options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) β 0.3 , b ) 0 , c ) 0.03 , d ) 1.08 , e ) 2.46 | c | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.8, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.8, 0.3)), const_1000), const_1000)) | what is the value of 3 x ^ 2 β 1.8 x + 0.3 for x = 0.3 ? | "3 x ^ 2 - 1.8 x + 0.3 for x = 0.6 = 3 ( 0.3 * 0.3 ) - 6 * 0.3 * ( 0.3 ) + 0.3 = - 3 * 0.3 * 0.3 + 0.3 = - 0.27 + 0.3 = 0.03 correct option : c" | a = 3 ** 2
b = 1 ** 8
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 1 ** 8
i = g - h
j = i / 1000
k = j / 1000
l = f - k
|
a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 86 | e | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(42, const_4)), 42) | the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 42 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 50 * 5 = 250 lowest is 42 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 42 = 168 . on the fifth day it is 250 - 168 = 281 range is therefore 128 - 42 = 86 answer e" | a = 2 + 3
b = 50 * a
c = 42 * 4
d = b - c
e = d - 42
|
a ) rs . 129 , b ) rs . 144 , c ) rs . 156 , d ) rs . 168 , e ) rs . 178 | b | divide(multiply(378, multiply(2, 4)), multiply(add(3, 4), 3)) | a , b and c enter into a partnership and theirs shares are in the ratio 1 / 2 : 1 / 3 : 1 / 4 . after two months , a withdraws half of his capital and after 10 months , a profit of rs . 378 is divided among them . what is b β s share ? | ratio of initial investments = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 . let their initial investments be 6 x , 2 x and 3 x respectively . a : b : c = ( 6 x * 2 + 3 x * 10 ) : ( 4 x * 12 ) : ( 3 x * 12 ) = 42 : 48 : 36 = 7 : 8 : 6 . therefore , b β s share = rs . 378 * 8 / 21 = rs . 144 . answer : b | a = 2 * 4
b = 378 * a
c = 3 + 4
d = c * 3
e = b / d
|
a ) 5 , b ) 50 , c ) 10 ^ 5 , d ) 5 ^ 10 , e ) 8 ^ 10 - 3 ^ 10 | c | power(10, subtract(8, 3)) | on a scale that measures the intensity of a certain phenomenon , a reading of d + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of d . on that scale , the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3 ? | to solve this problem we need to examine the information in the first sentence . we are told that β a reading of d + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of d . β let β s practice this idea with some real numbers . let β s say d is 2 . this means that d + 1 = 3 . with the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2 . furthermore , we can say that a reading of 4 is actually 10 x 10 = 10 ^ 2 times as great as the intensity of a reading of 2 . increasing one more unit , we can say that a reading of 5 is 10 x 10 x 10 = 10 ^ 3 times as great as the intensity of a reading of 2 . we have found a pattern , which can be applied to the problem presented in the stem : 3 is β one β unit away from 2 , and thus a reading of 3 is 10 ^ 1 times as great as the intensity of a reading of 2 . 4 is β two β units away from 2 , and thus a reading of 4 is 10 ^ 2 times as great as the intensity of a reading of 2 . 5 is β three β units away from 2 , and thus a reading of 5 is 10 ^ 3 times as great as the intensity of a measure of 2 . we can use this pattern to easily answer the question . here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3 . because 8 is 5 units greater than 3 , a reading of 8 is 10 ^ 5 times as great as the intensity corresponding to a reading of 3 . answer c . | a = 8 - 3
b = 10 ** a
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(add(2, const_3), const_3), const_4) | if 6 x ^ 2 + x - 12 = ( hx + b ) ( cx + d ) , then | h | + | b | + | c | + | d | = for a complete solution and more practice problems , see this blog : http : / / magoosh . com / gmat / 2012 / algebra - on . . . to - factor / | 6 x ^ 2 + x - 12 = 6 x ^ 2 + 9 x - 8 x - 12 = > 3 x ( 2 x + 3 ) - 4 ( 2 x + 3 ) = > ( 2 x + 3 ) ( 3 x - 4 ) = ( hx + b ) ( cx + d ) hence h = 2 , b = c = 3 , d = - 4 so , 2 + 3 + 3 + | - 4 | = 2 + 3 + 3 + 4 = 12 answer b . | a = 2 + 3
b = a + 3
c = b + 4
|
a ) 30 days , b ) 40 days , c ) 50 days , d ) 60 days , e ) 70 days | b | divide(add(multiply(divide(subtract(multiply(10, 7), multiply(8, 6)), subtract(multiply(8, 4), multiply(10, 3))), multiply(8, 4)), multiply(8, 6)), 10) | 4 men and 6 women finish a job in 8 days , while 3 men and 7 women finish it in 10 days . in how many days will 10 women working together finish it ? | explanation : let 1 man ' s 1 day work = x and 1 woman ' s 1 days work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving , we get y = 1 / 400 [ means work done by a woman in 1 day ] 10 women 1 day work = 10 / 400 = 1 / 40 10 women will finish the work in 40 days option b | a = 10 * 7
b = 8 * 6
c = a - b
d = 8 * 4
e = 10 * 3
f = d - e
g = c / f
h = 8 * 4
i = g * h
j = 8 * 6
k = i + j
l = k / 10
|
a ) 25 , b ) 8 , c ) 6 , d ) 5 , e ) 2 | c | add(sqrt(25), const_1) | how many trailing zeros will be there after the rightmost non - zero digit in the value of 25 ! ( factorial 25 ) ? | explanatory answer 25 ! means factorial 25 whose value = 25 * 24 * 23 * 22 * . . . . * 1 when a number that has 5 as its factor is multiplied by an even number , it results in a trailing zero . ( product of 5 and 2 is 10 and any number when multiplied with 10 or a power of 10 will have one or as many zeroes as the power of 10 with which it has been multiplied ) in 25 ! , the following numbers have 5 as their factor : 5 , 10 , 15 , 20 , and 25 . 25 is the square of 5 and hence it has two 5 s in it . in toto , it is equivalent of having six 5 s . there are at least 6 even numbers in 25 ! hence , the number 25 ! will have 6 trailing zeroes in it . choice c | a = math.sqrt(25)
b = a + 1
|
a ) 22 , b ) 25 , c ) 27 , d ) 38 , e ) 30 | b | add(multiply(divide(subtract(41, 5), add(5, 4)), 5), 5) | alice leaves her home and walks toward bob ' s house . one hour later , bob leaves his home and walks toward alice ' s house . the distance between their homes is 41 kilometers , alice ' s walking speed is 5 km / h , and bob ' s walking speed is 4 km / h . how many kilometers will alice walk before she meets bob ? | alice walks 5 km in the first hour so there are 36 km remaining . when bob starts walking , they complete a total of 9 km per hour . they will meet four hours after bob starts walking . since alice walks for 5 hours , she walks 25 km . the answer is b . | a = 41 - 5
b = 5 + 4
c = a / b
d = c * 5
e = d + 5
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | subtract(subtract(multiply(3, 6), add(subtract(17, 6), 3)), 6) | the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 5 . if the sum of the first and the last number is 17 . what is the last numbers ? | "a + b + c = 18 b + c + d = 15 a + d = 17 a β d = 3 a + d = 17 2 d = 14 d = 7 answer : d" | a = 3 * 6
b = 17 - 6
c = b + 3
d = a - c
e = d - 6
|
a ) 6.24 km , b ) 6 km , c ) 6.79 km , d ) 5.66 km , e ) 10 km | c | multiply(divide(multiply(add(7, 1.2), subtract(7, 1.2)), add(add(7, 1.2), subtract(7, 1.2))), const_2) | a man can row 7 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 7 s = 1.2 ds = 8.2 us = 5.8 x / 8.2 + x / 5.8 = 1 x = 3.40 d = 3.40 * 2 = 6.79 answer : c" | a = 7 + 1
b = 7 - 1
c = a * b
d = 7 + 1
e = 7 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 4 , b ) 5 , c ) 6 , d ) 3 , e ) 2 | c | multiply(const_4, 1) | how many prime numbers between 1 and 100 are factors of 67830 ? | "factor of 67830 = 2 * 3 * 5 * 7 * 17 * 19 - - - 6 prime numbers c" | a = 4 * 1
|
a ) 25 kmph , b ) 85 kmph , c ) 34 kmph , d ) 23 kmph , e ) 45 kmph | a | divide(add(20, 30), const_2) | the speed of a car is 20 km in the first hour and 30 km in the second hour . what is the average speed of the car ? | "s = ( 20 + 30 ) / 2 = 25 kmph answer : a" | a = 20 + 30
b = a / 2
|
a ) 33 , b ) 88 , c ) 27 , d ) 64 , e ) 99 | d | multiply(power(4, const_2), 4) | how many shots of 1 cm radius can be prepared from a sphere of 4 cm radius ? | "4 / 3 Γ― β¬ * 4 * 4 * 4 = 4 / 3 Γ― β¬ * 1 * 1 * 1 * x x = 64 answer : d" | a = 4 ** 2
b = a * 4
|
a ) 12 , b ) 15 , c ) 16 , d ) 18 , e ) 22 | a | subtract(add(7, 8), 3) | in a class , 7 students like to play basketball and 8 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 7 + 8 - 3 = 12 a )" | a = 7 + 8
b = a - 3
|
a ) $ 44 , b ) $ 48 , c ) $ 52 , d ) $ 56 , e ) $ 60 | d | divide(42, subtract(1, multiply(divide(1, 8), const_2))) | p has $ 42 more than what q and r together would have had if both b and c had 1 / 8 of what p has . how much does p have ? | "p = ( 2 / 8 ) * p + 42 ( 6 / 8 ) * p = 42 p = 56 the answer is d ." | a = 1 / 8
b = a * 2
c = 1 - b
d = 42 / c
|
a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 12 | e | multiply(divide(subtract(24, 3), add(3, 4)), 4) | one hour after yolanda started walking from x to y , a distance of 24 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob Ρ ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 24 , hence the distance at that time between them was 24 - 3 = 21 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 21 / 7 = 3 hours . in 6 hours b walked 3 * 4 = 12 miles . answer : e ." | a = 24 - 3
b = 3 + 4
c = a / b
d = c * 4
|
a ) 33 , b ) 11 , c ) 68 , d ) 36 , e ) 90 | e | divide(multiply(100, 10), const_100) | the cost of an article is decreased by 10 % . if the original cost is $ 100 , find the decrease cost . | "original cost = $ 100 decrease in it = 10 % of $ 100 = 10 / 100 Γ£ β 100 = 1000 / 100 = $ 10 therefore , decrease cost = $ 100 - $ 10 = $ 90 answer : e" | a = 100 * 10
b = a / 100
|
a ) 56 sec , b ) 1 minute , c ) 45 sec , d ) 58 sec , e ) 50 sec | a | divide(multiply(35, const_4), multiply(9, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 35 metres . if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 35 * 4 = 140 m time taken = 140 * 2 / 5 = 56 sec . answer is a" | a = 35 * 4
b = 1000 / 3600
c = 9 * b
d = a / c
|
a ) 6 days , b ) 5 days , c ) 3 days , d ) 7 days , e ) 12 days | c | floor(add(inverse(multiply(subtract(inverse(add(divide(const_1, 20), divide(const_1, 12))), 3), divide(const_1, 12))), const_1)) | a can do a piece of work in 20 days , b can do a piece of work in 12 days , both are working together for 3 days after a left the job , then how many days b requires to complete the remaining work ? | first take lcm of ( 20,12 ) = 60 ; total work = 60 ; a ' s one day work = 60 / 20 = 3 ; b ' s one day work = 60 / 12 = 5 ; one day work of both a + b = 3 + 5 = 8 ; total work completed in 3 days = 8 * 3 = 24 ; remaining work is = 60 - 24 = 36 ; remaining work completed by b in = 36 / 12 = 3 days . answer = c | a = 1 / 20
b = 1 / 12
c = a + b
d = 1/(c)
e = d - 3
f = 1 / 12
g = e * f
h = 1/(g)
i = h + 1
j = math.floor(i)
|
a ) 450 , b ) 375 , c ) 400 , d ) 425 , e ) 300 | d | subtract(divide(add(200, 800), const_2), divide(add(50, 300), const_2)) | the average ( arithmetic mean ) of the integers from 200 to 800 , inclusive , is how much greater than the average of the integers from 50 to 300 , inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 200 to 800 , inclusive = ( 200 + 800 ) / 2 = 500 average of numbers between 50 to 300 , inclusive = ( 50 + 300 ) / 2 = 175 difference = 500 - 175 = 425 answer is d" | a = 200 + 800
b = a / 2
c = 50 + 300
d = c / 2
e = b - d
|
a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum r of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : r = 2550 + ( 100 * 50 ) = 7550 . b" | a = 200 + 102
b = a / 2
c = 200 - 102
d = c / 2
e = d + 1
f = b * e
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | divide(add(multiply(10, const_2), 10), 6) | bipin is 6 times old as alok . bipin ' s age will be twice of chandan ' s age after 10 years . if chandan ' s 7 th birthday was celebrated 3 years ago , what is alok ' s present age ? | let the present ages of bipin , alok , chandan are b , a , c respectively given conditions are 1 ) b = 6 a 2 ) ( b + 10 ) = 2 * ( c + 10 ) 3 ) chandans present age = c = 10 years therefore b = 30 so a = 5 answer : d | a = 10 * 2
b = a + 10
c = b / 6
|
a ) 36 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 80 liters | e | divide(36, subtract(divide(3, 4), divide(30, const_100))) | a big container is 30 % full with water . if 36 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 36 liters of water is added , the container becomes 75 % full . hence these 36 liters account for 45 % of the container , which means that the capacity of it is 36 / 0.45 = 80 liters . or : if the capacity of the container is x liters then : 0.3 x + 36 = 0.75 x - - > x = 80 liters . answer : e ." | a = 3 / 4
b = 30 / 100
c = a - b
d = 36 / c
|
a ) 655.9 , b ) 659.75 , c ) 650.57 , d ) 655.9 , e ) 600.3 | b | multiply(circumface(divide(42, const_2)), 5) | find the cost of fencing around a circular field of diameter 42 m at the rate of rs . 5 a meter ? | "2 * 22 / 7 * 14 = 131.95 131.95 * 5 = rs . 659.75 answer : b" | a = 42 / 2
b = circumface * (
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a ) 46 m , b ) 60 m , c ) 58 m , d ) 78 m , e ) 80 m | e | divide(add(divide(5300, 26.50), multiply(const_2, 60)), const_4) | length of a rectangular plot is 60 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 60 ) metres . perimeter = 5300 / 26.5 m = 200 m . 2 [ ( x + 60 ) + x ] = 200 2 x + 60 = 100 2 x = 40 x = 20 . hence , length = x + 60 = 80 m e" | a = 5300 / 26
b = 2 * 60
c = a + b
d = c / 4
|
a ) 76 % , b ) 60 % , c ) 50 % , d ) 40 % , e ) 24 % | d | subtract(const_100, multiply(divide(divide(36, const_100), divide(subtract(const_100, 40), const_100)), const_100)) | exactly 36 % of the numbers in set a are even multiples of 3 . if 40 % of the even integers in set a are not multiples of 3 , what percent of the numbers in set a are not even integers ? | everything is correct except the red part with a typo : it should be n = 0.6 a - - > even numbers are 60 % of a - - > 40 % of a are not even integers . answer : d . | a = 36 / 100
b = 100 - 40
c = b / 100
d = a / c
e = d * 100
f = 100 - e
|
a ) 80 , b ) 90 , c ) 120 , d ) 150 , e ) 240 | a | add(divide(multiply(320, 3), 5), multiply(3, 5)) | the number of timeshare condos available at sunset beach is 3 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 320 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 3 / 5 x = 320 hence , x = 200 . so , number of timeshare condos available at playa del mar = 200 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 3 / 5 x = 2 / 5 x = 2 / 5 ( 200 ) = 80 the correct answer is a ." | a = 320 * 3
b = a / 5
c = 3 * 5
d = b + c
|
a ) - 324 , b ) 223 , c ) 324 , d ) - 223 , e ) - 410 | a | subtract(2, 3) | if the operation @ is defined for all a and b by the equation a @ b = ( a ^ 2 * b ) * 3 , then 2 @ ( 3 @ - 1 ) = | "work within the parenthesis first so solve ( 3 @ - 1 ) first ( 3 @ - 1 ) = ( ( 3 ^ 2 ) * - 1 ) * 3 = ( 9 * - 1 ) * 3 = - 9 * 3 = - 27 now take - 3 plug back into equation and solve the rest 2 @ ( - 27 ) = ( ( 2 ^ 2 ) * - 3 ) * 3 = ( 4 * - 27 ) * 3 = - 324 so - 324 is the answer . . . . this question is merely testing order of operations remember pemdas option a" | a = 2 - 3
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a ) 8 , b ) 12 , c ) 16 , d ) 20 , e ) 24 | b | divide(subtract(264, subtract(power(multiply(7, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 7 inches wide on each side , the total area would have been ( m + 264 ) square inches . what is the perimeter of the photograph , in inches ? | "let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 14 ) ( y + 14 ) = m and so ( 2 ) xy + 14 x + 14 y + 196 = m + 264 let ' s subtract equation ( 1 ) from equation ( 2 ) . 12 x + 12 y + 192 = 264 2 x + 2 y = 12 , which is the perimeter of the photograph . the answer is b ." | a = 7 * 2
b = a ** 2
c = 1 * 2
d = c ** 2
e = b - d
f = 264 - e
g = f / 2
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a ) 240 , b ) 75 , c ) 110 , d ) 130 , e ) 85 | e | add(divide(subtract(add(divide(subtract(240, 60), const_2), 60), 60), const_2), 60) | a picnic attracts 240 persons . there are 60 more men than women , and 60 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 60 i . e . y + ( y + 60 ) = 240 i . e . y = 90 i . e . adult = 90 + 20 = 110 adults include only men and women i . e . men + women = 110 let women , w = x then men , m = x + 60 i . e . x + ( x + 60 ) = 2 x + 60 = 110 i . e . x = 25 i . e . men , m = 25 + 60 = 85 answer : option e" | a = 240 - 60
b = a / 2
c = b + 60
d = c - 60
e = d / 2
f = e + 60
|
a ) $ 1.13 , b ) $ 1.14 , c ) $ 1.15 , d ) $ 1.16 , e ) $ 1.17 | d | divide(add(add(1.17, 1.3), multiply(const_2, 1.05)), add(add(const_1, const_1), const_2)) | the price for a loaf of bread in april was $ 1.17 , in may the price was $ 1.30 , and in june the price was $ 1.05 . if 2 / 3 as much bread was sold in april as in may , and twice as much was sold in june as in april , what was the average price for a loaf of bread sold during the 3 - month period ? | let x be the number of loaves sold in may . then 2 x / 3 loaves were sold in april and 4 x / 3 loaves were sold in june . the average price was ( 2 x / 3 ) ( 117 ) + 130 x + ( 4 x / 3 ) ( 105 ) / ( 2 x / 3 + x + 4 x / 3 ) = ( 78 + 130 + 140 ) / ( 3 ) = 348 / 3 = $ 1.16 the answer is d . | a = 1 + 17
b = 2 * 1
c = a + b
d = 1 + 1
e = d + 2
f = c / e
|
a ) 6 km , b ) 3 km , c ) 8 km , d ) 9 km , e ) 2 km | c | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 9), divide(const_1, 10)), divide(const_1, 11))), const_3), const_1000) | a person travels equal distances with speeds of 9 km / hr , 10 km / hr and 11 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 9 + x / 10 + x / 11 = 47 / 60 29 x / 96 = 47 / 60 = > x = 2.59 total distance = 3 * 2.59 = 7.78 km . answer : c" | a = 47 / const_60
b = 1 / 9
c = 1 / 10
d = b + c
e = 1 / 11
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150 | b | divide(multiply(22, 10), const_2) | if the sides of a triangle are 26 cm , 22 cm and 10 cm , what is its area ? | "the triangle with sides 26 cm , 22 cm and 10 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 22 * 10 = 110 cm 2 answer : option b" | a = 22 * 10
b = a / 2
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a ) 12 , b ) 13 , c ) 14 , d ) 16 , e ) 15 | b | divide(subtract(multiply(92, 5), multiply(5, 40)), subtract(92, 72)) | the average mark of the students of a class in a particular exam is 72 . if 5 students whose average mark in that exam is 40 are excluded , the average mark of the remaining will be 92 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 92 ( x - 5 ) 72 x - ( 5 * 40 ) = 92 ( x - 5 ) 260 = 20 x = > x = 13 answer : b" | a = 92 * 5
b = 5 * 40
c = a - b
d = 92 - 72
e = c / d
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a ) s . 8500 , b ) s . 6000 , c ) s . 6800 , d ) s . 9000 , e ) s . 7000 | c | divide(8160, add(add(multiply(divide(12, const_100), 3), multiply(divide(9, const_100), 5)), multiply(divide(13, const_100), subtract(8, 5)))) | suresh borrowed some money at the rate of 12 % p . a . for the first 3 years , 9 % p . a . for the next 5 years and 13 % p . a . for the period beyond 8 years . if the total interest paid by him at the end of 11 years is rs . 8160 , how much money did he borrow ? | using the formula s . i = p * t * r / 100 ( x * 12 * 3 / 100 ) + ( x * 9 * 5 / 100 ) + ( x * 13 * 3 / 100 ) = 8160 x = 6800 i . e money borrowed is rs . 6800 answer : c | a = 12 / 100
b = a * 3
c = 9 / 100
d = c * 5
e = b + d
f = 13 / 100
g = 8 - 5
h = f * g
i = e + h
j = 8160 / i
|
a ) 23 / 30 , b ) 11 / 15 , c ) 7 / 10 , d ) 1 / 6 , e ) 2 / 15 | b | divide(add(10, 4), multiply(10, const_2)) | in township k , 1 / 4 of the housing units are equiped with cable tv . if 1 / 10 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 4 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 4 ) = 1 / 12 therefore only video cassette recorder = 1 / 10 - 1 / 12 = 1 / 60 total = 1 / 4 + 1 / 60 + neither cable tv nor videocassette recorders 1 = 4 / 15 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = 1 - 4 / 15 = 11 / 15 hence b ." | a = 10 + 4
b = 10 * 2
c = a / b
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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) | 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 = 5 * 2
b = a * 2
c = 5 * 2
d = c * 2
e = b + d
f = 5 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 58 , b ) 14 , c ) 145 , d ) 69 , e ) 74 | d | subtract(multiply(add(divide(subtract(multiply(12, 4), 3), 2), 9), 3), 12) | what will be the result of 12 ^ 4 + 3 + 2 ^ 9 + 3 - 8 / 2 ? | "the multiplication ( e . g . 12 ^ 4 ) and division ( e . g . 8 / 2 ) is done first and then the sum ( e . g . 48 + 3 ) and subtraction ( e . g . 4 - 4 ) , and the answer is 69 , option d ." | a = 12 * 4
b = a - 3
c = b / 2
d = c + 9
e = d * 3
f = e - 12
|
a ) 41 / 50 , b ) 1 / 221 , c ) 1 / 216 , d ) 1 / 36 , e ) 1 / 42 | d | divide(const_1, power(subtract(divide(21, const_3), const_1), const_2)) | in a certain game of dice , the player β s score is determined as a sum of two throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if jim plays this game against 21 other players , what is the probability of the minimum score that will guarantee jim some monetary payoff ? | toguaranteethat jim will get some monetary payoff he must score the maximum score of 6 + 6 = 12 , because if he gets even one less than that so 11 , someone can get 12 and jim will get nothing . p ( 12 ) = 1 / 6 ^ 2 = 1 / 36 . answer : d | a = 21 / 3
b = a - 1
c = b ** 2
d = 1 / c
|
a ) 150 , b ) 280 , c ) 310 , d ) 420 , e ) 300 | e | multiply(multiply(multiply(1, const_10), const_10), const_3) | how many times will the digit 6 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 6 ) 3000 / 10 = 300 times . answer : e ." | a = 1 * 10
b = a * 10
c = b * 3
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | divide(multiply(60, const_2), add(divide(60, 20), const_1)) | an assembly line produces 20 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this whole time ? | "the time to produce the first 60 cogs is 60 / 20 = 3 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / 4 hours = 30 cogs per hour . the answer is b ." | a = 60 * 2
b = 60 / 20
c = b + 1
d = a / c
|
a ) 420 , b ) 484 , c ) 526 , d ) 578 , e ) 624 | e | divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 120 % of 13 / 24 of 960 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 40 = 624 the answer is e ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 120 * n
p = o / 100
|
a ) 45 % , b ) 56 % , c ) 57 % , d ) 70 % , e ) 72 % | c | multiply(divide(11628, add(add(1136, 7636), 11628)), const_100) | 3 candidates in an election and received 1136 , 7636 and 11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 1136 + 7636 + 11628 ) = 20400 so , required percentage = 11628 / 20400 * 100 = 57 % c" | a = 1136 + 7636
b = a + 11628
c = 11628 / b
d = c * 100
|
a ) 22.25 , b ) 31.25 , c ) 35 , d ) 40.5 , e ) none of these | d | add(80, const_1) | the average of first 80 non - zero positive numbers is : | "explanation : sum of first n natural numbers = n ( n + 1 ) / 2 so , average of first n natural numbers = n ( n + 1 ) / 2 n = ( n + 1 ) / 2 required average = ( 80 + 1 ) / 2 = 81 / 2 = 40.5 . answer : d" | a = 80 + 1
|
a ) 90 , b ) 120 , c ) 150 , d ) 210 , e ) 300 | d | divide(multiply(60, subtract(add(4, 3.5), const_1)), const_2) | yesterday it took robert 4 hours to drive from city a to city b . today it took robert 3.5 hours to drive back from city Π² to city a along the same route . if he had saved 15 minutes in both trips , the speed for the round trip would be 60 miles per hour . what is the distance between city a and city b ? | "2 d / 60 = 7 ( because time = 4 + 3.5 - 0.5 hrs ) = > d = 210 answer - d" | a = 4 + 3
b = a - 1
c = 60 * b
d = c / 2
|
a ) s . 147 , b ) s . 248 , c ) s . 244 , d ) s . 229 , e ) s . 216 | e | add(divide(153, subtract(const_1, divide(15, const_100))), multiply(divide(153, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | a shopkeeper loses 15 % , if an article is sold for rs . 153 . what should be the selling price of the article to gain 20 % ? | given that sp = rs . 153 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 153 ) / 85 = rs . 180 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 180 * 120 ) / 100 = rs . 216 answer : e | a = 15 / 100
b = 1 - a
c = 153 / b
d = 15 / 100
e = 1 - d
f = 153 / e
g = 20 / 100
h = f * g
i = c + h
|
a ) 339 , b ) 500 , c ) 288 , d ) 277 , e ) 112 | b | divide(multiply(45, 45), add(2.05, const_2)) | the sum of two numbers is 45 . sum of their quotient and reciprocal is 2.05 , find the product of the numbers . | sol : let a , b be the numbers . a + b = 45 ab + baab + ba = 2.05 β a 2 + b 2 ab β a 2 + b 2 ab = 2.05 β ( a + b ) 2 β 2 abab = 2.05 β ( a + b ) 2 β 2 abab = 2.05 β ( a + b ) 2 β ( a + b ) 2 = 2.05 ab + 2 ab = 4.05 ab β β ab = 4524 . 054524.05 = 500 answer : b | a = 45 * 45
b = 2 + 5
c = a / b
|
a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what q percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people q = 20 . . ans 5 / 20 or 25 % c" | a = 3 + 1
b = 3 + 2
c = a / b
d = 2 * c
e = 3 + 1
f = 3 + 2
g = e / f
h = 3 * g
i = 3 - h
j = d - i
k = 3 + 1
l = j / k
m = l * 100
|
a ) 391 , b ) 435 , c ) 410 , d ) 423 , e ) 445 | b | divide(500, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 500 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 500 j = 500 / 1.15 j = 50,000 / 115 = 435 thus b is the correct answer ." | a = 15 / 100
b = 1 + a
c = 500 / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(subtract(multiply(56, 10), multiply(52, 10)), subtract(60, 52)) | at a certain fruit stand , the price of each apple is 50 cents and the price of each orange is 60 cents . mary selects a total of 10 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 56 cents . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 52 cents ? | "let number of apples = a number of oranges = b a + b = 10 - - - 1 . 56 = ( . 5 a + . 6 b ) / 10 = > 56 = 5 a + 6 b - - - - 2 solving 1 and 2 , we get a = 4 b = 6 let the number of oranges put back = c 52 * ( 10 - c ) = 50 * 4 + 60 ( 6 - c ) = > c = 5 answer e" | a = 56 * 10
b = 52 * 10
c = a - b
d = 60 - 52
e = c / d
|
a ) 22 , b ) 24 , c ) 20 , d ) 25 , e ) 27 | c | divide(28, add(const_1, divide(40, const_100))) | mary can do a piece of work in 28 days . rosy is 40 % more efficient than mary . the number of days taken by rosy to do the same piece of work is ? | "ratio of times taken by mary and rosy = 140 : 100 = 14 : 10 suppose rosy takes x days to do the work . 14 : 10 : : 28 : x = > x = 20 days . hence , rosy takes 20 days to complete the work . answer : c" | a = 40 / 100
b = 1 + a
c = 28 / b
|
a ) 4 , b ) 9 , c ) 12 , d ) 20 , e ) 25 | b | divide(40, subtract(5, divide(const_1, const_2))) | company x provides bottled water to its employees in 5 liter bottles , $ 5 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 40 fixed monthly rate and $ 2 for 20 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 20 l bottles will be x / 4 ( to equate the vol . ) since the total cost will be equal , 5 x = 40 + 2 * x / 4 so x = 8.88 or 9 . answer is ( b ) . | a = 1 / 2
b = 5 - a
c = 40 / b
|
a ) 12 , b ) 16 , c ) 24 , d ) 28 , e ) 32 | c | divide(divide(multiply(800, 15), const_100), 5) | a reduction of 15 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 15 / 100 ) = 120 - - - - 5 ? - - - - 1 = > rs . 24 answer : c" | a = 800 * 15
b = a / 100
c = b / 5
|
a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 12 , e ) 3 / 28 | e | multiply(divide(4, add(add(4, 2), 2)), divide(2, subtract(add(add(4, 2), 2), const_1))) | the membership of a committee consists of 4 english teachers , 2 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee β s report , what is the probability that the two members selected will both be english teachers ? | "probability of first member an english teacher = 3 / 8 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 8 x 2 / 7 = 3 / 28 ( e )" | a = 4 + 2
b = a + 2
c = 4 / b
d = 4 + 2
e = d + 2
f = e - 1
g = 2 / f
h = c * g
|
a ) 40 , b ) 46 , c ) 88 , d ) 77 , e ) 21 | b | divide(360, multiply(subtract(45, 215), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 215 m long ? | ": speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 215 = 575 m required time = 575 * 2 / 25 = 46 sec answer : b" | a = 45 - 215
b = a * const_0_2778
c = 360 / b
|
a ) 3.33 days , b ) 3.99 days , c ) 2.99 days , d ) 2.98 days , e ) 2.44 days | a | inverse(add(inverse(5), inverse(10))) | a and b complete a work in 5 days . a alone can do it in 10 days . if both together can do the work in how many days ? | "1 / 5 + 1 / 10 = 3 / 10 10 / 3 = 3.33 days answer : a" | a = 1/(5)
b = 1/(10)
c = a + b
d = 1/(c)
|
a ) 44 % , b ) 100 % , c ) 105 % , d ) 225 % , e ) 125 % | e | multiply(divide(subtract(divide(3, 2), divide(2, 3)), divide(2, 3)), const_100) | when tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tom is what percent lesser than the quantifying chopped by tammy ? . | "tom chops 4 lbs in 6 minutes tammy chops 9 lbs in 6 minutes so in the same amount of time , tom chops 125 % less than tammy , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , and 9 x is still 125 % greater than 4 x . ans : e" | a = 3 / 2
b = 2 / 3
c = a - b
d = 2 / 3
e = c / d
f = e * 100
|
a ) 4 , b ) 5 , c ) 14 , d ) 12 , e ) none of these | c | multiply(subtract(const_1, divide(9, 15)), 35) | suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 35 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | "the part of job that suresh completes in 9 hours = 9 Γ’ Β β 15 = 3 Γ’ Β β 5 remaining job = 1 - 3 Γ’ Β β 5 = 2 Γ’ Β β 5 remaining job can be done by ashutosh in 2 Γ’ Β β 5 Γ£ β 35 = 14 hours answer c" | a = 9 / 15
b = 1 - a
c = b * 35
|
a ) 12 , b ) 25 , c ) 60 , d ) 30 , e ) 20 | a | divide(multiply(9, 40), subtract(40, 10)) | a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 9 days , a finished the remaining work . in how many days a alone can finish the job ? | "a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 9 days whole work will be done by a in 9 * 4 / 3 = 12 days answer is a" | a = 9 * 40
b = 40 - 10
c = a / b
|
a ) 53 / 10 , b ) 55 / 10 , c ) 56 / 10 , d ) 57 / 10 , e ) 58 / 10 | e | divide(250, multiply(add(64, 92), const_0_2778)) | two trains of length 90 m and 95 m are 250 m apart . they start moving towards each other on parallel tracks , at speeds 64 kmph and 92 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 = ( 64 + 92 ) * 5 / 18 = 43.3 mps . the time required = d / s = 250 / 43.3 = 58 / 10 sec . answer : e" | a = 64 + 92
b = a * const_0_2778
c = 250 / b
|
a ) 456780 , b ) 678910 , c ) 997479 , d ) 997918 , e ) 997920 | d | divide(multiply(const_1000, const_1000), 10) | what is the greatest 6 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 4 , 5 , 6 , 7 , and 8 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 8 when we divide by 10 , so d is the only possible answer . | a = 1000 * 1000
b = a / 10
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a ) 8.8 % , b ) 9 % , c ) 9.2 % , d ) 8.6 % , e ) 10.6 % | e | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(10, 35), const_100)), add(15, 35)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 10 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.12 * 15 + 0.10 * 35 = 5.3 kg , so the percentage is 5.3 / 50 * 100 = 10.6 % . answer : e ." | a = 12 * 15
b = a / 100
c = 10 * 35
d = c / 100
e = b + d
f = 15 + 35
g = e / f
h = g * 100
|
a ) 19 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 72 % | a | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | a baseball card decreased in value 10 % 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 ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.9 = 90 after second year price = 90 * 0.9 = 81 final decrease = [ ( 100 - 81 ) / 100 ] * 100 = 19 % correct answer - a" | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 2 , b ) 3 , c ) 9 , d ) 15 , e ) 30 | b | subtract(divide(subtract(120, 90), subtract(90, 80)), divide(subtract(120, const_100), const_100)) | at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 80 . if the restaurant serves 120 customers today , raising the average to 90 customers per day , what is the value of x ? | "withoutusing the formula , we can see that today the restaurant served 30 customers above the average . the total amount above the average must equal total amount below the average . this additional 30 customers must offset the β deficit β below the average of 90 created on the x days the restaurant served only 80 customers per day . 30 / 10 = 3 days . choice ( a ) . withthe formula , we can set up the following : 90 = ( 80 x + 120 ) / ( x + 1 ) 90 x + 90 = 80 x + 120 10 x = 30 x = 3 answer choice ( b )" | a = 120 - 90
b = 90 - 80
c = a / b
d = 120 - 100
e = d / 100
f = c - e
|
a ) 2287 , b ) 70 , c ) 128 , d ) 797 , e ) 120 | b | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(10, const_100)) | bhanu spends 30 % of his income on petrol on scooter 10 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 10 % ( 700 ) = rs . 70 answer : b" | a = 30 / 100
b = 300 / a
c = b - 300
d = 10 / 100
e = c * d
|
a ) 20 , b ) 150 , c ) 384 , d ) 300 , e ) 320 | c | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | how many 4 digit numbers are there , if it is known that the first digit is odd , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 1 can be used only once ? | "4 options for the first digit : 1 , 3 , 5 , 7 , 9 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 5 * 5 * 4 * 4 = 400 numbers with one 1 - s , 1 x 1 x 1 * 1 * 4 * 4 = 16 . thus there are 400 - 16 = 384 such numbers . answer : c ." | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 * 4
e = c * d
f = 4 + 1
g = 4 + 1
h = f * g
i = h * 4
j = e - i
|
a ) 5 : 1 , b ) 10 : 1 , c ) 25 : 1 , d ) 50 : 1 , e ) 150 : 1 | c | multiply(5, 5) | if the length of the sides of two cubes are in the ratio 5 : 1 , what is the ratio of their total surface area ? | "let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 5 x ) ^ 2 = 150 x ^ 2 . the ratio of surface areas is 25 : 1 . the answer is c ." | a = 5 * 5
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a ) rs . 11,300 , b ) rs . 11,500 , c ) rs . 11,302 , d ) rs . 11,301 , e ) rs . 11,303 | b | subtract(multiply(add(1500, 400), add(24, const_1)), multiply(1500, 24)) | the average monthly salary of 24 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 400 . what is the manager ' s monthly salary ? | explanation : manager ' s monthly salary = rs . ( 1900 x 25 - 1500 x 24 ) = rs . 11,500 answer : b | a = 1500 + 400
b = 24 + 1
c = a * b
d = 1500 * 24
e = c - d
|
a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 64 , e ) 1 : 2 | d | divide(power(1, 4), power(4, 4)) | the triplicate ratio of 1 : 4 is ? | "1 ^ 3 : 4 ^ 3 = 1 : 64 answer : d" | a = 1 ** 4
b = 4 ** 4
c = a / b
|
a ) 330000 , b ) 340000 , c ) 347000 , d ) 333200 , e ) 357000 | d | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(70, const_100)) | in an election , candidate a got 70 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 70 % therefore , the number of valid votes polled in favour of candidate a = 70 % of 476000 = 70 / 100 Γ 476000 = 33320000 / 100 = 333200 d )" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 70 / 100
e = c * d
|
a ) 122.9 m , b ) 127.5 m . , c ) 122.2 m , d ) 85 m , e ) 12289 m | d | subtract(400, divide(multiply(subtract(400, 100), subtract(800, const_100.0)), 800)) | in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 400 m ? | "when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 400 m , the number of meters by which a beats c = ( 400 * 212.5 ) / 1000 = 85 m . answer : d" | a = 400 - 100
b = 800 - 100
c = a * b
d = c / 800
e = 400 - d
|
a ) 18 , b ) 12 , c ) 11 , d ) 19 , e ) 114 | a | multiply(add(add(const_1, const_2), const_3), 3) | a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 3 days . b can do the work alone in ? | "suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 3 6 / x = 1 / 3 = > x = 18 so , b takes 9 hours to finish the work . answer : a" | a = 1 + 2
b = a + 3
c = b * 3
|
a ) 11 / 12 , b ) 13 / 17 , c ) 22 / 30 , d ) 12 / 19 , e ) 11 / 19 | c | divide(add(multiply(40, const_100), multiply(40, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 40))) | paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 60 % of his income to paul and 40 % of his income to quentin , quentin ' s new income would be what fraction of paul ' s new income ? | "make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get p = 12 and q 8.8 so 8.8 / 12 = = 2.2 / 3 ans : c" | a = 40 * 100
b = 100 - 20
c = 40 * b
d = a + c
e = 40 * 100
f = 40 + 20
g = f * 40
h = e + g
i = d / h
|
['a ) 1 : 4', 'b ) 2 : 4', 'c ) 3 : 4', 'd ) 4 : 4', 'e ) 5 : 4'] | a | divide(multiply(const_1, const_1), power(multiply(const_2, const_1), const_2)) | if the length of the side of a square is doubled , what is the ratio of the areas of the original square to the area of the new square ? | solution if x be the side of the original square , then its area is equal to x 2 if x is doubled to 2 x , then the new area is equal to ( 2 x ) 2 = 4 x 2 the ratio of the areas of the original square to the area of the new square x 2 / ( 4 x 2 ) = 1 / 4 or 1 : 4 answer is a | a = 1 * 1
b = 2 * 1
c = b ** 2
d = a / c
|
a ) $ 58.40 , b ) $ 40 . 00 , c ) $ 24.60 , d ) $ 18.40 , e ) $ 16.00 | d | multiply(subtract(730, 500), divide(8, const_100)) | in country x a returning tourist may import goods with a total value of $ 500 or less tax free , but must pay an 8 percent tax on the portion of the total value in excess of $ 500 . what tax must be paid by a returning tourist who imports goods with a total value of $ 730 ? | the tourist must pay tax on $ 730 - $ 500 = $ 230 . thus the amount to be paid is 0.08 * $ 230 = $ 18.40 . answer : d . | a = 730 - 500
b = 8 / 100
c = a * b
|
a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 18 | c | add(add(divide(3, 4), multiply(divide(3, 4), 15)), multiply(const_0_25, 15)) | total 45 cows 15 cow gives each 2 liter milk 15 cow gives each 3 / 4 liter milk 15 cow gives each 1 / 4 liter milk this is split into 3 son per each 15 cows & 15 liter milk how ? | "15 cow 2 liter each = 30 liter 15 cow 3 / 4 liter each = 3 / 4 = 0.75 * 15 = 11.25 15 cow 1 / 4 liter each = 1 / 4 = 0.25 * 15 = 3.75 add 30 + 11.25 + 3.75 = 45 milk split into 3 son each 15 liter then 45 / 3 = 15 answer : c" | a = 3 / 4
b = 3 / 4
c = b * 15
d = a + c
e = const_0_25 * 15
f = d + e
|
a ) 6 , b ) 9 , c ) 12 , d ) 15 , e ) 18 | b | add(add(add(add(add(multiply(multiply(2, 2), 2), multiply(multiply(2, 2), 2)), multiply(multiply(2, 2), 2)), 2), const_4), const_4) | if both 2 ^ 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 = 9 the answer is b ." | a = 2 * 2
b = a * 2
c = 2 * 2
d = c * 2
e = b + d
f = 2 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(multiply(4, divide(3, 4)), 12) | an outlet pipe can empty 3 / 4 of a cistern in 12 minutes . in 4 minutes , what part of the cistern will be emptied ? | "4 / 12 * 3 / 4 = 1 / 4 the answer is c ." | a = 3 / 4
b = 4 * a
c = b / 12
|
a ) 1 , b ) 9 / 64 , c ) 1 / 8 , d ) 3 / 16 , e ) 9 / 32 | b | subtract(3, multiply(divide(factorial(6), factorial(3)), power(divide(3, 3), 6))) | a couple decides to have 6 children . if they succeed in having 6 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 3 girls and 3 boys ? | "sample space = 2 ^ 6 = 64 . favourable events = { bbbggg } , { bgbgbg } , { bgbggb } , { bggbbg } { gggbbb } , { gbgbgb } , { gbbggb } . { gbgbbg } { gbbgbg } probability = 9 / 64 = 9 / 64 ans ( b ) ." | a = math.factorial(6)
b = math.factorial(3)
c = a / b
d = 3 / 3
e = d ** 6
f = c * e
g = 3 - f
|
a ) 15 seconds , b ) 12.1 seconds , c ) 10 seconds , d ) 8.1 seconds , e ) none of these | b | divide(add(110, 132), multiply(72, const_0_2778)) | how long does a train 110 meters long running at the speed of 72 km / hour take to cross a bridge 132 meters in length ? | "explanation : speed = 72 km / hour = 72 * ( 5 / 18 ) m / sec = 20 m / sec total distance to be covered = 110 + 132 = 142 meters time = distance / speed = 242 / 20 = 12.1 seconds option b" | a = 110 + 132
b = 72 * const_0_2778
c = a / b
|
a ) rs . 2000 , b ) rs . 3000 , c ) rs . 4000 , d ) rs . 4500 , e ) rs . 5000 | c | subtract(add(add(5050, 6250), 5200), multiply(6250, const_2)) | the averageincome of m and n is rs . 5050 . the average monthly income of n and o is rs . 6250 and the average monthly income of m and ois rs . 5200 . the income of m is ? | let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 5050 x 2 ) = 10100 . . . . ( i ) q + r = ( 6250 x 2 ) = 12500 . . . . ( ii ) m + o = ( 5200 x 2 ) = 10400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( m + n + o ) = 33000 or m + n + o = 16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) , we get m = 4000 . m ' s monthly income = rs . 4000 c | a = 5050 + 6250
b = a + 5200
c = 6250 * 2
d = b - c
|
a ) 330 % , b ) 390 % , c ) 200 % , d ) 250 % , e ) 300 % | e | divide(multiply(const_3.0, const_100), 4) | a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 1 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 1 he selling at 3 a rupee , his sp = 4 profit = sp - cp = 4 - 1 = 3 profit percent = 3 / 1 * 100 = 300 % answer : e" | a = 3 * 0
b = a / 4
|
a ) 21 Β½ days , b ) 22 Β½ days , c ) 23 Β½ days , d ) 12 Β½ days , e ) 7 Β½ days | e | add(divide(const_1, 5), divide(const_1, 15)) | a can do a job in 5 days and b can do it in 15 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "explanation : 1 / 5 + 1 / 15 = 4 / 15 = 4 / 15 15 / 4 = 15 / 4 * 2 = 7 Β½ days answer : e" | a = 1 / 5
b = 1 / 15
c = a + b
|
a ) 1.25 % , b ) 5 % , c ) 6.25 % , d ) 6.67 % , e ) 11.7 % | b | multiply(divide(multiply(multiply(const_100, const_100), divide(3, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 3 percent sodium chloride by volume . if 4,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the remaining solution will be approximately what percent sodium chloride ? means : what percent of the remaining solution is sodium chloride . now , since the remaining solution is 10,000 - 4,000 = 6,000 gallons and sodium chloride is 300 gallons ( 3 % of initial solution of 10,000 gallons ) then sodium chloride is 300 / 6,000 * 100 = 5 % of the remaining solution of 6,000 gallons . answer : b ." | a = 100 * 100
b = 3 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 1 / 2 , b ) 7 / 12 , c ) 3 / 7 , d ) 5 / 12 , e ) 6 / 17 | c | divide(subtract(8, multiply(const_2, const_3)), 8) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 8 | "total number of cases = 7 * 7 = 49 favourable cases = [ ( 2,7 ) , ( 3,6 ) , ( 3,7 ) , ( 4,5 ) , ( 4,6 ) , ( 4,7 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 5,7 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) , ( 6,7 ) , ( 7,2 ) , ( 7,3 ) , ( 7,4 ) , ( 7,5 ) , ( 7,6 ) , ( 7,7 ) ] = 21 so probability = 21 / 49 = 3 / 7 answer is c" | a = 2 * 3
b = 8 - a
c = b / 8
|
a ) 500 % , b ) 250 % , c ) 500 / 3 % , d ) 125 % , e ) 60 % | e | multiply(add(const_1, divide(20, const_100)), subtract(const_100, 50)) | if x is 20 percent more than y and y is 50 percent less than z , then x is what percent of z ? | "let ' s take values of x , y , z which can satisfy given equations . let z = 200 , so y becomes 100 ( since y is 50 % less than z ) , x becomes 120 ( since 20 % more than y ) now req is ( x / z ) * 100 = ( 120 / 200 ) * 100 = 60 % answer : e" | a = 20 / 100
b = 1 + a
c = 100 - 50
d = b * c
|
a ) rs . 1000 , b ) rs . 850 , c ) rs . 500 , d ) rs . 1200 , e ) rs . 900 | a | subtract(1072, divide(multiply(subtract(1144, 1072), 2), 4)) | a sum of money at simple interest amounts to rs . 1072 in 2 years and to rs . 1144 in 4 years . the sum is : | "s . i . for 2 years = rs . ( 1144 - 1072 ) = rs . 72 . s . i . for 1 year = rs . 72 / 2 = rs . 36 . principal = rs . ( 1072 - 72 ) = rs . 1000 . answer : option a" | a = 1144 - 1072
b = a * 2
c = b / 4
d = 1072 - c
|
a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 80.00 | d | multiply(subtract(10, divide(multiply(12, 8), const_100)), 10) | an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 12 % . also , if a customer will β buy at least 10 β items they will deduct an additional 8 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 20 % , thus he must pay slightly more than $ 80 . only answer choice d fits . answer : d ." | a = 12 * 8
b = a / 100
c = 10 - b
d = c * 10
|
a ) 5.2 days , b ) 9.2 days , c ) 8.2 days , d ) 6.2 days , e ) 7.2 days | e | inverse(add(inverse(add(add(add(add(const_2, const_3), 15), add(add(const_2, const_3), 15)), add(const_2, const_3))), add(inverse(15), inverse(add(add(const_2, const_3), 15))))) | a , b and c can do a work in 15 , 20,45 days respectively . in how many days they can complete the work together . | lcm = 180 no of days = [ 180 / ( 180 / 15 + 180 / 20 + 180 / 45 ) = [ 180 / ( 12 + 9 + 4 ) ] = [ 180 / 25 ] = 7.2 days answer : e | a = 2 + 3
b = a + 15
c = 2 + 3
d = c + 15
e = b + d
f = 2 + 3
g = e + f
h = 1/(g)
i = 1/(15)
j = 2 + 3
k = j + 15
l = 1/(k)
m = i + l
n = h + m
o = 1/(n)
|
a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) none | b | divide(subtract(1365, 15), subtract(6, const_1)) | the difference between two numbers is 1365 . when the larger number is divided by the smaller one , the quotient is 6 and the remainder is 15 . the smaller number is : | "solution let the numbers be x and ( x + 1365 ) . then , x + 1365 = 6 x + 15 β 5 x = 1350 β x = 270 . answer b" | a = 1365 - 15
b = 6 - 1
c = a / b
|
a ) 13.6 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 61.9 sec , e ) 47.98 sec | a | divide(add(100, 145), multiply(65, const_0_2778)) | how long does a train 100 m long running at the speed of 65 km / hr takes to cross a bridge 145 m length ? | "speed = 65 * 5 / 18 = 18 m / sec total distance covered = 100 + 145 = 245 m . required time = 245 / 18 = 13.6 sec . answer : a" | a = 100 + 145
b = 65 * const_0_2778
c = a / b
|
a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23 | c | add(10, 11) | lilly has 10 fish and rosy has 11 fish . in total , how many fish do they have in all ? | "10 + 11 = 21 the answer is c ." | a = 10 + 11
|
a ) 1.5 % , b ) 0.4 % , c ) 2.5 % , d ) 3.5 % , e ) 4 % | b | multiply(divide(4, 1), const_100) | what percent is 4 gm of 1 kg ? | "1 kg = 1000 gm 4 / 1000 Γ 100 = 400 / 1000 = 2 / 5 = 0.4 % b )" | a = 4 / 1
b = a * 100
|
['a ) 40 β cm Β³', 'b ) 60 β cm Β³', 'c ) 136 β cm Β³', 'd ) 138 β cm Β³', 'e ) none'] | c | multiply(const_pi, multiply(divide(16, const_2), sqrt(add(power(15, const_2), power(divide(16, const_2), const_2))))) | the curved surface of a right circular cone of height 15 cm and base diameter 16 cm is | sol . h = 14 cm , r = 7 cm . so , l = β r Β² + h Β² = β 8 Β² + ( 15 ) Β² = 17 cm . β΄ curved surface area = β rl = ( β * 8 * 17 ) cm Β² = 136 β cm Β² answer c | a = 16 / 2
b = 15 ** 2
c = 16 / 2
d = c ** 2
e = b + d
f = math.sqrt(e)
g = a * f
h = math.pi * g
|
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 4 | d | multiply(subtract(4, 3), divide(460, add(multiply(4, 11), multiply(3, 16)))) | we run a business that rents out canoes and kayaks . a canoe rental costs $ 11 per day , and a kayak rental costs $ 16 dollars per day . one day , our business rents out 4 canoes for every 3 kayaks and receives a total of $ 460 in revenue . how many more canoes than kayaks were rented out ? | "let x be the number of canoes . then 3 x / 4 is the number of kayaks . 11 x + ( 3 x / 4 ) * 16 = 460 11 x + 12 x = 460 23 x = 460 x = 20 ( canoes ) 3 x / 4 = 15 ( kayaks ) there were 20 - 15 = 5 more canoes rented out . the answer is d ." | a = 4 - 3
b = 4 * 11
c = 3 * 16
d = b + c
e = 460 / d
f = a * e
|
a ) 12 , b ) 13 , c ) 16.6 , d ) 17 , e ) 23 | a | subtract(subtract(25, const_2), add(10, const_1)) | set w consists of all the prime numbers between 10 and 25 . what is the range of set w ? | answer of set w is 12 . | a = 25 - 2
b = 10 + 1
c = a - b
|
a ) 238 , b ) 360 , c ) 377 , d ) 277 , e ) 291 | b | add(multiply(multiply(add(15, 15), const_3), const_3), multiply(add(15, 15), const_3)) | at 15 : 15 pm railway time , what will be the angle between minute and hour hand ? | at 15 : 15 both minute and hour hands will be in same position . hence the angle between them will be zero degrees . however 360 degree also means both minute and hour hands are in same position . hence correct answers are both zero and 360 degrees answer : b | a = 15 + 15
b = a * 3
c = b * 3
d = 15 + 15
e = d * 3
f = c + e
|
a ) 1 : 6 , b ) 6 : 1 , c ) 1 : 5 , d ) 5 : 1 , e ) 1 : 4 | a | divide(2000, 12000) | x and y start a business with rs . 2000 and rs . 12000 respectively . hoe should they share their profits at the end of one year ? | "they should share the profits in the ratio of their investments . the ratio of the investments made by x and y = 2000 : 12000 = > 1 : 6 . answer : a" | a = 2000 / 12000
|
a ) 150 , b ) 145 , c ) 165 , d ) 162 , e ) 152 | d | subtract(multiply(20, multiply(70, const_0_2778)), 130) | a train 130 m long crosses a platform with a speed of 70 kmph in 20 sec ; find the length of the platform ? | "d = 130 + x t = 20 s = 130 + x / 15 * 18 / 5 = 70 kmph . = > ( 130 + x ) * 18 = 5250 = > 130 + x = 292 = > x = 162 m answer : d" | a = 70 * const_0_2778
b = 20 * a
c = b - 130
|
['a ) 50', 'b ) 60', 'c ) 70', 'd ) 80', 'e ) 90'] | b | divide(288, add(add(divide(7, 5), divide(7, 5)), const_2)) | the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 288 meters , what is the width of the field ? | let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 288 , 2 ( 7 / 5 ) w + 2 w = 288 solve the above equation to find : w = 60 m and l = 84 m . correct answer b ) 60 | a = 7 / 5
b = 7 / 5
c = a + b
d = c + 2
e = 288 / d
|
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