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 ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | d | subtract(1000, subtract(add(divide(1000, 10), divide(1000, 35)), divide(1000, multiply(10, 35)))) | what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 10 nor by 35 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here s... | a = 1000 / 10
b = 1000 / 35
c = a + b
d = 10 * 35
e = 1000 / d
f = c - e
g = 1000 - f
|
a ) 8 , b ) 14 , c ) 15 , d ) 18 , e ) 19 | e | add(multiply(9, const_2), const_1) | the average age of applicants for a new job is 31 , with a standard deviation of 9 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ? | "within one standard deviation of the average age means 31 + / - 7 24 - - 31 - - 38 number of dif . ages - 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 total = 19 e" | a = 9 * 2
b = a + 1
|
a ) 300 , b ) 318 , c ) 322 , d ) 324 , e ) 332 | e | add(multiply(8, const_4), multiply(divide(50, 10), const_60)) | a man walks at a rate of 10 mph . after every ten miles , he rests for 8 minutes . how much time does he take to walk 50 miles ? | "to cover 50 miles the man needs ( time ) = ( distance ) / ( rate ) = 50 / 10 = 5 hours = 300 minutes . he will also rest 4 times ( after 10 , 20 , 30 and 40 miles ) , so total resting time = 4 * 8 = 32 minutes . total time = 300 + 32 = 332 minutes . answer : e ." | a = 8 * 4
b = 50 / 10
c = b * const_60
d = a + c
|
a ) 22 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 36 km | e | multiply(add(6, 3), 4) | roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 6 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ? | distance = relative speed * time = ( 6 + 3 ) * 4 = 36 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : e | a = 6 + 3
b = a * 4
|
a ) 10 % , b ) 20 % , c ) 35 % , d ) 40 % , e ) 50 % | b | multiply(divide(subtract(360, 300), 300), const_100) | the price of a book is increased from $ 300 to $ 360 . what is the % of increase in its price ? | "explanation : change in the price = rs 360 – rs 300 = rs 60 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 60 300 ) * 100 = 20 % b" | a = 360 - 300
b = a / 300
c = b * 100
|
a ) 1.2 , b ) 10.92 , c ) 12.98 , d ) 12.38 , e ) none | b | divide(multiply(multiply(26, 6), 7), const_100) | the simple interest on rs . 26 for 6 months at the rate of 7 paise per rupeeper month is | "sol . s . i . = rs . [ 26 * 7 / 100 * 6 ] = rs . 10.92 answer b" | a = 26 * 6
b = a * 7
c = b / 100
|
a ) $ 2100 , b ) $ 2222 , c ) $ 2320 , d ) $ 2083.33 , e ) $ 2183.33 | d | add(divide(subtract(multiply(divide(20, const_100), 25000), const_1000), const_2), add(add(subtract(const_100, 20), const_3), const_0_33)) | ms . mary sold two properties , x and y , for $ 25000 each . she sold property x for 20 % more than she paid for it and sold property y for 20 % less than she paid for it . if expenses are disregarded , what was her total net gain or loss , if any , on the two properties ? | there is a property to solve such questions withcommon selling priceandcommon % gain and loss . such cases always result in a loss and . . . total % loss = ( common gain % or loss % / 10 ) ^ 2 hence here loss % = ( 20 / 10 ) ^ 2 = 4 % which means he recovered only 96 % of his investment which amount to a total revenue ... | a = 20 / 100
b = a * 25000
c = b - 1000
d = c / 2
e = 100 - 20
f = e + 3
g = f + const_0_33
h = d + g
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | a | subtract(1, multiply(divide(1, 3), divide(1, 2))) | harold and millicent are getting married and need to combine their already - full libraries . if harold , who has 1 / 2 as many books as millicent , brings 1 / 3 of his books to their new home , then millicent will have enough room to bring 1 / 3 of her books to their new home . what fraction of millicent ' s old libra... | "because we see h willbring 1 / 3 of his booksto the new home - - > try to pick a number that isdivisible by 3 . before : assume h = 30 books h = 1 / 2 m - - > m = 60 books after : h ' = 1 / 3 h = 10 books m ' = 1 / 3 m = 20 books total = 30 books m ' = 30 = 1 / 2 * 60 ratio : 1 / 2 ans : a" | a = 1 / 3
b = 1 / 2
c = a * b
d = 1 - c
|
a ) 22 , b ) 30 , c ) 15 , d ) 18 , e ) 20 | b | subtract(divide(multiply(multiply(30, subtract(10, 2)), 50), multiply(2, subtract(150, 50))), 30) | an engineer undertakes a project to build a road 10 km long in 150 days and employs 30 men for the purpose . after 50 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 100 days 2 km done hence remaining is 8 km also , work has to be completed in next 100 days ( 150 - 50 = 100 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion he... | a = 10 - 2
b = 30 * a
c = b * 50
d = 150 - 50
e = 2 * d
f = c / e
g = f - 30
|
a ) 8 , b ) 7 , c ) 9 , d ) 7.5 , e ) 6 | b | divide(280, add(30, add(10, 20))) | a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 280 - mile journey ? assume that the train makes no intermediate stops . | "a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 280 - mile journey ? assume that the train makes no intermediate stops . i think the easiest way to solv... | a = 10 + 20
b = 30 + a
c = 280 / b
|
a ) 4 : 3 , b ) 12 : 11 , c ) 7 : 4 , d ) 6 : 5 , e ) 6 : 11 | d | divide(add(multiply(4, divide(28, add(4, 3))), 8), add(multiply(3, divide(28, add(4, 3))), 8)) | the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 28 years . the ratio of their ages after 8 years will be | "let mini ’ s age = 4 x and minakshi ’ s age = 3 x then 4 x + 3 x = 28 x = 4 mini ’ s age = 16 years and minakshi ’ s age = 12 years ratio of their ages after 8 years = ( 16 + 8 ) : ( 12 + 8 ) = 24 : 20 = 6 : 5 answer : d" | a = 4 + 3
b = 28 / a
c = 4 * b
d = c + 8
e = 4 + 3
f = 28 / e
g = 3 * f
h = g + 8
i = d / h
|
a ) 3 / 40000 , b ) 1 / 1000 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | b | divide(1, const_3) | a certain junior class has 1000 students and a certain senior class has 600 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let ' s see pick 60 / 1000 first then we can only pick 1 other pair from the 800 so total will be 60 / 600 * 1000 simplify and you get 1 / 10000 answer is b" | a = 1 / 3
|
a ) 99.999 , b ) 100.2 , c ) 134 , d ) 99.99 , e ) 99.9 | d | subtract(multiply(1, const_100), divide(1, const_100)) | what is the difference between the place values of two 1 ' s in the numeral 135.21 | required difference = 100 - 0.01 = 99.99 answer is d | a = 1 * 100
b = 1 / 100
c = a - b
|
a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 9 : 4 , e ) 36 : 8 | e | divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2)) | a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is four times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that ... | "water : soap = 3 : 2 soap : salt = 12 : 2 = > for 12 soap , salt = 2 = > for 2 soap , salt = ( 2 / 12 ) * 2 = 1 / 3 so , water : soap : salt = 3 : 2 : 1 / 3 = 36 : 24 : 4 after open container , water : soap : salt = 18 : 24 : 4 so , water : salt = 18 : 4 = 36 : 8 e" | a = 2 * 3
b = a * 3
c = 2 * 2
d = c * 2
e = b / d
|
a ) e = 21 , b ) e = 22 , c ) e = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | if x + ( 1 / x ) = 5 , what is the value of e = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a = 5 ** 2
b = a - 2
|
a ) 10368 , b ) 10638 , c ) 10836 , d ) 10846 , e ) none of them | a | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 24,36) | find the smallest number of five digits exactly divisible by 16 , 24,36 and 54 . | "smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 16,24 , 36,54 i . e 432 , on dividing 10000 by 432 , we get 64 as remainder . therefore , required number = 10000 + ( 432 – 64 ) = 10368 . answer is a ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 24
|
a ) 1 / 32 , b ) 1 / 4 , c ) 1 , d ) 4 , e ) 5 | b | divide(multiply(divide(4, 8), const_100), multiply(divide(8, 4), const_100)) | 8 is 4 % of a , and 4 is 8 % of b . c equals b / a . what is the value of c ? | "explanation : given , = > 4 a / 100 = 8 . = > a = 8 × ( 100 / 4 ) = 200 . - - - - - - - - - ( i ) and , = > ( 8 / 100 ) × b = 4 . = > b = 50 . - - - - - - - - - ( ii ) now , c = b / a ( from ( i ) and ( ii ) ) = > 50 / 200 . = > 1 / 4 . answer : b" | a = 4 / 8
b = a * 100
c = 8 / 4
d = c * 100
e = b / d
|
['a ) 2 : 5', 'b ) 2 : 9', 'c ) 2 : 2', 'd ) 2 : 9', 'e ) 2 : 1'] | a | divide(divide(1, 10), power(divide(1, 2), const_2)) | the volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ? | the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 12 h 1 / r 22 h 2 = > 1 / 10 = ( 1 ) 2 h 1 / ( 2 ) 2 h 2 = > h 1 / h 2 = 2 / 5 i . e . h 1 : h 2 = 2 : 5 answer : a | a = 1 / 10
b = 1 / 2
c = b ** 2
d = a / c
|
a ) 50 kg , b ) 60 kg , c ) 61 kg , d ) 62 kg , e ) 91 kg | a | divide(subtract(multiply(add(30, 30), 40), multiply(30, subtract(40, 10))), 30) | the average weight of a group of 30 friends increases by 10 kg when the weight of additional 30 friends was added . if average weight of the whole group after including the additional 30 members is 40 kg , what is the average weight of the additional friends ? | let a = avg . wt . of additional 30 friends original total weight = ( 30 friends ) ( 30 kg avge ) = 900 kg ( 900 + 30 a ) / ( 30 + 30 ) = 40 kg avge a = 50 kg answer - a | a = 30 + 30
b = a * 40
c = 40 - 10
d = 30 * c
e = b - d
f = e / 30
|
a ) 10 , 40 , b ) 20 , 30 , c ) 35 , 15 , d ) 30 , 20 , e ) 15 , 35 | d | subtract(50, divide(subtract(50, divide(20, const_2)), const_2)) | sum of two numbers is 50 . two times of the difference of first and seceond is 20 . then the numbers will be ? | explanation : x + y = 50 2 x ã ¢ â ‚ ¬ â € œ 2 y = 20 x = 30 y = 20 answer : d | a = 20 / 2
b = 50 - a
c = b / 2
d = 50 - c
|
a ) 28 % , b ) 67.5 % , c ) 64.8 % , d ) 70 % , e ) 72 % | b | add(multiply(divide(divide(25, const_100), subtract(1, divide(1, 10))), const_100), 2) | the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 25 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 25 % , price = 81 - 20.25 = 60.75 which is 60.75 / 90 * 100 of the sale price on day 1 = ~ 67.5 % answer b" | a = 25 / 100
b = 1 / 10
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) − 220 , b ) − 100 , c ) 100 , d ) 135 , e ) it can not be determined from the information given | b | subtract(multiply(const_60.0, const_2), multiply(110, const_2)) | if the average ( arithmetic mean ) of a and b is 110 , and the average of b and c is 160 , what is the value of a − c ? | "( a + b ) / 2 = 110 = = = > a + b = 220 ( b + c ) / 2 = 160 = = = > b + c = 320 ( a + b ) - ( b + c ) = 220 - 320 = = = > a + b - b - c = - 100 = = = > a - c = - 100 answer : b" | a = const_60 * 0
b = 110 * 2
c = a - b
|
a ) 35 , b ) 24 , c ) 17 , d ) 6 , e ) 5 | a | add(power(divide(subtract(7, sqrt(subtract(power(7, const_2), multiply(const_4, 7)))), const_2), 2), power(divide(add(7, sqrt(subtract(power(7, const_2), multiply(const_4, 7)))), const_2), 2)) | if a and b are the roots of the equation x 2 - 7 x + 7 = 0 , then the value of a 2 + b 2 is : | sol . ( b ) the sum of roots = a + b = 7 product of roots = ab = 8 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 49 - 14 = 35 answer a | a = 7 ** 2
b = 4 * 7
c = a - b
d = math.sqrt(c)
e = 7 - d
f = e / 2
g = f ** 2
h = 7 ** 2
i = 4 * 7
j = h - i
k = math.sqrt(j)
l = 7 + k
m = l / 2
n = m ** 2
o = g + n
|
a ) 12 , b ) 13 , c ) 14 , d ) 16 , e ) 19 | d | add(divide(33, add(1, 2)), divide(30, add(4, 2))) | each machine of type a has 4 steel parts and 2 chrome parts . each machine of type b has 2 steel parts and 1 chrome parts . if a certain group of type a and type b machines has a total of 30 steel parts and 33 chrome parts , how many machines are in the group | "look at the below representation of the problem : steel chrome total a 4 2 30 > > no . of type a machines = 30 / 6 = 5 b 2 1 33 > > no . of type b machines = 33 / 3 = 11 so the answer is 16 i . e d . hope its clear ." | a = 1 + 2
b = 33 / a
c = 4 + 2
d = 30 / c
e = b + d
|
a ) 75 , b ) 100 , c ) 125 , d ) 175 , e ) 225 | c | divide(subtract(multiply(divide(750, const_3), const_4), 750), const_2) | there are 750 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 750 - x x / 2 + 750 - x / 4 = 1 / 3 * ( 750 ) = 250 x = 250 x / 2 = 125 is supposed to be the answer answer : c" | a = 750 / 3
b = a * 4
c = b - 750
d = c / 2
|
a ) 8 , b ) 10 , c ) 6 , d ) 14 , e ) 16 | c | add(floor(divide(16, 3)), floor(divide(16, power(3, const_2)))) | if m = 3 ^ n , what is the greatest value of n for which m is a factor of 16 ! | "solution - consider multiples of 25 ! = > 3 , 6,9 , 12,15 count no . of 3 in each multiple . 3 = 3 x 1 - > 1 6 = 3 x 2 - > 1 9 = 3 x 3 - > 2 12 = 3 x 4 - > 1 15 = 3 x 5 - > 1 - - - - count 3 ' s = 6 so answer is 6" | a = 16 / 3
b = math.floor(a)
c = 3 ** 2
d = 16 / c
e = math.floor(d)
f = b + e
|
a ) 188 , b ) 180 , c ) 156 , d ) 840 , e ) 121 | b | divide(subtract(multiply(500, divide(10, const_100)), 5), divide(25, const_100)) | if 25 % of x is 5 less than 10 % of 500 , then x is ? | "25 % of x = x / 4 ; 10 % of 500 = 10 / 100 * 500 = 50 given that , x / 4 = 50 - 5 = > x / 4 = 45 = > x = 180 . answer : b" | a = 10 / 100
b = 500 * a
c = b - 5
d = 25 / 100
e = c / d
|
a ) 20 sec , b ) 15 sec , c ) 45 sec , d ) 50 sec , e ) 1 min | c | divide(675, add(8, 7)) | two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 675 meters , after what time will they meet at the starting point ? | "they meet every 675 / 7 + 8 = 45 sec answer is c" | a = 8 + 7
b = 675 / a
|
a ) - 16 , b ) - 10 , c ) 0 , d ) 14 , e ) 16 | b | add(sqrt(16), sqrt(64)) | if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 16 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | "( x + 1 ) ^ 2 < = 16 x < = 3 x > = - 5 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 5 × - 6 = 30 minimum possible value of xy is - 5 × 8 = - 40 - 40 + 30 = - 10 answer : b" | a = math.sqrt(16)
b = math.sqrt(64)
c = a + b
|
a ) 1 , b ) 10 , c ) 9 , d ) 0 , e ) 8 | a | subtract(add(10, 1), add(6, 4)) | what is the least possible value of expression e = ( x - 1 ) ( x - 3 ) ( x - 4 ) ( x - 6 ) + 10 for real values of x ? | explanation : e = ( x - 1 ) ( x - 6 ) ( x - 3 ) ( x - 4 ) + 10 e = ( x 2 - 7 x + 6 ) ( x 2 - 7 x + 12 ) + 10 let x 2 - 7 x + 6 = y e = y 2 + 6 y + 10 e = ( y + 3 ) 2 + 1 minimum value = 1 , when y = - 3 answer : a | a = 10 + 1
b = 6 + 4
c = a - b
|
a ) 76 days , b ) 30 days , c ) 98 days , d ) 31 days , e ) 22 days | b | inverse(subtract(inverse(12), inverse(20))) | a and b can finish a work in 12 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 12 – 1 / 20 = 2 / 60 = 1 / 30 = > 30 days answer : b" | a = 1/(12)
b = 1/(20)
c = a - b
d = 1/(c)
|
a ) 76 sec , b ) 67 sec , c ) 98 sec , d ) 36 sec , e ) 32 sec | e | divide(add(200, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 200 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 200 + 120 = 320 m . time taken = 320 / 10 = 32 sec . answer : e | a = 200 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 82 % , b ) 8.2 % , c ) 0.82 % , d ) 0.082 % , e ) 0.0082 % | a | multiply(divide(82, 100.00), 100.00) | a certain tax rate is $ 82 per $ 100.00 . what is the rate , expressed as a percent ? | "here in question it is asking $ . 82 is what percent of $ 100 . suppose $ . 82 is x % of 100 means 100 * ( x / 100 ) = 82 hence x = 82 % so answer is a" | a = 82 / 100
b = a * 100
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | a | divide(const_100.0, const_10) | how many integers from 101 to 600 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 5 = 5 6 is not possible here because it would result in a number greater than 6 ( i . e 606 , 616 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 5 * 10... | a = 100 / 0
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a ) 16 , b ) 18 , c ) 19 , d ) 20 , e ) 28 | e | subtract(divide(subtract(multiply(16, 5), 30), const_2), const_2) | the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 16 , and k < m < r < s < t . if t is 30 , what is the greatest possible value of the median of the 5 integers ? | "we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 16 which means that in effect , all numbers are equal to 16 . if the largest number is 30 , it is 14 more than 16 . we need r to be ma... | a = 16 * 5
b = a - 30
c = b / 2
d = c - 2
|
a ) 35 , b ) 30 , c ) 43 , d ) 60 , e ) 65 | c | add(add(subtract(add(multiply(6, 6), multiply(6, 10)), add(multiply(6, 10), 6)), 10), const_3) | x , a , z , and b are single digit positive integers . x = 1 / 6 a . z = 1 / 6 b . ( 10 a + b ) – ( 10 x + z ) could not equal | "a = 6 x , b = 6 z therefore ( 6 x . 10 + 6 z ) - ( 10 x + z ) = ( 6 - 1 ) ( 10 x + z ) = 5 . ( 10 x + z ) number should be divisible by 5 c" | a = 6 * 6
b = 6 * 10
c = a + b
d = 6 * 10
e = d + 6
f = c - e
g = f + 10
h = g + 3
|
a ) 520 , b ) 440 , c ) 260 , d ) 280 , e ) 120 | d | multiply(power(add(const_1, divide(5, const_100)), 8), 200) | rs . 200 amounts to rs . 800 in 8 years at simple interest . if the interest is increased by 5 % , it would amount to how much ? | "( 200 * 5 * 8 ) / 100 = 80 200 + 80 = 280 answer : d" | a = 5 / 100
b = 1 + a
c = b ** 8
d = c * 200
|
a ) 13 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | multiply(power(const_2, 336), factorial(336)) | the product of three consecutive numbers is 336 . then the sum of the smallest two numbers is ? | "product of three numbers = 336 336 = 6 * 7 * 8 . so , the three numbers are 6 , 7 and 8 . and sum of smallest of these two = 6 + 7 = 13 . answer : option a" | a = 2 ** 336
b = math.factorial(336)
c = a * b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(add(log(6), const_2), divide(const_1, const_2)) | a , b , c are positive numbers such that they are in increasing geometric progression then how many such numbers are there in ( loga + logb + logc ) / 6 = log 6 | given that a , b , c are in gp so b / a = c / b = > b ^ 2 = ac ( loga + logb + logc ) / 6 = log 6 = > log ( abc ) = 6 * log 6 = > log ( ac * b ) = log ( 6 ^ 6 ) = > log ( b ^ 3 ) = log ( 6 ^ 6 ) = > b ^ 3 = 6 ^ 6 = > b ^ 3 = ( 6 ^ 2 ) ^ 3 = > b = 6 ^ 2 = 36 it means a = 2 , c = 18 or a = 3 c = 12 or a = 4 c = 9 or a = ... | a = math.log(6)
b = a + 2
c = 1 / 2
d = b - c
|
a ) 15 , b ) 14 , c ) 13 , d ) 12 , e ) 11 | d | multiply(multiply(2, 4), multiply(3, 6)) | calculate the l . c . m of 2 / 5 , 4 / 7 , 3 / 7 , 6 / 13 is : | "required l . c . m = l . c . m . of 2 , 4 , 3 , 6 / h . c . f . of 5 , 7 , 7 , 13 = 12 / 1 = 12 answer is d" | a = 2 * 4
b = 3 * 6
c = a * b
|
a ) 23 , b ) 32 , c ) 34 , d ) 43 , e ) 48 | e | sqrt(divide(multiply(square_area(8), 18), inverse(const_2))) | the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 18 of the area of the field . what is the length of the field ? | "explanation : a / 18 = 8 * 8 = > a = 8 * 8 * 18 x * 2 x = 8 * 8 * 18 x = 24 = > 2 x = 48 answer : option e" | a = square_area * (
b = a / 18
c = 1/(2)
d = math.sqrt(b)
|
a ) 131 , b ) 135 , c ) 169 , d ) 196 , e ) 212 | d | add(multiply(divide(subtract(290, 8), const_3), const_2), 8) | if jake loses 8 pounds , he will weigh twice as much as his sister . together they now weigh 290 pounds . what is jake ’ s present weight , in pounds ? | "lets say j is the weight of jack and s is the wt of his sister . if he loses 8 pounds , he s twice as heavy as his sister . j - 8 = 2 * s also , together they weight 290 pounds j + s = 290 solvong the 2 equation , we get j = 196 pounds ! d" | a = 290 - 8
b = a / 3
c = b * 2
d = c + 8
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 20 | e | divide(59, const_10) | how many integers from 0 to 59 , inclusive , have a remainder of 1 when divided by 3 ? | "my ans is also c . 17 . explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 58 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term... | a = 59 / 10
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a ) 45,000 , b ) 40,000 , c ) 50,000 , d ) 55,000 , e ) 60,000 | b | multiply(multiply(power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33), 2,500), power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33)) | david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 2,500 and then he left the lab . he retu... | "let the rate be x , then population of the bacteria after each hour can be given as 2500 , 2500 x , 2500 ( x ^ 2 ) , 2500 ( x ^ 3 ) now population at 4 pm = 160,000 thus we have 2500 ( x ^ 3 ) = 160,000 = 64 thus x = 4 therefore population at 3 pm = 2500 ( 16 ) = 40,000 answer : b" | a = 2 * 4
b = a * 2
c = b / 2
d = c ** const_0_33
e = d * 2
f = 2 * 4
g = f * 2
h = g / 2
i = h ** const_0_33
j = e * i
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a ) $ 0.94 , b ) $ 0.96 , c ) $ 0.98 , d ) $ 1.02 , e ) $ 1.20 | d | multiply(add(const_1, divide(8, const_100)), divide(0.80, divide(subtract(const_100, 15), const_100))) | the manager of a produce market purchased a quantity of tomatoes for $ 0.80 per pound . due to improper handling , 15 percent of the tomatoes , by weight , were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal ... | "assume the manager bought 100 tomatoes . cost price = 80 given : 15 % are damaged - - > available tomatoes to sell = 85 85 * x - 80 = 0.08 * 80 85 x - 80 = 6.4 85 x = 86.64 x = 86.64 / 85 = 87 / 85 ( approx ) = 1.023 x is slightly under 1.023 = 1.02 answer : d" | a = 8 / 100
b = 1 + a
c = 100 - 15
d = c / 100
e = 0 / 80
f = b * e
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a ) 123 , b ) 127 , c ) 235 , d ) 4 , e ) 505 | d | gcd(subtract(1856, 4), subtract(1642, 6)) | the greatest number which on dividing 1642 and 1856 leaves remainders 6 and 4 respectively , is : | "explanation : required number = h . c . f . of ( 1642 - 6 ) and ( 1856 - 4 ) = h . c . f . of 1636 and 1852 = 4 . answer : d" | a = 1856 - 4
b = 1642 - 6
c = math.gcd(a, b)
|
a ) 22 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | a | subtract(27, add(floor(divide(9, const_2)), const_1)) | marcella has 27 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "marcella has 27 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 27 pairs of shoes minus the 5 ' destroyed ' pairs yields 22 p... | a = 9 / 2
b = math.floor(a)
c = b + 1
d = 27 - c
|
a ) 5 , b ) 4 , c ) 20 , d ) 25 , e ) 30 | c | multiply(5, const_4) | how many 3 - digit numbers can be formed from the digits 2 , 3 , 5 , 6 , 7 and 9 , which are divisible by 5 and none of the digits is repeated ? | since each desired number is divisible by 5 , so we must have 5 at the unit place . so , there is 1 way of doing it . the tens place can now be filled by any of the remaining 5 digits ( 2 , 3 , 6 , 7 , 9 ) . so , there are 5 ways of filling the tens place . the hundreds place can now be filled by any of the remaining 4... | a = 5 * 4
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a ) 35 : 44 , b ) 34 : 44 , c ) 22 : 44 , d ) 20 : 40 , e ) 50 : 45 | a | divide(multiply(105, const_2), multiply(88, const_3)) | a man invests some money partly in 12 % stock at 105 and partly in 8 % stock at 88 . to obtain equal dividends from both , he must invest the money in the ratio . | in case of stock 1 , if he invest rs . 105 , he will get a dividend of rs . 12 ( assume face value = 100 ) in case of stock 2 , if he invest rs . 88 , he will get a dividend of rs . 8 ( assume face value = 100 ) ie , if he invest rs . ( 88 * 12 ) / 8 , he will get a dividend of rs . 12 required ratio = 105 : ( 88 × 12 ... | a = 105 * 2
b = 88 * 3
c = a / b
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a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 900 , e ) $ 950 | a | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,055 . how much did she invest at 5 % ? | "et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1055 / 1000 ) - 1 = 0.055 = 5.5 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of ... | a = 5 * 2
b = a * 100
c = 5 * 2
d = c * 100
e = 5 * 2
f = e * 5
g = f + 4
h = d + g
i = 5 * 2
j = i * 100
k = h - j
l = k * 100
m = 5 * 2
n = m * 100
o = 5 * n
p = l - o
q = b - p
|
a ) 18 , b ) 24 , c ) 36 , d ) 42 , e ) 48 | b | multiply(2, 4) | three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 288 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 288 x = 24 hcf of 2 x , 3 x and 4 x = x = 24 the answer is b ." | a = 2 * 4
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a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | a | subtract(const_100, subtract(subtract(const_100, 10), 20)) | a merchant sells an item at a 10 % discount , but still makes a gross profit of 20 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | original sp = x cost = c current selling price = . 9 x ( 10 % discount ) . 9 x = 1.2 c ( 20 % profit ) x = 1.2 / . 9 * c x = 4 / 3 c original selling price is 1.3 c which is 30 % profit answer a | a = 100 - 10
b = a - 20
c = 100 - b
|
a ) . 7 , b ) . 07 , c ) . 05 , d ) 0.07 , e ) none of these | b | divide(70, const_1000) | what decimal fraction is 70 ml of a litre ? | "answer required fraction = 70 / 1000 = 7 / 100 = . 07 correct option : b" | a = 70 / 1000
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a ) 8 , b ) 16 , c ) 17 , d ) 18 , e ) 34 | c | add(subtract(add(20, 8), subtract(20, 8)), const_1) | the average age of applicants for a new job is 20 , with a standard deviation of 8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the max... | "minimum age = average - 1 standard deviation = 20 - 8 = 12 maximum age = average + 1 standard deviation = 20 + 8 = 28 maximum number of different ages of the applicants = 28 - 12 + 1 = 17 answer c" | a = 20 + 8
b = 20 - 8
c = a - b
d = c + 1
|
a ) 2800 , b ) 2403 , c ) 3998 , d ) 2539 , e ) 1930 | a | divide(multiply(7000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | p , q and r have $ 7000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | a 2800 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 7000 - r ) = > 3 r = 14000 - 2 r = > 5 r = 14000 = > r = 2800 . | a = 2 * 2
b = 7000 * a
c = 2 * 2
d = c / 3
e = d * 3
f = 1 * 3
g = e + f
h = 1 * 3
i = g + h
j = b / i
|
['a ) 750', 'b ) 850', 'c ) 950', 'd ) 1050', 'e ) none of the above'] | a | divide(volume_rectangular_prism(5, multiply(5, const_2), multiply(5, const_3)), volume_cube(const_1)) | unit size cubes are stacked inside a big rectangular box having dimensions corresponding to three consecutive multiples of 5 . choose exact number of cubes that can completely fill the box . | let the dimensions of the box be , length = 5 * a , breadth = 5 * ( a + 1 ) , height = 5 * ( a + 2 ) hence , volume = 5 * 5 * 5 * a * ( a + 1 ) * ( a + 2 ) among any 3 consecutive positive integers , we will either have ( a number that is divisible by both 23 ) or ( a number divisible by 2 and another number divisible ... | a = 5 * 2
b = 5 * 3
c = volume_rectangular_prism / (
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a ) 17 : 5 , b ) 17 : 3 , c ) 17 : 6 , d ) 17 : 7 , e ) 17 : 8 | b | divide(85000, 15000) | p and q started a business investing rs 85000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 85000 : 15000 = 85 : 15 = 17 : 3 option b" | a = 85000 / 15000
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a ) 8.8 % , b ) 9 % , c ) 10.1 % , d ) 8.6 % , e ) 8.4 % | c | multiply(divide(add(divide(multiply(15, 15), const_100), divide(multiply(8, 35), const_100)), add(15, 35)), const_100) | in one alloy there is 15 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.15 * 15 + 0.08 * 35 = 5.05 kg , so the percentage is 5.05 / 50 * 100 = 10.1 % . answer : c ." | a = 15 * 15
b = a / 100
c = 8 * 35
d = c / 100
e = b + d
f = 15 + 35
g = e / f
h = g * 100
|
a ) 1.5 , b ) 2 , c ) 2.4 , d ) 3 , e ) 3.6 | e | inverse(add(divide(const_1, 6), divide(const_1, multiply(divide(6, const_2), const_3)))) | working alone , sawyer finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . sawyer alone cleans the entire house in 6 hours . how many hours will it take nick and sawyer to clean the entire house if they work together ? | answer is 3.6 hours . sawyer does the complete house in 6 hours while nick does it in 9 hours . 1 / ( 1 / 6 + 1 / 9 ) = 3.6 answer is e | a = 1 / 6
b = 6 / 2
c = b * 3
d = 1 / c
e = a + d
f = 1/(e)
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a ) 68 , b ) 70.4 , c ) 86 , d ) 114.4 , e ) 108 | d | add(88, multiply(divide(30, const_100), 88)) | if x is 30 percent greater than 88 , then x = | "x = 88 * 1.3 = 114.4 so the answer is d ." | a = 30 / 100
b = a * 88
c = 88 + b
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a ) - 5 , b ) - 4 , c ) 4 , d ) 3 , e ) 2 | c | subtract(5, 1) | find value for x from below equation ? x + 1 = 5 | 1 . subtract 1 from both sides : x + 1 - 1 = 5 - 1 2 . simplify both sides : x = 4 c | a = 5 - 1
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a ) 4 , b ) 3 , c ) 2 , d ) 10 , e ) 5 | d | subtract(25, reminder(1015, 25)) | what least number should be added to 1015 , so that the sum is completely divisible by 25 ? | "1015 ã · 25 = 40 with remainder = 15 15 + 10 = 25 . hence 10 should be added to 1015 so that the sum will be divisible by 25 answer : option d" | a = 25 - reminder
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a ) 14.64 % , b ) 15.64 % , c ) 16.64 % , d ) 17.64 % , e ) 18.64 % | c | multiply(add(4, 4), 2) | the population of a city increases @ 4 % p . a . there is an additional annual increase of 4 % of the population due to the influx of job seekers , find the % increase in population after 2 years ? | total annual increament in population = 4 + 4 = 8 % let the population be x population after 2 years = 1.08 x + . 0864 x population increase = 1.08 x + . 0864 x - x % increase = ( ( 1.08 x + . 0864 x - x ) / x ) * 100 = ( 1.08 + . 0864 - 1 ) * 100 = . 1664 * 100 = 16.64 % answer : c | a = 4 + 4
b = a * 2
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a ) 240 , b ) 789 , c ) 520 , d ) 879 , e ) 456 | a | multiply(add(31, 1), divide(floor(divide(31, const_2)), const_2)) | the sum of even numbers between 1 and 31 is : | explanation : let sn = ( 2 + 4 + 6 + . . . + 30 ) . this is an a . p . in which a = 2 , d = 2 and l = 30 let the number of terms be n . then a + ( n - 1 ) d = 30 = > 2 + ( n - 1 ) x 2 = 30 n = 15 . sn = n / 2 ( a + l ) = 15 / 2 x ( 2 + 30 ) = ( 15 x 16 ) = 240 . answer : a | a = 31 + 1
b = 31 / 2
c = math.floor(b)
d = c / 2
e = a * d
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a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 62 | d | add(divide(450, 9), 3) | a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 450 ki... | "suppose food eaten in first hour : x the ap is : x , x + 3 , x + 6 , . . . . [ number of terms ' n ' = 9 ] therefore , 9 / 2 [ 2 x + ( 9 - 1 ) * 3 ] = 450 . solving for x , x = 38 . now , 6 th term will be : x + ( 6 - 1 ) * d = 38 + 5 * 3 = 53 . hence d ! !" | a = 450 / 9
b = a + 3
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a ) 56 kg , b ) 90 kg , c ) 75 kg , d ) data inadequate , e ) none of these | c | add(multiply(8, 2.5), 55) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 55 kg . what might be the weight of the new person ? | "c 75 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 55 + 20 ) kg = 75 kg ." | a = 8 * 2
b = a + 55
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a ) 3600 , b ) 3400 , c ) 3608 , d ) 3602 , e ) 3603 | b | subtract(17000, multiply(divide(4, 5), 17000)) | income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 17000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 17000 = > x = 3400 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3400 . answer : b" | a = 4 / 5
b = a * 17000
c = 17000 - b
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a ) 80.0 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 82.8 | e | multiply(multiply(divide(subtract(const_100, 8), const_100), divide(subtract(const_100, 10), const_100)), const_100) | a store reduced the price of all items in the store by 8 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | consider price of the all items as $ 100 after a initial reduction of 8 % price becomes = 0.92 * 100 = $ 92 after the final reduction of 10 % price becomes = 0.9 * 92 = $ 82.8 price of all items on second day is 82.8 % of price on first day correct answer option e | a = 100 - 8
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = e * 100
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a ) 80 , b ) 150 , c ) 100 , d ) 180 , e ) 220 | a | subtract(add(175, 325), subtract(470, 50)) | out of 470 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 470 - 50 = 420 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 420 = 80 . answer is a" | a = 175 + 325
b = 470 - 50
c = a - b
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a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54 | a | subtract(60, subtract(add(41, 22), 9)) | in a class of 60 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 60 54 + not = 60 not = 60 - 54 = 6 so answer is a" | a = 41 + 22
b = a - 9
c = 60 - b
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a ) 60 , b ) 72 , c ) 84 , d ) 90 , e ) 216 | e | multiply(subtract(divide(multiply(const_2, const_2), subtract(9, multiply(const_2, 4))), divide(const_2, subtract(9, 4))), const_60) | tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 4 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 9 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | e is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 4 t ) = 9 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 4 t ) = 9 t = = > 4 = = > 240 minutes difference , 216 minutes e | a = 2 * 2
b = 2 * 4
c = 9 - b
d = a / c
e = 9 - 4
f = 2 / e
g = d - f
h = g * const_60
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a ) 70 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | a | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(10, 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 10 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,07 r = x / 100 * 0.1 r answer a" | a = 30 / 100
b = 1 - a
c = 10 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
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a ) 8 / 15 , b ) 7 / 15 , c ) 1 / 15 , d ) 3 / 15 , e ) none of these | a | subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 4)) | p is able to do a piece of work in 15 days and q can do the same work in 20 days . if they can work together for 4 days , what is the fraction of work left ? | explanation : p ' s 1 - day work = 1 / 15 q ' s 1 - day work = 1 / 20 work done by ( p + q ) in 1 day = 1 / 15 + 1 / 20 = 7 / 60 . work done by them in 4 days = ( 7 / 60 ) * 4 = 7 / 15 . work left = 1 - ( 7 / 15 ) = 8 / 15 . answer is a | a = 1 / 20
b = 1 / 15
c = a + b
d = c * 4
e = 1 - d
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a ) 24440 , b ) 36440 , c ) 72440 , d ) 181440 , e ) 216440 | d | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | of the five - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "of the five - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for p - 9 ways ; choosing the digit for q - 8 ways ; choosing the digit for r - ... | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
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a ) 1.2 kg , b ) 1.5 kg , c ) 1.4 kg , d ) 1.9 kg , e ) none of these | c | divide(divide(subtract(const_100, 30), const_100), divide(50, const_100)) | when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 30 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 70 % ( 100 - 30 ) . therefore 0.5 x amount of flower - nectar = 0.70 x amount of honey = 0.70 x 1 kg therefore amount of flower - nectar needed = ( 0.70 / 0.51 ) kg = 1.4 kgs answer : c" | a = 100 - 30
b = a / 100
c = 50 / 100
d = b / c
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(subtract(add(add(add(1, 1), 2), 10), add(7, 7)), const_2) | for any integer m greater than 1 , $ m denotes the product of all the integers from 1 to m , inclusive . how many prime numbers are there between $ 7 + 2 and $ 7 + 10 , inclusive ? | $ is basically a factorial of a number . so , we are asked to find the number of primes between 7 ! + 2 and 7 ! + 10 , inclusive . from each number 7 ! + k were 2 ≤ k ≤ 102 ≤ k ≤ 10 we can factor out k , thus there are no pries in the given range . for example : 7 ! + 2 = 2 ( 3 * 4 * 5 * 6 * 7 + 1 ) - - > a multiple of... | a = 1 + 1
b = a + 2
c = b + 10
d = 7 + 7
e = c - d
f = e + 2
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a ) 82 , b ) 17 , c ) 12 , d ) 24 , e ) 18 | d | subtract(const_60, multiply(const_60, divide(30, 50))) | excluding stoppages , the speed of a train is 50 kmph and including stoppages it is 30 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 20 / 50 * 60 = 24 answer : option d" | a = 30 / 50
b = const_60 * a
c = const_60 - b
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a ) 11 , b ) 12 , c ) 15 , d ) 14 , e ) 19 | c | add(9, add(4, 2)) | find the constant k so that : - x 2 - ( k + 9 ) x - 8 = - ( x - 2 ) ( x - 4 ) | - x 2 - ( k + 9 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 9 ) x - 8 = - x 2 + 6 x - 8 - ( k + 9 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 15 : solve the above for k correct answer c | a = 4 + 2
b = 9 + a
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a ) 230 , b ) 270 , c ) 260 , d ) 256 , e ) 298 | b | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 230) | a 230 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 2340 ) / 9 = 500 / 9 = > x = 270 . answer : b" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 230
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a ) 100 marks , b ) 200 marks , c ) 160 marks , d ) 371 marks , e ) 827 marks | c | add(multiply(divide(add(40, 20), subtract(divide(60, const_100), divide(40, const_100))), divide(40, const_100)), 40) | a candidate who gets 40 % of the marks fails by 40 marks . but another candidate who gets 60 % marks gets 20 marks more than necessary for passing . find the number of marks for passing ? | "40 % - - - - - - - - - - - - 40 60 % - - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - - - - 60 40 % - - - - - - - - - - - - - - ? 120 + 40 = 160 marks answer : c" | a = 40 + 20
b = 60 / 100
c = 40 / 100
d = b - c
e = a / d
f = 40 / 100
g = e * f
h = g + 40
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a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 40 | e | multiply(1, 10) | two numbers are in the ratio of 1 : 2 . if 10 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 10 ) / ( y + 10 ) = 3 / 5 = > x = 20 and y = 40 answer : e" | a = 1 * 10
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a ) 288 , b ) 225 , c ) 277 , d ) 272 , e ) 150 | b | multiply(divide(multiply(90, const_1000), const_3600), 9) | a train running at the speed of 90 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 90 * 5 / 18 = 25 m / sec length of the train = speed * time = 25 * 9 = 225 m answer : b" | a = 90 * 1000
b = a / 3600
c = b * 9
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a ) s 400 , b ) s 500 , c ) s 850 , d ) s 540 , e ) s 170 | e | multiply(4, divide(850, add(add(4, 2), const_3))) | rs . 850 is divided so that 4 times the first share , twice the 2 nd share and twice the third share amount to the same . what is the value of the first share ? | "a + b + c = 850 4 a = 2 b = 2 c = x a : b : c = 1 / 4 : 1 / 2 : 1 / 2 = 1 : 2 : 2 1 / 5 * 850 = rs 170 answer : e" | a = 4 + 2
b = a + 3
c = 850 / b
d = 4 * c
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a ) a ) 36 , b ) b ) 40 , c ) c ) 41 , d ) d ) 42 , e ) e ) 44 | b | subtract(550, multiply(85, 6)) | we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 550 . how many green hats did we buy ? | "let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 550 . 6 ( 85 - g ) + 7 g = 550 . 510 - 6 g + 7 g = 550 . g = 550 - 510 = 40 . the answer is b ." | a = 85 * 6
b = 550 - a
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a ) 12 , b ) 15 , c ) 5 , d ) 9 , e ) none of these | d | multiply(6, divide(54, 6)) | solve for x and check : 6 x = 54 | "solution : dividing each side by 6 , we obtain ( 6 x / 6 ) = ( 54 / 6 ) therefore : x = 9 check : 6 x = 54 ( 6 * 9 ) = 54 54 = 54 answer : d" | a = 54 / 6
b = 6 * a
|
a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 42 | a | multiply(const_3, divide(60, const_10)) | jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 12 percent solution will be required ? | let a = amount of 2 % acid and b = amount of 12 % acid . now , the equation translates to , 0.02 a + . 12 b = . 05 ( a + b ) but a + b = 60 therefore . 02 a + . 12 b = . 05 ( 60 ) = > 2 a + 12 b = 300 but b = 60 - a therefore 2 a + 12 ( 60 - a ) = 300 = > 10 a = 420 hence a = 42 . , b = 60 - 42 = 18 answer : a | a = 60 / 10
b = 3 * a
|
a ) 16 , b ) 17 , c ) 12 , d ) 16 , e ) 16 | c | subtract(const_60, multiply(const_60, divide(36, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 9 / 45 * 60 12 answer : c" | a = 36 / 45
b = const_60 * a
c = const_60 - b
|
a ) 24 , b ) 42 , c ) 36 , d ) 48 , e ) 50 | a | divide(add(divide(const_2, const_10), 1), subtract(divide(1, const_4), divide(const_2, const_10))) | one - fourth of a number is greater than one - fifth of the number succeeding it by 1 . find the number . | number is 24 as 1 / 4 th of 24 = 6 1 / 5 th of 25 ( 24 + 1 ) = 5 6 = 5 + 1 answer : a | a = 2 / 10
b = a + 1
c = 1 / 4
d = 2 / 10
e = c - d
f = b / e
|
a ) 250 , b ) 425 , c ) 675 , d ) 700 , e ) 600 | e | multiply(add(divide(100, 20), divide(150, 10)), 30) | working at their respective constant rates , machine a makes 100 copies in 20 minutes and machine b makes 150 copies in 10 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ? | "machine a can produce 100 * 30 / 20 = 150 copies and , machine b can produce 150 * 30 / 10 = 450 copies total producing 600 copies . e is the answer" | a = 100 / 20
b = 150 / 10
c = a + b
d = c * 30
|
a ) 8 gallons , b ) 4 gallons , c ) 6 gallons , d ) 5.5 gallons , e ) 10 gallons | b | divide(160, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 160 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 160 kilometers ? 160 / 40 = 4 * 1 gallon = 4 gallons correct answer b" | a = 160 / 40
|
a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 51 % . | d | multiply(divide(add(multiply(divide(25, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 25 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | 25 % of 2 litres = 0.5 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.9 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 29 % answer : d | a = 25 / 100
b = a * 2
c = 40 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
|
a ) 84 , b ) 136 , c ) 172 , d ) 210 , e ) 478 | d | multiply(divide(multiply(105, 6), add(75, 105)), const_60) | cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 6 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 6 hrs therefore , 6 = d / 75 + d / 105 lcm of 75 and 105 = 525 6 = d / 75 + d / 105 = > 6 = 7 d / 525 + 5 d / ... | a = 105 * 6
b = 75 + 105
c = a / b
d = c * const_60
|
a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 10 | e | subtract(subtract(multiply(30, add(const_2, const_3)), multiply(20, const_4)), 20) | the average of temperatures at noontime from monday to friday is 30 ; the lowest one is 20 , what is the possible maximum range of the temperatures ? | "average = 30 , sum of temperatures = 30 * 5 = 150 as the min temperature is 20 , max would be 150 - 4 * 20 = 30 - - > the range = 30 ( max ) - 20 ( min ) = 10 answer : e" | a = 2 + 3
b = 30 * a
c = 20 * 4
d = b - c
e = d - 20
|
a ) 3010 , b ) 2010 , c ) 1000 , d ) 3100 , e ) 2100 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | find the c . i . on a sum of rs . 10000 for 6 months at 20 % per annum , interest being compounded halfyearly ? | "c . i . = 10000 ( 11 / 10 ) ^ 1 - 10000 = 1000 answer : c" | a = const_0_25 / 4
b = 1 + a
c = const_0_25 / 4
d = 1 + c
e = b * d
f = 10000 * e
g = f - 10000
|
a ) 80 , b ) 100 , c ) 160 , d ) 240 , e ) 300 | c | multiply(divide(const_60, 15), 40) | if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 40 minutes ? | "since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 40 minutes the population increases by 40 x 4 = 160 people . answer . c ." | a = const_60 / 15
b = a * 40
|
a ) 35 , b ) 37 , c ) 70 , d ) 75 , e ) 50 | c | subtract(add(multiply(7, 30), multiply(7, 80)), multiply(10, 42)) | the average of 10 result is 42 . average of the first 7 of them is 30 and that of the last 7 is 80 . find the 8 th result ? | "sum of all the 10 results = 10 * 42 = 420 sum of the first 7 of them = 7 * 30 = 210 sum of the last 7 of them = 7 * 80 = 560 so , the 8 th number = 420 + 210 - 560 = 70 . c" | a = 7 * 30
b = 7 * 80
c = a + b
d = 10 * 42
e = c - d
|
a ) $ 65,625 , b ) $ 105,000 , c ) $ 125,000 , d ) $ 183,750 , e ) $ 1 , 050,000 | c | subtract(subtract(divide(divide(26250, divide(multiply(7, 3), const_100)), const_1000), const_10), const_10) | a certain boxer has agreed to pay his opponent a fee of 3 % of his total purse for every pound over the specified weight limit he weighs in . if the boxer pays his opponent a fee of $ 26250 after weighing in 7 pounds over the specified limit , what was the boxer ' s purse ? | . 03 * 7 = . 21 26250 / . 21 = $ 125,000 answer : c | a = 7 * 3
b = a / 100
c = 26250 / b
d = c / 1000
e = d - 10
f = e - 10
|
['a ) 36', 'b ) 32', 'c ) 24', 'd ) 21', 'e ) 15'] | c | divide(volume_rectangular_prism(multiply(const_2, const_3), 9, 12), 27) | what is the maximum number q of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ? | 27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can ... | a = 2 * 3
b = volume_rectangular_prism / (
|
a ) 2436 , b ) 2801 , c ) - 2801 , d ) - 1601 , e ) none of them | d | multiply(subtract(const_1, const_2), subtract(multiply(64, 29), 165)) | - 64 x 29 + 165 = ? | "given exp . = - 64 x ( 30 - 1 ) + 165 = - ( 64 x 30 ) + 64 + 165 = - 1830 + 229 = - 1601 answer is d" | a = 1 - 2
b = 64 * 29
c = b - 165
d = a * c
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 2.5 | a | multiply(subtract(5, 4), 4) | if ( c - a ) / ( c - b ) = 1 , then ( 5 b - 4 a ) / ( c - a ) = | "let ' s say c = 3 , b = 1 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 0.5 answer a ( hopefully ) ) )" | a = 5 - 4
b = a * 4
|
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