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 |
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a ) 8 / 3 , b ) 3 / 8 , c ) 8 / 5 , d ) 4 / 1 , e ) 5 / 3 | d | divide(divide(900, 3), divide(300, 4)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 900 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 900 km time taken by eddy = 3 hours average speed of eddy = 900 / 3 = 300 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 300 / 75 = 4 / 1 answer d" | a = 900 / 3
b = 300 / 4
c = a / b
|
a ) 1 , b ) 3.5 , c ) 40 , d ) 49 , e ) 30 | c | divide(1, divide(1, 40)) | if 40 honey bees make 40 grams of honey in 40 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 40 : : 40 : x honey 40 : 1 = > 1 x 40 x x = 40 x 1 x 40 = > x = 40 . answer : c" | a = 1 / 40
b = 1 / a
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a ) 25 : 10 , b ) 25 : 8 , c ) 25 : 9 , d ) 8 : 25 , e ) 24 : 8 | b | divide(multiply(5, 5), 8) | a dog takes 5 leaps for every 8 leaps of a hare . if one leap of the dog is equal to 5 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 5 ) leaps of hare : 8 leaps of hare = 25 : 8 answer : b" | a = 5 * 5
b = a / 8
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a ) 27.12 % , b ) 23.12 % , c ) 25.55 % , d ) 31.22 % , e ) 56.22 % | a | multiply(const_100, divide(divide(multiply(add(15, const_100), 25), const_100), add(const_100, 6))) | of the families in city x in 2001 , 25 percent owned a personal computer . the number of families in city x owning a computer in 2008 was 15 percent greater than it was in 2001 , and the total number of families in city x was 6 percent greater in 2008 than it was in 2001 . what percent of the families in city x owned a personal computer in 2008 ? | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 25 number of families owning computer in 1998 = 25 * 115 / 100 = 28.75 number of families in 1998 = 106 the percentage = 28.75 / 106 * 100 = 27.12 % . answer : a" | a = 15 + 100
b = a * 25
c = b / 100
d = 100 + 6
e = c / d
f = 100 * e
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(multiply(1,000, 1,000), multiply(100, 1,000)) | the mass of 1 cubic meter of a substance is 100 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "100 kg - 1 cubic meter ; 100,000 g - 1 cubic meter ; 100,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 100,000 = 10 / 1 = 10 cubic centimeters . answer : c ." | a = 1 * 0
b = 100 * 1
c = a / b
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a ) 9 min , b ) 8 min , c ) 4 min , d ) 6 min , e ) none of these | b | add(multiply(12, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 24)), const_2)), 12), const_60)) | two pipes a and b can fill a tank in 12 and 24 minutes respectively . if both the pipes are used together , how long will it take to fill the tank ? | "explanation : tank can be filled by pipe a in 12 minutes and pipe b in 24 minutes . = > part filled by a in 1 minute = 1 / 12 = > part filled by b in 1 minute = 1 / 24 = > part filled by both in 1 minute = 1 / 12 + 1 / 24 = 3 / 24 = 1 / 8 time taken to fill the entire tank = 8 minutes answer b" | a = 12 * 100
b = 1 / 12
c = 1 / 24
d = b + c
e = d * 2
f = 1 - e
g = f * 12
h = g * const_60
i = a + h
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | subtract(subtract(99, 82), divide(99, add(const_1, const_10))) | what number is obtained by adding the units digits of 734 ^ 99 and 347 ^ 82 ? | "the units digit of 734 ^ 99 is 4 because 4 raised to the power of an odd integer ends in 4 . the units digit of 347 ^ 82 is 9 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 82 is in the form 4 n + 2 , the units digit is 9 . then 4 + 9 = 13 . the answer is d ." | a = 99 - 82
b = 1 + 10
c = 99 / b
d = a - c
|
a ) 24 , b ) 28 , c ) 32 , d ) 56 , e ) 58 | a | multiply(divide(divide(12880, 230), 7), 3) | the ratio between the number of sheep and the number of horses at the stewar farm is 3 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | et no of sheep and horses are 3 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 3 * 8 ) = 24 answer : a | a = 12880 / 230
b = a / 7
c = b * 3
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | divide(const_1, subtract(divide(const_1, 2), divide(const_1, add(2, divide(4, 5))))) | a pump can fill a tank with a water in 2 hours . because of a leak , it took 2 and 4 / 5 hours to fill the tank . the leak can drain all the water of the full tank in how many hours ? | the rate of the pump + leak = 5 / 14 1 / 2 - leak ' s rate = 5 / 14 leak ' s rate = 1 / 2 - 5 / 14 = 1 / 7 the leak will empty the tank in 7 hours . the answer is d . | a = 1 / 2
b = 4 / 5
c = 2 + b
d = 1 / c
e = a - d
f = 1 / e
|
a ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 70 | a | divide(100, add(divide(100, const_100), const_1)) | the sum of number of boys and girls in a school is 100 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 100 = 100 x + x / 100 * 100 = 100 2 * x = 100 x = 50 answer is a" | a = 100 / 100
b = a + 1
c = 100 / b
|
a ) 22 days , b ) 25 days , c ) 28 days , d ) 16 days , e ) 27 days | b | divide(multiply(20, 50), add(20, 20)) | the food in a camp lasts for 20 men for 50 days . if 20 more men join , how many days will the food last ? | one man can consume the same food in 20 * 50 = 1000 days . 20 more men join , the total number of men = 40 the number of days the food will last = 1000 / 40 = 25 days . answer : b | a = 20 * 50
b = 20 + 20
c = a / b
|
a ) 18 , b ) 22 , c ) 24 , d ) 28 , e ) 32 | b | multiply(2, 4) | three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 264 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 264 x = 22 hcf of 2 x , 3 x and 4 x = x = 22 the answer is b ." | a = 2 * 4
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a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 3 / 4 | a | divide(subtract(divide(1, 2), divide(1, 3)), divide(1, 3)) | in a class of students , 1 / 2 of the number of girls is equal to 1 / 3 of the total number of students . what is the ratio of boys to girls in the class ? | "( 1 / 2 ) g = ( 1 / 3 ) ( b + g ) 3 g = 2 b + 2 g g = 2 b b / g = 1 / 2 . the answer is a ." | a = 1 / 2
b = 1 / 3
c = a - b
d = 1 / 3
e = c / d
|
['a ) 3 : 4', 'b ) 1 : 4', 'c ) 1 : 2', 'd ) 2 : 3', 'e ) 1 : 3'] | a | divide(multiply(multiply(divide(45, const_360), 2), const_pi), multiply(multiply(divide(60, const_360), const_2), const_pi)) | two circles are there one with arc area 60 digree equal to 45 digree of another circle find the area of circle 1 : circle 2 ratio | area for first circle = ( 3.14 * r ^ 2 ) / 6 area of second circke = ( 3.14 * r ^ 2 ) / 8 therefore , ratio = 3 : 4 answer : a | a = 45 / 360
b = a * 2
c = b * math.pi
d = 60 / 360
e = d * 2
f = e * math.pi
g = c / f
|
a ) 2,000 , b ) 3,000 , c ) 4,000 , d ) 8,000 , e ) 9,000 | c | multiply(divide(divide(subtract(subtract(multiply(multiply(const_3, const_4), const_1000), multiply(multiply(multiply(const_3, const_4), const_1000), divide(1, 6))), multiply(subtract(multiply(multiply(const_3, const_4), const_1000), multiply(multiply(multiply(const_3, const_4), const_1000), divide(1, 6))), divide(3, 5))), const_1000), const_4), const_3) | in a recent head - to - head run - off election , 12,000 absentee ballets were cast . 1 / 6 of the absentee ballets were thrown out and 3 / 5 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | 5 / 6 * 2 / 5 ( total absentee votes ) = 1 / 3 ( total votes ) = 1 / 3 * 12000 = 4000 answer is c | a = 3 * 4
b = a * 1000
c = 3 * 4
d = c * 1000
e = 1 / 6
f = d * e
g = b - f
h = 3 * 4
i = h * 1000
j = 3 * 4
k = j * 1000
l = 1 / 6
m = k * l
n = i - m
o = 3 / 5
p = n * o
q = g - p
r = q / 1000
s = r / 4
t = s * 3
|
a ) 76 kmph , b ) 54 kmph , c ) 72 kmph , d ) 36 kmph , e ) 91 kmph | d | divide(divide(300, const_1000), divide(30, const_3600)) | a train 300 m long can cross an electric pole in 30 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 300 / 30 s = 10 m / sec speed = 10 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 36 kmph answer : d" | a = 300 / 1000
b = 30 / 3600
c = a / b
|
a ) 6.54 % , b ) 4.54 % , c ) 1.51 % , d ) 1.94 % , e ) 5.54 % | c | divide(const_100, 66) | at what rate percent of simple interest will a sum of money double itself in 66 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 66 ) = 100 / 66 = 1.51 % answer : c" | a = 100 / 66
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(7, 4) | an = 2 an - 1 + 4 and qn = 4 qn - 1 + 8 for all n > 1 . if a 5 = q 4 and a 7 = 316 , what is the first value of n for which qn is an integer ? | assuming i have understood the symbols used correctly answer is c - 3 a 7 = 2 s 6 + 4 = 316 a 6 = 312 / 2 = 156 = 2 a 5 + 4 a 5 = 152 / 2 = 76 = q 4 q 4 = 4 q 3 + 8 = 76 q 3 = 68 / 4 = 17 q 3 = 4 q 2 + 8 = 17 q 2 = 9 / 4 = 2.25 q 2 = 4 q 1 + 8 q 1 will not be integer = c | a = 7 - 4
|
a ) 22 , b ) 77 , c ) 877 , d ) 99 , e ) 88 | b | add(multiply(multiply(5.95, 3.78), const_3), const_10) | courtyard 3.78 meters long 5.95 meters wide is to be paved exactly with square tiles , all of the same size . what is the largest size of the tile which could be used for the purpose ? | 3.78 meters = 378 cm = 2 Γ 3 Γ 3 Γ 3 Γ 7 5.25 meters = 525 cm = 5 Γ 5 Γ 3 Γ 7 hence common factors are 3 and 7 hence lcm = 3 Γ 7 = 21 hence largest size of square tiles that can be paved exactly with square tiles is 77 cm . answer : b | a = 5 * 95
b = a * 3
c = b + 10
|
a ) 562 , b ) 356 , c ) 452 , d ) 416 , e ) 512 | d | add(multiply(divide(8, 5), 160), 160) | in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 160 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 160 x = 32 total number of students = 13 x = 13 * 32 = 416 answer is d" | a = 8 / 5
b = a * 160
c = b + 160
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | subtract(add(multiply(reminder(5, 500), 3), reminder(3, 500)), reminder(1, 500)) | x is a positive integer less than 500 . when x is divided by 5 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 5 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 500 n = 24.42 or n = 24 answer : - d" | a = reminder * (
b = a + 3
c = b - reminder
|
a ) 45 , b ) 62 , c ) 70 , d ) 77 , e ) 98 | b | add(add(add(divide(lcm(lcm(lcm(4, 1), 1), 3), 4), divide(lcm(lcm(lcm(4, 1), 1), 3), 1)), divide(lcm(lcm(lcm(4, 1), 1), 3), 1)), divide(lcm(lcm(lcm(4, 1), 1), 3), 3)) | john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 4 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | "rakesh : rahul : ranjan : rohit = 1 / 4 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 4 x 60 ) = 15 . rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : total number of pencils are ( 15 x + 20 x + 15 x + 12 x ) = 62 x . for minimum number of pencils x = 1 . the person should have at least 62 pencils . b )" | a = math.lcm(4, 1)
b = math.lcm(a, 1)
c = math.lcm(b, 3)
d = c / 4
e = math.lcm(4, 1)
f = math.lcm(e, 1)
g = math.lcm(f, 3)
h = g / 1
i = d + h
j = math.lcm(4, 1)
k = math.lcm(j, 1)
l = math.lcm(k, 3)
m = l / 1
n = i + m
o = math.lcm(4, 1)
p = math.lcm(o, 1)
q = math.lcm(p, 3)
r = q / 3
s = n + r
|
a ) $ 570 , b ) $ 586 , c ) $ 630 , d ) $ 686 , e ) $ 715 | b | add(500, subtract(500, divide(add(add(add(add(406, 413), 420), 436), 395), 5))) | tough and tricky questions : word problems . a salesman ' s income consists of commission and base salary . his weekly income totals over the past 5 weeks have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) income over the next 5 weeks be to increase his average weekly income to $ 500 over the 10 - week period ? | official solution : ( b ) first , we need to add up the wages over the past 5 weeks : $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 . to average $ 500 over 10 weeks , the salesman would need to earn : $ 500 Γ 10 = $ 5000 . subtract $ 2070 from $ 5000 to determine how much he would need to earn , in total , over the next 5 weeks to average $ 500 for the 10 weeks : $ 5000 β $ 2070 = $ 2930 . dividing $ 2930 by 5 will give us the amount he needs to earn on average over the next 5 weeks : $ 2930 / 5 = $ 586 . the correct answer is choice ( b ) . | a = 406 + 413
b = a + 420
c = b + 436
d = c + 395
e = d / 5
f = 500 - e
g = 500 + f
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a ) 8 / 27 , b ) 10 / 31 , c ) 12 / 37 , d ) 14 / 41 , e ) 16 / 49 | a | subtract(1, divide(const_2, 3)) | when a random experiment is conducted , the probability that event a occurs is 1 / 3 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 3 * 1 / 3 * 2 / 3 * 2 / 3 = 4 / 81 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 4 / 81 ) = 8 / 27 the answer is a ." | a = 2 / 3
b = 1 - a
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a ) 200 , b ) 225 , c ) 93.75 , d ) 320 , e ) none of these | c | divide(6, multiply(divide(16, const_100), divide(40, const_100))) | if 16 % of 40 % of a number is 6 , then the number is | explanation : let 16 / 100 Γ 40 / 100 Γ a = 6 a = 6 Γ 100 Γ 100 / 16 Γ 40 = 93.75 correct option : c | a = 16 / 100
b = 40 / 100
c = a * b
d = 6 / c
|
a ) 18 / 25 , b ) 17 / 20 , c ) 11 / 15 , d ) 7 / 10 , e ) 3 / 5 | a | multiply(divide(subtract(divide(multiply(3, multiply(multiply(3, 3), 5)), 5), divide(multiply(divide(multiply(3, multiply(multiply(3, 3), 5)), 5), 3), 3)), subtract(multiply(multiply(3, 3), 5), divide(multiply(2, multiply(multiply(3, 3), 5)), 3))), const_100) | one day a car rental agency rented 2 / 3 of its cars , including 3 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 2 / 5 ) ( 3 / 5 ) = 6 / 25 of all the cars . the cars which were not rented is 1 / 3 of all the cars . the percent of non - rented cars which had cd players is ( 6 / 25 ) / ( 1 / 3 ) = 18 / 25 the answer is a ." | a = 3 * 3
b = a * 5
c = 3 * b
d = c / 5
e = 3 * 3
f = e * 5
g = 3 * f
h = g / 5
i = h * 3
j = i / 3
k = d - j
l = 3 * 3
m = l * 5
n = 3 * 3
o = n * 5
p = 2 * o
q = p / 3
r = m - q
s = k / r
t = s * 100
|
a ) 4 % , b ) 5 % , c ) 6 % , d ) 8 % , e ) 10 % | e | multiply(divide(subtract(add(20, const_100), add(8, const_100)), add(20, const_100)), const_100) | if the price of gasoline increases by 20 % and a driver intends to spend only 8 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ? | "let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.2 * p ) y = 1.08 ( p * x ) y = ( 1.08 / 1.2 ) x = 0.9 x which is a reduction of 10 % . the answer is e ." | a = 20 + 100
b = 8 + 100
c = a - b
d = 20 + 100
e = c / d
f = e * 100
|
a ) 70 kg , b ) 80 kg , c ) 85 kg , d ) 90 kg , e ) 95 kg | a | add(multiply(8, 2.5), 50) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 50 + 20 ) kg = 70 kg . a )" | a = 8 * 2
b = a + 50
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a ) 4 / 3 , b ) 6 / 5 , c ) 8 / 7 , d ) 10 / 9 , e ) 12 / 11 | e | inverse(add(add(inverse(4), inverse(2)), inverse(6))) | machine a can finish a job in 4 hours , machine Π² can finish the job in 2 hours , and machine Ρ can finish the job in 6 hours . how many hours will it take for a , b , and Ρ together to finish the job ? | "the combined rate is 1 / 4 + 1 / 2 + 1 / 6 = 11 / 12 of the job per hour . the time to complete the job is 12 / 11 hours . the answer is e ." | a = 1/(4)
b = 1/(2)
c = a + b
d = 1/(6)
e = c + d
f = 1/(e)
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a ) 48 , b ) 32 , c ) 24 , d ) 18 , e ) 12 | c | multiply(3, multiply(const_2, const_4)) | a cubical block of metal weighs 3 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 3 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 3 pounds = 24 pounds so answer is c and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 3 * 8 = 24 so answer is c" | a = 2 * 4
b = 3 * a
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a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 , e ) 3 / 4 | d | divide(2, power(const_2, 2)) | two coins are tossed . find the probability of at most 2 tails ? | "n ( s ) = 2 ^ 2 = 4 . . . . ( 2 coins are tossed ) let e is the event of getting at most 2 tails n ( e ) = hh , tt , ht , th = 4 p ( e ) = n ( e ) / n ( s ) = 4 / 4 = 1 ans - d" | a = 2 ** 2
b = 2 / a
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a ) 21 , b ) 28 , c ) 90 , d ) 26 , e ) 70 | e | subtract(divide(8925, 85), 35) | a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 35 per metre of cloth . what is the cost price of one metre of cloth ? | "explanation : sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 35 = rs . 70 . answer : e" | a = 8925 / 85
b = a - 35
|
a ) 22 sec , b ) 35 sec , c ) 25 sec , d ) 18 sec , e ) 17 sec | b | divide(add(150, 200), multiply(36, const_0_2778)) | how many seconds will a train 200 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ? | "d = 200 + 150 = 350 s = 36 * 5 / 18 = 10 mps t = 350 / 10 = 35 sec answer : b" | a = 150 + 200
b = 36 * const_0_2778
c = a / b
|
a ) 82 , b ) 83 , c ) 84 , d ) 85 , e ) 86 | c | divide(427, add(divide(7, 7), add(divide(7, 3), divide(7, 4)))) | a sum of rs . 427 is to be divided among a , b and c in such a way that 3 times a β s share , 4 times b β s share and 7 times c β s share are all equal . the share of c is | "a + b + c = 427 3 a = 4 b = 7 c = > ( 7 c / 3 ) + ( 7 c / 4 ) + c = 427 = > c = 84 answer : c" | a = 7 / 7
b = 7 / 3
c = 7 / 4
d = b + c
e = a + d
f = 427 / e
|
a ) 10 , b ) 12.6 , c ) 22.5 , d ) 30 , e ) 40.8 | d | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(12, 4), subtract(12, const_2))), 4) | find the average of first 4 multiples of 12 ? | "average = ( 12 + 24 + 36 + 48 ) / 4 = 30 answer is d" | a = 4 + 1
b = 4 + 1
c = b + 2
d = a + c
e = 12 - 4
f = 12 - 2
g = e + f
h = d + g
i = h / 4
|
a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | e | add(multiply(power(divide(128, multiply(1, multiply(1, const_2))), inverse(const_3)), multiply(1, const_2)), multiply(1, const_2)) | there is a rectangular prism made of 1 in cubes that has been covered in tin foil . there are exactly 128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these 128 cubes is twice the length and twice the height , what is the p measure in inches of the width of the foil covered prism ? | "if the width is w , then length and height would be w / 2 . so , w * w / 2 * w / 2 = 128 = > w ^ 3 = ( 2 ^ 3 ) * 64 = ( 2 ^ 3 ) * ( 4 ^ 3 ) = > w = 2 * 4 = 8 in . along the width of the cuboid , 8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = 8 + 2 = p = 10 ans e ." | a = 1 * 2
b = 1 * a
c = 128 / b
d = 1/(3)
e = c ** d
f = 1 * 2
g = e * f
h = 1 * 2
i = g + h
|
a ) 83 % , b ) 83 1 β 2 % , c ) 84 % , d ) 85 % , e ) 86 2 β 3 % | c | divide(add(multiply(15, 80), multiply(10, 90)), 25) | if 15 students in a class average 80 % on an english exam and 10 students average 90 % on the same exam , what is the average in percent for all 25 students ? | the average percentage = ( 15 x 80 % + 10 x 90 % ) / 25 = ( 15 x 80 % + 10 x 80 % + 10 x 10 % ) / 25 . . . . . . ( as 90 % = 10 % + 80 % ) = ( 25 x 80 % + 10 x 10 % ) / 25 = 80 % + 10 x 10 % / 25 = 80 % + 4 % = 84 % answer : c | a = 15 * 80
b = 10 * 90
c = a + b
d = c / 25
|
a ) 52 kmph , b ) 56 kmph , c ) 58 kmph , d ) 62 kmph , e ) 72 kmph | d | subtract(multiply(divide(280, 9), const_3_6), 50) | a man sitting in a train which is traveling at 50 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed | "explanation : relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) Γ ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 50 ) kmph = 62 kmph answer : option d" | a = 280 / 9
b = a * const_3_6
c = b - 50
|
a ) 6 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(96, const_10), const_10) | how many positive factors do 160 and 96 have in common ? | "the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 160 and 96 is 32 number of factors of 32 = 6 answer : a" | a = 96 - 10
b = a / 10
|
a ) 15 , b ) 1 , c ) 2 , d ) 14 , e ) 17 | d | add(subtract(50, const_100), const_4) | how many odd prime numbers are there less than 50 ? | "odd prime number less than 50 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 there is 14 the odd prime number answer is d" | a = 50 - 100
b = a + 4
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a ) 13 % , b ) 80 % , c ) 20 % , d ) 17 % , e ) 12 % | a | subtract(15, const_1) | at a local appliance manufacturing facility , the workers received a 15 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | "let ' s say he works usually 10 hours and earns 100 per hour . 10 * 100 = 1000 10 * 115 = 1150 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 115 = 8.7 so he can reduce his work by 1.3 hours . which is > 13 % . answer a" | a = 15 - 1
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a ) 1.2 % , b ) 0.5 % , c ) 0.7 % , d ) 0.9 % , e ) 0.3 % | b | multiply(divide(divide(40, 4), 2000), const_100) | if the simple interest on $ 2000 increase by $ 40 , when the time increase by 4 years . find the rate percent per annum . | "applying the above formula , we have 40 = 2000 * r * 4 / 100 r = 40 * 100 / 2000 * 4 = 0.5 % . answer is b" | a = 40 / 4
b = a / 2000
c = b * 100
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a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 16 / 21 | e | add(multiply(divide(4, 7), divide(subtract(7, 2), subtract(7, 1))), multiply(divide(4, 7), divide(subtract(7, 4), subtract(7, 1)))) | in a room filled with 7 people , 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | "# of selections of 2 out of 7 - c 27 = 21 c 72 = 21 ; # of selections of 2 people which are not siblings - c 12 β c 12 c 21 β c 21 ( one from first pair of siblings * one from second pair of siblings ) + c 12 β c 13 c 21 β c 31 ( one from first pair of siblings * one from triple ) + c 1 ^ _ 2 * c ^ 1 _ 3 ( one from second pair of siblings * one from triple ) = 4 + 6 + 6 = 16 = 4 + 6 + 6 = 16 . p = 16 / 21 answer : e" | a = 4 / 7
b = 7 - 2
c = 7 - 1
d = b / c
e = a * d
f = 4 / 7
g = 7 - 4
h = 7 - 1
i = g / h
j = f * i
k = e + j
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a ) 250 , b ) 425 , c ) 550 , d ) 700 , e ) 750 | c | multiply(add(divide(100, 12), divide(150, 15)), 30) | working at their respective constant rates , machine a makes 100 copies in 12 minutes and machine b makes 150 copies in 15 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ? | machine a can produce 100 * 30 / 12 = 250 copies and , machine b can produce 150 * 30 / 15 = 300 copies total producing 550 copies . c is the answer | a = 100 / 12
b = 150 / 15
c = a + b
d = c * 30
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a ) 58 , b ) 66 , c ) 74 , d ) 82 , e ) 90 | d | subtract(456, reminder(897326, 456)) | which number can we add to 897326 to get a number exactly divisible by 456 ? | 897326 / 456 = 1967 with a remainder of 374 . we need to add 456 - 374 = 82 the answer is d . | a = 456 - reminder
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a ) 1 / 4 , b ) 1 / 9 , c ) 4 / 3 , d ) 3 / 9 , e ) 2 / 9 | e | divide(add(7, const_1), multiply(const_6, const_6)) | when two dice are rolled , what is the probability that the sum is either 7 or 11 ? | total number of outcomes possible = 36 probability of getting sum of 7 = 6 / 36 probability of getting sum of 11 = 2 / 36 probability that the sum is either 7 or 11 = 6 / 36 + 2 / 36 = 2 / 9 answer : e | a = 7 + 1
b = 6 * 6
c = a / b
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a ) 708 m , b ) 704 m , c ) 774 m , d ) 760.57 m , e ) 744 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 24.2), const_2), 500), const_100) | the radius of a wheel is 24.2 cm . what is the distance covered by the wheel in making 500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 24.2 = 76057 cm = 760.57 m answer : d" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 24
g = f * 2
h = g * 500
i = h / 100
|
a ) 2160 , b ) 2350 , c ) 2000 , d ) 2300 , e ) none of these | a | subtract(divide(multiply(add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 3940), const_100), subtract(const_100, 10)), add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 3940)) | after spending rs . 5000 on rent , rs . 1500 on milk , rs . 4500 on groceries , rs . 2500 on childrens education rs . 2000 on petrol and rs . 3940 on miscellaneous expenses , mr . kishore saved 10 % of his monthly salary . how much did he save in rs . ? | "explanation : total exp = 5000 + 1500 + 4500 + 2500 + 2000 + 3940 = 19440 exp in % = 100 - 10 = 90 % , 19440 = 90 % saving = 10 % = 19440 x 10 / 90 = rs . 2160 answer : a" | a = 5000 + 1500
b = a + 4500
c = b + 2500
d = c + 2000
e = d + 3940
f = e * 100
g = 100 - 10
h = f / g
i = 5000 + 1500
j = i + 4500
k = j + 2500
l = k + 2000
m = l + 3940
n = h - m
|
a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 51 | e | add(add(44, multiply(3, 2)), const_1) | the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 44 . what is a possible value for the mean of the distribution ? | "the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 44 : { mean } - 3 * { sd } > 44 ; { mean } - 6 > 44 ; { mean } > 50 . answer : e ." | a = 3 * 2
b = 44 + a
c = b + 1
|
a ) 7 / 16 , b ) 5 / 16 , c ) 3 / 16 , d ) 7 / 8 , e ) 5 / 8 | b | divide(add(floor(divide(78, 23)), const_2), subtract(subtract(floor(divide(910, 45)), floor(divide(78, 23))), const_1)) | the set x = { 78 , 910 } . the set y = { 23 , 45 } . we will randomly choose one element x from set x and one element y from set y . what is the probability that x / y will be an integer ? | the total number of x and y pairs is 4 * 4 = 16 . there are five pairs such that x / y is an integer . the probability that x / y is an integer is 5 / 16 . the answer is b . | a = 78 / 23
b = math.floor(a)
c = b + 2
d = 910 / 45
e = math.floor(d)
f = 78 / 23
g = math.floor(f)
h = e - g
i = h - 1
j = c / i
|
a ) 4 ^ 12 , b ) 4 ^ 35 , c ) 17 ( 4 ^ 5 ) , d ) 8 ^ 12 , e ) 65 ( 4 ^ 5 ) | e | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of 4 ^ 5 + 4 ^ 8 ? | "4 ^ 5 + 4 ^ 8 = 4 ^ 5 ( 1 + 4 ^ 3 ) = 4 ^ 5 * 65 answer e" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | multiply(divide(65, const_100), 12) | john bought a total of 12 mangoes and oranges . each mango costs 80 cents and each orange costs 50 cents . if the average price of the 12 mangoes and oranges that john originally purchased was 65 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | "let number of mangoes be x , number of oranges be 12 - x 0.80 x + ( 12 - x ) 0.50 / 12 = 0.65 solving for x , we get x = 6 - - > mangoes 6 , oranges 6 now , number of oranges to be returned be y [ 0.80 * 6 + ( 6 - y ) * 0.50 ] / 12 - y = 0.72 solving for y , y = 4 ans : a" | a = 65 / 100
b = a * 12
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a ) 3 : 1 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 15 : 4 | a | divide(add(multiply(4, divide(add(5, 1), add(4, 2))), 5), add(multiply(2, divide(add(5, 1), add(4, 2))), 1)) | two vessels contains equal number of mixtures milk and water in the ratio 4 : 2 and 5 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 4 / 6 + 5 / 6 ) : ( 2 / 6 + 1 / 6 ) = 9 / 6 : 3 / 6 = 3 : 1 answer is a" | a = 5 + 1
b = 4 + 2
c = a / b
d = 4 * c
e = d + 5
f = 5 + 1
g = 4 + 2
h = f / g
i = 2 * h
j = i + 1
k = e / j
|
a ) 7650 , b ) 9250 , c ) 12,650 , d ) 14,250 , e ) 15,150 | a | add(divide(subtract(subtract(201, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 201 ? | "100 + 102 + . . . + 200 = 100 * 51 + ( 2 + 4 + . . . + 100 ) = 100 * 51 + 2 * ( 1 + 2 + . . . + 50 ) = 100 * 51 + 2 ( 50 ) ( 51 ) / 2 = 100 * 51 + 50 * 51 = 150 ( 51 ) = 7650 the answer is a ." | a = 201 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 4.5 , e ) 5.5 | d | divide(90, multiply(72, const_0_2778)) | in what time will a train 90 m long cross an electric pole , it its speed be 72 km / hr ? | "speed = 72 * 5 / 18 = 20 m / sec time taken = 90 / 20 = 4.5 sec . answer : d" | a = 72 * const_0_2778
b = 90 / a
|
a ) 111.12 , b ) 250 , c ) 111.64 , d ) 111.11 , e ) 101.12 | b | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 400)), subtract(multiply(const_100, const_10), 200))) | a can give b 200 meters start and c 400 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 800 m and c runs 600 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 600 ) / 800 = 750 m . b can give c = 1000 - 750 = 250 m . answer : b" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 400
e = b * d
f = 100 * 10
g = f - 200
h = e / g
i = a - h
|
a ) 1.95 , b ) 2.95 , c ) 4.95 , d ) 3.95 , e ) 5.95 | b | multiply(const_12, divide(multiply(40, divide(40, const_100)), 65)) | a reduction of 40 % in the price of bananas would enable a man to obtain 65 more for rs . 40 , what is reduced price per dozen ? | "40 * ( 40 / 100 ) = 16 - - - 65 ? - - - 12 = > rs . 2.95 answer : b" | a = 40 / 100
b = 40 * a
c = b / 65
d = 12 * c
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a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 20 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 20), subtract(const_100, 10)), const_100), subtract(const_100, 10)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 20 % while buying and by 10 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 120 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 120 - 90 ) / 90 * 100 = 20 % answer : d ." | a = 100 + 20
b = 100 - 10
c = a - b
d = c * 100
e = 100 - 10
f = d / e
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | subtract(divide(subtract(115, 50), subtract(55, 50)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 115 units raises the average to 55 units per day , what is the value of n ? | the daily production was raised by 5 units for n days , which has a weighting of 5 n . 5 n = 115 - 55 = 60 n = 12 the answer is c . | a = 115 - 50
b = 55 - 50
c = a / b
d = c - 1
|
a ) 279 , b ) 283 , c ) 308 , d ) 318 , e ) 338 | c | multiply(11, 275) | the h . c . f . of two numbers is 11 and their l . c . m . is 7700 . if one of the numbers is 275 , then the other is : | other number = ( 11 x 7700 ) / 275 = 308 . answer : c | a = 11 * 275
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a ) 7 min , b ) 8 min , c ) 14.4 min , d ) 10 min , e ) 11 min | c | multiply(const_60, divide(subtract(54, 41), 54)) | excluding stoppages , the speed of a bus is 54 kmph and including stoppages , it is 41 kmph . for how many minutes does the bus stop per hour ? | due to stoppages , it covers 13 km less . time taken to cover 13 km = ( 13 / 54 ) x 60 = 14.4 min answer : c | a = 54 - 41
b = a / 54
c = const_60 * b
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a ) 60 sec , b ) 40 sec , c ) 30 sec , d ) 45 sec , e ) 1 min 10 sec | a | divide(add(add(100, 150), 50), subtract(15, 10)) | a train length 100 m going with speed 10 m / sec another train 150 m long going with speed 15 m / sec . the distance between two train is 50 m . then how much time second train will take to cross the first train ? | net speed = 15 - 10 = 5 m / sec net distance = 100 + 150 + 50 = 300 total time = 300 / 5 = 60 sec answer a | a = 100 + 150
b = a + 50
c = 15 - 10
d = b / c
|
a ) 20 minutes , b ) 30 minutes , c ) 40 minutes , d ) 50 minutes , e ) 60 minutes | c | divide(power(10, const_3), add(15, 10)) | one copy machine can make 15 copies a minute , and a second copy machine makes 10 copies a minute . if the two copiers work together , how long would it take them to make 1,000 copies ? | total work done by both machines in a minute = 15 + 10 = 25 copies total number of copies required = 1000 time = 1000 / 25 = 40 mins answer c | a = 10 ** 3
b = 15 + 10
c = a / b
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a ) 8 pm , b ) 6 pm , c ) 11 pm , d ) 10 pm , e ) 5 pm | c | add(6, divide(50, add(6, 4))) | a and b start walking towards each other at 6 pm at speed of 6 kmph and 4 kmph . they were initially 50 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 50 / 4 + 6 = 50 / 10 = 5 hrs after 6 pm = 11 pm answer is c" | a = 6 + 4
b = 50 / a
c = 6 + b
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | divide(add(multiply(factorial(33), factorial(2)), multiply(factorial(33), factorial(3))), 33) | what is the units digit of 33 ^ 2 * 17 ^ 3 * 49 ^ 3 ? | "the units digit of 33 ^ 2 is the units digit of 3 * 3 = 9 which is 9 . the units digit of 17 ^ 3 is the units digit of 7 * 7 * 7 = 343 which is 3 . the units digit of 49 ^ 3 is the units digit of 9 * 9 * 9 = 729 which is 9 . the units digit of 9 * 3 * 9 = 243 is 3 . the answer is b ." | a = math.factorial(33)
b = math.factorial(2)
c = a * b
d = math.factorial(33)
e = math.factorial(3)
f = d * e
g = c + f
h = g / 33
|
a ) 34 , b ) 35 , c ) 36 , d ) 37 1 / 25 , e ) 40 | c | multiply(const_100, divide(add(multiply(10, divide(30, const_100)), multiply(2, divide(30, const_100))), add(add(multiply(10, divide(30, const_100)), multiply(2, divide(30, const_100))), add(subtract(multiply(10, divide(70, const_100)), 2), multiply(2, divide(70, const_100)))))) | solution y is 30 percent liquid x and 70 percent water . if 2 kilograms of water evaporate from 10 kilograms of solution y and 2 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ? | "in 8 kilograms of solution y there are 0.3 * 10 = 3 kilograms of solution x ; after 2 kilograms of water are replaced by 2 kilograms of solution y , to the existing 2.4 kilograms of solution x , 0.3 * 2 = 0.6 kilograms of solution x are added , so in the new solution of 8 kilograms there are 3 + 0.6 = 3.6 kilograms of solution x , which is 3.6 / 10 * 100 = 36 % of this new solution . answer : c ." | a = 30 / 100
b = 10 * a
c = 30 / 100
d = 2 * c
e = b + d
f = 30 / 100
g = 10 * f
h = 30 / 100
i = 2 * h
j = g + i
k = 70 / 100
l = 10 * k
m = l - 2
n = 70 / 100
o = 2 * n
p = m + o
q = j + p
r = e / q
s = 100 * r
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | b | divide(volume_rectangular_prism(6, 4, 3), 4) | water is poured into a tank so that the tank is being filled at the rate of 4 cubic feet per hour . if the empty rectangular tank is 6 feet long , 4 feet wide , and 3 feet deep , approximately how many hours does it take to fill the tank ? | "the volume the tank is : length * width * depth = 6 * 4 * 3 = 72 cubic feet . 72 cubic feet / 4 cubic feet per hour = 18 hours . it will take 18 hours to fill the tank . the answer is b ." | a = volume_rectangular_prism / (
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a ) 197 min , b ) 100 min , c ) 177 min , d ) 176 min , e ) 186 min | b | inverse(subtract(add(divide(const_1, 60), divide(const_1, 75)), divide(const_1, 50))) | two pipes a and b can separately fill a cistern in 60 min and 75 min respectively . there is a third pipe in the bottom of the cistern to empty it . if all the 3 pipes are simultaneously opened , then the cistern is full in 50 min . in how much time , the third pipe alone can empty the cistern ? | work done by the third pipe in 1 min = 1 / 50 - ( 1 / 60 + 1 / 75 ) = - 1 / 100 . [ - ve sign means emptying ] the third pipe alone can empty the cistern in 100 min . answer : b | a = 1 / 60
b = 1 / 75
c = a + b
d = 1 / 50
e = c - d
f = 1/(e)
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a ) 250 , b ) 450 , c ) 750 , d ) 800 , e ) none of them | b | multiply(divide(const_1, add(add(const_4, 3), const_1)), 1200) | a and b undertake to do a piece of work for rs . 1200 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of b . | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a β s share = rs . ( 1200 * 4 / 8 ) = rs . 600 , b ' s share = rs . ( 1200 * 3 / 8 ) = rs . 450 c ' s share = rs . [ 1200 - ( 300 + 225 Β» ) = rs . 150 . answer is b" | a = 4 + 3
b = a + 1
c = 1 / b
d = c * 1200
|
a ) 22 , b ) 89.5 , c ) 76.9 , d ) 83.5 , e ) 11 | d | subtract(divide(9890, 92), 24) | a trader sells 92 meters of cloth for rs . 9890 at the profit of rs . 24 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 9890 / 92 = rs . 107.5 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 107.5 - rs . 24 = rs . 83.5 answer : d" | a = 9890 / 92
b = a - 24
|
a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4 | b | divide(divide(divide(lcm(21, 57), 57), const_4), const_4) | what is the least value of x . so that 21 x 57 is divisible by 3 . | "explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 1 + x + 5 + 7 = 15 + x least value of x may be 0 therefore 15 + 0 = 15 is divisible by 3 . answer : option b" | a = math.lcm(21, 57)
b = a / 57
c = b / 4
d = c / 4
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a ) 12 , b ) 16 , c ) 36 , d ) 48 , e ) 98 | c | lcm(multiply(3, 3), multiply(4, 3)) | the ratio of numbers is 3 : 4 and their h . c . f is 3 . their l . c . m is : | "let the numbers be 3 x and 4 x . then their h . c . f = x . so , x = 3 . so , the numbers are 9 and 12 . l . c . m of 9 and 12 = 36 . answer : c" | a = 3 * 3
b = 4 * 3
c = math.lcm(a, b)
|
a ) 67.4 kg , b ) 57.75 kg , c ) 58.85 kg , d ) 49.4 kg , e ) none of these | c | divide(add(multiply(58.4, 20), subtract(65, 56)), 20) | the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 65 kg . what is the correct weight ? | "actual total weight is ( 20 Γ£ β 58.4 - 56 + 65 ) = 1177 kgs actual average weight is 1177 / 20 = 58.85 kgs answer is c" | a = 58 * 4
b = 65 - 56
c = a + b
d = c / 20
|
a ) 297 , b ) 879 , c ) 342 , d ) 762 , e ) 224 | e | subtract(subtract(400, divide(multiply(400, 30), const_100)), divide(multiply(subtract(400, divide(multiply(400, 30), const_100)), 20), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 30 % and 20 % is ? | "400 * ( 70 / 100 ) * ( 80 / 100 ) = 224 answer : e" | a = 400 * 30
b = a / 100
c = 400 - b
d = 400 * 30
e = d / 100
f = 400 - e
g = f * 20
h = g / 100
i = c - h
|
a ) 196 m , b ) 194 m , c ) 186 m , d ) 296 m , e ) 106 m | a | multiply(divide(add(const_4, const_3), const_3), 84) | a runs 1 ΒΎ times as fast as b . if a gives b a start of 84 m , bow far must winning post be so that a and b might reach it at the same time ? | . ratio of the rates of a and b = 7 / 4 : 1 = 7 : 4 . so , in a race of 7 m , a gains 3 m over b . : . 3 m are gained by a in a race of 7 m . : . 84 m are gained by a in a race of ( 7 / 3 x 84 ) m = 196 m . : . winning post must be 196 m away from the starting point . answer is a . | a = 4 + 3
b = a / 3
c = b * 84
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(multiply(0.5, 12), const_1) | a box contains 13 apples , 12 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ? | "p ( choosing a red apple 7 times in a row ) = 12 / 13 * 11 / 12 * 10 / 11 * 9 / 10 * 8 / 9 * 7 / 8 * 6 / 7 = 6 / 13 < 0.5 the answer is e ." | a = 0 * 5
b = a + 1
|
a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | a | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 40 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the lowest possible sale price ? | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 20 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , a is the answer" | a = 80 / 2
|
a ) 15 , b ) 16 , c ) 19 , d ) 24 , e ) 36 | d | multiply(6, 4) | a = { 2 , 3 , 5 , 7 , 11 } b = { 2 , 4 , 6 , 19 } two integers will be randomly selected from sets a and b , one integer from set a and one from set b , and then multiplied together . how many different products can be obtained ? | step 1 : find out the number of products you get . 5 distinct numbers in set a and 4 distinct in set b so number of products = 5 * 4 = 20 step 2 : remove the products that appear more than once . notice that 5 , 7 , 11 and 13 are primes and none of their multiples are in either set . so ignore them . we just need to focus on 2 and 3 of set a and 2 , 4 and 6 of set b . 2 , 3 2 , 4 , 6 the only product repeated when you take a number from each set is 12 . ( 3 * 4 and 2 * 6 ) rest all are distinct . answer = 20 - 1 = 24 note here that the second step will involve manual calculation since it will depend on the specific numbers you have in the two sets . d | a = 6 * 4
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a ) 770 , b ) 870 , c ) 1825 , d ) 1925 , e ) 925 | d | divide(multiply(multiply(100, add(const_10, const_1)), add(const_3, const_4)), subtract(add(const_10, const_1), add(const_3, const_4))) | if one - seventh of a number exceeds its eleventh part by 100 then the number is β¦ | let no be : x 1 / 7 x - 1 / 11 x = 100 4 x / 77 = 100 x = 7700 / 4 = 1925 answer : d | a = 10 + 1
b = 100 * a
c = 3 + 4
d = b * c
e = 10 + 1
f = 3 + 4
g = e - f
h = d / g
|
['a ) 48 degrees', 'b ) 54 degrees', 'c ) 72 degrees', 'd ) 84 degrees', 'e ) 72 degrees'] | e | divide(subtract(const_180, divide(const_360, 10)), const_2) | a regular 10 sided area is inscribed in a circle . if a and b are adjacent vertices of the pentagon and o is the center of the circle , what is the value of β oab ? | its a regular 10 sided so divide the globe by 10 . = 360 / 10 = 36 36 would be the angle at o . angles at a and b are equivalent , so 36 + 2 x = 180 2 x = 144 x = 72 angles oab and oba will be 72 degrees each . e | a = 360 / 10
b = const_180 - a
c = b / 2
|
a ) 3.9 , b ) 4.2 , c ) 4.5 , d ) 4.8 , e ) 5.1 | c | divide(add(subtract(6, 3), add(6, 3)), const_2) | a rower whose speed is 6 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 3 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? | "time upstream = d / 3 time downstream = d / 9 total time = d / 3 + d / 9 = 4 d / 9 average speed = 2 d / ( 4 d / 9 ) = 4.5 km / hr the answer is c ." | a = 6 - 3
b = 6 + 3
c = a + b
d = c / 2
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | multiply(subtract(const_3, const_1), 3) | list i : 3 , 4 , 8 , 19 list ii : x , 3 , 4 , 8 , 19 | we start by calculating the median of the numbers of list i : 3 , 4 , 8 , 19 . we see that the numbers in the list are in order already and , since we have an even number of numbers , the median is the average of the two middle numbers . median = ( 4 + 8 ) / 2 median = 12 / 2 median = 6 the median of list i is 6 . looking at list ii : x , 3 , 4 , 8 , 19 , we see that we have an odd number of terms . thus , when the list is ordered from least to greatest the median must be the middle term . since the medians of the two lists must be equal , we know that the median of list ii must be 6 and therefore x is 6 . the answer is a . | a = 3 - 1
b = a * 3
|
a ) 8 / 12 , b ) 6 / 16 , c ) 4 / 16 , d ) 6 / 10 , e ) 4 / 10 | a | divide(divide(subtract(20, 4), add(const_1, const_1)), add(divide(subtract(20, 4), add(const_1, const_1)), 4)) | there are 4 more women than there are men on a local co - ed softball team . if there are a total of 20 players on the team , what is the ratio of men to women ? | "w = m + 4 w + m = 20 m + 4 + m = 20 2 m = 16 m = 8 w = 12 ratio : 8 : 12 ans : a" | a = 20 - 4
b = 1 + 1
c = a / b
d = 20 - 4
e = 1 + 1
f = d / e
g = f + 4
h = c / g
|
a ) 9 : 12 , b ) 9 : 18 , c ) 9 : 26 , d ) 9 : 32 , e ) 9 : 38 | d | divide(9, subtract(53, const_3)) | it is currently 9 : 53 pm . what time was it in the morning exactly 149,061 minutes ago ? | divide by 60 to convert 149,061 minutes to hours : 149,061 / 60 = 2,484 r 21 . that is 2,484 hours and 21 minutes . all of the answers are in the 9 am hour before : 53 , thus we can assume 2,484 previous must be 9 : 53 am . 21 minutes before that is 9 : 32 am . d | a = 53 - 3
b = 9 / a
|
a ) 1 kmph , b ) 7 kmph , c ) 98 kmph , d ) 6 kmph , e ) 8 kmph | e | divide(subtract(20, 4), const_2) | a man can row his boat with the stream at 20 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 20 us = 4 s = ? s = ( 20 - 4 ) / 2 = 8 kmph answer : e" | a = 20 - 4
b = a / 2
|
a ) 19 , b ) 20 , c ) 30 , d ) 33 , e ) 34 | c | multiply(6, subtract(6, const_1)) | 6 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "though 2 * ( 6 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 5 , so he ' ll play 10 matches . similarly , second guy has to play with the rest of 4 as his 2 games with the first guy are already played . so he plays 8 matches . this continues like this and the total matches are 10 + 8 + 6 . . . + 2 10 + 8 + . . . + 2 = 2 ( 5 + 4 + . . . + 1 ) = 2 ( ( 5 * 6 ) / 2 ) = 5 * 6 = 30 . answer : c" | a = 6 - 1
b = 6 * a
|
a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 17 kmph | e | subtract(51, divide(multiply(51, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 51 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 51 / 3 = 17 kmph answer : e" | a = 51 * 2
b = a / 3
c = 51 - b
|
a ) 2 , b ) 3 , c ) 4 , d ) 11 , e ) can not be determined from the information given . | d | subtract(multiply(4, 3), const_1) | the average ( arithmetic man ) of 3 integers a , b , and c is exactly 4 times the median . if a < b < c and a = 0 , what is the value of c / b ? | the average of three integers a , b , and c is exactly twice the median - - > ( a + b + c ) / 3 = 4 b - - > since a = 0 , then ( 0 + b + c ) / 3 = 4 b - - > c = 11 b - - > c / b = 11 . answer : d . | a = 4 * 3
b = a - 1
|
a ) 1960 , b ) 1965 , c ) 1970 , d ) 1975 , e ) 1980 | e | add(1950, add(1.75, add(4, 3))) | in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.75 times as old as robert ? | "in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 ro = 10 , ri = 40 now for each year we can calculate : 1960 : ri = 50 , ro = 20 1965 : ri = 55 , ro = 25 1980 : ri = 40 , ro = 70 thus 8 ans : e" | a = 4 + 3
b = 1 + 75
c = 1950 + b
|
a ) 75 , b ) 65 , c ) 45 , d ) 25 , e ) 35 | d | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 50), add(const_100, 50))), const_100) | raman ' s salary was decreased by 50 % and subsequently increased by 50 % . how much percent does he loss | "explanation : let the origianl salary = rs . 100 it will be 150 % of ( 50 % of 100 ) = ( 150 / 100 ) * ( 50 / 100 ) * 100 = 75 so new salary is 75 , it means his loss is 25 % answer : option d" | a = 100 * 100
b = 100 - 50
c = 100 + 50
d = b * c
e = a - d
f = e / 100
|
a ) 5.26 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.53 % | e | multiply(subtract(inverse(divide(930, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 930 gm . for a kg . his gain is β¦ % . | "his percentage gain is 100 * 7 / 930 as he is gaining 70 units for his purchase of 930 units . so 7.53 % . answer : e" | a = 4 + 1
b = a * 2
c = b * 100
d = 930 / c
e = 1/(d)
f = e - 1
g = f * 100
|
a ) $ 100,000 , b ) $ 125,000 , c ) $ 150,000 , d ) $ 200,000 , e ) $ 250,000 | a | divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2))) | the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 62,500 , how much profit did the business make ? | "majority owner + two partners = 0.25 * p + 2 * ( 0.25 ) * ( 0.75 ) * p = 0.25 * p + ( 0.5 ) * ( 0.75 ) * p = 0.25 * p + ( 0.375 ) * p = ( 0.625 ) * p this equals $ 62,500 . ( 0.625 ) * p = 62,500 - - - - - divide both sides by 625 ( 0.001 ) * p = 100 - - - - - - multiply by 1000 p = 100,000 answer = a" | a = 1 + 4
b = 100 * a
c = 100 * b
d = 25 / 100
e = 25 / 100
f = 25 / 100
g = 1 - f
h = e * g
i = h * 2
j = d + i
k = c / j
|
a ) 16 / 625 , b ) 56 / 625 , c ) 96 / 625 , d ) 126 / 625 , e ) 156 / 625 | c | subtract(1, divide(const_2, 5)) | when a random experiment is conducted , the probability that event a occurs is 1 / 5 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 5 * 1 / 5 * 4 / 5 * 4 / 5 = 16 / 625 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 16 / 625 ) = 96 / 625 the answer is c ." | a = 2 / 5
b = 1 - a
|
a ) 28 sec , b ) 36 sec , c ) 48 sec , d ) 52 sec , e ) 56 sec | c | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.9, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is : | "relative speed = ( 60 + 90 ) km / hr = 150 x 5 / 18 = 120 / 3 m / sec distance covered = ( 1.10 + 0.9 ) km = 2 km = 2000 m . required time = 2000 x 3 / 125 = 48 sec . answer : c ." | a = 1 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 0 * 9
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 1.8 , b ) 2.1 , c ) 2.4 , d ) 2.7 , e ) 3.0 | c | subtract(6, multiply(divide(multiply(6, 30), const_100), const_2)) | a 6 - liter solution is 30 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | "let x be the amount of pure alcohol required . 0.3 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 1.8 x = 2.4 liters the answer is c ." | a = 6 * 30
b = a / 100
c = b * 2
d = 6 - c
|
a ) 15 , b ) 18 , c ) 20 , d ) 26 , e ) none | d | divide(add(add(17, const_4), subtract(32, const_4)), const_2) | find the average of all the numbers between 17 and 32 which are divisible by 3 . | "sol . average = ( 21 + 24 + 27 + 30 / 4 ) = 102 / 4 = 25.5 answer d" | a = 17 + 4
b = 32 - 4
c = a + b
d = c / 2
|
a ) 2 hours , b ) 4 hours 12 minutes , c ) 4 hours , d ) 5 hours , e ) none | b | divide(72, add(13, 4)) | a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr . find the time taken by the boat to go 72 km downstream ? | "solution speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 72 km downstream = ( 72 / 17 ) hrs = 4 hrs 12 minutes . answer b" | a = 13 + 4
b = 72 / a
|
a ) 45 , b ) 50 , c ) 55 , d ) 100 , e ) 60 | a | multiply(divide(add(const_1, subtract(const_10, const_1)), const_2), subtract(const_10, const_1)) | if you add all the numbers on your mobile phone , what is the answer ? | 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 are in your mobile phone . therefore 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 answer is a | a = 10 - 1
b = 1 + a
c = b / 2
d = 10 - 1
e = c * d
|
a ) 0 , b ) 5 , c ) 383 , d ) 875 , e ) 965 | c | divide(multiply(190, 192), const_4) | what is the sum of the integers from - 190 to 192 inclusive ? | "sum / n = average . sum = ( average ) ( n ) average = a + b / 2 = 190 + 192 / 2 = 1 number of items ( n ) = b - a + 1 = 192 - ( - 190 ) + 1 = 192 + 191 = 383 . sum = average * n = 1 * 383 = 383 . answer is c" | a = 190 * 192
b = a / 4
|
a ) 22 % , b ) 23 % , c ) 24 % , d ) 26 % , e ) 34 % | d | divide(22, divide(subtract(const_100, 15), const_100)) | in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 15 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | since 15 % of the customers who sample the candyare notcaught , then 88 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.85 = 0.22 ; { % of customers who sample candy } = 0.259 answer : d . | a = 100 - 15
b = a / 100
c = 22 / b
|
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