Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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one week , a certain truck rental lot had a total of 20 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 10 trucks on the lot that saturday morning , what is ... | "n - not rented trucks ; r - rented trucks n + r = 20 n + r / 2 = 10 r = 20 b" | a ) 18 , b ) 20 , c ) 12 , d ) 8 , e ) 4 | b | add(10, multiply(20, divide(50, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n2,#1)| | general |
the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 47 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 48 - 47 ) = 1801 . correct mean = 1801 / 50 = 36.02 answer : d" | a ) 36.7 , b ) 36.1 , c ) 36.5 , d ) 36.02 , e ) 36.3 | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 47)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
hillary and eddy are climbing to the summit of mt . everest from a base camp 5000 ft from the summit . when they depart for the summit at 06 : 00 , hillary climbs at a rate of 800 ft / hr with eddy lagging behind at a slower rate of 500 ft / hr . if hillary stops 1000 ft short of the summit and then descends at a rate ... | solution : h stopped 1000 ft before reaching the final point , time taken to reach 4000 ft = 4000 / 800 = 5 hrs . this means she reached there at 11 : 00 . speed difference between them is 800 - 500 = 300 ft / hr so by the time h stops they have 1500 ft of distance so now here we use relative speed formula they both ar... | a ) 11.0 , b ) 12.0 , c ) 13.0 , d ) 14.0 , e ) 15.0 | b | add(add(6, divide(subtract(5000, 1000), 800)), const_1) | subtract(n0,n5)|divide(#0,n3)|add(n1,#1)|add(#2,const_1) | general |
$ 200 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 200 * 2 / 5 = $ 80 answer is a" | a ) $ 80 , b ) $ 50 , c ) $ 25 , d ) $ 200 , e ) $ 50 | a | multiply(divide(200, add(divide(2, 3), const_1)), divide(2, 3)) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general |
if a ( a + 6 ) = 27 and b ( b + 6 ) = 27 , where a β b , then a + b = | "a ( a + 6 ) = 27 and b ( b + 6 ) = 27 = > a , b must be integers and if a is - 9 or 3 , b will be 3 and - 9 respectively = > a + b = - 6 ans : b" | a ) β 48 , b ) β 6 , c ) 2 , d ) 46 , e ) 48 | b | add(divide(27, const_10), divide(27, divide(27, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
the base of a right triangle is 8 and hypotenuse is 10 . its area is ? | explanation : h 2 = ( 10 ) 2 - ( 8 ) 2 - ( 6 ) 2 - > h = 6 1 / 2 * 8 * 6 = 24 answer is d | ['a ) 12', 'b ) 80', 'c ) 59', 'd ) 24', 'e ) 25'] | d | divide(multiply(sqrt(subtract(multiply(10, 10), multiply(8, 8))), 8), const_2) | multiply(n1,n1)|multiply(n0,n0)|subtract(#0,#1)|sqrt(#2)|multiply(n0,#3)|divide(#4,const_2) | geometry |
jack and jill work at a hospital with 3 other workers . for an internal review , 2 of the 5 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | 1 / 5 c 2 = 1 / 10 . answer : b . | a ) 1 / 3 , b ) 1 / 10 , c ) 1 / 15 , d ) 3 / 8 , e ) 2 / 3 | b | inverse(divide(factorial(5), multiply(factorial(2), factorial(3)))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4) | physics |
a river 2 m deep and 45 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is ? | "explanation : ( 4000 * 2 * 45 ) / 60 = 6000 m 3 answer : option a" | a ) 6000 m 3 , b ) 4580 m 3 , c ) 18500 m 3 , d ) 4900 m 3 , e ) 4700 m 3 | a | divide(multiply(multiply(2, 45), multiply(4, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics |
the ratio of two quantities is 5 to 6 . if each of the quantities is increased by 15 , what is the ratio of these 2 new quantities ? | the ratio can not be straight way added any quantity . . . 5 : 6 means 5 x : 6 x . . . so when you add a quantity , it becomes 5 x + 15 : 6 x + 15 . . so value of x is must . . ans e | a ) 5 : 6 , b ) 25 : 27 , c ) 15 : 16 , d ) 20 : 21 , e ) it can not be determined from the information given | e | divide(add(5, 15), add(6, 15)) | add(n0,n2)|add(n1,n2)|divide(#0,#1) | other |
a sum of $ 1000 amounts to $ 1500 in 5 years at simple interest . if the interest rate is increased by 5 % it would amount to how much ? | "s . i = 1500 - 1000 = 500 p = $ 1000 t = 5 years r = 100 * 500 / 1000 * 5 = 10 % new rate = 10 + 5 = 15 % new s . i . = 1000 * 15 * 5 / 100 = $ 750 new amount = 1000 + 750 = $ 1750 answer is e" | a ) $ 1000 , b ) $ 750 , c ) $ 1541 , d ) $ 1478 , e ) $ 1750 | e | add(1000, multiply(multiply(1000, divide(add(divide(multiply(divide(subtract(1500, 1000), 5), const_100), 1000), 5), const_100)), 5)) | subtract(n1,n0)|divide(#0,n2)|multiply(#1,const_100)|divide(#2,n0)|add(n2,#3)|divide(#4,const_100)|multiply(n0,#5)|multiply(n2,#6)|add(n0,#7)| | gain |
if f ( x ) = ax ^ 5 β 3 x ^ 2 + ax ^ 2 β x , then f ( b ) β f ( - b ) will equal : | "f ( x ) = ax ^ 5 β 3 x ^ 2 + ax ^ 2 β x f ( b ) = ab ^ 5 β 3 b ^ 2 + ab ^ 2 β b f ( - b ) = - ab ^ 5 β 3 b ^ 2 + ab ^ 2 + b f ( b ) - f ( - b ) = ab ^ 5 β 3 b ^ 2 + ab ^ 2 β b + ab ^ 5 + 3 b ^ 2 - ab ^ 2 β b = 2 ab ^ 5 - 2 b answer d" | a ) 0 , b ) 2 ab , c ) 2 ab ^ 4 β 6 , d ) 2 ab ^ 5 - 2 b , e ) ab ^ 5 | d | multiply(2, const_10) | multiply(n2,const_10)| | general |
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 : | "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 : d" | a ) 2 : 5 , b ) 2 : 3 , c ) 2 : 4 , d ) 4 : 5 , e ) 2 : 9 | d | divide(subtract(15.8, 15.4), subtract(16.3, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
a man whose bowling average is 12.4 , takes 8 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 8 + x ) 12 solve equation x = 175 answer : c" | a ) 173 , b ) 174 , c ) 175 , d ) 176 , e ) 177 | c | divide(subtract(multiply(floor(12.4), 8), 26), subtract(12.4, floor(12.4))) | floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2)| | general |
a number when divided by 342 gives a remainder 47 . when the same number is divided by 19 what would be the remainder ? | number = 342 k + 47 = 19 * 18 k + 19 * 2 + 9 = 19 ( 18 k + 2 ) + 9 . the given number when divided by 19 gives 18 k + 2 as quotient and 9 as remainder . answer b 9 | a ) 0 , b ) 9 , c ) 19 , d ) 29 , e ) 18 | b | reminder(47, 19) | reminder(n1,n2) | general |
a train passes a station platform in 62 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 62 = 15 = > x = 444 m answer : d" | a ) 615 m , b ) 240 m , c ) 168 m , d ) 444 m , e ) 691 m | d | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
a , b and c enter into a partnership by investing $ 7000 , $ 11000 and $ 18000 respectively . at the end of 8 months , b receives $ 2200 as his share . find the share of a . | "the ratio of capital of a , b and c = 7000 : 11000 : 18000 = 7 : 11 : 18 a ' s share = ( 7 / 11 ) * 2200 = $ 1400 the answer is c ." | a ) $ 1200 , b ) $ 1300 , c ) $ 1400 , d ) $ 1500 , e ) $ 1600 | c | divide(multiply(multiply(7000, divide(2200, divide(multiply(11000, 8), const_100))), 8), const_100) | multiply(n1,n3)|divide(#0,const_100)|divide(n4,#1)|multiply(n0,#2)|multiply(n3,#3)|divide(#4,const_100)| | gain |
if 40 % of a certain number is 160 , then what is 10 % of that number ? | explanation : 40 % = 40 * 4 = 160 10 % = 10 * 4 = 40 answer : option b | a ) 100 , b ) 40 , c ) 80 , d ) 120 , e ) 110 | b | multiply(divide(160, divide(40, const_100)), divide(10, const_100)) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1) | gain |
a man took loan from a bank at the rate of 12 % p . a . s . i . after 20 years he had to pay rs . 2100 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 2100 ) / ( 12 * 20 ) = rs . 875 answer : d" | a ) 600 , b ) 715 , c ) 600 , d ) 875 , e ) 900 | d | divide(divide(multiply(2100, const_100), 20), 12) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)| | gain |
the average ( arithmetic mean ) of 24 , 35 , and 58 is 6 more than the average of 19 , 51 , and x . what is x ? | "the average of 24 , 35 , and 58 is 39 . the average of 19 , 51 and x is 33 . then 19 + 51 + x = 99 . x = 29 . the answer is d ." | a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | d | subtract(add(add(24, 35), 58), add(add(multiply(6, const_3), 19), 51)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general |
instead of multiplying a number by 5 , the number is divided by 5 . what is the percentage of error obtained ? | "let the number be x the right number is 5 x the wrong number is x / 5 error is ( 5 x - x / 5 ) = 24 x / 5 percentage of error is ( ( 24 x / 5 ) / 5 x ) * 100 = 96 % answer : c" | a ) 98 % , b ) 97 % , c ) 96 % , d ) 99 % , e ) 95 % | c | subtract(const_100, multiply(divide(divide(5, 5), multiply(5, 5)), const_100)) | divide(n1,n1)|multiply(n0,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | general |
a person crosses a 600 m long street in 5 minutes , what is his speed in km per hour ? | "explanation : speed = { \ color { blue } \ left ( \ frac { 600 } { 5 \ times 60 } \ right ) m / sec = 2 m / sec = \ left ( 2 \ times \ frac { 18 } { 5 } \ right ) km / hr = 7.2 km / hr } answer : b" | a ) 4.3 , b ) 7.2 , c ) 2.3 , d ) 3.2 , e ) 1.1 | b | divide(divide(600, const_1000), divide(multiply(5, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics |
the length of a train and that of a platform are equal . if with a speed of 72 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 72 * 5 / 18 ] m / sec = 20 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 20 = > x = 20 * 60 / 2 = 600 answer : e" | a ) 299 , b ) 266 , c ) 299 , d ) 750 , e ) 600 | e | divide(divide(multiply(72, const_1000), divide(const_60, const_1)), const_2) | divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)| | physics |
square a has an area of 81 square centimeters . square b has a perimeter of 32 centimeters . if square b is placed within square a and a random point is chosen within square a , what is the probability the point is not within square b ? | i guess it ' s mean that square b is placed within square aentirely . since , the perimeter of b is 32 , then its side is 32 / 4 = 8 and the area is 4 ^ 2 = 64 ; empty space between the squares is 81 - 64 = 17 square centimeters , so if a random point is in this area then it wo n ' t be within square b : p = favorable ... | ['a ) 17 / 64', 'b ) 17 / 81', 'c ) 16 / 25', 'd ) 3 / 5', 'e ) 6 / 25'] | b | divide(subtract(81, power(divide(32, const_4), const_2)), 81) | divide(n1,const_4)|power(#0,const_2)|subtract(n0,#1)|divide(#2,n0) | geometry |
one hour after yolanda started walking from x to y , a distance of 60 miles , bob started walking along the same road from y to x . if yolanda Γ’ s walking rate was 5 miles per hour and bob Γ’ s was 6 miles per hour , how many miles had bob walked when they met ? | "let t be the number of hours that bob had walked when he met yolanda . then , when they met , bob had walked 4 t miles and yolanda had walked 5 ( t + 1 ) miles . these distances must sum to 60 miles , so 6 t + 5 ( t + 1 ) = 60 , which may be solved for t as follows 6 t + 5 ( t + 1 ) = 60 6 t + 5 t + 5 = 60 11 t = 55 t... | a ) 40 , b ) 25 , c ) 28 , d ) 30 , e ) 30.5 | d | multiply(divide(subtract(60, 5), add(5, 6)), 6) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics |
in a class of 25 students , 15 play hockey and 16 play basketball . if there are 4 students who play neither sport , determine the number of students who play both hockey and basketball ? | let the number of who play both be x number of students who play sports 25 - 4 = 21 use venn diagram ( 15 - x ) + x + ( 16 - x ) = 21 which gives x = 10 answer is a 10 play both sport | a ) 10 , b ) 11 , c ) 13 , d ) 14 , e ) 15 | a | subtract(add(15, 16), subtract(25, 4)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1) | other |
what is the difference between 80 % of 170 and 35 % of 300 . | "( 80 / 100 ) * 170 Γ’ β¬ β ( 35 / 100 ) * 300 136 - 105 = 31 answer : e" | a ) 30 , b ) 29 , c ) 37 , d ) 36 , e ) 31 | e | subtract(divide(multiply(80, 170), const_100), divide(multiply(35, 300), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
a car travels first 160 km at 90 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 90 km / hr time taken to travel first 160 km = distancespeed = 160 / 90 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 160 / 80 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 160 / 90 + 160 / 80 average speed = total distance trave... | a ) 71.11 , b ) 84.7 , c ) 71.1 , d ) 71.17 , e ) 71.13 | b | multiply(320, divide(const_1, add(divide(160, 90), divide(160, 80)))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(const_1,#2)|multiply(n4,#3)| | general |
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 37 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 37 sec = 37 * 10 = 370 m . the length of the faster train = 370 m . answer : b" | a ) 270 , b ) 370 , c ) 266 , d ) 299 , e ) 126 | b | multiply(divide(subtract(72, 36), const_3_6), 37) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
a water tank which could be filled in 5 hours takes one more hour to be filled as a result of a leak in its bottom . if the tank is full calculate the time it will take for the leak empty it ? | part filled without leak in 1 hour = 1 / 5 part filled with leak in 1 hour = 1 / 6 work done by leak in 1 hour = 1 / 5 Γ’ Λ β 1 / 6 = 30 hours answer : b | a ) 32 hours , b ) 30 hours , c ) 34 hours , d ) 31 hours , e ) 33 hours | b | divide(5, divide(const_1, add(5, const_1))) | add(n0,const_1)|divide(const_1,#0)|divide(n0,#1) | physics |
country x taxes each of its citizens an amount equal to 10 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 ? | "equation is correct , so math must be a problem . 0.10 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 4,000 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 12,000 - - > x = 60,000 . answer : c ." | a ) $ 40,000 , b ) $ 56,000 , c ) $ 60,000 , d ) $ 66,667 , e ) $ 80,000 | c | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(10, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | 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)| | general |
two trains of length 100 m and 200 m are 70 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 72 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 54 + 72 ) * 5 / 18 = 7 * 5 = 35 mps . the time required = d / s = 70 / 35 = 2 sec . answer : a" | a ) 2 sec , b ) 32 / 3 sec , c ) 20 / 7 sec , d ) 32 / 3 sec , e ) 53 / 2 sec | a | divide(70, multiply(add(54, 72), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
the area of a rectangular plot is 24 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | l Γ b = 24 Γ b β΄ l = 24 m and l β b = 10 β΄ b = 24 β 10 = 14 m answer a | ['a ) 14 metres', 'b ) 5 metres', 'c ) 7.5 metres', 'd ) data inadequate', 'e ) none of these'] | a | subtract(24, 10) | subtract(n0,n1) | geometry |
peter ' s average ( arithmetic mean ) test score on 4 tests is 78 . what must be the student ' s score on a 5 th test for the peter ' s average score on the 5 tests to be 80 ? | e . 88 peter must score at least an 80 for sure . if he scores an 8 , then he will need to score 2 pots for each of the 4 other tests tomake upthe difference . they each were at 78 ( at least the average is , but this is a small point and does n ' t matter to the answer ) . so 4 tests that were each 2 points short of t... | a ) 80 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | e | subtract(multiply(5, 80), multiply(4, 78)) | multiply(n2,n4)|multiply(n0,n1)|subtract(#0,#1) | general |
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 14 sec . what is the length of the platform ? | "speed = 300 / 14 = 150 / 7 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 150 / 7 = > x = 535.71 m . answer : d" | a ) 287 m , b ) 350 m , c ) 267 m , d ) 535.71 m , e ) 656 m | d | subtract(multiply(speed(300, 14), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 21 gallons of gasoline ? | "so 12.2 * 21 = 256 . . imo option d is correct answer . ." | a ) 190 , b ) 284.6 , c ) 300 , d ) 256 , e ) 312 | d | multiply(12.2, 21) | multiply(n0,n2)| | general |
find the 15 th term of an arithmetic progression whose first term is 2 and the common difference is 3 . | "n th term of a . p = a + ( n - 1 ) * d = 2 + ( 15 - 1 ) * 3 , = 2 + 42 = 44 . answer : c" | a ) 45 , b ) 38 , c ) 44 , d ) 40 , e ) 46 | c | add(multiply(subtract(15, const_1), 3), 2) | subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1)| | general |
a fill pipe can fill 1 / 2 of cistern in 15 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 15 * 2 * 1 / 2 = 15 minutes answer is d" | a ) 1 min , b ) 5 min , c ) 10 min , d ) 15 min , e ) 20 min | d | divide(15, 1) | divide(n2,n0)| | physics |
a train passes a platform in 30 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 | "explanation : speed of the train = 54 km / hr = ( 54 Γ 10 ) / 30 m / s = 15 m / s length of the train = speed Γ time taken to cross the man = 15 Γ 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 30 = > 300 + l = 15 Γ 30 = 450 = > l = 450 - 300 = ... | a ) 150 meter , b ) 240 meter , c ) 200 meter , d ) 260 meter , e ) none of these | a | multiply(multiply(const_0_2778, 54), subtract(30, 20)) | multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)| | physics |
kiran has 85 currency notes in all , some of which were of rs . 100 denomination and the remaining of rs . 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount did she have in the denomination of rs . 50 ? | "let the number of 50 β rupee notes be x . then , the number of 100 - rupee notes = ( 85 β x ) 50 x + 100 ( 85 β x ) = 5000 = x + 2 ( 85 β x ) = 100 = x = 70 so , required amount = rs . ( 50 x 70 ) = rs . 3500 answer : option b" | a ) 1900 , b ) 3500 , c ) 4000 , d ) 4200 , e ) 4450 | b | multiply(subtract(85, divide(subtract(5000, multiply(50, 85)), 50)), 50) | multiply(n0,n2)|subtract(n3,#0)|divide(#1,n2)|subtract(n0,#2)|multiply(n2,#3)| | general |
a 250 meter long train crosses a platform in 40 seconds while it crosses a signal pole in 20 seconds . what is the length of the platform ? | "speed = [ 250 / 20 ] m / sec = 25 / 2 m / sec . let the length of the platform be x meters . then , x + 250 / 40 = 25 / 2 2 ( x + 250 ) = 1000 Γ¨ x = 250 m . answer : d" | a ) 500 , b ) 1000 , c ) 350 , d ) 250 , e ) 20 | d | subtract(multiply(divide(250, 20), 40), 250) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
in a 4 person race , medals are awarded to the fastest 3 runners . the first - place runner receives a gold medal , the second - place runner receives a silver medal , and the third - place runner receives a bronze medal . in the event of a tie , the tied runners receive the same color medal . ( for example , if there ... | gold can be awarded in 4 ways , then silver can be awarded in 3 ways and bronze in 2 ways . therefore the total number of ways are : 4 * 3 * 2 = 24 ways of awarding the medals and the same number of ways of forming the circle . this is when there is no tie . and if there is tie , for example all three receive the gold ... | a ) 24 , b ) 52 , c ) 96 , d ) 144 , e ) 648 | c | multiply(factorial(4), 4) | factorial(n0)|multiply(n0,#0) | general |
a goods train runs at the speed of 72 kmph and crosses a 350 m long platform in 26 seconds . what is the length of the goods train ? | explanation : speed = [ 72 x ( 5 / 18 ) ] m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , [ ( x + 350 ) / 26 ] = 20 = > x + 350 = 520 = > x = 170 . answer : d | a ) 230 m , b ) 140 m , c ) 160 m , d ) 170 m , e ) none of these | d | multiply(subtract(26, divide(350, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2) | physics |
the population of a town increased from 1 , 62,500 to 2 , 62,500 in a decade . the average percent increase of population per year is : | "solution increase in 10 year = ( 262500 - 162500 ) = 100000 . increase % = ( 100000 / 162500 x 100 ) % = 61.54 % Γ’ Λ Β΄ required average = ( 61.54 / 10 ) % = 6.154 % answer c" | a ) 4.37 % , b ) 5 % , c ) 6.154 % , d ) 8.75 % , e ) none of these | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
find the probability that a number selected from numbers 1 , 2 , 3 , . . . , 45 is a prime number , when each of the given numbers is equally likely to be selected ? | "let x be the event of selecting a prime number . x = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41,43 } n ( x ) = 14 , n ( s ) = 45 hence , the required probability is 14 / 45 . answer : b" | a ) 10 / 45 , b ) 14 / 45 , c ) 8 / 45 , d ) 3 / 45 , e ) 7 / 45 | b | divide(multiply(3, const_5), 45) | multiply(n2,const_5)|divide(#0,n3)| | probability |
on increasing the price of t . v . sets by 80 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ? | "explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 Γ 100 ) = rs . 10000 new sale value = rs . ( 180 Γ 80 ) = rs . 14400 increase % = 4400 / 10000 Γ 100 = 44 % answer : a" | a ) 44 , b ) 5 , c ) 56 , d ) 2 , e ) 7 | a | subtract(add(const_100, 80), add(divide(multiply(add(const_100, 80), 20), const_100), const_100)) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#2,const_100)|subtract(#0,#3)| | gain |
in a hostel , the number of students decreased by 15 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the prev... | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 15 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.85 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.85 * 1.2 ) = > f ( new ) = 0.9804 f therefore th... | a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 1.96 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 15), add(const_100, 20)))), const_100) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general |
a family pays $ 800 per year for an insurance plan that pays 70 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total . | "upfront payment for insurance plan = 800 $ family needs to pay 30 % of first 1000 $ in expense = 300 $ total amount paid by family when medical expenses are equal to or greater than 1000 $ = 800 + 300 = 1100 $ total amount paid by insurance plan for first 1000 $ = 800 $ total amount paid by family will equal amount pa... | a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,400 , d ) $ 1,800 , e ) $ 2,200 | c | subtract(1,000, 800) | subtract(n2,n0)| | general |
a bag contains 5 red , 6 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | p ( both are red ) , = 5 c 215 c 2 = 5 c 215 c 2 = 10 / 105 = 2 / 21 b | a ) 2 / 15 , b ) 2 / 21 , c ) 5 / 26 , d ) 3 / 29 , e ) 4 / 27 | b | divide(choose(5, 2), choose(add(add(5, 6), 4), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3) | other |
seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 27 percent ryegrass , what percent of the weight of this mixture is x ? | "- - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 27 % 0.4 x + ( m - x ) 0.25 = 0.27 m 0.15 x = 0.02 m x = 0.1333 m x = 13.33 % of m b" | a ) 10 % , b ) 13.33 % , c ) 40 % , d ) 50 % , e ) 66.66 % | b | divide(subtract(27, 25), subtract(divide(40, const_100), divide(25, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n2)|subtract(#0,#1)|divide(#2,#3)| | gain |
a , b , c , d , e , f are the only 6 families in indira nagar . a , b , c , d , e and f has 7 , 8 , 10 , 13 , 6 , and 10 member in their families respectively . if 1 member from all the 6 families left their respective families to accommodate themselves in the hostel of iim lucknow , then the average number of member n... | answer required average = ( ( 7 - 1 ) + ( 8 - 1 ) + ( 10 - 1 ) + ( 13 - 1 ) + ( 6 - 1 ) + ( 10 - 1 ) ) / 6 = ( 7 + 8 + 10 + 13 + 6 + 10 ) / 6 - ( 6 x 1 ) / 6 = 9 - 1 = 8 correct option : a | a ) 8 , b ) 9 , c ) 10 , d ) 13 , e ) 15 | a | subtract(10, const_2) | subtract(n3,const_2) | general |
a rectangular grass field is 75 m * 40 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 ? | area = ( l + b + 2 d ) 2 d = ( 75 + 40 + 2.5 * 2 ) 2 * 2.5 = > 600 600 * 2 = rs . 1200 answer : c | ['a ) s . 1350', 'b ) s . 1327', 'c ) s . 1200', 'd ) s . 1397', 'e ) s . 1927'] | c | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(40, multiply(2.5, const_2))), rectangle_area(75, 40)), 2) | multiply(n2,const_2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5) | geometry |
if - 2 x + 3 y = 8 and 4 x - 3 y = 20 , what is the sum of x and y ? | "given - 2 x + 3 y = 8 - - - eq 1 4 x - 3 y = 20 - - eq 2 sum both eqns we get 2 x = 28 = > x = 14 sub 2 x in eq 1 = > - 28 + 3 y = 8 . = > y = 12 now x + y = 12 + 14 = 26 option d is correct answer ." | a ) 24 , b ) 28 , c ) 18 , d ) 26 , e ) 32 | d | multiply(divide(add(20, multiply(divide(add(8, 20), 3), 3)), 2), divide(add(8, 20), 3)) | add(n2,n5)|divide(#0,n4)|multiply(n4,#1)|add(n5,#2)|divide(#3,n0)|multiply(#4,#1)| | general |
in a river flowing at 2 km / hr , a boat travels 72 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 72 km total time taken = time taken downstream + time taken upstream = ( 72 / 8 ) + ( 72 / 4 ) = 27 hr . answer : option d | a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 27 hours , e ) none of these | d | add(divide(72, subtract(6, 2)), divide(72, add(6, 2))) | add(n0,n2)|subtract(n2,n0)|divide(n1,#1)|divide(n1,#0)|add(#2,#3) | general |
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 . 3.50 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 β 6 = 666 * 3.5 = 2331 answer : e" | a ) s . 1014 , b ) s . 1140 , c ) s . 999 , d ) s . 1085 , e ) s . 2331 | e | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 3.50), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics |
if a - b = 5 and a 2 + b 2 = 29 , find the value of ab . | 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 29 - 25 = 4 ab = 2 . answer : e | a ) 6 , b ) 8 , c ) 10 , d ) 11 , e ) 2 | e | divide(subtract(29, power(5, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1) | general |
how many miles is 320 km ? | "200 miles 320 / 1.6 = 200 answer : c" | a ) 150 miles , b ) 175 miles , c ) 200 miles , d ) 225 miles , e ) 250 miles | c | divide(volume_cube(const_10), volume_cube(const_2.0)) | volume_cube(const_10)|volume_cube(const_2.0)|divide(#0,#1)| | physics |
60 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 420 1 child ' s 1 day work = 1 / 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 420 + 10 / 140 ) = 1 / 12 5 women and 10 children will complete the work in 12 days . answer : e" | a ) 8 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 12 days | e | inverse(add(divide(5, multiply(60, 7)), divide(60, multiply(60, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics |
how many seconds will it take for a car that is traveling at a constant rate of 14 miles per hour to travel a distance of 96 yards ? ( 1 mile = 1,160 yards ) | "speed = 14 miles / hr = 6.84 yard / s distance = 96 yards time = distance / speed = 96 / 6.84 = 14 sec ans - d" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 17 | d | divide(96, multiply(divide(14, const_3600), multiply(subtract(add(14, 14), const_4), const_10))) | add(n0,n0)|divide(n0,const_3600)|subtract(#0,const_4)|multiply(#2,const_10)|multiply(#1,#3)|divide(n1,#4)| | physics |
in how many years rs 200 will produce the same interest at 10 % as rs . 400 produce in 5 years at 12 % | "explanation : clue : firstly we need to calculate the si with prinical 400 , time 5 years and rate 12 % , it will be rs . 240 then we can get the time as time = ( 100 * 240 ) / ( 200 * 10 ) = 12 option c" | a ) 16 , b ) 13 , c ) 12 , d ) 14 , e ) 15 | c | divide(multiply(divide(multiply(400, 12), const_100), 5), multiply(divide(10, const_100), 200)) | divide(n1,const_100)|multiply(n2,n4)|divide(#1,const_100)|multiply(n0,#0)|multiply(n3,#2)|divide(#4,#3)| | gain |
a watch was sold at a loss of 10 % . if it was sold for rs . 196 more , there would have been a gain of 4 % . what is the cost price ? | "90 % 104 % - - - - - - - - 14 % - - - - 196 100 % - - - - ? = > rs : 1400 answer : e" | a ) s : 1000 , b ) s : 1067 , c ) s : 1278 , d ) s : 1028 , e ) s : 1400 | e | divide(multiply(196, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 37 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "37 % is 12 % - points above 25 % and 3 % - points below 40 % . thus the ratio of mixture y to mixture x is 1 : 4 . the percent of mixture x is 4 / 5 = 80 % . the answer is e ." | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 80 % | e | divide(subtract(37, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | add(n2,const_1)|divide(n0,const_100)|divide(#0,const_100)|subtract(n4,#0)|subtract(#1,#2)|divide(#3,#4)| | gain |
how many cubes of 10 cm edge can be put in a cubical box of 1 m edge . | "explanation : number of cubes = 100 β 100 β 100 / 10 β 10 β 10 = 1000 note : 1 m = 100 cm option b" | a ) 10000 cubes , b ) 1000 cubes , c ) 100 cubes , d ) 50 cubes , e ) none of these | b | divide(volume_cube(1), volume_cube(divide(10, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | physics |
the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 35 kms . what was the total distance travelled in 12 hours ? | "explanation : 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 , d = 2 . required distance = 12 / 2 [ 2 Γ 35 + { 12 - 1 ) Γ 2 ] = 6 ( 70 + 23 ) = 552 kms . answer : c" | a ) 456 kms , b ) 482 kms , c ) 552 kms , d ) 556 kms , e ) none of these | c | multiply(add(multiply(2, 35), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics |
a and b together can do a piece of work in 10 days . b alone can finish it in 20 days . in how many days can a alone finish the work ? | time taken by a to finish the work = xy / ( y - x ) = 10 x 20 / ( 20 - 10 ) = 200 / 10 = 20 days answer : a | a ) 20 , b ) 10 , c ) 15 , d ) 5 , e ) 7 | a | inverse(subtract(inverse(10), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2) | physics |
the ratio of pens to pencils is 5 to 6 . there are 7 more pencils than pens . how many pencils are there ? | "let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 7 x = 7 the number of pencils is 42 . the answer is b ." | a ) 36 , b ) 42 , c ) 48 , d ) 55 , e ) 64 | b | add(multiply(7, 5), 7) | multiply(n0,n2)|add(n2,#0)| | other |
the average score of a cricketer for 10 matches is 38.9 runs . if the average for the first 6 matches is 42 , then find the average for the last 4 matches . | solution required average = ( 38.9 x 10 ) - ( 42 x 6 ) / 4 = 137 / 4 = 34.25 answer c | a ) 33.5 , b ) 33.5 , c ) 34.25 , d ) 35 , e ) none of these | c | divide(subtract(multiply(10, 38.9), multiply(6, 42)), 4) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general |
the area of a square is equal to seven times the area of a rectangle of dimensions 25 cm * 343 cm . what is the perimeter of the square ? | "area of the square = s * s = 7 ( 343 * 25 ) = > s = 5 * 7 * 7 = 245 cm perimeter of the square = 4 * 245 = 980 cm . answer : c" | a ) 810 cm , b ) 800 cm , c ) 980 cm , d ) 718 cm , e ) 860 cm | c | multiply(sqrt(multiply(rectangle_area(25, 343), divide(343, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
the distance between a & b is 600 km . a person is traveling from a to b at 70 km / hr started at 10 am and another person is traveling from b to a at 80 km / hr and started at same time . then at what time they meet together . | let x hours be they will meet together distance covered by 1 st person + distance covered by 2 nd person = 600 km 70 x + 80 x = 600 x = 4 hr they will meet = 10 am + 4 hr = 2 pm answer is a | a ) 2 pm , b ) 4 pm , c ) 1 pm , d ) 3 pm , e ) 5 pm | a | divide(add(add(70, 80), 10), 80) | add(n1,n3)|add(n2,#0)|divide(#1,n3) | physics |
a train running at 1 / 4 of its own speed reached a place in 12 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 4 * 12 = 3 hrs time saved = 12 - 3 = 9 hrs answer : a" | a ) 9 hrs , b ) 10 hrs , c ) 12 hrs , d ) 15 hrs , e ) 6 hrs | a | multiply(divide(1, 4), 12) | divide(n0,n1)|multiply(n2,#0)| | physics |
when a merchant imported a certain item , he paid a 9 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 67.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.09 x = 67.5 x = $ 750 the total value was $ 750 + $ 1,000 = $ 1,750 . the answer is c ." | a ) $ 1,150 , b ) $ 1,450 , c ) $ 1,750 , d ) $ 2,150 , e ) $ 2,450 | c | floor(divide(add(divide(67.50, divide(9, const_100)), 1,000), 1,000)) | divide(n0,const_100)|divide(n2,#0)|add(#1,n1)|divide(#2,n1)|floor(#3)| | general |
the average weight of a group of boys is 30 kg . after a boy of weight 37 kg joins the group , the average weight of the group goes up by 1.00001 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 37 kg joins the group , total weight of boys = 30 x + 37 so 30 x + 37 = 31 ( x + 1 ) = > x = 6 . answer : c" | a ) a ) 4 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 1 | c | add(subtract(37, add(30, 1.00001)), 1.00001) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
of 55 players on a cricket team , 37 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 ? | "total = 55 thrower = 37 rest = 55 - 37 = 18 left handed = 18 / 3 = 6 right handed = 12 if all thrower are right handed then total right handed is 37 + 12 = 49 so a . 49 is the right answer" | a ) 49 , b ) 55 , c ) 59 , d ) 71 , e ) 92 | a | add(37, multiply(subtract(const_1, divide(const_1, const_3)), subtract(55, 37))) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general |
a person want to give his money of $ 900 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ? | "b ' s share = 900 * 3 / 9 = $ 300 answer is b" | a ) $ 400 , b ) $ 300 , c ) $ 500 , d ) $ 180 , e ) $ 200 | b | multiply(divide(3, add(add(2, 3), 4)), 900) | add(n1,n2)|add(n4,#0)|divide(n1,#1)|multiply(n0,#2)| | other |
a can do a piece of work in 21 days and b in 28 days . together they started the work and b left after 4 days . in how many days can a alone do the remaining work ? | "let a worked for x days . x / 21 + 4 / 28 = 1 = > x / 21 = 6 / 7 = > x = 18 a worked for 18 days . so , a can complete the remaining work in 18 - 4 = 14 days . answer : d" | a ) 22 , b ) 88 , c ) 27 , d ) 14 , e ) 99 | d | add(divide(subtract(const_1, add(multiply(4, divide(const_1, 21)), multiply(4, divide(const_1, 28)))), divide(const_1, 28)), 4) | divide(const_1,n0)|divide(const_1,n1)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)|add(n2,#6)| | physics |
if a and b are the two values of t that satisfy the equation t ^ 2 Γ’ β¬ β 12 t + 20 = 0 , with a > b , what is the value of a Γ’ β¬ β b ? | factor the left side of the equation : t ^ 2 Γ’ β¬ β 12 t + 20 = 0 ( t Γ’ β¬ β 2 ) ( t Γ’ β¬ β 10 ) = 0 t = 2 , t = 10 thus , a = 10 and b = 2 . so a Γ’ β¬ β b = 10 Γ’ β¬ β 2 = 8 . the answer is d . | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | subtract(divide(20, 2), 2) | divide(n2,n0)|subtract(#0,n0) | general |
the average ( arithmetic mean ) of 4 different integers is 74 . if the largest integer is 90 , what is the least possible value of the smallest integer ? | "total of integers = 74 * 4 = 296 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 296 - ( 90 + 89 + 8... | a ) 1 , b ) 19 , c ) 29 , d ) 30 , e ) 33 | c | subtract(subtract(subtract(multiply(74, 4), 90), subtract(90, const_1)), subtract(90, const_2)) | multiply(n0,n1)|subtract(n2,const_1)|subtract(n2,const_2)|subtract(#0,n2)|subtract(#3,#1)|subtract(#4,#2)| | general |
how many numbers from 2 to 11 are exactly divisible by 2 ? | "2 / 2 = 1 and 11 / 2 = 5 5 - 1 = 4 4 + 1 = 5 numbers . answer : c" | a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 7 , e ) e ) 8 | c | add(divide(subtract(multiply(floor(divide(11, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
if w is the set of all the integers between 35 and 64 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 63 , the smallest is 36 ) : 63 - 36 = 27 step 2 . divide by 3 : 27 / 3 = 9 step 3 . add 1 : 9 + 1 = 10 . so there are 10 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . ... | a ) 18 , b ) 19 , c ) 61 , d ) 31 , e ) 11 | b | subtract(add(floor(divide(subtract(64, 35), 3)), divide(subtract(64, 35), 2)), floor(divide(subtract(64, 35), multiply(2, 3)))) | 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)| | other |
4 milkmen rented a pasture . a grazed 24 cows for 3 months ; b 10 for 5 months ; c 35 cows for 4 months and d 21 cows for 3 months . if a ' s share of rent is rs . 1440 , find the total rent of the field . | ratio of shares of a , b , c , d = ( 24 x 3 ) : ( 10 x 5 ) : ( 35 x 4 ) : ( 21 x 3 ) = 72 : 50 : 140 : 63 . let total rent be rs . x . then , a β s share = rs . ( 72 x ) / 325 ( 72 x ) / 325 = 1440 = x = ( 1440 x 325 ) / 72 = 6500 hence , total rent of the field is rs . 6500 . answer is a . | a ) 6500 , b ) 3550 , c ) 4250 , d ) 2250 , e ) 3250 | a | multiply(divide(add(add(add(multiply(24, 3), multiply(10, 5)), multiply(35, 4)), multiply(21, 3)), multiply(24, 3)), 1440) | multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n5)|multiply(n2,n7)|add(#0,#1)|add(#4,#2)|add(#5,#3)|divide(#6,#0)|multiply(n9,#7) | general |
how many multiples of 4 are there between 12 and 108 , inclusive ? | "the multiples of 4 are from 4 * 3 up to 4 * 27 . 27 - 3 + 1 = 25 . the answer is e ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | e | add(divide(subtract(108, 12), 4), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
123 : 36 : : 221 : ? | "( 1 + 2 + 3 ) ^ 2 = 36 ( 2 + 2 + 1 ) ^ 2 = 25 answer : d" | a ) 52 , b ) 69 , c ) 72 , d ) 25 , e ) 32 | d | multiply(221, divide(123, 36)) | divide(n0,n1)|multiply(n2,#0)| | general |
an inspector rejects 0.02 % of the meters as defective . how many will he examine to reject 2 ? | "let the number of meters to be examined be x then , 0.02 % of x = 2 ( 2 / 100 ) * ( ( 1 / 100 ) * x = 2 x = 10000 answer is b" | a ) a ) 1500 , b ) b ) 10000 , c ) c ) 2500 , d ) d ) 3000 , e ) e ) 3100 | b | divide(multiply(2, const_100), 0.02) | multiply(n1,const_100)|divide(#0,n0)| | gain |
if a - b = 10 and a ^ 2 + b ^ 2 = 150 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 150 - 100 = 50 = > ab = 25 answer : c" | a ) 10 , b ) 12 , c ) 25 , d ) 18 , e ) 19 | c | multiply(multiply(add(10, divide(subtract(sqrt(150), 10), 2)), divide(subtract(sqrt(150), 10), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general |
eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 900 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy :... | "distance traveled by eddy = 900 km time taken by eddy = 3 hours average speed of eddy = 900 / 3 = 300 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 300 / 75 = 4 / 1 answer d" | a ) 8 / 3 , b ) 3 / 8 , c ) 8 / 5 , d ) 4 / 1 , e ) 5 / 3 | d | divide(divide(900, 3), divide(300, 4)) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)| | physics |
if 40 honey bees make 40 grams of honey in 40 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 40 : : 40 : x honey 40 : 1 = > 1 x 40 x x = 40 x 1 x 40 = > x = 40 . answer : c" | a ) 1 , b ) 3.5 , c ) 40 , d ) 49 , e ) 30 | c | divide(1, divide(1, 40)) | divide(n3,n0)|divide(n3,#0)| | physics |
a dog takes 5 leaps for every 8 leaps of a hare . if one leap of the dog is equal to 5 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 5 ) leaps of hare : 8 leaps of hare = 25 : 8 answer : b" | a ) 25 : 10 , b ) 25 : 8 , c ) 25 : 9 , d ) 8 : 25 , e ) 24 : 8 | b | divide(multiply(5, 5), 8) | multiply(n0,n2)|divide(#0,n1)| | other |
of the families in city x in 2001 , 25 percent owned a personal computer . the number of families in city x owning a computer in 2008 was 15 percent greater than it was in 2001 , and the total number of families in city x was 6 percent greater in 2008 than it was in 2001 . what percent of the families in city x owned a... | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 25 number of families owning computer in 1998 = 25 * 115 / 100 = 28.75 number of families in 1998 = 106 the percentage = 28.75 / 106 * 100 = 27.12 % . answer : a" | a ) 27.12 % , b ) 23.12 % , c ) 25.55 % , d ) 31.22 % , e ) 56.22 % | a | multiply(const_100, divide(divide(multiply(add(15, const_100), 25), const_100), add(const_100, 6))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general |
the mass of 1 cubic meter of a substance is 100 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 ) | "100 kg - 1 cubic meter ; 100,000 g - 1 cubic meter ; 100,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 100,000 = 10 / 1 = 10 cubic centimeters . answer : c ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(multiply(1,000, 1,000), multiply(100, 1,000)) | multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)| | geometry |
two pipes a and b can fill a tank in 12 and 24 minutes respectively . if both the pipes are used together , how long will it take to fill the tank ? | "explanation : tank can be filled by pipe a in 12 minutes and pipe b in 24 minutes . = > part filled by a in 1 minute = 1 / 12 = > part filled by b in 1 minute = 1 / 24 = > part filled by both in 1 minute = 1 / 12 + 1 / 24 = 3 / 24 = 1 / 8 time taken to fill the entire tank = 8 minutes answer b" | a ) 9 min , b ) 8 min , c ) 4 min , d ) 6 min , e ) none of these | b | add(multiply(12, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 24)), const_2)), 12), const_60)) | divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)| | physics |
what number is obtained by adding the units digits of 734 ^ 99 and 347 ^ 82 ? | "the units digit of 734 ^ 99 is 4 because 4 raised to the power of an odd integer ends in 4 . the units digit of 347 ^ 82 is 9 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 82 is in the form 4 n + 2 , the units digit is 9 . then 4 + 9 = 13 . the answer is d ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | subtract(subtract(99, 82), divide(99, add(const_1, const_10))) | add(const_1,const_10)|subtract(n1,n3)|divide(n1,#0)|subtract(#1,#2)| | general |
the ratio between the number of sheep and the number of horses at the stewar farm is 3 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | et no of sheep and horses are 3 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 3 * 8 ) = 24 answer : a | a ) 24 , b ) 28 , c ) 32 , d ) 56 , e ) 58 | a | multiply(divide(divide(12880, 230), 7), 3) | divide(n3,n2)|divide(#0,n1)|multiply(n0,#1) | other |
a pump can fill a tank with a water in 2 hours . because of a leak , it took 2 and 4 / 5 hours to fill the tank . the leak can drain all the water of the full tank in how many hours ? | the rate of the pump + leak = 5 / 14 1 / 2 - leak ' s rate = 5 / 14 leak ' s rate = 1 / 2 - 5 / 14 = 1 / 7 the leak will empty the tank in 7 hours . the answer is d . | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | divide(const_1, subtract(divide(const_1, 2), divide(const_1, add(2, divide(4, 5))))) | divide(const_1,n0)|divide(n2,n3)|add(n0,#1)|divide(const_1,#2)|subtract(#0,#3)|divide(const_1,#4) | physics |
the sum of number of boys and girls in a school is 100 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 100 = 100 x + x / 100 * 100 = 100 2 * x = 100 x = 50 answer is a" | a ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 70 | a | divide(100, add(divide(100, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
the food in a camp lasts for 20 men for 50 days . if 20 more men join , how many days will the food last ? | one man can consume the same food in 20 * 50 = 1000 days . 20 more men join , the total number of men = 40 the number of days the food will last = 1000 / 40 = 25 days . answer : b | a ) 22 days , b ) 25 days , c ) 28 days , d ) 16 days , e ) 27 days | b | divide(multiply(20, 50), add(20, 20)) | add(n0,n0)|multiply(n0,n1)|divide(#1,#0) | physics |
three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 264 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 264 x = 22 hcf of 2 x , 3 x and 4 x = x = 22 the answer is b ." | a ) 18 , b ) 22 , c ) 24 , d ) 28 , e ) 32 | b | multiply(2, 4) | multiply(n0,n2)| | other |
in a class of students , 1 / 2 of the number of girls is equal to 1 / 3 of the total number of students . what is the ratio of boys to girls in the class ? | "( 1 / 2 ) g = ( 1 / 3 ) ( b + g ) 3 g = 2 b + 2 g g = 2 b b / g = 1 / 2 . the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 3 / 4 | a | divide(subtract(divide(1, 2), divide(1, 3)), divide(1, 3)) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#1)| | general |
two circles are there one with arc area 60 digree equal to 45 digree of another circle find the area of circle 1 : circle 2 ratio | area for first circle = ( 3.14 * r ^ 2 ) / 6 area of second circke = ( 3.14 * r ^ 2 ) / 8 therefore , ratio = 3 : 4 answer : a | ['a ) 3 : 4', 'b ) 1 : 4', 'c ) 1 : 2', 'd ) 2 : 3', 'e ) 1 : 3'] | a | divide(multiply(multiply(divide(45, const_360), 2), const_pi), multiply(multiply(divide(60, const_360), const_2), const_pi)) | divide(n1,const_360)|divide(n0,const_360)|multiply(n3,#0)|multiply(#1,const_2)|multiply(#2,const_pi)|multiply(#3,const_pi)|divide(#4,#5) | geometry |
in a recent head - to - head run - off election , 12,000 absentee ballets were cast . 1 / 6 of the absentee ballets were thrown out and 3 / 5 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | 5 / 6 * 2 / 5 ( total absentee votes ) = 1 / 3 ( total votes ) = 1 / 3 * 12000 = 4000 answer is c | a ) 2,000 , b ) 3,000 , c ) 4,000 , d ) 8,000 , e ) 9,000 | c | multiply(divide(divide(subtract(subtract(multiply(multiply(const_3, const_4), const_1000), multiply(multiply(multiply(const_3, const_4), const_1000), divide(1, 6))), multiply(subtract(multiply(multiply(const_3, const_4), const_1000), multiply(multiply(multiply(const_3, const_4), const_1000), divide(1, 6))), divide(3, 5... | divide(n1,n2)|divide(n3,n4)|multiply(const_3,const_4)|multiply(#2,const_1000)|multiply(#0,#3)|subtract(#3,#4)|multiply(#1,#5)|subtract(#5,#6)|divide(#7,const_1000)|divide(#8,const_4)|multiply(#9,const_3) | general |
a train 300 m long can cross an electric pole in 30 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 300 / 30 s = 10 m / sec speed = 10 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 36 kmph answer : d" | a ) 76 kmph , b ) 54 kmph , c ) 72 kmph , d ) 36 kmph , e ) 91 kmph | d | divide(divide(300, const_1000), divide(30, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
at what rate percent of simple interest will a sum of money double itself in 66 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 66 ) = 100 / 66 = 1.51 % answer : c" | a ) 6.54 % , b ) 4.54 % , c ) 1.51 % , d ) 1.94 % , e ) 5.54 % | c | divide(const_100, 66) | divide(const_100,n0)| | gain |
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