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 ) 12 , b ) 20 , c ) 8 , d ) all of these , e ) none of these | c | multiply(divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3)) | find the smallest number in a gp whose sum is 38 and product 1728 | "in gp : - a / r + a + ar . . . . = 38 a ( 1 + r + r ^ 2 ) = 38 r . . . . . . eq 1 product a ^ 3 = 1728 a = 12 now put this value in eq 1 12 ( 1 + r + r ^ 2 ) = 38 r r = 2 / 3 and 3 / 2 . . so sallest number is 8 answer : c" | a = 38 / 1000
b = a / 3
c = b / 3
d = c / 3
e = 38 / 1000
f = e / 3
g = f / 3
h = g / 3
i = d * h
|
a ) 13.25 , b ) 38.25 , c ) 33.25 , d ) 34.25 , e ) 31.25 | c | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 76), const_2) | 76 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 76 = 33.25 answer : c" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 76
f = e / 2
|
a ) 2674 , b ) 2944 , c ) 2464 , d ) 2924 , e ) 2994 | b | add(multiply(40, 72), 64) | in a division sum , the quotient is 40 , the divisor 72 and the remainder 64 , find the dividend ? | explanation : 40 * 72 + 64 = 2944 answer : b | a = 40 * 72
b = a + 64
|
a ) 68 , b ) 84 , c ) 72 , d ) 60 , e ) 64 | d | divide(subtract(multiply(80, 3), multiply(80, divide(60, const_100))), add(divide(60, const_100), 1)) | a store sells chairs and tables . if the price of 3 chairs and 1 table is 60 % of the price of 1 chair and 3 tables , and the price of 1 table and 1 chair is $ 80 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . ) | "let c = chair ; t = table 3 c + 1 t = 0.6 ( 1 c + 3 t ) or c ( 3 - 0.6 ) = 1.8 t - 1 t or 2.4 c = 0.8 t therefore c = 0.8 / 2.4 t = 1 / 3 t ic + 1 t = 80 or 1 / 3 t + 1 t = 80 therefore t = 80 * 3 / 4 = 60 d" | a = 80 * 3
b = 60 / 100
c = 80 * b
d = a - c
e = 60 / 100
f = e + 1
g = d / f
|
a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 3 / 5 , e ) 3 / 8 | a | multiply(subtract(1, divide(1, 4)), subtract(1, divide(1, 3))) | the probability that a man will be alive for 10 more yrs is 1 / 4 & the probability that his wife will alive for 10 more yrs is 1 / 3 . the probability that none of them will be alive for 10 more yrs , is | sol . required probability = pg . ) x p ( b ) = ( 1 β d x ( 1 β i ) = : x 1 = 1 / 2 ans . ( a ) | a = 1 / 4
b = 1 - a
c = 1 / 3
d = 1 - c
e = b * d
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | c | multiply(4, const_2) | how many positive even integers less than 100 contain digits 4 or 7 ? | "one digit numbers : 4 two digit numbers : 4 at tens place : 40 , 42,44 , 46,48 7 at tens place : 70 , 72,74 , 76,78 4 at units place : 14 , 24,34 , 54,64 , 84,94 ( duplicates removed ) if 7 is at units place , the number cant be even total : 1 + 5 + 5 + 7 = 18 answer c" | a = 4 * 2
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a ) 24 . , b ) 32 . , c ) 36 , d ) 42 . , e ) 44 . | b | inverse(add(inverse(48), divide(inverse(48), const_2))) | a car traveled from san diego to san francisco at an average speed of 48 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 48 x total distance traveled = 2 * 48 x = 96 x total time taken = x + 2 x = 3 x average speed = 96 x / 3 x = 32 answer : b" | a = 1/(48)
b = 1/(48)
c = b / 2
d = a + c
e = 1/(d)
|
a ) 12.0 , b ) 12.1 , c ) 12.2 , d ) 10.0 , e ) 10.4 | d | sqrt(multiply(8, 12.4)) | amount of bacteria present time amount 1 : 00 p . m . 8.0 grams 4 : 00 p . m . x grams 7 : 00 p . m . 12.4 grams data for a certain biology experiment are given in the table above . if the amount of bacteria present increased by the same fraction during each of the two 3 - hour periods shown , how many grams of bacteria were present at 4 : 00 p . m . ? | the question says that bacteria increased by same fraction , not by same amount in 2 intervals of 3 hours . let x represent the amount of bacteria present at 4 : 00 pm . since the fractional increase must remain constant from 1 to 4 pm as it is from 4 pm to 7 pm : fractional increase from 1 pm to 4 pm = x / 8.0 fractional increase from 4 pm to 7 pm = 12.4 / x x \ 8 = 12.4 \ x x ^ 2 = 12.4 * 8 x = 10 answer : d | a = 8 * 12
b = math.sqrt(a)
|
a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 4 , d ) 7 : 9 , e ) none of these | d | divide(subtract(6.30, 5.60), subtract(7.20, 6.30)) | find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.60 a kg to produce a mixture worth rs . 6.30 a kg ? | "solution required ratio = 70 : 90 = 7 : 9 answer d" | a = 6 - 30
b = 7 - 20
c = a / b
|
a ) 175 , b ) 280 , c ) 189 , d ) 278 , e ) 268 | a | divide(multiply(300, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | i bought two books ; for rs . 300 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 300 - x ) 119 / 100 x = 175 answer : a" | a = 100 + 19
b = 300 * a
c = 100 - 15
d = 100 + 19
e = c + d
f = b / e
|
a ) 50 days , b ) 83.6 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(130, const_100), 60)) | p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 130 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 380 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 130 days - - > 130 / 250 work done = > 52 % 22 % work left . . . r alone will take 22 % * 380 = 83.6 days answer is ( b )" | a = 4 * 50
b = 130 / 100
c = b * 60
d = a - c
|
a ) 450 . , b ) 459 . , c ) 432 . , d ) 502 . , e ) 522 . | a | multiply(divide(multiply(49, 6), subtract(49, 15)), 15) | a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 15 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "pastry is divided in 15 customers equally . so , total number of pastry must be a multiple of 15 only option a satisfies the condition , and hence is the answer" | a = 49 * 6
b = 49 - 15
c = a / b
d = c * 15
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(8, 3), const_2), const_4), 8), add(multiply(multiply(multiply(8, 3), const_2), const_4), 3)), 1) | for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 8 + 3 and * 8 + 8 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 8 would be divisible by all numbers from 1 to 8 . = > * 8 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 8 is divisible by 3 ) in fact adding anyprimenumber between 1 to 8 to * 8 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 8 + 3 and * 8 + 11 then the answer would be 1 . for * 8 + 3 and * 8 + 13 , it is 2 and so on . . . a" | a = 8 * 3
b = a * 2
c = b * 4
d = c + 8
e = 8 * 3
f = e * 2
g = f * 4
h = g + 3
i = d - h
j = i - 1
|
a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 9 cm | b | sqrt(divide(multiply(multiply(const_pi, multiply(12, divide(12, const_2))), const_2), multiply(const_pi, const_4))) | the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each . the radius of the sphere is : | "explanation : curved surface area of sphere = 4 / Ο r 2 surface area of cylinder = 2 Ο rh = > 4 / Ο r 2 = 2 Ο rh = > r 2 = 6 β 12 / 2 = > r 2 = 36 = > r = 6 note : diameter of cylinder is 12 so radius is taken as 6 . option b" | a = 12 / 2
b = 12 * a
c = math.pi * b
d = c * 2
e = math.pi * 4
f = d / e
g = math.sqrt(f)
|
a ) 21000 , b ) 19800 , c ) 16500 , d ) 15800 , e ) 8700 | c | divide(multiply(subtract(25200, 5400), 5), add(const_2, const_4)) | during the first week of performances of a certain play , 5400 tickets were sold , all at reduced price . during the remaining weeks of performances , 5 times as many tickets were sold at full price as were sold at reduced price . if the total number of tickets e sold was 25200 , how many of them were sold at full price ? | given : the question tells us about the number of tickets sold in the first week at reduced price = 5400 . it also tells us that during the remaining weeks , the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price . the total tickets sold was 25200 , and we are asked to find the number of tickets that where sold at full price . approach : the question asks us to find the number of ticket sold at full price . we know that the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price in remaining weeks . hence , assuming tickets sold at reduced price in remaining weeks to be x would give us the no . of tickets sold at full price to be 5 x . tickets sold in first week + tickets sold in remaining weeks = total no . of tickets soldi . e . tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold we know the no . of tickets sold during first week and the total no . of tickets sold . we will use this relation to get to our answer . working out : tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold substituting values in the equation , we get 5400 + 5 x + x = 25200 x = 3300 tickets sold at full price = 5 x = 16500 answer : option c | a = 25200 - 5400
b = a * 5
c = 2 + 4
d = b / c
|
a ) 24 % , b ) 12.5 % , c ) 30 % , d ) 36 % , e ) 40 % | b | multiply(divide(100, subtract(900, 100)), const_100) | a cricket bat is sold for $ 900 , making a profit of $ 100 . the profit percentage would be | "100 / ( 900 - 100 ) = 100 / 800 = 0.125 = 12.5 % . answer : b" | a = 900 - 100
b = 100 / a
c = b * 100
|
a ) 186 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(279, subtract(20, 8)), 8) | a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 279 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 279 + l ) / 20 = l / 8 2232 + 8 l = 20 l l = 2232 / 12 = 186 m answer a" | a = 20 - 8
b = 279 / a
c = b * 8
|
a ) 144 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(60, const_100))) | mary ' s income is 60 % more than tim ' s income and tim ' s income is 10 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.9 = 90 m = 90 * 1.6 = 144 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 144 * 100 / 100 = 144 ans : a" | a = 100 - 10
b = 100 - 10
c = 60 / 100
d = b * c
e = a + d
|
a ) 597 , b ) 608 , c ) 613 , d ) 616 , e ) 621 | a | add(multiply(subtract(75, const_1), 8), 5) | let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 75 th number in this set ? | "the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 75 th number = 8 * ( 75 - 1 ) + 5 = 597 answer = a" | a = 75 - 1
b = a * 8
c = b + 5
|
a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 60 | e | divide(subtract(75, 54), subtract(const_1, divide(65, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 54 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "the total number of books is 75 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 54 books , thus the number of un - returned books is 75 - 54 = 21 books . 0.35 x = 21 x = 60 the answer is e ." | a = 75 - 54
b = 65 / 100
c = 1 - b
d = a / c
|
a ) 388 , b ) 2998 , c ) 277 , d ) 500 , e ) 400 | e | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 86 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 86 5 x / 100 + 90 β 6 x / 100 = 86 x / 100 = 4 = > x = 400 . answer : e" | a = 5 + 6
b = a * 100
|
a ) 2445 , b ) 2331 , c ) 2874 , d ) 2801 , e ) 2256 | e | subtract(subtract(add(add(multiply(multiply(add(const_2, const_3), const_3), const_10), multiply(subtract(multiply(9, const_3), const_3), const_100)), add(const_3, const_3)), const_100), const_100) | each of the positive integers a , b , and c is a 3 - digit integer . if each of the digits 1 through 9 appears in one of these 3 integers , what is the maximum possible value of the sum of a , b , and c ? | according to the stem we should use the digits 1 through 9 to construct 3 three - digit integers , so that their sum is as big as possible . to maximize the sum , maximize the hundreds digits of a , b , and c , so make them 9 , 8 , and 7 . next , minimize tens digits . make them 6 , 5 , and 4 . use the remaining digits ( 3 , 2 , and 1 ) for units digits . so , a would be 963 , b would be 852 and c would be 741 . 741 + 852 + 963 = 2256 . answer : e . | a = 2 + 3
b = a * 3
c = b * 10
d = 9 * 3
e = d - 3
f = e * 100
g = c + f
h = 3 + 3
i = g + h
j = i - 100
k = j - 100
|
a ) 15 % , b ) 32 % , c ) 40 % , d ) 62 % , e ) 80 % | e | multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(20, const_100)))), const_1)) | the profits of qrs company rose 50 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ? | "assume 100 in march , then 150 in april as 50 % increase , then 120 in may as 20 % decrease from april , and then 180 in june which is 150 % of 120 . so overall increase is from 100 to 180 is 80 % answer e" | a = 50 / 100
b = 1 + a
c = 50 / 100
d = 1 + c
e = 20 / 100
f = 1 - e
g = d * f
h = b * g
i = h - 1
j = 100 * i
|
a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | c | divide(subtract(100, add(37, 3)), 100) | a certain bag contains 100 balls Γ’ β¬ β 20 white , 30 green , 10 yellow , 37 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 20 + 30 + 10 ) / 100 = 60 / 100 = 0.6 . answer is c" | a = 37 + 3
b = 100 - a
c = b / 100
|
a ) 2977 , b ) 3837 , c ) 4097 , d ) 4847 , e ) 5387 | c | subtract(subtract(subtract(multiply(257, const_10), const_100), 2), 2) | what is the next number : 2 , 17 , 257 , __ | "4 ^ 0 + 1 = 2 4 ^ 2 + 1 = 17 4 ^ 4 + 1 = 257 4 ^ 6 + 1 = 4097 the answer is c ." | a = 257 * 10
b = a - 100
c = b - 2
d = c - 2
|
a ) 3.5 percent increase , b ) 9.5 percent increase , c ) 1.5 percent increase , d ) 3.9 percent increase , e ) 3.5 percent increase | e | multiply(divide(subtract(multiply(add(const_100, 15), subtract(const_100, 10)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100) | calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 15 % and its breadth is decreased by 10 % ? | "let l and b be 100 each 100 * 100 = 10000 l increase by 15 % = 115 b decrease by 10 % = 90 115 * 90 = 10350 3.5 percent increase answer : e" | a = 100 + 15
b = 100 - 10
c = a * b
d = 100 * 100
e = c - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) 5363 , b ) 6240 , c ) 4860 , d ) 1987 , e ) 2732 | c | multiply(divide(108, 10), 450) | in order to obtain an income of rs . 450 from 10 % stock at rs . 108 , one must make an investment of | "to obtain rs . 10 , investment = rs . 108 . to obtain rs . 450 , investment = = rs . 4860 . answer : c" | a = 108 / 10
b = a * 450
|
a ) 72 , b ) 85 , c ) 30 , d ) 51 , e ) 45 | c | add(20, divide(20, const_2)) | in a ratio which is equal to 4 : 6 , if the antecedent is 20 , then the consequent is ? | "we have 4 / 6 = 20 / x 4 x = 20 * 6 x = 30 consequent = 30 answer is c" | a = 20 / 2
b = 20 + a
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a ) 10 metres , b ) 5 metres , c ) 7.5 metres , d ) data inadequate , e ) none of these | a | subtract(20, 10) | the area of a rectangular plot is 20 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | "l Γ b = 20 Γ b β΄ l = 20 m and l β b = 10 β΄ b = 20 β 10 = 10 m answer a" | a = 20 - 10
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a ) 784598562 , b ) 765858254 , c ) 784569825 , d ) 786461346 , e ) none of these | d | multiply(divide(78654, 9999), const_100) | 78654 * 9999 | "explanation : 78654 * ( 10000 - 1 ) = 786540000 - 78654 = 786461346 option d" | a = 78654 / 9999
b = a * 100
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a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 2 / 3 | b | divide(subtract(60, 40), subtract(const_100, 40)) | the weight of a glass of jar is 40 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 40 g weight of coffee beans = 60 g weight of jar and remaining beans = 60 g weight of remaining beans = 20 g fraction remaining = 20 / 60 = 1 / 3 answer is b ." | a = 60 - 40
b = 100 - 40
c = a / b
|
a ) 80000 , b ) 60000 , c ) 50000 , d ) 70000 , e ) 90000 | d | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 75,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 175000 - x ) respectively . then , = 1 / 3 x = 1 / 2 ( 175000 - x ) = x / 3 + x / 2 = 87500 = 5 x / 6 = 87500 = x = 87500 x 6 / 5 = 105000 savings in public provident fund = rs . ( 175000 - 105000 ) = rs . 70000 answer is d ." | a = 100 * 10
b = a * 100
c = 100 * 10
d = c * 100
e = 2 * 100
f = e * 100
g = d - f
h = b + g
i = 3 / 2
j = i + 1
k = 1 / j
l = h * k
|
a ) 13 , b ) 19 , c ) 15 , d ) 16 , e ) 17 | b | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | how many positive integers less than 120 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 117 = 13 * 9 = 117 for 12 : 12 . . . 120 = 12 * 10 = 120 but there is one integer 13 * 12 . so n = ( 9 ) + ( 10 ) = 19 b" | a = 4 + 13
b = a - 1
c = math.factorial(b)
d = math.factorial(13)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 80 , b ) 720 , c ) 760 , d ) 840 , e ) 860 | b | multiply(multiply(2, 9), multiply(multiply(2, 4), 5)) | if the operation β¬ is defined for all x and y by the equation x β¬ y = 2 * x * y , then 9 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 2 * 4 * 5 = 40 9 β¬ 40 = 2 * 9 * 40 = 720 hence , answer is b" | a = 2 * 9
b = 2 * 4
c = b * 5
d = a * c
|
a ) 14 / 15 , b ) 43 / 45 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | b | divide(divide(43, 4.5), divide(43, 4.3)) | jack and jill are marathon runners . jack can finish a marathon ( 43 km ) in 4.5 hours and jill can run a marathon in 4.3 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 43 / ( 9 / 2 ) = 86 / 9 average speed of jill = 43 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 86 / 9 ) / 10 = 86 / 90 = 43 / 45 answer b" | a = 43 / 4
b = 43 / 4
c = a / b
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | add(multiply(6, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 6), divide(const_1, 12)), const_2)), 6), const_60)) | two pipes a and b can fill a tank in 6 and 12 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 1 / 6 part filled by b in 1 min . = 1 / 12 part filled by ( a + b ) in 1 min . = 1 / 6 + 1 / 12 = 1 / 4 . both the pipes can fill the tank in 4 minutes . answer : a" | a = 6 * 100
b = 1 / 6
c = 1 / 12
d = b + c
e = d * 2
f = 1 - e
g = f * 6
h = g * const_60
i = a + h
|
a ) 14 , b ) 16 , c ) 21 , d ) 24 , e ) 27 | d | divide(336, divide(subtract(420, 336), 6)) | a car traveled 420 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 420 miles will be covered in 420 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 420 / h = > 336 / c = 420 / ( c + 6 ) = > 336 c + 336 * 6 = 420 c = > c = 336 * 6 / 84 = 24 answer d ." | a = 420 - 336
b = a / 6
c = 336 / b
|
a ) 75 , b ) 65 , c ) 43 , d ) 52 , e ) 78 | a | multiply(divide(65, 13), 15) | 65 bags of nuts are to be divided among 13 students . each bag contains 15 nuts . how many nuts will each student receive ? | 65 Γ· 13 = 5 bags per student 5 x 15 = 75 nuts per student correct answer a | a = 65 / 13
b = a * 15
|
a ) 5 / 2 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 / 5 , e ) 0 | b | divide(add(1, 2), add(2, 4)) | if x = 1 - 4 t and y = 2 t - 2 , then for what value of t does x = y ? | we are given x = 1 β 4 t and y = 2 t β 2 , and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus , we can substitute 1 β 4 t for x and 2 t β 2 for y in the equation x = y . this gives us : 1 β 4 t = 2 t β 2 3 = 6 t 1 / 2 = t the answer is b . | a = 1 + 2
b = 2 + 4
c = a / b
|
a ) 4 , b ) 5 , c ) 6 , d ) 6 , e ) 1 | a | multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33)) | the sum of two numbers is 528 and their h . c . f is 33 . the number of pairs of numbers satisfying the above conditions is : | "let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 * 7 , 33 * 9 ) . the number of such pairs is 4 . answer : a" | a = 528 + 33
b = 1 + 1
c = a / b
d = 528 + 33
e = 1 + 1
f = d / e
g = f - 33
h = c * g
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a ) 5 , b ) 6 , c ) 4 , d ) 9 , e ) 5.5 | c | divide(divide(multiply(const_4, const_4), const_2), const_2) | what is the last digit in ( 7 ^ 95 - 3 ^ 58 ) ? | explanation : unit digit in 795 = unit digit in [ ( 74 ) 23 x 73 ] = unit digit in [ ( unit digit in ( 2401 ) ) 23 x ( 343 ) ] = unit digit in ( 123 x 343 ) = unit digit in ( 343 ) = 3 unit digit in 358 = unit digit in [ ( 34 ) 14 x 32 ] = unit digit in [ unit digit in ( 81 ) 14 x 32 ] = unit digit in [ ( 1 ) 14 x 32 ] = unit digit in ( 1 x 9 ) = unit digit in ( 9 ) = 9 unit digit in ( 795 - 358 ) = unit digit in ( 343 - 9 ) = unit digit in ( 334 ) = 4 . so , answer c | a = 4 * 4
b = a / 2
c = b / 2
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a ) 32 , b ) 28 , c ) 36 , d ) 40 , e ) 50 | a | divide(multiply(multiply(16, 10), 12), multiply(multiply(5, 4), 3)) | how many cuboids of length 5 m , width 4 m and height 3 m can be farmed from a cuboid of 16 m length , 10 m width and 12 m height . | ( 16 Γ£ β 10 Γ£ β 12 ) / ( 5 Γ£ β 4 Γ£ β 3 ) = 32 answer is a . | a = 16 * 10
b = a * 12
c = 5 * 4
d = c * 3
e = b / d
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a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | e | divide(divide(200, const_2), divide(10, const_100)) | one night 10 percent of the female officers on a police force were on duty . if 200 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 200 number of female officers on duty on that night = 200 / 2 = 100 ( 10 / 100 ) * f = 100 = > f = 1000 answer e" | a = 200 / 2
b = 10 / 100
c = a / b
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a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 90 , e ) $ 100 | b | divide(multiply(subtract(divide(140, subtract(const_1, divide(30, const_100))), 140), const_100), 140) | a furniture dealer purchased a desk for $ 140 and then set the selling price equal to the purchase price plus a markup that was 30 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 140 selling price = x 140 + 0.3 * x = x 0.7 * x = 140 x = 200 profit = 200 - 140 = 60 answer : b" | a = 30 / 100
b = 1 - a
c = 140 / b
d = c - 140
e = d * 100
f = e / 140
|
a ) 91 / 198 , b ) 91 / 12 , c ) 91 / 13 , d ) 91 / 10 , e ) 91 / 11 | b | divide(96, 97) | find 96 Γ Γ 97 | "here both numbers are less than 100 . so they are deficient of - 4 and - 3 compared with 100 . so answer : b" | a = 96 / 97
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a ) 18 , b ) 25 , c ) 20 , d ) 32 , e ) 33 | e | divide(multiply(77, 3), add(add(const_1, const_2), 4)) | of 3 numbers , the third is twice the first and the second is 4 times the first . if their average is 77 , the smallest of the 3 numbers is : | explanation : let first number be x . so , 2 nd no . = 4 x & 3 rd no . = 2 x . so , x + 4 x + 2 x = 77 Γ 3 = 231 7 x = 231 x = 231 / 7 hence , smallest number x = 33 answer : e | a = 77 * 3
b = 1 + 2
c = b + 4
d = a / c
|
a ) 330 , b ) 770 , c ) 660 , d ) 100 , e ) 600 | c | multiply(multiply(const_pi, 10), 21) | the slant height of a cone is 21 cm and radius of the base is 10 cm , find the curved surface of the cone . | "Ο * 21 * 10 = 660 answer : c" | a = math.pi * 10
b = a * 21
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a ) 2 % , b ) 17 % , c ) 25 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(400, 320), 320), const_100) | a store reported total sales of $ 400 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ? | "new value β old value ) / old value x 100 we are given : february sales this year = 400 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 400 million and the old value = 320 million . let β s plug them into our percent change formula . ( new value β old value ) / old value x 100 [ ( 400 β 320 ) / 320 ] x 100 = 25 % . the answer is c ." | a = 400 - 320
b = a / 320
c = b * 100
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a ) 3 : 5 , b ) 9 : 25 , c ) 5 : 2 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(20, const_2))) | two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 20 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 10 / speed b since the times are same , 50 pi / speed a = 20 pi / speed b speeda / speed b = 50 pi / 30 pi = 5 / 2 correct option : c" | a = 50 / 2
b = circumface / (
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a ) 45 % , b ) 50 % , c ) 55 % , d ) 60 % , e ) 65 % | e | multiply(const_100, divide(subtract(subtract(const_100, 56), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | in a company , 56 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | "the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 44 % of all employees are women , so non - union women are 44 % - 18 % = 26 % 40 % of all employees are non - union . the percent of non - union employees who are women is 26 % / 40 % = 65 % the answer is e ." | a = 100 - 56
b = 70 / 100
c = 60 * b
d = 60 - c
e = a - d
f = 100 - 60
g = e / f
h = 100 * g
|
a ) 20 , b ) 25 , c ) 30 , d ) 11 , e ) 35 | d | divide(subtract(140, 15), 10) | in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 15 and the other one , that is cheaper , costs $ 10 . the factory holds a $ 140 worth stock of 13 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 10 modules must be 11 so that the leftover 2 modules are of $ 15 which will give a total value $ 140 . 11 * 10 + 2 * 15 = 110 + 30 = 140 answer : d" | a = 140 - 15
b = a / 10
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | c | add(add(divide(subtract(sqrt(add(power(add(multiply(add(const_10, const_1), 4), multiply(4, const_2)), const_2), multiply(multiply(4, multiply(add(const_10, const_1), const_2)), subtract(208, multiply(const_4, const_4))))), add(multiply(add(const_10, const_1), 4), multiply(4, const_2))), multiply(add(const_10, const_1), 4)), 4), multiply(divide(subtract(sqrt(add(power(add(multiply(add(const_10, const_1), 4), multiply(4, const_2)), const_2), multiply(multiply(4, multiply(add(const_10, const_1), const_2)), subtract(208, multiply(const_4, const_4))))), add(multiply(add(const_10, const_1), 4), multiply(4, const_2))), multiply(add(const_10, const_1), 4)), 10)) | in a two digit number it is known that its unit ' s digit exceeds its 10 ' s digit by 4 and that the product of the given number and the sum of its digits is equal to 208 , what is the number ? | xy is 2 digit no . y - x = 4 ; ( 10 x + y ) * ( 2 x + 4 ) = 208 ; sub y = x + 4 ; we get 26 answer : c | a = 10 + 1
b = a * 4
c = 4 * 2
d = b + c
e = d ** 2
f = 10 + 1
g = f * 2
h = 4 * g
i = 4 * 4
j = 208 - i
k = h * j
l = e + k
m = math.sqrt(l)
n = 10 + 1
o = n * 4
p = 4 * 2
q = o + p
r = m - q
s = 10 + 1
t = s * 4
u = r / t
v = u + 4
w = 10 + 1
x = w * 4
y = 4 * 2
z = x + y
A = z ** 2
B = 10 + 1
C = B * 2
D = 4 * C
E = 4 * 4
F = 208 - E
G = D * F
H = A + G
I = math.sqrt(H)
J = 10 + 1
K = J * 4
L = 4 * 2
M = K + L
N = I - M
O = 10 + 1
P = O * 4
Q = N / P
R = Q * 10
S = v + R
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a ) 15.14 mins , b ) 16.14 mins , c ) 17.14 mins , d ) 18.14 mins , e ) 19.14 mins | c | divide(const_1, subtract(add(divide(const_1, 30), divide(const_1, 20)), divide(const_1, 40))) | if a fills a tank in 30 mins , and b fills the same tank in 20 mins , and c emptys it in 40 min , if a , b , c works together how much time will the tank take to fill ? | 1 / 30 + 1 / 20 - 1 / 40 = 7 / 120 so total time = 120 / 7 = 17.14 mins answer : c | a = 1 / 30
b = 1 / 20
c = a + b
d = 1 / 40
e = c - d
f = 1 / e
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a ) 36 % , b ) 38 % , c ) 16 % , d ) 39 % , e ) 12 % | d | subtract(const_100, divide(multiply(340, const_100), 560)) | an article is bought for rs . 560 and sold for rs . 340 , find the loss percent ? | 560 - - - - 220 100 - - - - ? = > 39 % answer : d | a = 340 * 100
b = a / 560
c = 100 - b
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a ) $ 4 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 24 | e | subtract(180, add(add(multiply(180, divide(1, 5)), multiply(180, divide(1, 6))), multiply(180, divide(1, 2)))) | jennifer had $ 180 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | 1 / 5 x $ 180 = $ 36 for sandwich 1 / 6 x $ 180 = $ 30 for museum 1 / 2 x $ 180 = $ 90 for book $ 36 + $ 30 + $ 90 = $ 156 spent $ 180 - $ 156 = $ 24 left over correct answer e | a = 1 / 5
b = 180 * a
c = 1 / 6
d = 180 * c
e = b + d
f = 1 / 2
g = 180 * f
h = e + g
i = 180 - h
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a ) 13 , b ) 13.5 , c ) 14.4 , d ) 14.5 , e ) 15 | c | divide(add(multiply(16, const_10), divide(multiply(16, const_10), const_2)), add(divide(multiply(16, const_10), 12), divide(divide(multiply(16, const_10), const_2), 16))) | 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 12 miles per gallon , and from town b to town c , the car averaged 16 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 ? | "ans is c given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 12 ) + ( x / 16 ) ) = 14.4 ( formula avg speed = total distance / total time )" | a = 16 * 10
b = 16 * 10
c = b / 2
d = a + c
e = 16 * 10
f = e / 12
g = 16 * 10
h = g / 2
i = h / 16
j = f + i
k = d / j
|
a ) 64 , b ) 68 , c ) 70 , d ) 79 , e ) 80 | a | multiply(const_2, divide(96, add(const_2, const_1))) | the greater of two numbers is twice the less , and the sum of the numbers is 96 . what are the numbers ? | let x = the less number 2 x = the greater number the sum of the numbers is 96 x + 2 x = 96 3 x = 96 x = 32 2 x = 64 option a | a = 2 + 1
b = 96 / a
c = 2 * b
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a ) 2 hours , b ) 2.1 hours , c ) 3 hours , d ) 3.5 hours , e ) 4.5 hours | b | inverse(add(divide(const_1, 24), add(divide(const_1, 3), divide(const_1, 9)))) | pipe p can fill a tank in 3 hours , pipe q in 9 hours and pipe r in 24 hours . if all the pipes are open , in how many hours will the tank be filled ? | "explanation : part filled by ( p + q + r ) in 1 hour = ( 1 / 3 + 1 / 9 + 1 / 24 ) = 35 / 72 all the three pipes together will fill the tank = 72 / 35 = 2.1 hours answer b" | a = 1 / 24
b = 1 / 3
c = 1 / 9
d = b + c
e = a + d
f = 1/(e)
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a ) 2 hours , b ) 1 hour , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | a | divide(8, divide(add(multiply(divide(1, 10), const_60), divide(2, 2)), const_2)) | a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 8 km in stationary water ? | "speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 2 + 6 ) = 4 kmhr time taken in stationary = 8 / 4 = 2 hrs answer : a" | a = 1 / 10
b = a * const_60
c = 2 / 2
d = b + c
e = d / 2
f = 8 / e
|
a ) 55 / 7 , b ) 1 , c ) 135 / 32 , d ) 222 / 37 , e ) 252 / 58 | c | inverse(add(divide(subtract(divide(add(inverse(add(add(const_4, const_1), const_4)), multiply(inverse(add(const_4, const_1)), const_2)), const_2), inverse(add(const_4, const_1))), divide(const_3, const_2)), inverse(add(const_4, const_1)))) | one woman and one man can build a wall together in five hours , but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together , it would take them nine hours to build the wall . assuming that rates for men , women and girls remain constant , how many hours would it take one woman , one man , and one girl , working together , to build the wall ? | solution : let work done by man , women and girl per hour be m , w , g respectively . then , m + w = 1 / 5 - - > ( 1 ) , w + 2 g = 1 / 5 - - > ( 2 ) and m + g = 1 / 9 - - > ( 3 ) . no . of hours it would take forone woman , one man , and one girl , working together , to build the wall , n = 1 / m + w + g from ( 1 ) and ( 2 ) , m = 2 g and from ( 3 ) g = 1 / 27 , m = 2 / 27 and w = 17 / 135 . so , n = 1 / ( 32 / 135 ) = 135 / 32 option , c | a = 4 + 1
b = a + 4
c = 1/(b)
d = 4 + 1
e = 1/(d)
f = e * 2
g = c + f
h = g / 2
i = 4 + 1
j = 1/(i)
k = h - j
l = 3 / 2
m = k / l
n = 4 + 1
o = 1/(n)
p = m + o
q = 1/(p)
|
a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1400 , e ) none of these | d | multiply(35, 40) | a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 35 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 β 35 = rs . 170 cp on 40 metres = 170 x 40 = rs . 6800 profit earned on 40 metres cloth = rs . 8200 β rs . 6800 = rs . 1400 . answer : option d" | a = 35 * 40
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a ) 2 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | a | multiply(divide(divide(multiply(25, 25), const_100), 36), const_12) | a reduction of 25 % in the price of apples would enable a man to obtain 36 more for rs . 25 , what is reduced price per dozen ? | "a 25 * ( 25 / 100 ) = 6 - - - 36 ? - - - 12 = > rs . 2" | a = 25 * 25
b = a / 100
c = b / 36
d = c * 12
|
a ) 8 : 7 , b ) 7 : 8 , c ) 8 : 6 , d ) 6 : 7 , e ) 6 : 8 | a | divide(add(4, add(4, divide(multiply(6, 4), subtract(7, 6)))), add(4, divide(multiply(6, 4), subtract(7, 6)))) | the ratio between the present age of p and q is 6 : 7 . if q is 4 years old than p , what will be the ratio of the ages of p and q after 4 years ? | let p and q ' s age is 6 x and 7 x years respectively . 7 x - 6 x = 4 x = 4 required ratio = ( 6 x + 4 ) : ( 7 x + 4 ) = 28 : 32 = 7 : 8 option a is answer | a = 6 * 4
b = 7 - 6
c = a / b
d = 4 + c
e = 4 + d
f = 6 * 4
g = 7 - 6
h = f / g
i = 4 + h
j = e / i
|
a ) 100 , b ) 110 , c ) 120 , d ) 105 , e ) 115 | c | subtract(100, subtract(power(2, divide(100, 60)), const_1)) | what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 60 ? | "put in simple words , we need to find the highest power of 2 in 100 ^ 60 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 60 = ( 2 ^ 2 * 5 ^ 2 ) ^ 60 = 2 ^ 120 * 5 ^ 120 answer : c" | a = 100 / 60
b = 2 ** a
c = b - 1
d = 100 - c
|
a ) 15 , b ) 45 , c ) 75 , d ) 125 , e ) 150 | a | add(add(add(add(add(multiply(multiply(5, 2), 3), multiply(multiply(5, 2), 3)), multiply(multiply(5, 2), 3)), 2), const_4), const_4) | if both 5 ^ 3 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ^ 2 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) has two appearances of 3 ( in 12 ^ 2 ) and two appearances of 5 ( in 10 ^ 2 ) . thus n must include at least 3 * 5 = 15 the answer is a ." | a = 5 * 2
b = a * 3
c = 5 * 2
d = c * 3
e = b + d
f = 5 * 2
g = f * 3
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | divide(add(8, 2), const_2) | in one hour , a boat goes 8 km along the stream and 2 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 8 + 2 ) kmph . = 5 kmph . answer b" | a = 8 + 2
b = a / 2
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a ) 187 , b ) 124 , c ) 172 , d ) 144 , e ) 175 | d | subtract(subtract(200, divide(multiply(200, 20), const_100)), divide(multiply(subtract(200, divide(multiply(200, 20), const_100)), 10), const_100)) | the sale price sarees listed for rs . 200 after successive discount is 20 % and 10 % is ? | "200 * ( 80 / 100 ) * ( 90 / 100 ) = 144 answer : d" | a = 200 * 20
b = a / 100
c = 200 - b
d = 200 * 20
e = d / 100
f = 200 - e
g = f * 10
h = g / 100
i = c - h
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a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | e | multiply(divide(7, subtract(9, 7)), 20) | sandy is younger than molly by 20 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 20 x = 10 sandy is 70 years old . the answer is e ." | a = 9 - 7
b = 7 / a
c = b * 20
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a ) 50 , b ) 10 , c ) 15 , d ) 25 , e ) 20 | e | divide(100, 5) | 5 times a number equals one 100 . the number is | since five times the number equals one hundred , then the number equals one hundred divided by five . therefore the number is 100 / 5 = 20 . correct answer e | a = 100 / 5
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a ) s 222 , b ) s 246 , c ) s 220 , d ) s 210 , e ) s 217 | b | add(add(200, divide(500, 100)), multiply(divide(20, 100), add(200, divide(500, 100)))) | the manufacturing cost of a shoe is rs . 200 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 200 + ( 500 / 100 ) = 205 . gain = 20 % = > sp = 1.2 cp = 1.2 x 205 = 246 answer : b" | a = 500 / 100
b = 200 + a
c = 20 / 100
d = 500 / 100
e = 200 + d
f = c * e
g = b + f
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a ) 824 , b ) 824.36 , c ) 824.32 , d ) 824.24 , e ) 824.26 | c | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 10,000), 10,000) | find the compound interest on $ 10,000 in 2 years at 4 % per annum , the interest being compounded half - yearly | "principal = $ 10000 ; rate = 2 % per half - year ; time = 2 years = 4 half - years . amount = $ [ 10000 * ( 1 + ( 2 / 100 ) ) 4 ] = $ ( 10000 * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) ) = $ 10824.32 . : . c . i . = $ ( 10824.32 - 10000 ) = $ 824.32 . answer c ." | a = 4 / 100
b = a / 2
c = b + 1
d = c ** 4
e = d * 10
f = e - 10
|
a ) 7 % , b ) 3 % , c ) 5 % , d ) 8 % , e ) 2 % | b | divide(const_100, 30) | in how many years will a sum of money doubles itself at 30 % per annum on simple interest ? | "p = ( p * 30 * r ) / 100 r = 3 % answer : b" | a = 100 / 30
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a ) rs . 186 , b ) rs . 258 , c ) rs . 486 , d ) rs . 586 , e ) none of these | a | multiply(divide(25, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 25 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) Γ£ β h ] + ( l Γ£ β b ) = [ 2 ( 25 + 12 ) Γ£ β 6 ] + ( 25 Γ£ β 12 ) = 744 sq m cost of plastering = 744 Γ£ β ( 25 / 100 ) = rs . 186 answer : a" | a = 25 / 100
b = 25 * 12
c = 25 * 6
d = 2 * c
e = 12 * 6
f = e * 2
g = d + f
h = b + g
i = a * h
|
a ) $ 144 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | multiply(2, divide(multiply(36, divide(2, 3)), divide(const_1, const_3))) | a collection of books went on sale , and 2 / 3 of them were sold for $ 2.00 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 2.00 ) or $ 144 . the best answer is therefore a . answer : a . | a = 2 / 3
b = 36 * a
c = 1 / 3
d = b / c
e = 2 * d
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a ) 25 , b ) 15 , c ) 30 , d ) 45 , e ) 20 | c | divide(multiply(30, 20), subtract(50, 30)) | in some quantity of ghee , 50 % is pure ghee and 50 % is vanaspati . if 20 kg of pure ghee is added , then the strength of vanaspati ghee becomes 30 % . the original quantity was ? | "let the original quantity be x then , vanaspati ghee in xkg = 50 x / 100 kg = x / 2 kg ( x / 2 ) / ( x + 20 ) = 30 / 100 x / ( 2 x + 40 ) = 3 / 10 x = 30 answer is c" | a = 30 * 20
b = 50 - 30
c = a / b
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a ) 38 . , b ) 40 . , c ) 42 . , d ) 48 . , e ) 46 . | d | subtract(multiply(sqrt(divide(1024, 4)), 4), sqrt(divide(1024, 4))) | the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 1024 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 1024 x ^ 2 = 256 x = 16 length = 4 * 16 = 64 difference = 64 - 16 = 48 d is the answer" | a = 1024 / 4
b = math.sqrt(a)
c = b * 4
d = 1024 / 4
e = math.sqrt(d)
f = c - e
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a ) 52.5 , b ) 50 , c ) 62.3 , d ) 62.1 , e ) 62.4 | b | divide(multiply(add(38, divide(multiply(38, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 38 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 38 sp = 38 * ( 125 / 100 ) = 38 mp * ( 95 / 100 ) = 47.5 mp = 50 answer : b" | a = 38 * 25
b = a / 100
c = 38 + b
d = c * 100
e = 100 - 5
f = d / e
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a ) 20 % , b ) 25 % , c ) 18 % , d ) 33 1 / 3 % , e ) none of these | d | multiply(divide(divide(6, const_100), divide(18, const_100)), const_100) | if the given two numbers are respectively 6 % and 18 % of a third number , then what percentage is the first of the second ? | "here , l = 6 and m = 18 therefore , first number = l / m x 100 % of second number = 6 / 18 x 100 % of second number = 33 1 / 3 % of second number answer : d" | a = 6 / 100
b = 18 / 100
c = a / b
d = c * 100
|
a ) rs . 125 , b ) rs . 80 , c ) rs . 99 , d ) rs . 109 , e ) rs . 101 | b | multiply(divide(const_100, 5), 4) | a 4 % stock yield 5 % . the market value of the stock is ? | "for an income of rs . 5 , investment = rs . 100 for an income of rs . 4 , investment = rs . ( 100 / 5 ) x 4 = rs . 80 answer : b" | a = 100 / 5
b = a * 4
|
a ) 48.29 , b ) 50 , c ) 26.8 , d ) 45.29 , e ) 30 | a | divide(add(288, 101), divide(multiply(29, const_1000), const_3600)) | a train is 288 meter long is running at a speed of 29 km / hour . in what time will it pass a bridge of 101 meter length ? | "speed = 29 km / hr = 29 * ( 5 / 18 ) m / sec = 145 / 18 m / sec total distance = 288 + 101 = 389 meter time = distance / speed = 389 * ( 18 / 145 ) = 48.29 seconds . answer : a" | a = 288 + 101
b = 29 * 1000
c = b / 3600
d = a / c
|
a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100) | rani bought more apples than oranges . she sells apples at βΉ 23 apiece and makes 15 % profit . she sells oranges at βΉ 10 apiece and makes 25 % profit . if she gets βΉ 653 after selling all the apples and oranges , find her profit percentage c . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % c = ( 97 / 556 ) * 100 = 17.4 % answer : b" | a = 2 * 10
b = 2 * 10
c = b + 1
d = a * c
e = 25 / 100
f = e + 1
g = 10 / f
h = 15 + 2
i = g * h
j = d + i
k = 653 - j
l = 2 * 10
m = 2 * 10
n = m + 1
o = l * n
p = 25 / 100
q = p + 1
r = 10 / q
s = 15 + 2
t = r * s
u = o + t
v = k / u
w = v * 100
|
a ) 50.4 % , b ) 63.4 % , c ) 66.5 % , d ) 34.9 % , e ) 43.5 % | b | multiply(divide(subtract(1100, add(add(add(75, 50), add(64, 78)), 135)), 1100), const_100) | john had a stock of 1100 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 1100 - n = 1100 - 402 = 698 percentage books not sold / total number of books = 698 / 1100 = 0.634 = 63.4 % correct answer b" | a = 75 + 50
b = 64 + 78
c = a + b
d = c + 135
e = 1100 - d
f = e / 1100
g = f * 100
|
a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 , e ) 12 | a | min(divide(108, const_3), divide(108, 6)) | in the third grade of windblown school there are 108 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "one third of them failed the math test = 1 / 3 of 108 = 36 . passed in math test = 82 . and 1 / 6 failed that literature test = 18 . it is a possible scenario that these 18 ( failed in literature ) were the ones who passed the math test . so in this case , no one failed in both . hence a ." | a = 108 / 3
b = 108 / 6
c = min(a)
|
a ) β 48 , b ) β 2 , c ) 4 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 21), subtract(4, 21)), const_1), const_1), const_1), const_1) | if a ( a - 4 ) = 21 and b ( b - 4 ) = 21 , where a β b , then a + b = | "a ( a - 4 ) = 21 and b ( b - 4 ) = 21 = > a , b must be integers and if a is - 3 or 7 , b will be 7 and - 3 respectively = > a + b = 4 ans : c" | a = 4 + 21
b = 4 - 21
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
['a ) 50.25 m', 'b ) 50.5 m', 'c ) 50.65 m', 'd ) 50.85', 'e ) none'] | b | sqrt(2550.25) | a housing society has been allotted a square piece of land measuring 2550.25 sq . m . what is the side of the plot ? | solution side = β 2550.25 = β 255025 / 100 = 505 / 10 = 50.5 m . answer b | a = math.sqrt(2550)
|
a ) 6.5 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | a | multiply(divide(subtract(add(10, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(10, add(const_0_25, const_0_25)), 5))), const_100) | in a certificate by mistake a candidate gave his height as 10 % more than actual height . in the interview panel , he clarified that his height was 5 feet 5 nches . find the % correction made by the candidate from his stated height to his actual height ? | "his height was = 5 feet 5 inch = 5 + 60 = 65 inch . required % correction = 65 * ( 1.10 - 1 ) = 6.5 a" | a = const_0_25 + const_0_25
b = 10 + a
c = b - 5
d = const_0_25 + const_0_25
e = 10 + d
f = e - 5
g = 100 + f
h = c / g
i = h * 100
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1064 , d ) data inadequate , e ) none of these | c | add(900, divide(multiply(multiply(900, add(divide(multiply(subtract(956, 900), const_100), multiply(900, 3)), 4)), 3), const_100)) | rs . 900 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 900 become in 3 years ? | "solution s . i . = rs . ( 956 - 900 ) = rs . 56 rate = ( 100 x 56 / 900 x 3 ) = 56 / 27 % new rate = ( 56 / 27 + 4 ) % = 164 / 27 % new s . i . = rs . ( 900 x 164 / 27 x 3 / 100 ) rs . 164 . β΄ new amount = rs . ( 900 + 164 ) = rs . 1064 . answer c" | a = 956 - 900
b = a * 100
c = 900 * 3
d = b / c
e = d + 4
f = 900 * e
g = f * 3
h = g / 100
i = 900 + h
|
a ) 55 , b ) 20 , c ) 96 , d ) 90 , e ) 40 | d | multiply(divide(const_60, 20), 30) | if the population of a certain country increases at the rate of one person every 20 seconds , by how many persons does the population increase in 30 minutes ? | answer = 3 * 30 = 90 answer = d | a = const_60 / 20
b = a * 30
|
a ) 1.17 kmph , b ) 1.67 kmph , c ) 0.69 kmph , d ) 1.45 kmph , e ) none of these | b | divide(multiply(50, 2), add(add(10, 2), 50)) | jake took a bus from home to market , that travels at 10 kmph . while walking back at 2 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 50 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 10 , by walking = x / 2 , by cycling = x / 50 hours : . average speed = total distance / total time = 2 x / x / 5 + x / 2 + x / 50 = 2 * 60 / 12 + 30 + 1.2 = 1.67 answer : b | a = 50 * 2
b = 10 + 2
c = b + 50
d = a / c
|
a ) $ 27,500 , b ) $ 35,000 , c ) $ 37,500 , d ) $ 42,500 , e ) $ 50,000 | d | divide(subtract(multiply(45000, 50), multiply(10, 55000)), 40) | a survey reveals that the average income of a company β s customers is $ 45000 per year . if 50 customers respond to the survey and the average income of the wealthiest 10 of those customers is $ 55000 , what is the average income of the other 40 customers ? is there a way to solve this using weighted average concept instead of doing tedious calculations ? | let x be the average of 40 customers 40 * x + 10 * 55000 = 50 * 45000 solving this we have x = 42500 answer is d . | a = 45000 * 50
b = 10 * 55000
c = a - b
d = c / 40
|
a ) 28 , b ) 31 , c ) 33 , d ) 35 , e ) 38 | c | divide(363, 11) | hcf of two numbers is 11 and the product of these numbers is 363 , what is the the greater number ? | the numbers be 11 a and 11 b 11 a Γ 11 b = 363 ab = 3 co - primes with product 3 are ( 1 , 3 ) hence the numbers with hcf 11 and product 363 = ( 11 Γ 1 , 11 Γ 3 ) = ( 11 , 33 ) numbers are 11 and 33 the greater number = 33 | a = 363 / 11
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | a | multiply(4, 2) | if n divided by 7 has a remainder of 2 , what is the remainder when 4 times n is divided by 7 ? | "as per question = > n = 7 p + 2 for some integer p hence 4 n = > 28 q + 8 = > remainder = > 6 for some integer q alternatively = > n = 2 > 3 n = > 8 = > 8 divided by 7 will leave a remainder 1 hence a" | a = 4 * 2
|
a ) 1344 , b ) 3360 , c ) 8400 , d ) 5040 , e ) 67200 | d | multiply(multiply(560, divide(12, 8)), 6) | running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 12 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 12 machine produce 560 * 6 * ( 12 / 8 ) in 6 minutes 560 * 6 * 12 / 8 = 5040 answer is d ." | a = 12 / 8
b = 560 * a
c = b * 6
|
a ) 4 , b ) 5 , c ) 8 , d ) 10 , e ) 12 | c | multiply(divide(multiply(8, 8), 8), divide(5, 5)) | if 8 workers can build 8 cars in 8 days , then how many days would it take 5 workers to build 5 cars ? | "8 workers can build 1 car per day on average . 1 worker can build 1 / 8 of a car per day . 5 workers can build 5 / 8 car per day . the time required to build 5 cars is 5 / ( 5 / 8 ) = 8 days the answer is c ." | a = 8 * 8
b = a / 8
c = 5 / 5
d = b * c
|
a ) 120 , b ) 256 , c ) 343 , d ) 441 , e ) 512 | d | add(add(add(add(add(multiply(multiply(7, 2), 2), multiply(multiply(7, 2), 2)), multiply(multiply(7, 2), 2)), 2), const_4), const_4) | if both 7 ^ 2 and 3 ^ 4 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 ^ 2 * 7 ^ 2 = 441 the answer is d ." | a = 7 * 2
b = a * 2
c = 7 * 2
d = c * 2
e = b + d
f = 7 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 648 , b ) 350 , c ) 252 , d ) 225 , e ) 26 | b | add(add(divide(subtract(1400, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | how many times digit 6 is used while writing numbers from 100 to 1400 ? | "there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' so far we have calculated 190 . the total now comes to 280 . the nearest to which is 350 . hence b ." | a = 1400 - 100
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 6 * 2
h = f + g
|
a ) 300 . , b ) 412 . , c ) 432 . , d ) 502 . , e ) 522 . | a | multiply(divide(multiply(49, 6), subtract(49, 25)), 25) | a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 25 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "pastry is divided in 25 customers equally . so , total number of pastry must be a multiple of 25 only option a satisfies the condition , and hence is the answer" | a = 49 * 6
b = 49 - 25
c = a / b
d = c * 25
|
a ) 131 , b ) 176 , c ) 274 , d ) 405 , e ) 495 | c | add(add(add(add(add(add(add(40, const_1), add(add(40, const_1), const_1)), add(add(add(40, const_1), const_1), const_2)), add(add(add(add(40, const_1), const_1), const_2), const_1)), add(add(add(add(add(40, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(40, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(40, const_1), const_1), const_2), const_1), const_1), const_1), const_1)) | the sum of the non - prime numbers between 40 and 50 , non - inclusive , is | "the sum of the non - prime numbers between 40 and 50 , non - inclusive , are { 42 , 44 , 45 , 46 , 48 , 49 } . 6 times 40 something is approx north of 240 hence the answer is ( c ) 274" | a = 40 + 1
b = 40 + 1
c = b + 1
d = a + c
e = 40 + 1
f = e + 1
g = f + 2
h = d + g
i = 40 + 1
j = i + 1
k = j + 2
l = k + 1
m = h + l
n = 40 + 1
o = n + 1
p = o + 2
q = p + 1
r = q + 1
s = m + r
t = 40 + 1
u = t + 1
v = u + 2
w = v + 1
x = w + 1
y = x + 1
z = s + y
A = 40 + 1
B = A + 1
C = B + 2
D = C + 1
E = D + 1
F = E + 1
G = F + 1
H = z + G
|
a ) 2 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these | c | multiply(divide(divide(multiply(subtract(const_100, 90), 20), const_100), subtract(const_100, 20)), const_100) | fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 20 kg of fresh grapes ? | "the weight of non - water in 20 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 20 β 0.1 = x β 0.8 - - > x = 2.5 answer : c ." | a = 100 - 90
b = a * 20
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
|
a ) 10 , b ) 30 , c ) 36 , d ) 60 , e ) 90 | c | divide(multiply(9, subtract(9, const_1)), const_2) | there are 9 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "9 players are there . two players play one game with one another . so 9 c 2 = 9 x 8 / 2 = 36 so option c is correct" | a = 9 - 1
b = 9 * a
c = b / 2
|
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