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 ) 2 hours , b ) 2 hours 20 minutes , c ) 2 hours 30 minutes , d ) 2 hours 45 minutes , e ) 3 hours | e | divide(add(13, 5), subtract(53, 47)) | truck x is 13 miles ahead of truck y , which is traveling the same direction along the same route as truck x . if truck x is traveling at an average speed of 47 miles per hour and truck y is traveling at an average speed of 53 miles per hour , how long will it take truck y to overtake and drive 5 miles ahead of truck x ? | relative speed = 53 - 47 = 6 miles per hour dist required = 13 + 5 = 18 miles time taken to overtake = 18 / 6 = 3 hours . e is the answer . | a = 13 + 5
b = 53 - 47
c = a / b
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a ) 1280 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3400 | a | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(125, 11.25), 6)) | how many bricks , each measuring 125 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 125 x 11.25 x 6 ) = 1280 answer : a" | a = 8 * 100
b = 6 * 100
c = a * b
d = c * 22
e = 125 * 11
f = e * 6
g = d / f
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a ) 27 , b ) 33 , c ) 54 , d ) 81 , e ) 120 | e | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 5) | the number of boxes in a warehouse can be divided evenly into 5 equal shipments by boat or 24 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | answer is the lcm of 5 and 24 = 120 answer e | a = 2 * 2
b = a * 2
c = b * 3
d = c * 5
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a ) 4360 , b ) 4460 , c ) 4560 , d ) 4660 , e ) 4760 | b | multiply(divide(subtract(10,035, 6690), subtract(multiply(6690, 6), multiply(3, 10,035))), const_100) | a sum of money amounts to rs . 6690 after 3 years and to rs . 10,035 after 6 years on compound interest . find the sum . | "let the sum be rs . p . then p ( 1 + r / 100 ) 3 = 6690 β¦ ( i ) p ( 1 + r / 100 ) 6 = 10035 β¦ ( ii ) on dividing , we get ( 1 + r / 100 ) 3 = 10025 / 6690 = 3 / 2 . substituting this value in ( i ) , we get : p * ( 3 / 2 ) = 6690 or p = ( 6690 * 2 / 3 ) = 4460 hence , the sum is rs . 4460 . answer : b" | a = 10 - 35
b = 6690 * 6
c = 3 * 10
d = b - c
e = a / d
f = e * 100
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a ) 166 , b ) 105 , c ) 175 , d ) 177 , e ) 169 | c | add(add(multiply(divide(const_100, 45), 45), multiply(divide(30, 45), 45)), 45) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 45 , what is the total amount ? | "x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 45 35 - - - ? = > 175 answer : c" | a = 100 / 45
b = a * 45
c = 30 / 45
d = c * 45
e = b + d
f = e + 45
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(const_60, multiply(divide(95, 100), const_60)) | without stoppages , a train travels certain distance with an average speed of 100 km / h , and with stoppages , it covers the same distance with an average speed of 95 km / h . how many minutes per hour the train stops ? | due to stoppages , it covers 5 km less . time taken to cover 5 km = 5 Γ’ Β β 100 h = 1 Γ’ Β β 20 h = 1 Γ’ Β β 20 Γ£ β 60 min = 3 min answer c | a = 95 / 100
b = a * const_60
c = const_60 - b
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a ) 558 , b ) 502 , c ) 516 , d ) 612 , e ) 700 | a | multiply(add(multiply(multiply(add(25, 12), const_2), 6), multiply(25, 12)), divide(75, const_100)) | a tank is 25 m long , 12 m wide and 6 m deep . what is the cost of plastering its walls and bottom at the rate of 75 paise per sq . m ? | "total surface s = 2 lw + 2 lh + 2 wh = ( 25 * 12 ) + ( 2 * 25 * 6 ) + ( 2 * 12 * 6 ) = 300 + 300 + 144 = 744 m 2 therefore cost of plastering = 744 * 75 = 5800 paise = 558 answer a" | a = 25 + 12
b = a * 2
c = b * 6
d = 25 * 12
e = c + d
f = 75 / 100
g = e * f
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a ) 9 , b ) 18.4 , c ) 15.5 , d ) 16.7 , e ) 13.4 | c | divide(add(divide(26, 2), divide(36, 2)), const_2) | a man swims downstream 36 km and upstream 26 km taking 2 hours each time , what is the speed of the man in still water ? | "36 - - - 2 ds = 18 ? - - - - 1 26 - - - - 2 us = 13 ? - - - - 1 m = ? m = ( 18 + 13 ) / 2 = 15.5 answer : c" | a = 26 / 2
b = 36 / 2
c = a + b
d = c / 2
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a ) 8876 , b ) 2765 , c ) 6400 , d ) 1298 , e ) 1261 | c | multiply(multiply(const_1, const_12), divide(19200, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 19200 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 19200 = 6400 answer : c" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 19200 / h
j = a * i
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a ) 90 , b ) 100 , c ) 110 , d ) 120 , e ) 130 | c | divide(add(110, 220), 3) | a student chose a number , multiplied it by 3 , then subtracted 220 from the result and got 110 . what was the number he chose ? | "let x be the number he chose , then 3 β
x β 220 = 110 3 x = 330 x = 110 correct answer c" | a = 110 + 220
b = a / 3
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a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 11 , d ) 2 ^ 35 , e ) 2 ^ 37 | c | divide(multiply(2, add(2, 2)), 2) | 2 + 2 + 2 Β² + 2 Β³ . . . + 2 ^ 10 | "2 + 2 = 2 ^ 2 2 ^ 2 + 2 ^ 2 = ( 2 ^ 2 ) * ( 1 + 1 ) = 2 ^ 3 2 ^ 3 + 2 ^ 3 = ( 2 ^ 3 ) * ( 1 + 1 ) = 2 ^ 4 so you can notice the pattern . . . in the end you will have 2 ^ 10 + 2 ^ 10 , which will give you 2 ^ 11 answer c" | a = 2 + 2
b = 2 * a
c = b / 2
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a ) 450 , b ) 600 , c ) 750 , d ) 2400 , e ) 3200 | b | multiply(3600, divide(1200, 7200)) | a football field is 7200 square yards . if 1200 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 a ) 9600 yards need 1200 lbs 1 yard will need 1200 / 7200 = 1 / 6 lbs 3600 yards will need 1 / 6 * 3600 yards = 600 lbs b" | a = 1200 / 7200
b = 3600 * a
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a ) 2135 , b ) 6551 , c ) 2704 , d ) 6466 , e ) 1113 | c | subtract(power(add(476, 424), 2), multiply(multiply(4, 476), 424)) | { ( 476 + 424 ) 2 - 4 x 476 x 424 } = ? | given exp . = [ ( a + b ) 2 - 4 ab ] , where a = 476 and b = 424 = [ ( 476 + 424 ) 2 - 4 x 476 x 424 ] = [ ( 900 ) 2 - 807296 ] = 810000 - 807296 = 2704 . correct option is c | a = 476 + 424
b = a ** 2
c = 4 * 476
d = c * 424
e = b - d
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(divide(18, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 18 - meter platform and caught it after it had traveled 53.4 meters . how many times did the ball bounce ? | "ans : 6 division of total diatance travelled will be 18 + 18 + 9 + 4.5 + 2.25 + 1.125 + 0.5 ans b" | a = 18 / 2
b = a / 2
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a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 66 2 / 3 % | e | multiply(divide(subtract(150, 90), 90), const_100) | a certain telescope increases the visual range at a particular location from 90 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | "original visual range = 90 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 90 ) / 90 * 100 % = 2 / 3 * 100 % = 66.67 % answer e" | a = 150 - 90
b = a / 90
c = b * 100
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a ) $ 180 , b ) $ 280 , c ) $ 420 , d ) $ 550 , e ) $ 700 | d | multiply(divide(35, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 35 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 35 0.07 = 35 p = 500 the original selling price was 500 * 1.1 = 550 . the correct answer is d ." | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 35 / h
j = 10 / 100
k = 1 + j
l = i * k
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a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | c | add(multiply(const_3, 32), divide(40, const_2)) | a car left town a towards town b driving at a speed of v = 32 km / hr . after 3 hours on the road the driver stopped for 15 min in town c . because of a closed road he had to change his route , making the trip 28 km longer . he increased his speed to v = 40 km / hr but still he was 30 min late . find the distance the car has covered . | the driver did not plan the stop at c . suppose it took x hours for him to get from c to b . then the distance is s = 40 β
x km . it took x β 3060 β 1560 = x β 4560 = x β 34 h to drive from c to b . the distance from c to b is 32 ( x β 34 ) km , which is 28 km shorter than 40 β
x , i . e . 32 ( x β 34 ) + 28 = 40 x 32 x β 24 + 28 = 40 x 4 = 8 x x = 12 hr β
x = 30 min . then the time of the trip from c to b was 30 min . the distance covered equals 3 β
32 + 12 β
40 = 96 + 20 = 116 km . so answer is c . | a = 3 * 32
b = 40 / 2
c = a + b
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a ) 145 , b ) 155 , c ) 165 , d ) 175 , e ) 185 | c | divide(333, add(add(multiply(divide(3, 5), divide(7, 11)), divide(7, 11)), const_1)) | total num of pupiles in 3 grades of scholl is 333 . the no of pupiles in grades 1 and 2 are un 3 : 5 ratio and 2 and 3 grades are 7 : 11 ratio . what is the strength of class that has highest no of pupils | ratio g 1 : g 2 = 3 : 5 and g 2 : g 3 = 7 : 11 so g 1 : g 2 : g 3 = 21 : 35 : 55 let the strength of three classes are 21 x , 35 x and 55 x respectively , then 21 x + 35 x + 55 x = 333 = > 111 x = 333 or x = 3 so strength of the class with highest number of pupils = 55 x = 55 * 3 = 165 answer : c | a = 3 / 5
b = 7 / 11
c = a * b
d = 7 / 11
e = c + d
f = e + 1
g = 333 / f
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a ) 20 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 40 sec | a | multiply(const_3600, divide(divide(150, const_1000), add(25, 2))) | a train 150 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 Γ 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 150 Γ 2 / 15 ) sec = 20 sec answer : a | a = 150 / 1000
b = 25 + 2
c = a / b
d = 3600 * c
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a ) 12 , b ) 14 , c ) 15 , d ) 24 , e ) 36 | d | multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3)) | initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many d women are currently in the room ? | the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d | a = 14 - 2
b = 5 * a
c = b / 4
d = c - 3
e = 2 * d
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a ) 2 β 2 , b ) 2 β 5 , c ) 6 , d ) 4 β 2 , e ) none of these | c | divide(divide(multiply(add(const_10, const_4), const_2), const_2), const_2) | twenty four meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ? | "area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 24 = > x / 360 * 2 * pi * r + 2 r = 24 = > 2 a / r + 2 r = 24 = > a = r 12 - r ^ 2 = r 12 - r ^ 2 we will now max using derivations max value of a will found at a = 0 i . e 12 - 2 r = 0 r = 6 c" | a = 10 + 4
b = a * 2
c = b / 2
d = c / 2
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a ) 277 , b ) 960 , c ) 277 , d ) 266 , e ) 768 | e | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(20, const_100))) | the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 20 % profit ? | "let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 120 % of rs . 640 = 120 / 100 * 640 = rs . 768 . answer : e" | a = 832 - 448
b = a / 2
c = 832 - b
d = 20 / 100
e = 1 + d
f = c * e
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a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50 | b | divide(subtract(multiply(multiply(2, multiply(2, multiply(2, multiply(2, 2)))), 2), const_4), const_10) | how many positive integers , from 2 to 100 , inclusive , are not divisible by odd integers greater than 1 ? | the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence answer : b | a = 2 * 2
b = 2 * a
c = 2 * b
d = 2 * c
e = d * 2
f = e - 4
g = f / 10
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a ) 10 days , b ) 30 days , c ) 60 days , d ) 70 days , e ) 50 days | a | divide(multiply(6, 15), subtract(15, 6)) | a and b finish the job in 15 days . while a , b and c can finish it in 6 days . c alone will finish the job in | "explanation : 6 = ( 15 * x ) / ( 15 + x ) 90 + 6 x = 15 x 9 x = 90 x = 10 answer : option a" | a = 6 * 15
b = 15 - 6
c = a / b
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a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50 | d | subtract(multiply(subtract(60, 10), const_3), multiply(subtract(60, 30), const_3)) | in x game of billiards , x can give y 10 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 100 ? | "x scores 60 while y score 50 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 50 = 60 . in x game of 100 points , y gives ( 100 - 60 ) = 40 points to c . d" | a = 60 - 10
b = a * 3
c = 60 - 30
d = c * 3
e = b - d
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a ) 22.8 , b ) 21.8 , c ) 20.8 , d ) 28.8 , e ) 29.8 | c | divide(multiply(multiply(const_2, 16), 30), add(16, 30)) | find avrg speed if a man travels at speed of 16 km / hr up and 30 km / hr dawn at an altitude of 200 m . | "avg speed = 2 * x * y / ( x + y ) = 2 * 16 * 30 / ( 16 + 30 ) = 20.8 answer : c" | a = 2 * 16
b = a * 30
c = 16 + 30
d = b / c
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a ) 20 days , b ) 5 1 / 3 days , c ) 6 days , d ) 8 days , e ) 7 days | b | divide(const_1, subtract(divide(const_1, 8), divide(const_1, 16))) | a can do a work in 16 days and b can do it in 8 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 16 b ' s 1 day ' s work = 1 / 8 they work together = 1 / 16 + 1 / 8 = 3 / 16 = 16 / 3 = 5 1 / 3 days answer : option b" | a = 1 / 8
b = 1 / 16
c = a - b
d = 1 / c
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a ) 150 , b ) 750 , c ) 1,250 , d ) 1,000 , e ) 2,500 | d | multiply(50, 5) | in a forest 100 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 100 tagged deers comprise 10 % of total # of deers - - > total # of deers = 100 * 10 = 1,000 . answer : d ." | a = 50 * 5
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a ) 1500 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2000 | e | multiply(multiply(divide(20, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 20 of money ? | "p = 20 ( 100 / 10 ) 2 = > p = 2000 answer : e" | a = 10 * 2
b = 20 / a
c = b * 100
d = 10 * 2
e = c * d
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a ) 45 , b ) 36 , c ) 40 , d ) 54 , e ) 48 | d | divide(original_price_before_loss(20, 90), divide(original_price_before_gain(20, 50), 20)) | a man sold 20 articles for $ 50 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 50 * ( 100 % - 20 % ) / 20 = $ 2.0 required production costs for a loss of 20 % : $ 90 * ( 100 % + 20 % ) = $ 108 number of articles to be sold for $ 108 to incur a 20 % loss : $ 108 / $ 2.0 = 54 thus , solution d is correct ." | a = original_price_before_loss / (
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a ) 3 , b ) 2 , c ) 5 , d ) 6 , e ) 1 | b | subtract(17, reminder(433124, 17)) | find the least number must be added to 433124 so that remaining no . is divisible by 17 ? | "on dividing 433124 by 17 we get the remainder 15 , so 2 should be added b" | a = 17 - reminder
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a ) s . 109 , b ) s . 108 , c ) s . 100 , d ) s . 103 , e ) s . 105 | e | multiply(5, divide(1554, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1554 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1554 74 x = 1554 = > x = 21 5 x = 105 rs . answer : e" | a = 6 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1554 / e
g = 5 * f
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a ) 85 kg , b ) 90 kg , c ) 83 kg , d ) 80 kg , e ) 82 kg | d | add(multiply(25, 1), 55) | the average weight of 25 girls increases by 1 kg when a new girl comes in place of one of them weighing 55 kg . what might be the weight of the new girl ? | "total weight increased = 25 x 1 kg = 25 kg . weight of new person = 55 + 25 kg = 80 kg answer : d" | a = 25 * 1
b = a + 55
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a ) a . 14 , b ) b . 12 , c ) c . 11 , d ) d . 15 , e ) e . 16 | e | divide(divide(multiply(120, const_2), const_3), const_4) | everyone shakes hands with everyone else in a room . total number of handshakes is 120 . number of persons = ? | "in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 120 or n ( n - 1 ) = 240 or n = 16 answer is ( e )" | a = 120 * 2
b = a / 3
c = b / 4
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a ) 400 , b ) 420 , c ) 430 , d ) 450 , e ) 525 | e | divide(subtract(multiply(15, 470), multiply(15, 120)), subtract(120, 110)) | the average salary of the employees in a office is rs . 120 / month . the avg salary of officers is rs . 470 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office . | "let no . of non - officers be x 15 * 470 + x * 110 = ( x + 15 ) 120 x = 525 e" | a = 15 * 470
b = 15 * 120
c = a - b
d = 120 - 110
e = c / d
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a ) 280 , b ) 282 , c ) 284 , d ) 286 , e ) 288 | b | add(3, lcm(31, 9)) | find the least number which when divided by 31 and 9 leaves a remainder of 3 in each case . | "the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 31 , 9 ) + 3 = 282 . answer : b" | a = math.lcm(31, 9)
b = 3 + a
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a ) 20 s , b ) 24 s , c ) 22 s , d ) 23 s , e ) 21 s | b | divide(500, multiply(add(45, 30), const_0_2778)) | two train each 500 m long , are running in opposite directions on parallel tracks . if their speeds are 45 km / hr and 30 km / hr respectively , the time taken by the slower train to pass the driver of the faster one is ? | relative speed = 45 + 30 = 75 km / hr = 750 / 36 m / s = 125 / 6 m / s we are calculating the time taken by the slower train to pass the driver of the faster one . hence the distance = length of the smaller train = 500 m time = distance / speed = 500 / ( 125 / 6 ) = 24 s answer is b . | a = 45 + 30
b = a * const_0_2778
c = 500 / b
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a ) 45 , b ) 43 , c ) 51 , d ) 53 , e ) 57 | b | add(32, const_1) | the average age of 32 students in a group is 10 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "age of the teacher = ( 33 * 11 - 32 * 10 ) = 43 years . answer : b" | a = 32 + 1
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a ) 16000 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17500 | e | multiply(divide(subtract(21000, divide(multiply(21000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 21000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 21000 - 10 / 100 ( 21000 ) = 21000 - 2100 = $ 18900 = > 1.08 x = 18900 = > x = 17500 e" | a = 21000 * 10
b = a / 100
c = 21000 - b
d = 100 + 8
e = c / d
f = e * 100
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a ) 15 , b ) 20 , c ) 30 , d ) 40 , e ) 45 | d | divide(subtract(subtract(200, 80), 60), add(3, const_1)) | a marketing firm determined that , of 200 households surveyed , 80 used neither brand k nor brand b soap , 60 used only brand k soap , and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 households surveyed used both brands of soap ? | solution for soap k and soap b ( d ) 40 | a = 200 - 80
b = a - 60
c = 3 + 1
d = b / c
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a ) 22 , b ) 21 , c ) 8 , d ) 9 , e ) 81 | b | add(multiply(10, const_2), const_1) | a shopkeeper cheats to the extent of 10 % while buying and selling , by using false weights . his total gain is . | answer : b ) 21 % | a = 10 * 2
b = a + 1
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a ) 3 : 5 , b ) 5 : 7 , c ) 7 : 11 , d ) 9 : 11 , e ) 11 : 15 | d | divide(subtract(const_100, 5), add(const_100, 15)) | the number of singles that a baseball player hit increased 15 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 5 percent . if the total number of singles and doubles that she hit increased 4 percent from her first season to her second season , what is the ratio of the number of singles she hit in her first season to the number of doubles she hit in her first season ? | soln : - 1.15 s + 0.95 d = 1.04 [ s + d ] 0.11 s = 0.09 d s / d = 9 / 11 answer : d | a = 100 - 5
b = 100 + 15
c = a / b
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a ) 18 , b ) 29 , c ) 37 , d ) 44 , e ) 51 | e | multiply(add(2, const_1), subtract(25, 2)) | the product x of two prime numbers is between 17 and 55 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25 , then x = | "option b & c can be ruled out as they themselves are prime numbers 18 = 2 * 9 = 3 * 6 > > ignore 44 = 2 * 22 = 4 * 11 > > ignore 51 = 3 * 17 > > answer answer e" | a = 2 + 1
b = 25 - 2
c = a * b
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a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 72.33 | e | add(add(add(60, const_3), const_4), const_4) | the average of first 6 prime numbers which are between 60 and 90 is | explanation : first six prime numbers which are between 60 and 90 = 61 , 67 , 71 , 73 , 79 , 83 average = ( 61 + 67 + 71 + 73 + 79 + 83 ) / 6 = 72.33 answer : e | a = 60 + 3
b = a + 4
c = b + 4
|
a ) 1350 , b ) 1550 , c ) 9676 , d ) 1679 , e ) 2691 | b | multiply(subtract(rectangle_area(add(95, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(95, 55)), 2) | a rectangular grass field is 95 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 95 + 55 + 2.5 * 2 ) 2 * 2.5 = > 775 775 * 2 = rs . 1550 answer : b" | a = 2 * 5
b = 95 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 266 , b ) 278 , c ) 342 , d ) 787 , e ) 191 | a | subtract(subtract(350, divide(multiply(350, 20), const_100)), divide(multiply(subtract(350, divide(multiply(350, 20), const_100)), 5), const_100)) | the sale price sarees listed for rs . 350 after successive discount is 20 % and 5 % is ? | "350 * ( 80 / 100 ) * ( 95 / 100 ) = 266 answer : a" | a = 350 * 20
b = a / 100
c = 350 - b
d = 350 * 20
e = d / 100
f = 350 - e
g = f * 5
h = g / 100
i = c - h
|
a ) 20.14 seconds , b ) 12.14 seconds , c ) 22.14 seconds , d ) 32.14 seconds , e ) none of these | d | divide(add(360, 140), divide(multiply(56, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 56 km / hour . in what time will it pass a bridge of 140 meter length . | "explanation : speed = 56 km / hr = 56 * ( 5 / 18 ) m / sec = 140 / 9 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 / 140 / 9 = 32.14 seconds answer : d" | a = 360 + 140
b = 56 * 1000
c = b / 3600
d = a / c
|
a ) s . 40 , b ) s . 46 , c ) s . 55 , d ) s . 41 , e ) s . 42 | c | divide(divide(multiply(1100, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 1100 , what is the reduced price for kg ? | "800 * ( 25 / 100 ) = 275 - - - - 5 ? - - - - 1 = > rs . 55 answer : c" | a = 1100 * 25
b = a / 100
c = b / 5
|
a ) 350 , b ) 500 , c ) 650 , d ) 235 , e ) 450 | b | add(subtract(subtract(subtract(add(add(280, 254), 280), 97), 138), 152), 73) | in a college , 280 students study hindi , 254 students study sanskrit and 280 students study english . 97 students study hindi as well as physics , 138 students study sanskrit as well as english and 152 students study hindi as well as english . 73 students study all the 3 languages . how many total number of students are there in that college ? | b | a = 280 + 254
b = a + 280
c = b - 97
d = c - 138
e = d - 152
f = e + 73
|
a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 230 | a | multiply(subtract(20, const_1), divide(20, const_2)) | 20 men shake hands with each other . maximum no of handshakes without cyclic handshakes . | "1 st person will shake hand with 19 people 2 nd person will shake hand with 18 people 3 rd person will shake hand with 17 people . . . . . . total no . of handshakes = 19 + 18 + 17 + . . . + 3 + 2 + 1 = 19 * ( 19 + 1 ) / 2 = 190 or , if there are n persons then no . of shakehands = nc 2 = 20 c 2 = 190 answer : a" | a = 20 - 1
b = 20 / 2
c = a * b
|
a ) 3.75 days , b ) 3.78 days , c ) 5.3 days , d ) 3.15 days , e ) 2.75 days | c | inverse(add(inverse(8), inverse(16))) | a and b complete a work in 8 days . a alone can do it in 16 days . if both together can do the work in how many days ? | "1 / 8 + 1 / 16 = 3 / 16 16 / 3 = 5.3 days answer : c" | a = 1/(8)
b = 1/(16)
c = a + b
d = 1/(c)
|
a ) 5 monkeys , b ) 6 monkeys , c ) 7 monkeys , d ) 8 monkeys , e ) 9 monkeys | d | divide(48, divide(48, 8)) | suppose 8 monkeys take 8 minutes to eat 8 bananas . how many monkeys would it take to eat 48 bananas in 48 minutes | "8 monkeys take 8 minutes to eat 8 bananas means each monkey takes 8 min for 1 banana in 48 mins one monkey can eat 6 bananas . . . . which means that 8 monkeys can eat 48 bananas in 48 minutes so answer is 8 monkeys answer : d" | a = 48 / 8
b = 48 / a
|
a ) a . 150 , b ) b . 300 , c ) c . 600 , d ) d . 120 , e ) e . 200 | d | multiply(0.2, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100))) | a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.2 per box per month for the record storage , what is the total amount that the company pays each month for record storage ? | volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.2 = 120 answer : d | a = 15 * 12
b = a * 10
c = b / 100
d = c / 100
e = d / 100
f = 1 / 8
g = 0 * 2
|
a ) 7 / 17 , b ) 13 / 33 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | b | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 1.3 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | "the easiest way for me is to plug in numbers . let the number of basic stereos produced be 40 , and number of delux stereos produced be 20 . total of 60 stereos . if it takes an hour to produce a basic stereo then it will take 1.3 hours to produce a deluxe stereo . 40 basic stereos = 40 hours . 20 delux stereos = 26 hours . total hours = 66 . then the fraction would be 26 / 66 = 13 / 33 . therefore answer b b ." | a = 3 + 2
b = a / 10
|
a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 240 | e | divide(multiply(528, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 528 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 528 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 528 β y [ 12 / 10 + 1 ] = 528 β 22 y / 10 = 528 β 22 y = 5280 β y = 5280 / 22 = 480 / 2 = rs . 240 e" | a = 1 + 4
b = a * 2
c = 528 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 2.25 hours , b ) 8.75 hours , c ) 12 hours , d ) 24 hours , e ) 16 hours | d | divide(multiply(6, 8), 2) | 8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 2 such machines to perform the same job ? | "let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 2 machines 2 units of work ( 2 machines ) - - - > 1 hour for 48 units of work 2 * 24 - - - > 1 * 24 hours d 24 hours" | a = 6 * 8
b = a / 2
|
a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 49 , e ) 47 | d | subtract(50, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 50 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "working backward from the day it ' s covered : day 50 : fully covered day 49 : half covered so 49 days answer : d" | a = 50 - 1
|
a ) 17 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | a | floor(multiply(const_100, divide(subtract(2564.36, 2200), 2200))) | a shopkeeper sold an article for rs 2564.36 . approximately what was his profit percent if the cost price of the article was rs 2200 | "explanation : gain % = ( 364.36 * 100 / 2200 ) = 16.56 % = 17 % approx option a" | a = 2564 - 36
b = a / 2200
c = 100 * b
d = math.floor(c)
|
a ) β 48 , b ) β 2 , c ) 46 , d ) 6 , e ) 48 | d | subtract(subtract(subtract(subtract(add(add(6, 27), subtract(6, 27)), const_1), const_1), const_1), const_1) | if a ( a - 6 ) = 27 and b ( b - 6 ) = 27 , where a β b , then a + b = ? | "i . e . if a = - 3 then b = 9 or if a = 9 then b = - 3 but in each case a + b = - 3 + 9 = 6 answer : option d" | a = 6 + 27
b = 6 - 27
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) 8 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 14 % | e | multiply(subtract(divide(22, const_100), divide(subtract(8.64, multiply(divide(22, const_100), multiply(18, const_2))), subtract(multiply(15, const_3), multiply(18, const_2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 8.64 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discount rates is 22 percent , what is the discount rate on pony jeans ? | let x be the discount on pony jeans . then 0.22 - x is the discount on fox jeans . 3 ( 0.22 - x ) ( 15 ) + 2 x ( 18 ) = 8.64 9.9 - 45 x + 36 x = 8.64 9 x = 1.26 x = 0.14 the answer is e . | a = 22 / 100
b = 22 / 100
c = 18 * 2
d = b * c
e = 8 - 64
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
|
a ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 144 pi , e ) 576 pi | d | multiply(multiply(4, 18), multiply(const_2, const_pi)) | a cylindrical can has a radius of 4 centimeters and a height of 18 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ? | "they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 4 * 18 = 144 pi answer is d ." | a = 4 * 18
b = 2 * math.pi
c = a * b
|
a ) 2 , b ) 6 , c ) 4 , d ) 5 , e ) 1 | e | subtract(divide(add(multiply(const_10, 5), 5), 3), divide(add(const_10, 3), 4)) | what is 5 2 / 3 - 3 5 / 4 divided by 5 / 2 - 7 / 6 ? | "5 2 / 3 - 3 5 / 4 = 17 / 3 - 20 / 4 = ( 68 - 60 ) / 12 = 8 / 12 = 4 / 3 5 / 2 - 7 / 6 = ( 15 - 7 ) / 6 = 8 / 6 = 4 / 3 so 4 / 3 / 4 / 3 = 1 answer - e" | a = 10 * 5
b = a + 5
c = b / 3
d = 10 + 3
e = d / 4
f = c - e
|
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | a | sqrt(add(multiply(131, const_2), 222)) | sum of the squares of 3 no . ' s 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)
|
a ) $ 150 , b ) $ 245 , c ) $ 365 , d ) $ 715 , e ) $ 265 | e | subtract(divide(subtract(multiply(500, 7), add(add(add(406, 413), add(436, 420)), 395)), const_2), 450) | a salesman ' s income consists of a commission and a base salary of $ 450 per week . over the past 5 weeks , his weekly income totals have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 over the 7 - week period ? | "total weekly income over 5 weeks = $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2070 = $ 1430 from this , we subtract base salary for 2 weeks i . e $ 450 * 2 = $ 900 therefore , commission = $ 1430 - $ 900 = $ 530 for 2 weeks avg weekly commission = $ 265 answer e" | a = 500 * 7
b = 406 + 413
c = 436 + 420
d = b + c
e = d + 395
f = a - e
g = f / 2
h = g - 450
|
a ) 360 , b ) 900 , c ) 540 , d ) 640 , e ) 720 | b | divide(divide(400, subtract(1, divide(1, 3))), subtract(1, divide(1, 3))) | an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 400 sculptures are not on display , how many pieces of art does the gallery have ? | "too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 / 3 of the pieces not on display are sculptures . 400 sculptures are not on display , so according to ( iii ) 2 / 3 * { not on display } = 400 - - > { not on display } = 600 . according to ( i ) 2 / 3 * { total } = 600 - - > { total } = 900 . answer : b ." | a = 1 / 3
b = 1 - a
c = 400 / b
d = 1 / 3
e = 1 - d
f = c / e
|
a ) 244140626 , b ) 243120617 , c ) 242580531 , d ) 241230438 , e ) 240760355 | a | divide(add(power(25, 7), 25), 25) | what is ( 25 ^ 7 + 25 ) / 25 ? | "( 25 ^ 7 + 25 ) / 25 = 25 * ( 25 ^ 6 + 1 ) / 25 = 25 ^ 6 + 1 clearly this is a number which ends with a 6 in the units place . the answer is a ." | a = 25 ** 7
b = a + 25
c = b / 25
|
a ) 5 / 17 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 7 | a | divide(5, add(multiply(4, 3), 5)) | chris mixed 5 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 3 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ? | "1 lbs of raisin = $ 1 5 lbs of raisin = $ 5 1 lbs of nuts = $ 3 4 lbs of nuts = $ 12 total value of mixture = 12 + 5 = 17 fraction of the value of raisin = 5 / 17 = 1 / 5 ans : a" | a = 4 * 3
b = a + 5
c = 5 / b
|
a ) 118 , b ) 150 , c ) 300 , d ) 258 , e ) 191 | c | multiply(divide(multiply(60, const_1000), const_3600), 18) | a train running at the speed of 60 km / hr crosses a pole in 18 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 18 = 300 m answer : c" | a = 60 * 1000
b = a / 3600
c = b * 18
|
a ) - 66 , b ) 13 , c ) - 15 , d ) - 31 , e ) 40 | c | multiply(subtract(const_2.0, const_4), 3) | find the value for x from below equation : x / 3 = - 5 ? | "1 . multiply both sides by 3 : x * 3 / 3 = - 5 / 3 simplify both sides : x = - 15 c" | a = 2 - 0
b = a * 3
|
a ) 256 , b ) 512 , c ) 768 , d ) 1024 , e ) 7168 | e | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.7, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.7), const_10) | if an average hard drive had a capacity of 0.7 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | 0.7 * 2 ^ 10 = 0.7 * 1024 = 716.8 the answer is e . | a = 0 * 7
b = a * 2
c = b * 2
d = c * 2
e = d * 2
f = e * 2
g = f * 2
h = g * 2
i = h * 2
j = i * 2
k = j + 0
l = k * 10
|
a ) 29.25 , b ) 29.28 , c ) 29.27 , d ) 29.22 , e ) 29.21 | a | add(multiply(8, 3), divide(add(multiply(9, 2), multiply(8, 3)), subtract(9, const_1))) | 9 men went to a hotel . 8 of them spent rs . 3 each over their meals and the ninth spent rs . 2 more than the average expenditure of all the 9 . determine the total money spent by them ? a . rs . 29.25 | average of 9 = x 9 x = 8 * 3 + x * 2 x = 3.25 total = 9 * 3.25 = 29.25 . answer : a | a = 8 * 3
b = 9 * 2
c = 8 * 3
d = b + c
e = 9 - 1
f = d / e
g = a + f
|
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 315 | e | multiply(multiply(10, 3), 7) | a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 c 7 * 2 c 10 = 7 * 45 = 315 the answer is ( e )" | a = 10 * 3
b = a * 7
|
a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 40 | b | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 6 and 34 which are divisible by 5 | "exp . multiples of 5 between 6 and 34 are 10 , 15,20 , 25,30 average = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 5 ( 10 + 30 ) / 2 * 5 = 40 / 2 = 20 answer : b" | a = 6 + 4
b = 34 - 4
c = a + b
d = c / 2
|
a ) 50 , b ) 99 , c ) 88 , d ) 76 , e ) 106 | e | divide(add(420, 640), multiply(subtract(72, 36), const_0_2778)) | how much time will a train of length 420 m moving at a speed of 72 kmph take to cross another train of length 640 m , moving at 36 kmph in the same direction ? | "the distance to be covered = sum of their lengths = 420 + 640 = 1060 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 1060 / 10 = 106 sec . answer : e" | a = 420 + 640
b = 72 - 36
c = b * const_0_2778
d = a / c
|
a ) 5 , b ) 15 , c ) 55 , d ) 95 , e ) 1 | e | divide(subtract(110, divide(110, add(divide(10, const_100), const_1))), 10) | machine p and machine q are each used to manufacture 110 sprockets . it takes machine p 10 hours longer to produce 110 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 10 / x = 110 / 1.1 x + 10 1.1 ( 110 ) = 110 + 11 x 11 x = 11 x = 1 the answer is e | a = 10 / 100
b = a + 1
c = 110 / b
d = 110 - c
e = d / 10
|
a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | b | divide(divide(subtract(const_100, 25), const_100), divide(50, const_100)) | when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 25 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 75 % ( 100 - 25 ) . therefore 0.5 x amount of flower - nectar = 0.75 x amount of honey = 0.75 x 1 kg therefore amount of flower - nectar needed = ( 0.75 / 0.51 ) kg = 1.5 kgs answer : b" | a = 100 - 25
b = a / 100
c = 50 / 100
d = b / c
|
a ) 44 % , b ) 40 % , c ) 50 % , d ) 56.25 % , e ) 37.5 % | e | multiply(subtract(divide(add(const_100, 30), subtract(const_100, 20)), const_1), const_100) | a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 30 % impurities to the product . find the net profit percentage of the dealer ? | "the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 30 % impurities to the product . it means that he is selling $ 5 worth of product for $ 5.5 . so his profit is $ 5.5 - $ 4 = $ 1.5 and his profit percent is ( 1.5 / 4 ) * 100 = 37.5 % answer : - e" | a = 100 + 30
b = 100 - 20
c = a / b
d = c - 1
e = d * 100
|
a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 126 | e | multiply(divide(144, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 144 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 144 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 144 = 16 / 30 x x = 270 270 - 144 = 126 answer e" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 144 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | c | subtract(95, 10) | 95 white and black tiles will be used to form a 10 x 10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | answer = c please refer diagram below 95 - 10 = 85 | a = 95 - 10
|
a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 2 sec | e | divide(20, multiply(36, const_0_2778)) | in what time will a railway train 20 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 60 / 36 * 18 / 5 = 2 sec answer : e" | a = 36 * const_0_2778
b = 20 / a
|
a ) 5 % , b ) 6 % , c ) 2 % , d ) 25 % , e ) 1 % | d | divide(multiply(const_100, 400), multiply(800, 2)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 400 in 2 years ? | interest for 2 yrs = 400 interest for 1 yr = 200 interest rate = 200 / 800 x 100 = 25 % answer : d | a = 100 * 400
b = 800 * 2
c = a / b
|
['a ) 4', 'b ) 27', 'c ) 12', 'd ) 16', 'e ) 18'] | b | divide(power(9, const_2), const_3) | if the area of a square with sides of length 9 centimeters is equal to the area of a rectangle with a width of 3 centimeters , what is the length of the rectangle , in centimeters ? | let length of rectangle = l 9 ^ 2 = l * 3 = > l = 81 / 3 = 27 answer b | a = 9 ** 2
b = a / 3
|
a ) 1 / 18 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 5 / 36 | e | divide(subtract(6, const_1), multiply(6, 6)) | taequan is playing a dice game where he rolls two , fair , 6 - sided dice twice . if he gets the same roll on his second try as his first , he wins . if he rolled an 8 on his first try , what is the probability that he will win with his second roll ? | there are 5 ways to roll an 8 : 2 and 6 , 6 and 2 , 3 and 5 , 5 and 3 , 4 and 4 ( 4 and 4 is counted just once ) . there are 6 * 6 = 36 ways to roll two six - sided dice . the dice are fair , so the probability of rolling an 8 is 5 / 36 . e | a = 6 - 1
b = 6 * 6
c = a / b
|
a ) 11 : 5 , b ) 11 : 8 , c ) 13 : 7 , d ) 25 : 24 , e ) 16 : 9 | d | divide(multiply(100, 5), multiply(80, 6)) | car a runs at the speed of 100 km / hr & reaches its destination in 5 hr . car b runs at the speed of 80 km / h & reaches its destination in 6 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 100 Γ 5 = 500 km distance travelled by car b = 80 Γ 6 = 480 km ratio = 500 / 480 = 25 : 24 answer : d" | a = 100 * 5
b = 80 * 6
c = a / b
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(subtract(subtract(100, multiply(100, divide(3, 7))), multiply(subtract(100, multiply(100, divide(3, 7))), divide(3, 7))), 3), subtract(subtract(subtract(100, multiply(100, divide(3, 7))), multiply(subtract(100, multiply(100, divide(3, 7))), divide(3, 7))), 3)) | george baked a total of 100 pizzas for 7 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ? | "3 / 5 of the 100 pizzas cooked on saturday = 60 pizzas 3 / 5 of the remaining pizzas on sunday = 24 pizzas we ' re left with ( 100 - 60 - 24 ) = 16 pizzas for the remaining 5 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w th f 6 4 3 2 1 = 16 w = 3 a" | a = 3 / 7
b = 100 * a
c = 100 - b
d = 3 / 7
e = 100 * d
f = 100 - e
g = 3 / 7
h = f * g
i = c - h
j = i - 3
k = 3 / 7
l = 100 * k
m = 100 - l
n = 3 / 7
o = 100 * n
p = 100 - o
q = 3 / 7
r = p * q
s = m - r
t = s - 3
u = j / t
|
a ) 7 hours , b ) 8 hours , c ) 9 hours , d ) 10 hours , e ) 11 hours | d | multiply(subtract(divide(const_1, 2), divide(const_1, add(2, divide(const_1, 2)))), const_100) | pipe a usually fills a tank in 2 hours . on account of a leak at the bottom of the tank , it takes pipe a 30 more minutes to fill the tank . how long will the leak take to empty a full tank if pipe a is shut ? | pipe a fills the tank normally in 2 hours . therefore , it will fill Β½ of the tank in an hour . let the leak take x hours to empty a full tank when pipe a is shut . therefore , the leak will empty of the tank in an hour . the net amount of water that gets filled in the tank in an hour when pipe a is open and when there is a leak = of the tank . β ( 1 ) when there is a leak , the problem states that pipe a takes two and a half hours to fill the tank . i . e . hours . therefore , in an hour , of the tank gets filled . β ( 2 ) equating ( 1 ) and ( 2 ) , we get = > = > x = 10 hours . the problem can also be mentally done as follows . pipe a takes 2 hours to fill the tank . therefore , it fills half the tank in an hour or 50 % of the tank in an hour . when there is a leak it takes 2 hours 30 minutes for the tank to fill . i . e hours to fill the tank or or 40 % of the tank gets filled . on account of the leak , ( 50 - 40 ) % = 10 % of the water gets wasted every hour . therefore , the leak will take 10 hours to drain a full tank . answer : d | a = 1 / 2
b = 1 / 2
c = 2 + b
d = 1 / c
e = a - d
f = e * 100
|
a ) 25 m , b ) 45 m , c ) 32 m , d ) 50 m , e ) 62 m | a | multiply(5, 5) | walking at 5 / 6 th of its usual speed a cab is 5 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 5 m ut / 5 = 5 m ut = 25 m answer is a" | a = 5 * 5
|
a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | b | divide(1, 37) | what is the 8 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 2 nd , 4 th , 6 th , . . . ) to the right of the decimal point is alternating by 2 , 0,7 , thus 8 th digit to the right of the decimal point is 2 . answer : b ." | a = 1 / 37
|
a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(add(6, 3), power(3, 2)) | find the value of x : 3 x ^ 2 - 6 x + 3 = 0 | a = 3 , b = - 6 , c = 3 x 1,2 = ( 6 Β± β ( ( - 6 ) 2 - 4 Γ 3 Γ 3 ) ) / ( 2 Γ 3 ) = ( 6 Β± β ( 36 - 36 ) ) / 6 = ( 6 Β± 0 ) / 6 x 1 = x 2 = 1 b | a = 6 + 3
b = 3 ** 2
c = a / b
|
a ) 199 sec , b ) 190 sec , c ) 726 sec , d ) 90 sec , e ) 128 sec | d | subtract(divide(multiply(const_1, const_1000), divide(50, 5)), 5) | in a kilometer race , a beats b by 50 meters or 5 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 5 ) / 50 = 100 sec . time taken by a = 100 - 10 = 90 sec . answer : d" | a = 1 * 1000
b = 50 / 5
c = a / b
d = c - 5
|
a ) 4722 , b ) 4725 , c ) 4728 , d ) 4731 , e ) 4733 | a | add(lcm(lcm(27, 35), lcm(25, 21)), 3) | what is the smallest number which when increased by 3 is divisible by 27 , 35 , 25 , & 21 ? | "here smallest number means lcm hence lcm of 27,35 , 25,21 = 3 ^ 3 x 5 ^ 2 x 7 = 4725 therefore required number is 4725 - 3 = 4722 ans - a" | a = math.lcm(27, 35)
b = math.lcm(25, 21)
c = math.lcm(a, b)
d = c + 3
|
a ) 50 , b ) 60 , c ) 55 , d ) 45 , e ) 53 | a | subtract(multiply(20, multiply(130, const_0_2778)), 150) | a train 150 m long crosses a platform 130 m long in 20 sec ; find the speed of the train ? | "d = 150 + 130 = 280 t = 20 s = 280 / 20 * 18 / 5 = 50 kmph . answer : a" | a = 130 * const_0_2778
b = 20 * a
c = b - 150
|
a ) 27 , b ) 48 , c ) 80 , d ) 72 , e ) 18 | c | subtract(multiply(multiply(5, 5), divide(245, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(245, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 245 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 49 x = 245 x = 5 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 5 = 80 answer c" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 245 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 245 / n
p = i * o
q = h - p
|
a ) 145 , b ) 185 , c ) 245 , d ) 135 , e ) 140 | d | add(multiply(multiply(5, 8), const_10), 6) | when 6 + 2 = 85 , 8 + 3 = 115 , 7 + 5 = 125 , then 5 + 8 = ? | "6 + 2 = > 6 + 2 = 8 = > 8 Γ£ β 10 + 5 = 85 8 + 3 = > 8 + 3 = 11 = > 11 Γ£ β 10 + 5 = 115 7 + 5 = > 7 + 5 = 12 = > 12 Γ£ β 10 + 5 = 125 then 5 + 8 = > 5 + 8 = 13 = > 13 Γ£ β 10 + 5 = 135 answer : d" | a = 5 * 8
b = a * 10
c = b + 6
|
a ) 13 , b ) 17 , c ) 18 , d ) 8 , e ) 12 | d | divide(multiply(8, 4), subtract(8, 4)) | pipe a can fill a tank in 4 hours . due to a leak at the bottom , it takes 8 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 4 - 1 / x = 1 / 8 = > 1 / x = 1 / 4 - 1 / 8 = ( 2 - 1 ) / 8 = 1 / 8 = > x = 8 . answer : d" | a = 8 * 4
b = 8 - 4
c = a / b
|
a ) 50803200 , b ) 50328000 , c ) 50003280 , d ) 50703200 , e ) 50603200 | a | multiply(factorial(7), factorial(7)) | in how many ways 7 boys and 7 girls can be seated in a row so that boys and girls are alternate ? | "number of ways = 2 * 7 ! * 7 ! = 50803200 answer is a" | a = math.factorial(7)
b = math.factorial(7)
c = a * b
|
a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 20 | e | subtract(23, const_3) | how many two - element subsets of < 1 , 23 , 45 , 67 > are there that do not contain the pair of elements 2 and 4 ? | 7 c 2 - 1 = 20 answer is e | a = 23 - 3
|
a ) 1 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | a certain junior class has 3000 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let ' s see pick 60 / 3000 first then we can only pick 1 other pair from the 800 so total will be 60 / 800 * 3000 simplify and you get 1 / 40000 answer is a" | a = 1 / 3
|
a ) 10 days , b ) 5 days , c ) 15 days , d ) 12 days , e ) 10 days | a | multiply(subtract(20, 15), const_2) | there is a sufficient food for 200 men for 20 days after 15 days 100 men leave the place . for how many days will the rest of the food last for the rest of the men ? | explanation : after 13 days , 100 men left , therefore 5 days * 200 men = x days * 100 men x = 10 days answer : option a | a = 20 - 15
b = a * 2
|
a ) 90 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | multiply(divide(divide(add(divide(120, const_2), 120), 30), const_4), divide(120, const_2)) | a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 30 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | "a - b = 120 miles b - c = 60 miles avg speed = 30 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 30 = 180 / 4 t t = 90 b - c = 90 mph answer a" | a = 120 / 2
b = a + 120
c = b / 30
d = c / 4
e = 120 / 2
f = d * e
|
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