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 ) 5 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min | c | multiply(33, divide(1, 11)) | a pipe can fill a cistern in 33 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 33 minutes 1 / 11 part filled in = 33 * 1 / 11 = 3 minutes answer is c" | a = 1 / 11
b = 33 * a
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a ) 46.3 % , b ) 50.4 % , c ) 54.2 % , d ) 57.8 % , e ) 61.3 % | d | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(8)))) | a wooden cube whose edge length is 8 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ? | "there are a total of 8 * 8 * 8 = 512 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 6 * 6 * 6 = 216 cubes . the number of cubes with at least one side painted red is 512 - 216 = 296 cubes the probability that a cube has at least one side painted red is 296 / 512 which is about 57.8 % the answer is d ." | a = 1 * 4
b = volume_cube / (
c = 1 - b
d = 100 * c
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a ) 45 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | a | multiply(36, inverse(subtract(const_1, divide(8, 40)))) | x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 36 days . how long will y take to complete the work ? | work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 36 days whole work will be done by y in 36 * 5 / 4 = 45 days answer is a | a = 8 / 40
b = 1 - a
c = 1/(b)
d = 36 * c
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(subtract(5, 3), const_1) | mr . hanson has a pool of money from which to give out christmas bonuses to his 5 employees . he wants all his employees to have identical bonuses , but when he divides them out , he finds he has 2 dollars left over , as the payroll system will only accept whole dollar amounts . the next christmas , mr . hanson discovers the pool for the bonuses is 3 times the size . this time , how much money will be left over ? | set x = to amount of money for bonuses , and y to be the amount of each individual bonus . x = 5 * y + 2 . 3 x = 15 * y + 6 3 x = 3 * 5 * y + 6 3 x = 5 ( 3 ( y + 1 ) ) + 1 remainder is then 1 , so option b . | a = 5 - 3
b = a - 1
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a ) 4586970843 , b ) 4686970743 , c ) 4690910862 , d ) 4586870843 , e ) none | c | multiply(469138, 9999) | calculate 469138 x 9999 = ? | "answer 469138 x 9999 = 469138 x ( 10000 - 1 ) = 4691380000 - 469138 = 4690910862 . option : c" | a = 469138 * 9999
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a ) 55 : 23 , b ) 52 : 33 , c ) 52 : 53 , d ) 27 : 29 , e ) 51 : 34 | d | divide(add(const_100, 8), add(const_100, 16)) | the cash difference between the selling prices of an book at a profit of 8 % and 16 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 108 % of x / 116 % of x = 108 / 116 = 27 / 29 = 27 : 29 d" | a = 100 + 8
b = 100 + 16
c = a / b
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a ) 4 , b ) 7 , c ) 8 , d ) 9 , e ) 26 | d | divide(subtract(divide(44, const_2), sqrt(subtract(power(divide(44, const_2), const_2), multiply(const_4, 117)))), const_2) | if a rectangular billboard has an area of 117 square feet and a perimeter of 44 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 117 and a perimeter of 44 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 117 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 117 , so the correct answer is let ' s test answer d : 9 if . . . the shorter side = 9 . . . the area = 117 . . . . 117 / 9 = 13 = the longer side perimeter = 9 + 9 + 13 + 13 = 44 d" | a = 44 / 2
b = 44 / 2
c = b ** 2
d = 4 * 117
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
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a ) 90 kms . , b ) 100 kms . , c ) 110 kms . , d ) 120 kms . , e ) 130 kms . | d | multiply(divide(120, add(60, 60)), 120) | bird is flying 120 km / hr between b to r . two trains at b to r at 60 kmph . the distance traveled by the bird before it is killed ? | it will depend on the distance between b and r . if distance between b and r is 120 kms , time taken by trains before collision = 120 / ( 60 + 60 ) = 1 hr distance traveled by bird in 1 hr = 120 kms . so the distance traveled by the bird before it is killed = 120 kms . answer : d | a = 60 + 60
b = 120 / a
c = b * 120
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a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(15, multiply(3, const_4)) | a = 5 ^ 15 - 125 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itself . on the other hand the stem says that x could get how many different number not must get different number ( this is very important issue ) as our friends say , if we simplify numerator more we can obtain : 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 ( 124 ) = 5 ^ 12 ( 31 * 2 * 2 ) divided by x and we are told that this fraction is an integer . so , x could be ( not must be ) 5 , 31 , or 2 ! ! ! so , x could get none different values and answer is a . . . ." | a = 3 * 4
b = 15 - a
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a ) 200 m , b ) 20 m , c ) 403 m , d ) 20 m , e ) 23 m | a | subtract(600, divide(multiply(subtract(600, 100), 4), 5)) | in a 600 m race , the ratio of the speeds of two contestants a and b is 5 : 4 . a has a start of 100 m . then , a wins by : | to reach the winning post a will have to cover a distance of ( 600 - 100 ) m , i . e . , 500 m . while a covers 5 m , b covers 4 m . while a covers 500 m , b covers 4 x 500 / 5 m = 400 m . thus , when a reaches the winning post , b covers 400 m and therefore remains 200 m behind . a wins by 200 m . answer : a | a = 600 - 100
b = a * 4
c = b / 5
d = 600 - c
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a ) 20 years , b ) 30 years , c ) 15 years , d ) 25 years , e ) 28 years | b | divide(multiply(subtract(25, 5), 3), 2) | at present the ratio between the ages of arun and deepak is 2 : 3 . after 5 years , arun ' s age will be 25 years . what is the age of deepak at present ? | "let the present ages of arun and deepak be 2 x years and 3 x years respectively 2 x + 5 = 25 2 x = 20 x = 10 deepak ' s age = 3 x = 30 years answer is b" | a = 25 - 5
b = a * 3
c = b / 2
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a ) 6 , b ) 5 , c ) 4 , d ) 10 , e ) 8 | b | divide(add(200, 100), 60) | the length of the bridge is 200 m , which a 100 m long train cross in 60 sec . then what is the speed of train ? | speed = length / time length = ( 200 + 100 ) = 300 m time = 60 sec speed = 300 / 60 = 5 m / sec answer b | a = 200 + 100
b = a / 60
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a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | if a = 105 and a ^ 3 = 21 × 25 × 45 × e , what is the value of e ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 × 25 × 45 × e = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x e = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x e = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x e e = 7 ^ 2 = 49 answer d" | a = 105 ** 3
b = 21 * 25
c = b * 45
d = a / c
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a ) $ 3 , b ) $ 7 , c ) $ 1 , d ) $ 5 , e ) $ 9 | d | divide(multiply(50, subtract(const_100, 10)), multiply(const_100, 9)) | if i earn a profit 50 $ in a day and i have 9 employees . after keeping 10 percent profit for me , and distribute remianing money equally among my employees then how much each employee would get ? | since i earned $ 50 sp 10 percent would be 5 $ and 45 $ would remain . dividing 45 by 9 , each employee would get 5 $ each and hince option d is correct . | a = 100 - 10
b = 50 * a
c = 100 * 9
d = b / c
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a ) rs . 175 , b ) rs . 92 , c ) rs . 112.50 , d ) rs . 116.50 , e ) none of these | a | multiply(divide(const_100, 8), 14) | a 14 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 14 , investment = rs . ( 100 / 8 x 14 ) = rs . 175 ∴ market value of rs . 100 stock = rs . 175 answer a" | a = 100 / 8
b = a * 14
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a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(84, const_3), divide(84, 6)) | in the third grade of windblown school there are 84 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "total = 84 failed in math = 84 / 3 = 28 failed in literature = 84 / 6 = 14 the least failed in both can be 0 while max can be 14 answer a" | a = 84 / 3
b = 84 / 6
c = min(a)
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a ) 1 : 9 , b ) 1 : 2 , c ) 1 : 8 , d ) 1 : 1 , e ) 1 : 6 | c | divide(1, const_4) | the diameters of two spheres are in the ratio 1 : 2 what is the ratio of their volumes ? | "1 : 8 answer : c" | a = 1 / 4
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a ) 237 , b ) 320 , c ) 177 , d ) 166 , e ) 111 | b | floor(divide(8200, add(25, divide(2.5, const_100)))) | find the number of shares that can be bought for rs . 8200 if the market value is rs . 25 each with brokerage being 2.5 % . | "explanation : cost of each share = ( 25 + 2.5 % of 25 ) = rs . 25.625 therefore , number of shares = 8200 / 25.625 = 320 answer : b" | a = 2 / 5
b = 25 + a
c = 8200 / b
d = math.floor(c)
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a ) 1 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | d | divide(multiply(30, divide(multiply(10, const_1000), const_60)), const_1000) | find the distance covered by a man walking for 30 min at a speed of 10 km / hr ? | "distance = 10 * 20 / 60 = 5 km answer is d" | a = 10 * 1000
b = a / const_60
c = 30 * b
d = c / 1000
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a ) 1 / 3 , b ) 2 / 3 , c ) 7 / 5 , d ) 5 / 7 , e ) 3 / 2 | c | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_2, const_3)), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))), subtract(divide(const_1, const_2), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))) | a certain ball team has an equal number of right - and left - handed players . on a certain day , two - thirds of the players were absent from practice . of the players at practice that day , one - third were right handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of left handed players who were not at practice ? | "say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 12 absent and 6 present . of the players at practice that day , one - third were right - handed - - > 6 * 1 / 3 = 2 were right - handed and 4 left - handed . the number of right - handed players who were not at practice that day is 9 - 2 = 7 . the number of left - handed players who were not at practice that days is 9 - 4 = 5 . the ratio = 7 / 5 . answer : c ." | a = 1 / 2
b = 2 / 3
c = 1 - b
d = 2 / 3
e = 1 - d
f = 2 / 3
g = e * f
h = c - g
i = a - h
j = 1 / 2
k = 2 / 3
l = 1 - k
m = 2 / 3
n = l * m
o = j - n
p = i / o
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a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 5 : 11 , e ) 7 : 3 | d | divide(add(multiply(12, divide(1, add(1, 2))), multiply(4, divide(1, add(1, 3)))), subtract(add(12, 4), add(multiply(12, divide(1, add(1, 2))), multiply(4, divide(1, add(1, 3)))))) | a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 4 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 4 ounces in the ratio 1 : 3 or , x + 3 x = 4 gives x ( alcohol ) = 1 and remaining water = 3 mixing both we have alcohol = 4 + 1 and water = 8 + 3 ratio thus alcohol / water = 5 / 11 answer d" | a = 1 + 2
b = 1 / a
c = 12 * b
d = 1 + 3
e = 1 / d
f = 4 * e
g = c + f
h = 12 + 4
i = 1 + 2
j = 1 / i
k = 12 * j
l = 1 + 3
m = 1 / l
n = 4 * m
o = k + n
p = h - o
q = g / p
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a ) 100 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(84, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 84 is : | "solution solution let the number be x . x - 16 % of x = 84 x - 16 / 100 x = 84 x - 4 / 25 x = 84 21 / 25 x = 84 x = ( 84 x 25 / 21 ) = 100 answer a" | a = 84 * 100
b = 100 - 16
c = a / b
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a ) rs . 520 , b ) rs . 480 , c ) rs . 420 , d ) rs . 200 , e ) rs . 600 | d | divide(multiply(subtract(21, 20), const_100), const_2) | find the principal which yields a simple interest of rs . 20 and compound interest of rs . 21 in two years , at the same percent rate per annum ? | "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 21 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 21 – 20 ) / 20 ] * 100 = 5 % p . a . let the principal be rs . x time = t = 2 years % rate = 5 % p . a . si = ( prt / 100 ) 20 = ( x * 5 * 2 ) / 100 x = rs . 200 answer : d" | a = 21 - 20
b = a * 100
c = b / 2
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a ) 14 , b ) 15 , c ) 20 , d ) 21 , e ) 24 | d | divide(multiply(subtract(32, 4), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 4 years rahul age will be 32 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 4 = 32 = > x = 7 so deepak age is = 3 ( 7 ) = 21 option d" | a = 32 - 4
b = a * 3
c = b / 4
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a ) 510 , b ) 450 , c ) 500 , d ) 520 , e ) 589 | b | divide(850, multiply(subtract(78, 1), const_0_2778)) | a train 850 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 850 + x ) / 60 = 65 / 3 x = 450 m . answer : option b" | a = 78 - 1
b = a * const_0_2778
c = 850 / b
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a ) 85 , b ) 95 , c ) 105 , d ) 120 , e ) 210 | a | multiply(divide(140, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 140 of the workers of that company are men , how many of the workers are women ? | "20 % of the workerswithoutretirement plan are women , this means 80 % of the workerswithoutretirement plan are men right ? makes 100 % = no retirement plan . then 40 % of those wo have a retirement plan are men , which means 60 % of those who have a retirement plan are women . these percentages are always refering to the subgroupsretirement plan = yesandno retirement plan . make the total workers equal x and work from there and it should be fine > x / 3 have no retirement plan at all . i struggled with the question a lot , it was hard for me although usually i have no problems with sets questions like this . i believe it must be a 700 + question . yes , you are right . i read the question incorrectly ! i glazed overnotfor some reason . if 20 % of the workers who do not have a retirement plan are women , 80 % who do not have retirement plan are men . also 40 % who do have retirement plan are men . since 1 / 3 do not have retirement plan and 2 / 3 do have , it becomes a simple weighted average question . cavg = ( 80 * 1 + 40 * 2 ) / 3 = 160 / 3 % are men total = 120 * 300 / 160 = 225 no . of women = 225 - 140 = 85 a" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 140 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 2 , d ) 4 / 5 , e ) 4 / 7 | c | divide(divide(6, 15), divide(2, 15)) | if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a ∪ b ) = 6 / 15 find p ( a | b ) | "p ( a | b ) = p ( a ∪ b ) / p ( b ) p ( a | b ) = ( 6 / 15 ) / ( 4 / 15 ) = 3 / 2 . c" | a = 6 / 15
b = 2 / 15
c = a / b
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a ) 20 / 9 , b ) 40 / 9 , c ) 50 / 9 , d ) 24 / 7 , e ) 80 / 9 | d | divide(const_1, add(divide(const_1, 8), divide(const_1, 6))) | worker a takes 8 hours to do a job . worker b takes 6 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "one day work of a = 1 / 8 one day work of b = 1 / 6 so one day work of a and b together = 1 / 8 + 1 / 6 = 7 / 24 so total days required = 24 / 7 answer : d" | a = 1 / 8
b = 1 / 6
c = a + b
d = 1 / c
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a ) 54 , b ) 32 , c ) 75 , d ) 28 , e ) 11 | d | sqrt(divide(multiply(square_area(7), 8), inverse(const_2))) | the length of the rectangular field is double its width . inside the field there is square shaped pond 7 m long . if the area of the pond is 1 / 8 of the area of the field . what is the length of the field ? | "a / 8 = 7 * 7 = > a = 7 * 7 * 8 x * 2 x = 7 * 7 * 8 x = 14 = > 2 x = 28 answer : d" | a = square_area * (
b = a / 8
c = 1/(2)
d = math.sqrt(b)
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a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | b | add(divide(subtract(multiply(floor(divide(97, 9)), 9), multiply(add(floor(divide(32, 9)), const_1), 9)), 9), const_1) | how many numbers from 32 to 97 are exactly divisible by 9 ? | "option ' b ' 32 / 9 = 3 and 97 / 9 = 10 = = > 10 - 3 = 7 numbers" | a = 97 / 9
b = math.floor(a)
c = b * 9
d = 32 / 9
e = math.floor(d)
f = e + 1
g = f * 9
h = c - g
i = h / 9
j = i + 1
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a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 12 | c | multiply(4, const_2) | the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 4 ) , how many units is p from the y - axis ? | "the x - axis is 4 units from the point p . thus the y - axis is 8 units from the point p . the answer is c ." | a = 4 * 2
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a ) 278.4 mtrs , b ) 378.4 mtrs , c ) 478.4 mtrs , d ) 488.4 mtrs , e ) 567.6 mtrs | e | multiply(66, divide(multiply(86, 8), multiply(20, 4))) | if 20 men can build a wall 66 metres long in 4 days , what length of a similar can be built by 86 men in 8 days ? | "if 20 men can build a wall 66 metres long in 4 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) / ( 4 * 20 ) = 567.6 mtrs answer : e" | a = 86 * 8
b = 20 * 4
c = a / b
d = 66 * c
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a ) 10 % , b ) 15 % , c ) 50 % , d ) 25 % , e ) 30 % | c | multiply(divide(250, 500), const_100) | the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 250 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 250 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 250 ) * 100 / 500 = 50 % answer : option c" | a = 250 / 500
b = a * 100
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a ) 440 m , b ) 500 m , c ) 260 m , d ) 430 m , e ) 450 m | b | subtract(multiply(divide(multiply(78, const_1000), const_3600), const_60), 800) | a train , 800 meter long is running with a speed of 78 km / hr . it crosses a tunnel in 1 minute . what is the length of the tunnel ( in meters ) ? | "explanation : distance = 800 + x meter where x is the length of the tunnel time = 1 minute = 60 seconds speed = 78 km / hr = 78 × 10 / 36 m / s = 130 / 6 = 65 / 3 m / s distance / time = speed ( 800 + x ) / 60 = 65 / 3 = > 800 + x = 20 × 65 = 1300 = > x = 1300 - 800 = 500 meter answer : option b" | a = 78 * 1000
b = a / 3600
c = b * const_60
d = c - 800
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a ) 50 , b ) 75 , c ) 150 , d ) 125 , e ) 100 | d | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 150), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 120)), 30) | a library has an average of 150 visitors on sundays and 120 on other days . the avearge number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begin with sunday , so there will be five sundays in the month required average = ( 150 * 5 + 120 * 25 ) / 30 = 3750 / 30 = 125 answer is d" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 150
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 120
l = e + k
m = l / 30
|
a ) 30 , b ) 120 , c ) 55 , d ) 35 , e ) none of these | e | divide(divide(divide(divide(divide(520, const_3), const_3), const_4), const_4), const_4) | find the smallest number which should be multiplied with 520 to make it a perfect square . | "520 = 26 * 20 = 2 * 13 * 22 * 5 = 23 * 13 * 5 required smallest number = 2 * 13 * 5 = 130 130 is the smallest number which should be multiplied with 520 to make it a perfect square . answer : e" | a = 520 / 3
b = a / 3
c = b / 4
d = c / 4
e = d / 4
|
a ) 60 , b ) 70 , c ) 80 , d ) 90 , e ) 100 | a | subtract(100, subtract(subtract(200, 120), 40)) | in a company of 200 employees , 120 are females . a total of 100 employees have advanced degrees and the rest have a college degree only . if 40 employees are males with a college degree only , how many employees are females with advanced degrees ? | "the number of males is 200 - 120 = 80 . the number of males with advanced degrees is 80 - 40 = 40 . the number of females with advanced degrees is 100 - 40 = 60 . the answer is a ." | a = 200 - 120
b = a - 40
c = 100 - b
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a ) 5 / 6 , b ) 5 / 9 , c ) 5 / 1 , d ) 5 / 3 , e ) 5 / 12 | e | divide(divide(subtract(8.75, 7.50), subtract(8.75, 4.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 4.5)))) | in what ratio should a variety of rice costing rs . 4.5 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 4.5 ) = 5 / 12 answer : e" | a = 8 - 75
b = 8 - 75
c = a / b
d = 8 - 75
e = 8 - 75
f = d / e
g = 1 - f
h = c / g
|
a ) 400 , b ) 625 , c ) 1818 , d ) 2500 , e ) 10 000 | c | divide(60, divide(2, 60)) | in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 60 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "the percent of tagged fish in the second catch is 2 / 60 * 100 = 3.33 % . we are told that 3.33 % approximates the percent of tagged fish in the pond . since there are 60 tagged fish , then we have 0.033 x = 60 - - > x = 1,818 . answer : c ." | a = 2 / 60
b = 60 / a
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a ) 500 m , b ) 1110 m , c ) 950 m , d ) 1000 m , e ) 1300 m | a | multiply(divide(multiply(15, const_1000), const_60), 10) | a man walking at a rate of 15 km / hr crosses a bridge in 10 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 10 minutes = 15 / 18 * 10 * 60 = 500 m answer is a" | a = 15 * 1000
b = a / const_60
c = b * 10
|
a ) 1361 , b ) 1362 , c ) 1950 , d ) 1364 , e ) 1365 | c | divide(multiply(61, add(61, 12)), 13) | 12 + 13 + 14 + . . . 61 + 62 + 63 = ? | "sum = 12 + 13 + 14 + . . . 61 + 62 + 63 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 63 positive integers = 63 * 64 / 2 sum of first 11 positive integers = 11 * 12 / 2 sum = 12 + 13 + 14 + . . . 61 + 62 + 63 = 63 * 64 / 2 - 11 * 12 / 2 = 1950 answer : c" | a = 61 + 12
b = 61 * a
c = b / 13
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | a | divide(divide(21, 3), 21) | tickets numbered from 1 to 21 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 ] p ( e ) = n ( e ) / n ( s ) = 7 / 21 = 1 / 3 the answer is a ." | a = 21 / 3
b = a / 21
|
a ) 2000 , b ) 2100 , c ) 2200 , d ) 2300 , e ) 2400 | b | multiply(divide(63, 3), const_100) | a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 63 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480 | "1 percent for 3 years = 63 1 percent for 1 year = 21 = > 100 percent = 2100 answer : b" | a = 63 / 3
b = a * 100
|
a ) 524.48 , b ) 556.02 , c ) 574.36 , d ) 585.64 , e ) 595.46 | d | multiply(128.5, power(add(const_4, const_1), const_4)) | ( 128.5 x 64 ) + ( 13.8 x 465 ) = ? x 25 | "explanation : ? = ( 128.5 x 64 ) + ( 13.8 x 465 ) / 25 = 8224 + 6417 / 25 = 585.64 answer : option d" | a = 4 + 1
b = a ** 4
c = 128 * 5
|
a ) 160 , b ) 190 , c ) 225 , d ) 250 , e ) 500 | a | sqrt(divide(multiply(128, const_100), divide(50, const_100))) | 128 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ? | let b be the number of boys in the school . 128 = xb / 100 b = 0.5 x 12800 = 0.5 x ^ 2 x ^ 2 = 25600 x = 160 the answer is a . | a = 128 * 100
b = 50 / 100
c = a / b
d = math.sqrt(c)
|
a ) 1 / 5 , b ) 3 / 10 , c ) 2 / 5 , d ) 1 / 2 , e ) 3 / 5 | c | divide(choose(subtract(5, 1), 1), choose(5, 2)) | a basket contains 5 apples , of which 1 is spoiled and the rest are good . if henry is to select 2 apples from the basket simultaneously and at random , what is the possibility that the 2 apples selected will include the spoiled apple ? | "p ( both apples are not spoiled ) = 4 c 2 / 5 c 2 = 4 * 3 / 5 * 4 = 3 / 5 p ( one of the apple spoiled ) = 1 - 3 / 5 = 2 / 5 answer is c ." | a = 5 - 1
b = math.comb(a, 1)
c = math.comb(5, 2)
d = b / c
|
a ) 40 / 3 , b ) 40 , c ) 120 , d ) 360 , e ) 420 | d | multiply(multiply(40, 3), 3) | the difference in compound interest earned on a deposit ( compounded annually ) in year 1 and year 2 is $ 40 . had the interest rate been 3 times its present value , the difference w would have been how much ? | case 1 : deposit = $ x ; rate of increase = r . interest yearned in 1 year = xr . deposit in 1 year = x + xr . interest yearned in 2 year = ( x + xr ) r . the difference w = ( x + xr ) r - xr = xr ^ 2 = 40 . case 2 : deposit = $ x ; rate of increase = 3 r . interest yearned in 1 year = x ( 3 r ) . deposit in 1 year = x + 3 xr . interest yearned in 2 year = ( x + 3 xr ) 3 r . the difference = ( x + 3 xr ) 3 r - 3 xr = 9 xr ^ 2 . since from case 1 we know that xr ^ 2 = 40 , then 9 xr ^ 2 = 9 * 40 = 360 . answer : d . | a = 40 * 3
b = a * 3
|
a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 87.50 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(7500, multiply(3, 4)))), multiply(4, sqrt(divide(7500, multiply(3, 4))))), 25), const_100) | the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 7500 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 7500 ⇔ 12 x 2 = 7500 ⇔ x 2 = 625 ⇔ x = 25 . so , length = 75 m and breadth = 100 m . perimeter = [ 2 ( 75 + 100 ) ] m = 350 m . ∴ cost of fencing = rs . ( 0.25 × 350 ) = rs . 87.50 . answer d" | a = 3 * 4
b = 7500 / a
c = math.sqrt(b)
d = 3 * c
e = 3 * 4
f = 7500 / e
g = math.sqrt(f)
h = 4 * g
i = rectangle_perimeter * (
j = i / 25
|
a ) 38 % , b ) 40 % , c ) 44 % , d ) 48 % , e ) 50 % | c | multiply(subtract(multiply(divide(add(const_100, 20), const_100), divide(add(const_100, 20), const_100)), const_1), const_100) | the percentage increase in the area of a rectangle , if each of its sides is increased by 20 % is | let original length = x metres and original breadth = y metres . original area = ( xy ) m 2 . new length = 120 x m = 6 x m . 100 5 new breadth = 120 y m = 6 y m . 100 5 new area = 6 x x 6 y m 2 = 36 xy m 2 . 5 5 25 the difference between the original area = xy and new - area 36 / 25 xy is = ( 36 / 25 ) xy - xy = xy ( 36 / 25 - 1 ) = xy ( 11 / 25 ) or ( 11 / 25 ) xy increase % = 11 xy x 1 x 100 % = 44 % . 25 xy c ) | a = 100 + 20
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 4 / 3 , b ) 3 / 2 , c ) 2 / 6 , d ) 7 / 8 , e ) 8 / 7 | b | divide(multiply(add(add(const_100, const_60), const_1), 5), const_100) | what is the value of ( p + q ) / ( p - q ) if p / q is 5 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 5 + 1 ) / ( 5 - 1 ) = 6 / 4 = 3 / 2 answer : b" | a = 100 + const_60
b = a + 1
c = b * 5
d = c / 100
|
a ) 140 cm 2 , b ) 150 cm 2 , c ) 168 cm 2 , d ) 182 cm 2 , e ) 200 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(2, 8)), 28) | find the area of the quadrilateral of one of its diagonals is 28 cm and its off sets 8 cm and 2 cm ? | "1 / 2 * 28 ( 8 + 2 ) = 140 cm 2 answer : a" | a = 1 / 2
b = 2 + 8
c = a * b
d = c * 28
|
a ) 6 , b ) 9 , c ) 12 , d ) 11 , e ) 48 | d | subtract(44, divide(multiply(3, 44), add(3, const_1))) | 3 machines operating independently , simultaneously , and at the same constant rate can fill a certain production order in 44 hours . if one additional machine were used under the same operating conditions , in how manyfewerhours of simultaneous operation could the production order be fulfilled ? | if 3 machines can do the work in 44 hr then 4 machines can do the work in 3 / 4 * 44 = 33 hrs . hence time saved will be 11 hr option d is the correct answer | a = 3 * 44
b = 3 + 1
c = a / b
d = 44 - c
|
a ) 3 , b ) 5 , c ) 8 , d ) 11 , e ) 17 | a | divide(1365, multiply(multiply(add(const_2, const_3), add(add(const_2, const_3), const_2)), add(const_10, const_3))) | in a certain business school class , p students are accounting majors , q students are finance majors , r students are marketing majors , and s students are strategy majors . if pqrs = 1365 , and if 1 < p < q < r < s , how many students in the class are accounting majors ? | pqrs = 1365 = 3 * 5 * 7 * 13 since 1 < p < q < r < s , the number of students who are accounting majors is p = 3 . the answer is a . | a = 2 + 3
b = 2 + 3
c = b + 2
d = a * c
e = 10 + 3
f = d * e
g = 1365 / f
|
a ) 5 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 15 sec | d | divide(250, multiply(subtract(58, 8), const_0_2778)) | a train 250 m long is running at a speed of 58 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "answer : d . speed of the train relative to man = ( 58 - 8 ) kmph = ( 50 * 5 / 18 ) m / sec = ( 125 / 9 ) m / sec time taken by the train to cross the man = time taken by it to cover 250 m at 125 / 9 m / sec = 250 * 9 / 125 sec = 18 sec" | a = 58 - 8
b = a * const_0_2778
c = 250 / b
|
a ) - 3 , b ) 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | b | divide(add(divide(subtract(4, 10), 2), 2), divide(add(2, 6), 2)) | line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 10 ) . what is the slope of line m ? | "ans : b solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 10 ) . midpoint is ( 4 , - 3 ) as we can see that the y axis of intercept point is ( 0 , - 2 ) means line m is parallel to x axis slope m = 1 ans : b" | a = 4 - 10
b = a / 2
c = b + 2
d = 2 + 6
e = d / 2
f = c / e
|
a ) 33.3 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 66.6 % | b | divide(20, divide(subtract(const_100, 50), const_100)) | on a certain transatlantic crossing , 20 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 50 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ? | "0.20 p = rt + c 0.5 ( rt ) = no c = > 0.50 ( rt ) had c 0.20 p = 0.50 ( rt ) rt / p = 40 % answer - b" | a = 100 - 50
b = a / 100
c = 20 / b
|
a ) 69 . , b ) 73 . , c ) 81 , d ) 91 , e ) 108 | b | subtract(subtract(420, 312), subtract(250, divide(multiply(250, 86), const_100))) | in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | total students = 420 boys = 312 , girls = 108 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 108 - 35 = 73 . correct option : b | a = 420 - 312
b = 250 * 86
c = b / 100
d = 250 - c
e = a - d
|
a ) 145 , b ) 150 , c ) 175 , d ) 180 , e ) 130 | d | multiply(divide(60, 20), const_60) | a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 60 percent . | 1 hr = 20 percent . thus 15 min = 5 percent . now to charge 60 percent 180 min . answer : d | a = 60 / 20
b = a * const_60
|
a ) 3 , b ) 0 , c ) 1 , d ) 2 , e ) 4 | a | subtract(subtract(14, add(1, 4)), 6) | if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 14 . what is the value of f ( 6 ) ? | "since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 14 then f ( 5 ) = 14 - 5 = 9 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 9 - 6 = 3 . answer : a . questions on funtions to practice :" | a = 1 + 4
b = 14 - a
c = b - 6
|
a ) 210 , b ) 280 , c ) 560 , d ) 450 , e ) none | c | divide(multiply(80, 70), subtract(80, 70)) | the banker â € ™ s discount of a certain sum of money is rs . 80 and the true discount on the same sum for the same time is rs . 70 . the sum due is | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 80 * 70 / 80 - 70 ] = rs . [ 80 * 70 / 10 ] = rs . 560 answer c" | a = 80 * 70
b = 80 - 70
c = a / b
|
a ) 2.5 sec , b ) 1.9 sec , c ) 3.8 sec , d ) 6.9 sec , e ) 2.9 sec | c | divide(150, multiply(144, const_0_2778)) | in what time will a train 150 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 150 / 40 = 3.8 sec . answer : c" | a = 144 * const_0_2778
b = 150 / a
|
a ) 50 % , b ) 60 % , c ) 55 % , d ) 75 % , e ) 81 % | e | multiply(power(divide(subtract(const_100, 10), const_100), const_3), const_100) | from a container having pure milk , 10 % is replaced by water and the process is repeated twice . at the end of the second operation , the milk is ? | "let the total quantity of original milk = 1000 gm milk after 1 st operation = 90 % of 1000 = 900 gm milk after second operation = 90 % of 900 = 810 gm strength of final mixture = 81 % answer is e" | a = 100 - 10
b = a / 100
c = b ** 3
d = c * 100
|
a ) 237 , b ) 246 , c ) 255 , d ) 264 , e ) 273 | c | subtract(lcm(24, 11), 9) | what is the smallest number which , when increased by 9 , is divisible by 8 , 11 , and 24 ? | lcm ( 8 , 11,24 ) = 24 x 11 = 264 so the least divisible number is 264 , and the number we are looking for is 264 - 9 = 255 . the answer is c . | a = math.lcm(24, 11)
b = a - 9
|
a ) 0.05 % , b ) 0.5 % , c ) 1 % , d ) 5 % , e ) 10 % | d | subtract(const_100, multiply(divide(19, const_4), const_100)) | on an order of 19 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 20 dozens free . equivalent discount = 1 / 20 * 100 = 5 % . answer d ) 5 %" | a = 19 / 4
b = a * 100
c = 100 - b
|
a ) 0.004 % , b ) 0.02 % , c ) 0.2 % , d ) 4 % , e ) 40 % | d | multiply(divide(multiply(0.02, 20), 10), const_100) | a glass was filled with 10 ounces of water , and 0.02 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ? | we are given that 0.02 ounces of water evaporated each day . furthermore , we know that this process happened over a 20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply 0.02 by 20 . this gives us : 0.02 x 20 = 0.4 ounces finally , we are asked for “ what percent ” of the original amount of water evaporated during this period . to determine this percentage , we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated / original amount ) x 100 % ( 0.4 / 10 ) x 100 % ( 4 / 100 ) x 100 % = 4 % answer d | a = 0 * 2
b = a / 10
c = b * 100
|
a ) $ 2000 , b ) $ 2500 , c ) $ 3000 , d ) $ 2800 , e ) $ 1540 | d | divide(multiply(divide(multiply(140, 10), subtract(15, 10)), const_100), 10) | i sold a book at a profit of 10 % . had i sold it for $ 140 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 140 5 % of cost = $ 140 cost = 140 * 100 / 5 = $ 2800 answer is d" | a = 140 * 10
b = 15 - 10
c = a / b
d = c * 100
e = d / 10
|
a ) 0.005 % , b ) 0.05 % , c ) 0.5 % , d ) 5 % , e ) 25 % | d | multiply(divide(multiply(0.05, 15), 15), const_100) | a glass was filled with 15 ounces of water , and 0.05 ounce of the water evaporated each day during a 15 - day period . what percent of the original amount of water evaporated during this period ? | "in 15 days 15 * 0.05 = 0.75 ounces of water evaporated , which is 0.75 / 15 â ˆ — 100 = 5 of the original amount of water . answer : d ." | a = 0 * 5
b = a / 15
c = b * 100
|
a ) 5 days , b ) 15 days , c ) 12 days , d ) 9 days , e ) 10 days | b | inverse(multiply(inverse(10), subtract(const_1, multiply(5, inverse(15))))) | a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 10 days . in how many days b alone can do the work ? | "explanation : a ’ s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 10 x = 15 days answer : option b" | a = 1/(10)
b = 1/(15)
c = 5 * b
d = 1 - c
e = a * d
f = 1/(e)
|
a ) 7500 , b ) 5000 , c ) 5400 , d ) 7000 , e ) 8500 | a | multiply(subtract(divide(add(multiply(add(100, 25), 10), 500), 25), 10), add(100, 25)) | in a hostel there were 100 students . to accommodate 25 more students the average is decreased by rupees 10 . but total expenditure increased by rs . 500 . find the total expenditure of the hostel now ? | let average is x 100 x + 500 = 125 ( x – 10 ) x = 70 100 * 70 + 500 = 7500 answer : a | a = 100 + 25
b = a * 10
c = b + 500
d = c / 25
e = d - 10
f = 100 + 25
g = e * f
|
a ) 5 : 28 , b ) 5 : 19 , c ) 7 : 12 , d ) 5 : 13 , e ) 7 : 22 | e | divide(subtract(sqrt(4356), 24), multiply(sqrt(4356), const_2)) | the area of a square is 4356 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4356 a = 66 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 42 : 132 = 7 : 22 answer : e" | a = math.sqrt(4356)
b = a - 24
c = math.sqrt(4356)
d = c * 2
e = b / d
|
a ) 18 , b ) 80 , c ) 72 , d ) 90 , e ) 108 | b | subtract(divide(144, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(144, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2))) | pat , kate , and mark charged a total of 144 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much time as mark , how many more hours did mark charge to the project than kate ? | "80 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 144 = > x = 16 32 : 16 : 96 96 - 16 = 80 answer b" | a = 1 / 3
b = 1 + a
c = 3 * 2
d = 1 / c
e = b + d
f = 144 / e
g = 1 / 3
h = 1 + g
i = 3 * 2
j = 1 / i
k = h + j
l = 144 / k
m = 3 * 2
n = l / m
o = f - n
|
a ) 15.3 % , b ) 25 % , c ) 30 % , d ) 33 % , e ) 40 % | a | multiply(subtract(divide(add(const_100, 50), add(const_100, 30)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 30 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 50 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | "another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 50 = 30 + b + 30 b / 100 b = 15.3 answer ( a )" | a = 100 + 50
b = 100 + 30
c = a / b
d = c - 1
e = d * 100
|
a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(50, const_100), divide(subtract(const_100, 20), const_100))) | in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 50 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | percent of students who are 25 years old or older is 0.5 * 48 + 0.2 * 52 = ~ 34 , so percent of people who are less than 25 years old is 100 - 3 = 66 . answer : b . | a = 50 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 1 - d
|
a ) rs . 7070 , b ) rs . 5991 , c ) rs . 6020 , d ) rs . 6850 , e ) none of these | a | subtract(multiply(add(5, const_1), 6200), add(add(add(add(5420, 5660), 6200), 6350), 6500)) | a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6200 ? | "explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6200 * 6 ) – 30,130 ] = rs . ( 37200 – 30,130 ) = rs . 7070 answer a" | a = 5 + 1
b = a * 6200
c = 5420 + 5660
d = c + 6200
e = d + 6350
f = e + 6500
g = b - f
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 20 | e | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), const_3.0) | the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 1 , 000,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 1000 ( 1 , 000,000 / 1,000 = 1000 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 1000 , where x represents the number of times the population doubles . a lot of people remember that 2 ^ 10 = 1,024 . i . e . the population has to double 10 times . since it takes the population 2 minutes to double once it takes 10 * 2 minutes = 20 minutes to double nine times . thus , solution e = 20 is correct ." | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 3
|
a ) rs . 150.50 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 1093.75 , e ) none | d | divide(1183, power(add(divide(4, const_100), const_1), 2)) | the present worth of rs . 1183 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 1183 / ( 1 + 4 / 100 ) ² ] = rs . ( 1183 x 25 / 26 x 25 / 26 ) = rs . 1093.75 answer d" | a = 4 / 100
b = a + 1
c = b ** 2
d = 1183 / c
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | d | multiply(divide(subtract(20, 12), 20), const_100) | the c . p of 20 pens is equal to the s . p of 12 pens . find his gain % or loss % ? | "20 cp = 12 sp 20 - - - 8 cpgain 100 - - - ? = > 40 % answer : d" | a = 20 - 12
b = a / 20
c = b * 100
|
a ) a ) 5.8 , b ) b ) 1.3 , c ) c ) 9.6 , d ) d ) 12.5 , e ) e ) 7.9 | a | divide(subtract(15, 10), subtract(const_1, divide(15, 100))) | how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 15 % solution ? | amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 15 / 100 200 + 20 x = 300 + 3 x 17 x = 100 x = 5.8 answer is a | a = 15 - 10
b = 15 / 100
c = 1 - b
d = a / c
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | e | add(divide(subtract(112, 16), 4), const_1) | how many multiples of 4 are there between 16 and 112 , inclusive ? | "the multiples of 4 are from 4 * 4 up to 4 * 28 . 28 - 4 + 1 = 25 . the answer is e ." | a = 112 - 16
b = a / 4
c = b + 1
|
a ) 3 , b ) 60 , c ) 9 , d ) 12 , e ) 14 | b | multiply(divide(35, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3) | jill has 35 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ? | let the number of each size of jar = wthen 1 / 4 w + 1 / 2 w + w = 35 1 3 / 4 w = 35 w = 20 the total number of jars = 3 w = 60 answer : b | a = 1 / 2
b = const_0_25 + a
c = 1 + b
d = 35 / c
e = d * 3
|
a ) 22 , b ) 27 , c ) 99 , d ) 88 , e ) 11 | b | multiply(divide(subtract(1080, 850), 850), const_100) | a cycle is bought for rs . 850 and sold for rs . 1080 , find the gain percent ? | 850 - - - - 180 100 - - - - ? = > 27 % answer : b | a = 1080 - 850
b = a / 850
c = b * 100
|
a ) 4 : 3 , b ) 9 : 11 , c ) 3 : 1 , d ) 2 : 1 , e ) 11 : 3 | a | divide(subtract(const_100, 2), add(const_100, 5)) | the number of singles that a baseball player hit increased 5 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 2 percent . if the total number of singles and doubles that she hit increased 2 percent from her first season to her second season , what is the ratio of the number of singles she hit in her first season to the number of doubles she hit in her first season ? | "soln : - 1.05 s + 0.98 d = 1.02 [ s + d ] 0.03 s = 0.04 d s / d = 4 / 3 answer : a" | a = 100 - 2
b = 100 + 5
c = a / b
|
a ) 100 km , b ) 95 km , c ) 135 km , d ) 80 km , e ) 125 km | e | multiply(25, 5) | a person is traveling at 25 km / hr and reached his destiny in 5 hr find the distance ? | speed = 25 km / hr time = 5 hr distance = 25 * 5 = 125 km answer is e | a = 25 * 5
|
a ) 404 , b ) 408 , c ) 412 , d ) 416 , e ) 420 | d | multiply(divide(add(14, 50), const_2), divide(add(subtract(50, 14), 3), 3)) | in a theater , the first row has 14 seats and each row has 3 more seats than previous row . if the last row has 50 seats , what is the total number of seats in the theater ? | the number of seats in the theater is 14 + ( 14 + 3 ) + . . . + ( 14 + 36 ) = 13 ( 14 ) + 3 ( 1 + 2 + . . . + 12 ) = 13 ( 14 ) + 3 ( 12 ) ( 13 ) / 2 = 13 ( 14 + 18 ) = 13 ( 32 ) = 416 the answer is d . | a = 14 + 50
b = a / 2
c = 50 - 14
d = c + 3
e = d / 3
f = b * e
|
a ) 85 kmph , b ) 115 kmph , c ) 105 kmph , d ) 95 kmph , e ) 125 kmph | b | subtract(divide(divide(200, 6), const_0_2778), 5) | a train 200 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ? | "let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 200 / [ ( x + 5 ) * 5 / 18 ] = 6 30 ( x + 5 ) = 3600 x = 115 kmph answer is b" | a = 200 / 6
b = a / const_0_2778
c = b - 5
|
a ) 8 , b ) 10 , c ) 16 , d ) 17 , e ) 18 | c | subtract(17, 8) | 8 ^ 100 is divisible by 17 then find the remainder ? | "this is an extremely difficult problem to solve with out fermat ' s little theorem . by applying fermat ' s little theorem , we know that 816 when divided by 17 , the remainder is 1 . so divide 100 by 16 and find the remainder . remainder = 4 therefore , 100 = ( 16 × 6 ) + 4 now this problem can be written as 810017 = 816 × 6 + 417 = ( 816 ) 6 × 8417 now this problem simply boils down to ( 1 ) 6 × 8417 = 8417 84 = 82 × 82 , we need to find the remainder when 64 × 64 is divisible by 17 . or 13 × 13 = 169 . when 169 is divided by 17 , remainder is 16 . c" | a = 17 - 8
|
a ) 21 , b ) 27 , c ) 26 , d ) 80 , e ) 28 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(80, 3)) | a student scored an average of 80 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 80 * 3 = 240 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 240 = 80 . answer : d" | a = 90 * 2
b = 70 * 2
c = a + b
d = 80 * 3
e = c - d
|
a ) 25 % , b ) 28 % , c ) 32 % , d ) 38 % , e ) 40 % | d | multiply(divide(subtract(divide(54, const_100), multiply(divide(62, const_100), divide(2, const_3))), divide(1, const_3)), const_100) | in an election , candidate douglas won 54 percent of the total vote in counties x and y . he won 62 percent of the vote in county x . if the ratio of people who voted in county x to county y is 2 : 1 , what percent of the vote did candidate douglas win in county y ? | "given voters in ratio 2 : 1 let x has 200 votersy has 100 voters for x 62 % voted means 62 * 200 = 124 votes combined for xy has 300 voters and voted 54 % so total votes = 162 balance votes = 162 - 124 = 38 as y has 100 voters so 38 votes means 38 % of votes required ans d" | a = 54 / 100
b = 62 / 100
c = 2 / 3
d = b * c
e = a - d
f = 1 / 3
g = e / f
h = g * 100
|
a ) 754 m , b ) 704 m , c ) 90.524 m , d ) 706 m , e ) 91.688 m | e | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 14.6), const_2), 100), const_100) | the radius of a wheel is 14.6 cm . what is the distance covered by the wheel in making 100 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 100 resolutions . = 100 * 2 * 22 / 7 * 14.6 = 9168.8 cm = 91.688 m answer : e" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 14
g = f * 2
h = g * 100
i = h / 100
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | e | subtract(const_1, multiply(divide(1, 3), divide(1, 2))) | harold and millicent are getting married and need to combine their already - full libraries . if harold , who has 1 / 2 as many books as millicent , brings 1 / 3 of his books to their new home , then millicent will have enough room to bring 1 / 2 of her books to their new home . what fraction of millicent ' s old library capacity is the new home ' s library capacity | because we see h willbring 1 / 3 of his booksto the new home - - > try to pick a number that isdivisible by 3 . before : assume h = 30 books h = 1 / 2 m - - > m = 60 books after : h ' = 1 / 3 h = 10 books m ' = 1 / 2 m = 30 books total = 40 books m ' = 40 = 5 / 6 * 60 ratio : 5 / 6 ans : e | a = 1 / 3
b = 1 / 2
c = a * b
d = 1 - c
|
a ) 86 , b ) 76 , c ) 56 , d ) 36 , e ) 16 | a | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 448 * 348 ) ^ 2 | "( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 448 * 348 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 8,8 ; 5 is missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 48 * 48 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 56 * 48 ) ) ^ 2 = ( 44 * 88 ) ^ 2 = 72 ^ 2 = 84 hence answer is a" | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) 33 : 25 , b ) 47 : 20 , c ) 47 : 25 , d ) 22 : 25 , e ) 91 : 50 | e | divide(add(82, const_100), const_100) | the number of sailors on a ship is 82 % more than the number of officers . the ratio of sailors to officers would be | "sailor = 1.82 * officer sailor / officer = 1.82 / 1 = 182 / 100 = 91 / 50 answer will be e" | a = 82 + 100
b = a / 100
|
a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 45 | c | subtract(50, multiply(const_2, divide(divide(50, const_2), add(const_1, const_4)))) | let f ( x , y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = 50 , what is the maximum value of y for which f ( x , y ) = 0 ? | the question is finding y such that ( 50 - y ) ! is a multiple of 50 . that means we need to have 2 * 5 ^ 2 in ( 50 - y ) ! 10 ! is the smallest factorial number with 2 * 5 ^ 2 as a factor . 50 - y = 10 y = 40 the answer is c . | a = 50 / 2
b = 1 + 4
c = a / b
d = 2 * c
e = 50 - d
|
a ) 5 , b ) 9 , c ) 10 , d ) 18 , e ) 30 | d | multiply(subtract(9, 9), 9) | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 9 ? | "what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 9 ? 9 ^ 9 = ( 3 ^ 2 ) ^ 9 = 3 ^ 18 d . 18" | a = 9 - 9
b = a * 9
|
a ) 19 , b ) 18 , c ) 16 , d ) 17 , e ) 14 | d | power(divide(84, divide(subtract(84, 24), sqrt(9))), const_2) | the speed of a railway engine is 84 km per hour when no compartment is attached , and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached , the maximum number of compartments that can be carried by the engine is : | "the reduction in speed is directly proportional to the square root of the number of compartments attached doesreductionmean amount subtracted ? or percentage decrease ? there are at least two interpretations , and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words , if s = speed , and n = number of compartments , then s = 84 - k * sqrt ( n ) wherekis a constant of the proportionality . in general , if a is directly proportional to b , we can write a = k * b and solve for k . if n = 9 , then s = 24 24 = 84 - k * sqrt ( 9 ) = 84 - 3 k k = 20 now , we need to know : what value of n makes s go to zero ? 0 = 84 - 20 * sqrt ( n ) 20 * sqrt ( n ) = 84 sqrt ( n ) = 4.2 n = 4.2 ^ 2 > 17 with 18 compartments , the train does not budge . therefore , it would budge if there were one fewer cars . thus , 17 is the maximum number of cars the engine can pull and still move . d" | a = 84 - 24
b = math.sqrt(9)
c = a / b
d = 84 / c
e = d ** 2
|
a ) 675 , b ) 680 , c ) 685 , d ) 690 , e ) 695 | a | divide(90, subtract(divide(90, 20), 9)) | a train covers a distance of 90 km in 20 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 90 / 20 * 60 ) km / hr = ( 270 * 5 / 18 ) m / sec = 75 m / sec . length of the train = 75 * 9 = 675 m . answer : option a" | a = 90 / 20
b = a - 9
c = 90 / b
|
a ) 4 / 7 , b ) 3 / 10 , c ) 3 / 18 , d ) 13 / 16 , e ) 4 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 3), 2), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | 3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 3 men and 2 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 3 m + 2 w = 8 w answer is 8 / 14 = 4 / 7 answer : a" | a = 8 - 2
b = 6 - 3
c = a / b
d = c * 3
e = d + 2
f = 8 - 2
g = 6 - 3
h = f / g
i = 6 * h
j = i + 2
k = e / j
|
a ) 120,60 , b ) 240,120 , c ) 280,90 , d ) 220,140 , e ) none of them | b | add(divide(subtract(360, 20), add(const_1, divide(const_1, const_2))), 20) | a third of arun â € ™ s marks in mathematics exceed a half of his marks in english by 20 . if he got 360 marks in two subjects together how many marks did he got in english ? | "let arun â € ™ s marks in mathematics and english be x and y then ( 1 / 3 ) x - ( 1 / 2 ) y = 20 2 x - 3 y = 120 â € ¦ â € ¦ > ( 1 ) x + y = 360 â € ¦ â € ¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 240 and y = 120 answer is b ." | a = 360 - 20
b = 1 / 2
c = 1 + b
d = a / c
e = d + 20
|
a ) 11 , b ) 82 , c ) 10 , d ) 28 , e ) 23 | c | subtract(20, 10) | in an ngo the daily average wages of 20 illiterate employees is decreased from rs . 25 to rs . 10 , thus the average salary of all the literate and illiterate employees is decreased rs . 10 per day . the no . of educated employees working in the ngo is ? | "explanation : total employees = \ inline \ frac { ( 25 - 10 ) \ times 20 } { 10 } = 30 hence number of educated employees = 30 - 20 = 10 answer : c" | a = 20 - 10
|
a ) 69 . , b ) 73 . , c ) 81 . , d ) 91 . , e ) 115 . | e | subtract(subtract(500, 350), subtract(250, divide(multiply(250, 86), const_100))) | in the fifth grade at parkway elementary school there are 500 students . 350 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "total students = 500 boys = 350 , girls = 150 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 150 - 35 = 115 . correct option : e" | a = 500 - 350
b = 250 * 86
c = b / 100
d = 250 - c
e = a - d
|
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