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a student gets 55 % in one subject , 65 % in the other . to get an overall of 55 % how much should get in third subject . | "let the 3 rd subject % = x 55 + 65 + x = 3 * 55 120 + x = 165 x = 165 - 120 = 45 answer : c" | a ) 75 % , b ) 25 % , c ) 45 % , d ) 55 % , e ) 65 % | c | subtract(multiply(55, const_3), add(55, 65)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | gain |
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 ? | "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" | a ) 16000 , b ) 10000 , c ) 40000 , d ) 12000 , e ) 5000 | c | divide(4000, divide(10, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
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 ? | "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" | a ) 800 , b ) 820 , c ) 850 , d ) 870 , e ) 900 | d | multiply(add(28, const_2), subtract(add(28, const_2), const_1)) | add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)| | physics |
what percent is 7 gm of 1 kg ? | "1 kg = 1000 gm 7 / 1000 Γ 100 = 700 / 1000 = 7 / 10 = 0.7 % a )" | a ) 0.7 % , b ) 0.5 % , c ) 2.5 % , d ) 3.5 % , e ) 4 % | a | multiply(divide(7, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
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 ? | "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" | a ) 14 % , b ) 23 % , c ) 30 % , d ) 50 % , e ) 100 % | a | multiply(divide(subtract(add(const_100, 60), add(const_100, 40)), add(const_100, 40)), const_100) | add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | general |
25 is subtracted from 75.00001 % of a number , the result is 50 . find the number ? | "( 75 / 100 ) * x Γ’ β¬ β 25 = 50 7.5 x = 750 x = 100 answer : e" | a ) 150 , b ) 75 , c ) 125 , d ) 95 , e ) 100 | e | divide(add(25, 50), divide(75.00001, const_100)) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1)| | gain |
an association of mathematics teachers has 1,500 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? | "total umber of members = 1500 number of members that cast votes = 525 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315 / 1500 ) * 100 = 21 % answer e" | a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 21 % | e | multiply(divide(multiply(divide(60, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | divide(n2,const_100)|power(const_4,const_2)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
half a number plus 7 is 11 . what is the number ? | "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" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | multiply(subtract(11, 7), const_2) | subtract(n1,n0)|multiply(#0,const_2)| | general |
if 5 % more is gained by selling an article for rs . 1000 than by selling it for rs . 20 , the cost of the article is ? | "let c . p . be rs . x . then , 5 % of x = 1000 - 20 = 80 x / 20 = 80 = > x = 1600 answer : b" | a ) 127 , b ) 1600 , c ) 1200 , d ) 1680 , e ) 1800 | b | divide(subtract(1000, 20), divide(5, const_100)) | divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)| | gain |
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 ? | "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" | a ) 3 , b ) 4.5 , c ) 4 , d ) 3.5 , e ) 5 | d | multiply(divide(add(multiply(25, divide(10, const_100)), multiply(50, divide(2, const_100))), const_100), const_100) | 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)| | general |
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 ? | "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 ." | a ) 400 , b ) 625 , c ) 1250 , d ) 1528 , e ) 10 000 | d | divide(55, divide(2, 55)) | divide(n2,n1)|divide(n0,#0)| | gain |
how many 1 / 6 s are there in 37 1 / 2 ? | required number = ( 75 / 2 ) / ( 1 / 6 ) = ( 75 / 2 x 6 / 1 ) = 225 . answer : a | a ) 225 , b ) 425 , c ) 520 , d ) 600 , e ) 700 | a | divide(add(37, divide(1, 2)), divide(1, 6)) | divide(n0,n4)|divide(n0,n1)|add(n2,#0)|divide(#2,#1) | general |
what is the smallest positive perfect square that is divisible by 4 , 9 , and 18 ? | "the number needs to be divisible by 2 ^ 2 , 3 ^ 2 , and 2 * 3 ^ 2 . the smallest such perfect square is 2 ^ 2 * 3 ^ 2 = 36 the answer is a ." | a ) 36 , b ) 81 , c ) 144 , d ) 625 , e ) 900 | a | add(multiply(multiply(multiply(4, power(const_3, const_2)), 9), const_2), multiply(9, 18)) | multiply(n1,n2)|power(const_3,const_2)|multiply(n0,#1)|multiply(n1,#2)|multiply(#3,const_2)|add(#4,#0)| | geometry |
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 ? | "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" | a ) 3 : 4 , b ) 3 : 0 , c ) 9 : 1 , d ) 9 : 2 , e ) 3 : 9 | c | divide(add(multiply(7, divide(add(7, 7), subtract(7, 3))), 7), subtract(multiply(3, divide(add(7, 7), subtract(7, 3))), 7)) | add(n0,n0)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n0,#3)|subtract(#4,n0)|divide(#5,#6)| | other |
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 ? | "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" | a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 11 , d ) 7 : 3 , e ) 15 : 4 | d | divide(add(multiply(3, divide(add(4, 1), add(3, 2))), 4), add(multiply(2, divide(add(4, 1), add(3, 2))), 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)| | other |
from a pack of 52 cards , 1 card is drawn at random . what is the probability that a red king is drawn ? | "the total number of cards is 52 . the number of red kings is 2 . p ( red king ) = 2 / 52 = 1 / 26 the answer is d ." | a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 13 , d ) 1 / 26 , e ) 1 / 52 | d | divide(multiply(const_4, const_3), 52) | multiply(const_3,const_4)|divide(#0,n0)| | probability |
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 ? | soln : - 460 / 115 ) 110 = 385 in this case long division does not take much time . ( 4 / 1 ) 110 = rs . 440 answer : b | a ) rs . 400 , b ) rs . 440 , c ) rs . 150 , d ) rs . 460 , e ) rs . 520 | b | add(divide(460, add(const_1, divide(15, const_100))), multiply(divide(10, const_100), divide(460, add(const_1, divide(15, const_100))))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4) | general |
a fruit seller had some oranges . he sells 10 % oranges and still has 360 oranges . how many oranges he had originally ? | "explanation : he sells 10 % of oranges and still there are 360 oranges remaining = > 90 % of oranges = 360 β ( 90 Γ total oranges ) / 100 = 360 β total oranges / 100 = 4 β total oranges = 4 Γ 100 = 400 answer : option d" | a ) 420 , b ) 700 , c ) 220 , d ) 400 , e ) none of these | d | add(360, multiply(360, divide(10, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain |
because he β 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 ? | "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 %" | a ) 22.5 % , b ) 24 % , c ) 30 % , d ) 33 % , e ) 20 % | b | multiply(const_100, divide(add(divide(40, const_100), multiply(4, divide(20, const_100))), add(const_1, 4))) | 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)| | gain |
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 ) | "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" | a ) 10 min . , b ) 41 min . , c ) 17 min . , d ) 23 min . , e ) 25 min . | c | divide(multiply(10, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)| | physics |
the length of the bridge , which a train 180 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "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" | a ) 876 m , b ) 157 m , c ) 195 m , d ) 156 m , e ) 167 m | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 180) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
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 ? | let the number be xy . given xy β yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten β s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) β ( 10 y + x ) = 36 = > x β y = 4 = > 2 y β y = 4 . hence , ( x + y ) β ( x β y ) = 3 y β y = 2 y = 8 b | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | b | multiply(divide(36, subtract(multiply(subtract(const_10, const_1), multiply(2, const_1)), subtract(const_10, const_1))), const_2) | multiply(n0,const_1)|subtract(const_10,const_1)|multiply(#0,#1)|subtract(#2,#1)|divide(n1,#3)|multiply(#4,const_2) | general |
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 ? | "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" | a ) 424 . , b ) 428 . , c ) 430 . , d ) 432 . , e ) 436 | d | add(subtract(312, divide(112, const_2)), subtract(232, divide(112, const_2))) | divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|add(#1,#2)| | other |
if the average ( arithmetic mean ) of a and b is 100 , and the average of b and c is 160 , what is the value of a β c ? | "question : a - c = ? ( a + b ) / 2 = 100 = = = > a + b = 200 ( b + c ) / 2 = 160 = = = > b + c = 320 ( a + b ) - ( b + c ) = 200 - 320 = = = > a + b - b - c = - 120 = = = > a - c = - 120 answer : a" | a ) β 120 , b ) β 100 , c ) 100 , d ) 135 , e ) it can not be determined from the information given | a | subtract(multiply(160, const_2), multiply(100, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
in a company of 11 employees , 5 employees earn $ 32,000 , 4 employees earn $ 45,000 , and the 2 highest - paid employees earn the same amount . if the average annual salary for the 11 employees is $ 48,000 , what is the annual salary for each of the highest - paid employees ? | "5 * 32,000 + 4 * 45,000 + 2 x = 11 * 48,000 2 x = 528,000 - 160,000 - 180,000 2 x = 188,000 x = 94,000 the answer is b ." | a ) $ 90,000 , b ) $ 94,000 , c ) $ 98,000 , d ) $ 102,000 , e ) $ 106,000 | b | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(2, 2), add(4, const_1)), const_2), divide(const_1, const_2)), 11), multiply(multiply(11, 2), add(4, const_1))), multiply(multiply(multiply(2, 2), add(4, const_1)), 4)), 2), const_1000), const_1000), 5) | add(const_1,n3)|divide(const_1,const_2)|multiply(n5,n5)|multiply(n0,n5)|multiply(#0,#2)|multiply(#0,#3)|add(#4,const_2)|multiply(n3,#4)|add(#6,#1)|multiply(n0,#8)|subtract(#9,#5)|subtract(#10,#7)|divide(#11,n5)|multiply(#12,const_1000)|divide(#13,const_1000)|subtract(#14,n1)| | general |
if n = 2 ^ 0.20 and n ^ b = 16 , b must equal | "20 / 100 = 1 / 5 n = 2 ^ 1 / 5 n ^ b = 2 ^ 4 ( 2 ^ 1 / 5 ) ^ b = 2 ^ 4 b = 20 answer : c" | a ) 3 / 80 , b ) 3 / 5 , c ) 20 , d ) 5 / 3 , e ) 80 / 3 | c | divide(log(16), log(power(2, 0.20))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
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 | "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" | a ) rs . 640 , b ) rs . 760 , c ) rs . 860 , d ) rs . 960 , e ) none of these | a | multiply(800, subtract(const_1, divide(20, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|multiply(n0,#1)| | gain |
find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 14 per cubic meter ? | "22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 14 = 1386 answer : a" | a ) 1386 , b ) 2799 , c ) 2890 , d ) 1485 , e ) 2780 | a | multiply(volume_cylinder(divide(3, const_2), 14), 14) | divide(n1,const_2)|volume_cylinder(#0,n0)|multiply(n2,#1)| | physics |
rani bought more apples than oranges . she sells apples at βΉ 23 apiece and makes 15 % profit . she sells oranges at βΉ 10 apiece and makes 25 % profit . if she gets βΉ 653 after selling all the apples and oranges , find her profit percentage z . | "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" | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | 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) | 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)| | gain |
the sum of digits of a two digit number is 13 , the difference between the digits is 5 . find the number | "description : = > x + y = 13 , x - y = 5 adding these 2 x = 18 = > x = 9 , y = 4 . thus the number is 94 answer b" | a ) 85 , b ) 94 , c ) 83 , d ) 72 , e ) none | b | add(multiply(divide(add(13, 5), const_2), 13), subtract(13, divide(add(13, 5), const_2))) | add(n0,n1)|divide(#0,const_2)|multiply(n0,#1)|subtract(n0,#1)|add(#2,#3)| | general |
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 ? | "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" | a ) 10 days , b ) 11 days , c ) 13 days , d ) 15 days , e ) 17 days | d | inverse(add(divide(6, multiply(8, 25)), divide(11, multiply(12, 25)))) | multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)| | physics |
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 . | "explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) Γ 5 / 18 m / sec . therefore 140 / ( ( x + 5 ) Γ 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 2520 < = > x = 79 speed of the train is 79 kmph . answer : option e" | a ) 45 kmph , b ) 50 kmph , c ) 55 kmph , d ) 60 kmph , e ) 79 kmph | e | subtract(divide(140, multiply(6, const_0_2778)), 5) | multiply(n1,const_0_2778)|divide(n0,#0)|subtract(#1,n2)| | physics |
a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 45 = = > x = 50 km / hr . answer : b" | a ) 10 km / hr , b ) 50 km / hr , c ) 14 km / hr , d ) 17 km / hr , e ) 77 km / hr | b | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n2)|divide(#3,const_0_2778)| | physics |
0.002 x 0.6 = ? | "explanation : 2 x 6 = 12 . sum of decimal places = 4 0.002 x 0.6 = 0.0012 answer - a" | a ) 0.0012 , b ) 0.001 , c ) 0.01 , d ) 0.1 , e ) none of these | a | multiply(divide(0.002, 0.6), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
find the average of all the number between 6 and 34 which are divisible by 5 . | "solution average = ( 10 + 15 + 20 + 25 + 30 / 5 ) = 100 / 2 = 20 answer b" | a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 32 | b | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
if the wheel is 14 cm then the number of revolutions to cover a distance of 880 cm is ? | "2 * 22 / 7 * 14 * x = 880 = > x = 10 answer : b" | a ) a ) 15 , b ) b ) 10 , c ) c ) 14 , d ) d ) 12 , e ) e ) 11 | b | divide(880, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | 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)| | physics |
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 ? | "2 c = 1 / 8 + 1 / 9 β 1 / 7 = 47 / 504 c = 47 / 1008 = > 1008 / 47 = 21.4 days the answer is c ." | a ) 14.1 days , b ) 18.8 days , c ) 21.4 days , d ) 24.3 days , e ) 26.5 days | c | 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))) | 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)| | physics |
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 ? | "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" | a ) $ 25 , b ) $ 26 , c ) $ 27 , d ) $ 28 , e ) $ 30 | a | multiply(power(5, 1), 5) | power(n0,n1)|multiply(n0,#0)| | general |
what is the smallest positive integer x , such that 1,152 x is a perfect cube ? | "we need to make 1152 x a perfect cube , hence we need to have the powers a multiple of 3 1152 = 2 ^ 7 * 3 ^ 2 the minimum value of x for which 1152 x is a perfect cube = 2 ^ 2 * 3 = 12 correct option : d" | a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | d | add(const_3, const_4) | add(const_3,const_4)| | geometry |
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 ? | "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 ." | a ) 105 , b ) 140 , c ) 175 , d ) 210 , e ) 245 | e | multiply(35, divide(multiply(divide(36, const_60), 35), subtract(38, 35))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
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 ? | "0.85 * { original price } = $ 51 - - > { original price } = $ 60 . { final price } = $ 51 * 1.25 = $ 63.75 . the difference = $ 63.75 - 60 $ = $ 3.75 . answer : c ." | a ) $ 0.00 , b ) $ 1.00 , c ) $ 3.75 , d ) $ 5.00 , e ) $ 6.80 | c | subtract(multiply(51, divide(add(const_100, 25), const_100)), divide(51, divide(subtract(const_100, 15), const_100))) | 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)| | general |
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 ? | in same time , they cover 110 km & 90 km respectively so ratio of their speed = 110 : 90 = 11 : 9 answer : a | a ) 11 : 9 , b ) 11 : 2 , c ) 91 : 9 , d ) 11 : 1 , e ) 11 : 5 | a | inverse(divide(subtract(200, 110), 110)) | subtract(n2,n3)|divide(#0,n3)|inverse(#1) | physics |
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 ? | "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" | a ) 1991 , b ) 2991 , c ) 3991 , d ) 4521 , e ) 5991 | b | subtract(multiply(add(5, const_1), 4500), add(add(add(add(4435, 4927), 4855), 5230), 4562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
the set s consists of 5 numbers : { 1 , 2,3 , 4,5 } . if all possible subsets including the null set are created and one subset is chosen at random , what is the probability that the subset has 4 or 5 as its largest number ? | "the set s has 2 ^ 5 = 32 subsets . the number 5 is in half of these subsets . thus 5 is the largest number in 16 subsets of s . of the remaining 16 subsets , 4 is an element in 8 of them . thus 4 is the largest number in 8 subsets of s . the probability that 4 or 5 is the largest number is 24 / 32 = 3 / 4 . the answer is c ." | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 5 / 8 , e ) 11 / 16 | c | divide(multiply(5, 4), power(const_2, 5)) | multiply(n4,n5)|power(const_2,n0)|divide(#0,#1)| | probability |
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 $ 211 ? | "i think it should be e . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 9 20 - pack for 3.05 * 9 = $ 15.25 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 9 = 2180" | a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,124 , e ) 2,180 | e | multiply(divide(211, 22.95), 250) | divide(n6,n5)|multiply(n4,#0)| | general |
what is the sum of all 3 digit integers formed using the digits 34 and 5 ( repetition is allowed ) | n = 3 * 3 * 3 = 27 = ( 555 + 333 ) / 2 = 444 sum = number of integers x average value n * = 27 * 444 = 11988 answer = d | a ) 11982 , b ) 11984 , c ) 11985 , d ) 11988 , e ) 11986 | d | multiply(multiply(add(add(const_100, const_10), const_1), add(add(3, 5), const_4)), power(const_3, const_2)) | 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) | general |
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 ? | 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 | a ) 95 minutes , b ) 90 minutes , c ) 88 minutes , d ) 95 minutes , e ) none of these | b | divide(multiply(multiply(const_60, 10), divide(18, const_2)), const_60) | divide(n2,const_2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,const_60) | physics |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 35 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 35 4 / x = 1 / 35 = > x = 140 min . answer : e" | a ) 229 , b ) 787 , c ) 144 , d ) 128 , e ) 140 | e | multiply(add(const_1, const_4), 35) | add(const_1,const_4)|multiply(n0,#0)| | physics |
if x ^ 2 + y ^ 2 = 20 and xy = 3 , then ( x β y ) ^ 2 = | "but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 20 - 2 * 3 = 14 . . answer c" | a ) 8 , b ) 11 , c ) 14 , d ) 17 , e ) 20 | c | power(3, 2) | power(n3,n0)| | general |
what is the smallest positive integer x such that 108 x is the cube of a positive integer ? | "given 108 x is a perfect cube so we will take 216 = 6 * 6 * 6 108 x = 216 x = 216 / 108 = 2 correct option is a" | a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 7 | a | add(const_3, const_4) | add(const_3,const_4)| | geometry |
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 . | "solution average of 7 people after the last one enters = 151 . Γ’ Λ Β΄ required weight = ( 7 x 151 ) - ( 6 x 150 ) = 1057 - 900 = 157 . answer a" | a ) 157 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 150)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
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 ? | "0.2 x = 4 ( 860 - 800 ) 0.2 x = 240 x = 1200 answer e" | a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | e | divide(multiply(subtract(860, 800), 4), divide(20, const_100)) | divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0)| | general |
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 ? | "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" | a ) 38 kg , b ) 25.6 kg , c ) 28 kg , d ) 26.5 kg , e ) none of these | a | divide(multiply(10, 42.75), 11.25) | multiply(n1,n2)|divide(#0,n0)| | physics |
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 : | explanation : 1 work done = 8 Γ 20 1 man = 3 / 2 boys 1 boy = 2 / 3 men then , 9 boys = 9 Γ 2 / 3 men = 6 men then , 4 men + 9 boys = 10 men then , 8 Γ 20 = 10 Γ ? days ? days = 8 Γ 20 / 10 = 16 days . answer : option d | a ) 10 days , b ) 12 days , c ) 15 days , d ) 16 days , e ) 20 days | d | divide(multiply(multiply(8, divide(const_3, const_2)), 20), add(multiply(4, divide(const_3, const_2)), 9)) | divide(const_3,const_2)|multiply(n0,#0)|multiply(n2,#0)|add(n3,#2)|multiply(n1,#1)|divide(#4,#3) | physics |
john makes $ 40 a week from his job . he earns a raise and now makes $ 70 a week . what is the % increase ? | "increase = ( 30 / 40 ) * 100 = ( 3 / 4 ) * 100 = 75 % . e" | a ) 16 % , b ) 16.66 % , c ) 17.9 % , d ) 18.12 % , e ) 75 % | e | multiply(divide(subtract(70, 40), 40), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
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 | "explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 410 + 140 = 550 meter time = distance / speed = 550 β 2 / 25 = 44 seconds option b" | a ) 20 seconds , b ) 44 seconds , c ) 40 seconds , d ) 50 seconds , e ) none of these | b | divide(add(410, 140), divide(multiply(45, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 15 terms of the arithmetic progression ? | "the sum of the first n terms of a g . p . is given by , where ' a ' is the first term of the g . p . , ' r ' is the common ratio and ' n ' is the number of terms in the g . p . therefore , the sum of the first 6 terms of the g . p will be equal to and sum of the first 3 terms of the g . p . will be equal to use the ratio between these two sums to find ' r ' the ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1 i . e . or r 3 + 1 = 9 r 3 = 8 r = 2 the answer is c" | a ) 3 , b ) 1 / 3 , c ) 2 , d ) 9 , e ) 1 / 9 | c | divide(multiply(20, 15), const_2) | multiply(n0,n1)|divide(#0,const_2)| | general |
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 ? | "a + b + c + d = 200 a + b = 80 c + d = 120 greatest possible = 119 ( just less than 1 ) answer = c" | a ) 55 , b ) 65 , c ) 119 , d ) 109 , e ) 115 | c | subtract(multiply(50, 4), multiply(40, 2)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
if f ( x ) = 3 x ^ 4 - 4 x ^ 3 - 2 x ^ 2 + 6 x , then f ( - 1 ) = | "f ( - 1 ) = 3 ( - 1 ) ^ 4 - 4 ( - 1 ) ^ 3 - 2 ( - 1 ) ^ 2 + 6 ( - 1 ) = 3 + 4 - 2 - 6 = - 1 the answer is b ." | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | b | add(subtract(subtract(multiply(3, power(negate(1), 4)), multiply(4, power(negate(1), 3))), multiply(3, power(negate(1), 2))), multiply(6, negate(1))) | negate(n7)|multiply(n6,#0)|power(#0,n1)|power(#0,n0)|power(#0,n5)|multiply(n0,#2)|multiply(n1,#3)|multiply(n0,#4)|subtract(#5,#6)|subtract(#8,#7)|add(#1,#9)| | general |
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 ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 150 . s . p . = 75 % of rs . 150 = rs . 112.50 . β΄ gain % = 12.50 % . answer a" | a ) 12.50 % , b ) 13.50 % , c ) 14 % , d ) 14.50 % , e ) none | a | 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) | 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)| | gain |
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 ) | "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 ." | a ) 3.33 , b ) 3.34 , c ) 3.53 , d ) 5.32 , e ) 3.92 | a | divide(multiply(1,000, 1,000), multiply(300, 1,000)) | multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)| | geometry |
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 ? | "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 )" | a ) 1 : 20 , b ) 1 : 10 , c ) 3 : 8 , d ) 3 : 4 , e ) 3 : 2 | c | divide(subtract(33, divide(36, divide(add(const_100, 50), const_100))), divide(36, divide(add(const_100, 50), const_100))) | add(n1,const_100)|divide(#0,const_100)|divide(n2,#1)|subtract(n0,#2)|divide(#3,#2)| | gain |
a train 520 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "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" | a ) 88 kmph , b ) 94 kmph , c ) 72 kmph , d ) 16 kmph , e ) 18 kmph | b | divide(divide(520, const_1000), divide(20, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
in a throw of dice what is the probability of ge Γ¦ Β« ng number greater than 4 | "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" | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these | b | divide(subtract(const_6, 4), const_6) | subtract(const_6,n0)|divide(#0,const_6)| | probability |
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 ? | "60 + 65 + 25 = 150 % the number is 50 % above 100 % because 50 % of the counties were counted twice . the answer is c ." | a ) 12.5 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 67.5 % | c | subtract(add(60, 65), subtract(const_100, 25)) | add(n0,n1)|subtract(const_100,n2)|subtract(#0,#1)| | gain |
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 ? | "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" | a ) 17 hr , b ) 14 hr , c ) 12 hr , d ) 19 hr , e ) 23 hr | a | divide(8.5, subtract(5.5, 5)) | subtract(n1,n0)|divide(n2,#0)| | gain |
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 ? | "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 ." | a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25 | d | divide(divide(300, 5), 3) | divide(n0,n2)|divide(#0,n1)| | physics |
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 . | "let no . of non - officers be x 15 * 460 + x * 110 = ( x + 15 ) 120 x = 510 e" | a ) 400 , b ) 420 , c ) 430 , d ) 450 , e ) 510 | e | divide(subtract(multiply(15, 460), multiply(15, 120)), subtract(120, 110)) | multiply(n1,n3)|multiply(n0,n3)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)| | general |
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 : | "clearly , the numbers are ( 10 x 13 ) and ( 10 x 14 ) . larger number = ( 10 x 14 ) = 140 . answer : option d" | a ) 100 , b ) 120 , c ) 180 , d ) 140 , e ) 60 | d | multiply(10, 14) | multiply(n0,n2)| | other |
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 . | "explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 / 21 * 672 = 256 answer : a" | a ) 256 , b ) 388 , c ) 379 , d ) 277 , e ) 122 | a | 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)))) | 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)| | gain |
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 ? | 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 | a ) 17 , b ) 98 , c ) 88 , d ) 85 , e ) 83 | d | divide(subtract(multiply(5, subtract(12.4, 0.4)), 26), 0.4) | subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3) | general |
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 ? | 27 * 12 : 54 * x = 2 : 1 x = 3 12 - 3 = 9 answer : a | a ) 9 , b ) 6 , c ) 7 , d ) 8 , e ) 2 | a | subtract(multiply(const_4, const_3), divide(divide(multiply(27000, multiply(const_4, const_3)), 54000), 2)) | multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3) | other |
in a recent election , james received 2.1 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 46 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 2.1 / 100 ) * 2000 = 42 votes to win = ( 46 / 100 ) * total votes + 1 = ( 46 / 100 ) * 2000 + 1 = 921 remaining voted needed to win election = 921 - 42 = 879 answer : option b" | a ) 901 , b ) 879 , c ) 990 , d ) 991 , e ) 1,001 | b | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 2.1)) | add(const_1000,const_1000)|multiply(n0,#0)|subtract(#0,#1)| | general |
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 ? | "( 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" | a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | a | divide(20, divide(add(add(divide(20, 11), divide(20, 20)), add(divide(20, 11), divide(20, 20))), const_2)) | 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)| | physics |
let c be defined as the sum of all prime numbers between 0 and 38 . what is c / 3 | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 33 , 37 sum , c = 230 c / 3 = 76.6 answer b" | a ) 155 , b ) 76.6 , c ) 61 , d ) 47 , e ) 43 | b | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)| | general |
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 ? | "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" | a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | add(n1,n4)|subtract(n0,#0)|subtract(n2,#1)| | general |
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 . | 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 ) β 5 a + 2 b = 2 / 3 ( 6 a + 8 b ) β 15 a + 6 b = 12 a + 16 b β 3 a = 10 b β ab = 10 / 3 β a : b = 10 : 3 answer : option d | a ) 2 : 1 , b ) 1 : 2 , c ) 4 : 3 , d ) 10 : 3 , e ) 3 : 2 | d | 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))) | 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) | general |
a man invests some money partly in 9 % stock at 96 and partly in 12 % stock at 120 . to obtain equal dividends from both , he must invest the money in the ratio ? | "for an income of re . 1 in 9 % stock at 96 , investment = rs . 96 / 9 = rs . 32 / 3 for an income re . 1 in 12 % stock at 120 , investment = rs . 120 / 12 = rs . 10 . ratio of investments = ( 32 / 3 ) : 10 = 32 : 30 = 16 : 15 answer : c" | a ) 16 : 18 , b ) 16 : 13 , c ) 16 : 15 , d ) 16 : 12 , e ) 16 : 11 | c | divide(multiply(96, const_2), multiply(120, const_3)) | multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)| | other |
20 men do a work in 20 days . how many men are needed to finish the work in 10 days ? | "men required to finish the work in 10 days = 20 * 20 / 10 = 40 answer is e" | a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 40 | e | divide(multiply(20, 20), 10) | multiply(n0,n1)|divide(#0,n2)| | physics |
if 5 a = 6 b and ab β 0 , what is the ratio of a / 6 to b / 5 ? | "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" | a ) 36 / 25 , b ) 10 / 12 , c ) 1 , d ) 1 / 5 , e ) 25 / 36 | c | divide(multiply(5, 6), multiply(6, 5)) | multiply(n0,n1)|divide(#0,#0)| | general |
a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 33 per litre so as to make a profit of 25 % on cost when he sells the resulting milk and water mixture for rs . 36 in what ratio should he mix the water and milk ? | "water = w ( liter ) milk = m ( liter ) = = > cost = price x quantity = 0.33 m = = > revenue = price x quantity = 0.36 ( m + w ) = = > profit = 0.36 ( m + w ) - 0.33 m = 0.25 * ( 0.33 m ) [ 25 % of cost ] = = > 0.36 m + 0.36 w - 0.33 m = 0.0825 m = = > 0.0525 m = 0.36 w = = > m / w = 0.36 / 0.0525 = 124 / 21 - - or - - w / m = 21 / 124 e is correct ." | a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 21 : 124 | e | divide(const_1, divide(25, const_2)) | divide(n1,const_2)|divide(const_1,#0)| | gain |
gary β s gas station serves an average of 15 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary β s station serve over the course of a typical week ? | 6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 15 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 15 + 10 ) = 16 * 70 = 1120 answer : a | a ) 1,120 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(15, 10), multiply(9, add(const_4, const_1)))), const_1000)) | add(n1,n3)|add(n0,n1)|add(const_1,const_4)|multiply(n2,#2)|add(#1,#3)|multiply(#0,#4)|divide(#5,const_1000)|floor(#6) | physics |
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 ? | "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" | a ) 10 , b ) 12 , c ) 20 , d ) 30 , e ) 15 | b | inverse(add(divide(const_1, 20), divide(const_1, 30))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics |
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 ) | 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 | a ) 14 / 15 , b ) 15 / 14 , c ) 4 / 5 , d ) 6 / 5 , e ) can not be determined | d | divide(divide(42, 3.5), divide(42, 4.2)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1) | physics |
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 ? | 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 | a ) 970.3 liters , b ) 1000.45 liters , c ) 879.65 liters , d ) 1020.56 liters , e ) 910.95 liters | a | subtract(subtract(subtract(1000, 10), 10), 10) | subtract(n0,n1)|subtract(#0,n1)|subtract(#1,n1) | general |
the workforce of company x is 60 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 50 % . how many employees did the company have after hiring the additional male workers ? | "let x be the total worker then 0.6 x = female worker and 0.4 x is male worker then 20 male worker added 06 x / ( 0.4 x + 20 ) = 50 / 100 or 60 x = 50 * ( 0.4 x + 100 ) = 20 x + 5000 or 40 x = 5000 , x = 500 / 4 = 125 total worker = 125 + 20 = 145 a" | a ) 145 , b ) 188 , c ) 189 , d ) 190 , e ) 191 | a | add(divide(multiply(divide(50, const_100), 20), subtract(divide(60, const_100), divide(50, const_100))), 20) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain |
the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 216 sq . cm , what is the width of the rectangle ? | "2 l + 2 w = 5 w l = 3 w / 2 w * l = 216 3 w ^ 2 / 2 = 216 w ^ 2 = 144 w = 12 the answer is a ." | a ) 12 cm , b ) 16 cm , c ) 18 cm , d ) 20 cm , e ) 22 cm | a | divide(216, const_10) | divide(n2,const_10)| | geometry |
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 ? | "total circuit = 18 / 3 = 6 total feet squirrel travels = 6 * 3 = 18 feet answer : e" | a ) 10 feet , b ) 12 feet , c ) 13 feet , d ) 15 feet , e ) 18 feet | e | multiply(divide(18, 3), 3) | divide(n1,n0)|multiply(n2,#0)| | geometry |
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 ? | 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 . | ['a ) 15 ohms', 'b ) 15 / 8 ohms', 'c ) 1 / 8 ohms', 'd ) 8 / 15 ohms', 'e ) 8 ohms'] | b | divide(multiply(3, 5), add(3, 5)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0) | geometry |
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 | 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 . β΄ difference of b and c β s shares = rs . ( 26400 x 9 / 22 - 26400 x 6 / 22 ) = rs . 3600 . answer c | a ) rs . 2400 , b ) rs . 3000 , c ) rs . 3600 , d ) rs . 4800 , e ) none of these | c | 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)))))) | 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) | general |
in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.815 per pound ? | "using weighted average method : let x be the proportion in which the $ 0.8 per pound flour is mixed with $ 0.9 per pound flour . thus 0.8 * x + 0.9 * ( 1 - x ) = 0.815 0.9 β 0.1 x = 0.815 x = 0.85 thus ratio of both flours is 6 : 1 c" | a ) 1 : 3 , b ) 1 : 2 , c ) 6 : 1 , d ) 2 : 1 , e ) 3 : 1 | c | divide(0.815, add(0.8, 0.9)) | add(n0,n1)|divide(n2,#0)| | general |
if the cost price of 20 articles is equal to the selling price of 25 articles , what is the % profit or % loss made by the merchant ? | "explanatory answer approach : assume a value for cost price . compute cost price and selling price for the same number of articles let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 25 articles = cost price of 20 articles = $ 20 . let us find the cost price of 25 articles . cost price of 25 articles = 25 * 1 = $ 25 . therefore , profit made on sale of 25 articles = selling price of 25 articles - cost price of 25 articles = 20 - 25 = - $ 5 . because the profit is in the negative , the merchant has made a loss of $ 5 . therefore , % loss = loss / cost price β 100 % loss = 5 / 25 β 100 = 20 % loss . choice c" | a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 5 % profit | c | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 20)), divide(multiply(const_100, 25), 20))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
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 : | "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" | a ) 60 m , b ) 20 m , c ) 40 m , d ) 20 m , e ) 23 m | c | subtract(500, divide(multiply(subtract(500, 155), 4), 3)) | subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)| | physics |
dan ' s age after 16 years will be 6 times his age 4 years ago . what is the present age of dan ? | "let dan ' s present age be x . x + 16 = 6 ( x - 4 ) 5 x = 40 x = 8 the answer is a ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | divide(add(16, multiply(4, 6)), subtract(6, const_1)) | multiply(n1,n2)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)| | general |
the ratio of two quantities is 10 : 7 . if each of the quantities is increased by 2 , their ratio changes to 15 : 11 then the greatest number is ? | "let the numbers be 10 x and 7 x then 10 x + 2 / 7 x + 2 = 15 / 11 110 x + 22 = 105 x + 30 5 x = 8 x = 1.6 greatest number = 10 * 1.6 = 16 answer is d" | a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 20 | d | divide(add(10, 2), add(7, 2)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | general |
by selling 20 pencils for a rupee a man loses 60 % . how many for a rupee should he sell in order to gain 60 % ? | "40 % - - - 20 160 % - - - ? 40 / 160 * 20 = 5 answer : e" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | e | multiply(divide(const_1, multiply(add(const_100, 60), divide(const_1, subtract(const_100, 60)))), 20) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
after decreasing 24 % in the price of an article costs rs . 1216 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 1216 cp = 16 * 100 = > cp = 1600 answer : d" | a ) 1667 , b ) 6789 , c ) 1200 , d ) 1600 , e ) 1421 | d | divide(1216, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
if ( 18 ^ a ) * 9 ^ ( 3 a β 1 ) = ( 2 ^ 3 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | 18 ^ a ) * 9 ^ ( 3 a β 1 ) = ( 2 ^ 3 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a β 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 | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 3 | e | multiply(3, 1) | multiply(n2,n3) | general |
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 ? | 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 . | a ) 75 , b ) 100 , c ) 125 , d ) 150 , e ) 175 | a | divide(30, subtract(subtract(const_1, inverse(10)), divide(const_1, const_2))) | divide(const_1,const_2)|inverse(n0)|subtract(const_1,#1)|subtract(#2,#0)|divide(n1,#3) | physics |
a bag marked at $ 250 is sold for $ 120 . the rate of discount is ? | "rate of discount = 130 / 250 * 100 = 52 % answer is d" | a ) 10 % , b ) 25 % , c ) 20 % , d ) 52 % , e ) 45 % | d | multiply(divide(subtract(250, 120), 250), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |