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 ) 66 , b ) 26 , c ) 42 , d ) 27 , e ) 14 | e | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 32), const_2) | 32 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 32 = 14 answer : e" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 32
f = e / 2
|
a ) 33 , b ) 77 , c ) 18 , d ) 99 , e ) 66 | c | divide(multiply(9, 6), subtract(9, 6)) | pipe a can fill a tank in 6 hours . due to a leak at the bottom , it takes 9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 6 - 1 / x = 1 / 9 = > 1 / x = 1 / 6 - 1 / 9 = ( 3 - 2 ) / 18 = 1 / 18 = > x = 18 . answer : c" | a = 9 * 6
b = 9 - 6
c = a / b
|
a ) 15 / 2 , b ) - 2 , c ) 1 , d ) 2 , e ) 3 | a | divide(multiply(3, 3), 2) | if x / y = 3 and ( 2 a - x ) / ( 5 b - y ) = 3 , then the value of a / b is ? | "x = 3 y 2 a - 3 y = 3 ( 5 b - y ) 2 a - 3 y = 15 b - 3 y 2 a = 15 b a / b = 15 / 2 answer : a" | a = 3 * 3
b = a / 2
|
a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 70 | e | multiply(7, const_4) | what could be the range of a set consisting of odd multiples of 7 ? | "a set of odd multiples of 7 would be something like 7 , 21 , 35 , 49 . . . . the range would always be an even number multiple of 7 . answer : e" | a = 7 * 4
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a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 5 , d ) 6 : 5 , e ) none | d | divide(multiply(48, const_2), multiply(110, const_3)) | a man invests some money partly in 4 % stock at 48 and partly in 10 % stock at 110 . to obtain equal dividends from both , he must invest the money in the ratio : | "solution for an income of rs . 1 in 9 % stock at 96 , investment = rs . ( 48 / 4 ) = rs . 12 . for an income of rs . 1 in 12 % stock at 120 , investment = rs . ( 110 / 10 ) = rs . 10 . β΄ ratio of investments = 12 : 10 = 6 : 5 answer d" | a = 48 * 2
b = 110 * 3
c = a / b
|
a ) 2 , b ) 1 , c ) 3 , d ) 5 , e ) 8 | e | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1096) | what least number must be added to 1096 , so that the sum is completely divisible by 23 ? | "1096 / 23 = 47.6 - - that is 23 * 47 = 1081 now , 1096 - 1081 = 15 to get completely divided by 23 add 8 to 1096 ans - e" | a = 4 * 10
b = a + 2
c = b * 23
d = c - 1096
|
a ) $ 60 , b ) $ 90 , c ) $ 120 , d ) $ 150 , e ) $ 180 | e | divide(45, subtract(divide(3, 2), divide(5, 4))) | the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 5 : 4 . if the oil bill for february had been $ 45 more , the corresponding ratio would have been 3 : 2 . how much was the oil bill for january ? | 5 : 4 = 15 : 12 and 3 : 2 = 18 : 12 . an increase in $ 45 increases the ratio by 3 : 12 . therefore an increase in $ 15 increases the ratio by 1 : 12 . therefore , january ' s bill was 12 ( $ 15 ) = $ 180 . the answer is e . | a = 3 / 2
b = 5 / 4
c = a - b
d = 45 / c
|
a ) 112 , b ) 133 , c ) 150 , d ) 167 , e ) 225 | e | multiply(50, divide(450, add(50, 50))) | trains a and b start simultaneously from stations 450 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 50 miles per hour and 50 miles per hour , respectively , how many miles will train a have traveled when the trains pass each other , to the nearest mile ? | "since we know the distance ( 450 ) and the combined rate ( 100 ) , we plug it into the formula : distance = rate * time 450 = 100 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 450 / 100 from dividing 100 on both sides to isolate time in the equation above . time will be 4.5 hours so now you can plug that in for train a β s distance . distance = rate * time distance = 50 * 4.5 distance = 225 according to answer choice e ." | a = 50 + 50
b = 450 / a
c = 50 * b
|
a ) 1 : 2 , b ) 1 : 4 , c ) 3 : 7 , d ) 7 : 12 , e ) 4 : 5 | b | divide(subtract(64, 60), subtract(80, 64)) | a theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 80 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ? | "deviation from the mean for the daytime staff = 80 - 64 = 16 . deviation from the mean for the evening staff = 64 - 60 = 4 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 4 : 16 = 1 : 4 . the answer is b ." | a = 64 - 60
b = 80 - 64
c = a / b
|
a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | c | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(12, 3)), 12)), subtract(12, 9)), divide(const_1, multiply(12, 3))) | 12 welders work at a constant rate they complete an order in 3 days . if after the first day , 9 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | 1 . we need to find out the time taken by 3 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 12 * 3 = 36 days 4 . but on day 1 twelve workers had already worked finishing 1 / 3 of the job . so 3 workers have to finish only 2 / 3 of the job . 5 . total time taken by 3 workers can be got from formula used at ( 2 ) . i . e . , 3 * total time taken = 36 . total time taken by 3 workers to finish the complete job is 36 / 3 = 12 days . 6 . time taken by 6 workers to finish 2 / 3 of the job is 2 / 3 * 12 = 8 days . the answer is choice c | a = 12 * 3
b = 1 / a
c = b * 12
d = 1 - c
e = 12 - 9
f = d / e
g = 12 * 3
h = 1 / g
i = f / h
|
a ) 7 , b ) 5 , c ) 9 , d ) 6 , e ) 4 | c | floor(divide(surface_rectangular_prism(const_3.0, 12, 2), multiply(12, 15))) | a rectangular cube has sides measuring 15 inches long by 12 inches wide by 2 inches high . if the surface area of the rectangle is the same as a cube , what do the sides / walls of the cube measure ? round to the nearest whole number . | first calculate the surface area of the rectangle by multiplying the length and width of the rectangle together , then multiply by 2 to get both sides of the rectangle . this calculates to 360 inches . find the surface area of the sides of the rectangular cube . multiply the height by the length of the rectangle . multiply the answer by 2 ( for the 2 sides on the rectangular cube ) . this calculates to 60 inches . then multiply the width of the rectangle by the height . multiply the answer by 2 ( for the 2 sides on the rectangular cube ) . this calculates to 48 inches . add the totals together to get the rectangular cube surface area of 468 inches . since a square cube has 6 sides - divide the surface area by 6 . this calculates to 78 inches as the surface area of each side of the cube . the square root of 78 is 8.831 . rounding to the nearest whole number that makes each side of the cube measuring 9 inches . the correct answer is ( c ) . | a = surface_rectangular_prism / (
b = 12 * 15
c = math.floor(a, b)
|
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9 | c | add(add(const_4, const_3), const_2) | what is the units digit of the solution to 177 ^ 28 β 133 ^ 23 ? | "let ' s reduce the clutter and make it easy to solve so , 7 ^ 28 - 3 ^ 23 will have the same units digit as the big numbers above both 7 and 3 have a cyclicity of 4 , i . e . their powers repeat the units digit after every 4 th power so , 7 ^ 28 has the same units digit as 7 ^ 4 , which is 1 similarly , 3 ^ 23 has the same units digit as 3 ^ 3 , which is 7 now , we get xx . . . xx 1 - xx . . . . xx 7 = xx . . . xx 4 thus , the units digit of the solution to 177 ^ 28 - 133 ^ 23 is 4 correct option : c" | a = 4 + 3
b = a + 2
|
a ) 1000 , b ) 2876 , c ) 1077 , d ) 2778 , e ) 2711 | c | divide(multiply(140, const_100), subtract(add(const_100, 3), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 3 % . what is the cost price ? | "90 % 103 % - - - - - - - - 13 % - - - - 140 100 % - - - - ? = > rs . 1077 answer : c" | a = 140 * 100
b = 100 + 3
c = 100 - 10
d = b - c
e = a / d
|
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 3 / 16 , e ) 3 / 5 | d | divide(const_2, 6) | if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 6 children have two children of the same sex and one of the opposite sex ? | "no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 6 = 6 c 2 total possible outcomes - 2 ^ 6 ( each child can be either a girl or a boy ) probability = 2 c 1 * 6 c 2 / 2 ^ 6 = 2 * 6 / 2 * 2 * 2 * 2 * 2 * 2 = 12 / 64 = 3 / 16 ans = d" | a = 2 / 6
|
a ) 0.625 , b ) 0.0625 , c ) 6.25 , d ) 0.00625 , e ) none | d | divide(multiply(25, 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 25 % of 25 % equal to | "solution 25 % of 25 % = 25 / 100 x 25 / 100 = 1 / 16 = 0.0625 answer d" | 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 = 25 * n
p = o / 100
|
a ) 3 , b ) 2 , c ) - 1 , d ) - 2 , e ) - 3 | e | subtract(multiply(21, divide(subtract(multiply(2, 3), 6), subtract(multiply(11, 3), multiply(2, 12)))), add(multiply(12, divide(subtract(multiply(2, 3), 6), subtract(multiply(11, 3), multiply(2, 12)))), 3)) | when positive integer x is divided by 12 , the quotient is y and the remainder is 3 . when 2 x is divided by 11 , the quotient is 3 y and the remainder is 6 . what is the value of 21 y β x ? | "( 1 ) x = 12 y + 3 ( 2 ) 2 x = 33 y + 6 ( 2 ) - ( 1 ) : x = 21 y + 3 21 y - x = - 3 the answer is e ." | a = 2 * 3
b = a - 6
c = 11 * 3
d = 2 * 12
e = c - d
f = b / e
g = 21 * f
h = 2 * 3
i = h - 6
j = 11 * 3
k = 2 * 12
l = j - k
m = i / l
n = 12 * m
o = n + 3
p = g - o
|
a ) 26 , b ) 452 , c ) 450 , d ) 440 , e ) 28 | b | multiply(multiply(const_pi, 8), 18) | the slant height of a cone is 18 cm and radius of the base is 8 cm , find the curved surface of the cone ? | "Ο * 18 * 8 = 452 answer : b" | a = math.pi * 8
b = a * 18
|
a ) 10 , b ) 30 , c ) 20 , d ) 40 , e ) 50 | d | add(add(add(const_3, const_2), divide(500, 20)), const_10) | a car has a 20 - gallon fuel to travel 500 miles . at this rate , how many gallons of fuel are needed for a 1000 mile car ? | number of gallons of fuel per mile = 500 / 20 = 25 gallons per mile number of gallons of fuel for a 1000 mile car = 1000 / 25 = 40 answer : d | a = 3 + 2
b = 500 / 20
c = a + b
d = c + 10
|
a ) 40 , b ) 50 , c ) 30 , d ) 60 , e ) 45 | c | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | nicky and cristina are running a 300 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . c" | a = 3 * 12
b = a / 3
c = 3 * 12
d = 5 - 3
e = c / d
f = b + e
|
a ) 216 cm 2 , b ) 238 cm 2 , c ) 240 cm 2 , d ) 247 cm 2 , e ) 250 cm 2 | d | divide(multiply(13, add(20, 18)), const_2) | nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 13 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 13 ) = 247 cm 2 answer : d" | a = 20 + 18
b = 13 * a
c = b / 2
|
a ) 1218 , b ) 2777 , c ) 1000 , d ) 2688 , e ) 1991 | c | divide(760, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 760 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 760 cp = 10 * 100 = > cp = 1000 answer : c" | a = 24 / 100
b = 1 - a
c = 760 / b
|
a ) $ 1000 , b ) $ 5250 , c ) $ 2500 , d ) $ 4500 , e ) $ 1200 | b | subtract(7000, divide(7000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | the salaries of a and b together amount to $ 7000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 7000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 7000 - x ) x = $ 5250 answer is b" | a = 100 - 85
b = 100 - 95
c = a / b
d = c + 1
e = 7000 / d
f = 7000 - e
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 8 | e | subtract(multiply(add(floor(divide(1056, 28)), const_1), 28), 1056) | what is the least number should be added to 1056 , so the sum of the number is completely divisible by 28 ? | "( 1056 / 28 ) gives remainder 20 20 + 8 = 28 , so we need to add 8 e" | a = 1056 / 28
b = math.floor(a)
c = b + 1
d = c * 28
e = d - 1056
|
a ) 2.51 % , b ) 4.51 % , c ) 3.51 % , d ) 6.51 % , e ) 7.51 % | a | subtract(multiply(multiply(add(const_100, 34), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | shopkeeper rise price by 34 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ? | let d initial price be 100 34 % rise now price = 134 / 100 * 100 = 134 10 % discount then price = 134 * 90 / 100 = 120.6 15 % discount then price = 120.6 * 85 / 100 = 102.51 so gain = 102.51 - 100 = 2.51 gain % = gain * 100 / cp = = > 2.51 * 100 / 100 = 2.51 % answer : a | a = 100 + 34
b = 100 - 10
c = b / 100
d = a * c
e = 100 - 15
f = e / 100
g = d * f
h = g - 100
|
['a ) 2', 'b ) 1', 'c ) 3', 'd ) 5', 'e ) 6'] | a | divide(divide(divide(divide(divide(1152, const_3), const_3), const_4), const_4), const_4) | find the smallest number which should be multiplied with 1152 to make it a perfect square . | 1152 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 required smallest number = 2 2 is the smallest number which should be multiplied with 1152 to make it a perfect square . answer : a | a = 1152 / 3
b = a / 3
c = b / 4
d = c / 4
e = d / 4
|
a ) 91.5 , b ) 92 , c ) 93 , d ) 94 , e ) 58.5 | e | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 45 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 45 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 58.5 cm answer : option e" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 82 , b ) 27 , c ) 12 , d ) 82 , e ) 18 | b | subtract(const_60, multiply(const_60, divide(30, 55))) | excluding stoppages , the speed of a train is 55 kmph and including stoppages it is 30 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 25 / 55 * 60 = 27 answer : option b" | a = 30 / 55
b = const_60 * a
c = const_60 - b
|
a ) 10 , b ) 20 , c ) 28 , d ) 30 , e ) 15 | c | divide(add(add(16, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 16 and 34 which are divisible by 5 ? | "average = ( 20 + 25 + 30 ) 3 = 85 / 3 = 28 answer is c" | a = 16 + 4
b = 34 - 4
c = a + b
d = c / 2
|
a ) 6.24 km , b ) 6 km , c ) 9.86 km , d ) 5.66 km , e ) 10 km | c | multiply(divide(multiply(add(10, 1.2), subtract(10, 1.2)), add(add(10, 1.2), subtract(10, 1.2))), const_2) | a man can row 10 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 = 10 s = 1.2 ds = 11.2 us = 8.8 x / 11.2 + x / 8.8 = 1 x = 4.93 d = 4.93 * 2 = 9.86 answer : c | a = 10 + 1
b = 10 - 1
c = a * b
d = 10 + 1
e = 10 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | floor(divide(reminder(power(36, reminder(10, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 36 ^ 10 ? | "36 ^ 10 = 6 ^ 20 if you type powers of six they end in 6 36 16 96 76 56 the pattern is 3 - - > 1 - - > 9 - - > 7 - - > 5 so for 36 we start with 3 - - > 9 - - > 5 - - > 1 - - > 7 and repeat . 36 ^ 10 will come at 7 . answer d ." | a = 4 + 1
b = 36 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
a ) 8 / 5 , b ) 5 / 8 , c ) 4 , d ) 5 , e ) it can not be determined | e | divide(multiply(8, 5), multiply(5, 8)) | the ratio between x and y is 8 / 5 ; x and y are decreased by 5 , what is the ratio between the new values of x and y ? | "ratio = 8 k / 5 k = 8 / 5 , 16 / 10 , etc . x and y are decreased by 5 - - > ( 8 k - 5 ) / ( 5 k - 5 ) new ratio can be 3 / 0 , 11 / 5 , etc . answer : e" | a = 8 * 5
b = 5 * 8
c = a / b
|
a ) β 108 , b ) β 44 , c ) 10 , d ) - 18 , e ) 18 | d | divide(add(add(add(multiply(multiply(20, const_2), 2), multiply(20, const_2)), 20), 4), multiply(4, const_2)) | # p is defined as 2 p + 20 for any number p . what is p , if # ( # ( # p ) ) = - 4 ? | # p = 2 p + 20 - - - > # ( # p ) = 2 ( 2 p + 20 ) + 20 = 4 p + 60 and thus # ( 4 p + 60 ) = 2 ( 4 p + 60 ) + 20 = 8 p + 140 = - 4 - - - > 8 p = - 144 - - - > p = - 18 , d is the correct answer . | a = 20 * 2
b = a * 2
c = 20 * 2
d = b + c
e = d + 20
f = e + 4
g = 4 * 2
h = f / g
|
a ) 32 , b ) 41 , c ) 45 , d ) 48 , e ) 50 | b | subtract(multiply(30, 11), subtract(multiply(30, 10), 11)) | the average student age of a certain class which has 30 students is 10 . if one student aged 11 go to another class and the age of the class teacher is included the average changes to 11 . what is the age of class teacher ? | "sum of ages of class before replacement = 30 x 10 = 300 sum of ages of class without teacher = 300 - 11 = 289 sum of ages of class after replacement = 30 x 11 = 330 age of teacher = 330 - 289 = 41 answer : b" | a = 30 * 11
b = 30 * 10
c = b - 11
d = a - c
|
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | c | add(const_2, const_2) | 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 = 2 + 2
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a ) 2 : 12 , b ) 4 : 25 , c ) 5 : 31 , d ) 2 : 19 , e ) 3 : 17 | b | divide(const_4, power(5, 2)) | the diagonals of two squares are in the ratio of 2 : 5 then find the ratio of its areas ? | "let the diagonals of the squares be 2 x and 5 x ratio of their areas = 1 / 2 ( 2 x ) ^ 2 : 1 / 2 ( 5 x ) ^ 2 = 4 x ^ 2 : 25 x ^ 2 = 4 : 25 answer is b" | a = 5 ** 2
b = 4 / a
|
a ) 1 , b ) 10 , c ) 30 , d ) 20 , e ) 21 | c | subtract(multiply(divide(60, 2), add(divide(60, 2), 1)), multiply(divide(add(59, 1), 2), add(divide(subtract(59, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 60 , inclusive , andbequals the sum of the odd integers from 1 to 59 , inclusive , what is the value of a - b ? | "this is a solution from beatthegmat : even numbers : ( 60 - 2 ) / 2 + 1 = 30 even integers . ( 60 + 2 ) / 2 = 31 is the average of the even set . sum = avg * ( # of elements ) = 31 * 30 = 930 = a odd numbers : ( 59 - 1 ) / 2 + 1 = 30 odd integers . ( 59 + 1 ) / 2 = 30 is the average of the odd set . sum = avg * ( # of elements ) = 30 * 30 = 900 = b a - b = 930 - 900 = 30 . answer : c" | a = 60 / 2
b = 60 / 2
c = b + 1
d = a * c
e = 59 + 1
f = e / 2
g = 59 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
|
a ) 50 , b ) 78 , c ) 267 , d ) 29 , e ) 10 | e | subtract(add(200, 310), 500) | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 310 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 310 - - - - - - - - - - - - - - a + b + 2 c = 510 a + b + c = 500 - - - - - - - - - - - - - - - - c = 10 answer : e" | a = 200 + 310
b = a - 500
|
a ) 50 kmph , b ) 60 kmph , c ) 32 kmph , d ) 70 kmph , e ) none | c | divide(288, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 288 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 288 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 288 / 9 = 32 kmph c )" | a = 6 * 3
b = a / 2
c = 288 / b
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a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 % | c | multiply(divide(add(multiply(50, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2 tax on others = 2.4 percentage = 4.4 / 100 4.4 % answer : c" | a = 4 / 100
b = 50 * a
c = 8 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 65 , b ) 110 , c ) 115 , d ) 130 , e ) 135 | b | add(multiply(30, const_2), 70) | of the 200 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 30 percent of the employees prefer city y and 70 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ? | "140 prefer x ( group 1 ) ; 60 prefer y ( group 2 ) . city y needs 150 people : letall 60 who prefer y ( entire group 2 ) be relocated there , the rest 90 will be those who prefer x from group 1 ; city x needs 50 people : 140 - 90 = 50 from group 1 will be relocated to x , which they prefer . so , the highest possible number of employees who will be relocated to the city they prefer is 60 + 50 = 110 . answer : b ." | a = 30 * 2
b = a + 70
|
a ) 25 minutes , b ) 10 minutes , c ) 20 minutes , d ) 80 minutes , e ) 60 minutes | d | divide(subtract(12.00, 8.00), subtract(0.25, 0.20)) | united telephone charges a base rate of $ 8.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ? | "lets take number of minutesx . given that , 8 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 80 minutes ans d" | a = 12 - 0
b = 0 - 25
c = a / b
|
a ) 1.74 % , b ) 1.94 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(600, 0.8), multiply(700, 1.8)), const_1000) | by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 600 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 600 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 600 ) / 100 + ( 1.8 * 700 ) / 100 = 17.4 gms % of liquid x in resultant mixture = ( 17.4 / 1000 ) * 100 = 1.74 % a" | a = 600 * 0
b = 700 * 1
c = a + b
d = c / 1000
|
a ) 3660 , b ) 2881 , c ) 2887 , d ) 9977 , e ) 2212 | a | multiply(divide(6300, add(add(6300, 4200), 10500)), 12200) | 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 . 12200 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12200 = 3660 . answer : a" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 12200
|
a ) 20 % , b ) 40 % , c ) 60 % , d ) 80 % , e ) 100 % | e | divide(20, subtract(const_1, divide(80, const_100))) | on a certain transatlantic crossing , 20 percent of a ship β s passengers held round - trip tickets and also took their cars abroad the ship . if 80 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship β s passengers held round - trip tickets ? | "let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.2 x had round trip tickets and took their cars . 0.2 x = 0.2 t x = t the answer is e ." | a = 80 / 100
b = 1 - a
c = 20 / b
|
a ) s . 2890 , b ) s . 4046 , c ) s . 1190 , d ) s . 1620 , e ) s . 2680 | b | multiply(divide(multiply(divide(686, 7), 17), 7), 17) | the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 686 , krishan has ? | "ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 686 : n & there n = 289 x 686 / 49 = rs . 4046 answer : b" | a = 686 / 7
b = a * 17
c = b / 7
d = c * 17
|
a ) 5 , b ) 6 , c ) 9 , d ) 11 , e ) 15 | c | divide(add(220, 43), 17) | a no . when divided by 220 gives a remainder 43 , what remainder will be obtainedby dividingthe same no . 17 ? | "221 + 43 = 264 / 17 = 9 ( remainder ) c" | a = 220 + 43
b = a / 17
|
a ) 11 , b ) 10 , c ) 8 , d ) 12 , e ) 15 | a | add(8, divide(subtract(110, 20), add(25, 20))) | two stations a and b are 110 km apart on a straight line . one train starts from a at 8 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 8 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 110 45 x = 135 x = 3 . so , they meet at 11 a . m . answer : a" | a = 110 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(2, subtract(divide(multiply(const_2, 2), 5), 1)) | it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 2 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 2 * x / 2 + x = 5 4 x = 5 x + 10 = > x = 10 a" | a = 2 * 2
b = a / 5
c = b - 1
d = 2 / c
|
a ) 60 , b ) w = 70 , c ) w = 75 , d ) w = 80 , e ) w = 100 | c | divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2)) | a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days w will the work get over ? | "we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 therefore total no . of days of job = 750 man - day / 8 days = 375 / 4 = 94 days ( approx . ) now since this is total and ques . is asking for additional no . of days , so 94 - 20 = 74 days the nearest approx . to answer is 75 ans : c ( 75 days )" | a = 10 * 20
b = a / const_0_25
c = 1 - const_0_25
d = b * c
e = 10 - 2
f = d / e
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 10 | a | divide(const_1, add(inverse(24), inverse(40))) | ravi can do a piece of work in 24 days while prakash can do it in 40 days . in how many days will they finish it together ? | "1 / 24 + 1 / 40 = 1 / 15 15 days answer : a" | a = 1/(24)
b = 1/(40)
c = a + b
d = 1 / c
|
a ) 18 % , b ) 72 % , c ) 32 % , d ) 16 % , e ) 52 % | d | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 30)), const_100)) | the tax on a commodity is diminished by 30 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ? | 100 * 100 = 10000 70 * 120 = 8400 10000 - - - - - - - 1600 100 - - - - - - - ? = 16 % answer : d | a = 100 + 20
b = 100 - 30
c = a * b
d = c / 100
e = 100 - d
|
a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 3 , e ) 32 / 36 | b | divide(divide(2, const_2), add(5, const_1)) | when throwing 2 dices , and looking at the sum of numbers on the dices - what is the probability to have a sum which is smaller than 5 ? | a dice is composed of 6 numbers - 12 , 34 , 56 . when trowing 2 dices - there are 36 options ( 6 x 6 = 36 ) . if we want the sum of two dices to be < 5 there are those options : ( 11 ) , ( 12 ) , ( 21 ) , ( 13 ) , ( 31 ) , ( 22 ) - total of 6 options . therefore - the probability to have a sum which is smaller than 5 is 6 / 36 = 1 / 6 . answer : b | a = 2 / 2
b = 5 + 1
c = a / b
|
a ) 22 , b ) 28 , c ) 32 , d ) 48 , e ) 92 | a | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 = > x = 22 . answer is a ." | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
|
a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | d | multiply(divide(160, 4), const_3) | in a mixed college 160 students are there in one class . out of this 160 students 1 / 4 students are girls . how many boys are there ? | total number of students : 160 total girls : 160 * 1 / 4 = 40 total boys : 160 - 40 = 120 answer is d | a = 160 / 4
b = a * 3
|
a ) 39 : 40 , b ) 39 : 49 , c ) 15 : 16 , d ) 14 : 18 , e ) 39 : 41 | c | divide(add(300, multiply(divide(300, add(subtract(500, const_100), 300)), add(200, multiply(const_2, add(const_3, const_2))))), subtract(500, const_100)) | a and b put in rs . 300 and rs . 500 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 200 where as b does not . in what ratio should they divide the second year ' s profit ? | explanation : 3 : 5 a = 3 / 8 * 200 = 75 375 : 400 39 : 40 answer : c | a = 500 - 100
b = a + 300
c = 300 / b
d = 3 + 2
e = 2 * d
f = 200 + e
g = c * f
h = 300 + g
i = 500 - 100
j = h / i
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.18, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.18, subtract(const_0_25, divide(const_1, const_100))), 14)) | if 42.18 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ? | "42.18 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.18 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.18 = km / 50 . . . . . . 18 / 100 = 3 m / 50 . . . . . . m = 3 k + m = 3 + 3 = 6 answer : a" | a = 1 / 100
b = const_0_25 - a
c = b * 50
d = 1 / 100
e = const_0_25 - d
f = 42 - 18
g = f / 14
h = c / g
i = 1 / 100
j = const_0_25 - i
k = 42 - 18
l = k / 14
m = h + l
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 20 | e | divide(5, subtract(divide(multiply(const_2, 5), 8), 1)) | it takes joey the postman 1 hours to run a 5 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 8 mile / hour , what is the speed with which joey returns ? | let his speed for one half of the journey be 5 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 5 * x / 5 + x = 8 10 x = 8 x + 40 = > 2 x = 40 = > x = 20 e | a = 2 * 5
b = a / 8
c = b - 1
d = 5 / c
|
a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 4 / 20 , e ) 3 / 10 | d | divide(subtract(20, 16), 20) | a β s speed is 20 / 16 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 ? | "we have the ratio of a β s speed and b β s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 16 meters in that time . so if the race is 20 meters long , when a reaches the finish line , b would be 4 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 4 meters so that he doesn β t need to cover that . in that case , the time required by a ( to cover 20 meters ) would be the same as the time required by b ( to cover 16 meters ) to reach the finish line . so b should get a head start of 4 / 20 th of the race . answer ( d )" | a = 20 - 16
b = a / 20
|
a ) 16 , b ) 14 , c ) 18 , d ) 10 , e ) 20 | a | add(add(4, 3), add(add(4, 4), 1)) | for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 980 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 980 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 1000 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 980 so , 3 y < 468 = = > y < 156 so , y can take any value which is less than 162 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . a" | a = 4 + 3
b = 4 + 4
c = b + 1
d = a + c
|
a ) 88 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(55, const_4), multiply(9, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 55 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 55 * 4 = 220 m time taken = 220 * 2 / 5 = 88 sec answer is a" | a = 55 * 4
b = 1000 / 3600
c = 9 * b
d = a / c
|
a ) 25 , b ) 20 , c ) 19 , d ) 23 , e ) 18 | b | divide(multiply(20, 5), subtract(20, 15)) | a contractor undertook to do a piece of work in 15 days . he employed certain number of laboures but 5 of them were absent from the very first day and the rest could finish the work in only 20 days . find the number of men originally employed ? | "let the number of men originally employed be x . 15 x = 20 ( x Γ’ β¬ β 5 ) or x = 20 answer b" | a = 20 * 5
b = 20 - 15
c = a / b
|
a ) 65 , b ) 60.003662 , c ) 80.2 , d ) 85 , e ) 90 | b | add(multiply(power(2, multiply(divide(60, 2), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 2 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 30 minutes after it was poured was how many degrees fahrenheit ? | "first , we have to find a . we know that after t = 2 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 2 a ) + 60 60 = 120 * ( 2 ^ - 2 a ) 60 / 120 = 2 ^ - 2 a 1 / 2 = 2 ^ - 2 a 2 ^ - 1 = 2 ^ - 2 a - 1 = - 2 a 1 / 2 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 2 * 30 ) + 60 f = 120 * ( 2 ^ - 15 ) + 60 f = 120 * ( 1 / 2 ^ 15 ) + 60 f = 120 * 1 / 32768 + 60 f = 0.003662 + 60 = 60.003662 answer b !" | a = 60 / 2
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
|
a ) $ 15 , b ) $ 20 , c ) $ 25 , d ) $ 30 , e ) $ 22 | c | divide(multiply(divide(multiply(500, 50), const_100), 10), const_100) | 10 % of ( 50 % of $ 500 ) is ? | "10 % of ( 50 % of $ 500 ) = 10 / 100 ( 50 / 100 * 500 ) = $ 25 answer is c" | a = 500 * 50
b = a / 100
c = b * 10
d = c / 100
|
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 36 | e | multiply(subtract(divide(60, const_2), divide(multiply(60, 20), const_100)), const_2) | a chemical supply company has 60 liters of a 20 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 20 % hno 3 solution means hno 3 = 12 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 12 + x = 50 % of ( 60 + x ) or x = 36 . hence , e" | a = 60 / 2
b = 60 * 20
c = b / 100
d = a - c
e = d * 2
|
a ) 3 days , b ) 1 day , c ) 2 days , d ) 4 days , e ) 5 days | b | divide(subtract(const_1, add(multiply(divide(const_1, 5), const_2), multiply(divide(const_1, 25), const_2))), divide(const_1, 25)) | a can finish a piece of work in 5 days . b can do it in 25 days . they work together for four days and then a goes away . in how many days will b finish the work ? | "4 / 5 + ( 4 + x ) / 25 = 1 = > x = 1 day answer : b" | a = 1 / 5
b = a * 2
c = 1 / 25
d = c * 2
e = b + d
f = 1 - e
g = 1 / 25
h = f / g
|
a ) 4 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 1 days | a | inverse(add(divide(5, multiply(4, 7)), divide(4, multiply(4, 14)))) | 4 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 28 1 child ' s 1 day work = 1 / 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 28 + 10 / 140 ) = 1 / 4 5 women and 10 children will complete the work in 4 days . answer : a" | a = 4 * 7
b = 5 / a
c = 4 * 14
d = 4 / c
e = b + d
f = 1/(e)
|
a ) 6 , b ) 1 , c ) 0 , d ) 2 , e ) 7 | d | subtract(add(13, power(add(const_1, const_4), 2)), multiply(12, 3)) | if p is a prime number greater than 3 , find the remainder when p ^ 2 + 13 is divided by 12 . | "every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 13 = 36 n ^ 2 + 12 n + 1 + 13 = 36 n ^ 2 + 12 n + 12 + 2 if p = 6 n - 1 , then p ^ 2 + 13 = 36 n ^ 2 - 12 n + 1 + 13 = 36 n ^ 2 - 12 n + 12 + 2 when divided by 12 , it must leave a remainder of 2 . the answer is d ." | a = 1 + 4
b = a ** 2
c = 13 + b
d = 12 * 3
e = c - d
|
a ) 484 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,440 | e | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | yesterday ' s closing prices of 2,640 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ? | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2640 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 1440 . e is my answer ." | a = 20 * 100
b = a - 10
c = b - 10
d = 20 / 100
e = 1 + d
f = e + 1
g = c / f
h = 20 / 100
i = 1 + h
j = g * i
|
a ) 7 / 10 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | divide(add(subtract(20, add(6, 9)), 9), 20) | a bag consists of 20 marbles , of which 6 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | "bag consists of 20 marbles , of which 6 are blue , 9 are red remainder are white . so , white = 20 - 6 - 9 = 5 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 5 / 20 = 14 / 20 = 7 / 10 hence , answer will be a ." | a = 6 + 9
b = 20 - a
c = b + 9
d = c / 20
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(72, divide(72, 12)) | suppose 12 monkeys take 12 minutes to eat 12 bananas . how many monkeys would it take to eat 72 bananas in 72 minutes ? | "one monkey takes 12 min to eat 1 banana , so in 72 mins 1 monkey will eat 6 bananas , so for 72 bananas in 72 min we need 72 / 6 = 12 monkeys answer : d" | a = 72 / 12
b = 72 / a
|
a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | e | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(5, add(add(5, 2), 2))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(2, add(add(5, 2), 2)))) | a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 5 : 2 : 2 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 2 : 5 , respectively . how many more acres of land were planted with tobacco under the new system ? | originally ( 2 / 9 ) * 1350 = 300 acres were planted with tobacco . in the new system ( 5 / 9 ) * 1350 = 750 acres were planted with tobacco . thus 750 - 300 = 450 more acres were planted with tobacco . the answer is e . | a = 100 * 10
b = 3 * 100
c = a + b
d = 5 * 10
e = c + d
f = 5 + 2
g = f + 2
h = 5 / g
i = e * h
j = 100 * 10
k = 3 * 100
l = j + k
m = 5 * 10
n = l + m
o = 5 + 2
p = o + 2
q = 2 / p
r = n * q
s = i - r
|
a ) 44 , b ) 52 , c ) 57 , d ) 65 , e ) 80 | a | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 9) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of charlie 9 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of charlie = 7 * 5 = 35 charlie ' s age 9 years hence = 35 + 9 = 44 answer = a" | a = 8 * 3
b = a + 76
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 9
|
['a ) 32.9', 'b ) 32.4', 'c ) 22.4', 'd ) 32.8', 'e ) 32.1'] | a | add(divide(circumface(6.4), const_2), multiply(6.4, const_2)) | the radius of a semi circle is 6.4 cm then its perimeter is ? | 36 / 7 r = 6.4 = 32.9 answer : a | a = circumface / (
b = a + 2
|
a ) 81 , b ) 82 , c ) 83 , d ) 84 , e ) 85 | c | divide(subtract(multiply(87, const_4), 95), subtract(const_4, const_1)) | the grade point average of the entire class is 87 . if the average of one third of the class is 95 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 87 x = ( 1 / 3 ) 95 x + ( 2 / 3 ) ( p ) x 261 = 95 + 2 p 2 p = 166 p = 83 . the answer is c ." | a = 87 * 4
b = a - 95
c = 4 - 1
d = b / c
|
a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | a | add(triangle_perimeter(30, 30, 30), triangle_perimeter(30, 30, 30)) | an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 30 cm , find the sum of the perimeters of all the triangles . | "we have 30 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 15 and continues . so we have 30,15 , 7.5 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 30 / 1 - ( 1 / 2 ) = 60 equilateral triangle perimeter is 3 a = 3 * 60 = 180 . so option a ." | a = triangle_perimeter + (
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 25 , e ) 6 / 25 | e | divide(6, multiply(4, 5)) | john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of 2 , 3 , 4 , 5 , or 6 miles per hour and amanda runs at a constant rate of 3 , 4 , 5 , 6 , or 7 miles per hour . what is the probability that john has traveled farther than amanda by the time they meet ? | "john will run farther if he runs at 6 mph and amanda runs at 5 mph , 4 mph , or 3 mph . in this case , p ( john runs farther ) = 1 / 5 * 3 / 5 = 3 / 25 john will run farther if he runs at 5 mph and amanda runs at 4 mph or 3 mph . in this case , p ( john runs farther ) = 1 / 5 * 2 / 5 = 2 / 25 john will run farther if he runs at 4 mph and amanda runs at 3 mph . in this case , p ( john runs farther ) = 1 / 5 * 1 / 5 = 1 / 25 overall , p ( john runs farther ) = 3 / 25 + 2 / 25 + 1 / 25 = 6 / 25 the answer is e ." | a = 4 * 5
b = 6 / a
|
a ) 33 , b ) 88 , c ) 76 , d ) 123 , e ) 12 | c | add(add(power(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2))) | the sum of four consecutive even numbers is 292 . what would be the largest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 292 = > x = 37 smallest number is : 2 ( x + 1 ) = 76 . answer : c" | a = 292 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 292 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 292 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 292 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | a | floor(divide(127, 5)) | on dividing 127 by a number , the quotient is 5 and the remainder is 2 . find the divisor . | "d = ( d - r ) / q = ( 127 - 2 ) / 5 = 125 / 5 = 25 a" | a = 127 / 5
b = math.floor(a)
|
a ) 90 min , b ) 160 min , c ) 110 min , d ) 120 min , e ) 130 min | b | inverse(subtract(add(inverse(80), inverse(160)), inverse(40))) | two pipes a and b can separately fill a cistern in 80 minutes and 160 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 40 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 40 - ( 1 / 80 + 1 / 160 ) = - 1 / 160 third pipe can empty in 160 minutes answer : b" | a = 1/(80)
b = 1/(160)
c = a + b
d = 1/(40)
e = c - d
f = 1/(e)
|
a ) 77 % , b ) 78 % , c ) 79 % , d ) 80 % , e ) 81 % | d | divide(add(multiply(15, 70), multiply(10, 95)), 25) | if 15 students in a class average 70 % on an exam and 10 students average 95 % on the same exam , what is the average in percent for all 25 students ? | "( 15 * 70 + 10 * 95 ) / 25 = 80 % the answer is d ." | a = 15 * 70
b = 10 * 95
c = a + b
d = c / 25
|
a ) rs . 55 , b ) rs . 60 , c ) rs . 65 , d ) rs . 70 , e ) rs . 75 | b | divide(divide(multiply(1800, 30), const_100), 9) | a reduction of 30 % in the price of oil enables a house wife to obtain 9 kgs more for rs . 1800 , what is the reduced price for kg ? | "explanation : 1800 * ( 30 / 100 ) = 540 - - - - 9 ? - - - - 1 = > rs . 60 answer : b" | a = 1800 * 30
b = a / 100
c = b / 9
|
a ) 182 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(273, subtract(20, 8)), 8) | a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 273 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 273 + l ) / 20 = l / 8 2184 + 8 l = 20 l l = 2184 / 12 = 182 m answer a" | a = 20 - 8
b = 273 / a
c = b * 8
|
a ) 2,858 , b ) 18,667 , c ) 21,429 , d ) 35,000 , e ) 56,000 | e | add(divide(divide(divide(add(150000, 130000), subtract(15, 10)), const_1000), const_3), const_3) | a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130000 and variables costs averaging $ 10 per item . if the manufacturer β s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150000 ? | let the items manufactured or sold bex 130000 + 10 x = 15 x - 150000 5 x = 280000 x = 56000 ans : e | a = 150000 + 130000
b = 15 - 10
c = a / b
d = c / 1000
e = d / 3
f = e + 3
|
a ) 17 kmph , b ) 19 kmph , c ) 15 kmph , d ) 12 kmph , e ) 16 kmph | d | divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), add(const_4, const_1)) | robert is travelling on his cycle and has calculated to reach point a at 2 p . m . if he travels at 10 km / hr ; he will reach there at 12 noon if he travels at 15 km / hr . at what speed must he travel to reach a at 1 p . m . ? | d 12 kmph let the distance traveled be x km . then , x / 10 - x / 15 = 2 3 x - 2 x = 60 = > x = 60 km . time taken to travel 60 km at 10 km / hr = 60 / 10 = 6 hrs . so , robert started 6 hours before 2 . p . m . i . e . , at 8 a . m . required speed = 60 / 5 = 12 kmph . | a = 1 / 10
b = 1 / 15
c = a - b
d = 2 / c
e = 4 + 1
f = d / e
|
a ) 10 % , b ) 33.33 % , c ) 46.66 % , d ) 50 % , e ) 66.66 % | c | divide(subtract(32, 25), subtract(divide(40, const_100), divide(25, const_100))) | seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 32 percent ryegrass , what percent of the weight of this mixture is x ? | "- - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 32 % 0.4 x + ( m - x ) 0.25 = 0.32 m 0.15 x = 0.07 m x = 0.4666 m x = 46.66 % of m c" | a = 32 - 25
b = 40 / 100
c = 25 / 100
d = b - c
e = a / d
|
a ) 287,000 , b ) 290,000 , c ) 284,000 , d ) 285,000 , e ) 286,000 | c | divide(subtract(multiply(multiply(multiply(add(divide(150, const_100), const_1), 2000), add(subtract(2007, 2003), const_1)), const_12), multiply(2000, const_12)), const_1000) | during 2003 , a company produced an average of 2000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ? | company produced 12 * 2000 = 24,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 24,000 . the gives the average of ( x + 24,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 + 150 % ( 36,000 ) = 80,000 . so : ( x + 36,000 ) / 4 = 80,000 x + 36,000 = 320,000 x = 284,000 the answer is c . | a = 150 / 100
b = a + 1
c = b * 2000
d = 2007 - 2003
e = d + 1
f = c * e
g = f * 12
h = 2000 * 12
i = g - h
j = i / 1000
|
a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 26048 | e | subtract(power(add(5, const_3), 5), divide(factorial(add(5, const_3)), factorial(const_3))) | in how many ways can a 5 - letter password be chosen , using the letters a , b , c , d , e , f , g , and / or h such that at least one letter is repeated within the password ? | total number of four letter passwords = 8 * 8 * 8 * 8 * 8 = 32768 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 8 c 5 * 5 ! = 56 * 120 = 6720 - - - - - - ( 2 ) therefore required value = ( 1 ) - ( 2 ) = 32768 - 6720 = 26048 e | a = 5 + 3
b = a ** 5
c = 5 + 3
d = math.factorial(c)
e = math.factorial(3)
f = d / e
g = b - f
|
a ) 25 , b ) 30 , c ) 42 , d ) 45 , e ) 50 | c | divide(700, multiply(const_0_2778, subtract(63, 3))) | how many seconds will a 700 metre long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "relative speed of the train = 63 - 3 = 60 kmph = 60 * 5 / 18 = 50 / 3 m / sec t = 700 * 3 / 50 = 42 sec answer : c" | a = 63 - 3
b = const_0_2778 * a
c = 700 / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | e | multiply(const_60, divide(multiply(divide(2, const_60), subtract(15, 5)), 5)) | a man walking at a constant rate of 5 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 15 miles per hour . the woman stops to wait for the man 2 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? | "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 2 * 15 / 60 = 1 / 2 miles and the man walks the distance mm , which is 2 * 5 / 60 = 1 / 6 mile . the difference of 1 / 2 - 1 / 6 = 1 / 3 miles ( mw ) will be covered by the man in ( 1 / 3 ) / 5 = 1 / 15 of an hour , which is 4 minutes . answer e ." | a = 2 / const_60
b = 15 - 5
c = a * b
d = c / 5
e = const_60 * d
|
a ) 800 , b ) 1,250 , c ) 8,000 , d ) 4,000 , e ) 80,000 | d | multiply(divide(multiply(divide(multiply(2, 10), 50), power(10, const_4)), const_1000), 2) | a certain galaxy is known to comprise approximately 2 x 10 ^ 11 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | total no . of stars on galaxy = 2 * 10 ^ 11 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 2 * 10 ^ 11 / 50 * 10 ^ 6 = 20 * 10 ^ 3 / 5 = 4000 therefore answer is d | a = 2 * 10
b = a / 50
c = 10 ** 4
d = b * c
e = d / 1000
f = e * 2
|
a ) 191 , b ) 355 , c ) 737 , d ) 846 , e ) 1,560 | d | divide(multiply(550, const_100), subtract(const_100, 35)) | a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 550 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 550 = > x = ( 550 * 100 ) / 65 = 846 answer d" | a = 550 * 100
b = 100 - 35
c = a / b
|
a ) 54 , b ) 1440 , c ) 2,160 , d ) 2,916 , e ) 148,824 | b | multiply(multiply(5, subtract(5, const_1)), multiply(9, 5)) | right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 5 β€ x β€ 2 and 4 β€ y β€ 9 , then how many different triangles can be drawn that will meet these conditions ? | "we have the rectangle with dimensions 9 * 5 ( 9 horizontal dots and 5 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 5 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot is already occupied by a ) ; choose the y coordinate for vertex b ( as x coordinate is fixed by a ) : 4 c 1 , ( 5 - 1 = 4 as 1 vertical dot is already occupied by a ) . 9 c 1 * 5 c * 8 c 1 * 4 c 1 = 1440 . answer : b ." | a = 5 - 1
b = 5 * a
c = 9 * 5
d = b * c
|
a ) a ) 3 , b ) b ) 5 , c ) c ) 7 , d ) d ) 9 , e ) e ) 11 | a | add(reminder(multiply(5, const_4), const_10), const_1) | one half of a two digit number exceeds its one third by 5 . what is the sum of the digits of the number ? | "x / 2 β x / 3 = 5 = > x = 30 3 + 0 = 3 answer : a" | a = 5 * 4
b = reminder + (
|
a ) 1 : 0 , b ) 1 : 8 , c ) 1 : 7 , d ) 1 : 512 , e ) 1 : 1 | d | divide(power(const_2.0, 8), power(const_3.0, 8)) | the triplicate ratio of 1 : 8 is ? | "13 : 83 = 1 : 512 answer : d" | a = 2 ** 0
b = 3 ** 0
c = a / b
|
['a ) 96', 'b ) 196', 'c ) 128', 'd ) 144', 'e ) 168'] | b | multiply(12, multiply(8, 2)) | a white paper 2 inches wide is placed around a rectangular black paper with dimensions 8 inches by 12 inches . what is the area of the white paper , in square inches ? | this question is an example of a ' punch out ' question - we have to find the area of everything , then ' punch out ' the part that we do n ' t want . we ' re told that a white paper 2 inches wide is placed around a rectangular black paper with dimensions 8 inches by 12 inches . we ' re asked for the area of the frame , in square inches . area of a rectangle = ( length ) ( width ) so the area of the black paper is . . . ( 8 ) ( 12 ) = 96 the white paper ' adds ' 2 inches to the top , bottom , left and right ' sides ' of the picture , so the area of everything is . . . ( 8 + 2 + 2 ) ( 12 + 2 + 2 ) = ( 12 ) ( 16 ) = 196 when we ' punch out ' the area of the black paper , we ' ll be left with the area of the white paper : 192 - 96 = 96 final answer : b | a = 8 * 2
b = 12 * a
|
a ) 125 , b ) 111 , c ) 129 , d ) 101 , e ) 141 | d | gcd(subtract(4351, 8), subtract(5161, 10)) | find the greatest number which on dividing 4351 and 5161 , leaves a reminder of 8 and 10 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 4351 - 8 ) and ( 5161 - 10 ) = hcf ( 4343 , 5151 ) = 101 option d" | a = 4351 - 8
b = 5161 - 10
c = math.gcd(a, b)
|
a ) a ) 265 , b ) b ) 280 , c ) c ) 290 , d ) d ) 300 , e ) e ) 310 | a | divide(add(1340, 15), subtract(6, const_1)) | the difference of two numbers is 1340 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1340 ) . x + 1340 = 6 x + 15 5 x = 1325 x = 265 smaller number = 265 . a )" | a = 1340 + 15
b = 6 - 1
c = a / b
|
a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | b | divide(multiply(choose(5, 2), choose(1, 1)), choose(add(add(5, 4), 1), 1)) | a jar contains only red , yellow , and orange marbles . if there are 1 red , 5 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 1 / 45 . this helped me realise that for the 3 possible orderings the probabbility is the same . so , it should be ( 1 / 45 ) * ( 1 ) , which indeed is 1 / 45 . b" | a = math.comb(5, 2)
b = math.comb(1, 1)
c = a * b
d = 5 + 4
e = d + 1
f = math.comb(e, 1)
g = c / f
|
a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 66 inches | a | divide(add(multiply(6, const_12), 8), 4) | a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : a" | a = 6 * 12
b = a + 8
c = b / 4
|
a ) 27 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | a | subtract(add(floor(divide(subtract(99, 59), 3)), divide(subtract(99, 59), 2)), floor(divide(subtract(99, 59), multiply(const_2, const_3)))) | if w is the set of all the integers between 59 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | multiples of 2 from 59 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 58 = [ 99 / 2 ] - [ 58 / 2 ] = 49 - 29 = 20 multiples of 3 from 59 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 58 = [ 99 / 3 ] - [ 58 / 3 ] = 33 - 169 = 14 multiples of 2 and 3 bothi . e . 6 from 59 to 99 = multiples of 6 from 1 to 99 - multiples of 6 from 1 to 58 = [ 99 / 6 ] - [ 58 / 6 ] = 16 - 9 = 7 these 8 numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3 i . e . total numbers in w = 20 + 14 - 7 = 27 answer option a | a = 99 - 59
b = a / 3
c = math.floor(b)
d = 99 - 59
e = d / 2
f = c + e
g = 99 - 59
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
|
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