options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(subtract(multiply(24, add(4, 2)), add(multiply(24, 4), multiply(10, 4))), 2), 2) | 10 years ago , the average age of a family of 4 members was 24 years . two children having been born ( with age diference of 2 years ) , the present average age of the family is the same . the present age of the youngest child is : | "explanation : total age of 4 members , 10 years ago = ( 24 x 4 ) years = 96 years . total age of 4 members now = [ 96 + ( 10 x 4 ) ] years = 136 years . total age of 6 members now = ( 24 x 6 ) years = 144 years . sum of the ages of 2 children = ( 144 - 136 ) years = 8 years . let the age of the younger child be x years . then , age of the elder child = ( x + 2 ) years . so , x + ( x + 2 ) = 8 ⇔ x = 3 age of younger child = 3 years . answer : c" | a = 4 + 2
b = 24 * a
c = 24 * 4
d = 10 * 4
e = c + d
f = b - e
g = f - 2
h = g / 2
|
a ) 40 / 3 , b ) 11 , c ) 7 , d ) 7 / 3 , e ) 9 / 7 | c | divide(multiply(70, 3), 30) | if it takes 70 workers 3 hours to disassemble the exhibition rides at a small amusement park , how many hours would it take 30 workers to do this same job ? | 70 workers = 3 hours then , 1 worker = 3 * 70 hours 30 workers = ( 3 * 70 ) / ( 30 ) = 7 answer : c | a = 70 * 3
b = a / 30
|
a ) 45 % , b ) 46.67 % , c ) 600 / 11 , d ) 55 % , e ) 35 % | b | multiply(divide(subtract(150, add(multiply(5, 10), multiply(10, 5))), 150), const_100) | a batsman scored 150 runs which included 5 boundaries and 10 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 150 − ( 5 × 4 + 10 × 6 ) . = > 150 − 80 = > 70 hence , the required percentage is : - = > 70 / 150 * 100 = > 46.67 % answer : b" | a = 5 * 10
b = 10 * 5
c = a + b
d = 150 - c
e = d / 150
f = e * 100
|
a ) 233 , b ) 600 , c ) 510 , d ) 771 , e ) 191 | c | subtract(multiply(390, 9), subtract(multiply(430, 9), 870)) | the average monthly salary of 8 workers and one supervisor in a factory was 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . whenthesupervisor , whosesalarywas 430 . when the supervisor , whose salary was 870 per month , retired , a new person was appointed and then the average salary of 9 people was $ $ 390 per month . the salary of the new supervisor is : | "explanation : total salary of 8 workers and supervisor together = 9 ã — 430 = 3870 now total salary of 8 workers = 3870 â ˆ ’ 870 = 3000 total salary of 9 workers including the new supervisor = 9 ã — 390 = 3510 salary of the new supervisor = 3510 â ˆ ’ 3000 = 510 answer : c" | a = 390 * 9
b = 430 * 9
c = b - 870
d = a - c
|
a ) $ 100 , b ) $ 75 , c ) $ 20 , d ) $ 120 , e ) $ 80 | e | multiply(100, subtract(const_1, divide(divide(200, const_2), add(400, divide(200, const_2))))) | a invested $ 400 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find a shares ? | a : b = 400 * 12 : 200 * 6 a : b = 4 : 1 a ' s share = 100 * 4 / 5 = $ 80 answer is e | a = 200 / 2
b = 200 / 2
c = 400 + b
d = a / c
e = 1 - d
f = 100 * e
|
a ) 65 , b ) 38 , c ) 20 , d ) 28 , e ) 21 | b | subtract(subtract(150, 62), add(25, 25)) | two cars start from the opposite places of a main road , 150 km apart . first car runs for 25 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run only 62 km along the main road . what would be the distance between two cars at this point ? | answer : b ) 38 km | a = 150 - 62
b = 25 + 25
c = a - b
|
a ) 25 , b ) 30 , c ) 20 , d ) 16 , e ) 12 | c | multiply(subtract(subtract(const_1, divide(10, const_100)), divide(10, 50)), 50) | george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 10 % of the money for taxi fare and buys 10 mangoes . how many oranges can he buy ? | "let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 10 % of 200 = 20 for taxi fare , so he is left with 180 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 80 ( 180 - 100 ) no of oranges he can buy = 80 / 4 = > 20 so , george can buy 10 oranges . c" | a = 10 / 100
b = 1 - a
c = 10 / 50
d = b - c
e = d * 50
|
a ) 1 / 16 , b ) 2 / 15 , c ) 2 / 17 , d ) 1 / 8 , e ) 1 / 7 | c | divide(subtract(17, 15), 17) | a ’ s speed is 17 / 15 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 17 v / 15 ) = xd / v ( 15 / 17 ) * d / v = x * d / v x = 15 / 17 b should have a head start of 2 / 17 of the full distance . the answer is c ." | a = 17 - 15
b = a / 17
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(multiply(0.5, const_10), const_1) | a box contains 8 apples , 7 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ? | "p ( choosing a red apple 5 times in a row ) = 7 / 8 * 6 / 7 * 5 / 6 * 4 / 5 * 3 / 4 = 3 / 8 < 0.5 the answer is c ." | a = 0 * 5
b = a + 1
|
a ) 4436 toys , b ) 5487 toys , c ) 2000 toys , d ) 2354 toys , e ) 1375 toys | c | divide(8000, 4) | a factory produces 8000 toys per week . if the workers at this factory work 4 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 4 days ) by 4 . 8000 / 4 = 2000 toys correct answer c" | a = 8000 / 4
|
a ) 1 / 7 , b ) 1 / 12 , c ) 1 / 14 , d ) 1 / 18 , e ) 1 / 21 | c | divide(choose(6, 2), choose(10, 6)) | a store has 10 bottles of juice , including 6 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 2 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought . | the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 2 bottles of apple juice is 6 c 2 * 4 c 4 = 15 * 1 = 15 p ( selling 2 bottles of apple juice ) = 15 / 210 = 5 / 70 = 1 / 14 the answer is c . | a = math.comb(6, 2)
b = math.comb(10, 6)
c = a / b
|
a ) 6150 , b ) 6200 , c ) 6050 , d ) 6075 , e ) 5075 | d | multiply(subtract(divide(multiply(89, add(89, 1)), 2), multiply(divide(subtract(89, 1), 2), add(divide(subtract(89, 1), 2), 1))), 3) | if 1 + 2 + 3 + . . . + n = n ( n + 1 ) , then 3 ( 1 + 3 + 5 + . . . . + 89 ) = ? | "explanation : to solve this use the formula of ap , sn = ( n / 2 ) ( a + l ) . . . . . . . . . . . . . . . . ( 1 ) to find n , use = > tn = a + ( n - 1 ) d = > 89 = 1 + ( n - 1 ) 2 = > n = 45 use value of n in ( 1 ) then , sn = ( 45 / 2 ) ( 1 + 89 ) = 2025 ans : - 3 ( sn ) = 6075 answer : d" | a = 89 + 1
b = 89 * a
c = b / 2
d = 89 - 1
e = d / 2
f = 89 - 1
g = f / 2
h = g + 1
i = e * h
j = c - i
k = j * 3
|
a ) 12 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 10 sec | a | divide(120, multiply(36, const_0_2778)) | in what time will a railway train 120 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 120 / 36 * 18 / 5 = 12 sec answer : a" | a = 36 * const_0_2778
b = 120 / a
|
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent q of the distribution is less than m + d ? | "d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , q = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d ." | a = 100 - 68
b = a / 2
c = 100 - b
|
a ) rs . 15,000 , b ) rs . 15,500 , c ) rs . 15,600 , d ) rs . 28,875 , e ) none | d | multiply(1400, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 1400 per sq . metre . | "solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1400 x 20.625 ) = rs . 28875 . answer d" | a = 5 * 5
b = 1400 * a
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(multiply(multiply(const_2, const_3), 105), 105) | find the smallest positive integer n such that 11 n - 1 is divisible by 105 . | the last digit of 11 n − 1 is 0 , so 5 always divides 11 n − 1 . also , checking the remainders when the powers of 11 are divided by 3 and 7 we get that 3 divides 11 n − 1 exactly when n is even , and 7 divides 11 n − 1 exactly when 3 divides n . correct answer d | a = 2 * 3
b = a * 105
c = b / 105
|
a ) 2 , b ) 4 , c ) 8 , d ) 59 , e ) 32 | d | add(subtract(89, add(add(add(15, 10), 5), 3)), 3) | there are 89 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | "lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 89 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 89 = 15 + 10 + 5 + 3 + x + y + s we need to know total snakes = x + y + s + 3 = 59 answer : d" | a = 15 + 10
b = a + 5
c = b + 3
d = 89 - c
e = d + 3
|
a ) 1629 , b ) 1656 , c ) 1277 , d ) 6298 , e ) 1279 | b | subtract(2691, divide(multiply(multiply(3, 5), 2691), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2691 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2691 - x ) * 3 * 5 ) / 100 24 x / 100 = 40365 / 100 - 15 x / 100 39 x = 40365 = > x = 1035 second sum = 2691 â € “ 1035 = 1656 answer : b" | a = 3 * 5
b = a * 2691
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2691 - f
|
a ) $ 155 , b ) $ 225 , c ) $ 510 , d ) $ 850 , e ) $ 1250 | b | multiply(divide(multiply(2.16, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.50) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 2.16 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 160,000 / 4800 = 450 boxes total cost = 450 × $ 0.5 = $ 225 the answer is b ." | a = 1000 * 1000
b = 2 * 16
c = 20 * 20
d = c * 12
e = b / d
f = e * 0
|
a ) 8 days , b ) 10 days , c ) 15 days , d ) 11 3 / 7 days , e ) none of these | d | add(divide(subtract(const_1, add(multiply(5, divide(const_1, 15)), multiply(5, divide(const_1, 20)))), divide(const_1, 20)), 5) | a can do a piece of work in 15 days and b in 20 days . they began the work together but 5 days before the completion of the work , a leaves . the work was completed in ? | "explanation : ( x â € “ 5 ) / 15 + x / 20 = 1 x = 11 3 / 7 days answer is d" | a = 1 / 15
b = 5 * a
c = 1 / 20
d = 5 * c
e = b + d
f = 1 - e
g = 1 / 20
h = f / g
i = h + 5
|
a ) 23 , b ) 22 , c ) 15 , d ) 36 , e ) 48 | c | multiply(10, divide(const_3, const_2)) | a is half good a work man as b and together they finish a job in 10 days . in how many days working alone b finish the job ? | "c 15 wc = 1 : 2 2 x + x = 1 / 10 = > x = 1 / 30 2 x = 1 / 15 = > 15 days" | a = 3 / 2
b = 10 * a
|
a ) 35 m east , b ) 35 m north , c ) 30 m west , d ) 45 m east , e ) 55 m east | a | add(20, 15) | rohit walked 25 m towards south . then he turned to his left and walked 20 m . he then turned to his left and walked 25 m . he again turned to his right and walked 15 m . at what distance is he from the starting point and in which direction ? | ! 20 ! - - - > 15 ! ! ! - - - > ! 20 + 15 = 35 answer : a | a = 20 + 15
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a ) 90 , b ) 45 , c ) 110 , d ) 40 , e ) 50 | b | divide(multiply(10, divide(3, const_3.0)), subtract(divide(5, add(6, 3)), multiply(divide(3, add(6, 3)), divide(3, 6)))) | a mixture contains milk and water in the ratio 6 : 3 . on adding 10 litters of water , the ratio of milk to water becomes 6 : 5 . the quantity of milk in the original mixture is ? | "explanation : milk : water = 6 : 3 6 x : 3 x + 10 = 6 : 5 5 [ 6 x ] = 6 [ 3 x + 10 ] 30 x = 18 x + 60 30 x - 18 x = 60 12 x = 60 x = 5 the quantity of milk in the original mixture is = 6 : 3 = 6 + 3 = 9 9 x = 45 short cut method - 2 : for only milk problems milk : water 6 : 3 6 : 5 milk ratio same but water ratio 2 part incress per 10 liters 2 part of ratio - - - - - - - > 10 liters 9 part of ratio - - - - - - - > 45 liters answer : option b" | a = 3 / 3
b = 10 * a
c = 6 + 3
d = 5 / c
e = 6 + 3
f = 3 / e
g = 3 / 6
h = f * g
i = d - h
j = b / i
|
a ) 1 / 2 , b ) 3 / 5 , c ) 4 / 7 , d ) 5 / 7 , e ) 7 / 9 | d | divide(multiply(subtract(const_100, 10), 20), multiply(add(const_100, 20), 21)) | if the numerator of a fraction be increased by 20 % and its denominator be diminished by 10 % , the value of the fraction is 20 / 21 . find the original fraction ? | "let the original fraction be x / y then , 120 % of x / 90 % of y = 20 / 21 120 x / 90 y = 20 / 21 x / y = 5 / 7 answer is d" | a = 100 - 10
b = a * 20
c = 100 + 20
d = c * 21
e = b / d
|
a ) 1 / 5 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 4 , e ) 3 / 2 | a | multiply(divide(subtract(const_100, 80), add(const_100, 25)), divide(5, 4)) | the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 25 percent of a , and m equals b decreased by 80 percent of b , what is the value of m / x ? | "a / b = 4 / 5 m / x = ( 1 / 5 ) * 5 / ( 5 / 4 ) * 4 = 1 / 5 the answer is a ." | a = 100 - 80
b = 100 + 25
c = a / b
d = 5 / 4
e = c * d
|
a ) rs . 22,000 , b ) rs . 21,000 , c ) rs . 20,000 , d ) rs . 23,000 , e ) rs . 24,000 | c | divide(multiply(200, const_100), subtract(const_100, add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(10, const_100))))) | a man saves 10 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 200 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 90 savings = rs . 10 present expenditure 90 + 90 * ( 10 / 100 ) = rs . 99 present savings = 100 – 99 = rs . 1 if savings is rs . 1 , salary = rs . 100 if savings is rs . 200 , salary = 100 / 1 * 200 = 20000 answer : c" | a = 200 * 100
b = 100 - 10
c = 100 - 10
d = 10 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) $ 50000 , b ) $ 60000 , c ) $ 60500 , d ) $ 65000 , e ) can not be determined | e | log(subtract(divide(10, 2), 10)) | the median annual household income in a certain community of 21 households is $ 50000 . if the mean w income of a household increases by 10 % per year over the next 2 years , what will the median income in the community be in 2 years ? | answer is e , because there are different numbers in the set and we are not sure which side of the numbers in the set will be increased so the mean w is increase by 10 % . it could be the case that small number of higher end incomes increased a little or many low end incomes increased - it can not be identified . | a = 10 / 2
b = a - 10
c = math.log(b)
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(9, 8) | 4 different children have jelly beans : aaron has 5 , bianca has 7 , callie has 8 , and dante has 9 . how many jelly beans must dante give to aaron to ensure that no child has more than 1 fewer jelly beans than any other child ? | since bianca and callie are both within 1 jelly bean of each other and aaron has 5 , dante must provide 3 of his 9 jelly beans so each child has no more than 1 fewer jelly bean than any other child . dante + aaron = 9 + 5 = 14 / 2 = 7 9 - 8 = 1 so dante must provide 3 jelly beans to aaron . answer ( a ) | a = 9 - 8
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(divide(12, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 12 - meter platform and caught it after it had traveled 35.65 meters . how many times did the ball bounce ? | "ans : 6 division of total diatance travelled will be 12 + 12 + 6 + 3 + 1.5 + . 75 + 0.4 ans b" | a = 12 / 2
b = a / 2
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a ) 18 , b ) 27 , c ) 26 , d ) 16 , e ) 12 | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 20), divide(const_1, 30))), 10) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 10 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 10 days , a can do 1 / 2 of the project . thus a and b must complete 1 / 2 of the project , which takes 6 days . the total number of days is 6 + 10 = 16 . the answer is d ." | a = 1 / 30
b = a * 10
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 10
|
a ) $ 54000 , b ) $ 48000 , c ) $ 36520 , d ) $ 32450 , e ) $ 42500 | a | multiply(128000, power(subtract(const_1, divide(25, const_100)), 3)) | a present value of a machine is $ 128000 . its value depletiation rate is 25 % per annum then find the machine value after 3 years ? | "p = $ 128000 r = 25 % t = 3 years machine value after 3 years = p / ( 1 - r / 100 ) ^ t = 128000 * 3 / 4 * 3 / 4 * 3 / 4 = $ 54000 answer is a" | a = 25 / 100
b = 1 - a
c = b ** 3
d = 128000 * c
|
a ) 11 / 14 , b ) 13 / 18 , c ) 4 / 7 , d ) 3 / 7 , e ) 3 / 14 | a | divide(subtract(divide(2, add(1, 2)), subtract(divide(2, 9), multiply(divide(2, 9), divide(3, 4)))), divide(subtract(9, 2), 9)) | when 2 / 9 of the votes on a certain resolution have been counted , 3 / 4 of those counted are in favor of the resolution . what fraction f of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution ? | "if we use variable for total votes there will be too many fractions to manipulate with , so pick some smart # : let set total # of votes is 18 . 2 / 9 of the votes on a certain resolution have been counted - - > 4 counted and 18 - 4 = 14 votes left to be counted ; 3 / 4 of those counted are in favor of the resolution - - > 3 in favor and 1 against ; ratio of those who voted against to those who voted for to be 2 to 1 there should be total of 18 * 2 / 3 = 12 people who voted against , so in the remaining 14 votes there should be 12 - 1 = 11 people who voted against . thus f = 11 / 14 of the remaining votes must be against . answer : a ." | a = 1 + 2
b = 2 / a
c = 2 / 9
d = 2 / 9
e = 3 / 4
f = d * e
g = c - f
h = b - g
i = 9 - 2
j = i / 9
k = h / j
|
a ) 50 , b ) 200 , c ) 380 , d ) 798 , e ) 400 | d | subtract(divide(8000, 10), divide(8000, multiply(20, const_100))) | when 1 / 20 % of 8000 is subtracted from 1 / 10 of 8000 , the difference is | 1 / 20 % of 8000 = 4 1 / 10 of 8000 = 800 800 - 2 = 798 ans : d | a = 8000 / 10
b = 20 * 100
c = 8000 / b
d = a - c
|
a ) 1 , b ) 3 , c ) 13 , d ) 5 , e ) 14 | e | subtract(divide(const_100.0, const_2), multiply(47, 47)) | what is the remainder when 47 * 50 is divided by 8 ? | "we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 47 * 50 ) / 8 } } = remainder of ( 47 / 8 ) * remainder of ( 50 / 8 ) = 7 * 2 = 14 answer : e" | a = 100 / 0
b = 47 * 47
c = a - b
|
a ) 43 , b ) 36 , c ) 86 , d ) 129 , e ) 11 | c | multiply(subtract(const_1, divide(14, 100)), const_100) | there are 100 students in a class . if 14 % are absent on a particular day , find the number of students present in the class . | number of students absent on a particular day = 14 % of 100 i . e . , 14 / 100 × 100 = 14 therefore , the number of students present = 100 - 14 = 86 students . answer : c | a = 14 / 100
b = 1 - a
c = b * 100
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 72 % | a | multiply(divide(72, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 72 ° of the circle , what percentage of megatek employees are in manufacturing ? | "72 ° divided by 360 ° equals 0.2 , therefore the sector is equal to 20 % of the total answer : a" | a = 3600 / 10
b = 72 / a
c = b * 100
|
a ) 33 , b ) 28 , c ) 27 , d ) 25 , e ) 23 | d | divide(add(24, 26), const_2) | if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 28 , 26 , 19 , 33 and 35 ? | "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 28 , 26 , 19 , 33 and 35 19 , 24 , 26 , 28 , 30 , 33 , 35 , 42 . . . numbers less than median are 19 , 24 , 26 , 28 . . the median of these numbers = center of 24 and 26 = 25 d" | a = 24 + 26
b = a / 2
|
a ) 12 kg , b ) 90 kg , c ) 72 kg , d ) 96 kg , e ) none | b | multiply(multiply(multiply(3, 2), divide(1.5, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1.5 cm when a man gets on it . the mass of man is | "solution volume of water displaced = ( 3 x 2 x 0.015 ) m 3 = 0.09 m 3 . mass of man = volume of water displaced × density of water = ( 0.09 × 1000 ) kg = 90 kg . answer b" | a = 3 * 2
b = 1 / 5
c = a * b
d = c * 1000
|
a ) 180 , b ) 98 , c ) 27 , d ) 21 , e ) 22 | a | add(multiply(multiply(1, 6), const_100), multiply(3, 6)) | three numbers are in the ratio 1 : 3 : 6 and their average is 100 . the largest number is : | "explanation : let the numbers be x , 3 x and 6 x , then , ( x + 3 x + 6 x ) / 3 = 100 = > 10 x = 100 * 3 = > x = 30 largest number 6 x = 6 * 30 = 180 answer : a" | a = 1 * 6
b = a * 100
c = 3 * 6
d = b + c
|
a ) 128 , b ) 142 , c ) 143 , d ) 141 , e ) 129 | e | divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7) | how many 3 digit positive integers r exist that when divided by 7 leave a remainder of 5 ? | "minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e ." | a = 100 * 10
b = a - 1
c = 10 + 4
d = c * 7
e = d + 5
f = b - e
g = f / 7
|
a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 8 , e ) 8 / 7 | b | divide(divide(divide(20, const_100), divide(30, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(30, const_100))) | a total of 30 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 20 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ] | "let there be g geese , of which 0.3 g are male and 0.7 g are female lets say m geese migrated , 0.2 m are male and 0.8 m are female migration rate male = 0.2 m / 0.3 g = ( 2 / 3 ) * ( m / g ) migration rate female = 0.8 m / 0.7 g = ( 8 / 7 ) * ( m / g ) ratio of migration rates = ( 2 / 3 ) / ( 8 / 7 ) = 7 / 12 answer is ( b )" | a = 20 / 100
b = 30 / 100
c = a / b
d = 2 * 4
e = d * 10
f = e / 100
g = 30 / 100
h = f / g
i = c / h
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | d | divide(subtract(add(88, 93), multiply(51, const_2)), subtract(51, 48)) | the average weight of a group of persons increased from 48 kg to 51 kg , when two persons weighing 88 kg and 93 kg join the group . find the initial number of members in the group ? | "let the initial number of members in the group be n . initial total weight of all the members in the group = n ( 48 ) from the data , 48 n + 88 + 93 = 51 ( n + 2 ) = > 51 n - 48 n = 79 = > n = 26 therefore there were 26 members in the group initially . answer : d" | a = 88 + 93
b = 51 * 2
c = a - b
d = 51 - 48
e = c / d
|
a ) 60 % , b ) 50 % , c ) 42 % , d ) 25 % , e ) 10 % | c | multiply(divide(60, 25), const_100) | what percent of 60 is 25 ? | "the phrasing of the question is difficult for some students . ` ` what percent of 60 is 25 ? ' ' is the same question as ` ` 25 is what percent of 60 ? ' ' many students find this later way of phrasing the question easier to work with . if x is y percent of z , then x / z = y / 100 ; for example , 10 is 50 % of 20 ; x = 10 , y = 50 , and z = 20 similarly , if you wanted to know what percent of 100 is 50 , you would intuitively know that it is 50 % since it is 50 / 100 or 50 per cent ( literally , per 100 ) . applying this logic to the problem at hand : 25 / 60 = . 42 = 42 % thus , 25 is 42 % of 60 if this way of solving the problem is difficult to conceptualize , consider another approach . it should be clear that 30 is 50 % ( or 1 / 2 ) or 60 . since 25 is less than 30 , 25 must be less than 50 % of 60 . this means that any answer that is not less than 50 % is wrong . since 25 % is 1 / 4 and 1 / 4 of 60 is 15 ( since 15 * 4 = 60 ) , 25 % is too small . by process of elimination , the answer is 42 % answer c" | a = 60 / 25
b = a * 100
|
a ) 7.19 , b ) 6.8 , c ) 7.2 , d ) 7.15 , e ) 7.11 | b | divide(add(121, 153), multiply(add(80, 65), const_0_2778)) | two trains 121 meters and 153 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 121 + 153 ) / ( 80 + 65 ) * 18 / 5 t = 6.8 answer : b" | a = 121 + 153
b = 80 + 65
c = b * const_0_2778
d = a / c
|
a ) 10 , b ) 12.6 , c ) 22.5 , d ) 31.3 , e ) 37.5 | e | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(15, 4), subtract(15, const_2))), 4) | find the average of first 4 multiples of 15 ? | "average = ( 15 + 30 + 45 + 60 ) / 4 = 37.5 answer is e" | a = 4 + 1
b = 4 + 1
c = b + 2
d = a + c
e = 15 - 4
f = 15 - 2
g = e + f
h = d + g
i = h / 4
|
a ) 500 , b ) 1000 , c ) 1200 , d ) 1300 , e ) 1500 | a | multiply(const_100, 5) | the price of 3 chudi and 6 tops is rs . 1500 . with the same money one can buy 1 chudi and 12 tops . if one wants to buy 5 tops , how much shall she have to pay ? | let the price of a chudi and a top be rs . x and rs . y respectively . then , 3 x + 6 y = 1500 . . . . ( i ) and x + 12 y = 1500 . . . . ( ii ) divide equation ( i ) by 3 , we get the below equation . = x + 2 y = 500 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 12 y = 1500 ( - ) x + 2 y = 500 - - - - - - - - - - - - - - - - 10 y = 1000 - - - - - - - - - - - - - - - - cost of 5 tops = 5 * 100 = 500 answer : a | a = 100 * 5
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a ) 17 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | a | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 10), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 10), const_100))), const_100) | the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 10 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 10 % then output equals 121 to restore new holidy demand output to original 100 . final - initial / final * 100 = 21 / 121 * 100 = 8 / 33 * 100 = 17 % approxiamately . option a is correct ." | a = 100 + 10
b = a / 100
c = 100 + 10
d = c / 100
e = b * d
f = e - 1
g = 100 + 10
h = g / 100
i = 100 + 10
j = i / 100
k = h * j
l = f / k
m = l * 100
|
a ) 2 / 15 , b ) 2 / 21 , c ) 3 / 55 , d ) 3 / 29 , e ) 4 / 27 | c | divide(choose(3, 2), choose(add(add(3, 4), 4), 2)) | a bag contains 3 red , 4 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 3 c 211 c 2 = 3 c 211 c 2 = 3 / 55 c" | a = math.comb(3, 2)
b = 3 + 4
c = b + 4
d = math.comb(c, 2)
e = a / d
|
a ) 6 : 06 , b ) 6 : 36 , c ) 7 : 06 , d ) 7 : 36 , e ) 8 : 06 | c | divide(add(5, divide(multiply(add(subtract(5, 4), divide(subtract(06, 30), const_60)), 30), subtract(39, 30))), 06) | tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 39 pages per hour . if tom starts reading a novel at 4 : 30 , and jan begins reading an identical copy of the same book at 5 : 06 , at what time will they be reading the same page ? | "since tom reads an average of 1 page every 2 minutes , tom will read 18 pages in the first 36 minutes . jan can catch tom at a rate of 9 pages per hour , so it will take 2 hours to catch tom . the answer is c ." | a = 5 - 4
b = 6 - 30
c = b / const_60
d = a + c
e = d * 30
f = 39 - 30
g = e / f
h = 5 + g
i = h / 6
|
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 21 | d | subtract(24, multiply(const_2, divide(divide(24, const_2), add(const_1, const_4)))) | let f ( x , y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = 24 , what is the maximum value of y for which f ( x , y ) = 0 ? | "the question is finding y such that ( 24 - y ) ! is a multiple of 24 . that means we need to have 2 ^ 3 * 3 in ( 24 - y ) ! 4 ! is the smallest factorial number with 2 ^ 3 * 3 as a factor . 24 - y = 4 y = 20 the answer is d ." | a = 24 / 2
b = 1 + 4
c = a / b
d = 2 * c
e = 24 - d
|
a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 15.7 . , e ) 16.4 . | d | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.5) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.5 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.5 = 6 * 0 + 2 * 7.2 + 8 - 6.5 = 0 + 14.4 + 1.5 15.7 answer d" | a = 2 * 7
b = 4 / 10
c = 8 - 4
d = a + c
e = d - 6
|
a ) 6 , b ) 7 , c ) 6.5 , d ) 6 2 / 3 , e ) 6 1 / 4 | d | inverse(add(divide(const_1, 15), divide(const_1, 12))) | 15 girls can complete a rangoli in 12 days , and 10 boys can complete the same work in 15 days . to complete the work earlier 10 boys and 15 girls are employed , in how many days will this work get completed ? | work done by 15 girls per day = 1 / 12 work done by 10 boys per day = 1 / 15 work done by 15 girls and 10 boys per day = 1 / 12 + 1 / 15 = 3 / 20 = 6 2 / 3 10 boys and 15 girls will complete the work in = 6 2 / 3 days d | a = 1 / 15
b = 1 / 12
c = a + b
d = 1/(c)
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | b | floor(divide(100, 9)) | on dividing 100 by a number , the quotient is 9 and the remainder is 1 . find the divisor ? | "d = ( d - r ) / q = ( 100 - 1 ) / 9 = 99 / 9 = 11 b )" | a = 100 / 9
b = math.floor(a)
|
a ) 1 / 3 , b ) 7 / 12 , c ) 5 / 12 , d ) 2 / 3 , e ) 1 / 2 | c | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 5)) | a can do a work in 15 days and b in 20 days . if they work on it together for 5 days , then the fraction of the work that is left is : | "ans is : c a ' s 1 day ' s work = 1 / 15 b ' s 1 day ' s work = 1 / 20 ( a + b ) ' s 1 day ' s work = ( 1 / 15 + 1 / 20 ) = 7 / 60 ( a + b ) ' s 5 day ' s work = ( 7 / 60 * 5 ) = 7 / 12 therefore , remaining work = ( 1 - 7 / 12 ) = 5 / 12 . answer : c" | a = 1 / 15
b = 1 / 20
c = a + b
d = c * 5
e = 1 - d
|
a ) 9 th minute , b ) 21 st minute , c ) 11 th minute , d ) 22 nd minute , e ) 13 th minute | a | add(multiply(multiply(const_4, 2), 2), 1) | a monkey ascends a greased pole 6 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 4 meters . i mean this will continue for first 4 * 2 = 8 mins . he would have reached 4 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 8 + 1 = 9 mins . so , asnwer will be a" | a = 4 * 2
b = a * 2
c = b + 1
|
a ) 1 , b ) 63 , c ) 66 , d ) 67 , e ) none of these | c | subtract(add(power(67, 67), 67), multiply(68, floor(divide(add(power(67, 67), 67), 68)))) | what will be the reminder when ( 67 ^ 67 + 67 ) is divided by 68 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 67 ^ 67 + 1 ) will be divisible by ( 67 + 1 ) ; ( 67 ^ 67 + 1 ) + 66 when divided by 68 will give 66 as remainder . correct option : c" | a = 67 ** 67
b = a + 67
c = 67 ** 67
d = c + 67
e = d / 68
f = math.floor(e)
g = 68 * f
h = b - g
|
a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | a | divide(subtract(multiply(140, divide(add(const_100, 50), const_100)), 10), add(divide(add(const_100, 50), const_100), const_1)) | if leo gains 10 pounds , he will weigh 50 % more than his sister kendra . currently their combined weight is 140 pounds . what is leo ' s current weight ? | "l + k = 140 and so k = 140 - l l + 10 = 1.5 k = 1.5 ( 140 - l ) 2.5 l = 200 l = 80 the answer is a ." | a = 100 + 50
b = a / 100
c = 140 * b
d = c - 10
e = 100 + 50
f = e / 100
g = f + 1
h = d / g
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(subtract(20, 6), multiply(add(divide(6, 2), 2), 2)) | aunt marge is giving candy to each of her nephews and nieces . she has 20 pieces of candy and she gives all the candy to the children according to her wish . if robert gets 2 more pieces than kate , bill gets 6 less than mary , mary gets 2 more pieces than robert , and kate gets 2 more pieces than bill , how many pieces of candy does kate get ? | use the initial letter of each name as a variable : r = k + 2 b = m − 6 m = r + 2 k = 2 + bk = = r − 2 b = k − 2 = ( r − 2 ) − 2 = r − 4 m = r + 2 r + b + m + k = 20 r + r − 4 + r + 2 + r − 2 = 20 r = 6 k = r − 2 = 6 − 2 = 4 answer : b | a = 20 - 6
b = 6 / 2
c = b + 2
d = c * 2
e = a - d
|
a ) 2 / 9 , b ) 1 / 3 , c ) 4 / 11 , d ) 2 / 3 , e ) 7 / 9 | c | divide(4, 11) | in the list 3 , 3 , 4 , 4 , 5 , 5 , 5 , 5 , 7 , 11 , 21 , what fraction of the data is less than the mode ? | mode : the mode of any set is the term which has the highest frequency ( occurrence ) highest frequent term in the set is 5 ( with frequency 4 ) hence mode = 5 two terms ( 3 , 3 , 4 , 4 ) out of a total of 11 terms are less than mode of the set . fraction of set that are less than mode of set = 4 / 11 answer : option c | a = 4 / 11
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a ) 104 , b ) 200 , c ) 625 , d ) 832 , e ) 833 | b | add(divide(subtract(16, 3200), 4), const_1) | how many multiples of 4 are less than 3200 , and also multiples of 16 ? | the lcm of 4 and 16 is 16 . divide 3200 / 16 = 200 . xxx . so b is your answer . | a = 16 - 3200
b = a / 4
c = b + 1
|
a ) 8 hrs , b ) 5 hrs , c ) 15 hrs , d ) 15 hrs , e ) 20 hrs | b | divide(90, add(10, 8)) | a man can row a boat at 10 kmph in still water and the speed of the stream is 8 kmph . what is then time taken to row a distance of 90 km down the stream ? | speed in down stream = 10 + 8 = 18 time taken to cover 90 km down stream = 90 / 18 = 5 hrs . answer : b | a = 10 + 8
b = 90 / a
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a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 55 | e | add(add(power(add(add(divide(subtract(subtract(109, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(109, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(109, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(109, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive number is 109 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 109 2 x + 1 = 109 2 x = 108 x = 54 therefore larger number is x + 1 = 55 answer : e" | a = 109 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 109 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 109 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 109 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 105 , b ) 160 , c ) 200 , d ) 71 , e ) 120 | e | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 1480) | what is the least number to be added to 1480 to make it a perfect square ? | "the numbers greater than 1480 and are square of number is 1600 . the least number that should be added 1480 to make it perfect square = 1600 - 1480 = 120 . answer : e" | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 1480
|
a ) $ 960 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,100 , e ) $ 2,250 | d | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.00), const_1000) | a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.00 . if the company ’ s latest order required a truck with at least 288,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "total volume is 288000 given lbh = 8 * 10 * 12 . the number of hats inside it = 288000 / 10 * 8 * 12 = 300 . price of each hat is 7.5 $ then total value is 300 * 7.0 = 2100 . imo option d is correct answer . ." | a = 3 * 100
b = 8 * 10
c = a + b
d = c + 4
e = d * 1000
f = 8 * 10
g = f * 12
h = e / g
i = h * 7
j = i / 1000
|
a ) rs . 30000 , b ) rs . 37000 , c ) rs . 37500 , d ) rs . 40000 , e ) rs . 40500 | c | multiply(divide(200, 80), 15000) | in order to obtain an income of rs . 15000 from 80 % stock at rs . 200 , one must make an investment of | "explanation : market value = rs . 200 . required income = rs . 15000 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 80 ( ie , 80 % of the face value 100 ) , investment = rs . 200 to obtain rs . 15000 , investment = 200 / 80 ã — 15000 = rs . 37500 answer : option c" | a = 200 / 80
b = a * 15000
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a ) 53 m 2 , b ) 51 m 2 , c ) 50.2 m 2 , d ) 45 m 2 , e ) 40 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(40, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 40 ° is ? | "40 / 360 * 22 / 7 * 12 * 12 = 50.2 m 2 answer : c" | a = 12 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 40 / g
i = f * h
|
a ) 31.23 , b ) 25.12 , c ) 36.25 , d ) 23.17 , e ) 27.66 | d | divide(add(multiply(28, 22), multiply(15, 13)), add(22, 13)) | the average runs scored by a batsman in 22 matches is 28 . in the next 13 matches the batsman scored an average of 15 runs . find his average in all the 35 matches ? | total score of the batsman in 22 matches = 616 . total score of the batsman in the next 13 matches = 195 . total score of the batsman in the 35 matches = 811 . average score of the batsman = 811 / 35 = 23.17 . answer : d | a = 28 * 22
b = 15 * 13
c = a + b
d = 22 + 13
e = c / d
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a ) 48 , b ) 54 , c ) 72 , d ) 84 , e ) 90 | e | divide(multiply(330, const_3), add(const_10, const_1)) | the sum of the numbers is 330 . if the first number be twice the second and third number be one - third of the first , then the second number is : | "let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 330 11 x / 3 = 330 x = 90 answer : e" | a = 330 * 3
b = 10 + 1
c = a / b
|
a ) 1250 , b ) 625 , c ) 600 , d ) 750 , e ) 375 | d | multiply(5, 150) | a trailer carries 3 , 4 and 5 crates on a trip . each crate weighs no less than 150 kg . what is the maximum weight of the crates on a single trip ? | "max no . of crates = 5 . max weight = 150 kg max . weight carried = 5 * 150 = 750 kg = d" | a = 5 * 150
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a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 2 / 3 | d | divide(subtract(60, 20), subtract(const_100, 20)) | the weight of a glass of jar is 20 % 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 = 20 g weight of coffee beans = 80 g weight of jar and remaining beans = 60 g weight of remaining beans = 40 g fraction remaining = 40 / 80 = 1 / 2 answer is d ." | a = 60 - 20
b = 100 - 20
c = a / b
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a ) 80 , b ) 75 , c ) 70 , d ) 60 , e ) 65 | e | divide(multiply(subtract(multiply(divide(add(4, 5), 5), 4), 5), add(multiply(divide(add(4, 5), 5), subtract(4, 5)), 1)), 5) | the points a ( 0 , 0 ) , b ( 0 , 4 a - 5 ) and c ( 2 a + 1 , 2 a + 4 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ? | "1 / 2 bh = 1 / 2 ( 2 a + 1 ) ( 2 a + 4 ) now 4 a - 5 = 2 a + 4 2 a = 9 therefore , a ( 0,0 ) ; b ( 0,13 ) ; c ( 10,13 ) 1 / 2 * 10 * 13 = 65 answer : e" | a = 4 + 5
b = a / 5
c = b * 4
d = c - 5
e = 4 + 5
f = e / 5
g = 4 - 5
h = f * g
i = h + 1
j = d * i
k = j / 5
|
a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.4 feet , e ) 3.5 feet | c | divide(add(add(multiply(add(2, 10), 2), 10), 6), add(2, 10)) | carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 6 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 6 = 40 40 / 12 = 3.3 feet answer : c" | a = 2 + 10
b = a * 2
c = b + 10
d = c + 6
e = 2 + 10
f = d / e
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 256 | e | power(2, multiply(const_4, 2)) | xy = 2 then what is ( 2 ^ ( x + y ) ^ 2 ) / ( 2 ^ ( x - y ) ^ 2 ) | ( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 8 2 ^ 8 = 256 answer e | a = 4 * 2
b = 2 ** a
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a ) 20 / 7 , b ) 20 / 3 , c ) 23 / 6 , d ) 10 / 3 , e ) 10 / 7 | d | multiply(divide(subtract(divide(7, 6), const_1), 5), const_100) | a sum of money becomes 7 / 6 of itself in 5 years at a certain rate of simple interest . the rate per annum is ? | "let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 5 years . rate = ( 100 * x ) / ( x * 6 * 5 ) = 10 / 3 % . answer : d" | a = 7 / 6
b = a - 1
c = b / 5
d = c * 100
|
a ) 17 years , b ) 19 years , c ) 29 years , d ) 8 years , e ) 12 years | d | divide(multiply(subtract(22, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 22 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 22 5 x = 20 = > x = 4 hence , b ' s age = 2 x = 8 years . answer : d" | a = 22 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | multiply(subtract(16, 6), const_2) | all the students of class are told to sit in circle shape . here the boy at the 6 th position is exactly opposite to 16 th boy . total number of boys in the class ? | as half the circle shape consist of 16 - 6 = 10 boys , so total number of boys in full circle = 2 * 10 = 20 answer : c | a = 16 - 6
b = a * 2
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a ) 5 : 28 , b ) 5 : 19 , c ) 5 : 12 , d ) 5 : 13 , e ) 11 : 34 | e | divide(subtract(sqrt(4624), 24), multiply(sqrt(4624), const_2)) | the area of a square is 4624 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4624 a 68 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 44 : 136 = 11 : 34 answer : e" | a = math.sqrt(4624)
b = a - 24
c = math.sqrt(4624)
d = c * 2
e = b / d
|
a ) 8.5 seconds , b ) 2.22 seconds , c ) 3.5 seconds , d ) 2.5 seconds , e ) 2.6 seconds | b | divide(100, multiply(162, const_0_2778)) | in what time will a train 100 meters long cross an electric pole , if its speed is 162 km / hr | "first convert speed into m / sec speed = 162 * ( 5 / 18 ) = 45 m / sec time = distance / speed = 100 / 45 = 2.22 seconds answer : b" | a = 162 * const_0_2778
b = 100 / a
|
a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint t , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , lets further reduce 0.75 : 1.25 gives 1 + 1.5 = 2.5 gallons of mixture . this looks ok , but lets reduce further just to be sure 0.375 : 0.625 gives 0.5 + 1 = 1.5 gallons of mixture , thats less than 2 gallons of mixture , so not acceptable . so correct ans is 2.5 gallons . b" | a = const_0_25 * 2
b = 2 * 5
c = a + b
|
a ) rs . 1000.15 , b ) rs . 1100.95 , c ) rs . 1892.85 , d ) rs . 1050.85 , e ) rs . 1200.25 | c | divide(2120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 2120 ? | "2120 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1892.85 answer : c" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 2120 / f
|
a ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | divide(subtract(multiply(8, subtract(40, 4)), multiply(8, 32)), 4) | the average age of an adult class is 40 years . 8 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 8 x 32 = ( y + 8 ) x 36 â ‡ ’ 40 y + 256 = 36 y + 288 â ‡ ’ 4 y = 32 â ˆ ´ y = 8 a" | a = 40 - 4
b = 8 * a
c = 8 * 32
d = b - c
e = d / 4
|
a ) 2 : 5 , b ) 3 : 4 , c ) 5 : 8 , d ) 2 : 3 , e ) 1 : 2 | c | divide(5, const_60) | what is the ratio of two no . 5 xy and 8 xy . | "5 xy / 8 xy = 5 : 8 answer c" | a = 5 / const_60
|
a ) 14 , b ) 13 , c ) 12 , d ) 11 , e ) 10 | d | subtract(subtract(subtract(17, 2), const_4), const_1) | how many positive integers less than 17 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | "the number = 2 a + 3 b < 17 when a = 1 , b = 1 , 2 , 3 , 4 , 5 - > 2 a = 2 ; 3 b = 3 , 6 , 9 , 12 , 15 - > the number = 5 , 8 , 11 , 14 - - > 4 numbers when a = 2 , b = 1,2 , 3,4 - > . . . . - - > 4 numbers when a = 3 , b = 1 , 2,3 - - > . . . . - - > 3 numbers total number is already 11 . look at the answer there is no number greater than 11 - - > we dont need to try any more answer must be d" | a = 17 - 2
b = a - 4
c = b - 1
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(23, reminder(1056, 23)) | what least number should be added to 1056 , so that the sum is completely divisible by 23 ? | "that is 1056 / 23 = 45 remainder = 21 = > 21 + 2 = 23 hence 2 should be added to 1056 so that the sum will be divisible by 23 answer is option b" | a = 23 - reminder
|
a ) 96 , b ) 240 , c ) t = 256 , d ) t = 384 , e ) t = 480 | e | multiply(multiply(16, const_2), 15) | at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses t at the restaurant ? ( assume that all boxes are filled to capacity . ) | "most test takers would recognize the system of equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e" | a = 16 * 2
b = a * 15
|
a ) $ 112.00 , b ) $ 145.60 , c ) $ 163.80 , d ) $ 182.00 , e ) $ 100.00 | a | multiply(200, divide(add(30, 20), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 200 before any discount ? | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we ' re told to maximize the discount ( thus , 30 % off the original price and then 20 % off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions , decimals , etc . ) : 30 % off = ( 1 - . 3 ) = ( 1 - 30 / 100 ) = ( . 7 ) and the same can be done with the 20 % additional discount . . . the final price of an item that originally cost $ 200 would be . . . . . ( $ 200 ) ( . 7 ) ( . 8 ) = ( $ 200 ) ( . 56 ) = 112 final answer : a" | a = 30 + 20
b = a / 100
c = 200 * b
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 30 | e | divide(550, multiply(add(60, 6), const_0_2778)) | a train 550 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 550 * 3 / 55 = 30 sec . answer : option e" | a = 60 + 6
b = a * const_0_2778
c = 550 / b
|
a ) 5 : 2 , b ) 5 : 1 , c ) 5 : 3 , d ) 4 : 1 , e ) 3 : 1 | c | divide(subtract(180, 170), subtract(186, 180)) | students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 186 cms . what was the ratio of men to women ? | "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 186 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 186 m ) / ( w + m ) = 180 170 w + 186 m = 180 w + 180 m 6 m = 10 w 3 m = 5 w m / w = 5 / 3 the ratio of men to women is 5 to 3 . c" | a = 180 - 170
b = 186 - 180
c = a / b
|
a ) 1 km , b ) 2 km , c ) 3.19 km , d ) 4 km , e ) 5 km | c | multiply(multiply(divide(divide(50, 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 50 minutes . the total distance is ? | "c 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 50 / 60 47 x / 60 = 50 / 60 = > x = 1.06 . total distance = 3 * 1.06 = 3.19 km ." | a = 50 / 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
|
a ) 694 , b ) 792 , c ) 698 , d ) 784 , e ) 789 | b | add(add(multiply(300, const_2), const_100), subtract(subtract(const_100, const_4), const_4)) | how many digits are used to number a book containg 300 pages . | explanation : number of digits in 1 digit numbers = 1 to 9 = 9 number of digits in 2 digit numbers = 10 to 99 = 90 ( 2 ) = 180 number of digits in 3 digit numbers = 100 to 300 = 201 ( 3 ) = 603 total = 792 answer : option b | a = 300 * 2
b = a + 100
c = 100 - 4
d = c - 4
e = b + d
|
a ) 1200 , b ) 3000 , c ) 1000 , d ) 1400 , e ) 2400 | d | divide(subtract(multiply(divide(6, const_100), 2800), multiply(2800, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | barbata invests $ 2800 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | "let the additional invested amount for 8 % interest be x ; equation will be ; 2800 + 0.05 * 2800 + x + 0.08 x = 2800 + x + 0.06 ( 2800 + x ) 0.05 * 2800 + 0.08 x = 0.06 x + 0.06 * 2800 0.02 x = 2800 ( 0.06 - 0.05 ) x = 2800 * 0.01 / 0.02 = 1400 ans : ` ` d ' '" | a = 6 / 100
b = a * 2800
c = 5 / 100
d = 2800 * c
e = b - d
f = 8 / 100
g = 6 / 100
h = f - g
i = e / h
|
a ) 3.33 % , b ) 5.93 % , c ) 4.33 % , d ) 9.33 % , e ) 11.67 % | e | multiply(divide(divide(subtract(950, 600), 600), 5), const_100) | at what rate percent on simple interest will rs . 600 amount to rs . 950 in 5 years ? | "350 = ( 600 * 5 * r ) / 100 r = 11.67 % answer : e" | a = 950 - 600
b = a / 600
c = b / 5
d = c * 100
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 11 | a | inverse(add(divide(const_1, 6), add(divide(const_1, 90), divide(const_1, 45)))) | a , b and c can do a work in 90 , 45 and 6 days respectively . if they work together , in how many days will they complete the work ? | one days ' s work of a , b and c = 1 / 90 + 1 / 45 + 1 / 6 = ( 1 + 2 + 15 ) / 90 = 1 / 5 a , b and c together can do the work in 5 days . answer : a | a = 1 / 6
b = 1 / 90
c = 1 / 45
d = b + c
e = a + d
f = 1/(e)
|
a ) 33 , b ) 67 , c ) 30 , d ) 15 , e ) 17 | a | divide(add(25, 41), const_2) | a man can row upstream at 25 kmph and downstream at 41 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 41 m = ( 41 + 25 ) / 2 = 33 answer : a" | a = 25 + 41
b = a / 2
|
a ) 337 , b ) 262.5 , c ) 299.5 , d ) 266.5 , e ) 299 | b | divide(multiply(450, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | i bought two books ; for rs . 450 . 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 ) = ( 450 - x ) 119 / 100 x = 262.5 answer : b" | a = 100 + 19
b = 450 * a
c = 100 - 15
d = 100 + 19
e = c + d
f = b / e
|
a ) 1.2 , b ) 20.8 , c ) 2.3 , d ) 6.7 , e ) 8.2 | b | add(divide(subtract(100, 1), 5), const_1) | how many multiples of 5 are there between 1 and 100 ( both are inclusive ) ? | "the answer is ( 100 - 1 ) / 5 + 1 = 20.8 answer is b" | a = 100 - 1
b = a / 5
c = b + 1
|
a ) 3 / 10 , b ) 2 / 5 , c ) 1 / 2 , d ) 4 , e ) 6 / 5 | d | divide(4, subtract(multiply(subtract(6, 4), 5), 4)) | when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 2 / 5 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 2 / 5 ) ( 4 + 4 x ) or , x = 4 answer d" | a = 6 - 4
b = a * 5
c = b - 4
d = 4 / c
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 98 % | e | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(14, const_100)), divide(10, const_100)), const_100) | in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 14 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | 0,098 r = x / 100 * 0.1 r answer e | a = 30 / 100
b = 1 - a
c = 14 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 15 kmph , b ) 11 kmph , c ) 88 kmph , d ) 20 kmph , e ) 12 kmph | d | multiply(divide(11, const_60), 110) | the speed of a train is 110 kmph . what is the distance covered by it in 11 minutes ? | "110 * 11 / 60 = 20 kmph answer : d" | a = 11 / const_60
b = a * 110
|
a ) rs . 1500 , b ) rs . 2000 , c ) rs . 1000 , d ) rs . 1200 , e ) rs . 1800 | b | subtract(2240, divide(multiply(subtract(2600, 2240), 2), 5)) | a sum of money at simple interest amounts to rs . 2240 in 2 years and to rs . 2600 in 5 years . the sum is : | "s . i . for 3 years = rs . ( 2600 - 2240 ) = rs . 360 . s . i . for 1 year = rs . 360 / 3 = rs . 120 . s . i . for 2 years = rs . ( 120 x 2 ) = rs . 240 . principal = rs . ( 2240 - 240 ) = rs . 2000 . answer : option b" | a = 2600 - 2240
b = a * 2
c = b / 5
d = 2240 - c
|
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