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 |
|---|---|---|---|---|---|
a ) 388,620 , b ) 388,260 , c ) 362,880 , d ) 352,880 , e ) 342,880 | c | factorial(const_3) | how many words , with or without meaning , can be formed using all letters of the word beautiful using each letter exactly once ? | "the word beautiful has exactly 9 letters which are all different . therefore the number of words that can be formed = number of permutations of 9 letters taken all at a time . = p ( 9 , 9 ) = 9 ! = 9 x 8 × 7 x 6 × 5 x 4 x 3 x 2 × 1 = 362,880 answer : c" | a = math.factorial(3)
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a ) 7.38 , b ) 5.38 , c ) 4.38 , d ) 6.38 , e ) 3.38 | d | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 940), 940), const_100) | a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 940 grams . find the profit percent . | explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 940 100 + g / 100 + 0 = 1000 / 940 100 + g = 100 / 94 * 100 g = 6.38 answer : d | a = 4 + 1
b = 4 + 1
c = a + b
d = c * 100
e = d - 940
f = e / 940
g = f * 100
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a ) 14 , b ) 9 , c ) 12 , d ) 15 , e ) 17 | a | add(4, const_1) | the average of first six multiples of 4 is : | "solution average = 4 ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 84 / 6 = 14 answer a" | a = 4 + 1
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a ) 24 km , b ) 30 km , c ) 96 km , d ) 12 km , e ) 15 km | c | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 20), add(subtract(10, 2), add(10, 2))) | a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 20 hour to row to a place and come back , how far is the place ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 20 2 x + 3 x = 480 x = 96 km answer is c" | a = 10 - 2
b = 10 + 2
c = a * b
d = c * 20
e = 10 - 2
f = 10 + 2
g = e + f
h = d / g
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a ) 9 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 25 % | e | divide(multiply(20, const_100), subtract(const_100, 20)) | during a sale , the price of a pair of shoes is marked down 20 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 80 price after sale = 100 percent increase = 20 / 80 * 100 = 25 % approx . correct option : e" | a = 20 * 100
b = 100 - 20
c = a / b
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a ) 2.5 sec , b ) 8.0 sec , c ) 3.5 sec , d ) 2.9 sec , e ) 9.5 sec | b | divide(320, multiply(144, const_0_2778)) | in what time will a train 320 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 320 / 40 = 8.0 sec . answer : b" | a = 144 * const_0_2778
b = 320 / a
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a ) 60 , b ) 90 , c ) 120 , d ) 172 , e ) 188 | a | subtract(200, subtract(add(multiply(divide(200, const_100), 56), multiply(divide(200, const_100), 44)), multiply(divide(200, const_100), 30))) | among 200 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 60 using your analysis : we see that the total number of students who study either maths or sociology = 88 + 112 - 60 = 140 so , in the image we know that the number of students in the zone with the black boundary = 140 let ' s assume the number of students who studyonlybiology to beb ( this is the number that we have to maximize ) and , let ' s assume the number of students who study none of the three subjects , that is the number of students in the white space = w since the total number of students = 200 , we can write : 140 + b + w = 200 or , b + w = 200 - 140 = 60 that is , b = 60 - w so , the maximum value ofbwill happen forw = 0 this is how we get , the maximum value ofb = 60 a" | a = 200 / 100
b = a * 56
c = 200 / 100
d = c * 44
e = b + d
f = 200 / 100
g = f * 30
h = e - g
i = 200 - h
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a ) 24 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 15 km | a | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 5), add(subtract(10, 2), add(10, 2))) | a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 5 hour to row to a place and come back , how far is the place ? | speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 5 2 x + 3 x = 120 x = 24 km answer is a | a = 10 - 2
b = 10 + 2
c = a * b
d = c * 5
e = 10 - 2
f = 10 + 2
g = e + f
h = d / g
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a ) s . 2400 , b ) s . 2016 , c ) s . 1400 , d ) s . 3480 , e ) s . 2000 | b | multiply(multiply(84, 24), divide(70, 10)) | what is the cost of leveling the field in the form of parallelogram at the rate of rs . 70 / 10 sq . metre , whose base & perpendicular distance from the other side being 84 m & 24 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 84 * 24 = 2016 m . total cost of levelling = rs . 2016 b" | a = 84 * 24
b = 70 / 10
c = a * b
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a ) a / 14 , b ) 5 a / 7 , c ) 7 a / 5 , d ) 14 a , e ) 35 a | d | floor(divide(70, add(5, const_1))) | during a certain two - week period , 70 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 5 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | "total movies = 100 . comedies = 70 . action + drama = 30 . since there were 5 times as many dramas as action movies , then action + 5 * action = 30 - - > action = a = 5 . comedies = 70 = 14 a . answer : d ." | a = 5 + 1
b = 70 / a
c = math.floor(b)
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a ) $ 40,000 , b ) $ 56,000 , c ) $ 64,000 , d ) $ 58,000 , e ) $ 80,000 | d | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(11, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | country x taxes each of its citizens an amount equal to 11 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ? | "equation is correct , so math must be a problem . 0.11 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 4,400 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 11,600 - - > x = 58,000 . answer : d ." | a = 4 * 10
b = a * 1000
c = 4 * 2
d = c * 1000
e = 11 / 100
f = 4 * 10
g = f * 1000
h = e * g
i = d - h
j = 20 / 100
k = i / j
l = b + k
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 12 | d | subtract(subtract(subtract(14, 4), const_1), const_1) | list k consists of 14 consecutive integers . if - 4 is the least integer in list k , what is the range of the positive integers in list k ? | answer = d = 8 if least = - 4 , then largest = 9 range = 9 - 1 = 8 | a = 14 - 4
b = a - 1
c = b - 1
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a ) 1 km , b ) 2 km , c ) 3 km , d ) 4 km , e ) 2.55 km | e | multiply(multiply(divide(divide(40, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 40 minutes . the total distance is ? | "e 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 40 / 60 47 x / 60 = 40 / 60 = > x = 0.85 . total distance = 3 * 0.85 = 2.55 km ." | a = 40 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = g * 3
i = h * 1000
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a ) 30 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(20, 65), multiply(65, 20)), add(20, 65)) | the average of 20 results is 65 and the average of other 65 results is 20 . what is the average of all the results ? | "answer sum of 85 result = sum of 20 result + sum of 65 result . = 20 x 65 + 65 x 20 = 2600 correct option : a" | a = 20 * 65
b = 65 * 20
c = a + b
d = 20 + 65
e = c / d
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a ) 46 , b ) 96 , c ) 35 , d ) 87 , e ) 13 | b | multiply(multiply(divide(1620, 135), divide(8, const_100)), const_100) | by investing rs . 1620 in 8 % stock , michael earns rs . 135 . the stock is then quoted at | "to earn rs . 135 , investment = rs . 1620 . to earn rs . 8 , investment = rs . 96 . market value of rs . 100 stock = rs . 96 . answer : b" | a = 1620 / 135
b = 8 / 100
c = a * b
d = c * 100
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a ) 430 , b ) 438 , c ) 436 , d ) 480 , e ) 422 | d | divide(192, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 192 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 192 = > 4 x / 10 = 192 = > 4 x = 1920 = > x = 480 . answer : d" | a = 100 - 70
b = 70 - a
c = b / 100
d = 192 / c
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a ) 45 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 15 | a | subtract(50, multiply(divide(50, const_100), 20)) | how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 20 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 20 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 20 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 20 x = 30 since you need a 15 l solution , you must evaporate 45 of the original 50 l solution to get a 20 % solution . answer is a ." | a = 50 / 100
b = a * 20
c = 50 - b
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a ) 1,000 , b ) 500 , c ) 250 , d ) 50 , e ) 0 | a | divide(divide(multiply(subtract(multiply(10, const_2), 10), const_1000), 10), const_1000) | a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 10 minutes , a combustion reaction creates 10,000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | solution : 10,000 carbon dioxide molecules are created over a period of 10 minutes . therefore 10,000 / 10 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 × 1,000 = 1,000 carbon atoms and 2 × 1,000 = 2,000 oxygen atoms . the difference is 2,000 – 1,000 = 1,000 . the correct answer is a . | a = 10 * 2
b = a - 10
c = b * 1000
d = c / 10
e = d / 1000
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a ) 15 , b ) 18 , c ) 13 , d ) 19 , e ) 20 | e | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 6 and 34 which is divisible by 5 | "avg = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 100 / 5 = > 20 answer e" | a = 6 + 4
b = 34 - 4
c = a + b
d = c / 2
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a ) 369.63 , b ) 372.33 , c ) 702.33 , d ) 702 , e ) none of them | a | add(add(add(33, 333), 3.33), divide(const_3, const_10)) | . 3 + 33 + 333 + 3.33 = ? | . 3 33 333 3.33 - - - - - - - - - - 369.63 - - - - - - - - - - answer is a | a = 33 + 333
b = a + 3
c = 3 / 10
d = b + c
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a ) 2 , b ) 1 , c ) 3 , d ) 4 , e ) 2.1 | a | divide(subtract(divide(45, 5), divide(25, 5)), const_2) | an woman swims downstream 45 km and upstream 25 km taking 5 hours each time ; what is the speed of the current ? | "45 - - - 5 ds = 9 ? - - - - 1 25 - - - - 5 us = 5 ? - - - - 1 s = ? s = ( 9 - 5 ) / 2 = 2 answer : a" | a = 45 / 5
b = 25 / 5
c = a - b
d = c / 2
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a ) 25 , b ) 28 , c ) 30 , d ) 35 , e ) 41 | e | subtract(multiply(add(6, 6), const_4), add(6, const_1)) | find a sum for first 6 prime number ' ss ? | required sum = ( 2 + 3 + 5 + 7 + 11 + 13 ) = 41 note : 1 is not a prime number option e | a = 6 + 6
b = a * 4
c = 6 + 1
d = b - c
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a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | e | divide(divide(8400, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 8400 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 8400 / ( 28 * 24 * 2 ) = 6.25 answer e" | a = 8400 / 28
b = 28 - 4
c = b * 2
d = a / c
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a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 15 | b | subtract(3,381, add(multiply(multiply(multiply(9, 11), 17), const_2), 8)) | what least number should be subtracted from 3,381 so that the remainder when divided by 9 , 11 , and 17 will leave in each case the same remainder 8 ? | "the lcm of 9 , 11 and 17 is 1,683 . the next multiple is 2 * 1,683 = 3,366 . 3,366 + { remainder } = 3,366 + 8 = 3,374 , which is 7 less than 3,381 . answer : b ." | a = 9 * 11
b = a * 17
c = b * 2
d = c + 8
e = 3 - 381
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a ) 10 % , b ) 7.8 % , c ) 11 % , d ) 12.5 % , e ) none | b | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), const_1), add(const_100, 40))) | a trader marked the selling price of an article at 40 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | sol . let c . p . = rs . 100 . then , marked price = rs . 140 , s . p . = rs . 99 . ∴ discount % = [ 11 / 140 * 100 ] % = 7.8 % answer b | a = 2 + 3
b = a * 2
c = b + 1
d = 100 + 40
e = c / d
f = 100 * e
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a ) 7 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | a | divide(divide(7, 16), divide(1, 16)) | diana is painting statues . she has 7 / 16 of a gallon of paint remaining . each statue requires 1 / 16 gallon of paint . how many statues can she paint ? | "number of statues = all the paint ÷ amount used per statue = 7 / 16 ÷ 1 / 16 = 7 / 16 * 16 / 1 = 7 / 1 = 7 answer is a ." | a = 7 / 16
b = 1 / 16
c = a / b
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a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6500 | e | divide(1300, divide(subtract(60, subtract(const_100, 60)), const_100)) | in an election a candidate who gets 60 % of the votes is elected by a majority of 1300 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1300 20 x / 100 = 1300 x = 1300 * 100 / 20 = 6500 answer is e" | a = 100 - 60
b = 60 - a
c = b / 100
d = 1300 / c
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a ) 10 kg , b ) 15 kg , c ) 20 kg , d ) 25 kg , e ) 40 kg | b | multiply(divide(divide(multiply(subtract(const_100, 60), 30), const_100), subtract(const_100, 20)), const_100) | fresh grapes contain 60 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 30 kg of fresh grapes ? | "from the question we know : 30 kg * 60 % = 18 kg of water in the fresh grapes 30 kg - 18 kg of water = 12 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 20 % of the weight of the dry graps is water and we know that 12 kg is non - water mass we can set up the following equation : x = 1 / 5 ( x ) + 12 kg 4 / 5 ( x ) = 12 kg x = 15 kg answer - b" | a = 100 - 60
b = a * 30
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
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a ) 15 , b ) 18 , c ) 10 , d ) 16 , e ) none of these | a | subtract(const_60, multiply(divide(60, 80), const_60)) | without stoppages , a train travels certain distance with an average speed of 80 km / h , and with stoppages , it covers the same distance with an average speed of 60 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 20 km less . time taken to cover 20 km = 20 ⁄ 80 h = 1 ⁄ 4 h = 1 ⁄ 4 × 60 min = 15 min answer a" | a = 60 / 80
b = a * const_60
c = const_60 - b
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a ) 3 , b ) 1 , c ) 1 / 3 , d ) - 1 / 3 , e ) - 3 | b | divide(1, subtract(5, 4)) | if 5 / ( 4 + 1 / x ) = 1 , then x = | "the expression 5 / ( 4 + 1 / x ) = 1 should have been equal to something . if 5 / ( 4 + 1 / x ) = 1 = > 5 x / ( 4 x + 1 ) = 1 = > 5 x = 4 x + 1 = > x = 1 correct option : b" | a = 5 - 4
b = 1 / a
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a ) 44 , b ) 48 , c ) 45 , d ) 43 , e ) 50 | a | subtract(multiply(add(22, 1), add(21, 1)), multiply(22, 21)) | the average age of a class of 22 students is 21 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "total age = 22 * 21 tot age of all stu + age of the teacher = 23 * 22 age of the teacher = 23 * 22 - 22 * 21 = 44 answer a" | a = 22 + 1
b = 21 + 1
c = a * b
d = 22 * 21
e = c - d
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a ) 25 % , b ) 26 % , c ) 27 % , d ) 28 % , e ) 29 % | a | multiply(divide(22, 88), const_100) | by selling 88 pens , a trader gains the cost of 22 pens . find his gain percentage ? | let the cp of each pen be rs . 1 . cp of 88 pens = rs . 88 profit = cost of 22 pens = rs . 22 profit % = 22 / 88 * 100 = 25 % answer : a | a = 22 / 88
b = a * 100
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a ) 512 , b ) 1024 , c ) 768 , d ) 2048 , e ) 4096 | b | subtract(power(2, add(9, const_1)), const_1) | the population of a bacteria colony doubles every day . if it was started 9 days ago with 2 bacteria and each bacteria lives for 13 days , how large is the colony today ? | "9 days ago - 2 8 days ago - 4 7 days ago - 8 6 days ago - 16 5 days ago - 32 4 days ago - 64 3 days ago - 128 2 days ago - 256 yesterday - 512 today - 1024 ans : b" | a = 9 + 1
b = 2 ** a
c = b - 1
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a ) 5 , b ) 7 , c ) 10 , d ) 15 , e ) 24 | b | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 35), divide(9, 55))), const_4) | due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "9 / 35 - 9 / 55 = 9 / 5 * ( 11 - 7 ) / 77 = 9 / 5 * 4 / 77 * 60 min = 9 * 12 * 4 / 77 = 432 / 77 ~ 5.6 answer - b" | a = 55 + 9
b = a - 1
c = 9 / 35
d = 9 / 55
e = c - d
f = b * e
g = max(f)
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a ) 0 , b ) 1 , c ) 36 , d ) 118 , e ) 420 | b | subtract(power(6, subtract(const_1, const_1)), power(subtract(const_1, const_1), 6)) | if k is a non - negative integer and 18 ^ k is a divisor of 624,938 then 6 ^ k - k ^ 6 = | "6 + 2 + 4 + 9 + 3 + 8 = 32 , so this number is not divisible by 3 and thus not divisible by 18 . therefore , k = 0 6 ^ k - k ^ 6 = 1 - 0 = 1 the answer is b ." | a = 1 - 1
b = 6 ** a
c = 1 - 1
d = c ** 6
e = b - d
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a ) 7 : 18 , b ) 3 : 7 , c ) 1 : 2 , d ) 2 : 5 , e ) 4 : 9 | c | divide(add(1, 2), add(3, 3)) | in the first m games of a team ' s season , the ratio of the team ' s wins to its losses was 1 : 3 . in the subsequent n games , the ratio of the team ´ s wins to losses was 2 : 3 . if m : n = 4 : 5 , what was the ratio of the team ' s wins to its losses for all m + n games ? | m = 4 / 9 of total games n = 5 / 9 of total games wins = 1 / 4 * 4 / 9 + 2 / 5 * 5 / 9 = 1 / 9 + 2 / 9 = 1 / 3 losses = 1 - 1 / 3 = 2 / 3 the ratio of wins to losses is 1 : 2 . the answer is c . | a = 1 + 2
b = 3 + 3
c = a / b
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a ) 540 km , b ) 495 km , c ) 276 km , d ) 178 km , e ) 176 km | b | multiply(add(20, 25), divide(55, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 55 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 55 km more than the other , 25 x â € “ 20 x = 55 5 x = 55 x = 11 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 11 = 495 km . answer : b" | a = 20 + 25
b = 25 - 20
c = 55 / b
d = a * c
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a ) 1 : 8 , b ) 1 : 6 , c ) 1 : 2 , d ) 1 : 1 , e ) 1 : 7 | d | divide(divide(multiply(5, 3), multiply(6, 2)), divide(multiply(3, 4), multiply(2, 5))) | the compound ratio of 5 : 6 , 3 : 2 and 4 : 5 ? | "5 / 6 * 3 / 2 * 4 / 5 = 1 / 1 1 : 1 answer : d" | a = 5 * 3
b = 6 * 2
c = a / b
d = 3 * 4
e = 2 * 5
f = d / e
g = c / f
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a ) 74 , b ) 146 , c ) 86 , d ) 92 , e ) 98 | b | add(multiply(add(multiply(6, const_3), 17), divide(add(multiply(6, const_3), 17), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 17 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 17 = 35 5 * quotient = 35 quotient = 7 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 7 ) + 6 = 146 . b )" | a = 6 * 3
b = a + 17
c = 6 * 3
d = c + 17
e = d / 5
f = b * e
g = f + 6
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a ) 643 , b ) 652 , c ) 454 , d ) 460 , e ) 490 | c | multiply(divide(495, add(const_1, divide(36, const_100))), add(const_1, divide(25, const_100))) | if albert ’ s monthly earnings rise by 36 % , he would earn $ 495 . if , instead , his earnings rise by only 25 % , how much ( in $ ) would he earn this month ? | "= 495 / 1.36 ∗ 1.25 = 454 = 454 answer is c" | a = 36 / 100
b = 1 + a
c = 495 / b
d = 25 / 100
e = 1 + d
f = c * e
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a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | d | add(10, 9) | lilly has 10 fish and rosy has 9 fish . in total , how many fish do they have in all ? | "10 + 9 = 19 the answer is d ." | a = 10 + 9
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a ) . 50 , b ) . 45 , c ) . 40 , d ) . 48 , e ) . 52 | c | divide(subtract(add(add(1000, divide(1000, divide(const_100, 30))), 100), 1000), 1000) | after giving a discount of rs . 100 the shopkeeper still gets a profit of 30 % , if the cost price is rs . 1000 . find the markup % ? | "cost price = 1000 s . p = 1000 * 130 / 100 = 1300 disc = 100 so . . . mark price = 1300 + 100 = 1400 . . . . . . mark up % = 1400 - 1000 / 1000 = 400 / 1000 = . 40 ( or ) 40 % answer : c" | a = 100 / 30
b = 1000 / a
c = 1000 + b
d = c + 100
e = d - 1000
f = e / 1000
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a ) 1025 , b ) 1075 , c ) 1125 , d ) 1175 , e ) 1225 | c | divide(divide(divide(120, subtract(const_1, divide(3, 5))), divide(3, 5)), divide(1, 3)) | of the goose eggs laid at a certain pond , 1 / 3 hatched and 4 / 5 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 120 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "let x be the number of eggs that were laid . ( 2 / 5 ) ( 4 / 5 ) ( 1 / 3 ) x = 120 ( 8 / 75 ) x = 120 x = 1125 the answer is c ." | a = 3 / 5
b = 1 - a
c = 120 / b
d = 3 / 5
e = c / d
f = 1 / 3
g = e / f
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a ) 40 , b ) 30 , c ) 42 , d ) 33 , e ) 35 | b | divide(add(multiply(24, const_10), divide(multiply(24, const_10), const_2)), add(divide(multiply(24, const_10), 20), divide(divide(multiply(24, const_10), const_2), 24))) | a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 20 miles per gallon , and from town b to town c , the car averaged 24 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ? | "step 1 ) took lcm of 20 and 24 . . came as 120 . step 2 ) 120 distance between b to c . . . do 120 / 24 hence 5 gallons used step 3 ) twice distance . . hence 120 * 2 = 240 . . . do as above . . 240 / 20 = 7 gallons used step 4 ) total gallons . . 5 + 7 = 12 gallons step ) total miles = 120 + 240 = 360 miles hence . . average of whole journey = 360 / 12 which comes to 30 answer : b" | a = 24 * 10
b = 24 * 10
c = b / 2
d = a + c
e = 24 * 10
f = e / 20
g = 24 * 10
h = g / 2
i = h / 24
j = f + i
k = d / j
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a ) 3 , b ) 13 , c ) 14 , d ) 17 , e ) 20 | b | divide(subtract(multiply(add(10, 23), divide(50, const_100)), 10), divide(50, const_100)) | a bowl of fruit contains 10 apples and 23 oranges . how many oranges must be removed so that 50 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 10 number of oranges = 23 let number of oranges that must be removed so that 50 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 10 + ( 23 - x ) = 33 - x 10 / ( 33 - x ) = 5 / 10 = > 20 = 33 - x = > x = 13 answer b" | a = 10 + 23
b = 50 / 100
c = a * b
d = c - 10
e = 50 / 100
f = d / e
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | add(add(3, 3), 5) | given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 5 ? | "p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit of p + 5 is 6 + 5 = 11 . answer : e ." | a = 3 + 3
b = a + 5
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a ) 2 a , b ) a / 2 , c ) 6 a , d ) 7 a , e ) 9 a | d | add(4, 3) | if a / x = 1 / 3 and a / y = 1 / 4 , then ( x + y ) = | ratio 1 : 3 a = x ratio 2 : 4 a = y x + y = 7 a answer is d | a = 4 + 3
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a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 3 | e | subtract(divide(subtract(90, 70), subtract(75, 70)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 70 units . if today ' s production of 90 units raises the average to 75 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 70 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 70 n + 90 = 75 * ( n + 1 ) - - > n = 3 . answer : e ." | a = 90 - 70
b = 75 - 70
c = a / b
d = c - 1
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a ) 3 , b ) 9 , c ) 7 , d ) 2 , e ) 6 | b | inverse(divide(inverse(6), add(const_2, const_1))) | a is twice as good a work man as b and together they finish the work in 6 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 6 = > x = 1 / 18 2 x = 1 / 9 a can do the work in 9 days . answer : b" | a = 1/(6)
b = 2 + 1
c = a / b
d = 1/(c)
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a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 9 : 4 , e ) 45 : 8 | e | divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2)) | a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is five times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that time , what is the ratio of water to salt in the solution ? | "water : soap = 3 : 2 soap : salt = 15 : 2 = > for 15 soap , salt = 2 = > for 2 soap , salt = ( 2 / 15 ) * 2 = 4 / 15 so , water : soap : salt = 3 : 2 : 4 / 15 = 45 : 30 : 4 after open container , water : soap : salt = 22.5 : 30 : 4 so , water : salt = 22.5 : 4 = 45 : 8 e" | a = 2 * 3
b = a * 3
c = 2 * 2
d = c * 2
e = b / d
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a ) 2 , b ) 12 , c ) 3 , d ) 8 , e ) 4 | b | multiply(4, power(9, divide(1, 2))) | if a @ b = a * b ^ ( 1 / 2 ) then 4 @ 9 = ? self made | 4 * ( 9 ) ^ 1 / 2 = 4 * 3 = 12 b is the answer | a = 1 / 2
b = 9 ** a
c = 4 * b
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a ) 976374 , b ) 979923 , c ) 980241 , d ) 1178125 , e ) 1083875 | d | multiply(1000000, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1000000 . if in 2005 there isan increment of 25 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 25 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1178125 d" | a = 25 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1000000 * h
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a ) $ 600 , b ) $ 1000 , c ) $ 1200 , d ) $ 1500 , e ) $ 2000 | c | subtract(subtract(multiply(power(add(divide(divide(2, const_2), const_100), const_1), const_2), 20), 20), divide(multiply(20, 2), const_100)) | the difference between simple interest and compound interest of a certain sum of money at 20 % per annum for 2 years is $ 48 . then the sum is | "20 % of ( year 1 interest ) = 20 % of ( 20 % of principal ) = 48 principal = $ 1200 answer ( c )" | a = 2 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 20
f = e - 20
g = 20 * 2
h = g / 100
i = f - h
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a ) 640 , b ) 500 , c ) 3000 , d ) 1000 , e ) 2000 | a | divide(144, divide(subtract(multiply(15, 5), multiply(15, 3.5)), const_100)) | the equal amounts of money are deposited in two banks each at 15 % per annum for 3.5 years and 5 years respectively . if the difference between their interests is rs . 144 , find the each sum ? | ( p * 5 * 15 ) / 100 - ( p * 3.5 * 15 ) / 100 = 144 75 p / 100 â € “ 52.5 p / 100 = 144 22.5 p = 144 * 100 = > p = rs . 640 answer a | a = 15 * 5
b = 15 * 3
c = a - b
d = c / 100
e = 144 / d
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a ) 20 , b ) 30 , c ) 40 , d ) 60 , e ) 90 | a | divide(add(multiply(10, 9), multiply(10, 7)), add(9, 7)) | the number of stamps that p and q had were in the ratio of 5 : 3 respectively . after p gave q 10 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 9 : 7 . as a result of the gift , p had how many more stamps than q ? | "p started with 5 k stamps and q started with 3 k stamps . ( 5 k - 10 ) / ( 3 k + 10 ) = 9 / 7 35 k - 27 k = 160 k = 20 p has 5 ( 20 ) - 10 = 90 stamps and q has 3 ( 20 ) + 10 = 70 stamps . the answer is a ." | a = 10 * 9
b = 10 * 7
c = a + b
d = 9 + 7
e = c / d
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a ) 21 , b ) 10 , c ) 61 , d ) 72 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 31,360 multiplied by y is the square of an integer , then y must be | "31360 = 8 * 8 * 7 * 7 * 5 * 2 , so we need one 5 and one 2 to make it a square of a number . so 5 * 2 = 10 ans : b" | a = 2 + 3
b = a * 2
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a ) 35 % , b ) 37 % , c ) 41.3 % , d ) 42 % , e ) 45 % | c | multiply(const_100, divide(add(multiply(8, divide(30, const_100)), multiply(3, divide(30, const_100))), add(add(multiply(8, divide(30, const_100)), multiply(3, divide(30, const_100))), add(subtract(multiply(8, divide(70, const_100)), 3), multiply(3, divide(70, const_100)))))) | solution y is 30 percent liquid x and 70 percent water . if 3 kilograms of water evaporate from 8 kilograms of solution y and 3 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ? | in 8 kilograms of solution y there are 0.3 * 8 = 2.4 kilograms of solution x ; after 3 kilograms of water are replaced by 3 kilograms of solution y , to the existing 2.4 kilograms of solution x , 0.3 * 3 = 0.9 kilograms of solution x are added , so in the new solution of 8 kilograms there are 2.4 + 0.9 = 3.3 kilograms of solution x , which is 3.3 / 8 * 100 = 41.3 % of this new solution . answer : c . | a = 30 / 100
b = 8 * a
c = 30 / 100
d = 3 * c
e = b + d
f = 30 / 100
g = 8 * f
h = 30 / 100
i = 3 * h
j = g + i
k = 70 / 100
l = 8 * k
m = l - 3
n = 70 / 100
o = 3 * n
p = m + o
q = j + p
r = e / q
s = 100 * r
|
a ) 22 kmph , b ) 108 kmph , c ) 54 kmph , d ) 71 kmph , e ) 88 kmph | b | multiply(const_3_6, divide(480, 16)) | a train 480 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 480 / 16 * 18 / 5 = 108 kmph answer : b" | a = 480 / 16
b = const_3_6 * a
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a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 5 hours , e ) none | e | divide(80, add(13, 4)) | a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr . find the time taken by the boat to go 80 km downstream ? | "solution speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 80 km downstream = ( 80 / 17 ) hrs = 4 hrs 42 minutes . answer e" | a = 13 + 4
b = 80 / a
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a ) 187 , b ) 279 , c ) 280 , d ) 285 , e ) 262 | d | divide(add(multiply(add(const_4, const_1), 510), multiply(multiply(add(const_4, const_1), add(const_4, const_1)), 240)), 30) | a library has an average of 510 visitors on sunday and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begins with a sunday , so there will be five sundays in the month . required average = [ ( 510 * 5 ) + ( 240 * 25 ) ] / 30 = 8550 / 30 = 285 . answer : d" | a = 4 + 1
b = a * 510
c = 4 + 1
d = 4 + 1
e = c * d
f = e * 240
g = b + f
h = g / 30
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a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | d | add(divide(add(35, 7), const_2), 7) | the sum of present age of abe and the age before 7 years is 35 . find the present age of abe . what will be his age after 7 years ? | present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 35 2 x - 7 = 35 2 x = 35 + 7 x = 42 / 2 x = 21 z = x + 7 = 21 + 7 = 28 answer : d | a = 35 + 7
b = a / 2
c = b + 7
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | divide(multiply(6, 6), const_2) | the two lines y = x and x = - 6 intersect on the coordinate plane . what is the value of the area of the figure formed by the intersecting lines and the x - axis ? | "the point of intersection is ( - 6 , - 6 ) . the triangle has a base of length 6 and a height of 6 . area = ( 1 / 2 ) * base * height = ( 1 / 2 ) * 6 * 6 = 18 the answer is d ." | a = 6 * 6
b = a / 2
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a ) 287 , b ) 600 , c ) 289 , d ) 300 , e ) 207 | d | multiply(multiply(36, const_0_2778), 30) | what distance will be covered by a bus moving at 36 kmph in 30 seconds ? | "36 kmph = 36 * 5 / 18 = 10 mps d = speed * time = 10 * 30 = 300 m . answer : d" | a = 36 * const_0_2778
b = a * 30
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a ) 2 : 3 , b ) 3 : 4 , c ) 3 : 10 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.75, const_100), multiply(2.5, const_100)) | if 2.5 of a number is equal to 0.75 of another number , the ratio of the numbers is : | "2.5 a = 0.75 b - > a / b = 0.75 / 2.5 = 75 / 250 = 3 / 10 : . a : b = 3 : 10 answer : c" | a = 0 * 75
b = 2 * 5
c = a / b
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a ) 22 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 13 kg , e ) none of these | d | subtract(multiply(add(29, const_1), 27.5), multiply(29, 28)) | the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.5 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.5 weight of the new student = ( 30 * 27.5 – 29 * 28 ) = 825 - 812 = 13 answer : d" | a = 29 + 1
b = a * 27
c = 29 * 28
d = b - c
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a ) 15 / 2 , b ) 6 / 5 , c ) 5 / 6 , d ) 3 / 10 , e ) 2 / 1 | e | add(divide(4, 4), divide(3, 4)) | if 4 tic equals 3 tacs and 6 tacs equal 4 tocs , what is the ratio of one tic to one toc ? | "4 tic = 3 * tac and 6 * tac = 4 * toc ; 24 * tic = 18 * tac and 18 * tac = 12 * toc - - > 24 * tic = 12 * toc - - > tic / toc = 24 / 12 = 2 / 1 . answer : e ." | a = 4 / 4
b = 3 / 4
c = a + b
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a ) 11 kmph , b ) 10 kmph , c ) 12 kmph , d ) 16 kmph , e ) 15 kmph | b | divide(30, add(const_1, const_2)) | the speed of a boat in still water is 30 kmph . what is the speed of the stream if the boat can cover 80 km downstream or 40 km upstream in the same time ? | x = the speed of the stream ( 30 + x ) / ( 30 - x ) = 2 / 1 30 + x = 60 - 2 x 3 x = 30 x = 10 km / hour if the speed of the stream is 10 km / hour , then the ' downstream ' speed of the boat is 30 + 10 = 40 km / hour and the ' upstream ' speed of the boat is 30 - 10 = 20 km / hour . in that way , if the boat traveled for 2 hours , it would travel 2 x 40 = 80 km downstream and 2 x 20 = 40 km / hour upstream . answer : b | a = 1 + 2
b = 30 / a
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a ) 22 % , b ) 18 % , c ) 24 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(150, divide(multiply(105, const_100), subtract(const_100, 12.55))), const_100), 150) | after getting 2 successive discounts , a shirt with a list price of rs 150 is available at rs 105 . if the second discount is 12.55 , find the first discount . | "let the first discount be x % then , 87.5 % of ( 100 - x ) % of 150 = 105 = 87.5 / 100 * ( 100 - x ) / 100 * 450 = 150 = > 105 = > 100 - x = ( 105 * 100 * 100 ) / ( 150 * 87.5 ) = 80 x = ( 100 - 80 ) = 20 first discount = 20 % answer is e ." | a = 105 * 100
b = 100 - 12
c = a / b
d = 150 - c
e = d * 100
f = e / 150
|
['a ) 44 m', 'b ) 66 m', 'c ) 26 m', 'd ) 56 m', 'e ) 25 m'] | a | divide(multiply(add(60, 50), const_2), 5) | a rectangular plot measuring 60 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | perimeter of the plot = 2 ( 60 + 50 ) = 220 m no of poles = 220 / 5 = 44 m answer : a | a = 60 + 50
b = a * 2
c = b / 5
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a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | d | subtract(15, multiply(2, 1.5)) | the arithmetic mean and standard deviation of a certain normal distribution are 15 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 15 - 2 * 1.5 = 12 . answer : d . | a = 2 * 1
b = 15 - a
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a ) 15 % , b ) 20 % , c ) 10 % , d ) 8 % , e ) 22 % | d | divide(const_100, 12) | in how many years will a sum of money doubles itself at 12 % per annum on simple interest ? | "p = ( p * 12 * r ) / 100 r = 8 % answer : d" | a = 100 / 12
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a ) 288 , b ) 278 , c ) 800 , d ) 1600 , e ) 121 | d | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 20) | 20 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 20 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 5 x = 320 x = 1600 answer : d" | a = 650 * 20
b = a / 100
c = b + 190
d = c * 100
e = d / 20
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a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | b | divide(add(3, divide(const_1, const_2)), divide(9, 18)) | it will take 18 days for cameron to complete a certain task alone . he worked for 9 days before she was joined by sandra . both of them completed the remaining task in 3 and half days . how many days will it take both of them to complete the entire job ? | explanation : cameron and sandra completed half work in 3.5 days = > they can complete whole work in 7 days answer : option b | a = 1 / 2
b = 3 + a
c = 9 / 18
d = b / c
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a ) $ 37.5 , b ) $ 64 , c ) $ 75 , d ) $ 96 , e ) can not be determined | a | divide(add(70, 80), const_2.0) | if greg buys 5 shirts , 4 trousers and 2 ties , the total cost is $ 80 . if greg buys 7 shirts , 4 trousers and 2 ties , the total cost is $ 70 . how much will it cost him to buy 2 trousers , 3 shirts and 1 ties ? | "solution : 5 x + 4 y + 2 z = 80 7 x + 4 y + 2 z = 70 adding both the equations = 12 x + 8 y + 4 z = 150 3 x + 2 y + 1 z = 37.5 ans a" | a = 70 + 80
b = a / 2
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a ) 190 , b ) 200 , c ) 185 , d ) 166 , e ) 213 | a | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 1), add(const_2, const_4)) | what is the sum of all the numbers between 1 and 16 , inclusive ? | "all you have to do is add 1 + 2 + 3 + 4 . . . + 14 + 15 + 16 , which is 190 . final answer : a" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 1
t = 2 + 4
u = s + t
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a ) 19200 , b ) 17606 , c ) 17604 , d ) 17600 , e ) 117601 | a | divide(multiply(add(const_100, 20), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 20 % if no discount was offered ? | "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 20 % if no discount was offered = 16000 * 120 / 100 = rs . 19200 . answer : a" | a = 100 + 20
b = 12500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
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a ) 11 , b ) 77 , c ) 8 3 / 4 , d ) 8 1 / 2 , e ) 8 | d | add(8, divide(100, add(50, 45))) | the distance between delhi and mathura is 100 kms . a starts from delhi with a speed of 50 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 45 kmph at 8 p . m . from delhi . when will they meet ? | "d = 100 – 50 = 50 rs = 45 + 50 = 95 t = 90 / 45 = 0.5 hours 8 a . m . + 0.5 = 8 1 / 2 a . m . . answer : d" | a = 50 + 45
b = 100 / a
c = 8 + b
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['a ) 375', 'b ) 2570', 'c ) 2800', 'd ) 3315', 'e ) none of these'] | d | divide(rectangle_area(multiply(16.25, const_100), multiply(12.75, const_100)), multiply(add(add(const_10, const_10), add(const_3, const_2)), add(add(const_10, const_10), add(const_3, const_2)))) | the length and breadth of a rectangular floor are 16.25 metre and 12.75 metre respectively . find how many minimum number of square tiles would be required to cover it completely ? | explanation : since we require minimum number of square tiles , the size of the tile is given as the h . c . f . of two sides of the room . the h . c . f . of 1625 cm & 1275 cm . is 25 cm . hence , we get , required number = ( 1625 * 1275 ) / ( 25 * 25 ) = 3315 answer : d | a = 16 * 25
b = 12 * 75
c = rectangle_area / (
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a ) 86 , b ) 67 , c ) 31 , d ) 15 , e ) 17 | c | divide(add(25, 37), const_2) | a man can row upstream at 25 kmph and downstream at 37 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 37 m = ( 37 + 25 ) / 2 = 31 answer : c" | a = 25 + 37
b = a / 2
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a ) 24 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | a | divide(24, subtract(divide(const_1, divide(50, const_100)), const_1)) | walking at 50 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 50 % of speed meand 1 / 2 s takes 2 t . it takes 24 minutes extra to cover the distance . then 2 t = t + 24 t = 24 . option a is correct" | a = 50 / 100
b = 1 / a
c = b - 1
d = 24 / c
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a ) 40000 , b ) 24000 , c ) 26682 , d ) 29973 , e ) 12312 | a | multiply(divide(const_100, 80), 32000) | 80 % of the population of a village is 32000 . the total population of the village is ? | "x * ( 80 / 100 ) = 32000 x = 400 * 100 x = 40000 answer : a" | a = 100 / 80
b = a * 32000
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a ) 12.5 % , b ) 75 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(4, const_1), 4), const_100) | a number x is 4 times another number y . the percentage that y is less than x is | "say y = 1 and x = 4 . then y = 1 is less than x = 4 by ( 4 - 1 ) / 4 * 100 = 3 / 4 * 100 = 75 % . answer : b ." | a = 4 - 1
b = a / 4
c = b * 100
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a ) 1000 , b ) 1200 , c ) 1400 , d ) 1450 , e ) 1500 | b | divide(subtract(multiply(1600, 11), multiply(1600, 8)), subtract(12, 8)) | a trader has 1600 kg of sugar . he sells a part at 8 % profit and the rest at 12 % profit . if he gains 11 % on the whole , find the quantity sold at 12 % ? | by rule of alligation , % profit by selling part 1 % profit by selling part 2 8 12 net % profit 11 12 - 11 = 1 11 - 8 = 3 = > quantity of part 1 : quantity of part 2 = 1 : 3 given that total quantity = 1600 kg hence , quantity of part 2 ( quantity sold at 12 % profit ) = 1600 × 3 4 = 1200 b ) | a = 1600 * 11
b = 1600 * 8
c = a - b
d = 12 - 8
e = c / d
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a ) 30 % , b ) 32 % , c ) 45 % , d ) 68 % , e ) 70 % | b | subtract(const_100, add(multiply(divide(subtract(const_100, 30), const_100), multiply(const_100, divide(40, const_100))), multiply(const_100, divide(40, const_100)))) | john spent 40 percent of his earning last month on rent and 30 percent less than what he spent on rent to purchase a new dishwasher . what percent of last month ' s earning did john have left over ? | say john ' s earning last month was $ 100 . john spent 40 percent of his earning last month on rent - - > $ 40 on rent ; 30 percent less than what he spent on rent to purchase a new dishwasher - - > $ 40 * 0.7 = $ 28 on the dishwasher . left over amount 100 - ( 40 + 28 ) = $ 32 . answer : b . | a = 100 - 30
b = a / 100
c = 40 / 100
d = 100 * c
e = b * d
f = 40 / 100
g = 100 * f
h = e + g
i = 100 - h
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a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 10 | e | inverse(add(divide(5, multiply(10, 7)), divide(10, multiply(10, 14)))) | 10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 4 children take to complete the work ? | "1 women ' s 1 day work = 1 / 70 1 child ' s 1 day work = 1 / 140 ( 5 women + 4 children ) ' s 1 day work = ( 5 / 10 + 4 / 140 ) = 1 / 10 5 women and 4 children will complete the work in 10 days . e" | a = 10 * 7
b = 5 / a
c = 10 * 14
d = 10 / c
e = b + d
f = 1/(e)
|
a ) $ 33.75 , b ) $ 47.25 , c ) $ 60.75 , d ) $ 54.00 , e ) $ 70.00 | c | add(multiply(7.5, 4.50), multiply(multiply(subtract(10.5, 7.5), 2.0), 4.50)) | lloyd normally works 7.5 hours per day and earns $ 4.50 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 2.0 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ? | "daily working hour * regular rate + overtime * increased rate 7.5 * 4.5 + 3 * 4.5 * 2.0 = 60.75 answer c" | a = 7 * 5
b = 10 - 5
c = b * 2
d = c * 4
e = a + d
|
a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 25 % , e ) 30 % | d | multiply(divide(20, subtract(const_100, 20)), const_100) | on a certain road , 20 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 20 % of them exceeded the speed limit and received the ticket , i . e . x / 5 . again , suppose total no . of motorists who exceeded the speed limit are y . 20 % of y exceeded the speed limit but did n ' t received the ticket , i . e . y / 5 . it means 4 y / 5 received the ticket . hence , 4 y / 5 = x / 5 or y / x = 1 / 4 or y / x * 100 = 1 / 4 * 100 = 25 % d" | a = 100 - 20
b = 20 / a
c = b * 100
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | b | subtract(multiply(16, 2), 30) | if the remainder is 16 when the integer n is divided by 30 , what is the remainder when 2 n is divided by 15 ? | n = 30 k + 16 2 n = 2 ( 30 k + 16 ) = 4 k * 15 + 32 = 4 k * 15 + 2 * 15 + 2 = 15 j + 2 the answer is b . | a = 16 * 2
b = a - 30
|
a ) 48 , b ) 50 , c ) 52 , d ) 54 , e ) 56 | e | add(divide(factorial(7), factorial(5)), multiply(7, const_2)) | to apply for the position of photographer at a local magazine , a photographer needs to include 4 or 5 photos in an envelope accompanying the application . if the photographer has pre - selected 7 photos representative of her work , how many choices does she have to provide the photos for the magazine ? | 7 c 4 + 7 c 5 = 35 + 21 = 56 the answer is e . | a = math.factorial(7)
b = math.factorial(5)
c = a / b
d = 7 * 2
e = c + d
|
a ) 47 , b ) 76 , c ) 28 , d ) 66 , e ) 11 | d | multiply(divide(30, subtract(6, const_1)), subtract(12, const_1)) | at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 30 seconds . how long does it tick at 12 ′ o clock | explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 30 / 5 = 6 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 6 = 66 secs . answer : d | a = 6 - 1
b = 30 / a
c = 12 - 1
d = b * c
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 11 | e | divide(subtract(1100, multiply(5, add(20, 20))), add(add(20, 20), 20)) | john purchased some shirts and trousers for $ 1100 . he paid $ 550 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ? | "given that the total purchase of two items cost 1100 . so the average purchase of one item will cost 1100 / 2 = 550 . its given as total shirt cost 100 $ less . hence total shirt cost = 550 - 275 and total trouser cost = 300 + 275 5 shirts = 275 $ = = > one shirt = 55 $ one trouser = 55 + 20 = 75 $ total trousers = 825 / 75 = 11 . e" | a = 20 + 20
b = 5 * a
c = 1100 - b
d = 20 + 20
e = d + 20
f = c / e
|
a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent t of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is t = . 25 / 1 * 100 = 25 % d is the answer ." | a = 9 / 12
b = a * 100
c = 100 - b
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 3 , e ) 2 / 5 | a | divide(const_1, const_2) | in a throw of a coin find the probability of getting a head ? | s = { h , t } e = { h } p ( e ) = 1 / 2 answer is a | a = 1 / 2
|
a ) 5 days , b ) 7 days , c ) 12 days , d ) 9 days , e ) 18 days | e | inverse(multiply(inverse(12), subtract(const_1, multiply(5, inverse(15))))) | a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 12 days . in how many days b alone can do the work ? | explanation : a ’ s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 12 x = 18 days answer : option e | a = 1/(12)
b = 1/(15)
c = 5 * b
d = 1 - c
e = a * d
f = 1/(e)
|
a ) 878 , b ) 545 , c ) 434 , d ) 442 , e ) 452 | e | subtract(460, multiply(multiply(12, 3), 2)) | evaluate : 460 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 460 - 12 * 3 * 2 = 460 - 8 = 452 correct answer e" | a = 12 * 3
b = a * 2
c = 460 - b
|
a ) 4 min , b ) 3 min 3 sec , c ) 4 min 4 sec , d ) 5 min 3 sec , e ) 4 min 3 sec | e | subtract(subtract(multiply(divide(7, 28), const_1000), 7), multiply(const_2, const_100)) | in a km race , a beats bby 28 metres or 7 seconds . find a ' s timeoverthe course | clearly , b covers 28 m in 7 seconds . : . b ' s time over the course = ( 278 x 1000 ) sec = 250 seconds . : . a ' s time over the course = ( 250 - 7 - ) sec = 243 sec = 4 min . 3 sec . answer e 4 min 3 sec | a = 7 / 28
b = a * 1000
c = b - 7
d = 2 * 100
e = c - d
|
a ) 350 m , b ) 200 m , c ) 400 m , d ) 800 m , e ) none of them | d | divide(multiply(200, 8), subtract(10, 8)) | a thief is spotted by a policeman from a distance of 200 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 200 m / 1000 ) x 1 / 2 hr = 1 / 10 hr . in 1 / 10 hrs , the thief covers a distance of 8 x 1 / 10 km = 4 / 5 km = 800 m answer is d ." | a = 200 * 8
b = 10 - 8
c = a / b
|
a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50 | e | add(30, 20) | set x consists of 10 integers and has median of 30 and a range of 20 . what is the value of the greatest possible integer that can be present in the set ? | note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 50 this is possible . hence it is the greatest such number . answer ( e ) | a = 30 + 20
|
a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 7.25 hrs , e ) 11.25 hrs | e | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 9))) | a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 5 - 1 / 9 = 4 / 45 therefore the cistern will be filled in 45 / 4 hours or 11.25 hours . answer : e" | a = 1 / 5
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 221 , b ) 287 , c ) 400 , d ) 589 , e ) 171 | d | divide(650, power(add(const_1, divide(5, const_100)), 2)) | find the sum lend at c . i . at 5 p . c per annum will amount to rs . 650 in 2 years ? | "explanation : 650 = p ( 21 / 20 ) 2 p = 589.56 answer : d" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 650 / c
|
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