{"problem":"tough and tricky questions : word problems . operation # is defined as : a # b = 4 a ^ 2 + 4 b ^ 2 + 8 ab for all non - negative integers . what is the value of ( a + b ) + 8 , when a # b = 100 ?","rationale":"official solution : ( b ) we know that a # b = 100 and a # b = 4 a ² + 4 b ² + 8 ab . so 4 a ² + 4 b ² + 8 ab = 100 we can see that 4 a ² + 4 b ² + 8 ab is a well - known formula for ( 2 a + 2 b ) ² . therefore ( 2 a + 2 b ) ² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbers . so we can conclude that 2 ( a + b ) = 10 . ( a + b ) + 8 = 10 \/ 2 + 8 = 13 . the correct answer is d","correct":"d","options":{"a":"5 ","b":"8 ","c":"10 ","d":"13","e":"17"},"options_float":{"a":5.0,"b":8.0,"c":10.0,"d":13.0,"e":17.0},"annotated_formula":"add(sqrt(divide(100, 4)), 8)","linear_formula":"divide(n6,n0)|sqrt(#0)|add(n4,#1)","chain":"100 \/ 4<\/gadget>\n25<\/output>\n25 ** (1\/2)<\/gadget>\n5<\/output>\n5 + 8<\/gadget>\n13<\/output>\n13<\/result>","index":1419} +{"problem":"according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 240 6 ounces servings of orange juice ?","rationale":"\"its e . total juice rquired = 240 * 6 = 1440 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1440 \/ 48 = 30 . answer e\"","correct":"e","options":{"a":"25 ","b":"34 ","c":"50 ","d":"67","e":"30"},"options_float":{"a":25.0,"b":34.0,"c":50.0,"d":67.0,"e":30.0},"annotated_formula":"divide(divide(multiply(240, 6), 12), const_4)","linear_formula":"multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)|","chain":"240 * 6<\/gadget>\n1_440<\/output>\n1_440 \/ 12<\/gadget>\n120<\/output>\n120 \/ 4<\/gadget>\n30<\/output>\n30<\/result>","index":1421} +{"problem":"mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 38400 , how much amount per month would he have spent on clothes and transport ?","rationale":"\"∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 38400 ⁄ 12 = 3200 answer c\"","correct":"c","options":{"a":"4038 ","b":"8076 ","c":"3200 ","d":"4845.6","e":"none of these"},"options_float":{"a":4038.0,"b":8076.0,"c":3200.0,"d":4845.6,"e":null},"annotated_formula":"multiply(divide(divide(38400, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100))","linear_formula":"multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)|","chain":"100 - 60<\/gadget>\n40<\/output>\n40 * 50<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n38_400 \/ (1\/5)<\/gadget>\n192_000<\/output>\n3 * 4<\/gadget>\n12<\/output>\n192_000 \/ 12<\/gadget>\n16_000<\/output>\n16_000 * (1\/5)<\/gadget>\n3_200<\/output>\n3_200<\/result>","index":1422} +{"problem":"daniel went to a shop and bought things worth rs . 50 , out of which 90 paise went on sales tax on taxable purchases . if the tax rate was 15 % , then what was the cost of the tax free items ?","rationale":"total cost of the items he purchased = rs . 50 given that out of this rs . 50 , 90 paise is given as tax = > total tax incurred = 90 paise = rs . 90 \/ 100 let the cost of the tax free items = x given that tax rate = 15 % ∴ ( 50 − 90 \/ 100 − x ) 15 \/ 100 = 90 \/ 100 ⇒ 15 ( 50 − 0.9 − x ) = 90 ⇒ ( 50 − 0.9 − x ) = 6 ⇒ x = 50 − 0.9 − 6 = 43.1 c","correct":"c","options":{"a":"19.7 ","b":"20 ","c":"43.1 ","d":"21.5","e":"22"},"options_float":{"a":19.7,"b":20.0,"c":43.1,"d":21.5,"e":22.0},"annotated_formula":"subtract(subtract(50, divide(90, const_100)), divide(90, 15))","linear_formula":"divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)","chain":"90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n50 - (9\/10)<\/gadget>\n491\/10 = around 49.1<\/output>\n90 \/ 15<\/gadget>\n6<\/output>\n(491\/10) - 6<\/gadget>\n431\/10 = around 43.1<\/output>\n431\/10 = around 43.1<\/result>","index":1423} +{"problem":"if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 3410 , 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 = 3410 * ( 6 \/ 31 ) = 110 * 6 = 660 . answer : option a\"","correct":"a","options":{"a":"660 ","b":"700 ","c":"800 ","d":"900","e":"none of these"},"options_float":{"a":660.0,"b":700.0,"c":800.0,"d":900.0,"e":null},"annotated_formula":"multiply(3410, 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>\n3_410 * (6\/31)<\/gadget>\n660<\/output>\n660<\/result>","index":1425} +{"problem":"running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 \/ 4 w widgets in 3 days , how many days would it take machine x alone to produce 1 w widgets ?","rationale":"let y produce w widgets in y days hence , in 1 day y will produce w \/ y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w \/ y + 2 widgets . hence together x and y in 1 day will produce { w \/ y + w \/ y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w \/ y + w \/ y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 \/ 4 ) w widgets . equating , 3 * [ { w \/ y + w \/ y + 2 } ] = ( 5 \/ 4 ) w take out w common and move 3 to denominator of rhs w { 1 \/ y + 1 \/ ( y + 2 ) } = ( 5 \/ 12 ) w canceling w from both sides { 1 \/ y + 1 \/ ( y + 2 ) } = 5 \/ 12 2 y + 2 \/ y ( y + 2 ) = 5 \/ 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 \/ 5 or y = 4 discarding y = - 6 \/ 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 1 * 6 = 6 days to produce 1 w widgets . answer : b","correct":"b","options":{"a":"4 ","b":"6 ","c":"8 ","d":"10","e":"12"},"options_float":{"a":4.0,"b":6.0,"c":8.0,"d":10.0,"e":12.0},"annotated_formula":"divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2)","linear_formula":"multiply(n2,n3)|multiply(n0,#0)|subtract(#1,#0)|divide(#2,n0)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24 - 12<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6<\/result>","index":1426} +{"problem":"a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 175 , how many coins of each type are there ?","rationale":"\"let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 175 ⇒ 1.75 x = 175 ⇒ x = 100 coins answer d\"","correct":"d","options":{"a":"20 coins ","b":"30 coins ","c":"50 coins ","d":"100 coins","e":"none of these"},"options_float":{"a":20.0,"b":30.0,"c":50.0,"d":100.0,"e":null},"annotated_formula":"divide(175, 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>\n175 \/ (7\/4)<\/gadget>\n100<\/output>\n100<\/result>","index":1427} +{"problem":"x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 6 . if x invested rs . 5000 . the amount invested by y is","rationale":"solution suppose y invested rs . y then , 5000 \/ y = 2 \/ 6 â € ¹ = â € º y = ( 5000 ã — 6 \/ 2 ) . â € ¹ = â € º y = 15000 . answer d","correct":"d","options":{"a":"rs . 45,000 ","b":"rs . 50,000 ","c":"rs . 60,000 ","d":"rs . 15,000","e":"none"},"options_float":{"a":45000.0,"b":50000.0,"c":60000.0,"d":15000.0,"e":null},"annotated_formula":"divide(multiply(5000, 6), 2)","linear_formula":"multiply(n1,n2)|divide(#0,n0)","chain":"5_000 * 6<\/gadget>\n30_000<\/output>\n30_000 \/ 2<\/gadget>\n15_000<\/output>\n15_000<\/result>","index":1428} +{"problem":"little john had $ 16.10 . he spent $ 3.25 on sweets and gave to his two friends $ 2.20 each . how much money was left ?","rationale":"john spent and gave to his two friends a total of 3.25 + 2.20 + 2.20 = $ 7.65 money left 16.10 - 7.65 = $ 8.45 answer : e","correct":"e","options":{"a":"$ 6.45 ","b":"$ 8.35 ","c":"$ 8.75 ","d":"$ 8.85","e":"$ 8.45"},"options_float":{"a":6.45,"b":8.35,"c":8.75,"d":8.85,"e":8.45},"annotated_formula":"subtract(16.1, add(3.25, add(2.2, 2.2)))","linear_formula":"add(n2,n2)|add(n1,#0)|subtract(n0,#1)","chain":"2.2 + 2.2<\/gadget>\n4.4<\/output>\n3.25 + 4.4<\/gadget>\n7.65<\/output>\n16.1 - 7.65<\/gadget>\n8.45<\/output>\n8.45<\/result>","index":1433} +{"problem":"3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 10 pumps work to empty the tank in 1 day ?","rationale":"\"3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 10 pumps then 48 \/ 10 = 4.8 hrs . answer : a\"","correct":"a","options":{"a":"4.8 ","b":"3.6 ","c":"1.1 ","d":"1.2","e":"1.3"},"options_float":{"a":4.8,"b":3.6,"c":1.1,"d":1.2,"e":1.3},"annotated_formula":"divide(multiply(multiply(3, 8), 2), 10)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|","chain":"3 * 8<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\n48 \/ 10<\/gadget>\n24\/5 = around 4.8<\/output>\n24\/5 = around 4.8<\/result>","index":1434} +{"problem":"the surface area of a sphere is 4 π r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area e of that hemisphere ?","rationale":"given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 \/ pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 \/ pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 \/ 2 = 6 ( curved part , not including the flat rounded base ) . but the total surface area = 6 + area of the base of a hemisphere . = 6 + 3 = 9 . answer is d ! !","correct":"d","options":{"a":"6 \/ π ","b":"9 \/ π ","c":"6 ","d":"9","e":"12"},"options_float":{"a":6.0,"b":9.0,"c":6.0,"d":9.0,"e":12.0},"annotated_formula":"add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi)))","linear_formula":"divide(n2,const_pi)|multiply(n0,const_pi)|multiply(#0,#1)|multiply(#0,const_pi)|divide(#2,n1)|add(#4,#3)","chain":"4 * pi<\/gadget>\n4*pi = around 12.566371<\/output>\n3 \/ pi<\/gadget>\n3\/pi = around 0.95493<\/output>\n(4*pi) * (3\/pi)<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\npi * (3\/pi)<\/gadget>\n3<\/output>\n6 + 3<\/gadget>\n9<\/output>\n9<\/result>","index":1435} +{"problem":"if n is a positive integer such that n ! \/ ( n - 2 ) ! = 342 , find n .","rationale":"we write n ! = n * ( n - 1 ) ( n - 2 ! ) therefore n ! \/ ( n - 2 ) ! = n ( n - 1 ) * ( n - 2 ) ! \/ ( n - 2 ) ! = n ( n - 1 ) . - - > n ( n - 1 ) = 342 - - > n ^ 2 - n - 342 = 0 - - > n ^ 2 - 19 n + 18 n - 342 = 0 - - > n ( n - 19 ) + 18 ( n - 19 ) = 0 - - > ( n - 19 ) ( n + 18 ) = 0 therefore n - 19 = 0 ; n + 18 = 0 ; ( i . e ) n = 19 ; n = - 18 we want positive integer . so , n = 19 . answer : c","correct":"c","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":"sqrt(add(342, divide(const_1, const_4)))","linear_formula":"divide(const_1,const_4)|add(n1,#0)|sqrt(#1)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n342 + (1\/4)<\/gadget>\n1_369\/4 = around 342.25<\/output>\n(1_369\/4) ** (1\/2)<\/gadget>\n37\/2 = around 18.5<\/output>\n37\/2 = around 18.5<\/result>","index":1437} +{"problem":"a person has 100 $ in 10 $ and 5 $ bill . if the 5 $ bill quantity is twice that of 10 $ bill . what is quantity of 10 $ .","rationale":"let amount of 10 $ be x . then amount of 5 $ be 2 x . now 5 * 2 x + 10 * x = 100 . thus x = 5 . answer : e","correct":"e","options":{"a":"2 ","b":"6 ","c":"7 ","d":"8","e":"5"},"options_float":{"a":2.0,"b":6.0,"c":7.0,"d":8.0,"e":5.0},"annotated_formula":"divide(divide(100, 10), const_2)","linear_formula":"divide(n0,n1)|divide(#0,const_2)","chain":"100 \/ 10<\/gadget>\n10<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":1438} +{"problem":"for a group of n people , k of whom are of the same sex , the ( n - k ) \/ n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 5 of whom are females , by how much does the index for the females exceed the index for the males in the group ?","rationale":"\"index for females = ( 20 - 5 ) \/ 20 = 3 \/ 4 = 0.75 index for males = ( 20 - 15 \/ 20 = 1 \/ 4 = 0.25 index for females exceeds males by 0.75 - 0.25 = 0.5 answer : c\"","correct":"c","options":{"a":"0.05 ","b":"0.0625 ","c":"0.5 ","d":"0.25","e":"0.6"},"options_float":{"a":0.05,"b":0.0625,"c":0.5,"d":0.25,"e":0.6},"annotated_formula":"subtract(divide(subtract(20, 5), 20), divide(5, 20))","linear_formula":"divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)|","chain":"20 - 5<\/gadget>\n15<\/output>\n15 \/ 20<\/gadget>\n3\/4 = around 0.75<\/output>\n5 \/ 20<\/gadget>\n1\/4 = around 0.25<\/output>\n(3\/4) - (1\/4)<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":1439} +{"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 12 meters , and the interior of tank b has a height of 12 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 12 \/ ( 2 * pi ) . the capacity of tank a is 10 * pi * 144 \/ ( 4 * pi ^ 2 ) = 360 \/ ( pi ) the radius of tank b is 10 \/ ( 2 * pi ) . the capacity of tank b is 12 * pi * 100 \/ ( 4 * pi ^ 2 ) = 300 \/ ( pi ) tank a \/ tank b = 360 \/ 300 = 12 \/ 10 = 120 % the answer is e .\"","correct":"e","options":{"a":"80 % ","b":"90 % ","c":"100 % ","d":"110 %","e":"120 %"},"options_float":{"a":80.0,"b":90.0,"c":100.0,"d":110.0,"e":120.0},"annotated_formula":"multiply(multiply(power(divide(12, 10), const_2), divide(10, 12)), const_100)","linear_formula":"divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)|","chain":"12 \/ 10<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) ** 2<\/gadget>\n36\/25 = around 1.44<\/output>\n10 \/ 12<\/gadget>\n5\/6 = around 0.833333<\/output>\n(36\/25) * (5\/6)<\/gadget>\n6\/5 = around 1.2<\/output>\n(6\/5) * 100<\/gadget>\n120<\/output>\n120<\/result>","index":1440} +{"problem":"eric throws 2 dice , and his score is the sum of the values shown . sandra throws one dice and her score is the square of the value shown . what is the probabilty that sandras score will be strictly higher than erics score ? ?","rationale":"sandra score can be like 1,4 , 9,16 , 25,36 eric score less then 1 - - > 0 eric score less then 4 = ( 1,1 ) , ( 1,2 ) ( 2,1 ) - - > 3 eric score less then 9 are ( 1,1 ) ( 1,2 ) ( 1,3 ) ( 1,4 ) ( 1,5 ) ( 1,6 ) ( 2,1 ) ( 2,2 ) ( 2,3 ) ( 2,4 ) ( 2,5 ) ( 2,6 ) ( 3,1 ) ( 3,2 ) ( 3,3 ) ( 3,4 ) ( 3,5 ) ( 4,1 ) ( 4,2 ) ( 4,3 ) ( 4,4 ) ( 5,1 ) ( 5,2 ) ( 5,3 ) ( 6,1 ) ( 6,2 ) - - > 26 eric score will always be less then 16 - - - > 36 eric score will always be less then 25 - - - > 36 eric score will always be less then 36 - - - > 36 total favorable outcomes = 3 + 26 + 36 + 36 + 36 = 137 total possible outcomes = 216 ( 36 * 6 ) probability = 137 \/ 216 answer : a","correct":"a","options":{"a":"137 \/ 216 ","b":"137 \/ 218 ","c":"137 \/ 217 ","d":"136 \/ 216","e":"138 \/ 216"},"options_float":{"a":0.6342592593,"b":0.628440367,"c":0.6313364055,"d":0.6296296296,"e":0.6388888889},"annotated_formula":"divide(add(add(add(add(const_3, subtract(power(multiply(2, const_3), const_2), const_10)), power(multiply(2, const_3), const_2)), power(multiply(2, const_3), const_2)), power(multiply(2, const_3), const_2)), multiply(power(multiply(2, const_3), const_2), multiply(2, const_3)))","linear_formula":"multiply(n0,const_3)|power(#0,const_2)|multiply(#0,#1)|subtract(#1,const_10)|add(#3,const_3)|add(#4,#1)|add(#5,#1)|add(#6,#1)|divide(#7,#2)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 ** 2<\/gadget>\n36<\/output>\n36 - 10<\/gadget>\n26<\/output>\n3 + 26<\/gadget>\n29<\/output>\n29 + 36<\/gadget>\n65<\/output>\n65 + 36<\/gadget>\n101<\/output>\n101 + 36<\/gadget>\n137<\/output>\n36 * 6<\/gadget>\n216<\/output>\n137 \/ 216<\/gadget>\n137\/216 = around 0.634259<\/output>\n137\/216 = around 0.634259<\/result>","index":1441} +{"problem":"if $ 910 are divided between worker a and worker b in the ratio 5 : 9 , what is the share that worker b will get ?","rationale":"worker b will get 9 \/ 14 = 64.29 % the answer is c .","correct":"c","options":{"a":"62.27 % ","b":"63.28 % ","c":"64.29 % ","d":"65.31 %","e":"66.32 %"},"options_float":{"a":62.27,"b":63.28,"c":64.29,"d":65.31,"e":66.32},"annotated_formula":"divide(910, add(5, 9))","linear_formula":"add(n1,n2)|divide(n0,#0)","chain":"5 + 9<\/gadget>\n14<\/output>\n910 \/ 14<\/gadget>\n65<\/output>\n65<\/result>","index":1442} +{"problem":"someone on a skateboard is traveling 8 miles per hour . how many feet does she travel in 5 seconds ? ( 1 mile = 5280 feet )","rationale":"\"per second = > 8 * 5280 ft \/ 60 * 60 = 11.73 ft 5 seconds = > 11.73 * 5 = 58.65 ft answer : e\"","correct":"e","options":{"a":"60 ft ","b":"52 ft ","c":"53 ft ","d":"55 ft","e":"58.65 ft"},"options_float":{"a":60.0,"b":52.0,"c":53.0,"d":55.0,"e":58.65},"annotated_formula":"multiply(5, divide(multiply(8, 5280), const_3600))","linear_formula":"multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"8 * 5_280<\/gadget>\n42_240<\/output>\n42_240 \/ 3_600<\/gadget>\n176\/15 = around 11.733333<\/output>\n5 * (176\/15)<\/gadget>\n176\/3 = around 58.666667<\/output>\n176\/3 = around 58.666667<\/result>","index":1443} +{"problem":"a car traveling at a certain constant speed takes 30 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 30 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?","rationale":"\"30 * t = 1 km = > t = 1 \/ 30 km \/ h v * ( t + 30 \/ 3600 ) = 1 v ( 1 \/ 30 + 30 \/ 3600 ) = 1 v ( 150 \/ 3600 ) = 1 v = 24 km \/ h the answer is e .\"","correct":"e","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(1, divide(add(multiply(const_3600, divide(1, 30)), 30), const_3600))","linear_formula":"divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)|","chain":"1 \/ 30<\/gadget>\n1\/30 = around 0.033333<\/output>\n3_600 * (1\/30)<\/gadget>\n120<\/output>\n120 + 30<\/gadget>\n150<\/output>\n150 \/ 3_600<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ (1\/24)<\/gadget>\n24<\/output>\n24<\/result>","index":1444} +{"problem":"of the families in city x in 1998 , 30 percent owned a personal computer . the number of families in city x owning a computer in 2002 was 20 percent greater than it was in 1998 , and the total number of families in city x was 8 percent greater in 2002 than it was in 1998 . what percent of the families in city x owned a personal computer in 2002 ?","rationale":"say a 100 families existed in 1998 then the number of families owning a computer in 1998 - 30 number of families owning computer in 2002 = 30 * 120 \/ 100 = 36 number of families in 2002 = 108 the percentage = 36 \/ 108 * 100 = 33.33 % . option : d","correct":"d","options":{"a":"50.12 % ","b":"52.66 % ","c":"56.33 % ","d":"33.33 %","e":"74.12 %"},"options_float":{"a":50.12,"b":52.66,"c":56.33,"d":33.33,"e":74.12},"annotated_formula":"multiply(const_100, divide(divide(multiply(add(20, const_100), 30), const_100), add(const_100, 8)))","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 + 8<\/gadget>\n108<\/output>\n36 \/ 108<\/gadget>\n1\/3 = around 0.333333<\/output>\n100 * (1\/3)<\/gadget>\n100\/3 = around 33.333333<\/output>\n100\/3 = around 33.333333<\/result>","index":1445} +{"problem":"recently , i decided to walk down an escalator of a tube station . i did some quick calculation in my mind . i found that if i walk down 20 ` ` 6 steps , i require thirty seconds to reach the bottom . however , if i am able to step down thirty ` ` 4 stairs , i would only require eighteen seconds to get to the bottom . if the time is measured from the moment the top step begins to descend to the time i step off the last step at the bottom ?","rationale":"26 steps 30 seconds and for 34 steps only 18 seconds left to reach botto . means he covered 8 steps ( i . e . 34 - 26 ) in 12 ( i . e 30 - 18 ) seconds the spped of the boy is 8 steps in 12 seconds after further simplyfy . . 2 steps in 3 seconds after 34 steps only 18 seconds , means 12 more steps are left total steps are 34 + 12 = 46 answer : e","correct":"e","options":{"a":"43 ","b":"44 ","c":"45 ","d":"40","e":"46"},"options_float":{"a":43.0,"b":44.0,"c":45.0,"d":40.0,"e":46.0},"annotated_formula":"add(add(multiply(const_3, const_10), 4), multiply(divide(subtract(add(multiply(const_3, const_10), 4), add(20, 6)), subtract(multiply(const_3, const_10), multiply(6, const_3))), multiply(6, const_3)))","linear_formula":"add(n0,n1)|multiply(const_10,const_3)|multiply(n1,const_3)|add(n2,#1)|subtract(#1,#2)|subtract(#3,#0)|divide(#5,#4)|multiply(#6,#2)|add(#3,#7)","chain":"3 * 10<\/gadget>\n30<\/output>\n30 + 4<\/gadget>\n34<\/output>\n20 + 6<\/gadget>\n26<\/output>\n34 - 26<\/gadget>\n8<\/output>\n6 * 3<\/gadget>\n18<\/output>\n30 - 18<\/gadget>\n12<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 18<\/gadget>\n12<\/output>\n34 + 12<\/gadget>\n46<\/output>\n46<\/result>","index":1446} +{"problem":"if y > 0 , ( 2 y ) \/ 20 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"soln : - can be reduced to y \/ 10 + 3 y \/ 10 = 2 y \/ 5 = 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":"multiply(const_100, add(divide(2, 20), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"2 \/ 20<\/gadget>\n1\/10 = around 0.1<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/10) + (3\/10)<\/gadget>\n2\/5 = around 0.4<\/output>\n100 * (2\/5)<\/gadget>\n40<\/output>\n40<\/result>","index":1447} +{"problem":"at a loading dock , each worker on the night crew loaded 2 \/ 3 as many boxes as each worker on the day crew . if the night crew has 5 \/ 6 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ?","rationale":"let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 5 x \/ 6 ) ( 2 y \/ 3 ) = 5 xy \/ 9 the total number of boxes is xy + 5 xy \/ 9 = 14 xy \/ 9 the fraction loaded by the day crew is xy \/ ( 14 xy \/ 9 ) = 9 \/ 14 the answer is d .","correct":"d","options":{"a":"6 \/ 11 ","b":"7 \/ 12 ","c":"8 \/ 13 ","d":"9 \/ 14","e":"11 \/ 15"},"options_float":{"a":0.5454545455,"b":0.5833333333,"c":0.6153846154,"d":0.6428571429,"e":0.7333333333},"annotated_formula":"divide(multiply(6, 3), add(multiply(6, 3), multiply(2, 5)))","linear_formula":"multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)","chain":"6 * 3<\/gadget>\n18<\/output>\n2 * 5<\/gadget>\n10<\/output>\n18 + 10<\/gadget>\n28<\/output>\n18 \/ 28<\/gadget>\n9\/14 = around 0.642857<\/output>\n9\/14 = around 0.642857<\/result>","index":1448} +{"problem":"if a train , travelling at a speed of 90 kmph , crosses a pole in 6 sec , then the length of train is ?","rationale":"\"e e = 90 * 5 \/ 18 * 6 = 150 m\"","correct":"e","options":{"a":"281 m ","b":"112 m ","c":"117 m ","d":"125 m","e":"150 m"},"options_float":{"a":281.0,"b":112.0,"c":117.0,"d":125.0,"e":150.0},"annotated_formula":"multiply(multiply(90, const_0_2778), 6)","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 * 6<\/gadget>\n150<\/output>\n150<\/result>","index":1449} +{"problem":"find the number of different prime factors of 25650","rationale":"explanation : l . c . m of 25650 = 2 x 3 x 3 x 3 x 5 x 5 x 19 3 , 2 , 5,19 number of different prime factors is 4 . answer : option a","correct":"a","options":{"a":"4 ","b":"2 ","c":"3 ","d":"5","e":"6"},"options_float":{"a":4.0,"b":2.0,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"add(const_2, const_2)","linear_formula":"add(const_2,const_2)","chain":"2 + 2<\/gadget>\n4<\/output>\n4<\/result>","index":1450} +{"problem":"the mean of 50 observations was 36 . it was found later that an observation 58 was wrongly taken as 43 . the corrected new mean is ?","rationale":"\"correct sum = ( 36 * 50 + 58 - 43 ) = 1815 . correct mean = 1815 \/ 50 = 36.3 answer : e\"","correct":"e","options":{"a":"36.7 ","b":"36.1 ","c":"36.5 ","d":"36.9","e":"36.3"},"options_float":{"a":36.7,"b":36.1,"c":36.5,"d":36.9,"e":36.3},"annotated_formula":"divide(add(multiply(36, 50), subtract(subtract(50, const_2), 43)), 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 - 43<\/gadget>\n5<\/output>\n1_800 + 5<\/gadget>\n1_805<\/output>\n1_805 \/ 50<\/gadget>\n361\/10 = around 36.1<\/output>\n361\/10 = around 36.1<\/result>","index":1451} +{"problem":"in a school of 450 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 % 450 * 18 \/ 100 = 81 answer : d\"","correct":"d","options":{"a":"72 ","b":"75 ","c":"80 ","d":"81","e":"90"},"options_float":{"a":72.0,"b":75.0,"c":80.0,"d":81.0,"e":90.0},"annotated_formula":"divide(multiply(450, 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>\n450 * 18<\/gadget>\n8_100<\/output>\n8_100 \/ 100<\/gadget>\n81<\/output>\n81<\/result>","index":1452} +{"problem":"find the volume and surface area of a cuboid 16 m long , 14 m broad and 7 m high .","rationale":"volume = ( 16 x 14 x 7 ) m ^ 3 = 1568 m ^ 3 . surface area = [ 2 ( 16 x 14 + 14 x 7 + 16 x 7 ) ] cm ^ 2 = ( 2 x 434 ) cm ^ 2 = 868 cm ^ 2 . answer is d","correct":"d","options":{"a":"878 cm ^ 2 ","b":"858 cm ^ 2 ","c":"838 cm ^ 2 ","d":"868 cm ^ 2","e":"none of them"},"options_float":{"a":878.0,"b":858.0,"c":838.0,"d":868.0,"e":null},"annotated_formula":"multiply(add(multiply(16, 7), add(multiply(16, 14), multiply(14, 7))), const_2)","linear_formula":"multiply(n0,n1)|multiply(n1,n2)|multiply(n0,n2)|add(#0,#1)|add(#3,#2)|multiply(#4,const_2)","chain":"16 * 7<\/gadget>\n112<\/output>\n16 * 14<\/gadget>\n224<\/output>\n14 * 7<\/gadget>\n98<\/output>\n224 + 98<\/gadget>\n322<\/output>\n112 + 322<\/gadget>\n434<\/output>\n434 * 2<\/gadget>\n868<\/output>\n868<\/result>","index":1455} +{"problem":"3 - twentieths of the members of a social club are retirees who are also bridge players , 5 - twentieths of the members are retirees , and one - half of the members are bridge players . if 120 of the members are neither retirees nor bridge players , what is the total number of members in the social club ?","rationale":"{ total } = { retirees } + { bridge players } - { both } + { neither } x = 5 \/ 20 * x + x \/ 2 - 3 \/ 20 * x + 120 20 x = 5 x + 10 x - 3 x + 120 * 20 ( multiply by 20 ) 12 x = 120 * 20 x = 200 . answer : c","correct":"c","options":{"a":"240 ","b":"300 ","c":"200 ","d":"400","e":"480"},"options_float":{"a":240.0,"b":300.0,"c":200.0,"d":400.0,"e":480.0},"annotated_formula":"multiply(subtract(120, multiply(5, const_4)), const_2)","linear_formula":"multiply(n1,const_4)|subtract(n2,#0)|multiply(#1,const_2)","chain":"5 * 4<\/gadget>\n20<\/output>\n120 - 20<\/gadget>\n100<\/output>\n100 * 2<\/gadget>\n200<\/output>\n200<\/result>","index":1457} +{"problem":"the total marks obtained by a student in mathematics and physics is 80 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 = 80 and c - p = 20 m + c \/ 2 = [ ( m + p ) + ( c - p ) ] \/ 2 = ( 80 + 20 ) \/ 2 = 50 . answer : d","correct":"d","options":{"a":"40 ","b":"30 ","c":"25 ","d":"50","e":"none of these ."},"options_float":{"a":40.0,"b":30.0,"c":25.0,"d":50.0,"e":null},"annotated_formula":"divide(add(80, 20), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"80 + 20<\/gadget>\n100<\/output>\n100 \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":1459} +{"problem":"the perimeter of a triangle is 44 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 * 44 \/ 2 = 55 cm 2 answer : b\"","correct":"b","options":{"a":"38 cm 2 ","b":"55 cm 2 ","c":"65 cm 2 ","d":"45 cm 2","e":"35 cm 2"},"options_float":{"a":38.0,"b":55.0,"c":65.0,"d":45.0,"e":35.0},"annotated_formula":"triangle_area(2.5, 44)","linear_formula":"triangle_area(n0,n1)|","chain":"(2.5 * 44) \/ 2<\/gadget>\n55<\/output>\n55<\/result>","index":1460} +{"problem":"the principal that amounts to rs . 4903 in 3 years at 6 1 \/ 4 % per annum c . i . compounded annually , is ?","rationale":"\"principal = [ 4913 \/ ( 1 + 25 \/ ( 4 * 100 ) ) 3 ] = 4903 * 16 \/ 17 * 16 \/ 17 * 16 \/ 17 = rs . 4076 . answer : b\"","correct":"b","options":{"a":"s . 3096 ","b":"s . 4076 ","c":"s . 4085 ","d":"s . 4096","e":"s . 5096"},"options_float":{"a":3096.0,"b":4076.0,"c":4085.0,"d":4096.0,"e":5096.0},"annotated_formula":"divide(4903, 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_903 \/ (4_913\/4_096)<\/gadget>\n20_082_688\/4_913 = around 4_087.662935<\/output>\n20_082_688\/4_913 = around 4_087.662935<\/result>","index":1461} +{"problem":"in a group of 95 students , 36 are taking history , and 32 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ?","rationale":"\"number of students taking history = h = 36 number of students taking statistics = s = 32 total number of students = t = 90 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 95 - 59 = 36 letxbe the number of students taking both history and statistics . then t = h + s + n - x or 95 = 36 + 32 + 36 - x or x = 9 now , number of students taking only history will be h - x or 36 - 9 = 27 answer : - e\"","correct":"e","options":{"a":"9 ","b":"19 ","c":"23 ","d":"45","e":"27"},"options_float":{"a":9.0,"b":19.0,"c":23.0,"d":45.0,"e":27.0},"annotated_formula":"subtract(36, subtract(add(36, 32), 59))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n1,#1)|","chain":"36 + 32<\/gadget>\n68<\/output>\n68 - 59<\/gadget>\n9<\/output>\n36 - 9<\/gadget>\n27<\/output>\n27<\/result>","index":1462} +{"problem":"the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 80 cents more than the commodity y ?","rationale":"\"the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 80 = $ 1.00 cents on commodity y , which will take 10 years . the answer is b .\"","correct":"b","options":{"a":"2010 . ","b":"2011 . ","c":"2012 . ","d":"2013 .","e":"2014 ."},"options_float":{"a":2010.0,"b":2011.0,"c":2012.0,"d":2013.0,"e":2014.0},"annotated_formula":"add(2001, divide(add(divide(80, const_100), subtract(4.40, 4.20)), subtract(divide(30, const_100), subtract(4.40, 4.20))))","linear_formula":"divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n4.4 - 4.2<\/gadget>\n0.2<\/output>\n(4\/5) + 0.2<\/gadget>\n1<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) - 0.2<\/gadget>\n0.1<\/output>\n1 \/ 0.1<\/gadget>\n10<\/output>\n2_001 + 10<\/gadget>\n2_011<\/output>\n2_011<\/result>","index":1463} +{"problem":"there are 13 clubs in a full deck of 52 cards . in a certain game , you pick a card from a standard deck of 52 cards . if the card is a club , you win . if the card is not a club , the person replaces the card to the deck , reshuffles , and draws again . the person keeps repeating that process until he picks a club , and the point is to measure how many draws it took before the person picked a club and , thereby , won . what is the probability that one will pick the first club on the forth draw or later ?","rationale":"favorable case = the club is picked in the third draw or later unfavorable cases = the club is picked in either first draw , second draw or third draws probability = favorable outcomes \/ total out comes also probability = 1 - ( unfavorable outcomes \/ total out comes ) unfavorable case 1 : probability of club picked in first draw = 13 \/ 52 = 1 \/ 4 unfavorable case 2 : probability of club picked in second draw ( i . e . first draw is not club ) = ( 39 \/ 52 ) * ( 13 \/ 52 ) = ( 3 \/ 4 ) * ( 1 \/ 4 ) = 3 \/ 16 unfavorable case 3 : probability of club picked in third draw ( i . e . first and 2 nd draws are not clubs ) = ( 39 \/ 52 ) * ( 39 \/ 52 ) * ( 13 \/ 52 ) = ( 3 \/ 4 ) * ( 3 \/ 4 ) * ( 1 \/ 4 ) = 9 \/ 64 total unfavorable probability = ( 1 \/ 4 ) + ( 3 \/ 16 ) + ( 9 \/ 64 ) = ( 16 \/ 64 ) + ( 12 \/ 64 ) + ( 9 \/ 64 ) = 37 \/ 64 i . e . , favorable probability = 1 - ( 37 \/ 64 ) = 27 \/ 64 answer : option e","correct":"e","options":{"a":"1 \/ 2 ","b":"3 \/ 4 ","c":"7 \/ 8 ","d":"27 \/ 32","e":"27 \/ 64"},"options_float":{"a":0.5,"b":0.75,"c":0.875,"d":0.84375,"e":0.421875},"annotated_formula":"multiply(multiply(divide(subtract(52, 13), 52), divide(subtract(52, 13), 52)), divide(subtract(52, 13), 52))","linear_formula":"subtract(n1,n0)|divide(#0,n1)|multiply(#1,#1)|multiply(#1,#2)","chain":"52 - 13<\/gadget>\n39<\/output>\n39 \/ 52<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * (3\/4)<\/gadget>\n9\/16 = around 0.5625<\/output>\n(9\/16) * (3\/4)<\/gadget>\n27\/64 = around 0.421875<\/output>\n27\/64 = around 0.421875<\/result>","index":1464} +{"problem":"the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 70 , and what number ?","rationale":"\"a 1 = 120 \/ 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 80 = 25 answer : b\"","correct":"b","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":"subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 70))","linear_formula":"add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)|","chain":"20 + 40<\/gadget>\n60<\/output>\n60 + 60<\/gadget>\n120<\/output>\n5 * 3<\/gadget>\n15<\/output>\n15 + 10<\/gadget>\n25<\/output>\n25 + 70<\/gadget>\n95<\/output>\n120 - 95<\/gadget>\n25<\/output>\n25<\/result>","index":1465} +{"problem":"if 28 % of a number exceeds 18 % of it by 7.2 , then find the number ?","rationale":"use the elimination method to find the correct option . of all the options only 90 fits 28 % of 90 = 25.2 18 % of 90 = 16.2 25.2 - 16.2 = 7.2 required number is 90 . answer : e","correct":"e","options":{"a":"50 ","b":"34 ","c":"55 ","d":"60","e":"90"},"options_float":{"a":50.0,"b":34.0,"c":55.0,"d":60.0,"e":90.0},"annotated_formula":"add(divide(7.2, divide(subtract(28, 18), const_100)), 18)","linear_formula":"subtract(n0,n1)|divide(#0,const_100)|divide(n2,#1)|add(n1,#2)","chain":"28 - 18<\/gadget>\n10<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n7.2 \/ (1\/10)<\/gadget>\n72<\/output>\n72 + 18<\/gadget>\n90<\/output>\n90<\/result>","index":1467} +{"problem":"how many factors of 60 are odd numbers greater than 1 ?","rationale":"prime factors of 60 are 2 ^ 2,3 ^ 1,5 ^ 1 total divisors = ( power if a prime factor + 1 ) total no . of odd factors ( 3,5 , ) = ( 1 + 1 ) ( 1 + 1 ) = 4 since we need odd divisors other than 1 = > 4 - 1 = 3 odd divisors d is the answer","correct":"d","options":{"a":"3 ","b":"4 ","c":"5 ","d":"3","e":"7"},"options_float":{"a":3.0,"b":4.0,"c":5.0,"d":3.0,"e":7.0},"annotated_formula":"divide(60, multiply(const_10, const_2))","linear_formula":"multiply(const_10,const_2)|divide(n0,#0)","chain":"10 * 2<\/gadget>\n20<\/output>\n60 \/ 20<\/gadget>\n3<\/output>\n3<\/result>","index":1468} +{"problem":"a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent e of the distribution is less than m + d ?","rationale":"d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , e = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d . d","correct":"d","options":{"a":"16 % ","b":"32 % ","c":"48 % ","d":"84 %","e":"92 %"},"options_float":{"a":16.0,"b":32.0,"c":48.0,"d":84.0,"e":92.0},"annotated_formula":"subtract(const_100, divide(subtract(const_100, 68), const_2))","linear_formula":"subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)","chain":"100 - 68<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n100 - 16<\/gadget>\n84<\/output>\n84<\/result>","index":1469} +{"problem":"village p ’ s population is 1150 greater than village q ' s population . if village q ’ s population were reduced by 200 people , then village p ’ s population would be 4 times as large as village q ' s population . what is village q ' s current population ?","rationale":"p = q + 1150 . p = 4 ( q - 200 ) . 4 ( q - 200 ) = q + 1150 . 3 q = 1950 . q = 650 . the answer is c .","correct":"c","options":{"a":"600 ","b":"625 ","c":"650 ","d":"675","e":"700"},"options_float":{"a":600.0,"b":625.0,"c":650.0,"d":675.0,"e":700.0},"annotated_formula":"divide(add(1150, multiply(200, 4)), const_3)","linear_formula":"multiply(n1,n2)|add(n0,#0)|divide(#1,const_3)","chain":"200 * 4<\/gadget>\n800<\/output>\n1_150 + 800<\/gadget>\n1_950<\/output>\n1_950 \/ 3<\/gadget>\n650<\/output>\n650<\/result>","index":1470} +{"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 hours e 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 e 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":1474} +{"problem":"jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 50,000 after completion of one year . what will be madhu ' s share of profit ?","rationale":"\"30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 \/ 2 * 50,000 i . e . rs . 25,000 answer : c\"","correct":"c","options":{"a":"rs . 27,000 ","b":"rs . 24,000 ","c":"rs . 25,000 ","d":"rs . 36,000","e":"none of these"},"options_float":{"a":27000.0,"b":24000.0,"c":25000.0,"d":36000.0,"e":null},"annotated_formula":"multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3)))","linear_formula":"add(n1,const_3)|multiply(n1,const_4)|multiply(const_3,const_4)|multiply(#0,n1)|multiply(#0,const_100)|divide(#3,n1)|multiply(#1,#3)|multiply(#4,const_100)|divide(#5,#2)|multiply(#6,const_100)|add(#9,#7)|multiply(#10,#8)|","chain":"4 * 2<\/gadget>\n8<\/output>\n2 + 3<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n8 * 10<\/gadget>\n80<\/output>\n80 * 100<\/gadget>\n8_000<\/output>\n5 * 100<\/gadget>\n500<\/output>\n500 * 100<\/gadget>\n50_000<\/output>\n8_000 + 50_000<\/gadget>\n58_000<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n4 * 3<\/gadget>\n12<\/output>\n5 \/ 12<\/gadget>\n5\/12 = around 0.416667<\/output>\n58_000 * (5\/12)<\/gadget>\n72_500\/3 = around 24_166.666667<\/output>\n72_500\/3 = around 24_166.666667<\/result>","index":1478} +{"problem":"in a certain school , the ratio of boys to girls is 5 to 13 . if there are 160 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 160 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 160 . so 1 part corresponds to 160 \/ 8 = 20 and so 5 parts correspond to 5 * 10 = 100 . a\"","correct":"a","options":{"a":"100 ","b":"36 ","c":"45 ","d":"72","e":"117"},"options_float":{"a":100.0,"b":36.0,"c":45.0,"d":72.0,"e":117.0},"annotated_formula":"subtract(divide(160, subtract(const_1, divide(5, 13))), 160)","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>\n160 \/ (8\/13)<\/gadget>\n260<\/output>\n260 - 160<\/gadget>\n100<\/output>\n100<\/result>","index":1480} +{"problem":"ajay can walk 4 km in 1 hour . in how many hours he can walk 40 km ?","rationale":"\"1 hour he walk 4 km he walk 40 km in = 40 \/ 4 * 1 = 10 hours answer is b\"","correct":"b","options":{"a":"5 hrs ","b":"10 hrs ","c":"15 hrs ","d":"20 hrs","e":"30 hrs"},"options_float":{"a":5.0,"b":10.0,"c":15.0,"d":20.0,"e":30.0},"annotated_formula":"divide(40, 4)","linear_formula":"divide(n2,n0)|","chain":"40 \/ 4<\/gadget>\n10<\/output>\n10<\/result>","index":1481} +{"problem":"ramesh has solved 108 questions in an examination . if he got only ‘ 0 ’ marks , then how many questions were wrong when one mark is given for each one correct answer and 1 \/ 3 mark is subtracted on each wrong answer .","rationale":"if ramesh attempts ' x ' questions correct and ' y ' questions wrong , then x + y = 108 - - - ( i ) & x - ( 1 \/ 3 ) y = 0 - - - ( ii ) on solving x = 27 , y = 81 answer : d","correct":"d","options":{"a":"78 ","b":"79 ","c":"80 ","d":"81","e":"82"},"options_float":{"a":78.0,"b":79.0,"c":80.0,"d":81.0,"e":82.0},"annotated_formula":"subtract(108, divide(multiply(divide(1, 3), 108), add(const_1, divide(1, 3))))","linear_formula":"divide(n2,n3)|add(#0,const_1)|multiply(n0,#0)|divide(#2,#1)|subtract(n0,#3)","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 108<\/gadget>\n36<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n36 \/ (4\/3)<\/gadget>\n27<\/output>\n108 - 27<\/gadget>\n81<\/output>\n81<\/result>","index":1482} +{"problem":"what is the measure of the angle x 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 x = 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":1483} +{"problem":"after decreasing 15 % in the price of an article costs rs . 915 . find the actual cost of an article ?","rationale":"\"cp * ( 85 \/ 100 ) = 915 cp = 10.76 * 100 = > cp = 1076 answer : c\"","correct":"c","options":{"a":"915 ","b":"1000 ","c":"1076 ","d":"1067","e":"1760"},"options_float":{"a":915.0,"b":1000.0,"c":1076.0,"d":1067.0,"e":1760.0},"annotated_formula":"divide(915, subtract(const_1, divide(15, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|","chain":"15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 - (3\/20)<\/gadget>\n17\/20 = around 0.85<\/output>\n915 \/ (17\/20)<\/gadget>\n18_300\/17 = around 1_076.470588<\/output>\n18_300\/17 = around 1_076.470588<\/result>","index":1484} +{"problem":"dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 12 hours and then stops , how many hours will it take annie , working alone , to complete the job ?","rationale":"\"dan can complete 1 \/ 15 of the job per hour . in 12 hours , dan completes 12 ( 1 \/ 15 ) = 4 \/ 5 of the job . annie can complete 1 \/ 10 of the job per hour . to complete the job , annie will take 1 \/ 5 \/ 1 \/ 10 = 2 hours . the answer is 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":"multiply(subtract(const_1, divide(12, 15)), 10)","linear_formula":"divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)|","chain":"12 \/ 15<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 10<\/gadget>\n2<\/output>\n2<\/result>","index":1485} +{"problem":"a circle graph shows how the megatech corporation allocates its research and development budget : 17 % microphotonics ; 24 % home electronics ; 15 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ?","rationale":"here all percentage when summed we need to get 100 % . as per data 17 + 24 + 15 + 29 + 8 = 93 % . so remaining 7 % is the balance for the astrophysics . since this is a circle all percentage must be equal to 360 degrees . 100 % - - - - 360 degrees then 7 % will be 26 degrees . . imo option c .","correct":"c","options":{"a":"8 ° ","b":"10 ° ","c":"26 ° ","d":"36 °","e":"52 °"},"options_float":{"a":8.0,"b":10.0,"c":26.0,"d":36.0,"e":52.0},"annotated_formula":"divide(multiply(subtract(const_100, add(add(add(add(17, 24), 15), 29), 8)), divide(const_3600, const_10)), const_100)","linear_formula":"add(n0,n1)|divide(const_3600,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|subtract(const_100,#4)|multiply(#1,#5)|divide(#6,const_100)","chain":"17 + 24<\/gadget>\n41<\/output>\n41 + 15<\/gadget>\n56<\/output>\n56 + 29<\/gadget>\n85<\/output>\n85 + 8<\/gadget>\n93<\/output>\n100 - 93<\/gadget>\n7<\/output>\n3_600 \/ 10<\/gadget>\n360<\/output>\n7 * 360<\/gadget>\n2_520<\/output>\n2_520 \/ 100<\/gadget>\n126\/5 = around 25.2<\/output>\n126\/5 = around 25.2<\/result>","index":1486} +{"problem":"peter invests a sum of money and gets back an amount of $ 830 in 3 years . david invests an equal amount of money and gets an amount of $ 854 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 = 854 - 830 = 24 interest earned for 3 years = 24 * 3 = 72 amount invested = 830 - 72 = 758 answer : d\"","correct":"d","options":{"a":"670 ","b":"664 ","c":"698 ","d":"758","e":"700"},"options_float":{"a":670.0,"b":664.0,"c":698.0,"d":758.0,"e":700.0},"annotated_formula":"subtract(830, multiply(divide(subtract(854, 830), 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":"854 - 830<\/gadget>\n24<\/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>\n24 \/ (1\/100)<\/gadget>\n2_400<\/output>\n2_400 * (3\/100)<\/gadget>\n72<\/output>\n830 - 72<\/gadget>\n758<\/output>\n758<\/result>","index":1489} +{"problem":"the simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the rate of 8 per cent in 8 months . how much amount was borrowed ?","rationale":"let the amount be x . from the question , x × 14 × 6 \/ 1200 − x × 8 × 8 \/ 1200 = 250 ∴ x = 15000 answer a","correct":"a","options":{"a":"15000 ","b":"25000 ","c":"7500 ","d":"14500","e":"none of these"},"options_float":{"a":15000.0,"b":25000.0,"c":7500.0,"d":14500.0,"e":null},"annotated_formula":"divide(250, subtract(multiply(divide(14, const_12), divide(6, const_100)), multiply(divide(8, const_12), divide(8, const_100))))","linear_formula":"divide(n0,const_12)|divide(n1,const_100)|divide(n3,const_12)|divide(n3,const_100)|multiply(#0,#1)|multiply(#2,#3)|subtract(#4,#5)|divide(n2,#6)","chain":"14 \/ 12<\/gadget>\n7\/6 = around 1.166667<\/output>\n6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n(7\/6) * (3\/50)<\/gadget>\n7\/100 = around 0.07<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/3) * (2\/25)<\/gadget>\n4\/75 = around 0.053333<\/output>\n(7\/100) - (4\/75)<\/gadget>\n1\/60 = around 0.016667<\/output>\n250 \/ (1\/60)<\/gadget>\n15_000<\/output>\n15_000<\/result>","index":1490} +{"problem":"if $ 120 invested at a certain rate of simple interest amounts to $ 240 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ?","rationale":"\"120 amounts to 240 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 240 120 + 120 * ( r \/ 100 ) * ( 3 ) = 240 = > r = 100 \/ 3 150 in 6 years = principal + interest = 150 + 150 * ( r \/ 100 ) * ( 6 ) = 450 answer is e .\"","correct":"e","options":{"a":"$ 190 ","b":"$ 180 ","c":"$ 200 ","d":"$ 340","e":"$ 450"},"options_float":{"a":190.0,"b":180.0,"c":200.0,"d":340.0,"e":450.0},"annotated_formula":"add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(240, 120), 120), 120), 3)), 120))","linear_formula":"multiply(n3,n4)|subtract(n1,n0)|multiply(#1,n0)|divide(#2,n0)|divide(#3,n2)|multiply(#4,#0)|divide(#5,n0)|add(n3,#6)|","chain":"150 * 6<\/gadget>\n900<\/output>\n240 - 120<\/gadget>\n120<\/output>\n120 * 120<\/gadget>\n14_400<\/output>\n14_400 \/ 120<\/gadget>\n120<\/output>\n120 \/ 3<\/gadget>\n40<\/output>\n900 * 40<\/gadget>\n36_000<\/output>\n36_000 \/ 120<\/gadget>\n300<\/output>\n150 + 300<\/gadget>\n450<\/output>\n450<\/result>","index":1492} +{"problem":"a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 9 : 6 : 5 , respectively , by weight . if 40 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ?","rationale":"we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 6 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : 9 x + 6 x + 5 x = 40 20 x = 40 x = 2 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( 5 ) ( 2 ) = 10 pounds of apples = ( 9 ) ( 2 ) = 18 thus we know that there are 18 – 10 = 8 more pounds of apples than grapes . answer is c .","correct":"c","options":{"a":"15 ","b":"12 ","c":"8 ","d":"6","e":"4"},"options_float":{"a":15.0,"b":12.0,"c":8.0,"d":6.0,"e":4.0},"annotated_formula":"subtract(multiply(9, const_2), multiply(5, const_2))","linear_formula":"multiply(n0,const_2)|multiply(n2,const_2)|subtract(#0,#1)","chain":"9 * 2<\/gadget>\n18<\/output>\n5 * 2<\/gadget>\n10<\/output>\n18 - 10<\/gadget>\n8<\/output>\n8<\/result>","index":1494} +{"problem":"the population of a town increases 20 % and 25 % respectively in two consecutive years . after the growth the present population of the town is 1500 . then what is the population of the town 2 years ago ?","rationale":"\"explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1500 * 100 \/ 120 * 100 \/ 125 = 1000 answer : option a\"","correct":"a","options":{"a":"1000 ","b":"1100 ","c":"1200 ","d":"1300","e":"1400"},"options_float":{"a":1000.0,"b":1100.0,"c":1200.0,"d":1300.0,"e":1400.0},"annotated_formula":"divide(1500, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n(6\/5) * (5\/4)<\/gadget>\n3\/2 = around 1.5<\/output>\n1_500 \/ (3\/2)<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1495} +{"problem":"find the average of all the numbers between 6 and 34 which are divisible by 5 .","rationale":"\"solution average = ( 10 + 15 + 20 + 25 + 30 \/ 5 ) = 100 \/ 5 = 20 . answer b\"","correct":"b","options":{"a":"18 ","b":"20 ","c":"24 ","d":"30","e":"32"},"options_float":{"a":18.0,"b":20.0,"c":24.0,"d":30.0,"e":32.0},"annotated_formula":"divide(add(add(6, const_4), subtract(34, const_4)), const_2)","linear_formula":"add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|","chain":"6 + 4<\/gadget>\n10<\/output>\n34 - 4<\/gadget>\n30<\/output>\n10 + 30<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":1498} +{"problem":"in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.24 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]","rationale":"\"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 24 cents ; a half - liter of kerosene costs as much as 8 eggs - - > 8 eggs = 1 \/ 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 16 eggs = 16 \/ 12 * 24 = 32 cents . answer : c .\"","correct":"c","options":{"a":"0.32 ","b":"0.44 ","c":"32 ","d":"44","e":"55"},"options_float":{"a":0.32,"b":0.44,"c":32.0,"d":44.0,"e":55.0},"annotated_formula":"multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.24, 100))","linear_formula":"divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n8 \/ (1\/2)<\/gadget>\n16<\/output>\n16 \/ 12<\/gadget>\n4\/3 = around 1.333333<\/output>\n0.24 * 100<\/gadget>\n24<\/output>\n(4\/3) * 24<\/gadget>\n32<\/output>\n32<\/result>","index":1499} +{"problem":"a train running at the speed of 162 km \/ hr crosses a pole in 9 seconds . find the length of the train .","rationale":"\"speed = 162 * ( 5 \/ 18 ) m \/ sec = 45 m \/ sec length of train ( distance ) = speed * time ( 45 ) * 9 = 405 meter answer : d\"","correct":"d","options":{"a":"150 meter ","b":"286 meter ","c":"186 meter ","d":"405 meter","e":"265 meter"},"options_float":{"a":150.0,"b":286.0,"c":186.0,"d":405.0,"e":265.0},"annotated_formula":"multiply(divide(multiply(162, const_1000), const_3600), 9)","linear_formula":"multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|","chain":"162 * 1_000<\/gadget>\n162_000<\/output>\n162_000 \/ 3_600<\/gadget>\n45<\/output>\n45 * 9<\/gadget>\n405<\/output>\n405<\/result>","index":1500} +{"problem":"a student has to obtain 30 % of the total marks to pass . he got 150 marks and failed by 30 marks . the maximum marks are ?","rationale":"\"let the maximum marks be x then , 30 % of x = 150 + 30 30 x \/ 100 = 180 30 x = 180 * 100 = 18000 x = 600 answer is b\"","correct":"b","options":{"a":"750 ","b":"600 ","c":"650 ","d":"550","e":"500"},"options_float":{"a":750.0,"b":600.0,"c":650.0,"d":550.0,"e":500.0},"annotated_formula":"divide(add(150, 30), divide(30, const_100))","linear_formula":"add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|","chain":"150 + 30<\/gadget>\n180<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n180 \/ (3\/10)<\/gadget>\n600<\/output>\n600<\/result>","index":1501} +{"problem":"a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 5 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ?","rationale":"\"1 \/ x - 1 \/ 6 = - 1 \/ 8 x = 24 hrs 24 * 60 * 5 = 7200 . answer : c\"","correct":"c","options":{"a":"5729 ","b":"5760 ","c":"7200 ","d":"2870","e":"2799"},"options_float":{"a":5729.0,"b":5760.0,"c":7200.0,"d":2870.0,"e":2799.0},"annotated_formula":"divide(multiply(5, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1))","linear_formula":"multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)|","chain":"8 * 60<\/gadget>\n480<\/output>\n5 * 480<\/gadget>\n2_400<\/output>\n6 * 60<\/gadget>\n360<\/output>\n480 \/ 360<\/gadget>\n4\/3 = around 1.333333<\/output>\n(4\/3) - 1<\/gadget>\n1\/3 = around 0.333333<\/output>\n2_400 \/ (1\/3)<\/gadget>\n7_200<\/output>\n7_200<\/result>","index":1502} +{"problem":"we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 560 . how many green hats did we buy ?","rationale":"\"let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 560 . 6 ( 85 - g ) + 7 g = 560 . 510 - 6 g + 7 g = 560 . g = 560 - 510 = 50 . the answer is b .\"","correct":"b","options":{"a":"36 ","b":"50 ","c":"40 ","d":"42","e":"44"},"options_float":{"a":36.0,"b":50.0,"c":40.0,"d":42.0,"e":44.0},"annotated_formula":"subtract(560, multiply(85, 6))","linear_formula":"multiply(n0,n1)|subtract(n3,#0)|","chain":"85 * 6<\/gadget>\n510<\/output>\n560 - 510<\/gadget>\n50<\/output>\n50<\/result>","index":1503} +{"problem":"the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1498 respectively . what is the rate of interest ?","rationale":"\"1498 - 1400 = 98 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 98 ) \/ ( 1400 ) = 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":"divide(multiply(subtract(1498, 1400), const_100), 1400)","linear_formula":"subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)|","chain":"1_498 - 1_400<\/gadget>\n98<\/output>\n98 * 100<\/gadget>\n9_800<\/output>\n9_800 \/ 1_400<\/gadget>\n7<\/output>\n7<\/result>","index":1505} +{"problem":"a man can row 9 kmph in still water . when the river is running at 3.1 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?","rationale":"\"m = 9 s = 3.1 ds = 12.1 us = 5.9 x \/ 12.1 + x \/ 5.9 = 1 x = 3.97 d = 3.97 * 2 = 7.94 answer : e\"","correct":"e","options":{"a":"2.21 ","b":"2.48 ","c":"9.24 ","d":"7.29","e":"7.94"},"options_float":{"a":2.21,"b":2.48,"c":9.24,"d":7.29,"e":7.94},"annotated_formula":"multiply(divide(multiply(add(9, 3.1), subtract(9, 3.1)), add(add(9, 3.1), subtract(9, 3.1))), const_2)","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|","chain":"9 + 3.1<\/gadget>\n12.1<\/output>\n9 - 3.1<\/gadget>\n5.9<\/output>\n12.1 * 5.9<\/gadget>\n71.39<\/output>\n12.1 + 5.9<\/gadget>\n18<\/output>\n71.39 \/ 18<\/gadget>\n3.966111<\/output>\n3.966111 * 2<\/gadget>\n7.932222<\/output>\n7.932222<\/result>","index":1506} +{"problem":"a man invests some money partly in 12 % stock at 105 and partly in 8 % stock at 88 . to obtain equal dividends from both , he must invest the money in the ratio :","rationale":"\"n case of stock 1 , if he invest rs . 105 , he will get a dividend of rs . 12 ( assume face value = 100 ) in case of stock 2 , if he invest rs . 88 , he will get a dividend of rs . 8 ( assume face value = 100 ) ie , if he invest rs . ( 88 * 12 ) \/ 8 , he will get a dividend of rs . 12 required ratio = 105 : ( 88 × 12 ) \/ 8 = 105 : ( 11 × 12 ) = 35 : ( 11 × 4 ) = 35 : 44 answer : option d\"","correct":"d","options":{"a":"31 : 44 ","b":"31 : 27 ","c":"16 : 15 ","d":"35 : 44","e":"35 : 27"},"options_float":{"a":0.7045454545,"b":1.1481481481,"c":1.0666666667,"d":0.7954545455,"e":1.2962962963},"annotated_formula":"divide(multiply(105, const_2), multiply(88, const_3))","linear_formula":"multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)|","chain":"105 * 2<\/gadget>\n210<\/output>\n88 * 3<\/gadget>\n264<\/output>\n210 \/ 264<\/gadget>\n35\/44 = around 0.795455<\/output>\n35\/44 = around 0.795455<\/result>","index":1507} +{"problem":"a man cycling along the road noticed that every 15 minutes a bus overtakes him and every 5 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ?","rationale":"\"let ' s say the distance between the buses is d . we want to determine interval = \\ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 15 minutes a bus overtakes cyclist : \\ frac { d } { b - c } = 15 , d = 15 b - 15 c ; every 5 minutes cyclist meets an oncoming bus : \\ frac { d } { b + c } = 4 , d = 4 b + 4 c ; d = 15 b - 15 c = 5 b + 5 c , - - > b = 2 c , - - > d = 15 b - 15 b \/ 2 = 15 b \/ 2 . interval = \\ frac { d } { b } = \\ frac { 15 \/ 2 b } { b } = 15 \/ 2 answer : e ( 15 \/ 2 minutes ) .\"","correct":"e","options":{"a":"5 minutes ","b":"6 minutes ","c":"8 minutes ","d":"9 minutes","e":"15 \/ 2 minutes"},"options_float":{"a":5.0,"b":6.0,"c":8.0,"d":9.0,"e":7.5},"annotated_formula":"divide(subtract(15, divide(15, divide(add(5, 15), subtract(15, 5)))), const_1)","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|divide(n0,#2)|subtract(n0,#3)|divide(#4,const_1)|","chain":"5 + 15<\/gadget>\n20<\/output>\n15 - 5<\/gadget>\n10<\/output>\n20 \/ 10<\/gadget>\n2<\/output>\n15 \/ 2<\/gadget>\n15\/2 = around 7.5<\/output>\n15 - (15\/2)<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) \/ 1<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":1508} +{"problem":"rakesh ' s mathematics test had 75 problems , 10 arithmetic , 30 algebra , 35 geometry problems . although he answered 70 % of arithmetic , 40 % of arithmetic and 60 % of geometry problems correctly , still he got less than 60 % problems right . how many more questions he would have to answer more to get passed ?","rationale":"explanation : number of questions attempted correctly = ( 70 % of 10 + 40 % of 30 + 60 % of 35 ) = 7 + 12 + 21 = 40 . questions to be answered correctly for 60 % = 60 % of total quotations = 60 % of 75 = 45 . he would have to answer 45 - 40 = 5 answer : 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":"subtract(divide(multiply(75, 60), const_100), add(add(divide(multiply(10, 70), const_100), divide(multiply(30, 40), const_100)), divide(multiply(35, 60), const_100)))","linear_formula":"multiply(n0,n6)|multiply(n1,n4)|multiply(n2,n5)|multiply(n3,n6)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|add(#5,#6)|add(#8,#7)|subtract(#4,#9)","chain":"75 * 60<\/gadget>\n4_500<\/output>\n4_500 \/ 100<\/gadget>\n45<\/output>\n10 * 70<\/gadget>\n700<\/output>\n700 \/ 100<\/gadget>\n7<\/output>\n30 * 40<\/gadget>\n1_200<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n7 + 12<\/gadget>\n19<\/output>\n35 * 60<\/gadget>\n2_100<\/output>\n2_100 \/ 100<\/gadget>\n21<\/output>\n19 + 21<\/gadget>\n40<\/output>\n45 - 40<\/gadget>\n5<\/output>\n5<\/result>","index":1509} +{"problem":"a number x is multiplied by 7 , and this product is then divided by 3 . 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 ( 7 x \/ 3 ) to be perfect square x has to 7 \/ 3 ans : b\"","correct":"b","options":{"a":"25 \/ 9 ","b":"7 \/ 3 ","c":"5 \/ 3 ","d":"3 \/ 5","e":"9 \/ 25"},"options_float":{"a":2.7777777778,"b":2.3333333333,"c":1.6666666667,"d":0.6,"e":0.36},"annotated_formula":"divide(7, 3)","linear_formula":"divide(n0,n1)|","chain":"7 \/ 3<\/gadget>\n7\/3 = around 2.333333<\/output>\n7\/3 = around 2.333333<\/result>","index":1510} +{"problem":"one day , connie plays a game with a fair 6 - sided die . connie rolls the die until she rolls a 6 , at which point the game ends . if she rolls a 6 on her first turn , connie wins 6 dollars . for each subsequent turn , connie wins 1 6 of the amount she would have won the previous turn . what is connie ' s expected earnings from the game ?","rationale":"connie has a 1 6 chance of winning 6 dollars her first turn . she has a 5 \/ 6 1 \/ 6 chance of winning 1 dollar her second turn . next , she has a 25 36 1 \/ 6 chance of winning 1 \/ 6 dollars her third turn . generalizing , connie ' s expected earnings form a geometric series with initial term 1 \/ 6 * 6 = 1 and common ratio 5 \/ 6 * 1 \/ 6 = 5 \/ 36 . hence , connie ' s expected earnings are 1 \/ 1 - 5 \/ 36 = 36 \/ 31 correct answer d","correct":"d","options":{"a":"32 \/ 31 ","b":"33 \/ 31 ","c":"34 \/ 31 ","d":"36 \/ 31","e":"0 \/ 31"},"options_float":{"a":1.0322580645,"b":1.064516129,"c":1.0967741935,"d":1.1612903226,"e":0.0},"annotated_formula":"divide(const_1, subtract(const_1, divide(divide(subtract(6, 1), 6), 6)))","linear_formula":"subtract(n0,n4)|divide(#0,n0)|divide(#1,n0)|subtract(const_1,#2)|divide(const_1,#3)","chain":"6 - 1<\/gadget>\n5<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(5\/6) \/ 6<\/gadget>\n5\/36 = around 0.138889<\/output>\n1 - (5\/36)<\/gadget>\n31\/36 = around 0.861111<\/output>\n1 \/ (31\/36)<\/gadget>\n36\/31 = around 1.16129<\/output>\n36\/31 = around 1.16129<\/result>","index":1512} +{"problem":"in the manufacture of a certain product , 6 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":"\"0.06 * 0.05 = 0.003 = 0.3 % the answer is b .\"","correct":"b","options":{"a":"0.125 % ","b":"0.3 % ","c":"0.8 % ","d":"1.25 %","e":"2.0 %"},"options_float":{"a":0.125,"b":0.3,"c":0.8,"d":1.25,"e":2.0},"annotated_formula":"multiply(6, divide(5, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n6 * (1\/20)<\/gadget>\n3\/10 = around 0.3<\/output>\n3\/10 = around 0.3<\/result>","index":1513} +{"problem":"in a group of ducks and cows , the total number of legs are 26 more than twice the no . of heads . find the total no . of buffaloes .","rationale":"\"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 26 = > 2 x = 26 = > x = 13 c\"","correct":"c","options":{"a":"11 ","b":"12 ","c":"13 ","d":"16","e":"18"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":16.0,"e":18.0},"annotated_formula":"divide(26, const_2)","linear_formula":"divide(n0,const_2)|","chain":"26 \/ 2<\/gadget>\n13<\/output>\n13<\/result>","index":1514} +{"problem":"a man walking at 3 \/ 4 th of the speed , reaches his office late by 2 hours . what is the usual time ?","rationale":"at 3 \/ 4 th of speed he is late by ' 2 hrs ' x - 3 \/ 4 ( x ) = 2 x = 8 so 8 - 2 = 6 hrs ( since 2 hrs late ) answer : c","correct":"c","options":{"a":"5 hours ","b":"3 hours ","c":"6 hours ","d":"12 hours","e":"15 hours"},"options_float":{"a":5.0,"b":3.0,"c":6.0,"d":12.0,"e":15.0},"annotated_formula":"divide(multiply(multiply(multiply(divide(3, 4), 2), divide(3, 4)), 2), subtract(multiply(divide(3, 4), 2), multiply(multiply(divide(3, 4), 2), divide(3, 4))))","linear_formula":"divide(n0,n1)|multiply(n2,#0)|multiply(#0,#1)|multiply(n2,#2)|subtract(#1,#2)|divide(#3,#4)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * (3\/4)<\/gadget>\n9\/8 = around 1.125<\/output>\n(9\/8) * 2<\/gadget>\n9\/4 = around 2.25<\/output>\n(3\/2) - (9\/8)<\/gadget>\n3\/8 = around 0.375<\/output>\n(9\/4) \/ (3\/8)<\/gadget>\n6<\/output>\n6<\/result>","index":1515} +{"problem":"given a + b = 1 , find the value of 2 a + 2 b . two solutions are presented below . only one is correct , even though both yield the correct answer .","rationale":"because a + b = 1 , 2 a + 2 b = 2 ( a + b ) = 2 × 1 = 2 . correct answer d","correct":"d","options":{"a":"3 ","b":"5 ","c":"4 ","d":"2","e":"1"},"options_float":{"a":3.0,"b":5.0,"c":4.0,"d":2.0,"e":1.0},"annotated_formula":"subtract(add(add(2, 1), 2), add(2, 1))","linear_formula":"add(n0,n1)|add(n1,#0)|subtract(#1,#0)","chain":"2 + 1<\/gadget>\n3<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 - 3<\/gadget>\n2<\/output>\n2<\/result>","index":1516} +{"problem":"a sum fetched total simple interest of 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?","rationale":"\"let the sums be p . now , 45 % of p = 4016.25 or , p = 8925 answer a\"","correct":"a","options":{"a":"8925 ","b":"8032.5 ","c":"4462.5 ","d":"8900","e":"none of these"},"options_float":{"a":8925.0,"b":8032.5,"c":4462.5,"d":8900.0,"e":null},"annotated_formula":"divide(multiply(const_100, 4016.25), multiply(9, 5))","linear_formula":"multiply(n0,const_100)|multiply(n1,n2)|divide(#0,#1)|","chain":"100 * 4_016.25<\/gadget>\n401_625<\/output>\n9 * 5<\/gadget>\n45<\/output>\n401_625 \/ 45<\/gadget>\n8_925<\/output>\n8_925<\/result>","index":1517} +{"problem":"a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 7 , ab = 5 and ac = 11 , what is the length of ae ?","rationale":"\"ac = 11 and ab = 5 , so bc = 6 . bc = 2 cd so cd = 3 . the length of ae is ab + bc + cd + de = 5 + 6 + 3 + 7 = 21 the answer is 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":"add(add(11, divide(subtract(11, 5), 2)), 7)","linear_formula":"subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)|","chain":"11 - 5<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n11 + 3<\/gadget>\n14<\/output>\n14 + 7<\/gadget>\n21<\/output>\n21<\/result>","index":1521} +{"problem":"each of the integers from 1 to 17 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ?","rationale":"\"out of the 17 integers : 9 are odd and 8 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 9 odd numbers first 2 . 10 th number will be the first even number so we need to withdraw at least 10 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 10 . ( d )\"","correct":"d","options":{"a":"19 ","b":"12 ","c":"11 ","d":"10","e":"3"},"options_float":{"a":19.0,"b":12.0,"c":11.0,"d":10.0,"e":3.0},"annotated_formula":"add(divide(17, const_2), 1)","linear_formula":"divide(n1,const_2)|add(n0,#0)|","chain":"17 \/ 2<\/gadget>\n17\/2 = around 8.5<\/output>\n(17\/2) + 1<\/gadget>\n19\/2 = around 9.5<\/output>\n19\/2 = around 9.5<\/result>","index":1523} +{"problem":"the averge score of a cricketer for 10 matches is 45 runs . if the average for the first 6 matches is 48 . then find the average for the last 4 matches ?","rationale":"sum of last 4 matches = ( ( 10 × 45 ) – ( 6 × 48 ) = 162 average = 162 \/ 4 = 40.5 answer : d","correct":"d","options":{"a":"43.25 ","b":"43 ","c":"38 ","d":"40.5","e":"36"},"options_float":{"a":43.25,"b":43.0,"c":38.0,"d":40.5,"e":36.0},"annotated_formula":"divide(subtract(multiply(45, 10), multiply(6, 48)), 4)","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)","chain":"45 * 10<\/gadget>\n450<\/output>\n6 * 48<\/gadget>\n288<\/output>\n450 - 288<\/gadget>\n162<\/output>\n162 \/ 4<\/gadget>\n81\/2 = around 40.5<\/output>\n81\/2 = around 40.5<\/result>","index":1524} +{"problem":"a shopkeeper fixes the marked price of an item 40 % above its cost price . the percentage of discount allowed to gain 8 % is","rationale":"\"explanation : let the cost price = rs 100 then , marked price = rs 140 required gain = 8 % , so selling price = rs 108 discount = 140 - 108 = 32 discount % = ( 32 \/ 140 ) * 100 = 22.85 % option b\"","correct":"b","options":{"a":"23.85 % ","b":"22.85 % ","c":"21.85 % ","d":"20.85 %","e":"none of these"},"options_float":{"a":23.85,"b":22.85,"c":21.85,"d":20.85,"e":null},"annotated_formula":"subtract(const_100, multiply(divide(add(8, const_100), add(40, const_100)), const_100))","linear_formula":"add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|","chain":"8 + 100<\/gadget>\n108<\/output>\n40 + 100<\/gadget>\n140<\/output>\n108 \/ 140<\/gadget>\n27\/35 = around 0.771429<\/output>\n(27\/35) * 100<\/gadget>\n540\/7 = around 77.142857<\/output>\n100 - (540\/7)<\/gadget>\n160\/7 = around 22.857143<\/output>\n160\/7 = around 22.857143<\/result>","index":1525} +{"problem":"how long does a train 60 m long travelling at 60 kmph takes to cross a bridge of 80 m in length ?","rationale":"\"b 16.8 sec d = 60 + 80 = 140 m s = 60 * 5 \/ 18 = 50 \/ 3 t = 140 * 3 \/ 50 = 8.4 sec answer is b\"","correct":"b","options":{"a":"5.8 sec ","b":"8.4 sec ","c":"12.4 sec ","d":"6.8 sec","e":"1.8 sec"},"options_float":{"a":5.8,"b":8.4,"c":12.4,"d":6.8,"e":1.8},"annotated_formula":"divide(add(60, 80), multiply(60, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"60 + 80<\/gadget>\n140<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n140 \/ (50\/3)<\/gadget>\n42\/5 = around 8.4<\/output>\n42\/5 = around 8.4<\/result>","index":1526} +{"problem":"a circle graph shows how the budget of a certain company was spent : 55 percent for salaries , 9 percent for research and development , 5 percent for utilities , 4 percent for equipment , 2 percent for supplies , and the remainder for transportation . if the area of each sector of the graph is proportional to the percent of the budget it represents , how many degrees of the circle are used to represent transportation ?","rationale":"\"the percent of the budget for transportation is 100 - ( 55 + 9 + 5 + 4 + 2 ) = 25 % 100 % of the circle is 360 degrees . then ( 25 % \/ 100 % ) * 360 = 90 degrees the answer is e .\"","correct":"e","options":{"a":"18 ° ","b":"36 ° ","c":"54 ° ","d":"72 °","e":"90 °"},"options_float":{"a":18.0,"b":36.0,"c":54.0,"d":72.0,"e":90.0},"annotated_formula":"divide(multiply(const_360, subtract(const_100, add(add(add(add(55, 9), 5), 4), 2))), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|subtract(const_100,#3)|multiply(#4,const_360)|divide(#5,const_100)|","chain":"55 + 9<\/gadget>\n64<\/output>\n64 + 5<\/gadget>\n69<\/output>\n69 + 4<\/gadget>\n73<\/output>\n73 + 2<\/gadget>\n75<\/output>\n100 - 75<\/gadget>\n25<\/output>\n360 * 25<\/gadget>\n9_000<\/output>\n9_000 \/ 100<\/gadget>\n90<\/output>\n90<\/result>","index":1527} +{"problem":"which is the least number that must be subtracted from 1856 so that the remainder when divided by 7 , 12 , 10 is 4 ?","rationale":"first we need to figure out what numbers are exactly divisible by 7 , 12,10 . this will be the set { lcm , lcmx 2 , lcmx 3 , . . . } lcm ( 7 , 12,10 ) = 42 * 10 = 420 the numbers which will leave remainder 4 will be { 420 + 4 , 420 x 2 + 4 , , . . . } the largest such number less than or equal to 1856 is 420 * 4 + 4 or 1684 to obtain this you need to subtract 172 . b","correct":"b","options":{"a":"168 ","b":"172 ","c":"182 ","d":"140","e":"160"},"options_float":{"a":168.0,"b":172.0,"c":182.0,"d":140.0,"e":160.0},"annotated_formula":"subtract(1856, add(4, multiply(gcd(1856, lcm(lcm(7, 12), 10)), lcm(lcm(7, 12), 10))))","linear_formula":"lcm(n1,n2)|lcm(n3,#0)|gcd(n0,#1)|multiply(#2,#1)|add(n4,#3)|subtract(n0,#4)","chain":"lcm(7, 12)<\/gadget>\n84<\/output>\nlcm(84, 10)<\/gadget>\n420<\/output>\ngcd(1_856, 420)<\/gadget>\n4<\/output>\n4 * 420<\/gadget>\n1_680<\/output>\n4 + 1_680<\/gadget>\n1_684<\/output>\n1_856 - 1_684<\/gadget>\n172<\/output>\n172<\/result>","index":1531} +{"problem":"how many bricks , each measuring 25 cm * 11.25 cm * 6 cm , will be needed to build a wall 8 m * 6 m * 22.5 m","rationale":"\"to solve this type of question , simply divide the volume of wall with the volume of brick to get the numbers of required bricks so lets solve this number of bricks = volume of wall \/ volume of 1 brick = 800 ∗ 600 ∗ 22.5 \/ 25 ∗ 11.25 ∗ 6 = 6400 answer : a\"","correct":"a","options":{"a":"6400 ","b":"3777 ","c":"2679 ","d":"2667","e":"1997"},"options_float":{"a":6400.0,"b":3777.0,"c":2679.0,"d":2667.0,"e":1997.0},"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":1534} +{"problem":"a soccer team played 160 games and won 65 percent of them . how many games did it win ?","rationale":"\"65 % of 160 = x 0.65 * 160 = x 104 = x answer : c\"","correct":"c","options":{"a":"84 ","b":"94 ","c":"104 ","d":"114","e":"124"},"options_float":{"a":84.0,"b":94.0,"c":104.0,"d":114.0,"e":124.0},"annotated_formula":"divide(multiply(65, 160), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|","chain":"65 * 160<\/gadget>\n10_400<\/output>\n10_400 \/ 100<\/gadget>\n104<\/output>\n104<\/result>","index":1535} +{"problem":"a trader sells 80 meters of cloth for rs . 9000 at the profit of rs . 23.5 per metre of cloth . what is the cost price of one metre of cloth ?","rationale":"\"sp of 1 m of cloth = 9000 \/ 80 = rs . 112.5 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 112.5 - rs . 23.5 = rs . 89 . answer : b\"","correct":"b","options":{"a":"22 ","b":"89 ","c":"90 ","d":"78","e":"80"},"options_float":{"a":22.0,"b":89.0,"c":90.0,"d":78.0,"e":80.0},"annotated_formula":"subtract(divide(9000, 80), 23.5)","linear_formula":"divide(n1,n0)|subtract(#0,n2)|","chain":"9_000 \/ 80<\/gadget>\n225\/2 = around 112.5<\/output>\n(225\/2) - 23.5<\/gadget>\n89<\/output>\n89<\/result>","index":1537} +{"problem":"a farm has chickens , cows and sheep . there are 6 times the number of chickens and cows than sheep . if there are more cows than chickens or sheep , and together , cows and chickens have a total of 100 feet and heads , how many sheep live at the farm ?","rationale":"chicken - ch cows - c sheep - s ch + c = 6 s c > ch and c > s each cow has 4 legs and 1 head each chicken has 2 legs and 1 head so 5 c + 3 ch = 100 ( sum of legs and head ) there are 2 possible solutions to this equation c = 11 and ch = 9 or c = 14 and ch = 10 since from first equation where ch + c = 6 s the sum of ch and c should be divisbile by 6 . 20 is not so the only possible solution is c = 14 and ch = 10 . so s = 4 answer : d","correct":"d","options":{"a":"5 ","b":"8 ","c":"10 ","d":"4","e":"17"},"options_float":{"a":5.0,"b":8.0,"c":10.0,"d":4.0,"e":17.0},"annotated_formula":"subtract(6, const_2)","linear_formula":"subtract(n0,const_2)","chain":"6 - 2<\/gadget>\n4<\/output>\n4<\/result>","index":1538} +{"problem":"in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 75 percent of the registered voters who are democrats and 25 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ?","rationale":"\"say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.75 = 45 democrats are expected to vote for candidate a ; 40 * 0.25 = 10 republicans are expected to vote for candidate a . thus total of 45 + 10 = 55 registered voters are expected to vote for candidate a , which is 55 % of the total number of registered voters . answer : d .\"","correct":"d","options":{"a":"50 % ","b":"53 % ","c":"54 % ","d":"55 %","e":"57 %"},"options_float":{"a":50.0,"b":53.0,"c":54.0,"d":55.0,"e":57.0},"annotated_formula":"add(multiply(60, divide(75, const_100)), multiply(subtract(const_100, 60), divide(25, const_100)))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4)|","chain":"75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n60 * (3\/4)<\/gadget>\n45<\/output>\n100 - 60<\/gadget>\n40<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n40 * (1\/4)<\/gadget>\n10<\/output>\n45 + 10<\/gadget>\n55<\/output>\n55<\/result>","index":1539} +{"problem":"a rower can row 5 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 \/ 3 + x \/ 7 = 1 x = 21 \/ 10 = 2.1 km the answer is c .\"","correct":"c","options":{"a":"1.5 ","b":"1.8 ","c":"2.1 ","d":"2.4","e":"2.7"},"options_float":{"a":1.5,"b":1.8,"c":2.1,"d":2.4,"e":2.7},"annotated_formula":"multiply(divide(subtract(5, 2), add(add(5, 2), subtract(5, 2))), add(5, 2))","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)|","chain":"5 - 2<\/gadget>\n3<\/output>\n5 + 2<\/gadget>\n7<\/output>\n7 + 3<\/gadget>\n10<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 7<\/gadget>\n21\/10 = around 2.1<\/output>\n21\/10 = around 2.1<\/result>","index":1540} +{"problem":"a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 24 m * 2 m * 0.75 m ?","rationale":"\"24 * 2 * 0.75 = 20 \/ 100 * 10 \/ 100 * 7.5 \/ 100 * x 24 = 1 \/ 100 * x = > x = 24000 answer : c\"","correct":"c","options":{"a":"29798 ","b":"27908 ","c":"24000 ","d":"25000","e":"27991"},"options_float":{"a":29798.0,"b":27908.0,"c":24000.0,"d":25000.0,"e":27991.0},"annotated_formula":"divide(divide(divide(multiply(multiply(multiply(24, 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":"24 * 100<\/gadget>\n2_400<\/output>\n2 * 100<\/gadget>\n200<\/output>\n2_400 * 200<\/gadget>\n480_000<\/output>\n0.75 * 100<\/gadget>\n75<\/output>\n480_000 * 75<\/gadget>\n36_000_000<\/output>\n36_000_000 \/ 20<\/gadget>\n1_800_000<\/output>\n1_800_000 \/ 10<\/gadget>\n180_000<\/output>\n180_000 \/ 7.5<\/gadget>\n24_000<\/output>\n24_000<\/result>","index":1541} +{"problem":"in a group of ducks and cows , the total number of legs are 28 more than twice the no . of heads . find the total no . of buffaloes .","rationale":"\"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 28 = > 2 x = 28 = > x = 14 c\"","correct":"c","options":{"a":"11 ","b":"12 ","c":"14 ","d":"16","e":"18"},"options_float":{"a":11.0,"b":12.0,"c":14.0,"d":16.0,"e":18.0},"annotated_formula":"divide(28, const_2)","linear_formula":"divide(n0,const_2)|","chain":"28 \/ 2<\/gadget>\n14<\/output>\n14<\/result>","index":1545} +{"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 26 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 26 ) = rs . 208 . profit = 208 - 150 = 58 so , profit % = ( 58 \/ 150 ) * 100 = 39 % option b\"","correct":"b","options":{"a":"12 % ","b":"39 % ","c":"14 % ","d":"15 %","e":"16 %"},"options_float":{"a":12.0,"b":39.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"divide(multiply(subtract(multiply(26, 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>\n26 * 8<\/gadget>\n208<\/output>\n5 * 18<\/gadget>\n90<\/output>\n3 * 20<\/gadget>\n60<\/output>\n90 + 60<\/gadget>\n150<\/output>\n208 - 150<\/gadget>\n58<\/output>\n58 * 100<\/gadget>\n5_800<\/output>\n5_800 \/ 150<\/gadget>\n116\/3 = around 38.666667<\/output>\n116\/3 = around 38.666667<\/result>","index":1546} +{"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 \/ 4 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 \/ 4 ) * 270 = 67 gallons total paint used = 157 gallons . option b\"","correct":"b","options":{"a":"18 ","b":"157 ","c":"175 ","d":"216","e":"250"},"options_float":{"a":18.0,"b":157.0,"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)), 4))","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 \/ 4<\/gadget>\n135\/2 = around 67.5<\/output>\n90 + (135\/2)<\/gadget>\n315\/2 = around 157.5<\/output>\n315\/2 = around 157.5<\/result>","index":1548} +{"problem":"a man has some hens and cows . if the number of heads be 50 and the number of feet equals 160 , then the number of hens will be :","rationale":"\"explanation : let the number of hens be x and the number of cows be y . then , x + y = 50 . . . . ( i ) and 2 x + 4 y = 160 x + 2 y = 80 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = 20 , y = 30 the required answer = 20 . answer : d\"","correct":"d","options":{"a":"22 ","b":"23 ","c":"24 ","d":"20","e":"28"},"options_float":{"a":22.0,"b":23.0,"c":24.0,"d":20.0,"e":28.0},"annotated_formula":"divide(subtract(multiply(50, const_4), 160), const_2)","linear_formula":"multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|","chain":"50 * 4<\/gadget>\n200<\/output>\n200 - 160<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20<\/result>","index":1551} +{"problem":"if a population of women in a town is 50 % of men . what is the population of men as a percentage 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 . men = 10 women = 5 we ' re asked for the number of men , as a percentage of the number of women . m \/ w % = 10 \/ 5 = 200 % answer is c\"","correct":"c","options":{"a":"100 % ","b":"120 % ","c":"200 % ","d":"150 %","e":"180 %"},"options_float":{"a":100.0,"b":120.0,"c":200.0,"d":150.0,"e":180.0},"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":1552} +{"problem":"how many even multiples of 65 are there between 649 and 1301 ?","rationale":"\"650 = 10 * 65 1300 = 20 * 65 the even multiples are 65 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b .\"","correct":"b","options":{"a":"5 ","b":"6 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":5.0,"b":6.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"add(divide(subtract(1301, 649), multiply(65, const_2)), const_1)","linear_formula":"multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)|","chain":"1_301 - 649<\/gadget>\n652<\/output>\n65 * 2<\/gadget>\n130<\/output>\n652 \/ 130<\/gadget>\n326\/65 = around 5.015385<\/output>\n(326\/65) + 1<\/gadget>\n391\/65 = around 6.015385<\/output>\n391\/65 = around 6.015385<\/result>","index":1553} +{"problem":"two trains are moving in opposite directions with speed of 70 km \/ hr and 90 km \/ hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds","rationale":"\"explanation : relative speed = 70 + 90 = 160 km \/ hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 \/ 160 hr = 1 \/ 80 hr = 3600 \/ 80 seconds = = 45 seconds answer : option b\"","correct":"b","options":{"a":"56 ","b":"45 ","c":"47 ","d":"26","e":"25"},"options_float":{"a":56.0,"b":45.0,"c":47.0,"d":26.0,"e":25.0},"annotated_formula":"multiply(divide(add(1.10, 0.9), add(70, 90)), const_3600)","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600)|","chain":"1.1 + 0.9<\/gadget>\n2<\/output>\n70 + 90<\/gadget>\n160<\/output>\n2 \/ 160<\/gadget>\n1\/80 = around 0.0125<\/output>\n(1\/80) * 3_600<\/gadget>\n45<\/output>\n45<\/result>","index":1554} +{"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 10 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?","rationale":"\"0,08 r = x \/ 100 * 0.1 r answer a\"","correct":"a","options":{"a":"80 % ","b":"105 % ","c":"120 % ","d":"124.2 %","e":"138 %"},"options_float":{"a":80.0,"b":105.0,"c":120.0,"d":124.2,"e":138.0},"annotated_formula":"multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(10, 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>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(4\/5) * (1\/10)<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) \/ (1\/10)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":1555} +{"problem":"if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 5 ) , what is the value of x ?","rationale":"\"( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 5 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 5 hence x = 7 . answer is a\"","correct":"a","options":{"a":"7 ","b":"11 ","c":"13 ","d":"15","e":"17"},"options_float":{"a":7.0,"b":11.0,"c":13.0,"d":15.0,"e":17.0},"annotated_formula":"add(5, 2)","linear_formula":"add(n0,n5)|","chain":"5 + 2<\/gadget>\n7<\/output>\n7<\/result>","index":1557} +{"problem":"in an examination , the percentage of students qualified to the students appeared from school ' p ' is 70 % . in school ' q ' , the number of students appeared is 30 % more than the students appeared from school ' p ' and the number of students qualified from school ' q ' is 50 % more than the students qualified from school ' p ' . what is the % of students qualified to the number of students appeared from school ' q ' ?","rationale":"explanation : number of students appeared from school ' p ' = 100 , say number of students qualified from school ' p ' = 70 and number of students appeared from school ' q ' = 130 number of students qualified from school ' q ' = 50 % more than those qualified from school ' p ' . = 70 + 35 = 105 % of students qualified to the number of students appeared from school b = 105 \/ 130 * 100 = 80.76 % answer : b","correct":"b","options":{"a":"80.78 % ","b":"80.76 % ","c":"80.72 % ","d":"80.79 %","e":"80.74 %"},"options_float":{"a":80.78,"b":80.76,"c":80.72,"d":80.79,"e":80.74},"annotated_formula":"multiply(divide(multiply(divide(add(50, const_100), const_100), divide(70, const_100)), divide(add(30, const_100), const_100)), const_100)","linear_formula":"add(n2,const_100)|add(n1,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#3,#2)|divide(#5,#4)|multiply(#6,const_100)","chain":"50 + 100<\/gadget>\n150<\/output>\n150 \/ 100<\/gadget>\n3\/2 = around 1.5<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(3\/2) * (7\/10)<\/gadget>\n21\/20 = around 1.05<\/output>\n30 + 100<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n(21\/20) \/ (13\/10)<\/gadget>\n21\/26 = around 0.807692<\/output>\n(21\/26) * 100<\/gadget>\n1_050\/13 = around 80.769231<\/output>\n1_050\/13 = around 80.769231<\/result>","index":1559} +{"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 75 . 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 ( 75 ) = 300 x = 71 and the fourth person has a height of 71 + 10 = 81 inches the answer is e .","correct":"e","options":{"a":"73 ","b":"75 ","c":"77 ","d":"79","e":"81"},"options_float":{"a":73.0,"b":75.0,"c":77.0,"d":79.0,"e":81.0},"annotated_formula":"add(divide(subtract(multiply(75, 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":"75 * 4<\/gadget>\n300<\/output>\n4 + 6<\/gadget>\n10<\/output>\n6 + 10<\/gadget>\n16<\/output>\n300 - 16<\/gadget>\n284<\/output>\n284 \/ 4<\/gadget>\n71<\/output>\n71 + 10<\/gadget>\n81<\/output>\n81<\/result>","index":1560} +{"problem":"if the angles of an n sided polygon are in a . p and a > = 20 and d > = 20 then wat is the maximum possible value of n ?","rationale":"sum of interior angles of a polygon = ( n - 2 ) × 180 ° where n = number of sides there will be n angles which are in a . p . therefore , since we need to find maximum value of n , put minimum value for a and d . i . e . , take a = 20 and d = 20 then sum of the angles = n \/ 2 [ 2 × 20 + ( n − 1 ) 20 ] therefore , 10 n ( n + 1 ) = ( n − 2 ) 180 n = 14.52 since number of sides must be a positive integer , maximum value of n = 14 answer : b","correct":"b","options":{"a":"12 ","b":"14 ","c":"21 ","d":"25","e":"cant determine"},"options_float":{"a":12.0,"b":14.0,"c":21.0,"d":25.0,"e":null},"annotated_formula":"divide(add(sqrt(subtract(power(subtract(20, const_3), const_2), multiply(const_4, multiply(divide(add(multiply(const_10, multiply(const_4, const_2)), const_100), const_10), const_2)))), subtract(20, const_3)), const_2)","linear_formula":"multiply(const_2,const_4)|subtract(n0,const_3)|multiply(#0,const_10)|power(#1,const_2)|add(#2,const_100)|divide(#4,const_10)|multiply(#5,const_2)|multiply(#6,const_4)|subtract(#3,#7)|sqrt(#8)|add(#9,#1)|divide(#10,const_2)","chain":"20 - 3<\/gadget>\n17<\/output>\n17 ** 2<\/gadget>\n289<\/output>\n4 * 2<\/gadget>\n8<\/output>\n10 * 8<\/gadget>\n80<\/output>\n80 + 100<\/gadget>\n180<\/output>\n180 \/ 10<\/gadget>\n18<\/output>\n18 * 2<\/gadget>\n36<\/output>\n4 * 36<\/gadget>\n144<\/output>\n289 - 144<\/gadget>\n145<\/output>\n145 ** (1\/2)<\/gadget>\nsqrt(145) = around 12.041595<\/output>\n(sqrt(145)) + 17<\/gadget>\nsqrt(145) + 17 = around 29.041595<\/output>\n(sqrt(145) + 17) \/ 2<\/gadget>\nsqrt(145)\/2 + 17\/2 = around 14.520797<\/output>\nsqrt(145)\/2 + 17\/2 = around 14.520797<\/result>","index":1561} +{"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 82 ?","rationale":"\"four wheeler = 20 * 4 = 80 ( max ) 2 wheel = 1 so no of 4 wheeler = 20 answer : d\"","correct":"d","options":{"a":"11 ","b":"12 ","c":"13 ","d":"20","e":"25"},"options_float":{"a":11.0,"b":12.0,"c":13.0,"d":20.0,"e":25.0},"annotated_formula":"divide(subtract(82, 2), 4)","linear_formula":"subtract(n3,n0)|divide(#0,n1)|","chain":"82 - 2<\/gadget>\n80<\/output>\n80 \/ 4<\/gadget>\n20<\/output>\n20<\/result>","index":1565} +{"problem":"if 50 % of x is 25 less than 25 % of 2500 , then x is ?","rationale":"\"50 % of x = x \/ 2 ; 25 % of 2500 = 25 \/ 100 * 2500 = 625 given that , x \/ 2 = 625 - 25 = > x \/ 2 = 600 = > x = 1200 . answer : c\"","correct":"c","options":{"a":"1880 ","b":"2160 ","c":"1200 ","d":"8400","e":"1210"},"options_float":{"a":1880.0,"b":2160.0,"c":1200.0,"d":8400.0,"e":1210.0},"annotated_formula":"divide(subtract(multiply(2500, divide(25, const_100)), 25), divide(50, const_100))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n2_500 * (1\/4)<\/gadget>\n625<\/output>\n625 - 25<\/gadget>\n600<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n600 \/ (1\/2)<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":1566} +{"problem":"a contractor undertakes to built a walls in 50 days . he employs 30 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":"\"30 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 , 30 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 30 × 25 × 0.6 \/ 25 × 0.4 = 45 men answerc\"","correct":"c","options":{"a":"25 ","b":"30 ","c":"45 ","d":"20","e":"none of these"},"options_float":{"a":25.0,"b":30.0,"c":45.0,"d":20.0,"e":null},"annotated_formula":"divide(multiply(30, 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>\n30 * (3\/5)<\/gadget>\n18<\/output>\n4 \/ 10<\/gadget>\n2\/5 = around 0.4<\/output>\n18 \/ (2\/5)<\/gadget>\n45<\/output>\n45<\/result>","index":1567} +{"problem":"in an examination , 30 % of total students failed in hindi , 35 % failed in english and 35 % in both . the percentage of these who passed in both the subjects is :","rationale":"\"pass percentage = 100 - ( 30 + 35 - 35 ) = 100 - 30 = 70 answer : d\"","correct":"d","options":{"a":"10 % ","b":"20 % ","c":"30 % ","d":"70 %","e":"50 %"},"options_float":{"a":10.0,"b":20.0,"c":30.0,"d":70.0,"e":50.0},"annotated_formula":"subtract(const_100, subtract(add(30, 35), 35))","linear_formula":"add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|","chain":"30 + 35<\/gadget>\n65<\/output>\n65 - 35<\/gadget>\n30<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70<\/result>","index":1568} +{"problem":"in a shop , the profit is 320 % of the cost . if the cost increases by 25 % but the selling price remains constant , find out approximately what percentage of the selling price is the profit ?","rationale":"let the cp = 100 profit = ( 320 \/ 100 ) × 100 = 320 sp = cp + profit = 100 + 320 = 420 if the cost increases by 25 % , new cp = ( 125 \/ 100 ) × 100 = 125 selling price is constant , hence new sp = 420 profit = sp – cp = 420 – 125 = 295 required percentage = ( 295 \/ 420 ) × 100 = 2950 \/ 42 = 1475 \/ 21 ≈ 70 answer : e","correct":"e","options":{"a":"180 % ","b":"120 % ","c":"90 % ","d":"80 %","e":"70 %"},"options_float":{"a":180.0,"b":120.0,"c":90.0,"d":80.0,"e":70.0},"annotated_formula":"multiply(divide(subtract(add(320, const_100), add(25, const_100)), add(320, const_100)), const_100)","linear_formula":"add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)","chain":"320 + 100<\/gadget>\n420<\/output>\n25 + 100<\/gadget>\n125<\/output>\n420 - 125<\/gadget>\n295<\/output>\n295 \/ 420<\/gadget>\n59\/84 = around 0.702381<\/output>\n(59\/84) * 100<\/gadget>\n1_475\/21 = around 70.238095<\/output>\n1_475\/21 = around 70.238095<\/result>","index":1569} +{"problem":"a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 10 % on conveyance . if his savings at the end of the month is rs . 1200 , then his salary per month in rupees is :","rationale":"total expenditure = 40 + 25 + 15 + 10 = 90 % saving = ( 100 - 90 ) = 10 % 10 \/ 100 × salary = 1200 , salary = 12000 rs . answer : a","correct":"a","options":{"a":"12000 ","b":"6000 ","c":"8000 ","d":"10000","e":"none of these"},"options_float":{"a":12000.0,"b":6000.0,"c":8000.0,"d":10000.0,"e":null},"annotated_formula":"divide(multiply(1200, const_100), 10)","linear_formula":"multiply(n4,const_100)|divide(#0,n3)","chain":"1_200 * 100<\/gadget>\n120_000<\/output>\n120_000 \/ 10<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":1570} +{"problem":"the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 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 * 7 ) + 6000 ( v - 7 ) < = > 2000 v = 42000 < = > v = 21 . answer : b\"","correct":"b","options":{"a":"76 ","b":"21 ","c":"26 ","d":"28","e":"11"},"options_float":{"a":76.0,"b":21.0,"c":26.0,"d":28.0,"e":11.0},"annotated_formula":"add(7, divide(multiply(7, 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>\n7 * 4_000<\/gadget>\n28_000<\/output>\n8_000 - 6_000<\/gadget>\n2_000<\/output>\n28_000 \/ 2_000<\/gadget>\n14<\/output>\n7 + 14<\/gadget>\n21<\/output>\n21<\/result>","index":1571} +{"problem":"a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 108 . what was the number he chose ?","rationale":"\"let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 108 x = 123 answer : a\"","correct":"a","options":{"a":"123 ","b":"267 ","c":"277 ","d":"267","e":"120"},"options_float":{"a":123.0,"b":267.0,"c":277.0,"d":267.0,"e":120.0},"annotated_formula":"divide(add(108, 138), 2)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"108 + 138<\/gadget>\n246<\/output>\n246 \/ 2<\/gadget>\n123<\/output>\n123<\/result>","index":1572} +{"problem":"express 30 mps in kmph ?","rationale":"\"30 * 18 \/ 5 = 108 kmph answer : b\"","correct":"b","options":{"a":"122 ","b":"108 ","c":"110 ","d":"150","e":"100"},"options_float":{"a":122.0,"b":108.0,"c":110.0,"d":150.0,"e":100.0},"annotated_formula":"multiply(divide(30, const_1000), const_3600)","linear_formula":"divide(n0,const_1000)|multiply(#0,const_3600)|","chain":"30 \/ 1_000<\/gadget>\n3\/100 = around 0.03<\/output>\n(3\/100) * 3_600<\/gadget>\n108<\/output>\n108<\/result>","index":1574} +{"problem":"in a group of ducks and cows , the total number of legs are 8 more than twice the no . of heads . find the total no . of buffaloes .","rationale":"\"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 8 = > 2 x = 8 = > x = 4 b\"","correct":"b","options":{"a":"5 ","b":"4 ","c":"6 ","d":"3","e":"2"},"options_float":{"a":5.0,"b":4.0,"c":6.0,"d":3.0,"e":2.0},"annotated_formula":"divide(8, const_2)","linear_formula":"divide(n0,const_2)|","chain":"8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":1575} +{"problem":"if log 8 x + log 8 1 \/ 6 = 1 \/ 3 , then the value of x is :","rationale":"\"log 8 x + log 8 ( 1 \/ 6 ) = 1 \/ 3 = > ( log x \/ log 8 ) + ( log 1 \/ 6 \/ log 8 ) = log ( 81 \/ 3 ) = log 2 = > log x = log 2 – log 1 \/ 6 = log ( 2 * 6 \/ 1 ) = log 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":"multiply(power(8, divide(1, 3)), 6)","linear_formula":"divide(n2,n5)|power(n0,#0)|multiply(n3,#1)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n8 ** (1\/3)<\/gadget>\n2<\/output>\n2 * 6<\/gadget>\n12<\/output>\n12<\/result>","index":1577} +{"problem":"if the sum of two numbers is 12 and the sum of their squares is 124 , then the product of the numbers is","rationale":"\"sol . let the numbers be x and y . then , ( x + y ) = 12 and x 2 + y 2 = 124 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 12 ) 2 - 124 = 144 - 124 = 20 xy = 10 . answer a\"","correct":"a","options":{"a":"10 ","b":"44 ","c":"80 ","d":"88","e":"90"},"options_float":{"a":10.0,"b":44.0,"c":80.0,"d":88.0,"e":90.0},"annotated_formula":"divide(subtract(power(12, const_2), 124), const_2)","linear_formula":"power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|","chain":"12 ** 2<\/gadget>\n144<\/output>\n144 - 124<\/gadget>\n20<\/output>\n20 \/ 2<\/gadget>\n10<\/output>\n10<\/result>","index":1578} +{"problem":"the ratio between the sale price and the cost price of an article is 8 : 5 . what is the ratio between the profit and the cost price of that article ?","rationale":"\"c . p . = rs . 5 x and s . p . = rs . 8 x . then , gain = rs . 3 x required ratio = 3 x : 5 x = 3 : 5 d\"","correct":"d","options":{"a":"23 ","b":"1 : 2 ","c":"2 : 5 ","d":"3 : 5","e":"25"},"options_float":{"a":23.0,"b":0.5,"c":0.4,"d":0.6,"e":25.0},"annotated_formula":"divide(subtract(8, 5), 5)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|","chain":"8 - 5<\/gadget>\n3<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n3\/5 = around 0.6<\/result>","index":1579} +{"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 9 even numbers . his mother immediately worked out the answer . how much had he scored in math ?","rationale":"\"b 90 sum = ( n x n ) + n hence , 9 x 9 = 81 + 9 = 90\"","correct":"b","options":{"a":"80 ","b":"90 ","c":"30 ","d":"70","e":"60"},"options_float":{"a":80.0,"b":90.0,"c":30.0,"d":70.0,"e":60.0},"annotated_formula":"multiply(add(9, const_1), 9)","linear_formula":"add(n0,const_1)|multiply(n0,#0)|","chain":"9 + 1<\/gadget>\n10<\/output>\n10 * 9<\/gadget>\n90<\/output>\n90<\/result>","index":1580} +{"problem":"out of 400 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ?","rationale":"\"n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 400 - 50 = 350 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 350 = 150 . answer is b\"","correct":"b","options":{"a":"120 ","b":"150 ","c":"100 ","d":"180","e":"220"},"options_float":{"a":120.0,"b":150.0,"c":100.0,"d":180.0,"e":220.0},"annotated_formula":"subtract(add(175, 325), subtract(400, 50))","linear_formula":"add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|","chain":"175 + 325<\/gadget>\n500<\/output>\n400 - 50<\/gadget>\n350<\/output>\n500 - 350<\/gadget>\n150<\/output>\n150<\/result>","index":1583} +{"problem":"50 men shake hands with each other . maximum no of handshakes without cyclic handshakes .","rationale":"total no . of handshakes = 49 + 48 + 47 + . . . + 3 + 2 + 1 = 19 * ( 19 + 1 ) \/ 2 = 1225 or , if there are n persons then no . of shakehands = nc 2 = 50 c 2 = 1225 answer : c","correct":"c","options":{"a":"190 ","b":"200 ","c":"1225 ","d":"220","e":"230"},"options_float":{"a":190.0,"b":200.0,"c":1225.0,"d":220.0,"e":230.0},"annotated_formula":"multiply(subtract(50, const_1), divide(50, const_2))","linear_formula":"divide(n0,const_2)|subtract(n0,const_1)|multiply(#0,#1)","chain":"50 - 1<\/gadget>\n49<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n49 * 25<\/gadget>\n1_225<\/output>\n1_225<\/result>","index":1585} +{"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 10 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 10 minutes the population increases by 10 x 4 = 40 people . answer . a .\"","correct":"a","options":{"a":"40 ","b":"100 ","c":"150 ","d":"240","e":"300"},"options_float":{"a":40.0,"b":100.0,"c":150.0,"d":240.0,"e":300.0},"annotated_formula":"multiply(divide(const_60, 15), 10)","linear_formula":"divide(const_60,n0)|multiply(n1,#0)|","chain":"60 \/ 15<\/gadget>\n4<\/output>\n4 * 10<\/gadget>\n40<\/output>\n40<\/result>","index":1586} +{"problem":"if p ( a ) = 3 \/ 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 ) = 3 \/ 5 . 2 \/ 5 p ( a n b ) = 6 \/ 25 . a\"","correct":"a","options":{"a":"6 \/ 25 ","b":"3 \/ 25 ","c":"8 \/ 25 ","d":"2 \/ 13","e":"3 \/ 17"},"options_float":{"a":0.24,"b":0.12,"c":0.32,"d":0.1538461538,"e":0.1764705882},"annotated_formula":"multiply(divide(3, 5), divide(2, 5))","linear_formula":"divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(3\/5) * (2\/5)<\/gadget>\n6\/25 = around 0.24<\/output>\n6\/25 = around 0.24<\/result>","index":1588} +{"problem":"if a certain coin is flipped , the probability that the coin will land heads is 1 \/ 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips ?","rationale":"\"on the first three flips , you must get heads . whats the probability of getting heads ? its 1 \/ 2 so for the first three flips , your probability is ( 1 \/ 2 ) ^ 3 = 1 \/ 8 now for the last two , you want to get tails only . whats the prob of getting tails ? well , its the same as prob of getting a heads , namely , 1 \/ 2 for the last two flips , your probability is ( 1 \/ 2 ) ^ 2 = 1 \/ 4 so your overall probability for the event in question is 1 \/ 8 * 1 \/ 4 = 1 \/ 32 answer : e\"","correct":"e","options":{"a":"3 \/ 5 ","b":"1 \/ 2 ","c":"1 \/ 5 ","d":"1 \/ 8","e":"1 \/ 32"},"options_float":{"a":0.6,"b":0.5,"c":0.2,"d":0.125,"e":0.03125},"annotated_formula":"power(divide(1, 2), 5)","linear_formula":"divide(n0,n1)|power(#0,n2)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) ** 5<\/gadget>\n1\/32 = around 0.03125<\/output>\n1\/32 = around 0.03125<\/result>","index":1589} +{"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 q 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 ] q = 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":1590} +{"problem":"a piece of work can finish by a certain number of men in 100 days . if however , 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 ) : . ( x - 10 ) : x : : 100 : 110 or x - 10 \/ x = 100 \/ 110 or 11 x - 110 = 10 x or x = 110 so , originally there were 110 men . answer : d","correct":"d","options":{"a":"75 ","b":"82 ","c":"100 ","d":"110","e":"120"},"options_float":{"a":75.0,"b":82.0,"c":100.0,"d":110.0,"e":120.0},"annotated_formula":"divide(multiply(divide(add(100, 10), 10), 10), subtract(divide(add(100, 10), 10), 10))","linear_formula":"add(n0,n1)|divide(#0,n1)|multiply(n1,#1)|subtract(#1,n1)|divide(#2,#3)","chain":"100 + 10<\/gadget>\n110<\/output>\n110 \/ 10<\/gadget>\n11<\/output>\n11 * 10<\/gadget>\n110<\/output>\n11 - 10<\/gadget>\n1<\/output>\n110 \/ 1<\/gadget>\n110<\/output>\n110<\/result>","index":1591} +{"problem":"how many seconds will a train 150 meters long take to cross a bridge 200 meters long if the speed of the train is 54 kmph ?","rationale":"d = 150 + 200 = 350 s = 54 * 5 \/ 18 = 15 mps t = 350 \/ 15 = 23.3 sec c ) 23.3 sec","correct":"c","options":{"a":"17 sec ","b":"21 sec ","c":"23.3 sec ","d":"27.5 sec","e":"29 sec"},"options_float":{"a":17.0,"b":21.0,"c":23.3,"d":27.5,"e":29.0},"annotated_formula":"divide(add(200, 150), multiply(54, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)","chain":"200 + 150<\/gadget>\n350<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n54 * (5\/18)<\/gadget>\n15<\/output>\n350 \/ 15<\/gadget>\n70\/3 = around 23.333333<\/output>\n70\/3 = around 23.333333<\/result>","index":1592} +{"problem":"an error 6 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ?","rationale":"\"percentage error in calculated area = ( 6 + 6 + ( 6 ã — 6 ) \/ 100 ) % = 12.36 % answer : e\"","correct":"e","options":{"a":"14.05 % ","b":"14.02 % ","c":"14 % ","d":"13 %","e":"12.36 %"},"options_float":{"a":14.05,"b":14.02,"c":14.0,"d":13.0,"e":12.36},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 6)), 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 + 6<\/gadget>\n106<\/output>\n106 ** 2<\/gadget>\n11_236<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n11_236 - 10_000<\/gadget>\n1_236<\/output>\n1_236 * 100<\/gadget>\n123_600<\/output>\n123_600 \/ 10_000<\/gadget>\n309\/25 = around 12.36<\/output>\n309\/25 = around 12.36<\/result>","index":1593} +{"problem":"liz drove from point a to point b at 40 km \/ h . on her way back she drove at 50 km \/ h and therefore her way back lasted one hour less . what is the distance ( in km ) between a and b ?","rationale":"distance is same s 1 t 1 = s 2 t 2 40 t = 50 ( t - 1 ) t = 5 distance = speed * time 40 * 5 = 200 answer : b","correct":"b","options":{"a":"150 ","b":"200 ","c":"450 ","d":"500","e":"600"},"options_float":{"a":150.0,"b":200.0,"c":450.0,"d":500.0,"e":600.0},"annotated_formula":"multiply(divide(50, subtract(50, 40)), 40)","linear_formula":"subtract(n1,n0)|divide(n1,#0)|multiply(n0,#1)","chain":"50 - 40<\/gadget>\n10<\/output>\n50 \/ 10<\/gadget>\n5<\/output>\n5 * 40<\/gadget>\n200<\/output>\n200<\/result>","index":1594} +{"problem":"the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 75 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 \/ ( 7 \/ 4 ) * 4 = 2 \/ 7 the answer is e .\"","correct":"e","options":{"a":"2 \/ 5 ","b":"3 \/ 4 ","c":"4 \/ 5 ","d":"5 \/ 4","e":"2 \/ 7"},"options_float":{"a":0.4,"b":0.75,"c":0.8,"d":1.25,"e":0.2857142857},"annotated_formula":"multiply(divide(subtract(const_100, 60), add(const_100, 75)), 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 + 75<\/gadget>\n175<\/output>\n40 \/ 175<\/gadget>\n8\/35 = around 0.228571<\/output>\n5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n(8\/35) * (5\/4)<\/gadget>\n2\/7 = around 0.285714<\/output>\n2\/7 = around 0.285714<\/result>","index":1596} +{"problem":"in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 700 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 2,100 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?","rationale":"ratio of 2 bedroom apartment : 1 bedroom apartment = 700 : 2100 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 \/ 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 \/ 4 ) * 100 - - - > 25 % answer : a","correct":"a","options":{"a":"25 % ","b":"15 % ","c":"20 % ","d":"40 %","e":"45 %"},"options_float":{"a":25.0,"b":15.0,"c":20.0,"d":40.0,"e":45.0},"annotated_formula":"divide(multiply(700, const_100), add(add(multiply(const_2, const_1000), const_100), 700))","linear_formula":"multiply(n0,const_100)|multiply(const_1000,const_2)|add(#1,const_100)|add(n0,#2)|divide(#0,#3)","chain":"700 * 100<\/gadget>\n70_000<\/output>\n2 * 1_000<\/gadget>\n2_000<\/output>\n2_000 + 100<\/gadget>\n2_100<\/output>\n2_100 + 700<\/gadget>\n2_800<\/output>\n70_000 \/ 2_800<\/gadget>\n25<\/output>\n25<\/result>","index":1597} +{"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 8 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 \/ 4 = 12 miles \/ gallon . current engine miles \/ gallon is 8 miles \/ gallon . additional 4 miles \/ gallon is required to match the actual mileage . answer : b\"","correct":"b","options":{"a":"2 ","b":"4 ","c":"12 ","d":"40","e":"160"},"options_float":{"a":2.0,"b":4.0,"c":12.0,"d":40.0,"e":160.0},"annotated_formula":"subtract(divide(480, 40), 8)","linear_formula":"divide(n1,n0)|subtract(#0,n2)|","chain":"480 \/ 40<\/gadget>\n12<\/output>\n12 - 8<\/gadget>\n4<\/output>\n4<\/result>","index":1598} +{"problem":"a square is drawn inside a right - angled triangle with the two perpendicular sides as 12 cm and 8 cm . what is the side of the largest possible square that can be drawn ?","rationale":"area of triangle is 1 \/ 2 * 12 * 8 = 48 side of square = x the entire triangle split into two right angled triangle and one square with dimensions as follows i ) square with side x ii ) right angled triangle with perpendicular sides x and 12 - x iii ) right angled triangle with perpendicular sides 8 - x and x sum of area of all three = 48 = x 2 + 1 \/ 2 * x * ( 12 - x ) + 1 \/ 2 * x * ( 8 - x ) = 48 = x = 4.8 cm answer : a","correct":"a","options":{"a":"4.8 cm ","b":"4.4 cm ","c":"4.9 cm ","d":"5.0 cm","e":"5.2 cm"},"options_float":{"a":4.8,"b":4.4,"c":4.9,"d":5.0,"e":5.2},"annotated_formula":"divide(divide(multiply(12, 8), const_2), const_10)","linear_formula":"multiply(n0,n1)|divide(#0,const_2)|divide(#1,const_10)","chain":"12 * 8<\/gadget>\n96<\/output>\n96 \/ 2<\/gadget>\n48<\/output>\n48 \/ 10<\/gadget>\n24\/5 = around 4.8<\/output>\n24\/5 = around 4.8<\/result>","index":1600} +{"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 social 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 ( b )\"","correct":"b","options":{"a":"2 \/ 3 ","b":"1 \/ 12 ","c":"2 \/ 9 ","d":"1 \/ 2","e":"1 \/ 24"},"options_float":{"a":0.6666666667,"b":0.0833333333,"c":0.2222222222,"d":0.5,"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":1601} +{"problem":"in a 160 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 160 m in 7 \/ 56 * 160 = 20 seconds since a beats b by 7 seconds , a runs 160 m in ( 20 - 7 ) = 13 seconds hence , a ' s time over the course = 13 seconds answer : c","correct":"c","options":{"a":"22 seconds ","b":"12 seconds ","c":"13 seconds ","d":"18 seconds","e":"28 seconds"},"options_float":{"a":22.0,"b":12.0,"c":13.0,"d":18.0,"e":28.0},"annotated_formula":"subtract(multiply(divide(7, 56), 160), 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) * 160<\/gadget>\n20<\/output>\n20 - 7<\/gadget>\n13<\/output>\n13<\/result>","index":1602} +{"problem":"by how much is 70 % of 120 greater than 35 % of 200 .","rationale":"( 70 \/ 100 ) * 120 â € “ ( 35 \/ 100 ) * 200 84 - 70 = 14 answer : b","correct":"b","options":{"a":"15 ","b":"14 ","c":"13 ","d":"16","e":"17"},"options_float":{"a":15.0,"b":14.0,"c":13.0,"d":16.0,"e":17.0},"annotated_formula":"subtract(multiply(120, divide(70, const_100)), multiply(divide(35, const_100), 200))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)","chain":"70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n120 * (7\/10)<\/gadget>\n84<\/output>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n(7\/20) * 200<\/gadget>\n70<\/output>\n84 - 70<\/gadget>\n14<\/output>\n14<\/result>","index":1605} +{"problem":"in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 5 years older than b , the present age of b is","rationale":"\"explanation : let b ' s age = x years . then , as age = ( x + 5 ) years . ( x + 5 + 10 ) = 2 ( x — 10 ) hence x = 35 . present age of b = 35 years answer : option a\"","correct":"a","options":{"a":"35 ","b":"37 ","c":"39 ","d":"41","e":"42"},"options_float":{"a":35.0,"b":37.0,"c":39.0,"d":41.0,"e":42.0},"annotated_formula":"add(multiply(const_2, 10), add(5, 10))","linear_formula":"add(n0,n3)|multiply(n0,const_2)|add(#0,#1)|","chain":"2 * 10<\/gadget>\n20<\/output>\n5 + 10<\/gadget>\n15<\/output>\n20 + 15<\/gadget>\n35<\/output>\n35<\/result>","index":1607} +{"problem":"if 2 + 7 = 57 ; 3 + 6 = 63 ; 5 + 9 = 206 then 5 + 8 = ?","rationale":"\"2 ^ 3 + 7 ^ 2 = 57 3 ^ 3 + 6 ^ 2 = 63 5 ^ 3 + 9 ^ 2 = 206 and 5 ^ 3 + 8 ^ 2 = 189 answer : e\"","correct":"e","options":{"a":"185 ","b":"186 ","c":"177 ","d":"168","e":"189"},"options_float":{"a":185.0,"b":186.0,"c":177.0,"d":168.0,"e":189.0},"annotated_formula":"add(power(5, 3), power(8, 2))","linear_formula":"power(n9,n3)|power(n10,n0)|add(#0,#1)|","chain":"5 ** 3<\/gadget>\n125<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n125 + 64<\/gadget>\n189<\/output>\n189<\/result>","index":1608} +{"problem":"if a man lost 7 % by selling oranges at the rate of 21 a rupee at how many a rupee must he sell them to gain 42 % ?","rationale":"\"93 % - - - - 21 142 % - - - - ? 93 \/ 142 * 21 = 13.75 answer : e\"","correct":"e","options":{"a":"12.75 ","b":"11.75 ","c":"8.75 ","d":"15.75","e":"13.75"},"options_float":{"a":12.75,"b":11.75,"c":8.75,"d":15.75,"e":13.75},"annotated_formula":"divide(multiply(subtract(const_100, 7), 21), add(const_100, 42))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|","chain":"100 - 7<\/gadget>\n93<\/output>\n93 * 21<\/gadget>\n1_953<\/output>\n100 + 42<\/gadget>\n142<\/output>\n1_953 \/ 142<\/gadget>\n1_953\/142 = around 13.753521<\/output>\n1_953\/142 = around 13.753521<\/result>","index":1609} +{"problem":"a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.4 m and its walls are 5 cm thick . the thickness of the bottom is :","rationale":"\"explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 140 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 140 - x ) = 8000 × 1000 = > ( 140 - x ) = 8000 × 1000 \/ 320 = 100 = > x = 40 cm = 4 dm . answer : b\"","correct":"b","options":{"a":"90 cm ","b":"4 dm ","c":"1 m ","d":"1.1 cm","e":"none of these"},"options_float":{"a":90.0,"b":4.0,"c":1.0,"d":1.1,"e":null},"annotated_formula":"subtract(multiply(multiply(3.3, 2.6), 1.4), divide(8000, const_1000))","linear_formula":"divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0)|","chain":"3.3 * 2.6<\/gadget>\n8.58<\/output>\n8.58 * 1.4<\/gadget>\n12.012<\/output>\n8_000 \/ 1_000<\/gadget>\n8<\/output>\n12.012 - 8<\/gadget>\n4.012<\/output>\n4.012<\/result>","index":1611} +{"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 16 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 = 16 x = 24 days answer : option d\"","correct":"d","options":{"a":"5 days ","b":"7 days ","c":"12 days ","d":"24 days","e":"10 days"},"options_float":{"a":5.0,"b":7.0,"c":12.0,"d":24.0,"e":10.0},"annotated_formula":"inverse(multiply(inverse(16), 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 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/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\/16) * (2\/3)<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ (1\/24)<\/gadget>\n24<\/output>\n24<\/result>","index":1614} +{"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 140 meter length .","rationale":"\"speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 360 + 140 = 500 meter time = distance \/ speed = 500 ∗ 2 \/ 25 = 40 seconds answer : d\"","correct":"d","options":{"a":"20 seconds ","b":"27 seconds ","c":"30 seconds ","d":"40 seconds","e":"50 seconds"},"options_float":{"a":20.0,"b":27.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(add(360, 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":"360 + 140<\/gadget>\n500<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n500 \/ (25\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":1616} +{"problem":"if a 2 - b 2 = 9 and a * b = 4 , find a 4 + b 4 .","rationale":"\"a 2 - b 2 = 9 : given a 4 + b 4 - 2 a 2 b 2 = 92 : square both sides and expand . a * b = 4 : given a 2 b 2 = 42 : square both sides . a 4 + b 4 - 2 ( 16 ) = 81 : substitute a 4 + b 4 = 113 correct answer c\"","correct":"c","options":{"a":"32 ","b":"90 ","c":"113 ","d":"92","e":"81"},"options_float":{"a":32.0,"b":90.0,"c":113.0,"d":92.0,"e":81.0},"annotated_formula":"add(power(9, 2), multiply(power(4, 2), 2))","linear_formula":"power(n3,n0)|power(n2,n0)|multiply(#0,n0)|add(#2,#1)|","chain":"9 ** 2<\/gadget>\n81<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n16 * 2<\/gadget>\n32<\/output>\n81 + 32<\/gadget>\n113<\/output>\n113<\/result>","index":1617} +{"problem":"a chemist mixes one liter of pure water with x liters of a 50 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ?","rationale":"\"concentration of salt in pure solution = 0 concentration of salt in salt solution = 50 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 50 - 15 ) \/ ( 15 - 0 ) = 7 \/ 3 1 \/ x = 7 \/ 3 x = 3 \/ 7 answer : e\"","correct":"e","options":{"a":"1 \/ 4 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"1","e":"3 \/ 7"},"options_float":{"a":0.25,"b":0.3333333333,"c":0.5,"d":1.0,"e":0.4285714286},"annotated_formula":"divide(15, subtract(50, 15))","linear_formula":"subtract(n0,n1)|divide(n1,#0)|","chain":"50 - 15<\/gadget>\n35<\/output>\n15 \/ 35<\/gadget>\n3\/7 = around 0.428571<\/output>\n3\/7 = around 0.428571<\/result>","index":1621} +{"problem":"the cyclist going at a constant rate of 18 miles per hour is passed by a motor - cyclist traveling in the same direction along the same path at 48 miles per hour . the motor - cyclist stops to wait for the cyclist 15 minutes after passing cyclist , while the cyclist continues to travel at constant rate , how many minutes must the motor - cyclist wait until the cyclist catches up ?","rationale":"for the 15 minutes the motor - cyclist continues to overtake the cyclist , she is going at 30 miles per hour faster than the cyclist . once the motor - cyclist stops , the cyclist is going at 18 miles per hour while the motor - cyclist is at rest so the amount of time the cyclist will take to cover the distance between them is going to be in the ratio of the relative speeds . 30 \/ 18 * 15 or 25 minutes answer is ( a )","correct":"a","options":{"a":"25 ","b":"30 ","c":"35 ","d":"40","e":"45"},"options_float":{"a":25.0,"b":30.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"divide(multiply(subtract(divide(48, const_4), divide(18, const_4)), const_60), 18)","linear_formula":"divide(n1,const_4)|divide(n0,const_4)|subtract(#0,#1)|multiply(#2,const_60)|divide(#3,n0)","chain":"48 \/ 4<\/gadget>\n12<\/output>\n18 \/ 4<\/gadget>\n9\/2 = around 4.5<\/output>\n12 - (9\/2)<\/gadget>\n15\/2 = around 7.5<\/output>\n(15\/2) * 60<\/gadget>\n450<\/output>\n450 \/ 18<\/gadget>\n25<\/output>\n25<\/result>","index":1622} +{"problem":"a number is doubled and 9 is added . if the resultant is trebled , it becomes 81 . what is that number ?","rationale":"\"let the number be x . then , 3 ( 2 x + 9 ) = 81 2 x = 18 = > x = 9 answer : e\"","correct":"e","options":{"a":"3.5 ","b":"6 ","c":"8 ","d":"7","e":"9"},"options_float":{"a":3.5,"b":6.0,"c":8.0,"d":7.0,"e":9.0},"annotated_formula":"divide(subtract(81, 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>\n81 - 27<\/gadget>\n54<\/output>\n3 * 2<\/gadget>\n6<\/output>\n54 \/ 6<\/gadget>\n9<\/output>\n9<\/result>","index":1623} +{"problem":"in a renowned city , the average birth rate is 9 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day .","rationale":"\"every 2 seconds , 6 persons are added ( 9 - 3 ) . every second 3 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 3 = 259200 option e\"","correct":"e","options":{"a":"32,300 ","b":"172,800 ","c":"468,830 ","d":"338,200","e":"259,200"},"options_float":{"a":32300.0,"b":172800.0,"c":468830.0,"d":338200.0,"e":259200.0},"annotated_formula":"multiply(multiply(subtract(9, 3), const_3600), const_12)","linear_formula":"subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)|","chain":"9 - 3<\/gadget>\n6<\/output>\n6 * 3_600<\/gadget>\n21_600<\/output>\n21_600 * 12<\/gadget>\n259_200<\/output>\n259_200<\/result>","index":1625} +{"problem":"the squared value of the diagonal of a rectangle is ( 64 + b 2 ) sq cm , where b is less than 8 cm . what is the breadth of that rectangle ?","rationale":"diagonal 2 = 64 + b 2 or , 10 ( 2 ) = 64 + 6 ( 2 ) answer a","correct":"a","options":{"a":"6 cm ","b":"10 cm ","c":"8 cm ","d":"data inadequate","e":"none of these"},"options_float":{"a":6.0,"b":10.0,"c":8.0,"d":null,"e":null},"annotated_formula":"subtract(sqrt(64), const_2)","linear_formula":"sqrt(n0)|subtract(#0,const_2)","chain":"64 ** (1\/2)<\/gadget>\n8<\/output>\n8 - 2<\/gadget>\n6<\/output>\n6<\/result>","index":1630} +{"problem":"the cost price of 60 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 60 articles = 60 y the selling price of x articles = 1.20 y * x 1.20 y * x = 60 y x = 60 \/ 1.2 = 50 the answer is d .\"","correct":"d","options":{"a":"42 ","b":"45 ","c":"48 ","d":"50","e":"54"},"options_float":{"a":42.0,"b":45.0,"c":48.0,"d":50.0,"e":54.0},"annotated_formula":"divide(multiply(60, const_4), add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|","chain":"60 * 4<\/gadget>\n240<\/output>\n4 + 1<\/gadget>\n5<\/output>\n240 \/ 5<\/gadget>\n48<\/output>\n48<\/result>","index":1631} +{"problem":"on a certain transatlantic crossing , 40 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 20 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":"\"let t be the total number of passengers . let x be the number of people with round trip tickets . 0.4 t had round trip tickets and took their cars . 0.8 x had round trip tickets and took their cars . 0.8 x = 0.4 t x = 0.5 t the answer is c .\"","correct":"c","options":{"a":"20 % ","b":"40 % ","c":"50 % ","d":"60 %","e":"80 %"},"options_float":{"a":20.0,"b":40.0,"c":50.0,"d":60.0,"e":80.0},"annotated_formula":"divide(40, subtract(const_1, divide(20, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n40 \/ (4\/5)<\/gadget>\n50<\/output>\n50<\/result>","index":1633} +{"problem":"in a certain game , a large container is filled with red , yellow , green , and blue beads worth , respectively , 7 , 5 , 3 , and 2 points each . a number of beads are then removed from the container . if the product of the point values of the removed beads is 30 , 870000 , how many red beads were removed ?","rationale":"30 , 870,000 = 2 ^ 4 * 5 ^ 4 * 3087 = 2 ^ 4 * 3 * 5 ^ 4 * 1029 = 2 ^ 4 * 3 ^ 2 * 5 ^ 4 * 343 = 2 ^ 4 * 3 ^ 2 * 5 ^ 4 * 7 ^ 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":"divide(multiply(3, const_1), const_1)","linear_formula":"multiply(n2,const_1)|divide(#0,const_1)","chain":"3 * 1<\/gadget>\n3<\/output>\n3 \/ 1<\/gadget>\n3<\/output>\n3<\/result>","index":1635} +{"problem":"# 88 a necklace is made by stringing q no individual beads together in the repeating pattern red bead , green bead , white bead , blue bead , and yellow bead . if the necklace design begins with a red bead and ends with a white bead , then q could equal","rationale":"you can just write out the pattern and count : rgwbyrgwbyrgwby . . . but to save time a good test taker will just look for a pattern . min # is 3 , because w is the third one . then every 5 beads another white comes , so it must be 3 + 5 + 5 + 5 . . and so on . . . 3 + 5 = 8 3 + 5 + 5 = 13 3 + 5 + 5 + 5 = 18 3 + 5 + 5 + 5 + 5 = 23 so you see it ends in either 8 or 3 . pick an answer that ends in either 8 or 3 . only one answer does , b .","correct":"b","options":{"a":"16 ","b":"28 ","c":"41 ","d":"54","e":"65"},"options_float":{"a":16.0,"b":28.0,"c":41.0,"d":54.0,"e":65.0},"annotated_formula":"add(add(add(add(add(add(divide(88, 88), const_2), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3))","linear_formula":"add(const_2,const_3)|divide(n0,n0)|add(#1,const_2)|add(#2,#0)|add(#3,#0)|add(#4,#0)|add(#5,#0)|add(#6,#0)","chain":"88 \/ 88<\/gadget>\n1<\/output>\n1 + 2<\/gadget>\n3<\/output>\n2 + 3<\/gadget>\n5<\/output>\n3 + 5<\/gadget>\n8<\/output>\n8 + 5<\/gadget>\n13<\/output>\n13 + 5<\/gadget>\n18<\/output>\n18 + 5<\/gadget>\n23<\/output>\n23 + 5<\/gadget>\n28<\/output>\n28<\/result>","index":1636} +{"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 $ 16 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 ( 16 ) = 0.8 x x = 1.2 ( 16 ) \/ 0.8 = 1.5 ( 16 ) = $ 24 the answer is e .","correct":"e","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(add(multiply(divide(20, const_100), 16), 16), 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) * 16<\/gadget>\n16\/5 = around 3.2<\/output>\n(16\/5) + 16<\/gadget>\n96\/5 = around 19.2<\/output>\n100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(96\/5) \/ (4\/5)<\/gadget>\n24<\/output>\n24<\/result>","index":1637} +{"problem":"rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 3000 . then his monthly salary is","rationale":"\"sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 3000 â ‡ ” 20 \/ 100 x = 3000 â ‡ ” x = 3000 ã — 5 = 15000 . answer a\"","correct":"a","options":{"a":"rs . 15000 ","b":"rs . 12000 ","c":"rs . 9000 ","d":"rs . 6000","e":"rs . 3000"},"options_float":{"a":15000.0,"b":12000.0,"c":9000.0,"d":6000.0,"e":3000.0},"annotated_formula":"multiply(3000, add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n4,#0)|","chain":"4 + 1<\/gadget>\n5<\/output>\n3_000 * 5<\/gadget>\n15_000<\/output>\n15_000<\/result>","index":1638} +{"problem":"last year a certain bond price with a face value of 5000 yielded 9 % 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 = 0.09 * 5000 = 0.065 * selling price - - > selling price = 0.09 * 5000 \/ 0.065 - - > selling price = ~ 6,923 answer : e .\"","correct":"e","options":{"a":"4063 ","b":"5325 ","c":"5351 ","d":"6000","e":"6923"},"options_float":{"a":4063.0,"b":5325.0,"c":5351.0,"d":6000.0,"e":6923.0},"annotated_formula":"divide(multiply(5000, divide(9, const_100)), divide(6.5, const_100))","linear_formula":"divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)|","chain":"9 \/ 100<\/gadget>\n9\/100 = around 0.09<\/output>\n5_000 * (9\/100)<\/gadget>\n450<\/output>\n6.5 \/ 100<\/gadget>\n0.065<\/output>\n450 \/ 0.065<\/gadget>\n6_923.076923<\/output>\n6_923.076923<\/result>","index":1640} +{"problem":"a train of 40 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 = 40 * 60 + 1500 = 3900 m t = 3900 \/ 60 * 18 \/ 5 = 234 sec = 3.9 mins answer : d","correct":"d","options":{"a":"6 ","b":"3 ","c":"4 ","d":"3.9","e":"3.6"},"options_float":{"a":6.0,"b":3.0,"c":4.0,"d":3.9,"e":3.6},"annotated_formula":"add(divide(multiply(add(40, const_1), 60), const_1000), 1.5)","linear_formula":"add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)","chain":"40 + 1<\/gadget>\n41<\/output>\n41 * 60<\/gadget>\n2_460<\/output>\n2_460 \/ 1_000<\/gadget>\n123\/50 = around 2.46<\/output>\n(123\/50) + 1.5<\/gadget>\n3.96<\/output>\n3.96<\/result>","index":1641} +{"problem":"( 0.15 ) ( power 3 ) - ( 0.1 ) ( power 3 ) \/ ( 0.15 ) ( power 2 ) + 0.015 + ( 0.1 ) ( power 2 ) is :","rationale":"\"given expression = ( 0.15 ) ( power 3 ) - ( 0.1 ) ( power 3 ) \/ ( 0.15 ) ( power 2 ) + ( 0.15 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) \/ a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.15 - 0.1 ) = 0.05 answer is c .\"","correct":"c","options":{"a":"0.68 ","b":"0.08 ","c":"0.05 ","d":"0.06","e":"none of them"},"options_float":{"a":0.68,"b":0.08,"c":0.05,"d":0.06,"e":null},"annotated_formula":"divide(subtract(power(0.15, 3), power(0.1, 3)), add(add(power(0.15, 2), 0.015), power(0.1, 2)))","linear_formula":"power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)|","chain":"0.15 ** 3<\/gadget>\n0.003375<\/output>\n0.1 ** 3<\/gadget>\n0.001<\/output>\n0.003375 - 0.001<\/gadget>\n0.002375<\/output>\n0.15 ** 2<\/gadget>\n0.0225<\/output>\n0.0225 + 0.015<\/gadget>\n0.0375<\/output>\n0.1 ** 2<\/gadget>\n0.01<\/output>\n0.0375 + 0.01<\/gadget>\n0.0475<\/output>\n0.002375 \/ 0.0475<\/gadget>\n0.05<\/output>\n0.05<\/result>","index":1643} +{"problem":"a retailer sells 7 shirts . the first 2 he sells for $ 38 and $ 42 . if the retailer wishes to sell the 7 shirts for an overall average price of over $ 50 , what must be the minimum average price of the remaining 5 shirts ?","rationale":"first 2 shirts are sold for $ 38 and $ 42 = $ 80 . to get average price of $ 50 , total sale should be 7 * $ 50 = $ 350 so remaining 5 shirts to be sold for $ 350 - $ 80 = $ 270 answer should be 270 \/ 5 = $ 54.00 that is a","correct":"a","options":{"a":"$ 54.00 ","b":"$ 57.00 ","c":"$ 58.00 ","d":"$ 60.50","e":"$ 63.00"},"options_float":{"a":54.0,"b":57.0,"c":58.0,"d":60.5,"e":63.0},"annotated_formula":"divide(subtract(multiply(7, 50), add(38, 42)), subtract(7, 2))","linear_formula":"add(n2,n3)|multiply(n0,n5)|subtract(n0,n1)|subtract(#1,#0)|divide(#3,#2)","chain":"7 * 50<\/gadget>\n350<\/output>\n38 + 42<\/gadget>\n80<\/output>\n350 - 80<\/gadget>\n270<\/output>\n7 - 2<\/gadget>\n5<\/output>\n270 \/ 5<\/gadget>\n54<\/output>\n54<\/result>","index":1644} +{"problem":"a person crosses a 500 m long street in 4 minutes . what is his speed in km per hour ?","rationale":"\"distance = 500 meter time = 4 minutes = 4 x 60 seconds = 240 seconds speed = distance \/ time = 500 \/ 240 = 2.08 m \/ s = 2.08 ã — 18 \/ 5 km \/ hr = 7.5 km \/ hr answer : a\"","correct":"a","options":{"a":"7.5 ","b":"2.6 ","c":"3.9 ","d":"8.2","e":"2.7"},"options_float":{"a":7.5,"b":2.6,"c":3.9,"d":8.2,"e":2.7},"annotated_formula":"divide(divide(500, const_1000), divide(multiply(4, const_60), const_3600))","linear_formula":"divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)|","chain":"500 \/ 1_000<\/gadget>\n1\/2 = around 0.5<\/output>\n4 * 60<\/gadget>\n240<\/output>\n240 \/ 3_600<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/2) \/ (1\/15)<\/gadget>\n15\/2 = around 7.5<\/output>\n15\/2 = around 7.5<\/result>","index":1645} +{"problem":"if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 160 , how much more remains to be paid ?","rationale":"\"10 \/ 100 p = 160 > > p = 160 * 100 \/ 10 = 1600 1600 - 160 = 1440 answer : e\"","correct":"e","options":{"a":"$ 880 ","b":"$ 990 ","c":"$ 1,000 ","d":"$ 1,100","e":"$ 1,440"},"options_float":{"a":880.0,"b":990.0,"c":1000.0,"d":1100.0,"e":1440.0},"annotated_formula":"subtract(multiply(160, divide(const_100, 10)), 160)","linear_formula":"divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)|","chain":"100 \/ 10<\/gadget>\n10<\/output>\n160 * 10<\/gadget>\n1_600<\/output>\n1_600 - 160<\/gadget>\n1_440<\/output>\n1_440<\/result>","index":1646} +{"problem":"if the perimeter of a rectangular garden is 600 m , its length when its breadth is 100 m is ?","rationale":"\"2 ( l + 100 ) = 600 = > l = 200 m answer : c\"","correct":"c","options":{"a":"227 ","b":"247 ","c":"200 ","d":"277","e":"121"},"options_float":{"a":227.0,"b":247.0,"c":200.0,"d":277.0,"e":121.0},"annotated_formula":"subtract(divide(600, const_2), 100)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"600 \/ 2<\/gadget>\n300<\/output>\n300 - 100<\/gadget>\n200<\/output>\n200<\/result>","index":1647} +{"problem":"you need to pick any number from ' 1 , 3 , 5 , 7 , 9 , 11 , 13 and 15 ' to make below equation true . ( ) + ( ) + ( ) = 30 can you solve it ?","rationale":"solution : 3 ! + 15 + 9 = 30 explanation : 3 ! = 3 * 2 * 1 = 6 6 + 15 + 9 = 30 answer b","correct":"b","options":{"a":"29 ","b":"30 ","c":"31 ","d":"32","e":"33"},"options_float":{"a":29.0,"b":30.0,"c":31.0,"d":32.0,"e":33.0},"annotated_formula":"add(add(11, factorial(3)), 13)","linear_formula":"factorial(n1)|add(n5,#0)|add(n6,#1)","chain":"factorial(3)<\/gadget>\n6<\/output>\n11 + 6<\/gadget>\n17<\/output>\n17 + 13<\/gadget>\n30<\/output>\n30<\/result>","index":1649} +{"problem":"ram and shyam start a two - length swimming race at the same moment but from opposite ends of the pool . they swim in lanes at uniform speeds , but hardy is faster than andy . they 1 st pass at a point 18.5 m from the deep end and having completed one length each 1 is allowed to rest on the edge for exactly 45 sec . after setting off on the return length , the swimmers pass for the 2 nd time just 10.5 m from the shallow end . how long is the pool ?","rationale":"let x = length of pool at first meeting , combined distance = x at second meeting , combined distance = 3 x if andy swims 18.5 m of x , then he will swim 3 * 18.5 = 55.5 m of 3 x andy ' s total distance to second meeting = x + 10.5 m x + 10.5 = 55.5 m x = 45 m e","correct":"e","options":{"a":"65 ","b":"60 ","c":"55 ","d":"50","e":"45"},"options_float":{"a":65.0,"b":60.0,"c":55.0,"d":50.0,"e":45.0},"annotated_formula":"subtract(add(multiply(18.5, const_2), 18.5), 10.5)","linear_formula":"multiply(n1,const_2)|add(n1,#0)|subtract(#1,n5)","chain":"18.5 * 2<\/gadget>\n37<\/output>\n37 + 18.5<\/gadget>\n55.5<\/output>\n55.5 - 10.5<\/gadget>\n45<\/output>\n45<\/result>","index":1650} +{"problem":"what least number must besubtracted from 427398 so that remaining no . is divisible by 15 ?","rationale":"on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted answer : option a","correct":"a","options":{"a":"725117481 ","b":"343564689 ","c":"454564690 ","d":"759900434","e":"656590009"},"options_float":{"a":725117481.0,"b":343564689.0,"c":454564690.0,"d":759900434.0,"e":656590009.0},"annotated_formula":"subtract(subtract(subtract(multiply(multiply(multiply(427398, const_100), const_10), const_2), 427398), multiply(427398, const_100)), multiply(multiply(multiply(const_100, const_100), const_100), const_100))","linear_formula":"multiply(n0,const_100)|multiply(const_100,const_100)|multiply(#0,const_10)|multiply(#1,const_100)|multiply(#2,const_2)|multiply(#3,const_100)|subtract(#4,n0)|subtract(#6,#0)|subtract(#7,#5)","chain":"427_398 * 100<\/gadget>\n42_739_800<\/output>\n42_739_800 * 10<\/gadget>\n427_398_000<\/output>\n427_398_000 * 2<\/gadget>\n854_796_000<\/output>\n854_796_000 - 427_398<\/gadget>\n854_368_602<\/output>\n854_368_602 - 42_739_800<\/gadget>\n811_628_802<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n10_000 * 100<\/gadget>\n1_000_000<\/output>\n1_000_000 * 100<\/gadget>\n100_000_000<\/output>\n811_628_802 - 100_000_000<\/gadget>\n711_628_802<\/output>\n711_628_802<\/result>","index":1652} +{"problem":"in covering a distance of 48 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is :","rationale":"\"let abhay ' s speed be x km \/ hr . then , 48 \/ x - 48 \/ 2 x = 3 6 x = 48 x = 8 km \/ hr . answer : option e\"","correct":"e","options":{"a":"5 kmph ","b":"6 kmph ","c":"6.25 kmph ","d":"7.5 kmph","e":"8 kmph"},"options_float":{"a":5.0,"b":6.0,"c":6.25,"d":7.5,"e":8.0},"annotated_formula":"divide(subtract(48, divide(48, 2)), add(1, 2))","linear_formula":"add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0)|","chain":"48 \/ 2<\/gadget>\n24<\/output>\n48 - 24<\/gadget>\n24<\/output>\n1 + 2<\/gadget>\n3<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n8<\/result>","index":1655} +{"problem":"the speed of a bus increases by 2 kmph after every one hour . if the distance travelled in the first one hour was 35 km , what was the total distance travelled in 12 hours ?","rationale":"dist 1 st hr = 35 km speed of bus by 2 kmph 2 nd hr = 37 km 3 rd hr = 39 km tot = 35 + 37 + 39 + . . . . ( 12 terms ) 12 \/ 2 ( 2 * 35 + ( 12 - 1 ) 2 ] = 6 * 92 = 552 answer c","correct":"c","options":{"a":"550 ","b":"500 ","c":"552 ","d":"560","e":"580"},"options_float":{"a":550.0,"b":500.0,"c":552.0,"d":560.0,"e":580.0},"annotated_formula":"multiply(divide(12, 2), add(multiply(subtract(12, const_1), 2), multiply(2, 35)))","linear_formula":"divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#3,#1)|multiply(#4,#0)","chain":"12 \/ 2<\/gadget>\n6<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n2 * 35<\/gadget>\n70<\/output>\n22 + 70<\/gadget>\n92<\/output>\n6 * 92<\/gadget>\n552<\/output>\n552<\/result>","index":1657} +{"problem":"harkamal purchased 8 kg of grapes at the rate of 90 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 = 90 × 8 = 720 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 720 + 495 = 1215 . b )\"","correct":"b","options":{"a":"1055 ","b":"1215 ","c":"1065 ","d":"1070","e":"1080"},"options_float":{"a":1055.0,"b":1215.0,"c":1065.0,"d":1070.0,"e":1080.0},"annotated_formula":"add(multiply(8, 90), multiply(9, 55))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"8 * 90<\/gadget>\n720<\/output>\n9 * 55<\/gadget>\n495<\/output>\n720 + 495<\/gadget>\n1_215<\/output>\n1_215<\/result>","index":1658} +{"problem":"the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 16 terms of the arithmetic progression ?","rationale":"\"n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n \/ 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 16 . . . we get 15 \/ 2 [ 2 a + 14 d ] = 16 [ a + 7 d ] = 16 * 10 = 160 . . . answer is d . . .\"","correct":"d","options":{"a":"300 ","b":"120 ","c":"150 ","d":"160","e":"270"},"options_float":{"a":300.0,"b":120.0,"c":150.0,"d":160.0,"e":270.0},"annotated_formula":"divide(multiply(20, 16), const_2)","linear_formula":"multiply(n0,n1)|divide(#0,const_2)|","chain":"20 * 16<\/gadget>\n320<\/output>\n320 \/ 2<\/gadget>\n160<\/output>\n160<\/result>","index":1659} +{"problem":"there are 3 red shoes & 7 green shoes . if two of red shoes are drawn what is the probability of getting red shoes","rationale":"\"taking 2 red shoe the probablity is 3 c 2 from 10 shoes probablity of taking 2 red shoe is 3 c 2 \/ 10 c 2 = 1 \/ 15 answer : d\"","correct":"d","options":{"a":"1 \/ 13 ","b":"1 \/ 14 ","c":"1 \/ 12 ","d":"1 \/ 15","e":"1 \/ 16"},"options_float":{"a":0.0769230769,"b":0.0714285714,"c":0.0833333333,"d":0.0666666667,"e":0.0625},"annotated_formula":"divide(choose(3, const_2), choose(add(3, 7), const_2))","linear_formula":"add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)|","chain":"binomial(3, 2)<\/gadget>\n3<\/output>\n3 + 7<\/gadget>\n10<\/output>\nbinomial(10, 2)<\/gadget>\n45<\/output>\n3 \/ 45<\/gadget>\n1\/15 = around 0.066667<\/output>\n1\/15 = around 0.066667<\/result>","index":1660} +{"problem":"it will take 16 days for mary to complete a certain task alone . she worked for 8 days before she was joined by her sister . both of them completed the remaining task in 2 and half days . if her sister had joined her when she started the task , how many days would it have taken ?","rationale":"explanation : mary and her sister complete half work in 2.5 days = > they can complete whole work in 5 days answer : option d","correct":"d","options":{"a":"6 ","b":"8 ","c":"2 ","d":"5","e":"4"},"options_float":{"a":6.0,"b":8.0,"c":2.0,"d":5.0,"e":4.0},"annotated_formula":"add(divide(divide(const_1, 8), divide(const_1, 16)), const_3)","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|divide(#0,#1)|add(#2,const_3)","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n(1\/8) \/ (1\/16)<\/gadget>\n2<\/output>\n2 + 3<\/gadget>\n5<\/output>\n5<\/result>","index":1661} +{"problem":"on dividing 73 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?","rationale":"\"d = ( d - r ) \/ q = ( 73 - 1 ) \/ 9 = 72 \/ 9 = 8 a )\"","correct":"a","options":{"a":"8 ","b":"15 ","c":"16 ","d":"17","e":"18"},"options_float":{"a":8.0,"b":15.0,"c":16.0,"d":17.0,"e":18.0},"annotated_formula":"floor(divide(73, 9))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"73 \/ 9<\/gadget>\n73\/9 = around 8.111111<\/output>\nfloor(73\/9)<\/gadget>\n8<\/output>\n8<\/result>","index":1662} +{"problem":"shruti purchased several number of 3 articles p , q and r in the proportion 3 : 2 : 3 . if the unit costs of the articles p , q and r are 200 , rs . 90 and rs . 60 respectively , how many articles of q must have been purchased in the total purchases of rs . 4800 ?","rationale":"explanation : let the number of articles of types p , q and r be 3 a , 2 a and 3 a respectively . thus , we get , ( 200 x 3 a ) + ( 90 x 2 a ) + ( 60 x 3 a ) = 4800 960 a = 4800 a = 5 hence , the number of articles of type “ q ” = 2 x 5 = 10 answer b","correct":"b","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(divide(4800, add(add(multiply(3, 200), multiply(2, 90)), multiply(3, 60))), 2)","linear_formula":"multiply(n0,n4)|multiply(n2,n5)|multiply(n0,n6)|add(#0,#1)|add(#3,#2)|divide(n7,#4)|multiply(n2,#5)","chain":"3 * 200<\/gadget>\n600<\/output>\n2 * 90<\/gadget>\n180<\/output>\n600 + 180<\/gadget>\n780<\/output>\n3 * 60<\/gadget>\n180<\/output>\n780 + 180<\/gadget>\n960<\/output>\n4_800 \/ 960<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10<\/result>","index":1663} +{"problem":"a boy multiplied 987 by a certain number and obtained 559981 as his answer . if in the answer both 9 are wrong and the other digits are correct , then the correct answer would be :","rationale":"987 = 3 x 7 x 47 so , the required number must be divisible by each one of 3 , 7 , 47 553681 - > ( sum of digits = 28 , not divisible by 3 ) 555181 - > ( sum of digits = 25 , not divisible by 3 ) 555681 is divisible by 3 , 7 , 47 answer c","correct":"c","options":{"a":"553681 ","b":"555181 ","c":"555681 ","d":"556581","e":"556881"},"options_float":{"a":553681.0,"b":555181.0,"c":555681.0,"d":556581.0,"e":556881.0},"annotated_formula":"multiply(subtract(subtract(divide(559981, 987), const_4), const_0_33), 987)","linear_formula":"divide(n1,n0)|subtract(#0,const_4)|subtract(#1,const_0_33)|multiply(n0,#2)","chain":"559_981 \/ 987<\/gadget>\n559_981\/987 = around 567.356636<\/output>\n(559_981\/987) - 4<\/gadget>\n556_033\/987 = around 563.356636<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(556_033\/987) - (1\/3)<\/gadget>\n555_704\/987 = around 563.023303<\/output>\n(555_704\/987) * 987<\/gadget>\n555_704<\/output>\n555_704<\/result>","index":1665} +{"problem":"the least number which when increased by 5 each divisible by each one of 24 , 32 , 36 and 54 is :","rationale":"solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 5 = 864 - 5 = 859 . answer b","correct":"b","options":{"a":"427 ","b":"859 ","c":"869 ","d":"4320","e":"none of these"},"options_float":{"a":427.0,"b":859.0,"c":869.0,"d":4320.0,"e":null},"annotated_formula":"subtract(lcm(lcm(lcm(24, 32), 36), 54), 5)","linear_formula":"lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)","chain":"lcm(24, 32)<\/gadget>\n96<\/output>\nlcm(96, 36)<\/gadget>\n288<\/output>\nlcm(288, 54)<\/gadget>\n864<\/output>\n864 - 5<\/gadget>\n859<\/output>\n859<\/result>","index":1666} +{"problem":"a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 80 % 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 180 , so from 2003 to 2005 it increased by ( 180 - 140 ) \/ 140 = 40 \/ 140 = 2 \/ 7 = ~ 29 % . answer : e .\"","correct":"e","options":{"a":"50 % ","b":"40 % ","c":"35 % ","d":"32 %","e":"29 %"},"options_float":{"a":50.0,"b":40.0,"c":35.0,"d":32.0,"e":29.0},"annotated_formula":"multiply(divide(subtract(add(const_1, divide(80, 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":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 + (4\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 + (2\/5)<\/gadget>\n7\/5 = around 1.4<\/output>\n(9\/5) - (7\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) \/ (7\/5)<\/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":1668} +{"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 65 , what is the common ratio of the g . p ?","rationale":"65 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) \/ ( a 1 + a 2 + a 3 ) factorize the same terms 65 = 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 . . . . . . . . . 65 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) \/ ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 64 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) \/ ( ( 1 + r ^ 1 + r ^ 2 ) ) 64 = r ^ 3 r = 4 a","correct":"a","options":{"a":"4 ","b":"1 \/ 4 ","c":"2 ","d":"9","e":"1 \/ 9"},"options_float":{"a":4.0,"b":0.25,"c":2.0,"d":9.0,"e":0.1111111111},"annotated_formula":"power(subtract(65, const_1), divide(const_1, const_3))","linear_formula":"divide(const_1,const_3)|subtract(n2,const_1)|power(#1,#0)","chain":"65 - 1<\/gadget>\n64<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n64 ** (1\/3)<\/gadget>\n4<\/output>\n4<\/result>","index":1670} +{"problem":"when jessica withdrew $ 200 from her bank account , her account balance decreased by 2 \/ 5 . if she deposits an amount equal to 1 \/ 3 of the remaining balance , what will be the final balance in her bank account ?","rationale":"\"as per the question 200 = 2 a \/ 5 thus - a which is the total amount = 500 the amount thus left = 300 she then deposited 1 \/ 3 of 300 = 100 total amount in her account = 400 answer c\"","correct":"c","options":{"a":"300 ","b":"375 ","c":"400 ","d":"500","e":"575"},"options_float":{"a":300.0,"b":375.0,"c":400.0,"d":500.0,"e":575.0},"annotated_formula":"multiply(subtract(divide(200, subtract(1, divide(const_3, 5))), 200), add(1, divide(1, 3)))","linear_formula":"divide(n3,n4)|divide(const_3,n2)|add(n3,#0)|subtract(n3,#1)|divide(n0,#3)|subtract(#4,n0)|multiply(#2,#5)|","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n200 \/ (2\/5)<\/gadget>\n500<\/output>\n500 - 200<\/gadget>\n300<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n300 * (4\/3)<\/gadget>\n400<\/output>\n400<\/result>","index":1671} +{"problem":"a man traveled a total distance of 1200 km . he traveled one - third of the whole trip by plane and the distance traveled by train is two - thirds of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ?","rationale":"total distance traveled = 1200 km . distance traveled by plane = 400 km . distance traveled by bus = x distance traveled by train = 2 x \/ 3 x + 2 x \/ 3 + 400 = 1200 5 x \/ 3 = 800 x = 480 km the answer is c .","correct":"c","options":{"a":"400 ","b":"440 ","c":"480 ","d":"520","e":"560"},"options_float":{"a":400.0,"b":440.0,"c":480.0,"d":520.0,"e":560.0},"annotated_formula":"divide(multiply(divide(multiply(1200, const_2), const_3), const_3), add(const_2, const_3))","linear_formula":"add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)","chain":"1_200 * 2<\/gadget>\n2_400<\/output>\n2_400 \/ 3<\/gadget>\n800<\/output>\n800 * 3<\/gadget>\n2_400<\/output>\n2 + 3<\/gadget>\n5<\/output>\n2_400 \/ 5<\/gadget>\n480<\/output>\n480<\/result>","index":1672} +{"problem":"in town x , 64 percent of the population are employed , and 40 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 40 are employed males , hence 24 % are employed females . ( employed females ) \/ ( total employed people ) = 24 \/ 64 = 38 % answer : a .\"","correct":"a","options":{"a":"38 % ","b":"25 % ","c":"32 % ","d":"40 %","e":"52 %"},"options_float":{"a":38.0,"b":25.0,"c":32.0,"d":40.0,"e":52.0},"annotated_formula":"multiply(divide(subtract(64, 40), 64), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"64 - 40<\/gadget>\n24<\/output>\n24 \/ 64<\/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":1673} +{"problem":"a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 237 for it , the cost price of the cricket bat for a is :","rationale":"\"125 % of 120 % of a = 237 125 \/ 100 * 120 \/ 100 * a = 237 a = 237 * 2 \/ 3 = 158 . answer c\"","correct":"c","options":{"a":"150 ","b":"120 ","c":"158 ","d":"160","e":"210"},"options_float":{"a":150.0,"b":120.0,"c":158.0,"d":160.0,"e":210.0},"annotated_formula":"divide(237, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n(6\/5) * (5\/4)<\/gadget>\n3\/2 = around 1.5<\/output>\n237 \/ (3\/2)<\/gadget>\n158<\/output>\n158<\/result>","index":1675} +{"problem":"how many numbers between 100 and 756 are divisible by 2 , 3 , and 7 together ?","rationale":"\"explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 756 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 756 : 756 \/ 42 = 18 hence , 42 * 16 = 756 is the last number divisible by 42 within 756 hence , total numbers divisible by 2 , 3 , 7 together are ( 18 â € “ 2 ) = 16 answer : d\"","correct":"d","options":{"a":"112 ","b":"77 ","c":"267 ","d":"16","e":"99"},"options_float":{"a":112.0,"b":77.0,"c":267.0,"d":16.0,"e":99.0},"annotated_formula":"subtract(divide(756, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7)))","linear_formula":"multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)|","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 7<\/gadget>\n42<\/output>\n756 \/ 42<\/gadget>\n18<\/output>\n100 \/ 42<\/gadget>\n50\/21 = around 2.380952<\/output>\n18 - (50\/21)<\/gadget>\n328\/21 = around 15.619048<\/output>\n328\/21 = around 15.619048<\/result>","index":1676} +{"problem":"if the sides of a cube are in the ratio 9 : 5 . what is the ratio of their diagonals ?","rationale":"\"explanation : diagonal of a cube = a √ 3 where a is side a 1 : a 2 = 9 : 5 d 1 : d 2 = 9 : 5 where √ 3 cancelled both side answer : a\"","correct":"a","options":{"a":"9 : 5 ","b":"16 : 9 ","c":"4 : ","d":"3 : 4","e":"3 : 8"},"options_float":{"a":1.8,"b":1.7777777778,"c":4.0,"d":0.75,"e":0.375},"annotated_formula":"divide(9, 5)","linear_formula":"divide(n0,n1)|","chain":"9 \/ 5<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":1677} +{"problem":"a and b together do a work in 20 days . b and c together in 15 days and c and a in 12 days . so a , b and c together finish same work in how many days ?","rationale":"( a + b ) work in 1 day = 1 \/ 20 , ( b + c ) work in 1 days = 1 \/ 15 . , ( c + a ) work in 1 days = 1 \/ 12 ( 1 ) adding = 2 [ a + b + c ] in 1 day work = [ 1 \/ 20 + 1 \/ 15 + 1 \/ 12 ] = 1 \/ 5 ( a + b + c ) work in 1 day = 1 \/ 10 so , all three together finish work in 10 days answer d","correct":"d","options":{"a":"12 ","b":"15 ","c":"8 ","d":"10","e":"11"},"options_float":{"a":12.0,"b":15.0,"c":8.0,"d":10.0,"e":11.0},"annotated_formula":"inverse(divide(add(inverse(12), add(inverse(20), inverse(15))), const_2))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|inverse(#5)","chain":"1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/20) + (1\/15)<\/gadget>\n7\/60 = around 0.116667<\/output>\n(1\/12) + (7\/60)<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) \/ 2<\/gadget>\n1\/10 = around 0.1<\/output>\n1 \/ (1\/10)<\/gadget>\n10<\/output>\n10<\/result>","index":1678} +{"problem":"if a train , travelling at a speed of 180 kmph , crosses a pole in 6 sec , then the length of train is ?","rationale":"d = 180 * 5 \/ 18 * 6 = 300 m answer : a","correct":"a","options":{"a":"300 ","b":"125 ","c":"288 ","d":"266","e":"121"},"options_float":{"a":300.0,"b":125.0,"c":288.0,"d":266.0,"e":121.0},"annotated_formula":"multiply(multiply(180, const_0_2778), 6)","linear_formula":"multiply(n0,const_0_2778)|multiply(n1,#0)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n180 * (5\/18)<\/gadget>\n50<\/output>\n50 * 6<\/gadget>\n300<\/output>\n300<\/result>","index":1679} +{"problem":"a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 36 . 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 . 180 ( 36 * 5 ) profit percentage = ( 180 - 100 ) \/ 100 * 100 = 80 % answer : a\"","correct":"a","options":{"a":"80 % ","b":"50 % ","c":"59 % ","d":"40 %","e":"53 %"},"options_float":{"a":80.0,"b":50.0,"c":59.0,"d":40.0,"e":53.0},"annotated_formula":"subtract(multiply(36, 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>\n36 * 5<\/gadget>\n180<\/output>\n25 * 4<\/gadget>\n100<\/output>\n180 - 100<\/gadget>\n80<\/output>\n80<\/result>","index":1680} +{"problem":"a can do a piece of work 60 days . b can do work in 90 days . in how many days they will complete the work together ?","rationale":"lcm = 180 , ratio = 60 : 90 = 2 : 3 no of days = 180 \/ ( 2 + 3 ) = 180 \/ 5 = 36 days answer : a","correct":"a","options":{"a":"36 days ","b":"32 days ","c":"19 days ","d":"17 days","e":"18 days"},"options_float":{"a":36.0,"b":32.0,"c":19.0,"d":17.0,"e":18.0},"annotated_formula":"divide(const_1, add(divide(const_1, 60), divide(const_1, 90)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)","chain":"1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n1 \/ 90<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/60) + (1\/90)<\/gadget>\n1\/36 = around 0.027778<\/output>\n1 \/ (1\/36)<\/gadget>\n36<\/output>\n36<\/result>","index":1681} +{"problem":"what least number should be added to 1536 , so that the sum is completely divisible by 21 ?","rationale":"\"1536 ÷ 21 = 73 reminder - 3 3 + 18 = 21 hence 18 should be added to 1536 so that the sum will be divisible by 21 answer : 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":"subtract(21, reminder(1536, 21))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|","chain":"1_536 % 21<\/gadget>\n3<\/output>\n21 - 3<\/gadget>\n18<\/output>\n18<\/result>","index":1682} +{"problem":"gary drove from point a to point b at 60 km \/ h . on his way back he took a train travelling at 110 km \/ h and therefore his trip back lasted 5 hours less . what is the distance ( in km ) between a and b ?","rationale":"distance = speed * time d 1 = s 1 t 1 d 2 = s 2 t 2 the distance from point a to point b is the same for each trip so , d 1 = d 2 and t 2 = t 1 - 5 thus , s 1 t 1 = s 2 t 2 60 t 1 = s 2 ( t 1 - 5 ) t 1 = 11 60 * 11 = 660 answer : c","correct":"c","options":{"a":"600 . ","b":"630 . ","c":"660 . ","d":"690 .","e":"720 ."},"options_float":{"a":600.0,"b":630.0,"c":660.0,"d":690.0,"e":720.0},"annotated_formula":"multiply(60, divide(multiply(110, 5), subtract(110, 60)))","linear_formula":"multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)","chain":"110 * 5<\/gadget>\n550<\/output>\n110 - 60<\/gadget>\n50<\/output>\n550 \/ 50<\/gadget>\n11<\/output>\n60 * 11<\/gadget>\n660<\/output>\n660<\/result>","index":1684} +{"problem":"when n is divided by 19 , the remainder is 7 . find thee difference between previous remainder and the remainder when 18 n is divided by 9 ?","rationale":"let n = 7 ( leaves a remainder of 7 when divided by 19 ) 18 n = 18 ( 7 ) = 126 , which leaves a remainder of 0 when divided by 9 . difference = 7 - 0 = 7 . answer a","correct":"a","options":{"a":"7 ","b":"5 ","c":"0 ","d":"3","e":"9"},"options_float":{"a":7.0,"b":5.0,"c":0.0,"d":3.0,"e":9.0},"annotated_formula":"subtract(7, reminder(18, 9))","linear_formula":"reminder(n2,n3)|subtract(n1,#0)","chain":"18 % 9<\/gadget>\n0<\/output>\n7 - 0<\/gadget>\n7<\/output>\n7<\/result>","index":1685} +{"problem":"solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 75 percent carbonated water , what percent of the volume of the mixture is p ?","rationale":"\"75 % is 5 % - points below 80 % and 20 % - points above 55 % . so the ratio of solution p to solution q is 4 : 1 . mixture p is 4 \/ 5 = 80 % of the volume of mixture pq . the answer is d .\"","correct":"d","options":{"a":"40 % ","b":"50 % ","c":"60 % ","d":"80 %","e":"90 %"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":80.0,"e":90.0},"annotated_formula":"multiply(divide(subtract(divide(75, const_100), divide(55, const_100)), add(subtract(divide(75, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(75, const_100)))), const_100)","linear_formula":"divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#0)|add(#3,#4)|divide(#3,#5)|multiply(#6,const_100)|","chain":"75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n55 \/ 100<\/gadget>\n11\/20 = around 0.55<\/output>\n(3\/4) - (11\/20)<\/gadget>\n1\/5 = around 0.2<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) - (3\/4)<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/5) + (1\/20)<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/5) \/ (1\/4)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":1686} +{"problem":"the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km \/ hr and 15 km \/ hr respectively . they will meet for the first time in ?","rationale":"\"clearly , the two will meet when they are 1000 m apart to be 20 + 15 = 35 km apart , they take 1 hour to be 1000 m apart , they take 35 * 1000 \/ 1000 = 35 min . answer is c\"","correct":"c","options":{"a":"50 min ","b":"40 min ","c":"35 min ","d":"25 min","e":"20 min"},"options_float":{"a":50.0,"b":40.0,"c":35.0,"d":25.0,"e":20.0},"annotated_formula":"add(20, 15)","linear_formula":"add(n1,n2)|","chain":"20 + 15<\/gadget>\n35<\/output>\n35<\/result>","index":1688} +{"problem":"the average weight of a class of 24 students is 35 kg . if the weight of the teacher be included , the average rises by 400 g . the weight of the teacher is :","rationale":"weight of the teacher = ( 35.4 x 25 - 35 x 24 ) kg = 45 kg . answer : a","correct":"a","options":{"a":"45 ","b":"46 ","c":"47 ","d":"48","e":"49"},"options_float":{"a":45.0,"b":46.0,"c":47.0,"d":48.0,"e":49.0},"annotated_formula":"subtract(multiply(add(35, divide(400, const_1000)), add(24, const_1)), multiply(24, 35))","linear_formula":"add(n0,const_1)|divide(n2,const_1000)|multiply(n0,n1)|add(n1,#1)|multiply(#3,#0)|subtract(#4,#2)","chain":"400 \/ 1_000<\/gadget>\n2\/5 = around 0.4<\/output>\n35 + (2\/5)<\/gadget>\n177\/5 = around 35.4<\/output>\n24 + 1<\/gadget>\n25<\/output>\n(177\/5) * 25<\/gadget>\n885<\/output>\n24 * 35<\/gadget>\n840<\/output>\n885 - 840<\/gadget>\n45<\/output>\n45<\/result>","index":1689} +{"problem":"the contents of a certain box consist of 14 apples and 24 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ?","rationale":"\"the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 \/ 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 24 oranges , so to get to 6 oranges , we should remove 24 - 6 = 18 oranges . answer a\"","correct":"a","options":{"a":"18 ","b":"6 ","c":"14 ","d":"17","e":"20"},"options_float":{"a":18.0,"b":6.0,"c":14.0,"d":17.0,"e":20.0},"annotated_formula":"subtract(add(14, 24), divide(14, divide(70, const_100)))","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)|","chain":"14 + 24<\/gadget>\n38<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n14 \/ (7\/10)<\/gadget>\n20<\/output>\n38 - 20<\/gadget>\n18<\/output>\n18<\/result>","index":1690} +{"problem":"mary ' s income is 60 percent more than tim ' s income , and tim ' s income is 40 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 = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 96 ( 60 percent more than tim ' s income ) . thus , mary ' s income ( 96 ) is 96 % of juan ' s income ( 100 ) . answer : c\"","correct":"c","options":{"a":"124 % ","b":"120 % ","c":"96 % ","d":"80 %","e":"64 %"},"options_float":{"a":124.0,"b":120.0,"c":96.0,"d":80.0,"e":64.0},"annotated_formula":"multiply(multiply(subtract(const_1, divide(40, 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":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n(3\/5) * (8\/5)<\/gadget>\n24\/25 = around 0.96<\/output>\n(24\/25) * 100<\/gadget>\n96<\/output>\n96<\/result>","index":1691} +{"problem":"a and b are two circles . the radius of a is four times as large as the diameter of b . what is the ratio between the areas of the circles ?","rationale":"given : the radius of a is 4 times as large as the diameter of b . = > r ( a ) = 4 * d ( b ) = 4 * 2 * r ( b ) = 8 r ( b ) . the radius are in ratio of 1 : 8 thus the area will be in the ratio of square of radius . 1 : 64 . hence d .","correct":"d","options":{"a":"1 : 8 . ","b":"1 : 2 . ","c":"1 : 24 . ","d":"1 : 64 .","e":"1 : 6 ."},"options_float":{"a":0.125,"b":0.5,"c":0.0416666667,"d":0.015625,"e":0.1666666667},"annotated_formula":"divide(power(const_1, const_2), power(multiply(const_2, const_4), const_2))","linear_formula":"multiply(const_2,const_4)|power(const_1,const_2)|power(#0,const_2)|divide(#1,#2)","chain":"1 ** 2<\/gadget>\n1<\/output>\n2 * 4<\/gadget>\n8<\/output>\n8 ** 2<\/gadget>\n64<\/output>\n1 \/ 64<\/gadget>\n1\/64 = around 0.015625<\/output>\n1\/64 = around 0.015625<\/result>","index":1692} +{"problem":"a can run 192 metre in 28 seconds and b in 32 seconds . by what distance a beat b ?","rationale":"\"clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance \/ time taken by b = 192 \/ 32 = 6 m \/ s distance covered by b in 4 seconds = speed ã — time = 6 ã — 4 = 24 metre i . e . , a beat b by 24 metre answer is c\"","correct":"c","options":{"a":"38 metre ","b":"28 metre ","c":"24 metre ","d":"15 metre","e":"28 metre"},"options_float":{"a":38.0,"b":28.0,"c":24.0,"d":15.0,"e":28.0},"annotated_formula":"subtract(192, multiply(divide(192, 32), 28))","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|","chain":"192 \/ 32<\/gadget>\n6<\/output>\n6 * 28<\/gadget>\n168<\/output>\n192 - 168<\/gadget>\n24<\/output>\n24<\/result>","index":1693} +{"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 , 585.75 . 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 ) = 1585.75 = > x = 791 e\"","correct":"e","options":{"a":"0 ","b":"233 ","c":"500 ","d":"695","e":"791"},"options_float":{"a":0.0,"b":233.0,"c":500.0,"d":695.0,"e":791.0},"annotated_formula":"divide(add(const_1000, 585.75), const_2)","linear_formula":"add(n4,const_1000)|divide(#0,const_2)|","chain":"1_000 + 585.75<\/gadget>\n1_585.75<\/output>\n1_585.75 \/ 2<\/gadget>\n792.875<\/output>\n792.875<\/result>","index":1694} +{"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 550 ?","rationale":"\"slope of 2 and a y - intercept of 2 y - coordinate is 550 y = 2 x + 2 548 = 2 x x = 274 answer : e . 274\"","correct":"e","options":{"a":"249 ","b":"498 ","c":"676 ","d":"823","e":"274"},"options_float":{"a":249.0,"b":498.0,"c":676.0,"d":823.0,"e":274.0},"annotated_formula":"divide(subtract(550, 2), 2)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|","chain":"550 - 2<\/gadget>\n548<\/output>\n548 \/ 2<\/gadget>\n274<\/output>\n274<\/result>","index":1697} +{"problem":"if xy = 4 , x \/ y = 16 , for positive numbers x and y , y = ?","rationale":"\"very easy question . 2 variables and 2 easy equations . xy = 4 - - - > x = 4 \/ y - ( i ) x \/ y = 16 - - - > replacing ( i ) here - - - > 4 \/ ( y ^ 2 ) = 16 - - - > y ^ 2 = 4 \/ 16 = 1 \/ 4 - - - > y = 1 \/ 2 or - 1 \/ 2 the question states that x and y are positive integers . therefore , y = 1 \/ 2 is the answer . answer a .\"","correct":"a","options":{"a":"1 \/ 2 ","b":"2 ","c":"1 \/ 3 ","d":"3","e":"1 \/ 6"},"options_float":{"a":0.5,"b":2.0,"c":0.3333333333,"d":3.0,"e":0.1666666667},"annotated_formula":"sqrt(divide(4, 16))","linear_formula":"divide(n0,n1)|sqrt(#0)|","chain":"4 \/ 16<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) ** (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":1700} +{"problem":"if a boat is rowed downstream for 24 km in 4 hours and upstream for 48 km in 24 hours , what is the speed of the boat and the river ?","rationale":"\"explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) \/ 2 y = ( ds â € “ us ) \/ 2 in the above problem ds = 6 ; us = 2 x = ( 6 + 2 ) \/ 2 = 8 \/ 2 = 4 km \/ hr y = ( 6 - 2 ) \/ 2 = 4 \/ 2 = 2 km \/ hr answer : e\"","correct":"e","options":{"a":"4 , 3 ","b":"4 , 4 ","c":"3 , 3 ","d":"4 , 5","e":"4 , 2"},"options_float":{"a":4.0,"b":4.0,"c":3.0,"d":4.0,"e":4.0},"annotated_formula":"divide(add(divide(48, 24), divide(24, 4)), const_2)","linear_formula":"divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2)|","chain":"48 \/ 24<\/gadget>\n2<\/output>\n24 \/ 4<\/gadget>\n6<\/output>\n2 + 6<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":1701} +{"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 86 , 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 = 86 thus m = 688 - - - - - - - - 2 also , f \/ f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 688 + 92 f ) \/ ( 8 + f ) = 90 solving this we get f = 16 ans : 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":"divide(subtract(multiply(90, 8), multiply(86, 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>\n86 * 8<\/gadget>\n688<\/output>\n720 - 688<\/gadget>\n32<\/output>\n92 - 90<\/gadget>\n2<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16<\/result>","index":1704} +{"problem":"a cycle is bought for rs . 800 and sold for rs . 1080 , find the gain percent ?","rationale":"\"800 - - - - 180 100 - - - - ? = > 35 % answer : c\"","correct":"c","options":{"a":"22 ","b":"20 ","c":"35 ","d":"88","e":"11"},"options_float":{"a":22.0,"b":20.0,"c":35.0,"d":88.0,"e":11.0},"annotated_formula":"multiply(divide(subtract(1080, 800), 800), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_080 - 800<\/gadget>\n280<\/output>\n280 \/ 800<\/gadget>\n7\/20 = around 0.35<\/output>\n(7\/20) * 100<\/gadget>\n35<\/output>\n35<\/result>","index":1705} +{"problem":"a train running at the speed of 60 km \/ hr crosses a pole in 6 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 ) * 6 = 100 meter answer : c\"","correct":"c","options":{"a":"150 ","b":"278 ","c":"100 ","d":"776","e":"191"},"options_float":{"a":150.0,"b":278.0,"c":100.0,"d":776.0,"e":191.0},"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":1707} +{"problem":"laura took out a charge account at the general store and agreed to pay 8 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ?","rationale":"\"principal that is amount taken by laura at year beginning = 35 $ rate of interest = 8 % interest = ( 8 \/ 100 ) * 35 = 2.8 $ total amount that laura owes a year later = 35 + 2.8 = 37.8 $ answer d\"","correct":"d","options":{"a":"$ 2.10 ","b":"$ 37.10 ","c":"$ 37.16 ","d":"$ 37.8","e":"$ 38.80"},"options_float":{"a":2.1,"b":37.1,"c":37.16,"d":37.8,"e":38.8},"annotated_formula":"add(multiply(divide(8, const_100), 35), 35)","linear_formula":"divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|","chain":"8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) * 35<\/gadget>\n14\/5 = around 2.8<\/output>\n(14\/5) + 35<\/gadget>\n189\/5 = around 37.8<\/output>\n189\/5 = around 37.8<\/result>","index":1708} +{"problem":"a man purchased 3 blankets @ rs . 100 each , 4 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ?","rationale":"explanation : 10 * 150 = 1500 3 * 100 + 4 * 150 = 900 1500 – 900 = 600 a","correct":"a","options":{"a":"600 ","b":"350 ","c":"450 ","d":"470","e":"500"},"options_float":{"a":600.0,"b":350.0,"c":450.0,"d":470.0,"e":500.0},"annotated_formula":"subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(4, 150)))","linear_formula":"multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)","chain":"10 * 150<\/gadget>\n1_500<\/output>\n3 * 100<\/gadget>\n300<\/output>\n4 * 150<\/gadget>\n600<\/output>\n300 + 600<\/gadget>\n900<\/output>\n1_500 - 900<\/gadget>\n600<\/output>\n600<\/result>","index":1709} +{"problem":"rates for having a manuscript typed at a certain typing service are $ 6 per page for the first time a page is typed and $ 4 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ?","rationale":"\"50 pages typed 1 x 40 pages typed 2 x ( original + one revision ) 10 pages typed 3 x ( original + two revisions ) 50 ( 6 ) + 40 ( 6 + 4 ) + 10 ( 6 + 4 + 4 ) = 300 + 400 + 140 = 840 answer - b\"","correct":"b","options":{"a":"$ 850 ","b":"$ 840 ","c":"$ 860 ","d":"$ 870","e":"$ 880"},"options_float":{"a":850.0,"b":840.0,"c":860.0,"d":870.0,"e":880.0},"annotated_formula":"add(add(multiply(100, 6), multiply(40, 4)), multiply(multiply(10, 4), const_2))","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)|","chain":"100 * 6<\/gadget>\n600<\/output>\n40 * 4<\/gadget>\n160<\/output>\n600 + 160<\/gadget>\n760<\/output>\n10 * 4<\/gadget>\n40<\/output>\n40 * 2<\/gadget>\n80<\/output>\n760 + 80<\/gadget>\n840<\/output>\n840<\/result>","index":1710} +{"problem":"a gambler has won 40 % of his 30 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 ( 30 ) + 0.7 x = 0.6 ( x + 30 ) 0.1 x = 6 x = 60 the answer is e .","correct":"e","options":{"a":"36 ","b":"42 ","c":"48 ","d":"54","e":"60"},"options_float":{"a":36.0,"b":42.0,"c":48.0,"d":54.0,"e":60.0},"annotated_formula":"divide(subtract(multiply(30, divide(60, const_100)), multiply(30, 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>\n30 * (3\/5)<\/gadget>\n18<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n30 * (2\/5)<\/gadget>\n12<\/output>\n18 - 12<\/gadget>\n6<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) - (3\/5)<\/gadget>\n1\/10 = around 0.1<\/output>\n6 \/ (1\/10)<\/gadget>\n60<\/output>\n60<\/result>","index":1711} +{"problem":"a is 30 % more efficient than b . how much time they will working together take to complete a job which a alone could have done in 23 days ?","rationale":"\"the ratio of times taken by a and b = 100 : 130 = 10 : 13 suppose b can do work in x days then 10 : 13 : : 23 : x x = ( 23 * 13 ) \/ 10 x = 299 \/ 10 a ' s 1 day ' s work = 1 \/ 23 b ' s 1 day ' s work = 10 \/ 299 ( a + b ) ' s 1 day ' s work = 1 \/ 23 + 10 \/ 299 = 23 \/ 299 = 1 \/ 13 a and b together can do work in 13 days answer ( b )\"","correct":"b","options":{"a":"25 days ","b":"13 days ","c":"14 days ","d":"20 days","e":"15 days"},"options_float":{"a":25.0,"b":13.0,"c":14.0,"d":20.0,"e":15.0},"annotated_formula":"inverse(add(divide(const_1, 23), divide(const_1, multiply(add(divide(30, const_100), const_1), 23))))","linear_formula":"divide(const_1,n1)|divide(n0,const_100)|add(#1,const_1)|multiply(n1,#2)|divide(const_1,#3)|add(#0,#4)|inverse(#5)|","chain":"1 \/ 23<\/gadget>\n1\/23 = around 0.043478<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) + 1<\/gadget>\n13\/10 = around 1.3<\/output>\n(13\/10) * 23<\/gadget>\n299\/10 = around 29.9<\/output>\n1 \/ (299\/10)<\/gadget>\n10\/299 = around 0.033445<\/output>\n(1\/23) + (10\/299)<\/gadget>\n1\/13 = around 0.076923<\/output>\n1 \/ (1\/13)<\/gadget>\n13<\/output>\n13<\/result>","index":1713} +{"problem":"this year , mbb consulting fired 6 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ?","rationale":"\"100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 6 % reduction in employees lead to 94 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 94 employees is 94 * 1100 = 103400 now the new salary is more than previous salary by x % . x = ( 103400 \/ 100000 ) * 100 = 103.4 % so the answer is d\"","correct":"d","options":{"a":"98.5 % ","b":"100.0 % ","c":"102.8 % ","d":"103.4 %","e":"105.0 %"},"options_float":{"a":98.5,"b":100.0,"c":102.8,"d":103.4,"e":105.0},"annotated_formula":"divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 6), const_4)), multiply(const_100, 10))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n100 + 1_000<\/gadget>\n1_100<\/output>\n100 - 6<\/gadget>\n94<\/output>\n94 + 4<\/gadget>\n98<\/output>\n1_100 * 98<\/gadget>\n107_800<\/output>\n107_800 \/ 1_000<\/gadget>\n539\/5 = around 107.8<\/output>\n539\/5 = around 107.8<\/result>","index":1714} +{"problem":"find the cost of fencing around a circular field of diameter 12 m at the rate of rs . 3.50 a meter ?","rationale":"\"2 * 22 \/ 7 * 6 = 37.7 37.7 * 3 1 \/ 2 = rs . 131.95 answer : a\"","correct":"a","options":{"a":"131.95 ","b":"132.9 ","c":"140.33 ","d":"123.4","e":"190.4"},"options_float":{"a":131.95,"b":132.9,"c":140.33,"d":123.4,"e":190.4},"annotated_formula":"multiply(circumface(divide(12, const_2)), 3.50)","linear_formula":"divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|","chain":"12 \/ 2<\/gadget>\n6<\/output>\n2 * pi * 6<\/gadget>\n12*pi = around 37.699112<\/output>\n(12*pi) * 3.5<\/gadget>\n42.0*pi = around 131.946891<\/output>\n42.0*pi = around 131.946891<\/result>","index":1717} +{"problem":"a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 28 adults have their meal , find the total number of children that can be catered with the remaining food .","rationale":"\"explanation : as there is meal for 70 adults and 28 have their meal , the meal left can be catered to 42 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 42 adults = 54 children hence , the meal can be catered to 54 children . answer : b\"","correct":"b","options":{"a":"33 ","b":"54 ","c":"18 ","d":"17","e":"01"},"options_float":{"a":33.0,"b":54.0,"c":18.0,"d":17.0,"e":1.0},"annotated_formula":"multiply(subtract(70, 28), divide(90, 70))","linear_formula":"divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1)|","chain":"70 - 28<\/gadget>\n42<\/output>\n90 \/ 70<\/gadget>\n9\/7 = around 1.285714<\/output>\n42 * (9\/7)<\/gadget>\n54<\/output>\n54<\/result>","index":1718} +{"problem":"a side of beef lost 25 percent of its weight in processing . if the side of beef weighed 540 pounds after processing , how many pounds did it weigh before processing ?","rationale":"\"let weight of side of beef before processing = x ( 75 \/ 100 ) * x = 540 = > x = ( 540 * 100 ) \/ 75 = 720 answer c\"","correct":"c","options":{"a":"191 ","b":"355 ","c":"720 ","d":"840","e":"1,560"},"options_float":{"a":191.0,"b":355.0,"c":720.0,"d":840.0,"e":1560.0},"annotated_formula":"divide(multiply(540, const_100), subtract(const_100, 25))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|","chain":"540 * 100<\/gadget>\n54_000<\/output>\n100 - 25<\/gadget>\n75<\/output>\n54_000 \/ 75<\/gadget>\n720<\/output>\n720<\/result>","index":1719} +{"problem":"when a random experiment is conducted , the probability that event a occurs is 1 \/ 3 . if the random experiment is conducted 5 independent times , what is the probability that event a occurs exactly twice ?","rationale":"one case is : 1 \/ 3 * 1 \/ 3 * 2 \/ 3 * 2 \/ 3 * 2 \/ 3 = 2 ^ 3 \/ 3 ^ 5 we have 5 ! \/ 2 ! * 3 ! = 10 such cases so , 2 ^ 3 * 10 \/ 3 ^ 5 = 80 \/ 243 answer : d .","correct":"d","options":{"a":"5 \/ 243 ","b":"25 \/ 243 ","c":"64 \/ 243 ","d":"80 \/ 243","e":"16 \/ 17"},"options_float":{"a":0.0205761317,"b":0.1028806584,"c":0.2633744856,"d":0.329218107,"e":0.9411764706},"annotated_formula":"subtract(1, divide(const_2, const_3))","linear_formula":"divide(const_2,const_3)|subtract(n0,#0)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":1720} +{"problem":"a reduction of 30 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 940 , what is the reduced price for kg ?","rationale":"\"940 * ( 30 \/ 100 ) = 282 - - - - 6 ? - - - - 1 = > rs . 47 answer : b\"","correct":"b","options":{"a":"72 ","b":"47 ","c":"40 ","d":"28","e":"20"},"options_float":{"a":72.0,"b":47.0,"c":40.0,"d":28.0,"e":20.0},"annotated_formula":"divide(divide(multiply(940, 30), const_100), 6)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|","chain":"940 * 30<\/gadget>\n28_200<\/output>\n28_200 \/ 100<\/gadget>\n282<\/output>\n282 \/ 6<\/gadget>\n47<\/output>\n47<\/result>","index":1721} +{"problem":"a certain company reported that the revenue on sales increased 30 % from 2000 to 2003 , and increased 80 % 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 130 and and in 2005 180 , so from 2003 to 2005 it increased by ( 180 - 130 ) \/ 130 = 50 \/ 130 = 39 % answer : e .","correct":"e","options":{"a":"50 % ","b":"40 % ","c":"35 % ","d":"32 %","e":"39 %"},"options_float":{"a":50.0,"b":40.0,"c":35.0,"d":32.0,"e":39.0},"annotated_formula":"multiply(divide(subtract(add(const_1, divide(80, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(30, 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":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 + (4\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n(9\/5) - (13\/10)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) \/ (13\/10)<\/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":1723} +{"problem":"if a man lost 5 % by selling oranges at the rate of 8 a rupee at how many a rupee must he sell them to gain 52 % ?","rationale":"\"95 % - - - - 8 152 % - - - - ? 95 \/ 152 * 8 = 5 answer : a\"","correct":"a","options":{"a":"5 ","b":"8 ","c":"7 ","d":"4","e":"2"},"options_float":{"a":5.0,"b":8.0,"c":7.0,"d":4.0,"e":2.0},"annotated_formula":"divide(multiply(subtract(const_100, 5), 8), add(const_100, 52))","linear_formula":"add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|","chain":"100 - 5<\/gadget>\n95<\/output>\n95 * 8<\/gadget>\n760<\/output>\n100 + 52<\/gadget>\n152<\/output>\n760 \/ 152<\/gadget>\n5<\/output>\n5<\/result>","index":1724} +{"problem":"there are 32 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ?","rationale":"the total number of stations = 34 from 34 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 34 p 2 ways . 34 p 2 = 34 * 33 = 1122 answer : d","correct":"d","options":{"a":"1800 ","b":"1820 ","c":"1150 ","d":"1122","e":"1900"},"options_float":{"a":1800.0,"b":1820.0,"c":1150.0,"d":1122.0,"e":1900.0},"annotated_formula":"multiply(add(32, const_2), subtract(add(32, const_2), const_1))","linear_formula":"add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)","chain":"32 + 2<\/gadget>\n34<\/output>\n34 - 1<\/gadget>\n33<\/output>\n34 * 33<\/gadget>\n1_122<\/output>\n1_122<\/result>","index":1725} +{"problem":"a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 5 , ab = 5 and ac = 11 , what is the length of ae ?","rationale":"\"ac = 11 and ab = 5 , so bc = 6 . bc = 2 cd so cd = 3 . the length of ae is ab + bc + cd + de = 5 + 6 + 3 + 5 = 19 the answer is a .\"","correct":"a","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":"add(add(11, divide(subtract(11, 5), 2)), 5)","linear_formula":"subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)|","chain":"11 - 5<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n11 + 3<\/gadget>\n14<\/output>\n14 + 5<\/gadget>\n19<\/output>\n19<\/result>","index":1728} +{"problem":"a can run 288 metre in 28 seconds and b in 32 seconds . by what distance a beat b ?","rationale":"\"clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance \/ time taken by b = 288 \/ 32 = 9 m \/ s distance covered by b in 4 seconds = speed ã — time = 9 ã — 4 = 36 metre i . e . , a beat b by 36 metre answer is e\"","correct":"e","options":{"a":"38 metre ","b":"28 metre ","c":"23 metre ","d":"15 metre","e":"36 metre"},"options_float":{"a":38.0,"b":28.0,"c":23.0,"d":15.0,"e":36.0},"annotated_formula":"subtract(288, multiply(divide(288, 32), 28))","linear_formula":"divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|","chain":"288 \/ 32<\/gadget>\n9<\/output>\n9 * 28<\/gadget>\n252<\/output>\n288 - 252<\/gadget>\n36<\/output>\n36<\/result>","index":1729} +{"problem":"if x \/ y = 7 \/ 4 , then ( x + y ) \/ ( x - y ) = ?","rationale":"\"any x and y satisfying x \/ y = 7 \/ 4 should give the same value for ( x + y ) \/ ( x - y ) . say x = 7 and y = 4 , then ( x + y ) \/ ( x - y ) = ( 7 + 4 ) \/ ( 7 - 4 ) = 11 \/ 3 . answer : b .\"","correct":"b","options":{"a":"5 ","b":"11 \/ 3 ","c":"- 1 \/ 6 ","d":"- 1 \/ 5","e":"- 5"},"options_float":{"a":5.0,"b":3.6666666667,"c":-0.1666666667,"d":-0.2,"e":-5.0},"annotated_formula":"divide(add(7, 4), subtract(7, 4))","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"7 + 4<\/gadget>\n11<\/output>\n7 - 4<\/gadget>\n3<\/output>\n11 \/ 3<\/gadget>\n11\/3 = around 3.666667<\/output>\n11\/3 = around 3.666667<\/result>","index":1731} +{"problem":"of the families in city x in 1956 , 25 percent owned a personal computer . the number of families in city x owning a computer in 1960 was 15 percent greater than it was in 1956 , and the total number of families in city x was 5 percent greater in 1956 than it was in 1960 . what percent of the families in city x owned a personal computer in 1960 ?","rationale":"\"say a 100 families existed in 1956 then the number of families owning a computer in 1956 - 25 number of families owning computer in 1960 = 25 * 115 \/ 100 = 28.75 number of families in 1960 = 105 the percentage = 28.75 \/ 105 * 100 = 27.38 % . option : e\"","correct":"e","options":{"a":"50 % ","b":"51.22 % ","c":"5.26 % ","d":"7.20 %","e":"27.38 %"},"options_float":{"a":50.0,"b":51.22,"c":5.26,"d":7.2,"e":27.38},"annotated_formula":"multiply(const_100, divide(divide(multiply(add(15, const_100), 25), const_100), add(const_100, 5)))","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":"15 + 100<\/gadget>\n115<\/output>\n115 * 25<\/gadget>\n2_875<\/output>\n2_875 \/ 100<\/gadget>\n115\/4 = around 28.75<\/output>\n100 + 5<\/gadget>\n105<\/output>\n(115\/4) \/ 105<\/gadget>\n23\/84 = around 0.27381<\/output>\n100 * (23\/84)<\/gadget>\n575\/21 = around 27.380952<\/output>\n575\/21 = around 27.380952<\/result>","index":1734} +{"problem":"if the cost price is 91 % of sp then what is the profit %","rationale":"\"sol . sp = rs 100 : then cp = rs 91 : profit = rs 9 . profit = { ( 9 \/ 91 ) * 100 } % = 9.89 % answer is a .\"","correct":"a","options":{"a":"9.89 % ","b":"8.90 % ","c":"9.00 % ","d":"8.00 %","e":"9.27 %"},"options_float":{"a":9.89,"b":8.9,"c":9.0,"d":8.0,"e":9.27},"annotated_formula":"multiply(divide(subtract(const_100, 91), 91), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"100 - 91<\/gadget>\n9<\/output>\n9 \/ 91<\/gadget>\n9\/91 = around 0.098901<\/output>\n(9\/91) * 100<\/gadget>\n900\/91 = around 9.89011<\/output>\n900\/91 = around 9.89011<\/result>","index":1735} +{"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 . 24 , find the sum of money ?","rationale":"\"a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 24 41 - - - - ? = > rs . 123 answer : d\"","correct":"d","options":{"a":"288 ","b":"262 ","c":"72 ","d":"123","e":"267"},"options_float":{"a":288.0,"b":262.0,"c":72.0,"d":123.0,"e":267.0},"annotated_formula":"multiply(divide(24, 40), add(add(const_100, 65), 40))","linear_formula":"add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)|","chain":"24 \/ 40<\/gadget>\n3\/5 = around 0.6<\/output>\n100 + 65<\/gadget>\n165<\/output>\n165 + 40<\/gadget>\n205<\/output>\n(3\/5) * 205<\/gadget>\n123<\/output>\n123<\/result>","index":1736} +{"problem":"tough and tricky questions : combinations . 8 contestants representing 4 different countries advance to the finals of a fencing championship . assuming all competitors have an equal chance of winning , how many possibilities are there with respect to how a first - place and second - place medal can be awarded ?","rationale":"number of ways first - place medal can be awarded to four contestants = 8 number of ways second - place medal can be awarded to contestants after awarding first - place medal = 3 therefore number of possibilities = 8 * 3 = 24 answer : e","correct":"e","options":{"a":"6 ","b":"7 ","c":"12 ","d":"16","e":"24"},"options_float":{"a":6.0,"b":7.0,"c":12.0,"d":16.0,"e":24.0},"annotated_formula":"multiply(8, subtract(4, const_1))","linear_formula":"subtract(n1,const_1)|multiply(n0,#0)","chain":"4 - 1<\/gadget>\n3<\/output>\n8 * 3<\/gadget>\n24<\/output>\n24<\/result>","index":1738} +{"problem":"a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 \/ 3 of what c invests . at the end of the year , the profit earned is rs . 44000 . what is the share of c ?","rationale":"\"explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x \/ 3 ) the inverstment of a = rs . ( 3 × ( 2 \/ 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x \/ 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 \/ 11 ) × 44000 ] = rs . 12000 answer : option a\"","correct":"a","options":{"a":"rs . 12000 ","b":"rs . 13375 ","c":"rs . 11750 ","d":"rs . 11625","e":"none of these"},"options_float":{"a":12000.0,"b":13375.0,"c":11750.0,"d":11625.0,"e":null},"annotated_formula":"multiply(44000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1)))","linear_formula":"divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)|","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 3<\/gadget>\n2<\/output>\n(2\/3) + 2<\/gadget>\n8\/3 = around 2.666667<\/output>\n(8\/3) + 1<\/gadget>\n11\/3 = around 3.666667<\/output>\n1 \/ (11\/3)<\/gadget>\n3\/11 = around 0.272727<\/output>\n44_000 * (3\/11)<\/gadget>\n12_000<\/output>\n12_000<\/result>","index":1739} +{"problem":"a sum of rs . 100 is lent at simple interest of 3 % p . a . for the first month , 9 % p . a . for the second month , 27 % p . a . for the third month and so on . what is the total amount of interest earned at the end of the year approximately","rationale":"total amount of interest is i = p \/ 100 * 1 [ 3 \/ 12 + 9 \/ 12 + 27 \/ 12 … . 312 \/ 12 where p = 100 ; i = 1 \/ 12 ( 3 + 9 + … . . 312 ) i = 1 \/ 12 ( 3 ( 312 - 1 ) ) \/ 3 - 1 = 531440 * 3 \/ 12 * 2 = rs . 66430 answer : d","correct":"d","options":{"a":"rs . 797160 ","b":"rs . 791160 ","c":"rs . 65930 ","d":"rs . 66430","e":"rs . 67430"},"options_float":{"a":797160.0,"b":791160.0,"c":65930.0,"d":66430.0,"e":67430.0},"annotated_formula":"divide(multiply(subtract(power(3, const_12), const_1), 3), multiply(const_12, const_2))","linear_formula":"multiply(const_12,const_2)|power(n1,const_12)|subtract(#1,const_1)|multiply(n1,#2)|divide(#3,#0)","chain":"3 ** 12<\/gadget>\n531_441<\/output>\n531_441 - 1<\/gadget>\n531_440<\/output>\n531_440 * 3<\/gadget>\n1_594_320<\/output>\n12 * 2<\/gadget>\n24<\/output>\n1_594_320 \/ 24<\/gadget>\n66_430<\/output>\n66_430<\/result>","index":1741} +{"problem":"suppose you have three identical prisms with congruent equilateral triangles as the end - polygons . suppose you attach them by the rectangular faces so they are perfectly aligned . there will be some large faces created by two or more co - planar faces of the individual prisms : count each such large face as one . given that , how many faces does the resultant solid have ?","rationale":"the top and the bottom are each single faces formed by three equilateral triangles joining , as in the diagram on the left , to make an isosceles trapezoid . top = 1 face , and bottom = 1 face . this is a four - sided figure , so there are four rectangles extending from the bottom of this prism to the congruent figure at the top . notice , in particular , the larger vertical face in the “ back ” of the diagram to the right is formed by two faces of the original triangular prisms lying next to each other and smoothly joining . total = 1 top + 1 bottom + 4 sides = 6 faces . answer = b .","correct":"b","options":{"a":"4 ","b":"6 ","c":"9 ","d":"10","e":"12"},"options_float":{"a":4.0,"b":6.0,"c":9.0,"d":10.0,"e":12.0},"annotated_formula":"add(add(const_4, const_1), const_1)","linear_formula":"add(const_1,const_4)|add(#0,const_1)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n6<\/result>","index":1742} +{"problem":"a man bought 10 crates of mangoes for $ 40 total . if he lost 2 of the crates , at what price would he have to sell each of the remaining crates in order to earn a total profit of 20 percent of the total cost ?","rationale":"as given , after lost , the remaining 8 crates total cost = $ 40 so , 1 crate cost = 40 \/ 8 = 5 to get 20 % profit , 1 crate cost should be = 5 + 5 * 20 \/ 100 = $ 6 answer : a","correct":"a","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(add(40, multiply(40, divide(20, const_100))), subtract(10, 2))","linear_formula":"divide(n3,const_100)|subtract(n0,n2)|multiply(n1,#0)|add(n1,#2)|divide(#3,#1)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n40 * (1\/5)<\/gadget>\n8<\/output>\n40 + 8<\/gadget>\n48<\/output>\n10 - 2<\/gadget>\n8<\/output>\n48 \/ 8<\/gadget>\n6<\/output>\n6<\/result>","index":1745} +{"problem":"a number increased by 30 % gives 650 . the number is ?","rationale":"\"formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 30 % = 130 % 130 % - - - - - - - > 650 ( 130 ã — 5 = 650 ) 100 % - - - - - - - > 400 ( 100 ã — 5 = 500 ) option ' d '\"","correct":"d","options":{"a":"200 ","b":"300 ","c":"400 ","d":"500","e":"600"},"options_float":{"a":200.0,"b":300.0,"c":400.0,"d":500.0,"e":600.0},"annotated_formula":"divide(650, add(const_1, divide(30, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n650 \/ (13\/10)<\/gadget>\n500<\/output>\n500<\/result>","index":1749} +{"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 20 days .","rationale":"\"they together completed 4 \/ 10 work in 4 days . balance 6 \/ 10 work will be completed by arun alone in 20 * 6 \/ 10 = 12 days . answer : a\"","correct":"a","options":{"a":"12 days . ","b":"17 days . ","c":"18 days . ","d":"19 days .","e":"20 days ."},"options_float":{"a":12.0,"b":17.0,"c":18.0,"d":19.0,"e":20.0},"annotated_formula":"subtract(20, multiply(divide(20, 10), 4))","linear_formula":"divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)|","chain":"20 \/ 10<\/gadget>\n2<\/output>\n2 * 4<\/gadget>\n8<\/output>\n20 - 8<\/gadget>\n12<\/output>\n12<\/result>","index":1751} +{"problem":"if ' x ' is a positive integer exactly divisible by 6 or 15 but not divisible by 12 . what could possibly be the value of ' x ' ?","rationale":"120 and 60 are both divisible by 6 and 15 but also by 12 . so they are not the right answer . 36 and 54 are both clearly not divisible by 15 ( not correct ) 90 is both divisible by 6 and 15 but not by 12 . answer : ( d ) 90","correct":"d","options":{"a":"120 ","b":"60 ","c":"36 ","d":"90","e":"54"},"options_float":{"a":120.0,"b":60.0,"c":36.0,"d":90.0,"e":54.0},"annotated_formula":"multiply(6, 15)","linear_formula":"multiply(n0,n1)","chain":"6 * 15<\/gadget>\n90<\/output>\n90<\/result>","index":1752} +{"problem":"on a sum of money , the simple interest for 2 years is rs . 324 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is","rationale":"\"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 - 324 = 16 ( r ã — si ) \/ ( 2 ã — 100 ) = 16 ( r ã — 324 ) \/ ( 2 ã — 100 ) = 16 r = 9.87 % answer : option c\"","correct":"c","options":{"a":"15 % ","b":"14.25 % ","c":"9.87 % ","d":"10.5 %","e":"11.5 %"},"options_float":{"a":15.0,"b":14.25,"c":9.87,"d":10.5,"e":11.5},"annotated_formula":"divide(multiply(const_100, subtract(subtract(340, divide(324, 2)), divide(324, 2))), divide(324, 2))","linear_formula":"divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)|","chain":"324 \/ 2<\/gadget>\n162<\/output>\n340 - 162<\/gadget>\n178<\/output>\n178 - 162<\/gadget>\n16<\/output>\n100 * 16<\/gadget>\n1_600<\/output>\n1_600 \/ 162<\/gadget>\n800\/81 = around 9.876543<\/output>\n800\/81 = around 9.876543<\/result>","index":1754} +{"problem":"a taxi leaves point a 2 hours after a bus left the same spot . the bus is traveling 20 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in two hours .","rationale":"\"let the speed of bus be v - 20 , speed of taxi be v the bus travelled a total of 4 hrs and taxi a total of 2 hrs . hence 4 * ( v - 20 ) = 2 v 4 v - 80 = 2 v 2 v = 80 v = 40 mph e\"","correct":"e","options":{"a":"42 ","b":"48 ","c":"44 ","d":"50","e":"40"},"options_float":{"a":42.0,"b":48.0,"c":44.0,"d":50.0,"e":40.0},"annotated_formula":"divide(add(multiply(2, 20), multiply(2, 20)), subtract(add(2, 2), 2))","linear_formula":"add(n0,n0)|multiply(n0,n1)|add(#1,#1)|subtract(#0,n0)|divide(#2,#3)|","chain":"2 * 20<\/gadget>\n40<\/output>\n40 + 40<\/gadget>\n80<\/output>\n2 + 2<\/gadget>\n4<\/output>\n4 - 2<\/gadget>\n2<\/output>\n80 \/ 2<\/gadget>\n40<\/output>\n40<\/result>","index":1756} +{"problem":"monica planned her birthday party . she prepared 5 muffins for each of her guests and kept aside two additional muffins in case someone will want extra . after the party , it turned out that one of the guests did n ' t come but every one of the guests that did come ate 6 muffins and 6 muffins remained . how many guests did monica plan on ?","rationale":"let x be the number of guests . number of muffins prepared = 5 x + 2 number of muffins eaten + number of muffins remaining = number of muffins prepared 6 ( x - 1 ) + 6 = 5 x + 2 6 x = 5 x + 2 x = 2 answer : a","correct":"a","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":"divide(subtract(add(6, 6), const_2), 5)","linear_formula":"add(n1,n1)|subtract(#0,const_2)|divide(#1,n0)","chain":"6 + 6<\/gadget>\n12<\/output>\n12 - 2<\/gadget>\n10<\/output>\n10 \/ 5<\/gadget>\n2<\/output>\n2<\/result>","index":1762} +{"problem":"a boy rides his bicycle 10 km at an average speed of 12 km \/ hr and again travels 12 km at an average speed of 10 km \/ hr . his average speed for the entire trip is approximately ?","rationale":"total distance traveled = 10 + 12 = 22 km \/ hr . total time taken = 10 \/ 12 + 12 \/ 10 = 61 \/ 30 hrs . average speed = 22 * 30 \/ 61 = 10.8 km \/ hr . answer : b","correct":"b","options":{"a":"10.7 km \/ hr ","b":"10.8 km \/ hr ","c":"17.8 km \/ hr ","d":"10.5 km \/ hr","e":"30.8 km \/ hr"},"options_float":{"a":10.7,"b":10.8,"c":17.8,"d":10.5,"e":30.8},"annotated_formula":"divide(add(12, 10), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"12 + 10<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n11<\/result>","index":1763} +{"problem":"if 3 ^ x = 2 , then 3 ^ ( 4 x + 3 ) =","rationale":"3 ^ x = 2 3 ^ 4 x = 2 ^ 4 3 ^ 4 x = 16 3 ^ ( 4 x + 3 ) = 3 ^ 4 x * 3 ^ 3 = 16 * 27 = 432 answer : c","correct":"c","options":{"a":"429 ","b":"454 ","c":"432 ","d":"438","e":"108"},"options_float":{"a":429.0,"b":454.0,"c":432.0,"d":438.0,"e":108.0},"annotated_formula":"power(3, add(multiply(4, divide(log(const_2), log(const_3))), 3))","linear_formula":"log(const_2)|log(const_3)|divide(#0,#1)|multiply(n3,#2)|add(n0,#3)|power(n0,#4)","chain":"log(2)<\/gadget>\nlog(2) = around 0.693147<\/output>\nlog(3)<\/gadget>\nlog(3) = around 1.098612<\/output>\nlog(2) \/ log(3)<\/gadget>\nlog(2)\/log(3) = around 0.63093<\/output>\n4 * (log(2)\/log(3))<\/gadget>\n4*log(2)\/log(3) = around 2.523719<\/output>\n(4*log(2)\/log(3)) + 3<\/gadget>\n4*log(2)\/log(3) + 3 = around 5.523719<\/output>\n3 ** (4*log(2)\/log(3) + 3)<\/gadget>\n3**(4*log(2)\/log(3) + 3) = around 432<\/output>\n3**(4*log(2)\/log(3) + 3) = around 432<\/result>","index":1764} +{"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 360 meters , after what time will they meet at the starting point ?","rationale":"\"they meet every 360 \/ 7 + 8 = 24 sec answer is d\"","correct":"d","options":{"a":"20 sec ","b":"15 sec ","c":"30 sec ","d":"24 sec","e":"1 min"},"options_float":{"a":20.0,"b":15.0,"c":30.0,"d":24.0,"e":1.0},"annotated_formula":"divide(360, add(8, 7))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"8 + 7<\/gadget>\n15<\/output>\n360 \/ 15<\/gadget>\n24<\/output>\n24<\/result>","index":1765} +{"problem":"a van takes 6 hours to cover a distance of 540 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 = 6 distence = 540 3 \/ 2 of 6 hours = 6 * 3 \/ 2 = 9 hours required speed = 540 \/ 9 = 60 kmph c )\"","correct":"c","options":{"a":"40 kmph ","b":"50 kmph ","c":"60 kmph ","d":"75 kmph","e":"860 kmph"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":75.0,"e":860.0},"annotated_formula":"divide(540, multiply(divide(3, 2), 6))","linear_formula":"divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 6<\/gadget>\n9<\/output>\n540 \/ 9<\/gadget>\n60<\/output>\n60<\/result>","index":1768} +{"problem":"if an object travels 90 feet in 2 seconds , what is the object ’ s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet )","rationale":"\"90 feet \/ 2 seconds = 45 feet \/ second ( 45 feet \/ second ) * ( 3600 seconds \/ hour ) * ( 1 mile \/ 5280 feet ) = 30.68 ( approximately ) the answer is a .\"","correct":"a","options":{"a":"30.68 ","b":"32.34 ","c":"34.56 ","d":"36.72","e":"38.95"},"options_float":{"a":30.68,"b":32.34,"c":34.56,"d":36.72,"e":38.95},"annotated_formula":"divide(divide(90, 5280), multiply(2, divide(1, const_3600)))","linear_formula":"divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)|","chain":"90 \/ 5_280<\/gadget>\n3\/176 = around 0.017045<\/output>\n1 \/ 3_600<\/gadget>\n1\/3_600 = around 0.000278<\/output>\n2 * (1\/3_600)<\/gadget>\n1\/1_800 = around 0.000556<\/output>\n(3\/176) \/ (1\/1_800)<\/gadget>\n675\/22 = around 30.681818<\/output>\n675\/22 = around 30.681818<\/result>","index":1770} +{"problem":"the population of a village is 13400 . it increases annually at the rate of 21 % p . a . what will be its population after 2 years ?","rationale":"\"formula : ( after = 100 denominator ago = 100 numerator ) 13400 × 121 \/ 100 × 121 \/ 100 = 19619 c\"","correct":"c","options":{"a":"10000 ","b":"12000 ","c":"19619 ","d":"18989","e":"14400"},"options_float":{"a":10000.0,"b":12000.0,"c":19619.0,"d":18989.0,"e":14400.0},"annotated_formula":"multiply(13400, power(add(const_1, divide(21, const_100)), 2))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|","chain":"21 \/ 100<\/gadget>\n21\/100 = around 0.21<\/output>\n1 + (21\/100)<\/gadget>\n121\/100 = around 1.21<\/output>\n(121\/100) ** 2<\/gadget>\n14_641\/10_000 = around 1.4641<\/output>\n13_400 * (14_641\/10_000)<\/gadget>\n980_947\/50 = around 19_618.94<\/output>\n980_947\/50 = around 19_618.94<\/result>","index":1773} +{"problem":"a box measuring 49 inches long by 21 inches wide by 14 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 . 7 is the biggest number that could divide all three , 49 , 21 and 14 . thus side of cube must be 7 , and total number of cubes = 49 \/ 7 * 21 \/ 7 * 14 \/ 7 = 42 ans b it is .\"","correct":"b","options":{"a":"17 ","b":"42 ","c":"54 ","d":"108","e":"864"},"options_float":{"a":17.0,"b":42.0,"c":54.0,"d":108.0,"e":864.0},"annotated_formula":"divide(multiply(multiply(49, 21), 14), volume_cube(divide(14, const_2)))","linear_formula":"divide(n2,const_2)|multiply(n0,n1)|multiply(n2,#1)|volume_cube(#0)|divide(#2,#3)|","chain":"49 * 21<\/gadget>\n1_029<\/output>\n1_029 * 14<\/gadget>\n14_406<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7 ** 3<\/gadget>\n343<\/output>\n14_406 \/ 343<\/gadget>\n42<\/output>\n42<\/result>","index":1775} +{"problem":"sonika deposited rs . 7000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ?","rationale":"\"( 7000 * 3 * 2 ) \/ 100 = 420 9200 - - - - - - - - 9620 answer : c\"","correct":"c","options":{"a":"9680 ","b":"2277 ","c":"9620 ","d":"2774","e":"1212"},"options_float":{"a":9680.0,"b":2277.0,"c":9620.0,"d":2774.0,"e":1212.0},"annotated_formula":"add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 7000), 3), 7000)), 7000), 3), 7000)","linear_formula":"divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)|","chain":"2 \/ 100<\/gadget>\n1\/50 = around 0.02<\/output>\n9_200 - 7_000<\/gadget>\n2_200<\/output>\n2_200 \/ 3<\/gadget>\n2_200\/3 = around 733.333333<\/output>\n(2_200\/3) \/ 7_000<\/gadget>\n11\/105 = around 0.104762<\/output>\n(1\/50) + (11\/105)<\/gadget>\n131\/1_050 = around 0.124762<\/output>\n(131\/1_050) * 7_000<\/gadget>\n2_620\/3 = around 873.333333<\/output>\n(2_620\/3) * 3<\/gadget>\n2_620<\/output>\n2_620 + 7_000<\/gadget>\n9_620<\/output>\n9_620<\/result>","index":1777} +{"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 the race ends in a dead heat ?","rationale":"\"we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 13 meters in that time . so if the race is 20 meters long , when a reaches the finish line , b would be 7 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 7 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 20 meters ) would be the same as the time required by b ( to cover 13 meters ) to reach the finish line . so b should get a head start of 7 \/ 20 th of the race . answer ( d )\"","correct":"d","options":{"a":"1 \/ 17 ","b":"3 \/ 17 ","c":"1 \/ 10 ","d":"7 \/ 20","e":"3 \/ 10"},"options_float":{"a":0.0588235294,"b":0.1764705882,"c":0.1,"d":0.35,"e":0.3},"annotated_formula":"divide(subtract(20, 13), 20)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|","chain":"20 - 13<\/gadget>\n7<\/output>\n7 \/ 20<\/gadget>\n7\/20 = around 0.35<\/output>\n7\/20 = around 0.35<\/result>","index":1778} +{"problem":"what is the largest number of 4 digits which is divisible by 15 , 25 , 40 and 75 ?","rationale":"explanation : largest number of four digits = 9999 lcm of 15 , 25 , 40 and 75 = 600 9999 ÷ 600 = 16 , remainder = 399 hence , largest number of four digits which is divisible by 15 , 25 , 40 and 75 = 9999 - 399 = 9600 answer : a","correct":"a","options":{"a":"9600 ","b":"5200 ","c":"362 ","d":"958","e":"258"},"options_float":{"a":9600.0,"b":5200.0,"c":362.0,"d":958.0,"e":258.0},"annotated_formula":"subtract(subtract(multiply(const_100, const_100), const_1), reminder(subtract(multiply(const_100, const_100), const_1), lcm(lcm(lcm(15, 25), 40), 75)))","linear_formula":"lcm(n1,n2)|multiply(const_100,const_100)|lcm(n3,#0)|subtract(#1,const_1)|lcm(n4,#2)|reminder(#3,#4)|subtract(#3,#5)","chain":"100 * 100<\/gadget>\n10_000<\/output>\n10_000 - 1<\/gadget>\n9_999<\/output>\nlcm(15, 25)<\/gadget>\n75<\/output>\nlcm(75, 40)<\/gadget>\n600<\/output>\nlcm(600, 75)<\/gadget>\n600<\/output>\n9_999 % 600<\/gadget>\n399<\/output>\n9_999 - 399<\/gadget>\n9_600<\/output>\n9_600<\/result>","index":1780} +{"problem":"what will be the cost of building a fence around a square plot with area equal to 64 sq ft , if the price per foot of building the fence is rs . 58 ?","rationale":"\"let the side of the square plot be a ft . a 2 = 64 = > a = 8 length of the fence = perimeter of the plot = 4 a = 32 ft . cost of building the fence = 32 * 58 = rs . 1856 . answer : d\"","correct":"d","options":{"a":"3944 ","b":"2882 ","c":"2999 ","d":"1856","e":"2121"},"options_float":{"a":3944.0,"b":2882.0,"c":2999.0,"d":1856.0,"e":2121.0},"annotated_formula":"multiply(square_perimeter(sqrt(64)), 58)","linear_formula":"sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|","chain":"64 ** (1\/2)<\/gadget>\n8<\/output>\n4 * 8<\/gadget>\n32<\/output>\n32 * 58<\/gadget>\n1_856<\/output>\n1_856<\/result>","index":1781} +{"problem":"a 6 litre sol is 20 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?","rationale":"\"20 % of 6 = 1.2 50 % of 6 = 3 shortage is 1.8 so we need to have 1.8 \/ 50 % to get 50 % alcohol content . = 3.6 d\"","correct":"d","options":{"a":"a . 0.6 ","b":"b . 1 ","c":"c . 2.1 ","d":"d . 3.6","e":"e . 5.4"},"options_float":{"a":0.6,"b":1.0,"c":2.1,"d":3.6,"e":5.4},"annotated_formula":"subtract(6, multiply(const_2, multiply(divide(20, const_100), 6)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 6<\/gadget>\n6\/5 = around 1.2<\/output>\n2 * (6\/5)<\/gadget>\n12\/5 = around 2.4<\/output>\n6 - (12\/5)<\/gadget>\n18\/5 = around 3.6<\/output>\n18\/5 = around 3.6<\/result>","index":1782} +{"problem":"if 45 % of z is 60 % of y and y is 75 % of x , what percent of x is z ?","rationale":"\"( 45 \/ 100 ) z = ( 60 \/ 100 ) y and y = ( 75 \/ 100 ) x i . e . y = ( 3 \/ 4 ) x i . e . ( 45 \/ 100 ) z = ( 60 \/ 100 ) * ( 3 \/ 4 ) x i . e . z = ( 60 * 3 ) x \/ ( 45 * 4 ) i . e . z = ( 1 ) x = ( 100 \/ 100 ) x i . e . z is 100 % of x answer : option c\"","correct":"c","options":{"a":"200 ","b":"160 ","c":"100 ","d":"65","e":"50"},"options_float":{"a":200.0,"b":160.0,"c":100.0,"d":65.0,"e":50.0},"annotated_formula":"multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(60, const_100))), const_100)","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)|","chain":"75 \/ 100<\/gadget>\n3\/4 = around 0.75<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(9\/20) \/ (3\/5)<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) \/ (3\/4)<\/gadget>\n1<\/output>\n1 * 100<\/gadget>\n100<\/output>\n100<\/result>","index":1783} +{"problem":"how many odd 3 - digit integers greater than 800 are there such that all their digits are different ?","rationale":"in the range 800 - 900 : 1 choice for the first digit : 8 ; 5 choices for the third digit : 1 , 3 , 5 , 7 , 9 ( as the number is odd ) ; 8 choices for the second digit : 10 digits - first digit - third digit = 8 digits . 1 * 5 * 8 = 40 . in the range 900 - 999 : 1 choice for the first digit : 9 ; 4 choices for the third digit : 1 , 3 , 5 , 7 ( 9 is out as it ' s the first digit ) ; 8 choices for the second digit : 10 digits - first digit - third digit = 8 digits . 1 * 4 * 8 = 32 . total : 40 + 32 = 72 . answer : c","correct":"c","options":{"a":"40 ","b":"56 ","c":"72 ","d":"81","e":"104"},"options_float":{"a":40.0,"b":56.0,"c":72.0,"d":81.0,"e":104.0},"annotated_formula":"add(multiply(add(3, add(const_3, const_2)), const_4), multiply(multiply(const_1, add(const_3, const_2)), add(3, add(const_3, const_2))))","linear_formula":"add(const_2,const_3)|add(n0,#0)|multiply(#0,const_1)|multiply(#1,const_4)|multiply(#1,#2)|add(#3,#4)","chain":"3 + 2<\/gadget>\n5<\/output>\n3 + 5<\/gadget>\n8<\/output>\n8 * 4<\/gadget>\n32<\/output>\n1 * 5<\/gadget>\n5<\/output>\n5 * 8<\/gadget>\n40<\/output>\n32 + 40<\/gadget>\n72<\/output>\n72<\/result>","index":1784} +{"problem":"the ratio of spinsters to cats is 2 to 7 . if there are 40 more cats than spinsters , how many spinsters are there ?","rationale":"\"let 2 x be the number of spinsters . then 7 x is the number of cats . 7 x - 2 x = 40 x = 8 and the number of spinsters is 2 ( 8 ) = 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":"multiply(divide(40, subtract(7, 2)), 2)","linear_formula":"subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|","chain":"7 - 2<\/gadget>\n5<\/output>\n40 \/ 5<\/gadget>\n8<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16<\/result>","index":1785} +{"problem":"33 1 \/ 3 % of 240 ?","rationale":"\"33 1 \/ 3 % = 1 \/ 3 1 \/ 3 × 240 = 80 a )\"","correct":"a","options":{"a":"80 ","b":"90 ","c":"110 ","d":"120","e":"130"},"options_float":{"a":80.0,"b":90.0,"c":110.0,"d":120.0,"e":130.0},"annotated_formula":"divide(multiply(add(33, divide(1, 3)), 240), 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) * 240<\/gadget>\n8_000<\/output>\n8_000 \/ 100<\/gadget>\n80<\/output>\n80<\/result>","index":1786} +{"problem":"one pump drains one - half of a pond in 7 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ?","rationale":"\"first pump drains 1 \/ 2 of the tank in 7 hours so 14 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 \/ 14 + 1 \/ a ) part in 1 hour son in 1 \/ 2 hour they drain 1 \/ 2 * ( 1 \/ 14 + 1 \/ a ) part of the tank given that in 1 \/ 2 hour they drain 1 \/ 2 of the tank hence we can say 1 \/ 2 * ( 1 \/ 14 + 1 \/ a ) = 1 \/ 2 solving u get a = 14 \/ 13 = 1.1 hence answer is b\"","correct":"b","options":{"a":"1 hour ","b":"1.1 hour ","c":"3 hours ","d":"5 hours","e":"6 hours"},"options_float":{"a":1.0,"b":1.1,"c":3.0,"d":5.0,"e":6.0},"annotated_formula":"divide(const_1, subtract(const_1, divide(const_1, multiply(7, const_2))))","linear_formula":"multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)|","chain":"7 * 2<\/gadget>\n14<\/output>\n1 \/ 14<\/gadget>\n1\/14 = around 0.071429<\/output>\n1 - (1\/14)<\/gadget>\n13\/14 = around 0.928571<\/output>\n1 \/ (13\/14)<\/gadget>\n14\/13 = around 1.076923<\/output>\n14\/13 = around 1.076923<\/result>","index":1789} +{"problem":"what is the area of square field whose side of length 18 m ?","rationale":"\"18 * 18 = 324 sq m answer : d\"","correct":"d","options":{"a":"225 sq m ","b":"186 sq m ","c":"586 sq m ","d":"324 sq m","e":"296 sq m"},"options_float":{"a":225.0,"b":186.0,"c":586.0,"d":324.0,"e":296.0},"annotated_formula":"square_area(18)","linear_formula":"square_area(n0)|","chain":"18 ** 2<\/gadget>\n324<\/output>\n324<\/result>","index":1790} +{"problem":"how much is 80 % of 40 is greater than 4 \/ 5 of 30 ?","rationale":"\"( 80 \/ 100 ) * 40 â € “ ( 4 \/ 5 ) * 30 32 - 24 = 8 answer : c\"","correct":"c","options":{"a":"12 ","b":"27 ","c":"8 ","d":"12","e":"81"},"options_float":{"a":12.0,"b":27.0,"c":8.0,"d":12.0,"e":81.0},"annotated_formula":"subtract(multiply(40, divide(80, 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":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n40 * (4\/5)<\/gadget>\n32<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 30<\/gadget>\n24<\/output>\n32 - 24<\/gadget>\n8<\/output>\n8<\/result>","index":1791} +{"problem":"how many unique positive odd integers less than 60 are equal to the product of a positive multiple of 5 and an odd number ?","rationale":"\"the odd numbers less than 60 are 1 , 3,5 , . . . 59 some of them are equal to product of 5 and odd number hence odd multiples of 5 which are less than 60 are 5,15 , 25,35 , 45,55 ( total = 6 ) ; answer : b\"","correct":"b","options":{"a":"4 ","b":"6 ","c":"11 ","d":"12","e":"15"},"options_float":{"a":4.0,"b":6.0,"c":11.0,"d":12.0,"e":15.0},"annotated_formula":"divide(divide(60, 5), const_2)","linear_formula":"divide(n0,n1)|divide(#0,const_2)|","chain":"60 \/ 5<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6<\/result>","index":1792} +{"problem":"in a sports club with 30 members , 17 play badminton and 18 play tennis and 2 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 ) + ( 18 - x ) + x + 2 = 30 37 - 2 x + x = 30 37 - x = 30 x = 7 so 7 members play both badminton and tennis . 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(add(add(17, 18), 2), 30)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)|","chain":"17 + 18<\/gadget>\n35<\/output>\n35 + 2<\/gadget>\n37<\/output>\n37 - 30<\/gadget>\n7<\/output>\n7<\/result>","index":1793} +{"problem":"working at their respective constant rates , machine a makes 100 copies in 15 minutes and machine b makes 150 copies in 10 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ?","rationale":"\"machine a can produce 100 * 30 \/ 15 = 200 copies and , machine b can produce 150 * 30 \/ 10 = 450 copies total producing 650 copies . c is the answer\"","correct":"c","options":{"a":"250 ","b":"425 ","c":"650 ","d":"700","e":"750"},"options_float":{"a":250.0,"b":425.0,"c":650.0,"d":700.0,"e":750.0},"annotated_formula":"multiply(add(divide(100, 15), divide(150, 10)), 30)","linear_formula":"divide(n0,n1)|divide(n2,n3)|add(#0,#1)|multiply(n4,#2)|","chain":"100 \/ 15<\/gadget>\n20\/3 = around 6.666667<\/output>\n150 \/ 10<\/gadget>\n15<\/output>\n(20\/3) + 15<\/gadget>\n65\/3 = around 21.666667<\/output>\n(65\/3) * 30<\/gadget>\n650<\/output>\n650<\/result>","index":1798} +{"problem":"one side of a rectangular field is 15 m and one of its diagonals is 17 m . find the area of the field .","rationale":"\"other side = ( ( 17 ) 2 - ( 15 ) 2 ) ( 1 \/ 2 ) = ( 289 - 225 ) ( 1 \/ 2 ) = ( 64 ) ( 1 \/ 2 ) = 8 m . area = ( 15 x 8 ) m 2 = 120 m 2 . ans : a\"","correct":"a","options":{"a":"120 ","b":"147 ","c":"251 ","d":"451","e":"258"},"options_float":{"a":120.0,"b":147.0,"c":251.0,"d":451.0,"e":258.0},"annotated_formula":"rectangle_area(15, sqrt(subtract(power(17, 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":"17 ** 2<\/gadget>\n289<\/output>\n15 ** 2<\/gadget>\n225<\/output>\n289 - 225<\/gadget>\n64<\/output>\n64 ** (1\/2)<\/gadget>\n8<\/output>\n15 * 8<\/gadget>\n120<\/output>\n120<\/result>","index":1805} +{"problem":"a certain quantity is measured on two different scales , the t - scale and the s - scale , that are related linearly . measurements on the t - scale of 6 and 24 correspond to measurements on the s - scale of 30 and 60 , respectively . what measurement on the t - scale corresponds to a measurement of 100 on the s - scale ?","rationale":"first , we have to understand what linearly means . it ' s not a straight ratio ( since 6 : 30 does not equal 24 : 60 ) . we need to look at the increases in each measurement to see what the scalar actually is . from 6 to 24 we have an increase of 18 . from 30 to 60 we have an increase of 30 . therefore , the increase ratio is 18 : 30 or 3 : 5 . in other words , for every 3 that t increases , s increases by 5 . we know that s is 100 . to get from 60 to 100 , we went up by 40 , or 8 jumps of 5 ; therefore , t will go up by 8 jumps of 3 . 24 + 8 ( 3 ) = 24 + 24 = 48 = c","correct":"c","options":{"a":"20 ","b":"36 ","c":"48 ","d":"60","e":"84"},"options_float":{"a":20.0,"b":36.0,"c":48.0,"d":60.0,"e":84.0},"annotated_formula":"add(multiply(subtract(24, 6), divide(subtract(100, 60), subtract(60, 30))), 24)","linear_formula":"subtract(n4,n3)|subtract(n3,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(#3,#2)|add(n1,#4)","chain":"24 - 6<\/gadget>\n18<\/output>\n100 - 60<\/gadget>\n40<\/output>\n60 - 30<\/gadget>\n30<\/output>\n40 \/ 30<\/gadget>\n4\/3 = around 1.333333<\/output>\n18 * (4\/3)<\/gadget>\n24<\/output>\n24 + 24<\/gadget>\n48<\/output>\n48<\/result>","index":1807} +{"problem":"the sum of two numbers is 184 . if one - third of the one exceeds one - seventh of the other by 8 , find the smaller number .","rationale":"let the numbers be x and ( 184 - x ) . then , ( x \/ 3 ) - ( 184 - x ) \/ 7 = 8 7 x - 3 ( 184 - x ) = 168 10 x = 720 , x = 72 . hence the correct answer is option a ) 72 .","correct":"a","options":{"a":"72 ","b":"64 ","c":"84 ","d":"12","e":"92"},"options_float":{"a":72.0,"b":64.0,"c":84.0,"d":12.0,"e":92.0},"annotated_formula":"divide(add(multiply(184, const_3), multiply(multiply(add(const_3, const_4), const_3), 8)), add(add(const_3, const_4), const_3))","linear_formula":"add(const_3,const_4)|multiply(n0,const_3)|add(#0,const_3)|multiply(#0,const_3)|multiply(n1,#3)|add(#1,#4)|divide(#5,#2)","chain":"184 * 3<\/gadget>\n552<\/output>\n3 + 4<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21 * 8<\/gadget>\n168<\/output>\n552 + 168<\/gadget>\n720<\/output>\n7 + 3<\/gadget>\n10<\/output>\n720 \/ 10<\/gadget>\n72<\/output>\n72<\/result>","index":1808} +{"problem":"the sum of the first 50 positive even integers is 2550 . what is the sum q of even integers from 102 to 200 inclusive ?","rationale":"\"my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : q = 2550 + ( 100 * 50 ) = 7550 . b\"","correct":"b","options":{"a":"5100 ","b":"7550 ","c":"10100 ","d":"15500","e":"20100"},"options_float":{"a":5100.0,"b":7550.0,"c":10100.0,"d":15500.0,"e":20100.0},"annotated_formula":"multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1))","linear_formula":"add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)|","chain":"200 + 102<\/gadget>\n302<\/output>\n302 \/ 2<\/gadget>\n151<\/output>\n200 - 102<\/gadget>\n98<\/output>\n98 \/ 2<\/gadget>\n49<\/output>\n49 + 1<\/gadget>\n50<\/output>\n151 * 50<\/gadget>\n7_550<\/output>\n7_550<\/result>","index":1809} +{"problem":"how many litres of pure acid are there in 15 litres of a 20 % solution","rationale":"explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 15 * 20 \/ 100 = 3 answer : option d","correct":"d","options":{"a":"4 ","b":"5 ","c":"2 ","d":"3","e":"1"},"options_float":{"a":4.0,"b":5.0,"c":2.0,"d":3.0,"e":1.0},"annotated_formula":"multiply(divide(20, const_100), 15)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 15<\/gadget>\n3<\/output>\n3<\/result>","index":1810} +{"problem":"rs . 385 were divided among x , y , z in such a way that x had rs . 20 more than y and z had rs 15 more than x . how much was y ’ s share ?","rationale":"let y gets rs x . then we can say x gets rs ( x + 20 ) and y gets rs ( x + 35 ) . x + 20 + x + x + 35 = 385 3 x = 330 x = 110 . r ’ s share = rs ( 110 + 35 ) = rs . 145 b","correct":"b","options":{"a":"rs . 130 ","b":"rs . 145 ","c":"rs . 154 ","d":"rs . 160","e":"rs . 164"},"options_float":{"a":130.0,"b":145.0,"c":154.0,"d":160.0,"e":164.0},"annotated_formula":"add(15, add(divide(subtract(385, add(15, add(20, 20))), const_3), 20))","linear_formula":"add(n1,n1)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|add(n1,#3)|add(n2,#4)","chain":"20 + 20<\/gadget>\n40<\/output>\n15 + 40<\/gadget>\n55<\/output>\n385 - 55<\/gadget>\n330<\/output>\n330 \/ 3<\/gadget>\n110<\/output>\n110 + 20<\/gadget>\n130<\/output>\n15 + 130<\/gadget>\n145<\/output>\n145<\/result>","index":1811} +{"problem":"when n divided by 3 , the remainder is 2 . when n divided by 4 , the remainder is 1 what is the the remainder when divided by 16","rationale":"case 1 n = 5,8 , 11,14 , 17,20 case 2 m = 5 , 9,13 , 17,21 therefore n = 17 remainder of 17 \/ 16 will be 1 a","correct":"a","options":{"a":"1 ","b":"3 ","c":"4 ","d":"5","e":"2"},"options_float":{"a":1.0,"b":3.0,"c":4.0,"d":5.0,"e":2.0},"annotated_formula":"floor(divide(add(add(const_12, const_3), add(const_2, const_4)), 16))","linear_formula":"add(const_12,const_3)|add(const_2,const_4)|add(#0,#1)|divide(#2,n4)|floor(#3)","chain":"12 + 3<\/gadget>\n15<\/output>\n2 + 4<\/gadget>\n6<\/output>\n15 + 6<\/gadget>\n21<\/output>\n21 \/ 16<\/gadget>\n21\/16 = around 1.3125<\/output>\nfloor(21\/16)<\/gadget>\n1<\/output>\n1<\/result>","index":1812} +{"problem":"a group of men decided to do a work in 20 days , but 8 of them became absent . if the rest of the group did the work in 28 days , find the original number of men ?","rationale":"\"original number of men = 8 * 28 \/ ( 28 - 20 ) = 28 answer is b\"","correct":"b","options":{"a":"60 ","b":"28 ","c":"40 ","d":"50","e":"25"},"options_float":{"a":60.0,"b":28.0,"c":40.0,"d":50.0,"e":25.0},"annotated_formula":"divide(multiply(8, 28), subtract(28, 20))","linear_formula":"multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|","chain":"8 * 28<\/gadget>\n224<\/output>\n28 - 20<\/gadget>\n8<\/output>\n224 \/ 8<\/gadget>\n28<\/output>\n28<\/result>","index":1813} +{"problem":"three walls have wallpaper covering a combined area of 280 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 36 square meters . what is the area that is covered with three layers of wallpaper ?","rationale":"\"280 - 180 = 100 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 100 - 36 = 64 sq m of the wallpaper overlaps in three layers . 64 sq m makes two extra layers hence the area over which it makes two extra layers is 32 sq m . answer ( b ) .\"","correct":"b","options":{"a":"5 square meters ","b":"32 square meters ","c":"42 square meters ","d":"83.3 square meters","e":"120 square meters"},"options_float":{"a":5.0,"b":32.0,"c":42.0,"d":83.3,"e":120.0},"annotated_formula":"divide(subtract(subtract(280, 180), 36), const_2)","linear_formula":"subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|","chain":"280 - 180<\/gadget>\n100<\/output>\n100 - 36<\/gadget>\n64<\/output>\n64 \/ 2<\/gadget>\n32<\/output>\n32<\/result>","index":1814} +{"problem":"mr . x , a mathematician , defines a number as ' connected with 6 if it is divisible by 6 or if the sum of its digits is 6 , or if 6 is one of the digits of the number . other numbers are all ' not connected with 6 ' . as per this definition , the number of integers from 1 to 60 ( both inclusive ) which are not connected with 6 is","rationale":"explanation : numbers from 1 to 60 , which are divisible by 6 are : 6 , 1218 , 24 , 30 , 3642 , 48 , 54 , 60 . there are 10 such numbers . numbers from 1 to 60 , the sum of whose digits is 6 are : 6 , 15 , 24 , 33 , 42 , 51 , 60 . there are 7 such numbers of which 4 are common to the above ones . so , there are 3 such uncommon numbers . numbers from 1 to 60 , which have 6 as one of the digits are 6 , 16 , 26 , 36 , 46 , 56 , 60 . clearly , there are 4 such uncommon numbers . so , numbers ' not connected with 6 ' = 60 - ( 10 + 3 + 4 ) = 43 . answer : d","correct":"d","options":{"a":"18 ","b":"22 ","c":"42 ","d":"43","e":"45"},"options_float":{"a":18.0,"b":22.0,"c":42.0,"d":43.0,"e":45.0},"annotated_formula":"subtract(60, add(add(const_10, const_3), const_4))","linear_formula":"add(const_10,const_3)|add(#0,const_4)|subtract(n6,#1)","chain":"10 + 3<\/gadget>\n13<\/output>\n13 + 4<\/gadget>\n17<\/output>\n60 - 17<\/gadget>\n43<\/output>\n43<\/result>","index":1815} +{"problem":"a camera lens filter kit containing 5 filters sells for $ 67.50 . if the filters are purchased individually , 2 of them are priced at $ 7.45 each , 2 at $ 10.05 each , 1 at $ 14.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ?","rationale":"\"cost of kit = $ 67.50 if filters are purchased individually - $ 7.45 * 2 + $ 10.05 * 2 + $ 14.50 = $ 49.50 amount saved = $ 67.50 - $ 49.50 = $ 18 required % age = ( $ 18 \/ $ 67.50 ) * 100 = 26.66 % so , the correct answer is b .\"","correct":"b","options":{"a":"25.66 % ","b":"26.66 % ","c":"27.66 % ","d":"28.66 %","e":"29.66 %"},"options_float":{"a":25.66,"b":26.66,"c":27.66,"d":28.66,"e":29.66},"annotated_formula":"divide(multiply(subtract(67.50, add(14.50, add(multiply(2, 7.45), multiply(2, 10.05)))), const_100), 67.50)","linear_formula":"multiply(n2,n3)|multiply(n2,n5)|add(#0,#1)|add(n7,#2)|subtract(n1,#3)|multiply(#4,const_100)|divide(#5,n1)|","chain":"2 * 7.45<\/gadget>\n14.9<\/output>\n2 * 10.05<\/gadget>\n20.1<\/output>\n14.9 + 20.1<\/gadget>\n35<\/output>\n14.5 + 35<\/gadget>\n49.5<\/output>\n67.5 - 49.5<\/gadget>\n18<\/output>\n18 * 100<\/gadget>\n1_800<\/output>\n1_800 \/ 67.5<\/gadget>\n26.666667<\/output>\n26.666667<\/result>","index":1816} +{"problem":"if n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2 ^ k is a factor of n ? .","rationale":"20 ! = 1 * 2 * 3 * 4 * 5 . . . * 19 * 20 ( this is 20 factorial written as 20 ! ) n = 1 * 2 * 3 * 4 * 5 * 6 * 7 . . . . . * 19 * 20 how many 2 s are there in n ? one 2 from 2 two 2 s from 4 one two from 6 three 2 s from 8 and so on . . . when you count them all , you get 18 . answer : d","correct":"d","options":{"a":"10 ","b":"12 ","c":"15 ","d":"18","e":"20"},"options_float":{"a":10.0,"b":12.0,"c":15.0,"d":18.0,"e":20.0},"annotated_formula":"add(add(add(divide(20, 2), divide(20, power(2, const_2))), floor(divide(20, power(power(2, const_2), 2)))), floor(divide(20, power(2, const_3))))","linear_formula":"divide(n1,n2)|power(n2,const_2)|power(n2,const_3)|divide(n1,#1)|divide(n1,#2)|power(#1,n2)|add(#0,#3)|divide(n1,#5)|floor(#4)|floor(#7)|add(#6,#9)|add(#10,#8)","chain":"20 \/ 2<\/gadget>\n10<\/output>\n2 ** 2<\/gadget>\n4<\/output>\n20 \/ 4<\/gadget>\n5<\/output>\n10 + 5<\/gadget>\n15<\/output>\n4 ** 2<\/gadget>\n16<\/output>\n20 \/ 16<\/gadget>\n5\/4 = around 1.25<\/output>\nfloor(5\/4)<\/gadget>\n1<\/output>\n15 + 1<\/gadget>\n16<\/output>\n2 ** 3<\/gadget>\n8<\/output>\n20 \/ 8<\/gadget>\n5\/2 = around 2.5<\/output>\nfloor(5\/2)<\/gadget>\n2<\/output>\n16 + 2<\/gadget>\n18<\/output>\n18<\/result>","index":1818} +{"problem":"a man performs 1 \/ 2 of the total journey by rail , 1 \/ 4 by bus and the remaining 4 km on foot . his total journey is","rationale":"\"explanation : let the journey be x km then , 1 x \/ 2 + 1 x \/ 4 + 4 = x 3 x + 16 = 4 x x = 16 km answer : option b\"","correct":"b","options":{"a":"18 km ","b":"16 km ","c":"12 km ","d":"24 km","e":"25 km"},"options_float":{"a":18.0,"b":16.0,"c":12.0,"d":24.0,"e":25.0},"annotated_formula":"multiply(4, 4)","linear_formula":"multiply(n3,n4)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16<\/result>","index":1819} +{"problem":"what is the probability that a two digit number selected at random is a multiple of 3 and not a multiple of 5 ?","rationale":"\"explanation : since every third number starting from 10 will be divisible by 3 , so total number of numbers divisible by 3 are 90 \/ 3 = 30 numbers which are divisible by 3 and 5 both are numbers which are multiple of 15 . for the range 10 to 99 , 15 is the first number divisible by 15 and 90 is the last number . so total number of numbers divisible by 15 are : ( 90 - 15 ) \/ 15 + 1 = 5 + 1 = 6 number of numbers which are divisible by 3 are 30 and number of numbers which are divisible by 3 and 5 both are 6 . so number of numbers divisible by 3 and not by 5 are : 30 - 6 = 24 so total probability = 24 \/ 90 = 4 \/ 15 answer : c\"","correct":"c","options":{"a":"1 \/ 15 ","b":"2 \/ 15 ","c":"4 \/ 15 ","d":"7 \/ 15","e":"none of these"},"options_float":{"a":0.0666666667,"b":0.1333333333,"c":0.2666666667,"d":0.4666666667,"e":null},"annotated_formula":"divide(subtract(multiply(multiply(5, const_2), 3), multiply(3, const_2)), multiply(multiply(5, const_2), multiply(3, 3)))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|multiply(n0,n0)|multiply(n0,#0)|multiply(#0,#2)|subtract(#3,#1)|divide(#5,#4)|","chain":"5 * 2<\/gadget>\n10<\/output>\n10 * 3<\/gadget>\n30<\/output>\n3 * 2<\/gadget>\n6<\/output>\n30 - 6<\/gadget>\n24<\/output>\n3 * 3<\/gadget>\n9<\/output>\n10 * 9<\/gadget>\n90<\/output>\n24 \/ 90<\/gadget>\n4\/15 = around 0.266667<\/output>\n4\/15 = around 0.266667<\/result>","index":1820} +{"problem":"bhanu spends 30 % of his income on petrol on scooter 21 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ?","rationale":"\"given 30 % ( income ) = 300 ⇒ ⇒ income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 21 % ( 700 ) = rs . 147 answer : e\"","correct":"e","options":{"a":"2287 ","b":"140 ","c":"128 ","d":"797","e":"147"},"options_float":{"a":2287.0,"b":140.0,"c":128.0,"d":797.0,"e":147.0},"annotated_formula":"multiply(subtract(divide(300, divide(30, const_100)), 300), divide(21, const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n300 \/ (3\/10)<\/gadget>\n1_000<\/output>\n1_000 - 300<\/gadget>\n700<\/output>\n21 \/ 100<\/gadget>\n21\/100 = around 0.21<\/output>\n700 * (21\/100)<\/gadget>\n147<\/output>\n147<\/result>","index":1821} +{"problem":"one - third less than 25 % of 180 is equal to :","rationale":"\"lots of ways to tackle this . 25 % of 180 = 45 1 \/ 3 of 45 = 15 so , 1 \/ 3 less than 45 is equal to 45 - 15 = 30 answer : b\"","correct":"b","options":{"a":"15 ","b":"30 ","c":"35 ","d":"40","e":"45"},"options_float":{"a":15.0,"b":30.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"subtract(multiply(divide(25, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(25, const_100), 180)))","linear_formula":"divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 180<\/gadget>\n45<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 45<\/gadget>\n15<\/output>\n45 - 15<\/gadget>\n30<\/output>\n30<\/result>","index":1823} +{"problem":"in arun ’ s opinion , his weight is greater than 65 kg but less than 72 kg . his brother doest not agree with arun and he thinks that arun ’ s weight is greater than 60 kg but less than 70 kg . his mother ’ s view is that his weight can not be greater than 68 kg . if all are them are correct in their estimation , what is the average of different probable weights of arun ?","rationale":"explanation : let arun ’ s weight by x kg . according to arun , 65 < x < 72 according to arun ’ s brother , 60 < x < 70 . according to arun ’ s mother , x < = 68 the values satisfying all the above conditions are 66 , 67 and 68 . required average = [ 66 + 67 + 68 \/ 3 ] = [ 201 \/ 3 ] = 67 kg answer b","correct":"b","options":{"a":"66 kg ","b":"67 kg ","c":"68 kg ","d":"69 kg","e":"none of these"},"options_float":{"a":66.0,"b":67.0,"c":68.0,"d":69.0,"e":null},"annotated_formula":"divide(add(add(subtract(68, const_1), subtract(subtract(68, const_1), const_1)), 68), const_3)","linear_formula":"subtract(n4,const_1)|subtract(#0,const_1)|add(#0,#1)|add(n4,#2)|divide(#3,const_3)","chain":"68 - 1<\/gadget>\n67<\/output>\n67 - 1<\/gadget>\n66<\/output>\n67 + 66<\/gadget>\n133<\/output>\n133 + 68<\/gadget>\n201<\/output>\n201 \/ 3<\/gadget>\n67<\/output>\n67<\/result>","index":1825} +{"problem":"there are 8 books on a shelf , of which 3 are paperbacks and 5 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ?","rationale":"approach 1 at - least 1 paper back = total - no paper back 8 c 4 - 5 c 4 = 65 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 3 c 1 * 5 c 3 + 3 c 2 * 5 c 2 + 3 c 3 * 5 c 1 = 65 answer is e","correct":"e","options":{"a":"40 ","b":"45 ","c":"50 ","d":"55","e":"65"},"options_float":{"a":40.0,"b":45.0,"c":50.0,"d":55.0,"e":65.0},"annotated_formula":"subtract(choose(8, 4), choose(5, 4))","linear_formula":"choose(n0,n3)|choose(n2,n3)|subtract(#0,#1)","chain":"binomial(8, 4)<\/gadget>\n70<\/output>\nbinomial(5, 4)<\/gadget>\n5<\/output>\n70 - 5<\/gadget>\n65<\/output>\n65<\/result>","index":1826} +{"problem":"rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 500 . then his monthly salary is","rationale":"\"sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 500 â ‡ ” 20 \/ 100 x = 500 â ‡ ” x = 500 ã — 5 = 2500 . answer a\"","correct":"a","options":{"a":"rs . 2500 ","b":"rs . 3500 ","c":"rs . 1500 ","d":"rs . 500","e":"rs . 2000"},"options_float":{"a":2500.0,"b":3500.0,"c":1500.0,"d":500.0,"e":2000.0},"annotated_formula":"multiply(500, add(const_4, const_1))","linear_formula":"add(const_1,const_4)|multiply(n4,#0)|","chain":"4 + 1<\/gadget>\n5<\/output>\n500 * 5<\/gadget>\n2_500<\/output>\n2_500<\/result>","index":1827} +{"problem":"robert left from a pvt company . management hold his salary rs . 15000 \/ - for one month . earlier robert borrowed rs . 7280 \/ - from company . but robert forget that . after one month robert asked his salary and accountant gives rs . 18500 \/ - to him . what is the incentive amount given to robert ?","rationale":"\"total salary = rs . 15000 \/ - borrowed money = 7280 \/ - balance salary = 15000 - 7280 = 7720 paid amount = 18500 \/ - incentive amount = 18500 - 7720 = 10780 \/ - answer is c\"","correct":"c","options":{"a":"9500 ","b":"12500 ","c":"10780 ","d":"10500","e":"8600"},"options_float":{"a":9500.0,"b":12500.0,"c":10780.0,"d":10500.0,"e":8600.0},"annotated_formula":"subtract(18500, 7280)","linear_formula":"subtract(n2,n1)|","chain":"18_500 - 7_280<\/gadget>\n11_220<\/output>\n11_220<\/result>","index":1829} +{"problem":"the food in a camp lasts for 40 men for 20 days . if 10 more men join , how many days will the food last ?","rationale":"one man can consume the same food in 40 * 20 = 800 days . 10 more men join , the total number of men = 50 the number of days the food will last = 800 \/ 50 = 16 days . answer : d","correct":"d","options":{"a":"22 days ","b":"30 days ","c":"23 days ","d":"16 days","e":"17 days"},"options_float":{"a":22.0,"b":30.0,"c":23.0,"d":16.0,"e":17.0},"annotated_formula":"divide(multiply(40, 20), add(40, 10))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)","chain":"40 * 20<\/gadget>\n800<\/output>\n40 + 10<\/gadget>\n50<\/output>\n800 \/ 50<\/gadget>\n16<\/output>\n16<\/result>","index":1831} +{"problem":"a trader bought a car at 30 % 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 = 70 s = 70 * ( 150 \/ 100 ) = 105 100 - 105 = 5 % answer : e\"","correct":"e","options":{"a":"7 % ","b":"62 % ","c":"12 % ","d":"19 %","e":"5 %"},"options_float":{"a":7.0,"b":62.0,"c":12.0,"d":19.0,"e":5.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 30), 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 - 30<\/gadget>\n70<\/output>\n100 + 50<\/gadget>\n150<\/output>\n70 * 150<\/gadget>\n10_500<\/output>\n10_500 \/ 100<\/gadget>\n105<\/output>\n105 \/ 100<\/gadget>\n21\/20 = around 1.05<\/output>\n(21\/20) - 1<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 100<\/gadget>\n5<\/output>\n5<\/result>","index":1832} +{"problem":"in a weight - lifting competition , the total weight of joe ' s two lifts was 450 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ?","rationale":"\"this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 450 pounds . we sum the two variables to obtain : f + s = 450 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f – 300 = s we can now plug in ( 2 f – 300 ) for s into the first equation , so we have : f + 2 f – 300 = 450 3 f = 750 f = 250 answer is a .\"","correct":"a","options":{"a":"250 ","b":"275 ","c":"325 ","d":"350","e":"400"},"options_float":{"a":250.0,"b":275.0,"c":325.0,"d":350.0,"e":400.0},"annotated_formula":"divide(add(450, 300), const_3)","linear_formula":"add(n0,n1)|divide(#0,const_3)|","chain":"450 + 300<\/gadget>\n750<\/output>\n750 \/ 3<\/gadget>\n250<\/output>\n250<\/result>","index":1834} +{"problem":"a particular library has 150 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 108 books in the special collection at that time , how many books of the special collection were loaned out during that month ?","rationale":"\"the total number of books is 150 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 108 books , thus the number of un - returned books is 150 - 108 = 42 books . 0.35 x = 42 x = 120 the answer is e .\"","correct":"e","options":{"a":"40 ","b":"60 ","c":"80 ","d":"100","e":"120"},"options_float":{"a":40.0,"b":60.0,"c":80.0,"d":100.0,"e":120.0},"annotated_formula":"divide(subtract(150, 108), subtract(const_1, divide(65, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"150 - 108<\/gadget>\n42<\/output>\n65 \/ 100<\/gadget>\n13\/20 = around 0.65<\/output>\n1 - (13\/20)<\/gadget>\n7\/20 = around 0.35<\/output>\n42 \/ (7\/20)<\/gadget>\n120<\/output>\n120<\/result>","index":1835} +{"problem":"ramu rides his bike at an average speed of 45 km \/ hr and reaches his desitination in 4 hours . somu covers the same distance in 6 hours . if ramu covered his journey at an average speed which was 9 km \/ hr less and somu covered his journey at an average speed which was 10 km \/ hr more , then the difference in their times taken to reach the destination would be ( in minutes ) .","rationale":"distance travelled by ramu = 45 * 4 = 180 km somu travelled the same distance in 6 hours . his speed = 180 \/ 6 = 30 km \/ hr hence in the conditional case , ramu ' s speed = 45 - 9 = 36 km \/ hr and somu ' s speed = 30 + 10 = 40 km \/ hr . therefore travel time of ramu and somu would be 5 hours and 4.5 hours respectively . hence difference in the time taken = 0.5 hours = 30 minutes . answer : b","correct":"b","options":{"a":"23 minutes ","b":"30 minutes ","c":"43 minutes ","d":"23 minutes","e":"33 minutes"},"options_float":{"a":23.0,"b":30.0,"c":43.0,"d":23.0,"e":33.0},"annotated_formula":"multiply(subtract(divide(multiply(45, 4), subtract(45, 9)), divide(multiply(45, 4), add(divide(multiply(45, 4), 6), 10))), const_60)","linear_formula":"multiply(n0,n1)|subtract(n0,n3)|divide(#0,#1)|divide(#0,n2)|add(n4,#3)|divide(#0,#4)|subtract(#2,#5)|multiply(#6,const_60)","chain":"45 * 4<\/gadget>\n180<\/output>\n45 - 9<\/gadget>\n36<\/output>\n180 \/ 36<\/gadget>\n5<\/output>\n180 \/ 6<\/gadget>\n30<\/output>\n30 + 10<\/gadget>\n40<\/output>\n180 \/ 40<\/gadget>\n9\/2 = around 4.5<\/output>\n5 - (9\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 60<\/gadget>\n30<\/output>\n30<\/result>","index":1836} +{"problem":"elena purchased brand x pens for $ 5.00 apiece and brand y for $ 2.80 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ?","rationale":"\"4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 6 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : 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":"subtract(multiply(const_4.0, 12), 42.00)","linear_formula":"multiply(const_4.0,n2)|subtract(#0,n3)|","chain":"4 * 12<\/gadget>\n48<\/output>\n48 - 42<\/gadget>\n6<\/output>\n6<\/result>","index":1838} +{"problem":"in order to fence a square manish fixed 48 poles . if the distance between two poles , is 2 metres then what will be the area of the square so formed ?","rationale":"let the side of the square be x m . ∴ perimeter of the square = 48 × 2 = 4 x ∴ x = 24 m ∴ area = ( 24 ) 2 = 576 m 2 answer d","correct":"d","options":{"a":"133 cm 2 ","b":"276 cm 2 ","c":"2500 cm 2 ","d":"576 cm 2","e":"none of these"},"options_float":{"a":133.0,"b":276.0,"c":2500.0,"d":576.0,"e":null},"annotated_formula":"square_area(divide(48, 2))","linear_formula":"divide(n0,n1)|square_area(#0)","chain":"48 \/ 2<\/gadget>\n24<\/output>\n24 ** 2<\/gadget>\n576<\/output>\n576<\/result>","index":1840} +{"problem":"find the area of a parallelogram with base 20 cm and height 10 cm ?","rationale":"area of a parallelogram = base * height = 20 * 10 = 200 cm 2 answer : e","correct":"e","options":{"a":"290 cm 2 ","b":"380 cm 2 ","c":"270 cm 2 ","d":"280 cm 2","e":"200 cm 2"},"options_float":{"a":290.0,"b":380.0,"c":270.0,"d":280.0,"e":200.0},"annotated_formula":"multiply(20, 10)","linear_formula":"multiply(n0,n1)","chain":"20 * 10<\/gadget>\n200<\/output>\n200<\/result>","index":1841} +{"problem":"a certain drink of type a is prepared by mixing 4 parts milk with 3 parts fruit juice . another drink of type b is prepared by mixing 4 parts of fruit juice and 3 parts of milk . how many liters of fruit juice must be added to 84 liters of drink a to convert it to drink b ?","rationale":"\"in 84 liters of drink a , there are 48 liters of milk and 36 liters of juice . with 48 liters of milk , we need a total of 64 liters of juice to make drink b . we need to add 28 liters of juice . 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":"subtract(divide(multiply(multiply(divide(4, add(4, 3)), 84), 4), 3), multiply(divide(3, add(4, 3)), 84))","linear_formula":"add(n0,n1)|divide(n0,#0)|divide(n1,#0)|multiply(n4,#1)|multiply(n4,#2)|multiply(n0,#3)|divide(#5,n1)|subtract(#6,#4)|","chain":"4 + 3<\/gadget>\n7<\/output>\n4 \/ 7<\/gadget>\n4\/7 = around 0.571429<\/output>\n(4\/7) * 84<\/gadget>\n48<\/output>\n48 * 4<\/gadget>\n192<\/output>\n192 \/ 3<\/gadget>\n64<\/output>\n3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n(3\/7) * 84<\/gadget>\n36<\/output>\n64 - 36<\/gadget>\n28<\/output>\n28<\/result>","index":1842} +{"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 r difference 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 r 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":1844} +{"problem":"in a school with 620 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 = ( 620 - x ) . then , ( 11 3 \/ 4 × 620 ) ⇔ 11 x + 12 ( 620 - x ) ⇔ x = 7440 - 7285 ⇔ 155 . answer c\"","correct":"c","options":{"a":"150 ","b":"200 ","c":"155 ","d":"350","e":"none"},"options_float":{"a":150.0,"b":200.0,"c":155.0,"d":350.0,"e":null},"annotated_formula":"subtract(multiply(12, 620), multiply(add(11, divide(9, 12)), 620))","linear_formula":"divide(n4,n1)|multiply(n0,n1)|add(n2,#0)|multiply(n0,#2)|subtract(#1,#3)|","chain":"12 * 620<\/gadget>\n7_440<\/output>\n9 \/ 12<\/gadget>\n3\/4 = around 0.75<\/output>\n11 + (3\/4)<\/gadget>\n47\/4 = around 11.75<\/output>\n(47\/4) * 620<\/gadget>\n7_285<\/output>\n7_440 - 7_285<\/gadget>\n155<\/output>\n155<\/result>","index":1845} +{"problem":"two trains are moving in opposite directions with speed of 60 km \/ hr and 90 km \/ hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds","rationale":"\"relative speed = 60 + 90 = 150 km \/ hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 \/ 150 hr = 1 \/ \/ 75 hr = 3600 \/ 75 seconds = 1200 \/ 25 = 240 \/ 5 = 48 seconds answer is e .\"","correct":"e","options":{"a":"40 ","b":"42 ","c":"44 ","d":"46","e":"48"},"options_float":{"a":40.0,"b":42.0,"c":44.0,"d":46.0,"e":48.0},"annotated_formula":"multiply(divide(add(1.10, 0.9), add(60, 90)), const_3600)","linear_formula":"add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600)|","chain":"1.1 + 0.9<\/gadget>\n2<\/output>\n60 + 90<\/gadget>\n150<\/output>\n2 \/ 150<\/gadget>\n1\/75 = around 0.013333<\/output>\n(1\/75) * 3_600<\/gadget>\n48<\/output>\n48<\/result>","index":1846} +{"problem":"the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km \/ hr and 14 km \/ hr respectively . they will meet for the first time in ?","rationale":"clearly , the two will meet when they are 1000 m apart to be 20 + 14 = 34 km apart , they take 1 hour to be 1000 m apart , they take 34 * 1000 \/ 1000 = 34 min . answer is c","correct":"c","options":{"a":"50 min ","b":"40 min ","c":"34 min ","d":"25 min","e":"20 min"},"options_float":{"a":50.0,"b":40.0,"c":34.0,"d":25.0,"e":20.0},"annotated_formula":"add(20, 14)","linear_formula":"add(n1,n2)","chain":"20 + 14<\/gadget>\n34<\/output>\n34<\/result>","index":1847} +{"problem":"the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 90 and rs . 80 respectively . the sum is :","rationale":"\"sol . sum = s . i . * t . d . \/ ( s . i ) - ( t . d . ) = 90 * 80 \/ ( 90 - 80 ) = rs . 720 . answer c\"","correct":"c","options":{"a":"1360 ","b":"1450 ","c":"720 ","d":"1800","e":"none"},"options_float":{"a":1360.0,"b":1450.0,"c":720.0,"d":1800.0,"e":null},"annotated_formula":"divide(multiply(90, 80), subtract(90, 80))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"90 * 80<\/gadget>\n7_200<\/output>\n90 - 80<\/gadget>\n10<\/output>\n7_200 \/ 10<\/gadget>\n720<\/output>\n720<\/result>","index":1848} +{"problem":"if shreehari walks in the speed of 4.5 km \/ hr from his house , in what time will he reach his school which is 750 m long from his house ?","rationale":"speed = 4.5 * 5 \/ 18 = 1.25 m \/ sec time taken = 750 \/ 1.25 = 600 sec ie . 10 mins . answer : c","correct":"c","options":{"a":"5 ","b":"30 ","c":"10 ","d":"12","e":"15"},"options_float":{"a":5.0,"b":30.0,"c":10.0,"d":12.0,"e":15.0},"annotated_formula":"multiply(divide(divide(750, const_1000), 4.5), const_60)","linear_formula":"divide(n1,const_1000)|divide(#0,n0)|multiply(#1,const_60)","chain":"750 \/ 1_000<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) \/ 4.5<\/gadget>\n0.166667<\/output>\n0.166667 * 60<\/gadget>\n10.00002<\/output>\n10.00002<\/result>","index":1850} +{"problem":"if x is a number such that x ^ 2 + 2 x - 24 = 0 and x ^ 2 - 5 x + 4 = 0 , then x =","rationale":"x ^ 2 + 2 x - 24 = ( x + 6 ) ( x - 4 ) = 0 then x = - 6 or x = 4 . x ^ 2 - 5 x + 4 = ( x - 4 ) ( x - 1 ) = 0 then x = 4 or x = 1 . thus x = 4 . the answer is a .","correct":"a","options":{"a":"4 ","b":"- 4 ","c":"- 3 ","d":"- 6","e":"1"},"options_float":{"a":4.0,"b":-4.0,"c":-3.0,"d":-6.0,"e":1.0},"annotated_formula":"divide(add(4, 24), add(2, 5))","linear_formula":"add(n2,n6)|add(n0,n5)|divide(#0,#1)","chain":"4 + 24<\/gadget>\n28<\/output>\n2 + 5<\/gadget>\n7<\/output>\n28 \/ 7<\/gadget>\n4<\/output>\n4<\/result>","index":1851} +{"problem":"a library has an average of 510 visitors on sunday and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is","rationale":"sol . since the month begins with a sunday , so there will be five sundays in the month . ∴ required average = [ 510 x 5 + 240 x 25 \/ 30 ] = 8550 \/ 30 = 285 answer d","correct":"d","options":{"a":"250 ","b":"276 ","c":"280 ","d":"285","e":"none"},"options_float":{"a":250.0,"b":276.0,"c":280.0,"d":285.0,"e":null},"annotated_formula":"divide(add(multiply(add(const_4, const_1), 510), multiply(multiply(add(const_4, const_1), add(const_4, const_1)), 240)), 30)","linear_formula":"add(const_1,const_4)|multiply(n0,#0)|multiply(#0,#0)|multiply(n1,#2)|add(#1,#3)|divide(#4,n2)","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 510<\/gadget>\n2_550<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25 * 240<\/gadget>\n6_000<\/output>\n2_550 + 6_000<\/gadget>\n8_550<\/output>\n8_550 \/ 30<\/gadget>\n285<\/output>\n285<\/result>","index":1852} +{"problem":"a horse chases a pony 8 hours after the pony runs . horse takes 12 hours to reach the pony . if the average speed of the horse is 320 kmph , what s the average speed of the pony ?","rationale":"pony take 20 hours and horse take 12 hours . . . then distance chased by them is 320 * 12 . so speed of pony is ( 320 * 12 ) \/ 20 = 192 kmph . answer is b","correct":"b","options":{"a":"182 kmph ","b":"192 kmph ","c":"193 kmph ","d":"196 kmph","e":"190 kmph"},"options_float":{"a":182.0,"b":192.0,"c":193.0,"d":196.0,"e":190.0},"annotated_formula":"divide(multiply(12, 320), add(12, 8))","linear_formula":"add(n0,n1)|multiply(n1,n2)|divide(#1,#0)","chain":"12 * 320<\/gadget>\n3_840<\/output>\n12 + 8<\/gadget>\n20<\/output>\n3_840 \/ 20<\/gadget>\n192<\/output>\n192<\/result>","index":1853} +{"problem":"a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 50 $ for the meal . assuming that they will pay 10 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ?","rationale":"\"let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.15 * 1.10 x = 50 x = 39.52 hence , the correct option is a\"","correct":"a","options":{"a":"39.52 $ ","b":"40.63 $ ","c":"41.63 $ ","d":"42.15 $","e":"41.15 $"},"options_float":{"a":39.52,"b":40.63,"c":41.63,"d":42.15,"e":41.15},"annotated_formula":"divide(50, add(divide(add(10, 15), const_100), const_1))","linear_formula":"add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2)|","chain":"10 + 15<\/gadget>\n25<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n50 \/ (5\/4)<\/gadget>\n40<\/output>\n40<\/result>","index":1855} +{"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 150 ?","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 = 150 \/ 2 = 75 answer is d\"","correct":"d","options":{"a":"10 ","b":"25 ","c":"50 ","d":"75","e":"100"},"options_float":{"a":10.0,"b":25.0,"c":50.0,"d":75.0,"e":100.0},"annotated_formula":"divide(150, const_2)","linear_formula":"divide(n0,const_2)|","chain":"150 \/ 2<\/gadget>\n75<\/output>\n75<\/result>","index":1857} +{"problem":"a chemist mixes one liter of pure water with x liters of a 60 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ?","rationale":"\"concentration of salt in pure solution = 0 concentration of salt in salt solution = 60 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 60 - 15 ) \/ ( 15 - 0 ) = 3 \/ 1 1 \/ x = 3 \/ 1 x = 1 \/ 3 answer : b\"","correct":"b","options":{"a":"1 \/ 4 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"1","e":"3"},"options_float":{"a":0.25,"b":0.3333333333,"c":0.5,"d":1.0,"e":3.0},"annotated_formula":"divide(15, subtract(60, 15))","linear_formula":"subtract(n0,n1)|divide(n1,#0)|","chain":"60 - 15<\/gadget>\n45<\/output>\n15 \/ 45<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":1859} +{"problem":"a certain car uses one gallon of gasoline every 32 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ?","rationale":"\"4 miles on the highway = 4 \/ 32 gallons ; 4 miles in the city = 4 \/ 20 gallons ; total = 4 \/ 32 + 4 \/ 20 = 13 \/ 40 gallons . 8 miles on the highway = 8 \/ 30 gallons . the % change = ( 13 \/ 40 - 8 \/ 32 ) \/ ( 8 \/ 32 ) = 0.30 . answer : e .\"","correct":"e","options":{"a":"15 % ","b":"20 % ","c":"22.5 % ","d":"25 %","e":"30 %"},"options_float":{"a":15.0,"b":20.0,"c":22.5,"d":25.0,"e":30.0},"annotated_formula":"multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 32))), multiply(8, divide(const_1, 32))), multiply(8, divide(const_1, 32))), const_100)","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n4,#1)|add(#2,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)|","chain":"1 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 4<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 32<\/gadget>\n1\/32 = around 0.03125<\/output>\n4 * (1\/32)<\/gadget>\n1\/8 = around 0.125<\/output>\n(1\/5) + (1\/8)<\/gadget>\n13\/40 = around 0.325<\/output>\n8 * (1\/32)<\/gadget>\n1\/4 = around 0.25<\/output>\n(13\/40) - (1\/4)<\/gadget>\n3\/40 = around 0.075<\/output>\n(3\/40) \/ (1\/4)<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":1860} +{"problem":"a retailer sold an appliance for $ 100 . if the retailer ' s gross profit on the appliance was 25 percent of the retailer ' s cost for the appliance , how many dollars was the retailer ' s gross profit ?","rationale":"let p be the original price paid by the retailer . 1.25 * p = 100 p = 80 the profit is $ 20 . the answer is c .","correct":"c","options":{"a":"$ 10 ","b":"$ 16 ","c":"$ 20 ","d":"$ 24","e":"$ 25"},"options_float":{"a":10.0,"b":16.0,"c":20.0,"d":24.0,"e":25.0},"annotated_formula":"subtract(100, multiply(divide(const_1, add(divide(25, 100), const_1)), 100))","linear_formula":"divide(n1,n0)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2)|subtract(n0,#3)","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n1 \/ (5\/4)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n20<\/result>","index":1861} +{"problem":"24 oz of juice p and 25 oz of juice t are mixed to make smothies x and y . the ratio of p to t in smothie x is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie x ?","rationale":"easy way to solve this question is start from the answer and then conform the information provided in the question . we can start from option d i . e 20 . . . as a quantity of juice p in x because it is the only one option that gets divided by 4 is 20 . . . since in the x the juice p to t ratio is 4 : 1 this gives us that quantity of juice p in x = 20 therefore quantity of juice t will be 5 . . . hence ratio = 4 : 1 this will lead to quantity of juice p in x = 4 and quantity of juice t = 20 . . . hence ratio 1 : 5 if we calculate total juice p = 24 and total of juice v = 25 it fits because totals are same as what mentioned in the question . . . thus ans is 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":"subtract(24, divide(subtract(multiply(4, 25), 24), subtract(multiply(4, 5), 1)))","linear_formula":"multiply(n1,n2)|multiply(n2,n5)|subtract(#0,n0)|subtract(#1,n3)|divide(#2,#3)|subtract(n0,#4)","chain":"4 * 25<\/gadget>\n100<\/output>\n100 - 24<\/gadget>\n76<\/output>\n4 * 5<\/gadget>\n20<\/output>\n20 - 1<\/gadget>\n19<\/output>\n76 \/ 19<\/gadget>\n4<\/output>\n24 - 4<\/gadget>\n20<\/output>\n20<\/result>","index":1862} +{"problem":"if 85 percent of the test takers taking an old paper and pencil gmat exam answered the first question on a given math section correctly , and 75 percent of the test takers answered the second question correctly , and 15 percent of the test takers answered neither question correctly , what percent answered both correctly ?","rationale":"{ total } = { first correctly } + { second correctly } - { both correctly } + { neither correctly } 100 = 85 + 75 - { both correctly } + 15 { both correctly } = 75 . answer : d .","correct":"d","options":{"a":"60 % ","b":"65 % ","c":"70 % ","d":"75 %","e":"80 %"},"options_float":{"a":60.0,"b":65.0,"c":70.0,"d":75.0,"e":80.0},"annotated_formula":"subtract(add(add(85, 75), 15), const_100)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,const_100)","chain":"85 + 75<\/gadget>\n160<\/output>\n160 + 15<\/gadget>\n175<\/output>\n175 - 100<\/gadget>\n75<\/output>\n75<\/result>","index":1863} +{"problem":"find the least number of complete years in which a sum of money put out at 25 % compound interest will be more than double of itself ?","rationale":"\"4 years answer : d\"","correct":"d","options":{"a":"6 years ","b":"9 years ","c":"5 years ","d":"4 years","e":"6 years"},"options_float":{"a":6.0,"b":9.0,"c":5.0,"d":4.0,"e":6.0},"annotated_formula":"floor(add(divide(log(const_2), log(add(const_1, divide(25, 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>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\nlog(5\/4)<\/gadget>\nlog(5\/4) = around 0.223144<\/output>\nlog(2) \/ log(5\/4)<\/gadget>\nlog(2)\/log(5\/4) = around 3.106284<\/output>\n(log(2)\/log(5\/4)) + 1<\/gadget>\n1 + log(2)\/log(5\/4) = around 4.106284<\/output>\nfloor(1 + log(2)\/log(5\/4))<\/gadget>\n4<\/output>\n4<\/result>","index":1865} +{"problem":"mrs . evans gave a test to her freshmen economics class , which has 29 students enrolled and 24 of them answered question 1 correctly . if 22 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 = 22 . number of students who answered both q 1 & q 2 correctly = 22 . answer : c\"","correct":"c","options":{"a":"18 ","b":"19 ","c":"22 ","d":"20","e":"19"},"options_float":{"a":18.0,"b":19.0,"c":22.0,"d":20.0,"e":19.0},"annotated_formula":"subtract(add(add(24, 22), 5), 29)","linear_formula":"add(n1,n3)|add(n5,#0)|subtract(#1,n0)|","chain":"24 + 22<\/gadget>\n46<\/output>\n46 + 5<\/gadget>\n51<\/output>\n51 - 29<\/gadget>\n22<\/output>\n22<\/result>","index":1866} +{"problem":"7 m - 20 = 2 m , then m + 7 is equal to ?","rationale":"7 m - 20 = 2 m so , 5 m = 20 so , m + 7 = 11 answer : c","correct":"c","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":"add(divide(20, subtract(7, 2)), 7)","linear_formula":"subtract(n0,n2)|divide(n1,#0)|add(n0,#1)","chain":"7 - 2<\/gadget>\n5<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n4 + 7<\/gadget>\n11<\/output>\n11<\/result>","index":1867} +{"problem":"the smallest value of n , for which 2 n + 1 is not a prime number is","rationale":"\"sol . = ( 2 × 1 + 1 ) = 3 , = ( 2 × 3 + 1 ) = 7 , = ( 2 × 4 + 1 ) = 9 , which is not prime . therefore n = 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":"add(2, 2)","linear_formula":"add(n0,n0)|","chain":"2 + 2<\/gadget>\n4<\/output>\n4<\/result>","index":1869} +{"problem":"two 100 multiples of 7 are chosen at random , and 300 multiples of 8 are chosen at random . approximately what percentage of the 500 selected numbers are odd ?","rationale":"two hundred multiples of seven will have 100 even and 100 odd numbers 300 multiples of eight will have all even ( being multiple of 8 ) probability of number selected being odd = total odd numbers \/ total available numbers probability = 100 \/ 500 = 0.2 = 20 % answer : option a","correct":"a","options":{"a":"20 % ","b":"25 % ","c":"40 % ","d":"50 %","e":"80 %"},"options_float":{"a":20.0,"b":25.0,"c":40.0,"d":50.0,"e":80.0},"annotated_formula":"multiply(divide(100, 500), const_100)","linear_formula":"divide(n0,n4)|multiply(#0,const_100)","chain":"100 \/ 500<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":1870} +{"problem":"in traveling from a dormitory to a certain city , a student went 1 \/ 3 of the way by foot , 1 \/ 4 of the way by bus , and the remaining 5 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 \/ 2 x + 3 \/ 5 x + 5 x - 1 \/ 2 x - 3 \/ 5 x = 5 x = 12 km option : e\"","correct":"e","options":{"a":"13 ","b":"17 ","c":"42 ","d":"15","e":"12"},"options_float":{"a":13.0,"b":17.0,"c":42.0,"d":15.0,"e":12.0},"annotated_formula":"multiply(5, inverse(subtract(1, add(divide(1, 3), divide(1, 4)))))","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>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/3) + (1\/4)<\/gadget>\n7\/12 = around 0.583333<\/output>\n1 - (7\/12)<\/gadget>\n5\/12 = around 0.416667<\/output>\n1 \/ (5\/12)<\/gadget>\n12\/5 = around 2.4<\/output>\n5 * (12\/5)<\/gadget>\n12<\/output>\n12<\/result>","index":1871} +{"problem":"tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 10 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 + 10 mph ) \/ 2 = 5.5 mph correct option is : b","correct":"b","options":{"a":"8 mph ","b":"5.5 mph ","c":"3.5 mph ","d":"4 mph","e":"0.5 mph"},"options_float":{"a":8.0,"b":5.5,"c":3.5,"d":4.0,"e":0.5},"annotated_formula":"divide(add(1, 10), const_2)","linear_formula":"add(n0,n1)|divide(#0,const_2)","chain":"1 + 10<\/gadget>\n11<\/output>\n11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\n11\/2 = around 5.5<\/result>","index":1873} +{"problem":"j is 25 % less than p and 20 % less than t . t is x % less than p . what is the value of x ?","rationale":"\"let , p = 400 then j = ( 75 \/ 100 ) * 400 = 300 also j = ( 80 \/ 100 ) * t i . e . t = 300 * 100 \/ 80 = 375 and t = [ 1 - ( x \/ 100 ) ] * p i . e . 100 - x = 100 * t \/ p = 100 * 375 \/ 400 = 93.75 i . e . x = 6.25 answer : option d\"","correct":"d","options":{"a":"93.5 ","b":"90 ","c":"6.75 ","d":"6.25","e":"2"},"options_float":{"a":93.5,"b":90.0,"c":6.75,"d":6.25,"e":2.0},"annotated_formula":"divide(multiply(25, 25), const_100)","linear_formula":"multiply(n0,n0)|divide(#0,const_100)|","chain":"25 * 25<\/gadget>\n625<\/output>\n625 \/ 100<\/gadget>\n25\/4 = around 6.25<\/output>\n25\/4 = around 6.25<\/result>","index":1874} +{"problem":"denise is trying to open a safe whose combination she does not know . if the safe has 4000 possible combinations , and she can try 75 different possibilities , what is the probability that she does not pick the one correct combination .","rationale":"when trying the first time the probability denise does n ' t pick the correct combination = 3999 \/ 4000 second time , as the total number of possible combinations reduced by one , not picking the right one would be 3998 \/ 3999 . third time 3997 \/ 3998 . . . and the same 75 times . so we get : 3999 \/ 4000 ∗ 3998 \/ 3999 ∗ . . . ∗ 3925 \/ 39263999 \/ 4000 ∗ 3998 \/ 3999 ∗ . . . ∗ 3925 \/ 3926 every denominator but the first will cancel out and every nominator but the last will cancel out as well . we ' ll get 3925 \/ 4000 = 157 \/ 160 . answer : c .","correct":"c","options":{"a":"1 ","b":"159 \/ 160 ","c":"157 \/ 160 ","d":"4 3 \/ 160","e":"0"},"options_float":{"a":1.0,"b":0.99375,"c":0.98125,"d":4.0,"e":0.0},"annotated_formula":"divide(subtract(4000, 75), 4000)","linear_formula":"subtract(n0,n1)|divide(#0,n0)","chain":"4_000 - 75<\/gadget>\n3_925<\/output>\n3_925 \/ 4_000<\/gadget>\n157\/160 = around 0.98125<\/output>\n157\/160 = around 0.98125<\/result>","index":1875} +{"problem":"in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many q cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]","rationale":"\"main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer q = 44 . d\"","correct":"d","options":{"a":"0.33 ","b":"0.44 ","c":"0.55 ","d":"44","e":"55"},"options_float":{"a":0.33,"b":0.44,"c":0.55,"d":44.0,"e":55.0},"annotated_formula":"multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100))","linear_formula":"divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n8 \/ (1\/2)<\/gadget>\n16<\/output>\n16 \/ 12<\/gadget>\n4\/3 = around 1.333333<\/output>\n0.33 * 100<\/gadget>\n33<\/output>\n(4\/3) * 33<\/gadget>\n44<\/output>\n44<\/result>","index":1877} +{"problem":"two spheres of their radios in the ratio 4 : 3 . find its volumes ratio ?","rationale":"sphere volume ( v ) = 4 \/ 3 π r ( power 3 ) : 4 \/ 3 π r ( power 3 ) = 4 ( power 3 ) : 3 ( power 3 ) = 64 : 27 answer is d .","correct":"d","options":{"a":"64 : 13 ","b":"13 : 64 ","c":"27 : 64 ","d":"64 : 27","e":"none of them"},"options_float":{"a":4.9230769231,"b":0.203125,"c":0.421875,"d":2.3703703704,"e":null},"annotated_formula":"divide(volume_sphere(4), volume_sphere(3))","linear_formula":"volume_sphere(n0)|volume_sphere(n1)|divide(#0,#1)","chain":"4\/3 * pi * (4 ** 3)<\/gadget>\n256*pi\/3 = around 268.082573<\/output>\n4\/3 * pi * (3 ** 3)<\/gadget>\n36*pi = around 113.097336<\/output>\n(256*pi\/3) \/ (36*pi)<\/gadget>\n64\/27 = around 2.37037<\/output>\n64\/27 = around 2.37037<\/result>","index":1878} +{"problem":"danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 1536 , what is the area of the side face of the box ?","rationale":"\"lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 \/ 2 w * d ( l = 1 \/ 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 1536 l * 1.5 l * 2 l = 1536 l = 8 side face area = l * d = l * 2 l = 8 * 2 * 8 = 128 e is the answer\"","correct":"e","options":{"a":"34 ","b":"65 ","c":"88 ","d":"90","e":"128"},"options_float":{"a":34.0,"b":65.0,"c":88.0,"d":90.0,"e":128.0},"annotated_formula":"divide(power(multiply(const_3, power(1536, const_2)), divide(const_1, const_3)), 1.5)","linear_formula":"divide(const_1,const_3)|power(n1,const_2)|multiply(#1,const_3)|power(#2,#0)|divide(#3,n0)|","chain":"1_536 ** 2<\/gadget>\n2_359_296<\/output>\n3 * 2_359_296<\/gadget>\n7_077_888<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n7_077_888 ** (1\/3)<\/gadget>\n192<\/output>\n192 \/ 1.5<\/gadget>\n128<\/output>\n128<\/result>","index":1880} +{"problem":"sum of the squares of 3 no . is 276 and the sum of their products taken two at a time is 150 . find the sum ?","rationale":"\"( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 276 + 2 * 150 a + b + c = â ˆ š 576 = 24 answer d\"","correct":"d","options":{"a":"22 ","b":"18 ","c":"26 ","d":"24","e":"32"},"options_float":{"a":22.0,"b":18.0,"c":26.0,"d":24.0,"e":32.0},"annotated_formula":"sqrt(add(276, multiply(150, const_2)))","linear_formula":"multiply(n2,const_2)|add(n1,#0)|sqrt(#1)|","chain":"150 * 2<\/gadget>\n300<\/output>\n276 + 300<\/gadget>\n576<\/output>\n576 ** (1\/2)<\/gadget>\n24<\/output>\n24<\/result>","index":1881} +{"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 14 questions , if sterling had answered 2 more questions correctly , he would have earned 31 pieces of candy . how many of the 14 questions did zachary answer correctly ?","rationale":"i got two equations : 3 x - 2 y = 25 x + y = 14 3 x - 2 ( 14 - x ) = 25 3 x - 28 + 2 x = 25 5 x = 53 x = 10.6 or between 10 and 11 . ( ans 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(add(subtract(multiply(2, subtract(14, 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":"14 - 2<\/gadget>\n12<\/output>\n2 * 12<\/gadget>\n24<\/output>\n3 * 2<\/gadget>\n6<\/output>\n24 - 6<\/gadget>\n18<\/output>\n18 + 31<\/gadget>\n49<\/output>\n3 + 2<\/gadget>\n5<\/output>\n49 \/ 5<\/gadget>\n49\/5 = around 9.8<\/output>\n49\/5 = around 9.8<\/result>","index":1882} +{"problem":"bhaman travelled for 15 hours . he covered the first half of the distance at 40 kmph and remaining half of the distance at 10 kmph . find the distance travelled by bhaman ?","rationale":"let the distance travelled be x km . total time = ( x \/ 2 ) \/ 40 + ( x \/ 2 ) \/ 10 = 15 = > x \/ 80 + x \/ 20 = 15 = > ( x + 4 x ) \/ 80 = 15 = > x = 240 km answer : a","correct":"a","options":{"a":"240 ","b":"230 ","c":"260 ","d":"220","e":"340"},"options_float":{"a":240.0,"b":230.0,"c":260.0,"d":220.0,"e":340.0},"annotated_formula":"multiply(divide(15, add(divide(multiply(const_2, 40), const_10), divide(multiply(const_2, 10), const_10))), multiply(multiply(divide(multiply(const_2, 40), const_10), divide(multiply(const_2, 10), const_10)), const_10))","linear_formula":"multiply(n1,const_2)|multiply(n2,const_2)|divide(#0,const_10)|divide(#1,const_10)|add(#2,#3)|multiply(#2,#3)|divide(n0,#4)|multiply(#5,const_10)|multiply(#6,#7)","chain":"2 * 40<\/gadget>\n80<\/output>\n80 \/ 10<\/gadget>\n8<\/output>\n2 * 10<\/gadget>\n20<\/output>\n20 \/ 10<\/gadget>\n2<\/output>\n8 + 2<\/gadget>\n10<\/output>\n15 \/ 10<\/gadget>\n3\/2 = around 1.5<\/output>\n8 * 2<\/gadget>\n16<\/output>\n16 * 10<\/gadget>\n160<\/output>\n(3\/2) * 160<\/gadget>\n240<\/output>\n240<\/result>","index":1883} +{"problem":"in a survey of parents , exactly 7 \/ 8 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 \/ 8 * 2 \/ 5 = 1 \/ 20 of all the parents surveyed . the percent of parents without full - time jobs is 3 \/ 20 + 1 \/ 20 = 1 \/ 5 = 20 % the answer is a .","correct":"a","options":{"a":"20 % ","b":"23 % ","c":"15 % ","d":"18 %","e":"16 %"},"options_float":{"a":20.0,"b":23.0,"c":15.0,"d":18.0,"e":16.0},"annotated_formula":"add(subtract(subtract(const_100, 40), multiply(divide(3, 4), subtract(const_100, 40))), subtract(40, multiply(divide(7, 8), 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>\n7 \/ 8<\/gadget>\n7\/8 = around 0.875<\/output>\n(7\/8) * 40<\/gadget>\n35<\/output>\n40 - 35<\/gadget>\n5<\/output>\n15 + 5<\/gadget>\n20<\/output>\n20<\/result>","index":1884} +{"problem":"three pipes of same capacity can fill a tank in 8 hours . if there are only two pipes of same capacity , the tank can be filled in .","rationale":"the part of the tank filled by three pipes in one hour = 1 \/ 8 = > the part of the tank filled by two pipes in 1 hour = 2 \/ 3 * 1 \/ 8 = 1 \/ 12 . the tank can be filled in 12 hours . answer : b","correct":"b","options":{"a":"11 hours ","b":"12 hours ","c":"15 hours ","d":"16 hours","e":"17 hours"},"options_float":{"a":11.0,"b":12.0,"c":15.0,"d":16.0,"e":17.0},"annotated_formula":"inverse(multiply(divide(const_2, const_3), divide(const_1, 8)))","linear_formula":"divide(const_2,const_3)|divide(const_1,n0)|multiply(#0,#1)|inverse(#2)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n(2\/3) * (1\/8)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":1885} +{"problem":"john makes $ 40 a week from his job . he earns a raise andnow makes $ 70 a week . what is the % increase ?","rationale":"\"increase = ( 30 \/ 40 ) * 100 = ( 3 \/ 4 ) * 100 = 75 % . e\"","correct":"e","options":{"a":"16 % ","b":"16.66 % ","c":"76.69 % ","d":"76.98 %","e":"75 %"},"options_float":{"a":16.0,"b":16.66,"c":76.69,"d":76.98,"e":75.0},"annotated_formula":"multiply(divide(subtract(70, 40), 40), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"70 - 40<\/gadget>\n30<\/output>\n30 \/ 40<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/4) * 100<\/gadget>\n75<\/output>\n75<\/result>","index":1887} +{"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 . 6500 ?","rationale":"\"let the sale in the sixth month = x then ( 6435 + 6927 + 6855 + 7230 + 6562 + x ) \/ 6 = 6500 = > 6435 + 6927 + 6855 + 7230 + 6562 + x = 6 × 6500 = > 34009 + x = 39000 = > x = 39000 − 34009 = 4991 answer is c .\"","correct":"c","options":{"a":"4551 ","b":"4771 ","c":"4991 ","d":"4881","e":"4661"},"options_float":{"a":4551.0,"b":4771.0,"c":4991.0,"d":4881.0,"e":4661.0},"annotated_formula":"subtract(multiply(add(5, const_1), 6500), 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_500<\/gadget>\n39_000<\/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>\n39_000 - 34_009<\/gadget>\n4_991<\/output>\n4_991<\/result>","index":1889} +{"problem":"calculate the number of bricks , each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 cm x 6 m when 10 % of its volume is occupied by mortar ?","rationale":"explanation : let the number of bricks be ' n ' 10 x 4 \/ 100 x 6 x 90 \/ 100 = 25 \/ 100 x 15 \/ 100 x 8 \/ 100 x n 10 x 4 x 6 x 90 = 15 x 2 x n = > n = 720 . answer is a","correct":"a","options":{"a":"720 ","b":"600 ","c":"660 ","d":"6000","e":"none of these"},"options_float":{"a":720.0,"b":600.0,"c":660.0,"d":6000.0,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(10, divide(4, const_100)), 6), subtract(const_1, divide(10, const_100))), multiply(multiply(divide(25, const_100), divide(15, const_100)), divide(8, const_100)))","linear_formula":"divide(n4,const_100)|divide(n3,const_100)|divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|multiply(n3,#0)|multiply(#3,#4)|subtract(const_1,#1)|multiply(n5,#5)|multiply(#2,#6)|multiply(#8,#7)|divide(#10,#9)","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n10 * (1\/25)<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 6<\/gadget>\n12\/5 = around 2.4<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n(12\/5) * (9\/10)<\/gadget>\n54\/25 = around 2.16<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(1\/4) * (3\/20)<\/gadget>\n3\/80 = around 0.0375<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(3\/80) * (2\/25)<\/gadget>\n3\/1_000 = around 0.003<\/output>\n(54\/25) \/ (3\/1_000)<\/gadget>\n720<\/output>\n720<\/result>","index":1891} +{"problem":"the lcm and hcf of two numbers are 8 and 48 respectively . if one of them is 24 , find the other ?","rationale":"hcf x lcm = product of numbers 8 x 48 = 24 x the other number other number = ( 8 x 48 ) \/ 24 other number = 16 answer : d","correct":"d","options":{"a":"12 ","b":"14 ","c":"15 ","d":"16","e":"20"},"options_float":{"a":12.0,"b":14.0,"c":15.0,"d":16.0,"e":20.0},"annotated_formula":"divide(multiply(8, 48), 24)","linear_formula":"multiply(n0,n1)|divide(#0,n2)","chain":"8 * 48<\/gadget>\n384<\/output>\n384 \/ 24<\/gadget>\n16<\/output>\n16<\/result>","index":1893} +{"problem":"one night a certain hotel rented 3 \/ 5 of its rooms , including 2 \/ 3 of their air conditioned rooms . if 3 \/ 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ?","rationale":"the rooms which were not rented is 2 \/ 5 the ac rooms which were not rented is ( 1 \/ 3 ) * ( 3 \/ 5 ) = 1 \/ 5 the percentage of unrented rooms which were ac rooms is ( 1 \/ 5 ) \/ ( 2 \/ 5 ) = 1 \/ 2 = 50 % the answer is a .","correct":"a","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":"multiply(divide(multiply(subtract(const_1, divide(2, 3)), multiply(divide(3, 5), const_100)), subtract(const_100, multiply(divide(3, 5), const_100))), const_100)","linear_formula":"divide(n0,n1)|divide(n2,n0)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_100,#2)|divide(#4,#5)|multiply(#6,const_100)","chain":"2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n(1\/3) * 60<\/gadget>\n20<\/output>\n100 - 60<\/gadget>\n40<\/output>\n20 \/ 40<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50<\/result>","index":1894} +{"problem":"in what time will a train 180 m long cross an electric pole , it its speed be 144 km \/ hr ?","rationale":"\"speed = 144 * 5 \/ 18 = 40 m \/ sec time taken = 180 \/ 40 = 4.5 sec . answer : d\"","correct":"d","options":{"a":"2.5 sec ","b":"1.9 sec ","c":"8.9 sec ","d":"4.5 sec","e":"2.9 sec"},"options_float":{"a":2.5,"b":1.9,"c":8.9,"d":4.5,"e":2.9},"annotated_formula":"divide(180, 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>\n180 \/ 40<\/gadget>\n9\/2 = around 4.5<\/output>\n9\/2 = around 4.5<\/result>","index":1895} +{"problem":"the average monthly salary of 25 employees in an organisation is rs . 2500 . if the manager ' s salary is added , then the average salary increases by rs . 200 . what is the manager ' s monthly salary ?","rationale":"\"manager ' s monthly salary = rs . ( 2700 * 26 - 2500 * 25 ) = rs . 7700 answer : a\"","correct":"a","options":{"a":"rs . 7700 ","b":"rs . 3618 ","c":"rs . 3600 ","d":"rs . 3619","e":"rs . 3610"},"options_float":{"a":7700.0,"b":3618.0,"c":3600.0,"d":3619.0,"e":3610.0},"annotated_formula":"subtract(multiply(add(2500, 200), add(25, const_1)), multiply(2500, 25))","linear_formula":"add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"2_500 + 200<\/gadget>\n2_700<\/output>\n25 + 1<\/gadget>\n26<\/output>\n2_700 * 26<\/gadget>\n70_200<\/output>\n2_500 * 25<\/gadget>\n62_500<\/output>\n70_200 - 62_500<\/gadget>\n7_700<\/output>\n7_700<\/result>","index":1897} +{"problem":"we had $ 1400 left after spending 30 % of the money that we took for shopping . how much money did we start with ?","rationale":"\"let x be the amount of money we started with . 0.7 x = 1400 x = 2000 the answer is a .\"","correct":"a","options":{"a":"$ 2000 ","b":"$ 2100 ","c":"$ 2200 ","d":"$ 2300","e":"$ 2400"},"options_float":{"a":2000.0,"b":2100.0,"c":2200.0,"d":2300.0,"e":2400.0},"annotated_formula":"divide(1400, subtract(const_1, divide(30, const_100)))","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 - (3\/10)<\/gadget>\n7\/10 = around 0.7<\/output>\n1_400 \/ (7\/10)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":1899} +{"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 94 % 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 94 % water and 6 % of non - water , so now 1 pound of non - water composes 6 % of cucucmbers , which means that the new weight of cucumbers is 1 \/ 0.06 = 16 pounds . answer : b .\"","correct":"b","options":{"a":"2 ","b":"16 ","c":"92 ","d":"96","e":"98"},"options_float":{"a":2.0,"b":16.0,"c":92.0,"d":96.0,"e":98.0},"annotated_formula":"multiply(divide(subtract(100, 99), subtract(100, 94)), 100)","linear_formula":"subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|","chain":"100 - 99<\/gadget>\n1<\/output>\n100 - 94<\/gadget>\n6<\/output>\n1 \/ 6<\/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":1900} +{"problem":"in a party every person shakes hands with every other person . if there are 105 hands shakes , find the number of person in the party .","rationale":"\"let n be the number of persons in the party . number of hands shake = 105 ; total number of hands shake is given by nc 2 . now , according to the question , nc 2 = 105 ; or , n ! \/ [ 2 ! * ( n - 2 ) ! ] = 105 ; or , n * ( n - 1 ) \/ 2 = 105 ; or , n 2 - n = 210 ; or , n 2 - n - 210 = 0 ; or , n = 15 , - 14 ; but , we can not take negative value of n . so , n = 15 i . e . number of persons in the party = 15 . option d\"","correct":"d","options":{"a":"14 ","b":"12 ","c":"13 ","d":"15","e":"16"},"options_float":{"a":14.0,"b":12.0,"c":13.0,"d":15.0,"e":16.0},"annotated_formula":"divide(add(sqrt(add(multiply(multiply(105, 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":"105 * 2<\/gadget>\n210<\/output>\n210 * 4<\/gadget>\n840<\/output>\n840 + 1<\/gadget>\n841<\/output>\n841 ** (1\/2)<\/gadget>\n29<\/output>\n29 + 1<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15<\/result>","index":1902} +{"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 5 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 + 5 x = 20.5 minutes answer : a\"","correct":"a","options":{"a":"20.5 ","b":"15 ","c":"25.5 ","d":"30","e":"60"},"options_float":{"a":20.5,"b":15.0,"c":25.5,"d":30.0,"e":60.0},"annotated_formula":"subtract(subtract(divide(const_60, const_2), 5), 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 - 5<\/gadget>\n25<\/output>\n20 \/ 1.5<\/gadget>\n13.333333<\/output>\n60 \/ 13.333333<\/gadget>\n4.5<\/output>\n25 - 4.5<\/gadget>\n20.5<\/output>\n20.5<\/result>","index":1904} +{"problem":"a lawn is in the form of a rectangle having its sides in the ratio 2 : 3 . the area of the lawn is ( 1 \/ 6 ) hectares . find the length and breadth of the lawn .","rationale":"let length = 2 x meters and breadth = 3 x meter . now , area = ( 1 \/ 6 ) x 1000 m 2 = 5000 \/ 3 m 2 so , 2 x * 3 x = 5000 \/ 3 < = > x 2 = 2500 \/ 9 < = > x = 50 \/ 3 therefore length = 2 x = ( 100 \/ 3 ) m = 33 ( 1 \/ 3 ) m and breadth = 3 x = 3 ( 50 \/ 3 ) m = 50 m . answer is a","correct":"a","options":{"a":"50 ","b":"30 ","c":"20 ","d":"40","e":"10"},"options_float":{"a":50.0,"b":30.0,"c":20.0,"d":40.0,"e":10.0},"annotated_formula":"divide(multiply(divide(multiply(const_10, const_1000), 6), 3), const_100)","linear_formula":"multiply(const_10,const_1000)|divide(#0,n3)|multiply(n1,#1)|divide(#2,const_100)","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 6<\/gadget>\n5_000\/3 = around 1_666.666667<\/output>\n(5_000\/3) * 3<\/gadget>\n5_000<\/output>\n5_000 \/ 100<\/gadget>\n50<\/output>\n50<\/result>","index":1907} +{"problem":"12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 10 hours a day , the number of men required ?","rationale":"\"that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 10 × 8 ? ( i . e . no . of men required ) = 12 × 8 × 10 \/ 10 × 8 = 12 days b )\"","correct":"b","options":{"a":"10 days ","b":"12 days ","c":"13 days ","d":"11 days","e":"9 days"},"options_float":{"a":10.0,"b":12.0,"c":13.0,"d":11.0,"e":9.0},"annotated_formula":"divide(multiply(multiply(12, 10), 8), multiply(8, 10))","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 * 10<\/gadget>\n80<\/output>\n960 \/ 80<\/gadget>\n12<\/output>\n12<\/result>","index":1908} +{"problem":"there are 240 doctors and nurses at a hospital . if the ratio of doctors to nurses is 3 to 7 , how many nurses are at the hospital ?","rationale":"\"the number of nurses at the hospital is ( 7 \/ 10 ) * 240 = 168 . the answer is a .\"","correct":"a","options":{"a":"168 ","b":"172 ","c":"176 ","d":"180","e":"184"},"options_float":{"a":168.0,"b":172.0,"c":176.0,"d":180.0,"e":184.0},"annotated_formula":"multiply(divide(240, add(3, 7)), 7)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"3 + 7<\/gadget>\n10<\/output>\n240 \/ 10<\/gadget>\n24<\/output>\n24 * 7<\/gadget>\n168<\/output>\n168<\/result>","index":1909} +{"problem":"what annual payment will discharge a debt of rs . 1125 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 = 1125 820 x + 1125 * 441 x = 605.03 so , value of each installment = rs . 605.03 answer : option d\"","correct":"d","options":{"a":"993.2 ","b":"551.25 ","c":"534.33 ","d":"605.03","e":"646.33"},"options_float":{"a":993.2,"b":551.25,"c":534.33,"d":605.03,"e":646.33},"annotated_formula":"divide(multiply(power(add(divide(5, const_100), const_1), 2), 1125), 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_125<\/gadget>\n19_845\/16 = around 1_240.3125<\/output>\n(19_845\/16) \/ 2<\/gadget>\n19_845\/32 = around 620.15625<\/output>\n19_845\/32 = around 620.15625<\/result>","index":1910} +{"problem":"if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 2 ¤ √ 2 =","rationale":"\"x = √ 2 and y also = √ 2 applying the function ( √ 2 + √ 2 ) ^ 2 - ( √ 2 - √ 2 ) ^ 2 = ( 2 √ 2 ) ^ 2 - 0 = 4 x 2 = 8 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = √ 2 and y = √ 2 you get the answer 8 . answer b\"","correct":"b","options":{"a":"0 ","b":"8 ","c":"10 ","d":"15","e":"20"},"options_float":{"a":0.0,"b":8.0,"c":10.0,"d":15.0,"e":20.0},"annotated_formula":"power(add(sqrt(2), sqrt(2)), 2)","linear_formula":"sqrt(n2)|add(#0,#0)|power(#1,n0)|","chain":"2 ** (1\/2)<\/gadget>\nsqrt(2) = around 1.414214<\/output>\n(sqrt(2)) + (sqrt(2))<\/gadget>\n2*sqrt(2) = around 2.828427<\/output>\n(2*sqrt(2)) ** 2<\/gadget>\n8<\/output>\n8<\/result>","index":1912} +{"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 4 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 * 4 = > 1 \/ 12 working alone taimour can do the job in 1 = 1 \/ 12 * hours = > 12 answer c","correct":"c","options":{"a":"6 hours ","b":"8 hours ","c":"12 hours ","d":"24 hours","e":"32 hours"},"options_float":{"a":6.0,"b":8.0,"c":12.0,"d":24.0,"e":32.0},"annotated_formula":"multiply(4, const_3)","linear_formula":"multiply(n1,const_3)","chain":"4 * 3<\/gadget>\n12<\/output>\n12<\/result>","index":1913} +{"problem":"find √ ? \/ 20 = 4 ?","rationale":"\"answer let √ n \/ 20 = 4 then √ n = 20 x 4 = 80 ∴ n = 80 x 80 = 6400 . correct option : b\"","correct":"b","options":{"a":"76 ","b":"6400 ","c":"304 ","d":"1296","e":"none"},"options_float":{"a":76.0,"b":6400.0,"c":304.0,"d":1296.0,"e":null},"annotated_formula":"power(multiply(4, 20), const_2)","linear_formula":"multiply(n0,n1)|power(#0,const_2)|","chain":"4 * 20<\/gadget>\n80<\/output>\n80 ** 2<\/gadget>\n6_400<\/output>\n6_400<\/result>","index":1916} +{"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 45 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 = 45 \/ 3 = 15 kmph answer : d\"","correct":"d","options":{"a":"12 kmph ","b":"13 kmph ","c":"14 kmph ","d":"15 kmph","e":"16 kmph"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":16.0},"annotated_formula":"subtract(45, divide(multiply(45, const_2), const_3))","linear_formula":"multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)|","chain":"45 * 2<\/gadget>\n90<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n45 - 30<\/gadget>\n15<\/output>\n15<\/result>","index":1917} +{"problem":"a box contains 14 pairs of shoes ( 28 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 28 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 \/ 27 ( as after taking one at random there are 27 shoes left and only one is the pair of the first one ) = 1 \/ 27 answer : d\"","correct":"d","options":{"a":"1 \/ 190 ","b":"1 \/ 20 ","c":"1 \/ 19 ","d":"1 \/ 27","e":"1 \/ 9"},"options_float":{"a":0.0052631579,"b":0.05,"c":0.0526315789,"d":0.037037037,"e":0.1111111111},"annotated_formula":"divide(const_1, subtract(28, const_1))","linear_formula":"subtract(n1,const_1)|divide(const_1,#0)|","chain":"28 - 1<\/gadget>\n27<\/output>\n1 \/ 27<\/gadget>\n1\/27 = around 0.037037<\/output>\n1\/27 = around 0.037037<\/result>","index":1918} +{"problem":"there are 1000 students in a school and among them 10 % of them attends chess class . 10 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ?","rationale":"\"10 % of 1000 gives 100 . so 100 attends chess and 10 % of 100 gives 10 . so 10 enrolled for swimming answer : b\"","correct":"b","options":{"a":"1 ","b":"10 ","c":"100 ","d":"50","e":"20"},"options_float":{"a":1.0,"b":10.0,"c":100.0,"d":50.0,"e":20.0},"annotated_formula":"divide(multiply(divide(multiply(10, 1000), const_100), 10), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)|","chain":"10 * 1_000<\/gadget>\n10_000<\/output>\n10_000 \/ 100<\/gadget>\n100<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n1_000 \/ 100<\/gadget>\n10<\/output>\n10<\/result>","index":1920} +{"problem":"a furniture dealer purchased a desk for $ 150 and then set the selling price equal to the purchase price plus a markup that was 20 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ?","rationale":"anyway , in this question , there is no discount but the mark up is given as 40 % of the selling price . so it is not 20 % of $ 150 but instead , 20 % of selling price which is obtained by adding mark up to $ 150 . so if selling price is s , 150 + 20 % of s = s s = 150 \/ 0.8 profit = 150 \/ 0.8 - 150 = which is calculated on cost price in % terms . so 37.5 \/ 150 * 100 = 25 % is profit . e","correct":"e","options":{"a":"45 % ","b":"40 % ","c":"35 % ","d":"30 %","e":"25 %"},"options_float":{"a":45.0,"b":40.0,"c":35.0,"d":30.0,"e":25.0},"annotated_formula":"divide(multiply(subtract(divide(150, subtract(const_1, divide(20, const_100))), 150), const_100), 150)","linear_formula":"divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n150 \/ (4\/5)<\/gadget>\n375\/2 = around 187.5<\/output>\n(375\/2) - 150<\/gadget>\n75\/2 = around 37.5<\/output>\n(75\/2) * 100<\/gadget>\n3_750<\/output>\n3_750 \/ 150<\/gadget>\n25<\/output>\n25<\/result>","index":1921} +{"problem":"a and b start walking towards each other at 4 pm at speed of 2 kmph and 3 kmph . they were initially 15 km apart . at what time do they meet ?","rationale":"time of meeting = distance \/ relative speed = 15 \/ 3 + 2 = 15 \/ 5 = 3 hrs after 4 pm = 7 pm answer is c","correct":"c","options":{"a":"8 pm ","b":"6 pm ","c":"7 pm ","d":"10 pm","e":"5 pm"},"options_float":{"a":8.0,"b":6.0,"c":7.0,"d":10.0,"e":5.0},"annotated_formula":"add(4, divide(15, add(2, 3)))","linear_formula":"add(n1,n2)|divide(n3,#0)|add(n0,#1)","chain":"2 + 3<\/gadget>\n5<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\n4 + 3<\/gadget>\n7<\/output>\n7<\/result>","index":1922} +{"problem":"in a certain parallelogram the degree measure of one angle exceeds that of the other by 20 what is the degree measure of the smaller angle ?","rationale":"\"in a parallelogram opposite angles are equal and the angles at each side are supplementary to each other ( supplementary angles are two angles that add up to 180 ° ) . given : x + ( x + 20 ) = 180 - - > x = 80 . answer : b .\"","correct":"b","options":{"a":"75 ","b":"80 ","c":"85 ","d":"90","e":"95"},"options_float":{"a":75.0,"b":80.0,"c":85.0,"d":90.0,"e":95.0},"annotated_formula":"divide(subtract(divide(const_3600, const_10), multiply(20, const_2)), const_4)","linear_formula":"divide(const_3600,const_10)|multiply(n0,const_2)|subtract(#0,#1)|divide(#2,const_4)|","chain":"3_600 \/ 10<\/gadget>\n360<\/output>\n20 * 2<\/gadget>\n40<\/output>\n360 - 40<\/gadget>\n320<\/output>\n320 \/ 4<\/gadget>\n80<\/output>\n80<\/result>","index":1924} +{"problem":"3 years ago , the average age of a , b and c was 27 years and that of b and c 5 years ago was 20 years . a ’ s present age is :","rationale":"explanation : sum of the present ages of a , b and c = ( 27 × 3 + 3 × 3 ) years = 90 years . sum of the present ages of b and c = ( 20 × 2 + 5 × 2 ) years = 50 years . a ' s present age = 90 – 50 = 40 years . answer : c","correct":"c","options":{"a":"22 ","b":"88 ","c":"40 ","d":"87","e":"17"},"options_float":{"a":22.0,"b":88.0,"c":40.0,"d":87.0,"e":17.0},"annotated_formula":"subtract(add(multiply(27, 3), multiply(3, 3)), add(multiply(20, const_2), multiply(5, const_2)))","linear_formula":"multiply(n0,n1)|multiply(n0,n0)|multiply(n3,const_2)|multiply(n2,const_2)|add(#0,#1)|add(#2,#3)|subtract(#4,#5)","chain":"27 * 3<\/gadget>\n81<\/output>\n3 * 3<\/gadget>\n9<\/output>\n81 + 9<\/gadget>\n90<\/output>\n20 * 2<\/gadget>\n40<\/output>\n5 * 2<\/gadget>\n10<\/output>\n40 + 10<\/gadget>\n50<\/output>\n90 - 50<\/gadget>\n40<\/output>\n40<\/result>","index":1925} +{"problem":"if the perimeter of a rectangular garden is 600 m , its length when its breadth is 120 m is ?","rationale":"\"2 ( l + 120 ) = 600 = > l = 180 m answer : b\"","correct":"b","options":{"a":"286 m ","b":"180 m ","c":"200 m ","d":"166 m","e":"187 m"},"options_float":{"a":286.0,"b":180.0,"c":200.0,"d":166.0,"e":187.0},"annotated_formula":"subtract(divide(600, const_2), 120)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"600 \/ 2<\/gadget>\n300<\/output>\n300 - 120<\/gadget>\n180<\/output>\n180<\/result>","index":1928} +{"problem":"the average of 5 consecutive even numbers a , b , c , d and e is 20 . what percent of e is d ?","rationale":"explanation : in such a case the middle number ( c ) is the average ∴ c = 20 and d = 22 and e = 24 required percentage = 22 \/ 24 x 100 = 91.7 answer : option b","correct":"b","options":{"a":"90.1 ","b":"91.7 ","c":"97.1 ","d":"101.1","e":"107.1"},"options_float":{"a":90.1,"b":91.7,"c":97.1,"d":101.1,"e":107.1},"annotated_formula":"multiply(divide(add(add(add(subtract(20, divide(add(add(add(const_2, multiply(const_2, const_2)), multiply(const_2, const_3)), multiply(const_2, const_4)), 5)), const_2), const_2), const_2), add(add(add(add(subtract(20, divide(add(add(add(const_2, multiply(const_2, const_2)), multiply(const_2, const_3)), multiply(const_2, const_4)), 5)), const_2), const_2), const_2), const_2)), const_100)","linear_formula":"multiply(const_2,const_2)|multiply(const_2,const_3)|multiply(const_2,const_4)|add(#0,const_2)|add(#3,#1)|add(#4,#2)|divide(#5,n0)|subtract(n1,#6)|add(#7,const_2)|add(#8,const_2)|add(#9,const_2)|add(#10,const_2)|divide(#10,#11)|multiply(#12,const_100)","chain":"2 * 2<\/gadget>\n4<\/output>\n2 + 4<\/gadget>\n6<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6 + 6<\/gadget>\n12<\/output>\n2 * 4<\/gadget>\n8<\/output>\n12 + 8<\/gadget>\n20<\/output>\n20 \/ 5<\/gadget>\n4<\/output>\n20 - 4<\/gadget>\n16<\/output>\n16 + 2<\/gadget>\n18<\/output>\n18 + 2<\/gadget>\n20<\/output>\n20 + 2<\/gadget>\n22<\/output>\n22 + 2<\/gadget>\n24<\/output>\n22 \/ 24<\/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":1930} +{"problem":"the length of rectangle is thrice its breadth and its perimeter is 72 m , find the area of the rectangle ?","rationale":"\"2 ( 3 x + x ) = 72 l = 27 b = 9 lb = 27 * 9 = 243 answer : c\"","correct":"c","options":{"a":"432 ","b":"212 ","c":"243 ","d":"992","e":"212"},"options_float":{"a":432.0,"b":212.0,"c":243.0,"d":992.0,"e":212.0},"annotated_formula":"multiply(multiply(divide(72, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(72, 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>\n72 \/ 8<\/gadget>\n9<\/output>\n9 * 3<\/gadget>\n27<\/output>\n27 * 9<\/gadget>\n243<\/output>\n243<\/result>","index":1931} +{"problem":"there are 5 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 = 5 and for second = 4 so total methods = 5 * 4 = 20 answer : b\"","correct":"b","options":{"a":"19 ","b":"20 ","c":"30 ","d":"32","e":"25"},"options_float":{"a":19.0,"b":20.0,"c":30.0,"d":32.0,"e":25.0},"annotated_formula":"add(choose(5, 2), choose(5, 2))","linear_formula":"choose(n0,n1)|add(#0,#0)|","chain":"binomial(5, 2)<\/gadget>\n10<\/output>\n10 + 10<\/gadget>\n20<\/output>\n20<\/result>","index":1932} +{"problem":"if m is a positive integer and m ^ 2 is divisible by 36 , 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 12 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is d but if m = 48 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":"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(36, const_3), const_3))","linear_formula":"divide(n1,const_3)|divide(#0,const_3)|multiply(#1,const_3)|","chain":"36 \/ 3<\/gadget>\n12<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12<\/result>","index":1933} +{"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 $ 0.60 per glass on the first day , what was the price per f 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 * 0.6 = 3 * x - - > x = $ 0.4 . answer : d .\"","correct":"d","options":{"a":"$ 015 ","b":"$ 0.20 ","c":"$ 0.30 ","d":"$ 0.40","e":"$ 0.45"},"options_float":{"a":15.0,"b":0.2,"c":0.3,"d":0.4,"e":0.45},"annotated_formula":"divide(multiply(add(const_1, const_1), 0.60), 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 * 0.6<\/gadget>\n1.2<\/output>\n1 + 2<\/gadget>\n3<\/output>\n1.2 \/ 3<\/gadget>\n0.4<\/output>\n0.4<\/result>","index":1936} +{"problem":"in what time will a train 110 m long cross an electric pole , it its speed be 128 km \/ hr ?","rationale":"\"speed = 128 * 5 \/ 18 = 36 m \/ sec time taken = 110 \/ 36 = 3.1 sec . answer : c\"","correct":"c","options":{"a":"2.5 sec ","b":"2.9 sec ","c":"3.1 sec ","d":"8.7 sec","e":"8.5 sec"},"options_float":{"a":2.5,"b":2.9,"c":3.1,"d":8.7,"e":8.5},"annotated_formula":"divide(110, multiply(128, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n128 * (5\/18)<\/gadget>\n320\/9 = around 35.555556<\/output>\n110 \/ (320\/9)<\/gadget>\n99\/32 = around 3.09375<\/output>\n99\/32 = around 3.09375<\/result>","index":1937} +{"problem":"two trains 160 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 = 160 + 160 = 320 m . required time = 320 * 9 \/ 250 = 288 \/ 25 = 11.52 sec . answer : e\"","correct":"e","options":{"a":"10.52 sec ","b":"18.8 sec ","c":"14.52 sec ","d":"10.8 sec","e":"11.52 sec"},"options_float":{"a":10.52,"b":18.8,"c":14.52,"d":10.8,"e":11.52},"annotated_formula":"divide(add(160, 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":"160 + 160<\/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":1939} +{"problem":"two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 48 kmph respectively . in how much time will they cross each other , if they are running in the same direction ?","rationale":"\"solution relative speed = ( 48 - 40 ) kmph = 8 kmph = ( 8 x 5 \/ 18 ) m \/ sec = ( 40 \/ 18 ) m \/ sec time taken = ( 350 x 18 \/ 40 ) sec = 157.5 sec . answer c\"","correct":"c","options":{"a":"72 sec ","b":"132 sec ","c":"157.5 sec ","d":"252 sec","e":"none"},"options_float":{"a":72.0,"b":132.0,"c":157.5,"d":252.0,"e":null},"annotated_formula":"multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(48, 40)))","linear_formula":"add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)|","chain":"200 + 150<\/gadget>\n350<\/output>\n350 \/ 1_000<\/gadget>\n7\/20 = around 0.35<\/output>\n48 - 40<\/gadget>\n8<\/output>\n(7\/20) \/ 8<\/gadget>\n7\/160 = around 0.04375<\/output>\n3_600 * (7\/160)<\/gadget>\n315\/2 = around 157.5<\/output>\n315\/2 = around 157.5<\/result>","index":1942} +{"problem":"a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 10 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 10 % of that is added to the weight , the weight claimed by the trader is obtained . if he charges the cost price of the weight that he claims , find his profit percentage .","rationale":"\"lets say the indicated weight is x for $ 10 but the trader actually get 1.1 x for $ 10 now he tells to the customer that its 1.1 ( 1.1 x ) = 1.21 x and charges the price accordingly i . e . 1.21 x * 10 \/ x = 12.1 so profit % = ( 12.1 - 10 ) \/ 10 = 0.21 = 21 % answer : b\"","correct":"b","options":{"a":"20 % ","b":"21 % ","c":"21.33 % ","d":"22.109 %","e":"23 %"},"options_float":{"a":20.0,"b":21.0,"c":21.33,"d":22.109,"e":23.0},"annotated_formula":"subtract(multiply(divide(add(const_100, 10), const_100), add(const_100, 10)), const_100)","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)|","chain":"100 + 10<\/gadget>\n110<\/output>\n110 \/ 100<\/gadget>\n11\/10 = around 1.1<\/output>\n(11\/10) * 110<\/gadget>\n121<\/output>\n121 - 100<\/gadget>\n21<\/output>\n21<\/result>","index":1943} +{"problem":"if the wheel is 14 cm then the number of revolutions to cover a distance of 2288 cm is ?","rationale":"\"2 * 22 \/ 7 * 14 * x = 2288 = > x = 26 answer : a\"","correct":"a","options":{"a":"26 ","b":"28 ","c":"17 ","d":"12","e":"88"},"options_float":{"a":26.0,"b":28.0,"c":17.0,"d":12.0,"e":88.0},"annotated_formula":"divide(2288, 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>\n2_288 \/ (2_198\/25)<\/gadget>\n28_600\/1_099 = around 26.023658<\/output>\n28_600\/1_099 = around 26.023658<\/result>","index":1944} +{"problem":"in the seaside 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 seaside 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(subtract(50, divide(multiply(90, 50), const_100)), const_100), 5), 50)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,const_100)|divide(#3,n2)|subtract(#4,n0)","chain":"90 * 50<\/gadget>\n4_500<\/output>\n4_500 \/ 100<\/gadget>\n45<\/output>\n50 - 45<\/gadget>\n5<\/output>\n5 * 100<\/gadget>\n500<\/output>\n500 \/ 5<\/gadget>\n100<\/output>\n100 - 50<\/gadget>\n50<\/output>\n50<\/result>","index":1946} +{"problem":"the average price of 3 items of furniture is rs . 15000 . if their prices are in the ratio 2 : 4 : 8 , the price of the cheapest item is ?","rationale":"let their prices be 3 x , 5 x and 7 x . then , 2 x + 6 x + 8 x = ( 15000 * 3 ) or x = 2812.5 . cost of cheapest item = 2 x = rs . 5625 . answer : c","correct":"c","options":{"a":"2379 ","b":"2889 ","c":"5625 ","d":"9000","e":"28311"},"options_float":{"a":2379.0,"b":2889.0,"c":5625.0,"d":9000.0,"e":28311.0},"annotated_formula":"divide(multiply(3, 15000), 8)","linear_formula":"multiply(n0,n1)|divide(#0,n4)","chain":"3 * 15_000<\/gadget>\n45_000<\/output>\n45_000 \/ 8<\/gadget>\n5_625<\/output>\n5_625<\/result>","index":1947} +{"problem":"in a group of 15 people , 8 read english , 7 read french while 3 of them read none of these two . how many of them read french and english both ?","rationale":"in the following venn diagram , f and e represent people who read french and english respectively . now , [ f + ( { f ∩ e } ) + e ] = 15 - 3 ( or ) f + e + ( f ∩ e ) = 12 . . . . . . ( 1 ) also , f + ( f ∩ e ) = 7 ; e + ( f ∩ e ) = 8 . by adding , f + e + 2 ( f ∩ e ) = 15 - - - - - - - - - - ( 2 ) by subtracting ( 1 ) from ( 2 ) , we get ( f ∩ e ) = 3 . ∴ 3 of them read both french and english . answer : b","correct":"b","options":{"a":"2 ","b":"3 ","c":"4 ","d":"7","e":"5"},"options_float":{"a":2.0,"b":3.0,"c":4.0,"d":7.0,"e":5.0},"annotated_formula":"subtract(add(8, 7), subtract(15, 3))","linear_formula":"add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)","chain":"8 + 7<\/gadget>\n15<\/output>\n15 - 3<\/gadget>\n12<\/output>\n15 - 12<\/gadget>\n3<\/output>\n3<\/result>","index":1951} +{"problem":"if a person walks at 14 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 traveled be x km . then , x \/ 10 = ( x + 20 ) \/ 14 4 x - 200 = > x = 50 km . answer : a\"","correct":"a","options":{"a":"50 km ","b":"16 km ","c":"18 km ","d":"29 km","e":"19 km"},"options_float":{"a":50.0,"b":16.0,"c":18.0,"d":29.0,"e":19.0},"annotated_formula":"multiply(10, divide(20, subtract(14, 10)))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|","chain":"14 - 10<\/gadget>\n4<\/output>\n20 \/ 4<\/gadget>\n5<\/output>\n10 * 5<\/gadget>\n50<\/output>\n50<\/result>","index":1952} +{"problem":"sum of 36 odd numbers is ?","rationale":"\"sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 36 odd numbers = 36 ^ 2 = 1296 answer : c\"","correct":"c","options":{"a":"1294 ","b":"1295 ","c":"1296 ","d":"1297","e":"1298"},"options_float":{"a":1294.0,"b":1295.0,"c":1296.0,"d":1297.0,"e":1298.0},"annotated_formula":"multiply(multiply(36, const_2), divide(36, const_2))","linear_formula":"divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1)|","chain":"36 * 2<\/gadget>\n72<\/output>\n36 \/ 2<\/gadget>\n18<\/output>\n72 * 18<\/gadget>\n1_296<\/output>\n1_296<\/result>","index":1953} +{"problem":"15 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ?","rationale":"\"though 2 * ( 15 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 14 , so he ' ll play 28 matches . similarly , second guy has to play with the rest of 13 as his 2 games with the first guy are already played . so he plays 26 matches . this continues like this and the total matches are 28 + 26 + 24 . . . + 2 28 + 26 + . . . + 2 = 2 ( 14 + 13 + . . . + 1 ) = 2 ( ( 14 * 15 ) \/ 2 ) = 14 * 15 = 210 . answer : c\"","correct":"c","options":{"a":"190 ","b":"200 ","c":"210 ","d":"220","e":"225"},"options_float":{"a":190.0,"b":200.0,"c":210.0,"d":220.0,"e":225.0},"annotated_formula":"multiply(15, subtract(15, const_1))","linear_formula":"subtract(n0,const_1)|multiply(n0,#0)|","chain":"15 - 1<\/gadget>\n14<\/output>\n15 * 14<\/gadget>\n210<\/output>\n210<\/result>","index":1954} +{"problem":"the height of the wall is 6 times its width and length of the wall is 8 times its height . if the volume of the wall be 36000 cu . m . its width is","rationale":"\"explanation : let width = x then , height = 6 x and length = 48 x 48 x ã — 6 x ã — x = 36000 x = 5 answer : b\"","correct":"b","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(36000, multiply(multiply(6, 8), 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 * 8<\/gadget>\n48<\/output>\n48 * 6<\/gadget>\n288<\/output>\n36_000 \/ 288<\/gadget>\n125<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n125 ** (1\/3)<\/gadget>\n5<\/output>\n5<\/result>","index":1955} +{"problem":"one - third of 600 is what percent of 120 ?","rationale":"answer let one - third of 600 is n % of 120 . ∵ 600 \/ 3 = ( n x 120 ) \/ 100 ∴ n = ( 200 x 100 ) \/ 120 = 166.6 correct option : e","correct":"e","options":{"a":"313.2 ","b":"30.1 ","c":"12.24 ","d":"none of these","e":"166.6"},"options_float":{"a":313.2,"b":30.1,"c":12.24,"d":null,"e":166.6},"annotated_formula":"multiply(const_100, divide(divide(600, const_3), 120))","linear_formula":"divide(n0,const_3)|divide(#0,n1)|multiply(#1,const_100)","chain":"600 \/ 3<\/gadget>\n200<\/output>\n200 \/ 120<\/gadget>\n5\/3 = around 1.666667<\/output>\n100 * (5\/3)<\/gadget>\n500\/3 = around 166.666667<\/output>\n500\/3 = around 166.666667<\/result>","index":1956} +{"problem":"how many diagonals does a 58 - sided convex polygon have ?","rationale":"\"a 58 - sided convex polygon has 58 vertices . if we examine a single vertex , we can see that we can connect it with 55 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 58 vertices can be connected with 55 vertices to create a diagonal then the total number of diagonals would be ( 58 ) ( 55 ) = 3190 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3190 by 2 to get 1595 . the answer is c .\"","correct":"c","options":{"a":"870 ","b":"1125 ","c":"1595 ","d":"2560","e":"3190"},"options_float":{"a":870.0,"b":1125.0,"c":1595.0,"d":2560.0,"e":3190.0},"annotated_formula":"divide(factorial(58), multiply(factorial(subtract(58, 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(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>\n58 - 2<\/gadget>\n56<\/output>\nfactorial(56)<\/gadget>\n710_998_587_804_863_451_854_045_647_463_724_949_736_497_978_881_168_458_687_447_040_000_000_000_000<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n710_998_587_804_863_451_854_045_647_463_724_949_736_497_978_881_168_458_687_447_040_000_000_000_000 * 2<\/gadget>\n1_421_997_175_609_726_903_708_091_294_927_449_899_472_995_957_762_336_917_374_894_080_000_000_000_000<\/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 \/ 1_421_997_175_609_726_903_708_091_294_927_449_899_472_995_957_762_336_917_374_894_080_000_000_000_000<\/gadget>\n1_653<\/output>\n1_653<\/result>","index":1957} +{"problem":"a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 545 pounds after processing , how many pounds did it weigh before processing ?","rationale":"let weight of side of beef before processing = x ( 65 \/ 100 ) * x = 545 = > x = ( 545 * 100 ) \/ 65 = 838 answer d","correct":"d","options":{"a":"191 ","b":"355 ","c":"737 ","d":"838","e":"1,560"},"options_float":{"a":191.0,"b":355.0,"c":737.0,"d":838.0,"e":1560.0},"annotated_formula":"divide(multiply(545, const_100), subtract(const_100, 35))","linear_formula":"multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)","chain":"545 * 100<\/gadget>\n54_500<\/output>\n100 - 35<\/gadget>\n65<\/output>\n54_500 \/ 65<\/gadget>\n10_900\/13 = around 838.461538<\/output>\n10_900\/13 = around 838.461538<\/result>","index":1958} +{"problem":"miller street begins at baker street and runs directly east for 4.5 kilometers until it ends when it meets turner street . miller street is intersected every 250 meters by a perpendicular street , and each of those streets other than baker street and turner street is given a number beginning at 1 st street ( one block east of baker street ) and continuing consecutively ( 2 nd street , 3 rd street , etc . . . ) until the highest - numbered street one block west of turner street . what is the highest - numbered street that intersects miller street ?","rationale":"4.5 km \/ 250 m = 18 . however , the street at the 4.5 - km mark is not 18 th street ; it is turner street . therefore , the highest numbered street is 17 th street . the answer is c .","correct":"c","options":{"a":"15 th ","b":"16 th ","c":"17 th ","d":"18 th","e":"19 th"},"options_float":{"a":15.0,"b":16.0,"c":17.0,"d":18.0,"e":19.0},"annotated_formula":"subtract(divide(4.5, divide(250, const_1000)), 1)","linear_formula":"divide(n1,const_1000)|divide(n0,#0)|subtract(#1,n2)","chain":"250 \/ 1_000<\/gadget>\n1\/4 = around 0.25<\/output>\n4.5 \/ (1\/4)<\/gadget>\n18<\/output>\n18 - 1<\/gadget>\n17<\/output>\n17<\/result>","index":1959} +{"problem":"the pressure someone experiences as he or she dives deeper and deeper in the ocean increases linearly . on the surface , the pressure is close to 15 pounds per square inch . 33 feet below the surface , the pressure is 30 pounds . if 25000 pounds per sq inch can crush your bones , what depth is extremely dangerous for humans ?","rationale":"solution : first , model the pressure ( p ) in terms of depth ( d ) with a linear equation . we will find the equation p = md + b use ( 0 , 15 ) and ( 33 , 30 ) to find m m = 30 - 15 \/ 33 - 0 m = 15 \/ 33 = 0.45 p = 0.45 d + b use ( 0 , 15 ) to find b 15 = 0.45 × 0 + b 15 = b p = 0.45 d + 15 25000 = 0.45 d + 15 25000 - 15 = 0.45 d + 15 - 15 24985 = 0.45 d d = 24985 \/ 0.45 = 55522 feet answer a","correct":"a","options":{"a":"55522 feet ","b":"45522 feet ","c":"35522 feet ","d":"25522 feet","e":"none"},"options_float":{"a":55522.0,"b":45522.0,"c":35522.0,"d":25522.0,"e":null},"annotated_formula":"divide(divide(subtract(25000, 15), divide(subtract(30, 15), multiply(33, multiply(const_4, const_3)))), multiply(const_4, const_3))","linear_formula":"multiply(const_3,const_4)|subtract(n3,n0)|subtract(n2,n0)|multiply(n1,#0)|divide(#2,#3)|divide(#1,#4)|divide(#5,#0)","chain":"25_000 - 15<\/gadget>\n24_985<\/output>\n30 - 15<\/gadget>\n15<\/output>\n4 * 3<\/gadget>\n12<\/output>\n33 * 12<\/gadget>\n396<\/output>\n15 \/ 396<\/gadget>\n5\/132 = around 0.037879<\/output>\n24_985 \/ (5\/132)<\/gadget>\n659_604<\/output>\n659_604 \/ 12<\/gadget>\n54_967<\/output>\n54_967<\/result>","index":1960} +{"problem":"in a class total 34 students , 16 are have a brother , 15 are have sisters , 9 students do n ' t have either brothers or sisters . find the number of students having both brother and sisters .","rationale":"total number of students = 34 let a be the number of students have a brothers . let b be the number of students have a sisters . aub = number of students have either brothers or sisters = 34 - 9 = 25 n ( aub ) = n ( a ) + n ( b ) - n ( anb ) 25 = 16 + 15 - n ( anb ) n ( anb ) = 31 - 25 n ( anb ) = 6 the number of students having both brother and sisters = 6 answer : 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":"subtract(add(16, 15), subtract(34, 9))","linear_formula":"add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)","chain":"16 + 15<\/gadget>\n31<\/output>\n34 - 9<\/gadget>\n25<\/output>\n31 - 25<\/gadget>\n6<\/output>\n6<\/result>","index":1961} +{"problem":"find the area of a parallelogram with base 26 cm and height 12 cm ?","rationale":"\"area of a parallelogram = base * height = 26 * 12 = 312 cm 2 answer : e\"","correct":"e","options":{"a":"281 ","b":"284 ","c":"288 ","d":"255","e":"312"},"options_float":{"a":281.0,"b":284.0,"c":288.0,"d":255.0,"e":312.0},"annotated_formula":"multiply(26, 12)","linear_formula":"multiply(n0,n1)|","chain":"26 * 12<\/gadget>\n312<\/output>\n312<\/result>","index":1962} +{"problem":"a can do a piece of work in 90 days & y can do it in 80 days . they began working together but a leaves after some days and then b completed the remaining work in 46 days . the number of days after which a left the work was ?","rationale":"a and b did the work for some days . after that , b completed the remaining work in 46 days . in 46 days , word done by b = 1 80 x 46 = 23 45 remaining work = 1 – 23 \/ 45 = 17 \/ 45 remaining work is done by both a and b together = 17 \/ 45 x 90 x 80 \/ 90 + 80 = 16 days c","correct":"c","options":{"a":"12 days ","b":"18 days ","c":"16 days ","d":"19 days","e":"21 days"},"options_float":{"a":12.0,"b":18.0,"c":16.0,"d":19.0,"e":21.0},"annotated_formula":"subtract(divide(subtract(const_1, multiply(divide(const_1, 80), 46)), add(divide(const_1, 80), divide(const_1, 90))), const_2)","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|subtract(#5,const_2)","chain":"1 \/ 80<\/gadget>\n1\/80 = around 0.0125<\/output>\n(1\/80) * 46<\/gadget>\n23\/40 = around 0.575<\/output>\n1 - (23\/40)<\/gadget>\n17\/40 = around 0.425<\/output>\n1 \/ 90<\/gadget>\n1\/90 = around 0.011111<\/output>\n(1\/80) + (1\/90)<\/gadget>\n17\/720 = around 0.023611<\/output>\n(17\/40) \/ (17\/720)<\/gadget>\n18<\/output>\n18 - 2<\/gadget>\n16<\/output>\n16<\/result>","index":1964} +{"problem":"by selling 10 pencils for a rupee a man loses 25 % . how many for a rupee should he sell in order to gain 25 % ?","rationale":"\"75 % - - - 10 125 % - - - ? 75 \/ 125 * 10 = 6 answer : a\"","correct":"a","options":{"a":"6 ","b":"9 ","c":"10 ","d":"89","e":"81"},"options_float":{"a":6.0,"b":9.0,"c":10.0,"d":89.0,"e":81.0},"annotated_formula":"multiply(divide(const_1, multiply(add(const_100, 25), divide(const_1, subtract(const_100, 25)))), 10)","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 + 25<\/gadget>\n125<\/output>\n100 - 25<\/gadget>\n75<\/output>\n1 \/ 75<\/gadget>\n1\/75 = around 0.013333<\/output>\n125 * (1\/75)<\/gadget>\n5\/3 = around 1.666667<\/output>\n1 \/ (5\/3)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 10<\/gadget>\n6<\/output>\n6<\/result>","index":1966} +{"problem":"1 \/ 0.05 is equal to","rationale":"\"explanation : 1 \/ 0.05 = ( 1 * 100 ) \/ 5 = 100 \/ 5 = 20 option d\"","correct":"d","options":{"a":"25.5 ","b":"2.5 ","c":"25 ","d":"20","e":"none of these"},"options_float":{"a":25.5,"b":2.5,"c":25.0,"d":20.0,"e":null},"annotated_formula":"divide(1, 0.05)","linear_formula":"divide(n0,n1)|","chain":"1 \/ 0.05<\/gadget>\n20<\/output>\n20<\/result>","index":1967} +{"problem":"amit and ananthu can do a work in 15 days and 60 days respectively . amit started the work and left after 3 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 \/ 15 amit ’ s 3 day ’ s work = 1 \/ 15 * 3 = 1 \/ 5 work left = 1 - 1 \/ 5 = 4 \/ 5 ananthu ’ s one day ’ s work = 1 \/ 60 ananthu can do work in = 4 \/ 5 * 60 = 48 days so total days = 48 + 3 = 51 days answer : e\"","correct":"e","options":{"a":"38 days ","b":"40 days ","c":"43 days ","d":"45 days","e":"51 days"},"options_float":{"a":38.0,"b":40.0,"c":43.0,"d":45.0,"e":51.0},"annotated_formula":"add(divide(subtract(const_1, multiply(inverse(15), 3)), inverse(60)), 3)","linear_formula":"inverse(n0)|inverse(n1)|multiply(#0,n2)|subtract(const_1,#2)|divide(#3,#1)|add(#4,n2)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/15) * 3<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n(4\/5) \/ (1\/60)<\/gadget>\n48<\/output>\n48 + 3<\/gadget>\n51<\/output>\n51<\/result>","index":1968} +{"problem":"a and b together can do a work in 7 days . if a alone can do it in 56 days . in how many days can b alone do it ?","rationale":"\"b 8 1 \/ 7 â € “ 1 \/ 56 = 1 \/ 8 = > 8\"","correct":"b","options":{"a":"11 ","b":"8 ","c":"21 ","d":"20","e":"25"},"options_float":{"a":11.0,"b":8.0,"c":21.0,"d":20.0,"e":25.0},"annotated_formula":"add(inverse(subtract(divide(const_1, 7), divide(const_1, 56))), 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 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ 56<\/gadget>\n1\/56 = around 0.017857<\/output>\n(1\/7) - (1\/56)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ (1\/8)<\/gadget>\n8<\/output>\n2 + 3<\/gadget>\n5<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n8 + (2\/5)<\/gadget>\n42\/5 = around 8.4<\/output>\n42\/5 = around 8.4<\/result>","index":1970} +{"problem":"a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 180 kilometers ?","rationale":"\"each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 180 kilometers ? 180 \/ 20 = 9 * 1 gallon = 9 gallons correct answer b\"","correct":"b","options":{"a":"8.5 gallons ","b":"9 gallons ","c":"6.5 gallons ","d":"5.5 gallons","e":"6 gallons"},"options_float":{"a":8.5,"b":9.0,"c":6.5,"d":5.5,"e":6.0},"annotated_formula":"divide(180, 20)","linear_formula":"divide(n1,n0)|","chain":"180 \/ 20<\/gadget>\n9<\/output>\n9<\/result>","index":1972} +{"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 680 sq . feet , how many feet of fencing will be required ?","rationale":"\"explanation : we have : l = 20 ft and lb = 680 sq . ft . so , b = 34 ft . length of fencing = ( l + 2 b ) = ( 20 + 68 ) ft = 88 ft . answer : d\"","correct":"d","options":{"a":"34 ","b":"40 ","c":"68 ","d":"88","e":"98"},"options_float":{"a":34.0,"b":40.0,"c":68.0,"d":88.0,"e":98.0},"annotated_formula":"add(multiply(divide(680, 20), const_2), 20)","linear_formula":"divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|","chain":"680 \/ 20<\/gadget>\n34<\/output>\n34 * 2<\/gadget>\n68<\/output>\n68 + 20<\/gadget>\n88<\/output>\n88<\/result>","index":1973} +{"problem":"the average salary of all the workers in a workshop is rs . 8000 . the average salary of 9 technicians is rs . 18000 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 = ( 18000 * 9 ) + 6000 ( v - 9 ) < = > 2000 v = 108000 < = > v = 54 answer : b ) 54\"","correct":"b","options":{"a":"22 ","b":"54 ","c":"88 ","d":"37","e":"29"},"options_float":{"a":22.0,"b":54.0,"c":88.0,"d":37.0,"e":29.0},"annotated_formula":"add(9, divide(multiply(9, subtract(18000, 8000)), subtract(8000, 6000)))","linear_formula":"subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)|","chain":"18_000 - 8_000<\/gadget>\n10_000<\/output>\n9 * 10_000<\/gadget>\n90_000<\/output>\n8_000 - 6_000<\/gadget>\n2_000<\/output>\n90_000 \/ 2_000<\/gadget>\n45<\/output>\n9 + 45<\/gadget>\n54<\/output>\n54<\/result>","index":1974} +{"problem":"a shopkeeper sold an book offering a discount of 5 % and earned a profit of 25 % . what would have been the percentage of profit earned if no discount was offered ?","rationale":"\"let c . p . be $ 100 . then , s . p . = $ 125 let marked price be $ x . then , 95 \/ 100 x = 125 x = 12500 \/ 95 = $ 131.6 now , s . p . = $ 131.6 , c . p . = $ 100 profit % = 31.6 % . a\"","correct":"a","options":{"a":"131.6 ","b":"120 ","c":"130 ","d":"110","e":"150"},"options_float":{"a":131.6,"b":120.0,"c":130.0,"d":110.0,"e":150.0},"annotated_formula":"multiply(const_100, divide(add(const_100, 25), subtract(const_100, 5)))","linear_formula":"add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|","chain":"100 + 25<\/gadget>\n125<\/output>\n100 - 5<\/gadget>\n95<\/output>\n125 \/ 95<\/gadget>\n25\/19 = around 1.315789<\/output>\n100 * (25\/19)<\/gadget>\n2_500\/19 = around 131.578947<\/output>\n2_500\/19 = around 131.578947<\/result>","index":1976} +{"problem":"find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 10 of money ?","rationale":"\"p = 10 ( 100 \/ 10 ) 2 = > p = 1000 answer : d\"","correct":"d","options":{"a":"1500 ","b":"1992 ","c":"9921 ","d":"1000","e":"2789"},"options_float":{"a":1500.0,"b":1992.0,"c":9921.0,"d":1000.0,"e":2789.0},"annotated_formula":"multiply(multiply(divide(10, multiply(10, 2)), const_100), multiply(10, 2))","linear_formula":"multiply(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|multiply(#2,#0)|","chain":"10 * 2<\/gadget>\n20<\/output>\n10 \/ 20<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 100<\/gadget>\n50<\/output>\n50 * 20<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":1978} +{"problem":"a room is 7 meters 68 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room .","rationale":"\"let us calculate both the length and width of the room in centimeters . length = 7 meters and 68 centimeters = 768 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 768 and 432 = 48 thus , the number of square tiles required = ( 768 x 432 ) \/ ( 48 x 48 ) = 16 x 9 = 144 answer : b\"","correct":"b","options":{"a":"107 ","b":"144 ","c":"175 ","d":"165","e":"130"},"options_float":{"a":107.0,"b":144.0,"c":175.0,"d":165.0,"e":130.0},"annotated_formula":"divide(multiply(add(multiply(7, const_100), 68), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(7, const_100), 68), add(multiply(4, const_100), 32)), gcd(add(multiply(7, const_100), 68), add(multiply(4, const_100), 32))))","linear_formula":"multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)|","chain":"7 * 100<\/gadget>\n700<\/output>\n700 + 68<\/gadget>\n768<\/output>\n4 * 100<\/gadget>\n400<\/output>\n400 + 32<\/gadget>\n432<\/output>\n768 * 432<\/gadget>\n331_776<\/output>\ngcd(768, 432)<\/gadget>\n48<\/output>\n48 * 48<\/gadget>\n2_304<\/output>\n331_776 \/ 2_304<\/gadget>\n144<\/output>\n144<\/result>","index":1980} +{"problem":"there is enough provisions for 600 men in an army camp for 25 days . if there were 300 men less , how long will the provision last ?","rationale":"\"exp : we have , m 1 d 1 = m 2 d 2 600 * 25 = 300 * d 2 d 2 = 600 * 25 \/ 300 = 50 days . answer : d\"","correct":"d","options":{"a":"35 days ","b":"40 days ","c":"45 days ","d":"50 days","e":"55 days"},"options_float":{"a":35.0,"b":40.0,"c":45.0,"d":50.0,"e":55.0},"annotated_formula":"divide(multiply(600, 25), 300)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"600 * 25<\/gadget>\n15_000<\/output>\n15_000 \/ 300<\/gadget>\n50<\/output>\n50<\/result>","index":1981} +{"problem":"two cars start from the opposite places of a main road , 140 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 : d ) 55 km","correct":"d","options":{"a":"65 ","b":"38 ","c":"20 ","d":"55","e":"21"},"options_float":{"a":65.0,"b":38.0,"c":20.0,"d":55.0,"e":21.0},"annotated_formula":"subtract(subtract(140, 35), add(25, 25))","linear_formula":"add(n1,n1)|subtract(n0,n4)|subtract(#1,#0)|","chain":"140 - 35<\/gadget>\n105<\/output>\n25 + 25<\/gadget>\n50<\/output>\n105 - 50<\/gadget>\n55<\/output>\n55<\/result>","index":1982} +{"problem":"1 ÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 1 ÷ 1 ) } ] = ?","rationale":"explanation : 1 ÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 1 ÷ 1 ) } ] = 1 ÷ [ 1 + 1 ÷ { 1 + 1 ÷ 1 } ] = 1 ÷ [ 1 + 1 ÷ { 1 + 1 } ] = 1 ÷ [ 1 + 1 ÷ 2 ] = 1 ÷ [ 1 + ( 1 \/ 2 ) ] = 1 ÷ 3 \/ 2 = 1 × 3 \/ 2 = 1 × 2 \/ 3 = 2 \/ 3 answer : option c","correct":"c","options":{"a":"5 \/ 3 ","b":"4 \/ 3 ","c":"2 \/ 3 ","d":"1 \/ 3","e":"1 \/ 5"},"options_float":{"a":1.6666666667,"b":1.3333333333,"c":0.6666666667,"d":0.3333333333,"e":0.2},"annotated_formula":"divide(1, add(1, divide(1, add(1, divide(1, 1)))))","linear_formula":"divide(n0,n0)|add(n0,#0)|divide(n0,#1)|add(n0,#2)|divide(n0,#3)","chain":"1 \/ 1<\/gadget>\n1<\/output>\n1 + 1<\/gadget>\n2<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n1 \/ (3\/2)<\/gadget>\n2\/3 = around 0.666667<\/output>\n2\/3 = around 0.666667<\/result>","index":1983} +{"problem":"john had a stock of 1000 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ?","rationale":"\"let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 1000 - n = 1000 - 402 = 598 percentage books not sold \/ total number of books = 598 \/ 1000 = 59.8 = 59.8 % correct answer a\"","correct":"a","options":{"a":"59.8 % ","b":"23.8 % ","c":"66.5 % ","d":"34.9 %","e":"43.5 %"},"options_float":{"a":59.8,"b":23.8,"c":66.5,"d":34.9,"e":43.5},"annotated_formula":"multiply(divide(subtract(1000, add(add(add(75, 50), add(64, 78)), 135)), 1000), const_100)","linear_formula":"add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)|","chain":"75 + 50<\/gadget>\n125<\/output>\n64 + 78<\/gadget>\n142<\/output>\n125 + 142<\/gadget>\n267<\/output>\n267 + 135<\/gadget>\n402<\/output>\n1_000 - 402<\/gadget>\n598<\/output>\n598 \/ 1_000<\/gadget>\n299\/500 = around 0.598<\/output>\n(299\/500) * 100<\/gadget>\n299\/5 = around 59.8<\/output>\n299\/5 = around 59.8<\/result>","index":1985} +{"problem":"a line that passes through ( – 1 , – 4 ) and ( 2 , k ) has a slope = k . what is the value of k ?","rationale":"\"slope = ( y 2 - y 1 ) \/ ( x 2 - x 1 ) = > k = ( k + 4 ) \/ ( 2 + 1 ) = > 3 k = k + 4 = > k = 2 ans d it is !\"","correct":"d","options":{"a":"3 \/ 4 ","b":"1 ","c":"4 \/ 3 ","d":"2","e":"7 \/ 2"},"options_float":{"a":0.75,"b":1.0,"c":1.3333333333,"d":2.0,"e":3.5},"annotated_formula":"divide(4, 2)","linear_formula":"divide(n1,n2)|","chain":"4 \/ 2<\/gadget>\n2<\/output>\n2<\/result>","index":1986} +{"problem":"p , q and r have $ 8000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ?","rationale":"\"c 3200 let the amount with r be $ r r = 2 \/ 3 ( total amount with p and q ) r = 2 \/ 3 ( 8000 - r ) = > 3 r = 16000 - 2 r = > 5 r = 16000 = > r = 3200 .\"","correct":"c","options":{"a":"2400 ","b":"2403 ","c":"3200 ","d":"2539","e":"1930"},"options_float":{"a":2400.0,"b":2403.0,"c":3200.0,"d":2539.0,"e":1930.0},"annotated_formula":"divide(multiply(8000, 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>\n8_000 * 4<\/gadget>\n32_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>\n32_000 \/ 10<\/gadget>\n3_200<\/output>\n3_200<\/result>","index":1987} +{"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 bil 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(1, 2))), floor(divide(16, add(divide(3, 4), 2)))), divide(8, add(divide(1, 3), 1)))","linear_formula":"divide(n1,n4)|divide(n1,n0)|divide(n4,n5)|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":1988} +{"problem":"of the 3,600 employees of company x , 16 \/ 25 are clerical . if the clerical staff were to be reduced by 1 \/ 4 , what percent of the total number of the remaining employees would then be clerical ?","rationale":"\"let ' s see , the way i did it was 16 \/ 25 are clerical out of 3600 so 2304 are clerical 2304 reduced by 1 \/ 4 is 2304 * 1 \/ 4 so it reduced 576 people , so there is 1728 clerical people left but since 576 people left , it also reduced from the total of 3600 so there are 3024 people total since 1728 clerical left \/ 3024 people total you get ( a ) 57 %\"","correct":"a","options":{"a":"57 % ","b":"22.2 % ","c":"20 % ","d":"12.5 %","e":"11.1 %"},"options_float":{"a":57.0,"b":22.2,"c":20.0,"d":12.5,"e":11.1},"annotated_formula":"multiply(divide(multiply(divide(16, 25), subtract(1, divide(1, 4))), add(multiply(divide(16, 25), subtract(1, divide(1, 4))), subtract(1, divide(16, 25)))), const_100)","linear_formula":"divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)|","chain":"16 \/ 25<\/gadget>\n16\/25 = around 0.64<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n(16\/25) * (3\/4)<\/gadget>\n12\/25 = around 0.48<\/output>\n1 - (16\/25)<\/gadget>\n9\/25 = around 0.36<\/output>\n(12\/25) + (9\/25)<\/gadget>\n21\/25 = around 0.84<\/output>\n(12\/25) \/ (21\/25)<\/gadget>\n4\/7 = around 0.571429<\/output>\n(4\/7) * 100<\/gadget>\n400\/7 = around 57.142857<\/output>\n400\/7 = around 57.142857<\/result>","index":1989} +{"problem":"the difference between two numbers is 1365 . when the larger number is divided by the smaller one , the quotient is 6 and the remainder is 15 . the smaller number is","rationale":"\"solution let the numbers be x and x ( x + 1365 ) . then , x + 1365 = 6 x + 15 ‹ = › 5 x = 1350 . ‹ = › x = 270 . answer d\"","correct":"d","options":{"a":"240 ","b":"250 ","c":"260 ","d":"270","e":"none"},"options_float":{"a":240.0,"b":250.0,"c":260.0,"d":270.0,"e":null},"annotated_formula":"divide(subtract(1365, 15), subtract(6, const_1))","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|","chain":"1_365 - 15<\/gadget>\n1_350<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_350 \/ 5<\/gadget>\n270<\/output>\n270<\/result>","index":1990} +{"problem":"the sum of present age of abe and the age before 7 years is 33 . find the present age of abe . what will be his age after 7 years ?","rationale":"\"present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 33 2 x - 7 = 33 2 x = 33 + 7 x = 40 \/ 2 x = 20 z = x + 7 = 20 + 7 = 27 answer : 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":"add(divide(add(33, 7), const_2), 7)","linear_formula":"add(n0,n1)|divide(#0,const_2)|add(n0,#1)|","chain":"33 + 7<\/gadget>\n40<\/output>\n40 \/ 2<\/gadget>\n20<\/output>\n20 + 7<\/gadget>\n27<\/output>\n27<\/result>","index":1991} +{"problem":"a train running at a speed of 60 kmph crosses a pole in 18 seconds . what is the length of the train ?","rationale":"60 kmph = 50 \/ 3 m \/ sec 50 \/ 3 * 18 = 300 m answer : b","correct":"b","options":{"a":"120 m ","b":"300 m ","c":"190 m ","d":"150 m","e":"160 m"},"options_float":{"a":120.0,"b":300.0,"c":190.0,"d":150.0,"e":160.0},"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":1994} +{"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 w 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":"80 ","c":"w . 90 ","d":"w . 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":1995} +{"problem":"when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of t ?","rationale":"\"a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . t * 23 + 39 = 25 * ( t + 1 ) t = 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":"subtract(divide(subtract(39, 15), add(2, 1)), 1)","linear_formula":"add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,n3)|","chain":"39 - 15<\/gadget>\n24<\/output>\n2 + 1<\/gadget>\n3<\/output>\n24 \/ 3<\/gadget>\n8<\/output>\n8 - 1<\/gadget>\n7<\/output>\n7<\/result>","index":1996} +{"problem":"the average earning of a person for the first 4 days of a week is rs 18 and for the last 4 days is rs 22 . if he earns rs 20 on the fourth day , his average earning for the whole week is ?","rationale":"total earning for the week = rs ( 4 × 18 + 4 × 22 - 20 ) = rs 140 average earning = rs 140 \/ 7 = rs 20 . answer : c","correct":"c","options":{"a":"rs 18.95 ","b":"rs 16 ","c":"rs 20 ","d":"rs 25.71","e":"none of these"},"options_float":{"a":18.95,"b":16.0,"c":20.0,"d":25.71,"e":null},"annotated_formula":"divide(subtract(add(multiply(4, 18), multiply(4, 22)), 20), add(const_4, const_3))","linear_formula":"add(const_3,const_4)|multiply(n0,n1)|multiply(n0,n3)|add(#1,#2)|subtract(#3,n4)|divide(#4,#0)","chain":"4 * 18<\/gadget>\n72<\/output>\n4 * 22<\/gadget>\n88<\/output>\n72 + 88<\/gadget>\n160<\/output>\n160 - 20<\/gadget>\n140<\/output>\n4 + 3<\/gadget>\n7<\/output>\n140 \/ 7<\/gadget>\n20<\/output>\n20<\/result>","index":1997} +{"problem":"mary can do a piece of work in 12 days . rosy is 50 % 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 = 150 : 100 = 3 : 2 suppose rosy takes x days to do the work . 3 : 2 : : 12 : x = > x = 8 days . hence , rosy takes 8 days to complete the work . answer : b","correct":"b","options":{"a":"7 ","b":"8 ","c":"9 ","d":"11","e":"10"},"options_float":{"a":7.0,"b":8.0,"c":9.0,"d":11.0,"e":10.0},"annotated_formula":"divide(12, add(const_1, divide(50, const_100)))","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n12 \/ (3\/2)<\/gadget>\n8<\/output>\n8<\/result>","index":1998} +{"problem":"points a , b , and , c have xy - coordinates ( 20 ) , ( 812 ) , and ( 140 ) , respectively . points x , y , and z have xy - coordinates ( 60 ) , ( 84 ) , and ( 100 ) , 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 ( 812 ) ) 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(multiply(const_4, const_4), const_2), multiply(subtract(20, multiply(const_4, const_2)), divide(subtract(20, multiply(const_4, const_2)), const_2)))","linear_formula":"multiply(const_4,const_4)|multiply(const_2,const_4)|divide(#0,const_2)|subtract(n0,#1)|divide(#3,const_2)|multiply(#4,#3)|divide(#2,#5)","chain":"4 * 4<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n4 * 2<\/gadget>\n8<\/output>\n20 - 8<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n12 * 6<\/gadget>\n72<\/output>\n8 \/ 72<\/gadget>\n1\/9 = around 0.111111<\/output>\n1\/9 = around 0.111111<\/result>","index":2000} +{"problem":"a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 174 , how many hours overtime did he work ?","rationale":"\"at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 174 - $ 120 = $ 54 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 54 \/ $ 6 = 9 ans is c\"","correct":"c","options":{"a":"8 ","b":"5 ","c":"9 ","d":"6","e":"10"},"options_float":{"a":8.0,"b":5.0,"c":9.0,"d":6.0,"e":10.0},"annotated_formula":"divide(subtract(174, multiply(3, 40)), multiply(3, const_2))","linear_formula":"multiply(n0,n1)|multiply(n0,const_2)|subtract(n2,#0)|divide(#2,#1)|","chain":"3 * 40<\/gadget>\n120<\/output>\n174 - 120<\/gadget>\n54<\/output>\n3 * 2<\/gadget>\n6<\/output>\n54 \/ 6<\/gadget>\n9<\/output>\n9<\/result>","index":2001} +{"problem":"the price of rice falls by 25 % . how much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously ?","rationale":"solution : let rs . 100 be spend on rice initially for 20 kg . as the price falls by 20 % , new price for 20 kg rice , = ( 100 - 25 % of 100 ) = 75 new price of rice = 75 \/ 20 = rs . 3.75 per kg . rice can bought now at = 100 \/ 3.75 = 26.67 kg . answer : option c","correct":"c","options":{"a":"5 kg ","b":"15 kg ","c":"26.67 kg ","d":"30 kg","e":"none"},"options_float":{"a":5.0,"b":15.0,"c":26.67,"d":30.0,"e":null},"annotated_formula":"divide(const_100, divide(subtract(const_100, 25), 20))","linear_formula":"subtract(const_100,n0)|divide(#0,n1)|divide(const_100,#1)","chain":"100 - 25<\/gadget>\n75<\/output>\n75 \/ 20<\/gadget>\n15\/4 = around 3.75<\/output>\n100 \/ (15\/4)<\/gadget>\n80\/3 = around 26.666667<\/output>\n80\/3 = around 26.666667<\/result>","index":2002} +{"problem":"john purchased 1370 large bottles at $ 1.99 per bottle and 690 small bottles at $ 1.50 per bottle . what was the approximate average price paid per bottle ?","rationale":"\"( 1370 * 1.99 + 690 * 1.50 ) \/ ( 1370 + 690 ) = ~ 1.83 - option ( e )\"","correct":"e","options":{"a":"$ 1.63 ","b":"$ 1.64 ","c":"$ 1.68 ","d":"$ 1.72","e":"$ 1.83"},"options_float":{"a":1.63,"b":1.64,"c":1.68,"d":1.72,"e":1.83},"annotated_formula":"divide(add(multiply(1370, 1.99), multiply(690, 1.50)), add(1370, 690))","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|","chain":"1_370 * 1.99<\/gadget>\n2_726.3<\/output>\n690 * 1.5<\/gadget>\n1_035<\/output>\n2_726.3 + 1_035<\/gadget>\n3_761.3<\/output>\n1_370 + 690<\/gadget>\n2_060<\/output>\n3_761.3 \/ 2_060<\/gadget>\n1.825874<\/output>\n1.825874<\/result>","index":2004} +{"problem":"find the ratio in which rice at rs . 7.30 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 : 100 = 3 : 5 answer : c\"","correct":"c","options":{"a":"2 : 0 ","b":"2 : 3 ","c":"3 : 5 ","d":"2 : 2","e":"2 : 8"},"options_float":{"a":null,"b":0.6666666667,"c":0.6,"d":1.0,"e":0.25},"annotated_formula":"divide(subtract(6.30, 5.70), subtract(7.30, 6.30))","linear_formula":"subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"6.3 - 5.7<\/gadget>\n0.6<\/output>\n7.3 - 6.3<\/gadget>\n1<\/output>\n0.6 \/ 1<\/gadget>\n0.6<\/output>\n0.6<\/result>","index":2005} +{"problem":"3 different containers contain 50 litres , 100 litres and 150 litres of mixture of milk and water respectively . what is the biggest measure can measure all the different quantities exactly ?","rationale":"m 1 = 50 litres m 2 = 100 litres m 3 = 150 litres required measurement = h . c . f . of m 1 , m 2 , m 3 = 50 litres answer is d","correct":"d","options":{"a":"120 litres ","b":"57 litres ","c":"60 litres ","d":"50 litres","e":"100 litres"},"options_float":{"a":120.0,"b":57.0,"c":60.0,"d":50.0,"e":100.0},"annotated_formula":"gcd(gcd(50, 100), 150)","linear_formula":"gcd(n1,n2)|gcd(n3,#0)","chain":"gcd(50, 100)<\/gadget>\n50<\/output>\ngcd(50, 150)<\/gadget>\n50<\/output>\n50<\/result>","index":2006} +{"problem":"2034 - ( 1002 \/ 20.04 ) = ?","rationale":"2034 - 50 = 1984 answer : e","correct":"e","options":{"a":"2984 ","b":"2983 ","c":"2982 ","d":"2981","e":"1984"},"options_float":{"a":2984.0,"b":2983.0,"c":2982.0,"d":2981.0,"e":1984.0},"annotated_formula":"subtract(2034, divide(1002, 20.04))","linear_formula":"divide(n1,n2)|subtract(n0,#0)","chain":"1_002 \/ 20.04<\/gadget>\n50<\/output>\n2_034 - 50<\/gadget>\n1_984<\/output>\n1_984<\/result>","index":2007} +{"problem":"x , y , and z are all unique numbers . if x is chosen randomly from the set { 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ?","rationale":"p ( x is prime ) = 1 \/ 2 p ( y is prime ) = 1 \/ 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 \/ 2 * 1 \/ 4 = 1 \/ 8 the answer is c .","correct":"c","options":{"a":"1 \/ 5 ","b":"3 \/ 20 ","c":"1 \/ 8 ","d":"3 \/ 10","e":"1 \/ 10"},"options_float":{"a":0.2,"b":0.15,"c":0.125,"d":0.3,"e":0.1},"annotated_formula":"multiply(divide(const_1, const_2), divide(const_1, const_4))","linear_formula":"divide(const_1,const_2)|divide(const_1,const_4)|multiply(#0,#1)","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>\n1\/8 = around 0.125<\/result>","index":2008} +{"problem":"marty ' s pizza shop guarantees that their pizzas all have at least 75 % of the surface area covered with toppings , with a crust of uniform width surrounding them . if you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust ?","rationale":"total area = 8 * 8 * pi radius = 64 pi surface = . 75 * 64 * pi = 48 pi radius of surface = 4 sqrt ( 3 ) ~ 6.8 radius width = 8 - 6.8 = 1.2 answer : b","correct":"b","options":{"a":"0.8 inches ","b":"1.1 inches ","c":"1.6 inches ","d":"2.0 inches","e":"2.5 inches"},"options_float":{"a":0.8,"b":1.1,"c":1.6,"d":2.0,"e":2.5},"annotated_formula":"divide(subtract(16, multiply(sqrt(divide(divide(multiply(circle_area(divide(16, const_2)), 75), const_100), const_pi)), const_2)), const_2)","linear_formula":"divide(n1,const_2)|circle_area(#0)|multiply(n0,#1)|divide(#2,const_100)|divide(#3,const_pi)|sqrt(#4)|multiply(#5,const_2)|subtract(n1,#6)|divide(#7,const_2)","chain":"16 \/ 2<\/gadget>\n8<\/output>\npi * (8 ** 2)<\/gadget>\n64*pi = around 201.06193<\/output>\n(64*pi) * 75<\/gadget>\n4800*pi = around 15_079.644737<\/output>\n(4800*pi) \/ 100<\/gadget>\n48*pi = around 150.796447<\/output>\n(48*pi) \/ pi<\/gadget>\n48<\/output>\n48 ** (1\/2)<\/gadget>\n4*sqrt(3) = around 6.928203<\/output>\n(4*sqrt(3)) * 2<\/gadget>\n8*sqrt(3) = around 13.856406<\/output>\n16 - (8*sqrt(3))<\/gadget>\n16 - 8*sqrt(3) = around 2.143594<\/output>\n(16 - 8*sqrt(3)) \/ 2<\/gadget>\n8 - 4*sqrt(3) = around 1.071797<\/output>\n8 - 4*sqrt(3) = around 1.071797<\/result>","index":2009} +{"problem":"compute all real solutions to 16 x + 4 x + 1 - 96 = 0","rationale":"if we substitute y = 4 x , we have y 2 + 4 y - 96 = 0 , so y = - 4 or y = 8 . the first does not map to a real solution , while the second maps to x = 3 \/ 2 correct answer a","correct":"a","options":{"a":"3 \/ 2 ","b":"3 \/ 3 ","c":"2 \/ 4 ","d":"4 \/ 4","e":"4 \/ 5"},"options_float":{"a":1.5,"b":1.0,"c":0.5,"d":1.0,"e":0.8},"annotated_formula":"divide(subtract(4, 1), subtract(subtract(4, 1), 1))","linear_formula":"subtract(n1,n2)|subtract(#0,n2)|divide(#0,#1)","chain":"4 - 1<\/gadget>\n3<\/output>\n3 - 1<\/gadget>\n2<\/output>\n3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":2010} +{"problem":"virginia , adrienne , and dennis have taught history for a combined total of 87 years . if virginia has taught for 9 more years than adrienne and for 9 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 - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 87 = > 3 d = 87 + 27 = 114 = > d = 38 answer c\"","correct":"c","options":{"a":"23 ","b":"32 ","c":"38 ","d":"41","e":"44"},"options_float":{"a":23.0,"b":32.0,"c":38.0,"d":41.0,"e":44.0},"annotated_formula":"add(divide(subtract(87, add(add(9, 9), 9)), const_3), add(9, 9))","linear_formula":"add(n1,n2)|add(n1,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)|","chain":"9 + 9<\/gadget>\n18<\/output>\n18 + 9<\/gadget>\n27<\/output>\n87 - 27<\/gadget>\n60<\/output>\n60 \/ 3<\/gadget>\n20<\/output>\n20 + 18<\/gadget>\n38<\/output>\n38<\/result>","index":2011} +{"problem":"the edges of a cuboid are 4 cm , 5 cm and 7 cm . find the volume of the cuboid ?","rationale":"\"4 * 5 * 7 = 120 answer : a\"","correct":"a","options":{"a":"140 ","b":"278 ","c":"378 ","d":"368","e":"367"},"options_float":{"a":140.0,"b":278.0,"c":378.0,"d":368.0,"e":367.0},"annotated_formula":"volume_rectangular_prism(4, 5, 7)","linear_formula":"volume_rectangular_prism(n0,n1,n2)|","chain":"4 * 5 * 7<\/gadget>\n140<\/output>\n140<\/result>","index":2012} +{"problem":"what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 20 nor by 35 ?","rationale":"\"in 1000 consecutive numbers , number of multiples of 20 = 1000 \/ 20 = 50 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 \/ 35 = 28 number of multiples of 20 * 35 i . e . 700 = 1000 \/ 700 = 1 number of integers from 1 to 1000 that are divisible by neither 20 nor by 35 = 1000 - ( 50 + 28 - 1 ) { using the concept of sets here ) = 923 answer is d\"","correct":"d","options":{"a":"567 ","b":"850 ","c":"560 ","d":"923","e":"240"},"options_float":{"a":567.0,"b":850.0,"c":560.0,"d":923.0,"e":240.0},"annotated_formula":"subtract(1000, subtract(add(divide(1000, 20), divide(1000, 35)), divide(1000, multiply(20, 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 \/ 20<\/gadget>\n50<\/output>\n1_000 \/ 35<\/gadget>\n200\/7 = around 28.571429<\/output>\n50 + (200\/7)<\/gadget>\n550\/7 = around 78.571429<\/output>\n20 * 35<\/gadget>\n700<\/output>\n1_000 \/ 700<\/gadget>\n10\/7 = around 1.428571<\/output>\n(550\/7) - (10\/7)<\/gadget>\n540\/7 = around 77.142857<\/output>\n1_000 - (540\/7)<\/gadget>\n6_460\/7 = around 922.857143<\/output>\n6_460\/7 = around 922.857143<\/result>","index":2013} +{"problem":"the population of a town is 10000 . it decreases annually at the rate of 20 % p . a . what will be its population after 2 years ?","rationale":"\"10000 × 80 \/ 100 × 80 \/ 100 = 6400 answer : a\"","correct":"a","options":{"a":"6400 ","b":"4500 ","c":"5120 ","d":"5230","e":"5366"},"options_float":{"a":6400.0,"b":4500.0,"c":5120.0,"d":5230.0,"e":5366.0},"annotated_formula":"subtract(subtract(10000, multiply(10000, divide(20, const_100))), multiply(subtract(10000, multiply(10000, divide(20, const_100))), divide(20, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n10_000 * (1\/5)<\/gadget>\n2_000<\/output>\n10_000 - 2_000<\/gadget>\n8_000<\/output>\n8_000 * (1\/5)<\/gadget>\n1_600<\/output>\n8_000 - 1_600<\/gadget>\n6_400<\/output>\n6_400<\/result>","index":2014} +{"problem":"in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 4 , 1 ) , ( 1 , 1 ) , ( 1 , - 3 ) and ( - 4 , - 3 ) ?","rationale":"\"length of side 1 = 4 + 1 = 5 length of side 2 = 3 + 1 = 4 area of rectangle = 5 * 4 = 20 b is the answer\"","correct":"b","options":{"a":"16 ","b":"20 ","c":"24 ","d":"25","e":"30"},"options_float":{"a":16.0,"b":20.0,"c":24.0,"d":25.0,"e":30.0},"annotated_formula":"multiply(add(4, 1), add(1, 3))","linear_formula":"add(n0,n1)|add(n1,n5)|multiply(#0,#1)|","chain":"4 + 1<\/gadget>\n5<\/output>\n1 + 3<\/gadget>\n4<\/output>\n5 * 4<\/gadget>\n20<\/output>\n20<\/result>","index":2015} +{"problem":"each person in a group of 110 investors has investments in either equities or securities or both . exactly 25 of the investors in equities have investments in securities , and exactly 40 of the investors in securities have investments in equities . how many have investments in equities ?","rationale":"explanation : the investors can be categorized into three groups : ( 1 ) those who have investments in equities only . ( 2 ) those who have investments in securities only . ( 3 ) those who have investments in both equities and securities . let x , y , and z denote the number of people in the respective categories . since the total number of investors is 110 , we have : - = > x + y + z = 110 - - - - - - - - - - - - - ( 1 ) . also , the number of people with investments in equities is x + z and the number of people with investments in securities is y + z . since exactly 25 % of the investors in equities have investments in securities , we have the equation = > ( 25 \/ 100 ) × ( x + z ) = z . = > ( 25 \/ 100 ) × x = 75 z \/ 100 . = > x = 3 z . - - - - - - - - - - - - - - ( 2 ) since exactly 40 % of the investors in securities have investments in equities , we have the equation = > ( 40 \/ 100 ) × ( y + z ) = z . = > ( y + z ) = 5 z \/ 2 . = > y = 3 z \/ 2 . - - - - - - - - - - - - - - - - - ( 3 ) substituting equations ( 2 ) and ( 3 ) into equation ( 1 ) gives : - = > 3 z + ( 3 z \/ 2 ) + z = 110 . = > 11 z \/ 2 = 110 . = > z = 110 × 2 \/ 11 = 20 . hence , the number of people with investments in equities is : = > x + z = 3 z + z = 3 × 20 + 20 = 60 + 20 = 80 . answer : b","correct":"b","options":{"a":"65 ","b":"80 ","c":"120 ","d":"180","e":"190"},"options_float":{"a":65.0,"b":80.0,"c":120.0,"d":180.0,"e":190.0},"annotated_formula":"multiply(divide(110, add(add(multiply(divide(divide(40, const_100), divide(25, const_100)), divide(25, const_100)), subtract(const_1, multiply(divide(divide(40, const_100), divide(25, const_100)), divide(25, const_100)))), subtract(divide(divide(40, const_100), divide(25, const_100)), multiply(divide(divide(40, const_100), divide(25, const_100)), divide(25, const_100))))), divide(divide(40, const_100), divide(25, const_100)))","linear_formula":"divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,#1)|subtract(const_1,#3)|subtract(#2,#3)|add(#3,#4)|add(#6,#5)|divide(n0,#7)|multiply(#8,#2)","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(2\/5) \/ (1\/4)<\/gadget>\n8\/5 = around 1.6<\/output>\n(8\/5) * (1\/4)<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(2\/5) + (3\/5)<\/gadget>\n1<\/output>\n(8\/5) - (2\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n1 + (6\/5)<\/gadget>\n11\/5 = around 2.2<\/output>\n110 \/ (11\/5)<\/gadget>\n50<\/output>\n50 * (8\/5)<\/gadget>\n80<\/output>\n80<\/result>","index":2016} +{"problem":"a certain number of two digits is 3 times the sum of its digits and if 45 be added to it , the degits are reversed . the number is .","rationale":"let the ten ' s digit be x and unit ' s digit be y . then 10 x + y = 3 ( x + y ) = = > 7 x - 2 y = 0 = = > 7 x - 2 y = 0 - - - - > ( i ) 10 + y + 45 = 10 y + x = = > y - x = 5 = = = > - 2 x + 2 y = 10 - - - - - - ( ii ) by adding ( i ) and ( ii ) we get 5 x = 10 so x = 2 and y = 7 so the required number is 27 . so the correct option is b ) 27 .","correct":"b","options":{"a":"23 ","b":"27 ","c":"32 ","d":"72","e":"46"},"options_float":{"a":23.0,"b":27.0,"c":32.0,"d":72.0,"e":46.0},"annotated_formula":"add(multiply(subtract(divide(subtract(45, const_10), subtract(subtract(const_10, 3), const_2)), subtract(subtract(const_10, 3), const_2)), const_10), divide(subtract(45, const_10), subtract(subtract(const_10, 3), const_2)))","linear_formula":"subtract(n1,const_10)|subtract(const_10,n0)|subtract(#1,const_2)|divide(#0,#2)|subtract(#3,#2)|multiply(#4,const_10)|add(#3,#5)","chain":"45 - 10<\/gadget>\n35<\/output>\n10 - 3<\/gadget>\n7<\/output>\n7 - 2<\/gadget>\n5<\/output>\n35 \/ 5<\/gadget>\n7<\/output>\n7 - 5<\/gadget>\n2<\/output>\n2 * 10<\/gadget>\n20<\/output>\n20 + 7<\/gadget>\n27<\/output>\n27<\/result>","index":2018} +{"problem":"tom , working alone , can paint a room in 16 hours . peter and john , working independently , can paint the same room in 8 hours and 4 hours , respectively . tom starts painting the room and works on his own for two hour . he is then joined by peter and they work together for two hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?","rationale":"\"let the time when all three were working together be t hours . then : tom worked for t + 4 hour and has done 1 \/ 16 * ( t + 4 ) part of the job ; peter worked for t + 2 hour and has done 1 \/ 8 * ( t + 2 ) part of the job ; john worked for t hours and has done 1 \/ 4 * t part of the job : 1 \/ 16 * ( t + 4 ) + 1 \/ 8 * ( t + 2 ) + 1 \/ 4 * t = 1 - - > multiply by 16 - - > ( t + 4 ) + ( 2 t + 2 ) + 4 t = 16 - - > t = 10 \/ 7 ; hence peter has done 1 \/ 8 * ( 10 \/ 7 + 2 ) = 1 \/ 8 * 24 \/ 7 = 3 \/ 7 answer : e\"","correct":"e","options":{"a":"4 \/ 7 ","b":"5 \/ 7 ","c":"2 \/ 7 ","d":"1 \/ 7","e":"3 \/ 7"},"options_float":{"a":0.5714285714,"b":0.7142857143,"c":0.2857142857,"d":0.1428571429,"e":0.4285714286},"annotated_formula":"divide(const_4, add(multiply(const_4, const_2), const_1))","linear_formula":"multiply(const_2,const_4)|add(#0,const_1)|divide(const_4,#1)|","chain":"4 * 2<\/gadget>\n8<\/output>\n8 + 1<\/gadget>\n9<\/output>\n4 \/ 9<\/gadget>\n4\/9 = around 0.444444<\/output>\n4\/9 = around 0.444444<\/result>","index":2020} +{"problem":"an angry arjun carried some arrows for fighting with bheeshm . with half the arrows , he cut down the arrows thrown by bheeshm on him and with 6 other arrows he killed the chariot driver of bheeshm . with one arrow each he knocked down respectively the chariot , the flag and the bow of bheeshm . finally , with one more than 4 times the square root of arrows he laid bheeshm unconscious on an arrow bed . find the total number of arrows arjun had .","rationale":"x \/ 2 + 6 + 3 + 1 + 4 sqrt ( x ) = x x \/ 2 + 10 + 4 sqrt ( x ) = x 4 sqrt ( x ) = x \/ 2 - 10 squaring on both sides 16 x = x ² \/ 4 + 100 - 10 x simplifying x ² - 104 x + 400 = 0 x = 100 , 4 x = 4 is not possible therefore x = 100 answer : 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":"power(add(6, 4), const_2)","linear_formula":"add(n0,n1)|power(#0,const_2)","chain":"6 + 4<\/gadget>\n10<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n100<\/result>","index":2021} +{"problem":"find the distance covered by a man walking for 10 min at a speed of 6 km \/ hr ?","rationale":"\"distance = 6 * 10 \/ 60 = 1 km answer is a\"","correct":"a","options":{"a":"1 km ","b":"2 km ","c":"3 km ","d":"1.5 km","e":"4 km"},"options_float":{"a":1.0,"b":2.0,"c":3.0,"d":1.5,"e":4.0},"annotated_formula":"divide(multiply(10, divide(multiply(6, const_1000), const_60)), const_1000)","linear_formula":"multiply(n1,const_1000)|divide(#0,const_60)|multiply(n0,#1)|divide(#2,const_1000)|","chain":"6 * 1_000<\/gadget>\n6_000<\/output>\n6_000 \/ 60<\/gadget>\n100<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n1_000 \/ 1_000<\/gadget>\n1<\/output>\n1<\/result>","index":2022} +{"problem":"in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 15 , what is the greatest possible value of b x a ?","rationale":"\"i also was confused when i was looking forabove number : d as far as i understood , 15 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 15 x 1 ( a = 1 , b = 5 respectively we have = 5 ) . to ensure , let ' s check ( avoid even multiples of 2,4 , 6,8 etc ( we will have 0 in units thus making our result 0 ) 15 x 3 = 45 ( a = 4 b = 5 respectively . hey ! that ' s 20 . but we do n ' t have this number in answer choices , move on ) . 15 x 5 = 75 ( a = 7 , b = 5 respectively . increasing trend , we have 35 now ) 15 x 7 = 105 ( a = 0 , b = 5 . have 0 now . can create a pattern ) imo e .\"","correct":"e","options":{"a":"0 ","b":"15 ","c":"25 ","d":"30","e":"35"},"options_float":{"a":0.0,"b":15.0,"c":25.0,"d":30.0,"e":35.0},"annotated_formula":"multiply(add(const_3, const_4), add(const_2, const_3))","linear_formula":"add(const_3,const_4)|add(const_2,const_3)|multiply(#0,#1)|","chain":"3 + 4<\/gadget>\n7<\/output>\n2 + 3<\/gadget>\n5<\/output>\n7 * 5<\/gadget>\n35<\/output>\n35<\/result>","index":2023} +{"problem":"let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 79 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 79 th number = 8 * ( 79 - 1 ) + 5 = 629 answer = e\"","correct":"e","options":{"a":"605 ","b":"608 ","c":"613 ","d":"616","e":"629"},"options_float":{"a":605.0,"b":608.0,"c":613.0,"d":616.0,"e":629.0},"annotated_formula":"add(multiply(subtract(79, const_1), 8), 5)","linear_formula":"subtract(n2,const_1)|multiply(n0,#0)|add(n1,#1)|","chain":"79 - 1<\/gadget>\n78<\/output>\n78 * 8<\/gadget>\n624<\/output>\n624 + 5<\/gadget>\n629<\/output>\n629<\/result>","index":2024} +{"problem":"the area of a sector of a circle of radius 4 cm formed by an arc of length 4.5 cm is ?","rationale":"\"( 4 * 4.5 ) \/ 2 = 9 answer : e\"","correct":"e","options":{"a":"8.78 ","b":"8.67 ","c":"8.75 ","d":"8.98","e":"9"},"options_float":{"a":8.78,"b":8.67,"c":8.75,"d":8.98,"e":9.0},"annotated_formula":"multiply(divide(const_1, const_2), multiply(4, 4.5))","linear_formula":"divide(const_1,const_2)|multiply(n0,n1)|multiply(#0,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n4 * 4.5<\/gadget>\n18<\/output>\n(1\/2) * 18<\/gadget>\n9<\/output>\n9<\/result>","index":2025} +{"problem":"the speed at which a man can row a boat in still water is 21 kmph . if he rows downstream , where the speed of current is 5 kmph , what time will he take to cover 90 metres ?","rationale":"speed of the boat downstream = 21 + 5 = 26 kmph = 26 * 5 \/ 18 = 7.22 m \/ s hence time taken to cover 90 m = 90 \/ 7.22 = 12.46 seconds . answer : d","correct":"d","options":{"a":"23.46 ","b":"27.46 ","c":"28.46 ","d":"12.46","e":"25.46"},"options_float":{"a":23.46,"b":27.46,"c":28.46,"d":12.46,"e":25.46},"annotated_formula":"divide(90, multiply(add(21, 5), const_0_2778))","linear_formula":"add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|","chain":"21 + 5<\/gadget>\n26<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n26 * (5\/18)<\/gadget>\n65\/9 = around 7.222222<\/output>\n90 \/ (65\/9)<\/gadget>\n162\/13 = around 12.461538<\/output>\n162\/13 = around 12.461538<\/result>","index":2028} +{"problem":"if a man lost 4 % by selling oranges at the rate of 48 a rupee at how many a rupee must he sell them to gain 44 % ?","rationale":"\"96 % - - - - 48 144 % - - - - ? 96 \/ 144 * 48 = 32 answer : e\"","correct":"e","options":{"a":"1 ","b":"8 ","c":"9 ","d":"4","e":"32"},"options_float":{"a":1.0,"b":8.0,"c":9.0,"d":4.0,"e":32.0},"annotated_formula":"divide(multiply(subtract(const_100, 4), 48), 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 * 48<\/gadget>\n4_608<\/output>\n100 + 44<\/gadget>\n144<\/output>\n4_608 \/ 144<\/gadget>\n32<\/output>\n32<\/result>","index":2029} +{"problem":"an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 28 metre of cloth ?","rationale":"\"explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 28 : x = > x = { \\ color { blue } \\ frac { 28 \\ times 1 } { 0.128 } } = > x = 218.75 so time will be approx 219 seconds answer : b\"","correct":"b","options":{"a":"175 seconds ","b":"219 seconds ","c":"155 seconds ","d":"115 seconds","e":"115 seconds"},"options_float":{"a":175.0,"b":219.0,"c":155.0,"d":115.0,"e":115.0},"annotated_formula":"divide(28, 0.128)","linear_formula":"divide(n1,n0)|","chain":"28 \/ 0.128<\/gadget>\n218.75<\/output>\n218.75<\/result>","index":2030} +{"problem":"the radius of a circular wheel is 1.75 m , how many revolutions will it make in traveling 1 km ?","rationale":"2 * 22 \/ 7 * 1.75 * x = 11000 x = 1000 answer : a","correct":"a","options":{"a":"1000 ","b":"2788 ","c":"2677 ","d":"2899","e":"2771"},"options_float":{"a":1000.0,"b":2788.0,"c":2677.0,"d":2899.0,"e":2771.0},"annotated_formula":"divide(multiply(multiply(multiply(const_pi, const_2), 1.75), const_1000), add(1, const_10))","linear_formula":"add(n1,const_10)|multiply(const_2,const_pi)|multiply(n0,#1)|multiply(#2,const_1000)|divide(#3,#0)","chain":"pi * 2<\/gadget>\n2*pi = around 6.283185<\/output>\n(2*pi) * 1.75<\/gadget>\n3.5*pi = around 10.995574<\/output>\n(3.5*pi) * 1_000<\/gadget>\n3500.0*pi = around 10_995.574288<\/output>\n1 + 10<\/gadget>\n11<\/output>\n(3500.0*pi) \/ 11<\/gadget>\n318.181818181818*pi = around 999.597663<\/output>\n318.181818181818*pi = around 999.597663<\/result>","index":2031} +{"problem":"a 340 - liter solution of kola is made from 88 % water , 5 % concentrated kola and the rest is made from sugar . if 4.2 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 + 4.2 + 6.8 = 361 numerator : 340 ( 1 - . 88 - . 05 ) + 4.2 340 ( 0.07 ) + 4.2 23.8 + 4.2 28 ratio : 28 \/ 361 = 0.077 answer : b\"","correct":"b","options":{"a":"6 % . ","b":"7.7 % . ","c":"9.2 % . ","d":"10.5 % .","e":"11 % ."},"options_float":{"a":6.0,"b":7.7,"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))), 4.2), add(add(add(340, 4.2), 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) + 4.2<\/gadget>\n28<\/output>\n340 + 4.2<\/gadget>\n344.2<\/output>\n344.2 + 10<\/gadget>\n354.2<\/output>\n354.2 + 6.8<\/gadget>\n361<\/output>\n28 \/ 361<\/gadget>\n28\/361 = around 0.077562<\/output>\n(28\/361) * 100<\/gadget>\n2_800\/361 = around 7.756233<\/output>\n2_800\/361 = around 7.756233<\/result>","index":2032} +{"problem":"if the radius of a cylinder is doubled and so is the height , what is the new volume of the cylinder divided by the old one ?","rationale":"let the radius be r and the the height be h . new radius = 2 r and height = 2 h . area ( new ) : area ( old ) = pi ∗ ( 2 r ) ^ 2 ∗ 2 h \/ pi ∗ r ^ 2 ∗ h = 8 : 1 . hence the answer is a .","correct":"a","options":{"a":"8 . ","b":"2 ","c":"6 ","d":"4","e":"10"},"options_float":{"a":8.0,"b":2.0,"c":6.0,"d":4.0,"e":10.0},"annotated_formula":"divide(volume_cylinder(multiply(const_1, const_2), multiply(const_1, const_2)), volume_cylinder(const_1, const_1))","linear_formula":"multiply(const_1,const_2)|volume_cylinder(const_1,const_1)|volume_cylinder(#0,#0)|divide(#2,#1)","chain":"1 * 2<\/gadget>\n2<\/output>\npi * (2 ** 2) * 2<\/gadget>\n8*pi = around 25.132741<\/output>\npi * (1 ** 2) * 1<\/gadget>\npi = around 3.141593<\/output>\n(8*pi) \/ pi<\/gadget>\n8<\/output>\n8<\/result>","index":2033} +{"problem":"oil is poured into a tank so that the tank is being filled at the rate of 4 cubic feet per hour . if the empty rectangular tank is 9 feet long , 8 feet wide , and 5 feet deep , approximately how many hours does it take to fill the tank ?","rationale":"the volume the tank is : length * width * depth = 9 * 8 * 5 = 360 cubic feet . 360 cubic feet \/ 4 cubic feet per hour = 90 hours . it will take 90 hours to fill the tank . the answer is d .","correct":"d","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":"divide(volume_rectangular_prism(9, 8, 5), 4)","linear_formula":"volume_rectangular_prism(n1,n2,n3)|divide(#0,n0)","chain":"9 * 8 * 5<\/gadget>\n360<\/output>\n360 \/ 4<\/gadget>\n90<\/output>\n90<\/result>","index":2037} +{"problem":"the speed of a boat in still water in 37 km \/ hr and the rate of current is 13 km \/ hr . the distance travelled downstream in 10 minutes is :","rationale":"\"explanation : speed downstream = ( 37 + 13 ) = 50 kmph time = 24 minutes = 10 \/ 60 hour = 1 \/ 6 hour distance travelled = time × speed = ( 1 \/ 6 ) × 50 = 8.33 km answer : option e\"","correct":"e","options":{"a":"10.44 km ","b":"10.6 km ","c":"11.4 km ","d":"11.22 km","e":"8.33 km"},"options_float":{"a":10.44,"b":10.6,"c":11.4,"d":11.22,"e":8.33},"annotated_formula":"multiply(add(37, 13), divide(10, const_60))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"37 + 13<\/gadget>\n50<\/output>\n10 \/ 60<\/gadget>\n1\/6 = around 0.166667<\/output>\n50 * (1\/6)<\/gadget>\n25\/3 = around 8.333333<\/output>\n25\/3 = around 8.333333<\/result>","index":2039} +{"problem":"a train 600 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 = 600 \/ 20 s = 30 m \/ sec speed = 30 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 108 kmph answer : b\"","correct":"b","options":{"a":"76 kmph ","b":"108 kmph ","c":"72 kmph ","d":"34 kmph","e":"91 kmph"},"options_float":{"a":76.0,"b":108.0,"c":72.0,"d":34.0,"e":91.0},"annotated_formula":"divide(divide(600, const_1000), divide(20, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"600 \/ 1_000<\/gadget>\n3\/5 = around 0.6<\/output>\n20 \/ 3_600<\/gadget>\n1\/180 = around 0.005556<\/output>\n(3\/5) \/ (1\/180)<\/gadget>\n108<\/output>\n108<\/result>","index":2040} +{"problem":"the average of 30 numbers is 25 . if each number is multiplied by 5 , find the new average ?","rationale":"\"sum of the 30 numbers = 30 * 25 = 750 if each number is multiplied by 5 , the sum also gets multiplied by 5 and the average also gets multiplied by 5 . thus , the new average = 25 * 5 = 125 . answer : b\"","correct":"b","options":{"a":"115 ","b":"125 ","c":"135 ","d":"145","e":"155"},"options_float":{"a":115.0,"b":125.0,"c":135.0,"d":145.0,"e":155.0},"annotated_formula":"multiply(25, 5)","linear_formula":"multiply(n1,n2)|","chain":"25 * 5<\/gadget>\n125<\/output>\n125<\/result>","index":2041} +{"problem":"√ ( 21 ) ^ 2","rationale":"\"explanation √ ( 21 ) ^ 2 = ? or , ? = 21 answer d\"","correct":"d","options":{"a":"7 ","b":"14 ","c":"49 ","d":"21","e":"none of these"},"options_float":{"a":7.0,"b":14.0,"c":49.0,"d":21.0,"e":null},"annotated_formula":"sqrt(power(21, 2))","linear_formula":"power(n0,n1)|sqrt(#0)|","chain":"21 ** 2<\/gadget>\n441<\/output>\n441 ** (1\/2)<\/gadget>\n21<\/output>\n21<\/result>","index":2042} +{"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 . 798 find the share of a .","rationale":"\"explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 \/ 21 * 798 = 304 answer : d\"","correct":"d","options":{"a":"240 ","b":"388 ","c":"379 ","d":"304","e":"122"},"options_float":{"a":240.0,"b":388.0,"c":379.0,"d":304.0,"e":122.0},"annotated_formula":"multiply(divide(798, 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>\n798 \/ 84_000<\/gadget>\n19\/2_000 = around 0.0095<\/output>\n(19\/2_000) * 32_000<\/gadget>\n304<\/output>\n304<\/result>","index":2043} +{"problem":"the speed of a boat in still water is 15 km \/ hr and the rate of the current is 3 km \/ hr . the distance travelled downstream in 12 minutes is","rationale":"\"speed downstream = ( 15 + 3 ) km \/ hr = 18 km \/ hr . distance travelled = ( 18 x 12 \/ 60 ) hours = 3.6 km . answer d\"","correct":"d","options":{"a":"1.2 km ","b":"1.8 km ","c":"2.4 km ","d":"3.6 km","e":"none"},"options_float":{"a":1.2,"b":1.8,"c":2.4,"d":3.6,"e":null},"annotated_formula":"multiply(divide(12, const_60), add(15, 3))","linear_formula":"add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|","chain":"12 \/ 60<\/gadget>\n1\/5 = around 0.2<\/output>\n15 + 3<\/gadget>\n18<\/output>\n(1\/5) * 18<\/gadget>\n18\/5 = around 3.6<\/output>\n18\/5 = around 3.6<\/result>","index":2045} +{"problem":"a certain animal in the zoo has consumed 39 pounds of food in 6 days . if it continues to eat at the same rate , in how many more days will its total consumption be 117 pounds ?","rationale":"ans is c : 39 pounds - - > 6 days 117 pounds - - > x days x = 117 * 6 \/ 39 = 18 the animal has already consumed food in 6 days so the the number of days for it ' s total consumption be 117 pounds is 18 - 6 = 12","correct":"c","options":{"a":"8 ","b":"7 ","c":"12 ","d":"9","e":"none of the above"},"options_float":{"a":8.0,"b":7.0,"c":12.0,"d":9.0,"e":null},"annotated_formula":"subtract(divide(117, divide(39, 6)), 6)","linear_formula":"divide(n0,n1)|divide(n2,#0)|subtract(#1,n1)","chain":"39 \/ 6<\/gadget>\n13\/2 = around 6.5<\/output>\n117 \/ (13\/2)<\/gadget>\n18<\/output>\n18 - 6<\/gadget>\n12<\/output>\n12<\/result>","index":2046} +{"problem":"the length of a rectangular floor is more than its breadth by 200 % . if rs . 441 is required to paint the floor at the rate of rs . 3 \/ sq m , 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 = 441 \/ 3 = 147 sq m l b = 147 i . e . , l * l \/ 3 = 147 l ^ 2 = 441 = > l = 21 . d\"","correct":"d","options":{"a":"12 ","b":"18 ","c":"20 ","d":"21","e":"24"},"options_float":{"a":12.0,"b":18.0,"c":20.0,"d":21.0,"e":24.0},"annotated_formula":"multiply(sqrt(divide(divide(441, 3), const_3)), const_3)","linear_formula":"divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)|","chain":"441 \/ 3<\/gadget>\n147<\/output>\n147 \/ 3<\/gadget>\n49<\/output>\n49 ** (1\/2)<\/gadget>\n7<\/output>\n7 * 3<\/gadget>\n21<\/output>\n21<\/result>","index":2047} +{"problem":"in a class of 42 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 = 42 avg # of books per student = 2 total # of books = 42 * 2 = 84 # of student borrowed at least 3 books = 42 - 2 - 12 - 10 = 18 # of books borrowed by above 18 students = 84 - ( 12 * 1 ) + ( 10 * 2 ) = 52 considering that 17 out of above 18 students borrowed only 3 books , # of books borrowed = 17 * 3 = 51 so maximum # of books borrowed by any single student = 52 - 51 = 1 option a\"","correct":"a","options":{"a":"1 ","b":"5 ","c":"8 ","d":"13","e":"15"},"options_float":{"a":1.0,"b":5.0,"c":8.0,"d":13.0,"e":15.0},"annotated_formula":"subtract(multiply(42, 2), add(multiply(subtract(subtract(42, 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":"42 * 2<\/gadget>\n84<\/output>\n12 * 1<\/gadget>\n12<\/output>\n12 + 10<\/gadget>\n22<\/output>\n22 + 2<\/gadget>\n24<\/output>\n42 - 24<\/gadget>\n18<\/output>\n18 - 1<\/gadget>\n17<\/output>\n17 * 3<\/gadget>\n51<\/output>\n10 * 2<\/gadget>\n20<\/output>\n12 + 20<\/gadget>\n32<\/output>\n51 + 32<\/gadget>\n83<\/output>\n84 - 83<\/gadget>\n1<\/output>\n1<\/result>","index":2048} +{"problem":"a set of consecutive positive integers beginning with 1 is written on the blackboard . a student came along and erased one number . the average of the remaining numbers is 35 * 7 \/ 17 . what was the number erased ?","rationale":"explanation : let the higher number be n and x be the number erased . then ( ( n ( n + 1 ) \/ 2 ) + x ) \/ ( n + 1 ) = 35 * 7 \/ 17 = 602 \/ 17 hence , n = 69 and x = 7 satisfy the above conditions . answer : a","correct":"a","options":{"a":"7 ","b":"8 ","c":"6 ","d":"5","e":"4"},"options_float":{"a":7.0,"b":8.0,"c":6.0,"d":5.0,"e":4.0},"annotated_formula":"subtract(multiply(divide(floor(multiply(add(35, divide(7, 17)), const_2)), const_2), subtract(floor(multiply(add(35, divide(7, 17)), const_2)), const_1)), multiply(add(35, divide(7, 17)), subtract(subtract(floor(multiply(add(35, divide(7, 17)), const_2)), const_1), 1)))","linear_formula":"divide(n2,n3)|add(n1,#0)|multiply(#1,const_2)|floor(#2)|divide(#3,const_2)|subtract(#3,const_1)|multiply(#4,#5)|subtract(#5,n0)|multiply(#1,#7)|subtract(#6,#8)","chain":"7 \/ 17<\/gadget>\n7\/17 = around 0.411765<\/output>\n35 + (7\/17)<\/gadget>\n602\/17 = around 35.411765<\/output>\n(602\/17) * 2<\/gadget>\n1_204\/17 = around 70.823529<\/output>\nfloor(1_204\/17)<\/gadget>\n70<\/output>\n70 \/ 2<\/gadget>\n35<\/output>\n70 - 1<\/gadget>\n69<\/output>\n35 * 69<\/gadget>\n2_415<\/output>\n69 - 1<\/gadget>\n68<\/output>\n(602\/17) * 68<\/gadget>\n2_408<\/output>\n2_415 - 2_408<\/gadget>\n7<\/output>\n7<\/result>","index":2049} +{"problem":"find the number which is nearest to 3105 and is exactly divisible by 21 ?","rationale":"on dividing 3105 by 21 , we get 18 as remainder . number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . answer e","correct":"e","options":{"a":"3100 ","b":"2500 ","c":"2545 ","d":"5800","e":"3108"},"options_float":{"a":3100.0,"b":2500.0,"c":2545.0,"d":5800.0,"e":3108.0},"annotated_formula":"add(3105, subtract(21, reminder(3105, 21)))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|add(n0,#1)","chain":"3_105 % 21<\/gadget>\n18<\/output>\n21 - 18<\/gadget>\n3<\/output>\n3_105 + 3<\/gadget>\n3_108<\/output>\n3_108<\/result>","index":2050} +{"problem":"the lengths of the diagonals of a rhombus are 20 and 48 meters . find the perimeter of the rhombus ?","rationale":"below is shown a rhombus with the given diagonals . consider the right triangle boc and apply pythagora ' s theorem as follows bc 2 = 10 ^ 2 + 24 ^ 2 and evaluate bc bc = 26 meters . we now evaluate the perimeter p as follows : p = 4 * 26 = 104 meters . answer is d","correct":"d","options":{"a":"150 merters ","b":"125 meters ","c":"96 meters ","d":"104 meters","e":"152 meters"},"options_float":{"a":150.0,"b":125.0,"c":96.0,"d":104.0,"e":152.0},"annotated_formula":"power(divide(20, const_2), const_2)","linear_formula":"divide(n0,const_2)|power(#0,const_2)","chain":"20 \/ 2<\/gadget>\n10<\/output>\n10 ** 2<\/gadget>\n100<\/output>\n100<\/result>","index":2051} +{"problem":"a courtyard is 21 meter long and 14 meter board is to be paved with bricks of dimensions 14 cm by 8 cm . the total number of bricks required is :","rationale":"\"explanation : number of bricks = courtyard area \/ 1 brick area = ( 2100 ã — 1400 \/ 14 ã — 8 ) = 26250 option d\"","correct":"d","options":{"a":"16000 ","b":"18000 ","c":"20000 ","d":"26250","e":"none of these"},"options_float":{"a":16000.0,"b":18000.0,"c":20000.0,"d":26250.0,"e":null},"annotated_formula":"divide(multiply(multiply(21, const_100), multiply(14, const_100)), multiply(14, 8))","linear_formula":"multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)|","chain":"21 * 100<\/gadget>\n2_100<\/output>\n14 * 100<\/gadget>\n1_400<\/output>\n2_100 * 1_400<\/gadget>\n2_940_000<\/output>\n14 * 8<\/gadget>\n112<\/output>\n2_940_000 \/ 112<\/gadget>\n26_250<\/output>\n26_250<\/result>","index":2053} +{"problem":"the average age of students of a class is 15.8 years . the average age of boys in the class is 16.3 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 * 16.3 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.3 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 \/ 0.5 = 4 \/ 5 required ratio = 4 \/ 5 : 1 = 4 : 5 . answer : c\"","correct":"c","options":{"a":"2 : 6 ","b":"2 : 3 ","c":"4 : 5 ","d":"2 : 1","e":"2 : 4"},"options_float":{"a":0.3333333333,"b":0.6666666667,"c":0.8,"d":2.0,"e":0.5},"annotated_formula":"divide(subtract(15.8, 15.4), subtract(16.3, 15.8))","linear_formula":"subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|","chain":"15.8 - 15.4<\/gadget>\n0.4<\/output>\n16.3 - 15.8<\/gadget>\n0.5<\/output>\n0.4 \/ 0.5<\/gadget>\n0.8<\/output>\n0.8<\/result>","index":2054} +{"problem":"during the first two weeks of june , the total rainfall in springdale was 40 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall in inches during the second week of june ?","rationale":"let x be the rainfall in the first week . then 1.5 x was the rainfall in the second week . 2.5 x = 40 x = 16 the rainfall during the second week was 1.5 * 16 = 24 inches 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(40, add(const_1, 1.5)), 1.5)","linear_formula":"add(n1,const_1)|divide(n0,#0)|multiply(n1,#1)","chain":"1 + 1.5<\/gadget>\n2.5<\/output>\n40 \/ 2.5<\/gadget>\n16<\/output>\n16 * 1.5<\/gadget>\n24<\/output>\n24<\/result>","index":2062} +{"problem":"the owner of a furniture shop charges his customer 18 % more than the cost price . if a customer paid rs . 6000 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 6000 ( 100 \/ 118 ) = rs . 5084 . answer : d\"","correct":"d","options":{"a":"rs . 6289 ","b":"rs . 6298 ","c":"rs . 6290 ","d":"rs . 5084","e":"rs . 6708"},"options_float":{"a":6289.0,"b":6298.0,"c":6290.0,"d":5084.0,"e":6708.0},"annotated_formula":"divide(6000, add(const_1, divide(18, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"18 \/ 100<\/gadget>\n9\/50 = around 0.18<\/output>\n1 + (9\/50)<\/gadget>\n59\/50 = around 1.18<\/output>\n6_000 \/ (59\/50)<\/gadget>\n300_000\/59 = around 5_084.745763<\/output>\n300_000\/59 = around 5_084.745763<\/result>","index":2063} +{"problem":"find the simple interest on $ 10000 at 7 % per annum for 12 months ?","rationale":"\"p = $ 10000 r = 7 % t = 12 \/ 12 years = 1 year s . i . = p * r * t \/ 100 = 10000 * 7 * 1 \/ 100 = $ 700 answer is d\"","correct":"d","options":{"a":"$ 410 ","b":"$ 500 ","c":"$ 650 ","d":"$ 700","e":"$ 1000"},"options_float":{"a":410.0,"b":500.0,"c":650.0,"d":700.0,"e":1000.0},"annotated_formula":"multiply(10000, divide(7, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"7 \/ 100<\/gadget>\n7\/100 = around 0.07<\/output>\n10_000 * (7\/100)<\/gadget>\n700<\/output>\n700<\/result>","index":2064} +{"problem":"two men and 7 children complete a certain piece of work in 4 days while 4 men and 4 children complete the same work in only 3 days . the number of days required by 1 man to complete the work is","rationale":"two men and 7 children complete a certain piece of work in 4 days or 8 men and 28 children complete a certain piece of work in 1 days 4 men and 4 children complete the same work in only 3 days . or 12 men and 12 children complete the same work in only 3 days . so 8 men + 28 children = 12 men + 12 children 1 man = 4 children 4 men and 4 children complete the same work in only 3 days or 4 men and 1 man ( in place of 4 children ) complete the same work in only 3 days or 5 men complete the same work in 3 days or 1 man will complete the same work in 5 * 3 = 15 days answer : b","correct":"b","options":{"a":"60 days ","b":"15 days ","c":"6 days ","d":"51 days","e":"50 days"},"options_float":{"a":60.0,"b":15.0,"c":6.0,"d":51.0,"e":50.0},"annotated_formula":"divide(subtract(multiply(7, 4), add(4, 4)), subtract(divide(7, 3), 1))","linear_formula":"add(n1,n1)|divide(n0,n4)|multiply(n0,n1)|subtract(#2,#0)|subtract(#1,n5)|divide(#3,#4)","chain":"7 * 4<\/gadget>\n28<\/output>\n4 + 4<\/gadget>\n8<\/output>\n28 - 8<\/gadget>\n20<\/output>\n7 \/ 3<\/gadget>\n7\/3 = around 2.333333<\/output>\n(7\/3) - 1<\/gadget>\n4\/3 = around 1.333333<\/output>\n20 \/ (4\/3)<\/gadget>\n15<\/output>\n15<\/result>","index":2065} +{"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 15 % . 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 = 0.85 * 150 x = 127.5 x hence total profit = 27.5 % option a","correct":"a","options":{"a":"27.5 % ","b":"30 % ","c":"35 % ","d":"37.5 %","e":"40 %"},"options_float":{"a":27.5,"b":30.0,"c":35.0,"d":37.5,"e":40.0},"annotated_formula":"subtract(multiply(divide(subtract(const_100, 15), 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 - 15<\/gadget>\n85<\/output>\n85 \/ 100<\/gadget>\n17\/20 = around 0.85<\/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(17\/20) * 150<\/gadget>\n255\/2 = around 127.5<\/output>\n(255\/2) - 100<\/gadget>\n55\/2 = around 27.5<\/output>\n55\/2 = around 27.5<\/result>","index":2066} +{"problem":"the total surface area of a solid hemisphere of diameter 14 cm , is :","rationale":"sol . total surface area = 3 ∏ r ² = [ 3 * 22 \/ 7 * 7 * 7 ] cm ² = 462 cm ² answer a","correct":"a","options":{"a":"462 cm ² ","b":"530 cm ² ","c":"1345 cm ² ","d":"1788 cm ²","e":"none"},"options_float":{"a":462.0,"b":530.0,"c":1345.0,"d":1788.0,"e":null},"annotated_formula":"multiply(multiply(const_3, const_pi), power(divide(14, const_2), const_2))","linear_formula":"divide(n0,const_2)|multiply(const_3,const_pi)|power(#0,const_2)|multiply(#1,#2)","chain":"3 * pi<\/gadget>\n3*pi = around 9.424778<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7 ** 2<\/gadget>\n49<\/output>\n(3*pi) * 49<\/gadget>\n147*pi = around 461.81412<\/output>\n147*pi = around 461.81412<\/result>","index":2067} +{"problem":"a certain company ’ s profit in 1996 was 25 percent greater than its profit in 1995 , and its profit in 1997 was 30 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ?","rationale":"\"profit in 1995 - 100 profit in 1996 - 125 % increae profit in 1997 in comparison to 1995 = 25 + 125 * 30 % = 62.5 correct option : b\"","correct":"b","options":{"a":"5 % ","b":"62.5 % ","c":"33 % ","d":"35 %","e":"38 %"},"options_float":{"a":5.0,"b":62.5,"c":33.0,"d":35.0,"e":38.0},"annotated_formula":"multiply(subtract(multiply(add(divide(30, const_100), const_1), add(const_1, divide(25, 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":"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>\n1 + (1\/4)<\/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":2068} +{"problem":"15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container h ?","rationale":"if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 \/ 37.5 ; expressed as a fraction this becomes ( 45 \/ 2 ) \/ ( 75 \/ 2 ) , or 45 \/ 2 * 2 \/ 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 \/ 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 \/ 15 of a means there must be 6 \/ 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 \/ 16 ; 13.5 \/ 24 = ( 27 \/ 2 ) * ( 1 \/ 24 ) = 9 \/ 16 . choice c is correct .","correct":"c","options":{"a":"a : 45 ","b":"b : 25 ","c":"c : 37.5 ","d":"d : 36","e":"e : 42"},"options_float":{"a":45.0,"b":25.0,"c":37.5,"d":36.0,"e":42.0},"annotated_formula":"divide(15, subtract(const_1, sqrt(divide(9, add(9, 16)))))","linear_formula":"add(n2,n3)|divide(n2,#0)|sqrt(#1)|subtract(const_1,#2)|divide(n0,#3)","chain":"9 + 16<\/gadget>\n25<\/output>\n9 \/ 25<\/gadget>\n9\/25 = around 0.36<\/output>\n(9\/25) ** (1\/2)<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n15 \/ (2\/5)<\/gadget>\n75\/2 = around 37.5<\/output>\n75\/2 = around 37.5<\/result>","index":2069} +{"problem":"after decreasing 90 % in the price of an article costs rs . 320 . find the actual cost of an article ?","rationale":"\"cp * ( 10 \/ 100 ) = 320 cp = 32 * 100 = > cp = 3200 answer : c\"","correct":"c","options":{"a":"2777 ","b":"2987 ","c":"3200 ","d":"9977","e":"1671"},"options_float":{"a":2777.0,"b":2987.0,"c":3200.0,"d":9977.0,"e":1671.0},"annotated_formula":"divide(320, subtract(const_1, divide(90, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|","chain":"90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n1 - (9\/10)<\/gadget>\n1\/10 = around 0.1<\/output>\n320 \/ (1\/10)<\/gadget>\n3_200<\/output>\n3_200<\/result>","index":2070} +{"problem":"the perimeter of a triangle is 44 cm and the in radius 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 in radius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 44 \/ 2 = 55 cm 2 answer : d\"","correct":"d","options":{"a":"76 ","b":"88 ","c":"66 ","d":"55","e":"35"},"options_float":{"a":76.0,"b":88.0,"c":66.0,"d":55.0,"e":35.0},"annotated_formula":"triangle_area(2.5, 44)","linear_formula":"triangle_area(n0,n1)|","chain":"(2.5 * 44) \/ 2<\/gadget>\n55<\/output>\n55<\/result>","index":2071} +{"problem":"if 0.45 : x : : 4 : 2 , then x is equal to","rationale":"\"sol . ( x × 4 ) = ( 0.45 × 2 ) ⇒ x = 0.9 \/ 4 = 0.225 . answer a\"","correct":"a","options":{"a":"0.225 ","b":"0.228 ","c":"0.254 ","d":"0.256","e":"none"},"options_float":{"a":0.225,"b":0.228,"c":0.254,"d":0.256,"e":null},"annotated_formula":"divide(multiply(0.45, 2), 4)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"0.45 * 2<\/gadget>\n0.9<\/output>\n0.9 \/ 4<\/gadget>\n0.225<\/output>\n0.225<\/result>","index":2073} +{"problem":"a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability e that both selected dogs are not littermates ?","rationale":"\"we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one dog of type b or none ( we can not have two dogs both of type b ) . the probability of choosing one dog of type b and one of type a is 3 \/ 9 * 6 \/ 8 * 2 = 1 \/ 2 ( the factor of 2 for the two possibilities ba and ab ) . the probability e of choosing two dogs of type a which are not littermates is 6 \/ 9 * 4 \/ 8 = 1 \/ 3 ( choose one a , then another a which is n ' t the previous one ' s littermate ) . the required probability is 1 \/ 2 + 1 \/ 3 = 5 \/ 6 . find the probability for the complementary event : choose aa or bb . probability of choosing two dogs of type a who are littermates is 6 \/ 9 * 1 \/ 8 = 1 \/ 12 . probability of choosing two dogs of type b ( who necessarily are littermates ) is 3 \/ 9 * 2 \/ 8 = 1 \/ 12 . again , we obtain 1 - ( 1 \/ 12 + 1 \/ 12 ) = 5 \/ 6 . answer : c\"","correct":"c","options":{"a":"1 \/ 6 ","b":"2 \/ 9 ","c":"5 \/ 6 ","d":"7 \/ 9","e":"8 \/ 9"},"options_float":{"a":0.1666666667,"b":0.2222222222,"c":0.8333333333,"d":0.7777777778,"e":0.8888888889},"annotated_formula":"divide(const_5, 6)","linear_formula":"divide(const_5,n1)|","chain":"5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n5\/6 = around 0.833333<\/result>","index":2074} +{"problem":"if 0.75 : x : : 5 : 12 , then x is equal to :","rationale":"\"explanation : ( x * 5 ) = ( 0.75 * 12 ) x = 9 \/ 5 = 1.80 answer : c\"","correct":"c","options":{"a":"1.12 ","b":"1.16 ","c":"1.8 ","d":"1.3","e":"none of these"},"options_float":{"a":1.12,"b":1.16,"c":1.8,"d":1.3,"e":null},"annotated_formula":"divide(multiply(0.75, 12), 5)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"0.75 * 12<\/gadget>\n9<\/output>\n9 \/ 5<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":2075} +{"problem":"two employees x and y are paid a total of rs . 650 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 = 650 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 ∴ 12 y \/ 10 + y = 650 ⇒ y [ 12 \/ 10 + 1 ] = 650 ⇒ 22 y \/ 10 = 650 ⇒ 22 y = 6500 ⇒ y = 6500 \/ 22 = rs . 295.45 c )\"","correct":"c","options":{"a":"s . 200.45 ","b":"s . 250.45 ","c":"s . 295.45 ","d":"s . 300.45","e":"s . 310.45"},"options_float":{"a":200.45,"b":250.45,"c":295.45,"d":300.45,"e":310.45},"annotated_formula":"divide(multiply(650, 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>\n650 * 10<\/gadget>\n6_500<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n6_500 \/ 22<\/gadget>\n3_250\/11 = around 295.454545<\/output>\n3_250\/11 = around 295.454545<\/result>","index":2077} +{"problem":"the average speed of a car is 1 3 \/ 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 bike in an hour = 25 × 2 = 50 km the speed of a car in an hour = 8 \/ 5 * 50 = 80 km so , the distance covered by car in 4 h is 80 × 4 = 320 km ans . ( b )","correct":"b","options":{"a":"400 km ","b":"320 km ","c":"360 km ","d":"550 km","e":"600 km"},"options_float":{"a":400.0,"b":320.0,"c":360.0,"d":550.0,"e":600.0},"annotated_formula":"multiply(multiply(add(1, divide(3, 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":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n575 \/ 23<\/gadget>\n25<\/output>\n2 * 25<\/gadget>\n50<\/output>\n(8\/5) * 50<\/gadget>\n80<\/output>\n80 * 4<\/gadget>\n320<\/output>\n320<\/result>","index":2078} +{"problem":"in a factory , there are 40 % technicians and 60 % non - technicians . if the 60 % of the technicians and 40 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ?","rationale":"total = 100 t = 40 nt = 60 40 * ( 60 \/ 100 ) = 24 60 * ( 40 \/ 100 ) = 24 24 + 24 = 48 = > 100 - 48 = 52 % answer : c","correct":"c","options":{"a":"62 % ","b":"57 % ","c":"52 % ","d":"22 %","e":"42 %"},"options_float":{"a":62.0,"b":57.0,"c":52.0,"d":22.0,"e":42.0},"annotated_formula":"subtract(const_100, add(multiply(40, divide(60, const_100)), multiply(divide(40, const_100), 60)))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(n0,#0)|multiply(n1,#1)|add(#2,#3)|subtract(const_100,#4)","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>\n(2\/5) * 60<\/gadget>\n24<\/output>\n24 + 24<\/gadget>\n48<\/output>\n100 - 48<\/gadget>\n52<\/output>\n52<\/result>","index":2081} +{"problem":"harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 75 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 = 75 × 9 = 675 . total cost he has to pay = 560 + 675 = 1235 . a )\"","correct":"a","options":{"a":"1235 ","b":"1055 ","c":"1065 ","d":"1075","e":"1080"},"options_float":{"a":1235.0,"b":1055.0,"c":1065.0,"d":1075.0,"e":1080.0},"annotated_formula":"add(multiply(8, 70), multiply(9, 75))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"8 * 70<\/gadget>\n560<\/output>\n9 * 75<\/gadget>\n675<\/output>\n560 + 675<\/gadget>\n1_235<\/output>\n1_235<\/result>","index":2083} +{"problem":"if the area of a square with sides of length 11 centimeters is equal to the area of a rectangle with a width of 11 centimeters , what is the length of the rectangle , in centimeters ?","rationale":"\"let length of rectangle = l 11 ^ 2 = l * 11 = > l = 121 \/ 11 = 11 answer c\"","correct":"c","options":{"a":"4 ","b":"8 ","c":"11 ","d":"16","e":"18"},"options_float":{"a":4.0,"b":8.0,"c":11.0,"d":16.0,"e":18.0},"annotated_formula":"divide(power(11, const_2), 11)","linear_formula":"power(n0,const_2)|divide(#0,n1)|","chain":"11 ** 2<\/gadget>\n121<\/output>\n121 \/ 11<\/gadget>\n11<\/output>\n11<\/result>","index":2084} +{"problem":"tanks p and b are each in the shape of a right circular cylinder . the interior of tank p 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 p is what percent of the capacity of tank b ?","rationale":"b . for p , r = 8 \/ 2 pi . its capacity = ( 4 pi ) ^ 2 * 10 = 160 pi for b , r = 10 \/ pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi p \/ b = 160 pi \/ 200 pi = 0.8","correct":"b","options":{"a":"75 % ","b":"80 % ","c":"100 % ","d":"120 %","e":"125 %"},"options_float":{"a":75.0,"b":80.0,"c":100.0,"d":120.0,"e":125.0},"annotated_formula":"multiply(divide(volume_cylinder(divide(divide(8, const_2), const_pi), 10), volume_cylinder(divide(divide(10, const_2), const_pi), 8)), const_100)","linear_formula":"divide(n1,const_2)|divide(n0,const_2)|divide(#0,const_pi)|divide(#1,const_pi)|volume_cylinder(#2,n0)|volume_cylinder(#3,n1)|divide(#4,#5)|multiply(#6,const_100)","chain":"8 \/ 2<\/gadget>\n4<\/output>\n4 \/ pi<\/gadget>\n4\/pi = around 1.27324<\/output>\npi * ((4\/pi) ** 2) * 10<\/gadget>\n160\/pi = around 50.929582<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 \/ pi<\/gadget>\n5\/pi = around 1.591549<\/output>\npi * ((5\/pi) ** 2) * 8<\/gadget>\n200\/pi = around 63.661977<\/output>\n(160\/pi) \/ (200\/pi)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":2085} +{"problem":"x and y are both integers . if x \/ y = 59.40 , then what is the sum of all the possible two digit remainders of x \/ y ?","rationale":"remainder = 0.40 - - > 40 \/ 100 - - > can be written as ( 40 \/ 4 ) \/ ( 100 \/ 4 ) = 10 \/ 25 so remainders can be 10 , 20 , 30 , 40 , . . . . . 90 . we need the sum of only 2 digit remainders - - > 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 = 450 answer : a","correct":"a","options":{"a":"450 ","b":"616 ","c":"672 ","d":"900","e":"1024"},"options_float":{"a":450.0,"b":616.0,"c":672.0,"d":900.0,"e":1024.0},"annotated_formula":"add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 59.4), const_3))","linear_formula":"add(const_2,const_3)|divide(const_3,const_2)|multiply(n0,#0)|multiply(#1,const_100)|add(#2,const_3)|add(#4,#3)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 100<\/gadget>\n150<\/output>\n2 + 3<\/gadget>\n5<\/output>\n5 * 59.4<\/gadget>\n297<\/output>\n297 + 3<\/gadget>\n300<\/output>\n150 + 300<\/gadget>\n450<\/output>\n450<\/result>","index":2086} +{"problem":"what quantity of water should be added to reduce 20 liters of 80 % acidic liquid to 20 % acidic liquid ?","rationale":"\"acid in 20 liters = 80 % of 20 = 16 liters suppose x liters of water be added . then 16 liters of acid in 20 + x liters of diluted solution 20 % of 20 + x = 16 20 + x = 80 x = 60 liters answer is c\"","correct":"c","options":{"a":"30 liters ","b":"50 liters ","c":"60 liters ","d":"70 liters","e":"80 liters"},"options_float":{"a":30.0,"b":50.0,"c":60.0,"d":70.0,"e":80.0},"annotated_formula":"subtract(divide(multiply(multiply(20, divide(80, const_100)), const_100), 20), 20)","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_100)|divide(#2,n2)|subtract(#3,n0)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n20 * (4\/5)<\/gadget>\n16<\/output>\n16 * 100<\/gadget>\n1_600<\/output>\n1_600 \/ 20<\/gadget>\n80<\/output>\n80 - 20<\/gadget>\n60<\/output>\n60<\/result>","index":2088} +{"problem":"if the compound interest on a certain sum of money for 5 years at 10 % per annum be rs . 993 , what would be the simple interest ?","rationale":"\"let p = principal a - amount we have a = p ( 1 + r \/ 100 ) 3 and ci = a - p atq 993 = p ( 1 + r \/ 100 ) 3 - p ? p = 3000 \/ - now si @ 10 % on 3000 \/ - for 5 yrs = ( 3000 x 10 x 5 ) \/ 100 = 1500 \/ - answer : a .\"","correct":"a","options":{"a":"rs . 1500 ","b":"rs . 890 ","c":"rs . 895 ","d":"rs . 900","e":"none"},"options_float":{"a":1500.0,"b":890.0,"c":895.0,"d":900.0,"e":null},"annotated_formula":"divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 5), const_100)","linear_formula":"multiply(const_3.0,const_100)|multiply(n1,#0)|multiply(n1,#1)|multiply(n0,#2)|divide(#3,const_100)|","chain":"3 * 100<\/gadget>\n300<\/output>\n300 * 10<\/gadget>\n3_000<\/output>\n3_000 * 10<\/gadget>\n30_000<\/output>\n30_000 * 5<\/gadget>\n150_000<\/output>\n150_000 \/ 100<\/gadget>\n1_500<\/output>\n1_500<\/result>","index":2089} +{"problem":"the percentage profit earned by selling an article for rs . 1620 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 , ( 1620 - x ) \/ x * 100 = ( x - 1280 ) \/ x * 100 1620 - x = x - 1280 2 x = 2900 = > x = 1450 required s . p . = 125 % of rs . 1450 = 125 \/ 100 * 1450 = rs . 1812.5 . answer b\"","correct":"b","options":{"a":"3000 ","b":"1812.5 ","c":"2000 ","d":"5600","e":"3400"},"options_float":{"a":3000.0,"b":1812.5,"c":2000.0,"d":5600.0,"e":3400.0},"annotated_formula":"multiply(divide(add(const_100, 25), const_100), divide(add(1620, 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_620 + 1_280<\/gadget>\n2_900<\/output>\n2_900 \/ 2<\/gadget>\n1_450<\/output>\n(5\/4) * 1_450<\/gadget>\n3_625\/2 = around 1_812.5<\/output>\n3_625\/2 = around 1_812.5<\/result>","index":2090} +{"problem":"a 750 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 = 750 \/ 18 = 125 \/ 3 m \/ sec . let the length of the platform be x meters . then , ( x + 750 ) \/ 39 = 125 \/ 3 = > x = 1625 m . l = 1625 - 750 = 875 answer : option b\"","correct":"b","options":{"a":"300 ","b":"875 ","c":"360 ","d":"770","e":"380"},"options_float":{"a":300.0,"b":875.0,"c":360.0,"d":770.0,"e":380.0},"annotated_formula":"subtract(multiply(speed(750, 18), 39), 750)","linear_formula":"speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|","chain":"750 \/ 18<\/gadget>\n125\/3 = around 41.666667<\/output>\n(125\/3) * 39<\/gadget>\n1_625<\/output>\n1_625 - 750<\/gadget>\n875<\/output>\n875<\/result>","index":2092} +{"problem":"when a certain number x is divided by 54 , the remainder is 22 . what is the remainder when x is divided by 18 ?","rationale":"\"let possible value of x is 76 least possible value of x \/ 18 is 76 \/ 18 = > 4 quotient with remainder 4 thus answer is ( d ) 4\"","correct":"d","options":{"a":"3 ","b":"1 ","c":"2 ","d":"4","e":"5"},"options_float":{"a":3.0,"b":1.0,"c":2.0,"d":4.0,"e":5.0},"annotated_formula":"reminder(22, 18)","linear_formula":"reminder(n1,n2)|","chain":"22 % 18<\/gadget>\n4<\/output>\n4<\/result>","index":2095} +{"problem":"marketing executives for a certain chewing gum company projected a 20 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 35 % . what percent of the projected revenue was the actual revenue ?","rationale":"\"last year revenue = 100 ( assume ) ; this year revenue = 65 ; projected revenue = 120 . actual \/ projected * 100 = 65 \/ 120 * 100 = 54.2 % . answer : a .\"","correct":"a","options":{"a":"54.2 % ","b":"58 % ","c":"62.5 % ","d":"64 %","e":"75 %"},"options_float":{"a":54.2,"b":58.0,"c":62.5,"d":64.0,"e":75.0},"annotated_formula":"multiply(divide(subtract(const_100, 35), add(20, const_100)), const_100)","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)|multiply(#2,const_100)|","chain":"100 - 35<\/gadget>\n65<\/output>\n20 + 100<\/gadget>\n120<\/output>\n65 \/ 120<\/gadget>\n13\/24 = around 0.541667<\/output>\n(13\/24) * 100<\/gadget>\n325\/6 = around 54.166667<\/output>\n325\/6 = around 54.166667<\/result>","index":2098} +{"problem":"mr . shah decided to walk down the escalator of a tube station . he found â that if he walks down 26 steps , he requires 30 seconds to reach the bottom . however , if he steps down 34 stairs he would only require 18 seconds to get to the bottom . if the time is measured from the moment the top step begins â to descend to the time he steps off the last step at the bottom , find out the height of the stair way in steps ?","rationale":"( s 1 * t 2 ~ s 2 * t 1 ) \/ ( t 2 ~ t 1 ) = ( 26 * 18 ~ 34 * 30 ) \/ ( 18 ~ 30 ) = 46 answer : c","correct":"c","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":"add(26, multiply(divide(subtract(34, 26), subtract(30, 18)), 30))","linear_formula":"subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n1,#2)|add(n0,#3)","chain":"34 - 26<\/gadget>\n8<\/output>\n30 - 18<\/gadget>\n12<\/output>\n8 \/ 12<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 30<\/gadget>\n20<\/output>\n26 + 20<\/gadget>\n46<\/output>\n46<\/result>","index":2099} +{"problem":"the average age of 18 students of a class is 18 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 18 th student is","rationale":"\"explanation : age of the 18 th student = [ 18 * 18 - ( 14 * 5 + 16 * 9 ) ] = ( 324 - 214 ) = 110 years . answer : c\"","correct":"c","options":{"a":"101 ","b":"66 ","c":"110 ","d":"160","e":"12"},"options_float":{"a":101.0,"b":66.0,"c":110.0,"d":160.0,"e":12.0},"annotated_formula":"subtract(multiply(18, 18), 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":"18 * 18<\/gadget>\n324<\/output>\n5 * 14<\/gadget>\n70<\/output>\n9 * 16<\/gadget>\n144<\/output>\n70 + 144<\/gadget>\n214<\/output>\n324 - 214<\/gadget>\n110<\/output>\n110<\/result>","index":2100} +{"problem":"on dividing 123 by a number , the quotient is 7 and the remainder is 4 . find the divisor .","rationale":"\"d = ( d - r ) \/ q = ( 123 - 4 ) \/ 7 = 119 \/ 7 = 17 c\"","correct":"c","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":"floor(divide(123, 7))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"123 \/ 7<\/gadget>\n123\/7 = around 17.571429<\/output>\nfloor(123\/7)<\/gadget>\n17<\/output>\n17<\/result>","index":2102} +{"problem":"a reduction of 10 % in the price of tea enables a dealer to purchase 25 kg more tea for rs . 22500 . what is the reduced price per kg of tea ?","rationale":"solution : 1 st method : let the original price of tea be rs . x \/ kg . after reduction the price becomes = x - 10 % of x = 9 x \/ 10 per kg . now , ( 22500 \/ ( 9 x \/ 10 ) ) - 22500 \/ x = 25 or , 22500 [ 10 \/ 9 x - 1 \/ x ] = 25 or , 25 * 9 x = 22500 ; or , x = ( 22500 \/ 2589 ) = rs . 100 . hence , new price = 90 per kg . thought process method : let the original price be rs . 100 per kg , he get tea = 22500 \/ 100 = 225 kg . after reduction the price becomes = 90 per kg , he get tea = 22500 \/ 90 = 250 kg . so , reduction price is rs . 90 per kg as it enables him to buy 25 kg of more tea . answer : option c","correct":"c","options":{"a":"rs . 70 ","b":"rs . 80 ","c":"rs . 90 ","d":"rs . 100","e":"none"},"options_float":{"a":70.0,"b":80.0,"c":90.0,"d":100.0,"e":null},"annotated_formula":"divide(divide(multiply(22500, subtract(divide(const_100, subtract(const_100, 10)), const_1)), 25), divide(const_100, subtract(const_100, 10)))","linear_formula":"subtract(const_100,n0)|divide(const_100,#0)|subtract(#1,const_1)|multiply(n2,#2)|divide(#3,n1)|divide(#4,#1)","chain":"100 - 10<\/gadget>\n90<\/output>\n100 \/ 90<\/gadget>\n10\/9 = around 1.111111<\/output>\n(10\/9) - 1<\/gadget>\n1\/9 = around 0.111111<\/output>\n22_500 * (1\/9)<\/gadget>\n2_500<\/output>\n2_500 \/ 25<\/gadget>\n100<\/output>\n100 \/ (10\/9)<\/gadget>\n90<\/output>\n90<\/result>","index":2103} +{"problem":"two pipes a and b can fill a tank in 36 hours and 45 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ?","rationale":"\"1 \/ ( 1 \/ 36 + 1 \/ 45 ) = 180 \/ 9 = 20 hrs answer : a\"","correct":"a","options":{"a":"20 hrs ","b":"22 hrs ","c":"23 hrs ","d":"24 hrs","e":"21 hrs"},"options_float":{"a":20.0,"b":22.0,"c":23.0,"d":24.0,"e":21.0},"annotated_formula":"divide(const_1, add(divide(const_1, 36), divide(const_1, 45)))","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 \/ 45<\/gadget>\n1\/45 = around 0.022222<\/output>\n(1\/36) + (1\/45)<\/gadget>\n1\/20 = around 0.05<\/output>\n1 \/ (1\/20)<\/gadget>\n20<\/output>\n20<\/result>","index":2107} +{"problem":"a certain country is divided into 6 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 \/ 18 the total number of progressives in the entire country , what fraction of the country is traditionalist ?","rationale":"\"let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p \/ 18 the total number of traditionalists is 6 p \/ 18 = p \/ 3 . the total population is p + p \/ 3 = 4 p \/ 3 p \/ ( 4 p \/ 3 ) = 3 \/ 4 the answer is e .\"","correct":"e","options":{"a":"1 \/ 5 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"2 \/ 3","e":"3 \/ 4"},"options_float":{"a":0.2,"b":0.3333333333,"c":0.5,"d":0.6666666667,"e":0.75},"annotated_formula":"subtract(1, divide(divide(6, 18), add(divide(6, 18), 1)))","linear_formula":"divide(n0,n2)|add(n1,#0)|divide(#0,#1)|subtract(n1,#2)|","chain":"6 \/ 18<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) + 1<\/gadget>\n4\/3 = around 1.333333<\/output>\n(1\/3) \/ (4\/3)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":2108} +{"problem":"in order to obtain an income of rs . 1250 from 10 % stock at rs . 210 , one must make an investment of","rationale":"\"to obtain rs . 10 , investment = rs . 210 . to obtain rs . 1250 , investment = = rs . 26250 . answer : b\"","correct":"b","options":{"a":"5363 ","b":"26250 ","c":"28678 ","d":"29002","e":"2732"},"options_float":{"a":5363.0,"b":26250.0,"c":28678.0,"d":29002.0,"e":2732.0},"annotated_formula":"multiply(divide(210, 10), 1250)","linear_formula":"divide(n2,n1)|multiply(n0,#0)|","chain":"210 \/ 10<\/gadget>\n21<\/output>\n21 * 1_250<\/gadget>\n26_250<\/output>\n26_250<\/result>","index":2110} +{"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 . 400 per sq . metre .","rationale":"\"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 400 x 20.625 ) = rs . 8,250 . answer b\"","correct":"b","options":{"a":"rs . 15,000 ","b":"rs . 8,250 ","c":"rs . 15,600 ","d":"rs . 16,500","e":"none"},"options_float":{"a":15000.0,"b":8250.0,"c":15600.0,"d":16500.0,"e":null},"annotated_formula":"multiply(400, multiply(5.5, 3.75))","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|","chain":"5.5 * 3.75<\/gadget>\n20.625<\/output>\n400 * 20.625<\/gadget>\n8_250<\/output>\n8_250<\/result>","index":2111} +{"problem":"given that 268 x 74 = 19632 , find the value of 2.68 x . 74 .","rationale":"\"solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9632 answer a\"","correct":"a","options":{"a":"1.9632 ","b":"1.0025 ","c":"1.5693 ","d":"1.0266","e":"none"},"options_float":{"a":1.9632,"b":1.0025,"c":1.5693,"d":1.0266,"e":null},"annotated_formula":"multiply(divide(268, const_100), divide(74, const_100))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|","chain":"268 \/ 100<\/gadget>\n67\/25 = around 2.68<\/output>\n74 \/ 100<\/gadget>\n37\/50 = around 0.74<\/output>\n(67\/25) * (37\/50)<\/gadget>\n2_479\/1_250 = around 1.9832<\/output>\n2_479\/1_250 = around 1.9832<\/result>","index":2112} +{"problem":"a student committee on academic integrity has 56 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 = 56 x ^ 2 - x - 56 = 0 ( x - 8 ) ( x + 7 ) = 0 x = 8 , - 7 - 7 ca n ' t possible . hence 8 should be the answer b","correct":"b","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(add(const_1, sqrt(add(multiply(const_4, 56), 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 * 56<\/gadget>\n224<\/output>\n-1<\/gadget>\n-1<\/output>\n(-1) ** 2<\/gadget>\n1<\/output>\n224 + 1<\/gadget>\n225<\/output>\n225 ** (1\/2)<\/gadget>\n15<\/output>\n1 + 15<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2113} +{"problem":"a businessman earns $ 26800 in december , thus decreasing his average annual ( january to december ) earnings by $ 1200 . his average annual earnings would be source : cmat preparation","rationale":"( x - 26,800 ) \/ 11 - x \/ 12 = 1,200 x = 480,000 x \/ 12 = 40,000 answer : c .","correct":"c","options":{"a":"$ 29000 ","b":"$ 33500 ","c":"$ 40000 ","d":"$ 41000","e":"$ 42300"},"options_float":{"a":29000.0,"b":33500.0,"c":40000.0,"d":41000.0,"e":42300.0},"annotated_formula":"add(26800, multiply(const_12, 1200))","linear_formula":"multiply(n1,const_12)|add(n0,#0)","chain":"12 * 1_200<\/gadget>\n14_400<\/output>\n26_800 + 14_400<\/gadget>\n41_200<\/output>\n41_200<\/result>","index":2114} +{"problem":"a can complete a certain job in 16 days . b is 60 % more efficient than a . in how many days can b complete the same job ?","rationale":"let , total work unit = 160 units a can finish in 16 days = 160 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 60 \/ 100 ) * 10 = 16 unit work days in which b will complete the work alone = 160 \/ 16 = 10 days answer : option e","correct":"e","options":{"a":"6 ","b":"6.25 ","c":"7 ","d":"7.5","e":"10"},"options_float":{"a":6.0,"b":6.25,"c":7.0,"d":7.5,"e":10.0},"annotated_formula":"divide(multiply(16, 60), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)","chain":"16 * 60<\/gadget>\n960<\/output>\n960 \/ 100<\/gadget>\n48\/5 = around 9.6<\/output>\n48\/5 = around 9.6<\/result>","index":2115} +{"problem":"a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 7 % of bananas were rotten . find the percentage of fruits in good condition ?","rationale":"\"total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 \/ 100 × 600 = 9000 \/ 100 = 90 number of rotten bananas = 7 % of 400 = 28 therefore , total number of rotten fruits = 90 + 28 = 118 therefore number of fruits in good condition = 1000 - 118 = 882 therefore percentage of fruits in good condition = ( 882 \/ 1000 × 100 ) % = ( 88200 \/ 1000 ) % = 88.2 % answer : b\"","correct":"b","options":{"a":"92.5 % ","b":"88.2 % ","c":"85.2 % ","d":"96.8 %","e":"78.9 %"},"options_float":{"a":92.5,"b":88.2,"c":85.2,"d":96.8,"e":78.9},"annotated_formula":"multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(7, const_100)))), add(600, 400)), const_100)","linear_formula":"add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)|","chain":"600 + 400<\/gadget>\n1_000<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n600 * (3\/20)<\/gadget>\n90<\/output>\n7 \/ 100<\/gadget>\n7\/100 = around 0.07<\/output>\n400 * (7\/100)<\/gadget>\n28<\/output>\n90 + 28<\/gadget>\n118<\/output>\n1_000 - 118<\/gadget>\n882<\/output>\n882 \/ 1_000<\/gadget>\n441\/500 = around 0.882<\/output>\n(441\/500) * 100<\/gadget>\n441\/5 = around 88.2<\/output>\n441\/5 = around 88.2<\/result>","index":2116} +{"problem":"sandy had $ 231 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ?","rationale":"\"let the money sandy took for shopping be x . 0.7 x = 231 x = 330 the answer is d .\"","correct":"d","options":{"a":"$ 270 ","b":"$ 290 ","c":"$ 310 ","d":"$ 330","e":"$ 350"},"options_float":{"a":270.0,"b":290.0,"c":310.0,"d":330.0,"e":350.0},"annotated_formula":"divide(231, divide(subtract(const_100, 30), const_100))","linear_formula":"subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|","chain":"100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n231 \/ (7\/10)<\/gadget>\n330<\/output>\n330<\/result>","index":2117} +{"problem":"3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 12 pumps work to empty the tank in 1 day ?","rationale":"\"3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 12 pumps then 48 \/ 12 = 4 hrs . answer : a\"","correct":"a","options":{"a":"4 ","b":"10 ","c":"11 ","d":"12","e":"13"},"options_float":{"a":4.0,"b":10.0,"c":11.0,"d":12.0,"e":13.0},"annotated_formula":"divide(multiply(multiply(3, 8), 2), 12)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|","chain":"3 * 8<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\n48 \/ 12<\/gadget>\n4<\/output>\n4<\/result>","index":2118} +{"problem":"the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 70 \/ kg . how much did a kg of coffee cost in june ?","rationale":"\"let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 \/ 2 kg ( 700 gm ) of tea and 1 \/ 2 kg ( 700 gm ) of coffee ( equal quantities ) = 70 1.2 x ( 1 \/ 2 ) + 0.8 x ( 1 \/ 2 ) = 70 = > x = 70 thus proved . . . option a .\"","correct":"a","options":{"a":"70 ","b":"60 ","c":"80 ","d":"100","e":"120"},"options_float":{"a":70.0,"b":60.0,"c":80.0,"d":100.0,"e":120.0},"annotated_formula":"divide(70, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1)))","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":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(1\/5) + 1<\/gadget>\n6\/5 = around 1.2<\/output>\n(4\/5) * (6\/5)<\/gadget>\n24\/25 = around 0.96<\/output>\n70 \/ (24\/25)<\/gadget>\n875\/12 = around 72.916667<\/output>\n875\/12 = around 72.916667<\/result>","index":2119} +{"problem":"a sum of rs . 66000 is divided into 3 parts such that the simple interests accrued on them for 6 , two and 11 years respectively may be equal . find the amount deposited for 11 years .","rationale":"let the amounts be x , y , z in ascending order of value . as the interest rate and interest accrued are same for 2 years 6 years and 11 years i . e . 2 x = 6 y = 11 z = k . l . c . m . of 2 , 611 = 66 so x : y : z : = 33000 : 11000 : 6000 the amount deposited for 11 years = 6000 answer : e","correct":"e","options":{"a":"6500 ","b":"2000 ","c":"4500 ","d":"3000","e":"6000"},"options_float":{"a":6500.0,"b":2000.0,"c":4500.0,"d":3000.0,"e":6000.0},"annotated_formula":"multiply(multiply(6, const_10), const_100)","linear_formula":"multiply(n2,const_10)|multiply(#0,const_100)","chain":"6 * 10<\/gadget>\n60<\/output>\n60 * 100<\/gadget>\n6_000<\/output>\n6_000<\/result>","index":2120} +{"problem":"in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what percent z of all the people in smithtown are left - handed women ?","rationale":"looking at the ratio we can take total number of people = 20 . . ans 5 \/ 20 or 25 % c","correct":"c","options":{"a":"50 % ","b":"40 % ","c":"25 % ","d":"20 %","e":"10 %"},"options_float":{"a":50.0,"b":40.0,"c":25.0,"d":20.0,"e":10.0},"annotated_formula":"multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100)","linear_formula":"add(n0,n1)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|multiply(n0,#2)|subtract(n0,#4)|subtract(#3,#5)|divide(#6,#0)|multiply(#7,const_100)","chain":"3 + 1<\/gadget>\n4<\/output>\n3 + 2<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n2 * (4\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n3 * (4\/5)<\/gadget>\n12\/5 = around 2.4<\/output>\n3 - (12\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(8\/5) - (3\/5)<\/gadget>\n1<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2122} +{"problem":"the average age of 20 men in the class is 15.6 years . 5 new men join and the new average becomes 14.56 years . what was the average age of 5 new men ?","rationale":"total age of 20 men = 15.6 x 20 = 312 now , total age of 25 men = 364 . total age of five men added later = 364 - 312 = 52 . hence , the total average of five men = 52 \/ 5 = 10.4 answer : d","correct":"d","options":{"a":"15.5 ","b":"15.4 ","c":"15.25 ","d":"10.4","e":"15.6"},"options_float":{"a":15.5,"b":15.4,"c":15.25,"d":10.4,"e":15.6},"annotated_formula":"divide(subtract(multiply(add(20, 5), 14.56), multiply(20, 15.6)), 5)","linear_formula":"add(n0,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2)","chain":"20 + 5<\/gadget>\n25<\/output>\n25 * 14.56<\/gadget>\n364<\/output>\n20 * 15.6<\/gadget>\n312<\/output>\n364 - 312<\/gadget>\n52<\/output>\n52 \/ 5<\/gadget>\n52\/5 = around 10.4<\/output>\n52\/5 = around 10.4<\/result>","index":2124} +{"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 . tee age of the 15 th student is ?","rationale":"\"age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = ( 225 - 214 ) = 11 years . answer : a\"","correct":"a","options":{"a":"11 years ","b":"17 years ","c":"67 years ","d":"14 years","e":"12 years"},"options_float":{"a":11.0,"b":17.0,"c":67.0,"d":14.0,"e":12.0},"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":2128} +{"problem":"rob also compared the empire state building and the petronas towers . what is the height difference between the two if the empire state building is 435 m tall and the petronas towers is 458 m tall ?","rationale":"458 - 435 = 23 . answer is c .","correct":"c","options":{"a":"9 ","b":"17 ","c":"23 ","d":"45","e":"12"},"options_float":{"a":9.0,"b":17.0,"c":23.0,"d":45.0,"e":12.0},"annotated_formula":"subtract(458, 435)","linear_formula":"subtract(n1,n0)","chain":"458 - 435<\/gadget>\n23<\/output>\n23<\/result>","index":2129} +{"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 $ 1000 . if the amount of the import tax that the merchant paid was $ 109.90 , what was the total value of the item ?","rationale":"let x be the value of the item . 0.07 * ( x - 1000 ) = 109.90 x = 2570 the answer is d .","correct":"d","options":{"a":"$ 1940 ","b":"$ 2150 ","c":"$ 2360 ","d":"$ 2570","e":"$ 2780"},"options_float":{"a":1940.0,"b":2150.0,"c":2360.0,"d":2570.0,"e":2780.0},"annotated_formula":"add(1000, divide(109.9, divide(7, const_100)))","linear_formula":"divide(n0,const_100)|divide(n2,#0)|add(n1,#1)","chain":"7 \/ 100<\/gadget>\n7\/100 = around 0.07<\/output>\n109.9 \/ (7\/100)<\/gadget>\n1_570<\/output>\n1_000 + 1_570<\/gadget>\n2_570<\/output>\n2_570<\/result>","index":2131} +{"problem":"find the area , diameter = 11 m .","rationale":"diameter = 11 meter . radius = diameter \/ 2 . = 11 \/ 2 . = 5.5 meter . area of a circle = ï € r 2 . here , pi ( ï € ) = 3.14 meter , radius ( r ) = 5.5 . area of a circle = 3.14 ã — 5.5 ã — 5.5 . . = 3.14 ã — 30.25 . = 95.07 square meter answer : b","correct":"b","options":{"a":"113.00 square meter ","b":"95.07 square meter ","c":"93.08 square meter ","d":"93.24 square meter","e":"113.43 square meter"},"options_float":{"a":113.0,"b":95.07,"c":93.08,"d":93.24,"e":113.43},"annotated_formula":"circle_area(divide(11, const_2))","linear_formula":"divide(n0,const_2)|circle_area(#0)","chain":"11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\npi * ((11\/2) ** 2)<\/gadget>\n121*pi\/4 = around 95.033178<\/output>\n121*pi\/4 = around 95.033178<\/result>","index":2133} +{"problem":"when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 5 , the remainder is 4 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 11 ?","rationale":"when positive integer n is divided by 3 , the remainder is 1 i . e . , n = 3 x + 1 values of n can be one of { 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 . . . . . . . . . . . . . . 49 , 52 , 59 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 5 , the remainder is 5 . . i . e . , n = 5 y + 4 values of n can be one of { 4 , 9 , 14 , 19 , . . . } combining both the sets we get n = { 4,19 , 52 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 11 or 11 x in case of n = 4 p = 7 so for min value of p , we take min value of n . d is the answer .","correct":"d","options":{"a":"1 ","b":"2 ","c":"5 ","d":"7","e":"20"},"options_float":{"a":1.0,"b":2.0,"c":5.0,"d":7.0,"e":20.0},"annotated_formula":"subtract(11, reminder(4, 5))","linear_formula":"reminder(n3,n2)|subtract(n4,#0)","chain":"4 % 5<\/gadget>\n4<\/output>\n11 - 4<\/gadget>\n7<\/output>\n7<\/result>","index":2134} +{"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 . 12400 after a year ?","rationale":"\"6300 : 4200 : 10500 3 : 2 : 5 3 \/ 10 * 12400 = 3720 . answer : c\"","correct":"c","options":{"a":"3630 ","b":"2881 ","c":"3720 ","d":"9977","e":"2212"},"options_float":{"a":3630.0,"b":2881.0,"c":3720.0,"d":9977.0,"e":2212.0},"annotated_formula":"multiply(divide(6300, add(add(6300, 4200), 10500)), 12400)","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_400<\/gadget>\n3_720<\/output>\n3_720<\/result>","index":2135} +{"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 100 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?","rationale":"sol . speed of the first train = [ 100 \/ 10 ] m \/ sec = 10 m \/ sec . speed of the second train = [ 100 \/ 15 ] m \/ sec = 6.7 m \/ sec . relative speed = ( 10 + 6.7 ) = m \/ sec = 16.7 m \/ sec . ∴ required time = ( 100 + 100 ) \/ 16.7 secc = 11.9 sec . answer b","correct":"b","options":{"a":"12 ","b":"11.9 ","c":"16 ","d":"20","e":"18"},"options_float":{"a":12.0,"b":11.9,"c":16.0,"d":20.0,"e":18.0},"annotated_formula":"divide(add(100, 100), add(divide(100, 15), divide(100, 10)))","linear_formula":"add(n2,n2)|divide(n2,n1)|divide(n2,n0)|add(#1,#2)|divide(#0,#3)","chain":"100 + 100<\/gadget>\n200<\/output>\n100 \/ 15<\/gadget>\n20\/3 = around 6.666667<\/output>\n100 \/ 10<\/gadget>\n10<\/output>\n(20\/3) + 10<\/gadget>\n50\/3 = around 16.666667<\/output>\n200 \/ (50\/3)<\/gadget>\n12<\/output>\n12<\/result>","index":2136} +{"problem":"if 40 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ?","rationale":"\"40 % of ( x - y ) = 20 % of ( x + y ) 40 \/ 100 ( x - y ) = 20 \/ 100 ( x + y ) x = 3 y required percentage = y \/ x * 100 = y \/ 3 y * 100 = 33.3 % answer is d\"","correct":"d","options":{"a":"50.5 % ","b":"44.4 % ","c":"22.2 % ","d":"33.3 %","e":"25 %"},"options_float":{"a":50.5,"b":44.4,"c":22.2,"d":33.3,"e":25.0},"annotated_formula":"multiply(divide(subtract(40, 20), add(40, 20)), const_100)","linear_formula":"add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|","chain":"40 - 20<\/gadget>\n20<\/output>\n40 + 20<\/gadget>\n60<\/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":2137} +{"problem":"in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 170 m . then , a wins by :","rationale":"\"to reach the winning post a will have to cover a distance of ( 500 - 170 ) m , i . e . , 330 m . while a covers 3 m , b covers 4 m . while a covers 330 m , b covers 4 x 330 \/ 3 m = 440 m . thus , when a reaches the winning post , b covers 440 m and therefore remains 60 m behind . a wins by 60 m . answer : a\"","correct":"a","options":{"a":"60 m ","b":"20 m ","c":"43 m ","d":"20 m","e":"23 m"},"options_float":{"a":60.0,"b":20.0,"c":43.0,"d":20.0,"e":23.0},"annotated_formula":"subtract(500, divide(multiply(subtract(500, 170), 4), 3))","linear_formula":"subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)|","chain":"500 - 170<\/gadget>\n330<\/output>\n330 * 4<\/gadget>\n1_320<\/output>\n1_320 \/ 3<\/gadget>\n440<\/output>\n500 - 440<\/gadget>\n60<\/output>\n60<\/result>","index":2138} +{"problem":"from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a red king ?","rationale":"p ( club ) = 1 \/ 4 p ( red king ) = 1 \/ 26 p ( club then a red king ) = 1 \/ 4 * 1 \/ 26 = 1 \/ 104 the answer is e .","correct":"e","options":{"a":"1 \/ 13 ","b":"1 \/ 15 ","c":"1 \/ 26 ","d":"1 \/ 52","e":"1 \/ 104"},"options_float":{"a":0.0769230769,"b":0.0666666667,"c":0.0384615385,"d":0.0192307692,"e":0.0096153846},"annotated_formula":"multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52))","linear_formula":"divide(const_2,const_52)|multiply(const_3,const_4)|add(#1,const_1)|divide(#2,const_52)|multiply(#3,#0)","chain":"3 * 4<\/gadget>\n12<\/output>\n12 + 1<\/gadget>\n13<\/output>\n13 \/ 52<\/gadget>\n1\/4 = around 0.25<\/output>\n2 \/ 52<\/gadget>\n1\/26 = around 0.038462<\/output>\n(1\/4) * (1\/26)<\/gadget>\n1\/104 = around 0.009615<\/output>\n1\/104 = around 0.009615<\/result>","index":2139} +{"problem":"a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many cubes are painted on only 2 faces ?","rationale":"the mini - cubes with 2 painted sides are all on the edge of the cube , in the ` ` middle ' ' of the edge . there are 4 in front , 4 in back and 4 more on the ` ` strip ' ' that runs around the left \/ top \/ right \/ bottom of the cube . 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":2140} +{"problem":"a man can do a piece of work in 6 days , but with the help of his son he can do it in 3 days . in what time can the son do it alone ?","rationale":"\"explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 \/ 3 − 1 \/ 6 ) = ( 6 − 3 ) \/ 18 = 1 \/ 6 so son will do whole work in 6 days answer : b\"","correct":"b","options":{"a":"7 days ","b":"6 days ","c":"5 days ","d":"4 days","e":"none of these"},"options_float":{"a":7.0,"b":6.0,"c":5.0,"d":4.0,"e":null},"annotated_formula":"divide(multiply(6, 3), subtract(6, 3))","linear_formula":"multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|","chain":"6 * 3<\/gadget>\n18<\/output>\n6 - 3<\/gadget>\n3<\/output>\n18 \/ 3<\/gadget>\n6<\/output>\n6<\/result>","index":2142} +{"problem":"if 3 < x < 6 < y < 9 , then what is the greatest possible positive integer difference of x and y ?","rationale":"\"3 < x < 6 < y < 9 ; 3 < x y < 9 3 + y < x + 9 y - x < 6 . positive integer difference is 5 ( for example y = 8.5 and x = 3.5 ) answer : 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":"subtract(subtract(9, 3), const_1)","linear_formula":"subtract(n2,n0)|subtract(#0,const_1)|","chain":"9 - 3<\/gadget>\n6<\/output>\n6 - 1<\/gadget>\n5<\/output>\n5<\/result>","index":2144} +{"problem":"if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 150 % of j ?","rationale":"\"imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 20 % of 350 , comes out to be 70 . . . . 150 % of 10 is 15 . . . . ( 70 * 100 ) \/ 15 = 466.66 . . . . ans : b\"","correct":"b","options":{"a":"0.35 ","b":"466 ","c":"35 ","d":"350","e":"3500"},"options_float":{"a":0.35,"b":466.0,"c":35.0,"d":350.0,"e":3500.0},"annotated_formula":"multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 20), 150), const_100)","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100)|","chain":"125 * 150<\/gadget>\n18_750<\/output>\n18_750 * 175<\/gadget>\n3_281_250<\/output>\n25 * 50<\/gadget>\n1_250<\/output>\n1_250 * 75<\/gadget>\n93_750<\/output>\n3_281_250 \/ 93_750<\/gadget>\n35<\/output>\n35 * 20<\/gadget>\n700<\/output>\n700 \/ 150<\/gadget>\n14\/3 = around 4.666667<\/output>\n(14\/3) * 100<\/gadget>\n1_400\/3 = around 466.666667<\/output>\n1_400\/3 = around 466.666667<\/result>","index":2145} +{"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 22 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 22 ) = rs . 176 . profit = 176 - 150 = 26 so , profit % = ( 26 \/ 150 ) * 100 = 17 % option e\"","correct":"e","options":{"a":"12 % ","b":"13 % ","c":"14 % ","d":"15 %","e":"17 %"},"options_float":{"a":12.0,"b":13.0,"c":14.0,"d":15.0,"e":17.0},"annotated_formula":"divide(multiply(subtract(multiply(22, 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>\n22 * 8<\/gadget>\n176<\/output>\n5 * 18<\/gadget>\n90<\/output>\n3 * 20<\/gadget>\n60<\/output>\n90 + 60<\/gadget>\n150<\/output>\n176 - 150<\/gadget>\n26<\/output>\n26 * 100<\/gadget>\n2_600<\/output>\n2_600 \/ 150<\/gadget>\n52\/3 = around 17.333333<\/output>\n52\/3 = around 17.333333<\/result>","index":2146} +{"problem":"the cost price of 20 articles is the same as the selling price of x articles . if the profit is 25 % , find out the value of x","rationale":"\"explanation : let the cost price of one article = rs . 1 cp of x articles = rs . x cp of 20 articles = 20 selling price of x articles = 20 profit = 25 % [ given ] ⇒ ( sp − cp \/ cp ) = 25 \/ 100 = 1 \/ 4 ⇒ ( 20 − x ) \/ x = 1 \/ 4 ⇒ 80 − 4 x = x ⇒ 5 x = 80 option d ⇒ x = 805 = 16\"","correct":"d","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":"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":2147} +{"problem":"an error 5 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ?","rationale":"\"percentage error in calculated area = ( 5 + 5 + ( 5 ã — 5 ) \/ 100 ) % = 10.25 % answer : d\"","correct":"d","options":{"a":"4.05 % ","b":"4.02 % ","c":"4 % ","d":"10.28 %","e":"2 %"},"options_float":{"a":4.05,"b":4.02,"c":4.0,"d":10.28,"e":2.0},"annotated_formula":"divide(multiply(subtract(square_area(add(const_100, 5)), 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 + 5<\/gadget>\n105<\/output>\n105 ** 2<\/gadget>\n11_025<\/output>\n100 ** 2<\/gadget>\n10_000<\/output>\n11_025 - 10_000<\/gadget>\n1_025<\/output>\n1_025 * 100<\/gadget>\n102_500<\/output>\n102_500 \/ 10_000<\/gadget>\n41\/4 = around 10.25<\/output>\n41\/4 = around 10.25<\/result>","index":2149} +{"problem":"a can give b 100 meters start and c 170 meters start in a kilometer race . how much start can b give c in a kilometer race ?","rationale":"\"explanation : a runs 1000 meters while b runs 900 meters and c runs 830 meters . therefore , b runs 900 meters while c runs 830 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 830 ) \/ 900 = 922.22 meters thus , b can give c ( 1000 - 922.22 ) = 77.77 meters start answer : c\"","correct":"c","options":{"a":"11.77 meters ","b":"55.77 meters ","c":"77.77 meters ","d":"113.77 meters","e":"none of these"},"options_float":{"a":11.77,"b":55.77,"c":77.77,"d":113.77,"e":null},"annotated_formula":"subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 170)), 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 - 170<\/gadget>\n830<\/output>\n1_000 * 830<\/gadget>\n830_000<\/output>\n1_000 - 100<\/gadget>\n900<\/output>\n830_000 \/ 900<\/gadget>\n8_300\/9 = around 922.222222<\/output>\n1_000 - (8_300\/9)<\/gadget>\n700\/9 = around 77.777778<\/output>\n700\/9 = around 77.777778<\/result>","index":2151} +{"problem":"company t produces two kinds of stereos : basic and deluxe . of the stereos produced by company t last month , 2 \/ 3 were basic and the rest were deluxe . if it takes 7 \/ 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ?","rationale":"# of basic stereos was 2 \/ 3 of total and # of deluxe stereos was 1 \/ 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 7 \/ 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 7 \/ 5 = 7 - - > total time for both of them would be 10 + 7 = 17 - - > deluxe \/ total = 7 \/ 17 . b","correct":"b","options":{"a":"5 \/ 17 ","b":"7 \/ 17 ","c":"4 \/ 17 ","d":"3 \/ 17","e":"5"},"options_float":{"a":0.2941176471,"b":0.4117647059,"c":0.2352941176,"d":0.1764705882,"e":5.0},"annotated_formula":"divide(multiply(5, divide(7, 5)), add(multiply(multiply(2, 5), const_1), multiply(5, divide(7, 5))))","linear_formula":"divide(n2,n3)|multiply(n0,n3)|multiply(n3,#0)|multiply(#1,const_1)|add(#3,#2)|divide(#2,#4)","chain":"7 \/ 5<\/gadget>\n7\/5 = around 1.4<\/output>\n5 * (7\/5)<\/gadget>\n7<\/output>\n2 * 5<\/gadget>\n10<\/output>\n10 * 1<\/gadget>\n10<\/output>\n10 + 7<\/gadget>\n17<\/output>\n7 \/ 17<\/gadget>\n7\/17 = around 0.411765<\/output>\n7\/17 = around 0.411765<\/result>","index":2152} +{"problem":"by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 45 % ?","rationale":"\"sp = 800 profit = 25 % cp = ( sp ) * [ 100 \/ ( 100 + p ) ] = 800 * [ 100 \/ 125 ] = 640 loss = 45 % = 45 % of 640 = rs . 288 sp = cp - loss = 640 - 288 = rs . 352 answer : b\"","correct":"b","options":{"a":"s . 429 ","b":"s . 352 ","c":"s . 429 ","d":"s . 128","e":"s . 419"},"options_float":{"a":429.0,"b":352.0,"c":429.0,"d":128.0,"e":419.0},"annotated_formula":"subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 45), 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":"800 * 100<\/gadget>\n80_000<\/output>\n25 + 100<\/gadget>\n125<\/output>\n80_000 \/ 125<\/gadget>\n640<\/output>\n640 * 45<\/gadget>\n28_800<\/output>\n28_800 \/ 100<\/gadget>\n288<\/output>\n640 - 288<\/gadget>\n352<\/output>\n352<\/result>","index":2155} +{"problem":"at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 5 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ?","rationale":"\"i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 5 ) = $ 200 total revenue : 1000 + 200 = $ 1200 the answer is b .\"","correct":"b","options":{"a":"$ 1220 ","b":"$ 1200 ","c":"$ 1300 ","d":"$ 1340","e":"$ 1880"},"options_float":{"a":1220.0,"b":1200.0,"c":1300.0,"d":1340.0,"e":1880.0},"annotated_formula":"add(multiply(add(10, 5), 30), multiply(subtract(100, 30), 10))","linear_formula":"add(n0,n1)|subtract(n2,n3)|multiply(n3,#0)|multiply(n0,#1)|add(#2,#3)|","chain":"10 + 5<\/gadget>\n15<\/output>\n15 * 30<\/gadget>\n450<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 * 10<\/gadget>\n700<\/output>\n450 + 700<\/gadget>\n1_150<\/output>\n1_150<\/result>","index":2156} +{"problem":"pat , kate , and mark charged a total of 126 hours to a certain project . if pat charged twice as much time to the project as kate and 1 \/ 3 as much time as mark , how many more hours did mark charge to the project than kate ?","rationale":"\"70 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 126 = > x = 14 28 : 14 : 84 84 - 14 = 70 answer d\"","correct":"d","options":{"a":"18 ","b":"36 ","c":"72 ","d":"70","e":"108"},"options_float":{"a":18.0,"b":36.0,"c":72.0,"d":70.0,"e":108.0},"annotated_formula":"subtract(divide(126, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(126, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2)))","linear_formula":"divide(n1,n2)|multiply(n2,const_2)|add(n1,#0)|divide(n1,#1)|add(#2,#3)|divide(n0,#4)|divide(#5,#1)|subtract(#5,#6)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 + (1\/3)<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 * 2<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(4\/3) + (1\/6)<\/gadget>\n3\/2 = around 1.5<\/output>\n126 \/ (3\/2)<\/gadget>\n84<\/output>\n84 \/ 6<\/gadget>\n14<\/output>\n84 - 14<\/gadget>\n70<\/output>\n70<\/result>","index":2157} +{"problem":"calculate the area of a triangle , if the sides of are 52 cm , 48 cm and 20 cm , what is its area ?","rationale":"\"the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 \/ 2 * 48 * 20 = 480 cm 2 answer : a\"","correct":"a","options":{"a":"480 cm 2 ","b":"580 cm 2 ","c":"380 cm 2 ","d":"180 cm 2","e":"280 cm 2"},"options_float":{"a":480.0,"b":580.0,"c":380.0,"d":180.0,"e":280.0},"annotated_formula":"multiply(divide(48, const_2), 20)","linear_formula":"divide(n1,const_2)|multiply(n2,#0)|","chain":"48 \/ 2<\/gadget>\n24<\/output>\n24 * 20<\/gadget>\n480<\/output>\n480<\/result>","index":2158} +{"problem":"the amount of time that three people worked on a special project was in the ratio of 2 to 4 to 6 . if the project took 144 hours , how many more hours did the hardest working person work than the person who worked the least ?","rationale":"let the persons be a , b , c . hours worked : a = 2 * 144 \/ 12 = 24 hours b = 4 * 144 \/ 12 = 48 hours c = 6 * 144 \/ 12 = 72 hours c is the hardest worker and a worked for the least number of hours . so the difference is 72 - 24 = 48 hours . answer : c","correct":"c","options":{"a":"47 hours ","b":"45 hours ","c":"48 hours ","d":"49 hours","e":"50 hours"},"options_float":{"a":47.0,"b":45.0,"c":48.0,"d":49.0,"e":50.0},"annotated_formula":"subtract(multiply(divide(144, add(add(2, 4), 6)), 6), multiply(divide(144, add(add(2, 4), 6)), 2))","linear_formula":"add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|multiply(n0,#2)|subtract(#3,#4)","chain":"2 + 4<\/gadget>\n6<\/output>\n6 + 6<\/gadget>\n12<\/output>\n144 \/ 12<\/gadget>\n12<\/output>\n12 * 6<\/gadget>\n72<\/output>\n12 * 2<\/gadget>\n24<\/output>\n72 - 24<\/gadget>\n48<\/output>\n48<\/result>","index":2160} +{"problem":"a car is purchased on hire - purchase . the cash price is $ 24 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ?","rationale":"\"explanation : cash price = $ 24 000 deposit = 10 % ã — $ 24 000 = $ 2400 loan amount = $ 24000 â ˆ ’ $ 2400 number of payments = 60 = $ 21600 i = p * r * t \/ 100 i = 12960 total amount = 21600 + 12960 = $ 34560 regular payment = total amount \/ number of payments = 576 answer : d\"","correct":"d","options":{"a":"$ 503 ","b":"$ 504 ","c":"$ 555 ","d":"$ 576","e":"$ 587"},"options_float":{"a":503.0,"b":504.0,"c":555.0,"d":576.0,"e":587.0},"annotated_formula":"add(divide(multiply(multiply(24, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(24, const_1000), subtract(const_1, divide(10, const_100)))))","linear_formula":"divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)|","chain":"24 * 1_000<\/gadget>\n24_000<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n24_000 * (9\/10)<\/gadget>\n21_600<\/output>\n21_600 \/ 60<\/gadget>\n360<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n(3\/25) \/ 12<\/gadget>\n1\/100 = around 0.01<\/output>\n(1\/100) * 21_600<\/gadget>\n216<\/output>\n360 + 216<\/gadget>\n576<\/output>\n576<\/result>","index":2162} +{"problem":"a truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes . what is the ratio of their speed ?","rationale":"\"explanation : speed of the truck = distance \/ time = 550 \/ 1 = 550 meters \/ minute speed of the train = distance \/ time = 33 \/ 45 km \/ minute = 33000 \/ 45 meters \/ minut speed of the truck \/ speed of the train = 550 \/ ( 33000 \/ 45 ) = ( 550 × 45 ) \/ 33000 = ( 55 × 45 ) \/ 3300 = ( 11 × 45 ) \/ 660 = ( 11 × 9 ) \/ 132 = 9 \/ 12 = 34 hence , speed of the truck : speed of the train = 3 : 4 answer : option d\"","correct":"d","options":{"a":"3 : 7 ","b":"4 : 7 ","c":"1 : 4 ","d":"3 : 4","e":"2 : 5"},"options_float":{"a":0.4285714286,"b":0.5714285714,"c":0.25,"d":0.75,"e":0.4},"annotated_formula":"divide(550, multiply(divide(33, 45), const_1000))","linear_formula":"divide(n2,n3)|multiply(#0,const_1000)|divide(n0,#1)|","chain":"33 \/ 45<\/gadget>\n11\/15 = around 0.733333<\/output>\n(11\/15) * 1_000<\/gadget>\n2_200\/3 = around 733.333333<\/output>\n550 \/ (2_200\/3)<\/gadget>\n3\/4 = around 0.75<\/output>\n3\/4 = around 0.75<\/result>","index":2163} +{"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 380 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 - 380 ) = 2 + 2 ( extra ) = > 2 x - 760 = x = > x = 760 . answer : c\"","correct":"c","options":{"a":"237 ","b":"287 ","c":"760 ","d":"287","e":"720"},"options_float":{"a":237.0,"b":287.0,"c":760.0,"d":287.0,"e":720.0},"annotated_formula":"multiply(380, const_2)","linear_formula":"multiply(n0,const_2)|","chain":"380 * 2<\/gadget>\n760<\/output>\n760<\/result>","index":2165} +{"problem":"in town p , 70 percent of the population are employed , and 42 percent of the population are employed males . what percent of the employed people in town p are females ?","rationale":"\"the percent of the population who are employed females is 70 - 42 = 28 % the percent of employed people who are female is 28 % \/ 70 % = 40 % . the answer is d .\"","correct":"d","options":{"a":"25 % ","b":"30 % ","c":"35 % ","d":"40 %","e":"45 %"},"options_float":{"a":25.0,"b":30.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"multiply(divide(subtract(70, 42), 70), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"70 - 42<\/gadget>\n28<\/output>\n28 \/ 70<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":2166} +{"problem":"andy solves problems 74 to 135 inclusive in a math exercise . how many problems does he solve ?","rationale":"\"135 - 74 + 1 = 62 ' b ' is the answer\"","correct":"b","options":{"a":"53 ","b":"62 ","c":"51 ","d":"50","e":"49"},"options_float":{"a":53.0,"b":62.0,"c":51.0,"d":50.0,"e":49.0},"annotated_formula":"add(subtract(135, 74), const_1)","linear_formula":"subtract(n1,n0)|add(#0,const_1)|","chain":"135 - 74<\/gadget>\n61<\/output>\n61 + 1<\/gadget>\n62<\/output>\n62<\/result>","index":2170} +{"problem":"a number increased by 25 % gives 520 . the number is ?","rationale":"\"formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 25 % = 125 % 125 % - - - - - - - > 520 ( 120 ã — 4.16 = 520 ) 100 % - - - - - - - > 416 ( 100 ã — 4.16 = 416 ) option ' e '\"","correct":"e","options":{"a":"216 ","b":"316 ","c":"616 ","d":"516","e":"416"},"options_float":{"a":216.0,"b":316.0,"c":616.0,"d":516.0,"e":416.0},"annotated_formula":"divide(520, 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>\n520 \/ (5\/4)<\/gadget>\n416<\/output>\n416<\/result>","index":2172} +{"problem":"joe drives 240 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ?","rationale":"\"t 1 = 240 \/ 60 = 4 hours t 2 = 120 \/ 40 = 3 hours t = t 1 + t 2 = 7 hours avg speed = total distance \/ t = 360 \/ 7 = 51 mph = b\"","correct":"b","options":{"a":"42 ","b":"51 ","c":"50 ","d":"54","e":"56"},"options_float":{"a":42.0,"b":51.0,"c":50.0,"d":54.0,"e":56.0},"annotated_formula":"divide(add(240, 120), add(divide(240, 60), divide(120, 40)))","linear_formula":"add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)|","chain":"240 + 120<\/gadget>\n360<\/output>\n240 \/ 60<\/gadget>\n4<\/output>\n120 \/ 40<\/gadget>\n3<\/output>\n4 + 3<\/gadget>\n7<\/output>\n360 \/ 7<\/gadget>\n360\/7 = around 51.428571<\/output>\n360\/7 = around 51.428571<\/result>","index":2174} +{"problem":"what is the sum of the multiples of 4 between 38 and 127 inclusive ?","rationale":"\"the fastest way in an ap is to find the average and multiply with total integers . . between 38 and 127 , the smallest multiple of 4 is 40 and largest = 124 . . average = ( 40 + 124 ) \/ 2 = 164 \/ 2 = 82 . . total numbers = ( 124 - 40 ) \/ 4 + 1 = = 84 \/ 4 + 1 = 27 + 1 = 22 . . sum = 82 * 22 = 1804 ans a\"","correct":"a","options":{"a":"1804 ","b":"1816 ","c":"1824 ","d":"1828","e":"1832"},"options_float":{"a":1804.0,"b":1816.0,"c":1824.0,"d":1828.0,"e":1832.0},"annotated_formula":"multiply(divide(add(subtract(127, const_3), add(38, const_2)), const_2), add(divide(subtract(subtract(127, const_3), add(38, 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":"127 - 3<\/gadget>\n124<\/output>\n38 + 2<\/gadget>\n40<\/output>\n124 + 40<\/gadget>\n164<\/output>\n164 \/ 2<\/gadget>\n82<\/output>\n124 - 40<\/gadget>\n84<\/output>\n84 \/ 4<\/gadget>\n21<\/output>\n21 + 1<\/gadget>\n22<\/output>\n82 * 22<\/gadget>\n1_804<\/output>\n1_804<\/result>","index":2175} +{"problem":"a train 110 m long is running with a speed of 30 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 = 30 + 6 = 36 km \/ hr . = 36 * 5 \/ 18 = 10 m \/ sec . time taken to pass the men = 110 \/ 10 = 11 sec . answer : d\"","correct":"d","options":{"a":"7 sec ","b":"6 sec ","c":"8 sec ","d":"11 sec","e":"2 sec"},"options_float":{"a":7.0,"b":6.0,"c":8.0,"d":11.0,"e":2.0},"annotated_formula":"divide(110, multiply(add(30, 6), const_0_2778))","linear_formula":"add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|","chain":"30 + 6<\/gadget>\n36<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n110 \/ 10<\/gadget>\n11<\/output>\n11<\/result>","index":2177} +{"problem":"there are 50 boys in a class . their average weight is 45 kg . when one boy leaves the class , the average reduces by 100 g . find the weight of the boy who left the class .","rationale":"here one boy is excluded and final average of the group decreases . ∴ change in average is ( – ) ve = – 0.1 kg . using the formula sum of the quantities excluded = ( changein no . ofquantities × origina laverage ) + ( changeinaverage × final no . ofquantities ) ⇒ weight of the boy who left = ( 1 × 45 ) – ( – 0.1 × 49 ) = 49.9 kg answer c","correct":"c","options":{"a":"40.9 kg ","b":"42.9 kg ","c":"49.9 kg ","d":"39.9 kg","e":"none of these"},"options_float":{"a":40.9,"b":42.9,"c":49.9,"d":39.9,"e":null},"annotated_formula":"add(45, divide(multiply(subtract(50, const_1), 100), const_1000))","linear_formula":"subtract(n0,const_1)|multiply(n2,#0)|divide(#1,const_1000)|add(n1,#2)","chain":"50 - 1<\/gadget>\n49<\/output>\n49 * 100<\/gadget>\n4_900<\/output>\n4_900 \/ 1_000<\/gadget>\n49\/10 = around 4.9<\/output>\n45 + (49\/10)<\/gadget>\n499\/10 = around 49.9<\/output>\n499\/10 = around 49.9<\/result>","index":2178} +{"problem":"if w \/ x = 1 \/ 3 and w \/ y = 4 \/ 15 , then ( x + y ) \/ y =","rationale":"\"w \/ x = 1 \/ 3 = > x = 3 w and w \/ y = 4 \/ 15 = > y = 15 \/ 4 w ( x + y ) \/ y = ( 3 w + 15 \/ 4 w ) \/ ( 15 \/ 4 w ) = ( 27 \/ 4 w ) \/ ( 15 \/ 4 w ) = 9 \/ 5 correct option : e\"","correct":"e","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(4, 1), divide(15, 3)), const_1)","linear_formula":"divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)|","chain":"4 \/ 1<\/gadget>\n4<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) + 1<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":2181} +{"problem":"tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 52 people bought tickets to the concert , what was the total revenue from ticket sales ?","rationale":"\"price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 \/ 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 \/ 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 22 tickets = 20 * 22 = 440 revenue generated from sales of 52 tickets = 120 + 340 + 440 = 900 $ answer d\"","correct":"d","options":{"a":"$ 600 ","b":"$ 740 ","c":"$ 850 ","d":"$ 900","e":"$ 1,140"},"options_float":{"a":600.0,"b":740.0,"c":850.0,"d":900.0,"e":1140.0},"annotated_formula":"multiply(add(add(subtract(subtract(52, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20)","linear_formula":"divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n0)|subtract(const_1,#0)|subtract(#2,n1)|subtract(const_1,#1)|multiply(n1,#3)|multiply(n0,#5)|add(#6,#4)|add(#8,#7)|multiply(n0,#9)|","chain":"52 - 20<\/gadget>\n32<\/output>\n32 - 10<\/gadget>\n22<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 10<\/gadget>\n6<\/output>\n22 + 6<\/gadget>\n28<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n1 - (3\/20)<\/gadget>\n17\/20 = around 0.85<\/output>\n(17\/20) * 20<\/gadget>\n17<\/output>\n28 + 17<\/gadget>\n45<\/output>\n45 * 20<\/gadget>\n900<\/output>\n900<\/result>","index":2182} +{"problem":"a university cafeteria offers 6 flavors of pizza - pork , gobi - manjurian , pepperoni , chicken , hawaiian and vegetarian . if a customer has an option ( but not the obligation ) to add extra cheese , mushrooms or both to any kind of pizza , how many different pizza varieties are available ?","rationale":"6 flavours * 6 choices = 6 c 1 * 6 c 1 = 6 * 6 = 36 = d","correct":"d","options":{"a":"4 ","b":"8 ","c":"12 ","d":"36","e":"32"},"options_float":{"a":4.0,"b":8.0,"c":12.0,"d":36.0,"e":32.0},"annotated_formula":"multiply(6, 6)","linear_formula":"multiply(n0,n0)","chain":"6 * 6<\/gadget>\n36<\/output>\n36<\/result>","index":2185} +{"problem":"in a zoo , the ratio of the number of cheetahs to the number 4 then what is the increase in the number of pandas ?","rationale":"one short cut to solve the problem is c : p = 1 : 3 c increased to 5 = > 1 : 3 = 5 : x = > x = 15 = > p increased by 12 b is the answer","correct":"b","options":{"a":"2 ","b":"12 ","c":"5 ","d":"10","e":"15"},"options_float":{"a":2.0,"b":12.0,"c":5.0,"d":10.0,"e":15.0},"annotated_formula":"subtract(multiply(4, 4), const_4)","linear_formula":"multiply(n0,n0)|subtract(#0,const_4)","chain":"4 * 4<\/gadget>\n16<\/output>\n16 - 4<\/gadget>\n12<\/output>\n12<\/result>","index":2187} +{"problem":"a train 250 m long running at 72 kmph crosses a platform in 50 sec . what is the length of the platform ?","rationale":"\"d = 72 * 5 \/ 18 = 50 = 1000 â € “ 250 = 750 m answer : d\"","correct":"d","options":{"a":"150 m ","b":"200 m ","c":"250 m ","d":"750 m","e":"300 m"},"options_float":{"a":150.0,"b":200.0,"c":250.0,"d":750.0,"e":300.0},"annotated_formula":"subtract(multiply(50, 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>\n50 * 20<\/gadget>\n1_000<\/output>\n1_000 - 250<\/gadget>\n750<\/output>\n750<\/result>","index":2188} +{"problem":"in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 12 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture .","rationale":"\"given that milk \/ water = 4 x \/ x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 \/ ( 9 + 12 ) = 36 \/ 21 = 12 \/ 7 . answer : a .\"","correct":"a","options":{"a":"12 \/ 7 ","b":"4 \/ 1 ","c":"2 \/ 3 ","d":"3 \/ 4","e":"3 \/ 2"},"options_float":{"a":1.7142857143,"b":4.0,"c":0.6666666667,"d":0.75,"e":1.5},"annotated_formula":"divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 12))","linear_formula":"add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)|","chain":"4 + 1<\/gadget>\n5<\/output>\n45 \/ 5<\/gadget>\n9<\/output>\n45 - 9<\/gadget>\n36<\/output>\n9 + 12<\/gadget>\n21<\/output>\n36 \/ 21<\/gadget>\n12\/7 = around 1.714286<\/output>\n12\/7 = around 1.714286<\/result>","index":2190} +{"problem":"in 1979 approximately 1 \/ 3 of the 32.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 \/ 2 the number that used kennedy airport and 4 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ?","rationale":"\"number of passengers using kennedy airport = 32 \/ 3 = ~ 10.67 passengers using miami airport = 10.67 \/ 2 = ~ 5.34 passengers using logan airport = 5.34 \/ 4 = ~ 1.33 so d\"","correct":"d","options":{"a":"18.6 ","b":"9.3 ","c":"6.2 ","d":"1.33","e":"1.6"},"options_float":{"a":18.6,"b":9.3,"c":6.2,"d":1.33,"e":1.6},"annotated_formula":"divide(divide(32.3, 3), multiply(4, 2))","linear_formula":"divide(n3,n2)|multiply(n5,n6)|divide(#0,#1)|","chain":"32.3 \/ 3<\/gadget>\n10.766667<\/output>\n4 * 2<\/gadget>\n8<\/output>\n10.766667 \/ 8<\/gadget>\n1.345833<\/output>\n1.345833<\/result>","index":2191} +{"problem":"what is the greater of the two numbers whose product is 2496 , given that the sum of the two numbers exceeds their difference by 64 ?","rationale":"\"let the greater and the smaller number be g and s respectively . gs = 2496 g + s exceeds g - s by 64 i . e . , g + s - ( g - s ) = 64 i . e . , 2 s = 64 = > s = 32 . g = 2496 \/ s = 78 . answer : d\"","correct":"d","options":{"a":"96 ","b":"108 ","c":"110 ","d":"78","e":"of these"},"options_float":{"a":96.0,"b":108.0,"c":110.0,"d":78.0,"e":null},"annotated_formula":"divide(2496, multiply(power(const_2, const_4), const_2))","linear_formula":"power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)|","chain":"2 ** 4<\/gadget>\n16<\/output>\n16 * 2<\/gadget>\n32<\/output>\n2_496 \/ 32<\/gadget>\n78<\/output>\n78<\/result>","index":2192} +{"problem":"of the goose eggs laid at a certain pond , 2 \/ 3 hatched and 3 \/ 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 \/ 5 did not survive the first year . if 125 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ?","rationale":"of the goose eggs laid at a certain pond , 2 \/ 3 hatched and 3 \/ 4 of the geese that hatched from those eggs survived the first month : 2 \/ 3 * 3 \/ 4 = 1 \/ 2 survived the first month . of the geese that survived the first month , 3 \/ 5 did not survive the first year : ( 1 - 3 \/ 5 ) * 1 \/ 2 = 1 \/ 5 survived the first year . 120 geese survived the first year : 1 \/ 5 * ( total ) = 125 - - > ( total ) = 625 . answer : d .","correct":"d","options":{"a":"280 ","b":"400 ","c":"540 ","d":"625","e":"840"},"options_float":{"a":280.0,"b":400.0,"c":540.0,"d":625.0,"e":840.0},"annotated_formula":"divide(divide(divide(125, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(const_2, const_3))","linear_formula":"divide(n1,n5)|divide(n1,n3)|divide(const_2,const_3)|subtract(const_1,#0)|divide(n6,#3)|divide(#4,#1)|divide(#5,#2)","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n125 \/ (2\/5)<\/gadget>\n625\/2 = around 312.5<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(625\/2) \/ (3\/4)<\/gadget>\n1_250\/3 = around 416.666667<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n(1_250\/3) \/ (2\/3)<\/gadget>\n625<\/output>\n625<\/result>","index":2194} +{"problem":"the average salary per month of 55 employees in a company is rs 8500 . if the managers salary is added , the average salary increases to rs 8800 , what is the salary of the manager ?","rationale":"explanation : salary of the manager = ( 56 * 8800 - 55 * 8500 ) = 25300 answer : d","correct":"d","options":{"a":"10000 ","b":"12000 ","c":"23000 ","d":"25300","e":"45000"},"options_float":{"a":10000.0,"b":12000.0,"c":23000.0,"d":25300.0,"e":45000.0},"annotated_formula":"subtract(multiply(add(55, const_1), 8800), multiply(55, 8500))","linear_formula":"add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)","chain":"55 + 1<\/gadget>\n56<\/output>\n56 * 8_800<\/gadget>\n492_800<\/output>\n55 * 8_500<\/gadget>\n467_500<\/output>\n492_800 - 467_500<\/gadget>\n25_300<\/output>\n25_300<\/result>","index":2195} +{"problem":"there are 24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted 3 roses , every three boys planted 1 birch . by the end of the day they planted 2424 plants . how many birches were planted ?","rationale":"\"let x be the number of roses . then the number of birches is 24 − x , and the number of boys is 3 × ( 24 − x ) . if each girl planted 3 roses , there are x 3 girls in the class . we know that there are 24 students in the class . therefore x 3 + 3 ( 24 − x ) = 24 x + 9 ( 24 − x ) = 3 ⋅ 24 x + 216 − 9 x = 72 216 − 72 = 8 x 1448 = x 1 x = 18 so , students planted 18 roses and 24 - x = 24 - 18 = 6 birches . correct answer is d ) 6\"","correct":"d","options":{"a":"2 ","b":"5 ","c":"8 ","d":"6","e":"4"},"options_float":{"a":2.0,"b":5.0,"c":8.0,"d":6.0,"e":4.0},"annotated_formula":"divide(subtract(multiply(3, 24), 24), subtract(multiply(3, 3), 1))","linear_formula":"multiply(n0,n1)|multiply(n1,n1)|subtract(#0,n0)|subtract(#1,n2)|divide(#2,#3)|","chain":"3 * 24<\/gadget>\n72<\/output>\n72 - 24<\/gadget>\n48<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 - 1<\/gadget>\n8<\/output>\n48 \/ 8<\/gadget>\n6<\/output>\n6<\/result>","index":2196} +{"problem":"in a simultaneous throw of pair of dice . find the probability of getting the total more than 7","rationale":"here n ( s ) = ( 6 * 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } p ( e ) = n ( e ) \/ n ( s ) = 15 \/ 36 = 5 \/ 12 option c","correct":"c","options":{"a":"5 \/ 7 ","b":"4 \/ 7 ","c":"5 \/ 12 ","d":"4 \/ 7","e":"1 \/ 6"},"options_float":{"a":0.7142857143,"b":0.5714285714,"c":0.4166666667,"d":0.5714285714,"e":0.1666666667},"annotated_formula":"divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2)))","linear_formula":"add(n0,const_4)|add(const_2,const_4)|add(#0,const_4)|multiply(#1,#1)|divide(#2,#3)","chain":"7 + 4<\/gadget>\n11<\/output>\n11 + 4<\/gadget>\n15<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n15 \/ 36<\/gadget>\n5\/12 = around 0.416667<\/output>\n5\/12 = around 0.416667<\/result>","index":2198} +{"problem":"a green grocer received a boxful of tomatoes and on opening the box found that several had gone bad . he then counted them up so that he could make a formal complaint and found that 68 were mouldy , which was 16 per cent of the total contents of the box . how many tomatoes were in the box ?","rationale":"b 425 ( 68 ã · 16 ) ã — 100","correct":"b","options":{"a":"336 ","b":"425 ","c":"275 ","d":"235","e":"689"},"options_float":{"a":336.0,"b":425.0,"c":275.0,"d":235.0,"e":689.0},"annotated_formula":"subtract(multiply(multiply(68, const_4), const_2), const_100)","linear_formula":"multiply(n0,const_4)|multiply(#0,const_2)|subtract(#1,const_100)","chain":"68 * 4<\/gadget>\n272<\/output>\n272 * 2<\/gadget>\n544<\/output>\n544 - 100<\/gadget>\n444<\/output>\n444<\/result>","index":2200} +{"problem":"how many boxes do we need if we have to carry 250 apples into boxes that each hold 25 apples ?","rationale":"sol . apples 250 each carries 25 = 250 \/ 25 = 10 answer : d","correct":"d","options":{"a":"9 ","b":"5 ","c":"7 ","d":"10","e":"none of the above"},"options_float":{"a":9.0,"b":5.0,"c":7.0,"d":10.0,"e":null},"annotated_formula":"divide(250, 25)","linear_formula":"divide(n0,n1)","chain":"250 \/ 25<\/gadget>\n10<\/output>\n10<\/result>","index":2202} +{"problem":"how many seconds does sandy take to cover a distance of 600 meters , if sandy runs at a speed of 15 km \/ hr ?","rationale":"\"15 km \/ hr = 15000 m \/ 3600 s = ( 150 \/ 36 ) m \/ s = ( 25 \/ 6 ) m \/ s time = 600 \/ ( 25 \/ 6 ) = 144 seconds 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(600, multiply(15, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n15 * (5\/18)<\/gadget>\n25\/6 = around 4.166667<\/output>\n600 \/ (25\/6)<\/gadget>\n144<\/output>\n144<\/result>","index":2203} +{"problem":"if shares of two persons in profits are rs . 600 and rs . 300 then ratio of their capitals is","rationale":"\"total profit = 1000 ratio = 600 \/ 300 = 2 : 1 answer : e\"","correct":"e","options":{"a":"3 : 4 ","b":"2 : 3 ","c":"4 : 3 ","d":"1 : 3","e":"2 : 1"},"options_float":{"a":0.75,"b":0.6666666667,"c":1.3333333333,"d":0.3333333333,"e":2.0},"annotated_formula":"divide(600, 300)","linear_formula":"divide(n0,n1)|","chain":"600 \/ 300<\/gadget>\n2<\/output>\n2<\/result>","index":2204} +{"problem":"the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is :","rationale":"\"correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 \/ 50 = 36.5 answer a\"","correct":"a","options":{"a":"36.5 ","b":"35 ","c":"34 ","d":"33","e":"32.5"},"options_float":{"a":36.5,"b":35.0,"c":34.0,"d":33.0,"e":32.5},"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":2205} +{"problem":"a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 500 each from people already solicited . people already solicited represent 50 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ?","rationale":"\"let us suppose there are 100 people . 50 % of them donated $ 25000 ( 500 * 50 ) $ 25000 is 60 % of total amount . so total amount = 25000 * 100 \/ 60 remaining amount is 40 % of total amount . 40 % of total amount = 25000 * ( 100 \/ 60 ) * ( 40 \/ 100 ) = 50000 \/ 3 this amount has to be divided by 50 ( remaining people are 50 ) so per head amount is 50000 \/ 3 \/ 50 = 32000 \/ 180 = 333.33 ; answer : b\"","correct":"b","options":{"a":"$ 200 ","b":"$ 333.33 ","c":"$ 100.25 ","d":"$ 277.78","e":"$ 377.00"},"options_float":{"a":200.0,"b":333.33,"c":100.25,"d":277.78,"e":377.0},"annotated_formula":"divide(multiply(divide(multiply(divide(50, const_100), 500), divide(60, const_100)), divide(50, const_100)), divide(60, const_100))","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)|","chain":"50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) * 500<\/gadget>\n250<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n250 \/ (3\/5)<\/gadget>\n1_250\/3 = around 416.666667<\/output>\n(1_250\/3) * (1\/2)<\/gadget>\n625\/3 = around 208.333333<\/output>\n(625\/3) \/ (3\/5)<\/gadget>\n3_125\/9 = around 347.222222<\/output>\n3_125\/9 = around 347.222222<\/result>","index":2206} +{"problem":"a certain bacteria colony doubles in size every day for 19 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 20 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 18 times . answer : d . similar questions to practice : hope it helps .\"","correct":"d","options":{"a":"6.33 ","b":"7.5 ","c":"10 ","d":"18","e":"19"},"options_float":{"a":6.33,"b":7.5,"c":10.0,"d":18.0,"e":19.0},"annotated_formula":"subtract(19, divide(19, 19))","linear_formula":"divide(n0,n0)|subtract(n0,#0)|","chain":"19 \/ 19<\/gadget>\n1<\/output>\n19 - 1<\/gadget>\n18<\/output>\n18<\/result>","index":2207} +{"problem":"the diameter of a circle is 4 \/ π . find the circumference of the circle .","rationale":"circumference = 2 * pi * r = 2 * pi * 4 \/ pi = > 8 a","correct":"a","options":{"a":"8 ","b":"4 π ","c":"4 ","d":"6","e":"5"},"options_float":{"a":8.0,"b":4.0,"c":4.0,"d":6.0,"e":5.0},"annotated_formula":"circumface(divide(4, const_pi))","linear_formula":"divide(n0,const_pi)|circumface(#0)","chain":"4 \/ pi<\/gadget>\n4\/pi = around 1.27324<\/output>\n2 * pi * (4\/pi)<\/gadget>\n8<\/output>\n8<\/result>","index":2208} +{"problem":"simplify : 0.3 * 0.3 + 0.3 * 0.3","rationale":"\"given exp . = 0.3 * 0.3 + ( 0.3 * 0.3 ) = 0.09 + 0.09 = 0.18 answer is c .\"","correct":"c","options":{"a":"0.52 ","b":"0.42 ","c":"0.18 ","d":"0.64","e":"0.46"},"options_float":{"a":0.52,"b":0.42,"c":0.18,"d":0.64,"e":0.46},"annotated_formula":"add(multiply(0.3, 0.3), multiply(0.3, 0.3))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"0.3 * 0.3<\/gadget>\n0.09<\/output>\n0.09 + 0.09<\/gadget>\n0.18<\/output>\n0.18<\/result>","index":2209} +{"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 will be a solution to the equation ( x - 4 ) ( x + 9 ) ( 2 x + 5 ) = 0 ?","rationale":"x = - 2.5 prob = 1 \/ 12 answer - a","correct":"a","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(1, multiply(6, 2))","linear_formula":"multiply(n1,n14)|divide(n5,#0)","chain":"6 * 2<\/gadget>\n12<\/output>\n1 \/ 12<\/gadget>\n1\/12 = around 0.083333<\/output>\n1\/12 = around 0.083333<\/result>","index":2212} +{"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 300 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 300 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 * 300 = 9000 . d = st 9000 \/ 1.5 = t , i think we can take 9000 meters as 9 km , then only we get t = 6 . ( 1000 meters = 1 km ) d )\"","correct":"d","options":{"a":"1 ","b":"2 ","c":"4 ","d":"6","e":"8"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":6.0,"e":8.0},"annotated_formula":"divide(divide(multiply(300, 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>\n300 * 30<\/gadget>\n9_000<\/output>\n9_000 \/ 1_000<\/gadget>\n9<\/output>\n15 \/ 10<\/gadget>\n3\/2 = around 1.5<\/output>\n9 \/ (3\/2)<\/gadget>\n6<\/output>\n6<\/result>","index":2213} +{"problem":"if 30 % of a class averages 95 % on a test , 50 % of the class averages 79 % 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 % - 30 % - 50 % = 20 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.30 x 95 + 0.50 x 79 + 0.20 x 60 = 80 the class average ( rounded ) is 80 % final answer e ) 80 %\"","correct":"e","options":{"a":"76 % ","b":"77 % ","c":"78 % ","d":"79 %","e":"80 %"},"options_float":{"a":76.0,"b":77.0,"c":78.0,"d":79.0,"e":80.0},"annotated_formula":"divide(add(add(multiply(30, 95), multiply(50, 79)), multiply(subtract(const_100, add(30, 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":"30 * 95<\/gadget>\n2_850<\/output>\n50 * 79<\/gadget>\n3_950<\/output>\n2_850 + 3_950<\/gadget>\n6_800<\/output>\n30 + 50<\/gadget>\n80<\/output>\n100 - 80<\/gadget>\n20<\/output>\n20 * 60<\/gadget>\n1_200<\/output>\n6_800 + 1_200<\/gadget>\n8_000<\/output>\n8_000 \/ 100<\/gadget>\n80<\/output>\n80<\/result>","index":2216} +{"problem":"if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ?","rationale":"if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? a . 2 b . 3 c . 4 d . 5 e . 6 - > if m = y = 60 , 2 ( m + y ) = 240 is derived . if m = y = 69 , 2 ( m + y ) = 276 is derived , which makes 4,5 , 6,7 possible for the tens digit . therefore , the answer is c .","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":"subtract(6, 2)","linear_formula":"subtract(n0,n1)","chain":"6 - 2<\/gadget>\n4<\/output>\n4<\/result>","index":2218} +{"problem":"the maximum number of students among them 848 pens and 630 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ?","rationale":"\"number of pens = 848 number of pencils = 630 required number of students = h . c . f . of 848 and 630 = 2 answer is c\"","correct":"c","options":{"a":"10 ","b":"4 ","c":"2 ","d":"14","e":"16"},"options_float":{"a":10.0,"b":4.0,"c":2.0,"d":14.0,"e":16.0},"annotated_formula":"gcd(848, 630)","linear_formula":"gcd(n0,n1)|","chain":"gcd(848, 630)<\/gadget>\n2<\/output>\n2<\/result>","index":2219} +{"problem":"find the ratio of the curved surfaces of two cylinders of same heights if their radii are in the ratio 1 : 2 ?","rationale":"1 : 2 answer : a","correct":"a","options":{"a":"1 : 2 ","b":"2 : 3 ","c":"2 : 9 ","d":"2 : 1","e":"2 : 2"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.2222222222,"d":2.0,"e":1.0},"annotated_formula":"divide(1, 2)","linear_formula":"divide(n0,n1)","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1\/2 = around 0.5<\/result>","index":2220} +{"problem":"the captain of a cricket team of 11 members is 29 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 age of the team ?","rationale":"\"explanation : let the average age of the whole team by x years . 11 x â € “ ( 29 + 32 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer e\"","correct":"e","options":{"a":"20 years ","b":"21 years ","c":"22 years ","d":"23 years","e":"26 years"},"options_float":{"a":20.0,"b":21.0,"c":22.0,"d":23.0,"e":26.0},"annotated_formula":"divide(subtract(add(29, add(29, 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":"29 + 3<\/gadget>\n32<\/output>\n29 + 32<\/gadget>\n61<\/output>\n3 * 3<\/gadget>\n9<\/output>\n61 - 9<\/gadget>\n52<\/output>\n52 \/ 2<\/gadget>\n26<\/output>\n26<\/result>","index":2222} +{"problem":"if the sides of a triangle are 31 cm , 29 cm and 15 cm , what is its area ?","rationale":"\"the triangle with sides 31 cm , 29 cm and 15 cm is right angled , where the hypotenuse is 31 cm . area of the triangle = 1 \/ 2 * 29 * 15 = 217.5 cm 2 answer : e\"","correct":"e","options":{"a":"220.75 cm 2 ","b":"258 cm 2 ","c":"225.50 cm 2 ","d":"222.25 cm 2","e":"217.5 cm 2"},"options_float":{"a":220.75,"b":258.0,"c":225.5,"d":222.25,"e":217.5},"annotated_formula":"divide(multiply(29, 15), const_2)","linear_formula":"multiply(n1,n2)|divide(#0,const_2)|","chain":"29 * 15<\/gadget>\n435<\/output>\n435 \/ 2<\/gadget>\n435\/2 = around 217.5<\/output>\n435\/2 = around 217.5<\/result>","index":2223} +{"problem":"in what ratio should a variety of rice costing rs . 6.5 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ?","rationale":"\"let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x \/ y = ( 8.75 - 7.50 ) \/ ( 7.50 - 6.5 ) = 5 \/ 4 answer : c\"","correct":"c","options":{"a":"5 \/ 6 ","b":"5 \/ 9 ","c":"5 \/ 4 ","d":"5 \/ 3","e":"7 \/ 6"},"options_float":{"a":0.8333333333,"b":0.5555555556,"c":1.25,"d":1.6666666667,"e":1.1666666667},"annotated_formula":"divide(divide(subtract(8.75, 7.50), subtract(8.75, 6.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6.5))))","linear_formula":"subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|","chain":"8.75 - 7.5<\/gadget>\n1.25<\/output>\n8.75 - 6.5<\/gadget>\n2.25<\/output>\n1.25 \/ 2.25<\/gadget>\n0.555556<\/output>\n1 - 0.555556<\/gadget>\n0.444444<\/output>\n0.555556 \/ 0.444444<\/gadget>\n1.250002<\/output>\n1.250002<\/result>","index":2224} +{"problem":"a bag contains 7 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 seven green 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 : e\"","correct":"e","options":{"a":"7 \/ 16 ","b":"7 \/ 12 ","c":"7 \/ 19 ","d":"7 \/ 12","e":"7 \/ 15"},"options_float":{"a":0.4375,"b":0.5833333333,"c":0.3684210526,"d":0.5833333333,"e":0.4666666667},"annotated_formula":"add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 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":"7 + 8<\/gadget>\n15<\/output>\n8 \/ 15<\/gadget>\n8\/15 = around 0.533333<\/output>\n8 - 1<\/gadget>\n7<\/output>\n15 - 1<\/gadget>\n14<\/output>\n7 \/ 14<\/gadget>\n1\/2 = around 0.5<\/output>\n(8\/15) * (1\/2)<\/gadget>\n4\/15 = around 0.266667<\/output>\n7 \/ 15<\/gadget>\n7\/15 = around 0.466667<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6 \/ 14<\/gadget>\n3\/7 = around 0.428571<\/output>\n(7\/15) * (3\/7)<\/gadget>\n1\/5 = around 0.2<\/output>\n(4\/15) + (1\/5)<\/gadget>\n7\/15 = around 0.466667<\/output>\n7\/15 = around 0.466667<\/result>","index":2225} +{"problem":"the average monthly salary of 20 employees in an organisation is rs . 1400 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?","rationale":"\"explanation : manager ' s monthly salary rs . ( 1500 * 21 - 1400 * 20 ) = rs . 3500 . answer : e\"","correct":"e","options":{"a":"3600 ","b":"3890 ","c":"88798 ","d":"2789","e":"3500"},"options_float":{"a":3600.0,"b":3890.0,"c":88798.0,"d":2789.0,"e":3500.0},"annotated_formula":"subtract(multiply(add(1400, 100), add(20, const_1)), multiply(1400, 20))","linear_formula":"add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"1_400 + 100<\/gadget>\n1_500<\/output>\n20 + 1<\/gadget>\n21<\/output>\n1_500 * 21<\/gadget>\n31_500<\/output>\n1_400 * 20<\/gadget>\n28_000<\/output>\n31_500 - 28_000<\/gadget>\n3_500<\/output>\n3_500<\/result>","index":2226} +{"problem":"a train 400 m long can cross an electric pole in 10 sec and then find the speed of the train ?","rationale":"\"length = speed * time speed = l \/ t s = 400 \/ 10 s = 40 m \/ sec speed = 40 * 18 \/ 5 ( to convert m \/ sec in to kmph multiply by 18 \/ 5 ) speed = 144 kmph answer : b\"","correct":"b","options":{"a":"165 kmph ","b":"144 kmph ","c":"172 kmph ","d":"175 kmph","e":"178 kmph"},"options_float":{"a":165.0,"b":144.0,"c":172.0,"d":175.0,"e":178.0},"annotated_formula":"divide(divide(400, const_1000), divide(10, const_3600))","linear_formula":"divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|","chain":"400 \/ 1_000<\/gadget>\n2\/5 = around 0.4<\/output>\n10 \/ 3_600<\/gadget>\n1\/360 = around 0.002778<\/output>\n(2\/5) \/ (1\/360)<\/gadget>\n144<\/output>\n144<\/result>","index":2229} +{"problem":"a man can row his boat with the stream at 30 km \/ h and against the stream in 14 km \/ h . the man ' s rate is ?","rationale":"\"ds = 30 us = 14 s = ? s = ( 30 - 14 ) \/ 2 = 8 kmph answer : e\"","correct":"e","options":{"a":"1 kmph ","b":"4 kmph ","c":"5 kmph ","d":"7 kmph","e":"8 kmph"},"options_float":{"a":1.0,"b":4.0,"c":5.0,"d":7.0,"e":8.0},"annotated_formula":"divide(subtract(30, 14), const_2)","linear_formula":"subtract(n0,n1)|divide(#0,const_2)|","chain":"30 - 14<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2230} +{"problem":"3 men and 7 women can complete a work in 10 days . but 4 men and 6 women need 8 days to complete the same work . in how many days will 10 women complete the same work ?","rationale":"explanation : work done by 4 men and 6 women in 1 day = 1 \/ 8 work done by 3 men and 7 women in 1 day = 1 \/ 10 let 1 man does m work in 1 day and 1 woman does w work in 1 day . the above equations can be written as 4 m + 6 w = 1 \/ 8 - - - ( 1 ) 3 m + 7 w = 1 \/ 10 - - - ( 2 ) solving equation ( 1 ) and ( 2 ) , we get m = 11 \/ 400 and w = 1 \/ 400 amount of work 10 women can do in a day = 10 × ( 1 \/ 400 ) = 1 \/ 40 ie , 10 women can complete the work in 40 days answer : option b","correct":"b","options":{"a":"50 ","b":"40 ","c":"30 ","d":"20","e":"10"},"options_float":{"a":50.0,"b":40.0,"c":30.0,"d":20.0,"e":10.0},"annotated_formula":"inverse(multiply(divide(subtract(divide(const_1, 8), multiply(4, divide(subtract(divide(const_1, 10), multiply(divide(7, 6), divide(const_1, 8))), subtract(3, multiply(4, divide(7, 6)))))), 6), 10))","linear_formula":"divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)","chain":"1 \/ 8<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n7 \/ 6<\/gadget>\n7\/6 = around 1.166667<\/output>\n(7\/6) * (1\/8)<\/gadget>\n7\/48 = around 0.145833<\/output>\n(1\/10) - (7\/48)<\/gadget>\n-11\/240 = around -0.045833<\/output>\n4 * (7\/6)<\/gadget>\n14\/3 = around 4.666667<\/output>\n3 - (14\/3)<\/gadget>\n-5\/3 = around -1.666667<\/output>\n(-11\/240) \/ (-5\/3)<\/gadget>\n11\/400 = around 0.0275<\/output>\n4 * (11\/400)<\/gadget>\n11\/100 = around 0.11<\/output>\n(1\/8) - (11\/100)<\/gadget>\n3\/200 = around 0.015<\/output>\n(3\/200) \/ 6<\/gadget>\n1\/400 = around 0.0025<\/output>\n(1\/400) * 10<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":2231} +{"problem":"what is the characteristic of the logarithm of 0.0000134 ?","rationale":"log ( 0.0000134 ) . since there are four zeros between the decimal point and the first significant digit , the characteristic is – 5 . answer : b","correct":"b","options":{"a":"5 ","b":"- 5 ","c":"6 ","d":"- 6","e":"7"},"options_float":{"a":5.0,"b":-5.0,"c":6.0,"d":-6.0,"e":7.0},"annotated_formula":"floor(divide(log(0.0000134), log(const_10)))","linear_formula":"log(n0)|log(const_10)|divide(#0,#1)|floor(#2)","chain":"log(0.000013)<\/gadget>\n-11.250561<\/output>\nlog(10)<\/gadget>\nlog(10) = around 2.302585<\/output>\n(-11.250561) \/ log(10)<\/gadget>\n-11.250561\/log(10) = around -4.886057<\/output>\nfloor(-11.250561\/log(10))<\/gadget>\n-5<\/output>\n-5<\/result>","index":2232} +{"problem":"in the game of dubblefud , red chips , blue chips and green chips are each worth 2 , 4 and 5 points respectively . in a certain selection of chips , the product of the point values of the chips is 16000 . if the number of blue chips in this selection doubles the number of green chips , how many red chips are in the selection ?","rationale":"this is equivalent to : - 2 x * 4 y * 5 z = 16000 y \/ 2 = z ( given ) 2 x * 4 y * 5 y \/ 2 = 16000 2 x * y ^ 2 = 16000 \/ 10 2 x * y ^ 2 = 1600 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 2 as 2 * 2 * y ^ 2 = 1600 y ^ 2 = 400 y = 20 number of red chips = 2 hence 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":"divide(multiply(multiply(power(2, 4), power(2, const_3)), power(5, const_3)), multiply(power(const_2, multiply(2, const_3)), power(5, const_3)))","linear_formula":"multiply(n0,const_3)|power(n0,n1)|power(n0,const_3)|power(n2,const_3)|multiply(#1,#2)|power(const_2,#0)|multiply(#4,#3)|multiply(#5,#3)|divide(#6,#7)","chain":"2 ** 4<\/gadget>\n16<\/output>\n2 ** 3<\/gadget>\n8<\/output>\n16 * 8<\/gadget>\n128<\/output>\n5 ** 3<\/gadget>\n125<\/output>\n128 * 125<\/gadget>\n16_000<\/output>\n2 * 3<\/gadget>\n6<\/output>\n2 ** 6<\/gadget>\n64<\/output>\n64 * 125<\/gadget>\n8_000<\/output>\n16_000 \/ 8_000<\/gadget>\n2<\/output>\n2<\/result>","index":2233} +{"problem":"compound interest of rs . 2000 at 10 % per annum for 1 1 \/ 2 years will be ( interest compounded half yearly ) .","rationale":"\"10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 \/ 2 years ) so compound interest = 2000 [ 1 + ( 5 \/ 100 ) ] ^ 3 - 2000 = 2000 [ ( 21 \/ 20 ) ^ 3 - 1 ] = 2000 ( 9261 - 8000 ) \/ 8000 = 2 * 1261 \/ 8 = 315 answer : d\"","correct":"d","options":{"a":"rs . 473 ","b":"rs . 374 ","c":"rs . 495 ","d":"rs . 315","e":"none of the above"},"options_float":{"a":473.0,"b":374.0,"c":495.0,"d":315.0,"e":null},"annotated_formula":"subtract(multiply(2000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 2000)","linear_formula":"divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)|","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>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 2<\/gadget>\n3<\/output>\n(21\/20) ** 3<\/gadget>\n9_261\/8_000 = around 1.157625<\/output>\n2_000 * (9_261\/8_000)<\/gadget>\n9_261\/4 = around 2_315.25<\/output>\n(9_261\/4) - 2_000<\/gadget>\n1_261\/4 = around 315.25<\/output>\n1_261\/4 = around 315.25<\/result>","index":2234} +{"problem":"on average , the boys in the class have 20 pencils and the girls have 38 pencils . if the overall class average is 30 pencils , what is the ratio of boys to girls in the class ?","rationale":"( 38 g + 20 b ) \/ ( g + b ) = 30 38 g + 20 b = 30 ( g + b ) 8 g = 10 b b \/ g = 4 \/ 5 the answer is d .","correct":"d","options":{"a":"1 \/ 2 ","b":"2 \/ 3 ","c":"3 \/ 4 ","d":"4 \/ 5","e":"5 \/ 6"},"options_float":{"a":0.5,"b":0.6666666667,"c":0.75,"d":0.8,"e":0.8333333333},"annotated_formula":"divide(30, 38)","linear_formula":"divide(n2,n1)","chain":"30 \/ 38<\/gadget>\n15\/19 = around 0.789474<\/output>\n15\/19 = around 0.789474<\/result>","index":2238} +{"problem":"in a certain village , 200 litres of water are required per household per month . at this rate , if there are 5 households in the village , how long ( in months ) will 2000 litres of water last ?","rationale":"\"i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 5 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 5 = 1000 ( m * n ) water available is 2000 ( p ) it will last 2 months ( p \/ m * n ) ans : d\"","correct":"d","options":{"a":"9 ","b":"5 ","c":"6 ","d":"2","e":"4"},"options_float":{"a":9.0,"b":5.0,"c":6.0,"d":2.0,"e":4.0},"annotated_formula":"divide(2000, multiply(200, 5))","linear_formula":"multiply(n0,n1)|divide(n2,#0)|","chain":"200 * 5<\/gadget>\n1_000<\/output>\n2_000 \/ 1_000<\/gadget>\n2<\/output>\n2<\/result>","index":2239} +{"problem":"rs . 925 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 925 become in 3 years ?","rationale":"\"solution s . i . = rs . ( 956 - 925 ) = rs . 31 rate = ( 100 x 31 \/ 925 x 3 ) = 124 \/ 111 % new rate = ( 124 \/ 111 + 4 ) % = 568 \/ 111 % new s . i . = rs . ( 925 x 568 \/ 111 x 3 \/ 100 ) rs . 142 ∴ new amount = rs . ( 925 + 142 ) = rs . 1067 . answer c\"","correct":"c","options":{"a":"rs . 1020.80 ","b":"rs . 1025 ","c":"rs . 1067 ","d":"data inadequate","e":"none of these"},"options_float":{"a":1020.8,"b":1025.0,"c":1067.0,"d":null,"e":null},"annotated_formula":"add(925, divide(multiply(multiply(925, add(divide(multiply(subtract(956, 925), const_100), multiply(925, 3)), 4)), 3), const_100))","linear_formula":"multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|","chain":"956 - 925<\/gadget>\n31<\/output>\n31 * 100<\/gadget>\n3_100<\/output>\n925 * 3<\/gadget>\n2_775<\/output>\n3_100 \/ 2_775<\/gadget>\n124\/111 = around 1.117117<\/output>\n(124\/111) + 4<\/gadget>\n568\/111 = around 5.117117<\/output>\n925 * (568\/111)<\/gadget>\n14_200\/3 = around 4_733.333333<\/output>\n(14_200\/3) * 3<\/gadget>\n14_200<\/output>\n14_200 \/ 100<\/gadget>\n142<\/output>\n925 + 142<\/gadget>\n1_067<\/output>\n1_067<\/result>","index":2242} +{"problem":"exactly 3 \/ 7 of the people in the room are under the age of 21 , and exactly 5 \/ 12 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 12 ( in order 3 \/ 7 and 5 \/ 12 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 12 , which is 84 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 84 people in the room . therefore there are 3 \/ 7 * 84 = 36 people in the room under the age of 21 . answer : c .\"","correct":"c","options":{"a":"21 ","b":"35 ","c":"36 ","d":"60","e":"65"},"options_float":{"a":21.0,"b":35.0,"c":36.0,"d":60.0,"e":65.0},"annotated_formula":"divide(multiply(multiply(7, 12), 3), 7)","linear_formula":"multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)|","chain":"7 * 12<\/gadget>\n84<\/output>\n84 * 3<\/gadget>\n252<\/output>\n252 \/ 7<\/gadget>\n36<\/output>\n36<\/result>","index":2243} +{"problem":"if the average of r , b , c , 14 and 15 is 12 . what is the average value of r , b , c and 29","rationale":"r + b + c + 14 + 15 = 12 * 5 = 60 = > r + b + c = 60 - 29 = 31 r + b + c + 29 = 31 + 29 = 60 average = 60 \/ 4 = 15 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":"divide(add(subtract(multiply(add(const_4, const_1), 12), add(14, 15)), 29), const_4)","linear_formula":"add(const_1,const_4)|add(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|add(n3,#3)|divide(#4,const_4)","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 12<\/gadget>\n60<\/output>\n14 + 15<\/gadget>\n29<\/output>\n60 - 29<\/gadget>\n31<\/output>\n31 + 29<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15<\/result>","index":2246} +{"problem":"if 15 % of a is the same as 30 % of b , then a : b is :","rationale":"expl : 15 % of a i = 30 % of b = 15 a \/ 100 = 30 b \/ 100 = 2 \/ 1 = 2 : 1 answer : e","correct":"e","options":{"a":"1 : 4 ","b":"4 : 3 ","c":"6 : 7 ","d":"3 : 5","e":"2 : 1"},"options_float":{"a":0.25,"b":1.3333333333,"c":0.8571428571,"d":0.6,"e":2.0},"annotated_formula":"divide(divide(30, const_100), divide(15, const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/10) \/ (3\/20)<\/gadget>\n2<\/output>\n2<\/result>","index":2247} +{"problem":"two friends c and d leave point c and point d simultaneously and travel towards point d and point c 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 d take to cover the entire journey between point d and point c ?","rationale":"let x per minute be the speed of c and y per minute be the speed of d . after meeting at a point , c travels for 32 mins and d travels for 50 mins . so distance covered by each of them post point of crossing c = 32 x and d = 50 y the distance covered by c and d before they cross each would be distance covered by d and c post crossing respectively . therefore distance covered by d before he meets c = 32 x time taken by d cover 32 x distance = 32 x \/ y mins therefore total time taken by d = 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 c and d = x + y therefore time taken before c and d meet en - route = ( 32 x + 50 y ) \/ ( x + y ) time taken by d reach destination after meeting c = 50 mins total travel time for d = [ ( 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 a","correct":"a","options":{"a":"90 ","b":"80 ","c":"75 ","d":"60","e":"65"},"options_float":{"a":90.0,"b":80.0,"c":75.0,"d":60.0,"e":65.0},"annotated_formula":"add(sqrt(multiply(50, 32)), 50)","linear_formula":"multiply(n0,n1)|sqrt(#0)|add(n1,#1)","chain":"50 * 32<\/gadget>\n1_600<\/output>\n1_600 ** (1\/2)<\/gadget>\n40<\/output>\n40 + 50<\/gadget>\n90<\/output>\n90<\/result>","index":2250} +{"problem":"a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 225 for it , the cost price of the cricket bat for a is :","rationale":"\"a 150 125 % of 120 % of a = 225 125 \/ 100 * 120 \/ 100 * a = 225 a = 225 * 2 \/ 3 = 150 .\"","correct":"a","options":{"a":"150 ","b":"120 ","c":"130 ","d":"160","e":"210"},"options_float":{"a":150.0,"b":120.0,"c":130.0,"d":160.0,"e":210.0},"annotated_formula":"divide(225, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 + (1\/5)<\/gadget>\n6\/5 = around 1.2<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 + (1\/4)<\/gadget>\n5\/4 = around 1.25<\/output>\n(6\/5) * (5\/4)<\/gadget>\n3\/2 = around 1.5<\/output>\n225 \/ (3\/2)<\/gadget>\n150<\/output>\n150<\/result>","index":2251} +{"problem":"x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 5 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 \/ 5 ^ 2 = 9 \/ 25 answer : e\"","correct":"e","options":{"a":"3 ","b":"6 ","c":"1 \/ 9 ","d":"1 \/ 3","e":"9 \/ 25"},"options_float":{"a":3.0,"b":6.0,"c":0.1111111111,"d":0.3333333333,"e":0.36},"annotated_formula":"divide(multiply(1, power(3, const_2)), power(5, 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>\n5 ** 2<\/gadget>\n25<\/output>\n9 \/ 25<\/gadget>\n9\/25 = around 0.36<\/output>\n9\/25 = around 0.36<\/result>","index":2253} +{"problem":"in a class of 50 students , 20 play baseball , 15 play cricket and 11 play football . 7 play both baseball and cricket , 4 play cricket and football and 5 play baseball and football . if 18 students do not play any of these given sports , how many students play exactly two of these sports ?","rationale":"notice that 7 play both baseball and cricket does not mean that out of those 7 , some does not play football too . the same for cricket \/ football and baseball \/ football . [ color = # ffff 00 ] { total } = { baseball } + { cricket } + { football } - { hc + ch + hf } + { all three } + { neither } for more checkadvanced overlapping sets problems [ \/ color ] 50 = 20 + 15 + 11 - ( 7 + 4 + 5 ) + { all three } + 18 - - > { all three } = 2 ; those who play only baseball and cricket are 7 - 2 = 5 ; those who play only cricket and football are 4 - 2 = 2 ; those who play only baseball and football are 5 - 2 = 3 ; hence , 5 + 2 + 3 = 10 students play exactly two of these sports . answer : a .","correct":"a","options":{"a":"10 ","b":"46 ","c":"67 ","d":"68","e":"446"},"options_float":{"a":10.0,"b":46.0,"c":67.0,"d":68.0,"e":446.0},"annotated_formula":"add(subtract(5, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), add(subtract(7, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), subtract(4, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18)))))","linear_formula":"add(n1,n2)|add(n4,n5)|add(n3,#0)|add(n6,#1)|subtract(#2,#3)|add(n7,#4)|subtract(n0,#5)|subtract(n4,#6)|subtract(n5,#6)|subtract(n6,#6)|add(#7,#8)|add(#10,#9)","chain":"20 + 15<\/gadget>\n35<\/output>\n35 + 11<\/gadget>\n46<\/output>\n7 + 4<\/gadget>\n11<\/output>\n11 + 5<\/gadget>\n16<\/output>\n46 - 16<\/gadget>\n30<\/output>\n30 + 18<\/gadget>\n48<\/output>\n50 - 48<\/gadget>\n2<\/output>\n5 - 2<\/gadget>\n3<\/output>\n7 - 2<\/gadget>\n5<\/output>\n4 - 2<\/gadget>\n2<\/output>\n5 + 2<\/gadget>\n7<\/output>\n3 + 7<\/gadget>\n10<\/output>\n10<\/result>","index":2254} +{"problem":"the speed of a train is 90 kmph . what is the distance covered by it in 10 minutes ?","rationale":"\"90 * 10 \/ 60 = 15 kmph answer : a\"","correct":"a","options":{"a":"15 ","b":"87 ","c":"99 ","d":"77","e":"55"},"options_float":{"a":15.0,"b":87.0,"c":99.0,"d":77.0,"e":55.0},"annotated_formula":"multiply(divide(10, const_60), 90)","linear_formula":"divide(n1,const_60)|multiply(n0,#0)|","chain":"10 \/ 60<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 90<\/gadget>\n15<\/output>\n15<\/result>","index":2255} +{"problem":"a positive number x is multiplied by 5 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ?","rationale":"\"sq rt ( 5 x \/ 3 ) = x = > 5 x \/ 3 = x ^ 2 = > x = 5 \/ 3 ans - d\"","correct":"d","options":{"a":"9 \/ 4 ","b":"3 \/ 2 ","c":"4 \/ 3 ","d":"5 \/ 3","e":"1 \/ 2"},"options_float":{"a":2.25,"b":1.5,"c":1.3333333333,"d":1.6666666667,"e":0.5},"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":2257} +{"problem":"the price of commodity x increases by 40 paise every year , while the price of commodity y increases by 15 paise every year . if in 2001 , the price of commodity x was rs . 4.20 and that of y was rs . 6.30 , in which year commodity x will cost 40 paise more than the commodity y ?","rationale":"\"suppose commodity x will cost 40 paise more than y after z years . then , ( 4.20 + 0.40 z ) - ( 6.30 + 0.15 z ) = 0.40 0.25 z = 0.40 + 2.10 z = 2.50 \/ 0.25 = 250 \/ 25 = 10 . therefore , x will cost 40 paise more than y 10 years after 2001 i . e . , 2011 . answer is d .\"","correct":"d","options":{"a":"2010 ","b":"2001 ","c":"2012 ","d":"2011","e":"2009"},"options_float":{"a":2010.0,"b":2001.0,"c":2012.0,"d":2011.0,"e":2009.0},"annotated_formula":"add(2001, divide(add(divide(40, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), divide(15, const_100))))","linear_formula":"divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n6.3 - 4.2<\/gadget>\n2.1<\/output>\n(2\/5) + 2.1<\/gadget>\n2.5<\/output>\n15 \/ 100<\/gadget>\n3\/20 = around 0.15<\/output>\n(2\/5) - (3\/20)<\/gadget>\n1\/4 = around 0.25<\/output>\n2.5 \/ (1\/4)<\/gadget>\n10<\/output>\n2_001 + 10<\/gadget>\n2_011<\/output>\n2_011<\/result>","index":2258} +{"problem":"if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 13 . what is the value of f ( 6 ) ?","rationale":"\"since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 13 then f ( 5 ) = 13 - 5 = 8 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 8 - 6 = 2 . answer : d . questions on funtions to practice :\"","correct":"d","options":{"a":"- 1 ","b":"0 ","c":"1 ","d":"2","e":"4"},"options_float":{"a":-1.0,"b":0.0,"c":1.0,"d":2.0,"e":4.0},"annotated_formula":"subtract(subtract(13, add(1, 4)), 6)","linear_formula":"add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)|","chain":"1 + 4<\/gadget>\n5<\/output>\n13 - 5<\/gadget>\n8<\/output>\n8 - 6<\/gadget>\n2<\/output>\n2<\/result>","index":2263} +{"problem":"what will be the area of a semi - circle of 14 metres diameter ?","rationale":"area of semicircle = ½ π r 2 = ½ × 22 ⁄ 7 × 7 × 7 = 77 m 2 answer b","correct":"b","options":{"a":"154 sq metres ","b":"77 sq metres ","c":"308 sq metres ","d":"22 sq metres","e":"none of these"},"options_float":{"a":154.0,"b":77.0,"c":308.0,"d":22.0,"e":null},"annotated_formula":"divide(circle_area(divide(14, const_2)), const_2)","linear_formula":"divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)","chain":"14 \/ 2<\/gadget>\n7<\/output>\npi * (7 ** 2)<\/gadget>\n49*pi = around 153.93804<\/output>\n(49*pi) \/ 2<\/gadget>\n49*pi\/2 = around 76.96902<\/output>\n49*pi\/2 = around 76.96902<\/result>","index":2264} +{"problem":"the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 2200 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 2200 ( 100 \/ 110 ) = rs . 2000 answer : b\"","correct":"b","options":{"a":"2299 ","b":"2000 ","c":"2670 ","d":"6725","e":"2601"},"options_float":{"a":2299.0,"b":2000.0,"c":2670.0,"d":6725.0,"e":2601.0},"annotated_formula":"divide(2200, 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>\n2_200 \/ (11\/10)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2265} +{"problem":"one fourth of a solution that was 10 % salt 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 salt by weight ?","rationale":"\"say the second solution ( which was 1 \/ 4 th of total ) was x % salt , then 3 \/ 4 * 0.1 + 1 \/ 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : b .\"","correct":"b","options":{"a":"24 % ","b":"34 % ","c":"22 % ","d":"18 %","e":"8.5 %"},"options_float":{"a":24.0,"b":34.0,"c":22.0,"d":18.0,"e":8.5},"annotated_formula":"multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(10, const_100), const_4), divide(10, const_100))), const_100)","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)|","chain":"16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n(4\/25) * 4<\/gadget>\n16\/25 = around 0.64<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 4<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) - (1\/10)<\/gadget>\n3\/10 = around 0.3<\/output>\n(16\/25) - (3\/10)<\/gadget>\n17\/50 = around 0.34<\/output>\n(17\/50) * 100<\/gadget>\n34<\/output>\n34<\/result>","index":2266} +{"problem":"if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 85 , 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 = 170 - - 2 subtracting 1 from 2 we have c - a = 80 ; answer : b\"","correct":"b","options":{"a":"25 ","b":"80 ","c":"90 ","d":"140","e":"it can not be determined from the information given"},"options_float":{"a":25.0,"b":80.0,"c":90.0,"d":140.0,"e":null},"annotated_formula":"subtract(multiply(85, const_2), multiply(45, const_2))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|","chain":"85 * 2<\/gadget>\n170<\/output>\n45 * 2<\/gadget>\n90<\/output>\n170 - 90<\/gadget>\n80<\/output>\n80<\/result>","index":2267} +{"problem":"3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 4 pumps work to empty the tank in 1 day ?","rationale":"the required number of working hours per day x , more pumps , less working hours per day ( indirect ) less days , more working hours per day ( indirect ) pumps 4 : 3 , days 1 : 2 } : : 8 : x therefore 4 * 1 * x = 3 * 2 * 8 , x = ( 3 * 2 * 8 ) \/ 4 x = 12 correct answer ( d )","correct":"d","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":"divide(multiply(multiply(3, 8), 2), 4)","linear_formula":"multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)","chain":"3 * 8<\/gadget>\n24<\/output>\n24 * 2<\/gadget>\n48<\/output>\n48 \/ 4<\/gadget>\n12<\/output>\n12<\/result>","index":2268} +{"problem":"peter invests a sum of money and gets back an amount of $ 810 in 3 years . david invests an equal amount of money and gets an amount of $ 854 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 = 854 - 810 = 44 interest earned for 3 years = 44 * 3 = 132 amount invested = 815 - 132 = 683 answer : b","correct":"b","options":{"a":"670 ","b":"683 ","c":"698 ","d":"744","e":"700"},"options_float":{"a":670.0,"b":683.0,"c":698.0,"d":744.0,"e":700.0},"annotated_formula":"subtract(810, multiply(divide(subtract(854, 810), 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":"854 - 810<\/gadget>\n44<\/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>\n44 \/ (1\/100)<\/gadget>\n4_400<\/output>\n4_400 * (3\/100)<\/gadget>\n132<\/output>\n810 - 132<\/gadget>\n678<\/output>\n678<\/result>","index":2269} +{"problem":"a envelop weight 9.2 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ?","rationale":"\"800 * 9.2 7360.0 gm 7.36 kg answer : e\"","correct":"e","options":{"a":"6.6 kg ","b":"6.8 kg ","c":"6.7 kg ","d":"6.9 kg","e":"7.36 kg"},"options_float":{"a":6.6,"b":6.8,"c":6.7,"d":6.9,"e":7.36},"annotated_formula":"divide(multiply(9.2, 800), const_1000)","linear_formula":"multiply(n0,n1)|divide(#0,const_1000)|","chain":"9.2 * 800<\/gadget>\n7_360<\/output>\n7_360 \/ 1_000<\/gadget>\n184\/25 = around 7.36<\/output>\n184\/25 = around 7.36<\/result>","index":2270} +{"problem":"what least number should be added to 1022 , so that the sum is completely divisible by 25 ?","rationale":"\"1022 ã · 25 = 40 with remainder = 22 22 + 3 = 25 . hence 3 should be added to 1022 so that the sum will be divisible by 25 answer : option b\"","correct":"b","options":{"a":"4 ","b":"3 ","c":"2 ","d":"0","e":"5"},"options_float":{"a":4.0,"b":3.0,"c":2.0,"d":0.0,"e":5.0},"annotated_formula":"subtract(25, reminder(1022, 25))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|","chain":"1_022 % 25<\/gadget>\n22<\/output>\n25 - 22<\/gadget>\n3<\/output>\n3<\/result>","index":2271} +{"problem":"a hostel had provisions for 250 men for 44 days . if 50 men left the hostel , how long will the food last at the same rate ?","rationale":"\"a hostel had provisions for 250 men for 44 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 44 250 × 44 = 200 x 5 × 44 = 4 x x = 5 × 11 = 55 answer a\"","correct":"a","options":{"a":"55 ","b":"40 ","c":"50 ","d":"60","e":"65"},"options_float":{"a":55.0,"b":40.0,"c":50.0,"d":60.0,"e":65.0},"annotated_formula":"divide(multiply(250, 44), subtract(250, 50))","linear_formula":"multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)|","chain":"250 * 44<\/gadget>\n11_000<\/output>\n250 - 50<\/gadget>\n200<\/output>\n11_000 \/ 200<\/gadget>\n55<\/output>\n55<\/result>","index":2272} +{"problem":"the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8339 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8339 ( 100 \/ 124 ) = rs . 6725 . answer : c\"","correct":"c","options":{"a":"rs . 6825 ","b":"rs . 6721 ","c":"rs . 6725 . ","d":"rs . 4298","e":"rs . 6729"},"options_float":{"a":6825.0,"b":6721.0,"c":6725.0,"d":4298.0,"e":6729.0},"annotated_formula":"divide(8339, add(const_1, divide(24, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"24 \/ 100<\/gadget>\n6\/25 = around 0.24<\/output>\n1 + (6\/25)<\/gadget>\n31\/25 = around 1.24<\/output>\n8_339 \/ (31\/25)<\/gadget>\n6_725<\/output>\n6_725<\/result>","index":2273} +{"problem":"63 + 5 * 12 \/ ( 180 \/ 3 ) = ?","rationale":"\"63 + 5 * 12 \/ ( 180 \/ 3 ) = 63 + 5 * 12 \/ ( 60 ) = 63 + ( 5 * 12 ) \/ 60 = 63 + 1 = 64 . answer : d\"","correct":"d","options":{"a":"22 ","b":"77 ","c":"29 ","d":"64","e":"21"},"options_float":{"a":22.0,"b":77.0,"c":29.0,"d":64.0,"e":21.0},"annotated_formula":"add(63, 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>\n63 + 1<\/gadget>\n64<\/output>\n64<\/result>","index":2274} +{"problem":"a hen leaps 3 leaps for every 2 leaps of a duck , but 4 leaps of the duck are equal to 3 leaps of the hen . what is the ratio of the speed of the hen to that of the duck ?","rationale":"\"given ; 4 duck = 3 hen ; or , duck \/ hen = 3 \/ 4 ; let hen ' s 1 leap = 4 meter and ducks 1 leap = 3 meter . then , ratio of speed of hen and duck = 4 * 3 \/ 3 * 2 = 2 : 1 ' ' answer : 2 : 1 ;\"","correct":"a","options":{"a":"2 : 1 ","b":"3 : 4 ","c":"4 : 3 ","d":"1 : 4","e":"5 : 6"},"options_float":{"a":2.0,"b":0.75,"c":1.3333333333,"d":0.25,"e":0.8333333333},"annotated_formula":"divide(divide(3, 2), divide(3, 4))","linear_formula":"divide(n0,n1)|divide(n3,n2)|divide(#0,#1)|","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n(3\/2) \/ (3\/4)<\/gadget>\n2<\/output>\n2<\/result>","index":2277} +{"problem":"praveen starts business with rs . 3640 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 , 3640 * 12 \/ 7 x = 2 \/ 3 = > 14 x = 131040 = > x = 9360 . answer : e\"","correct":"e","options":{"a":"s . 7500 ","b":"s . 8000 ","c":"s . 8500 ","d":"s . 9000","e":"s . 9360"},"options_float":{"a":7500.0,"b":8000.0,"c":8500.0,"d":9000.0,"e":9360.0},"annotated_formula":"divide(divide(3640, 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_640 \/ (7\/12)<\/gadget>\n6_240<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n6_240 \/ (2\/3)<\/gadget>\n9_360<\/output>\n9_360<\/result>","index":2278} +{"problem":"how many positive integers q between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ?","rationale":"1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e .","correct":"e","options":{"a":"3 ","b":"16 ","c":"75 ","d":"24","e":"26"},"options_float":{"a":3.0,"b":16.0,"c":75.0,"d":24.0,"e":26.0},"annotated_formula":"subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), floor(add(subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(add(const_10, 5), 2)), const_1))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5)))","linear_formula":"add(n4,const_10)|divide(n1,n2)|divide(n0,n2)|divide(n1,n3)|divide(n0,n3)|divide(n1,n4)|divide(n0,n4)|subtract(n1,n0)|add(#7,const_1)|add(n2,#0)|subtract(#1,#2)|subtract(#3,#4)|subtract(#5,#6)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|subtract(#8,#13)|subtract(#14,#9)|subtract(#15,#0)|add(#17,const_1)|floor(#19)|subtract(#16,#20)|subtract(#21,#18)","chain":"300 - 200<\/gadget>\n100<\/output>\n100 + 1<\/gadget>\n101<\/output>\n300 \/ 2<\/gadget>\n150<\/output>\n200 \/ 2<\/gadget>\n100<\/output>\n150 - 100<\/gadget>\n50<\/output>\n50 + 1<\/gadget>\n51<\/output>\n101 - 51<\/gadget>\n50<\/output>\n300 \/ 3<\/gadget>\n100<\/output>\n200 \/ 3<\/gadget>\n200\/3 = around 66.666667<\/output>\n100 - (200\/3)<\/gadget>\n100\/3 = around 33.333333<\/output>\n(100\/3) + 1<\/gadget>\n103\/3 = around 34.333333<\/output>\n10 + 5<\/gadget>\n15<\/output>\n15 + 2<\/gadget>\n17<\/output>\n(103\/3) - 17<\/gadget>\n52\/3 = around 17.333333<\/output>\n(52\/3) + 1<\/gadget>\n55\/3 = around 18.333333<\/output>\nfloor(55\/3)<\/gadget>\n18<\/output>\n50 - 18<\/gadget>\n32<\/output>\n300 \/ 5<\/gadget>\n60<\/output>\n200 \/ 5<\/gadget>\n40<\/output>\n60 - 40<\/gadget>\n20<\/output>\n20 + 1<\/gadget>\n21<\/output>\n21 - 15<\/gadget>\n6<\/output>\n32 - 6<\/gadget>\n26<\/output>\n26<\/result>","index":2280} +{"problem":"rs . 850 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 850 become in 3 years ?","rationale":"\"solution s . i . = rs . ( 956 - 850 ) = rs . 106 rate = ( 100 x 106 \/ 850 x 3 ) = 212 \/ 51 % new rate = ( 212 \/ 51 + 4 ) % = 416 \/ 51 % new s . i . = rs . ( 850 x 416 \/ 51 x 3 \/ 100 ) rs . 208 . ∴ new amount = rs . ( 850 + 208 ) = rs . 1058 . answer c\"","correct":"c","options":{"a":"rs . 1020.80 ","b":"rs . 1025 ","c":"rs . 1058 ","d":"data inadequate","e":"none of these"},"options_float":{"a":1020.8,"b":1025.0,"c":1058.0,"d":null,"e":null},"annotated_formula":"add(850, divide(multiply(multiply(850, add(divide(multiply(subtract(956, 850), const_100), multiply(850, 3)), 4)), 3), const_100))","linear_formula":"multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|","chain":"956 - 850<\/gadget>\n106<\/output>\n106 * 100<\/gadget>\n10_600<\/output>\n850 * 3<\/gadget>\n2_550<\/output>\n10_600 \/ 2_550<\/gadget>\n212\/51 = around 4.156863<\/output>\n(212\/51) + 4<\/gadget>\n416\/51 = around 8.156863<\/output>\n850 * (416\/51)<\/gadget>\n20_800\/3 = around 6_933.333333<\/output>\n(20_800\/3) * 3<\/gadget>\n20_800<\/output>\n20_800 \/ 100<\/gadget>\n208<\/output>\n850 + 208<\/gadget>\n1_058<\/output>\n1_058<\/result>","index":2281} +{"problem":"it takes ten minutes to load a certain video on a cellphone , and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates , how many seconds would it take them to load the video , rounded to the nearest hundredth ?","rationale":"\"the laptop can load the video at a rate of 1 \/ 15 of the video per second . the phone can load the video at a rate of 1 \/ ( 60 * 10 ) = 1 \/ 600 of the video per second . the combined rate is 1 \/ 15 + 1 \/ 600 = 41 \/ 600 of the video per second . the time required to load the video is 600 \/ 41 = 14.63 seconds . the answer is d .\"","correct":"d","options":{"a":"13.42 ","b":"13.86 ","c":"14.25 ","d":"14.63","e":"14.88"},"options_float":{"a":13.42,"b":13.86,"c":14.25,"d":14.63,"e":14.88},"annotated_formula":"subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))","linear_formula":"add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)|","chain":"2 + 3<\/gadget>\n5<\/output>\n5 + 4<\/gadget>\n9<\/output>\n9 * 60<\/gadget>\n540<\/output>\n1 \/ 540<\/gadget>\n1\/540 = around 0.001852<\/output>\n3 * 4<\/gadget>\n12<\/output>\n12 + 3<\/gadget>\n15<\/output>\n1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/540) + (1\/15)<\/gadget>\n37\/540 = around 0.068519<\/output>\n1 \/ (37\/540)<\/gadget>\n540\/37 = around 14.594595<\/output>\n4 * 4<\/gadget>\n16<\/output>\n16 * 3<\/gadget>\n48<\/output>\n48 - 2<\/gadget>\n46<\/output>\n100 * 100<\/gadget>\n10_000<\/output>\n46 \/ 10_000<\/gadget>\n23\/5_000 = around 0.0046<\/output>\n(540\/37) - (23\/5_000)<\/gadget>\n2_699_149\/185_000 = around 14.589995<\/output>\n2_699_149\/185_000 = around 14.589995<\/result>","index":2283} +{"problem":"a certain scholarship committee awarded scholarships in the amounts of $ 1250 , $ 2500 and $ 4000 . the committee awarded twice as many $ 2500 scholarships as $ 4000 and it awarded 3 times as many $ 1250 scholarships as $ 2500 scholarships . if the total of $ 75000 was awarded in $ 1250 scholarships , how many $ 4000 scholarships were awarded ?","rationale":"since the starting point is given as the $ 4000 scholarship , assume $ 4000 scholarships to be x by the given information , $ 2500 scholarships = 2 x and $ 1250 scholarships = 6 x gievn : total $ 1250 scholarships = $ 75000 6 x * 1250 = 75000 solve for x = 10 option d","correct":"d","options":{"a":"5 ","b":"6 ","c":"9 ","d":"10","e":"15"},"options_float":{"a":5.0,"b":6.0,"c":9.0,"d":10.0,"e":15.0},"annotated_formula":"divide(divide(75000, 1250), multiply(const_2, 3))","linear_formula":"divide(n8,n0)|multiply(n5,const_2)|divide(#0,#1)","chain":"75_000 \/ 1_250<\/gadget>\n60<\/output>\n2 * 3<\/gadget>\n6<\/output>\n60 \/ 6<\/gadget>\n10<\/output>\n10<\/result>","index":2284} +{"problem":"in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 6 eggs . if the cost of each pound of rice is $ 0.33 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]","rationale":"\"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 \/ 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 \/ 12 * 33 = 33 cents . answer : c .\"","correct":"c","options":{"a":"0.33 ","b":"0.44 ","c":"33 ","d":"44","e":"55"},"options_float":{"a":0.33,"b":0.44,"c":33.0,"d":44.0,"e":55.0},"annotated_formula":"multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.33, 100))","linear_formula":"divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n6 \/ (1\/2)<\/gadget>\n12<\/output>\n12 \/ 12<\/gadget>\n1<\/output>\n0.33 * 100<\/gadget>\n33<\/output>\n1 * 33<\/gadget>\n33<\/output>\n33<\/result>","index":2285} +{"problem":"a pupil ' s marks were wrongly entered as 73 instead of 40 . due to 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 = ( 73 - 40 ) = > x \/ 2 = 33 = > x = 66 . answer : c\"","correct":"c","options":{"a":"18 ","b":"82 ","c":"66 ","d":"27","e":"77"},"options_float":{"a":18.0,"b":82.0,"c":66.0,"d":27.0,"e":77.0},"annotated_formula":"multiply(subtract(73, 40), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)|","chain":"73 - 40<\/gadget>\n33<\/output>\n33 * 2<\/gadget>\n66<\/output>\n66<\/result>","index":2286} +{"problem":"a baseball card decreased in value 25 % 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":"\"let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 25 \/ 100 ) * 100 = 75 after second year , value of baseball card = ( 1 - 10 \/ 100 ) * 75 = 67.5 total percent decrease of the card ' s value over the two years = ( 100 - 67.5 ) \/ 100 * 100 % = 31.5 % answer c\"","correct":"c","options":{"a":"28 % ","b":"30 % ","c":"32.5 % ","d":"36 %","e":"72 %"},"options_float":{"a":28.0,"b":30.0,"c":32.5,"d":36.0,"e":72.0},"annotated_formula":"subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(25, 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>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n(9\/10) * (3\/4)<\/gadget>\n27\/40 = around 0.675<\/output>\n(27\/40) * 100<\/gadget>\n135\/2 = around 67.5<\/output>\n100 - (135\/2)<\/gadget>\n65\/2 = around 32.5<\/output>\n65\/2 = around 32.5<\/result>","index":2287} +{"problem":"if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?","rationale":"\"total nos of ways in which we can choose n = 96 n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? case 1 : n = odd then n + 2 = odd & n + 1 will be even i . e this needs get divided by 8 , hence is a multiple of 8 so we have 8 . . 96 = 12 multiples to fill the n + 1 pos hence 12 ways case 2 : n is even then n + 2 will be even & the product will be divisible by 24 & thus 8 so nos of values that can be used for n = 2 . . . . 96 ( all even nos ) i . e 48 nos total = 48 + 12 = 60 ways so reqd p = 60 \/ 96 = 5 \/ 8 ; answer : d\"","correct":"d","options":{"a":"1 \/ 4 ","b":"3 \/ 8 ","c":"1 \/ 2 ","d":"5 \/ 8","e":"3 \/ 4"},"options_float":{"a":0.25,"b":0.375,"c":0.5,"d":0.625,"e":0.75},"annotated_formula":"divide(add(divide(96, 2), divide(96, 8)), 96)","linear_formula":"divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|","chain":"96 \/ 2<\/gadget>\n48<\/output>\n96 \/ 8<\/gadget>\n12<\/output>\n48 + 12<\/gadget>\n60<\/output>\n60 \/ 96<\/gadget>\n5\/8 = around 0.625<\/output>\n5\/8 = around 0.625<\/result>","index":2288} +{"problem":"a dealer offers a cash discount of 16 % and still makes a profit of 25 % when he further allows 60 articles to be sold at the cost price of 50 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ?","rationale":"\"given cash discount - 16 % profit - 25 % items sold - 60 price sold at = list price of 50 assume list price = $ 10 total invoice = $ 500 - 16 % cash discount = $ 420 let cost price of 60 items be x so total cost = 60 * x given the shopkeeper had a profit of 25 % 60 * x * 125 \/ 100 = 420 or x = $ 7 * 4 \/ 5 = $ 28 \/ 5 which means his products were listed at $ 10 which is a 78 + ( 4 \/ 7 ) % markup over $ 28 \/ 5 answer e\"","correct":"e","options":{"a":"50 % ","b":"60 % ","c":"70 % ","d":"75 %","e":"78 + ( 4 \/ 7 ) %"},"options_float":{"a":50.0,"b":60.0,"c":70.0,"d":75.0,"e":78.0},"annotated_formula":"multiply(subtract(divide(divide(divide(add(const_100, 25), const_100), subtract(const_1, divide(subtract(60, 50), 60))), divide(subtract(const_100, 16), const_100)), const_1), const_100)","linear_formula":"add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)|","chain":"100 + 25<\/gadget>\n125<\/output>\n125 \/ 100<\/gadget>\n5\/4 = around 1.25<\/output>\n60 - 50<\/gadget>\n10<\/output>\n10 \/ 60<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 - (1\/6)<\/gadget>\n5\/6 = around 0.833333<\/output>\n(5\/4) \/ (5\/6)<\/gadget>\n3\/2 = around 1.5<\/output>\n100 - 16<\/gadget>\n84<\/output>\n84 \/ 100<\/gadget>\n21\/25 = around 0.84<\/output>\n(3\/2) \/ (21\/25)<\/gadget>\n25\/14 = around 1.785714<\/output>\n(25\/14) - 1<\/gadget>\n11\/14 = around 0.785714<\/output>\n(11\/14) * 100<\/gadget>\n550\/7 = around 78.571429<\/output>\n550\/7 = around 78.571429<\/result>","index":2291} +{"problem":"if n divided by 8 has a remainder of 1 , what is the remainder when 3 times n is divided by 8 ?","rationale":"as per question = > n = 8 p + 1 for some integer p hence 3 n = > 24 q + 3 = > remainder = > 3 for some integer q hence b","correct":"b","options":{"a":"1 ","b":"3 ","c":"7 ","d":"5","e":"6"},"options_float":{"a":1.0,"b":3.0,"c":7.0,"d":5.0,"e":6.0},"annotated_formula":"multiply(3, 1)","linear_formula":"multiply(n1,n2)","chain":"3 * 1<\/gadget>\n3<\/output>\n3<\/result>","index":2292} +{"problem":"the volume of a cube is 2197 cc . find its surface .","rationale":"a 3 = 2197 = > a = 13 6 a 2 = 6 * 13 * 13 = 1014 answer : b","correct":"b","options":{"a":"864 ","b":"1014 ","c":"1299 ","d":"1268","e":"1191"},"options_float":{"a":864.0,"b":1014.0,"c":1299.0,"d":1268.0,"e":1191.0},"annotated_formula":"surface_cube(cube_edge_by_volume(2197))","linear_formula":"cube_edge_by_volume(n0)|surface_cube(#0)|","chain":"2_197 ** (1\/3)<\/gadget>\n13<\/output>\n6 * (13 ** 2)<\/gadget>\n1_014<\/output>\n1_014<\/result>","index":2293} +{"problem":"15 business executives and 3 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 15 business exec and in each handshake 2 business execs are involved . hence 15 c 2 = 105 also , each of 15 exec will shake hand with every 3 other chairmen for total of 45 handshake . total = 45 + 105 = 150 ans : a\"","correct":"a","options":{"a":"150 ","b":"131 ","c":"115 ","d":"90","e":"45"},"options_float":{"a":150.0,"b":131.0,"c":115.0,"d":90.0,"e":45.0},"annotated_formula":"add(divide(multiply(15, subtract(15, const_1)), const_2), multiply(15, 3))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)|","chain":"15 - 1<\/gadget>\n14<\/output>\n15 * 14<\/gadget>\n210<\/output>\n210 \/ 2<\/gadget>\n105<\/output>\n15 * 3<\/gadget>\n45<\/output>\n105 + 45<\/gadget>\n150<\/output>\n150<\/result>","index":2294} +{"problem":"a is twice as fast as b . if b alone can do a piece of work in 18 days , in what time can a and b together complete the work ?","rationale":"\"a can do the work in 18 \/ 2 i . e . , 9 days . a and b ' s one day ' s work = 1 \/ 9 + 1 \/ 18 = ( 2 + 1 ) \/ 18 = 1 \/ 6 so a and b together can do the work in 6 days . answer : d\"","correct":"d","options":{"a":"10 ","b":"16 ","c":"18 ","d":"6","e":"12"},"options_float":{"a":10.0,"b":16.0,"c":18.0,"d":6.0,"e":12.0},"annotated_formula":"inverse(add(divide(const_1, 18), multiply(divide(const_1, 18), const_2)))","linear_formula":"divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 18<\/gadget>\n1\/18 = around 0.055556<\/output>\n(1\/18) * 2<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/18) + (1\/9)<\/gadget>\n1\/6 = around 0.166667<\/output>\n1 \/ (1\/6)<\/gadget>\n6<\/output>\n6<\/result>","index":2296} +{"problem":"the area of a triangle is with base 5.5 m and height 6 m ?","rationale":"\"1 \/ 2 * 5.5 * 6 = 16.5 m 2 answer : b\"","correct":"b","options":{"a":"11 m 2 ","b":"16.5 m 2 ","c":"18.5 m 2 ","d":"19.5 m 2","e":"12 m 2"},"options_float":{"a":11.0,"b":16.5,"c":18.5,"d":19.5,"e":12.0},"annotated_formula":"triangle_area(5.5, 6)","linear_formula":"triangle_area(n0,n1)|","chain":"(5.5 * 6) \/ 2<\/gadget>\n16.5<\/output>\n16.5<\/result>","index":2297} +{"problem":"the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 8400 for a computer table , then what was the cost price of the computer table ?","rationale":"\"cp = sp * ( 100 \/ ( 100 + profit % ) ) = 8400 ( 100 \/ 125 ) = rs . 6720 . answer : d\"","correct":"d","options":{"a":"rs . 5725 ","b":"rs . 5275 ","c":"rs . 6275 ","d":"rs . 6720","e":"none of these"},"options_float":{"a":5725.0,"b":5275.0,"c":6275.0,"d":6720.0,"e":null},"annotated_formula":"divide(8400, 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>\n8_400 \/ (5\/4)<\/gadget>\n6_720<\/output>\n6_720<\/result>","index":2299} +{"problem":"the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 10 and 11 . the larger of the two numbers is :","rationale":"\"clearly , the numbers are ( 23 x 10 ) and ( 23 x 11 ) . larger number = ( 23 x 11 ) = 253 . answer : option b\"","correct":"b","options":{"a":"276 ","b":"253 ","c":"322 ","d":"345","e":"395"},"options_float":{"a":276.0,"b":253.0,"c":322.0,"d":345.0,"e":395.0},"annotated_formula":"multiply(23, 11)","linear_formula":"multiply(n0,n2)|","chain":"23 * 11<\/gadget>\n253<\/output>\n253<\/result>","index":2301} +{"problem":"tim and é lan are 60 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ?","rationale":"tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice c meet the 2 : 1 ( tim : elan = 40 : 20 ) ratio of maintaining total distance traveled 60 miles so correct answer c","correct":"c","options":{"a":"30 miles . ","b":"35 miles . ","c":"40 miles . ","d":"60 miles .","e":"65 miles ."},"options_float":{"a":30.0,"b":35.0,"c":40.0,"d":60.0,"e":65.0},"annotated_formula":"multiply(divide(10, add(5, 10)), 60)","linear_formula":"add(n1,n2)|divide(n1,#0)|multiply(n0,#1)","chain":"5 + 10<\/gadget>\n15<\/output>\n10 \/ 15<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 60<\/gadget>\n40<\/output>\n40<\/result>","index":2303} +{"problem":"if y is 90 % greater than x , than x is what % less than y ?","rationale":"\"y = 1.9 x x = y \/ 1.9 = 10 y \/ 19 x is 9 \/ 19 less which is 47.4 % less than y . the answer is d .\"","correct":"d","options":{"a":"35.7 % ","b":"39.8 % ","c":"43.2 % ","d":"47.4 %","e":"51.2 %"},"options_float":{"a":35.7,"b":39.8,"c":43.2,"d":47.4,"e":51.2},"annotated_formula":"multiply(divide(90, add(90, const_100)), const_100)","linear_formula":"add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)|","chain":"90 + 100<\/gadget>\n190<\/output>\n90 \/ 190<\/gadget>\n9\/19 = around 0.473684<\/output>\n(9\/19) * 100<\/gadget>\n900\/19 = around 47.368421<\/output>\n900\/19 = around 47.368421<\/result>","index":2304} +{"problem":"what is the length of a bridge ( in meters ) , which a train 166 meters long and travelling at 45 km \/ h can cross in 40 seconds ?","rationale":"\"speed = 45 km \/ h = 45000 m \/ 3600 s = 25 \/ 2 m \/ s in 40 seconds , the train can travel 25 \/ 2 * 40 = 500 meters 500 = length of train + length of bridge length of bridge = 500 - 166 = 334 meters the answer is d .\"","correct":"d","options":{"a":"310 ","b":"318 ","c":"326 ","d":"334","e":"342"},"options_float":{"a":310.0,"b":318.0,"c":326.0,"d":334.0,"e":342.0},"annotated_formula":"subtract(multiply(multiply(45, const_0_2778), 40), 166)","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) * 40<\/gadget>\n500<\/output>\n500 - 166<\/gadget>\n334<\/output>\n334<\/result>","index":2305} +{"problem":"the average weight of 10 persons increases by 4.2 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":"\"solution total weight increased = ( 10 x 4.2 ) kg = 42 kg . weight of new person = ( 65 + 42 ) kg = 107 kg . answer a\"","correct":"a","options":{"a":"107 kg ","b":"80 kg ","c":"120 kg ","d":"90 kg","e":"100"},"options_float":{"a":107.0,"b":80.0,"c":120.0,"d":90.0,"e":100.0},"annotated_formula":"add(65, multiply(10, 4.2))","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"10 * 4.2<\/gadget>\n42<\/output>\n65 + 42<\/gadget>\n107<\/output>\n107<\/result>","index":2306} +{"problem":"a certain class of students is being divided into teams . the class can either be divided into 8 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?","rationale":"\"let total no of students in the class be n so , we are told that n is divisible by both 8 24 so , lets find the least common multiple of 8 24 , ie 24 so our answer is ( d ) 24\"","correct":"d","options":{"a":"6 ","b":"36 ","c":"48 ","d":"24","e":"72"},"options_float":{"a":6.0,"b":36.0,"c":48.0,"d":24.0,"e":72.0},"annotated_formula":"lcm(8, 24)","linear_formula":"lcm(n0,n1)|","chain":"lcm(8, 24)<\/gadget>\n24<\/output>\n24<\/result>","index":2308} +{"problem":"in a games hour 4 different types of players came to the ground ? cricket 11 , hokey 15 , football 21 , softball 15 . how many players are present in the ground ?","rationale":"\"total number of players = 11 + 15 + 21 + 15 = 62 answer is c\"","correct":"c","options":{"a":"70 ","b":"52 ","c":"62 ","d":"49","e":"50"},"options_float":{"a":70.0,"b":52.0,"c":62.0,"d":49.0,"e":50.0},"annotated_formula":"add(add(11, 15), add(21, 15))","linear_formula":"add(n1,n2)|add(n3,n4)|add(#0,#1)|","chain":"11 + 15<\/gadget>\n26<\/output>\n21 + 15<\/gadget>\n36<\/output>\n26 + 36<\/gadget>\n62<\/output>\n62<\/result>","index":2318} +{"problem":"a caterer ordered 125 ice - cream bars and 125 sundaes . if the total price was $ 275.00 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 = 275 = > 125 * s = 200 = > s = 1.6 answer e\"","correct":"e","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(275.00, multiply(125, 0.60)), 125)","linear_formula":"multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)|","chain":"125 * 0.6<\/gadget>\n75<\/output>\n275 - 75<\/gadget>\n200<\/output>\n200 \/ 125<\/gadget>\n8\/5 = around 1.6<\/output>\n8\/5 = around 1.6<\/result>","index":2321} +{"problem":"a batsman makes a score of 50 runs in the 6 th inning and thus increases his average by 2 . find his average after 6 th inning .","rationale":"\"let the average after 6 th inning = x then , average after 5 th inning = x - 2 5 ( x - 2 ) + 50 = 6 x x = 10 - 50 = 40 answer is 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":"add(subtract(50, multiply(6, 2)), 2)","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|","chain":"6 * 2<\/gadget>\n12<\/output>\n50 - 12<\/gadget>\n38<\/output>\n38 + 2<\/gadget>\n40<\/output>\n40<\/result>","index":2322} +{"problem":"the distance between two cities a and b is 330 km . a train starts from a at 8 a . m . and travels 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":"\"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 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":"12 ","b":"10 ","c":"11 ","d":"09","e":"03"},"options_float":{"a":12.0,"b":10.0,"c":11.0,"d":9.0,"e":3.0},"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":2326} +{"problem":"a high school has 360 students 1 \/ 2 attend the arithmetic club , 5 \/ 8 attend the biology club and 3 \/ 4 attend the chemistry club . 3 \/ 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 180 members ( 1 \/ 2 of 360 ) b - club has 225 members ( 5 \/ 8 of 360 ) c - club has 270 members ( 3 \/ 4 of 360 ) we can create an equation to solve this : 180 + 225 + 270 = 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 ( 135 ) . 180 + 225 + 270 = 360 + x + 270 x = 405 - 360 = 45 a\"","correct":"a","options":{"a":"45 ","b":"40 ","c":"35 ","d":"50","e":"55"},"options_float":{"a":45.0,"b":40.0,"c":35.0,"d":50.0,"e":55.0},"annotated_formula":"subtract(subtract(add(add(divide(multiply(360, 1), 2), divide(multiply(360, 5), 8)), divide(multiply(360, 3), 4)), multiply(divide(multiply(360, 3), 8), 2)), 360)","linear_formula":"multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|divide(#0,n2)|divide(#1,n4)|divide(#2,n6)|divide(#2,n4)|add(#3,#4)|multiply(n2,#6)|add(#7,#5)|subtract(#9,#8)|subtract(#10,n0)|","chain":"360 * 1<\/gadget>\n360<\/output>\n360 \/ 2<\/gadget>\n180<\/output>\n360 * 5<\/gadget>\n1_800<\/output>\n1_800 \/ 8<\/gadget>\n225<\/output>\n180 + 225<\/gadget>\n405<\/output>\n360 * 3<\/gadget>\n1_080<\/output>\n1_080 \/ 4<\/gadget>\n270<\/output>\n405 + 270<\/gadget>\n675<\/output>\n1_080 \/ 8<\/gadget>\n135<\/output>\n135 * 2<\/gadget>\n270<\/output>\n675 - 270<\/gadget>\n405<\/output>\n405 - 360<\/gadget>\n45<\/output>\n45<\/result>","index":2327} +{"problem":"the cash difference between the selling prices of an article at a profit of 8 % and 6 % is rs . 3 . the ratio of the two selling prices is ?","rationale":"\"let c . p . of the article be rs . x . then , required ratio = 108 % of x \/ 106 % of x = 108 \/ 106 = 54 \/ 53 = 54 : 53 answer : d\"","correct":"d","options":{"a":"52 : 56 ","b":"52 : 53 ","c":"52 : 50 ","d":"54 : 53","e":"52 : 51"},"options_float":{"a":0.9285714286,"b":0.9811320755,"c":1.04,"d":1.0188679245,"e":1.0196078431},"annotated_formula":"divide(add(const_100, 8), add(const_100, 6))","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|","chain":"100 + 8<\/gadget>\n108<\/output>\n100 + 6<\/gadget>\n106<\/output>\n108 \/ 106<\/gadget>\n54\/53 = around 1.018868<\/output>\n54\/53 = around 1.018868<\/result>","index":2328} +{"problem":"reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 588 as interest at the end of the loan period , what was the rate of interest ?","rationale":"\"let rate = r % and time = r years . then , 1200 x r x r \/ 100 = 588 12 r 2 = 588 r 2 = 49 r = 7 . answer : b\"","correct":"b","options":{"a":"3.6 ","b":"7 ","c":"18 ","d":"can not be determined","e":"none of these"},"options_float":{"a":3.6,"b":7.0,"c":18.0,"d":null,"e":null},"annotated_formula":"sqrt(divide(multiply(588, const_100), 1200))","linear_formula":"multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)|","chain":"588 * 100<\/gadget>\n58_800<\/output>\n58_800 \/ 1_200<\/gadget>\n49<\/output>\n49 ** (1\/2)<\/gadget>\n7<\/output>\n7<\/result>","index":2329} +{"problem":"what is x if x + 5 y = 24 and y = 2 ?","rationale":"\"substitute y by 2 in x + 5 y = 24 x + 5 ( 2 ) = 24 x + 10 = 24 if we substitute x by 14 in x + 10 = 24 , we have 14 + 10 = 24 . hence x = 14 correct answer e\"","correct":"e","options":{"a":"1 ","b":"3 ","c":"5 ","d":"7","e":"14"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":7.0,"e":14.0},"annotated_formula":"subtract(24, multiply(5, 2))","linear_formula":"multiply(n0,n2)|subtract(n1,#0)|","chain":"5 * 2<\/gadget>\n10<\/output>\n24 - 10<\/gadget>\n14<\/output>\n14<\/result>","index":2330} +{"problem":"the wages earned by robin is 20 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ?","rationale":"\"let wage of erica = 10 wage of robin = 1.2 * 10 = 12 wage of charles = 1.6 * 10 = 16 percentage by which wage earned by charles is more than that earned by robin = ( 16 - 12 ) \/ 12 * 100 % = 4 \/ 12 * 100 % = 33 % answer b\"","correct":"b","options":{"a":"18.75 % ","b":"33 % ","c":"30 % ","d":"50 %","e":"100 %"},"options_float":{"a":18.75,"b":33.0,"c":30.0,"d":50.0,"e":100.0},"annotated_formula":"multiply(divide(subtract(add(const_100, 60), add(const_100, 20)), add(const_100, 20)), const_100)","linear_formula":"add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"100 + 60<\/gadget>\n160<\/output>\n100 + 20<\/gadget>\n120<\/output>\n160 - 120<\/gadget>\n40<\/output>\n40 \/ 120<\/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":2331} +{"problem":"there are two numbers . if 10 % of the first number is added to the second number , then the second number increases to its 6 - fifth . what is the ratio of the first number to the second number ?","rationale":"let the two numbers be x and y . ( 1 \/ 10 ) * x + y = ( 6 \/ 5 ) * y ( 1 \/ 10 ) * x = ( 1 \/ 5 ) * y x \/ y = 2 \/ 1 = 2 \/ 1 the answer is e .","correct":"e","options":{"a":"3 : 2 ","b":"4 : 3 ","c":"8 : 7 ","d":"5 : 8","e":"2 : 1"},"options_float":{"a":1.5,"b":1.3333333333,"c":1.1428571429,"d":0.625,"e":2.0},"annotated_formula":"divide(divide(const_1, divide(10, const_2)), divide(const_1, 10))","linear_formula":"divide(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|divide(#2,#1)","chain":"10 \/ 2<\/gadget>\n5<\/output>\n1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n1 \/ 10<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/5) \/ (1\/10)<\/gadget>\n2<\/output>\n2<\/result>","index":2332} +{"problem":"each child has 2 pencils and 13 skittles . if there are 6 children , how many pencils are there in total ?","rationale":"2 * 6 = 12 . answer is b .","correct":"b","options":{"a":"16 ","b":"12 ","c":"18 ","d":"22","e":"08"},"options_float":{"a":16.0,"b":12.0,"c":18.0,"d":22.0,"e":8.0},"annotated_formula":"multiply(2, 6)","linear_formula":"multiply(n0,n2)|","chain":"2 * 6<\/gadget>\n12<\/output>\n12<\/result>","index":2333} +{"problem":"a reduction of 40 % in the price of bananas would enable a man to obtain 60 more for rs . 40 , what is reduced price per dozen ?","rationale":"\"40 * ( 40 \/ 100 ) = 16 - - - 60 ? - - - 12 = > rs . 3.2 answer : c\"","correct":"c","options":{"a":"1.2 ","b":"2.2 ","c":"3.2 ","d":"4.2","e":"5"},"options_float":{"a":1.2,"b":2.2,"c":3.2,"d":4.2,"e":5.0},"annotated_formula":"multiply(const_12, divide(multiply(40, divide(40, const_100)), 60))","linear_formula":"divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n40 * (2\/5)<\/gadget>\n16<\/output>\n16 \/ 60<\/gadget>\n4\/15 = around 0.266667<\/output>\n12 * (4\/15)<\/gadget>\n16\/5 = around 3.2<\/output>\n16\/5 = around 3.2<\/result>","index":2334} +{"problem":"in a fuel station the service costs $ 2.05 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 70 liters and a truck ' s tank is 120 % bigger and they are all empty - ?","rationale":"\"service cost of 3 van and 2 truck = 2.05 * ( 3 + 2 ) = 10.5 fuel in 3 van = 3 * 70 = 210 litre fuel in 2 trucks = 2 * 70 ( 1 + 120 \/ 100 ) = 308 total fuel ( van + truck ) = 518 litre total fuel cost = 518 * 0.6 = 310.8 total cost = fuel + service = 310.8 + 10.25 = 321.05 answer is b\"","correct":"b","options":{"a":"312.6 $ ","b":"321.05 $ ","c":"343.7 $ ","d":"398.85 $","e":"412.12 $"},"options_float":{"a":312.6,"b":321.05,"c":343.7,"d":398.85,"e":412.12},"annotated_formula":"add(multiply(multiply(add(70, divide(multiply(70, 120), const_100)), 2), 0.6), multiply(multiply(70, 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":"70 * 120<\/gadget>\n8_400<\/output>\n8_400 \/ 100<\/gadget>\n84<\/output>\n70 + 84<\/gadget>\n154<\/output>\n154 * 2<\/gadget>\n308<\/output>\n308 * 0.6<\/gadget>\n184.8<\/output>\n70 * 3<\/gadget>\n210<\/output>\n210 * 0.6<\/gadget>\n126<\/output>\n184.8 + 126<\/gadget>\n310.8<\/output>\n310.8<\/result>","index":2335} +{"problem":"a man two flats for $ 675958 each . on one he gains 16 % while on the other he loses 16 % . how much does he gain or lose in the whole transaction ?","rationale":"\"in such a case there is always a loss loss % = ( 16 \/ 10 ) ^ 2 = 64 \/ 25 = 2.56 % answer is b\"","correct":"b","options":{"a":"2 % ","b":"2.56 % ","c":"3.12 % ","d":"4.65 %","e":"5.12 %"},"options_float":{"a":2.0,"b":2.56,"c":3.12,"d":4.65,"e":5.12},"annotated_formula":"multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 16)), 675958), multiply(divide(const_100, subtract(const_100, 16)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 16)), 675958), multiply(divide(const_100, subtract(const_100, 16)), 675958))), const_100)","linear_formula":"add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100)|","chain":"100 + 16<\/gadget>\n116<\/output>\n100 \/ 116<\/gadget>\n25\/29 = around 0.862069<\/output>\n(25\/29) * 675_958<\/gadget>\n16_898_950\/29 = around 582_722.413793<\/output>\n100 - 16<\/gadget>\n84<\/output>\n100 \/ 84<\/gadget>\n25\/21 = around 1.190476<\/output>\n(25\/21) * 675_958<\/gadget>\n16_898_950\/21 = around 804_711.904762<\/output>\n(16_898_950\/29) + (16_898_950\/21)<\/gadget>\n844_947_500\/609 = around 1_387_434.318555<\/output>\n675_958 + 675_958<\/gadget>\n1_351_916<\/output>\n(844_947_500\/609) - 1_351_916<\/gadget>\n21_630_656\/609 = around 35_518.318555<\/output>\n(21_630_656\/609) \/ (844_947_500\/609)<\/gadget>\n16\/625 = around 0.0256<\/output>\n(16\/625) * 100<\/gadget>\n64\/25 = around 2.56<\/output>\n64\/25 = around 2.56<\/result>","index":2336} +{"problem":"a rectangular lawn of dimensions 120 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 . 3 per sq m ?","rationale":"\"area = ( l + b – d ) d ( 120 + 60 – 10 ) 10 = > 1700 m 2 1700 * 3 = rs . 5100 answer : a\"","correct":"a","options":{"a":"s . 5100 ","b":"s . 3900 ","c":"s . 3228 ","d":"s . 3922","e":"s . 3928"},"options_float":{"a":5100.0,"b":3900.0,"c":3228.0,"d":3922.0,"e":3928.0},"annotated_formula":"multiply(multiply(subtract(add(120, 60), 10), 10), 3)","linear_formula":"add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)|","chain":"120 + 60<\/gadget>\n180<\/output>\n180 - 10<\/gadget>\n170<\/output>\n170 * 10<\/gadget>\n1_700<\/output>\n1_700 * 3<\/gadget>\n5_100<\/output>\n5_100<\/result>","index":2338} +{"problem":"the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 502 to 600 inclusive ?","rationale":"\"2 + 4 + 6 + 8 + . . . + 100 = 2550 502 + 504 + . . . + 600 = 50 ( 500 ) + ( 2 + 4 + . . . + 100 ) = 25,000 + 2550 = 27,550 the answer is b .\"","correct":"b","options":{"a":"23,550 ","b":"27,550 ","c":"31,550 ","d":"35,550","e":"39,550"},"options_float":{"a":23550.0,"b":27550.0,"c":31550.0,"d":35550.0,"e":39550.0},"annotated_formula":"multiply(divide(add(600, 502), const_2), add(divide(subtract(600, 502), const_2), const_1))","linear_formula":"add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)|","chain":"600 + 502<\/gadget>\n1_102<\/output>\n1_102 \/ 2<\/gadget>\n551<\/output>\n600 - 502<\/gadget>\n98<\/output>\n98 \/ 2<\/gadget>\n49<\/output>\n49 + 1<\/gadget>\n50<\/output>\n551 * 50<\/gadget>\n27_550<\/output>\n27_550<\/result>","index":2339} +{"problem":"when x is multiplied by 3 , the result is 18 more than the result of subtracting x from 70 . what is the value of x ?","rationale":"\"the equation that can be formed is : 3 x - 18 = 70 - x or , 4 x = 88 or , x = 22 . e answer .\"","correct":"e","options":{"a":"- 4 ","b":"- 2 ","c":"11 ","d":"13","e":"22"},"options_float":{"a":-4.0,"b":-2.0,"c":11.0,"d":13.0,"e":22.0},"annotated_formula":"divide(add(70, 18), add(3, const_1))","linear_formula":"add(n1,n2)|add(const_1,n0)|divide(#0,#1)|","chain":"70 + 18<\/gadget>\n88<\/output>\n3 + 1<\/gadget>\n4<\/output>\n88 \/ 4<\/gadget>\n22<\/output>\n22<\/result>","index":2341} +{"problem":"10 men can complete a work in 7 days . but 10 women need 14 days to complete the same work . how many days will 5 men and 10 women need to complete the work ?","rationale":"work done by 10 men in 1 day = 1 \/ 7 work done by 1 man in 1 day = ( 1 \/ 7 ) \/ 10 = 1 \/ 70 work done by 10 women in 1 day = 1 \/ 14 work done by 1 woman in 1 day = 1 \/ 140 work done by 5 men and 10 women in 1 day = 5 × ( 1 \/ 70 ) + 10 × ( 1 \/ 140 ) = 5 \/ 70 + 10 \/ 140 = 1 \/ 7 = 5 men and 10 women can complete the work in 7 days answer : option c","correct":"c","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":"inverse(add(multiply(10, inverse(multiply(14, 10))), multiply(5, divide(inverse(7), 10))))","linear_formula":"inverse(n1)|multiply(n0,n3)|divide(#0,n0)|inverse(#1)|multiply(n0,#3)|multiply(n4,#2)|add(#4,#5)|inverse(#6)","chain":"14 * 10<\/gadget>\n140<\/output>\n1 \/ 140<\/gadget>\n1\/140 = around 0.007143<\/output>\n10 * (1\/140)<\/gadget>\n1\/14 = around 0.071429<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n(1\/7) \/ 10<\/gadget>\n1\/70 = around 0.014286<\/output>\n5 * (1\/70)<\/gadget>\n1\/14 = around 0.071429<\/output>\n(1\/14) + (1\/14)<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ (1\/7)<\/gadget>\n7<\/output>\n7<\/result>","index":2343} +{"problem":"vijay sells a cupboard at 10 % below cost price . had he got rs . 1500 more , he would have made a profit of 10 % . what is the cost price of the cupboard ?","rationale":"\"explanation : cost price = 1500 \/ ( 0.10 + 0.10 ) = 1500 \/ 0.20 = rs . 7500 answer b\"","correct":"b","options":{"a":"7450 ","b":"7500 ","c":"7400 ","d":"7500","e":"none of these"},"options_float":{"a":7450.0,"b":7500.0,"c":7400.0,"d":7500.0,"e":null},"annotated_formula":"divide(1500, divide(subtract(add(const_100, 10), subtract(const_100, 10)), const_100))","linear_formula":"add(n0,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#2,const_100)|divide(n1,#3)|","chain":"100 + 10<\/gadget>\n110<\/output>\n100 - 10<\/gadget>\n90<\/output>\n110 - 90<\/gadget>\n20<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1_500 \/ (1\/5)<\/gadget>\n7_500<\/output>\n7_500<\/result>","index":2344} +{"problem":"there has been successive increases of 20 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ?","rationale":"\"let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.2 * p = x * p y = x \/ ( 1.1 * 1.2 ) which is about 0.76 x which is a decrease of about 24 % . the answer is c .\"","correct":"c","options":{"a":"16 % ","b":"20 % ","c":"24 % ","d":"28 %","e":"32 %"},"options_float":{"a":16.0,"b":20.0,"c":24.0,"d":28.0,"e":32.0},"annotated_formula":"multiply(subtract(const_1, divide(const_100, add(add(const_100, 20), divide(multiply(add(const_100, 20), 10), const_100)))), const_100)","linear_formula":"add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100)|","chain":"100 + 20<\/gadget>\n120<\/output>\n120 * 10<\/gadget>\n1_200<\/output>\n1_200 \/ 100<\/gadget>\n12<\/output>\n120 + 12<\/gadget>\n132<\/output>\n100 \/ 132<\/gadget>\n25\/33 = around 0.757576<\/output>\n1 - (25\/33)<\/gadget>\n8\/33 = around 0.242424<\/output>\n(8\/33) * 100<\/gadget>\n800\/33 = around 24.242424<\/output>\n800\/33 = around 24.242424<\/result>","index":2345} +{"problem":"the length of a rectangle is increased by 35 % and its breadth is decreased by 20 % . what is the effect on its area ?","rationale":"\"100 * 100 = 10000 135 * 80 = 10800 answer : a\"","correct":"a","options":{"a":"10800 ","b":"1299 ","c":"1000 ","d":"10000","e":"2887"},"options_float":{"a":10800.0,"b":1299.0,"c":1000.0,"d":10000.0,"e":2887.0},"annotated_formula":"multiply(add(35, const_100), subtract(const_100, 20))","linear_formula":"add(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|","chain":"35 + 100<\/gadget>\n135<\/output>\n100 - 20<\/gadget>\n80<\/output>\n135 * 80<\/gadget>\n10_800<\/output>\n10_800<\/result>","index":2346} +{"problem":"express 25 mps in kmph ?","rationale":"\"25 * 18 \/ 5 = 90 kmph answer : c\"","correct":"c","options":{"a":"23 ","b":"88 ","c":"90 ","d":"27","e":"21"},"options_float":{"a":23.0,"b":88.0,"c":90.0,"d":27.0,"e":21.0},"annotated_formula":"multiply(divide(25, const_1000), const_3600)","linear_formula":"divide(n0,const_1000)|multiply(#0,const_3600)|","chain":"25 \/ 1_000<\/gadget>\n1\/40 = around 0.025<\/output>\n(1\/40) * 3_600<\/gadget>\n90<\/output>\n90<\/result>","index":2347} +{"problem":"a can finish a work in 24 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ?","rationale":"\"b ' s 10 day ' s work = ( 1 x 10 ) = 2 . 15 3 remaining work = ( 1 - 2 ) = 1 . 3 3 now , 1 work is done by a in 1 day . 24 therefore 1 work is done by a in ( 24 x 1 ) = 8 days . e\"","correct":"e","options":{"a":"6 ","b":"5 ","c":"5.5 ","d":"7","e":"8"},"options_float":{"a":6.0,"b":5.0,"c":5.5,"d":7.0,"e":8.0},"annotated_formula":"divide(multiply(multiply(divide(const_1, 15), 10), 24), const_2)","linear_formula":"divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)|","chain":"1 \/ 15<\/gadget>\n1\/15 = around 0.066667<\/output>\n(1\/15) * 10<\/gadget>\n2\/3 = around 0.666667<\/output>\n(2\/3) * 24<\/gadget>\n16<\/output>\n16 \/ 2<\/gadget>\n8<\/output>\n8<\/result>","index":2348} +{"problem":"a satellite is composed of 30 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 \/ 6 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ?","rationale":"\"let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x \/ 6 . the number of non - upgraded sensors on the whole satellite is 30 ( x \/ 6 ) = 5 x . the fraction of sensors which have been upgraded is x \/ ( x + 5 x ) = x \/ 6 x = 1 \/ 6 the answer is c .\"","correct":"c","options":{"a":"5 \/ 6 ","b":"1 \/ 5 ","c":"1 \/ 6 ","d":"1 \/ 7","e":"1 \/ 24"},"options_float":{"a":0.8333333333,"b":0.2,"c":0.1666666667,"d":0.1428571429,"e":0.0416666667},"annotated_formula":"divide(30, add(30, multiply(30, multiply(divide(1, 6), 30))))","linear_formula":"divide(n1,n2)|multiply(n0,#0)|multiply(n0,#1)|add(n0,#2)|divide(n0,#3)|","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 30<\/gadget>\n5<\/output>\n30 * 5<\/gadget>\n150<\/output>\n30 + 150<\/gadget>\n180<\/output>\n30 \/ 180<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":2349} +{"problem":"a certain class of students is being divided into teams . the class can either be divided into 16 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?","rationale":"\"let total no of students in the class be n so , we are told that n is divisible by both 16 24 so , lets find the least common multiple of 16 24 , ie 48 so our answer is ( c ) 48\"","correct":"c","options":{"a":"6 ","b":"36 ","c":"48 ","d":"60","e":"72"},"options_float":{"a":6.0,"b":36.0,"c":48.0,"d":60.0,"e":72.0},"annotated_formula":"lcm(16, 24)","linear_formula":"lcm(n0,n1)|","chain":"lcm(16, 24)<\/gadget>\n48<\/output>\n48<\/result>","index":2350} +{"problem":"if 5 % more is gained by selling an article for rs . 350 than by selling it for rs . 320 , the cost of the article is","rationale":"\"explanation : let c . p . be rs . x . then , 5 % of x = 350 - 320 = 30 x \/ 20 = 30 = > x = 600 answer : e\"","correct":"e","options":{"a":"289 ","b":"231 ","c":"200 ","d":"288","e":"600"},"options_float":{"a":289.0,"b":231.0,"c":200.0,"d":288.0,"e":600.0},"annotated_formula":"divide(subtract(350, 320), divide(5, const_100))","linear_formula":"divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)|","chain":"350 - 320<\/gadget>\n30<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n30 \/ (1\/20)<\/gadget>\n600<\/output>\n600<\/result>","index":2351} +{"problem":"a hall 36 m long and 15 m broad is to be paved with stones , each measuring 5 dm by 5 dm . the number of stones required is :","rationale":"\"area of the hall = 3600 * 1500 area of each stone = ( 50 * 50 ) therefore , number of stones = ( 3600 * 1500 \/ 50 * 50 ) = 2160 answer : d\"","correct":"d","options":{"a":"180 ","b":"1800 ","c":"18 ","d":"2160","e":"1.8"},"options_float":{"a":180.0,"b":1800.0,"c":18.0,"d":2160.0,"e":1.8},"annotated_formula":"divide(multiply(36, 15), divide(multiply(5, 5), const_100))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#0,#2)|","chain":"36 * 15<\/gadget>\n540<\/output>\n5 * 5<\/gadget>\n25<\/output>\n25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n540 \/ (1\/4)<\/gadget>\n2_160<\/output>\n2_160<\/result>","index":2352} +{"problem":"before leaving home for the town of madison , pete checks a map which shows that madison is 6 inches from his current location , gardensquare . pete arrives in madison 2.5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ?","rationale":"\"pete covered 2.5 * 60 = 150 miles which correspond to 6 inches on the map - - > scale in inches per mile is 6 \/ 150 = 1 \/ 25 . answer : a .\"","correct":"a","options":{"a":"1 \/ 25 ","b":"1 \/ 30 ","c":"1 \/ 10 ","d":"2","e":"30"},"options_float":{"a":0.04,"b":0.0333333333,"c":0.1,"d":2.0,"e":30.0},"annotated_formula":"divide(const_1, multiply(divide(2.5, 6), 60))","linear_formula":"divide(n1,n0)|multiply(#0,n2)|divide(const_1,#1)|","chain":"2.5 \/ 6<\/gadget>\n0.416667<\/output>\n0.416667 * 60<\/gadget>\n25.00002<\/output>\n1 \/ 25.00002<\/gadget>\n0.04<\/output>\n0.04<\/result>","index":2354} +{"problem":"a person lent a certain sum of money at 5 % per annum at simple interest and in 8 years the interest amounted to $ 480 less than the sum lent . what was the sum lent ?","rationale":"\"p - 480 = ( p * 5 * 8 ) \/ 100 p = 800 the answer is b .\"","correct":"b","options":{"a":"700 ","b":"800 ","c":"900 ","d":"1000","e":"1100"},"options_float":{"a":700.0,"b":800.0,"c":900.0,"d":1000.0,"e":1100.0},"annotated_formula":"divide(480, subtract(const_1, divide(multiply(5, 8), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|","chain":"5 * 8<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n480 \/ (3\/5)<\/gadget>\n800<\/output>\n800<\/result>","index":2355} +{"problem":"all numbers from 1 to 200 ( in decimal system ) are written in base 6 and base 7 systems . how many of the numbers will have a non - zero units digit in both base 6 and base 7 notations ?","rationale":"detailed solution if a number written in base 6 ends with a zero , it should be a multiple of 6 . in other words , the question wants us to find all numbers from 1 to 200 that are not multiples of 6 or 7 . there are 33 multiples of 6 less than 201 . there are 28 multiples of 7 less than 201 . there are 4 multiples of 6 & 7 ( or multiple of 42 ) from 1 to 200 . so , total multiples of 6 or 7 less than 201 = 33 + 28 - 4 = 57 . number of numbers with non - zero units digit = 200 - 57 = 143 . correct answer : a","correct":"a","options":{"a":"143 ","b":"200 ","c":"157 ","d":"122","e":"132"},"options_float":{"a":143.0,"b":200.0,"c":157.0,"d":122.0,"e":132.0},"annotated_formula":"subtract(200, subtract(add(divide(200, 6), divide(200, 7)), divide(200, multiply(6, 7))))","linear_formula":"divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)","chain":"200 \/ 6<\/gadget>\n100\/3 = around 33.333333<\/output>\n200 \/ 7<\/gadget>\n200\/7 = around 28.571429<\/output>\n(100\/3) + (200\/7)<\/gadget>\n1_300\/21 = around 61.904762<\/output>\n6 * 7<\/gadget>\n42<\/output>\n200 \/ 42<\/gadget>\n100\/21 = around 4.761905<\/output>\n(1_300\/21) - (100\/21)<\/gadget>\n400\/7 = around 57.142857<\/output>\n200 - (400\/7)<\/gadget>\n1_000\/7 = around 142.857143<\/output>\n1_000\/7 = around 142.857143<\/result>","index":2356} +{"problem":"an urn contains 6 red , 5 blue and 2 green marbles . if 2 marbles are picked at random , what is the probability that both are red ?","rationale":"option ( b ) is correct p ( both are red ) , 6 c 2 \/ 13 c 2 = 5 \/ 26 answer b","correct":"b","options":{"a":"6 \/ 13 ","b":"5 \/ 26 ","c":"6 \/ 26 ","d":"9 \/ 26","e":"10 \/ 27"},"options_float":{"a":0.4615384615,"b":0.1923076923,"c":0.2307692308,"d":0.3461538462,"e":0.3703703704},"annotated_formula":"divide(divide(multiply(6, 5), const_2), divide(multiply(add(add(6, 5), 2), subtract(add(add(6, 5), 2), const_1)), const_2))","linear_formula":"add(n0,n1)|multiply(n0,n1)|add(n2,#0)|divide(#1,const_2)|subtract(#2,const_1)|multiply(#2,#4)|divide(#5,const_2)|divide(#3,#6)","chain":"6 * 5<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n6 + 5<\/gadget>\n11<\/output>\n11 + 2<\/gadget>\n13<\/output>\n13 - 1<\/gadget>\n12<\/output>\n13 * 12<\/gadget>\n156<\/output>\n156 \/ 2<\/gadget>\n78<\/output>\n15 \/ 78<\/gadget>\n5\/26 = around 0.192308<\/output>\n5\/26 = around 0.192308<\/result>","index":2358} +{"problem":"if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 30 % of m is equal to what percent of 200 % of j ?","rationale":"\"imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 30 % of 350 , comes out to be 105 . . . . 200 % of 10 is 20 . . . . ( 105 * 100 ) \/ 20 = 525 . . . . ans : c\"","correct":"c","options":{"a":"0.35 ","b":"3.5 ","c":"525 ","d":"350","e":"3500"},"options_float":{"a":0.35,"b":3.5,"c":525.0,"d":350.0,"e":3500.0},"annotated_formula":"multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 30), 200), const_100)","linear_formula":"multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100)|","chain":"125 * 150<\/gadget>\n18_750<\/output>\n18_750 * 175<\/gadget>\n3_281_250<\/output>\n25 * 50<\/gadget>\n1_250<\/output>\n1_250 * 75<\/gadget>\n93_750<\/output>\n3_281_250 \/ 93_750<\/gadget>\n35<\/output>\n35 * 30<\/gadget>\n1_050<\/output>\n1_050 \/ 200<\/gadget>\n21\/4 = around 5.25<\/output>\n(21\/4) * 100<\/gadget>\n525<\/output>\n525<\/result>","index":2359} +{"problem":"solve below question 2 x + 1 = - 15","rationale":"\"2 x + 1 = - 15 x = - 8 a\"","correct":"a","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(15, 1)), 2)","linear_formula":"add(n1,n2)|negate(#0)|divide(#1,n0)|","chain":"15 + 1<\/gadget>\n16<\/output>\n-16<\/gadget>\n-16<\/output>\n(-16) \/ 2<\/gadget>\n-8<\/output>\n-8<\/result>","index":2360} +{"problem":"a salesman ' s income consists of a commission and a base salary of $ 350 per week . over the past 5 weeks , his weekly income totals have been $ 556 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 over the 7 - week period ?","rationale":"\"total weekly income over 5 weeks = $ 556 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2220 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2220 = $ 1280 from this , we subtract base salary for 2 weeks i . e $ 350 * 2 = $ 700 therefore , commission = $ 1280 - $ 700 = $ 580 for 2 weeks avg weekly commission = $ 290 answer b\"","correct":"b","options":{"a":"$ 150 ","b":"$ 290 ","c":"$ 365 ","d":"$ 715","e":"$ 730"},"options_float":{"a":150.0,"b":290.0,"c":365.0,"d":715.0,"e":730.0},"annotated_formula":"subtract(divide(subtract(multiply(500, 7), add(add(add(556, 413), add(436, 420)), 395)), const_2), 350)","linear_formula":"add(n2,n3)|add(n4,n5)|multiply(n7,n8)|add(#0,#1)|add(n6,#3)|subtract(#2,#4)|divide(#5,const_2)|subtract(#6,n0)|","chain":"500 * 7<\/gadget>\n3_500<\/output>\n556 + 413<\/gadget>\n969<\/output>\n436 + 420<\/gadget>\n856<\/output>\n969 + 856<\/gadget>\n1_825<\/output>\n1_825 + 395<\/gadget>\n2_220<\/output>\n3_500 - 2_220<\/gadget>\n1_280<\/output>\n1_280 \/ 2<\/gadget>\n640<\/output>\n640 - 350<\/gadget>\n290<\/output>\n290<\/result>","index":2361} +{"problem":"a man can row 4.8 km \/ hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ?","rationale":"\"speed of boat in still water ( b ) = 4.8 km \/ hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b \/ 3 = 1.6 km \/ hr . answer : e\"","correct":"e","options":{"a":"1.9 ","b":"1.7 ","c":"1.2 ","d":"1.5","e":"1.6"},"options_float":{"a":1.9,"b":1.7,"c":1.2,"d":1.5,"e":1.6},"annotated_formula":"divide(subtract(multiply(4.8, const_2), 4.8), const_3)","linear_formula":"multiply(n0,const_2)|subtract(#0,n0)|divide(#1,const_3)|","chain":"4.8 * 2<\/gadget>\n9.6<\/output>\n9.6 - 4.8<\/gadget>\n4.8<\/output>\n4.8 \/ 3<\/gadget>\n1.6<\/output>\n1.6<\/result>","index":2362} +{"problem":"125 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ?","rationale":"\"milk = 3 \/ 5 * 125 = 75 liters water = 50 liters 75 : ( 50 + p ) = 3 : 4 150 + 3 p = 400 = > p = 50 50 liters of water are to be added for the ratio become 3 : 4 . answer : d\"","correct":"d","options":{"a":"12 liters ","b":"32 liters ","c":"41 liters ","d":"50 liters","e":"34 liters"},"options_float":{"a":12.0,"b":32.0,"c":41.0,"d":50.0,"e":34.0},"annotated_formula":"multiply(divide(125, add(3, 2)), 2)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"3 + 2<\/gadget>\n5<\/output>\n125 \/ 5<\/gadget>\n25<\/output>\n25 * 2<\/gadget>\n50<\/output>\n50<\/result>","index":2363} +{"problem":"a pupil ' s marks were wrongly entered as 35 instead of 23 . 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 = ( 35 - 23 ) = > x \/ 2 = 12 = > x = 24 . answer : e","correct":"e","options":{"a":"30 ","b":"80 ","c":"20 ","d":"25","e":"24"},"options_float":{"a":30.0,"b":80.0,"c":20.0,"d":25.0,"e":24.0},"annotated_formula":"multiply(subtract(35, 23), const_2)","linear_formula":"subtract(n0,n1)|multiply(#0,const_2)","chain":"35 - 23<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24<\/result>","index":2364} +{"problem":"the radius of a wheel is 22.4 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 500 resolutions . = 400 * 2 * 22 \/ 7 * 22.4 = 56320 cm = 563.2 m answer : e\"","correct":"e","options":{"a":"708 m ","b":"704 m ","c":"774 m ","d":"714 m","e":"563.2 m"},"options_float":{"a":708.0,"b":704.0,"c":774.0,"d":714.0,"e":563.2},"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), 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) * 22.4<\/gadget>\n70.4<\/output>\n70.4 * 2<\/gadget>\n140.8<\/output>\n140.8 * 400<\/gadget>\n56_320<\/output>\n56_320 \/ 100<\/gadget>\n2_816\/5 = around 563.2<\/output>\n2_816\/5 = around 563.2<\/result>","index":2367} +{"problem":"q is as much younger than r as he is older than t . if the sum of the ages of r and t is 50 years , what is definitely the difference between r and q ' s age ?","rationale":"explanation : given r – q = q – t and r + t = 50 which gives q = 25 as the difference between r & q and q & t is same so answer is 25 years answer : c","correct":"c","options":{"a":"23 ","b":"28 ","c":"25 ","d":"19","e":"11"},"options_float":{"a":23.0,"b":28.0,"c":25.0,"d":19.0,"e":11.0},"annotated_formula":"divide(50, const_2)","linear_formula":"divide(n0,const_2)","chain":"50 \/ 2<\/gadget>\n25<\/output>\n25<\/result>","index":2368} +{"problem":"if x and y are integers , what is the least positive number of 24 x + 21 y ?","rationale":"\"24 x + 21 y = 3 ( 8 x + 7 y ) which will be a minimum positive number when 8 x + 7 y = 1 . 8 ( 1 ) + 7 ( - 1 ) = 1 then 3 ( 8 x + 7 y ) can have a minimum positive value of 3 . the answer is 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":"subtract(24, 21)","linear_formula":"subtract(n0,n1)|","chain":"24 - 21<\/gadget>\n3<\/output>\n3<\/result>","index":2369} +{"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 3 ?","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 + 3 = 35 runs total runs = new avg x new no . of innings = 35 * 21 = 735 runs made in the 11 th inning = 735 - 640 = 95 answer : a\"","correct":"a","options":{"a":"95 ","b":"106 ","c":"122 ","d":"116","e":"122"},"options_float":{"a":95.0,"b":106.0,"c":122.0,"d":116.0,"e":122.0},"annotated_formula":"subtract(multiply(add(20, const_1), add(3, 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>\n3 + 32<\/gadget>\n35<\/output>\n21 * 35<\/gadget>\n735<\/output>\n20 * 32<\/gadget>\n640<\/output>\n735 - 640<\/gadget>\n95<\/output>\n95<\/result>","index":2370} +{"problem":"a sum of money deposited at c . i . amounts to rs . 500 in 3 years and to rs . 650 in 4 years . find the rate percent ?","rationale":"\"500 - - - 150 100 - - - ? = > 30 % answer : b\"","correct":"b","options":{"a":"25 % ","b":"30 % ","c":"35 % ","d":"40 %","e":"45 %"},"options_float":{"a":25.0,"b":30.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"multiply(divide(subtract(650, 500), 500), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"650 - 500<\/gadget>\n150<\/output>\n150 \/ 500<\/gadget>\n3\/10 = around 0.3<\/output>\n(3\/10) * 100<\/gadget>\n30<\/output>\n30<\/result>","index":2371} +{"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 $ 36400 , 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 36400 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 364 so we have 1 , 3 , 9 , 27 . . . . . 36400 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 . . . 364 = 1 - 3 ^ n \/ 1 - 3 ; 1 - 3 ^ n \/ - 2 cross multiplying we get 364 x - 2 = 1 - 3 ^ n - 728 = 1 - 3 ^ n - 729 = - 3 ^ n 729 = 3 ^ n ( negatives cancel out ) 729 can also be re written as 3 ^ 6 therefore ; 3 ^ 6 = 3 ^ n thus n = 6 ( a )\"","correct":"a","options":{"a":"6 ","b":"8 ","c":"10 ","d":"11","e":"13"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":11.0,"e":13.0},"annotated_formula":"divide(log(add(divide(multiply(36400, 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":"36_400 * 2<\/gadget>\n72_800<\/output>\n72_800 \/ 100<\/gadget>\n728<\/output>\n728 + 1<\/gadget>\n729<\/output>\nlog(729)<\/gadget>\nlog(729) = around 6.591674<\/output>\nlog(3)<\/gadget>\nlog(3) = around 1.098612<\/output>\nlog(729) \/ log(3)<\/gadget>\nlog(729)\/log(3) = around 6<\/output>\nlog(729)\/log(3) = around 6<\/result>","index":2372} +{"problem":"a big container is 35 % full with water . if 16 liters of water is added , the container becomes 3 \/ 4 full . what is the capacity of the big container in liters ?","rationale":"\"16 liters is 40 % of the capacity c . 16 = 0.4 c c = 16 \/ 0.4 = 40 liters . the answer is b .\"","correct":"b","options":{"a":"32 ","b":"40 ","c":"48 ","d":"54","e":"60"},"options_float":{"a":32.0,"b":40.0,"c":48.0,"d":54.0,"e":60.0},"annotated_formula":"divide(16, subtract(divide(3, 4), divide(35, 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>\n35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n(3\/4) - (7\/20)<\/gadget>\n2\/5 = around 0.4<\/output>\n16 \/ (2\/5)<\/gadget>\n40<\/output>\n40<\/result>","index":2374} +{"problem":"in objective test a correct ans score 4 marks and on a wrong ans 2 marks are - - - . a student score 480 marks from 150 question . how many ans were correct ?","rationale":"let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = 150 = > 4 x - 2 y = 480 = > 6 x = 780 hence x = 130 answer : b","correct":"b","options":{"a":"120 ","b":"130 ","c":"110 ","d":"150","e":"180"},"options_float":{"a":120.0,"b":130.0,"c":110.0,"d":150.0,"e":180.0},"annotated_formula":"divide(add(480, multiply(150, 2)), add(4, 2))","linear_formula":"add(n0,n1)|multiply(n1,n3)|add(n2,#1)|divide(#2,#0)","chain":"150 * 2<\/gadget>\n300<\/output>\n480 + 300<\/gadget>\n780<\/output>\n4 + 2<\/gadget>\n6<\/output>\n780 \/ 6<\/gadget>\n130<\/output>\n130<\/result>","index":2375} +{"problem":"a number when divided by a certain divisor left remainder 251 , when twice the number was divided by the same divisor , the remainder was 112 . find the divisor ?","rationale":"\"easy solution : n = dq 1 + 251 2 n = 2 dq 1 + 502 - ( 1 ) 2 n = dq 2 + 112 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 390 d * some integer = 390 checking all options only ( c ) syncs with it . answer c\"","correct":"c","options":{"a":"370 ","b":"365 ","c":"390 ","d":"456","e":"460"},"options_float":{"a":370.0,"b":365.0,"c":390.0,"d":456.0,"e":460.0},"annotated_formula":"subtract(multiply(251, const_2), 112)","linear_formula":"multiply(n0,const_2)|subtract(#0,n1)|","chain":"251 * 2<\/gadget>\n502<\/output>\n502 - 112<\/gadget>\n390<\/output>\n390<\/result>","index":2376} +{"problem":"excluding stoppages , the average speed of a bus is 120 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 120 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 120 - 40 ) km i . e 80 km at 120 km \/ hr . stoppage time = 80 \/ 120 hrs = 40 min answer : e\"","correct":"e","options":{"a":"15 min ","b":"18 min ","c":"16 min ","d":"20 min","e":"40 min"},"options_float":{"a":15.0,"b":18.0,"c":16.0,"d":20.0,"e":40.0},"annotated_formula":"subtract(multiply(const_1, const_60), multiply(divide(40, 120), 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 \/ 120<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) * 60<\/gadget>\n20<\/output>\n60 - 20<\/gadget>\n40<\/output>\n40<\/result>","index":2377} +{"problem":"consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 9900 interest only for the period . the principal amount borrowed by her was","rationale":"\"explanation : principal = rs . ( 100 × 9900 \/ 12 × 3 ) = > rs . 27500 . answer : b\"","correct":"b","options":{"a":"rs . 2000 ","b":"rs . 27500 ","c":"rs . 15000 ","d":"rs . 20000","e":"none of these"},"options_float":{"a":2000.0,"b":27500.0,"c":15000.0,"d":20000.0,"e":null},"annotated_formula":"divide(9900, divide(multiply(3, 12), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|divide(n2,#1)|","chain":"3 * 12<\/gadget>\n36<\/output>\n36 \/ 100<\/gadget>\n9\/25 = around 0.36<\/output>\n9_900 \/ (9\/25)<\/gadget>\n27_500<\/output>\n27_500<\/result>","index":2379} +{"problem":"if the cost price is 96 % of the selling price , then what is the profit percent ?","rationale":"\"let s . p . = $ 100 c . p . = $ 96 profit = $ 4 profit % = 4 \/ 96 * 100 = 25 \/ 6 = 4.17 % answer is a\"","correct":"a","options":{"a":"4.17 % ","b":"5 % ","c":"6.12 % ","d":"3.25 %","e":"5.75 %"},"options_float":{"a":4.17,"b":5.0,"c":6.12,"d":3.25,"e":5.75},"annotated_formula":"multiply(divide(subtract(const_100, 96), 96), const_100)","linear_formula":"subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"100 - 96<\/gadget>\n4<\/output>\n4 \/ 96<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/24) * 100<\/gadget>\n25\/6 = around 4.166667<\/output>\n25\/6 = around 4.166667<\/result>","index":2380} +{"problem":"how many multiples of 5 are there between 70 and 358 ?","rationale":"\"5 * 14 = 70 5 * 71 = 355 total no of multiples = ( 71 - 14 ) + 1 = 57 + 1 = 58 answer is e .\"","correct":"e","options":{"a":"54 ","b":"55 ","c":"56 ","d":"57","e":"58"},"options_float":{"a":54.0,"b":55.0,"c":56.0,"d":57.0,"e":58.0},"annotated_formula":"add(divide(subtract(358, 70), 5), const_1)","linear_formula":"subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|","chain":"358 - 70<\/gadget>\n288<\/output>\n288 \/ 5<\/gadget>\n288\/5 = around 57.6<\/output>\n(288\/5) + 1<\/gadget>\n293\/5 = around 58.6<\/output>\n293\/5 = around 58.6<\/result>","index":2381} +{"problem":"a sum of money deposited at c . i . amounts to rs . 5000 in 5 years and to rs . 5750 in 6 years . find the rate percent ?","rationale":"5000 - - - 750 100 - - - ? = > 15 % answer : d","correct":"d","options":{"a":"1 % ","b":"5 % ","c":"10 % ","d":"15 %","e":"20 %"},"options_float":{"a":1.0,"b":5.0,"c":10.0,"d":15.0,"e":20.0},"annotated_formula":"multiply(divide(subtract(5750, 5000), 5000), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)","chain":"5_750 - 5_000<\/gadget>\n750<\/output>\n750 \/ 5_000<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 100<\/gadget>\n15<\/output>\n15<\/result>","index":2383} +{"problem":"renu can do a piece of work in 6 days , but with the help of her friend suma , she can do it in 5 days . in what time suma can do it alone ?","rationale":"renu â € ™ s one day â € ™ s work = 1 \/ 6 suma â € ™ s one day â € ™ s work = 1 \/ 5 - 1 \/ 6 = 1 \/ 30 suma can do it alone in 30 days . answer : e","correct":"e","options":{"a":"10 ","b":"12 ","c":"14 ","d":"15","e":"30"},"options_float":{"a":10.0,"b":12.0,"c":14.0,"d":15.0,"e":30.0},"annotated_formula":"inverse(subtract(divide(const_1, 5), divide(const_1, 6)))","linear_formula":"divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#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>\n1\/30 = around 0.033333<\/output>\n1 \/ (1\/30)<\/gadget>\n30<\/output>\n30<\/result>","index":2384} +{"problem":"if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 160 units ?","rationale":"\"here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units \/ 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours \/ 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 160 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 160 units is ' 8 times ' 20 , we need ' 8 times ' more time . ( 2.5 hours ) ( 8 times ) = 20 hours final answer : [ reveal ] spoiler : c\"","correct":"c","options":{"a":"50 hours ","b":"40 hours ","c":"20 hours ","d":"12 hours","e":"8 hours"},"options_float":{"a":50.0,"b":40.0,"c":20.0,"d":12.0,"e":8.0},"annotated_formula":"divide(160, multiply(divide(divide(20, 10), 5), 20))","linear_formula":"divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)|","chain":"20 \/ 10<\/gadget>\n2<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 20<\/gadget>\n8<\/output>\n160 \/ 8<\/gadget>\n20<\/output>\n20<\/result>","index":2385} +{"problem":"a policeman noticed a criminal from a distance of 200 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 3 minutes ?","rationale":"\"explanation : solution : relative speed = ( 9 - 8 ) = 1 km \/ hr . distance covered in 3 minutes = ( 1 * 3 \/ 60 ) km = 1 \/ 20 km = 50 m . . ' . distance between the criminal and policeman = ( 200 - 50 ) m = 150 m . answer : d\"","correct":"d","options":{"a":"100 m ","b":"120 m ","c":"130 m ","d":"150 m","e":"none of these"},"options_float":{"a":100.0,"b":120.0,"c":130.0,"d":150.0,"e":null},"annotated_formula":"subtract(200, multiply(divide(3, const_60), const_1000))","linear_formula":"divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)|","chain":"3 \/ 60<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 1_000<\/gadget>\n50<\/output>\n200 - 50<\/gadget>\n150<\/output>\n150<\/result>","index":2386} +{"problem":"the average of 6 no . ' s is 3.95 . the average of 2 of them is 4 , 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 ) + ( 3.85 * 2 ) ] = 8 required average = ( 8 \/ 2 ) = 4 e\"","correct":"e","options":{"a":"4.2 ","b":"4.4 ","c":"4.6 ","d":"5.6","e":"4"},"options_float":{"a":4.2,"b":4.4,"c":4.6,"d":5.6,"e":4.0},"annotated_formula":"divide(subtract(multiply(6, 3.95), add(multiply(2, 4), 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<\/gadget>\n8<\/output>\n2 * 3.85<\/gadget>\n7.7<\/output>\n8 + 7.7<\/gadget>\n15.7<\/output>\n23.7 - 15.7<\/gadget>\n8<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":2387} +{"problem":"the population of a town increased from 50000 to 80000 in a decade . the average percent increase of population per year is :","rationale":"solution increase in 10 year = ( 80000 - 50000 ) = 30000 . increase % = ( 30000 \/ 50000 x 100 ) % = 60 % â ˆ ´ required average = ( 60 \/ 10 ) % = 6 % answer c","correct":"c","options":{"a":"4.37 % ","b":"5 % ","c":"6 % ","d":"8.75 %","e":"none of these"},"options_float":{"a":4.37,"b":5.0,"c":6.0,"d":8.75,"e":null},"annotated_formula":"divide(multiply(divide(subtract(80000, 50000), 50000), const_100), const_10)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|divide(#2,const_10)","chain":"80_000 - 50_000<\/gadget>\n30_000<\/output>\n30_000 \/ 50_000<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 100<\/gadget>\n60<\/output>\n60 \/ 10<\/gadget>\n6<\/output>\n6<\/result>","index":2388} +{"problem":"carina has 130 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ?","rationale":"\"lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 130 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 130 - > y = 120 \/ 15 . = 8 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(130, multiply(5, 2)), add(10, 5))","linear_formula":"add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0)|","chain":"5 * 2<\/gadget>\n10<\/output>\n130 - 10<\/gadget>\n120<\/output>\n10 + 5<\/gadget>\n15<\/output>\n120 \/ 15<\/gadget>\n8<\/output>\n8<\/result>","index":2392} +{"problem":"if x = 1 \/ q and y = ( 2 \/ q ) - 6 , then for what value of q , x is equal to y ?","rationale":"explanation : x = y < = > 1 \/ q = ( 2 \/ q ) - 6 < = > 1 \/ q = 6 < = > q = 1 \/ 6 . answer : b","correct":"b","options":{"a":"1 \/ 3 ","b":"1 \/ 6 ","c":"6 ","d":"3","e":"2"},"options_float":{"a":0.3333333333,"b":0.1666666667,"c":6.0,"d":3.0,"e":2.0},"annotated_formula":"divide(subtract(2, 1), 6)","linear_formula":"subtract(n1,n0)|divide(#0,n2)","chain":"2 - 1<\/gadget>\n1<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":2393} +{"problem":"a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ?","rationale":"\"explanation : 90 % 104 % - - - - - - - - 14 % - - - - 140 100 % - - - - ? = > rs . 1000 a )\"","correct":"a","options":{"a":"rs . 1000 ","b":"rs . 1100 ","c":"rs . 1200 ","d":"rs . 1250","e":"rs . 1500"},"options_float":{"a":1000.0,"b":1100.0,"c":1200.0,"d":1250.0,"e":1500.0},"annotated_formula":"divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 10)))","linear_formula":"add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)|","chain":"140 * 100<\/gadget>\n14_000<\/output>\n100 + 4<\/gadget>\n104<\/output>\n100 - 10<\/gadget>\n90<\/output>\n104 - 90<\/gadget>\n14<\/output>\n14_000 \/ 14<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":2394} +{"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 50 days .","rationale":"they together completed 4 \/ 10 work in 4 days . balance 6 \/ 10 work will be completed by arun alone in 50 * 6 \/ 10 = 30 days . answer : d","correct":"d","options":{"a":"16 days . ","b":"17 days . ","c":"18 days . ","d":"30 days .","e":"20 days ."},"options_float":{"a":16.0,"b":17.0,"c":18.0,"d":30.0,"e":20.0},"annotated_formula":"subtract(50, multiply(divide(50, 10), 4))","linear_formula":"divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)","chain":"50 \/ 10<\/gadget>\n5<\/output>\n5 * 4<\/gadget>\n20<\/output>\n50 - 20<\/gadget>\n30<\/output>\n30<\/result>","index":2395} +{"problem":"a certain telescope increases the visual range at a particular location from 70 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ?","rationale":"\"original visual range = 70 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 70 ) \/ 70 * 100 % = 8 \/ 7 * 100 % = 114.28 % answer e\"","correct":"e","options":{"a":"30 % ","b":"33 1 \/ 2 % ","c":"40 % ","d":"60 %","e":"114.28 %"},"options_float":{"a":30.0,"b":33.0,"c":40.0,"d":60.0,"e":114.28},"annotated_formula":"multiply(divide(subtract(150, 70), 70), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"150 - 70<\/gadget>\n80<\/output>\n80 \/ 70<\/gadget>\n8\/7 = around 1.142857<\/output>\n(8\/7) * 100<\/gadget>\n800\/7 = around 114.285714<\/output>\n800\/7 = around 114.285714<\/result>","index":2396} +{"problem":"find the simple interest on rs . 68,000 at 16 2 \/ 3 % per annum for 9 months .","rationale":"\"p = rs . 68000 , r = 50 \/ 3 % p . a and t = 9 \/ 12 years = 3 \/ 4 years . s . i . = ( p * r * t ) \/ 100 = rs . ( 68,000 * ( 50 \/ 3 ) * ( 3 \/ 4 ) * ( 1 \/ 100 ) ) = rs . 8500 answer is a .\"","correct":"a","options":{"a":"rs . 8500 ","b":"rs . 8000 ","c":"rs . 7500 ","d":"rs . 7000","e":"rs . 6500"},"options_float":{"a":8500.0,"b":8000.0,"c":7500.0,"d":7000.0,"e":6500.0},"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":2398} +{"problem":"three cubes of iron whose edges are 6 cm , 8 cm and 10 cm respectively are melted and formed into a single cube . the edge of the new cube formed is","rationale":"sol . volume of the new cube = ( 63 + 83 + 103 ) cm 3 = 1728 cm 3 . let the edge of the new cube be a cm . ∴ a 3 = 1728 ⇒ a = 12 . answer b","correct":"b","options":{"a":"10 cm ","b":"12 cm ","c":"16 cm ","d":"18 cm","e":"none"},"options_float":{"a":10.0,"b":12.0,"c":16.0,"d":18.0,"e":null},"annotated_formula":"cube_edge_by_volume(add(volume_cube(10), add(volume_cube(6), volume_cube(8))))","linear_formula":"volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|cube_edge_by_volume(#4)","chain":"10 ** 3<\/gadget>\n1_000<\/output>\n6 ** 3<\/gadget>\n216<\/output>\n8 ** 3<\/gadget>\n512<\/output>\n216 + 512<\/gadget>\n728<\/output>\n1_000 + 728<\/gadget>\n1_728<\/output>\n1_728 ** (1\/3)<\/gadget>\n12<\/output>\n12<\/result>","index":2399} +{"problem":"after 6 games , team b had an average of 75 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 * 75 ) + 47 + x > 500 450 + 47 + x > 500 497 + x > 500 = > x > 3 option d\"","correct":"d","options":{"a":"5 ","b":"4 ","c":"7 ","d":"3","e":"8"},"options_float":{"a":5.0,"b":4.0,"c":7.0,"d":3.0,"e":8.0},"annotated_formula":"subtract(500, add(multiply(6, 75), 47))","linear_formula":"multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)|","chain":"6 * 75<\/gadget>\n450<\/output>\n450 + 47<\/gadget>\n497<\/output>\n500 - 497<\/gadget>\n3<\/output>\n3<\/result>","index":2404} +{"problem":"if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) , what is the smallest possible positive value of n ?","rationale":"( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) has one appearance of 3 ( in the 6 ) and no appearances of 5 . thus n must include at least 3 ^ 2 * 5 ^ 2 = 9 * 25 = 225 the answer is e .","correct":"e","options":{"a":"75 ","b":"125 ","c":"145 ","d":"175","e":"225"},"options_float":{"a":75.0,"b":125.0,"c":145.0,"d":175.0,"e":225.0},"annotated_formula":"add(add(add(add(add(multiply(multiply(5, 7), 2), multiply(multiply(5, 7), 2)), multiply(multiply(5, 7), 2)), 7), const_4), const_4)","linear_formula":"multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)","chain":"5 * 7<\/gadget>\n35<\/output>\n35 * 2<\/gadget>\n70<\/output>\n70 + 70<\/gadget>\n140<\/output>\n140 + 70<\/gadget>\n210<\/output>\n210 + 7<\/gadget>\n217<\/output>\n217 + 4<\/gadget>\n221<\/output>\n221 + 4<\/gadget>\n225<\/output>\n225<\/result>","index":2405} +{"problem":"there are 20 poles with a constant distance between each pole . a car takes 26 second to reach the 12 th pole . how much will it take to reach the last pole .","rationale":"\"assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 26 seconds 1 pole ( 26 \/ 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 26 \/ 11 ) seconds = 44.9091 seconds answer : b\"","correct":"b","options":{"a":"44.4543 ","b":"44.9091 ","c":"44.95128 ","d":"44.91288","e":"44.91222"},"options_float":{"a":44.4543,"b":44.9091,"c":44.95128,"d":44.91288,"e":44.91222},"annotated_formula":"multiply(divide(26, 12), 20)","linear_formula":"divide(n1,n2)|multiply(n0,#0)|","chain":"26 \/ 12<\/gadget>\n13\/6 = around 2.166667<\/output>\n(13\/6) * 20<\/gadget>\n130\/3 = around 43.333333<\/output>\n130\/3 = around 43.333333<\/result>","index":2407} +{"problem":"p has $ 21 more than what q and r together would have had if both b and c had 1 \/ 5 of what p has . how much does p have ?","rationale":"\"p = ( 2 \/ 5 ) * p + 21 ( 3 \/ 5 ) * p = 21 p = 35 the answer is a .\"","correct":"a","options":{"a":"$ 35 ","b":"$ 36 ","c":"$ 37 ","d":"$ 38","e":"$ 39"},"options_float":{"a":35.0,"b":36.0,"c":37.0,"d":38.0,"e":39.0},"annotated_formula":"divide(21, subtract(1, multiply(divide(1, 5), const_2)))","linear_formula":"divide(n1,n2)|multiply(#0,const_2)|subtract(n1,#1)|divide(n0,#2)|","chain":"1 \/ 5<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 2<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n21 \/ (3\/5)<\/gadget>\n35<\/output>\n35<\/result>","index":2408} +{"problem":"a large box contains 18 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ?","rationale":"\"the number of chocolate bars is equal to 18 ? 25 = 450 correct answer c\"","correct":"c","options":{"a":"350 ","b":"250 ","c":"450 ","d":"550","e":"650"},"options_float":{"a":350.0,"b":250.0,"c":450.0,"d":550.0,"e":650.0},"annotated_formula":"multiply(18, 25)","linear_formula":"multiply(n0,n1)|","chain":"18 * 25<\/gadget>\n450<\/output>\n450<\/result>","index":2409} +{"problem":"because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 5 times as much as mork did , what was their combined tax rate ?","rationale":"\"say morks income is - 100 so tax paid will be 40 say mindys income is 5 * 100 = 500 so tax paid is 30 % * 500 = 150 total tax paid = 40 + 150 = 190 . combined tax % will be 190 \/ 100 + 500 = 31.67 %\"","correct":"d","options":{"a":"32.5 % ","b":"34 % ","c":"35 % ","d":"31.67 %","e":"37.5 %"},"options_float":{"a":32.5,"b":34.0,"c":35.0,"d":31.67,"e":37.5},"annotated_formula":"multiply(const_100, divide(add(divide(40, const_100), multiply(5, divide(30, const_100))), add(const_1, 5)))","linear_formula":"add(n2,const_1)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|multiply(#5,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n5 * (3\/10)<\/gadget>\n3\/2 = around 1.5<\/output>\n(2\/5) + (3\/2)<\/gadget>\n19\/10 = around 1.9<\/output>\n1 + 5<\/gadget>\n6<\/output>\n(19\/10) \/ 6<\/gadget>\n19\/60 = around 0.316667<\/output>\n100 * (19\/60)<\/gadget>\n95\/3 = around 31.666667<\/output>\n95\/3 = around 31.666667<\/result>","index":2410} +{"problem":"find the number which is nearest to 3105 and is exactly divisible by 21 .","rationale":"\"sol . on dividing 3105 by 21 , we get 18 as remainder . number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . option b\"","correct":"b","options":{"a":"1208 ","b":"3108 ","c":"241 ","d":"217","e":"3147"},"options_float":{"a":1208.0,"b":3108.0,"c":241.0,"d":217.0,"e":3147.0},"annotated_formula":"add(3105, subtract(21, reminder(3105, 21)))","linear_formula":"reminder(n0,n1)|subtract(n1,#0)|add(n0,#1)|","chain":"3_105 % 21<\/gadget>\n18<\/output>\n21 - 18<\/gadget>\n3<\/output>\n3_105 + 3<\/gadget>\n3_108<\/output>\n3_108<\/result>","index":2413} +{"problem":"how many terminating zeroes r 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":2415} +{"problem":"a girl walking at the rate of 9 km per hour crosses a square field diagonally in 12 seconds . the area of the field is :","rationale":"distance covered in ( 9 × 1000 ) \/ ( 3600 ) × 12 = 30 m diagonal of squarre field = 30 m . area of square field = 30 ( power ) 2 \/ 2 = 900 \/ 2 = 450 sq . m answer is c .","correct":"c","options":{"a":"430 sq . m ","b":"425 sq . m ","c":"450 sq . m ","d":"475 sq . m","e":"350 sq . m"},"options_float":{"a":430.0,"b":425.0,"c":450.0,"d":475.0,"e":350.0},"annotated_formula":"divide(multiply(multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10))), multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10)))), const_2)","linear_formula":"multiply(n0,const_1000)|multiply(const_10,const_360)|divide(#0,#1)|multiply(n1,#2)|multiply(#3,#3)|divide(#4,const_2)","chain":"9 * 1_000<\/gadget>\n9_000<\/output>\n360 * 10<\/gadget>\n3_600<\/output>\n9_000 \/ 3_600<\/gadget>\n5\/2 = around 2.5<\/output>\n12 * (5\/2)<\/gadget>\n30<\/output>\n30 * 30<\/gadget>\n900<\/output>\n900 \/ 2<\/gadget>\n450<\/output>\n450<\/result>","index":2416} +{"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":"\"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 * 40 \/ 2 = 50 cm 2 answer : option c\"","correct":"c","options":{"a":"72 ","b":"828 ","c":"50 ","d":"34","e":"35"},"options_float":{"a":72.0,"b":828.0,"c":50.0,"d":34.0,"e":35.0},"annotated_formula":"triangle_area(2.5, 40)","linear_formula":"triangle_area(n0,n1)|","chain":"(2.5 * 40) \/ 2<\/gadget>\n50<\/output>\n50<\/result>","index":2419} +{"problem":"the probability that event b occurs is 0.6 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.4 , what is the probability that event a will occur ?","rationale":"p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.4 = 0.6 + p ( a ) - 0.25 p ( a ) = 0.05 ans : a","correct":"a","options":{"a":"0.05 ","b":"0.15 ","c":"0.45 ","d":"0.5","e":"0.55"},"options_float":{"a":0.05,"b":0.15,"c":0.45,"d":0.5,"e":0.55},"annotated_formula":"subtract(add(0.25, 0.4), 0.6)","linear_formula":"add(n1,n2)|subtract(#0,n0)","chain":"0.25 + 0.4<\/gadget>\n0.65<\/output>\n0.65 - 0.6<\/gadget>\n0.05<\/output>\n0.05<\/result>","index":2421} +{"problem":"if x and y are both odd prime numbers and x < y , how many distinct positive integer e factors does 2 xy have ?","rationale":"since 2 xy prime e 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":2422} +{"problem":"in a certain group of 10 developers , 4 developers code only in python and the rest program in either ruby on rails or php - but not both . if a developer organization is to choose a 3 - member team , which must have at least 1 developer who codes in python , how many different programming teams can be chosen ?","rationale":"two ways . . . 1 ) total ways = 10 c 3 = 10 ! \/ 7 ! 3 ! = 120 . . ways without python developer = 6 c 3 = 6 ! \/ 3 ! 3 ! = 20 . . ways of at least one python developer = 120 - 20 = 100 . . 2 ) ways of selecting only one = 4 * 6 c 2 = 4 * 15 = 60 . . ways of selecting only two = 4 c 2 * 6 c 1 = 6 * 6 = 36 . . ways of selecting all three = 4 c 3 = 4 = 4 . . total = 60 + 36 + 4 = 100 . . . answer : a","correct":"a","options":{"a":"100 ","b":"40 ","c":"66 ","d":"80","e":"75"},"options_float":{"a":100.0,"b":40.0,"c":66.0,"d":80.0,"e":75.0},"annotated_formula":"subtract(divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3))), divide(factorial(subtract(10, 4)), multiply(factorial(3), factorial(3))))","linear_formula":"factorial(n0)|factorial(n2)|subtract(n0,n2)|subtract(n0,n1)|factorial(#2)|factorial(#3)|multiply(#1,#1)|divide(#5,#6)|multiply(#4,#1)|divide(#0,#8)|subtract(#9,#7)","chain":"factorial(10)<\/gadget>\n3_628_800<\/output>\n10 - 3<\/gadget>\n7<\/output>\nfactorial(7)<\/gadget>\n5_040<\/output>\nfactorial(3)<\/gadget>\n6<\/output>\n5_040 * 6<\/gadget>\n30_240<\/output>\n3_628_800 \/ 30_240<\/gadget>\n120<\/output>\n10 - 4<\/gadget>\n6<\/output>\nfactorial(6)<\/gadget>\n720<\/output>\n6 * 6<\/gadget>\n36<\/output>\n720 \/ 36<\/gadget>\n20<\/output>\n120 - 20<\/gadget>\n100<\/output>\n100<\/result>","index":2423} +{"problem":"a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 920 gm . for a kg . his gain is … % .","rationale":"\"his percentage gain is 100 * 80 \/ 920 as he is gaining 80 units for his purchase of 920 units . so 8.69 % . answer : e\"","correct":"e","options":{"a":"5.26 % ","b":"5.36 % ","c":"4.26 % ","d":"6.26 %","e":"8.69 %"},"options_float":{"a":5.26,"b":5.36,"c":4.26,"d":6.26,"e":8.69},"annotated_formula":"multiply(subtract(inverse(divide(920, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100)","linear_formula":"add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)|","chain":"4 + 1<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n10 * 100<\/gadget>\n1_000<\/output>\n920 \/ 1_000<\/gadget>\n23\/25 = around 0.92<\/output>\n1 \/ (23\/25)<\/gadget>\n25\/23 = around 1.086957<\/output>\n(25\/23) - 1<\/gadget>\n2\/23 = around 0.086957<\/output>\n(2\/23) * 100<\/gadget>\n200\/23 = around 8.695652<\/output>\n200\/23 = around 8.695652<\/result>","index":2424} +{"problem":"5.40 can be expressed in terms of percentage as","rationale":"\"explanation : while calculation in terms of percentage we need to multiply by 100 , so 5.40 * 100 = 540 answer : option d\"","correct":"d","options":{"a":"5.04 % ","b":"50.4 % ","c":"209 % ","d":"540 %","e":"none of these"},"options_float":{"a":5.04,"b":50.4,"c":209.0,"d":540.0,"e":null},"annotated_formula":"multiply(5.40, const_100)","linear_formula":"multiply(n0,const_100)|","chain":"5.4 * 100<\/gadget>\n540<\/output>\n540<\/result>","index":2425} +{"problem":"a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer ’ s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer ’ s suggested retail price o f $ 25.00 ?","rationale":"\"for retail price = $ 25 first maximum discounted price = 25 - 30 % of 25 = 25 - 7.5 = 17.5 price after additional discount of 20 % = 17.5 - 20 % of 17.5 = 17.5 - 3.5 = 14 answer : option a\"","correct":"a","options":{"a":"$ 14.00 ","b":"$ 11.20 ","c":"$ 14.40 ","d":"$ 16.00","e":"$ 18.00"},"options_float":{"a":14.0,"b":11.2,"c":14.4,"d":16.0,"e":18.0},"annotated_formula":"multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 25.00))","linear_formula":"subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(n3,#3)|multiply(#2,#4)|","chain":"100 - 20<\/gadget>\n80<\/output>\n80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n(7\/10) * 25<\/gadget>\n35\/2 = around 17.5<\/output>\n(4\/5) * (35\/2)<\/gadget>\n14<\/output>\n14<\/result>","index":2426} +{"problem":"a number is mistakenly divided by 5 instead of being multiplied by 5 . find the percentage change in the result due t this mistake .","rationale":"lets take a number 20 20 \/ 5 = 4 20 * 5 = 100 diff = 100 - 4 = 96 % answer : a","correct":"a","options":{"a":"96 % ","b":"95 % ","c":"2400 % ","d":"200 %","e":"400 %"},"options_float":{"a":96.0,"b":95.0,"c":2400.0,"d":200.0,"e":400.0},"annotated_formula":"multiply(subtract(multiply(5, 5), const_1), divide(const_100, multiply(5, 5)))","linear_formula":"multiply(n0,n0)|divide(const_100,#0)|subtract(#0,const_1)|multiply(#1,#2)","chain":"5 * 5<\/gadget>\n25<\/output>\n25 - 1<\/gadget>\n24<\/output>\n100 \/ 25<\/gadget>\n4<\/output>\n24 * 4<\/gadget>\n96<\/output>\n96<\/result>","index":2427} +{"problem":"the first , second and third terms of the proportion are 56 , 16 , 49 . find the fourth term .","rationale":"explanation : let the fourth term be x . thus 56 , 16 , 49 , x are in proportion . product of extreme terms = 56 x product of mean terms = 16 x 49 since , the numbers make up a proportion therefore , 56 x = 16 49 or , x = ( 16 49 ) \/ 56 or , x = 14 therefore , the fourth term of the proportion is 14 . answer : b","correct":"b","options":{"a":"10 ","b":"14 ","c":"40 ","d":"50","e":"60"},"options_float":{"a":10.0,"b":14.0,"c":40.0,"d":50.0,"e":60.0},"annotated_formula":"divide(multiply(49, 16), 56)","linear_formula":"multiply(n1,n2)|divide(#0,n0)","chain":"49 * 16<\/gadget>\n784<\/output>\n784 \/ 56<\/gadget>\n14<\/output>\n14<\/result>","index":2428} +{"problem":"during a sale , the price of a pair of shoes is marked down 10 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ?","rationale":"\"assume the price = 100 price during sale = 90 price after sale = 100 percent increase = 10 \/ 90 * 100 = 11 % approx . correct option : c\"","correct":"c","options":{"a":"9 % ","b":"10 % ","c":"11 % ","d":"15 %","e":"90 %"},"options_float":{"a":9.0,"b":10.0,"c":11.0,"d":15.0,"e":90.0},"annotated_formula":"divide(multiply(10, const_100), subtract(const_100, 10))","linear_formula":"multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)|","chain":"10 * 100<\/gadget>\n1_000<\/output>\n100 - 10<\/gadget>\n90<\/output>\n1_000 \/ 90<\/gadget>\n100\/9 = around 11.111111<\/output>\n100\/9 = around 11.111111<\/result>","index":2429} +{"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 $ 324 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 = 324 . hourly wage = 324 \/ 36 = 9 answer : c ) 9\"","correct":"c","options":{"a":"2 ","b":"8 ","c":"9 ","d":"1","e":"2"},"options_float":{"a":2.0,"b":8.0,"c":9.0,"d":1.0,"e":2.0},"annotated_formula":"divide(324, 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>\n324 \/ 36<\/gadget>\n9<\/output>\n9<\/result>","index":2430} +{"problem":"determine the value of 3 * 27 \/ 31 + 81 \/ 93","rationale":"solution : both fractions should be reduced before performing arithmetic operations . we get 3 * 27 \/ 31 + 3.27 \/ 3.31 = 3 * 27 \/ 31 + 27 \/ 31 = 4 * 27 \/ 31 = 151 \/ 31 answer d","correct":"d","options":{"a":"0 ","b":"156 \/ 31 ","c":"123 \/ 31 ","d":"151 \/ 31","e":"none"},"options_float":{"a":0.0,"b":5.0322580645,"c":3.9677419355,"d":4.8709677419,"e":null},"annotated_formula":"divide(add(subtract(add(81, multiply(27, 3)), subtract(93, 81)), const_1), 31)","linear_formula":"multiply(n0,n1)|subtract(n4,n3)|add(n3,#0)|subtract(#2,#1)|add(#3,const_1)|divide(#4,n2)","chain":"27 * 3<\/gadget>\n81<\/output>\n81 + 81<\/gadget>\n162<\/output>\n93 - 81<\/gadget>\n12<\/output>\n162 - 12<\/gadget>\n150<\/output>\n150 + 1<\/gadget>\n151<\/output>\n151 \/ 31<\/gadget>\n151\/31 = around 4.870968<\/output>\n151\/31 = around 4.870968<\/result>","index":2431} +{"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 e in the nature reserve are not hawks , paddyfield - warblers , or kingfishers ?","rationale":"\"1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if non hawks are 700 ( this is 70 % of the total , so total = 1000 ) , then paddy warbs are 2 \/ 5 x 700 = 1400 \/ 5 = 280 . kingfishers , therefore , are 280 \/ 4 = 70 . lets add them up . 300 hawks + 280 peddy warbs + 70 kingsifhers = 650 . so all others are 1000 - 650 = 350 or 35 % of total birds . the main job here to to identify the smart number to start the question with . this can be time consuming , but once identified , this question can be solved fairly quickly . 2 . another method : if x is total - - > non hawks = 0.7 x - - > warbs = 0.4 ( 0.7 x ) - - > kfs = 0.25 ( 0.4 ( 0.7 x ) ) . our job is to find out e : ( 0.3 x + 0.28 x + 0.07 x ) \/ x . or 0.65 x \/ x = 0.65 . we need to find 1 - 0.65 = 0.35 or 35 % . 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":2432} +{"problem":"what is the smallest integer t greater than 1 that leaves a remainder of 1 when divided by any of the integers 6 , 8 , and 10 ?","rationale":"or u can just use the answer choices here . since the answers are already arranged in ascending order , the first number which gives remainder t as 1 for all three is the correct answer . in the given question , the first number which gives a remainder of 1 for 6,8 and 10 is 121 . c","correct":"c","options":{"a":"21 ","b":"41 ","c":"t = 121 ","d":"241","e":"481"},"options_float":{"a":21.0,"b":41.0,"c":121.0,"d":241.0,"e":481.0},"annotated_formula":"add(lcm(lcm(6, 8), 10), 1)","linear_formula":"lcm(n2,n3)|lcm(n4,#0)|add(n0,#1)","chain":"lcm(6, 8)<\/gadget>\n24<\/output>\nlcm(24, 10)<\/gadget>\n120<\/output>\n120 + 1<\/gadget>\n121<\/output>\n121<\/result>","index":2433} +{"problem":"a train 150 m long running at 72 kmph crosses a platform in 20 sec . what is the length of the platform ?","rationale":"\"e 250 e = 72 * 5 \/ 18 = 20 = 400 â € “ 150 = 250\"","correct":"e","options":{"a":"443 m ","b":"354 m ","c":"450 m ","d":"350 m","e":"250 m"},"options_float":{"a":443.0,"b":354.0,"c":450.0,"d":350.0,"e":250.0},"annotated_formula":"subtract(multiply(20, multiply(72, const_0_2778)), 150)","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>\n20 * 20<\/gadget>\n400<\/output>\n400 - 150<\/gadget>\n250<\/output>\n250<\/result>","index":2434} +{"problem":"the smallest number when increased by ` ` 1 ` ` is exactly divisible by 2 , 8 , 24 , 36 is :","rationale":"lcm = 72 72 - 1 = 71 answer : a","correct":"a","options":{"a":"71 ","b":"70 ","c":"72 ","d":"73","e":"36"},"options_float":{"a":71.0,"b":70.0,"c":72.0,"d":73.0,"e":36.0},"annotated_formula":"subtract(lcm(24, 36), 1)","linear_formula":"lcm(n3,n4)|subtract(#0,n0)","chain":"lcm(24, 36)<\/gadget>\n72<\/output>\n72 - 1<\/gadget>\n71<\/output>\n71<\/result>","index":2436} +{"problem":"a person bought 135 glass bowls at a rate of rs . 15 per bowl . he sold 115 of them at rs . 18 and the remaining broke . what is the percentage gain for a ?","rationale":"\"cp = 135 * 15 = 2025 and sp = 115 * 18 = 2070 gain % = 100 * ( 2070 - 2025 ) \/ 2025 = 20 \/ 9 answer : c\"","correct":"c","options":{"a":"40 ","b":"30 \/ 11 ","c":"20 \/ 9 ","d":"27 \/ 11","e":"29 \/ 8"},"options_float":{"a":40.0,"b":2.7272727273,"c":2.2222222222,"d":2.4545454545,"e":3.625},"annotated_formula":"multiply(divide(subtract(multiply(115, 18), multiply(135, 15)), multiply(135, 15)), const_100)","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|","chain":"115 * 18<\/gadget>\n2_070<\/output>\n135 * 15<\/gadget>\n2_025<\/output>\n2_070 - 2_025<\/gadget>\n45<\/output>\n45 \/ 2_025<\/gadget>\n1\/45 = around 0.022222<\/output>\n(1\/45) * 100<\/gadget>\n20\/9 = around 2.222222<\/output>\n20\/9 = around 2.222222<\/result>","index":2437} +{"problem":"the maximum number of students among them 1200 pens and 820 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ?","rationale":"\"number of pens = 1200 number of pencils = 820 required number of students = h . c . f . of 1200 and 820 = 20 answer is b\"","correct":"b","options":{"a":"40 ","b":"20 ","c":"60 ","d":"80","e":"65"},"options_float":{"a":40.0,"b":20.0,"c":60.0,"d":80.0,"e":65.0},"annotated_formula":"gcd(1200, 820)","linear_formula":"gcd(n0,n1)|","chain":"gcd(1_200, 820)<\/gadget>\n20<\/output>\n20<\/result>","index":2439} +{"problem":"in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 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":"this is a rather straight forward ratio problem . 1 . 80 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 \/ 50 2 \/ 50 = 80 \/ x thus , x = 2000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . d","correct":"d","options":{"a":"400 ","b":"625 ","c":"1,250 ","d":"2,000","e":"10,000"},"options_float":{"a":400.0,"b":625.0,"c":1250.0,"d":2000.0,"e":10000.0},"annotated_formula":"divide(80, divide(2, 50))","linear_formula":"divide(n2,n1)|divide(n0,#0)","chain":"2 \/ 50<\/gadget>\n1\/25 = around 0.04<\/output>\n80 \/ (1\/25)<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2440} +{"problem":"a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what w percent of the list price is the lowest possible sale price ?","rationale":"\"let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which w is 40 % of 80 hence , d is the answer\"","correct":"d","options":{"a":"20 ","b":"25 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(80, const_2)","linear_formula":"divide(n3,const_2)|","chain":"80 \/ 2<\/gadget>\n40<\/output>\n40<\/result>","index":2442} +{"problem":"a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 28 m * 2 m * 0.75 m ?","rationale":"\"28 * 2 * 0.75 = 20 \/ 100 * 10 \/ 100 * 7.5 \/ 100 * x 28 = 1 \/ 100 * x = > x = 28000 answer : a\"","correct":"a","options":{"a":"28000 ","b":"27908 ","c":"78902 ","d":"25000","e":"27991"},"options_float":{"a":28000.0,"b":27908.0,"c":78902.0,"d":25000.0,"e":27991.0},"annotated_formula":"divide(divide(divide(multiply(multiply(multiply(28, 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":"28 * 100<\/gadget>\n2_800<\/output>\n2 * 100<\/gadget>\n200<\/output>\n2_800 * 200<\/gadget>\n560_000<\/output>\n0.75 * 100<\/gadget>\n75<\/output>\n560_000 * 75<\/gadget>\n42_000_000<\/output>\n42_000_000 \/ 20<\/gadget>\n2_100_000<\/output>\n2_100_000 \/ 10<\/gadget>\n210_000<\/output>\n210_000 \/ 7.5<\/gadget>\n28_000<\/output>\n28_000<\/result>","index":2443} +{"problem":"an amount at compound interest sums to rs . 17640 \/ - in 2 years and to rs . 20286 \/ - in 3 years at the same rate of interest . find the rate percentage ?","rationale":"\"explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 20286 \/ - - 17640 \/ - = rs . 2646 \/ - rate of interest = ( 2646 \/ 17640 ) × ( 100 \/ 1 ) = > 15 % answer : option d\"","correct":"d","options":{"a":"5 % ","b":"7 % ","c":"9 % ","d":"15 %","e":"12 %"},"options_float":{"a":5.0,"b":7.0,"c":9.0,"d":15.0,"e":12.0},"annotated_formula":"multiply(divide(subtract(20286, 17640), 17640), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"20_286 - 17_640<\/gadget>\n2_646<\/output>\n2_646 \/ 17_640<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 100<\/gadget>\n15<\/output>\n15<\/result>","index":2444} +{"problem":"david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 600 and then he left the lab . he returned in time to take a reading at 4 p . m . , by which point the population had grown to 4,800 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ?","rationale":"let the rate be x , then population of the bacteria after each hour can be given as 600,600 x , 600 ( x ^ 2 ) , 600 ( x ^ 3 ) now population at 4 pm = 4800 thus we have 600 ( x ^ 3 ) = 4800 = 8 thus x = 2 therefore population at 3 pm = 600 ( 4 ) = 2400 answer : a","correct":"a","options":{"a":"2400 ","b":"3600 ","c":"3000 ","d":"2800","e":"2500"},"options_float":{"a":2400.0,"b":3600.0,"c":3000.0,"d":2800.0,"e":2500.0},"annotated_formula":"multiply(multiply(power(divide(multiply(multiply(2, 4), 600), 600), const_0_33), 600), power(divide(multiply(multiply(2, 4), 600), 600), const_0_33))","linear_formula":"multiply(n2,n4)|multiply(n1,#0)|divide(#1,n1)|power(#2,const_0_33)|multiply(n1,#3)|multiply(#4,#3)","chain":"2 * 4<\/gadget>\n8<\/output>\n8 * 600<\/gadget>\n4_800<\/output>\n4_800 \/ 600<\/gadget>\n8<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n8 ** (1\/3)<\/gadget>\n2<\/output>\n2 * 600<\/gadget>\n1_200<\/output>\n1_200 * 2<\/gadget>\n2_400<\/output>\n2_400<\/result>","index":2445} +{"problem":"5 years ago , the average age of a , b , c and d was 45 years . with e joining them now , the average of all the 5 is 50 years . the age of e is ?","rationale":"solution 5 years ago average age of a , b , c , d = 45 years = > 5 years ago total age of a , b , c , d = 45 x 4 = 180 years = > total present age of a , b , c , d = 180 + 5 x 4 = 200 years if e ' s present age is x years = 200 + x \/ 5 = 50 x = 50 years . answer a","correct":"a","options":{"a":"50 ","b":"47 ","c":"48 ","d":"49","e":"46"},"options_float":{"a":50.0,"b":47.0,"c":48.0,"d":49.0,"e":46.0},"annotated_formula":"subtract(multiply(50, 5), add(multiply(45, multiply(const_2, const_2)), multiply(5, const_4)))","linear_formula":"multiply(n0,n3)|multiply(const_2,const_2)|multiply(n0,const_4)|multiply(n1,#1)|add(#3,#2)|subtract(#0,#4)","chain":"50 * 5<\/gadget>\n250<\/output>\n2 * 2<\/gadget>\n4<\/output>\n45 * 4<\/gadget>\n180<\/output>\n5 * 4<\/gadget>\n20<\/output>\n180 + 20<\/gadget>\n200<\/output>\n250 - 200<\/gadget>\n50<\/output>\n50<\/result>","index":2446} +{"problem":"sandy bought 65 books for $ 1180 from one shop and 55 books for $ 860 from another shop . what is the average price that sandy paid per book ?","rationale":"\"average price per book = ( 1180 + 860 ) \/ ( 65 + 55 ) = 2040 \/ 120 = $ 17 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":"divide(add(1180, 860), add(65, 55))","linear_formula":"add(n1,n3)|add(n0,n2)|divide(#0,#1)|","chain":"1_180 + 860<\/gadget>\n2_040<\/output>\n65 + 55<\/gadget>\n120<\/output>\n2_040 \/ 120<\/gadget>\n17<\/output>\n17<\/result>","index":2447} +{"problem":"there are 15 slate rocks , 20 pumice rocks , and 10 granite rocks randomly distributed in a certain field . if 2 rocks are to be chosen at random and without replacement , what is the probability that both rocks will be slate rocks ?","rationale":"\"total no of rocks = 45 probability of choosing 1 st slate rock = 15 \/ 45 probability of choosing 2 nd slate rock = 14 \/ 44 ( without replacement ) so combined probability = 15 \/ 45 * 14 \/ 44 = 7 \/ 66 so , answer d .\"","correct":"d","options":{"a":"1 \/ 3 ","b":"7 \/ 22 ","c":"1 \/ 9 ","d":"7 \/ 66","e":"2 \/ 45"},"options_float":{"a":0.3333333333,"b":0.3181818182,"c":0.1111111111,"d":0.1060606061,"e":0.0444444444},"annotated_formula":"multiply(divide(15, add(add(15, 20), 10)), divide(subtract(15, const_1), subtract(add(add(15, 20), 10), const_1)))","linear_formula":"add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)|","chain":"15 + 20<\/gadget>\n35<\/output>\n35 + 10<\/gadget>\n45<\/output>\n15 \/ 45<\/gadget>\n1\/3 = around 0.333333<\/output>\n15 - 1<\/gadget>\n14<\/output>\n45 - 1<\/gadget>\n44<\/output>\n14 \/ 44<\/gadget>\n7\/22 = around 0.318182<\/output>\n(1\/3) * (7\/22)<\/gadget>\n7\/66 = around 0.106061<\/output>\n7\/66 = around 0.106061<\/result>","index":2448} +{"problem":"a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 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 \/ 4 - 1 \/ 9 = 5 \/ 36 the cistern will be filled in 36 \/ 5 hr = 7.2 hr answer is d\"","correct":"d","options":{"a":"6 hr ","b":"5.6 hr ","c":"9.5 hr ","d":"7.2 hr","e":"4 hr"},"options_float":{"a":6.0,"b":5.6,"c":9.5,"d":7.2,"e":4.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n(1\/4) - (1\/9)<\/gadget>\n5\/36 = around 0.138889<\/output>\n1 \/ (5\/36)<\/gadget>\n36\/5 = around 7.2<\/output>\n36\/5 = around 7.2<\/result>","index":2449} +{"problem":"approximately how many cubic feet of water are needed to fill a circular swimming pool that is 40 feet across and 7 feet deep ?","rationale":"\"answer should be e . v = \\ pir ^ 2 h = \\ pi * 20 ^ 2 * 7 = approximately 9000\"","correct":"e","options":{"a":"700 ","b":"1500 ","c":"3000 ","d":"5000","e":"9000"},"options_float":{"a":700.0,"b":1500.0,"c":3000.0,"d":5000.0,"e":9000.0},"annotated_formula":"volume_cylinder(divide(40, const_2), 7)","linear_formula":"divide(n0,const_2)|volume_cylinder(#0,n1)|","chain":"40 \/ 2<\/gadget>\n20<\/output>\npi * (20 ** 2) * 7<\/gadget>\n2800*pi = around 8_796.45943<\/output>\n2800*pi = around 8_796.45943<\/result>","index":2450} +{"problem":"( 3 x + 2 ) ( 2 x - 1 ) = ax ^ 2 + kx + n . what is the value of a - n + k ?","rationale":"\"expanding we have 6 x ^ 2 - 3 x + 4 x - 2 6 x ^ 2 + x - 2 taking coefficients , a = 6 , k = 1 , n = - 2 therefore a - n + k = 6 - ( - 2 ) + 1 = 8 + 1 = 9 the answer is c .\"","correct":"c","options":{"a":"5 ","b":"8 ","c":"9 ","d":"10","e":"11"},"options_float":{"a":5.0,"b":8.0,"c":9.0,"d":10.0,"e":11.0},"annotated_formula":"add(add(multiply(3, 2), multiply(1, 2)), subtract(multiply(2, 2), multiply(1, 3)))","linear_formula":"multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5)|","chain":"3 * 2<\/gadget>\n6<\/output>\n1 * 2<\/gadget>\n2<\/output>\n6 + 2<\/gadget>\n8<\/output>\n2 * 2<\/gadget>\n4<\/output>\n1 * 3<\/gadget>\n3<\/output>\n4 - 3<\/gadget>\n1<\/output>\n8 + 1<\/gadget>\n9<\/output>\n9<\/result>","index":2452} +{"problem":"what is the average ( arithmetic mean ) of 10 , 2030 , 4050 , 6070 , 8090 ?","rationale":"so addition of all term - 10 , 20 , 30 , . . . . . . . 90 so average = ( 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 ) \/ 9 = ( 450 ) \/ 9 = 50 hence , the correct answer is e .","correct":"e","options":{"a":"90 ","b":"95 ","c":"70 ","d":"85","e":"50"},"options_float":{"a":90.0,"b":95.0,"c":70.0,"d":85.0,"e":50.0},"annotated_formula":"subtract(divide(add(add(add(add(10, 2030), 4050), 6070), 8090), add(const_4, const_1)), multiply(multiply(const_100, const_10), const_4))","linear_formula":"add(n0,n1)|add(const_1,const_4)|multiply(const_10,const_100)|add(n2,#0)|multiply(#2,const_4)|add(n3,#3)|add(n4,#5)|divide(#6,#1)|subtract(#7,#4)","chain":"10 + 2_030<\/gadget>\n2_040<\/output>\n2_040 + 4_050<\/gadget>\n6_090<\/output>\n6_090 + 6_070<\/gadget>\n12_160<\/output>\n12_160 + 8_090<\/gadget>\n20_250<\/output>\n4 + 1<\/gadget>\n5<\/output>\n20_250 \/ 5<\/gadget>\n4_050<\/output>\n100 * 10<\/gadget>\n1_000<\/output>\n1_000 * 4<\/gadget>\n4_000<\/output>\n4_050 - 4_000<\/gadget>\n50<\/output>\n50<\/result>","index":2455} +{"problem":"the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 1323 sq m , then what is the breadth of the rectangular plot ?","rationale":"\"let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 1323 3 b 2 = 1323 b 2 = 441 = 21 ( b > 0 ) b = 21 m . answer : d\"","correct":"d","options":{"a":"11 ","b":"17 ","c":"18 ","d":"21","e":"1322"},"options_float":{"a":11.0,"b":17.0,"c":18.0,"d":21.0,"e":1322.0},"annotated_formula":"sqrt(divide(1323, const_3))","linear_formula":"divide(n0,const_3)|sqrt(#0)|","chain":"1_323 \/ 3<\/gadget>\n441<\/output>\n441 ** (1\/2)<\/gadget>\n21<\/output>\n21<\/result>","index":2457} +{"problem":"after an ice began to melt out from the freezer , in the first hour lost 3 \/ 4 , in the second hour lost 3 \/ 4 of its remaining . if after two hours , the volume is 0.3 cubic inches , what is the original volume of the cubic ice , in cubic inches ?","rationale":"\"let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 \/ 4 ) x - ( 3 \/ 4 * 1 \/ 4 * x ) = ( 1 \/ 16 ) x ( 1 \/ 16 ) x = 0.3 x = 4.8 alternate solution : try to backsolve . initial volume = 4.8 after one hour - - > ( 1 \/ 4 ) 4.8 = 1.2 after two hours - - > ( 1 \/ 4 ) 1.2 = 0.3 answer : c\"","correct":"c","options":{"a":"2.5 ","b":"3.0 ","c":"4.8 ","d":"6.5","e":"8.0"},"options_float":{"a":2.5,"b":3.0,"c":4.8,"d":6.5,"e":8.0},"annotated_formula":"divide(divide(0.3, const_0_25), const_0_25)","linear_formula":"divide(n4,const_0_25)|divide(#0,const_0_25)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n0.3 \/ (1\/4)<\/gadget>\n1.2<\/output>\n1.2 \/ (1\/4)<\/gadget>\n4.8<\/output>\n4.8<\/result>","index":2458} +{"problem":"the perimeter of a triangle is 22 cm and the inradius of the triangle is 3.5 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 = 3.5 * 22 \/ 2 = 38.5 cm 2 answer : e\"","correct":"e","options":{"a":"22 ","b":"35 ","c":"77 ","d":"54","e":"38"},"options_float":{"a":22.0,"b":35.0,"c":77.0,"d":54.0,"e":38.0},"annotated_formula":"triangle_area(3.5, 22)","linear_formula":"triangle_area(n0,n1)|","chain":"(3.5 * 22) \/ 2<\/gadget>\n38.5<\/output>\n38.5<\/result>","index":2459} +{"problem":"for a group of n people , k of whom are of the same sex , the ( n - k ) \/ n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 6 of whom are females , by how much does the index for the females exceed the index for the males in the group ?","rationale":"\"index for females = ( 20 - 6 ) \/ 20 = 7 \/ 10 = 0.7 index for males = ( 20 - 14 \/ 20 = 3 \/ 10 = 0.3 index for females exceeds males by 0.7 - 0.3 = 0.4 answer : a\"","correct":"a","options":{"a":"0.4 ","b":"0.0625 ","c":"0.2 ","d":"0.25","e":"0.6"},"options_float":{"a":0.4,"b":0.0625,"c":0.2,"d":0.25,"e":0.6},"annotated_formula":"subtract(divide(subtract(20, 6), 20), divide(6, 20))","linear_formula":"divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)|","chain":"20 - 6<\/gadget>\n14<\/output>\n14 \/ 20<\/gadget>\n7\/10 = around 0.7<\/output>\n6 \/ 20<\/gadget>\n3\/10 = around 0.3<\/output>\n(7\/10) - (3\/10)<\/gadget>\n2\/5 = around 0.4<\/output>\n2\/5 = around 0.4<\/result>","index":2460} +{"problem":"x does a work in 40 days . y does the same work in 60 days . in how many days they together will do the same work ?","rationale":"\"x ' s 1 day ' s work = 1 \/ 40 y ' s 1 day ' s work = 1 \/ 60 ( x + y ) ' s 1 day ' s work = ( 1 \/ 40 + 1 \/ 60 ) = 1 \/ 24 both together will finish the work in 24 days . correct option is c\"","correct":"c","options":{"a":"10 ","b":"12 ","c":"24 ","d":"30","e":"15"},"options_float":{"a":10.0,"b":12.0,"c":24.0,"d":30.0,"e":15.0},"annotated_formula":"inverse(add(divide(const_1, 40), divide(const_1, 60)))","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|","chain":"1 \/ 40<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ 60<\/gadget>\n1\/60 = around 0.016667<\/output>\n(1\/40) + (1\/60)<\/gadget>\n1\/24 = around 0.041667<\/output>\n1 \/ (1\/24)<\/gadget>\n24<\/output>\n24<\/result>","index":2461} +{"problem":"the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the sum and the number if the ratio between the digits of the number is 1 : 2 ?","rationale":"\"let the number be xy . given xy – yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten ’ s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) – ( 10 y + x ) = 36 = > x – y = 4 = > 2 y – y = 4 . hence , ( x + y ) – ( x – y ) = 3 y – y = 2 y = 8 b\"","correct":"b","options":{"a":"6 ","b":"8 ","c":"10 ","d":"14","e":"15"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":14.0,"e":15.0},"annotated_formula":"multiply(divide(36, subtract(multiply(subtract(const_10, 1), multiply(2, 1)), subtract(const_10, 1))), 2)","linear_formula":"multiply(n0,n2)|subtract(const_10,n2)|multiply(#0,#1)|subtract(#2,#1)|divide(n1,#3)|multiply(#4,n0)|","chain":"10 - 1<\/gadget>\n9<\/output>\n2 * 1<\/gadget>\n2<\/output>\n9 * 2<\/gadget>\n18<\/output>\n18 - 9<\/gadget>\n9<\/output>\n36 \/ 9<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n8<\/result>","index":2462} +{"problem":"7 carpet - weavers can weave 7 carpets in 7 days . at the same rate , how many carpets would be woven by 14 carpet - weavers in 14 days ?","rationale":"explanation : solution : let the required number of carpets be x . more weavers , more carpets ( direct proportion ) more days , more carpets ( direct proportion ) weavers 7 : 14 } : : 7 : x days 7 : 14 . ' . 7 * 7 * x = 14 * 14 * 7 < = > x = 14 * 14 * 7 \/ 7 * 7 = 28 . answer : b","correct":"b","options":{"a":"14 ","b":"28 ","c":"21 ","d":"35","e":"none of these"},"options_float":{"a":14.0,"b":28.0,"c":21.0,"d":35.0,"e":null},"annotated_formula":"add(14, add(7, 7))","linear_formula":"add(n0,n0)|add(n3,#0)","chain":"7 + 7<\/gadget>\n14<\/output>\n14 + 14<\/gadget>\n28<\/output>\n28<\/result>","index":2463} +{"problem":"what is the speed of the stream if a canoe rows upstream at 6 km \/ hr and downstream at 12 km \/ hr","rationale":"\"sol . speed of stream = 1 \/ 2 ( 12 - 6 ) kmph = 3 kmph . answer c\"","correct":"c","options":{"a":"1 kmph ","b":"4 kmph ","c":"3 kmph ","d":"2 kmph","e":"1.9 kmph"},"options_float":{"a":1.0,"b":4.0,"c":3.0,"d":2.0,"e":1.9},"annotated_formula":"divide(subtract(12, 6), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,const_2)|","chain":"12 - 6<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n3<\/result>","index":2464} +{"problem":"a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 40 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ?","rationale":"i year best is to give a number to his income , say 100 . . and let saving be x . . so expenditure = 100 - x next year - income = 140 savings = 2 x expenditure = 140 - 2 x . . now 140 - 2 x + 100 - x = 2 ( 100 - x ) . . . 240 - 3 x = 200 - 2 x . . . . . . . . . . . . . . . . x = 40 . . . saving % = 40 \/ 100 * 100 = 40 % answer : b","correct":"b","options":{"a":"45 % ","b":"40 % ","c":"25 % ","d":"28 %","e":"33.33 %"},"options_float":{"a":45.0,"b":40.0,"c":25.0,"d":28.0,"e":33.33},"annotated_formula":"multiply(divide(subtract(add(add(100, 40), 100), multiply(2, 100)), const_100), const_100)","linear_formula":"add(n0,n1)|multiply(n1,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_100)|multiply(#4,const_100)","chain":"100 + 40<\/gadget>\n140<\/output>\n140 + 100<\/gadget>\n240<\/output>\n2 * 100<\/gadget>\n200<\/output>\n240 - 200<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":2465} +{"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 , 587.75 . 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 ) = 1587.75 = > x = 787 e\"","correct":"e","options":{"a":"0 ","b":"233 ","c":"500 ","d":"695","e":"787"},"options_float":{"a":0.0,"b":233.0,"c":500.0,"d":695.0,"e":787.0},"annotated_formula":"divide(add(const_1000, 587.75), const_2)","linear_formula":"add(n4,const_1000)|divide(#0,const_2)|","chain":"1_000 + 587.75<\/gadget>\n1_587.75<\/output>\n1_587.75 \/ 2<\/gadget>\n793.875<\/output>\n793.875<\/result>","index":2468} +{"problem":"find large number 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 10 as remainder","rationale":"\"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 10 5 x = 1355 x = 271 large number = 271 + 1365 = 1636 a\"","correct":"a","options":{"a":"1636 ","b":"1346 ","c":"1378 ","d":"1635","e":"1489"},"options_float":{"a":1636.0,"b":1346.0,"c":1378.0,"d":1635.0,"e":1489.0},"annotated_formula":"multiply(divide(subtract(1365, 10), subtract(6, const_1)), 6)","linear_formula":"subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|","chain":"1_365 - 10<\/gadget>\n1_355<\/output>\n6 - 1<\/gadget>\n5<\/output>\n1_355 \/ 5<\/gadget>\n271<\/output>\n271 * 6<\/gadget>\n1_626<\/output>\n1_626<\/result>","index":2469} +{"problem":"two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fifth widget , starting with the fifth , and steven inspects every fourth , starting with the fourth , how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ?","rationale":"widgets inspected by lauren : ( ( 95 - 5 ) \/ 5 ) + 1 = 18 + 1 = 19 widgets inspected by steven : ( ( 96 - 4 ) \/ 4 ) + 1 = 23 + 1 = 24 widgets inspected by both : ( ( 96 \/ 12 ) + 1 = 9 total : 19 + 24 - 9 = 34 hence , widgets not inspected : 98 - 34 = 64 option d","correct":"d","options":{"a":"66 ","b":"68 ","c":"70 ","d":"64","e":"72"},"options_float":{"a":66.0,"b":68.0,"c":70.0,"d":64.0,"e":72.0},"annotated_formula":"subtract(98, subtract(add(floor(divide(98, add(const_4, const_1))), floor(divide(98, const_4))), floor(divide(98, add(const_10, add(const_4, const_1))))))","linear_formula":"add(const_1,const_4)|divide(n0,const_4)|add(#0,const_10)|divide(n0,#0)|floor(#1)|divide(n0,#2)|floor(#3)|add(#6,#4)|floor(#5)|subtract(#7,#8)|subtract(n0,#9)","chain":"4 + 1<\/gadget>\n5<\/output>\n98 \/ 5<\/gadget>\n98\/5 = around 19.6<\/output>\nfloor(98\/5)<\/gadget>\n19<\/output>\n98 \/ 4<\/gadget>\n49\/2 = around 24.5<\/output>\nfloor(49\/2)<\/gadget>\n24<\/output>\n19 + 24<\/gadget>\n43<\/output>\n10 + 5<\/gadget>\n15<\/output>\n98 \/ 15<\/gadget>\n98\/15 = around 6.533333<\/output>\nfloor(98\/15)<\/gadget>\n6<\/output>\n43 - 6<\/gadget>\n37<\/output>\n98 - 37<\/gadget>\n61<\/output>\n61<\/result>","index":2470} +{"problem":"the length of the bridge , which a train 130 metres long and travelling at 36 km \/ hr can cross in 45 seconds , is :","rationale":"\"speed = [ 36 x 5 \/ 18 ] m \/ sec = 10 m \/ sec time = 45 sec let the length of bridge be x metres . then , ( 130 + x ) \/ 45 = 10 = > 130 + x = 450 = > x = 320 m . answer : a\"","correct":"a","options":{"a":"320 m ","b":"225 m ","c":"245 m ","d":"250 m","e":"240 m"},"options_float":{"a":320.0,"b":225.0,"c":245.0,"d":250.0,"e":240.0},"annotated_formula":"subtract(multiply(divide(multiply(36, speed(const_1000, const_1)), speed(const_3600, const_1)), 45), 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>\n36 * 1_000<\/gadget>\n36_000<\/output>\n3_600 \/ 1<\/gadget>\n3_600<\/output>\n36_000 \/ 3_600<\/gadget>\n10<\/output>\n10 * 45<\/gadget>\n450<\/output>\n450 - 130<\/gadget>\n320<\/output>\n320<\/result>","index":2473} +{"problem":"the speed of a car increases by 2 kms after every one hour . if the distance travelled in the first one hour was 35 kms , what was the total distance travelled in 12 hours ?","rationale":"\"total distance travelled in 12 hours = ( 35 + 37 + 39 + . . . upto 12 terms ) . this is an a . p . with first term , a = 35 , number of terms , n = 12 , common difference d = 2 required distance = 12 \/ 2 ( 2 * 35 + ( 12 - 1 ) * 2 ) = 6 ( 70 + 22 ) = 552 km . correct option : c\"","correct":"c","options":{"a":"456 kms ","b":"482 kms ","c":"552 kms ","d":"556 kms","e":"none of these"},"options_float":{"a":456.0,"b":482.0,"c":552.0,"d":556.0,"e":null},"annotated_formula":"multiply(add(multiply(2, 35), 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 * 35<\/gadget>\n70<\/output>\n12 - 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n70 + 22<\/gadget>\n92<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n92 * 6<\/gadget>\n552<\/output>\n552<\/result>","index":2474} +{"problem":"a money lender lent rs . 1000 at 4 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years .","rationale":"( 1000 xtx 4 \/ 100 ) + ( 1400 xtx 5 \/ 100 ) = 350 â † ’ t = 3.2 answer a","correct":"a","options":{"a":"3.2 ","b":"3.75 ","c":"4 ","d":"4.25","e":"4.5"},"options_float":{"a":3.2,"b":3.75,"c":4.0,"d":4.25,"e":4.5},"annotated_formula":"divide(350, add(divide(multiply(4, 1000), const_100), divide(multiply(1400, 5), const_100)))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|divide(n4,#4)|","chain":"4 * 1_000<\/gadget>\n4_000<\/output>\n4_000 \/ 100<\/gadget>\n40<\/output>\n1_400 * 5<\/gadget>\n7_000<\/output>\n7_000 \/ 100<\/gadget>\n70<\/output>\n40 + 70<\/gadget>\n110<\/output>\n350 \/ 110<\/gadget>\n35\/11 = around 3.181818<\/output>\n35\/11 = around 3.181818<\/result>","index":2478} +{"problem":"for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 90 θ 33 ) θ 17 ) - ( 97 θ ( 33 θ 17 ) ) ?","rationale":"( ( 90 θ 33 ) θ 17 ) the remainder of 90 divided by 33 is 24 ; the remainder of 24 divided by 17 is 7 ; ( 97 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 97 divided by 16 is 1 . 7 - 1 = 6 . answer : 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":"subtract(reminder(reminder(90, 33), 17), reminder(97, reminder(33, 17)))","linear_formula":"reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3)","chain":"90 % 33<\/gadget>\n24<\/output>\n24 % 17<\/gadget>\n7<\/output>\n33 % 17<\/gadget>\n16<\/output>\n97 % 16<\/gadget>\n1<\/output>\n7 - 1<\/gadget>\n6<\/output>\n6<\/result>","index":2483} +{"problem":"if 1,000 microns = 1 decimeter , and 1,000 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ?","rationale":"\"given that 1,000 microns = 1 decimeter = 1,000 , 000,000 angstroms so , 1 micron = 1,000 , 000,000 \/ 1,000 = 1 , 000,000 answer : c\"","correct":"c","options":{"a":"1,000 ","b":"100 ","c":"1 , 000,000 ","d":"10","e":"10,000"},"options_float":{"a":1000.0,"b":100.0,"c":1.0,"d":10.0,"e":10000.0},"annotated_formula":"multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100))","linear_formula":"multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)|","chain":"100 * 100<\/gadget>\n10_000<\/output>\n1 \/ 10_000<\/gadget>\n1\/10_000 = around 0.0001<\/output>\n(1\/10_000) * 10_000<\/gadget>\n1<\/output>\n1<\/result>","index":2485} +{"problem":"if the sales tax reduced from 3 1 \/ 2 % to 3 1 \/ 3 % , then what difference does it make to a person who purchases an article with market price of $ 8400 ?","rationale":"\"required difference = [ 3 1 \/ 2 % of $ 8400 ] – [ 3 1 \/ 3 % of $ 8400 ] = [ ( 7 \/ 20 ) - ( 10 \/ 3 ) ] % of $ 8400 = 1 \/ 6 % of $ 8400 = $ [ ( 1 \/ 6 ) * ( 1 \/ 100 ) * 8400 ] = $ 14 . answer a .\"","correct":"a","options":{"a":"14 ","b":"24 ","c":"34 ","d":"12","e":"13"},"options_float":{"a":14.0,"b":24.0,"c":34.0,"d":12.0,"e":13.0},"annotated_formula":"divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 8400), const_100)","linear_formula":"divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100)|","chain":"1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(1\/2) + 3<\/gadget>\n7\/2 = around 3.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) + 3<\/gadget>\n10\/3 = around 3.333333<\/output>\n(7\/2) - (10\/3)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/6) * 8_400<\/gadget>\n1_400<\/output>\n1_400 \/ 100<\/gadget>\n14<\/output>\n14<\/result>","index":2486} +{"problem":"machine p and machine q are each used to manufacture 770 sprockets . it takes machine p 10 hours longer to produce 770 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?","rationale":"\"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 770 \/ x = 770 \/ 1.1 x + 10 1.1 ( 770 ) = 770 + 11 x 11 x = 77 x = 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":"divide(subtract(770, divide(770, add(divide(10, const_100), const_1))), 10)","linear_formula":"divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) + 1<\/gadget>\n11\/10 = around 1.1<\/output>\n770 \/ (11\/10)<\/gadget>\n700<\/output>\n770 - 700<\/gadget>\n70<\/output>\n70 \/ 10<\/gadget>\n7<\/output>\n7<\/result>","index":2488} +{"problem":"the c . p of 20 books is equal to the s . p of 30 books . find his gain % or loss % ?","rationale":"explanation : 20 cp = 30 sp 30 - - - 10 cp loss 100 - - - ? = > 33.33 % loss answer : c","correct":"c","options":{"a":"63.33 % ","b":"34.33 % ","c":"33.33 % ","d":"31.33 %","e":"36.33 %"},"options_float":{"a":63.33,"b":34.33,"c":33.33,"d":31.33,"e":36.33},"annotated_formula":"multiply(subtract(const_1, divide(20, 30)), const_100)","linear_formula":"divide(n0,n1)|subtract(const_1,#0)|multiply(#1,const_100)","chain":"20 \/ 30<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/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":2489} +{"problem":"weights of two friends ram and shyam are in the ratio 3 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ?","rationale":"\"solution : given ratio of ram and shayam ' s weight = 3 : 5 hence , ( x - 15 ) \/ ( 15 - 10 ) = 3 \/ 5 or , x = 18 % . answer : option a\"","correct":"a","options":{"a":"18 % ","b":"10 % ","c":"21 % ","d":"16 %","e":"none"},"options_float":{"a":18.0,"b":10.0,"c":21.0,"d":16.0,"e":null},"annotated_formula":"add(15, multiply(subtract(15, 10), divide(3, 5)))","linear_formula":"divide(n0,n1)|subtract(n4,n2)|multiply(#0,#1)|add(n4,#2)|","chain":"15 - 10<\/gadget>\n5<\/output>\n3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n5 * (3\/5)<\/gadget>\n3<\/output>\n15 + 3<\/gadget>\n18<\/output>\n18<\/result>","index":2490} +{"problem":"in the manufacture of a certain product , 7 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 = 7 % 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 ) * ( 7 \/ 100 ) * 100 % = ( 28 \/ 10000 ) * 100 % = ( 28 \/ 100 ) % = . 28 % answer b\"","correct":"b","options":{"a":"0.125 % ","b":"0.28 % ","c":"0.8 % ","d":"1.25 %","e":"2.0 %"},"options_float":{"a":0.125,"b":0.28,"c":0.8,"d":1.25,"e":2.0},"annotated_formula":"multiply(7, divide(4, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"4 \/ 100<\/gadget>\n1\/25 = around 0.04<\/output>\n7 * (1\/25)<\/gadget>\n7\/25 = around 0.28<\/output>\n7\/25 = around 0.28<\/result>","index":2491} +{"problem":"a trader bought a car at 30 % 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 = 70 s = 70 * ( 180 \/ 100 ) = 126 100 - 126 = 26 % answer : b\"","correct":"b","options":{"a":"18 % ","b":"26 % ","c":"12 % ","d":"32 %","e":"15 %"},"options_float":{"a":18.0,"b":26.0,"c":12.0,"d":32.0,"e":15.0},"annotated_formula":"multiply(subtract(divide(divide(multiply(subtract(const_100, 30), 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 - 30<\/gadget>\n70<\/output>\n100 + 80<\/gadget>\n180<\/output>\n70 * 180<\/gadget>\n12_600<\/output>\n12_600 \/ 100<\/gadget>\n126<\/output>\n126 \/ 100<\/gadget>\n63\/50 = around 1.26<\/output>\n(63\/50) - 1<\/gadget>\n13\/50 = around 0.26<\/output>\n(13\/50) * 100<\/gadget>\n26<\/output>\n26<\/result>","index":2493} +{"problem":"a boy goes to his school from his house at a speed of 3 km \/ hr and return at a speed of 2 km \/ hr . if he takes 5 hours in going and coming , the distance between his house and school is ?","rationale":"average speed = 2 * 3 * 2 \/ 3 + 2 = 12 \/ 5 km \/ hr distance traveled = 12 \/ 5 * 5 = 12 km distance between house and school = 12 \/ 2 = 6 km answer is b","correct":"b","options":{"a":"5 km ","b":"6 km ","c":"10 km ","d":"12 km","e":"8 km"},"options_float":{"a":5.0,"b":6.0,"c":10.0,"d":12.0,"e":8.0},"annotated_formula":"multiply(divide(5, add(divide(3, 2), const_1)), 3)","linear_formula":"divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2)","chain":"3 \/ 2<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) + 1<\/gadget>\n5\/2 = around 2.5<\/output>\n5 \/ (5\/2)<\/gadget>\n2<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6<\/result>","index":2494} +{"problem":"how many seconds will a 600 meter 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":"let length of tunnel is x meter distance = 600 + x meter time = 1 minute = 60 seconds speed = 78 km \/ hr = 78 * 5 \/ 18 m \/ s = 65 \/ 3 m \/ s distance = speed * time 600 + x = ( 65 \/ 3 ) * 60 600 + x = 20 * 65 = 1300 x = 1300 - 600 = 700 meters answer : a","correct":"a","options":{"a":"700 ","b":"288 ","c":"500 ","d":"277","e":"121"},"options_float":{"a":700.0,"b":288.0,"c":500.0,"d":277.0,"e":121.0},"annotated_formula":"multiply(multiply(subtract(divide(600, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2)","linear_formula":"subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2)","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>\n600 \/ (50\/3)<\/gadget>\n36<\/output>\n36 - 1<\/gadget>\n35<\/output>\n35 * 10<\/gadget>\n350<\/output>\n350 * 2<\/gadget>\n700<\/output>\n700<\/result>","index":2495} +{"problem":"a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ?","rationale":"a ' s 1 hour work = 1 \/ 4 ; ( b + c ) ' s 1 hour work = 1 \/ 3 ; ( a + b ) ' s 1 hour work = 1 \/ 2 ( a + b + c ) ' s 1 hour work = ( 1 \/ 4 + 1 \/ 3 ) = 7 \/ 12 c ' s 1 hour work = ( 7 \/ 12 - 1 \/ 2 ) = 1 \/ 12 c alone will take 12 hours to do the work . answer : a","correct":"a","options":{"a":"12 hours ","b":"10 hours ","c":"6 hours ","d":"8 hours","e":"4 hours"},"options_float":{"a":12.0,"b":10.0,"c":6.0,"d":8.0,"e":4.0},"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":2496} +{"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 , 60 percent of books that were loaned out are returned and there are 65 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 . 60 % of books that were loaned out are returned - - > 100 % - 60 % = 40 % of books that were loaned out are not returned . now , there are 68 books , thus 75 - 65 = 10 books are not returned . { loaned out } * 0.4 = 10 - - > { loaned out } = 25 . answer : b .\"","correct":"b","options":{"a":"20 ","b":"25 ","c":"35 ","d":"40","e":"55"},"options_float":{"a":20.0,"b":25.0,"c":35.0,"d":40.0,"e":55.0},"annotated_formula":"divide(subtract(75, 65), subtract(const_1, divide(60, const_100)))","linear_formula":"divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|","chain":"75 - 65<\/gadget>\n10<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n1 - (3\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n10 \/ (2\/5)<\/gadget>\n25<\/output>\n25<\/result>","index":2497} +{"problem":"cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost q is what percent of the original cost ?","rationale":"\"original cost 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":"q = 200 ","b":"q = 600 ","c":"q = 800 ","d":"q = 1600","e":"q = 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":2498} +{"problem":"albert is 2 times mary ’ s age and 4 times as old as betty . mary is 12 years younger than albert . how old is betty ?","rationale":"\"a = 2 m = m + 12 m = 12 a = 24 a = 4 b , and so b = 6 the answer is a .\"","correct":"a","options":{"a":"6 ","b":"12 ","c":"10 ","d":"15","e":"18"},"options_float":{"a":6.0,"b":12.0,"c":10.0,"d":15.0,"e":18.0},"annotated_formula":"divide(multiply(2, 12), 4)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"2 * 12<\/gadget>\n24<\/output>\n24 \/ 4<\/gadget>\n6<\/output>\n6<\/result>","index":2499} +{"problem":"what is the next number : 2 , 10 , 82 , __","rationale":"\"3 ^ 0 + 1 = 2 3 ^ 2 + 1 = 10 3 ^ 4 + 1 = 82 3 ^ 6 + 1 = 730 the answer is b .\"","correct":"b","options":{"a":"630 ","b":"730 ","c":"830 ","d":"848","e":"900"},"options_float":{"a":630.0,"b":730.0,"c":830.0,"d":848.0,"e":900.0},"annotated_formula":"subtract(subtract(subtract(multiply(82, const_10), const_100), 2), 2)","linear_formula":"multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)|","chain":"82 * 10<\/gadget>\n820<\/output>\n820 - 100<\/gadget>\n720<\/output>\n720 - 2<\/gadget>\n718<\/output>\n718 - 2<\/gadget>\n716<\/output>\n716<\/result>","index":2500} +{"problem":"marts income is 50 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 = ( 150 \/ 100 ) t t = ( 60 \/ 100 ) j = > m = ( 90 \/ 100 ) j answer d .","correct":"d","options":{"a":"124 % ","b":"120 % ","c":"96 % ","d":"90 %","e":"64 %"},"options_float":{"a":124.0,"b":120.0,"c":96.0,"d":90.0,"e":64.0},"annotated_formula":"multiply(divide(add(const_100, 50), 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 + 50<\/gadget>\n150<\/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>\n150 \/ (500\/3)<\/gadget>\n9\/10 = around 0.9<\/output>\n(9\/10) * 100<\/gadget>\n90<\/output>\n90<\/result>","index":2502} +{"problem":"what least number should be subtracted from 13601 such that the remainder is divisible by 87 ?","rationale":"\"13601 ÷ 87 = 156 , remainder = 29 hence 29 is the least number which can be subtracted from 13601 such that the remainder is divisible by 87 answer is b\"","correct":"b","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(13601, 87)","linear_formula":"reminder(n0,n1)|","chain":"13_601 % 87<\/gadget>\n29<\/output>\n29<\/result>","index":2503} +{"problem":"if the complement of a certain angle is 7 times the measure of that certain angle , then what is the measure of that certain angle ?","rationale":"thecomplementof angle a is the angle which , when added to angle a , gives 90 degrees . the two acute angles of a right triangle are complements , for example . the original angle is x , so the complement is 7 x , and together , these add up to 90 degrees . x + 7 x = 90 8 x = 90 x = 11.25 ° answer = ( e )","correct":"e","options":{"a":"45 ° ","b":"30 ° ","c":"22.5 ° ","d":"18 °","e":"11.25 °"},"options_float":{"a":45.0,"b":30.0,"c":22.5,"d":18.0,"e":11.25},"annotated_formula":"divide(subtract(const_100, const_10), add(7, const_1))","linear_formula":"add(n0,const_1)|subtract(const_100,const_10)|divide(#1,#0)","chain":"100 - 10<\/gadget>\n90<\/output>\n7 + 1<\/gadget>\n8<\/output>\n90 \/ 8<\/gadget>\n45\/4 = around 11.25<\/output>\n45\/4 = around 11.25<\/result>","index":2504} +{"problem":"the area of a triangle is with base 2 m and height 5 m ?","rationale":"\"1 \/ 2 * 2 * 5 = 5 m 2 answer : e\"","correct":"e","options":{"a":"11 ","b":"10 ","c":"787 ","d":"122","e":"5"},"options_float":{"a":11.0,"b":10.0,"c":787.0,"d":122.0,"e":5.0},"annotated_formula":"triangle_area(2, 5)","linear_formula":"triangle_area(n0,n1)|","chain":"(2 * 5) \/ 2<\/gadget>\n5<\/output>\n5<\/result>","index":2506} +{"problem":"ramu bought an old car for rs . 38000 . he spent rs . 12000 on repairs and sold it for rs . 64900 . what is his profit percent ?","rationale":"\"total cp = rs . 38000 + rs . 12000 = rs . 50000 and sp = rs . 64900 profit ( % ) = ( 64900 - 50000 ) \/ 50000 * 100 = 29.8 % answer : e\"","correct":"e","options":{"a":"17 % ","b":"19 % ","c":"18 % ","d":"14 %","e":"29.8 %"},"options_float":{"a":17.0,"b":19.0,"c":18.0,"d":14.0,"e":29.8},"annotated_formula":"multiply(divide(subtract(64900, add(38000, 12000)), add(38000, 12000)), const_100)","linear_formula":"add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|","chain":"38_000 + 12_000<\/gadget>\n50_000<\/output>\n64_900 - 50_000<\/gadget>\n14_900<\/output>\n14_900 \/ 50_000<\/gadget>\n149\/500 = around 0.298<\/output>\n(149\/500) * 100<\/gadget>\n149\/5 = around 29.8<\/output>\n149\/5 = around 29.8<\/result>","index":2509} +{"problem":"alfred buys an old scooter for $ 4700 and spends $ 800 on its repairs . if he sells the scooter for $ 6400 , his gain percent is ?","rationale":"\"c . p . = 4700 + 800 = $ 5500 s . p . = $ 6400 gain = 6400 - 5500 = $ 900 gain % = 900 \/ 5500 * 100 = 16.36 % answer is b\"","correct":"b","options":{"a":"5.45 % ","b":"16.36 % ","c":"7 % ","d":"8.12 %","e":"10 %"},"options_float":{"a":5.45,"b":16.36,"c":7.0,"d":8.12,"e":10.0},"annotated_formula":"multiply(divide(subtract(6400, add(4700, 800)), add(4700, 800)), const_100)","linear_formula":"add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|","chain":"4_700 + 800<\/gadget>\n5_500<\/output>\n6_400 - 5_500<\/gadget>\n900<\/output>\n900 \/ 5_500<\/gadget>\n9\/55 = around 0.163636<\/output>\n(9\/55) * 100<\/gadget>\n180\/11 = around 16.363636<\/output>\n180\/11 = around 16.363636<\/result>","index":2512} +{"problem":"3251 + 587 + 369 - ? = 3007","rationale":"let 4207 - x = 3007 then x = 4207 - 3007 = 1200 answer is d","correct":"d","options":{"a":"1250 ","b":"1300 ","c":"1375 ","d":"1200","e":"none of them"},"options_float":{"a":1250.0,"b":1300.0,"c":1375.0,"d":1200.0,"e":null},"annotated_formula":"subtract(add(add(3251, 587), 369), 3007)","linear_formula":"add(n0,n1)|add(n2,#0)|subtract(#1,n3)","chain":"3_251 + 587<\/gadget>\n3_838<\/output>\n3_838 + 369<\/gadget>\n4_207<\/output>\n4_207 - 3_007<\/gadget>\n1_200<\/output>\n1_200<\/result>","index":2513} +{"problem":"the manufacturer ’ s suggested retail price ( msrp ) of a certain item is $ 60 . store a sells the item for 20 percent more than the msrp . the regular price of the item at store b is 30 percent more than the msrp , but the item is currently on sale for 10 percent less than the regular price . if sales tax is 5 percent of the purchase price at both stores , what is the result when the total cost of the item at store b is subtracted from the total cost of the item at store a ?","rationale":"msrp = 60 price at store a = 60 ∗ 120100 = 72 = 60 ∗ 120100 = 72 price at store b = 60 ∗ 130100 ∗ 90100 = 70.2 = 60 ∗ 130100 ∗ 90100 = 70.2 difference = 72.0 - 70.2 = 1.8 sales tax applicable = 5 % on both = 1.8 + 0.09 = 1.89 answer = d","correct":"d","options":{"a":"$ 0 ","b":"$ 0.63 ","c":"$ 1.80 ","d":"$ 1.89","e":"$ 2.10"},"options_float":{"a":0.0,"b":0.63,"c":1.8,"d":1.89,"e":2.1},"annotated_formula":"subtract(multiply(60, divide(add(const_100, 20), const_100)), multiply(divide(subtract(const_100, 10), const_100), multiply(divide(add(const_100, 30), const_100), 60)))","linear_formula":"add(n1,const_100)|add(n2,const_100)|subtract(const_100,n3)|divide(#0,const_100)|divide(#2,const_100)|divide(#1,const_100)|multiply(n0,#3)|multiply(n0,#5)|multiply(#4,#7)|subtract(#6,#8)","chain":"100 + 20<\/gadget>\n120<\/output>\n120 \/ 100<\/gadget>\n6\/5 = around 1.2<\/output>\n60 * (6\/5)<\/gadget>\n72<\/output>\n100 - 10<\/gadget>\n90<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n100 + 30<\/gadget>\n130<\/output>\n130 \/ 100<\/gadget>\n13\/10 = around 1.3<\/output>\n(13\/10) * 60<\/gadget>\n78<\/output>\n(9\/10) * 78<\/gadget>\n351\/5 = around 70.2<\/output>\n72 - (351\/5)<\/gadget>\n9\/5 = around 1.8<\/output>\n9\/5 = around 1.8<\/result>","index":2514} +{"problem":"if the selling price of 100 articles is equal to the cost price of 63 articles , then the loss or gain percent is :","rationale":"\"let c . p . of each article be re . 1 . then , c . p . of 100 articles = rs . 100 ; s . p . of 100 articles = rs . 63 . loss % = 37 \/ 100 * 100 = 37 % answer : e\"","correct":"e","options":{"a":"39 % ","b":"36 % ","c":"35 % ","d":"40 %","e":"37 %"},"options_float":{"a":39.0,"b":36.0,"c":35.0,"d":40.0,"e":37.0},"annotated_formula":"subtract(100, 63)","linear_formula":"subtract(n0,n1)|","chain":"100 - 63<\/gadget>\n37<\/output>\n37<\/result>","index":2516} +{"problem":"rs . 1775 is divided amongst a , b , c so that 5 times a ' s share , 3 times b ' s share and 7 times c ' s share are all equal . find c ' s share ?","rationale":"a + b + c = 590 5 a = 3 b = 7 c = x a : b : c = 1 \/ 5 : 1 \/ 3 : 1 \/ 7 = 21 : 35 : 15 15 \/ 71 * 1775 = rs . 375 answer : e","correct":"e","options":{"a":"177 ","b":"150 ","c":"817 ","d":"716","e":"375"},"options_float":{"a":177.0,"b":150.0,"c":817.0,"d":716.0,"e":375.0},"annotated_formula":"divide(1775, add(add(divide(7, 5), divide(7, 3)), const_1))","linear_formula":"divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)|","chain":"7 \/ 5<\/gadget>\n7\/5 = around 1.4<\/output>\n7 \/ 3<\/gadget>\n7\/3 = around 2.333333<\/output>\n(7\/5) + (7\/3)<\/gadget>\n56\/15 = around 3.733333<\/output>\n(56\/15) + 1<\/gadget>\n71\/15 = around 4.733333<\/output>\n1_775 \/ (71\/15)<\/gadget>\n375<\/output>\n375<\/result>","index":2517} +{"problem":"n ^ ( n \/ 2 ) = 2 is true when n = 2 in the same way what is the value of n if n ^ ( n \/ 2 ) = 4 ?","rationale":"\"n ^ ( n \/ 2 ) = 4 apply log n \/ 2 logn = log 4 nlogn = 2 log 4 = log 4 ^ 2 = log 16 logn = log 16 now apply antilog n = 16 \/ n now n = 4 . answer : c\"","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":"divide(power(4, 2), 4)","linear_formula":"power(n4,n0)|divide(#0,n4)|","chain":"4 ** 2<\/gadget>\n16<\/output>\n16 \/ 4<\/gadget>\n4<\/output>\n4<\/result>","index":2518} +{"problem":"if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 23 , find the value of ab .","rationale":"\"2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 23 - 9 = 14 ab = 7 . answer is c .\"","correct":"c","options":{"a":"5 ","b":"8 ","c":"7 ","d":"10","e":"3"},"options_float":{"a":5.0,"b":8.0,"c":7.0,"d":10.0,"e":3.0},"annotated_formula":"divide(subtract(23, power(3, 2)), 2)","linear_formula":"power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|","chain":"3 ** 2<\/gadget>\n9<\/output>\n23 - 9<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":2520} +{"problem":"a person buys an article at $ 380 . at what price should he sell the article so as to make a profit of 25 % ?","rationale":"\"c 475 cost price = $ 380 profit = 25 % of 380 = $ 95 selling price = cost price + profit = 380 + 95 = 475\"","correct":"c","options":{"a":"445 ","b":"449 ","c":"475 ","d":"740","e":"460"},"options_float":{"a":445.0,"b":449.0,"c":475.0,"d":740.0,"e":460.0},"annotated_formula":"add(380, multiply(380, divide(25, const_100)))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n380 * (1\/4)<\/gadget>\n95<\/output>\n380 + 95<\/gadget>\n475<\/output>\n475<\/result>","index":2521} +{"problem":"in certain code ' twice ' is written as ' 34 $ 5 δ ' and ' wears ' is written as ' 4 δ 29 % ' . how is ' seat ' written in that code ?","rationale":"answer : option b","correct":"b","options":{"a":"22 ","b":"23 ","c":"697 ","d":"66 p","e":"82"},"options_float":{"a":22.0,"b":23.0,"c":697.0,"d":66.0,"e":82.0},"annotated_formula":"subtract(subtract(29, 5), const_1)","linear_formula":"subtract(n3,n1)|subtract(#0,const_1)","chain":"29 - 5<\/gadget>\n24<\/output>\n24 - 1<\/gadget>\n23<\/output>\n23<\/result>","index":2522} +{"problem":"the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 12.5 percent , what was the annual interest rate last year ?","rationale":"\"12.5 = 1.1 * x x = 11.36 % answer d )\"","correct":"d","options":{"a":"1 % ","b":"1.1 % ","c":"9.1 % ","d":"11.36 %","e":"10.8 %"},"options_float":{"a":1.0,"b":1.1,"c":9.1,"d":11.36,"e":10.8},"annotated_formula":"divide(multiply(12.5, const_100), add(12.5, const_100))","linear_formula":"add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)|","chain":"12.5 * 100<\/gadget>\n1_250<\/output>\n12.5 + 100<\/gadget>\n112.5<\/output>\n1_250 \/ 112.5<\/gadget>\n11.111111<\/output>\n11.111111<\/result>","index":2523} +{"problem":"two pipes a and b can fill a cistern in 10 and 15 minutes respectively . both fill pipes are opened together , but at the end of 3 minutes , ‘ b ’ is turned off . how much time will the cistern take to fill ?","rationale":"in one min , ( a + b ) fill the cistern = 1 ⁄ 10 + 1 ⁄ 15 = 1 ⁄ 6 th in 3 min , ( a + b ) fill the cistern = 3 ⁄ 6 = 1 ⁄ 2 th remaining part = 1 - 1 ⁄ 2 = 1 ⁄ 2 ∵ 1 ⁄ 10 th part filled by a in one min . ∴ 1 ⁄ 2 nd part filled by a in 10 × 1 ⁄ 2 = 5 min . ∴ total time = 3 + 5 = 8 min . answer b","correct":"b","options":{"a":"6 min ","b":"8 min ","c":"10 min ","d":"12 min","e":"none of these"},"options_float":{"a":6.0,"b":8.0,"c":10.0,"d":12.0,"e":null},"annotated_formula":"add(multiply(10, subtract(const_1, multiply(add(inverse(10), inverse(15)), const_3))), 3)","linear_formula":"inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,const_3)|subtract(const_1,#3)|multiply(n0,#4)|add(n2,#5)","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>\n(1\/6) * 3<\/gadget>\n1\/2 = around 0.5<\/output>\n1 - (1\/2)<\/gadget>\n1\/2 = around 0.5<\/output>\n10 * (1\/2)<\/gadget>\n5<\/output>\n5 + 3<\/gadget>\n8<\/output>\n8<\/result>","index":2524} +{"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 8 km ?","rationale":"\"the swimmer can swim against the current at a speed of 4 - 2 = 2 km \/ h . the time it will take is 8 \/ 2 = 4 hours . 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":"divide(8, subtract(4, 2))","linear_formula":"subtract(n0,n1)|divide(n2,#0)|","chain":"4 - 2<\/gadget>\n2<\/output>\n8 \/ 2<\/gadget>\n4<\/output>\n4<\/result>","index":2525} +{"problem":"find a sum for first 8 prime numbers ?","rationale":"required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 ) = 77 note : 1 is not a prime number option d","correct":"d","options":{"a":"22 ","b":"28 ","c":"59 ","d":"77","e":"84"},"options_float":{"a":22.0,"b":28.0,"c":59.0,"d":77.0,"e":84.0},"annotated_formula":"add(add(8, add(add(add(const_2, const_3), const_2), const_4)), add(add(add(add(add(add(add(add(add(add(add(const_2, const_3), const_2), const_4), const_2), const_4), add(add(add(add(const_2, const_3), const_2), const_4), const_2)), add(add(add(const_2, const_3), const_2), const_4)), add(add(const_2, const_3), const_2)), add(const_2, const_3)), const_3), const_2))","linear_formula":"add(const_2,const_3)|add(#0,const_2)|add(#1,const_4)|add(n0,#2)|add(#2,const_2)|add(#4,const_4)|add(#5,#4)|add(#6,#2)|add(#7,#1)|add(#8,#0)|add(#9,const_3)|add(#10,const_2)|add(#3,#11)","chain":"2 + 3<\/gadget>\n5<\/output>\n5 + 2<\/gadget>\n7<\/output>\n7 + 4<\/gadget>\n11<\/output>\n8 + 11<\/gadget>\n19<\/output>\n11 + 2<\/gadget>\n13<\/output>\n13 + 4<\/gadget>\n17<\/output>\n17 + 13<\/gadget>\n30<\/output>\n30 + 11<\/gadget>\n41<\/output>\n41 + 7<\/gadget>\n48<\/output>\n48 + 5<\/gadget>\n53<\/output>\n53 + 3<\/gadget>\n56<\/output>\n56 + 2<\/gadget>\n58<\/output>\n19 + 58<\/gadget>\n77<\/output>\n77<\/result>","index":2527} +{"problem":"a jogger running at 9 km \/ hr along side a railway track is 290 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 = 290 + 120 = 410 m . time taken = 410 \/ 10 = 41 sec . answer : e\"","correct":"e","options":{"a":"67 sec ","b":"89 sec ","c":"36 sec ","d":"87 sec","e":"41 sec"},"options_float":{"a":67.0,"b":89.0,"c":36.0,"d":87.0,"e":41.0},"annotated_formula":"divide(add(290, 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":"290 + 120<\/gadget>\n410<\/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>\n410 \/ 10<\/gadget>\n41<\/output>\n41<\/result>","index":2528} +{"problem":"working simultaneously and independently at an identical constant rate , 20 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 3 x units of product p in 8 days ?","rationale":"\"the rate of 20 machines is rate = job \/ time = x \/ 4 units per day - - > the rate of 1 machine 1 \/ 20 * ( x \/ 4 ) = x \/ 80 units per day ; now , again as { time } * { combined rate } = { job done } then 8 * ( m * x \/ 80 ) = 3 x - - > m = 30 . answer : a .\"","correct":"a","options":{"a":"30 ","b":"32 ","c":"35 ","d":"38","e":"40"},"options_float":{"a":30.0,"b":32.0,"c":35.0,"d":38.0,"e":40.0},"annotated_formula":"multiply(multiply(20, 3), divide(4, 8))","linear_formula":"divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)|","chain":"20 * 3<\/gadget>\n60<\/output>\n4 \/ 8<\/gadget>\n1\/2 = around 0.5<\/output>\n60 * (1\/2)<\/gadget>\n30<\/output>\n30<\/result>","index":2530} +{"problem":"the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 13 and 31 inclusive .","rationale":"we ' re dealing with a sequence of consecutive odd integers : 13 to 31 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 13 + 31 = 44 2 ) now look at the ' next smallest ' and the ' next biggest ' : 15 + 29 = 44 now we have proof that there is no middle term . we have 5 bunches of 44 5 ( 44 ) = 220 a","correct":"a","options":{"a":"220 ","b":"364 ","c":"410 ","d":"424","e":"450"},"options_float":{"a":220.0,"b":364.0,"c":410.0,"d":424.0,"e":450.0},"annotated_formula":"subtract(power(divide(add(31, const_1), 2), const_2), power(divide(add(subtract(13, 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,const_2)|power(#4,n0)|subtract(#5,#6)","chain":"31 + 1<\/gadget>\n32<\/output>\n32 \/ 2<\/gadget>\n16<\/output>\n16 ** 2<\/gadget>\n256<\/output>\n13 - 2<\/gadget>\n11<\/output>\n11 + 1<\/gadget>\n12<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6 ** 2<\/gadget>\n36<\/output>\n256 - 36<\/gadget>\n220<\/output>\n220<\/result>","index":2531} +{"problem":"if o is the center of the circle in the figure above and the area of the unshaded sector is 5 , what is the area of the shaded region ?","rationale":"60 \/ 360 = 1 \/ 6 1 \/ 6 of total area = 5 5 \/ 6 of total area = 5 * 5 = 25 answer : d","correct":"d","options":{"a":"25 \/ √ π ","b":"30 \/ √ π ","c":"20 ","d":"25","e":"30"},"options_float":{"a":25.0,"b":30.0,"c":20.0,"d":25.0,"e":30.0},"annotated_formula":"power(5, const_2)","linear_formula":"power(n0,const_2)","chain":"5 ** 2<\/gadget>\n25<\/output>\n25<\/result>","index":2533} +{"problem":"if a mixture is 5 ⁄ 9 alcohol by volume and 4 ⁄ 9 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 = { 5 \/ 9 } \/ { 4 \/ 9 } = 5 \/ 4 c\"","correct":"c","options":{"a":"1 \/ 4 ","b":"3 \/ 4 ","c":"5 \/ 4 ","d":"6 \/ 4","e":"7 \/ 4"},"options_float":{"a":0.25,"b":0.75,"c":1.25,"d":1.5,"e":1.75},"annotated_formula":"divide(divide(5, 9), divide(4, 9))","linear_formula":"divide(n0,n1)|divide(n2,n1)|divide(#0,#1)|","chain":"5 \/ 9<\/gadget>\n5\/9 = around 0.555556<\/output>\n4 \/ 9<\/gadget>\n4\/9 = around 0.444444<\/output>\n(5\/9) \/ (4\/9)<\/gadget>\n5\/4 = around 1.25<\/output>\n5\/4 = around 1.25<\/result>","index":2535} +{"problem":"two spherical balls lie on the ground touching . if one of the balls has a radius of 8 cm , and the point of contact is 12 cm above the ground , what is the radius of the other ball ( in centimeters ) ?","rationale":"\"a straight line will join the two centers and the point of contact , thus making similar triangles . 4 \/ 8 = ( r - 12 ) \/ r 4 r = 8 r - 96 r = 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":"add(add(const_4.0, 12), 8)","linear_formula":"add(const_4.0,n1)|add(#0,n0)|","chain":"4 + 12<\/gadget>\n16<\/output>\n16 + 8<\/gadget>\n24<\/output>\n24<\/result>","index":2536} +{"problem":"s = { 12 , 35 , 69 } t = { 45 , 67 , 13 } what is the probability that x chosen from s and y chosen from t will result x * y = even","rationale":"p : the probability that x * y is even , then p = 1 - p ( x * y is odd ) p ( x * y odd ) = p ( x odd ) * p ( y odd ) = 4 \/ 6 * 4 \/ 6 = 16 \/ 36 = 4 \/ 9 and p = 1 - 4 \/ 9 = 5 \/ 9 option : a","correct":"a","options":{"a":"5 \/ 9 ","b":"3 \/ 2 ","c":"1 \/ 2 ","d":"5 \/ 6","e":"6 \/ 5"},"options_float":{"a":0.5555555556,"b":1.5,"c":0.5,"d":0.8333333333,"e":1.2},"annotated_formula":"divide(add(divide(12, const_4), const_2), multiply(divide(12, const_4), divide(12, const_4)))","linear_formula":"divide(n0,const_4)|add(#0,const_2)|multiply(#0,#0)|divide(#1,#2)","chain":"12 \/ 4<\/gadget>\n3<\/output>\n3 + 2<\/gadget>\n5<\/output>\n3 * 3<\/gadget>\n9<\/output>\n5 \/ 9<\/gadget>\n5\/9 = around 0.555556<\/output>\n5\/9 = around 0.555556<\/result>","index":2537} +{"problem":"a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 110 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ?","rationale":"\"solution provided by stanford 2012 is correct : 110 \/ 800 choosing one of the cellos which has a pair viola , 1 \/ 600 choosing the viola which is the pair of chosen cello - - > p = 110 \/ 800 * 1 \/ 600 = 311 \/ 48,000 . answer : a .\"","correct":"a","options":{"a":"11 \/ 48,000 ","b":"1 \/ 8,100 ","c":"3 \/ 1,600 ","d":"1 \/ 90","e":"2 \/ 45"},"options_float":{"a":0.0002291667,"b":0.0001234568,"c":0.001875,"d":0.0111111111,"e":0.0444444444},"annotated_formula":"multiply(divide(110, 800), divide(const_1, 600))","linear_formula":"divide(n2,n0)|divide(const_1,n1)|multiply(#0,#1)|","chain":"110 \/ 800<\/gadget>\n11\/80 = around 0.1375<\/output>\n1 \/ 600<\/gadget>\n1\/600 = around 0.001667<\/output>\n(11\/80) * (1\/600)<\/gadget>\n11\/48_000 = around 0.000229<\/output>\n11\/48_000 = around 0.000229<\/result>","index":2538} +{"problem":"a certain roller coaster has 3 cars , and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 3 times , what is the probability that the passenger will ride in each of the 3 cars ?","rationale":"\"probability = ( favorable cases ) \/ ( total number of cases ) total number of ways in which a person can ride car = 3 * 3 * 3 = 27 ( in first ride he has 3 options to sit , in second right again he has 3 seats available to sit and so on ) number of favorable cases , i . e . , when he rides on different cars ; he can choose seat car in 3 ways in his 1 st ride . he can choose seat car in 2 ways in his 2 nd ride . he can choose seat car in 1 ways in his 3 rd ride . so , 3 * 2 * 1 = 6 ways thus , probability of choosing different seats = 6 \/ 27 = 2 \/ 9 answer : c\"","correct":"c","options":{"a":"0 ","b":"1 \/ 9 ","c":"2 \/ 9 ","d":"1 \/ 3","e":"1"},"options_float":{"a":0.0,"b":0.1111111111,"c":0.2222222222,"d":0.3333333333,"e":1.0},"annotated_formula":"multiply(factorial(3), power(divide(1, 3), 3))","linear_formula":"divide(n1,n0)|factorial(n0)|power(#0,n0)|multiply(#1,#2)|","chain":"factorial(3)<\/gadget>\n6<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/3) ** 3<\/gadget>\n1\/27 = around 0.037037<\/output>\n6 * (1\/27)<\/gadget>\n2\/9 = around 0.222222<\/output>\n2\/9 = around 0.222222<\/result>","index":2539} +{"problem":"a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 20 gallons of water and 40 gallons of salt are added , the resulting mixture is 33 1 \/ 3 % salt by volume . what is the value of x ?","rationale":"\"nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 \/ 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 40 gallons salt is added and 20 gallons of water is added . so salt now becomes - > ( 0.2 x + 40 ) and water - - > ( 0.6 x + 20 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 40 ) = ( 0.6 x + 20 ) \/ 2 = > 0.4 x + 80 = 0.6 x + 20 = > 0.2 x = 60 solving , x = 300 answer : a\"","correct":"a","options":{"a":"300 ","b":"75 ","c":"100 ","d":"150","e":"175"},"options_float":{"a":300.0,"b":75.0,"c":100.0,"d":150.0,"e":175.0},"annotated_formula":"divide(subtract(multiply(40, const_2), 20), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100))))","linear_formula":"divide(n0,const_100)|divide(n4,const_4)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#2,n1)|subtract(n4,#0)|multiply(#1,#5)|subtract(#5,#6)|subtract(#7,#3)|divide(#4,#8)|","chain":"40 * 2<\/gadget>\n80<\/output>\n80 - 20<\/gadget>\n60<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(4\/5) * (1\/4)<\/gadget>\n1\/5 = around 0.2<\/output>\n(4\/5) - (1\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n2 * (1\/5)<\/gadget>\n2\/5 = around 0.4<\/output>\n(3\/5) - (2\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n60 \/ (1\/5)<\/gadget>\n300<\/output>\n300<\/result>","index":2542} +{"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 w 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 w = 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":2543} +{"problem":"the diagonals of a rhombus are 18 cm and 22 cm . find its area ?","rationale":"1 \/ 2 * 18 * 22 = 198 answer : c","correct":"c","options":{"a":"277 ","b":"266 ","c":"198 ","d":"288","e":"212"},"options_float":{"a":277.0,"b":266.0,"c":198.0,"d":288.0,"e":212.0},"annotated_formula":"rhombus_area(18, 22)","linear_formula":"rhombus_area(n0,n1)","chain":"(18 * 22) \/ 2<\/gadget>\n198<\/output>\n198<\/result>","index":2545} +{"problem":"if 20 liters of chemical x are added to 80 liters of a mixture that is 25 % chemical x and 75 % chemical y , then what percentage of the resulting mixture is chemical x ?","rationale":"the amount of chemical x in the solution is 20 + 0.25 ( 80 ) = 40 liters . 40 liters \/ 100 liters = 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(20, multiply(divide(25, const_100), 80))","linear_formula":"divide(n2,const_100)|multiply(n1,#0)|add(n0,#1)","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 80<\/gadget>\n20<\/output>\n20 + 20<\/gadget>\n40<\/output>\n40<\/result>","index":2546} +{"problem":"a number increased by 20 % gives 600 . the number is","rationale":"\"formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 600 ( 120 × 5 = 600 ) 100 % - - - - - - - > 500 ( 100 × 5 = 500 ) d )\"","correct":"d","options":{"a":"250 ","b":"400 ","c":"450 ","d":"500","e":"520"},"options_float":{"a":250.0,"b":400.0,"c":450.0,"d":500.0,"e":520.0},"annotated_formula":"divide(600, 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>\n600 \/ (6\/5)<\/gadget>\n500<\/output>\n500<\/result>","index":2547} +{"problem":"when a laptop is sold for rs . 49,000 , the owner loses 30 % . at what price must that laptop be sold in order to gain 30 % ?","rationale":"70 : 49000 = 130 : x x = ( 49000 x 130 ) \/ 70 = 91000 . hence , s . p . = rs . 91,000 . answer : option e","correct":"e","options":{"a":"87000 ","b":"88000 ","c":"89000 ","d":"90000","e":"91000"},"options_float":{"a":87000.0,"b":88000.0,"c":89000.0,"d":90000.0,"e":91000.0},"annotated_formula":"multiply(divide(multiply(multiply(multiply(add(const_3, const_4), add(const_3, const_4)), const_100), multiply(add(const_3, const_2), const_2)), subtract(const_100, 30)), add(const_100, 30))","linear_formula":"add(n1,const_100)|add(const_3,const_4)|add(const_2,const_3)|subtract(const_100,n1)|multiply(#1,#1)|multiply(#2,const_2)|multiply(#4,const_100)|multiply(#6,#5)|divide(#7,#3)|multiply(#0,#8)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 7<\/gadget>\n49<\/output>\n49 * 100<\/gadget>\n4_900<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 * 2<\/gadget>\n10<\/output>\n4_900 * 10<\/gadget>\n49_000<\/output>\n100 - 30<\/gadget>\n70<\/output>\n49_000 \/ 70<\/gadget>\n700<\/output>\n100 + 30<\/gadget>\n130<\/output>\n700 * 130<\/gadget>\n91_000<\/output>\n91_000<\/result>","index":2548} +{"problem":"if a and b together can finish a work in 16 days . a can finish same work alone in 24 days then b alone can finish same work alone in how many days ?","rationale":"( a + b ) work in 1 day = 1 \/ 16 , a work in 1 day = 1 \/ 24 b work in 1 day = [ 1 \/ 16 - 1 \/ 24 ] = 1 \/ 48 . b alone can finish same work in 48 days . answer b","correct":"b","options":{"a":"30 days ","b":"48 days ","c":"40 days ","d":"36 days","e":"50 days"},"options_float":{"a":30.0,"b":48.0,"c":40.0,"d":36.0,"e":50.0},"annotated_formula":"inverse(subtract(inverse(16), inverse(24)))","linear_formula":"inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)","chain":"1 \/ 16<\/gadget>\n1\/16 = around 0.0625<\/output>\n1 \/ 24<\/gadget>\n1\/24 = around 0.041667<\/output>\n(1\/16) - (1\/24)<\/gadget>\n1\/48 = around 0.020833<\/output>\n1 \/ (1\/48)<\/gadget>\n48<\/output>\n48<\/result>","index":2549} +{"problem":"if 5 a + 7 b = m , where a and b are positive integers , what is the largest possible value of m for which exactly one pair of integers ( a , b ) makes the equation true ?","rationale":"5 * a 1 + 7 * b 1 = m 5 * a 2 + 7 * b 2 = m 5 * ( a 1 - a 2 ) = 7 * ( b 2 - b 1 ) since we are dealing with integers we can assume that a 1 - a 2 = 7 * q and b 2 - b 1 = 5 * q where q is integer , so whenever we get a pair for ( a ; b ) we can find another one by simply adding 7 to a and subtracting 5 from b or vice versa , subtracting 7 from a and adding 5 to b . lets check how it works for our numbers , starting from the largest : e ) 74 = 5 * 12 + 7 * 2 ( a 1 = 12 , b 1 = 2 ) , subtract 7 from a and add 5 to b respectively , so a 2 = 5 and b 2 = 7 , second pair - bad d ) 70 = 5 * 7 + 7 * 5 ( a 1 = 7 , b 1 = 5 ) , if we add 7 toawe will have to subtract 5 from b but b ca n ' t be 0 , so - no pair , if we subtract 7 froma , we ' ll get a = 0 which also is n ' t allowed - no pair , thus this is the only pair for ( a ; b ) that works , good ! , thus d is the answer","correct":"d","options":{"a":"35 ","b":"48 ","c":"69 ","d":"70","e":"74"},"options_float":{"a":35.0,"b":48.0,"c":69.0,"d":70.0,"e":74.0},"annotated_formula":"add(multiply(7, add(const_3, const_4)), multiply(5, const_4))","linear_formula":"add(const_3,const_4)|multiply(n0,const_4)|multiply(n1,#0)|add(#2,#1)","chain":"3 + 4<\/gadget>\n7<\/output>\n7 * 7<\/gadget>\n49<\/output>\n5 * 4<\/gadget>\n20<\/output>\n49 + 20<\/gadget>\n69<\/output>\n69<\/result>","index":2550} +{"problem":"the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 15 % ?","rationale":"\"let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 85 gain % = 21 \/ 64 * 100 = 32.8 % . answer : a\"","correct":"a","options":{"a":"32.8 % ","b":"37.6 % ","c":"38.5 % ","d":"17.5 %","e":"37.2 %"},"options_float":{"a":32.8,"b":37.6,"c":38.5,"d":17.5,"e":37.2},"annotated_formula":"multiply(subtract(divide(subtract(const_100, 15), 64), const_1), const_100)","linear_formula":"subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)|","chain":"100 - 15<\/gadget>\n85<\/output>\n85 \/ 64<\/gadget>\n85\/64 = around 1.328125<\/output>\n(85\/64) - 1<\/gadget>\n21\/64 = around 0.328125<\/output>\n(21\/64) * 100<\/gadget>\n525\/16 = around 32.8125<\/output>\n525\/16 = around 32.8125<\/result>","index":2551} +{"problem":"the calendar of the year 2028 can be used again in the year ?","rationale":"\"explanation : given year 2028 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2028 + 28 = 2056 answer : c\"","correct":"c","options":{"a":"2041 ","b":"1951 ","c":"2056 ","d":"1971","e":"1973"},"options_float":{"a":2041.0,"b":1951.0,"c":2056.0,"d":1971.0,"e":1973.0},"annotated_formula":"add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2028)","linear_formula":"multiply(const_4,const_4)|subtract(#0,const_2)|multiply(#1,const_2)|add(n0,#2)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16 - 2<\/gadget>\n14<\/output>\n14 * 2<\/gadget>\n28<\/output>\n28 + 2_028<\/gadget>\n2_056<\/output>\n2_056<\/result>","index":2552} +{"problem":"a man speaks truth 3 out of 4 times . he throws a die and reports it to be a 6 . what is the probability of it being a 6 ?","rationale":"explanation : there are two cases 1 ) he is telling truth that the die reports 6 , its probability = 3 \/ 4 * 1 \/ 6 = 1 \/ 8 2 ) he is telling lie that the die reports 6 , its probability = 1 \/ 4 * 5 \/ 6 = 5 \/ 24 so required probability = ( 1 \/ 8 ) \/ ( 1 \/ 8 ) + ( 5 \/ 24 ) = ( 1 \/ 8 ) \/ ( 1 \/ 3 ) = 3 \/ 8 hencer ( d ) is the correct answer answer : d","correct":"d","options":{"a":"3 \/ 5 ","b":"1 \/ 2 ","c":"3 \/ 4 ","d":"3 \/ 8","e":"3 \/ 6"},"options_float":{"a":0.6,"b":0.5,"c":0.75,"d":0.375,"e":0.5},"annotated_formula":"divide(multiply(divide(3, 4), divide(const_1, 6)), add(multiply(divide(3, 4), divide(const_1, 6)), multiply(divide(const_1, const_4), divide(const_5, 6))))","linear_formula":"divide(n0,n1)|divide(const_1,n2)|divide(const_1,const_4)|divide(const_5,n2)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5)|divide(#4,#6)","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(3\/4) * (1\/6)<\/gadget>\n1\/8 = around 0.125<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n(1\/4) * (5\/6)<\/gadget>\n5\/24 = around 0.208333<\/output>\n(1\/8) + (5\/24)<\/gadget>\n1\/3 = around 0.333333<\/output>\n(1\/8) \/ (1\/3)<\/gadget>\n3\/8 = around 0.375<\/output>\n3\/8 = around 0.375<\/result>","index":2553} +{"problem":"on a sum of money , simple interest for 2 years is rs 660 and compound interest is rs 696.30 , the rate of interest being the same in both cases .","rationale":"explanation : difference between c . i and s . i for 2 years = 36.30 s . i . for one year = 330 . s . i . on rs 330 for one year = 36.30 so r % = \\ frac { 100 * 36.30 } { 330 * 1 } = 11 % answer : d","correct":"d","options":{"a":"8 % ","b":"9 % ","c":"10 % ","d":"11 %","e":"none of these"},"options_float":{"a":8.0,"b":9.0,"c":10.0,"d":11.0,"e":null},"annotated_formula":"multiply(divide(multiply(subtract(696.3, 660), 2), 660), const_100)","linear_formula":"subtract(n2,n1)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_100)","chain":"696.3 - 660<\/gadget>\n36.3<\/output>\n36.3 * 2<\/gadget>\n72.6<\/output>\n72.6 \/ 660<\/gadget>\n0.11<\/output>\n0.11 * 100<\/gadget>\n11<\/output>\n11<\/result>","index":2555} +{"problem":"an amount at compound interest sums to rs . 17640 \/ - in 2 years and to rs . 19404 \/ - in 3 years at the same rate of interest . find the rate percentage ?","rationale":"explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 19404 \/ - - 17640 \/ - = rs . 1764 \/ - rate of interest = ( 1764 \/ 17640 ) × ( 100 \/ 1 ) = > 10 % answer : option d","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":"multiply(divide(subtract(19404, 17640), 17640), const_100)","linear_formula":"subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)","chain":"19_404 - 17_640<\/gadget>\n1_764<\/output>\n1_764 \/ 17_640<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 100<\/gadget>\n10<\/output>\n10<\/result>","index":2556} +{"problem":"a man is 22 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":"\"let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 22 ) + 2 = 2 ( x + 2 ) x + 24 = 2 x + 4 x = 20 . answer : c\"","correct":"c","options":{"a":"14 years ","b":"18 years ","c":"20 years ","d":"22 years","e":"16 years"},"options_float":{"a":14.0,"b":18.0,"c":20.0,"d":22.0,"e":16.0},"annotated_formula":"divide(subtract(22, 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>\n22 - 2<\/gadget>\n20<\/output>\n2 - 1<\/gadget>\n1<\/output>\n20 \/ 1<\/gadget>\n20<\/output>\n20<\/result>","index":2558} +{"problem":"if y > 0 , ( 10 y ) \/ 20 + ( 3 y ) \/ 10 is what percent of y ?","rationale":"\"can be reduced to y \/ 2 + 3 y \/ 10 = 4 y \/ 5 = 80 % e\"","correct":"e","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(10, 20), divide(3, 10)))","linear_formula":"divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|","chain":"10 \/ 20<\/gadget>\n1\/2 = around 0.5<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n(1\/2) + (3\/10)<\/gadget>\n4\/5 = around 0.8<\/output>\n100 * (4\/5)<\/gadget>\n80<\/output>\n80<\/result>","index":2561} +{"problem":"let the number which when multiplied by 11 is increased by 300 .","rationale":"solution let the number be x . then , 11 x - x = 300 ‹ = › 10 x = 300 x ‹ = › 30 . answer e","correct":"e","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(300, subtract(11, const_1))","linear_formula":"subtract(n0,const_1)|divide(n1,#0)","chain":"11 - 1<\/gadget>\n10<\/output>\n300 \/ 10<\/gadget>\n30<\/output>\n30<\/result>","index":2563} +{"problem":"in a division , a student took 78 as divisor instead of 36 . his answer was 24 . the correct answer is -","rationale":"\"x \/ 78 = 24 . x = 24 * 78 . so correct answer would be , ( 24 * 78 ) \/ 36 = 52 . answer : d\"","correct":"d","options":{"a":"42 ","b":"32 ","c":"48 ","d":"52","e":"38"},"options_float":{"a":42.0,"b":32.0,"c":48.0,"d":52.0,"e":38.0},"annotated_formula":"divide(multiply(78, 24), 36)","linear_formula":"multiply(n0,n2)|divide(#0,n1)|","chain":"78 * 24<\/gadget>\n1_872<\/output>\n1_872 \/ 36<\/gadget>\n52<\/output>\n52<\/result>","index":2564} +{"problem":"an athlete takes 10 seconds to run 100 m . what is his avg . speed in miles per hour ?","rationale":"his average speed is 10 m \/ s . which is 36 km \/ hr . but 36 km = 22.37 miles . the average speed of the athlete is 22.37 mph answer : a","correct":"a","options":{"a":"22.37 ","b":"26.66 ","c":"24.35 ","d":"36.0","e":"42.44"},"options_float":{"a":22.37,"b":26.66,"c":24.35,"d":36.0,"e":42.44},"annotated_formula":"divide(multiply(divide(100, const_1000), const_0_6), divide(10, const_3600))","linear_formula":"divide(n1,const_1000)|divide(n0,const_3600)|multiply(#0,const_0_6)|divide(#2,#1)","chain":"100 \/ 1_000<\/gadget>\n1\/10 = around 0.1<\/output>\n6 \/ 10<\/gadget>\n3\/5 = around 0.6<\/output>\n(1\/10) * (3\/5)<\/gadget>\n3\/50 = around 0.06<\/output>\n10 \/ 3_600<\/gadget>\n1\/360 = around 0.002778<\/output>\n(3\/50) \/ (1\/360)<\/gadget>\n108\/5 = around 21.6<\/output>\n108\/5 = around 21.6<\/result>","index":2565} +{"problem":"a pair of articles was bought for $ 50 at a discount of 50 % . what must be the marked price of each of the article ?","rationale":"\"s . p . of each of the article = 50 \/ 2 = $ 25 let m . p = $ x 50 % of x = 25 x = 25 * . 5 = $ 12.50 answer is b\"","correct":"b","options":{"a":"$ 25 ","b":"$ 12.50 ","c":"$ 29.65 ","d":"$ 35.95","e":"$ 45.62"},"options_float":{"a":25.0,"b":12.5,"c":29.65,"d":35.95,"e":45.62},"annotated_formula":"divide(multiply(subtract(const_100, 50), divide(50, const_2)), const_100)","linear_formula":"divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)|","chain":"100 - 50<\/gadget>\n50<\/output>\n50 \/ 2<\/gadget>\n25<\/output>\n50 * 25<\/gadget>\n1_250<\/output>\n1_250 \/ 100<\/gadget>\n25\/2 = around 12.5<\/output>\n25\/2 = around 12.5<\/result>","index":2567} +{"problem":"a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 40 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 6 hours ?","rationale":"the number of gallons in the tank is ( 1 \/ 4 ) 24 = 6 gallons the amount of sodium chloride is 0.4 ( 6 ) = 2.4 gallons at the start , the amount of water is 0.6 ( 6 ) = 3.6 gallons after 6 hours , the amount of water is 3.6 - 0.5 ( 6 ) = 0.6 gallons the concentration of water is 0.6 \/ ( 2.4 + 0.6 ) = 2 \/ 10 = 20 % the answer is b .","correct":"b","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":"multiply(divide(subtract(divide(multiply(6, subtract(const_100, 40)), const_100), multiply(0.5, 6)), subtract(6, multiply(0.5, 6))), const_100)","linear_formula":"multiply(n2,n3)|subtract(const_100,n0)|multiply(n3,#1)|subtract(n3,#0)|divide(#2,const_100)|subtract(#4,#0)|divide(#5,#3)|multiply(#6,const_100)","chain":"100 - 40<\/gadget>\n60<\/output>\n6 * 60<\/gadget>\n360<\/output>\n360 \/ 100<\/gadget>\n18\/5 = around 3.6<\/output>\n0.5 * 6<\/gadget>\n3<\/output>\n(18\/5) - 3<\/gadget>\n3\/5 = around 0.6<\/output>\n6 - 3<\/gadget>\n3<\/output>\n(3\/5) \/ 3<\/gadget>\n1\/5 = around 0.2<\/output>\n(1\/5) * 100<\/gadget>\n20<\/output>\n20<\/result>","index":2568} +{"problem":"in what proportion must rice at rs 3.10 per kg be mixed with rice at rs 3.75 per kg , so that the mixture be worth rs 3.25 a kg ?","rationale":"c . p of 1 kg of cheaper rice = rs 3.10 c . p of 1 kg of expensive rice = rs 3.75 the mixture be worth for 1 kg = rs 3.25 by the alligation rule : quantity of cheaper rice \/ quantity of expensive rice = ( 3.75 - 3.25 ) \/ ( 3.25 - 3.10 ) = ( 0.50 ) \/ ( 0.15 ) = 10 \/ 3 c","correct":"c","options":{"a":"7 \/ 3 ","b":"5 \/ 3 ","c":"10 \/ 3 ","d":"3 \/ 7","e":"11 \/ 5"},"options_float":{"a":2.3333333333,"b":1.6666666667,"c":3.3333333333,"d":0.4285714286,"e":2.2},"annotated_formula":"divide(subtract(3.75, 3.25), subtract(3.25, 3.1))","linear_formula":"subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)","chain":"3.75 - 3.25<\/gadget>\n0.5<\/output>\n3.25 - 3.1<\/gadget>\n0.15<\/output>\n0.5 \/ 0.15<\/gadget>\n3.333333<\/output>\n3.333333<\/result>","index":2569} +{"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 256 sq . feet , how many feet of fencing will be required ?","rationale":"\"given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 256 breadth = 12.8 feet area to be fenced = 2 b + l = 2 ( 12.8 ) + 20 = 45.6 feet answer : a\"","correct":"a","options":{"a":"45.6 ","b":"40 ","c":"68 ","d":"88","e":"92"},"options_float":{"a":45.6,"b":40.0,"c":68.0,"d":88.0,"e":92.0},"annotated_formula":"add(multiply(divide(256, 20), const_2), 20)","linear_formula":"divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|","chain":"256 \/ 20<\/gadget>\n64\/5 = around 12.8<\/output>\n(64\/5) * 2<\/gadget>\n128\/5 = around 25.6<\/output>\n(128\/5) + 20<\/gadget>\n228\/5 = around 45.6<\/output>\n228\/5 = around 45.6<\/result>","index":2572} +{"problem":"there are 180 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 ) * 180 = 108 . the answer is c .\"","correct":"c","options":{"a":"100 ","b":"104 ","c":"108 ","d":"112","e":"116"},"options_float":{"a":100.0,"b":104.0,"c":108.0,"d":112.0,"e":116.0},"annotated_formula":"multiply(divide(180, add(2, 3)), 3)","linear_formula":"add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|","chain":"2 + 3<\/gadget>\n5<\/output>\n180 \/ 5<\/gadget>\n36<\/output>\n36 * 3<\/gadget>\n108<\/output>\n108<\/result>","index":2573} +{"problem":"a reduction of 40 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ?","rationale":"\"800 * ( 40 \/ 100 ) = 320 - - - - 5 ? - - - - 1 = > rs . 64 answer : c\"","correct":"c","options":{"a":"80 ","b":"72 ","c":"64 ","d":"56","e":"48"},"options_float":{"a":80.0,"b":72.0,"c":64.0,"d":56.0,"e":48.0},"annotated_formula":"divide(divide(multiply(800, 40), const_100), 5)","linear_formula":"multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|","chain":"800 * 40<\/gadget>\n32_000<\/output>\n32_000 \/ 100<\/gadget>\n320<\/output>\n320 \/ 5<\/gadget>\n64<\/output>\n64<\/result>","index":2574} +{"problem":"what is the max number of rectangular boxes , each measuring 4 inches by 6 inches by 10 inches , that can be packed into a rectangular packing box measuring 16 inches by 18 inches by 30 inches , if all boxes are aligned in the same direction ?","rationale":"the 4 inch side should be aligned to the 16 inch side ( 4 layer ) 6 inch side should be aligned to the 18 inch side . ( 3 layer ) and 10 inch side should be aligned to the 30 inch side . ( 3 layer ) maximum number of rectangles = 4 * 3 * 3 = 36 answer is a","correct":"a","options":{"a":"36 ","b":"14 ","c":"12 ","d":"15","e":"11"},"options_float":{"a":36.0,"b":14.0,"c":12.0,"d":15.0,"e":11.0},"annotated_formula":"divide(multiply(multiply(16, 18), 30), multiply(multiply(4, 6), 10))","linear_formula":"multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)","chain":"16 * 18<\/gadget>\n288<\/output>\n288 * 30<\/gadget>\n8_640<\/output>\n4 * 6<\/gadget>\n24<\/output>\n24 * 10<\/gadget>\n240<\/output>\n8_640 \/ 240<\/gadget>\n36<\/output>\n36<\/result>","index":2578} +{"problem":"after decreasing 80 % in the price of an article costs rs . 320 . find the actual cost of an article ?","rationale":"\"cp * ( 20 \/ 100 ) = 320 cp = 16 * 100 = > cp = 1600 answer : e\"","correct":"e","options":{"a":"2777 ","b":"2987 ","c":"1200 ","d":"9977","e":"1600"},"options_float":{"a":2777.0,"b":2987.0,"c":1200.0,"d":9977.0,"e":1600.0},"annotated_formula":"divide(320, subtract(const_1, divide(80, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n1 - (4\/5)<\/gadget>\n1\/5 = around 0.2<\/output>\n320 \/ (1\/5)<\/gadget>\n1_600<\/output>\n1_600<\/result>","index":2579} +{"problem":"a batsman had a certain average of runs for 16 innings . in the 17 th innings , he made a score of 87 runs thereby increasing his average by 3 . what is his average after 17 innings ?","rationale":"explanation : assume his initial average = xx his total runs after 16 innings = 16 xx after scoring 87 runs his average got increased by 3 to xx + 3 so his total runs after 17 innings = 17 × ( xx + 3 ) but it was given that the difference in the total scores after 16 innings and 17 innings = 87 therefore 17 × ( x + 3 ) − 16 x = 87 ⇒ x = 3617 × ( x + 3 ) − 16 x = 87 ⇒ x = 36 his new average = 36 + 3 = 39 answer : a","correct":"a","options":{"a":"39 ","b":"88 ","c":"266 ","d":"278","e":"221"},"options_float":{"a":39.0,"b":88.0,"c":266.0,"d":278.0,"e":221.0},"annotated_formula":"add(subtract(87, multiply(17, 3)), 3)","linear_formula":"multiply(n1,n3)|subtract(n2,#0)|add(n3,#1)","chain":"17 * 3<\/gadget>\n51<\/output>\n87 - 51<\/gadget>\n36<\/output>\n36 + 3<\/gadget>\n39<\/output>\n39<\/result>","index":2581} +{"problem":"if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is - .","rationale":"\"explanation : p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) , where s denotes spade and k denotes king . p ( s ᴜ k ) = 13 \/ 52 + 4 \/ 52 - 1 \/ 52 = 4 \/ 13 answer : b\"","correct":"b","options":{"a":"2 \/ 10 ","b":"4 \/ 13 ","c":"3 \/ 5 ","d":"9 \/ 7","e":"1 \/ 4"},"options_float":{"a":0.2,"b":0.3076923077,"c":0.6,"d":1.2857142857,"e":0.25},"annotated_formula":"add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52))","linear_formula":"divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)|","chain":"3 \/ 52<\/gadget>\n3\/52 = around 0.057692<\/output>\n52 \/ 4<\/gadget>\n13<\/output>\n13 \/ 52<\/gadget>\n1\/4 = around 0.25<\/output>\n(3\/52) + (1\/4)<\/gadget>\n4\/13 = around 0.307692<\/output>\n4\/13 = around 0.307692<\/result>","index":2582} +{"problem":"a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 1 \/ 5 % his yearly income diminishes by rs . 61.50 . his capital is","rationale":"explanation : capital = rs . x , then 4 \/ 5 x = 61.5 x = 76.88 answer : c ) rs . 76.88","correct":"c","options":{"a":"22.378 ","b":"37.78 ","c":"76.88 ","d":"27.888","e":"12.771"},"options_float":{"a":22.378,"b":37.78,"c":76.88,"d":27.888,"e":12.771},"annotated_formula":"divide(61.5, divide(const_4, 5))","linear_formula":"divide(const_4,n3)|divide(n4,#0)","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n61.5 \/ (4\/5)<\/gadget>\n76.875<\/output>\n76.875<\/result>","index":2584} +{"problem":"thirty percent of the members of a swim club have passed the lifesaving test . among the members who havenotpassed the test , 19 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 + 19 = 49 members have not passed the test , thus 0.7 * total = 49 - - > total = 70 . answer : b .\"","correct":"b","options":{"a":"60 ","b":"70 ","c":"100 ","d":"120","e":"140"},"options_float":{"a":60.0,"b":70.0,"c":100.0,"d":120.0,"e":140.0},"annotated_formula":"divide(add(19, 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":"19 + 30<\/gadget>\n49<\/output>\n100 - 30<\/gadget>\n70<\/output>\n70 \/ 100<\/gadget>\n7\/10 = around 0.7<\/output>\n49 \/ (7\/10)<\/gadget>\n70<\/output>\n70<\/result>","index":2585} +{"problem":"the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 14 greater than the actual number and the second number added is 13 instead of 31 . find the correct average .","rationale":"\"sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 14 = 406 hence , new average = 406 ⁄ 10 = 40.6 answer c\"","correct":"c","options":{"a":"40.2 ","b":"40.4 ","c":"40.6 ","d":"40.8","e":"none of the above"},"options_float":{"a":40.2,"b":40.4,"c":40.6,"d":40.8,"e":null},"annotated_formula":"divide(subtract(add(multiply(40.2, 10), add(13, 14)), 31), 10)","linear_formula":"add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)|","chain":"40.2 * 10<\/gadget>\n402<\/output>\n13 + 14<\/gadget>\n27<\/output>\n402 + 27<\/gadget>\n429<\/output>\n429 - 31<\/gadget>\n398<\/output>\n398 \/ 10<\/gadget>\n199\/5 = around 39.8<\/output>\n199\/5 = around 39.8<\/result>","index":2590} +{"problem":"in 1990 the budgets for projects q and v were $ 500,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ?","rationale":"\"let the no of years it takes is x . 500 + 30 x = 780 - 10 x - - > 40 x = 280 and x = 7 . thus , it happens in 1997 . e .\"","correct":"e","options":{"a":"1992 ","b":"1993 ","c":"1994 ","d":"1995","e":"1997"},"options_float":{"a":1992.0,"b":1993.0,"c":1994.0,"d":1995.0,"e":1997.0},"annotated_formula":"add(1990, multiply(10, multiply(const_2, const_3)))","linear_formula":"multiply(const_2,const_3)|multiply(n3,#0)|add(n0,#1)|","chain":"2 * 3<\/gadget>\n6<\/output>\n10 * 6<\/gadget>\n60<\/output>\n1_990 + 60<\/gadget>\n2_050<\/output>\n2_050<\/result>","index":2591} +{"problem":"how many numbers between 100 and 798 are divisible by 2 , 3 , and 7 together ?","rationale":"explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 798 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 798 : 798 \/ 42 = 19 hence , 42 * 19 = 798 is the last number divisible by 42 within 798 hence , total numbers divisible by 2 , 3 , 7 together are ( 19 â € “ 2 ) = 17 answer : d","correct":"d","options":{"a":"112 ","b":"77 ","c":"267 ","d":"17","e":"99"},"options_float":{"a":112.0,"b":77.0,"c":267.0,"d":17.0,"e":99.0},"annotated_formula":"subtract(divide(798, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7)))","linear_formula":"multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)","chain":"2 * 3<\/gadget>\n6<\/output>\n6 * 7<\/gadget>\n42<\/output>\n798 \/ 42<\/gadget>\n19<\/output>\n100 \/ 42<\/gadget>\n50\/21 = around 2.380952<\/output>\n19 - (50\/21)<\/gadget>\n349\/21 = around 16.619048<\/output>\n349\/21 = around 16.619048<\/result>","index":2592} +{"problem":"a car traveled 340 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 340 miles will be covered in 340 \/ 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 = 340 \/ h = > 336 \/ c = 340 \/ ( c + 6 ) = > 336 c + 336 * 6 = 340 c = > c = 336 * 6 \/ 4 = 504 answer a .\"","correct":"a","options":{"a":"504 ","b":"416 ","c":"321 ","d":"220","e":"170"},"options_float":{"a":504.0,"b":416.0,"c":321.0,"d":220.0,"e":170.0},"annotated_formula":"divide(336, divide(subtract(340, 336), 6))","linear_formula":"subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)|","chain":"340 - 336<\/gadget>\n4<\/output>\n4 \/ 6<\/gadget>\n2\/3 = around 0.666667<\/output>\n336 \/ (2\/3)<\/gadget>\n504<\/output>\n504<\/result>","index":2593} +{"problem":"1850 men have provisions for 15 days . if 150 more men join them , for how many days will the provisions last now ?","rationale":"\"1850 * 15 = 2000 * x x = 13.9 answer : c\"","correct":"c","options":{"a":"12.9 ","b":"12.0 ","c":"13.9 ","d":"13.5","e":"12.1"},"options_float":{"a":12.9,"b":12.0,"c":13.9,"d":13.5,"e":12.1},"annotated_formula":"divide(multiply(15, 1850), add(1850, 150))","linear_formula":"add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|","chain":"15 * 1_850<\/gadget>\n27_750<\/output>\n1_850 + 150<\/gadget>\n2_000<\/output>\n27_750 \/ 2_000<\/gadget>\n111\/8 = around 13.875<\/output>\n111\/8 = around 13.875<\/result>","index":2594} +{"problem":"a number increased by 30 % gives 780 . the number is","rationale":"\"formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 30 % = 130 % 130 % - - - - - - - > 780 ( 130 × 6 = 780 ) 100 % - - - - - - - > 600 ( 100 × 6 = 600 ) d )\"","correct":"d","options":{"a":"250 ","b":"400 ","c":"450 ","d":"600","e":"520"},"options_float":{"a":250.0,"b":400.0,"c":450.0,"d":600.0,"e":520.0},"annotated_formula":"divide(780, add(const_1, divide(30, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n1 + (3\/10)<\/gadget>\n13\/10 = around 1.3<\/output>\n780 \/ (13\/10)<\/gadget>\n600<\/output>\n600<\/result>","index":2596} +{"problem":"if a > x > y > z on the number line , y is halfway between x and z , and x is halfway between w and z , then ( y - x ) \/ ( y - a ) =","rationale":"let y - z = t - - - > since y is halfway between x and z and x > y we have x - y = t . moreover x - z = ( x - y ) + ( y - z ) = 2 t . similarly since x is halfway between w and z , we have a - x = 2 t . so y - x = - t , y - a = - 3 t . - - - > ( y - x ) \/ ( y - a ) = 1 \/ 3 . the answer is ( b ) .","correct":"b","options":{"a":"1 \/ 4 ","b":"1 \/ 3 ","c":"1 \/ 2 ","d":"3 \/ 4","e":"1"},"options_float":{"a":0.25,"b":0.3333333333,"c":0.5,"d":0.75,"e":1.0},"annotated_formula":"divide(const_1, subtract(add(const_2, const_2), const_1))","linear_formula":"add(const_2,const_2)|subtract(#0,const_1)|divide(const_1,#1)","chain":"2 + 2<\/gadget>\n4<\/output>\n4 - 1<\/gadget>\n3<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1\/3 = around 0.333333<\/result>","index":2597} +{"problem":"3 friends james , david and charlie divide $ 1230 amongs them in such a way that if $ 5 , $ 10 and $ 15 are removed from the sums that james , david and charlie received respectively , then the share of the sums that they got will be in the ratio of 9 : 10 : 11 . how much did charlie receive ?","rationale":"a + b + c = 1230 given ratio 9 : 10 : 11 let us say the shares of a , b , c deducting 5 , 1015 be a , b , c a + b + c = 1230 - 30 = 1200 = 30 k c share = ( 1200 x 30 ) \/ 60 = 600 c = charlie share = 600 + 15 = 615 option e","correct":"e","options":{"a":"$ 600 ","b":"$ 575 ","c":"$ 550 ","d":"$ 580","e":"$ 615"},"options_float":{"a":600.0,"b":575.0,"c":550.0,"d":580.0,"e":615.0},"annotated_formula":"add(add(add(add(add(multiply(11, divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), 15), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11)))","linear_formula":"add(n2,n3)|add(n3,n5)|add(n4,#0)|add(n7,#1)|subtract(n1,#2)|divide(#4,#3)|multiply(n7,#5)|add(n4,#6)|add(#7,#5)|add(#8,#5)|add(#9,#5)|add(#10,#5)","chain":"5 + 10<\/gadget>\n15<\/output>\n15 + 15<\/gadget>\n30<\/output>\n1_230 - 30<\/gadget>\n1_200<\/output>\n9 + 10<\/gadget>\n19<\/output>\n19 + 11<\/gadget>\n30<\/output>\n1_200 \/ 30<\/gadget>\n40<\/output>\n11 * 40<\/gadget>\n440<\/output>\n440 + 15<\/gadget>\n455<\/output>\n455 + 40<\/gadget>\n495<\/output>\n495 + 40<\/gadget>\n535<\/output>\n535 + 40<\/gadget>\n575<\/output>\n575 + 40<\/gadget>\n615<\/output>\n615<\/result>","index":2598} +{"problem":"a rectangular grass field is 70 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ?","rationale":"\"area = ( l + b + 2 d ) 2 d = ( 70 + 55 + 2.5 * 2 ) 2 * 2.5 = > 650 650 * 2 = rs . 1300 answer : b\"","correct":"b","options":{"a":"s . 1350 ","b":"s . 1300 ","c":"s . 1328 ","d":"s . 1397","e":"s . 1927"},"options_float":{"a":1350.0,"b":1300.0,"c":1328.0,"d":1397.0,"e":1927.0},"annotated_formula":"multiply(subtract(rectangle_area(add(70, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(70, 55)), 2)","linear_formula":"multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)|","chain":"2.5 * 2<\/gadget>\n5<\/output>\n70 + 5<\/gadget>\n75<\/output>\n55 + 5<\/gadget>\n60<\/output>\n75 * 60<\/gadget>\n4_500<\/output>\n70 * 55<\/gadget>\n3_850<\/output>\n4_500 - 3_850<\/gadget>\n650<\/output>\n650 * 2<\/gadget>\n1_300<\/output>\n1_300<\/result>","index":2600} +{"problem":"calculate the area of a triangle , if the sides of are 39 cm , 36 cm and 15 cm , what is its area ?","rationale":"\"the triangle with sides 39 cm , 36 cm and 15 is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 \/ 2 * 36 * 15 = 270 cm 2 answer : e\"","correct":"e","options":{"a":"570 cm 2 ","b":"370 cm 2 ","c":"170 cm 2 ","d":"271 cm 2","e":"270 cm 2"},"options_float":{"a":570.0,"b":370.0,"c":170.0,"d":271.0,"e":270.0},"annotated_formula":"multiply(divide(36, const_2), 15)","linear_formula":"divide(n1,const_2)|multiply(n2,#0)|","chain":"36 \/ 2<\/gadget>\n18<\/output>\n18 * 15<\/gadget>\n270<\/output>\n270<\/result>","index":2601} +{"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 20 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 20 hours . he would have saved 20 â € “ 4 = 16 hours . answer : e\"","correct":"e","options":{"a":"2 ","b":"8 ","c":"1 ","d":"6","e":"16"},"options_float":{"a":2.0,"b":8.0,"c":1.0,"d":6.0,"e":16.0},"annotated_formula":"subtract(20, divide(20, 5))","linear_formula":"divide(n2,n0)|subtract(n3,#0)|","chain":"20 \/ 5<\/gadget>\n4<\/output>\n20 - 4<\/gadget>\n16<\/output>\n16<\/result>","index":2602} +{"problem":"a man can do a job in 15 days . his father takes 20 days and his son finishes it in 15 days . how long will they take to complete the job if they all work together ?","rationale":"\"1 day work of the three persons = ( 1 \/ 15 + 1 \/ 20 + 1 \/ 15 ) = 11 \/ 60 so , all three together will complete the work in 300 \/ 47 = 5.5 days . answer : c\"","correct":"c","options":{"a":"6.3 ","b":"6.9 ","c":"5.5 ","d":"6.1","e":"6.2"},"options_float":{"a":6.3,"b":6.9,"c":5.5,"d":6.1,"e":6.2},"annotated_formula":"divide(const_1, add(divide(const_1, 15), add(divide(const_1, 15), 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 \/ 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(1\/15) + (7\/60)<\/gadget>\n11\/60 = around 0.183333<\/output>\n1 \/ (11\/60)<\/gadget>\n60\/11 = around 5.454545<\/output>\n60\/11 = around 5.454545<\/result>","index":2604} +{"problem":"how long does a train 250 meters long running at the rate of 72 km \/ hr take to cross a bridge 150 meters in length ?","rationale":"\"distance = length of train + length of bridge = 250 + 150 = 400 speed = 72 km \/ hr = 72 * 5 \/ 18 = 20 m \/ s required time = 400 \/ 20 = 20 seconds answer is b\"","correct":"b","options":{"a":"10 sec ","b":"20 sec ","c":"25 sec ","d":"30 sec","e":"35 sec"},"options_float":{"a":10.0,"b":20.0,"c":25.0,"d":30.0,"e":35.0},"annotated_formula":"divide(add(250, 150), multiply(72, const_0_2778))","linear_formula":"add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|","chain":"250 + 150<\/gadget>\n400<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n400 \/ 20<\/gadget>\n20<\/output>\n20<\/result>","index":2605} +{"problem":"the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 10 cents less than the price of commodity y ?","rationale":"\"the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 10 cents less than commodity y . $ 2.00 \/ 25 cents = 8 years the answer is 2001 + 8 years = 2009 . the answer is b .\"","correct":"b","options":{"a":"2008 ","b":"2009 ","c":"2010 ","d":"2011","e":"2012"},"options_float":{"a":2008.0,"b":2009.0,"c":2010.0,"d":2011.0,"e":2012.0},"annotated_formula":"add(2001, divide(add(divide(10, const_100), subtract(7.30, 5.20)), subtract(divide(45, const_100), subtract(7.30, 5.20))))","linear_formula":"divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n7.3 - 5.2<\/gadget>\n2.1<\/output>\n(1\/10) + 2.1<\/gadget>\n2.2<\/output>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n(9\/20) - 2.1<\/gadget>\n-1.65<\/output>\n2.2 \/ (-1.65)<\/gadget>\n-1.333333<\/output>\n2_001 + (-1.333333)<\/gadget>\n1_999.666667<\/output>\n1_999.666667<\/result>","index":2606} +{"problem":"if a wholesaler distributes items to several outlets ( a , b , c and d ) in the ratio of 1 \/ 12 : 1 \/ 13 : 1 \/ 15 : 1 \/ 2 , then find the total number of items the wholesaler distributes ?","rationale":"here , a : b : c : d = 1 \/ 12 : 1 \/ 13 : 1 \/ 15 : 1 \/ 2 1 ) l . c . m of 12 : 13 : 15 : 2 is 780 2 ) find the number of books each friend received - - - - - - - - - ( to find no . of books each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 \/ 12 ) x 780 = 65 b = ( 1 \/ 13 ) x 780 = 60 c = ( 1 \/ 15 ) x 780 = 52 d = ( 1 \/ 2 ) x 780 = 390 3 ) total number of toys = ( 65 x + 60 x + 52 x + 390 x ) = 567 x minimum number of pens ( x ) = 1 therefore , total number of items = 567 items . correct option : a","correct":"a","options":{"a":"567 ","b":"167 ","c":"267 ","d":"467","e":"667"},"options_float":{"a":567.0,"b":167.0,"c":267.0,"d":467.0,"e":667.0},"annotated_formula":"add(add(multiply(const_100, const_4), const_100), add(multiply(15, const_4), add(const_4, const_3)))","linear_formula":"add(const_3,const_4)|multiply(const_100,const_4)|multiply(n5,const_4)|add(#1,const_100)|add(#0,#2)|add(#3,#4)","chain":"100 * 4<\/gadget>\n400<\/output>\n400 + 100<\/gadget>\n500<\/output>\n15 * 4<\/gadget>\n60<\/output>\n4 + 3<\/gadget>\n7<\/output>\n60 + 7<\/gadget>\n67<\/output>\n500 + 67<\/gadget>\n567<\/output>\n567<\/result>","index":2607} +{"problem":"john makes $ 50 a week from his job . he earns a raise and now makes $ 60 a week . what is the % increase ?","rationale":"\"increase = ( 10 \/ 50 ) * 100 = ( 1 \/ 5 ) * 100 = 20 % . e\"","correct":"e","options":{"a":"16 % ","b":"16.66 % ","c":"17.9 % ","d":"18.12 %","e":"20 %"},"options_float":{"a":16.0,"b":16.66,"c":17.9,"d":18.12,"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":2608} +{"problem":"in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 90 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ?","rationale":"\"total students = 420 boys = 312 , girls = 108 total playing soccer = 250 90 % of 250 = 225 are boys who play soccer . girls who play soccer = 25 . total girls who do not play soccer = 108 - 25 = 83 . correct option : b\"","correct":"b","options":{"a":"69 . ","b":"83 . ","c":"81 ","d":"91","e":"108"},"options_float":{"a":69.0,"b":83.0,"c":81.0,"d":91.0,"e":108.0},"annotated_formula":"subtract(subtract(420, 312), subtract(250, divide(multiply(250, 90), const_100)))","linear_formula":"multiply(n2,n3)|subtract(n0,n1)|divide(#0,const_100)|subtract(n2,#2)|subtract(#1,#3)|","chain":"420 - 312<\/gadget>\n108<\/output>\n250 * 90<\/gadget>\n22_500<\/output>\n22_500 \/ 100<\/gadget>\n225<\/output>\n250 - 225<\/gadget>\n25<\/output>\n108 - 25<\/gadget>\n83<\/output>\n83<\/result>","index":2609} +{"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 $ 14 , 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 14 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 14 solve for x x = $ 24.88 correct answer c\"","correct":"c","options":{"a":"$ 45.10 ","b":"$ 34.31 ","c":"$ 24.88 ","d":"$ 67.54","e":"$ 65.23"},"options_float":{"a":45.1,"b":34.31,"c":24.88,"d":67.54,"e":65.23},"annotated_formula":"divide(multiply(multiply(const_100, const_100), 14), 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 * 14<\/gadget>\n140_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>\n140_000 \/ 5_625<\/gadget>\n224\/9 = around 24.888889<\/output>\n224\/9 = around 24.888889<\/result>","index":2610} +{"problem":"the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1690 . the perimeter of the plot is ?","rationale":"\"sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1650 x = 60 required perimeter = 2 ( 60 + 70 ) = 260 e\"","correct":"e","options":{"a":"126 ","b":"156 ","c":"190 ","d":"321","e":"260"},"options_float":{"a":126.0,"b":156.0,"c":190.0,"d":321.0,"e":260.0},"annotated_formula":"multiply(add(divide(subtract(divide(divide(1690, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(1690, 6.5), const_2), 10), const_2), 10)), const_2)","linear_formula":"divide(n2,n1)|divide(#0,const_2)|subtract(#1,n0)|divide(#2,const_2)|add(#3,n0)|add(#4,#3)|multiply(#5,const_2)|","chain":"1_690 \/ 6.5<\/gadget>\n260<\/output>\n260 \/ 2<\/gadget>\n130<\/output>\n130 - 10<\/gadget>\n120<\/output>\n120 \/ 2<\/gadget>\n60<\/output>\n60 + 10<\/gadget>\n70<\/output>\n60 + 70<\/gadget>\n130<\/output>\n130 * 2<\/gadget>\n260<\/output>\n260<\/result>","index":2611} +{"problem":"the average weight of 4 person ' s increases by 1.5 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 1.5 ) kg = 6 kg . weight of new person = ( 95 + 6 ) kg = 101 kg . answer : option a\"","correct":"a","options":{"a":"101 kg ","b":"103.4 kg ","c":"105 kg ","d":"data inadequate","e":"none of these"},"options_float":{"a":101.0,"b":103.4,"c":105.0,"d":null,"e":null},"annotated_formula":"add(multiply(4, 1.5), 95)","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"4 * 1.5<\/gadget>\n6<\/output>\n6 + 95<\/gadget>\n101<\/output>\n101<\/result>","index":2612} +{"problem":"each of the three people individually can complete a certain job in 3 , 5 , 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 \/ 5 and 1 \/ 6 of the job per hour . the sum of these rates is 1 \/ 5 + 1 \/ 6 = 11 \/ 30 of the job per hour . the answer is c .\"","correct":"c","options":{"a":"4 \/ 15 ","b":"7 \/ 30 ","c":"11 \/ 30 ","d":"7 \/ 18","e":"5 \/ 18"},"options_float":{"a":0.2666666667,"b":0.2333333333,"c":0.3666666667,"d":0.3888888889,"e":0.2777777778},"annotated_formula":"add(divide(1, 5), divide(1, 6))","linear_formula":"divide(n3,n1)|divide(n3,n2)|add(#0,#1)|","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>\n11\/30 = around 0.366667<\/output>\n11\/30 = around 0.366667<\/result>","index":2613} +{"problem":"the average of 5 consecutive odd numbers a , b , c , d and e is 33 . what percent of a is d ?","rationale":"explanation : in such a case the middle number ( c ) is the average ∴ c = 33 and a = 31 and d = 35 required percentage = 31 \/ 35 x 100 = 88.6 answer : option b","correct":"b","options":{"a":"86.8 ","b":"88.6 ","c":"89.2 ","d":"90.1","e":"92.2"},"options_float":{"a":86.8,"b":88.6,"c":89.2,"d":90.1,"e":92.2},"annotated_formula":"multiply(const_100, divide(divide(multiply(33, 5), 5), add(add(add(divide(multiply(33, 5), 5), const_2), const_2), const_2)))","linear_formula":"multiply(n0,n1)|divide(#0,n0)|add(#1,const_2)|add(#2,const_2)|add(#3,const_2)|divide(#1,#4)|multiply(#5,const_100)","chain":"33 * 5<\/gadget>\n165<\/output>\n165 \/ 5<\/gadget>\n33<\/output>\n33 + 2<\/gadget>\n35<\/output>\n35 + 2<\/gadget>\n37<\/output>\n37 + 2<\/gadget>\n39<\/output>\n33 \/ 39<\/gadget>\n11\/13 = around 0.846154<\/output>\n100 * (11\/13)<\/gadget>\n1_100\/13 = around 84.615385<\/output>\n1_100\/13 = around 84.615385<\/result>","index":2614} +{"problem":"daniel went to a shop and bought things worth rs . 25 , out of which 60 paise 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 . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 60 paise = rs . 60 \/ 100 let the cost of the tax free items = x given that tax rate = 6 % ∴ ( 25 − 60 \/ 100 − x ) 6 \/ 100 = 60 \/ 100 ⇒ 6 ( 25 − 0.6 − x ) = 60 ⇒ ( 25 − 0.6 − x ) = 10 ⇒ x = 25 − 0.6 − 10 = 14.4 a\"","correct":"a","options":{"a":"14.4 ","b":"20 ","c":"21.3 ","d":"21.5","e":"22"},"options_float":{"a":14.4,"b":20.0,"c":21.3,"d":21.5,"e":22.0},"annotated_formula":"subtract(subtract(25, divide(60, const_100)), divide(60, 6))","linear_formula":"divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)|","chain":"60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n25 - (3\/5)<\/gadget>\n122\/5 = around 24.4<\/output>\n60 \/ 6<\/gadget>\n10<\/output>\n(122\/5) - 10<\/gadget>\n72\/5 = around 14.4<\/output>\n72\/5 = around 14.4<\/result>","index":2616} +{"problem":"the volumes of two cubes are in the ratio 27 : 125 , what shall be the ratio of their surface areas ?","rationale":"a 13 : a 23 = 27 : 125 a 1 : a 2 = 3 : 5 6 a 12 : 6 a 22 a 12 : a 22 = 9 : 25 answer : c","correct":"c","options":{"a":"6 : 25 ","b":"3 : 5 ","c":"9 : 25 ","d":"16 : 25","e":"19 : 25"},"options_float":{"a":0.24,"b":0.6,"c":0.36,"d":0.64,"e":0.76},"annotated_formula":"divide(surface_cube(divide(divide(27, const_3), const_3)), surface_cube(divide(125, divide(125, add(const_4, const_1)))))","linear_formula":"add(const_1,const_4)|divide(n0,const_3)|divide(#1,const_3)|divide(n1,#0)|divide(n1,#3)|surface_cube(#2)|surface_cube(#4)|divide(#5,#6)","chain":"27 \/ 3<\/gadget>\n9<\/output>\n9 \/ 3<\/gadget>\n3<\/output>\n6 * (3 ** 2)<\/gadget>\n54<\/output>\n4 + 1<\/gadget>\n5<\/output>\n125 \/ 5<\/gadget>\n25<\/output>\n125 \/ 25<\/gadget>\n5<\/output>\n6 * (5 ** 2)<\/gadget>\n150<\/output>\n54 \/ 150<\/gadget>\n9\/25 = around 0.36<\/output>\n9\/25 = around 0.36<\/result>","index":2617} +{"problem":"bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is - .","rationale":"explanation : drawing two balls of same color from seven green balls can be done in ⠁ · c â ‚ ‚ ways . similarly from eight white balls two can be drawn in ⠁ ¸ c â ‚ ‚ ways . p = ⠁ · c â ‚ ‚ \/ â ¹ ⠁ µ c â ‚ ‚ + ⠁ ¸ c â ‚ ‚ \/ â ¹ ⠁ µ c â ‚ ‚ = 7 \/ 15 a","correct":"a","options":{"a":"7 \/ 15 ","b":"2 \/ 8 ","c":"7 \/ 11 ","d":"13 \/ 5","e":"87"},"options_float":{"a":0.4666666667,"b":0.25,"c":0.6363636364,"d":2.6,"e":87.0},"annotated_formula":"divide(add(divide(factorial(7), multiply(factorial(subtract(7, const_2)), factorial(const_2))), divide(factorial(8), multiply(factorial(subtract(8, const_2)), factorial(const_2)))), divide(factorial(add(7, 8)), multiply(factorial(subtract(add(7, 8), const_2)), factorial(const_2))))","linear_formula":"add(n0,n1)|factorial(n0)|factorial(const_2)|factorial(n1)|subtract(n0,const_2)|subtract(n1,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|factorial(#9)|multiply(#6,#2)|multiply(#7,#2)|divide(#1,#11)|divide(#3,#12)|multiply(#10,#2)|add(#13,#14)|divide(#8,#15)|divide(#16,#17)","chain":"factorial(7)<\/gadget>\n5_040<\/output>\n7 - 2<\/gadget>\n5<\/output>\nfactorial(5)<\/gadget>\n120<\/output>\nfactorial(2)<\/gadget>\n2<\/output>\n120 * 2<\/gadget>\n240<\/output>\n5_040 \/ 240<\/gadget>\n21<\/output>\nfactorial(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>\n21 + 28<\/gadget>\n49<\/output>\n7 + 8<\/gadget>\n15<\/output>\nfactorial(15)<\/gadget>\n1_307_674_368_000<\/output>\n15 - 2<\/gadget>\n13<\/output>\nfactorial(13)<\/gadget>\n6_227_020_800<\/output>\n6_227_020_800 * 2<\/gadget>\n12_454_041_600<\/output>\n1_307_674_368_000 \/ 12_454_041_600<\/gadget>\n105<\/output>\n49 \/ 105<\/gadget>\n7\/15 = around 0.466667<\/output>\n7\/15 = around 0.466667<\/result>","index":2619} +{"problem":"two trains of equal length , running with the speeds of 60 and 40 kmph , take 75 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ?","rationale":"\"rs = 60 - 40 = 20 * 5 \/ 18 = 100 \/ 18 t = 75 d = 75 * 100 \/ 18 = 1250 \/ 3 rs = 60 + 50 = 100 * 5 \/ 18 t = 1250 \/ 3 * 18 \/ 500 = 15 sec answer : a\"","correct":"a","options":{"a":"15 sec ","b":"16 sec ","c":"14 sec ","d":"67 sec","e":"13 sec"},"options_float":{"a":15.0,"b":16.0,"c":14.0,"d":67.0,"e":13.0},"annotated_formula":"multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 75), inverse(multiply(const_0_2778, add(60, 40))))","linear_formula":"add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 - 40<\/gadget>\n20<\/output>\n(5\/18) * 20<\/gadget>\n50\/9 = around 5.555556<\/output>\n(50\/9) * 75<\/gadget>\n1_250\/3 = around 416.666667<\/output>\n60 + 40<\/gadget>\n100<\/output>\n(5\/18) * 100<\/gadget>\n250\/9 = around 27.777778<\/output>\n1 \/ (250\/9)<\/gadget>\n9\/250 = around 0.036<\/output>\n(1_250\/3) * (9\/250)<\/gadget>\n15<\/output>\n15<\/result>","index":2620} +{"problem":"if 40 % of a certain number is 160 , then what is 90 % of that number ?","rationale":"\"explanation : 40 % = 40 * 4 = 160 90 % = 90 * 4 = 360 answer : option d\"","correct":"d","options":{"a":"270 ","b":"380 ","c":"260 ","d":"360","e":"290"},"options_float":{"a":270.0,"b":380.0,"c":260.0,"d":360.0,"e":290.0},"annotated_formula":"multiply(divide(160, divide(40, const_100)), divide(90, const_100))","linear_formula":"divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n160 \/ (2\/5)<\/gadget>\n400<\/output>\n90 \/ 100<\/gadget>\n9\/10 = around 0.9<\/output>\n400 * (9\/10)<\/gadget>\n360<\/output>\n360<\/result>","index":2623} +{"problem":"subtracting 30 from a number , the remainder is one fourth of the number . find the number ?","rationale":"explanation : 3 \/ 4 x = 30 = > x = 40 answer : c","correct":"c","options":{"a":"29 ","b":"88 ","c":"40 ","d":"28","e":"27"},"options_float":{"a":29.0,"b":88.0,"c":40.0,"d":28.0,"e":27.0},"annotated_formula":"divide(30, subtract(const_1, divide(const_1, const_4)))","linear_formula":"divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1)","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n30 \/ (3\/4)<\/gadget>\n40<\/output>\n40<\/result>","index":2625} +{"problem":"anne bought doughnuts for a class breakfast party . she bought 12 chocolate doughnuts , 6 coconut doughnuts , and 8 jam - filled doughnuts . how many doughnuts did anne buy in all ?","rationale":"\"add the numbers of doughnuts . 12 + 6 + 8 = 26 . answer is b .\"","correct":"b","options":{"a":"25 ","b":"26 ","c":"39 ","d":"21","e":"11"},"options_float":{"a":25.0,"b":26.0,"c":39.0,"d":21.0,"e":11.0},"annotated_formula":"add(add(12, 6), 8)","linear_formula":"add(n0,n1)|add(n2,#0)|","chain":"12 + 6<\/gadget>\n18<\/output>\n18 + 8<\/gadget>\n26<\/output>\n26<\/result>","index":2629} +{"problem":"in what time will a train 100 m long cross an electric pole , it its speed be 90 km \/ hr ?","rationale":"\"speed = 90 * 5 \/ 18 = 25 m \/ sec time taken = 100 \/ 25 = 4 sec . answer : c\"","correct":"c","options":{"a":"2.5 ","b":"2.9 ","c":"4 sec ","d":"2.8","e":"2.1"},"options_float":{"a":2.5,"b":2.9,"c":4.0,"d":2.8,"e":2.1},"annotated_formula":"divide(100, multiply(90, const_0_2778))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n90 * (5\/18)<\/gadget>\n25<\/output>\n100 \/ 25<\/gadget>\n4<\/output>\n4<\/result>","index":2630} +{"problem":"if the number is divided by 3 , it reduced by 34 . the number is","rationale":"explanation : let the number be x . then , x - ( x \/ 3 ) = 34 = > 2 x \/ 3 = 34 = > x = 51 answer : option a","correct":"a","options":{"a":"51 ","b":"50 ","c":"45 ","d":"40","e":"36"},"options_float":{"a":51.0,"b":50.0,"c":45.0,"d":40.0,"e":36.0},"annotated_formula":"divide(multiply(34, 3), subtract(3, const_1))","linear_formula":"multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)","chain":"34 * 3<\/gadget>\n102<\/output>\n3 - 1<\/gadget>\n2<\/output>\n102 \/ 2<\/gadget>\n51<\/output>\n51<\/result>","index":2631} +{"problem":"if a train runs at 40 kmph , it reach its destination late by 11 minutes but if it runs at 50 kmph it is late by 5 minutes only . the correct time for a train to complete its journey is ? let the correct time to complete the journey be x min distance covered in ( x + 11 ) min . at 40 kmph distance covered in ( x + 5 ) min . at 50 kmph ( x + 11 ) \/ 60 * 40 = ( x + 5 ) \/ 60 * 50 x = 19 min","rationale":"let the correct time to complete the journey be x min distance covered in ( x + 11 ) min . at 40 kmph distance covered in ( x + 5 ) min . at 50 kmph ( x + 11 ) \/ 60 * 40 = ( x + 5 ) \/ 60 * 50 x = 19 min answer ( a )","correct":"a","options":{"a":"19 min ","b":"19 hrs ","c":"52 min ","d":"126 min","e":"52 min"},"options_float":{"a":19.0,"b":19.0,"c":52.0,"d":126.0,"e":52.0},"annotated_formula":"divide(subtract(multiply(multiply(60, 40), 11), multiply(multiply(60, 50), 5)), subtract(multiply(60, 50), multiply(60, 40)))","linear_formula":"multiply(n0,n9)|multiply(n2,n9)|multiply(n1,#0)|multiply(n3,#1)|subtract(#1,#0)|subtract(#2,#3)|divide(#5,#4)","chain":"60 * 40<\/gadget>\n2_400<\/output>\n2_400 * 11<\/gadget>\n26_400<\/output>\n60 * 50<\/gadget>\n3_000<\/output>\n3_000 * 5<\/gadget>\n15_000<\/output>\n26_400 - 15_000<\/gadget>\n11_400<\/output>\n3_000 - 2_400<\/gadget>\n600<\/output>\n11_400 \/ 600<\/gadget>\n19<\/output>\n19<\/result>","index":2632} +{"problem":"share rs . 5400 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ?","rationale":"\"amount received by sanjay . 4 \/ 12 x 5400 = 1800 = ( related ratio \/ sum of ratio ) x total amount so , the amount received by sanjay is 1800 . a\"","correct":"a","options":{"a":"1800 ","b":"980 ","c":"1200 ","d":"1240","e":"1400"},"options_float":{"a":1800.0,"b":980.0,"c":1200.0,"d":1240.0,"e":1400.0},"annotated_formula":"subtract(divide(5400, 2), divide(5400, 6))","linear_formula":"divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)|","chain":"5_400 \/ 2<\/gadget>\n2_700<\/output>\n5_400 \/ 6<\/gadget>\n900<\/output>\n2_700 - 900<\/gadget>\n1_800<\/output>\n1_800<\/result>","index":2634} +{"problem":"a room is a square of side 50 feet . a second room is of area 100 square yards . a third room is of area 200 square feet . which of these can seat maximum people ? ( hint : 1 yard = 3 feet )","rationale":"first room because area 50 * 50 = 2500 sq feet second room area 100 sq yard in feet 300 sq feet third room area 200 sq feet answer : a","correct":"a","options":{"a":"200 sq feet ","b":"300 sq feet ","c":"400 sq feet ","d":"500 sq feet","e":"600 sq feet"},"options_float":{"a":200.0,"b":300.0,"c":400.0,"d":500.0,"e":600.0},"annotated_formula":"multiply(100, const_2)","linear_formula":"multiply(n1,const_2)","chain":"100 * 2<\/gadget>\n200<\/output>\n200<\/result>","index":2635} +{"problem":"alex and brian start a business with rs . 7000 each , and after 8 months , brian withdraws half of his capital . how should they share the profits at the end of the 18 months ?","rationale":"alex invests rs . 7000 for 18 months , but brian invests rs . 7000 for the first 8 months and then withdraws rs . 3500 . so , the investment of brian for remaining 10 months is rs . 3500 only . alex : brian 7000 * 18 : ( 7000 * 8 ) + ( 3500 * 10 ) 126000 : 91000 alex : brian = 18 : 13 answer : e","correct":"e","options":{"a":"5 : 4 ","b":"4 : 3 ","c":"18 : 11 ","d":"3 : 2","e":"18 : 13"},"options_float":{"a":1.25,"b":1.3333333333,"c":1.6363636364,"d":1.5,"e":1.3846153846},"annotated_formula":"divide(18, add(const_12, const_1))","linear_formula":"add(const_1,const_12)|divide(n2,#0)","chain":"12 + 1<\/gadget>\n13<\/output>\n18 \/ 13<\/gadget>\n18\/13 = around 1.384615<\/output>\n18\/13 = around 1.384615<\/result>","index":2636} +{"problem":"of the people who responded to a market survey , 240 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of 6 to 1 , how many people responded to the survey ?","rationale":"\"ratio = 6 : 1 = > 6 x respondents preferred brand x and x preferred brand y since , no . of respondents who preferred brand x = 240 = > 6 x = 240 = > x = 40 hence total no . of respondents = 240 + 40 = 280 hence c is the answer .\"","correct":"c","options":{"a":"80 ","b":"160 ","c":"280 ","d":"360","e":"480"},"options_float":{"a":80.0,"b":160.0,"c":280.0,"d":360.0,"e":480.0},"annotated_formula":"add(divide(240, 6), 240)","linear_formula":"divide(n0,n1)|add(n0,#0)|","chain":"240 \/ 6<\/gadget>\n40<\/output>\n40 + 240<\/gadget>\n280<\/output>\n280<\/result>","index":2637} +{"problem":"two trains 119 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ?","rationale":"\"t = ( 119 + 165 ) \/ ( 80 + 65 ) * 18 \/ 5 t = 7.05 answer : e\"","correct":"e","options":{"a":"7.19 ","b":"7.17 ","c":"7.2 ","d":"7.15","e":"7.05"},"options_float":{"a":7.19,"b":7.17,"c":7.2,"d":7.15,"e":7.05},"annotated_formula":"divide(add(119, 165), multiply(add(80, 65), const_0_2778))","linear_formula":"add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|","chain":"119 + 165<\/gadget>\n284<\/output>\n80 + 65<\/gadget>\n145<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n145 * (5\/18)<\/gadget>\n725\/18 = around 40.277778<\/output>\n284 \/ (725\/18)<\/gadget>\n5_112\/725 = around 7.051034<\/output>\n5_112\/725 = around 7.051034<\/result>","index":2639} +{"problem":"if an integer e is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that e ( e + 1 ) ( e + 2 ) will be divisible by 8 ?","rationale":"for e total numbers 8 * 12 there are 12 numbers divisible by 8 - > 3 * 12 ( if 8 is an example - ( 6 , 78 ) , ( 7 , 89 ) , ( 8 , 910 ) ) and 12 numbers divisible by 4 but not divisible by 8 - > 2 * 12 ( if 4 is an example ( 2 , 34 ) and ( 4 , 56 ) ) the answer 5 \/ 8 - > d","correct":"d","options":{"a":"1 \/ 4 ","b":"3 \/ 8 ","c":"1 \/ 2 ","d":"5 \/ 8","e":"3 \/ 4"},"options_float":{"a":0.25,"b":0.375,"c":0.5,"d":0.625,"e":0.75},"annotated_formula":"divide(add(multiply(divide(divide(96, 8), 8), 2), 2), 8)","linear_formula":"divide(n1,n4)|divide(#0,n4)|multiply(n3,#1)|add(n3,#2)|divide(#3,n4)","chain":"96 \/ 8<\/gadget>\n12<\/output>\n12 \/ 8<\/gadget>\n3\/2 = around 1.5<\/output>\n(3\/2) * 2<\/gadget>\n3<\/output>\n3 + 2<\/gadget>\n5<\/output>\n5 \/ 8<\/gadget>\n5\/8 = around 0.625<\/output>\n5\/8 = around 0.625<\/result>","index":2640} +{"problem":"if 20 men can build a wall 66 metres long in 10 days , what length of a similar can be built by 86 men in 8 days ?","rationale":"\"if 20 men can build a wall 66 metres long in 10 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) \/ ( 10 * 20 ) = 227.04 mtrs answer : a\"","correct":"a","options":{"a":"227.04 mtrs ","b":"378.4 mtrs ","c":"478.4 mtrs ","d":"488.4 mtrs","e":"578.4 mtrs"},"options_float":{"a":227.04,"b":378.4,"c":478.4,"d":488.4,"e":578.4},"annotated_formula":"multiply(66, divide(multiply(86, 8), multiply(20, 10)))","linear_formula":"multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)|","chain":"86 * 8<\/gadget>\n688<\/output>\n20 * 10<\/gadget>\n200<\/output>\n688 \/ 200<\/gadget>\n86\/25 = around 3.44<\/output>\n66 * (86\/25)<\/gadget>\n5_676\/25 = around 227.04<\/output>\n5_676\/25 = around 227.04<\/result>","index":2641} +{"problem":"if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 9 ) , what is the value of x ?","rationale":"\"( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 9 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 9 hence x = 11 . 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":"add(9, 2)","linear_formula":"add(n0,n5)|","chain":"9 + 2<\/gadget>\n11<\/output>\n11<\/result>","index":2642} +{"problem":"by travelling at 60 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ?","rationale":"\"let the time taken to reach the destination be 3 x hours . total distance = 60 * 3 x = 180 x km he covered 2 \/ 3 * 180 x = 120 x km in 1 \/ 3 * 3 x = x hours so , the remaining 60 x km , he has to cover in 2 x hours . required speed = 60 x \/ 2 x = 30 kmph . answer : a\"","correct":"a","options":{"a":"30 kmph ","b":"28 kmph ","c":"26 kmph ","d":"24 kmph","e":"22 kmph"},"options_float":{"a":30.0,"b":28.0,"c":26.0,"d":24.0,"e":22.0},"annotated_formula":"divide(subtract(multiply(60, const_3), divide(multiply(multiply(60, const_3), const_2), const_3)), subtract(const_3, const_1))","linear_formula":"multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)|","chain":"60 * 3<\/gadget>\n180<\/output>\n180 * 2<\/gadget>\n360<\/output>\n360 \/ 3<\/gadget>\n120<\/output>\n180 - 120<\/gadget>\n60<\/output>\n3 - 1<\/gadget>\n2<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n30<\/result>","index":2643} +{"problem":"a candidate got 35 % of the votes polled and he lost to his rival by 2430 votes . how many votes were cast ?","rationale":"\"35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2430 100 % - - - - - - - - - ? = > 8100 answer : c\"","correct":"c","options":{"a":"7500 ","b":"3388 ","c":"8100 ","d":"2888","e":"2661"},"options_float":{"a":7500.0,"b":3388.0,"c":8100.0,"d":2888.0,"e":2661.0},"annotated_formula":"divide(2430, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100)))","linear_formula":"divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|","chain":"35 \/ 100<\/gadget>\n7\/20 = around 0.35<\/output>\n1 - (7\/20)<\/gadget>\n13\/20 = around 0.65<\/output>\n(13\/20) - (7\/20)<\/gadget>\n3\/10 = around 0.3<\/output>\n2_430 \/ (3\/10)<\/gadget>\n8_100<\/output>\n8_100<\/result>","index":2644} +{"problem":"if the average of 5 positive integers is 65 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 * 65 = 325 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 ) = 325 i . e . 5 a = 315 i . e . a = 63 i . e . largest e = 63 + 10 = 73 answer : option e\"","correct":"e","options":{"a":"50 ","b":"52 ","c":"59 ","d":"68","e":"73"},"options_float":{"a":50.0,"b":52.0,"c":59.0,"d":68.0,"e":73.0},"annotated_formula":"add(divide(subtract(multiply(65, 5), 10), 5), 10)","linear_formula":"multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)|","chain":"65 * 5<\/gadget>\n325<\/output>\n325 - 10<\/gadget>\n315<\/output>\n315 \/ 5<\/gadget>\n63<\/output>\n63 + 10<\/gadget>\n73<\/output>\n73<\/result>","index":2645} +{"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 20 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 \/ 20 work done by p , q and r in 1 day = 1 \/ 10 + 1 \/ 20 = 3 \/ 20 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 = 3 \/ 20 = > work done by p in 1 day = 3 \/ 40 = > work done by q and r in 1 day = 3 \/ 40 hence work done by q in 1 day = 3 \/ 40 â € “ 1 \/ 20 = 1 \/ 40 so q alone can do the work in 40 days answer is e .\"","correct":"e","options":{"a":"20 ","b":"22 ","c":"25 ","d":"27","e":"40"},"options_float":{"a":20.0,"b":22.0,"c":25.0,"d":27.0,"e":40.0},"annotated_formula":"divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 20)), const_2), divide(const_1, 20)))","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 \/ 20<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/10) + (1\/20)<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) \/ 2<\/gadget>\n3\/40 = around 0.075<\/output>\n(3\/40) - (1\/20)<\/gadget>\n1\/40 = around 0.025<\/output>\n1 \/ (1\/40)<\/gadget>\n40<\/output>\n40<\/result>","index":2646} +{"problem":"sale of rs 6835 , rs . 9927 , 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 , 6500 ?","rationale":"total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 39000 - 34009 ) = rs . 4966 answer : a","correct":"a","options":{"a":"4966 ","b":"2477 ","c":"2877 ","d":"2676","e":"1881"},"options_float":{"a":4966.0,"b":2477.0,"c":2877.0,"d":2676.0,"e":1881.0},"annotated_formula":"multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6500), 5)","linear_formula":"add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5)","chain":"6_835 + 9_927<\/gadget>\n16_762<\/output>\n16_762 + 6_855<\/gadget>\n23_617<\/output>\n23_617 + 7_230<\/gadget>\n30_847<\/output>\n30_847 + 6_562<\/gadget>\n37_409<\/output>\n37_409 \/ 5<\/gadget>\n37_409\/5 = around 7_481.8<\/output>\n(37_409\/5) - 6_500<\/gadget>\n4_909\/5 = around 981.8<\/output>\n(4_909\/5) * 5<\/gadget>\n4_909<\/output>\n4_909<\/result>","index":2647} +{"problem":"find the area of a parallelogram with base 15 cm and height 40 cm ?","rationale":"\"area of a parallelogram = base * height = 15 * 40 = 600 cm 2 answer : d\"","correct":"d","options":{"a":"200 cm 2 ","b":"100 cm 2 ","c":"42 cm 2 ","d":"600 cm 2","e":"230 cm 2"},"options_float":{"a":200.0,"b":100.0,"c":42.0,"d":600.0,"e":230.0},"annotated_formula":"multiply(15, 40)","linear_formula":"multiply(n0,n1)|","chain":"15 * 40<\/gadget>\n600<\/output>\n600<\/result>","index":2648} +{"problem":"a sum of money is distributed among a , b , c , d in the proportion of 1 : 3 : 4 : 2 . if c gets $ 500 more than d , what is the b ' s share ?","rationale":"let the shares of a , b , c , d are x , 3 x , 4 x , 2 x 4 x - 2 x = 500 x = 250 b ' s share = 3 x = $ 750 answer is c","correct":"c","options":{"a":"$ 450 ","b":"$ 500 ","c":"$ 750 ","d":"$ 800","e":"$ 840"},"options_float":{"a":450.0,"b":500.0,"c":750.0,"d":800.0,"e":840.0},"annotated_formula":"divide(multiply(divide(multiply(add(500, 500), 2), 4), 3), 2)","linear_formula":"add(n4,n4)|multiply(n3,#0)|divide(#1,n2)|multiply(n1,#2)|divide(#3,n3)","chain":"500 + 500<\/gadget>\n1_000<\/output>\n1_000 * 2<\/gadget>\n2_000<\/output>\n2_000 \/ 4<\/gadget>\n500<\/output>\n500 * 3<\/gadget>\n1_500<\/output>\n1_500 \/ 2<\/gadget>\n750<\/output>\n750<\/result>","index":2652} +{"problem":"the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 507 sq m , then what is the breadth of the rectangular plot ?","rationale":"\"let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 507 3 b 2 = 507 b 2 = 169 b = 13 m . answer : option c\"","correct":"c","options":{"a":"16 ","b":"17 ","c":"13 ","d":"19","e":"14"},"options_float":{"a":16.0,"b":17.0,"c":13.0,"d":19.0,"e":14.0},"annotated_formula":"sqrt(divide(507, const_3))","linear_formula":"divide(n0,const_3)|sqrt(#0)|","chain":"507 \/ 3<\/gadget>\n169<\/output>\n169 ** (1\/2)<\/gadget>\n13<\/output>\n13<\/result>","index":2653} +{"problem":"the number 341 is equal to the sum of the cubes of two integers . what is the product of those integers ?","rationale":"5 ^ 3 + 6 ^ 3 = 341 number is 5 * 6 = 30 d","correct":"d","options":{"a":"8 ","b":"15 ","c":"21 ","d":"30","e":"39"},"options_float":{"a":8.0,"b":15.0,"c":21.0,"d":30.0,"e":39.0},"annotated_formula":"multiply(floor(power(divide(341, const_2), divide(const_1, const_3))), power(subtract(341, power(floor(power(divide(341, 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":"341 \/ 2<\/gadget>\n341\/2 = around 170.5<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n(341\/2) ** (1\/3)<\/gadget>\n2**(2\/3)*341**(1\/3)\/2 = around 5.545084<\/output>\nfloor(2**(2\/3)*341**(1\/3)\/2)<\/gadget>\n5<\/output>\n5 ** 3<\/gadget>\n125<\/output>\n341 - 125<\/gadget>\n216<\/output>\n216 ** (1\/3)<\/gadget>\n6<\/output>\n5 * 6<\/gadget>\n30<\/output>\n30<\/result>","index":2654} +{"problem":"45 x ? = 25 % of 900","rationale":"\"answer let 45 x a = ( 25 x 900 ) \/ 100 ∴ a = ( 25 x 9 ) \/ 45 = 5 correct option : c\"","correct":"c","options":{"a":"16.2 ","b":"4 ","c":"5 ","d":"500","e":"none"},"options_float":{"a":16.2,"b":4.0,"c":5.0,"d":500.0,"e":null},"annotated_formula":"divide(multiply(divide(25, const_100), 900), 45)","linear_formula":"divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)|","chain":"25 \/ 100<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 900<\/gadget>\n225<\/output>\n225 \/ 45<\/gadget>\n5<\/output>\n5<\/result>","index":2656} +{"problem":"| x + 3 | – | 4 - x | = | 7 + x | how many solutions will this equation have ?","rationale":"you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , s ố x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , s ố í t becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 1 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":"divide(multiply(add(4, 3), const_2), 7)","linear_formula":"add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)","chain":"4 + 3<\/gadget>\n7<\/output>\n7 * 2<\/gadget>\n14<\/output>\n14 \/ 7<\/gadget>\n2<\/output>\n2<\/result>","index":2658} +{"problem":"a man is 24 years older than his son . in three years , his age will be twice the age of his son . the present age of the son is","rationale":"\"solution let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . then â € ¹ = â € º ( x + 24 ) + 3 = 2 ( x + 3 ) â € ¹ = â € º x + 27 = 2 x + 6 x = 21 . answer d\"","correct":"d","options":{"a":"14 years ","b":"18 years ","c":"20 years ","d":"21 years","e":"none"},"options_float":{"a":14.0,"b":18.0,"c":20.0,"d":21.0,"e":null},"annotated_formula":"divide(subtract(24, 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>\n24 - 2<\/gadget>\n22<\/output>\n2 - 1<\/gadget>\n1<\/output>\n22 \/ 1<\/gadget>\n22<\/output>\n22<\/result>","index":2659} +{"problem":"a man covers a certain distance q in a train . if the train moved 4 km \/ hr faster , it would take 30 min less . if it moved 2 km \/ hr slower , it would take 20 mins more . find the distance ?","rationale":"not really . when you solve the 2 equation above , you get , 6 t - 4 \/ 3 = 5 r \/ 6 from simplifying equation 1 4 t - 2 = r \/ 2 from simplifying equation 2 you can now multiply equation 2 by 5 to get 5 ( 4 t - 2 = r \/ 2 ) = 20 t - 10 = 5 r \/ 2 and then subtract this new equation from equation 1 to get t = 3 , followed by r = 20 to give you distance q = r * t = 20 * 3 = 60 km . d","correct":"d","options":{"a":"200 km ","b":"50 km ","c":"20 km ","d":"60 km","e":"80 km"},"options_float":{"a":200.0,"b":50.0,"c":20.0,"d":60.0,"e":80.0},"annotated_formula":"multiply(divide(subtract(multiply(4, 2), 4), const_2), 30)","linear_formula":"multiply(n0,n2)|subtract(#0,n0)|divide(#1,const_2)|multiply(n1,#2)","chain":"4 * 2<\/gadget>\n8<\/output>\n8 - 4<\/gadget>\n4<\/output>\n4 \/ 2<\/gadget>\n2<\/output>\n2 * 30<\/gadget>\n60<\/output>\n60<\/result>","index":2661} +{"problem":"if the a radio is sold for rs 490 and sold for rs 465.50 . find loss % .","rationale":"\"sol . cp = rs 490 , sp = 465.50 . loss = rs ( 490 - 465.50 ) = rs 24.50 . loss % = [ ( 24.50 \/ 490 ) * 100 ] % = 5 % answer is b .\"","correct":"b","options":{"a":"4 % ","b":"5 % ","c":"6 % ","d":"3 %","e":"5.5 %"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":3.0,"e":5.5},"annotated_formula":"multiply(divide(subtract(490, 465.50), 490), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|","chain":"490 - 465.5<\/gadget>\n24.5<\/output>\n24.5 \/ 490<\/gadget>\n0.05<\/output>\n0.05 * 100<\/gadget>\n5<\/output>\n5<\/result>","index":2662} +{"problem":"a cycle is bought for rs . 900 and sold for rs . 1160 , find the gain percent ?","rationale":"\"900 - - - - 260 100 - - - - ? = > 29 % answer : b\"","correct":"b","options":{"a":"11 ","b":"29 ","c":"99 ","d":"77","e":"18"},"options_float":{"a":11.0,"b":29.0,"c":99.0,"d":77.0,"e":18.0},"annotated_formula":"multiply(divide(subtract(1160, 900), 900), const_100)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|","chain":"1_160 - 900<\/gadget>\n260<\/output>\n260 \/ 900<\/gadget>\n13\/45 = around 0.288889<\/output>\n(13\/45) * 100<\/gadget>\n260\/9 = around 28.888889<\/output>\n260\/9 = around 28.888889<\/result>","index":2664} +{"problem":"the perimeter of a rectangular yard is completely surrounded by a fence that measures 12 meters . what is the length of the yard if the area of the yard is 9 meters squared ?","rationale":"perimeter of rectangular yard = 2 ( l + b ) = 12 - - > l + b = 6 area = l * b = 9 b = 6 - l l ( 6 - l ) = 9 6 l - l ^ 2 = 9 l ^ 2 - 6 l + 9 = 0 upon simplifying we get l = 3 . answer : b","correct":"b","options":{"a":"8 ","b":"1 ","c":"3 ","d":"4","e":"6"},"options_float":{"a":8.0,"b":1.0,"c":3.0,"d":4.0,"e":6.0},"annotated_formula":"subtract(const_4, const_3)","linear_formula":"subtract(const_4,const_3)","chain":"4 - 3<\/gadget>\n1<\/output>\n1<\/result>","index":2665} +{"problem":"x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 7 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 \/ 7 ^ 2 = 9 \/ 49 answer : e","correct":"e","options":{"a":"3 ","b":"6 ","c":"1 \/ 9 ","d":"1 \/ 3","e":"9 \/ 49"},"options_float":{"a":3.0,"b":6.0,"c":0.1111111111,"d":0.3333333333,"e":0.1836734694},"annotated_formula":"divide(multiply(1, power(3, const_2)), power(7, 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>\n7 ** 2<\/gadget>\n49<\/output>\n9 \/ 49<\/gadget>\n9\/49 = around 0.183673<\/output>\n9\/49 = around 0.183673<\/result>","index":2666} +{"problem":"a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 \/ 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 50 square feet of the wall , and no tiles overlap , what is the area of the entire wall ?","rationale":"\"the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 50 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 50 = 225 . total area = 50 + 225 = 275 . answer : b .\"","correct":"b","options":{"a":"160 ","b":"275 ","c":"360 ","d":"440","e":"560"},"options_float":{"a":160.0,"b":275.0,"c":360.0,"d":440.0,"e":560.0},"annotated_formula":"add(50, multiply(divide(multiply(50, 3), const_2), 3))","linear_formula":"multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)|","chain":"50 * 3<\/gadget>\n150<\/output>\n150 \/ 2<\/gadget>\n75<\/output>\n75 * 3<\/gadget>\n225<\/output>\n50 + 225<\/gadget>\n275<\/output>\n275<\/result>","index":2668} +{"problem":"a clothing store purchased a pair of pants for $ 90 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a clothing store owner decided to decrease the selling price by 20 percent . what was the clothing store ' s gross profit on this sale ?","rationale":"sale price ( sp ) = 90 + markup ( mp ) - - > mp = sp - 90 and given mp = sp \/ 4 ( 25 % is 1 \/ 4 th ) so sp \/ 4 = sp - 90 3 sp \/ 4 = 90 sp = 120 now a discount of 20 % is given so new sp is . 8 * 120 = 96 profit = 96 - 90 = 6.0 $ answer is d","correct":"d","options":{"a":"$ 14 ","b":"$ 5 ","c":"$ 10 ","d":"$ 6","e":"$ 8"},"options_float":{"a":14.0,"b":5.0,"c":10.0,"d":6.0,"e":8.0},"annotated_formula":"subtract(divide(multiply(subtract(const_100, 20), add(divide(90, const_3), 90)), const_100), 90)","linear_formula":"divide(n0,const_3)|subtract(const_100,n2)|add(n0,#0)|multiply(#2,#1)|divide(#3,const_100)|subtract(#4,n0)","chain":"100 - 20<\/gadget>\n80<\/output>\n90 \/ 3<\/gadget>\n30<\/output>\n30 + 90<\/gadget>\n120<\/output>\n80 * 120<\/gadget>\n9_600<\/output>\n9_600 \/ 100<\/gadget>\n96<\/output>\n96 - 90<\/gadget>\n6<\/output>\n6<\/result>","index":2669} +{"problem":"the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 6 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 6 ) = rs . 8915 . answer : option e\"","correct":"e","options":{"a":"4457 ","b":"4567 ","c":"4235 ","d":"4547","e":"8915"},"options_float":{"a":4457.0,"b":4567.0,"c":4235.0,"d":4547.0,"e":8915.0},"annotated_formula":"multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 6)","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)) * 6<\/gadget>\n5028.55844154167*sqrt(pi) = around 8_912.887774<\/output>\n5028.55844154167*sqrt(pi) = around 8_912.887774<\/result>","index":2671} +{"problem":"the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years q 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":"q = 20 ","b":"q = 25 ","c":"q = 30 ","d":"q = 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":2672} +{"problem":"a is twice as good a workman as b and they took 8 days together to do the work b alone can do it in .","rationale":"\"wc = 2 : 1 2 x + x = 1 \/ 8 x = 1 \/ 24 = > 24 days answer : a\"","correct":"a","options":{"a":"24 days ","b":"12 days ","c":"29 days ","d":"25 days","e":"27 days"},"options_float":{"a":24.0,"b":12.0,"c":29.0,"d":25.0,"e":27.0},"annotated_formula":"multiply(divide(multiply(8, 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>\n8 * 3<\/gadget>\n24<\/output>\n24 \/ 2<\/gadget>\n12<\/output>\n12 * 2<\/gadget>\n24<\/output>\n24<\/result>","index":2673} +{"problem":"how many unique positive odd integers less than 70 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 70 are odd multiples of 5 so we have 5,15 , 25,35 , 45,55 and 65 = 7 ans b\"","correct":"b","options":{"a":"4 ","b":"7 ","c":"11 ","d":"12","e":"15"},"options_float":{"a":4.0,"b":7.0,"c":11.0,"d":12.0,"e":15.0},"annotated_formula":"divide(divide(70, 5), const_2)","linear_formula":"divide(n0,n1)|divide(#0,const_2)|","chain":"70 \/ 5<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7<\/result>","index":2675} +{"problem":"find the value of 201834 x 99999 = m ?","rationale":"\"201834 x 99999 = 201834 x ( 100000 - 1 ) = 201834 x 100000 - 201834 x 1 = 20183400000 - 201834 = 20183198166 a\"","correct":"a","options":{"a":"20183198166 ","b":"20194181766 ","c":"20175292556 ","d":"20132191166","e":"20153198166"},"options_float":{"a":20183198166.0,"b":20194181766.0,"c":20175292556.0,"d":20132191166.0,"e":20153198166.0},"annotated_formula":"multiply(subtract(99999, const_4), 201834)","linear_formula":"subtract(n1,const_4)|multiply(#0,n0)|","chain":"99_999 - 4<\/gadget>\n99_995<\/output>\n99_995 * 201_834<\/gadget>\n20_182_390_830<\/output>\n20_182_390_830<\/result>","index":2676} +{"problem":"if a 5 percent deposit that has been paid toward the purchase of a certain product is $ 70 , how much more remains to be paid ?","rationale":"\"95 % remains to be paid so the remaining amount is 19 * 70 = $ 1330 . the answer is d .\"","correct":"d","options":{"a":"$ 1120 ","b":"$ 1190 ","c":"$ 1260 ","d":"$ 1330","e":"$ 1400"},"options_float":{"a":1120.0,"b":1190.0,"c":1260.0,"d":1330.0,"e":1400.0},"annotated_formula":"subtract(multiply(70, divide(const_100, 5)), 70)","linear_formula":"divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)|","chain":"100 \/ 5<\/gadget>\n20<\/output>\n70 * 20<\/gadget>\n1_400<\/output>\n1_400 - 70<\/gadget>\n1_330<\/output>\n1_330<\/result>","index":2677} +{"problem":"for all even integers n , h ( n ) is defined to be the sum of the even integers between 4 and n , inclusive . what is the value of h ( 18 ) \/ h ( 10 ) ?","rationale":"concept : when terms are in arithmetic progression ( a . p . ) i . e . terms are equally spaced then mean = median = ( first + last ) \/ 2 and sum = mean * number of terms h ( 18 ) = [ ( 4 + 18 ) \/ 2 ] * 8 = 88 h ( 10 ) = ( 4 + 10 ) \/ 2 ] * 4 = 28 h ( 18 ) \/ h ( 10 ) = ( 88 ) \/ ( 28 ) ~ 3 answer : a","correct":"a","options":{"a":"3 ","b":"1.8 ","c":"6 ","d":"18","e":"60"},"options_float":{"a":3.0,"b":1.8,"c":6.0,"d":18.0,"e":60.0},"annotated_formula":"divide(divide(multiply(add(18, 4), add(divide(subtract(18, 4), const_2), const_1)), const_2), divide(multiply(add(divide(subtract(10, 4), const_2), const_1), add(4, 10)), const_2))","linear_formula":"add(n0,n1)|add(n0,n2)|subtract(n1,n0)|subtract(n2,n0)|divide(#2,const_2)|divide(#3,const_2)|add(#4,const_1)|add(#5,const_1)|multiply(#0,#6)|multiply(#7,#1)|divide(#8,const_2)|divide(#9,const_2)|divide(#10,#11)","chain":"18 + 4<\/gadget>\n22<\/output>\n18 - 4<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n7 + 1<\/gadget>\n8<\/output>\n22 * 8<\/gadget>\n176<\/output>\n176 \/ 2<\/gadget>\n88<\/output>\n10 - 4<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n4 + 10<\/gadget>\n14<\/output>\n4 * 14<\/gadget>\n56<\/output>\n56 \/ 2<\/gadget>\n28<\/output>\n88 \/ 28<\/gadget>\n22\/7 = around 3.142857<\/output>\n22\/7 = around 3.142857<\/result>","index":2678} +{"problem":"sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 1.5 % more . she would get how much ?","rationale":"( 8000 * 3 * 1.5 ) \/ 100 = 360 9200 - - - - - - - - 9560 answer : a","correct":"a","options":{"a":"9560 ","b":"96288 ","c":"26667 ","d":"1662","e":"2882"},"options_float":{"a":9560.0,"b":96288.0,"c":26667.0,"d":1662.0,"e":2882.0},"annotated_formula":"add(multiply(multiply(add(divide(1.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000)","linear_formula":"divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)","chain":"1.5 \/ 100<\/gadget>\n0.015<\/output>\n9_200 - 8_000<\/gadget>\n1_200<\/output>\n1_200 \/ 3<\/gadget>\n400<\/output>\n400 \/ 8_000<\/gadget>\n1\/20 = around 0.05<\/output>\n0.015 + (1\/20)<\/gadget>\n0.065<\/output>\n0.065 * 8_000<\/gadget>\n520<\/output>\n520 * 3<\/gadget>\n1_560<\/output>\n1_560 + 8_000<\/gadget>\n9_560<\/output>\n9_560<\/result>","index":2681} +{"problem":"a horse is tethered to one corner of a rectangular grassy field 36 m by 20 m with a rope 12 m long . over how much area of the field can it graze ?","rationale":"\"area of the shaded portion = 1 ⁄ 4 × π × ( 12 ) 2 = 113 m 2 answer b\"","correct":"b","options":{"a":"154 cm 2 ","b":"113 m 2 ","c":"123 m 2 ","d":"115 m 2","e":"none of these"},"options_float":{"a":154.0,"b":113.0,"c":123.0,"d":115.0,"e":null},"annotated_formula":"divide(multiply(power(12, const_2), const_pi), const_4)","linear_formula":"power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)|","chain":"12 ** 2<\/gadget>\n144<\/output>\n144 * pi<\/gadget>\n144*pi = around 452.389342<\/output>\n(144*pi) \/ 4<\/gadget>\n36*pi = around 113.097336<\/output>\n36*pi = around 113.097336<\/result>","index":2682} +{"problem":"there are 6 people in the elevator . their average weight is 170 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 170 ) = 1057 - 1020 = 37 . answer a\"","correct":"a","options":{"a":"37 ","b":"168 ","c":"189 ","d":"190","e":"200"},"options_float":{"a":37.0,"b":168.0,"c":189.0,"d":190.0,"e":200.0},"annotated_formula":"subtract(multiply(151, 7), multiply(6, 170))","linear_formula":"multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|","chain":"151 * 7<\/gadget>\n1_057<\/output>\n6 * 170<\/gadget>\n1_020<\/output>\n1_057 - 1_020<\/gadget>\n37<\/output>\n37<\/result>","index":2683} +{"problem":"the mean of 50 observations is 100 . but later he found that there is decrements of 13 from each observations . what is the the updated mean is ?","rationale":"\"87 answer is a\"","correct":"a","options":{"a":"87 ","b":"97 ","c":"67 ","d":"57","e":"46"},"options_float":{"a":87.0,"b":97.0,"c":67.0,"d":57.0,"e":46.0},"annotated_formula":"subtract(100, 13)","linear_formula":"subtract(n1,n2)|","chain":"100 - 13<\/gadget>\n87<\/output>\n87<\/result>","index":2688} +{"problem":"a pupil ' s marks were wrongly entered as 83 instead of 63 . due to 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 = ( 83 - 63 ) = > x \/ 2 = 20 = > x = 40 answer : c\"","correct":"c","options":{"a":"13 ","b":"18 ","c":"40 ","d":"82","e":"43"},"options_float":{"a":13.0,"b":18.0,"c":40.0,"d":82.0,"e":43.0},"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":2690} +{"problem":"a trader sells 23 meters of cloth for rs . 529 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ?","rationale":"\"sp of 1 m of cloth = 529 \/ 23 = rs . 23 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 23 - rs . 5 = rs . 18 answer : e\"","correct":"e","options":{"a":"26 ","b":"88 ","c":"90 ","d":"42","e":"18"},"options_float":{"a":26.0,"b":88.0,"c":90.0,"d":42.0,"e":18.0},"annotated_formula":"subtract(divide(529, 23), 5)","linear_formula":"divide(n1,n0)|subtract(#0,n2)|","chain":"529 \/ 23<\/gadget>\n23<\/output>\n23 - 5<\/gadget>\n18<\/output>\n18<\/result>","index":2691} +{"problem":"if an article is sold at 18 % profit instead of 9 % profit , then the profit would be $ 54 more . what is the cost price ?","rationale":"9 % * cost price = $ 54 1 % * cost price = $ 54 \/ 9 = $ 6 the cost price is $ 600 . the answer is b .","correct":"b","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":"multiply(divide(54, 9), const_100)","linear_formula":"divide(n2,n1)|multiply(#0,const_100)","chain":"54 \/ 9<\/gadget>\n6<\/output>\n6 * 100<\/gadget>\n600<\/output>\n600<\/result>","index":2692} +{"problem":"if 8 cats can kill 8 rats in 8 minutes , how long will it take 100 cats to kill 100 rats ?","rationale":"it will take 8 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 8 minutes , so 100 cats can kill 100 rats in 8 minutes answer c","correct":"c","options":{"a":"6 minutes ","b":"7 minutes ","c":"8 minutes ","d":"9 minutes","e":"10 minutes"},"options_float":{"a":6.0,"b":7.0,"c":8.0,"d":9.0,"e":10.0},"annotated_formula":"multiply(8, const_1)","linear_formula":"multiply(n0,const_1)","chain":"8 * 1<\/gadget>\n8<\/output>\n8<\/result>","index":2693} +{"problem":"peter invested a certain sum of money in a simple interest bond whose value grew to $ 400 at the end of 3 years and to $ 600 at the end of another 2 years . what was the rate of interest in which he invested his sum ?","rationale":"\"lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of 3 years the principal p amounts to $ 400 and after a time of 5 years ( question says after another 5 years so 3 + 2 ) p becomes $ 600 . formulating the above data amount ( a 1 ) at end of 3 years a 1 = p ( 1 + 3 r \/ 100 ) = 400 amount ( a 2 ) at end of 8 years a 2 = p ( 1 + 5 r \/ 100 ) = 600 dividing a 2 by a 1 we get ( 1 + 5 r \/ 100 ) \/ ( 1 + 3 r \/ 100 ) = 6 \/ 8 after cross multiplication we are left with r = 100 option : a\"","correct":"a","options":{"a":"100 % ","b":"12.5 % ","c":"67 % ","d":"25 %","e":"33 %"},"options_float":{"a":100.0,"b":12.5,"c":67.0,"d":25.0,"e":33.0},"annotated_formula":"multiply(divide(divide(subtract(600, 400), 2), subtract(400, multiply(divide(subtract(600, 400), 2), 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":"600 - 400<\/gadget>\n200<\/output>\n200 \/ 2<\/gadget>\n100<\/output>\n100 * 3<\/gadget>\n300<\/output>\n400 - 300<\/gadget>\n100<\/output>\n100 \/ 100<\/gadget>\n1<\/output>\n1 * 100<\/gadget>\n100<\/output>\n100<\/result>","index":2694} +{"problem":"3 \/ 4 of 1 \/ 2 of 2 \/ 5 of 5080 = ?","rationale":"\"e 762 ? = 5080 * ( 2 \/ 5 ) * ( 1 \/ 2 ) * ( 3 \/ 4 ) = 762\"","correct":"e","options":{"a":"392 ","b":"229 ","c":"753 ","d":"493","e":"762"},"options_float":{"a":392.0,"b":229.0,"c":753.0,"d":493.0,"e":762.0},"annotated_formula":"multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5080)","linear_formula":"divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)|","chain":"3 \/ 4<\/gadget>\n3\/4 = around 0.75<\/output>\n1 \/ 2<\/gadget>\n1\/2 = around 0.5<\/output>\n(3\/4) * (1\/2)<\/gadget>\n3\/8 = around 0.375<\/output>\n2 \/ 5<\/gadget>\n2\/5 = around 0.4<\/output>\n(3\/8) * (2\/5)<\/gadget>\n3\/20 = around 0.15<\/output>\n(3\/20) * 5_080<\/gadget>\n762<\/output>\n762<\/result>","index":2695} +{"problem":"√ 4 percent of 4 √ 4 =","rationale":"√ 4 = 2 so , √ 4 percent of 4 √ 4 = 2 percent of ( 4 ) ( 2 ) = ( 2 \/ 100 ) ( 8 ) = 16 \/ 100 = 0.16 answer : a","correct":"a","options":{"a":"0.16 ","b":"0.17 ","c":"0.18 ","d":"0.2","e":"0.19"},"options_float":{"a":0.16,"b":0.17,"c":0.18,"d":0.2,"e":0.19},"annotated_formula":"divide(multiply(multiply(sqrt(4), sqrt(4)), 4), const_100)","linear_formula":"sqrt(n0)|multiply(#0,#0)|multiply(n0,#1)|divide(#2,const_100)","chain":"4 ** (1\/2)<\/gadget>\n2<\/output>\n2 * 2<\/gadget>\n4<\/output>\n4 * 4<\/gadget>\n16<\/output>\n16 \/ 100<\/gadget>\n4\/25 = around 0.16<\/output>\n4\/25 = around 0.16<\/result>","index":2697} +{"problem":"12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 12 hours a day , the number of men required ?","rationale":"\"that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 12 × 8 ? ( i . e . no . of men required ) = 12 × 8 × 10 \/ 12 × 8 = 8 days e )\"","correct":"e","options":{"a":"11 days ","b":"12 days ","c":"9 days ","d":"8 days","e":"10 days"},"options_float":{"a":11.0,"b":12.0,"c":9.0,"d":8.0,"e":10.0},"annotated_formula":"divide(multiply(multiply(12, 10), 8), multiply(8, 12))","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 * 12<\/gadget>\n96<\/output>\n960 \/ 96<\/gadget>\n10<\/output>\n10<\/result>","index":2699} +{"problem":"8 people decided to split the restaurant bill evenly . if the bill was $ 214.15 dollars , how much money did they 1 cent is the smallest unit ?","rationale":"\"if the last three digits of a whole number are divisible by 8 , then the entire number is divisible by 8 the last 3 digit 415 not divisible by a hence , we need to add 1 to this number for it to be divisible by 8 correct option : a\"","correct":"a","options":{"a":"$ 214.16 ","b":"$ 214.17 ","c":"$ 214.18 ","d":"$ 214.19","e":"$ 214.20"},"options_float":{"a":214.16,"b":214.17,"c":214.18,"d":214.19,"e":214.2},"annotated_formula":"add(214.15, divide(const_3, const_100))","linear_formula":"divide(const_3,const_100)|add(n1,#0)|","chain":"3 \/ 100<\/gadget>\n3\/100 = around 0.03<\/output>\n214.15 + (3\/100)<\/gadget>\n214.18<\/output>\n214.18<\/result>","index":2700} +{"problem":"a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what y percent of the list price is the lowest possible sale price ?","rationale":"\"let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which y is 40 % of 80 hence , d is the answer\"","correct":"d","options":{"a":"20 ","b":"25 ","c":"30 ","d":"40","e":"50"},"options_float":{"a":20.0,"b":25.0,"c":30.0,"d":40.0,"e":50.0},"annotated_formula":"divide(80, const_2)","linear_formula":"divide(n3,const_2)|","chain":"80 \/ 2<\/gadget>\n40<\/output>\n40<\/result>","index":2702} +{"problem":"each year for 4 years , a farmer increased the number of trees in a certain orchard by 1 \/ 4 of the number of trees in the orchard of the preceding year . if all of the trees thrived and there were 12500 trees in the orchard at the end of 4 year period , how many trees were in the orchard at the beginning of the 4 year period .","rationale":"\"trees increase by 1 \/ 4 the number of trees in preceding year . hence , correct answer must be divisible by 4 . based on divisibility rules , if last 2 digits are divisible by 4 then the number is divisible by 4 . thus , we can eliminate a , b , d , e the answer to be c again , trees increase by 1 \/ 4 the number of trees in preceding year . hence , the number of trees increase by 5 \/ 4 times the number of trees the preceding year . if x = initial number of trees = 5120 year 1 = 5 \/ 4 x year 2 = ( 5 \/ 4 ) ( 5 \/ 4 ) x year 3 = ( 5 \/ 4 ) ( 5 \/ 4 ) ( 5 \/ 4 ) x year 4 = ( 5 \/ 4 ) ( 5 \/ 4 ) ( 5 \/ 4 ) ( 5 \/ 4 ) x only for answer d : ( 5 \/ 4 ) ( 5 \/ 4 ) ( 5 \/ 4 ) ( 5 \/ 4 ) 5120 = 12500 hence , correct answer = c\"","correct":"c","options":{"a":"5113 ","b":"5117 ","c":"5120 ","d":"8119","e":"10115"},"options_float":{"a":5113.0,"b":5117.0,"c":5120.0,"d":8119.0,"e":10115.0},"annotated_formula":"divide(12500, power(add(divide(1, 4), 1), 4))","linear_formula":"divide(n1,n0)|add(n1,#0)|power(#1,n0)|divide(n3,#2)|","chain":"1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) + 1<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) ** 4<\/gadget>\n625\/256 = around 2.441406<\/output>\n12_500 \/ (625\/256)<\/gadget>\n5_120<\/output>\n5_120<\/result>","index":2703} +{"problem":"a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000.00001 more . at the end of the year , their profits amounted to rs . 714 find the share of a .","rationale":"explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 \/ 21 * 714 = 272 answer : b","correct":"b","options":{"a":"240 ","b":"272 ","c":"379 ","d":"277","e":"122"},"options_float":{"a":240.0,"b":272.0,"c":379.0,"d":277.0,"e":122.0},"annotated_formula":"multiply(divide(714, 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>\n714 \/ 84_000<\/gadget>\n17\/2_000 = around 0.0085<\/output>\n(17\/2_000) * 32_000<\/gadget>\n272<\/output>\n272<\/result>","index":2704} +{"problem":"a train 120 m long running at 60 kmph crosses a platform in 35 sec . what is the length of the platform ?","rationale":"\"d = 60 * 5 \/ 18 = 35 = 583 – 120 = 463 answer : c\"","correct":"c","options":{"a":"338 ","b":"277 ","c":"463 ","d":"456","e":"271"},"options_float":{"a":338.0,"b":277.0,"c":463.0,"d":456.0,"e":271.0},"annotated_formula":"subtract(multiply(35, multiply(60, const_0_2778)), 120)","linear_formula":"multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n60 * (5\/18)<\/gadget>\n50\/3 = around 16.666667<\/output>\n35 * (50\/3)<\/gadget>\n1_750\/3 = around 583.333333<\/output>\n(1_750\/3) - 120<\/gadget>\n1_390\/3 = around 463.333333<\/output>\n1_390\/3 = around 463.333333<\/result>","index":2705} +{"problem":"jar a has 6 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ?","rationale":"\"an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 106 marbles . now , for both jars to have equal marbles we should move 3 marbles from a to b , which is 3 \/ 106 = ~ 2.8 % of a . answer : a .\"","correct":"a","options":{"a":"2.8 % ","b":"3.0 % ","c":"3.2 % ","d":"3.4 %","e":"3.6 %"},"options_float":{"a":2.8,"b":3.0,"c":3.2,"d":3.4,"e":3.6},"annotated_formula":"multiply(divide(divide(6, const_2), add(6, const_100)), const_100)","linear_formula":"add(n0,const_100)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_100)|","chain":"6 \/ 2<\/gadget>\n3<\/output>\n6 + 100<\/gadget>\n106<\/output>\n3 \/ 106<\/gadget>\n3\/106 = around 0.028302<\/output>\n(3\/106) * 100<\/gadget>\n150\/53 = around 2.830189<\/output>\n150\/53 = around 2.830189<\/result>","index":2706} +{"problem":"what number has a 5 : 1 ratio to the number 10 ?","rationale":"\"answer : option b explanation : 5 : 1 = x : 10 x = 50 answer : option b\"","correct":"b","options":{"a":"74 ","b":"50 ","c":"94 ","d":"59","e":"48"},"options_float":{"a":74.0,"b":50.0,"c":94.0,"d":59.0,"e":48.0},"annotated_formula":"multiply(10, 5)","linear_formula":"multiply(n0,n2)|","chain":"10 * 5<\/gadget>\n50<\/output>\n50<\/result>","index":2710} +{"problem":"a , b and c can do a piece of work in 7 days , 14 days and 28 days respectively . how long will they take to finish the work , if all the three work together ?","rationale":"\"1 \/ 7 + 1 \/ 14 + 1 \/ 28 = 7 \/ 28 = 1 \/ 4 all three can finish the work in 4 days answer : a\"","correct":"a","options":{"a":"4 ","b":"9 ","c":"2 ","d":"11","e":"none"},"options_float":{"a":4.0,"b":9.0,"c":2.0,"d":11.0,"e":null},"annotated_formula":"inverse(add(inverse(28), add(inverse(7), inverse(14))))","linear_formula":"inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)|","chain":"1 \/ 28<\/gadget>\n1\/28 = around 0.035714<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1 \/ 14<\/gadget>\n1\/14 = around 0.071429<\/output>\n(1\/7) + (1\/14)<\/gadget>\n3\/14 = around 0.214286<\/output>\n(1\/28) + (3\/14)<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ (1\/4)<\/gadget>\n4<\/output>\n4<\/result>","index":2713} +{"problem":"in the youth summer village there are 200 people , 100 of them are not working , 75 of them have families and 125 of them like to sing in the shower . what is the largest possible number of people in the village , which are working , that do n ' t have families and that are singing in the shower ?","rationale":"total = 200 not working = 100 having family = 75 like to sing in shower = 125 working = 200 - 100 = 100 not having family = 200 - 75 = 125 like to sing in shower = 125 largest possible number is the lowest possible among the above thus 100 c","correct":"c","options":{"a":"125 ","b":"150 ","c":"100 ","d":"130","e":"140"},"options_float":{"a":125.0,"b":150.0,"c":100.0,"d":130.0,"e":140.0},"annotated_formula":"subtract(add(add(100, 75), 125), 200)","linear_formula":"add(n1,n2)|add(n3,#0)|subtract(#1,n0)","chain":"100 + 75<\/gadget>\n175<\/output>\n175 + 125<\/gadget>\n300<\/output>\n300 - 200<\/gadget>\n100<\/output>\n100<\/result>","index":2714} +{"problem":"nitin ranks 18 th in a class of 49 students . what is rank from the last ?","rationale":"\"explanation : number students behind the nitin in rank = ( 49 - 18 ) = 31 nitin is 32 nd from the last answer : c ) 32\"","correct":"c","options":{"a":"33 ","b":"38 ","c":"32 ","d":"28","e":"19"},"options_float":{"a":33.0,"b":38.0,"c":32.0,"d":28.0,"e":19.0},"annotated_formula":"subtract(49, 18)","linear_formula":"subtract(n1,n0)|","chain":"49 - 18<\/gadget>\n31<\/output>\n31<\/result>","index":2715} +{"problem":"lamp a flashes every 6 seconds , lamp b flashes every 8 seconds , lamp c flashes every 10 seconds . at a certain instant of time all 3 lamps flash simultaneously . during the period of 6 minutes after that how many times will exactly two lamps flash ? ( please include any flash of exactly two lights which occurs at the 6 minute mark . )","rationale":"6 minutes is 360 seconds . lamp a and lamp b will flash together every 24 seconds . 360 \/ 24 = 15 . in the time period , lamp a and lamp b will flash together 15 times . lamp a and lamp c will flash together every 30 seconds . 360 \/ 30 = 12 . in the time period , lamp a and lamp c will flash together 12 times . lamp b and lamp c will flash together every 40 seconds . 360 \/ 40 = 9 . in the time period , lamp b and lamp c will flash together 9 times . all three lights will flash together every 2 * 2 * 2 * 3 * 5 = 120 seconds . 360 \/ 120 = 3 . we have counted these triple flashes three times , so we need to subtract three times the number of times that all three lights flash together . the number of times that exactly two lights flash together is 15 + 12 + 9 - 9 = 27 times . the answer is d .","correct":"d","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":"subtract(add(add(divide(multiply(6, const_60), lcm(6, 8)), divide(multiply(6, const_60), lcm(6, 10))), divide(multiply(6, const_60), lcm(8, 10))), multiply(divide(multiply(6, const_60), lcm(lcm(6, 8), 10)), 3))","linear_formula":"lcm(n0,n1)|lcm(n0,n2)|lcm(n1,n2)|multiply(n0,const_60)|divide(#3,#0)|divide(#3,#1)|divide(#3,#2)|lcm(n2,#0)|add(#4,#5)|divide(#3,#7)|add(#8,#6)|multiply(n3,#9)|subtract(#10,#11)","chain":"6 * 60<\/gadget>\n360<\/output>\nlcm(6, 8)<\/gadget>\n24<\/output>\n360 \/ 24<\/gadget>\n15<\/output>\nlcm(6, 10)<\/gadget>\n30<\/output>\n360 \/ 30<\/gadget>\n12<\/output>\n15 + 12<\/gadget>\n27<\/output>\nlcm(8, 10)<\/gadget>\n40<\/output>\n360 \/ 40<\/gadget>\n9<\/output>\n27 + 9<\/gadget>\n36<\/output>\nlcm(24, 10)<\/gadget>\n120<\/output>\n360 \/ 120<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n36 - 9<\/gadget>\n27<\/output>\n27<\/result>","index":2716} +{"problem":"a man walking at a rate of 10 km \/ hr crosses a bridge in 12 minutes . the length of the bridge is ?","rationale":"speed = 10 * 5 \/ 18 = 50 \/ 18 m \/ sec distance covered in 10 minutes = 50 \/ 18 * 12 * 60 = 2000 m answer is a","correct":"a","options":{"a":"2000 ","b":"1492 ","c":"1667 ","d":"1254","e":"1112"},"options_float":{"a":2000.0,"b":1492.0,"c":1667.0,"d":1254.0,"e":1112.0},"annotated_formula":"multiply(divide(multiply(10, const_1000), const_60), 12)","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) * 12<\/gadget>\n2_000<\/output>\n2_000<\/result>","index":2718} +{"problem":"there are 1000 students in a school and among them 30 % of them attends chess class . 10 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ?","rationale":"30 % of 1000 gives 300 . so 300 attends chess and 10 % of 300 gives 30 . so 30 enrolled for swimming answer : d","correct":"d","options":{"a":"1 ","b":"10 ","c":"100 ","d":"30","e":"20"},"options_float":{"a":1.0,"b":10.0,"c":100.0,"d":30.0,"e":20.0},"annotated_formula":"divide(multiply(divide(multiply(30, 1000), const_100), 10), const_100)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)","chain":"30 * 1_000<\/gadget>\n30_000<\/output>\n30_000 \/ 100<\/gadget>\n300<\/output>\n300 * 10<\/gadget>\n3_000<\/output>\n3_000 \/ 100<\/gadget>\n30<\/output>\n30<\/result>","index":2719} +{"problem":"exactly 15 % of the reporters for a certain wire service cover local politics in country x . if 25 % 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 - - > 15 reporters cover local politics . now , as 25 % of the reporters who cover all politics do not cover local politics then the rest 75 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 75 % of them equal to 15 ( # of reporters who cover local politics ) : 0.75 x = 15 - - > x = 20 , hence 20 reporters cover politics and the rest 100 - 20 = 80 reporters do not cover politics at all . answer : d .\"","correct":"d","options":{"a":"20 % ","b":"42 % ","c":"44 % ","d":"80 %","e":"84 %"},"options_float":{"a":20.0,"b":42.0,"c":44.0,"d":80.0,"e":84.0},"annotated_formula":"multiply(subtract(const_1, divide(15, subtract(const_100, 25))), const_100)","linear_formula":"subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)|","chain":"100 - 25<\/gadget>\n75<\/output>\n15 \/ 75<\/gadget>\n1\/5 = around 0.2<\/output>\n1 - (1\/5)<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 100<\/gadget>\n80<\/output>\n80<\/result>","index":2721} +{"problem":"of 70 players on a football team , 43 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ?","rationale":"\"total = 70 thrower = 43 rest = 70 - 43 = 27 left handed = 27 \/ 3 = 9 right handed = 18 if all thrower are right handed then total right handed is 43 + 18 = 61 so c . 61 is the right answer\"","correct":"c","options":{"a":"54 ","b":"59 ","c":"61 ","d":"71","e":"92"},"options_float":{"a":54.0,"b":59.0,"c":61.0,"d":71.0,"e":92.0},"annotated_formula":"add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 43)), 43)","linear_formula":"divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 - (1\/3)<\/gadget>\n2\/3 = around 0.666667<\/output>\n70 - 43<\/gadget>\n27<\/output>\n(2\/3) * 27<\/gadget>\n18<\/output>\n18 + 43<\/gadget>\n61<\/output>\n61<\/result>","index":2723} +{"problem":"in how many seconds will a train 100 meters long pass an oak tree , if the speed of the train is 36 km \/ hr ?","rationale":"\"speed = 36 * 5 \/ 18 = 10 m \/ s time = 100 \/ 10 = 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(100, multiply(const_0_2778, 36))","linear_formula":"multiply(n1,const_0_2778)|divide(n0,#0)|","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(5\/18) * 36<\/gadget>\n10<\/output>\n100 \/ 10<\/gadget>\n10<\/output>\n10<\/result>","index":2724} +{"problem":"the bus fare for two persons for travelling between agra and aligarh id 4 - thirds the train fare between the same places for one person . the total fare paid by 6 persons travelling by bus and 8 persons travelling by train between the two places is rs . 1512 . find the train fare between the two places for one person ?","rationale":"let the train fare between the two places for one person be rs . t bus fare between the two places for two persons rs . 4 \/ 3 t = > 6 \/ 2 ( 4 \/ 3 t ) + 8 ( t ) = 1512 = > 12 t = 1512 = > t = 126 . answer : a","correct":"a","options":{"a":"rs . 126 ","b":"rs . 132 ","c":"rs . 120 ","d":"rs . 114","e":"none of these"},"options_float":{"a":126.0,"b":132.0,"c":120.0,"d":114.0,"e":null},"annotated_formula":"divide(1512, add(multiply(divide(6, const_2), divide(4, const_3)), 8))","linear_formula":"divide(n1,const_2)|divide(n0,const_3)|multiply(#0,#1)|add(n2,#2)|divide(n3,#3)","chain":"6 \/ 2<\/gadget>\n3<\/output>\n4 \/ 3<\/gadget>\n4\/3 = around 1.333333<\/output>\n3 * (4\/3)<\/gadget>\n4<\/output>\n4 + 8<\/gadget>\n12<\/output>\n1_512 \/ 12<\/gadget>\n126<\/output>\n126<\/result>","index":2727} +{"problem":"a goods bullet train runs at the speed of 72 km \/ hr and crosses a 250 m long platform in 26 seconds . what is the length of the goods bullet train ?","rationale":"e 270 m","correct":"e","options":{"a":"220 m ","b":"250 m ","c":"280 m ","d":"210 m","e":"270 m"},"options_float":{"a":220.0,"b":250.0,"c":280.0,"d":210.0,"e":270.0},"annotated_formula":"subtract(multiply(multiply(72, const_0_2778), 26), 250)","linear_formula":"multiply(n0,const_0_2778)|multiply(n2,#0)|subtract(#1,n1)","chain":"10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n72 * (5\/18)<\/gadget>\n20<\/output>\n20 * 26<\/gadget>\n520<\/output>\n520 - 250<\/gadget>\n270<\/output>\n270<\/result>","index":2729} +{"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 10 % 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 - 10 % of 140 = rs . 140 - 14 = rs . 126 . gain = rs . 126 â € “ rs . 100 = rs . 26 therefore , gain \/ profit percent = 26 % . answer : option a\"","correct":"a","options":{"a":"26 % ","b":"20 % ","c":"17 % ","d":"18 %","e":"none of these"},"options_float":{"a":26.0,"b":20.0,"c":17.0,"d":18.0,"e":null},"annotated_formula":"subtract(subtract(add(const_100, 40), multiply(add(const_100, 40), divide(10, 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>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n140 * (1\/10)<\/gadget>\n14<\/output>\n140 - 14<\/gadget>\n126<\/output>\n126 - 100<\/gadget>\n26<\/output>\n26<\/result>","index":2730} +{"problem":"evaluate : | 6 - 8 ( 3 - 12 ) | - | 5 - 11 | = ?","rationale":"\"according to order of operations , inner brackets first . hence | 6 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 6 - 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 = | 6 + 72 | - | 5 - 11 | = | 78 | - | - 6 | = 78 - 6 = 72 correct answer c ) 72\"","correct":"c","options":{"a":"40 ","b":"50 ","c":"72 ","d":"70","e":"80"},"options_float":{"a":40.0,"b":50.0,"c":72.0,"d":70.0,"e":80.0},"annotated_formula":"subtract(subtract(6, 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>\n6 - (-72)<\/gadget>\n78<\/output>\n5 - 11<\/gadget>\n-6<\/output>\n-(-6)<\/gadget>\n6<\/output>\n78 - 6<\/gadget>\n72<\/output>\n72<\/result>","index":2732} +{"problem":"on a trip , a cyclist averaged 11 miles per hour for the first 22 miles and 10 miles per hour for the remaining 20 miles . if the cyclist returned immediately via the same route and took a total of 9 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ?","rationale":"the time to go 42 miles was 22 \/ 11 + 20 \/ 10 = 2 + 2 = 4 hours . the average speed for the return trip was 42 miles \/ 5 hours = 8.4 mph . the answer is e .","correct":"e","options":{"a":"7.6 ","b":"7.8 ","c":"8 ","d":"8.2","e":"8.4"},"options_float":{"a":7.6,"b":7.8,"c":8.0,"d":8.2,"e":8.4},"annotated_formula":"divide(add(22, 20), subtract(9, add(divide(22, 11), divide(20, 10))))","linear_formula":"add(n1,n3)|divide(n1,n0)|divide(n3,n2)|add(#1,#2)|subtract(n4,#3)|divide(#0,#4)","chain":"22 + 20<\/gadget>\n42<\/output>\n22 \/ 11<\/gadget>\n2<\/output>\n20 \/ 10<\/gadget>\n2<\/output>\n2 + 2<\/gadget>\n4<\/output>\n9 - 4<\/gadget>\n5<\/output>\n42 \/ 5<\/gadget>\n42\/5 = around 8.4<\/output>\n42\/5 = around 8.4<\/result>","index":2735} +{"problem":"on increasing the number of lines in a page by 100 , they become 240 . what is the % of increase in the no . of lines in the page ?","rationale":"\"explanation : number of pages increased = 100 now , the number of pages of book = 240 number of pages of the books before increase = 240 – 100 = 140 % increase in the number of pages in the book = 100 \/ 140 x 100 % = 71.4 % d\"","correct":"d","options":{"a":"20 % ","b":"305 ","c":"50 % ","d":"71.4 %","e":"60 %"},"options_float":{"a":20.0,"b":305.0,"c":50.0,"d":71.4,"e":60.0},"annotated_formula":"subtract(multiply(divide(240, subtract(240, 100)), const_100), const_100)","linear_formula":"subtract(n1,n0)|divide(n1,#0)|multiply(#1,const_100)|subtract(#2,const_100)|","chain":"240 - 100<\/gadget>\n140<\/output>\n240 \/ 140<\/gadget>\n12\/7 = around 1.714286<\/output>\n(12\/7) * 100<\/gadget>\n1_200\/7 = around 171.428571<\/output>\n(1_200\/7) - 100<\/gadget>\n500\/7 = around 71.428571<\/output>\n500\/7 = around 71.428571<\/result>","index":2736} +{"problem":"find the sum 3 \/ 10 + 5 \/ 100 + 8 \/ 1000 in decimal form ?","rationale":"answer 3 \/ 10 + 5 \/ 100 + 8 \/ 1000 = 0.3 + 0.05 + 0.008 = 0.358 correct option : b","correct":"b","options":{"a":"0.853 ","b":"0.358 ","c":"3.58 ","d":"8.35","e":"none"},"options_float":{"a":0.853,"b":0.358,"c":3.58,"d":8.35,"e":null},"annotated_formula":"add(divide(8, 1000), add(divide(3, 10), divide(5, 100)))","linear_formula":"divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|add(#3,#2)","chain":"8 \/ 1_000<\/gadget>\n1\/125 = around 0.008<\/output>\n3 \/ 10<\/gadget>\n3\/10 = around 0.3<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(3\/10) + (1\/20)<\/gadget>\n7\/20 = around 0.35<\/output>\n(1\/125) + (7\/20)<\/gadget>\n179\/500 = around 0.358<\/output>\n179\/500 = around 0.358<\/result>","index":2737} +{"problem":"3 distinct single digit no a , b , c are in g . p . if abs ( x ) for real x is the absolute value of x ( x if x is positive or zero , and x if x is negative ) , then the no . of different possible values of abs ( a + b - c ) is","rationale":"if a = 1 , r = 2 then a = 1 , b = 2 , c = 4 then abs ( a + b - c ) = 1 if a = 1 , r = 3 then a = 1 , a = 3 , a = 9 then abs ( 1 + 3 - 9 ) = 5 if a = 2 , r = 2 , then a = 2 , b = 4 , c = 8 then abs ( 2 + 4 - 8 ) = 2 if a = 1 , r = - 2 then a = 1 , b = - 2 , c = 4 the abs ( 1 - 2 - 4 ) = 5 if a = 1 , r = - 3 then a = 1 , b = - 3 , c = 9 then abs ( 1 - 3 - 9 ) = 11 if a = 2 , r = - 2 then a = 2 , b = - 4 , c = - 8 then abs ( 2 - 4 - 8 ) = 10 so total 5 abs ( ) values answer : d","correct":"d","options":{"a":"6 ","b":"4 ","c":"3 ","d":"5","e":"2"},"options_float":{"a":6.0,"b":4.0,"c":3.0,"d":5.0,"e":2.0},"annotated_formula":"add(3, const_2)","linear_formula":"add(n0,const_2)","chain":"3 + 2<\/gadget>\n5<\/output>\n5<\/result>","index":2738} +{"problem":"sum of 19 odd numbers is ?","rationale":"\"sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 19 odd numbers = 19 ^ 2 = 361 answer : c\"","correct":"c","options":{"a":"341 ","b":"351 ","c":"361 ","d":"371","e":"381"},"options_float":{"a":341.0,"b":351.0,"c":361.0,"d":371.0,"e":381.0},"annotated_formula":"multiply(multiply(19, const_2), divide(19, const_2))","linear_formula":"divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1)|","chain":"19 * 2<\/gadget>\n38<\/output>\n19 \/ 2<\/gadget>\n19\/2 = around 9.5<\/output>\n38 * (19\/2)<\/gadget>\n361<\/output>\n361<\/result>","index":2739} +{"problem":"an engine moves at the speed of 90 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 78 kmph . what will be the speed of train when 25 coaches are attached .","rationale":"\"1 . no . of coaches = 9 sqr root = 3 speed decreases by 12 12 = k * 3 k = 4 no . of coaches = 25 swr root = 5 decrease = 5 * 4 = 20 new speed = 90 - 20 = 70 e\"","correct":"e","options":{"a":"90 ","b":"85 ","c":"80 ","d":"60","e":"70"},"options_float":{"a":90.0,"b":85.0,"c":80.0,"d":60.0,"e":70.0},"annotated_formula":"subtract(90, multiply(sqrt(25), divide(subtract(90, 78), sqrt(9))))","linear_formula":"sqrt(n1)|sqrt(n3)|subtract(n0,n2)|divide(#2,#0)|multiply(#3,#1)|subtract(n0,#4)|","chain":"25 ** (1\/2)<\/gadget>\n5<\/output>\n90 - 78<\/gadget>\n12<\/output>\n9 ** (1\/2)<\/gadget>\n3<\/output>\n12 \/ 3<\/gadget>\n4<\/output>\n5 * 4<\/gadget>\n20<\/output>\n90 - 20<\/gadget>\n70<\/output>\n70<\/result>","index":2740} +{"problem":"a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 264 metres long in 20 seconds . the length of the train ( in metres ) is :","rationale":"\"explanation : let the length of train be l m . acc . to question ( 264 + l ) \/ 20 = l \/ 8 2112 + 8 l = 20 l l = 2112 \/ 12 = 176 m answer b\"","correct":"b","options":{"a":"188 ","b":"176 ","c":"175 ","d":"96","e":"none of these"},"options_float":{"a":188.0,"b":176.0,"c":175.0,"d":96.0,"e":null},"annotated_formula":"multiply(divide(264, subtract(20, 8)), 8)","linear_formula":"subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)|","chain":"20 - 8<\/gadget>\n12<\/output>\n264 \/ 12<\/gadget>\n22<\/output>\n22 * 8<\/gadget>\n176<\/output>\n176<\/result>","index":2741} +{"problem":"a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 10 miles per gallon , and from town b to town c , the car averaged 12 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ?","rationale":"\"step 1 ) took lcm of 10 and 12 . . came as 30 . just multiplied by 10 . . . ( to make easy calculation ) step 2 ) 300 distance between b to c . . . do 300 \/ 12 hence 25 gallons used step 3 ) twice distance . . hence 300 * 2 = 600 . . . do as above . . 600 \/ 10 = 60 gallons used step 4 ) total gallons . . 25 + 60 = 85 gallons step ) total miles = 300 + 600 = 900 miles hence . . average of whole journey = 900 \/ 85 which comes to 10.6 answer : d\"","correct":"d","options":{"a":"11.5 ","b":"9.5 ","c":"13.5 ","d":"10.6","e":"14.5"},"options_float":{"a":11.5,"b":9.5,"c":13.5,"d":10.6,"e":14.5},"annotated_formula":"divide(add(multiply(12, const_10), divide(multiply(12, const_10), const_2)), add(divide(multiply(12, const_10), 10), divide(divide(multiply(12, const_10), const_2), 12)))","linear_formula":"multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)|","chain":"12 * 10<\/gadget>\n120<\/output>\n120 \/ 2<\/gadget>\n60<\/output>\n120 + 60<\/gadget>\n180<\/output>\n120 \/ 10<\/gadget>\n12<\/output>\n60 \/ 12<\/gadget>\n5<\/output>\n12 + 5<\/gadget>\n17<\/output>\n180 \/ 17<\/gadget>\n180\/17 = around 10.588235<\/output>\n180\/17 = around 10.588235<\/result>","index":2742} +{"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 16 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 = 16 . . 4 x + 12 = 16 . . x = 1 . . ans = x + 4 = 1 + 4 = 5 answer 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(divide(subtract(16, 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>\n16 - 12<\/gadget>\n4<\/output>\n4 \/ 4<\/gadget>\n1<\/output>\n1 + 4<\/gadget>\n5<\/output>\n5<\/result>","index":2745} +{"problem":"a train 450 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":"\"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 450 m at 50 \/ 3 m \/ sec = 450 * 3 \/ 50 sec = 27 sec answer : c .\"","correct":"c","options":{"a":"5 sec ","b":"39 sec ","c":"27 sec ","d":"15 sec","e":"18 sec"},"options_float":{"a":5.0,"b":39.0,"c":27.0,"d":15.0,"e":18.0},"annotated_formula":"divide(450, 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>\n450 \/ (50\/3)<\/gadget>\n27<\/output>\n27<\/result>","index":2747} +{"problem":"during a thanksgiving weekend , a car rental company rented 6 - tenths of their vehicles , including two - fifths of the 4 wds that it had . if 40 % of the vehicles are 4 wds , then what percent of the vehicles that were not rented were not 4 wds ?","rationale":"4 \/ 10 of all the vehicles were not rented . ( 3 \/ 5 ) ( 2 \/ 5 ) = 6 \/ 25 of all the vehicles are 4 wds that were not rented . ( 6 \/ 25 ) \/ ( 4 \/ 10 ) = 3 \/ 5 is the fraction of non - rented vehicles that were 4 wds 1 - 3 \/ 5 = 40 % of non - rented vehicles were not 4 wds . the answer is c .","correct":"c","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":"multiply(divide(divide(multiply(const_2, 40), add(const_3, const_2)), 40), const_100)","linear_formula":"add(const_2,const_3)|multiply(n2,const_2)|divide(#1,#0)|divide(#2,n2)|multiply(#3,const_100)","chain":"2 * 40<\/gadget>\n80<\/output>\n3 + 2<\/gadget>\n5<\/output>\n80 \/ 5<\/gadget>\n16<\/output>\n16 \/ 40<\/gadget>\n2\/5 = around 0.4<\/output>\n(2\/5) * 100<\/gadget>\n40<\/output>\n40<\/result>","index":2748} +{"problem":"if x < y < z and y - x > 7 , 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 > 5 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 can eliminate answer choices a , c and e 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 > 7 the least value for y - x must be 11 , the least value for x must be 2 , and , thus , the least possible value for y must be 11 ( y - 2 = 9 , y = 11 ) 2 < 11 < z , since z is odd , the least possible value for z is 13 z - x = 13 - 2 = 11 answer c\"","correct":"c","options":{"a":"6 ","b":"7 ","c":"11 ","d":"8","e":"10"},"options_float":{"a":6.0,"b":7.0,"c":11.0,"d":8.0,"e":10.0},"annotated_formula":"add(add(7, const_2), const_2)","linear_formula":"add(n0,const_2)|add(#0,const_2)|","chain":"7 + 2<\/gadget>\n9<\/output>\n9 + 2<\/gadget>\n11<\/output>\n11<\/result>","index":2749} +{"problem":"jim drove 1096 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 - 1096 = 104 miles correct answer a\"","correct":"a","options":{"a":"104 miles ","b":"432 miles ","c":"456 miles ","d":"887 miles","e":"767 miles"},"options_float":{"a":104.0,"b":432.0,"c":456.0,"d":887.0,"e":767.0},"annotated_formula":"subtract(1200, 1096)","linear_formula":"subtract(n1,n0)|","chain":"1_200 - 1_096<\/gadget>\n104<\/output>\n104<\/result>","index":2751} +{"problem":"ram , who is half as efficient as krish , will take 18 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 = 18 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 18 days is 18 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 = 18 w \/ 3 w = 6 days answer d\"","correct":"d","options":{"a":"16 days ","b":"12 days ","c":"8 days ","d":"6 days","e":"18 days"},"options_float":{"a":16.0,"b":12.0,"c":8.0,"d":6.0,"e":18.0},"annotated_formula":"inverse(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)|inverse(#3)|","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 \/ (1\/6)<\/gadget>\n6<\/output>\n6<\/result>","index":2752} +{"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":"\"external radius = 4 cm , internal radius = 3 cm . volume of iron = ( 22 \/ 7 x [ ( 4 ) ^ 2 - ( 3 ) ^ 2 ] x 21 ) cm ^ 3 ( 22 \/ 7 x 7 x 1 x 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":"3.06 kg"},"options_float":{"a":3.6,"b":3.696,"c":36.0,"d":36.9,"e":3.06},"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":2753} +{"problem":"the average weight of 7 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ?","rationale":"\"total weight increases = 7 × 1.5 = 10.5 kg so the weight of new person = 65 + 10.5 = 75.5 kg answer c\"","correct":"c","options":{"a":"76 kg ","b":"77 kg ","c":"75.5 kg ","d":"data inadequate","e":"none of these"},"options_float":{"a":76.0,"b":77.0,"c":75.5,"d":null,"e":null},"annotated_formula":"add(65, multiply(7, 1.5))","linear_formula":"multiply(n0,n1)|add(n2,#0)|","chain":"7 * 1.5<\/gadget>\n10.5<\/output>\n65 + 10.5<\/gadget>\n75.5<\/output>\n75.5<\/result>","index":2754} +{"problem":"mr . jones gave 40 % of the money he had to his wife . he also gave 20 % of the remaining amount to his 3 sons . and half of the amount now left was spent on miscellaneous items and the remaining amount of rs . 12000 was deposited in the bank . how much money did mr . jones have initially ?","rationale":"explanation : let the initial amount be x , amount given to his wife = ( 40 \/ 100 ) x = 2 x \/ 5 balance = ( x - ( 2 x \/ 5 ) ) = 3 x \/ 5 amount given to his wife = ( 20 \/ 100 ) * ( 3 x \/ 5 ) = 3 x \/ 25 balance = 3 x \/ 5 - 3 x \/ 25 = 12 x \/ 25 amountt spent on miscellaneous items = ( 1 \/ 2 ) * ( 12 x \/ 25 ) = 6 x \/ 25 which is equal to 12000 hence , = > 6 x \/ 25 = 12000 = > x = 50000 answer : c","correct":"c","options":{"a":"40000 ","b":"45000 ","c":"50000 ","d":"62000","e":"none of these"},"options_float":{"a":40000.0,"b":45000.0,"c":50000.0,"d":62000.0,"e":null},"annotated_formula":"divide(12000, multiply(divide(divide(const_100, const_2), const_100), multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100)))))","linear_formula":"divide(const_100,const_2)|divide(n0,const_100)|divide(n1,const_100)|divide(#0,const_100)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|divide(n3,#7)","chain":"100 \/ 2<\/gadget>\n50<\/output>\n50 \/ 100<\/gadget>\n1\/2 = around 0.5<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/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>\n12\/25 = around 0.48<\/output>\n(1\/2) * (12\/25)<\/gadget>\n6\/25 = around 0.24<\/output>\n12_000 \/ (6\/25)<\/gadget>\n50_000<\/output>\n50_000<\/result>","index":2755} +{"problem":"he total marks obtained by a student in physics , chemistry and mathematics is 170 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ?","rationale":"\"let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 170 + p c + m = 170 average mark obtained by the student in chemistry and mathematics = ( c + m ) \/ 2 = 170 \/ 2 = 85 . answer : d\"","correct":"d","options":{"a":"55 ","b":"65 ","c":"75 ","d":"85","e":"95"},"options_float":{"a":55.0,"b":65.0,"c":75.0,"d":85.0,"e":95.0},"annotated_formula":"divide(170, const_2)","linear_formula":"divide(n0,const_2)|","chain":"170 \/ 2<\/gadget>\n85<\/output>\n85<\/result>","index":2756} +{"problem":"25 onions on a scale weigh 5.12 kg . when 3 onions are removed from the scale , the average weight of the 22 onions is 200 grams . what is the average weight ( in grams ) of the 3 onions which were removed ?","rationale":"22 * 200 = 4400 . the other 3 onions weigh a total of 720 grams . the average weight is 720 \/ 3 = 240 grams . the answer is c .","correct":"c","options":{"a":"200 ","b":"220 ","c":"240 ","d":"260","e":"280"},"options_float":{"a":200.0,"b":220.0,"c":240.0,"d":260.0,"e":280.0},"annotated_formula":"divide(subtract(multiply(5.12, const_1000), multiply(22, 200)), 3)","linear_formula":"multiply(n1,const_1000)|multiply(n3,n4)|subtract(#0,#1)|divide(#2,n2)","chain":"5.12 * 1_000<\/gadget>\n5_120<\/output>\n22 * 200<\/gadget>\n4_400<\/output>\n5_120 - 4_400<\/gadget>\n720<\/output>\n720 \/ 3<\/gadget>\n240<\/output>\n240<\/result>","index":2758} +{"problem":"a profit of rs . 600 is divided between x and y in the ratio of 1 \/ 2 : 1 \/ 3 . what is the difference between their profit shares ?","rationale":"\"a profit of rs . 600 is divided between x and y in the ratio of 1 \/ 2 : 1 \/ 3 or 3 : 2 . so profits are 360 and 240 . difference in profit share = 360 - 240 = 120 answer : b\"","correct":"b","options":{"a":"s . 220 ","b":"s . 120 ","c":"s . 320 ","d":"s . 50","e":"s . 90"},"options_float":{"a":220.0,"b":120.0,"c":320.0,"d":50.0,"e":90.0},"annotated_formula":"subtract(divide(divide(600, add(divide(1, 2), divide(1, 3))), 2), divide(divide(600, add(divide(1, 2), divide(1, 3))), 3))","linear_formula":"divide(n1,n2)|divide(n1,n4)|add(#0,#1)|divide(n0,#2)|divide(#3,n2)|divide(#3,n4)|subtract(#4,#5)|","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>\n5\/6 = around 0.833333<\/output>\n600 \/ (5\/6)<\/gadget>\n720<\/output>\n720 \/ 2<\/gadget>\n360<\/output>\n720 \/ 3<\/gadget>\n240<\/output>\n360 - 240<\/gadget>\n120<\/output>\n120<\/result>","index":2759} +{"problem":"3 candidates contested an election and received 1000 , 2000 and 4000 votes respectively . what percentage of the total votes did the winning candidate got ?","rationale":"total number of votes polled = ( 1000 + 2000 + 4000 ) = 7000 required percentage = 4000 \/ 7000 * 100 = 57 % ( approximately ) answer : option c","correct":"c","options":{"a":"30 % ","b":"50 % ","c":"57 % ","d":"62 %","e":"75 %"},"options_float":{"a":30.0,"b":50.0,"c":57.0,"d":62.0,"e":75.0},"annotated_formula":"multiply(divide(4000, add(add(1000, 2000), 4000)), const_100)","linear_formula":"add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)","chain":"1_000 + 2_000<\/gadget>\n3_000<\/output>\n3_000 + 4_000<\/gadget>\n7_000<\/output>\n4_000 \/ 7_000<\/gadget>\n4\/7 = around 0.571429<\/output>\n(4\/7) * 100<\/gadget>\n400\/7 = around 57.142857<\/output>\n400\/7 = around 57.142857<\/result>","index":2761} +{"problem":"an old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 64 times the difference between the two numbers . how many coins did the old man have ?","rationale":"\"let the number of coins one son got be x and the number of coins another got be y . total = x + y . x ^ 2 - y ^ 2 = 64 ( x - y ) - - > x + y = 64 . answer : d .\"","correct":"d","options":{"a":"24 ","b":"26 ","c":"30 ","d":"64","e":"40"},"options_float":{"a":24.0,"b":26.0,"c":30.0,"d":64.0,"e":40.0},"annotated_formula":"floor(64)","linear_formula":"floor(n0)|","chain":"floor(64)<\/gadget>\n64<\/output>\n64<\/result>","index":2764} +{"problem":"20 litres of mixture contains 40 % alcohol and the rest water . if 8 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ?","rationale":"\"alcohol in the 20 litres of mix . = 40 % of 20 litres = ( 40 * 20 \/ 100 ) = 8 litres water in it = 20 - 8 = 12 litres new quantity of mix . = 20 + 8 = 28 litres quantity of alcohol in it = 8 litres percentage of alcohol in new mix . = 8 * 100 \/ 28 = 50 \/ 3 = 28.57 % answer is c\"","correct":"c","options":{"a":"26.32 % ","b":"35.14 % ","c":"28.57 % ","d":"25 %","e":"31.14 %"},"options_float":{"a":26.32,"b":35.14,"c":28.57,"d":25.0,"e":31.14},"annotated_formula":"multiply(divide(subtract(add(20, 8), add(multiply(divide(subtract(const_100, 40), const_100), 20), 8)), add(20, 8)), const_100)","linear_formula":"add(n0,n2)|subtract(const_100,n1)|divide(#1,const_100)|multiply(n0,#2)|add(n2,#3)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100)|","chain":"20 + 8<\/gadget>\n28<\/output>\n100 - 40<\/gadget>\n60<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * 20<\/gadget>\n12<\/output>\n12 + 8<\/gadget>\n20<\/output>\n28 - 20<\/gadget>\n8<\/output>\n8 \/ 28<\/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":2765} +{"problem":"12.036 divided by 0.04 gives :","rationale":"\"= 12.036 \/ 0.04 = 1203.6 \/ 4 = 300.9 answer is b .\"","correct":"b","options":{"a":"30.09 ","b":"300.9 ","c":"30.06 ","d":"100.9","e":"300.6"},"options_float":{"a":30.09,"b":300.9,"c":30.06,"d":100.9,"e":300.6},"annotated_formula":"divide(12.036, 0.04)","linear_formula":"divide(n0,n1)|","chain":"12.036 \/ 0.04<\/gadget>\n300.9<\/output>\n300.9<\/result>","index":2767} +{"problem":"if 20 liters of chemical x are added to 80 liters of a mixture that is 5 % chemical x and 95 % chemical y , then what percentage of the resulting mixture is chemical x ?","rationale":"\"the amount of chemical x in the solution is 20 + 0.05 ( 80 ) = 24 liters . 24 liters \/ 100 liters = 24 % the answer is a .\"","correct":"a","options":{"a":"24 % ","b":"26 % ","c":"28 % ","d":"30 %","e":"32 %"},"options_float":{"a":24.0,"b":26.0,"c":28.0,"d":30.0,"e":32.0},"annotated_formula":"add(20, multiply(divide(5, const_100), 80))","linear_formula":"divide(n2,const_100)|multiply(n1,#0)|add(n0,#1)|","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n(1\/20) * 80<\/gadget>\n4<\/output>\n20 + 4<\/gadget>\n24<\/output>\n24<\/result>","index":2768} +{"problem":"for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score r for the exam ?","rationale":"\"a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean r = 74 . answer : a .\"","correct":"a","options":{"a":"74 ","b":"76 ","c":"78 ","d":"80","e":"82"},"options_float":{"a":74.0,"b":76.0,"c":78.0,"d":80.0,"e":82.0},"annotated_formula":"divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3))","linear_formula":"add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)|","chain":"58 * 3<\/gadget>\n174<\/output>\n98 * 2<\/gadget>\n196<\/output>\n174 + 196<\/gadget>\n370<\/output>\n2 + 3<\/gadget>\n5<\/output>\n370 \/ 5<\/gadget>\n74<\/output>\n74<\/result>","index":2770} +{"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 240 meter length ?","rationale":"\"speed = 45 km \/ hr = 45 * ( 5 \/ 18 ) m \/ sec = 25 \/ 2 m \/ sec total distance = 360 + 240 = 600 meter time = distance \/ speed = 600 * ( 2 \/ 25 ) = 48 seconds answer : e\"","correct":"e","options":{"a":"65 seconds ","b":"46 seconds ","c":"40 seconds ","d":"97 seconds","e":"48 seconds"},"options_float":{"a":65.0,"b":46.0,"c":40.0,"d":97.0,"e":48.0},"annotated_formula":"divide(add(360, 240), 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 + 240<\/gadget>\n600<\/output>\n45 * 1_000<\/gadget>\n45_000<\/output>\n45_000 \/ 3_600<\/gadget>\n25\/2 = around 12.5<\/output>\n600 \/ (25\/2)<\/gadget>\n48<\/output>\n48<\/result>","index":2771} +{"problem":"in one year , the population , of a village increased by 10 % and in the next year , it decreased by 10 % . if at the end of 2 nd year , the population was 7920 , what was it in the beginning ?","rationale":"\"x * 110 \/ 100 * 90 \/ 100 = 7920 x * 0.99 = 7920 x = 7920 \/ 0.99 = > 8000 answer : b\"","correct":"b","options":{"a":"8008 ","b":"8000 ","c":"8022 ","d":"8021","e":"8022"},"options_float":{"a":8008.0,"b":8000.0,"c":8022.0,"d":8021.0,"e":8022.0},"annotated_formula":"divide(divide(7920, subtract(const_1, divide(10, const_100))), add(const_1, divide(10, const_100)))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|divide(n3,#2)|divide(#3,#1)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n7_920 \/ (9\/10)<\/gadget>\n8_800<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n8_800 \/ (11\/10)<\/gadget>\n8_000<\/output>\n8_000<\/result>","index":2772} +{"problem":"the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 35 % profit ?","rationale":"\"let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 135 % of $ 640 = $ 864 d\"","correct":"d","options":{"a":"$ 480 ","b":"$ 450 ","c":"$ 960 ","d":"$ 864","e":"$ 660"},"options_float":{"a":480.0,"b":450.0,"c":960.0,"d":864.0,"e":660.0},"annotated_formula":"add(divide(multiply(divide(add(832, 448), const_2), 35), const_100), divide(add(832, 448), const_2))","linear_formula":"add(n0,n1)|divide(#0,const_2)|multiply(n2,#1)|divide(#2,const_100)|add(#3,#1)|","chain":"832 + 448<\/gadget>\n1_280<\/output>\n1_280 \/ 2<\/gadget>\n640<\/output>\n640 * 35<\/gadget>\n22_400<\/output>\n22_400 \/ 100<\/gadget>\n224<\/output>\n224 + 640<\/gadget>\n864<\/output>\n864<\/result>","index":2773} +{"problem":"if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 3720 , 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 = 3720 * ( 6 \/ 31 ) = 120 * 6 = 720 . answer : option a\"","correct":"a","options":{"a":"720 ","b":"700 ","c":"800 ","d":"900","e":"none of these"},"options_float":{"a":720.0,"b":700.0,"c":800.0,"d":900.0,"e":null},"annotated_formula":"multiply(3720, 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>\n3_720 * (6\/31)<\/gadget>\n720<\/output>\n720<\/result>","index":2774} +{"problem":"it costs $ 2 for the first 15 minutes to use the bumper cars at a fair ground . after the first 15 minutes it costs $ 6 per hour . if a certain customer uses the bumper cars for 3 hours and 25 minutes , how much will it cost him ?","rationale":"3 hrs 25 min = 205 min first 15 min - - - - - - > $ 2 time left is 190 min . . . now , 60 min costs $ 6 1 min costs $ 6 \/ 60 190 min costs $ 6 \/ 60 * 190 = > $ 19 so , total cost will be $ 19 + $ 2 = > $ 21 the answer will be ( d ) $ 21","correct":"d","options":{"a":"$ 22 ","b":"$ 3 ","c":"$ 15 ","d":"$ 21","e":"$ 30"},"options_float":{"a":22.0,"b":3.0,"c":15.0,"d":21.0,"e":30.0},"annotated_formula":"add(multiply(divide(6, const_60), subtract(add(multiply(3, const_60), 25), 15)), 2)","linear_formula":"divide(n3,const_60)|multiply(n4,const_60)|add(n5,#1)|subtract(#2,n1)|multiply(#0,#3)|add(n0,#4)","chain":"6 \/ 60<\/gadget>\n1\/10 = around 0.1<\/output>\n3 * 60<\/gadget>\n180<\/output>\n180 + 25<\/gadget>\n205<\/output>\n205 - 15<\/gadget>\n190<\/output>\n(1\/10) * 190<\/gadget>\n19<\/output>\n19 + 2<\/gadget>\n21<\/output>\n21<\/result>","index":2775} +{"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 30 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 * 30 = 25500 p in rs = 25500 \/ 100 = rs 255 answer : a\"","correct":"a","options":{"a":"s 255 ","b":"s 780 ","c":"s 880 ","d":"s 480","e":"s 980"},"options_float":{"a":255.0,"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)))), 30), 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 * 30<\/gadget>\n25_500<\/output>\n25_500 \/ 100<\/gadget>\n255<\/output>\n255<\/result>","index":2776} +{"problem":"john and andrew can finish the work 9 days if they work together . they worked together for 6 days and then andrew left . john finished the remaining work in another 6 days . in how many days john alone can finish the work ?","rationale":"amount of work done by john and andrew in 1 day = 1 \/ 9 amount of work done by john and andrew in 6 days = 6 ã — ( 1 \/ 9 ) = 2 \/ 3 remaining work â € “ 1 â € “ 2 \/ 3 = 1 \/ 3 john completes 1 \/ 3 work in 6 days amount of work john can do in 1 day = ( 1 \/ 3 ) \/ 6 = 1 \/ 18 = > john can complete the work in 18 days answer : c","correct":"c","options":{"a":"30 days ","b":"60 days ","c":"18 days ","d":"80 days","e":"90 days"},"options_float":{"a":30.0,"b":60.0,"c":18.0,"d":80.0,"e":90.0},"annotated_formula":"divide(6, subtract(const_1, divide(6, 9)))","linear_formula":"divide(n1,n0)|subtract(const_1,#0)|divide(n1,#1)","chain":"6 \/ 9<\/gadget>\n2\/3 = around 0.666667<\/output>\n1 - (2\/3)<\/gadget>\n1\/3 = around 0.333333<\/output>\n6 \/ (1\/3)<\/gadget>\n18<\/output>\n18<\/result>","index":2777} +{"problem":"the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 160 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 € “ 160 x + ( x - 160 ) = 208 2 x = 368 x = 184 runs answer : a\"","correct":"a","options":{"a":"184 ","b":"367 ","c":"269 ","d":"177","e":"191"},"options_float":{"a":184.0,"b":367.0,"c":269.0,"d":177.0,"e":191.0},"annotated_formula":"divide(add(160, 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>\n160 + 208<\/gadget>\n368<\/output>\n368 \/ 2<\/gadget>\n184<\/output>\n184<\/result>","index":2780} +{"problem":"the total age of a and b is 14 years more than the total age of b and c . c is how many year younger than a","rationale":"\"explanation : given that a + b = 14 + b + c = > a ? c = 14 + b ? b = 14 = > c is younger than a by 14 years answer : option d\"","correct":"d","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(14, const_1)","linear_formula":"multiply(n0,const_1)|","chain":"14 * 1<\/gadget>\n14<\/output>\n14<\/result>","index":2781} +{"problem":"if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m , what is its area ?","rationale":"l - b = 23 . . . ( 1 ) perimeter = 206 2 ( l = b ) = 206 l + b = 103 . . . ( 2 ) ( 1 ) + ( 2 ) 2 l = 23 + 103 = 126 l = 126 \/ 2 = 63 metre substituting the value of l in ( 1 ) , we get 63 - b = 23 b = 63 - 23 = 40 metre area = lb = 63 ã — 40 = 2520 m 2 answer : a","correct":"a","options":{"a":"2520 ","b":"2510 ","c":"2525 ","d":"2025","e":"2020"},"options_float":{"a":2520.0,"b":2510.0,"c":2525.0,"d":2025.0,"e":2020.0},"annotated_formula":"rectangle_area(add(divide(subtract(206, multiply(const_2, 23)), const_4), 23), divide(subtract(206, 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>\n206 - 46<\/gadget>\n160<\/output>\n160 \/ 4<\/gadget>\n40<\/output>\n40 + 23<\/gadget>\n63<\/output>\n63 * 40<\/gadget>\n2_520<\/output>\n2_520<\/result>","index":2783} +{"problem":"a boat having a length 5 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 = ( 5 x 2 x 0.01 ) m 3 = 0.10 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.10 x 1000 ) kg = 100 kg . answer : a\"","correct":"a","options":{"a":"100 kg ","b":"60 kg ","c":"72 kg ","d":"96 kg","e":"none of these"},"options_float":{"a":100.0,"b":60.0,"c":72.0,"d":96.0,"e":null},"annotated_formula":"multiply(multiply(multiply(5, 2), divide(1, const_100)), const_1000)","linear_formula":"divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|","chain":"5 * 2<\/gadget>\n10<\/output>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n10 * (1\/100)<\/gadget>\n1\/10 = around 0.1<\/output>\n(1\/10) * 1_000<\/gadget>\n100<\/output>\n100<\/result>","index":2784} +{"problem":"if 100 cats kill 100 mice in 100 days , then 4 cats would kill 4 mice in how many days ?","rationale":"as 100 cats kill 100 mice in 100 days 1 cats kill 1 mouse in 100 days then 4 cats kill 4 mice in 100 days answer : d","correct":"d","options":{"a":"1 day ","b":"4 days ","c":"40 days ","d":"100 days","e":"50 days"},"options_float":{"a":1.0,"b":4.0,"c":40.0,"d":100.0,"e":50.0},"annotated_formula":"divide(multiply(multiply(4, 100), 100), multiply(100, 4))","linear_formula":"multiply(n0,n3)|multiply(n0,#0)|divide(#1,#0)","chain":"4 * 100<\/gadget>\n400<\/output>\n400 * 100<\/gadget>\n40_000<\/output>\n100 * 4<\/gadget>\n400<\/output>\n40_000 \/ 400<\/gadget>\n100<\/output>\n100<\/result>","index":2785} +{"problem":"of the 55 cars on a car lot , 40 have air - conditioning , 25 have power windows , and 12 have both air - conditioning and power windows . how many of the cars on the lot have neither air - conditioning nor power windows ?","rationale":"total - neither = all air conditioning + all power windows - both or 55 - neither = 40 + 25 - 12 = 53 . = > neither = 2 , hence d . answer : d","correct":"d","options":{"a":"15 ","b":"8 ","c":"10 ","d":"2","e":"18"},"options_float":{"a":15.0,"b":8.0,"c":10.0,"d":2.0,"e":18.0},"annotated_formula":"subtract(55, subtract(add(40, 25), 12))","linear_formula":"add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)","chain":"40 + 25<\/gadget>\n65<\/output>\n65 - 12<\/gadget>\n53<\/output>\n55 - 53<\/gadget>\n2<\/output>\n2<\/result>","index":2787} +{"problem":"a person lent a certain sum of money at 5 % per annum at simple interest and in 8 years the interest amounted to $ 420 less than the sum lent . what was the sum lent ?","rationale":"\"p - 420 = ( p * 5 * 8 ) \/ 100 p = 700 the answer is c .\"","correct":"c","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":"divide(420, subtract(const_1, divide(multiply(5, 8), const_100)))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|","chain":"5 * 8<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n1 - (2\/5)<\/gadget>\n3\/5 = around 0.6<\/output>\n420 \/ (3\/5)<\/gadget>\n700<\/output>\n700<\/result>","index":2790} +{"problem":"30 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 : 30 : : 18 : x [ please pay attention , we have written 27 : 30 rather than 30 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \\ color { blue } x = \\ frac { 30 \\ times 18 } { 27 } } x = 20 so 20 days will be required to get work done by 27 men . answer : a\"","correct":"a","options":{"a":"20 ","b":"77 ","c":"36 ","d":"25","e":"13"},"options_float":{"a":20.0,"b":77.0,"c":36.0,"d":25.0,"e":13.0},"annotated_formula":"divide(multiply(18, 30), 27)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"18 * 30<\/gadget>\n540<\/output>\n540 \/ 27<\/gadget>\n20<\/output>\n20<\/result>","index":2791} +{"problem":"if n is a positive integer and n ^ 2 is divisible by 200 , then what is the largest positive integer that must divide n ?","rationale":"200 = 2 ^ 3 * 5 ^ 2 if 200 divides n ^ 2 , then n must be divisible by 2 ^ 2 * 5 = 20 the answer is c .","correct":"c","options":{"a":"10 ","b":"15 ","c":"20 ","d":"36","e":"50"},"options_float":{"a":10.0,"b":15.0,"c":20.0,"d":36.0,"e":50.0},"annotated_formula":"multiply(sqrt(divide(200, 2)), 2)","linear_formula":"divide(n1,n0)|sqrt(#0)|multiply(n0,#1)","chain":"200 \/ 2<\/gadget>\n100<\/output>\n100 ** (1\/2)<\/gadget>\n10<\/output>\n10 * 2<\/gadget>\n20<\/output>\n20<\/result>","index":2792} +{"problem":"john paid a sum of money for purchasing 20 pens , which he recovered in full when he sold 10 of them . what was his percentage of profit or loss per pen ?","rationale":"a 100 % if the sum he paid whilst purchasing 20 pens = a , then the cost price of each pen = a \/ 20 . since the amount he got whilst selling 10 pens is also = a then the selling price of each pen = a \/ 10 . since selling price > cost price , he made a profit . profit per pen = selling price - cost price = a \/ 10 - a \/ 20 = a \/ 20 . profit percentage per pen = profit per pen \/ cost per pen x 100 = ( a \/ 20 ) \/ ( a \/ 20 ) x 100 = 100 %","correct":"a","options":{"a":"100 % ","b":"150 % ","c":"90 % ","d":"80 %","e":"95 %"},"options_float":{"a":100.0,"b":150.0,"c":90.0,"d":80.0,"e":95.0},"annotated_formula":"multiply(divide(subtract(20, 10), 10), const_100)","linear_formula":"subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)","chain":"20 - 10<\/gadget>\n10<\/output>\n10 \/ 10<\/gadget>\n1<\/output>\n1 * 100<\/gadget>\n100<\/output>\n100<\/result>","index":2793} +{"problem":"find the smallest number of 6 digits exactly divisible by 25 , 3545 and 15 .","rationale":"smallest number of six digits is 100000 . required number must be divisible by l . c . m . of 25,35 , 45,15 i . e 1575 , on dividing 100000 by 1575 , we get 800 as remainder . therefore , required number = 100000 + ( 1575 â € “ 800 ) = 100775 . answer is b .","correct":"b","options":{"a":"100555 ","b":"100775 ","c":"100885 ","d":"100995","e":"100665"},"options_float":{"a":100555.0,"b":100775.0,"c":100885.0,"d":100995.0,"e":100665.0},"annotated_formula":"multiply(power(const_100, const_2), const_10)","linear_formula":"power(const_100,const_2)|multiply(#0,const_10)","chain":"100 ** 2<\/gadget>\n10_000<\/output>\n10_000 * 10<\/gadget>\n100_000<\/output>\n100_000<\/result>","index":2795} +{"problem":"in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 700 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 2100 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?","rationale":"ratio of 2 bedroom apartment : 1 bedroom apartment = 700 : 2100 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 \/ 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 \/ 4 ) * 100 - - - > 25 % answer : a","correct":"a","options":{"a":"25 % ","b":"15 % ","c":"20 % ","d":"40 %","e":"45 %"},"options_float":{"a":25.0,"b":15.0,"c":20.0,"d":40.0,"e":45.0},"annotated_formula":"multiply(divide(const_1, add(const_3, const_1)), const_100)","linear_formula":"add(const_1,const_3)|divide(const_1,#0)|multiply(#1,const_100)","chain":"3 + 1<\/gadget>\n4<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2797} +{"problem":"if two - third of a bucket is filled in 6 minute then the time taken to fill the bucket completely will be .","rationale":"\"2 \/ 3 filled in 6 mint 1 \/ 3 filled in 3 mint thn 2 \/ 3 + 1 \/ 3 = 6 + 3 = 9 minutes answer : d\"","correct":"d","options":{"a":"90 seconds ","b":"70 seconds ","c":"60 seconds ","d":"9 minutes","e":"120 seconds"},"options_float":{"a":90.0,"b":70.0,"c":60.0,"d":9.0,"e":120.0},"annotated_formula":"multiply(divide(6, const_2), const_3)","linear_formula":"divide(n0,const_2)|multiply(#0,const_3)|","chain":"6 \/ 2<\/gadget>\n3<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9<\/result>","index":2798} +{"problem":"a batsman makes a score of 76 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning .","rationale":"\"let the average after 17 th inning = x . then , average after 16 th inning = ( x – 3 ) . ∴ 16 ( x – 3 ) + 76 = 17 x or x = ( 76 – 48 ) = 28 . answer b\"","correct":"b","options":{"a":"36 ","b":"28 ","c":"42 ","d":"45","e":"none of the above"},"options_float":{"a":36.0,"b":28.0,"c":42.0,"d":45.0,"e":null},"annotated_formula":"add(subtract(76, multiply(17, 3)), 3)","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|","chain":"17 * 3<\/gadget>\n51<\/output>\n76 - 51<\/gadget>\n25<\/output>\n25 + 3<\/gadget>\n28<\/output>\n28<\/result>","index":2799} +{"problem":"find the remainder of the division ( 2 ^ 14 ) \/ 7 .","rationale":"\"find the pattern of the remainders after each power : ( 2 ^ 1 ) \/ 7 remainder 2 ( 2 ^ 2 ) \/ 7 remainder 4 ( 2 ^ 3 ) \/ 7 remainder 1 - - > this is where the cycle ends ( 2 ^ 4 ) \/ 7 remainder 2 - - > this is where the cycle begins again ( 2 ^ 5 ) \/ 7 remainder 4 continuing the pattern to ( 2 ^ 14 ) \/ 7 gives us a remainder of 4 final answer : d ) 4\"","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":"reminder(power(2, 14), 7)","linear_formula":"power(n0,n1)|reminder(#0,n2)|","chain":"2 ** 14<\/gadget>\n16_384<\/output>\n16_384 % 7<\/gadget>\n4<\/output>\n4<\/result>","index":2800} +{"problem":"barbata invests $ 2600 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 ; 2600 + 0.05 * 2600 + x + 0.08 x = 2600 + x + 0.06 ( 2600 + x ) 0.05 * 2600 + 0.08 x = 0.06 x + 0.06 * 2600 0.02 x = 2600 ( 0.06 - 0.05 ) x = 2600 * 0.01 \/ 0.02 = 1300 ans : ` ` c ' '\"","correct":"c","options":{"a":"1200 ","b":"3000 ","c":"1300 ","d":"3600","e":"2400"},"options_float":{"a":1200.0,"b":3000.0,"c":1300.0,"d":3600.0,"e":2400.0},"annotated_formula":"divide(subtract(multiply(divide(6, const_100), 2600), multiply(2600, 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_600<\/gadget>\n156<\/output>\n5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n2_600 * (1\/20)<\/gadget>\n130<\/output>\n156 - 130<\/gadget>\n26<\/output>\n8 \/ 100<\/gadget>\n2\/25 = around 0.08<\/output>\n(2\/25) - (3\/50)<\/gadget>\n1\/50 = around 0.02<\/output>\n26 \/ (1\/50)<\/gadget>\n1_300<\/output>\n1_300<\/result>","index":2802} +{"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 w 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":"w = 256 ","d":"w = 384","e":"w = 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":2803} +{"problem":"if x is equal to the sum of the integers from 40 to 50 , inclusive , and y is the number of even integers from 40 to 50 , inclusive , what is the value of x + y ?","rationale":"\"sum s = n \/ 2 { 2 a + ( n - 1 ) d } = 11 \/ 2 { 2 * 40 + ( 11 - 1 ) * 1 } = 11 * 45 = 495 = x number of even number = ( 50 - 40 ) \/ 2 + 1 = 6 = y x + y = 495 + 6 = 501 d\"","correct":"d","options":{"a":"171 ","b":"281 ","c":"391 ","d":"501","e":"613"},"options_float":{"a":171.0,"b":281.0,"c":391.0,"d":501.0,"e":613.0},"annotated_formula":"add(multiply(divide(add(40, 50), const_2), add(subtract(50, 40), const_1)), add(divide(subtract(50, 40), const_2), const_1))","linear_formula":"add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)|","chain":"40 + 50<\/gadget>\n90<\/output>\n90 \/ 2<\/gadget>\n45<\/output>\n50 - 40<\/gadget>\n10<\/output>\n10 + 1<\/gadget>\n11<\/output>\n45 * 11<\/gadget>\n495<\/output>\n10 \/ 2<\/gadget>\n5<\/output>\n5 + 1<\/gadget>\n6<\/output>\n495 + 6<\/gadget>\n501<\/output>\n501<\/result>","index":2804} +{"problem":"a man swims downstream 28 km and upstream 16 km taking 4 hours each time , what is the speed of the man in still water ?","rationale":"\"28 - - - 4 ds = 7 ? - - - - 1 16 - - - - 4 us = 4 ? - - - - 1 m = ? m = ( 7 + 4 ) \/ 2 = 5.5 answer : e\"","correct":"e","options":{"a":"6.5 ","b":"8.6 ","c":"7.5 ","d":"9.2","e":"5.5"},"options_float":{"a":6.5,"b":8.6,"c":7.5,"d":9.2,"e":5.5},"annotated_formula":"divide(add(divide(16, 4), divide(28, 4)), const_2)","linear_formula":"divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|","chain":"16 \/ 4<\/gadget>\n4<\/output>\n28 \/ 4<\/gadget>\n7<\/output>\n4 + 7<\/gadget>\n11<\/output>\n11 \/ 2<\/gadget>\n11\/2 = around 5.5<\/output>\n11\/2 = around 5.5<\/result>","index":2805} +{"problem":"alice and bob drive at constant speeds toward each other on a highway . alice drives at a constant speed of 30 km per hour . at a certain time they pass by each other , and then keep driving away from each other , maintaining their constant speeds . if alice is 100 km away from bob at 7 am , and also 100 km away from bob at 11 am , then how fast is bob driving ( in kilometers per hour ) ?","rationale":"alice and bob complete 200 km \/ 4 hours = 50 km \/ hour bob ' s speed is 50 - 30 = 20 km \/ hour the answer is a .","correct":"a","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":"subtract(divide(add(100, 100), subtract(11, 7)), 30)","linear_formula":"add(n1,n1)|subtract(n4,n2)|divide(#0,#1)|subtract(#2,n0)","chain":"100 + 100<\/gadget>\n200<\/output>\n11 - 7<\/gadget>\n4<\/output>\n200 \/ 4<\/gadget>\n50<\/output>\n50 - 30<\/gadget>\n20<\/output>\n20<\/result>","index":2806} +{"problem":"the calendar of the year 2040 can be used again in the year ?","rationale":"\"explanation : given year 2040 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2040 + 28 = 2068 answer : e\"","correct":"e","options":{"a":"2063 ","b":"2061 ","c":"2111 ","d":"2191","e":"2068"},"options_float":{"a":2063.0,"b":2061.0,"c":2111.0,"d":2191.0,"e":2068.0},"annotated_formula":"add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2040)","linear_formula":"multiply(const_4,const_4)|subtract(#0,const_2)|multiply(#1,const_2)|add(n0,#2)|","chain":"4 * 4<\/gadget>\n16<\/output>\n16 - 2<\/gadget>\n14<\/output>\n14 * 2<\/gadget>\n28<\/output>\n28 + 2_040<\/gadget>\n2_068<\/output>\n2_068<\/result>","index":2807} +{"problem":"in what time will a train 100 metres long cross an electic pole , if its speed be 144 km \/ hr ?","rationale":"\"sol . 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":2809} +{"problem":"a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 25 hour to row to a place and come back , how far is the place ?","rationale":"\"speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x \/ 12 + x \/ 8 = 25 2 x + 3 x = 600 x = 120 km answer is d\"","correct":"d","options":{"a":"24 km ","b":"30 km ","c":"48 km ","d":"120 km","e":"15 km"},"options_float":{"a":24.0,"b":30.0,"c":48.0,"d":120.0,"e":15.0},"annotated_formula":"divide(multiply(multiply(subtract(10, 2), add(10, 2)), 25), add(subtract(10, 2), add(10, 2)))","linear_formula":"add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)|","chain":"10 - 2<\/gadget>\n8<\/output>\n10 + 2<\/gadget>\n12<\/output>\n8 * 12<\/gadget>\n96<\/output>\n96 * 25<\/gadget>\n2_400<\/output>\n8 + 12<\/gadget>\n20<\/output>\n2_400 \/ 20<\/gadget>\n120<\/output>\n120<\/result>","index":2810} +{"problem":"from a group of 4 boys and 4 girls , 4 children are to be randomly selected . what is the probability that 2 boys and 2 girls will be selected ?","rationale":"\"the total number of ways to choose 4 children from 8 is 8 c 4 = 70 the number of ways to choose 2 boys and 2 girls is 4 c 2 * 4 c 2 = 6 * 6 = 36 p ( 2 boys and 2 girls ) = 36 \/ 70 = 18 \/ 35 the answer is d .\"","correct":"d","options":{"a":"12 \/ 29 ","b":"14 \/ 31 ","c":"16 \/ 33 ","d":"18 \/ 35","e":"20 \/ 37"},"options_float":{"a":0.4137931034,"b":0.4516129032,"c":0.4848484848,"d":0.5142857143,"e":0.5405405405},"annotated_formula":"divide(multiply(choose(4, const_2), choose(4, const_2)), choose(add(4, 4), 4))","linear_formula":"add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)|","chain":"binomial(4, 2)<\/gadget>\n6<\/output>\n6 * 6<\/gadget>\n36<\/output>\n4 + 4<\/gadget>\n8<\/output>\nbinomial(8, 4)<\/gadget>\n70<\/output>\n36 \/ 70<\/gadget>\n18\/35 = around 0.514286<\/output>\n18\/35 = around 0.514286<\/result>","index":2811} +{"problem":"a can do a work in 8 days . b can do the same work in 24 days . if both a & b are working together in how many days they will finish the work ?","rationale":"\"a rate = 1 \/ 8 b rate = 1 \/ 24 ( a + b ) rate = ( 1 \/ 8 ) + ( 1 \/ 24 ) = 1 \/ 6 a & b finish the work in 6 days correct option is e\"","correct":"e","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(8, 24), add(8, 24))","linear_formula":"add(n0,n1)|multiply(n0,n1)|divide(#1,#0)|","chain":"8 * 24<\/gadget>\n192<\/output>\n8 + 24<\/gadget>\n32<\/output>\n192 \/ 32<\/gadget>\n6<\/output>\n6<\/result>","index":2813} +{"problem":"what is the remainder when 50 ! is divided by 16 ^ 8 ? ?","rationale":"\"16 raise to 8 = 2 raise to 32 , now highest power of 2 divisible by 50 ! is 25 + 12 + 6 + 3 + 1 = 47 since 2 raise to 47 is divisible , 2 raise to 32 also will be divisible answer : 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":"reminder(multiply(16, 50), 8)","linear_formula":"multiply(n0,n1)|reminder(#0,n2)|","chain":"16 * 50<\/gadget>\n800<\/output>\n800 % 8<\/gadget>\n0<\/output>\n0<\/result>","index":2814} +{"problem":"if the cost price of 140 pencils is equal to the selling price of 100 pencils , the gain percent is","rationale":"\"let c . p . of each pencil be re . 1 . then , c . p . of 100 pencils = rs . 100 ; s . p . of 100 pencils = rs . 140 . gain % = 40 \/ 100 * 100 = 40 % answer : e\"","correct":"e","options":{"a":"36 ","b":"37 ","c":"38 ","d":"39","e":"40"},"options_float":{"a":36.0,"b":37.0,"c":38.0,"d":39.0,"e":40.0},"annotated_formula":"divide(const_100, divide(100, subtract(140, 100)))","linear_formula":"subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)|","chain":"140 - 100<\/gadget>\n40<\/output>\n100 \/ 40<\/gadget>\n5\/2 = around 2.5<\/output>\n100 \/ (5\/2)<\/gadget>\n40<\/output>\n40<\/result>","index":2816} +{"problem":"mother , her daughter and her grand child weighs 140 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 \/ 5 th of her grand mother . what is the age of the daughter ?","rationale":"\"mother + daughter + child = 140 kg daughter + child = 60 kg mother = 140 - 60 = 80 kg child = 1 \/ 5 th of mother = ( 1 \/ 5 ) * 80 = 16 kg so now daughter = 140 - ( mother + child ) = 140 - ( 80 + 16 ) = 44 kg answer : a\"","correct":"a","options":{"a":"44 ","b":"47 ","c":"48 ","d":"49","e":"50"},"options_float":{"a":44.0,"b":47.0,"c":48.0,"d":49.0,"e":50.0},"annotated_formula":"subtract(60, divide(subtract(140, 60), 5))","linear_formula":"subtract(n0,n1)|divide(#0,n3)|subtract(n1,#1)|","chain":"140 - 60<\/gadget>\n80<\/output>\n80 \/ 5<\/gadget>\n16<\/output>\n60 - 16<\/gadget>\n44<\/output>\n44<\/result>","index":2817} +{"problem":"a mixture of sand and cement contains , 3 parts of sand and 5 parts of cement . how much of the mixture must be substituted with sand to make the mixture half sand and half cement ?","rationale":"we have total of 8 parts : 3 parts of sand and 5 parts of cement . in order there to be half sand and half cement ( 4 parts of sand and 4 parts of cement ) , we should remove 1 part of cement . with 1 part of cement comes 3 \/ 5 parts of sand , so we should remove 1 + 3 \/ 5 = 8 \/ 5 part of the mixture , which is ( 8 \/ 5 ) \/ 8 = 1 \/ 5 of the mixture . answer : c .","correct":"c","options":{"a":"1 \/ 3 ","b":"1 \/ 4 ","c":"1 \/ 5 ","d":"1 \/ 7","e":"1 \/ 8"},"options_float":{"a":0.3333333333,"b":0.25,"c":0.2,"d":0.1428571429,"e":0.125},"annotated_formula":"divide(add(const_1, divide(3, 5)), add(5, 3))","linear_formula":"add(n0,n1)|divide(n0,n1)|add(#1,const_1)|divide(#2,#0)","chain":"3 \/ 5<\/gadget>\n3\/5 = around 0.6<\/output>\n1 + (3\/5)<\/gadget>\n8\/5 = around 1.6<\/output>\n5 + 3<\/gadget>\n8<\/output>\n(8\/5) \/ 8<\/gadget>\n1\/5 = around 0.2<\/output>\n1\/5 = around 0.2<\/result>","index":2819} +{"problem":"a bookseller sells his books at a 20 % markup in price . if he sells a book for $ 24.00 , how much did he pay for it ?","rationale":"let the cost price of book = x selling price of book = 24 $ markup % = 20 ( 120 \/ 100 ) x = 24 = > x = 20 answer e","correct":"e","options":{"a":"$ 14.40 ","b":"$ 14.00 ","c":"$ 10.00 ","d":"$ 9.60","e":"$ 20.00"},"options_float":{"a":14.4,"b":14.0,"c":10.0,"d":9.6,"e":20.0},"annotated_formula":"subtract(24, multiply(divide(20, const_100), 24))","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) * 24<\/gadget>\n24\/5 = around 4.8<\/output>\n24 - (24\/5)<\/gadget>\n96\/5 = around 19.2<\/output>\n96\/5 = around 19.2<\/result>","index":2821} +{"problem":"in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 10 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ?","rationale":"\"the profit 0 f 2009 in terms of 2008 = 0.9 * 15 \/ 10 * 100 = 135 % c\"","correct":"c","options":{"a":"80 % ","b":"105 % ","c":"135 % ","d":"124.2 %","e":"138 %"},"options_float":{"a":80.0,"b":105.0,"c":135.0,"d":124.2,"e":138.0},"annotated_formula":"multiply(divide(multiply(15, subtract(const_1, divide(10, const_100))), 10), const_100)","linear_formula":"divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)|","chain":"10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n1 - (1\/10)<\/gadget>\n9\/10 = around 0.9<\/output>\n15 * (9\/10)<\/gadget>\n27\/2 = around 13.5<\/output>\n(27\/2) \/ 10<\/gadget>\n27\/20 = around 1.35<\/output>\n(27\/20) * 100<\/gadget>\n135<\/output>\n135<\/result>","index":2822} +{"problem":"find the simple interest on $ 10000 at 6 % per annum for 12 months ?","rationale":"\"p = $ 10000 r = 6 % t = 12 \/ 12 years = 1 year s . i . = p * r * t \/ 100 = 10000 * 6 * 1 \/ 100 = $ 600 answer is c\"","correct":"c","options":{"a":"$ 410 ","b":"$ 500 ","c":"$ 600 ","d":"$ 710","e":"$ 1000"},"options_float":{"a":410.0,"b":500.0,"c":600.0,"d":710.0,"e":1000.0},"annotated_formula":"multiply(10000, divide(6, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,#0)|","chain":"6 \/ 100<\/gadget>\n3\/50 = around 0.06<\/output>\n10_000 * (3\/50)<\/gadget>\n600<\/output>\n600<\/result>","index":2823} +{"problem":"of the 3,600 employees of company x , 12 \/ 25 are clerical . if the clerical staff were to be reduced by 1 \/ 4 , what percent of the total number of the remaining employees would then be clerical ?","rationale":"let ' s see , the way i did it was 12 \/ 25 are clerical out of 3600 so 1728 are clerical 1728 reduced by 1 \/ 4 is 1728 * 1 \/ 4 so it reduced 432 people , so there is 1296 clerical people left but since 432 people left , it also reduced from the total of 3600 so there are 3168 people total since 1296 clerical left \/ 3168 people total you get ( a ) 40 %","correct":"a","options":{"a":"40 % ","b":"22.2 % ","c":"20 % ","d":"12.5 %","e":"11.1 %"},"options_float":{"a":40.0,"b":22.2,"c":20.0,"d":12.5,"e":11.1},"annotated_formula":"multiply(divide(multiply(divide(12, 25), subtract(1, divide(1, 4))), add(multiply(divide(12, 25), subtract(1, divide(1, 4))), subtract(const_1, divide(12, 25)))), const_100)","linear_formula":"divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(const_1,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)","chain":"12 \/ 25<\/gadget>\n12\/25 = around 0.48<\/output>\n1 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 - (1\/4)<\/gadget>\n3\/4 = around 0.75<\/output>\n(12\/25) * (3\/4)<\/gadget>\n9\/25 = around 0.36<\/output>\n1 - (12\/25)<\/gadget>\n13\/25 = around 0.52<\/output>\n(9\/25) + (13\/25)<\/gadget>\n22\/25 = around 0.88<\/output>\n(9\/25) \/ (22\/25)<\/gadget>\n9\/22 = around 0.409091<\/output>\n(9\/22) * 100<\/gadget>\n450\/11 = around 40.909091<\/output>\n450\/11 = around 40.909091<\/result>","index":2825} +{"problem":"a man two flats for $ 675958 each . on one he gains 13 % while on the other he loses 13 % . how much does he gain or lose in the whole transaction ?","rationale":"\"in such a case there is always a loss loss % = ( 13 \/ 10 ) ^ 2 = 120 \/ 71 = 1.69 % answer is a\"","correct":"a","options":{"a":"1.69 % ","b":"2.56 % ","c":"3.12 % ","d":"4.65 %","e":"5.12 %"},"options_float":{"a":1.69,"b":2.56,"c":3.12,"d":4.65,"e":5.12},"annotated_formula":"multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 13)), 675958), multiply(divide(const_100, subtract(const_100, 13)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 13)), 675958), multiply(divide(const_100, subtract(const_100, 13)), 675958))), const_100)","linear_formula":"add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100)|","chain":"100 + 13<\/gadget>\n113<\/output>\n100 \/ 113<\/gadget>\n100\/113 = around 0.884956<\/output>\n(100\/113) * 675_958<\/gadget>\n67_595_800\/113 = around 598_192.920354<\/output>\n100 - 13<\/gadget>\n87<\/output>\n100 \/ 87<\/gadget>\n100\/87 = around 1.149425<\/output>\n(100\/87) * 675_958<\/gadget>\n67_595_800\/87 = around 776_963.218391<\/output>\n(67_595_800\/113) + (67_595_800\/87)<\/gadget>\n13_519_160_000\/9_831 = around 1_375_156.138745<\/output>\n675_958 + 675_958<\/gadget>\n1_351_916<\/output>\n(13_519_160_000\/9_831) - 1_351_916<\/gadget>\n228_473_804\/9_831 = around 23_240.138745<\/output>\n(228_473_804\/9_831) \/ (13_519_160_000\/9_831)<\/gadget>\n169\/10_000 = around 0.0169<\/output>\n(169\/10_000) * 100<\/gadget>\n169\/100 = around 1.69<\/output>\n169\/100 = around 1.69<\/result>","index":2826} +{"problem":"what is the greatest of 3 consecutive integers whose sum is 30 ?","rationale":"\"30 \/ 3 = 10 the three numbers are 9 , 10 , and 11 . the answer is d .\"","correct":"d","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":"add(divide(subtract(30, 3), 3), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)|","chain":"30 - 3<\/gadget>\n27<\/output>\n27 \/ 3<\/gadget>\n9<\/output>\n9 + 2<\/gadget>\n11<\/output>\n11<\/result>","index":2827} +{"problem":"students of 3 different classes appeared in common examination . pass average of 10 students of first class was 45 % , pass average of 15 students of second class was 60 % and pass average of 25 students of third class was 80 % then what will be the pass average of all students of 3 classes ?","rationale":"solution : sum of pass students of first , second and third class , = ( 45 % of 10 ) + ( 60 % of 15 ) + ( 80 % of 25 ) = 4.5 + 9 + 20 = 33.5 total students appeared , = 10 + 15 + 25 = 50 pass average , = 33.5 * 100 \/ 50 = 67 % . answer : option c","correct":"c","options":{"a":"74 % ","b":"75 % ","c":"67 % ","d":"72 %","e":"none"},"options_float":{"a":74.0,"b":75.0,"c":67.0,"d":72.0,"e":null},"annotated_formula":"divide(multiply(add(add(divide(multiply(10, 45), const_100), divide(multiply(15, 60), const_100)), divide(multiply(25, 80), const_100)), const_100), add(add(10, 15), 25))","linear_formula":"add(n1,n3)|multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(n5,#0)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|add(#5,#6)|add(#8,#7)|multiply(#9,const_100)|divide(#10,#4)","chain":"10 * 45<\/gadget>\n450<\/output>\n450 \/ 100<\/gadget>\n9\/2 = around 4.5<\/output>\n15 * 60<\/gadget>\n900<\/output>\n900 \/ 100<\/gadget>\n9<\/output>\n(9\/2) + 9<\/gadget>\n27\/2 = around 13.5<\/output>\n25 * 80<\/gadget>\n2_000<\/output>\n2_000 \/ 100<\/gadget>\n20<\/output>\n(27\/2) + 20<\/gadget>\n67\/2 = around 33.5<\/output>\n(67\/2) * 100<\/gadget>\n3_350<\/output>\n10 + 15<\/gadget>\n25<\/output>\n25 + 25<\/gadget>\n50<\/output>\n3_350 \/ 50<\/gadget>\n67<\/output>\n67<\/result>","index":2828} +{"problem":"if the sides of a triangle are 20 cm , 12 cm and 16 cm , what is its area ?","rationale":"\"the triangle with sides 20 cm , 12 cm and 16 cm is right angled , where the hypotenuse is 20 cm . area of the triangle = 1 \/ 2 * 12 * 16 = 96 cm 2 answer : option d\"","correct":"d","options":{"a":"70 ","b":"79 ","c":"85 ","d":"96","e":"92"},"options_float":{"a":70.0,"b":79.0,"c":85.0,"d":96.0,"e":92.0},"annotated_formula":"divide(multiply(12, 16), const_2)","linear_formula":"multiply(n1,n2)|divide(#0,const_2)|","chain":"12 * 16<\/gadget>\n192<\/output>\n192 \/ 2<\/gadget>\n96<\/output>\n96<\/result>","index":2829} +{"problem":"how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?","rationale":"\"d = 100 + 150 = 250 s = 36 * 5 \/ 18 = 10 mps t = 250 \/ 10 = 25 sec . answer : c\"","correct":"c","options":{"a":"2 ","b":"28 ","c":"25 ","d":"99","e":"12"},"options_float":{"a":2.0,"b":28.0,"c":25.0,"d":99.0,"e":12.0},"annotated_formula":"divide(add(150, 100), multiply(36, const_0_2778))","linear_formula":"add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|","chain":"150 + 100<\/gadget>\n250<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n36 * (5\/18)<\/gadget>\n10<\/output>\n250 \/ 10<\/gadget>\n25<\/output>\n25<\/result>","index":2830} +{"problem":"34.94 + 240.016 + 23.98 = ?","rationale":"34.94 240.016 + 23.98 - - - - - - - - 298.936 answer is a .","correct":"a","options":{"a":"298.936 ","b":"298.694 ","c":"289.496 ","d":"289.469","e":"298.964"},"options_float":{"a":298.936,"b":298.694,"c":289.496,"d":289.469,"e":298.964},"annotated_formula":"add(add(34.94, 240.016), 23.98)","linear_formula":"add(n0,n1)|add(n2,#0)","chain":"34.94 + 240.016<\/gadget>\n274.956<\/output>\n274.956 + 23.98<\/gadget>\n298.936<\/output>\n298.936<\/result>","index":2831} +{"problem":"meera purchased two 3 items from a shop . total price for 3 items is rs . 2000 \/ - she have given rs . 3000 \/ - what is the balance amount meera got ?","rationale":"total cost of items : 2000 \/ - amount paid : 3000 \/ - balance receivable : 3000 - 2000 = 1000 \/ - answer is b","correct":"b","options":{"a":"650 ","b":"1000 ","c":"1500 ","d":"800","e":"750"},"options_float":{"a":650.0,"b":1000.0,"c":1500.0,"d":800.0,"e":750.0},"annotated_formula":"subtract(3000, 2000)","linear_formula":"subtract(n3,n2)","chain":"3_000 - 2_000<\/gadget>\n1_000<\/output>\n1_000<\/result>","index":2832} +{"problem":"a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 36 instead of 86 then find the correct average marks up to two decimal places .","rationale":"\"correct average = 35 x 72 + ( 86 - 36 ) \/ 35 ≈ 72 + 1.43 = 73.43 answer d\"","correct":"d","options":{"a":"73.41 ","b":"74.31 ","c":"72.43 ","d":"73.43","e":"can not be determined"},"options_float":{"a":73.41,"b":74.31,"c":72.43,"d":73.43,"e":null},"annotated_formula":"divide(subtract(multiply(35, 72), subtract(86, 36)), 35)","linear_formula":"multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)|","chain":"35 * 72<\/gadget>\n2_520<\/output>\n86 - 36<\/gadget>\n50<\/output>\n2_520 - 50<\/gadget>\n2_470<\/output>\n2_470 \/ 35<\/gadget>\n494\/7 = around 70.571429<\/output>\n494\/7 = around 70.571429<\/result>","index":2833} +{"problem":"what is the smallest integer e greater than 1 that leaves a remainder of 1 when divided by any of the integers 6 , 8 , and 10 ?","rationale":"or u can just use the answer choices here . since the answers are already arranged in ascending order , the first number which gives remainder e as 1 for all three is the correct answer . in the given question , the first number which gives a remainder of 1 for 68 and 10 is 121 . c","correct":"c","options":{"a":"21 ","b":"41 ","c":"e = 121 ","d":"241","e":"481"},"options_float":{"a":21.0,"b":41.0,"c":121.0,"d":241.0,"e":481.0},"annotated_formula":"add(lcm(10, lcm(6, 8)), const_1)","linear_formula":"lcm(n2,n3)|lcm(n4,#0)|add(#1,const_1)","chain":"lcm(6, 8)<\/gadget>\n24<\/output>\nlcm(10, 24)<\/gadget>\n120<\/output>\n120 + 1<\/gadget>\n121<\/output>\n121<\/result>","index":2834} +{"problem":"a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 30 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 40 % of that is added to the weight , the weight claimed by the trader is obtained . if he charges the cost price of the weight that he claims , find his profit percentage .","rationale":"\"anyways , one can infer that he ' steals ' 30 % from suppliers and then charges 40 % extra to customers so basically 1.3 * 1.4 = 1.82 given that 1 is start point , we get 21 % more hence answer is b\"","correct":"b","options":{"a":"28 % ","b":"82 % ","c":"24.33 % ","d":"29.109 %","e":"78 %"},"options_float":{"a":28.0,"b":82.0,"c":24.33,"d":29.109,"e":78.0},"annotated_formula":"subtract(multiply(divide(add(const_100, 40), const_100), add(const_100, 30)), const_100)","linear_formula":"add(n0,const_100)|add(n1,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)|","chain":"100 + 40<\/gadget>\n140<\/output>\n140 \/ 100<\/gadget>\n7\/5 = around 1.4<\/output>\n100 + 30<\/gadget>\n130<\/output>\n(7\/5) * 130<\/gadget>\n182<\/output>\n182 - 100<\/gadget>\n82<\/output>\n82<\/result>","index":2835} +{"problem":"in a class of 37 students 26 play football and play 20 long tennis , if 17 play above , many play neither ?","rationale":"\"26 + 20 - 17 = 29 37 - 29 = 8 play neither answer is b\"","correct":"b","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(37, 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>\n37 - 29<\/gadget>\n8<\/output>\n8<\/result>","index":2838} +{"problem":"a boat can travel with a speed of 15 km \/ hr in still water . if the speed of the stream is 6 km \/ hr , find the time taken by the boat to go 86 km downstream .","rationale":"\"speed of boat in still water = 15 km \/ hr speed of the stream = 6 km \/ hr speed downstream = ( 15 + 6 ) = 21 km \/ hr time taken to travel 86 km downstream = 86 ⠁ „ 16 = 17 ⠁ „ 4 = 4.1 hours answer is a\"","correct":"a","options":{"a":"4.1 hr ","b":"5.25 hr ","c":"8.25 hr ","d":"2.25 hr","e":"2.50 hr"},"options_float":{"a":4.1,"b":5.25,"c":8.25,"d":2.25,"e":2.5},"annotated_formula":"divide(86, add(15, 6))","linear_formula":"add(n0,n1)|divide(n2,#0)|","chain":"15 + 6<\/gadget>\n21<\/output>\n86 \/ 21<\/gadget>\n86\/21 = around 4.095238<\/output>\n86\/21 = around 4.095238<\/result>","index":2839} +{"problem":"bhanu spends 30 % of his income on petrol on scooter 12 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ?","rationale":"given 30 % ( income ) = 300 ⇒ ⇒ income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 12 % ( 700 ) = rs . 84 answer : c","correct":"c","options":{"a":"62 ","b":"140 ","c":"84 ","d":"60","e":"123"},"options_float":{"a":62.0,"b":140.0,"c":84.0,"d":60.0,"e":123.0},"annotated_formula":"multiply(subtract(divide(300, divide(30, const_100)), 300), divide(12, const_100))","linear_formula":"divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)","chain":"30 \/ 100<\/gadget>\n3\/10 = around 0.3<\/output>\n300 \/ (3\/10)<\/gadget>\n1_000<\/output>\n1_000 - 300<\/gadget>\n700<\/output>\n12 \/ 100<\/gadget>\n3\/25 = around 0.12<\/output>\n700 * (3\/25)<\/gadget>\n84<\/output>\n84<\/result>","index":2841} +{"problem":"a train running at a speed of 36 km \/ h passes an electric pole in 15 seconds . in how many seconds will the whole train pass a 370 - meter long platform ?","rationale":"\"let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km \/ h = 36000 m \/ 3600 s = 10 m \/ s x = 15 * 10 = 150 m . the time taken to pass the platform = ( 150 + 370 ) \/ 10 = 52 seconds the answer is d .\"","correct":"d","options":{"a":"46 ","b":"48 ","c":"50 ","d":"52","e":"54"},"options_float":{"a":46.0,"b":48.0,"c":50.0,"d":52.0,"e":54.0},"annotated_formula":"divide(add(multiply(multiply(36, const_0_2778), 15), 370), multiply(36, const_0_2778))","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>\n36 * (5\/18)<\/gadget>\n10<\/output>\n10 * 15<\/gadget>\n150<\/output>\n150 + 370<\/gadget>\n520<\/output>\n520 \/ 10<\/gadget>\n52<\/output>\n52<\/result>","index":2842} +{"problem":"a pump can fill a tank with water in 3 hours . because of a leak , it took 3 1 \/ 3 hours to fill the tank . the leak can drain all the water of the tank in ?","rationale":"\"work done by the tank in 1 hour = ( 1 \/ 3 - 3 1 \/ 3 ) = 1 \/ 30 leak will empty the tank in 30 hrs . answer : c\"","correct":"c","options":{"a":"17 hr ","b":"19 hr ","c":"30 hr ","d":"14 hr","e":"16 hr"},"options_float":{"a":17.0,"b":19.0,"c":30.0,"d":14.0,"e":16.0},"annotated_formula":"inverse(subtract(divide(1, 3), inverse(divide(add(multiply(3, 3), 1), 3))))","linear_formula":"divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)|","chain":"1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 + 1<\/gadget>\n10<\/output>\n10 \/ 3<\/gadget>\n10\/3 = around 3.333333<\/output>\n1 \/ (10\/3)<\/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":2843} +{"problem":"from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 13 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ?","rationale":"\"soln : - 13 x = 45 - - > 87 \/ 77 * x = 45 - - > x = 45 * 77 \/ 87 = 677 \/ 17 = ~ 40 . answer : c .\"","correct":"c","options":{"a":"38 ","b":"39 ","c":"40 ","d":"41","e":"42"},"options_float":{"a":38.0,"b":39.0,"c":40.0,"d":41.0,"e":42.0},"annotated_formula":"multiply(divide(const_100, add(const_100, 13)), 45)","linear_formula":"add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)|","chain":"100 + 13<\/gadget>\n113<\/output>\n100 \/ 113<\/gadget>\n100\/113 = around 0.884956<\/output>\n(100\/113) * 45<\/gadget>\n4_500\/113 = around 39.823009<\/output>\n4_500\/113 = around 39.823009<\/result>","index":2844} +{"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 r 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 r 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":2846} +{"problem":"if ( a - b - c + d = 18 ) and ( a + b - c - d = 4 ) , what is the value of ( b - d ) ^ 2 ?","rationale":"\"eq 1 : a - b - c + d = 18 eq 2 : a + b - c - d = 4 ( 1 ) subtract eq 1 from eq 2 a - b - c + d = 18 - a + b - c - d = 4 - - - - - - - - - - - - - - - - - - - - - - - - - 2 b + 2 d = 14 ( 2 ) simplify - b + d = 7 b - d = - 7 ( b - d ) ^ 2 = ( - 7 ) ^ 2 = 49 my answer : a\"","correct":"a","options":{"a":"49 . ","b":"8 . ","c":"12 . ","d":"16 .","e":"64 ."},"options_float":{"a":49.0,"b":8.0,"c":12.0,"d":16.0,"e":64.0},"annotated_formula":"power(subtract(4, divide(add(18, 4), 2)), 2)","linear_formula":"add(n0,n1)|divide(#0,n2)|subtract(n1,#1)|power(#2,n2)|","chain":"18 + 4<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n4 - 11<\/gadget>\n-7<\/output>\n(-7) ** 2<\/gadget>\n49<\/output>\n49<\/result>","index":2847} +{"problem":"john purchased some shirts and trousers for $ 800 . he paid $ 400 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ?","rationale":"given that the total purchase of two items cost 800 . so the average purchase of one item will cost 800 \/ 2 = 400 . its given as total shirt cost 400 $ less . hence total shirt cost = 400 - 200 and total trouser cost = 400 + 200 5 shirts = 200 $ = = > one shirt = 40 $ one trouser = 40 + 20 = 60 $ total trousers = 600 \/ 60 = 10 . e","correct":"e","options":{"a":"4 ","b":"5 ","c":"6 ","d":"7","e":"10"},"options_float":{"a":4.0,"b":5.0,"c":6.0,"d":7.0,"e":10.0},"annotated_formula":"divide(subtract(800, multiply(5, add(20, 20))), add(add(20, 20), 20))","linear_formula":"add(n3,n3)|add(n3,#0)|multiply(n2,#0)|subtract(n0,#2)|divide(#3,#1)","chain":"20 + 20<\/gadget>\n40<\/output>\n5 * 40<\/gadget>\n200<\/output>\n800 - 200<\/gadget>\n600<\/output>\n40 + 20<\/gadget>\n60<\/output>\n600 \/ 60<\/gadget>\n10<\/output>\n10<\/result>","index":2848} +{"problem":"the average age of a class of 24 students is 23 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ?","rationale":"\"total age of all students = 24 ã — 23 total age of all students + age of the teacher = 25 ã — 24 age of the teacher = 25 ã — 24 â ˆ ’ 24 ã — 23 = 24 ( 25 â ˆ ’ 23 ) = 24 ã — 2 = 48 answer is c .\"","correct":"c","options":{"a":"40 ","b":"41 ","c":"48 ","d":"52","e":"43"},"options_float":{"a":40.0,"b":41.0,"c":48.0,"d":52.0,"e":43.0},"annotated_formula":"subtract(multiply(add(24, 1), add(23, 1)), multiply(24, 23))","linear_formula":"add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|","chain":"24 + 1<\/gadget>\n25<\/output>\n23 + 1<\/gadget>\n24<\/output>\n25 * 24<\/gadget>\n600<\/output>\n24 * 23<\/gadget>\n552<\/output>\n600 - 552<\/gadget>\n48<\/output>\n48<\/result>","index":2849} +{"problem":"45 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 : 45 : : 18 : x [ please pay attention , we have written 27 : 45 rather than 45 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \\ color { blue } x = \\ frac { 45 \\ times 18 } { 27 } } x = 30 so 30 days will be required to get work done by 27 men . answer : c\"","correct":"c","options":{"a":"24 ","b":"77 ","c":"30 ","d":"25","e":"13"},"options_float":{"a":24.0,"b":77.0,"c":30.0,"d":25.0,"e":13.0},"annotated_formula":"divide(multiply(18, 45), 27)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|","chain":"18 * 45<\/gadget>\n810<\/output>\n810 \/ 27<\/gadget>\n30<\/output>\n30<\/result>","index":2850} +{"problem":"27 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":"\"( 27 * 8 ) \/ 30 = ( x * 6 ) \/ 50 = > x = 60 60 – 27 = 33 answer : a\"","correct":"a","options":{"a":"33 ","b":"66 ","c":"88 ","d":"100","e":"281"},"options_float":{"a":33.0,"b":66.0,"c":88.0,"d":100.0,"e":281.0},"annotated_formula":"subtract(divide(multiply(divide(multiply(27, 8), 30), 50), 6), 27)","linear_formula":"multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)|","chain":"27 * 8<\/gadget>\n216<\/output>\n216 \/ 30<\/gadget>\n36\/5 = around 7.2<\/output>\n(36\/5) * 50<\/gadget>\n360<\/output>\n360 \/ 6<\/gadget>\n60<\/output>\n60 - 27<\/gadget>\n33<\/output>\n33<\/result>","index":2851} +{"problem":"n and m are each 3 - digit integers . each of the numbers 2 , 3 , 4,5 , 6 , and 7 is a digit of either n or m . what is the smallest possible positive difference between n and m ?","rationale":"\"you have 6 digits : 2 , 3 , 4 , 5 , 6 , 7 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 2 and 7 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 3 * * and 4 * * ) . make the 3 * * number as large as possible and make the 4 * * number as small as possible . 376 and 425 ( difference is 49 ) or try and make a pair with ( 5 * * and 6 * * ) . make the 5 * * number as large as possible and make the 6 * * number as small as possible . we get 574 and 623 ( difference is 49 ) b\"","correct":"b","options":{"a":"59 ","b":"49 ","c":"58 ","d":"113","e":"131"},"options_float":{"a":59.0,"b":49.0,"c":58.0,"d":113.0,"e":131.0},"annotated_formula":"subtract(subtract(const_100, multiply(subtract(7, 2), const_10)), const_1)","linear_formula":"subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)|","chain":"7 - 2<\/gadget>\n5<\/output>\n5 * 10<\/gadget>\n50<\/output>\n100 - 50<\/gadget>\n50<\/output>\n50 - 1<\/gadget>\n49<\/output>\n49<\/result>","index":2852} +{"problem":"15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 4 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 4 hrs a day will take ( 15 * 21 * 8 ) \/ ( 21 * 4 * ( 2 \/ 3 ) ) days = > 45 days answer : e\"","correct":"e","options":{"a":"30 ","b":"20 ","c":"15 ","d":"25","e":"45"},"options_float":{"a":30.0,"b":20.0,"c":15.0,"d":25.0,"e":45.0},"annotated_formula":"divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 4), 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 * 4<\/gadget>\n84<\/output>\n2 \/ 3<\/gadget>\n2\/3 = around 0.666667<\/output>\n84 * (2\/3)<\/gadget>\n56<\/output>\n2_520 \/ 56<\/gadget>\n45<\/output>\n45<\/result>","index":2854} +{"problem":"on dividing 21 by a number , the quotient is 10 and the remainder is 1 . find the divisor .","rationale":"\"d = ( d - r ) \/ q = ( 21 - 1 ) \/ 10 = 20 \/ 10 = 2 b\"","correct":"b","options":{"a":"1 ","b":"2 ","c":"4 ","d":"6","e":"7"},"options_float":{"a":1.0,"b":2.0,"c":4.0,"d":6.0,"e":7.0},"annotated_formula":"floor(divide(21, 10))","linear_formula":"divide(n0,n1)|floor(#0)|","chain":"21 \/ 10<\/gadget>\n21\/10 = around 2.1<\/output>\nfloor(21\/10)<\/gadget>\n2<\/output>\n2<\/result>","index":2856} +{"problem":"if length of a rectangle is equal to side of a square and breadth of rectangle is half of length . if area of square is 36 sq . m . calculate the area of rectangle ?","rationale":"side of square = √ 36 = 6 m . length = 6 m and breadth = 3 m area of rectangle = 6 * 3 = 18 sq . m answer a","correct":"a","options":{"a":"18 ","b":"20 ","c":"27 ","d":"32","e":"25"},"options_float":{"a":18.0,"b":20.0,"c":27.0,"d":32.0,"e":25.0},"annotated_formula":"multiply(sqrt(36), divide(sqrt(36), const_2))","linear_formula":"sqrt(n0)|divide(#0,const_2)|multiply(#1,#0)","chain":"36 ** (1\/2)<\/gadget>\n6<\/output>\n6 \/ 2<\/gadget>\n3<\/output>\n6 * 3<\/gadget>\n18<\/output>\n18<\/result>","index":2857} +{"problem":"the price of a t . v . set worth rs . 70000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ?","rationale":"\"money paid in cash = rs . 1000 balance payment = ( 70000 - 1000 ) = rs . 69000 answer : c\"","correct":"c","options":{"a":"22678 ","b":"26699 ","c":"69000 ","d":"19000","e":"26711"},"options_float":{"a":22678.0,"b":26699.0,"c":69000.0,"d":19000.0,"e":26711.0},"annotated_formula":"subtract(70000, 1000)","linear_formula":"subtract(n0,n2)|","chain":"70_000 - 1_000<\/gadget>\n69_000<\/output>\n69_000<\/result>","index":2858} +{"problem":"the original price of a suit is $ 200 . the price increased 20 % , and after this increase , the store published a 20 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 20 % off the increased price , how much did these consumers pay for the suit ?","rationale":"\"0.8 * ( 1.2 * 200 ) = $ 192 the answer is a .\"","correct":"a","options":{"a":"$ 192 ","b":"$ 198 ","c":"$ 200 ","d":"$ 208","e":"$ 216"},"options_float":{"a":192.0,"b":198.0,"c":200.0,"d":208.0,"e":216.0},"annotated_formula":"subtract(add(200, divide(multiply(200, 20), const_100)), divide(multiply(add(200, divide(multiply(200, 20), const_100)), 20), 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":"200 * 20<\/gadget>\n4_000<\/output>\n4_000 \/ 100<\/gadget>\n40<\/output>\n200 + 40<\/gadget>\n240<\/output>\n240 * 20<\/gadget>\n4_800<\/output>\n4_800 \/ 100<\/gadget>\n48<\/output>\n240 - 48<\/gadget>\n192<\/output>\n192<\/result>","index":2859} +{"problem":"in a basketball game , dhoni scored 30 points more than dravid , but only half as many points as shewag . if the 3 players scored a combined total of 150 points , how many points did dhoni score ?","rationale":"let dravid scored point = x then dhoni scored = x + 30 shewag scored = 2 * ( x + 30 ) = 2 x + 60 as given , x + x + 30 + 2 x + 60 = 150 points 4 x + 90 = 150 x = 150 - 90 \/ 4 = 15 so dhoni scored = x + 30 i . e ) 15 + 30 = 45 answer : e","correct":"e","options":{"a":"50 ","b":"52 ","c":"35 ","d":"40","e":"45"},"options_float":{"a":50.0,"b":52.0,"c":35.0,"d":40.0,"e":45.0},"annotated_formula":"divide(add(150, 30), add(add(const_2, const_1), const_1))","linear_formula":"add(n0,n2)|add(const_1,const_2)|add(#1,const_1)|divide(#0,#2)","chain":"150 + 30<\/gadget>\n180<\/output>\n2 + 1<\/gadget>\n3<\/output>\n3 + 1<\/gadget>\n4<\/output>\n180 \/ 4<\/gadget>\n45<\/output>\n45<\/result>","index":2862} +{"problem":"on a partly cloudy day , milton decides to walk back from work . when it is sunny , he walks at a speed of s miles \/ hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles \/ hr . if his average speed for the entire distance is 2.8 miles \/ hr , what fraction of the total distance did he cover while the sun was shining on him ?","rationale":"if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 \/ 4 of the total time . the formula for distance \/ rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is 2 \/ 14 = 1 \/ 7 . answer : d","correct":"d","options":{"a":"1 \/ 5 ","b":"1 \/ 6 ","c":"1 \/ 4 ","d":"1 \/ 7","e":"1 \/ 3"},"options_float":{"a":0.2,"b":0.1666666667,"c":0.25,"d":0.1428571429,"e":0.3333333333},"annotated_formula":"divide(1, divide(add(add(2.8, add(2.8, 2.8)), add(2.8, 2.8)), const_2))","linear_formula":"add(n1,n1)|add(n1,#0)|add(#1,#0)|divide(#2,const_2)|divide(n0,#3)","chain":"2.8 + 2.8<\/gadget>\n5.6<\/output>\n2.8 + 5.6<\/gadget>\n8.4<\/output>\n8.4 + 5.6<\/gadget>\n14<\/output>\n14 \/ 2<\/gadget>\n7<\/output>\n1 \/ 7<\/gadget>\n1\/7 = around 0.142857<\/output>\n1\/7 = around 0.142857<\/result>","index":2863} +{"problem":"if the sample interest on a sum of money 10 % per annum for 2 years is $ 1200 , find the compound interest on the same sum for the same period at the same rate ?","rationale":"rate = 10 % time = 2 years s . i . = $ 1200 principal = 100 * 1200 \/ 10 * 2 = $ 6000 amount = 6000 ( 1 + 10 \/ 100 ) ^ 2 = $ 7260 c . i . = 7260 - 6000 = $ 1260 answer is a","correct":"a","options":{"a":"$ 1260 ","b":"$ 1520 ","c":"$ 1356 ","d":"$ 1440","e":"$ 1210"},"options_float":{"a":1260.0,"b":1520.0,"c":1356.0,"d":1440.0,"e":1210.0},"annotated_formula":"subtract(add(divide(multiply(add(divide(multiply(1200, const_100), multiply(10, 2)), divide(multiply(divide(multiply(1200, const_100), multiply(10, 2)), 10), const_100)), 10), const_100), add(divide(multiply(1200, const_100), multiply(10, 2)), divide(multiply(divide(multiply(1200, const_100), multiply(10, 2)), 10), const_100))), divide(multiply(1200, const_100), multiply(10, 2)))","linear_formula":"multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_100)|add(#2,#4)|multiply(n0,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)","chain":"1_200 * 100<\/gadget>\n120_000<\/output>\n10 * 2<\/gadget>\n20<\/output>\n120_000 \/ 20<\/gadget>\n6_000<\/output>\n6_000 * 10<\/gadget>\n60_000<\/output>\n60_000 \/ 100<\/gadget>\n600<\/output>\n6_000 + 600<\/gadget>\n6_600<\/output>\n6_600 * 10<\/gadget>\n66_000<\/output>\n66_000 \/ 100<\/gadget>\n660<\/output>\n660 + 6_600<\/gadget>\n7_260<\/output>\n7_260 - 6_000<\/gadget>\n1_260<\/output>\n1_260<\/result>","index":2864} +{"problem":"if 35 % of a number is 12 less than 50 % of that number , then the number is ?","rationale":"\"let the number be x . then , 50 % of x - 35 % of x = 12 50 \/ 100 x - 35 \/ 100 x = 12 x = ( 12 * 100 ) \/ 15 = 80 . answer : d\"","correct":"d","options":{"a":"40 ","b":"50 ","c":"60 ","d":"80","e":"70"},"options_float":{"a":40.0,"b":50.0,"c":60.0,"d":80.0,"e":70.0},"annotated_formula":"divide(12, 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>\n12 \/ (3\/20)<\/gadget>\n80<\/output>\n80<\/result>","index":2865} +{"problem":"rs 50000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 8000 find money given by 10 % ?","rationale":"let first parrt is x and second part is y then x + y = 50000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 8000 = ( x * 10 * 1 ) \/ 100 + ( y * 20 * 1 ) \/ 100 80000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 80000 = 50000 + y so y = 30000 then x = 50000 - 30000 = 20000 first part = 20000 answer : a","correct":"a","options":{"a":"20000 ","b":"40000 ","c":"50000 ","d":"60000","e":"70000"},"options_float":{"a":20000.0,"b":40000.0,"c":50000.0,"d":60000.0,"e":70000.0},"annotated_formula":"divide(subtract(divide(multiply(50000, 20), const_100), 8000), divide(10, const_100))","linear_formula":"divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(#2,n3)|divide(#3,#0)","chain":"50_000 * 20<\/gadget>\n1_000_000<\/output>\n1_000_000 \/ 100<\/gadget>\n10_000<\/output>\n10_000 - 8_000<\/gadget>\n2_000<\/output>\n10 \/ 100<\/gadget>\n1\/10 = around 0.1<\/output>\n2_000 \/ (1\/10)<\/gadget>\n20_000<\/output>\n20_000<\/result>","index":2867} +{"problem":"a number when divided by a certain divisor left remainder 245 , when twice the number was divided by the same divisor , the remainder was 112 . find the divisor ?","rationale":"\"easy solution : n = dq 1 + 245 2 n = 2 dq 1 + 490 - ( 1 ) 2 n = dq 2 + 112 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 378 d * some integer = 378 checking all options only ( a ) syncs with it . answer a\"","correct":"a","options":{"a":"378 ","b":"365 ","c":"380 ","d":"456","e":"460"},"options_float":{"a":378.0,"b":365.0,"c":380.0,"d":456.0,"e":460.0},"annotated_formula":"subtract(multiply(245, const_2), 112)","linear_formula":"multiply(n0,const_2)|subtract(#0,n1)|","chain":"245 * 2<\/gadget>\n490<\/output>\n490 - 112<\/gadget>\n378<\/output>\n378<\/result>","index":2869} +{"problem":"85 white and black tiles will be used to form a 10 x 10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ?","rationale":"\"answer = a please refer diagram below 85 - 10 = 75\"","correct":"a","options":{"a":"75 ","b":"80 ","c":"85 ","d":"90","e":"95"},"options_float":{"a":75.0,"b":80.0,"c":85.0,"d":90.0,"e":95.0},"annotated_formula":"subtract(85, 10)","linear_formula":"subtract(n0,n1)|","chain":"85 - 10<\/gadget>\n75<\/output>\n75<\/result>","index":2871} +{"problem":"the parameter of a square is equal to the perimeter of a rectangle of length 16 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 . parameter of the rectangle = 2 ( 16 + 14 ) = 60 cm parameter of the square = 60 cm i . e . 4 a = 60 a = 15 diameter of the semicircle = 15 cm circimference of the semicircle = 1 \/ 2 ( ∏ ) ( 15 ) = 1 \/ 2 ( 22 \/ 7 ) ( 15 ) = 330 \/ 14 = 23.57 cm to two decimal places answer : option e\"","correct":"e","options":{"a":"34 ","b":"35 ","c":"56 ","d":"67","e":"23.57"},"options_float":{"a":34.0,"b":35.0,"c":56.0,"d":67.0,"e":23.57},"annotated_formula":"divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(16, 14)), const_2)), const_2)","linear_formula":"rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)|","chain":"2 * (16 + 14)<\/gadget>\n60<\/output>\n60 \/ 4<\/gadget>\n15<\/output>\n15 \/ 2<\/gadget>\n15\/2 = around 7.5<\/output>\n2 * pi * (15\/2)<\/gadget>\n15*pi = around 47.12389<\/output>\n(15*pi) \/ 2<\/gadget>\n15*pi\/2 = around 23.561945<\/output>\n15*pi\/2 = around 23.561945<\/result>","index":2872} +{"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 height of the building be x meters 40.25 : 28.75 : : 17.5 < = > 40.25 x x = 28.75 x 17.5 x = 28.75 x 17.5 \/ 40.25 x = 12.5 answer : option b\"","correct":"b","options":{"a":"10 m ","b":"12.5 m ","c":"17.5 m ","d":"21.25 m","e":"none"},"options_float":{"a":10.0,"b":12.5,"c":17.5,"d":21.25,"e":null},"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":2873} +{"problem":"x can do a piece of work in 4 hours ; y and z together can do it in 3 hours , while x and z together can do it in 2 hours . how long will y alone take to do it ?","rationale":"x 1 hour ' s work = 1 \/ 4 ; y + z ' s hour ' s work = 1 \/ 3 x + y + z ' s 1 hour ' s work = 1 \/ 4 + 1 \/ 3 = 7 \/ 12 y ' s 1 hour ' s work = ( 7 \/ 12 - 1 \/ 2 ) = 1 \/ 12 . y alone will take 12 hours to do the work . c","correct":"c","options":{"a":"5 hours ","b":"10 hours ","c":"12 hours ","d":"24 hours","e":"15 hours"},"options_float":{"a":5.0,"b":10.0,"c":12.0,"d":24.0,"e":15.0},"annotated_formula":"inverse(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)|inverse(#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":2874} +{"problem":"in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 40 % of all the employees are men . what % of all the employee went to the picnic ?","rationale":"\"total men in company 40 % means total women in company 60 % ( assume total people in company 100 % ) no of men employees attended picnic = 40 x ( 20 \/ 100 ) = 8 no of women employees attended picnic = 60 x ( 40 \/ 100 ) = 24 total percentage of employees attended the picnic = 8 + 24 = 32 % answer : a\"","correct":"a","options":{"a":"32 % ","b":"34 % ","c":"35 % ","d":"36 %","e":"37 %"},"options_float":{"a":32.0,"b":34.0,"c":35.0,"d":36.0,"e":37.0},"annotated_formula":"multiply(add(multiply(divide(40, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 40), const_100), divide(40, const_100))), const_100)","linear_formula":"divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100)|","chain":"40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n(2\/5) * (1\/5)<\/gadget>\n2\/25 = around 0.08<\/output>\n100 - 40<\/gadget>\n60<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) * (2\/5)<\/gadget>\n6\/25 = around 0.24<\/output>\n(2\/25) + (6\/25)<\/gadget>\n8\/25 = around 0.32<\/output>\n(8\/25) * 100<\/gadget>\n32<\/output>\n32<\/result>","index":2876} +{"problem":"excluding stoppages , the speed of a bus is 84 kmph and including stoppages , it is 70 kmph . for how many minutes does the bus stop per hour ?","rationale":"\"due to stoppages , it covers 14 km less . time taken to cover 14 km = ( 14 \/ 84 x 60 ) min = 10 min answer : b\"","correct":"b","options":{"a":"12 min ","b":"10 min ","c":"15 min ","d":"14 min","e":"13 min"},"options_float":{"a":12.0,"b":10.0,"c":15.0,"d":14.0,"e":13.0},"annotated_formula":"multiply(const_60, divide(subtract(84, 70), 84))","linear_formula":"subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)|","chain":"84 - 70<\/gadget>\n14<\/output>\n14 \/ 84<\/gadget>\n1\/6 = around 0.166667<\/output>\n60 * (1\/6)<\/gadget>\n10<\/output>\n10<\/result>","index":2878} +{"problem":"40 is what percent of 160 ?","rationale":"40 \/ 160 × 100 = 25 % answer : e","correct":"e","options":{"a":"35 % ","b":"40 % ","c":"45 % ","d":"50 %","e":"25 %"},"options_float":{"a":35.0,"b":40.0,"c":45.0,"d":50.0,"e":25.0},"annotated_formula":"multiply(divide(40, 160), const_100)","linear_formula":"divide(n0,n1)|multiply(#0,const_100)|","chain":"40 \/ 160<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2879} +{"problem":"in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 57 liters and they are all empty , how much money total will it cost to fuel all cars ?","rationale":"\"total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 57 ) = 465.60 hence answer will be ( e )\"","correct":"e","options":{"a":"320.50 $ ","b":"380.50 $ ","c":"425.50 $ ","d":"450.50 $","e":"465.60 $"},"options_float":{"a":320.5,"b":380.5,"c":425.5,"d":450.5,"e":465.6},"annotated_formula":"multiply(multiply(0.65, 57), 12)","linear_formula":"multiply(n1,n3)|multiply(n2,#0)|","chain":"0.65 * 57<\/gadget>\n37.05<\/output>\n37.05 * 12<\/gadget>\n444.6<\/output>\n444.6<\/result>","index":2880} +{"problem":"the l . c . m of 22 , 54 , 108 , 135 and 198 is","rationale":"answer : c ) 5940","correct":"c","options":{"a":"5942 ","b":"2887 ","c":"5940 ","d":"2888","e":"28881"},"options_float":{"a":5942.0,"b":2887.0,"c":5940.0,"d":2888.0,"e":28881.0},"annotated_formula":"multiply(multiply(multiply(multiply(const_2, const_2), multiply(multiply(const_3, const_3), const_3)), divide(divide(divide(135, const_3), const_3), const_3)), divide(22, const_2))","linear_formula":"divide(n0,const_2)|divide(n3,const_3)|multiply(const_2,const_2)|multiply(const_3,const_3)|divide(#1,const_3)|multiply(#3,const_3)|divide(#4,const_3)|multiply(#2,#5)|multiply(#6,#7)|multiply(#0,#8)","chain":"2 * 2<\/gadget>\n4<\/output>\n3 * 3<\/gadget>\n9<\/output>\n9 * 3<\/gadget>\n27<\/output>\n4 * 27<\/gadget>\n108<\/output>\n135 \/ 3<\/gadget>\n45<\/output>\n45 \/ 3<\/gadget>\n15<\/output>\n15 \/ 3<\/gadget>\n5<\/output>\n108 * 5<\/gadget>\n540<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n540 * 11<\/gadget>\n5_940<\/output>\n5_940<\/result>","index":2881} +{"problem":"the sale price sarees listed for rs . 280 after successive discount is 12 % and 8 % is ?","rationale":"\"280 * ( 88 \/ 100 ) * ( 92 \/ 100 ) = 226 answer : b\"","correct":"b","options":{"a":"288 ","b":"226 ","c":"250 ","d":"230","e":"262"},"options_float":{"a":288.0,"b":226.0,"c":250.0,"d":230.0,"e":262.0},"annotated_formula":"subtract(subtract(280, divide(multiply(280, 12), const_100)), divide(multiply(subtract(280, divide(multiply(280, 12), const_100)), 8), const_100))","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|","chain":"280 * 12<\/gadget>\n3_360<\/output>\n3_360 \/ 100<\/gadget>\n168\/5 = around 33.6<\/output>\n280 - (168\/5)<\/gadget>\n1_232\/5 = around 246.4<\/output>\n(1_232\/5) * 8<\/gadget>\n9_856\/5 = around 1_971.2<\/output>\n(9_856\/5) \/ 100<\/gadget>\n2_464\/125 = around 19.712<\/output>\n(1_232\/5) - (2_464\/125)<\/gadget>\n28_336\/125 = around 226.688<\/output>\n28_336\/125 = around 226.688<\/result>","index":2882} +{"problem":"a girl sitting in a train which is travelling at 40 kmph observes that a goods train travelling in a opposite direction , takes 12 seconds to pass him . if the goods train is 1120 m long , find its speed .","rationale":"relative speed = ( 1120 \/ 12 ) m \/ s = ( 1120 \/ 12 ) * ( 18 \/ 5 ) = 336 kmph speed of goods train = 336 - 40 = 296 kmph answer is b","correct":"b","options":{"a":"295 ","b":"296 ","c":"297 ","d":"298","e":"299"},"options_float":{"a":295.0,"b":296.0,"c":297.0,"d":298.0,"e":299.0},"annotated_formula":"subtract(divide(divide(1120, 12), const_0_2778), 40)","linear_formula":"divide(n2,n1)|divide(#0,const_0_2778)|subtract(#1,n0)","chain":"1_120 \/ 12<\/gadget>\n280\/3 = around 93.333333<\/output>\n10 \/ 36<\/gadget>\n5\/18 = around 0.277778<\/output>\n(280\/3) \/ (5\/18)<\/gadget>\n336<\/output>\n336 - 40<\/gadget>\n296<\/output>\n296<\/result>","index":2883} +{"problem":"a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction t of the sum of the 21 numbers in the list ?","rationale":"\"this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction t of n to the total would be 4 a \/ 24 a or 1 \/ 6 answer b\"","correct":"b","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, 4), 21), 4), const_2), 4), const_3))","linear_formula":"divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|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 - 4<\/gadget>\n9<\/output>\n9 - 3<\/gadget>\n6<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n1\/6 = around 0.166667<\/result>","index":2884} +{"problem":"a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 66 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480","rationale":"\"1 percent for 3 years = 66 1 percent for 1 year = 22 = > 100 percent = 2200 answer : c\"","correct":"c","options":{"a":"2000 ","b":"2100 ","c":"2200 ","d":"2300","e":"2400"},"options_float":{"a":2000.0,"b":2100.0,"c":2200.0,"d":2300.0,"e":2400.0},"annotated_formula":"multiply(divide(66, 3), const_100)","linear_formula":"divide(n2,n0)|multiply(#0,const_100)|","chain":"66 \/ 3<\/gadget>\n22<\/output>\n22 * 100<\/gadget>\n2_200<\/output>\n2_200<\/result>","index":2886} +{"problem":"andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.82 . what is the median gasoline price ?","rationale":"ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.82 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.82 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b","correct":"b","options":{"a":"$ 1 ","b":"$ 1.82 ","c":"$ 1.92 ","d":"$ 2.13","e":"$ 2.15"},"options_float":{"a":1.0,"b":1.82,"c":1.92,"d":2.13,"e":2.15},"annotated_formula":"min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.82), 7), 1.82)","linear_formula":"add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|add(n6,#3)|add(n7,#4)|divide(#5,n0)|min(n7,#6)","chain":"1.75 + 1.61<\/gadget>\n3.36<\/output>\n3.36 + 1.79<\/gadget>\n5.15<\/output>\n5.15 + 2.11<\/gadget>\n7.26<\/output>\n7.26 + 1.96<\/gadget>\n9.22<\/output>\n9.22 + 2.09<\/gadget>\n11.31<\/output>\n11.31 + 1.82<\/gadget>\n13.13<\/output>\n13.13 \/ 7<\/gadget>\n1.875714<\/output>\nmin(1.875714, 1.82)<\/gadget>\n1.82<\/output>\n1.82<\/result>","index":2889} +{"problem":"if 625 ( 5 ^ x ) = 1 then x =","rationale":"\"5 ^ x = 1 \/ 625 5 ^ x = 1 \/ 5 ^ 4 5 ^ x = 5 ^ - 4 x = - 4 b\"","correct":"b","options":{"a":"– 2 ","b":"- 4 ","c":"0 ","d":"- 1","e":"2"},"options_float":{"a":2.0,"b":-4.0,"c":0.0,"d":-1.0,"e":2.0},"annotated_formula":"divide(log(divide(1, 625)), log(5))","linear_formula":"divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|","chain":"1 \/ 625<\/gadget>\n1\/625 = around 0.0016<\/output>\nlog(1\/625)<\/gadget>\n-log(625) = around -6.437752<\/output>\nlog(5)<\/gadget>\nlog(5) = around 1.609438<\/output>\n(-log(625)) \/ log(5)<\/gadget>\n-log(625)\/log(5) = around -4<\/output>\n-log(625)\/log(5) = around -4<\/result>","index":2890} +{"problem":"find the least number of complete years in which a sum of money put out at 45 % 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(45, 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>\n45 \/ 100<\/gadget>\n9\/20 = around 0.45<\/output>\n1 + (9\/20)<\/gadget>\n29\/20 = around 1.45<\/output>\nlog(29\/20)<\/gadget>\nlog(29\/20) = around 0.371564<\/output>\nlog(2) \/ log(29\/20)<\/gadget>\nlog(2)\/log(29\/20) = around 1.865488<\/output>\n(log(2)\/log(29\/20)) + 1<\/gadget>\n1 + log(2)\/log(29\/20) = around 2.865488<\/output>\nfloor(1 + log(2)\/log(29\/20))<\/gadget>\n2<\/output>\n2<\/result>","index":2891} +{"problem":"the speed of the boat in still water in 12 kmph . it can travel downstream through 45 kms in 3 hrs . in what time would it cover the same distance upstream ?","rationale":"still water = 12 km \/ hr downstream = 45 \/ 3 = 15 km \/ hr upstream = > > still water = ( u + v \/ 2 ) = > > 12 = u + 15 \/ 2 = 9 km \/ hr so time taken in upstream = 45 \/ 9 = 5 hrs answer : d","correct":"d","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(45, subtract(12, subtract(divide(45, 3), 12)))","linear_formula":"divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)","chain":"45 \/ 3<\/gadget>\n15<\/output>\n15 - 12<\/gadget>\n3<\/output>\n12 - 3<\/gadget>\n9<\/output>\n45 \/ 9<\/gadget>\n5<\/output>\n5<\/result>","index":2892} +{"problem":"the avg weight of a , b & c is 55 kg . if d joins the group , the avg weight of the group becomes 60 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 58 kg . what is the weight of a ?","rationale":"\"a + b + c = 3 * 55 = 165 a + b + c + d = 4 * 60 = 240 - - - - ( i ) so , d = 75 & e = 75 + 3 = 78 b + c + d + e = 58 * 4 = 232 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 240 – 232 = 8 a = e + 8 = 78 + 8 = 86 answer : d\"","correct":"d","options":{"a":"56 ","b":"65 ","c":"75 ","d":"86","e":"90"},"options_float":{"a":56.0,"b":65.0,"c":75.0,"d":86.0,"e":90.0},"annotated_formula":"subtract(multiply(60, const_4), subtract(multiply(58, const_4), add(3, subtract(multiply(60, const_4), multiply(55, 3)))))","linear_formula":"multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,n2)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5)|","chain":"60 * 4<\/gadget>\n240<\/output>\n58 * 4<\/gadget>\n232<\/output>\n55 * 3<\/gadget>\n165<\/output>\n240 - 165<\/gadget>\n75<\/output>\n3 + 75<\/gadget>\n78<\/output>\n232 - 78<\/gadget>\n154<\/output>\n240 - 154<\/gadget>\n86<\/output>\n86<\/result>","index":2893} +{"problem":"if ' a ' completes a piece of work in 3 days , which ' b ' completes it in 5 days and ' c ' takes 10 days to complete the same work . how long will they take to complete the work , if they work together ?","rationale":"explanation : hint : a ' s one day work = 1 \/ 3 b ' s one day work = 1 \/ 5 c ' s one day work = 1 \/ 10 ( a + b + c ) ' s one day work = 1 \/ 3 + 1 \/ 5 + 1 \/ 10 = 1 \/ 1.5 hence , a , b & c together will take 1.5 days to complete the work . answer is a","correct":"a","options":{"a":"1.5 days ","b":"4.5 days ","c":"7 days ","d":"9.8 days","e":"9 days"},"options_float":{"a":1.5,"b":4.5,"c":7.0,"d":9.8,"e":9.0},"annotated_formula":"add(subtract(3, const_2), divide(5, 10))","linear_formula":"divide(n1,n2)|subtract(n0,const_2)|add(#0,#1)","chain":"3 - 2<\/gadget>\n1<\/output>\n5 \/ 10<\/gadget>\n1\/2 = around 0.5<\/output>\n1 + (1\/2)<\/gadget>\n3\/2 = around 1.5<\/output>\n3\/2 = around 1.5<\/result>","index":2896} +{"problem":"a certain drink of type a is prepared by mixing 4 parts milk with 3 parts fruit juice . another drink of type b is prepared by mixing 4 parts of fruit juice and 3 parts of milk . how many liters of fruit juice must be added to 105 liters of drink a to convert it to drink b ?","rationale":"\"in 105 liters of drink a , there are 60 liters of milk and 45 liters of juice . with 60 liters of milk , we need a total of 80 liters of juice to make drink b . we need to add 35 liters of juice . the answer is d .\"","correct":"d","options":{"a":"14 ","b":"21 ","c":"28 ","d":"35","e":"42"},"options_float":{"a":14.0,"b":21.0,"c":28.0,"d":35.0,"e":42.0},"annotated_formula":"subtract(divide(multiply(multiply(divide(4, add(4, 3)), 105), 4), 3), multiply(divide(3, add(4, 3)), 105))","linear_formula":"add(n0,n1)|divide(n0,#0)|divide(n1,#0)|multiply(n4,#1)|multiply(n4,#2)|multiply(n0,#3)|divide(#5,n1)|subtract(#6,#4)|","chain":"4 + 3<\/gadget>\n7<\/output>\n4 \/ 7<\/gadget>\n4\/7 = around 0.571429<\/output>\n(4\/7) * 105<\/gadget>\n60<\/output>\n60 * 4<\/gadget>\n240<\/output>\n240 \/ 3<\/gadget>\n80<\/output>\n3 \/ 7<\/gadget>\n3\/7 = around 0.428571<\/output>\n(3\/7) * 105<\/gadget>\n45<\/output>\n80 - 45<\/gadget>\n35<\/output>\n35<\/result>","index":2897} +{"problem":"at what rate percent per annum will a sum of money double in 9 years .","rationale":"\"let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) \/ ( p x 9 ) ] % = 11.1 % per annum . answer : d\"","correct":"d","options":{"a":"12.5 % ","b":"13.5 % ","c":"11.5 % ","d":"11.1 %","e":"21.5 %"},"options_float":{"a":12.5,"b":13.5,"c":11.5,"d":11.1,"e":21.5},"annotated_formula":"divide(const_100, 9)","linear_formula":"divide(const_100,n0)|","chain":"100 \/ 9<\/gadget>\n100\/9 = around 11.111111<\/output>\n100\/9 = around 11.111111<\/result>","index":2898} +{"problem":"if a * b = 2 a - 3 b + ab , then 3 * 5 + 5 * 3 is equal to :","rationale":"\"explanation : 3 * 5 + 5 * 3 = ( 2 * 3 - 3 * 5 + 3 * 5 ) + ( 2 * 5 - 3 * 3 + 5 * 3 ) = ( 6 + 10 - 9 + 15 ) = 22 . answer : a ) 22\"","correct":"a","options":{"a":"22 ","b":"37 ","c":"38 ","d":"398","e":"72"},"options_float":{"a":22.0,"b":37.0,"c":38.0,"d":398.0,"e":72.0},"annotated_formula":"add(multiply(2, 3), multiply(3, 5))","linear_formula":"multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)|","chain":"2 * 3<\/gadget>\n6<\/output>\n3 * 5<\/gadget>\n15<\/output>\n6 + 15<\/gadget>\n21<\/output>\n21<\/result>","index":2901} +{"problem":"noelle walks from point a to point b at an average speed of 3 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 4 kilometers per hour ?","rationale":"\"let ' s suppose that speed while returning was xkm \/ h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 \/ s 1 + s 2 4 = 2 * 3 * x \/ 3 + x x = 6 b is the answer\"","correct":"b","options":{"a":"5.75 ","b":"6 ","c":"7.25 ","d":"7.5","e":"7.75"},"options_float":{"a":5.75,"b":6.0,"c":7.25,"d":7.5,"e":7.75},"annotated_formula":"divide(multiply(4, 3), subtract(multiply(const_2, 3), 4))","linear_formula":"multiply(n0,n1)|multiply(n0,const_2)|subtract(#1,n1)|divide(#0,#2)|","chain":"4 * 3<\/gadget>\n12<\/output>\n2 * 3<\/gadget>\n6<\/output>\n6 - 4<\/gadget>\n2<\/output>\n12 \/ 2<\/gadget>\n6<\/output>\n6<\/result>","index":2902} +{"problem":"the product of x and y is a constant . if the value of x is increased by 40 % , by what percentage must the value of y be decreased ?","rationale":"x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.4 x ( 140 ) i . e . 140 * new ' y ' = 10000 i . e . new ' y ' = 10000 \/ 140 = 71.43 i . e . y decreases from 100 to 71.43 i . e . decrease of 28.57 % c","correct":"c","options":{"a":"34 % ","b":"36 % ","c":"28.57 % ","d":"30 %","e":"32 %"},"options_float":{"a":34.0,"b":36.0,"c":28.57,"d":30.0,"e":32.0},"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":2903} +{"problem":"if a and b are the roots of the equation x 2 - 6 x + 6 = 0 , then the value of a 2 + b 2 is :","rationale":"\"sol . ( b ) the sum of roots = a + b = 6 product of roots = ab = 6 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 36 - 12 = 24 answer b\"","correct":"b","options":{"a":"36 ","b":"24 ","c":"17 ","d":"6","e":"5"},"options_float":{"a":36.0,"b":24.0,"c":17.0,"d":6.0,"e":5.0},"annotated_formula":"add(power(divide(subtract(6, sqrt(subtract(power(6, 2), multiply(const_4, 6)))), 2), 2), power(divide(add(6, sqrt(subtract(power(6, 2), multiply(const_4, 6)))), 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":"6 ** 2<\/gadget>\n36<\/output>\n4 * 6<\/gadget>\n24<\/output>\n36 - 24<\/gadget>\n12<\/output>\n12 ** (1\/2)<\/gadget>\n2*sqrt(3) = around 3.464102<\/output>\n6 - (2*sqrt(3))<\/gadget>\n6 - 2*sqrt(3) = around 2.535898<\/output>\n(6 - 2*sqrt(3)) \/ 2<\/gadget>\n3 - sqrt(3) = around 1.267949<\/output>\n(3 - sqrt(3)) ** 2<\/gadget>\n(3 - sqrt(3))**2 = around 1.607695<\/output>\n6 + (2*sqrt(3))<\/gadget>\n2*sqrt(3) + 6 = around 9.464102<\/output>\n(2*sqrt(3) + 6) \/ 2<\/gadget>\nsqrt(3) + 3 = around 4.732051<\/output>\n(sqrt(3) + 3) ** 2<\/gadget>\n(sqrt(3) + 3)**2 = around 22.392305<\/output>\n((3 - sqrt(3))**2) + ((sqrt(3) + 3)**2)<\/gadget>\n(3 - sqrt(3))**2 + (sqrt(3) + 3)**2 = around 24<\/output>\n(3 - sqrt(3))**2 + (sqrt(3) + 3)**2 = around 24<\/result>","index":2904} +{"problem":"6 persons in an organization including x and y were to be divided in two groups of 3 members each . the total number of groups containing both x and y is what fraction of the total number of groups which can be formed ?","rationale":"the fraction is nothing but the probability . . number to choose 3 out of 6 = 6 c 3 number to choose x and y and 2 from remaining 4 = 4 c 2 . . prob of a and b choosen = 4 c 2 \/ 6 c 3 = 3 \/ 10 answer : d","correct":"d","options":{"a":"1 \/ 4 ","b":"1 \/ 70 ","c":"3 \/ 14 ","d":"3 \/ 10","e":"11 \/ 14"},"options_float":{"a":0.25,"b":0.0142857143,"c":0.2142857143,"d":0.3,"e":0.7857142857},"annotated_formula":"divide(factorial(3), multiply(subtract(6, const_1), const_4))","linear_formula":"factorial(n1)|subtract(n0,const_1)|multiply(#1,const_4)|divide(#0,#2)","chain":"factorial(3)<\/gadget>\n6<\/output>\n6 - 1<\/gadget>\n5<\/output>\n5 * 4<\/gadget>\n20<\/output>\n6 \/ 20<\/gadget>\n3\/10 = around 0.3<\/output>\n3\/10 = around 0.3<\/result>","index":2906} +{"problem":"having scored 94 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ?","rationale":"\"explanation : let the average score of the first 18 innings be n 18 n + 94 = 19 ( n + 4 ) = > n = 18 so , average score after 19 th innings = x + 4 = 22 . answer : d\"","correct":"d","options":{"a":"28 ","b":"27 ","c":"26 ","d":"22","e":"24"},"options_float":{"a":28.0,"b":27.0,"c":26.0,"d":22.0,"e":24.0},"annotated_formula":"add(subtract(94, multiply(19, 4)), 4)","linear_formula":"multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|","chain":"19 * 4<\/gadget>\n76<\/output>\n94 - 76<\/gadget>\n18<\/output>\n18 + 4<\/gadget>\n22<\/output>\n22<\/result>","index":2907} +{"problem":"a train 150 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 – 150 = 299 . answer : c","correct":"c","options":{"a":"288 ","b":"236 ","c":"350 ","d":"299","e":"266"},"options_float":{"a":288.0,"b":236.0,"c":350.0,"d":299.0,"e":266.0},"annotated_formula":"subtract(multiply(25, multiply(72, const_0_2778)), 150)","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 - 150<\/gadget>\n350<\/output>\n350<\/result>","index":2909} +{"problem":"the present worth of a certain bill due sometime hence is rs . 1296 and the true discount is rs . 72 . what is the banker ' s discount ?","rationale":"explanation : bg = ( td ) 2 \/ pw = 722 \/ 1296 = 72 × 72 \/ 1296 = 12 × 12 \/ 36 = 12 \/ 3 = rs . 4 bg = bd – td = > 4 = bd - 72 = > bd = 72 + 4 = rs . 76 answer : option a","correct":"a","options":{"a":"rs . 76 ","b":"rs . 72 ","c":"rs . 74 ","d":"rs . 4","e":"none of these"},"options_float":{"a":76.0,"b":72.0,"c":74.0,"d":4.0,"e":null},"annotated_formula":"add(72, divide(power(72, const_2), 1296))","linear_formula":"power(n1,const_2)|divide(#0,n0)|add(n1,#1)","chain":"72 ** 2<\/gadget>\n5_184<\/output>\n5_184 \/ 1_296<\/gadget>\n4<\/output>\n72 + 4<\/gadget>\n76<\/output>\n76<\/result>","index":2910} +{"problem":"the grade point average of the entire class is 88 . if the average of one third of the class is 94 , 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 . 88 x = ( 1 \/ 3 ) 94 x + ( 2 \/ 3 ) ( p ) x 264 = 94 + 2 p 2 p = 170 p = 85 . the answer is 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(88, const_4), 94), subtract(const_4, const_1))","linear_formula":"multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)|","chain":"88 * 4<\/gadget>\n352<\/output>\n352 - 94<\/gadget>\n258<\/output>\n4 - 1<\/gadget>\n3<\/output>\n258 \/ 3<\/gadget>\n86<\/output>\n86<\/result>","index":2911} +{"problem":"a truck covers a distance of 376 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?","rationale":"\"explanation : speed of the truck = distance \/ time = 376 \/ 8 = 47 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 47 + 18 ) = 65 kmph distance travelled by car = 376 + 14 = 390 km time taken by car = distance \/ speed = 390 \/ 65 = 6 hours . answer – a\"","correct":"a","options":{"a":"6 hours ","b":"5 hours ","c":"7 hours ","d":"8 hours","e":"none"},"options_float":{"a":6.0,"b":5.0,"c":7.0,"d":8.0,"e":null},"annotated_formula":"divide(add(376, 14), add(divide(376, 8), 18))","linear_formula":"add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2)|","chain":"376 + 14<\/gadget>\n390<\/output>\n376 \/ 8<\/gadget>\n47<\/output>\n47 + 18<\/gadget>\n65<\/output>\n390 \/ 65<\/gadget>\n6<\/output>\n6<\/result>","index":2912} +{"problem":"the ratio of the ages of maala and kala is 3 : 5 . the total of their ages is 3.2 decades . the proportion of their ages after 0.8 decades will be [ 1 decade = 10 years ]","rationale":"let , maala ’ s age = 3 a and kala ’ s age = 5 a then 3 a + 5 a = 32 a = 4 maala ’ s age = 12 years and kala ’ s age = 20 years proportion of their ages after 8 is = ( 12 + 8 ) : ( 20 + 8 ) = 20 : 28 = 5 : 7 answer : b","correct":"b","options":{"a":"6 : 5 ","b":"5 : 7 ","c":"4 : 5 ","d":"7 : 9","e":"3 : 6"},"options_float":{"a":1.2,"b":0.7142857143,"c":0.8,"d":0.7777777778,"e":0.5},"annotated_formula":"divide(add(multiply(divide(multiply(3.2, 10), add(3, 5)), 3), multiply(0.8, 10)), add(multiply(5, divide(multiply(3.2, 10), add(3, 5))), multiply(0.8, 10)))","linear_formula":"add(n0,n1)|multiply(n2,n5)|multiply(n3,n5)|divide(#1,#0)|multiply(n0,#3)|multiply(n1,#3)|add(#4,#2)|add(#5,#2)|divide(#6,#7)","chain":"3.2 * 10<\/gadget>\n32<\/output>\n3 + 5<\/gadget>\n8<\/output>\n32 \/ 8<\/gadget>\n4<\/output>\n4 * 3<\/gadget>\n12<\/output>\n0.8 * 10<\/gadget>\n8<\/output>\n12 + 8<\/gadget>\n20<\/output>\n5 * 4<\/gadget>\n20<\/output>\n20 + 8<\/gadget>\n28<\/output>\n20 \/ 28<\/gadget>\n5\/7 = around 0.714286<\/output>\n5\/7 = around 0.714286<\/result>","index":2913} +{"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 18 miles , how many miles was mike ' s ride ?","rationale":"\"the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 18 ) = $ 12 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 12 0.25 * x = 9.5 x = 38 miles the answer is c .\"","correct":"c","options":{"a":"30 ","b":"34 ","c":"38 ","d":"42","e":"48"},"options_float":{"a":30.0,"b":34.0,"c":38.0,"d":42.0,"e":48.0},"annotated_formula":"divide(subtract(add(add(2.50, 5.00), multiply(0.25, 18)), 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 * 18<\/gadget>\n4.5<\/output>\n7.5 + 4.5<\/gadget>\n12<\/output>\n12 - 2.5<\/gadget>\n9.5<\/output>\n9.5 \/ 0.25<\/gadget>\n38<\/output>\n38<\/result>","index":2914} +{"problem":"a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 55 percent .","rationale":"1 hr = 20 percent . thus 15 min = 5 percent . now to charge 55 percent 165 min . answer : d","correct":"d","options":{"a":"145 ","b":"150 ","c":"175 ","d":"165","e":"130"},"options_float":{"a":145.0,"b":150.0,"c":175.0,"d":165.0,"e":130.0},"annotated_formula":"multiply(divide(55, 20), const_60)","linear_formula":"divide(n2,n1)|multiply(#0,const_60)|","chain":"55 \/ 20<\/gadget>\n11\/4 = around 2.75<\/output>\n(11\/4) * 60<\/gadget>\n165<\/output>\n165<\/result>","index":2915} +{"problem":"in a box of 11 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 ) = 8 \/ 11 * 7 \/ 10 = 28 \/ 55 the answer is a .\"","correct":"a","options":{"a":"28 \/ 55 ","b":"19 \/ 33 ","c":"7 \/ 11 ","d":"4 \/ 7","e":"3 \/ 5"},"options_float":{"a":0.5090909091,"b":0.5757575758,"c":0.6363636364,"d":0.5714285714,"e":0.6},"annotated_formula":"multiply(divide(subtract(11, 3), 11), divide(subtract(subtract(11, 3), const_1), subtract(11, 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":"11 - 3<\/gadget>\n8<\/output>\n8 \/ 11<\/gadget>\n8\/11 = around 0.727273<\/output>\n8 - 1<\/gadget>\n7<\/output>\n11 - 1<\/gadget>\n10<\/output>\n7 \/ 10<\/gadget>\n7\/10 = around 0.7<\/output>\n(8\/11) * (7\/10)<\/gadget>\n28\/55 = around 0.509091<\/output>\n28\/55 = around 0.509091<\/result>","index":2916} +{"problem":"a rectangular box measures internally 1.6 m long , 1 m broad and 60 cm deep . the number of cubical box each of edge 20 cm that can be packed inside the box is :","rationale":"\"explanation : number of blocks = ( 160 x 100 x 60 \/ 20 x 20 x 20 ) = 120 answer : d\"","correct":"d","options":{"a":"30 ","b":"60 ","c":"90 ","d":"120","e":"140"},"options_float":{"a":30.0,"b":60.0,"c":90.0,"d":120.0,"e":140.0},"annotated_formula":"volume_rectangular_prism(divide(1.6, divide(20, const_100)), divide(1, divide(20, const_100)), divide(divide(60, const_100), divide(20, const_100)))","linear_formula":"divide(n3,const_100)|divide(n2,const_100)|divide(n0,#0)|divide(n1,#0)|divide(#1,#0)|volume_rectangular_prism(#2,#3,#4)|","chain":"20 \/ 100<\/gadget>\n1\/5 = around 0.2<\/output>\n1.6 \/ (1\/5)<\/gadget>\n8<\/output>\n1 \/ (1\/5)<\/gadget>\n5<\/output>\n60 \/ 100<\/gadget>\n3\/5 = around 0.6<\/output>\n(3\/5) \/ (1\/5)<\/gadget>\n3<\/output>\n8 * 5 * 3<\/gadget>\n120<\/output>\n120<\/result>","index":2919} +{"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 & q , having each turn together to fill the empty drum ?","rationale":"if caoacity of q is x units , then capacity of p is 3 x and capacity of drum is 60 * 3 x = 180 x . it will take 180 x \/ 4 x = 45 turns it will take for both the buckets p & q , having each turn together to fill the empty drum . answer : a","correct":"a","options":{"a":"45 ","b":"53 ","c":"54 ","d":"46","e":"63"},"options_float":{"a":45.0,"b":53.0,"c":54.0,"d":46.0,"e":63.0},"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":2921} +{"problem":"a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 40 now , the son age 5 years back was","rationale":"\"let the son ' s present age be x years . then , ( 40 - x ) = x x = 20 . son ' s age 5 years back = ( 20 - 5 ) = 15 years answer : c\"","correct":"c","options":{"a":"14 ","b":"17 ","c":"15 ","d":"19","e":"99"},"options_float":{"a":14.0,"b":17.0,"c":15.0,"d":19.0,"e":99.0},"annotated_formula":"subtract(divide(40, const_2), 5)","linear_formula":"divide(n0,const_2)|subtract(#0,n1)|","chain":"40 \/ 2<\/gadget>\n20<\/output>\n20 - 5<\/gadget>\n15<\/output>\n15<\/result>","index":2922} +{"problem":"the h . c . f . of two numbers is 30 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 ( 30 x 13 ) and ( 30 x 14 ) . larger number = ( 30 x 14 ) = 420 . answer : option c\"","correct":"c","options":{"a":"276 ","b":"299 ","c":"420 ","d":"345","e":"365"},"options_float":{"a":276.0,"b":299.0,"c":420.0,"d":345.0,"e":365.0},"annotated_formula":"multiply(30, 14)","linear_formula":"multiply(n0,n2)|","chain":"30 * 14<\/gadget>\n420<\/output>\n420<\/result>","index":2923} +{"problem":"find compound interest on rs . 7500 at 4 % per year for 2 years , compounded annually .","rationale":"\"amount = rs [ 7500 * ( 1 + ( 4 \/ 100 ) 2 ] = rs ( 7500 * ( 26 \/ 25 ) * ( 26 \/ 25 ) ) = rs . 8112 . therefore , compound interest = rs . ( 8112 - 7500 ) = rs . 612 . answer is e .\"","correct":"e","options":{"a":"812 ","b":"712 ","c":"412 ","d":"512","e":"612"},"options_float":{"a":812.0,"b":712.0,"c":412.0,"d":512.0,"e":612.0},"annotated_formula":"subtract(add(add(7500, divide(multiply(7500, 4), const_100)), divide(multiply(add(7500, divide(multiply(7500, 4), const_100)), 4), const_100)), 7500)","linear_formula":"multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)|","chain":"7_500 * 4<\/gadget>\n30_000<\/output>\n30_000 \/ 100<\/gadget>\n300<\/output>\n7_500 + 300<\/gadget>\n7_800<\/output>\n7_800 * 4<\/gadget>\n31_200<\/output>\n31_200 \/ 100<\/gadget>\n312<\/output>\n7_800 + 312<\/gadget>\n8_112<\/output>\n8_112 - 7_500<\/gadget>\n612<\/output>\n612<\/result>","index":2924} +{"problem":"in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 178 votes . find the total number of valid votes ?","rationale":"let the total number of valid votes be x . 70 % of x = 70 \/ 100 * x = 7 x \/ 10 number of votes secured by the other candidate = x - 7 x \/ 100 = 3 x \/ 10 given , 7 x \/ 10 - 3 x \/ 10 = 178 = > 4 x \/ 10 = 178 = > 4 x = 1780 = > x = 445 . answer : a","correct":"a","options":{"a":"445 ","b":"570 ","c":"480 ","d":"520","e":"550"},"options_float":{"a":445.0,"b":570.0,"c":480.0,"d":520.0,"e":550.0},"annotated_formula":"divide(178, divide(subtract(70, subtract(const_100, 70)), const_100))","linear_formula":"subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)","chain":"100 - 70<\/gadget>\n30<\/output>\n70 - 30<\/gadget>\n40<\/output>\n40 \/ 100<\/gadget>\n2\/5 = around 0.4<\/output>\n178 \/ (2\/5)<\/gadget>\n445<\/output>\n445<\/result>","index":2926} +{"problem":"6 computers , each working at the same constant rate , together can process a certain amount of data in 9 days . how many additional computers , each working at the same constant rate , will be needed to process the same amount of data in 6 days ?","rationale":"explanation : if six computers require 9 days to process the data , thats a total of 54 computer - days the product of 6 and 9 . if you change the number of computers or the number of days , 54 will have to remain the product , whether that means 54 days of one computer or one day with 54 computers . in 6 days , the number of computers is : 6 c = 54 c = 9 9 computers is 3 more than the 6 that it took to do the job in 9 days , so the correct choice is ( a ) .","correct":"a","options":{"a":"3 ","b":"5 ","c":"6 ","d":"9","e":"12"},"options_float":{"a":3.0,"b":5.0,"c":6.0,"d":9.0,"e":12.0},"annotated_formula":"subtract(divide(multiply(6, divide(const_1, 6)), divide(const_1, 9)), 6)","linear_formula":"divide(const_1,n0)|divide(const_1,n1)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,n0)","chain":"1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n6 * (1\/6)<\/gadget>\n1<\/output>\n1 \/ 9<\/gadget>\n1\/9 = around 0.111111<\/output>\n1 \/ (1\/9)<\/gadget>\n9<\/output>\n9 - 6<\/gadget>\n3<\/output>\n3<\/result>","index":2930} +{"problem":"the radius of a cylindrical vessel is 7 cm and height is 4 cm . find the whole surface of the cylinder ?","rationale":"\"r = 7 h = 4 2 π r ( h + r ) = 2 * 22 \/ 7 * 7 ( 11 ) = 484 answer : a\"","correct":"a","options":{"a":"484 ","b":"771 ","c":"440 ","d":"767","e":"1981"},"options_float":{"a":484.0,"b":771.0,"c":440.0,"d":767.0,"e":1981.0},"annotated_formula":"surface_cylinder(7, 4)","linear_formula":"surface_cylinder(n0,n1)|","chain":"2 * pi * 7 * (7 + 4)<\/gadget>\n154*pi = around 483.805269<\/output>\n154*pi = around 483.805269<\/result>","index":2931} +{"problem":"mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.00 . each pot cost 0.25 more than the next one below it in size . what was the cost , in dollars , of the largest pot ?","rationale":"\"this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 8.00 ( a 25 - cent increment ) and the the difference in sequential prices of the pots is 25 cents , the largest pot probably has a price that is a 25 - cent increment . from the answer choices , i would then test answer c first ( since answers b and d are not in 25 - cent increments ) . if . . . . the largest pot = $ 1.958 0.708 0.958 1.208 1.458 1.708 1.958 total = $ 8.00 so this must be the answer . b\"","correct":"b","options":{"a":"$ 1.75 ","b":"$ 1.96 ","c":"$ 2.00 ","d":"$ 2.15","e":"$ 2.30"},"options_float":{"a":1.75,"b":1.96,"c":2.0,"d":2.15,"e":2.3},"annotated_formula":"add(divide(subtract(8.00, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.25)), 6), multiply(subtract(6, const_1), 0.25))","linear_formula":"subtract(n0,const_1)|multiply(n0,#0)|multiply(n2,#0)|divide(#1,const_2)|multiply(n2,#3)|subtract(n1,#4)|divide(#5,n0)|add(#6,#2)|","chain":"6 - 1<\/gadget>\n5<\/output>\n5 * 6<\/gadget>\n30<\/output>\n30 \/ 2<\/gadget>\n15<\/output>\n15 * 0.25<\/gadget>\n3.75<\/output>\n8 - 3.75<\/gadget>\n4.25<\/output>\n4.25 \/ 6<\/gadget>\n0.708333<\/output>\n5 * 0.25<\/gadget>\n1.25<\/output>\n0.708333 + 1.25<\/gadget>\n1.958333<\/output>\n1.958333<\/result>","index":2932} +{"problem":"a rectangular grassy plot 110 m by 65 cm has a gravel path . 5 cm wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . mt","rationale":"\"explanation : area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = ( 110 - 5 ) * ( 65 - 5 ) = 6300 sq m area of the path = 7150 - 6300 = 850 sq m cost of gravelling the path = 850 * 80 \/ 100 = 680 rs answer : a ) 680 rs\"","correct":"a","options":{"a":"680 ","b":"378 ","c":"267 ","d":"299","e":"271"},"options_float":{"a":680.0,"b":378.0,"c":267.0,"d":299.0,"e":271.0},"annotated_formula":"multiply(divide(80, const_100), subtract(multiply(110, 65), multiply(subtract(110, 5), subtract(65, 5))))","linear_formula":"divide(n3,const_100)|multiply(n0,n1)|subtract(n0,n2)|subtract(n1,n2)|multiply(#2,#3)|subtract(#1,#4)|multiply(#0,#5)|","chain":"80 \/ 100<\/gadget>\n4\/5 = around 0.8<\/output>\n110 * 65<\/gadget>\n7_150<\/output>\n110 - 5<\/gadget>\n105<\/output>\n65 - 5<\/gadget>\n60<\/output>\n105 * 60<\/gadget>\n6_300<\/output>\n7_150 - 6_300<\/gadget>\n850<\/output>\n(4\/5) * 850<\/gadget>\n680<\/output>\n680<\/result>","index":2934} +{"problem":"in a dairy farm , 16 cows eat 16 bags of husk in 16 days . in how many days one cow will eat one bag of husk ?","rationale":"explanation : one bag of husk = 16 cows per day ⇒ 16 × 1 × 16 = 1 × 16 × x for one cow = 16 days answer : a","correct":"a","options":{"a":"16 ","b":"40 ","c":"20 ","d":"26","e":"30"},"options_float":{"a":16.0,"b":40.0,"c":20.0,"d":26.0,"e":30.0},"annotated_formula":"multiply(divide(16, 16), 16)","linear_formula":"divide(n0,n0)|multiply(n0,#0)","chain":"16 \/ 16<\/gadget>\n1<\/output>\n1 * 16<\/gadget>\n16<\/output>\n16<\/result>","index":2935} +{"problem":"in 12 pumps can raise 1218 tons of water in 11 days of 9 hrs each , how many pumps will raise 2030 tons of water in 12 days of 11 hrs each ?","rationale":"explanation : pumps work time 12 1218 99 x 2030 132 = > 1218 \/ ( 912 * 99 ) = 2020 \/ ( x × 132 ) = > x = 15 pumps answer : option b","correct":"b","options":{"a":"12 ","b":"15 ","c":"18 ","d":"21","e":"22"},"options_float":{"a":12.0,"b":15.0,"c":18.0,"d":21.0,"e":22.0},"annotated_formula":"divide(multiply(multiply(multiply(12, 11), 9), 2030), multiply(multiply(12, 11), 1218))","linear_formula":"multiply(n0,n2)|multiply(n3,#0)|multiply(n1,#0)|multiply(n4,#1)|divide(#3,#2)","chain":"12 * 11<\/gadget>\n132<\/output>\n132 * 9<\/gadget>\n1_188<\/output>\n1_188 * 2_030<\/gadget>\n2_411_640<\/output>\n132 * 1_218<\/gadget>\n160_776<\/output>\n2_411_640 \/ 160_776<\/gadget>\n15<\/output>\n15<\/result>","index":2936} +{"problem":"32.32 \/ 2000 is equal to :","rationale":"\"25.25 \/ 2000 = 2525 \/ 200000 = 0.01616 answer : d\"","correct":"d","options":{"a":"1.012526 ","b":"0.012625 ","c":"0.12526 ","d":"0.01616","e":"0.12725"},"options_float":{"a":1.012526,"b":0.012625,"c":0.12526,"d":0.01616,"e":0.12725},"annotated_formula":"divide(32.32, 2000)","linear_formula":"divide(n0,n1)|","chain":"32.32 \/ 2_000<\/gadget>\n0.01616<\/output>\n0.01616<\/result>","index":2940} +{"problem":"george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 5 % of the money for taxi fare and buys 25 mangoes . how many oranges can he buy ?","rationale":"let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 5 % of 200 = 10 for taxi fare , so he is left with 190 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 90 ( 190 - 100 ) no of oranges he can buy = 90 \/ 4 = > 22,5 so , george can buy 20 oranges . d","correct":"d","options":{"a":"25 ","b":"30 ","c":"20 ","d":"22.5","e":"12"},"options_float":{"a":25.0,"b":30.0,"c":20.0,"d":22.5,"e":12.0},"annotated_formula":"multiply(subtract(subtract(const_1, divide(5, const_100)), divide(25, 50)), 50)","linear_formula":"divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)","chain":"5 \/ 100<\/gadget>\n1\/20 = around 0.05<\/output>\n1 - (1\/20)<\/gadget>\n19\/20 = around 0.95<\/output>\n25 \/ 50<\/gadget>\n1\/2 = around 0.5<\/output>\n(19\/20) - (1\/2)<\/gadget>\n9\/20 = around 0.45<\/output>\n(9\/20) * 50<\/gadget>\n45\/2 = around 22.5<\/output>\n45\/2 = around 22.5<\/result>","index":2941} +{"problem":"if 4 spiders make 4 webs in 4 days , then 1 spider will make 1 web in how many days ?","rationale":"\"explanation : let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 4 | | : : 4 : 1 webs 4 : 1 | = > 1 * 4 * x = 4 * 1 * 4 = > x = 4 answer : d\"","correct":"d","options":{"a":"1 ","b":"3 ","c":"5 ","d":"4","e":"6"},"options_float":{"a":1.0,"b":3.0,"c":5.0,"d":4.0,"e":6.0},"annotated_formula":"multiply(1, 4)","linear_formula":"multiply(n0,n3)|","chain":"1 * 4<\/gadget>\n4<\/output>\n4<\/result>","index":2942} +{"problem":"in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 36 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ?","rationale":"\"girls = 900 , boys = 1100 passed = ( 36 % of 1100 ) + ( 32 % of 900 ) = 396 + 288 = 684 failed = 2000 - 684 = 1316 failed % = [ ( 1316 \/ 2000 ) x 100 ] % = 65.8 % . answer : b\"","correct":"b","options":{"a":"35.67 % ","b":"65.80 % ","c":"68.57 % ","d":"69.57 %","e":"none of these"},"options_float":{"a":35.67,"b":65.8,"c":68.57,"d":69.57,"e":null},"annotated_formula":"multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(36, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100)","linear_formula":"divide(n2,const_100)|divide(n3,const_100)|multiply(const_10,const_100)|multiply(#2,const_2)|multiply(n1,#1)|subtract(#3,n1)|multiply(#0,#5)|subtract(#3,#6)|subtract(#7,#4)|divide(#8,#3)|multiply(#9,const_100)|","chain":"100 * 10<\/gadget>\n1_000<\/output>\n2 * 1_000<\/gadget>\n2_000<\/output>\n36 \/ 100<\/gadget>\n9\/25 = around 0.36<\/output>\n2_000 - 900<\/gadget>\n1_100<\/output>\n(9\/25) * 1_100<\/gadget>\n396<\/output>\n2_000 - 396<\/gadget>\n1_604<\/output>\n32 \/ 100<\/gadget>\n8\/25 = around 0.32<\/output>\n(8\/25) * 900<\/gadget>\n288<\/output>\n1_604 - 288<\/gadget>\n1_316<\/output>\n1_316 \/ 2_000<\/gadget>\n329\/500 = around 0.658<\/output>\n(329\/500) * 100<\/gadget>\n329\/5 = around 65.8<\/output>\n329\/5 = around 65.8<\/result>","index":2943} +{"problem":"john left home and drove at the rate of 50 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ?","rationale":"\"the total distance d traveled by john is given by d = 50 * 2 + 3 * 55 = 265 miles . answer d\"","correct":"d","options":{"a":"235 miles . ","b":"245 miles . ","c":"255 miles . ","d":"265 miles .","e":"275 miles ."},"options_float":{"a":235.0,"b":245.0,"c":255.0,"d":265.0,"e":275.0},"annotated_formula":"add(multiply(50, 2), multiply(3, 55))","linear_formula":"multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|","chain":"50 * 2<\/gadget>\n100<\/output>\n3 * 55<\/gadget>\n165<\/output>\n100 + 165<\/gadget>\n265<\/output>\n265<\/result>","index":2944} +{"problem":"a boat having a length 3 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":"\"volume of water displaced = ( 3 x 2 x 0.01 ) m 3 = 0.06 m 3 . mass of man = volume of water displaced x density of water = ( 0.06 x 1000 ) kg = 60 kg . answer : option b\"","correct":"b","options":{"a":"12 kg ","b":"60 kg ","c":"72 kg ","d":"88 kg","e":"96 kg"},"options_float":{"a":12.0,"b":60.0,"c":72.0,"d":88.0,"e":96.0},"annotated_formula":"multiply(multiply(multiply(3, 2), divide(1, 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>\n1 \/ 100<\/gadget>\n1\/100 = around 0.01<\/output>\n6 * (1\/100)<\/gadget>\n3\/50 = around 0.06<\/output>\n(3\/50) * 1_000<\/gadget>\n60<\/output>\n60<\/result>","index":2945} +{"problem":"anne earned $ 3 an hour baby - sitting , and $ 4 an hour working in the garden . last week she did baby - sitting for 5 hours and garden work for 3 hours . how much more money does she need to buy a game that costs $ 35 ?","rationale":"5 x $ 3 = $ 15 for baby - sitting 3 x $ 4 = $ 12 for garden work $ 15 + $ 12 = $ 27 she has $ 35 - $ 27 = $ 8 more needed to buy the game correct answer a","correct":"a","options":{"a":"$ 8 ","b":"$ 12 ","c":"$ 6 ","d":"$ 21","e":"$ 10"},"options_float":{"a":8.0,"b":12.0,"c":6.0,"d":21.0,"e":10.0},"annotated_formula":"subtract(35, add(multiply(5, 3), multiply(3, 4)))","linear_formula":"multiply(n0,n2)|multiply(n0,n1)|add(#0,#1)|subtract(n4,#2)","chain":"5 * 3<\/gadget>\n15<\/output>\n3 * 4<\/gadget>\n12<\/output>\n15 + 12<\/gadget>\n27<\/output>\n35 - 27<\/gadget>\n8<\/output>\n8<\/result>","index":2947} +{"problem":"if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 80 , 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 = 160 - - 2 subtracting 1 from 2 we have c - a = 70 ; answer : b\"","correct":"b","options":{"a":"25 ","b":"70 ","c":"90 ","d":"140","e":"it can not be determined from the information given"},"options_float":{"a":25.0,"b":70.0,"c":90.0,"d":140.0,"e":null},"annotated_formula":"subtract(multiply(80, const_2), multiply(45, const_2))","linear_formula":"multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|","chain":"80 * 2<\/gadget>\n160<\/output>\n45 * 2<\/gadget>\n90<\/output>\n160 - 90<\/gadget>\n70<\/output>\n70<\/result>","index":2948} +{"problem":"a envelop weight 8.2 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ?","rationale":"\"800 * 8.2 6560.0 gm 6.56 kg answer : a\"","correct":"a","options":{"a":"6.56 kg ","b":"6.8 kg ","c":"6.7 kg ","d":"6.9 kg","e":"7.8 kg"},"options_float":{"a":6.56,"b":6.8,"c":6.7,"d":6.9,"e":7.8},"annotated_formula":"divide(multiply(8.2, 800), const_1000)","linear_formula":"multiply(n0,n1)|divide(#0,const_1000)|","chain":"8.2 * 800<\/gadget>\n6_560<\/output>\n6_560 \/ 1_000<\/gadget>\n164\/25 = around 6.56<\/output>\n164\/25 = around 6.56<\/result>","index":2949} +{"problem":"if a coin is flipped , the probability that the coin will land heads is 1 \/ 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 4 flips and not on the last flip ?","rationale":"( 1 \/ 2 ) * ( 1 \/ 2 ) * ( 1 \/ 2 ) * ( 1 \/ 2 ) * ( 1 \/ 2 ) = 1 \/ 32 answer : b","correct":"b","options":{"a":"1 \/ 8 ","b":"1 \/ 32 ","c":"1 \/ 4 ","d":"1 \/ 2","e":"1 \/ 16"},"options_float":{"a":0.125,"b":0.03125,"c":0.25,"d":0.5,"e":0.0625},"annotated_formula":"divide(const_1, power(2, 5))","linear_formula":"power(n1,n2)|divide(const_1,#0)","chain":"2 ** 5<\/gadget>\n32<\/output>\n1 \/ 32<\/gadget>\n1\/32 = around 0.03125<\/output>\n1\/32 = around 0.03125<\/result>","index":2950} +{"problem":"the cross - section of a water channel is a trapezium in shape . if the channel is 14 meters wide at the top and 8 meters wide at the bottom and the area of cross - section is 990 square meters , what is the depth of the channel ( in meters ) ?","rationale":"\"1 \/ 2 * d * ( 14 + 8 ) = 990 d = 90 the answer is e .\"","correct":"e","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(990, divide(add(14, 8), const_2))","linear_formula":"add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|","chain":"14 + 8<\/gadget>\n22<\/output>\n22 \/ 2<\/gadget>\n11<\/output>\n990 \/ 11<\/gadget>\n90<\/output>\n90<\/result>","index":2951} +{"problem":"income and expenditure of a person are in the ratio 5 : 4 . 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 . 5 x and rs . 4 x respectively . income , 5 x = 14000 = > x = 2800 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 2800 . answer : b\"","correct":"b","options":{"a":"3600 ","b":"2800 ","c":"3608 ","d":"3602","e":"3603"},"options_float":{"a":3600.0,"b":2800.0,"c":3608.0,"d":3602.0,"e":3603.0},"annotated_formula":"subtract(14000, multiply(divide(4, 5), 14000))","linear_formula":"divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)|","chain":"4 \/ 5<\/gadget>\n4\/5 = around 0.8<\/output>\n(4\/5) * 14_000<\/gadget>\n11_200<\/output>\n14_000 - 11_200<\/gadget>\n2_800<\/output>\n2_800<\/result>","index":2952} +{"problem":"two employees x and y are paid a total of rs . 750 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 = 750 but x = 120 % of y = 120 y \/ 100 = 12 y \/ 10 ∴ 12 y \/ 10 + y = 750 ⇒ y [ 12 \/ 10 + 1 ] = 750 ⇒ 22 y \/ 10 = 750 ⇒ 22 y = 7500 ⇒ y = 7500 \/ 22 = rs . 340.90 e )\"","correct":"e","options":{"a":"s . 200.90 ","b":"s . 250.90 ","c":"s . 290.90 ","d":"s . 300.90","e":"s . 340.90"},"options_float":{"a":200.9,"b":250.9,"c":290.9,"d":300.9,"e":340.9},"annotated_formula":"divide(multiply(750, 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>\n750 * 10<\/gadget>\n7_500<\/output>\n10 + 1<\/gadget>\n11<\/output>\n11 * 2<\/gadget>\n22<\/output>\n7_500 \/ 22<\/gadget>\n3_750\/11 = around 340.909091<\/output>\n3_750\/11 = around 340.909091<\/result>","index":2953} +{"problem":"the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 19 . find the sum :","rationale":"\"sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2100 = 10100 × 5 × 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 − 100100 ( 1 + 5100 ) 2 − 100 ∴ = 100 × 21 × 2120 × 20 − 100 = 414 = 100 × 21 × 2120 × 20 − 100 = 414 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 19 , the sum = 400 × 19 = 7600 answer b\"","correct":"b","options":{"a":"4500 ","b":"7600 ","c":"5000 ","d":"6000","e":"none of these"},"options_float":{"a":4500.0,"b":7600.0,"c":5000.0,"d":6000.0,"e":null},"annotated_formula":"divide(19, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100)))))","linear_formula":"divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5)|","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>\n2 * (1\/20)<\/gadget>\n1\/10 = around 0.1<\/output>\n1 + (1\/10)<\/gadget>\n11\/10 = around 1.1<\/output>\n(441\/400) - (11\/10)<\/gadget>\n1\/400 = around 0.0025<\/output>\n19 \/ (1\/400)<\/gadget>\n7_600<\/output>\n7_600<\/result>","index":2954} +{"problem":"machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 60 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 6 minutes ?","rationale":"\"machine b produces 100 part in 60 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 60 \/ 2 = 30 minutes . now , machine a produces 100 parts in 30 minutes which is 100 \/ 30 = 10 \/ 3 parts \/ minute . 10 \/ 3 parts x a total of 6 minutes = 20 a\"","correct":"a","options":{"a":"20 ","b":"30 ","c":"40 ","d":"10","e":"16"},"options_float":{"a":20.0,"b":30.0,"c":40.0,"d":10.0,"e":16.0},"annotated_formula":"multiply(multiply(divide(100, 60), const_2), 6)","linear_formula":"divide(n0,n2)|multiply(#0,const_2)|multiply(n3,#1)|","chain":"100 \/ 60<\/gadget>\n5\/3 = around 1.666667<\/output>\n(5\/3) * 2<\/gadget>\n10\/3 = around 3.333333<\/output>\n(10\/3) * 6<\/gadget>\n20<\/output>\n20<\/result>","index":2956} +{"problem":"if n is a prime number greater than 5 , what is the remainder when n ^ 2 is divided by 12 ?","rationale":"there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 5 , square it and see what would be the remainder upon division of it by 12 . n = 7 - - > n ^ 2 = 49 - - > remainder upon division 49 by 12 is 1 . 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":"subtract(power(add(5, const_2), 2), multiply(12, const_4))","linear_formula":"add(n0,const_2)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)","chain":"5 + 2<\/gadget>\n7<\/output>\n7 ** 2<\/gadget>\n49<\/output>\n12 * 4<\/gadget>\n48<\/output>\n49 - 48<\/gadget>\n1<\/output>\n1<\/result>","index":2958} +{"problem":"set j consists of 5 consecutive even numbers . if the smallest term in the set is - 2 , what is the range of the positive integers in set j ?","rationale":"since there are only 5 integers , another approach is the just list all 5 . we get : - 2 , 0,2 , 4 , 6 range of positive integers = 6 - 2 = 4 answer : c","correct":"c","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":"subtract(add(negate(2), multiply(subtract(5, const_1), 2)), 2)","linear_formula":"negate(n1)|subtract(n0,const_1)|multiply(n1,#1)|add(#2,#0)|subtract(#3,n1)","chain":"-2<\/gadget>\n-2<\/output>\n5 - 1<\/gadget>\n4<\/output>\n4 * 2<\/gadget>\n8<\/output>\n(-2) + 8<\/gadget>\n6<\/output>\n6 - 2<\/gadget>\n4<\/output>\n4<\/result>","index":2959} +{"problem":"the sum of all the integers g such that - 26 < g < 24 is","rationale":"\"easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 g = - 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":2961} +{"problem":"a student chose a number , multiplied it by 2 , then subtracted 180 from the result and got 104 . what was the number he chose ?","rationale":"\"solution : let x be the number he chose , then 2 * x * 180 = 104 2 x = 284 x = 142 correct answer d\"","correct":"d","options":{"a":"90 ","b":"100 ","c":"120 ","d":"142","e":"200"},"options_float":{"a":90.0,"b":100.0,"c":120.0,"d":142.0,"e":200.0},"annotated_formula":"divide(add(104, 180), 2)","linear_formula":"add(n1,n2)|divide(#0,n0)|","chain":"104 + 180<\/gadget>\n284<\/output>\n284 \/ 2<\/gadget>\n142<\/output>\n142<\/result>","index":2963} +{"problem":"two brothers take the same route to school on their bicycles , one gets to school in 25 minutes and the second one gets to school in 36 minutes . the ratio of their speeds is","rationale":"solution let us name the brothers as a and b . = ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 25 : â ˆ š 36 = 5 : 6 answer d","correct":"d","options":{"a":"4 : 5 ","b":"1 : 2 ","c":"6 : 7 ","d":"5 : 6","e":"none"},"options_float":{"a":0.8,"b":0.5,"c":0.8571428571,"d":0.8333333333,"e":null},"annotated_formula":"divide(sqrt(25), sqrt(36))","linear_formula":"sqrt(n0)|sqrt(n1)|divide(#0,#1)","chain":"25 ** (1\/2)<\/gadget>\n5<\/output>\n36 ** (1\/2)<\/gadget>\n6<\/output>\n5 \/ 6<\/gadget>\n5\/6 = around 0.833333<\/output>\n5\/6 = around 0.833333<\/result>","index":2964} +{"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 9 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 9 = 45 km imo , correct answer is ` ` c . ' '\"","correct":"c","options":{"a":"20 km ","b":"30 km ","c":"45 km ","d":"50 km","e":"60 km"},"options_float":{"a":20.0,"b":30.0,"c":45.0,"d":50.0,"e":60.0},"annotated_formula":"multiply(60, divide(multiply(5, 9), 60))","linear_formula":"multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|","chain":"5 * 9<\/gadget>\n45<\/output>\n45 \/ 60<\/gadget>\n3\/4 = around 0.75<\/output>\n60 * (3\/4)<\/gadget>\n45<\/output>\n45<\/result>","index":2965} +{"problem":"in a certain warehouse , 50 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 50 % of the packages weigh less than 75 pounds so , 50 % oft = number of packages that weigh less than 75 pounds 50 % of 240 = 120 , so 120 packages weigh less than 75 pounds of those 120 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 = 120 - 48 = 72 = c\"","correct":"c","options":{"a":"8 ","b":"64 ","c":"72 ","d":"102","e":"144"},"options_float":{"a":8.0,"b":64.0,"c":72.0,"d":102.0,"e":144.0},"annotated_formula":"subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 50), 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 * 50<\/gadget>\n12_000<\/output>\n12_000 \/ 100<\/gadget>\n120<\/output>\n120 - 48<\/gadget>\n72<\/output>\n72<\/result>","index":2966} +{"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 ( in km \/ hr ) 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":2967} +{"problem":"the ratio of the cost price and the selling price is 4 : 5 . the profit percent is ?","rationale":"\"let c . p . = rs . 4 x . then , s . p . = rs . 5 x gain = ( 5 x - 4 x ) = rs . x gain % = ( x * 100 ) \/ 4 x = 25 % . answer : c\"","correct":"c","options":{"a":"17 ","b":"56 ","c":"25 ","d":"28","e":"12"},"options_float":{"a":17.0,"b":56.0,"c":25.0,"d":28.0,"e":12.0},"annotated_formula":"multiply(subtract(divide(5, 4), const_1), const_100)","linear_formula":"divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)|","chain":"5 \/ 4<\/gadget>\n5\/4 = around 1.25<\/output>\n(5\/4) - 1<\/gadget>\n1\/4 = around 0.25<\/output>\n(1\/4) * 100<\/gadget>\n25<\/output>\n25<\/result>","index":2968} +{"problem":"if 45 % of a class averages 100 % on a test , 50 % of the class averages 78 % on the test , and the remainder of the class averages 65 % 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 100 + 0.50 x 78 + 0.05 x 63 = 87.15 the class average ( rounded ) is 87 % final answer b ) 87 %\"","correct":"b","options":{"a":"86 % ","b":"87 % ","c":"88 % ","d":"89 %","e":"90 %"},"options_float":{"a":86.0,"b":87.0,"c":88.0,"d":89.0,"e":90.0},"annotated_formula":"divide(add(add(multiply(45, 100), multiply(50, 78)), multiply(subtract(const_100, add(45, 50)), 65)), 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 * 100<\/gadget>\n4_500<\/output>\n50 * 78<\/gadget>\n3_900<\/output>\n4_500 + 3_900<\/gadget>\n8_400<\/output>\n45 + 50<\/gadget>\n95<\/output>\n100 - 95<\/gadget>\n5<\/output>\n5 * 65<\/gadget>\n325<\/output>\n8_400 + 325<\/gadget>\n8_725<\/output>\n8_725 \/ 100<\/gadget>\n349\/4 = around 87.25<\/output>\n349\/4 = around 87.25<\/result>","index":2969} +{"problem":"of the votes cast on a certain proposal , 62 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 + 62 0.2 x = 62 x = 310 the answer is d .","correct":"d","options":{"a":"280 ","b":"290 ","c":"300 ","d":"310","e":"320"},"options_float":{"a":280.0,"b":290.0,"c":300.0,"d":310.0,"e":320.0},"annotated_formula":"divide(62, 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>\n62 \/ (1\/5)<\/gadget>\n310<\/output>\n310<\/result>","index":2970} +{"problem":"how many of the integers between 20 and 80 are even ?","rationale":"\"number start between 20 to 80 is 60 numbers half of them is even . . which is 30 answer : b\"","correct":"b","options":{"a":"21 ","b":"30 ","c":"11 ","d":"10","e":"9"},"options_float":{"a":21.0,"b":30.0,"c":11.0,"d":10.0,"e":9.0},"annotated_formula":"divide(subtract(80, 20), const_2)","linear_formula":"subtract(n1,n0)|divide(#0,const_2)|","chain":"80 - 20<\/gadget>\n60<\/output>\n60 \/ 2<\/gadget>\n30<\/output>\n30<\/result>","index":2971} +{"problem":"if a and b are positive integers , and a = 5 b + 20 , the greatest common divisor of a and b can not be","rationale":"if b is 2 , 4 , 5 , or 10 , then gcd of a and b is 2 , 4 , 5 , and 10 respectively . so , by poe the answer must be d . still : if b is a multiple of 6 , then a is 20 greater than a multiple of 6 , so not a multiple of 6 , so both of them can not be divisive by 6 . answer : d .","correct":"d","options":{"a":"2 ","b":"4 ","c":"5 ","d":"6","e":"10"},"options_float":{"a":2.0,"b":4.0,"c":5.0,"d":6.0,"e":10.0},"annotated_formula":"add(divide(20, 5), const_2)","linear_formula":"divide(n1,n0)|add(#0,const_2)","chain":"20 \/ 5<\/gadget>\n4<\/output>\n4 + 2<\/gadget>\n6<\/output>\n6<\/result>","index":2972} +{"problem":"express 35 mps in kmph ?","rationale":"\"35 * 18 \/ 5 = 126 kmph answer : c\"","correct":"c","options":{"a":"122 ","b":"188 ","c":"126 ","d":"140","e":"124"},"options_float":{"a":122.0,"b":188.0,"c":126.0,"d":140.0,"e":124.0},"annotated_formula":"multiply(divide(35, const_1000), const_3600)","linear_formula":"divide(n0,const_1000)|multiply(#0,const_3600)|","chain":"35 \/ 1_000<\/gadget>\n7\/200 = around 0.035<\/output>\n(7\/200) * 3_600<\/gadget>\n126<\/output>\n126<\/result>","index":2976} +{"problem":"the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 2.50 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ?","rationale":"\"a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 – 6 = 666 * 2.5 = 1665 answer : d\"","correct":"d","options":{"a":"s . 1014 ","b":"s . 1140 ","c":"s . 999 ","d":"s . 1665","e":"s . 1020"},"options_float":{"a":1014.0,"b":1140.0,"c":999.0,"d":1665.0,"e":1020.0},"annotated_formula":"multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 2.50), 3)","linear_formula":"multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)|","chain":"3_136 ** (1\/2)<\/gadget>\n56<\/output>\n56 * 4<\/gadget>\n224<\/output>\n2 * 1<\/gadget>\n2<\/output>\n224 - 2<\/gadget>\n222<\/output>\n222 * 2.5<\/gadget>\n555<\/output>\n555 * 3<\/gadget>\n1_665<\/output>\n1_665<\/result>","index":2977} +{"problem":"there are 690 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ?","rationale":"female = x male = 690 - x x \/ 2 + 690 - x \/ 4 = 1 \/ 3 * ( 690 ) = 230 x = 230 x \/ 2 = 115 is supposed to be the answer m is missing something correct option c","correct":"c","options":{"a":"75 ","b":"100 ","c":"115 ","d":"175","e":"225"},"options_float":{"a":75.0,"b":100.0,"c":115.0,"d":175.0,"e":225.0},"annotated_formula":"divide(subtract(multiply(divide(690, const_3), const_4), 690), const_2)","linear_formula":"divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)","chain":"690 \/ 3<\/gadget>\n230<\/output>\n230 * 4<\/gadget>\n920<\/output>\n920 - 690<\/gadget>\n230<\/output>\n230 \/ 2<\/gadget>\n115<\/output>\n115<\/result>","index":2979} +{"problem":"find the area of a parallelogram with base 20 cm and height 40 cm ?","rationale":"\"area of a parallelogram = base * height = 20 * 40 = 800 cm 2 answer : c\"","correct":"c","options":{"a":"100 cm 2 ","b":"250 cm 2 ","c":"800 cm 2 ","d":"296 cm 2","e":"456 cm 2"},"options_float":{"a":100.0,"b":250.0,"c":800.0,"d":296.0,"e":456.0},"annotated_formula":"multiply(20, 40)","linear_formula":"multiply(n0,n1)|","chain":"20 * 40<\/gadget>\n800<\/output>\n800<\/result>","index":2980} +{"problem":"a can do a piece of work in 6 hours ; b and c together can do it in 4 hours , which a and c together can do it in 3 hours . how long will b alone take to do it ?","rationale":"\"a ' s 1 hour work = 1 \/ 6 ; ( b + c ) ' s 1 hour work = 1 \/ 4 ; ( a + c ) ' s 1 hour work = 1 \/ 3 ( a + b + c ) ' s 1 hour work = ( 1 \/ 4 + 1 \/ 6 ) = 5 \/ 12 b ' s 1 hour work = ( 5 \/ 12 - 1 \/ 3 ) = 1 \/ 12 b alone will take 12 hours to do the work . answer : d\"","correct":"d","options":{"a":"8 hours ","b":"6 hours ","c":"14 hours ","d":"12 hours","e":"5 hours"},"options_float":{"a":8.0,"b":6.0,"c":14.0,"d":12.0,"e":5.0},"annotated_formula":"divide(const_1, subtract(divide(const_1, 4), subtract(divide(const_1, 3), divide(const_1, 6))))","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 \/ 4<\/gadget>\n1\/4 = around 0.25<\/output>\n1 \/ 3<\/gadget>\n1\/3 = around 0.333333<\/output>\n1 \/ 6<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/3) - (1\/6)<\/gadget>\n1\/6 = around 0.166667<\/output>\n(1\/4) - (1\/6)<\/gadget>\n1\/12 = around 0.083333<\/output>\n1 \/ (1\/12)<\/gadget>\n12<\/output>\n12<\/result>","index":2982} +{"problem":"a train 250 m long running at 72 kmph crosses a platform in 30 sec . what is the length of the platform ?","rationale":"\"d = 72 * 5 \/ 18 = 30 = 600 â € “ 250 = 350 m answer : a\"","correct":"a","options":{"a":"350 m ","b":"200 m ","c":"250 m ","d":"270 m","e":"300 m"},"options_float":{"a":350.0,"b":200.0,"c":250.0,"d":270.0,"e":300.0},"annotated_formula":"subtract(multiply(30, 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>\n30 * 20<\/gadget>\n600<\/output>\n600 - 250<\/gadget>\n350<\/output>\n350<\/result>","index":2983} +{"problem":"4 shepherds were watching over the flocks and they were commenting on how many sheep they each had . if ram had 3 more sheep than he would have one less than rahul . wheras akar has the same number as the other 3 shepherds put togeher . if john had 3 less sheep he would have exactly trile the number of ram . if they were evenly distributed if they would each have 11 seep how many sheep did ram have ?","rationale":"akar has = ram + rahul + john after evenly distribution each has 11 . so , total no . is 44 so , akar has = 22 & ram + rahul + john = 22 also ram = rahul - 4 & john - 3 = 3 * ram solving these we get the sol . answer : 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":"divide(subtract(multiply(11, const_2), add(4, 3)), add(4, const_1))","linear_formula":"add(n0,n1)|add(n0,const_1)|multiply(n4,const_2)|subtract(#2,#0)|divide(#3,#1)","chain":"11 * 2<\/gadget>\n22<\/output>\n4 + 3<\/gadget>\n7<\/output>\n22 - 7<\/gadget>\n15<\/output>\n4 + 1<\/gadget>\n5<\/output>\n15 \/ 5<\/gadget>\n3<\/output>\n3<\/result>","index":2984}