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 ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) 0.03408 | e | multiply(0.016, 2.13) | if 213 × 16 = 3408 , then 0.016 × 2.13 is equal to : | "solution 0.016 × 2.13 = ( 16 / 1000 x 213 / 100 ) = ( 16 / 1000 x 213 / 100 ) = 3408 / 100000 = 0.03408 . answer e" | a = 0 * 16
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a ) 68 % , b ) 71 % , c ) 73 % , d ) 75 % , e ) 79 % | b | divide(add(multiply(70, 65), multiply(85, subtract(90, 70))), 90) | in a certain accounting class of 90 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 65 % , and the students on the make - up date had an average score of 85 % , what was the average score for the entire class ? | "70 % of the students took the final exam on the assigned day - - > 0.7 * 90 = 63 students took the final exam on the assigned day , with an average score of 65 . the rest of the students took the exam on a make - up date - - > 90 - 63 = 27 students took the exam on a make - up date , with an average score of 85 . ( average score ) = ( sum of the scores ) / ( total students ) = ( 63 * 65 + 27 * 85 ) / 90 = 71 . answer : b ." | a = 70 * 65
b = 90 - 70
c = 85 * b
d = a + c
e = d / 90
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a ) a ) 2 , b ) b ) - 13 , c ) c ) - 18 , d ) d ) - 17 , e ) e ) - 25 | e | multiply(subtract(3, 5), add(6, 2)) | 3 - 5 * 6 + 2 = | 3 - 5 * 6 + 2 = 5 - 30 = - 25 . . . so the answer is - 25 . . . e | a = 3 - 5
b = 6 + 2
c = a * b
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a ) 160 , b ) 170 , c ) 270 , d ) 190 , e ) 200 | c | divide(multiply(75, multiply(36, const_2)), 20) | 75 boys can complete a work in 36 days . how many men need to complete twice the work in 20 days | one man can complete the work in 36 * 75 = 2700 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 2700 m = 270 workers answer : c | a = 36 * 2
b = 75 * a
c = b / 20
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a ) 8 : 6 , b ) 8 : 9 , c ) 8 : 3 , d ) 8 : 7 , e ) 8 : 1 | d | divide(8, 7) | ratio of the earnings of a and b is 4 : 7 . if the earnings of a increases by 50 % and those of b decreased by 25 % , the new ratio of their earnings becomes 8 : 7 . what are a ' s earnings ? | let the original earnings of a and b be rs . 4 x and rs . 7 x . new earnings of a = 150 % 0 f rs . 4 x = ( 150 / 100 * 4 x ) = rs . 6 x new earnings of b = 75 % of rs . 7 x = ( 75 / 100 * 7 x ) = rs . 21 x / 4 6 x : 21 x / 4 = 8 : 7 this does not give x . so , the given data is inadequate . answer : d | a = 8 / 7
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 30 ? | "first , let ' s translate the expression 21 # 30 , using the definition given : 21 # 30 = 21 × 24 × 27 × 30 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 30 = 3 ^ 4 ( 7 × 8 × 9 × 10 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 30 = 3 ^ 4 ( 7 × 2 ^ 3 × 3 ^ 2 × ( 2 × 5 ) ) let ' s group the prime bases : 21 # 30 = 2 ^ 4 × 3 ^ 6 × 5 × 7 the sum of the exponents is 4 + 6 + 1 + 1 = 12 the answer is c ." | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
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a ) 9 , b ) 36 , c ) 2 , d ) 6 , e ) 1 | c | divide(sqrt(144), 6) | what is the square root of 144 , divided by 6 ? | "square root is a number times itself square root of 144 = 12 , 12 / 6 = 2 ( c ) 2" | a = math.sqrt(144)
b = a / 6
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a ) 21.21 , b ) 20.25 , c ) 32.25 , d ) 75.75 , e ) 62.61 | d | divide(divide(multiply(20, const_1000), const_60), multiply(multiply(divide(add(20, const_2), add(const_4, const_3)), const_2), divide(divide(140, const_100), const_2))) | the diameter of the driving wheel of a bus in 140 cm . how many revolutions per minute must the wheel make in order to keep a speed of 20 kmph ? | distance covered in 1 min = ( 20 * 1000 ) / 60 = 333.33 m circumference of the wheel = ( 2 * ( 22 / 7 ) * . 70 ) = 4.4 m no of revolution per min = 333.33 / 4.4 = 75.75 answer : d | a = 20 * 1000
b = a / const_60
c = 20 + 2
d = 4 + 3
e = c / d
f = e * 2
g = 140 / 100
h = g / 2
i = f * h
j = b / i
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a ) 28 % , b ) 30 % , c ) 36 % , d ) 36 % , e ) 40 % | d | multiply(divide(add(multiply(divide(30, 100), 400), multiply(100, divide(60, 100))), add(400, 100)), 100) | a grocer has 400 pounds of coffee in stock , 30 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer ’ s stock of coffee is decaffeinated ? | "1 . 30 % of 400 = 120 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 180 pounds of decaffeinated out of 500 pounds , that means 180 / 500 * 100 % = 36 % . the correct answer is d ." | a = 30 / 100
b = a * 400
c = 60 / 100
d = 100 * c
e = b + d
f = 400 + 100
g = e / f
h = g * 100
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a ) 77 kg , b ) 60 kg , c ) 77 kg , d ) 55 kg , e ) 47 kg | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets into it . the mass of the man is ? | "in this type of question , first we will calculate the volume of water displaces then will multiply with the density of water . volume of water displaced = 3 * 2 * 0.01 = 0.06 m cube mass of man = volume of water displaced * density of water = 0.06 * 1000 = 60 kg answer : b" | a = 3 * 2
b = 1 / 100
c = a * b
d = c * 1000
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a ) 2 , b ) 2.5 , c ) 3.2 , d ) 4.6 , e ) 2.8 | c | multiply(divide(500, add(multiply(4, const_60), 30)), const_3_6) | a cyclist covers a distance of 500 m in 4 min 30 sec . what is the speed in km / hr of the cyclist ? | distance = 500 m time = 4 min 10 sec = 250 sec speed = ( 500 / 250 ) m / s = 2 m / s speed in km / hr = 2 * 18 / 5 = 3.2 km / hr correct option is c | a = 4 * const_60
b = a + 30
c = 500 / b
d = c * const_3_6
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | subtract(9, reminder(add(multiply(76, const_100), 4), 9)) | the number n is 76 h 4 , where h represents the 10 ' s digit . if n is divisible by 9 , what is the value of h ? | integer is divisible by 9 - sum of digits is divisible by 9 answer : a | a = 76 * 100
b = a + 4
c = 9 - reminder
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a ) 12 , 20 , b ) 20 , 20 , c ) 20 , 11 , d ) 20 , 12 , e ) 12 , 12 | d | divide(add(divide(88, 11), divide(96, 3)), const_2) | if a boat is rowed downstream for 96 km in 3 hours and upstream for 88 km in 11 hours , what is the speed of the boat and the river ? | explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 32 ; us = 8 x = ( 32 + 8 ) / 2 = 40 / 2 = 20 km / hr y = ( 32 - 8 ) / 2 = 24 / 2 = 12 km / hr answer : d | a = 88 / 11
b = 96 / 3
c = a + b
d = c / 2
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a ) 79698 , b ) 20578 , c ) 80698 , d ) 81268 , e ) none of them | b | multiply(117, power(217, 83)) | 117 x 217 + 83 x 83 = ? | "= ( 117 ) ^ 2 + ( 83 ) ^ 2 = ( 100 + 17 ) ^ 2 + ( 100 - 17 ) ^ 2 = 2 [ ( 100 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 10000 + 289 ] = 2 x 10289 = 20578 answer is b" | a = 217 ** 83
b = 117 * a
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a ) 12.5 m , b ) 10 m , c ) 17.5 m , d ) 21.25 m , e ) 9.8 m | a | multiply(28.75, divide(17.5, 40.25)) | a flagstaff 17.5 m high casts a shadow of length 40.25 m . the height of the building , which casts a shadow of length 28.75 m under similar conditions will be | "let the height of the building x metres . less lengthy shadow , less in the height ( direct proportion ) 40.25 : 28.75 : : 17.5 : x 40.25 * x = 28.75 * 17.5 x = ( 28.75 x 17.5 ) / 40.25 x = 12.5 answer is a ." | a = 17 / 5
b = 28 * 75
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a ) 15 square meters , b ) 20 square meters , c ) 24 square meters , d ) 28 square meters , e ) 30 square meters | a | divide(subtract(subtract(196, 140), 24), const_2) | three rugs have a combined area of 196 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ? | "total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 194 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 15 . answer : a ." | a = 196 - 140
b = a - 24
c = b / 2
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a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | a line has a slope of 3 / 4 and intersects the point r ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point r of intersection is ( 40,0 ) . a is the correct answer ." | a = negate - (
b = negate * (
c = 3 / 4
d = a / b
e = negate * (
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a ) 25 , b ) 70 , c ) 50 , d ) 60 , e ) 75 | b | subtract(subtract(subtract(180, 60), 50), divide(subtract(subtract(180, 60), 50), const_3)) | 180 college students were asked in a survey if they preferred windows or mac brand computers . 50 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 60 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | "180 = 50 ( mac ) + x ( window ) + 60 ( both ) = > x = 70 answer : b" | a = 180 - 60
b = a - 50
c = 180 - 60
d = c - 50
e = d / 3
f = b - e
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a ) 15 , b ) 44 , c ) 40 , d ) 27 , e ) 12 | c | divide(add(360, 140), divide(multiply(45, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 2 / 25 ) = 40 seconds answer : c" | a = 360 + 140
b = 45 * 1000
c = b / 3600
d = a / c
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a ) a ) 130 , b ) b ) 131 , c ) c ) 148 , d ) d ) 158 , e ) e ) 160 | b | add(multiply(14, 9), 5) | what is the dividend . divisor 14 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 14 * 9 + 5 d = 126 + 5 d = 131 answer b" | a = 14 * 9
b = a + 5
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a ) 500 , b ) 800 , c ) 600 , d ) 700 , e ) 750 | b | divide(560, subtract(const_1, divide(30, const_100))) | shop offered 30 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | "sp * ( 70 / 100 ) = 560 sp = 8 * 100 = > cp = 800 answer : b" | a = 30 / 100
b = 1 - a
c = 560 / b
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a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | a | subtract(divide(add(18, 20), const_2), divide(add(add(add(add(add(1, 7), 18), 20), 29), 33), multiply(const_2, const_3))) | if q = { 1 , 7 , 18 , 20 , 29 , 33 } , how much less is the mean of the numbers in q than the median of the numbers in q ? | this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd , the middle term is the median . if the number of terms is even , the average of middle two terms is the median coming to this question , mean = ( 1 + 7 + 18 + 20 + 29 + 33 ) / 6 = 18 median = ( 18 + 20 ) / 2 = 19 difference = 1 option a | a = 18 + 20
b = a / 2
c = 1 + 7
d = c + 18
e = d + 20
f = e + 29
g = f + 33
h = 2 * 3
i = g / h
j = b - i
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a ) 66 % , b ) 75 % , c ) 86 % , d ) 116 % , e ) 150 % | b | multiply(divide(add(100, 20), add(100, divide(add(100, 20), const_2))), 100) | company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 20 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw would be approximately what percent of these combined assets ? | "let the price of company a ' s assets be 100 price of assets of kw is 20 % more than company a ' s assets which is 120 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 60 a + b = 160 kw = 120 kw / ( a + b ) * 100 = 120 / 160 * 100 = 75 % b" | a = 100 + 20
b = 100 + 20
c = b / 2
d = 100 + c
e = a / d
f = e * 100
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a ) 3 , b ) 6 , c ) 15 , d ) 17 , e ) 20 | c | divide(subtract(multiply(add(12, 23), divide(60, const_100)), 12), divide(60, const_100)) | a bowl of fruit contains 12 apples and 23 oranges . how many oranges must be removed so that 60 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 23 let number of oranges that must be removed so that 60 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 12 + ( 23 - x ) = 35 - x 12 / ( 35 - x ) = 6 / 10 = > 20 = 35 - x = > x = 15 answer c" | a = 12 + 23
b = 60 / 100
c = a * b
d = c - 12
e = 60 / 100
f = d / e
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a ) 26000 , b ) 27000 , c ) 28000 , d ) 29000 , e ) 30000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | if a town of 25,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | "let population in the beginning = 100 at the end of 1 st year population = 100 + 1 % of 100 = 100 + 1 = 101 at the end of 2 nd year population = 101 + 1 % of 101 = 101 + 1 approx = 102 approx ( on the lower side because 1 % of 101 is greater than 1 ) at the end of 3 rd year population = 102 + 1 % of 102 = 102 + 1 approx = 103 approx ( on the lower side because 1 % of 102 is greater than 1 ) at the end of 4 th year population = 103 + 1 % of 103 = 103 + 1 approx = 104 approx ( on the lower side because 1 % of 103 is greater than 1 ) at the end of 5 th year population = 104 + 1 % of 104 = 104 + 1 approx = 105 approx ( on the lower side because 1 % of 104 is greater than 1 ) i . e . 5 % greater than 25000 = 25000 + 5 % of 25000 = 25000 + 1250 = 26250 ( on the lower side ) i . e . population must be greater than 26250 but since the closest option is 26000 so answer : option a" | a = 3 + 4
b = 5 * a
c = b * 100
d = 3 + 4
e = d + 3
f = c * e
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a ) 1680 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2100), 16), add(multiply(16, 16), 4)) | one ton has 2100 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2100 pound = 13 * 2100 * 16 ounces . hence the answer is ( 13 * 2100 * 16 ) / 260 = 1680 . answer : a ." | a = 13 * 2100
b = a * 16
c = 16 * 16
d = c + 4
e = b / d
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a ) 3.25 % increase , b ) 3.25 % decrease , c ) 6.25 % increase , d ) 6.25 % decrease , e ) no change | d | subtract(const_100, subtract(add(25, const_100), divide(multiply(add(25, const_100), 25), const_100))) | the salary of a worker is first increased by 25 % and afterwards reduced by 25 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.75 ( 1.25 x ) = 0.9375 x the answer is d ." | a = 25 + 100
b = 25 + 100
c = b * 25
d = c / 100
e = a - d
f = 100 - e
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a ) 39 , b ) 33 , c ) 26 , d ) 21 , e ) 10 | a | subtract(divide(13, divide(2, 3)), 13) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 13 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 13 = > m = 39 answer a" | a = 2 / 3
b = 13 / a
c = b - 13
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a ) 120 , b ) 180 , c ) 240 , d ) 360 , e ) 480 | d | divide(multiply(30, 240), 20) | perimeter of front wheel = 30 , back wheel = 20 . if front wheel revolves 240 times . how many revolutions will the back wheel take ? | sice the distance covered by both the wheels should be same so let x be the no . of revolutions by the back wheel so 240 * 2 * pie * 30 = x * 2 * pie * 20 so x = 360 . answer : d | a = 30 * 240
b = a / 20
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a ) 20.83 % , b ) 23 % , c ) 18.3 % , d ) 19.75 % , e ) 21.23 % | a | multiply(divide(subtract(add(15, 3), add(multiply(divide(subtract(const_100, 25), const_100), 15), 3)), add(15, 3)), const_100) | 15 litres of mixture contains 25 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | alcohol in the 15 litres of mix . = 25 % of 15 litres = ( 25 * 15 / 100 ) = 3.75 litres water in it = 15 - 3.75 = 11.25 litres new quantity of mix . = 15 + 3 = 18 litres quantity of alcohol in it = 3.75 litres percentage of alcohol in new mix . = 3.75 * 100 / 18 = 20.83 % answer is a | a = 15 + 3
b = 100 - 25
c = b / 100
d = c * 15
e = d + 3
f = a - e
g = 15 + 3
h = f / g
i = h * 100
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a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1052 , d ) data inadequate , e ) none of these | c | add(800, divide(multiply(multiply(800, add(divide(multiply(subtract(956, 800), const_100), multiply(800, 3)), 4)), 3), const_100)) | rs . 800 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 800 become in 3 years ? | "solution s . i . = rs . ( 956 - 800 ) = rs . 156 rate = ( 100 x 156 / 800 x 3 ) = 6 1 / 2 % new rate = ( 6 1 / 2 + 4 ) % = 10 1 / 2 % new s . i . = rs . ( 800 x 21 / 2 x 3 / 100 ) rs . 252 . ∴ new amount = rs . ( 800 + 252 ) = rs . 1052 . answer c" | a = 956 - 800
b = a * 100
c = 800 * 3
d = b / c
e = d + 4
f = 800 * e
g = f * 3
h = g / 100
i = 800 + h
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a ) 17.5 % , b ) 24 % , c ) 30 % , d ) 35 % , e ) 38 % | e | subtract(multiply(add(const_100, 20), divide(add(const_100, 15), const_100)), const_100) | a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 15 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | "suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.15 = 138 increase in 1993 from 1991 = 138 - 100 = 38 answer : e" | a = 100 + 20
b = 100 + 15
c = b / 100
d = a * c
e = d - 100
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a ) 9680 , b ) 8130 , c ) 8134 , d ) 8556 , e ) 9808 | b | add(multiply(multiply(add(divide(3, const_100), divide(divide(subtract(7200, 6200), 5), 6200)), 6200), 5), 6200) | sonika deposited rs . 6200 which amounted to rs . 7200 after 5 years at simple interest . had the interest been 3 % more . she would get how much ? | "( 6200 * 5 * 3 ) / 100 = 930 7200 - - - - - - - - 8130 answer : b" | a = 3 / 100
b = 7200 - 6200
c = b / 5
d = c / 6200
e = a + d
f = e * 6200
g = f * 5
h = g + 6200
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a ) 20 lb , b ) 30 lb , c ) 10 lb , d ) 15 lb , e ) 5 lb | c | divide(50, const_1) | a bag of potatoes weighs 50 lbs divided by half of its weight . how much does the bag of potatoes weight ? | "sol . 50 ÷ 5 = 10 . answer : c" | a = 50 / 1
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a ) 5 , b ) 9 , c ) 12 , d ) 20 , e ) 30 | d | multiply(divide(const_10, const_2), const_4) | if x is a positive integer , then the least value of x for which x ! is divisible by 100,000 is ? can someone please explain intuitively what the question is asking ? | in order x ! to be divisible by 1,000 , it should have at least 3 trailing zeros . a trailing 0 in factorial of a number is produced by 2 and 5 in it : 2 * 5 = 10 . so , we need 10 to be in x ! at least in power of 3 . 5 ! = 120 has 1 trailing zeros . 10 ! will have 2 trailing zeros . 15 ! will have 3 trailing zeros . 20 ! will have 5 trailing zeros . answer : d . | a = 10 / 2
b = a * 4
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a ) 50 days , b ) 60 days , c ) 70 days , d ) 75 days , e ) 90 days | d | divide(multiply(600, 25), 200) | there is enough provisions for 600 men in an army camp for 25 days . if there were 200 men less , how long will the provision last ? | exp : we have , m 1 d 1 = m 2 d 2 600 * 25 = 200 * d 2 d 2 = 600 * 25 / 200 = 75 days . answer : d | a = 600 * 25
b = a / 200
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a ) 38,650 , b ) 28,650 , c ) 48,650 , d ) 68,650 , e ) 18,650 | b | divide(add(multiply(add(const_4, const_1), const_100), 40), sqrt(const_100)) | a person spent rs . 8,550 from his salary on food and 8,640 on house rent . after that he was left with 40 % of his monthly salary . what is his monthly salary ? | "total money spent on food and house rent = 8,550 + 8,640 = 17,190 which is 100 - 40 = 60 % of his monthly salary ∴ his salary = 17190 x 100 / 60 = 28650 answer : b" | a = 4 + 1
b = a * 100
c = b + 40
d = math.sqrt(100)
e = c / d
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a ) - 4 , b ) - 2 , c ) 11 , d ) 13 , e ) 22 | e | divide(add(62, 26), add(3, const_1)) | when x is multiplied by 3 , the result is 26 more than the result of subtracting x from 62 . what is the value of x ? | "the equation that can be formed is : 3 x - 26 = 62 - x or , 4 x = 88 or , x = 22 . answer : e" | a = 62 + 26
b = 3 + 1
c = a / b
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a ) 979 , b ) 1288 , c ) 2778 , d ) 2682 , e ) 1021 | a | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 6), const_100), const_1)) | find the principle on a certain sum of money at 6 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 6 * 12 / 5 ) / 100 ] p = 979 answer : a" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 6
e = d / 100
f = e + 1
g = 1120 / f
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a ) 300 , b ) 350 , c ) 390 , d ) 400 , e ) 450 | c | divide(multiply(divide(const_3600, const_10), add(5, const_60)), const_60) | my watch gains 5 min every hour . how many degrees the second hand moves in every min ? | gain 5 min every 1 hr degrees moves = = > 390 answer c | a = 3600 / 10
b = 5 + const_60
c = a * b
d = c / const_60
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a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 5 minutes | e | multiply(5, const_1) | if 5 cats can kill 5 rats in 5 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 5 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 5 minutes , so 100 cats can kill 100 rats in 5 minutes answer e" | a = 5 * 1
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a ) 2 , b ) 1 , c ) 3 , d ) 4 , e ) 8 | b | divide(subtract(multiply(4.2, 4.2), multiply(1.9, 1.9)), 2.3) | 4.2 x 4.2 - 1.9 x 1.9 / 2.3 x 6.1 is equal to : | "given expression = ( a 2 - b 2 ) / ( a + b ) ( a - b ) = ( a 2 - b 2 ) / ( a 2 - b 2 ) = 1 . answer is b ." | a = 4 * 2
b = 1 * 9
c = a - b
d = c / 2
|
a ) $ 288 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | multiply(const_2.0, divide(multiply(36, divide(2, 3)), divide(const_1, 3))) | a collection of books went on sale , and 2 / 3 of them were sold for $ 4 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | "since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 4 ) or $ 288 . the best answer is therefore a . answer : a ." | a = 2 / 3
b = 36 * a
c = 1 / 3
d = b / c
e = 2 * 0
|
a ) 6000 , b ) 5400 , c ) 3800 , d ) 4700 , e ) 2500 | a | divide(multiply(multiply(multiply(10, divide(4, const_100)), 5), subtract(const_1, divide(10, const_100))), multiply(multiply(divide(25, const_100), divide(15, const_100)), divide(8, const_100))) | calculate the number of bricks , each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 m x 5 m when 10 % of its volume is occupied by concrete ? | "explanation : let ' b ' be the nuber of bricks . = > 10 x 4 / 100 x 5 x 90 / 100 = 25 / 100 x 15 / 100 x 8 / 100 x b = > 10 x 20 x 90 = 15 x 2 x b = > b = 6000 answer is a" | a = 4 / 100
b = 10 * a
c = b * 5
d = 10 / 100
e = 1 - d
f = c * e
g = 25 / 100
h = 15 / 100
i = g * h
j = 8 / 100
k = i * j
l = f / k
|
a ) 28400 , b ) 28500 , c ) 28600 , d ) 28700 , e ) 28800 | e | multiply(factorial(5), factorial(5)) | in how many ways 5 boys and 5 girls can be seated in a row so that boys and girls are alternate ? | "number of ways = 2 * 5 ! * 5 ! = 28800 answer is e" | a = math.factorial(5)
b = math.factorial(5)
c = a * b
|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | e | subtract(subtract(multiply(divide(subtract(multiply(2450, 6), multiply(1250, 6)), subtract(1250, 850)), const_2), const_4), const_2) | the average monthly salary of laborers and supervisors in a factory is rs . 1250 per month ; where as the average monthly salary of 6 supervisors is rs . 2450 . if the average monthly salary of the laborers is rs . 850 find the number of laborers ? | "2450 850 \ / 1250 / \ 400 1200 1 : 3 1 - - > 6 3 - - > ? 18 answer : e" | a = 2450 * 6
b = 1250 * 6
c = a - b
d = 1250 - 850
e = c / d
f = e * 2
g = f - 4
h = g - 2
|
a ) 25 % , b ) 35 % , c ) 75 % , d ) 45 % , e ) 55 % | c | multiply(divide(const_3, const_4), const_100) | in a digital circuit which was to implement ( a b ) + ( a ) xor ( b ) , the designer implements ( a b ) ( a ) xor ( b ) what is the probability of error in it ? | a b ab a xor b ( ab ) + ( a xor b ) ( ab ) ( a xor b ) 0 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 1 0 so , we are getting 3 wrong answers out of 4 probability of error is 3 / 4 = 75 % answer : c | a = 3 / 4
b = a * 100
|
a ) 1.2 day , b ) 2.0 days , c ) 3 days , d ) 4.2 days , e ) 5 days | a | divide(36, multiply(divide(48, multiply(4, 2)), 5)) | if 4 men can colour 48 m long cloth in 2 days , then 5 men can colour 36 m long cloth in | the length of cloth painted by one man in one day = 48 / 4 × 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 × 5 = 1.2 day . a | a = 4 * 2
b = 48 / a
c = b * 5
d = 36 / c
|
a ) 130 % , b ) 120 % , c ) 150 % , d ) 160 % , e ) 170 % | b | add(multiply(subtract(multiply(add(const_1, divide(10, const_100)), const_2), const_1), const_100), const_100) | a man gains 10 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be . | "explanation : let the c . p . = x , then s . p . = ( 110 / 100 ) x = 11 x / 10 new s . p . = 2 ( 11 x / 10 ) = 22 x / 10 profit = 22 x / 10 - x = 12 x / 10 profit % = ( profit / c . p . ) * 100 = > ( 12 x / 10 ) * ( 1 / x ) * 100 = 120 % option b" | a = 10 / 100
b = 1 + a
c = b * 2
d = c - 1
e = d * 100
f = e + 100
|
a ) 10 days , b ) 5 days , c ) 6 days , d ) 7 days , e ) 8 days | c | divide(const_1, add(divide(const_1, 10), divide(const_1, 15))) | man 1 does a work in 10 days and man 2 does the same work in 15 days . in how many days they together will do the same work ? | man 1 day work = 1 / 10 man 2 day work = 1 / 15 both work = 1 / 10 + 1 / 15 = = = > 3 + 2 / 30 = = > 1 / 6 answer c | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / c
|
a ) 40 % pure , b ) 50 % pure , c ) 51.2 % pure , d ) 58.8 % pure , e ) none | c | multiply(power(divide(subtract(const_100, 20), const_100), const_3), const_100) | from a container having pure milk , 20 % is replaced by water and the process is repeated thrice . at the end of the third operation , the milk is | "solution let total quantity of original milk = 100 gm . milk after first operation = 80 % of 1000 ‹ = › 800 gm . milk after second operation = 80 % of 800 ‹ = › 640 gm . milk after third operation = 80 % of 640 ‹ = › 512 gm . strength of final mixtures = 51.2 % . answer c" | a = 100 - 20
b = a / 100
c = b ** 3
d = c * 100
|
a ) 5 , b ) 12 , c ) 10 , d ) 20 , e ) 30 | b | multiply(subtract(9, 6), 6) | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 6 ? | "what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 6 ? 9 ^ 6 = ( 3 ^ 2 ) ^ 6 = 3 ^ 12 b . 12" | a = 9 - 6
b = a * 6
|
a ) 1948816 , b ) 1981709 , c ) 18362619 , d ) 2031719 , e ) none of these | a | multiply(divide(1396, 1396), const_100) | 1396 x 1396 | "1396 x 1396 = ( 1396 ) 2 = ( 1400 - 2 ) 2 = ( 1400 ) 2 + ( 4 ) 2 - ( 4 x 1400 x 4 ) = 1948816 . answer : option a" | a = 1396 / 1396
b = a * 100
|
a ) 708 , b ) 718 , c ) 738 , d ) 748 , e ) 728 | e | multiply(divide(14, const_100), multiply(divide(4940, 9.50), 10)) | a man invested rs . 4940 in rs . 10 shares quoted at rs . 9.50 . if the rate of dividend be 14 % , his annual income is | "market value of a share = rs . 9.50 investment = rs . 4940 number of shares = 4940 / 9.50 = 520 face value of a share = rs . 10 dividend = 14 % dividend per share = ( 10 × 14 ) / 100 = rs . 1.4 his annual income = 520 × 1.4 = rs . 728 answer is e ." | a = 14 / 100
b = 4940 / 9
c = b * 10
d = a * c
|
a ) 14 % , b ) 25 % , c ) 28 % , d ) 34 % , e ) 40 % | e | multiply(divide(add(multiply(divide(10, const_100), 40), 20), add(40, 20)), const_100) | if 20 gallons of grape juice are added to 40 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ? | "official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 40 × 0.10 = 4 gallons of grape juice . 40 × 0.90 = 36 gallons of other components . if we add 20 gallons of grape juice , we will end up with 24 gallons of grape juice and 36 gallons of other components , and we will have a total of 60 gallons of the mixture . so 24 / 60 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 40 / 100 = 40 % . the correct answer is choice ( e )" | a = 10 / 100
b = a * 40
c = b + 20
d = 40 + 20
e = c / d
f = e * 100
|
a ) 25 , b ) 32 , c ) 40 , d ) 53 , e ) 54 | a | add(multiply(divide(30, multiply(2, 3)), 3), multiply(divide(30, multiply(2, 3)), 2)) | the l . c . m . of two numbers is 30 . the numbers are in the ratio 2 : 3 . then sum of the number is : | "let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 30 or x = 5 . the numbers are 10 and 15 . hence , required sum = ( 10 + 15 ) = 25 . answer : option a" | a = 2 * 3
b = 30 / a
c = b * 3
d = 2 * 3
e = 30 / d
f = e * 2
g = c + f
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | e | divide(subtract(100, multiply(12, const_3)), const_4) | a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 12 more blue balls than white balls , and twice as many red balls as blue balls , how many white balls does the box has ? | x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 12 more blue balls than white balls . . . equation # 2 : y = 12 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 2 y solve equation # 2 for z : z = y - 12 now , we can replace both x and z with y in equation # 1 2 y + y + ( y - 12 ) = 100 4 y - 12 = 100 4 y = 112 y = 28 there are 28 blue balls . this is 15 more than the number of white balls , so z = 16 . that ' s the answer . just as a check , x = 56 , and 56 + 28 + 16 = 100 . answer = 16 , ( e ) | a = 12 * 3
b = 100 - a
c = b / 4
|
a ) 15 / 36 , b ) 175 / 432 , c ) 575 / 1296 , d ) 583 / 1296 , e ) 1 / 2 | c | divide(subtract(1, divide(add(multiply(divide(multiply(multiply(subtract(6, 1), 6), add(multiply(subtract(6, 1), const_2), 1)), 6), const_2), multiply(6, 6)), power(multiply(6, 6), const_2))), const_2) | bill and jane play a simple game involving two fair dice , each of which has 6 sides numbered from 1 to 6 ( with an equal chance of landing on any side ) . bill rolls the dice and his score is the total of the two dice . jane then rolls the dice and her score is the total of her two dice . if jane ’ s score is higher than bill ’ s , she wins the game . what is the probability the jane will win the game ? | there are three possible outcomes in the game : 1 . b wins 2 . j wins 3 . draw so , 1 = p ( j ) + p ( b ) + p ( draw ) 1 = p ( j ) + p ( b ) + p ( draw ) the probability of b winning or j winning the game is same . so , 1 = 2 ∗ p ( j ) + p ( draw ) 1 = 2 ∗ p ( j ) + p ( draw ) so , we need to calculate the probability of a draw . a draw can happen when both will have the same sum . the possible sums are - 2 , 34 , 56 , 78 , 910 , 1112 prob . of each sum is - 2 - 11 - 1 / 36 3 - 1 2 , 2 1 - 2 / 36 4 - 1 3 , 2 2 , 3 1 - 3 / 36 5 - 1 4 , 2 3 , 3 2 , 41 - 4 / 36 6 - 1 5 , 2 4 , 3 3 , 4 2 , 5 1 - 5 / 36 7 - 1 6 , 2 5 , 3 4 , 4 3 , 5 2 , 61 - 6 / 36 8 - 2 6 , 3 5 , 4 4 , 5 3 , 6 2 , - 5 / 36 9 - 3 6 , 4 5 , 5 4 , 63 - 4 / 36 10 - 4 6 , 5 5 , 6 4 - 3 / 36 11 - 5 6 , 6 5 - 2 / 36 12 - 6 6 - 1 / 36 now , since both have to get the same sum , the probability will be same for both and the combined probability will be the square of the probabilities i . e . 2 - ( 1 / 36 ) ^ 2 3 - ( 2 / 36 ) ^ 2 etc . squaring the probabilities and adding them will give 146 / 1296 so , p ( d ) = 146 / 1296 1 = 2 * p ( j ) + p ( d ) 2 * p ( j ) = 1150 / 1296 p ( j ) = 575 / 1296 ans . c | a = 6 - 1
b = a * 6
c = 6 - 1
d = c * 2
e = d + 1
f = b * e
g = f / 6
h = g * 2
i = 6 * 6
j = h + i
k = 6 * 6
l = k ** 2
m = j / l
n = 1 - m
o = n / 2
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | multiply(11, const_1) | the total age of a and b is 11 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 11 + b + c = > a ? c = 11 + b ? b = 11 = > c is younger than a by 11 years answer : option a" | a = 11 * 1
|
a ) 400 , b ) 500 , c ) 550 , d ) 600 , e ) 700 | d | divide(multiply(40, 50), const_2) | if the sides of a triangle are 30 cm , 40 cm and 50 cm , what is its area ? | "the triangle with sides 30 cm , 40 cm and 50 cm is right angled , where the hypotenuse is 50 cm . area of the triangle = 1 / 2 * 40 * 30 = 600 cm 2 answer : option d" | a = 40 * 50
b = a / 2
|
['a ) 2', 'b ) 3', 'c ) 4', 'd ) 5', 'e ) 6'] | c | add(subtract(2, 4), multiply(2, const_3)) | the area of an equilateral triangle is subtracted from its base , and the perimeter is then added to this total , the result is 4 . what is the height of the equilateral triangle if its base is 2 ? | an eqilateral triangle means all sides are equal ; therefore , base = side . the equation is ; base - area + perimeter = b - a + p = b - 1 / 2 bh + 3 b , and b = 2 . so , 2 - 2 / 2 h + 6 = 4 , and 8 - h = 4 . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : c | a = 2 - 4
b = 2 * 3
c = a + b
|
a ) 29 / 5 , b ) 28 / 3 , c ) 27 / 6 , d ) 25 / 4 , e ) 29 / 6 | a | divide(subtract(factorial(6), factorial(4)), factorial(5)) | ( 6 ! - 4 ! ) / 5 ! = ? | ( 6 ! - 4 ! ) / 5 ! ( 4 ! * 5 * 6 - 4 ! ) / 5 ! 4 ! ( 5 * 6 - 1 ) / 4 ! * 5 29 / 5 answer a | a = math.factorial(6)
b = math.factorial(4)
c = a - b
d = math.factorial(5)
e = c / d
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(const_1, add(divide(const_1, 24), divide(const_1, 40))) | a sum of money is sufficient to pay p ' s wages for 24 days and q ' s wages for 40 days . the same money is sufficient to pay the wages of both for ? | let the total money be $ x p ' s 1 day work = $ x / 24 q ' s 1 day work = $ x / 40 p + q 1 day work = $ x / 15 money is sufficient to pay the wages of both for 15 days answer is d | a = 1 / 24
b = 1 / 40
c = a + b
d = 1 / c
|
a ) 1.99 km , b ) 1.55 km , c ) 1.82 km , d ) 2.87 km , e ) 1.87 km | e | multiply(divide(multiply(5, 3), subtract(3, 5)), divide(add(9, 6), const_60)) | if i walk at 5 km / h , i miss the bus by 9 minutes . if i walk at 3 km / h , i reach 6 minutes before the arrival of the bus . how far i walk to reach the bus stand ? | "d = product of speed difference of time / difference of speed d = 5 x 3 / 60 [ 9 â ˆ ’ ( â ˆ ’ 6 ) / 5 - 3 ] [ here , â € “ ve sign indicates before the schedule time ] â ‡ ’ d = 1.87 km answer e" | a = 5 * 3
b = 3 - 5
c = a / b
d = 9 + 6
e = d / const_60
f = c * e
|
a ) 20000 , b ) 15000 , c ) 25000 , d ) 34000 , e ) 30000 | b | multiply(27000, subtract(const_1, divide(multiply(30000, multiply(2, multiply(2, const_3))), add(multiply(45000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(30000, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 27000 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 27000 = 15000 b" | a = 2 * 3
b = 2 * a
c = 30000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 45000 * f
h = 2 * 3
i = 2 * h
j = 30000 * i
k = g + j
l = c / k
m = 1 - l
n = 27000 * m
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | e | add(divide(subtract(multiply(floor(divide(50, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1) | how many numbers from 10 to 50 are exactly divisible by 9 ? | "10 / 9 = 1 and 50 / 9 = 5 = = > 5 - 1 = 4 . therefore 4 answer : e" | a = 50 / 9
b = math.floor(a)
c = b * 9
d = 10 / 9
e = math.floor(d)
f = e + 1
g = f * 9
h = c - g
i = h / 9
j = i + 1
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 7 | e | subtract(add(6, 8), 7) | a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 7 years ago ? | "let ' s say age of richard is ` ` r ' ' age of david is ` ` d ' ' age of scott is ` ` s ' ' now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6 = ( s + 8 ) + 6 = s + 14 therefore , 2 s + 8 = s + 14 i . e . s = 6 i . e . r = 20 i . e . d = 14 now , how old was david 7 years ago ? i . e . d - 7 = 14 - 7 = 7 years answer : e" | a = 6 + 8
b = a - 7
|
a ) 7000 , b ) 8400 , c ) 6000 , d ) 5000 , e ) 4000 | b | divide(84, divide(subtract(7, 6), const_100)) | in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 84 more candidates got selected than a . what was the number of candidates appeared from each state ? | "state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 84 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 84 = > total appeared candidates in state b = 84 x 100 = 8400 = > total appeared candidates in state a = total appeared candidates in state b = 8400" | a = 7 - 6
b = a / 100
c = 84 / b
|
a ) 198 cm 2 , b ) 384 cm 2 , c ) 510 cm 2 , d ) 612 cm 2 , e ) 680 cm 2 | d | multiply(34, 18) | find the area of a parallelogram with base 34 cm and height 18 cm ? | "area of a parallelogram = base * height = 34 * 18 = 612 cm 2 answer : d" | a = 34 * 18
|
a ) 288 , b ) 390 , c ) 287 , d ) 266 , e ) 161 | b | divide(add(add(30, multiply(2, 10)), 30), 2) | a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey ? | "the total distance covered by the car in the first 10 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 = sum of 10 terms in ap whose first term is 30 and last term is 48 = 10 / 2 [ 30 + 48 ] = 390 km . answer : b" | a = 2 * 10
b = 30 + a
c = b + 30
d = c / 2
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | subtract(add(add(15, 18), 5), 35) | in a sports club with 35 members , 15 play badminton and 18 play tennis and 5 do not play either . how many members play both badminton and tennis ? | "15 + 18 = 33 but where as total number is 35 - 5 = 30 therefore answer is 33 - 30 = 3 hence answer is a" | a = 15 + 18
b = a + 5
c = b - 35
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(subtract(96, 48), subtract(48, 44)) | on day one , a store sold 96 pens . on each of the next d days , the company sold 44 pens . if the average daily sales of pens over this time period ( including day one ) is 48 pens , what is the value of d ? | 96 + 44 d = 48 ( d + 1 ) . 4 d = 48 . d = 12 . the answer is d . | a = 96 - 48
b = 48 - 44
c = a / b
|
a ) 22 , b ) 28 , c ) 99 , d ) 35 , e ) 66 | d | multiply(5, 7) | walking 7 / 6 of his usual rate , a boy reaches his school 5 min early . find his usual time to reach the school ? | "speed ratio = 1 : 7 / 6 = 6 : 7 time ratio = 7 : 6 1 - - - - - - - - 7 5 - - - - - - - - - ? = 35 m answer : d" | a = 5 * 7
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a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 201.3 , e ) $ 190 | d | floor(multiply(22, 9.15)) | carrie likes to buy t - shirts at the local clothing store . they cost $ 9.15 each . one day , she bought 22 t - shirts . how much money did she spend ? | $ 9.15 * 22 = $ 201.3 . answer is d . | a = 22 * 9
b = math.floor(a)
|
a ) 776 , b ) 1344 , c ) 299 , d ) 257 , e ) 125 | b | divide(multiply(divide(288, divide(subtract(56, subtract(const_100, 56)), const_100)), 56), const_100) | there were two candidates in an election . winner candidate received 56 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 56 % l = 44 % 56 % - 44 % = 12 % 12 % - - - - - - - - 288 56 % - - - - - - - - ? = > 1344 answer : b" | a = 100 - 56
b = 56 - a
c = b / 100
d = 288 / c
e = d * 56
f = e / 100
|
a ) 12628 , b ) 23488 , c ) 26688 , d ) 26667 , e ) 12672 | a | multiply(multiply(multiply(multiply(243, 3), 3), 3), const_2) | 243 has been divided into 3 parts such that half of the first part , one - third of the second part and one - fourth of the third part are equal . the largest part is ? | explanation : ratio of their shares = ( 35000 * 8 ) : ( 42000 * 10 ) = 2 : 3 . reena ' s share rs . 31570 * ( 2 / 5 ) = rs . 12628 . answer : a ) 12628 | a = 243 * 3
b = a * 3
c = b * 3
d = c * 2
|
['a ) 2', 'b ) 4', 'c ) 8', 'd ) 16', 'e ) none of the above'] | c | sqrt(power(8, const_2)) | in the rectangular coordinate system , points ( 8 , 0 ) and ( – 8 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ? | the answer is c it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 8 c | a = 8 ** 2
b = math.sqrt(a)
|
a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | a | divide(subtract(const_100, add(35, 20)), const_3) | a polling company surveyed a certain country , and it found that 35 % of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that 20 % had a favorable impression only of party e . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "s = 100 not ( e and b ) = 35 only e = 20 ( e and b ) / b = 1 / 2 let ( e and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans" | a = 35 + 20
b = 100 - a
c = b / 3
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a ) 3 ^ 8 , b ) 3 ^ 12 , c ) 3 ^ 16 , d ) 3 ^ 24 , e ) 3 ^ 32 | e | power(power(3, multiply(2, 4)), multiply(2, 2)) | if y ¤ w = y ^ ( 2 w ) for all positive integers , then ( 3 ¤ 4 ) ¤ 2 = | y ¤ w = y ^ ( 2 w ) 3 ¤ 4 = 3 ^ ( 8 ) ( 3 ¤ 4 ) ¤ 2 = ( 3 ^ 8 ) ¤ 2 = ( 3 ^ 8 ) ^ 4 = 3 ^ 32 correct option : e | a = 2 * 4
b = 3 ** a
c = 2 * 2
d = b ** c
|
['a ) 3 x / 2', 'b ) 4 x / 5', 'c ) 7 x / 5', 'd ) 5 x / 7', 'e ) 3 x / 4'] | a | divide(multiply(const_3, 45), const_2) | a circular rim a having a diameter of 45 inches is rotating at the rate of x inches / min . another circular rim b with a diameter of 30 inches is rotating at the rate of y inches / min . what is the value of y in terms of x , if both the rims reach their starting positions at the same time after every rotation . | t = s 1 / v 1 = s 2 / v 2 or , 45 / x = 30 / y or , y = 45 x / 30 = 3 x / 2 ( answer a ) | a = 3 * 45
b = a / 2
|
a ) 40 , b ) 60 , c ) 64 , d ) 72 , e ) none | c | multiply(divide(multiply(subtract(30, add(3, 3)), subtract(20, const_4)), multiply(20, 30)), const_100) | a typist uses a sheet measuring 20 cm by 30 cm lengthwise . if a margin of 2 cm is left on each side and a 3 cm margin on top and bottom , then percent of the page used for typing is | solution area of the sheet = ( 20 x 30 ) cm ² = 600 cm ² area used for typing = [ ( 20 - 4 ) x ( 30 - 6 ) ] cm ² = 384 cm ² required percentage = ( 384 / 600 x 100 ) % = 64 % . answer c | a = 3 + 3
b = 30 - a
c = 20 - 4
d = b * c
e = 20 * 30
f = d / e
g = f * 100
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a ) 66 , 78 , b ) 70 , 82 , c ) 94 , 106 , d ) 84 , 96 , e ) none of these | d | multiply(12, add(const_3, const_4)) | the h . c . f . of two numbers is 12 and their difference is 12 . the numbers are : | explanation : out of the given numbers , the two with h . c . f . 12 and difference 12 are 84 and 96 . answer is d | a = 3 + 4
b = 12 * a
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a ) 35000 , b ) 40000 , c ) 16000 , d ) 36000 , e ) 60000 | d | multiply(divide(80000, add(add(4, 7), 9)), 9) | john , james and jim ' investment ratio in a business is 4 : 7 : 9 . if total total investment is 80000 . find jim ' s investment ? | let the investment of john james and jim is 4 x 7 x and 9 x 4 x + 7 x + 9 x = 80000 20 x = 80000 x = 4000 jim ' investment = 9 * 4000 = 36000 answer : d | a = 4 + 7
b = a + 9
c = 80000 / b
d = c * 9
|
a ) 1 / 5 , b ) 1 / 1 , c ) 1 / 6 , d ) 1 / 3 , e ) 1 / 7 | c | multiply(3, add(divide(const_1, 30), divide(const_1, 45))) | two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 3 days ? | "a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 45 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 45 = 1 / 18 the part of the work completed in 3 days = 3 ( 1 / 18 ) = 1 / 6 . answer : c" | a = 1 / 30
b = 1 / 45
c = a + b
d = 3 * c
|
a ) 20 seconds , b ) 3 , c ) 4 , d ) 1 , e ) 5 | e | divide(8, subtract(3, 1.4)) | calculate how long it will take a swimmer to swim a distance of 8 km against the current of a river which flows at 1.4 km / hr , given that he can swim in still water at 3 km / h | swim in still water at = 3 speed of river = 1.4 us = 3 - 1.4 = 1.6 distance = 8 t = 8 / 1.6 = 5 answer : e | a = 3 - 1
b = 8 / a
|
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 8 , e ) 1 / 24 | a | divide(24, add(24, multiply(24, multiply(divide(1, 8), 24)))) | a satellite is composed of 24 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 8 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ? | "let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 8 . the number of non - upgraded sensors on the whole satellite is 24 ( x / 8 ) = 3 x . the fraction of sensors which have been upgraded is x / ( x + 3 x ) = x / 4 x = 1 / 4 the answer is a ." | a = 1 / 8
b = a * 24
c = 24 * b
d = 24 + c
e = 24 / d
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | c | divide(power(2, divide(power(2, 4), power(2, 2))), power(2, 1)) | if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 4 ø 2 ) ø 1 = ? | "4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 4 ø 1 = 2 ^ 4 / 2 = 8 the answer is c ." | a = 2 ** 4
b = 2 ** 2
c = a / b
d = 2 ** c
e = 2 ** 1
f = d / e
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(12, const_1) | if 3 ^ x * 4 ^ y = 531,441 and x – y = 12 , then x = ? | "since 531,441 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 12 and y = 0 , then x = 12 . the answer is c ." | a = 12 * 1
|
a ) s . 1690 , b ) s . 1610 , c ) s . 1890 , d ) s . 1602 , e ) s . 1092 | b | divide(multiply(subtract(const_100, 30), 2300), const_100) | a man buys a cycle for rs . 2300 and sells it at a loss of 30 % . what is the selling price of the cycle ? | "s . p . = 70 % of rs . 2300 = rs . 70 x 2300 / 100 = rs . 1610 answer : option b" | a = 100 - 30
b = a * 2300
c = b / 100
|
a ) 60 hours , b ) 6 hours , c ) 30 hours , d ) 11 hours , e ) 10 hours | d | divide(660, const_60) | how many hours are there in 660 minutes ? | "solution : number of minutes in 1 hour = 60 minutes so just divide 660 by 60 660 ã · 60 = 11 , so there are 11 hours in 660 minutes option d" | a = 660 / const_60
|
a ) 25 days , b ) 88 days , c ) 21 days , d ) 27 days , e ) 13 days | d | multiply(divide(multiply(9, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 9 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 9 x = 1 / 27 = > 27 days answer : d" | a = 2 + 1
b = 9 * a
c = b / 2
d = c * 2
|
a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 18 | c | divide(subtract(sqrt(add(multiply(multiply(16, 5), const_4), power(16, const_2))), 16), const_2) | tom read a book containing 480 pages by reading the same number of pages each day . if he would have finished the book 5 days earlier by reading 16 pages a day more , how many days did tom spend reading the book ? | "say the number of days tom spent reading the book was dd . then : the number of pages he read per day would be 480 / d the number of pages he would read per day with an increased speed would be 480 / d − 5 e are told that the number of pages for the second case was 16 pages per day more , so 480 / d = 480 / d − 5 − 16 at this point it ' s much better to plug answer choices rather than solve for dd . answer choice cc fits : 480 / 15 = 32 = 480 / 15 − 5 − 16 answer : c" | a = 16 * 5
b = a * 4
c = 16 ** 2
d = b + c
e = math.sqrt(d)
f = e - 16
g = f / 2
|
a ) 7 , b ) 5 , c ) 6 , d ) 4 , e ) 3 | b | divide(30, multiply(const_2, const_3)) | ajay and vijay have some marbles with them . ajay told vijay ` ` if you give me ' x ' marbles , both of us will have equal number of marbles ' ' . vijay then told ajay ` ` if you give me twice as many marbles , i will have 30 more marbles than you would ' ' . find ' x ' ? | if vijay gives ' x ' marbles to ajay then vijay and ajay would have v - x and a + x marbles . v - x = a + x - - - ( 1 ) if ajay gives 2 x marbles to vijay then ajay and vijay would have a - 2 x and v + 2 x marbles . v + 2 x - ( a - 2 x ) = 30 = > v - a + 4 x = 30 - - - ( 2 ) from ( 1 ) we have v - a = 2 x substituting v - a = 2 x in ( 2 ) 6 x = 30 = > x = 5 . answer : b | a = 2 * 3
b = 30 / a
|
a ) 15 , b ) 25 , c ) 40 , d ) 45 , e ) 75 | a | subtract(subtract(multiply(45, add(const_2, const_3)), multiply(42, const_4)), 42) | the average of temperatures at noontime from monday to friday is 45 ; the lowest one is 42 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 45 * 5 = 225 lowest is 42 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 42 = 168 . on the fifth day it is 225 - 168 = 57 range is therefore 57 - 42 = 15 answer a" | a = 2 + 3
b = 45 * a
c = 42 * 4
d = b - c
e = d - 42
|
a ) 9 , b ) 10 , c ) 11 , d ) 8 , e ) 12 | d | subtract(add(12, 4), multiply(const_2, 4)) | david is 12 years older than his daughter rosy . in 4 years david will be twice as old as rosy . what is rosy current age . | "now : rosy = x , david = x + 12 in 4 years rosy = x + 4 , david = x + 12 + 4 or 2 ( x + 4 ) x + 16 = 2 x + 8 x = 8 rosy is 8 years old answer : d" | a = 12 + 4
b = 2 * 4
c = a - b
|
a ) 169.49 , b ) 30 , c ) 120.32 , d ) 125.32 , e ) 15.152 | a | multiply(const_100, divide(divide(1200, const_3), 236)) | one - third of 1200 is what percent of 236 ? | "answer let one - third of 1200 is n % of 236 . ∵ 1200 / 3 = ( n x 236 ) / 100 ∴ n = ( 400 x 100 ) / 236 = 169.49 correct option : a" | a = 1200 / 3
b = a / 236
c = 100 * b
|
a ) 9000 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000 | e | subtract(subtract(8100, multiply(8100, divide(10, const_100))), multiply(subtract(8100, multiply(8100, divide(10, const_100))), divide(10, const_100))) | the population of a town is 8100 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8100 × 100 / 90 × 100 / 90 = 10000 e )" | a = 10 / 100
b = 8100 * a
c = 8100 - b
d = 10 / 100
e = 8100 * d
f = 8100 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 1200 , b ) 160 , c ) 360 , d ) 240 , e ) none of these | b | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(48, add(32, 4))), 4) | average age of students of an adult school is 48 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | explanation : let the original no . of students be x . according to situation , 48 x + 120 * 32 = ( x + 120 ) 36 ⇒ x = 40 so , required no . of students after joining the new students = x + 120 = 160 . answer : b | a = 32 + 4
b = a * 120
c = 120 * 32
d = b - c
e = 32 + 4
f = 48 - e
g = d / f
h = g * 4
|
a ) 15 , b ) 6 , c ) 1 , d ) 5 , e ) 12 | b | divide(add(8, const_4), subtract(9, 7)) | if 9 - 4 / x = 7 + 8 / x , then x = | we ' re given the equation 9 - 4 / x = 7 + 8 / x . we ' re asked for the value of x . the common - denominator of these 4 numbers is x , so we have to multiple 2 of the numbers to give them that denominator . . . 9 x - 4 x / x = 7 x + 8 x / x we can then eliminate that denominator , which gives us . . . . 9 x - 4 = 7 x + 8 12 = 2 x both 2 x and 12 have a common denominator of 2 , so we divide both sides by 2 for our final answer . . . 6 = x b | a = 8 + 4
b = 9 - 7
c = a / b
|
a ) $ 25 , b ) $ 32 , c ) $ 36 , d ) $ 40 . , e ) $ 42 . | e | add(add(multiply(12, 3), 4), multiply(divide(12, const_60), divide(25, 4))) | it costs $ 4 for the first 1 / 4 hour to use the laundry machine at the laundromat . after the first ¼ hour it costs $ 12 per hour . if a certain customer uses the laundry machine for 3 hours and 25 minutes , how much will it cost him ? | "3 hrs 25 min = 205 min first 15 min - - - - - - > $ 4 time left is 190 min . . . now , 60 min costs $ 12 1 min costs $ 12 / 60 190 min costs $ 12 / 60 * 190 = > $ 38 so , total cost will be $ 38 + $ 4 = > $ 42 hene answer will be ( e ) $ 42" | a = 12 * 3
b = a + 4
c = 12 / const_60
d = 25 / 4
e = c * d
f = b + e
|
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