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 ) 25 % , b ) 50 % , c ) 20 % , d ) 150 % , e ) 300 % | e | subtract(divide(30, divide(10, const_100)), const_100) | a man buys an article for $ 10 . and sells it for $ 30 . find the gain percent ? | "c . p . = $ 10 s . p . = $ 30 gain = $ 20 gain % = 20 / 10 * 100 = 300 % answer is e" | a = 10 / 100
b = 30 / a
c = b - 100
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a ) 7.6 % , b ) 7.7 % , c ) 20.52 % , d ) 23.62 % , e ) 27.82 % | c | multiply(const_100, divide(subtract(multiply(56, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | a man buys 56 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 56 pens = rs . 46 s . p . of 56 pens = 99 % of rs . 56 = rs . 55.44 profit % = ( profit / c . p . ) x 100 profit % = ( 9.44 ) x 100 = 20.52 % answer : c" | a = 100 - 1
b = 56 * a
c = 46 * 100
d = b - c
e = 46 * 100
f = d / e
g = 100 * f
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a ) 145 , b ) 272 , c ) 278 , d ) 277 , e ) 112 | a | multiply(divide(multiply(58, const_1000), const_3600), 9) | a train running at the speed of 58 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 58 * ( 5 / 18 ) m / sec = 145 / 9 m / sec length of train ( distance ) = speed * time ( 145 / 9 ) * 9 = 145 meter answer : a" | a = 58 * 1000
b = a / 3600
c = b * 9
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a ) 672 , b ) 682 , c ) 328 , d ) 964 , e ) 285 | a | add(subtract(812, multiply(const_2, const_100)), multiply(const_12, add(const_2, const_3))) | what is the least common multiple of 812 , 3214 | the given numbers are 8,12 , 32,14 thus the l . c . m is 2 * 2 * 2 * 1 * 3 * 4 * 7 = 672 the answer is a | a = 2 * 100
b = 812 - a
c = 2 + 3
d = 12 * c
e = b + d
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a ) 1.0 , b ) 1.2 , c ) 1.4 , d ) 1.6 , e ) 1.8 | a | multiply(divide(add(1.1, divide(100, const_1000)), 72), const_60) | a train with a length of 100 meters , is traveling at a speed of 72 km / hr . the train enters a tunnel 1.1 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | "72 km / hr = 1.2 km / min the total distance is 1.2 km . 1.2 / 1.2 = 1 minute the answer is a ." | a = 100 / 1000
b = 1 + 1
c = b / 72
d = c * const_60
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['a ) 32', 'b ) 36', 'c ) 40', 'd ) 44', 'e ) 48'] | c | multiply(const_2, 20) | the area of a square garden is q square feet and the perimeter is p feet . if q = 2 p + 20 , what is the perimeter of the garden in feet ? | let x be the length of one side of the square garden . x ^ 2 = 8 x + 20 x ^ 2 - 8 x - 20 = 0 ( x - 10 ) ( x + 2 ) = 0 x = 10 , - 2 p = 4 ( 10 ) = 40 the answer is c . | a = 2 * 20
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | c | subtract(add(13, 8), const_1) | a student is ranked 13 th from right and 8 th from left . how many students are there in totality ? | from right 13 , from left 8 total = 13 + 8 - 1 = 20 answer : c | a = 13 + 8
b = a - 1
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a ) 150 m , b ) 180 m , c ) 400 m , d ) 250 m , e ) 155 m | c | subtract(multiply(350, divide(20, divide(20, const_3))), multiply(200, divide(25, divide(20, const_3)))) | a train crosses a platform of 200 m in 20 sec , same train crosses another platform of length 350 m in 25 sec . then find the length of the train ? | "length of the train be β x β x + 200 / 20 = x + 350 / 25 5 x + 1000 = 4 x + 1400 x = 400 m answer : c" | a = 20 / 3
b = 20 / a
c = 350 * b
d = 20 / 3
e = 25 / d
f = 200 * e
g = c - f
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a ) 6 , b ) 8 , c ) 14 , d ) 16 , e ) 28 | b | divide(multiply(subtract(30, 20), 8), 10) | how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 30 percent butterfat to obtain milk that is 20 percent butterfat ? | "equate the fat : 0.1 x + 0.30 * 8 = 0.2 ( x + 8 ) - - > x = 8 . answer : b ." | a = 30 - 20
b = a * 8
c = b / 10
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a ) 5 , b ) 17 , c ) 16 , d ) 23 , e ) 19 | c | multiply(divide(8, 8), const_100) | 8 + 8 | c | a = 8 / 8
b = a * 100
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a ) 6 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | a | multiply(subtract(divide(power(40, const_2), 357), floor(divide(power(40, const_2), 357))), 357) | on dividing a number by 357 , we get 40 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 40 = ( 17 * 21 * y ) + ( 17 * 2 ) + 6 = 17 * ( 21 y + 2 ) + 6 . required number = 6 . answer is a" | a = 40 ** 2
b = a / 357
c = 40 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
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a ) 30 , b ) 45 , c ) 60 , d ) 90 , e ) 105 | e | multiply(14, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 14 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 5 x + 2 / 3 x + 14 x - 13 / 15 x = 14 2 / 15 x = 14 = > so x = ( 15 / 2 ) * 14 = 105 km answer e" | a = 1 / 5
b = 2 / 3
c = a + b
d = 1 - c
e = 1/(d)
f = 14 * e
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a ) 25 % , b ) 33 1 / 3 % , c ) 100 % , d ) 66 2 / 3 % , e ) 75 % | c | multiply(divide(subtract(60, 30), 30), const_100) | a certain protective additive increases from 30 days to 60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? | "general formula for percent increase or decrease , ( percent change ) : percent = change / original β 100 so , the time between maintenance checks increased by ( 60 β 30 ) / 30 β 100 = 100 answer : c" | a = 60 - 30
b = a / 30
c = b * 100
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a ) 1320 , b ) 1300 , c ) 1325 , d ) 120 , e ) none of these | c | add(divide(divide(1404, const_2), divide(add(const_100, 8), const_100)), divide(divide(1404, const_2), divide(add(const_100, divide(8, const_2)), const_100))) | the present worth of rs . 1404 due in two equal half - yearly instalments at 8 % per annum . simple interest is : | "explanation : required sum = pw of rs . 702 due 6 months hence + pw of rs . 702 due 1 year hence = rs . [ 100 Γ 702 / 100 + ( 8 Γ 12 ) + 100 Γ 702 / 100 + ( 8 Γ 1 ) ] = rs . 1325 answer : c" | a = 1404 / 2
b = 100 + 8
c = b / 100
d = a / c
e = 1404 / 2
f = 8 / 2
g = 100 + f
h = g / 100
i = e / h
j = d + i
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a ) $ 45.10 , b ) $ 34.31 , c ) $ 28.44 , d ) $ 67.54 , e ) $ 33.77 | e | divide(multiply(multiply(const_100, const_100), 19), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 19 , what was the price before the first discount ? | "let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 19 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 19 solve for x x = $ 33.77 correct answer e" | a = 100 * 100
b = a * 19
c = 100 - 25
d = c * 100
e = 100 - 25
f = e * 25
g = d - f
h = b / g
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a ) 26 , 14 , b ) 23 , 17 , c ) 17 , 23 , d ) 14 , 26 , e ) 15 , 25 | b | subtract(40, divide(subtract(40, divide(5, const_2)), const_2)) | sum of two numbers is 40 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 15 3 y Γ£ Β’ Γ’ β Β¬ Γ’ β¬ Ε 2 x = 5 x = 23 y = 17 answer : b" | a = 5 / 2
b = 40 - a
c = b / 2
d = 40 - c
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a ) $ 460 , b ) $ 480 , c ) $ 500 , d ) $ 520 , e ) $ 540 | c | divide(add(add(add(add(600, 250), 450), 400), 800), 5) | a cab driver 5 days income was $ 600 , $ 250 , $ 450 , $ 400 , $ 800 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 600 + 250 + 450 + 400 + 800 ) / 5 = 500 answer is c" | a = 600 + 250
b = a + 450
c = b + 400
d = c + 800
e = d / 5
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a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | c | add(add(const_4, const_3), const_2) | what is the units digit of ( 3 ^ 10 ) ( 4 ^ 13 ) ? | "- > the ones place of ( ~ 3 ) ^ n repeats after 4 times like 3 ο 9 ο 7 ο 1 ο 3 . the ones place of ( ~ 4 ) ^ n repeats after 2 times like 4 ο 6 ο 4 . then , 3 ^ 10 = 3 ^ 4 * 2 + 2 ο 3 ^ 2 . , 4 ^ 13 = 4 ^ 2 * 6 + 1 = 4 ^ 1 = ~ 4 which is ( 3 ^ 10 ) ( 4 ^ 13 ) ο ( 3 ^ 2 ) ( ~ 4 ) = ( ~ 9 ) ( ~ 4 ) = ~ 6 . therefore , the answer is c ." | a = 4 + 3
b = a + 2
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a ) 50 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | c | divide(multiply(36, divide(multiply(subtract(52, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 52 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 52 - 36 = 16 km / hr . = 16 * 5 / 18 = 40 / 9 m / sec . 2 x / 36 = 40 / 9 = > x = 80 . answer : c" | a = 52 - 36
b = a * 1000
c = b / 3600
d = 36 * c
e = d / 2
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a ) 2812 , b ) 8231 , c ) 2734 , d ) 4222 , e ) 2945 | e | divide(11780, 4) | a volume of 11780 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 11780 l can be kept in ( 11780 / 4 ) hemispheres ans . 2945 answer : e" | a = 11780 / 4
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a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0005 , e ) 0.0002 | e | divide(0.2, 1,000) | when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 0.2 centimeter . the actual diameter of the tissue , in centimeters , is | "it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 0.2 then x = 0.2 / 1000 = 0.0002 . the answer is e" | a = 0 / 2
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['a ) 16 β 2 Ο', 'b ) 8 β 2 Ο', 'c ) 22 β 2 Ο', 'd ) 12 β 2 Ο', 'e ) 9'] | a | multiply(diagonal(divide(64, const_4), divide(64, const_4)), const_pi) | square x is inscribed in circle y . if the perimeter of x is 64 , what is the circumference of y ? | square forms two right angled triangles . any time we have a right angle triangle inside a circle , the hypotenuse is the diameter . hypotenuse here = diagonal of the square = 16 sqrt ( 2 ) = diameter = > radius = 8 sqrt ( 2 ) circumference of the circle = 2 pi r = 16 pi sqrt ( 2 ) answer is a . | a = 64 / 4
b = 64 / 4
c = diagonal * (
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a ) 35 / 10 , b ) 31 / 10 , c ) 31 / 10 , d ) 36 / 10 , e ) 37 / 10 | a | divide(160, multiply(add(74, 92), const_0_2778)) | two trains of length 210 m and 120 m are 160 m apart . they start moving towards each other on parallel tracks , at speeds 74 kmph and 92 kmph . after how much time will the trains meet ? | they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 74 + 92 ) * 5 / 18 = 46.1 mps . the time required = d / s = 160 / 46.1 = 35 / 10 sec . answer : a | a = 74 + 92
b = a * const_0_2778
c = 160 / b
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a ) 5.98 , b ) 7.98 , c ) 7.96 , d ) 5.87 , e ) 5.82 | c | multiply(divide(multiply(add(8, 1.2), subtract(8, 1.2)), add(add(8, 1.2), subtract(8, 1.2))), const_2) | a man can row 8 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 8 s = 1.2 ds = 9.6 us = 6.8 x / 9.6 + x / 6.8 = 1 x = 3.98 d = 3.98 * 2 = 7.96 answer : c" | a = 8 + 1
b = 8 - 1
c = a * b
d = 8 + 1
e = 8 - 1
f = d + e
g = c / f
h = g * 2
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a ) 22 % , b ) 23 % , c ) 21 % , d ) 20 % , e ) 17 % | d | divide(multiply(subtract(150, divide(multiply(105, const_100), subtract(const_100, 12.55))), const_100), 150) | after getting 2 successive discounts , a shirt with a list price of rs 150 is available at rs 105 . if the second discount is 12.55 , find the first discount | let the first discount be x % then , 87.5 % of ( 100 - x ) % of 150 = 105 87.5 / 100 * ( 100 - x ) / 100 * 450 = 150 = > 105 = > 100 - x = ( 105 * 100 * 100 ) / ( 150 * 87.5 ) = 80 x = ( 100 - 80 ) = 20 first discount = 20 % option d | a = 105 * 100
b = 100 - 12
c = a / b
d = 150 - c
e = d * 100
f = e / 150
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a ) 102 / 102 , b ) 5 / 102 , c ) 1 / 102 , d ) 8 / 102 , e ) 13 / 102 | e | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | from a pack of 52 cards , two cards are drawn together at random . what is the probability of one is a spade and one is a heart ? | solution let s be the sample space . then , n ( s ) = 52 c 2 = ( 52 Γ 51 ) / ( 2 Γ 1 ) = 1326 . let e = event of getting 1 spade and 1 heart . n ( e ) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = ( 13 c 1 * 13 c 1 ) = 169 . p ( e ) = n ( e ) / n ( s ) = 169 / 1326 = 13 / 102 . answer e | a = 52 / 4
b = 52 / 4
c = a * b
d = math.comb(52, 2)
e = c / d
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a ) 21 , b ) 20 , c ) 25 , d ) 36 , e ) 45 | d | subtract(divide(multiply(add(divide(88, const_2), 2), 2), 2), 10) | 10 is added to a certain number , the sum is multiplied by 2 , the product is divided by 2 and 2 is subtracted from the quotient . the remainder left is half of 88 . what is the number ? | "let number is x . when 10 added to it , = ( x + 10 ) 2 multiplied to sum , = 2 * ( x + 10 ) now , = [ { 2 * ( x + 10 ) } / 2 ] and , = [ { 2 * ( x + 10 ) } / 2 ] - 2 according to question , [ { 2 * ( x + 10 ) } / 2 ] - 2 = half of 88 [ ( 2 x + 20 ) / 2 ) = 44 + 2 x + 10 = 46 x + 10 = 46 x = 46 - 10 x = 36 so , required number is : 36 . answer : d" | a = 88 / 2
b = a + 2
c = b * 2
d = c / 2
e = d - 10
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a ) 1 / 216 , b ) 1 / 215 , c ) 5 / 300 , d ) 1 / 415 , e ) 1 / 500 | a | multiply(divide(const_3, add(const_3, const_3)), divide(const_3, add(const_3, const_3))) | four dice are thrown simultaneously . find the probability that all of them show the same face . | "total no of elementary events = 6 ^ 4 n ( s ) = 6 ^ 4 x = { ( 1,1 , 1,1 ) ( 2,2 , 2,2 ) . . . . . ( 6,6 , 6,6 ) } n ( x ) = 6 n ( x ) / n ( s ) = 6 / 6 ^ 4 = 1 / 216 answer a" | a = 3 + 3
b = 3 / a
c = 3 + 3
d = 3 / c
e = b * d
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a ) 300 , b ) 375 , c ) 450 , d ) 500 , e ) 575 | c | multiply(subtract(divide(200, subtract(1, divide(const_3, 5))), 200), add(1, divide(1, 2))) | when jessica withdrew $ 200 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 2 of the remaining balance , what will be the final balance in her bank account ? | "as per the question 200 = 2 a / 5 thus - a which is the total amount = 500 the amount thus left = 300 she then deposited 1 / 2 of 300 = 150 total amount in her account = 450 answer c" | a = 3 / 5
b = 1 - a
c = 200 / b
d = c - 200
e = 1 / 2
f = 1 + e
g = d * f
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a ) 12 : 5 , b ) 12 : 7 , c ) 7 : 12 , d ) 5 : 7 , e ) 4 : 7 | a | divide(2, 3) | the simple form of the ratio 8 / 5 : 2 / 3 is ? | "8 / 5 : 2 / 3 = 12 : 5 answer : a" | a = 2 / 3
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a ) 3.5 % , b ) 2.4 % , c ) 10.2 % , d ) 5 % , e ) 2.6 % | c | subtract(subtract(16, 5), divide(multiply(16, 5), const_100)) | in measuring the sides of a rectangle , one side is taken 16 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements . | "say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . now , the area obtained with wrong measurements would be 116 * 95 = 11,020 , which is 10.2 % greater than the actual area . answer : c ." | a = 16 - 5
b = 16 * 5
c = b / 100
d = a - c
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a ) 600,800 , b ) 500,500 , c ) 300,700 , d ) 800,200 , e ) 550,450 | a | divide(multiply(6, 8), add(6, 8)) | a can do a work in 6 days . b can do the same work in 8 days . both a & b together will finish the work and they got $ 1000 from that work . find their shares ? | "ratio of their works a : b = 6 : 8 ratio of their wages a : b = 3 : 4 a ' s share = ( 3 / 5 ) 1000 = 600 b ' s share = ( 4 / 5 ) 1000 = 800 correct option is a" | a = 6 * 8
b = 6 + 8
c = a / b
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a ) 1648 , b ) 1632 , c ) 1626 , d ) 1612 , e ) 1800 | b | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 12), add(const_2, const_4)) | what is the sum of all the multiples of 12 between 30 and 200 ? | "you first have to know all the multiples of 12 between 30 and 200 . they are 12 , 24,36 , 48,60 , 72,84 , 96,108 , 120,132 , 144,156 , 168,180192 . if you add all these numbers together , you get 1632 . final answer : b" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 12
t = 2 + 4
u = s + t
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a ) 1 , b ) 2 , c ) 5 , d ) 4 , e ) 8 | d | subtract(multiply(add(floor(divide(1100, 23)), const_1), 23), 1100) | what is the least number should be added to 1100 , so the sum of the number is completely divisible by 23 ? | "( 1100 / 23 ) gives remainder 19 19 + 4 = 23 , so we need to add 4 answer : d" | a = 1100 / 23
b = math.floor(a)
c = b + 1
d = c * 23
e = d - 1100
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a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these | c | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | ritesh and co . generated revenue of rs . 1,250 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 1,250 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1250 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) Γ 1250 . = > 10,000 . hence , the total revenue by end of 2007 is rs . 10,000 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 10000 ) Γ 100 . = > 25 % . answer : c" | a = 1 + 4
b = 2 * a
c = b * 100
d = 2 * c
e = 1 + 4
f = e * 100
g = d + f
h = 1 + 4
i = 2 * h
j = i * 100
k = 2 + 4
l = k * 100
m = j + l
n = 12 / 5
o = m / n
p = g / o
q = p * 100
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a ) 8 / 25 , b ) 1 / 25 , c ) 2 / 25 , d ) 3 / 14 , e ) 25 / 8 | a | multiply(divide(4, 5), divide(2, 5)) | if p ( a ) = 4 / 5 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events . | "p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 4 / 5 . 2 / 5 p ( a n b ) = 8 / 25 . a" | a = 4 / 5
b = 2 / 5
c = a * b
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['a ) 98', 'b ) 99', 'c ) 100', 'd ) 101', 'e ) 102'] | a | subtract(multiply(multiply(5, 5), 5), multiply(multiply(subtract(5, const_2), subtract(5, const_2)), const_3)) | a making a cube with dimension 5 * 5 * 5 using 1 * 1 * 1 cubes . what is the number of cubes needed to make hollow cube looking of the same shape . | 125 1 * 1 * 1 small cubes = 5 * 5 * 5 cubes because hallow in inner we do n ' t have 3 * 3 * 3 cubes so 125 - 27 = 98 cubes required answer : a | a = 5 * 5
b = a * 5
c = 5 - 2
d = 5 - 2
e = c * d
f = e * 3
g = b - f
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a ) 14 , b ) 24 , c ) 28 , d ) 48 , e ) 96 | d | multiply(const_2, factorial(4)) | how many 4 digit numbers have no repeat digits , do not contain zero , and have a sum of digits f equal to 28 ? | first , look for all 4 digits without repeat that add up to 28 . to avoid repetition , start with the highest numbers first . start from the largest number possible 9874 . then the next largest number possible is 9865 . after this , you ' ll realize no other solution . clearly the solution needs to start with a 9 ( cuz otherwise 8765 is the largest possible , but only equals 26 ) . with a 9 , you also need an 8 ( cuz otherwise 9765 is the largest possible , but only equals 27 ) . with 98 __ only 74 and 65 work . so you have two solutions . each can be rearranged in 4 ! = 24 ways . so f = 24 + 24 = 48 . d | a = math.factorial(4)
b = 2 * a
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a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , how many q kilometers per hour did tammy walk on the second day ? | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its simple algebra for s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 but for some reason im getting 3.5 and not 4.5 . 8 s + 6 s + 3 = 52 14 s = 49 s = 3.5" | a = 14 + 2
b = a / 2
c = b - 2
d = c / 2
e = 52 - d
f = 14 + 2
g = f / 2
h = g - 2
i = 14 + 2
j = i / 2
k = h + j
l = e / k
m = 1 / 2
n = l + m
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a ) rs . 500 , b ) rs . 5100 , c ) rs . 530 , d ) rs . 540 , e ) none of these | a | divide(580, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 580 ? | "explanation : 580 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 500 answer : option a" | a = 3 * 5
b = a + 3
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 580 / f
|
a ) 120 , b ) 125 , c ) 130 , d ) 135 , e ) 150 | e | subtract(multiply(multiply(divide(300, subtract(add(divide(divide(50, const_100), divide(20, const_100)), const_1), divide(50, const_100))), divide(divide(50, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(300, subtract(add(divide(divide(50, const_100), divide(20, const_100)), const_1), divide(50, const_100))), subtract(const_1, divide(50, const_100)))) | of 300 surveyed students , 20 % of those who read book a also read book b and 50 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 50 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.5 b - - > a = 2.5 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 300 = 2.5 b + b - 0.5 b - - > b = 100 , a = 2.5 b = 250 and { both } = 0.5 b = 50 . the number of students who read only book a is { a } - { both } = 250 - 50 = 200 ; the number of students who read only book b is { b } - { both } = 100 - 50 = 50 ; the difference is 200 - 50 = 150 . answer : e ." | a = 50 / 100
b = 20 / 100
c = a / b
d = c + 1
e = 50 / 100
f = d - e
g = 300 / f
h = 50 / 100
i = 20 / 100
j = h / i
k = g * j
l = 20 / 100
m = 1 - l
n = k * m
o = 50 / 100
p = 20 / 100
q = o / p
r = q + 1
s = 50 / 100
t = r - s
u = 300 / t
v = 50 / 100
w = 1 - v
x = u * w
y = n - x
|
a ) 12 : 15 , b ) 1 : 4 , c ) 1 : 5 , d ) 2 : 5 , e ) 1 : 3 | a | divide(multiply(3, const_4), add(multiply(3, const_4), 3)) | a pet store holds cats and dogs . if the difference between the number of cats and the number of dogs is 3 . what could be the ratio of cats to dogs in the pet store ? | "say theratio of cats to dogs is a / b . then the numberof cats would be ax and the number of dogs bx , for some positive integer x . we are told that ax - bx = 3 - - > x ( a - b ) = 3 . since 3 is a prime number it could be broken into the product of two positive multiples only in one way : x ( a - b ) = 1 * 13 . the above implies that either x = 1 and ( a - b ) = 3 or x = 3 and ( a - b ) = 1 . therefore the correct answer should have the difference between numerator and denominator equal to 1 or 13 . for the original question only option which fits is e , 4 : 5 . cats = 3 * 4 = 12 and dogs = 3 * 5 = 15 . answer : a ." | a = 3 * 4
b = 3 * 4
c = b + 3
d = a / c
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | b | multiply(multiply(6, divide(8, 6)), divide(8, 6)) | 6 mat - weavers can weave 6 mats in 6 days . at the same rate , how many mats would be woven by 8 mat - weavers in 8 days ? | "let the required number of mats be x more mat - weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) hence we can write as ( mat - weavers ) 6 : 8 } : : 6 : x ( days ) 6 : 8 β 6 Γ 6 Γ x = 8 Γ 8 Γ 6 β x = 10 answer : b" | a = 8 / 6
b = 6 * a
c = 8 / 6
d = b * c
|
a ) 5 % , b ) 11 % , c ) 13 % , d ) 25 % , e ) 19 % | d | multiply(divide(subtract(const_100, 80), 80), const_100) | if the cost price is 80 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 80 profit = $ 20 profit % = 20 / 80 * 100 = 25 / 6 = 25 % answer is d" | a = 100 - 80
b = a / 80
c = b * 100
|
a ) 29 , b ) 38 , c ) 76 , d ) 37 , e ) 75 | c | divide(add(add(add(add(81, 65), 82), 67), 85), divide(const_10, const_2)) | dacid obtained 81 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 81 + 65 + 82 + 67 + 85 ) / 5 = 76 . answer : c" | a = 81 + 65
b = a + 82
c = b + 67
d = c + 85
e = 10 / 2
f = d / e
|
a ) 29 , b ) 1 , c ) 10 , d ) 30 , e ) 8 | a | subtract(30, const_1) | he average of 30 numbers is zero . of them , at the most , how many may be greater than zero ? | average of 30 numbers = 0 . sum of 30 numbers ( 0 x 30 ) = 0 . it is quite possible that 29 of these numbers may be positive and if their sum is a the 30 th number is ( - a ) . answer : option a | a = 30 - 1
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a ) 3 and 15 , b ) 3 and 20 , c ) 4 and 13 , d ) 4 and 14 , e ) 5 and 14 | a | add(multiply(add(floor(sqrt(11)), const_1), const_100), subtract(floor(sqrt(200)), const_1)) | in a certain deck of cards , each card has a positive integer written on it , in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer . if the each possible product is between 11 and 200 , then the least and greatest integer on the card would be | "given : 11 < x ( x + 1 ) < 200 . now , it ' s better to test the answer choices here rather than to solve : if x = 3 then x ( x + 1 ) = 12 > 11 - - > so , the least value is 4 . test for the largest value : if x = 15 then x ( x + 1 ) = 15 * 16 = 240 > 200 - - > discard b . answer : a ." | a = math.sqrt(11)
b = math.floor(a)
c = b + 1
d = c * 100
e = math.sqrt(200)
f = math.floor(e)
g = f - 1
h = d + g
|
a ) 1 , b ) 2 , c ) 5 , d ) 9 , e ) 11 | d | subtract(47, multiply(19, const_2)) | a number when divided by 342 gives a remainder 47 . when the same number if it divided by 19 , what would be the remainder ? | "on dividing the given number by 342 , let k be the quotient and 47 as remainder . then , number β 342 k + 47 = ( 19 x 18 k + 19 x 2 + 9 ) = 19 ( 18 k + 2 ) + 9 . ο the given number when divided by 19 , gives ( 18 k + 2 ) as quotient and 9 as remainder . answer d" | a = 19 * 2
b = 47 - a
|
a ) 6 , b ) 14 , c ) 36 , d ) 21 , e ) none of these | c | sqrt(power(36, 2)) | β ( 36 ) ^ 2 | explanation β ( 36 ) ^ 2 = ? or , ? = 36 answer c | a = 36 ** 2
b = math.sqrt(a)
|
a ) a ) 2520000 , b ) b ) 2620000 , c ) c ) 2820000 , d ) d ) 2920000 , e ) e ) 2420000 | b | add(multiply(multiply(divide(divide(100000, const_2), 10), const_3), const_4), 10000) | a and b start a business , with a investing the total capital of rs . 100000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 10000 per month from the total profit and any profit remaining is equally shared by both of them . at the end of the year , it was found that the income of a is twice that of b . find the total profit for the year ? | "interest received by a from b = 10 % of half of rs . 100000 = 10 % * 50000 = 5000 . amount received by a per annum for being a working partner = 10000 * 12 = rs . 120000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 5000 + 120000 + p ) total income of b = only his share from the remaining profit = ' p ' , as a and b share the remaining profit equally . income of a = twice the income of b ( 5000 + 120000 + p ) = 2 ( p ) p = 125000 total profit = 2 p + 120000 = 2 * 125000 + 120000 = 2620000 answer : b" | a = 100000 / 2
b = a / 10
c = b * 3
d = c * 4
e = d + 10000
|
a ) 0.003 , b ) 0.0005 , c ) 0.25 , d ) 0.005 , e ) none of these | d | multiply(divide(divide(1, const_100), 1), const_2) | double of quarter of 1 percent written as a decimal is : | "explanation : solution : ( 2 ) * ( 1 / 4 ) * 1 % = 2 * ( 1 / 4 * 1 / 100 ) = 1 / 400 = 2 * 0.0025 = 0.005 . answer : d" | a = 1 / 100
b = a / 1
c = b * 2
|
a ) 32 , b ) 87 , c ) 30 , d ) 99 , e ) 77 | c | add(floor(divide(multiply(multiply(21, 8), multiply(15, 3)), multiply(multiply(21, 2), 6))), const_1) | 15 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | "3 w = 2 m 15 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 15 * 21 * 8 = 14 * x * 6 x = 30 answer : c" | a = 21 * 8
b = 15 * 3
c = a * b
d = 21 * 2
e = d * 6
f = c / e
g = math.floor(f)
h = g + 1
|
a ) 11 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | a | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 49 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "i think brute force with some common sense should be used to solve this problem . write down all perfect squares less than 49 : 1 , 4 , 9 , 16 , 25 , 36 , 49 . . now , 49 should be the sum of 3 of those 7 numbers . also to simplify a little bit trial and error , we can notice that as 49 is an odd numbers then either all three numbers must be odd ( odd + odd + odd = odd ) or two must be even and one odd ( even + even + odd = odd ) . we can find that 49 equals to 4 + 9 + 36 = 2 ^ 2 + 3 ^ 2 + 6 ^ 2 = 49 - - > 2 + 3 + 6 = 11 . answer : a ." | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
|
a ) 55 , b ) 70 , c ) 75 , d ) 85 , e ) 95 | b | divide(140, const_2) | he total marks obtained by a student in physics , chemistry and mathematics is 140 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | "let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 140 + p c + m = 140 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 140 / 2 = 70 . answer : b" | a = 140 / 2
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a ) 8 , b ) 9 , c ) 6 , d ) 1 , e ) 5 | a | divide(multiply(2, multiply(multiply(4, 8), 5)), multiply(2, 20)) | 4 men can check exam papers in 8 days working 5 hours regularly . what is the total hours when 2 men will check the double of the papers in 20 days ? | let a man can do 1 unit of work in 1 hour . total units of work = 4 x 8 x 5 = 160 units . now work = 2 x 160 = 320 units . now 2 men work for 20 days . let in x hours they have to work per day . now total work = 2 Γ x Γ 20 = 40 x 40 x = 320 so x = 320 / 40 = 8 hours . answer : a | a = 4 * 8
b = a * 5
c = 2 * b
d = 2 * 20
e = c / d
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | add(add(3, 3), 1) | given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 1 ? | "p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > only in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . i . e 0 , 1,5 or 6 intersection of values is 6 , thus the units digit of p + 1 is 6 + 1 = 9 . answer : a ." | a = 3 + 3
b = a + 1
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a ) 14000 , b ) 14400 , c ) 14500 , d ) 14600 , e ) 16900 | e | add(10000, multiply(divide(multiply(10000, 30), const_100), 2)) | the population of a town is 10000 . it increases annually at the rate of 30 % p . a . what will be its population after 2 years ? | "formula : 10000 Γ 130 / 100 Γ 130 / 100 = 16900 answer : e" | a = 10000 * 30
b = a / 100
c = b * 2
d = 10000 + c
|
a ) 423 cm 2 , b ) 122 cm 2 , c ) 420 cm 2 , d ) 251 cm 2 , e ) 288 cm 2 | e | multiply(18, 16) | find the area of a parallelogram with base 18 cm and height 16 cm ? | "area of a parallelogram = base * height = 18 * 16 = 288 cm 2 answer : e" | a = 18 * 16
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a ) a ) 36 , b ) b ) 38 , c ) c ) 40 , d ) d ) 42 , e ) e ) 44 | b | subtract(50, multiply(multiply(12, 2), 2)) | evaluate : 50 - 12 Γ· 2 Γ 2 = | "according to order of operations , 12 Γ· 2 Γ 2 ( division and multiplication ) is done first from left to right 12 Γ· 2 Γ 2 = 6 Γ 2 = 12 hence 50 - 12 Γ· 6 Γ 2 = 50 - 12 = 38 correct answer is b ) 38" | a = 12 * 2
b = a * 2
c = 50 - b
|
a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(100, reminder(5, 5)) | when positive integer n is divided by 3 , the remainder is 2 . when n is divided by 5 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and next multiple of 7 will be 84 + 21 > 100 . . ans 0 . . a" | a = 100 - reminder
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a ) 4 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | a | divide(const_1, add(multiply(24, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 24 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 24 girls = 8 Γ£ β 6 / 4 + 12 = 12 + 24 = 36 girls 4 days to complete the work answer : option a" | a = 1 / 6
b = a / 24
c = 24 * b
d = 1 / 6
e = d / 16
f = 8 * e
g = c + f
h = 1 / g
|
a ) 150,287 , b ) 150,219 , c ) 150,200 , d ) 150,298 , e ) 150,217 | c | multiply(3, divide(100, 2)) | the incomes of two persons a and b are in the ratio 3 : 4 . if each saves rs . 100 per month , the ratio of their expenditures is 1 : 2 . find their incomes ? | the incomes of a and b be 3 p and 4 p . expenditures = income - savings ( 3 p - 100 ) and ( 4 p - 100 ) the ratio of their expenditure = 1 : 2 ( 3 p - 100 ) : ( 4 p - 100 ) = 1 : 2 2 p = 100 = > p = 50 their incomes = 150 , 200 answer : c | a = 100 / 2
b = 3 * a
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a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5 | e | subtract(1, divide(divide(6, 9), add(divide(6, 9), 1))) | a certain country is divided into 6 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 9 the total number of progressives in the entire country , what fraction of the country is traditionalist ? | let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 9 the total number of traditionalists is 6 p / 9 = 2 p / 3 . the total population is p + 2 p / 3 = 5 p / 3 p / ( 5 p / 3 ) = 3 / 5 the answer is e . | a = 6 / 9
b = 6 / 9
c = b + 1
d = a / c
e = 1 - d
|
a ) $ 150 , b ) $ 200 , c ) $ 250 , d ) $ 120 , e ) $ 100 | c | divide(50, divide(add(10, 10), const_100)) | a tradesman sold an article at a loss of 10 % . if the selling price had been increased by $ 50 , there would have been a gain of 10 % . what was the cost price of the article ? | "let c . p . be $ x then 110 % of x - 90 % of x = 50 20 % of x = 50 x / 5 = 50 x = 250 answer is c" | a = 10 + 10
b = a / 100
c = 50 / b
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a ) 399 , b ) 272 , c ) 799.2 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.20), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.20 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 Γ’ β¬ β 6 = 666 * 1.2 = 799.2 answer : c" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 1
f = e * 3
|
a ) 12653 , b ) 14587 , c ) 36975 , d ) 75000 , e ) 12874 | d | multiply(multiply(multiply(multiply(add(const_3, const_2), const_3), const_100), const_100), divide(3, add(add(1, 2), 3))) | in business , a and c invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by a and b was 2 : 3 , if rs 15,0000 was their profit , how much amount did b receive . | explanation : a : b = 2 : 3 = 2 : 3 = > a : c = 2 : 1 = 2 : 1 = > a : b : c = 2 : 3 : 1 b share = ( 3 / 6 ) * 150000 = 75000 option d | a = 3 + 2
b = a * 3
c = b * 100
d = c * 100
e = 1 + 2
f = e + 3
g = 3 / f
h = d * g
|
a ) 2016 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2015) | if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2015 ) = ? | "f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 β β f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 β β f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 β β f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2015 ) = 2016 ans : a" | a = 1 + 2015
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a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 450 , e ) $ 600 | d | divide(multiply(subtract(const_100, 10), divide(810, const_2)), const_100) | a pair of articles was bought for $ 810 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 810 / 2 = $ 405 let m . p = $ x 90 % of x = 405 x = 405 * 100 / 90 = $ 450 answer is d" | a = 100 - 10
b = 810 / 2
c = a * b
d = c / 100
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a ) 3.6 , b ) 7.2 , c ) 8.4 , d ) 10 , e ) 8 | b | divide(divide(600, const_1000), divide(multiply(5, const_60), const_3600)) | a person crosses a 600 m long street in 5 minutes . what is his speed in km per hour ? | "speed = 600 / ( 5 x 60 ) m / sec . = 2 m / sec . converting m / sec to km / hr = ( 2 x ( 18 / 5 ) ) km / hr = 7.2 km / hr . answer : b" | a = 600 / 1000
b = 5 * const_60
c = b / 3600
d = a / c
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a ) 1 / 9 , b ) 1 / 8 , c ) 1 / 11 , d ) 1 / 13 , e ) 2 / 5 | b | divide(subtract(subtract(add(3, add(4, 3)), 1), lcm(lcm(4, 2), 8)), lcm(lcm(4, 2), 8)) | if a and b are two events such that p ( a ) = 3 / 4 , p ( b ) = 1 / 2 and p ( a n b ) = 3 / 8 , find p ( not a and not b ) . | p ( not a and not b ) = 1 - ( p ( a ) + p ( b ) - p ( ab ) ) which you might find somewhere in your text . substituting in our probabilities we get : p ( not a and not b ) = 1 - ( 3 / 4 + 1 / 2 - 3 / 8 ) p ( not a and not b ) = 1 - ( 7 / 8 ) p ( not a and not b ) = 1 / 8 . b | a = 4 + 3
b = 3 + a
c = b - 1
d = math.lcm(4, 2)
e = math.lcm(d, 8)
f = c - e
g = math.lcm(4, 2)
h = math.lcm(g, 8)
i = f / h
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a ) 125 km , b ) 100 km , c ) 80 km , d ) 120 km , e ) 60 km | b | multiply(20, divide(25, subtract(25, 20))) | when sourav increases his speed from 20 km / hr to 25 km / hr , he takes one hour less than the usual time to cover a certain distance . what is the distance usually covered by him ? | detailed solution here , the distance travelled by sourav is constant in both the cases . so , 20 x t = 25 x ( t - 1 ) = d = ) 20 t = 25 t - 25 = ) 5 t = 25 = ) t = 5 hrs so , distance travelled = 20 x 5 = 100 km . correct answer b . | a = 25 - 20
b = 25 / a
c = 20 * b
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a ) 1443 , b ) 1554 , c ) 1665 , d ) 1776 , e ) 1887 | d | multiply(add(add(multiply(add(add(4, 3), 1), const_100), multiply(add(add(const_4.0, 3), 1), const_10)), add(add(4, 3), 1)), 3) | what is the sum of all possible 3 - digit numbers that can be constructed using the digits 1 , 3 , and 4 if each digit can be used only once in each number ? | "there are 6 possible arrangements of the three numbers . then each number will be in the hundreds , tens , and ones place two times each . the sum is 2 ( 111 ) + 2 ( 333 ) + 2 ( 444 ) = 1776 the answer is d ." | a = 4 + 3
b = a + 1
c = b * 100
d = 4 + 0
e = d + 1
f = e * 10
g = c + f
h = 4 + 3
i = h + 1
j = g + i
k = j * 3
|
a ) rs . 2500 , b ) rs . 2700 , c ) rs . 2800 , d ) rs . 5000 , e ) rs . 6000 | c | divide(subtract(multiply(divide(5, const_100), 4000), 144), subtract(divide(5, const_100), divide(3, const_100))) | 4000 was divided into two parts such a way that when first part was invested at 3 % and the second at 5 % , the whole annual interest from both the investments is rs . 144 , how much was put at 3 % ? | explanation : ( x * 3 * 1 ) / 100 + [ ( 4000 - x ) * 5 * 1 ] / 100 = 144 3 x / 100 + 200 Γ’ β¬ β 5 x / 100 = 144 2 x / 100 = 56 Γ£ Β¨ x = 2800 answer is c | a = 5 / 100
b = a * 4000
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
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a ) 20 , b ) 120 , c ) 360 , d ) 1200 , e ) 820 | d | divide(multiply(120, 200), 20) | if 20 % of a number = 200 , then 120 % of that number will be ? | "let the number x . then , 20 % of x = 200 x = ( 200 * 100 ) / 20 = 1000 120 % of x = ( 120 / 100 * 1000 ) = 1200 . answer : d" | a = 120 * 200
b = a / 20
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a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | d | divide(add(18, 10), add(14, 14)) | solve the equation for x : 14 ( - x + z ) + 18 = - 14 ( x - z ) - 10 | d 1 - 14 x + 14 z + 18 = - 14 x + 14 z - 10 - 28 x + 28 = 0 - 28 x = - 28 = > x = 1 | a = 18 + 10
b = 14 + 14
c = a / b
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a ) rs . 169.50 , b ) rs . 170 , c ) rs . 175.50 , d ) rs . 180 , e ) rs . 190 | c | add(add(add(add(153, const_10), const_10), 2), add(const_0_25, const_0_25)) | tea worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg will be : | explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . 126 + 135 = rs . 130.50 2 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 153 / 22.50 = 1 x - 153 = 22.50 x = 175.50 answer is c | a = 153 + 10
b = a + 10
c = b + 2
d = const_0_25 + const_0_25
e = c + d
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a ) 2200 , b ) 3600 , c ) 3270 , d ) 3700 , e ) 4000 | a | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(45, const_100))), 2000) | in an election between two candidates , one got 45 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 2000 , the number of valid votes that the other candidate got , was : | "number of valid votes = 80 % of 4000 = 3200 . valid votes polled by other candidate = 55 % of 4000 = ( 55 / 100 ) x 8000 = 2200 answer = a" | a = 20 / 100
b = 1 - a
c = 45 / 100
d = 1 - c
e = b * d
f = e * 2000
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a ) 1.69897 , b ) 2.69897 , c ) 1.99897 , d ) 0.69897 , e ) 5.69897 | a | divide(const_1, 0.30103) | if log 10 2 = 0.30103 , find the value of log 10 50 | "log 10 50 = log 10 ( 100 / 2 ) = log 100 - log 2 = 2 log 10 - log 2 = 2 ( 1 ) - 0.30103 = 1.69897 answer : a" | a = 1 / 0
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a ) $ 120 , b ) $ 500 , c ) $ 420 , d ) $ 610 , e ) $ 315 | c | divide(672, 5) | divide $ 672 among a , b in the ratio 5 : 3 . how many $ that a get ? | "sum of ratio terms = 5 + 3 = 8 a = 672 * 5 / 8 = $ 420 answer is c" | a = 672 / 5
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a ) 17 , b ) 25 , c ) 27 , d ) 35 , e ) 40 | b | add(multiply(divide(add(multiply(8, const_3), 75), add(add(5, 8), 7)), 8), 10) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 5 years ago , the sum of their ages was 75 , what will be the age of aman 10 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 5 + 8 x - 5 + 7 x - 5 = 75 x = 3 present age of aman = 5 * 3 = 15 aman ' s age 10 years hence = 15 + 10 = 25 answer = b" | a = 8 * 3
b = a + 75
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 10
|
a ) 240 , b ) 250 , c ) 280 , d ) 300 , e ) 320 | a | divide(42, subtract(1, add(add(add(divide(1, 3), divide(1, 8)), divide(1, 5)), divide(1, 6)))) | if 1 / 3 of the passengers on a ship are from north america , 1 / 8 are europeans , 1 / 5 are from africa , 1 / 6 are from asia and the remaining 42 people are citizens of other continents , then how many passengers are on board the ship ? | "1 / 3 + 1 / 8 + 1 / 5 + 1 / 6 = ( 40 + 15 + 24 + 20 ) / 120 = 99 / 120 let x be the number of passengers on the ship . 42 = ( 21 / 120 ) x = 7 x / 40 x = 240 the answer is a ." | a = 1 / 3
b = 1 / 8
c = a + b
d = 1 / 5
e = c + d
f = 1 / 6
g = e + f
h = 1 - g
i = 42 / h
|
a ) 136 , b ) 146 , c ) 166 , d ) 176 , e ) 216 | d | divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), power(divide(add(multiply(5, const_100), 44), power(const_2, const_4)), const_2)) | a room of 5 m 44 cm long and 3 m 74 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | area of the room = 544 * 374 sq cm size of largest square tile = h . c . f of 544 cm and 374 cm = 34 cm area of 1 tile = 34 * 34 sq cm no . of tiles required = ( 544 * 374 ) / ( 34 * 34 ) = 176 answer : d | a = 5 * 100
b = a + 44
c = 3 * 100
d = c + 74
e = b * d
f = 5 * 100
g = f + 44
h = 2 ** 4
i = g / h
j = i ** 2
k = e / j
|
a ) 350 , b ) 370 , c ) 380 , d ) 430 , e ) none | c | add(340, divide(multiply(4, subtract(420, 340)), add(3, 4))) | average expenditure of a person for the first 3 days of a week is rs . 340 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 340 total excess than assumed mean = 4 Γ ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 340 + rs . 40 = rs . 380 correct option : c" | a = 420 - 340
b = 4 * a
c = 3 + 4
d = b / c
e = 340 + d
|
a ) s . 575 , b ) s . 595 , c ) s . 590 , d ) s . 610 , e ) s . 585 | d | subtract(multiply(subtract(540, 400), 6), subtract(540, 400)) | if rs . 400 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | "80 = ( 400 * 4 * r ) / 100 r = 8.75 % i = ( 400 * 6 * 8.75 ) / 100 = 210 400 + 210 = 610 answer : d" | a = 540 - 400
b = a * 6
c = 540 - 400
d = b - c
|
a ) 1 , b ) 7 . , c ) 2 . , d ) 3 , e ) 9 | a | add(multiply(const_4, const_2), reminder(add(add(multiply(subtract(const_10, const_1), const_100), multiply(multiply(add(const_3, const_2), const_100), const_10)), multiply(add(const_12, add(const_3, const_2)), add(const_3, const_2))), 4)) | whats the reminder when 879,548 , 521,456 , 258,759 , 653,258 , 778,455 , 658,985 is divided by 4 | "a number ending in a 0 is divisible by 2 . a number ending in 2 zeroes is divisible by 4 . given the obscene number , you should immediately be convinced that you will need to focus on a very small part of it . 879,548 , 521,456 , 258,759 , 653,258 , 778,455 , 658,985 = 879,548 , 521,456 , 258,759 , 653,258 , 778,455 , 658,900 + 85 the first number is divisible by 16 . you just have to find the remainder when you divide 5287 by 16 . that will be the remainder when you divide the original number by 4 . 85 / 4 gives remainder 1 . answer ( a )" | a = 4 * 2
b = 10 - 1
c = b * 100
d = 3 + 2
e = d * 100
f = e * 10
g = c + f
h = 3 + 2
i = 12 + h
j = 3 + 2
k = i * j
l = g + k
m = a + reminder
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | divide(400, add(subtract(26, 2), const_1)) | 400 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 400 / 25 ) = 16 d" | a = 26 - 2
b = a + 1
c = 400 / b
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(divide(subtract(multiply(34, 6), multiply(30, 6)), subtract(46, 34)), 6) | a car averages 30 miles per hour for the first 6 hours of a trip and averages 46 miles per hour for each additional hour of travel time . if the average speed for the entire trip is 34 miles per hour , how many hours long is the trip ? | "let t be the total time of the trip . 30 * 6 + 46 ( t - 6 ) = 34 t 12 t = 276 - 180 t = 8 the answer is a ." | a = 34 * 6
b = 30 * 6
c = a - b
d = 46 - 34
e = c / d
f = e + 6
|
a ) 28 % , b ) 41 % , c ) 45 % , d ) 72 % , e ) 68 % | e | multiply(divide(multiply(divide(70, const_100), divide(subtract(8, 5), 8)), add(multiply(divide(subtract(const_100, 80), const_100), divide(5, 8)), multiply(divide(70, const_100), divide(subtract(8, 5), 8)))), const_100) | 5 / 8 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded if necessary to the nearest percent , the probability that a juliet supporter chosen at random resides in capulet is | "total population = 80 ( assume ) . 5 / 8 * 80 = 50 people from montague . 3 / 8 * 80 = 30 people from capulet . 0.2 * 50 = 10 people from montague support juliet . 0.7 * 30 = 21 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 21 / ( 10 + 21 ) = ~ 68 . answer : e" | a = 70 / 100
b = 8 - 5
c = b / 8
d = a * c
e = 100 - 80
f = e / 100
g = 5 / 8
h = f * g
i = 70 / 100
j = 8 - 5
k = j / 8
l = i * k
m = h + l
n = d / m
o = n * 100
|
a ) 3 , b ) 7 , c ) 21 , d ) 27 , e ) 189 | d | multiply(multiply(multiply(subtract(add(add(subtract(5, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(5, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(5, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | a β sophie germain β prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 5 is | "a prime number greater than 5 can have only the following four units digits : 1 , 3 , 7 , or 9 . if the units digit of p is 1 then the units digit of 2 p + 1 would be 3 , which is a possible units digit for a prime . for example consider p = 11 = prime - - > 2 p + 1 = 23 = prime ; if the units digit of p is 3 then the units digit of 2 p + 1 would be 7 , which is a possible units digit for a prime . for example consider p = 23 = prime - - > 2 p + 1 = 47 = prime ; if the units digit of p is 7 then the units digit of 2 p + 1 would be 5 , which is not a possible units digit for a prime ; if the units digit of p is 9 then the units digit of 2 p + 1 would be 9 , which is a possible units digit for a prime . for example consider p = 29 = prime - - > 2 p + 1 = 59 = prime . the product of all the possible units digits of sophie germain primes greater than 5 is 1 * 3 * 9 = 27 . answer : d ." | a = 5 - 1
b = 2 + 1
c = a + b
d = 2 + 1
e = c + d
f = e - 10
g = 5 - 1
h = 2 + 1
i = g + h
j = 2 * i
k = j + 1
l = k - 10
m = f * l
n = 2 + 1
o = m * n
p = 5 - 1
q = 2 + 1
r = p + q
s = 2 + 1
t = r + s
u = 2 + 1
v = t + u
w = 2 * v
x = w + 1
y = 2 * 10
z = x - y
A = o * z
|
a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.75 , e ) 1.0 | a | divide(multiply(divide(multiply(8, 5), const_100), 32), const_100) | 32 % of 5 / 8 = | "should be simple . 0.32 * 5 / 8 = 1.6 / 8 = 0.2 correct option : a" | a = 8 * 5
b = a / 100
c = b * 32
d = c / 100
|
a ) 18000 , b ) 22500 , c ) 24000 , d ) 26000 , e ) 32000 | b | multiply(multiply(subtract(const_1, divide(2, 5)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(1, 2), 1)) | in a mayoral election , candidate x received 1 / 2 more votes than candidate y , and candidate y received 2 / 5 fewer votes than z . if z received 25,000 votes how many votes did candidate x received ? | z = 25 - - > y received 2 / 5 fewer votes than z - - > y = z - 2 / 5 * z = 15 ; x received 1 / 2 more votes than y - - > x = y + 1 / 2 * y = 22,5 . answer : b . | a = 2 / 5
b = 1 - a
c = const_0_25 * 100
d = c * 100
e = d * 10
f = b * e
g = 1 / 2
h = g + 1
i = f * h
|
a ) 32 , b ) 33 , c ) 88 , d ) 66 , e ) 09 | b | divide(divide(add(150, 240), const_1000), divide(42, const_3600)) | a train 150 meters long completely crosses a 240 meters long bridge in 42 seconds . what is the speed of the train is ? | "s = ( 150 + 240 ) / 42 = 390 / 42 * 18 / 5 = 33 answer : b" | a = 150 + 240
b = a / 1000
c = 42 / 3600
d = b / c
|
a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salaries of all the employees at company e after july 1 last year was what percent t of thatbeforejuly 1 last year ? | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before t = [ 0.9 n ( 1.1 x ) ] / xn = 0.99 or 99 % . b" | a = 10 * 10
|
a ) 2 / 16 , b ) 1 / 16 , c ) 3 / 24 , d ) 5 / 16 , e ) 6 / 26 | d | divide(8, add(multiply(3, divide(8, 2)), multiply(4, divide(8, 3)))) | if 2 men or 3 women can reap a field in 8 days how long will 3 men and 4 women take to reap it ? | "explanation : 2 men reap 1 / 8 field in 1 day 1 man reap 1 / ( 2 x 8 ) 3 women reap 1 / 8 field in 1 day 1 woman reap 1 / ( 3 x 8 ) 3 men and 4 women reap ( 3 / ( 2 x 8 ) + 4 / ( 3 x 8 ) ) = 5 / 16 in 1 day 3 men and 4 women will reap the field in 5 / 16 days answer : option d" | a = 8 / 2
b = 3 * a
c = 8 / 3
d = 4 * c
e = b + d
f = 8 / e
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) none of them | b | divide(multiply(multiply(15, 9), 16), multiply(18, 8)) | if 15 men , working 9 hours a day , can reap a field in 16 days , in how many days will 18 men reap the field , working 8 hours a day ? | let the required number of days be x . more men , less days ( indirect proportion ) less hours per day , more days ( indirect proportion ) men 18 : 15 hours per day 8 : 9 } : : 16 : x ( 18 x 8 x x ) = ( 15 x 9 x 16 ) = x = ( 44 x 15 ) 144 = 15 hence , required number of days = 15 . answer is b . | a = 15 * 9
b = a * 16
c = 18 * 8
d = b / c
|
a ) 1256 , b ) 2000 , c ) 3600 , d ) 8400 , e ) 4566 | b | multiply(multiply(divide(200, multiply(50, 5)), const_100), multiply(50, 5)) | find the sum the difference between the compound and s . i . on a certain sum of money for 5 years at 50 % per annum is rs . 200 of money ? | "p = 200 ( 100 / 50 ) 5 = > p = 2000 answer : b" | a = 50 * 5
b = 200 / a
c = b * 100
d = 50 * 5
e = c * d
|
a ) 2 / 3 , b ) 2 / 5 , c ) 3 / 5 , d ) 3 / 10 , e ) 7 / 20 | d | divide(add(1, 2), multiply(2, 5)) | a small beaker is 1 / 2 filled with salt water . another beaker , which has 5 times the capacity of the small beaker , is 1 / 5 filled with fresh water . after dumping all of the salt water from the small beaker into the large beaker , to what fraction of its capacity will the large beaker be filled ? | 1 / 2 of the small beaker is 1 / 10 of the large beaker . 1 / 10 + 1 / 5 = 3 / 10 the answer is d . | a = 1 + 2
b = 2 * 5
c = a / b
|
a ) 450 , b ) 600 , c ) 750 , d ) 2400 , e ) 300 | e | multiply(3600, divide(800, 9600)) | a football field is 9600 square yards . if 800 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 3600 square yards ? | "answer e ) 9600 yards need 1200 lbs 1 yard will need 800 / 9600 = 1 / 12 lbs 3600 yards will need 1 / 12 * 3600 yards = 300 lbs" | a = 800 / 9600
b = 3600 * a
|
a ) 1.009 , b ) 100.09 , c ) 10.9 , d ) none of these , e ) can not be determined | a | divide(4.036, 0.04) | 4.036 divided by 0.04 gives | "solution 4.036 / 0.04 = 403.6 / 4 = 100.9 answer a" | a = 4 / 36
|
a ) 22 , b ) 25 , c ) 26 , d ) 28 , e ) 29 | a | sqrt(add(multiply(131, const_2), 222)) | sum of the squares of 3 numbers is 222 and the sum of their products taken two at a time is 131 . find the sum ? | ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 222 + 2 * 131 a + b + c = β 484 = 22 a | a = 131 * 2
b = a + 222
c = math.sqrt(b)
|
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