Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12600 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12600 = 3780 answer : e" | a ) 3630 , b ) 3637 , c ) 3630 , d ) 3631 , e ) 3780 | e | multiply(divide(6300, add(add(6300, 4200), 10500)), 12600) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain | E |
p is 3 times as old as q . in 11 years , p will be twice as old as q . how old was p 3 years ago ? | p = 3 q so q = p / 3 p + 11 = 2 ( q + 11 ) = 2 ( p / 3 + 11 ) p / 3 = 11 p = 33 three years ago , p was 30 . the answer is d . | a ) 21 , b ) 24 , c ) 27 , d ) 30 , e ) 33 | d | subtract(multiply(3, 11), 3) | multiply(n0,n1)|subtract(#0,n0) | general | D |
matt gets a $ 1,300 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 400 , how many sales has matt made ? | "let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1300 new average = ( xy + 1300 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1300 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1300 ) = ( xy + x + 150 y + 150 ) i . e . ( 1150 ) = ( x + 150 y ) new commission = 400 = 150 + x i . e . x = 250 i . e . y = 6 new sales = y + 1 = 7 answer : option d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(const_1000, const_100), subtract(400, subtract(400, 150))) | subtract(const_1000,const_100)|subtract(n2,n1)|subtract(n2,#1)|divide(#0,#2)| | general | D |
a grocer has a sale of rs . 800 , rs . 900 , rs . 1000 , rs . 700 and rs . 800 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 850 ? | "total sale for 5 months = rs . ( 800 + 900 + 1000 + 700 + 800 ) = rs . 4200 required sale = rs . [ ( 850 x 6 ) - 4200 ] = rs . ( 5100 - 4200 ) = rs . 900 . option d" | a ) s . 440 , b ) s . 850 , c ) s . 450 , d ) s . 900 , e ) s . 950 | d | subtract(multiply(add(5, const_1), 850), add(add(add(add(800, 900), 1000), 700), 800)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | D |
what is remainder of the division ( 1425 * 1427 * 1429 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : e" | a ) 5 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | e | subtract(multiply(multiply(1425, 1427), 1429), subtract(multiply(multiply(1425, 1427), 1429), const_3)) | multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)| | general | E |
for all numbers a and b , the operationis defined by ab = ( a + 2 ) ( b β 3 ) . if 3 x = β 25 , then x = | ( 3 + 2 ) ( x - 3 ) = - 25 . . x - 3 = - 5 . . x = - 2 b | a ) β 15 , b ) β 2 , c ) - 3 , d ) 6 , e ) 15 | b | subtract(3, divide(25, add(2, 3))) | add(n0,n1)|divide(n3,#0)|subtract(n1,#1) | general | B |
find the average of all the numbers between 4 and 32 which are divisible by 5 . | "solution average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 ) = 105 / 6 = 17.5 . answer a" | a ) 17.5 , b ) 20 , c ) 24 , d ) 30 , e ) 32 | a | divide(add(add(4, const_4), subtract(32, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general | A |
the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 7967 for a computer table , then what was the cost price of the computer table ? | "explanation : cp = sp * ( 100 / ( 100 + profit % ) ) = 7967 ( 100 / 124 ) = rs . 6425 . answer : b" | a ) 7297 , b ) 6425 , c ) 2871 , d ) 6725 , e ) 2981 | b | divide(7967, add(const_1, divide(24, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain | B |
perimeter of the back wheel = 9 feet , front wheel = 7 feet on a certain distance , the front wheel gets 10 revolutions more than the back wheel . what is the distance ? | for 1 revolution : front wheel goes 7 ft . and back wheel goes 9 feet . let distance is x feet . x / 7 = ( x / 9 ) + 10 or , 2 x / 63 = 10 or , x = ( 10 * 63 ) / 2 therefore , x = 315 answer : a | a ) 315 , b ) 325 , c ) 335 , d ) 345 , e ) 305 | a | multiply(9, divide(multiply(10, 7), subtract(9, 7))) | multiply(n1,n2)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2) | physics | A |
a baseball card decreased in value 60 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 60 / 100 ) * 100 = 40 after second year , value of baseball card = ( 1 - 10 / 100 ) * 40 = 36 total percent decrease of the card ' s value over the two years = ( 100 - 36 ) / 100 * 100 % = 64 % answer e | a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 64 % | e | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(60, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5) | gain | E |
how many different ways can 2 students be seated in a row of 6 desks , so that there is always at least one empty desk between the students ? | "10 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 10 x 2 = 20 d" | a ) 2 , b ) 3 , c ) 4 , d ) 20 , e ) 12 | d | permutation(subtract(6, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0)| | probability | D |
a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 36 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 36 pints will require 9 / 17 * 36 = 19 pints . . c" | a ) 9 , b ) 16 , c ) 19 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(36, divide(multiply(3, 2), multiply(4, 3))) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,#1)|multiply(n4,#2)| | general | C |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 120 games will be played during the tournament . find the number of participants . | "let p be the number of participants . pc 2 = 120 ( p ) ( p - 1 ) = 240 = 16 * 15 p = 16 the answer is b ." | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | b | divide(add(sqrt(add(multiply(multiply(120, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general | B |
the least number , which when divided by 12 , 15 , 20 and 63 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 12 , 15 , 20 , 63 ) + 8 = 1260 + 8 = 1268 . answer : d" | a ) 448 , b ) 488 , c ) 542 , d ) 1268 , e ) 560 | d | multiply(63, const_10) | multiply(n3,const_10)| | general | D |
the number 0.8 is how much greater than 1 / 8 ? | let x be the difference then . 8 - 1 / 8 = x 8 / 10 - 1 / 8 = x x = 27 / 40 ans b | a ) Β½ , b ) 27 / 40 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | b | subtract(0.8, divide(1, 8)) | divide(n1,n2)|subtract(n0,#0) | general | B |
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 | B |
24 oz of juice p and 25 oz of juice v are mixed to make smothies m and y . the ratio of p to v in smothie m is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie m ? | the ratio of p to v in smothie m is 4 is to 1 and that in y is 1 is to 5 . p 1 + p 2 = 24 v 1 + v 2 = 25 p 1 = 4 v 1 p 2 = v 2 / 5 4 v 1 + v 2 / 5 = 24 v 1 + v 2 = 25 4 v 2 - v 2 / 5 = 76 19 v 2 / 5 = 76 = > v 2 = 20 = > v 1 = 5 = > p 1 = 20 answer - d | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | multiply(divide(subtract(multiply(24, 5), 25), subtract(multiply(4, 5), 1)), 4) | multiply(n0,n5)|multiply(n2,n5)|subtract(#0,n1)|subtract(#1,n3)|divide(#2,#3)|multiply(n2,#4) | other | D |
a person want to give his money of $ 5400 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ? | "b ' s share = 5400 * 3 / 9 = $ 1800 answer is b" | a ) $ 2000 , b ) $ 1800 , c ) $ 2500 , d ) $ 1500 , e ) $ 1600 | b | multiply(divide(3, add(add(2, 3), 4)), 5400) | add(n1,n2)|add(n4,#0)|divide(n1,#1)|multiply(n0,#2)| | other | B |
a reduction of 35 % in the price of apples would enable a man to obtain 42 more for rs . 20 , what is reduced price per dozen ? | "a 20 * ( 35 / 100 ) = 6 - - - 42 ? - - - 12 = > rs . 1.7" | a ) 1.7 , b ) 3.5 , c ) 6.3 , d ) 7.1 , e ) 10.4 | a | multiply(divide(divide(multiply(35, 20), const_100), 42), const_12) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|multiply(#2,const_12)| | gain | A |
how many digits are required to number a book containing 260 pages ? | "9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 260 - ( 90 + 9 ) = 161 pages will require 161 * 3 = 483 digits . the total number of digits is 9 + 180 + 483 = 672 . the answer is b ." | a ) 784 , b ) 672 , c ) 492 , d ) 372 , e ) 300 | b | add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(260, const_100), const_1), const_3)) | subtract(const_10,const_1)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#4,#0)|add(#6,#5)| | general | B |
what is the greatest of 3 consecutive integers whose sum is 21 ? | "the sum of three consecutive integers can be written as n + ( n + 1 ) + ( n + 2 ) = 3 n + 3 if the sum is 24 , we need to solve the equation 3 n + 3 = 21 ; = > 3 n = 18 ; = > n = 6 the greatest of the three numbers is therefore 6 + 2 = 8 answer : c" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | c | add(divide(subtract(21, 3), 3), const_2) | subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)| | physics | C |
the average of 4 consecutive odd numbers is 24 . find the largest number | explanation : let the numbers are x , x + 2 , x + 4 , x + 6 , then = > x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) / 4 = 24 = > 4 x + 12 ) / 4 = 24 = > x + 3 = 24 = > x = 21 so largest number is 21 + 6 = 27 option b | a ) 25 , b ) 27 , c ) 29 , d ) 31 , e ) 32 | b | add(divide(multiply(4, 24), 4), const_3) | multiply(n0,n1)|divide(#0,n0)|add(#1,const_3) | general | B |
kim bought a total of $ 2.65 worth of postage stamps in 4 denominations . if she bought an equal number of 2 - cent and 20 - cent stamps and twice as many 10 - cent stamps as 5 - cent stamps , what is the least number of 1 - cent stamps she could have bought ? | let : # of 2 and 20 cents stamps = n # of 10 cent stamps = 2 n # of 1 cent stamps = m therefore : 2 n + 20 n + 10 ( 2 n ) + m = 265 cents 42 n + m = 265 forleastnumber of 1 cents stamps , maximize 50 n . hence , 42 * 5 + m = 265 m = 55 i . e . 55 stamps 1 cents each answer : c | a ) 5 , b ) 10 , c ) 55 , d ) 80 , e ) 95 | c | add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(reminder(multiply(2.65, const_100), add(20, 2)), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1) | add(n2,n3)|multiply(n0,const_100)|reminder(#1,#0)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,const_1)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|add(#13,const_1)|add(#14,const_1)|add(#15,const_1)|add(#16,const_1)|add(#17,const_1)|add(#18,const_1)|add(#19,const_1)|add(#20,const_1)|add(#21,const_1)|add(#22,const_1)|add(#23,const_1)|add(#24,const_1)|add(#25,const_1)|add(#26,const_1)|add(#27,const_1)|add(#28,const_1)|add(#29,const_1)|add(#30,const_1)|add(#31,const_1)|add(#32,const_1)|add(#33,const_1)|add(#34,const_1)|add(#35,const_1)|add(#36,const_1)|add(#37,const_1)|add(#38,const_1)|add(#39,const_1)|add(#40,const_1)|add(#41,const_1)|add(#42,const_1)|add(#43,const_1)|add(#44,const_1)|add(#45,const_1)|add(#46,const_1)|add(#47,const_1)|add(#48,const_1)|add(#49,const_1)|add(#50,const_1)|add(#51,const_1)|add(#52,const_1)|add(#53,const_1) | general | C |
in what time will a railway train 56 m long moving at the rate of 39 kmph pass a telegraph post on its way ? | "t = 56 / 39 * 18 / 5 = 5 sec answer : c" | a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec | c | divide(56, multiply(39, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
in the xy - plane the point ( - 2 - 3 ) is the centre of a circle , the point ( - 2 , 2 ) lies inside the circle and the point ( 5 , - 3 ) lies outside the circle . if the radius r of the circle r is an integer then r = | "can be solved without much calculations . you are given that ( - 2 , - 3 ) is the center of the circle . point ( 5 , - 3 ) lies inside the circle - - - > the radius is lesser than distance of ( - 2 , - 3 ) from ( 5 , - 3 ) - - - > lesser than 7 units but the radius will also be greater than the distance of ( - 2 , - 3 ) from ( - 2,2 ) - - - - > greater than 5 units . thus the radius is > 5 but < 7 and as it is an integer , the only possible value of radius = 6 units . a is the correct answer ." | a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | divide(add(subtract(5, negate(2)), 5), 2) | negate(n0)|subtract(n4,#0)|add(n4,#1)|divide(#2,n0)| | general | A |
find the area of a parallelogram with base 14 cm and height 24 cm . | "area of a parallelogram = base * height = 14 * 24 = 336 cm 2 answer : option a" | a ) 336 , b ) 384 , c ) 345 , d ) 244 , e ) 242 | a | multiply(14, 24) | multiply(n0,n1)| | geometry | A |
of the votes cast on a certain proposal , 68 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 68 0.2 x = 68 x = 340 the answer is b ." | a ) 300 , b ) 340 , c ) 380 , d ) 420 , e ) 460 | b | divide(68, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n0,#2)| | gain | B |
1 , 1 , 2 , 1 , 2 , 3 , 1 , 2 , 3 , 4 , 1 , 2 , __ ? | "1,1 , 2,1 , 2,3 , 1,2 , 3,4 , 1,2 , __ ? looking at the series 1 12 123 1234 12 ( 3 ) 45 so the answer is 3 . answer : c" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(const_3.0, 1) | multiply(n5,n0)| | general | C |
if a * b = 2 a β 3 b + ab , then 3 * 5 + 5 * 3 is equal to : | "solution 3 Γ 5 + 5 Γ 3 = ( 2 Γ 3 - 3 Γ 5 + 3 Γ 5 ) + 2 ( 2 Γ 5 - 3 Γ 3 + 5 Γ 3 ) = ( 6 + 10 - 9 + 15 ) = 22 . answer a" | a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 29 | a | multiply(3, 5) | multiply(n2,n3)| | general | A |
in a zoo there are 30 penguins , 22 zebras , and 8 tigers with some zookeepers . if the total number of heads are 132 less than the number of feet , the number of zookeepers is | explanation : let number of zookeepers be x . then , total number of feet = 2 x 30 + 4 x 22 + 4 x 8 + 2 x = 2 x + 180 . total number of heads = 30 + 22 + 8 + x = 60 + x . therefore ( 2 x + 180 ) - 132 = ( 60 + x ) + or x = 12 . answer : a | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 22 | a | subtract(multiply(30, const_2), subtract(add(add(multiply(30, const_2), multiply(22, const_4)), multiply(8, const_4)), 132)) | multiply(n0,const_2)|multiply(n1,const_4)|multiply(n2,const_4)|add(#0,#1)|add(#3,#2)|subtract(#4,n3)|subtract(#0,#5) | general | A |
a cube of edge 17 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ? | "increase in volume = volume of the cube = 17 * 17 * 17 cm ^ 3 rise in water level = volume / area = 17 * 17 * 17 / 20 * 15 = 16.37 cm answer is d" | a ) 6 cm , b ) 8.25 cm , c ) 11.25 cm , d ) 16.37 cm , e ) 20.62 cm | d | divide(volume_cube(17), multiply(20, 15)) | multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)| | geometry | D |
in business , a and c invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by a and b was 3 : 2 , if rs 157300 was their profit , how much amount did b receive . | "explanation : a : b = 3 : 2 = 6 : 4 = > a : c = 2 : 1 = 6 : 3 = > a : b : c = 6 : 4 : 3 b share = ( 4 / 13 ) * 157300 = 48400 option d" | a ) rs 48000 , b ) rs 47000 , c ) rs 47400 , d ) rs 48400 , e ) none of these | d | multiply(multiply(multiply(multiply(add(2, 2), 2), const_100), const_100), divide(2, add(add(1, 2), 2))) | add(n0,n1)|add(n0,n3)|add(n3,#0)|multiply(#1,n3)|divide(n3,#2)|multiply(#3,const_100)|multiply(#5,const_100)|multiply(#4,#6)| | gain | D |
find the value of 2.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 2.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 2.5 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 2.5 x 4 x 4 x 50 = 2000 correct option : a" | a ) 2000 , b ) 500 , c ) 900 , d ) 1600 , e ) none | a | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 2.5) | multiply(n1,n2)|multiply(n4,n5)|multiply(n3,#0)|multiply(n6,#1)|divide(#2,#3)|multiply(n0,#4)| | general | A |
a cistern is filled by pipe a in 16 hours and the full cistern can be leaked out by an exhaust pipe b in 20 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 16 - 1 / 20 ) hrs = 1 / 80 = 80 hrs answer : d" | a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | d | divide(const_1, subtract(divide(const_1, 16), divide(const_1, 20))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | D |
the price of a jacket is reduced by 15 % . during a special sale the price of the jacket is reduced another 30 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | 1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 15 % , therefore bringing down the price to $ 85 . 3 ) again it is further discounted by 30 % , therefore bringing down the price to $ 59.5 4 ) now 59.5 has to be added byx % in order to equal the original price . 59.5 + ( x % ) 59.5 = 100 . solving this eq for x , we get x = 68.25 ans is d | a ) 60 , b ) 65 , c ) 69 , d ) 68.25 , e ) 67.5 | d | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 15), multiply(subtract(const_100, 15), divide(30, const_100)))), subtract(subtract(const_100, 15), multiply(subtract(const_100, 15), divide(30, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100) | gain | D |
find the average of all the numbers between 6 and 34 which are divisible by 7 . | "solution average = ( 7 + 14 + 21 + 28 ) / 4 ) = 70 / 4 = 17.5 . answer c" | a ) 18 , b ) 20 , c ) 17.5 , d ) 30 , e ) 32 | c | 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 | C |
of the 17,210 employees of the anvil factory , 2 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ? | "the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 / 7 portion represents ( 14210 / 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 / 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 / 7 of the total workforce . if 1 / 7 of the workforce is journeymen , and 1 / 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 / 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 / 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 / 7 of it because 1 / 7 was eliminated . the remaining workers now account for 1 / 7 out of 6 / 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 28.6 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 / 12,180 is exactly 1 / 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you β re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . d" | a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 % | d | multiply(multiply(divide(divide(divide(2, 7), 2), add(divide(divide(2, 7), 2), subtract(const_1, divide(2, 7)))), const_100), const_3) | divide(n1,n2)|divide(#0,n1)|subtract(const_1,#0)|add(#1,#2)|divide(#1,#3)|multiply(#4,const_100)|multiply(#5,const_3)| | general | D |
if 7 a = 2 b = 15 , then 42 ab = | "7 a * 2 b = 15 * 15 = 225 14 ab = 225 i . e . 42 ab = 675 answer : option c" | a ) 75 , b ) 125 , c ) 675 , d ) 1100 , e ) 1250 | c | multiply(42, multiply(divide(15, 7), divide(15, 2))) | divide(n2,n0)|divide(n2,n1)|multiply(#0,#1)|multiply(n3,#2)| | general | C |
5 ^ 100 is divided by 18 . then what is the remainder ? | "here n = 18 = 2 Γ 32 Ο ( 18 ) = 18 ( 1 β 12 ) ( 1 β 13 ) = 6 so 56 when divided by 18 , remainder is 1 . so we can write the given expression 5100 = ( 56 ) 16 Γ 54 = ( 1 ) 16 Γ 54 = 52 Γ 52 = 7 Γ 7 = 49 now 49 when divided by 18 , remainder is 13 . c" | a ) 10 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | c | subtract(18, 5) | subtract(n2,n0)| | general | C |
n ^ ( n / 2 ) = 4 is true when n = 4 in the same way what is the value of n if n ^ ( n / 2 ) = 8 ? | n ^ ( n / 2 ) = 8 apply log n / 2 logn = log 8 nlogn = 2 log 8 = log 8 ^ 2 = log 64 logn = log 64 now apply antilog n = 64 / n now n = 8 . answer : b | a ) 4 , b ) 8 , c ) 2 , d ) 6 , e ) 10 | b | divide(power(8, 2), 8) | power(n4,n0)|divide(#0,n4)| | general | B |
we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 10 , and the difference between the total number of strawberries and the total number of blueberries would increase by 50 . each blue box contains how many blueberries ? | "let x be the number of blueberries in each blue box . then there are x + 10 strawberries in each red box . x + ( x + 10 ) = 50 x = 20 the answer is c ." | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | divide(subtract(50, 10), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general | C |
gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 45 and the turban . find the price of the turban . | "let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 45 . thus , 3434 ( 90 + x ) = 45 + x or 270 + 3 x = 180 + 4 x or x = 90 answer : d" | a ) 27 , b ) 36 , c ) 29 , d ) 90 , e ) 11 | d | subtract(multiply(divide(subtract(90, 45), subtract(const_12, 9)), const_12), 90) | subtract(n0,n2)|subtract(const_12,n1)|divide(#0,#1)|multiply(#2,const_12)|subtract(#3,n0)| | general | D |
the average age of a class is 15.8 years . the average age of boys in the class is 16.4 years while that of girls is 15.4 years . what is the ratio of boys to girls in the class ? | let x = number of girls y = number of boys sum of girls ages / x = 15.4 sum of boys ages / y = 16.4 the sum of the girls ' ages is 15.4 x , and boys ' sum is 16.4 y ( sum of girls ages + sum of boys ages ) / ( x + y ) = 15.8 ( 15.4 x + 16.4 y ) / ( x + y ) = 15.8 15.4 x + 16.4 y = 15.8 ( x + y ) 15.4 x + 16.4 y = 15.8 x + 15.8 y 0.6 y = 0.4 x 0.6 y / x = 0.4 y / x = 0.4 / 0.6 y / x = 2 / 3 which is the ratio of boys to girls answer : d | a ) 1 : 2 , b ) 3 : 4 , c ) 3 : 5 , d ) 2 : 3 , e ) none of these | d | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1) | general | D |
a furniture dealer purchased a desk for $ 140 and then set the selling price equal to the purchase price plus a markup that was 30 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 140 selling price = x 140 + 0.3 * x = x 0.7 * x = 140 x = 200 profit = 200 - 140 = 60 answer : b" | a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 90 , e ) $ 100 | b | divide(multiply(subtract(divide(140, subtract(const_1, divide(30, const_100))), 140), const_100), 140) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)| | gain | B |
the average of 35 numbers is 25 . if each number is multiplied by 5 , find the new average ? | "sum of the 35 numbers = 35 * 25 = 875 if each number is multiplied by 5 , the sum also gets multiplied by 5 and the average also gets multiplied by 5 . thus , the new average = 25 * 5 = 125 . answer : a" | a ) 125 , b ) 189 , c ) 297 , d ) 267 , e ) 298 | a | multiply(25, 5) | multiply(n1,n2)| | general | A |
a pipe takes a hours to fill the tank . but because of a leakage it took 3 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 180 x = 90 min answer : c" | a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 70 min | c | multiply(const_10, multiply(const_1, 3)) | multiply(n0,const_1)|multiply(#0,const_10)| | physics | C |
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 51 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 51 / 3 = 17 kmph answer : e" | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 17 kmph | e | subtract(51, divide(multiply(51, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics | E |
if xy > 0 , 1 / x + 1 / y = 15 , and 1 / xy = 5 , then ( x + y ) / 5 = ? | ( 1 / x + 1 / y ) = 15 canbe solved as { ( x + y ) / xy } = 5 . substituting for 1 / xy = 5 , we get x + y = 15 / 5 = = > ( x + y ) / 5 = 15 / ( 5 * 5 ) = 3 / 5 . c | a ) 1 / 25 , b ) 1 / 6 , c ) 3 / 5 , d ) 5 , e ) 6 | c | divide(divide(15, 5), 5) | divide(n3,n5)|divide(#0,n5) | general | C |
how many pounds of salt at 50 cents / lb must be mixed with 45 lbs of salt that costs 35 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 35 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 45 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt 1 ) = ( 50 - 45 ) / ( 45 - 35 ) = 1 / 2 we know that weight of salt 1 is 45 lbs . weight of salt 2 must be 90 lbs . answer ( e )" | a ) 20 , b ) 15 , c ) 40 , d ) 50 , e ) 90 | e | divide(subtract(multiply(48, 45), multiply(divide(add(const_100, 20), const_100), multiply(35, 45))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | add(n3,const_100)|multiply(n1,n4)|multiply(n1,n2)|divide(#0,const_100)|multiply(#3,#2)|multiply(n0,#3)|subtract(#1,#4)|subtract(#5,n4)|divide(#6,#7)| | gain | E |
in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 9 , 0 ) , ( 0 , - 3.5 ) , ( - 9 , 0 ) ? | "ares of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 9 + 9 = 18 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 18 * 7 = 63 a is the answer" | a ) 63 , b ) 88 , c ) 112 , d ) 116 , e ) 120 | a | rhombus_area(multiply(9, const_2), multiply(3.5, const_2)) | multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)| | geometry | A |
what is the sum of the multiples of 4 from 40 to 80 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 4 and add 1 . ( 80 / 4 ) - ( 40 / 4 ) + 1 multiply these together so what we show average * total numbers = sum ( 80 + 40 ) / 2 * ( 80 / 4 ) - ( 40 / 4 ) + 1 = sum 60 * 11 = 660 b" | a ) 560 , b ) 660 , c ) 800 , d ) 760 , e ) 480 | b | multiply(divide(add(subtract(80, const_3), add(40, const_2)), const_2), add(divide(subtract(subtract(80, const_3), add(40, const_2)), 4), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)| | general | B |
75 men working 8 hours per day dig 50 m deep . how many extra men should be put to dig to a depth of 70 m working 6 hours per day ? | ( 75 * 8 ) / 50 = ( x * 6 ) / 70 = > x = 140 140 Γ’ β¬ β 65 = 65 answer : d | a ) 75 , b ) 45 , c ) 55 , d ) 65 , e ) 35 | d | subtract(divide(multiply(divide(multiply(75, 8), 50), 70), 6), 75) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0) | physics | D |
mona and donald fly to rome for the weekend . they take cash only in notes of $ 10 and notes of β¬ 10 . mona carries 3 times the amount of euros donald carries . she also carries as many dollars as donald carries . the number of β¬ 10 notes they take is double the number of $ 10 notes they take . if donald carries a total of 39 notes ( of either $ 10 or β¬ 10 , ) then what is the total number of notes ( of either $ 10 or β¬ 10 , ) they take ? | let e 10 = x no . d 10 = y no . donald is having x + y notes mona carries 3 x + y again x = 2 y or donald x + y = 39 or 3 y = 39 y = 13 ; x = 26 , total notes they carry = 104 + 26 = 130 a | a ) 130 , b ) 80 , c ) 100 , d ) 120 , e ) 150 | a | add(multiply(39, 3), divide(39, 3)) | divide(n5,n2)|multiply(n2,n5)|add(#0,#1) | general | A |
mr . smith calculated the average of 10 3 digit numbers . but due to a mistake he reversed the digits of a number and thus his average increased by 19.8 the difference between the unit digit and hundreds digit of that number is : | since the average increased by 29.7 and there were a total of 10 numbers , it means the incorrect number was 198 greater than the correct number . say , the correct number was abc ( where a , b and c are the digits of the 3 digit number ) then the incorrect number was cba . 100 c + 10 b + a - ( 100 a + 10 b + c ) = 198 99 c - 99 a = 99 ( c - a ) = 198 198 = 99 * 2 = 99 ( c - a ) so c - a = 2 answer ( a ) | a ) 2 , b ) 6 , c ) 5 , d ) 3 , e ) 4 | a | divide(multiply(10, 19.8), subtract(const_100, const_1)) | multiply(n0,n2)|subtract(const_100,const_1)|divide(#0,#1) | general | A |
what is the area of a circle with the circumference of 88 cms ? | "circum = 2 * pi * r 88 = 2 * pi * r r = 44 / pi area of circle = pi * r 2 a = pi * ( 44 / pi ) 2 a = 616 answer : e" | a ) 618 sq cms , b ) 516 sq cms , c ) 600 sq cms , d ) 615 sq cms , e ) none of these | e | multiply(multiply(multiply(divide(divide(multiply(sqrt(88), const_4), const_pi), const_2), divide(divide(multiply(sqrt(88), const_4), const_pi), const_2)), const_pi), const_2) | sqrt(n0)|multiply(#0,const_4)|divide(#1,const_pi)|divide(#2,const_2)|multiply(#3,#3)|multiply(#4,const_pi)|multiply(#5,const_2)| | geometry | E |
the average of runs of a cricket player of 10 innings was 15 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | average after 11 innings = 19 required number of runs = ( 19 * 11 ) - ( 15 * 10 ) = 209 - 150 = 59 . answer : e | a ) 87 , b ) 16 , c ) 10 , d ) 76 , e ) 59 | e | subtract(multiply(add(10, const_1), add(4, 15)), multiply(10, 15)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2) | general | E |
x , y and z , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2350 , what will be z ' s share of the earnings ? | "the dollars earned will be in the same ratio as amount of work done 1 day work of z is 1 / 12 ( or 2 / 24 ) 1 day work of the combined workforce is ( 1 / 6 + 1 / 8 + 1 / 12 ) = 9 / 24 z ' s contribution is 2 / 9 of the combined effort translating effort to $ = 2 / 9 * 2350 = $ 522.3 hence : d" | a ) $ 525 , b ) $ 550.50 , c ) $ 1080.02 , d ) $ 522.3 , e ) $ 1020.2 | d | multiply(divide(2350, add(add(inverse(6), inverse(8)), inverse(12))), inverse(6)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|multiply(#5,#0)| | physics | D |
set j consists of 15 consecutive even numbers . if the smallest term in the set is - 10 , what is the range of the positive integers in set j ? | "since there are only 15 integers , another approach is the just list all 15 . we get : - 10 , - 8 , - 6 , - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 range of positive integers = 18 - 2 = 16 answer : d" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 20 | d | subtract(add(negate(10), multiply(subtract(15, const_1), const_2.0)), 10) | negate(n1)|subtract(n0,const_1)|multiply(const_2.0,#1)|add(#2,#0)|subtract(#3,n1)| | general | D |
find the average of first 4 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 ) / 4 = 12.5 answer is c" | a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 21 | c | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(5, 4), subtract(5, const_2))), 4) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general | C |
in one alloy there is 12 % chromium while in another alloy it is 8 % . 20 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 20 + 35 = 55 kg alloy is 0.12 * 20 + 0.08 * 35 = 5.2 kg , so the percentage is 5.2 / 55 * 100 = 9.2 % . answer : a ." | a ) 9.45 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | a | multiply(divide(add(divide(multiply(12, 20), const_100), divide(multiply(8, 35), const_100)), add(20, 35)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain | A |
the total number of digits used in numbering the pages of a book having 360 pages is | "total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 261 ) = ( 9 + 180 + 783 ) = 972 . answer : a" | a ) 972 , b ) 990 , c ) 1098 , d ) 1305 , e ) 1405 | a | subtract(subtract(multiply(360, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | multiply(n0,const_3)|subtract(const_100,const_1)|subtract(const_10,const_1)|subtract(#0,#1)|subtract(#3,#2)| | general | A |
find the number , difference between number and its 3 / 5 is 50 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 50 , = > ( 2 / 5 ) x = 50 = > 2 x = 50 * 5 , = > x = 125 option d" | a ) 120 , b ) 123 , c ) 124 , d ) 125 , e ) 359 | d | multiply(divide(50, const_2), 5) | divide(n2,const_2)|multiply(n1,#0)| | general | D |
find the number of divisors of 1728 . ? | "1728 = 2 ^ 6 * 3 ^ 3 hence the number of factors = ( 6 + 1 ) x ( 3 + 1 ) = 7 x 4 = 28 . we know that if a number represented in standard form ( a ^ m * b ^ n ) , then the number of factors is given by ( m + 1 ) ( n + 1 ) . answer is 28 answer : c" | a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | c | add(const_2, const_2) | add(const_2,const_2)| | general | C |
the average of temperatures at noontime from monday to friday is 25 ; the lowest one is 15 , what is the possible maximum range of the temperatures ? | average = 25 , sum of temperatures = 25 * 5 = 125 as the min temperature is 15 , max would be 125 - 4 * 15 = 65 - - > the range = 65 ( max ) - 15 ( min ) = 50 answer : c | a ) 20 , b ) 25 , c ) 50 , d ) 45 , e ) 75 | c | subtract(subtract(multiply(25, add(const_2, const_3)), multiply(15, const_4)), 15) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1) | general | C |
a sum of money at simple interest amounts to rs . 820 in 3 years and to rs . 854 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 854 - 820 ) = rs . 34 . s . i . for 3 years = rs . ( 34 x 3 ) = rs . 102 . principal = rs . ( 820 - 102 ) = rs . 718 . answer : b" | a ) 647 , b ) 718 , c ) 654 , d ) 847 , e ) 976 | b | subtract(820, divide(multiply(subtract(854, 820), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | B |
the number 500 can be written as the sum of the squares of 2 different positive integers . what is the difference of these 2 integers ? | "22 ^ 2 + 4 ^ 2 = 500 - - > 22 - 4 = 18 . d" | a ) 17 , b ) 16 , c ) 15 , d ) 18 , e ) 13 | d | add(add(add(const_4, 2), add(2, const_2)), 2) | add(n1,const_4)|add(const_2,n1)|add(#0,#1)|add(n1,#2)| | general | D |
the average age of a , b and c is 25 years . if the average age of a and c is 29 years , what is the age of b in years ? | "age of b = age of ( a + b + c ) Γ’ β¬ β age of ( a + c ) = 25 Γ£ β 3 Γ’ β¬ β 29 Γ£ β 2 = 75 Γ’ β¬ β 58 = 17 years a" | a ) 17 , b ) 35 , c ) 20 , d ) 32 , e ) 21 | a | subtract(multiply(25, const_3), multiply(29, const_2)) | multiply(n0,const_3)|multiply(n1,const_2)|subtract(#0,#1)| | general | A |
one woman and one man can build a wall together in three hours , but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together , it would take them five hours to build the wall . assuming that rates for men , women and girls remain constant , how many hours would it take one woman , one man , and one girl , working together , to build the wall ? | "solution : let work done by man , women and girl per hour be m , w , g respectively . then , m + w = 1 / 3 - - > ( 1 ) , w + 2 g = 1 / 3 - - > ( 2 ) and m + g = 1 / 5 - - > ( 3 ) . no . of hours it would take forone woman , one man , and one girl , working together , to build the wall , n = 1 / m + w + g from ( 1 ) and ( 2 ) , m = 2 g and from ( 3 ) g = 1 / 15 , m = 2 / 15 and w = 1 / 5 . so , n = 1 / ( 2 / 5 ) = 5 / 2 option , a" | a ) 5 / 2 , b ) 1 , c ) 10 / 7 , d ) 12 / 7 , e ) 22 / 7 | a | inverse(add(divide(subtract(divide(add(inverse(add(add(const_4, const_1), const_4)), multiply(inverse(add(const_4, const_1)), const_2)), const_2), inverse(add(const_4, const_1))), divide(const_3, const_2)), inverse(add(const_4, const_1)))) | add(const_1,const_4)|divide(const_3,const_2)|add(#0,const_4)|inverse(#0)|inverse(#2)|multiply(#3,const_2)|add(#4,#5)|divide(#6,const_2)|subtract(#7,#3)|divide(#8,#1)|add(#9,#3)|inverse(#10)| | physics | A |
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 ? | "let the cost price of 1 article be rs . 1 . therefore , cost price of 20 articles = 20 * 1 = rs . 20 the selling price of 25 articles = cost price of 20 articles = rs . 20 . let us find the cost price of 25 articles . cost price of 25 articles = 25 * 1 = rs . 25 . therefore , profit made on sale of 25 articles = selling price of 25 articles - cost price of 25 articles = 20 - 25 = - rs . 5 . because the profit is in the negative , the merchant has made a loss of rs . 5 . therefore , % loss = loss / cost price β 100 % loss = 5 / 25 β 100 = 20 % loss . answer : b" | a ) 25 % loss , b ) 20 % loss , c ) 25 % profit , d ) 20 % profit , e ) 5 % profit | b | 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 | B |
a man is 46 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 46 ) + 2 = 2 ( x + 2 ) x + 48 = 2 x + 4 x = 44 . answer : a" | a ) 44 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 16 years | a | divide(subtract(46, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general | A |
in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 79 more candidates got selected than a . what was the number of candidates appeared from each state ? | "state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 79 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 79 = > total appeared candidates in state b = 79 x 100 = 7900 = > total appeared candidates in state a = total appeared candidates in state b = 7900" | a ) 7000 , b ) 7900 , c ) 6000 , d ) 5000 , e ) 4000 | b | divide(79, divide(subtract(7, 6), const_100)) | subtract(n1,n0)|divide(#0,const_100)|divide(n2,#1)| | general | B |
the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 302 to 400 inclusive ? | "2 + 4 + 6 + 8 + . . . + 100 = 2550 302 + 304 + . . . + 400 = 50 ( 300 ) + ( 2 + 4 + . . . + 100 ) = 15,000 + 2550 = 17,550 the answer is c ." | a ) 11,550 , b ) 14,550 , c ) 17,550 , d ) 20,550 , e ) 23,550 | c | multiply(divide(add(400, 302), const_2), add(divide(subtract(400, 302), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)| | general | C |
if k and w are the dimensions of a rectangle that has area 20 and if k and w are integers such that k > w , what is the total number of possible values of k ? | kw = 20 = 20 * 1 = 10 * 2 = 5 * 4 - - > k can take 3 values , answer : b | ['a ) two', 'b ) three', 'c ) four', 'd ) five', 'e ) six'] | b | multiply(const_3, const_1) | multiply(const_1,const_3) | geometry | B |
a motorist travels to a place 150 km away at an average speed of 50 km / hr and returns at 30 km / hr . his average speed for the whole journey in km / hr is ? | explanation : average speed = ( 2 xy ) / ( x + y ) km / hr = ( 2 * 50 * 30 ) / ( 50 + 30 ) km / hr . 37.5 km / hr . answer : c | a ) 37.9 , b ) 37.53 , c ) 37.5 , d ) 37.1 , e ) 37.4 | c | divide(multiply(const_2, multiply(50, 30)), add(50, 30)) | add(n1,n2)|multiply(n1,n2)|multiply(#1,const_2)|divide(#2,#0) | general | C |
if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n . what is the value of n ? | "if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n what is the value of n 4 ^ 11 = ( 2 ^ 2 ) ^ 11 = 2 ^ 22 10 ^ n = ( 5 x 2 ) ^ n = 5 ^ n x 2 ^ n 5 ^ 21 x 2 ^ 22 = 2 x 2 ^ n x 5 ^ n = 2 ^ n + 1 x 5 ^ n = 2 ^ 22 x 5 ^ 21 n = 21 option b" | a ) 11 , b ) 21 , c ) 22 , d ) 23 , e ) 32 | b | divide(log(multiply(power(4, 11), power(5, 21))), log(2)) | log(n4)|power(n2,n3)|power(n0,n1)|multiply(#1,#2)|log(#3)|divide(#4,#0)| | general | B |
a light flashes every 6 seconds , how many times will it flash in ? of an hour ? | "1 flash = 6 sec for 1 min = 10 flashes so for 1 hour = 10 * 60 = 600 flashes . answer : b" | a ) 550 , b ) 600 , c ) 650 , d ) 700 , e ) 750 | b | divide(const_3600, 6) | divide(const_3600,n0)| | physics | B |
a man sitting in a train which is traveling at 55 kmph observes that a goods train , traveling in opposite direction , takes 10 seconds to pass him . if the goods train is 320 m long , find its speed | "explanation : relative speed = 320 / 10 m / sec = ( ( 320 / 10 ) Γ ( 18 / 5 ) ) kmph = 115 kmph . speed of goods train = ( 115 - 55 ) kmph = 60 kmph answer : option c" | a ) 52 kmph , b ) 56 kmph , c ) 60 kmph , d ) 62 kmph , e ) 72 kmph | c | subtract(multiply(divide(320, 10), const_3_6), 55) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics | C |
in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.835 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 ) = 0.835 0.9 β 0.1 x = 0.835 x = 0.65 thus ratio of both flours is 2 : 1 d" | a ) 1 : 3 , b ) 1 : 2 , c ) 1 : 1 , d ) 2 : 1 , e ) 3 : 1 | d | divide(0.835, add(0.8, 0.9)) | add(n0,n1)|divide(n2,#0)| | general | D |
a library has an average of 510 visitors on sunday and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begins with a sunday , so there will be five sundays in the month . required average = [ ( 510 * 5 ) + ( 240 * 25 ) ] / 30 = 8550 / 30 = 285 . answer : d" | a ) 187 , b ) 279 , c ) 280 , d ) 285 , e ) 262 | d | divide(add(multiply(add(const_4, const_1), 510), multiply(multiply(add(const_4, const_1), add(const_4, const_1)), 240)), 30) | add(const_1,const_4)|multiply(n0,#0)|multiply(#0,#0)|multiply(n1,#2)|add(#1,#3)|divide(#4,n2)| | general | D |
if a - b = 3 and a 2 + b 2 = 29 , then find the value of ab | explanation : 2 ab = ( a 2 + b 2 ) β ( a β b ) 2 = > 2 ab = 29 - 9 = 20 = > ab = 10 option c | a ) 7 , b ) 8.5 , c ) 10 , d ) 12 , e ) 14 | c | divide(subtract(29, power(3, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1) | general | C |
in kaya ' s teacher ' s desk there are 10 pink highlighters , 15 yellow highlighters , and 8 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 10 + 15 + 8 = 33 . answer is d ." | a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | d | add(add(10, 15), 8) | add(n0,n1)|add(n2,#0)| | general | D |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 4400 . what is the share of b ? | let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 4400 x ( 2 / 11 ) ) = rs . 800 . answer is b | a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | b | multiply(4400, divide(const_2, add(add(multiply(const_2, const_3), multiply(divide(const_2, const_3), const_3)), const_3))) | divide(const_2,const_3)|multiply(const_2,const_3)|multiply(#0,const_3)|add(#1,#2)|add(#3,const_3)|divide(const_2,#4)|multiply(n1,#5) | gain | B |
in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 190 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 190 x = 38 total number of students = 13 x = 13 * 38 = 494 answer is d" | a ) 562 , b ) 356 , c ) 452 , d ) 494 , e ) 512 | d | add(multiply(divide(8, 5), 190), 190) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)| | other | D |
a survey reveals that the average income of a company β s customers is $ 45000 per year . if 50 customers respond to the survey and the average income of the wealthiest 10 of those customers is $ 55000 , what is the average income of the other 40 customers ? is there a way to solve this using weighted average concept instead of doing tedious calculations ? | let x be the average of 40 customers 40 * x + 10 * 55000 = 50 * 45000 solving this we have x = 42500 answer is d . | a ) $ 27,500 , b ) $ 35,000 , c ) $ 37,500 , d ) $ 42,500 , e ) $ 50,000 | d | divide(subtract(multiply(45000, 50), multiply(10, 55000)), 40) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general | D |
the maximum numbers of students among them 781 pens and 710 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is | "olution required number of students . = h . c . f of 781 and 710 . Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ 71 . answer a" | a ) 71 , b ) 910 , c ) 1001 , d ) 1911 , e ) none | a | subtract(781, 710) | subtract(n0,n1)| | general | A |
the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 15 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the difference is 15 , the sum = 400 Γ 15 = 6000 answer d" | a ) 4500 , b ) 7500 , c ) 5000 , d ) 6000 , e ) none of these | d | divide(15, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5)| | gain | D |
a person crosses a 720 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 720 / ( 12 x 60 ) m / sec = 1 m / sec . converting m / sec to km / hr = 1 x ( 18 / 5 ) km / hr = 3.6 km / hr . answer : a" | a ) 3.6 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 5.5 | a | divide(divide(720, const_1000), divide(multiply(12, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics | A |
if both 5 ^ 2 and 3 ^ 4 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 ^ 2 * 5 ^ 2 = 225 the answer is d ." | a ) 80 , b ) 120 , c ) 150 , d ) 225 , e ) 300 | d | add(add(add(add(add(multiply(multiply(5, 2), const_2.0), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n4,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other | D |
what is the least number of square tiles required to pave the floor of a room 12 m 12 cm long and 1 m 20 cm broad ? | "solution length of largest tile = h . c . f . of 1212 cm & 120 cm = 12 cm . area of each tile = ( 12 x 12 ) cm 2 β΄ required number of tiles = [ 1212 x 120 / 12 x 12 ] = 1010 . answer a" | a ) 1010 , b ) 804 , c ) 814 , d ) 844 , e ) none | a | divide(multiply(add(multiply(1, const_100), 20), multiply(12, const_100)), power(20, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2)| | physics | A |
a β s speed is 21 / 19 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | let d be the full distance . let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 21 v / 19 ) = xd / v ( 19 / 21 ) * d / v = x * d / v x = 19 / 21 b should have a head start of 2 / 21 of the full distance . the answer is a . | a ) 2 / 21 , b ) 2 / 19 , c ) 1 / 11 , d ) 1 / 7 , e ) 1 / 5 | a | divide(subtract(21, 19), 21) | subtract(n0,n1)|divide(#0,n0) | general | A |
a can give b 80 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ? | a runs 1000 m while b runs 920 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) / 920 = 869.56 m b can give c = 1000 - 869.56 = 130.43 m . answer : b | a ) 130.12 , b ) 130.43 , c ) 130.22 , d ) 111.0 , e ) 111.12 | b | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 80))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4) | physics | B |
the ratio between x and y is 7 / 9 ; x is increased by 4 and y is decreased by 4 , what is the ratio between the new values of x and y ? | "ratio = 7 k / 9 k = 7 / 9 , 14 / 18 , etc . x is increased by 4 and y is decreased by 4 - - > ( 7 k + 4 ) / ( 9 k - 4 ) new ratio can be 11 / 15 , 18 / 14 , etc . answer : e" | a ) 7 / 9 , b ) 14 / 9 , c ) 7 / 18 , d ) 11 / 13 , e ) it can not be determined | e | divide(multiply(7, 9), multiply(9, 7)) | multiply(n0,n1)|divide(#0,#0)| | general | E |
if the sides of a triangle are 28 cm , 26 cm and 10 cm , what is its area ? | "the triangle with sides 28 cm , 26 cm and 10 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 26 * 10 = 130 cm 2 answer : d" | a ) 120 cm 2 , b ) 112 cm 2 , c ) 776 cm 2 , d ) 130 cm 2 , e ) 886 cm 2 | d | divide(multiply(26, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry | D |
what is the value of ( log 2 3 ) ( log 3 4 ) ( log 4 5 ) . . . ( log 63 64 ) ? | using the identity ( loga b ) ( logb c ) = loga c repeatedly , we obtain ( log 2 3 ) ( log 3 4 ) ( log 4 5 ) . . . ( log 63 64 ) = log 2 64 = 6 correct answer d | a ) 1 / 6 , b ) 2 , c ) 5 / 2 , d ) 6 , e ) 32 | d | multiply(add(5, const_1), divide(log(2), log(2))) | add(n5,const_1)|log(n0)|divide(#1,#1)|multiply(#0,#2) | other | D |
a bus travels first half distance between two places with a speed of 40 kmph and the rest half distance with a speed of 60 kmph . the average speed of the bus is ? | explanation : we know that speed = distance traveled / time taken let the total distance traveled by the car is 2 x km . then time taken by it to cover first half is hour . and for second half is hour . then average speed = total distance travelled / total time taken . i . e . = > = 48 kmph . answer : a | a ) 48 kmph , b ) 78 kmph , c ) 28 kmph , d ) 18 kmph , e ) 98 kmph | a | divide(add(40, 60), const_2) | add(n0,n1)|divide(#0,const_2) | physics | A |
rahul can done a work in 5 days and meena can done a work in 10 days . in how many days that both are work together ? | a ' s one day work = 1 / 5 b ' s one day work = 1 / 10 a + b = 1 / 5 + 1 / 10 = 10 + 5 / 50 = 10 / 3 days ans : option b | a ) 3 / 10 , b ) 10 / 3 , c ) 1 / 3 , d ) 5 , e ) 6 / 3 | b | inverse(add(divide(const_1, 5), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2) | physics | B |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 5500 , the number of valid votes that the other candidate got , was : | "b number of valid votes = 80 % of 5500 = 4400 . valid votes polled by other candidate = 45 % of 4400 = ( 45 / 100 x 4400 ) = 1980 ." | a ) 2800 , b ) 1980 , c ) 2900 , d ) 2200 , e ) 2300 | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 5500) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain | B |
what is the sum of all remainders obtained when the first 130 natural numbers are divided by 9 ? | "a positive integer can give only the following 9 remainders when divided by 9 : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 0 . 1 divided by 9 gives the remainder of 1 ; 2 divided by 9 gives the remainder of 2 ; . . . 8 divided by 9 gives the remainder of 8 ; 9 divided by 9 gives the remainder of 0 . we ' ll have 11 such blocks , since 99 / 9 = 11 . the last will be : 91 divided by 9 gives the remainder of 1 ; 92 divided by 9 gives the remainder of 2 ; . . . 98 divided by 9 gives the remainder of 8 ; 99 divided by 9 gives the remainder of 0 . the last number , 100 , gives the remainder of 1 when divided by 9 , thus the sum of all remainders will be : 11 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0 ) + 1 = 412 . answer : d ." | a ) 397 , b ) 401 , c ) 403 , d ) 412 , e ) 399 | d | add(const_3, multiply(divide(subtract(130, reminder(130, 9)), 9), add(add(add(const_1, const_2), const_3), add(add(const_1, const_4), add(const_4, add(add(const_1, const_2), const_3)))))) | add(const_1,const_2)|add(const_1,const_4)|reminder(n0,n1)|add(#0,const_3)|subtract(n0,#2)|add(#3,const_4)|divide(#4,n1)|add(#1,#5)|add(#3,#7)|multiply(#8,#6)|add(#9,const_3)| | general | D |
in a flight of 600 km , an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by 200 km / hr and the time of flight increased by 30 minutes . the duration of the flight is : | "let the duration of the flight be x hours then 600 / x - 600 / x ( 1 / 2 ) = 200 = 600 / x - 1200 / 2 x + 1 = 200 = x ( 2 x + 1 ) = 3 = 2 xsquare + x - 3 = 0 = ( 2 x + 3 ) ( x - 1 ) = 0 = x = 1 hr ( neglecting the - ve value of x ) answer : option a" | a ) 1 hour , b ) 2 hour , c ) 3 hour , d ) 4 hour , e ) 5 hour | a | divide(600, divide(multiply(200, 30), const_10)) | multiply(n1,n2)|divide(#0,const_10)|divide(n0,#1)| | general | A |
a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work in : | "explanation : ratio of rates of working of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 . b ' s 1 day ' s work = 1 / 12 a ' s 1 day ' s work = 1 / 6 ; ( 2 times of b ' s work ) ( a + b ) ' s 1 day ' s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer is a" | a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) 28 days | a | inverse(add(divide(const_1, 12), multiply(const_2, divide(const_1, 12)))) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)| | physics | A |
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