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 since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 910 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? d" | 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 = 25000 + 25000 = 50000 i . e . 96 % of cost = 40000 therefore , 4 % of cost ( loss ) = $ 2083.33 answer : d | 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 library capacity is the new home ' s library capacity ? | "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 hence , x = ( 30 * 8 * 50 ) / ( 2 * 100 ) thus , x = 60 thus , more men needed to finish the task = 60 - 30 = 30 answer : b" | 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 solve this problem would be simply to count the number of miles it travels per hour ( and in total ) hour miles / hour total miles 1 10 10 2 20 30 3 30 60 4 40 100 5 50 150 6 60 210 7 70 280 it takes a total of nine hours to cover the 280 mile distance . answer : b" | 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 time , what is the ratio of water to salt in the solution ? | "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 = 50 answer is a ." | a = 100 / 0
|
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 maximum so k and m should be as small as possible to get the average of 16 . since all the numbers are positive integers , k and m can not be less than 1 and 2 respectively . 1 is 15 less than 16 and 2 is 14 less than 16 which means k and m combined are 29 less than the average . 30 is already 14 more than 16 and hence we only have 29 - 14 = 15 extra to distribute between r and s . since s must be greater than r , r can be 16 + 12 = 28 and s can be 16 + 13 = 29 . so r is 28 . answer ( e )" | 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 = 6 c = 6 . hence there are four such numbers answer : d | 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 of an ap , a is the first term and d is the common difference . so , 58 = 1 + ( n - 1 ) 3 or , ( n - 1 ) 3 = 57 or , n - 1 = 19 or , n = 20 e" | a = 59 / 10
|
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 returned in time to take a reading at 4 p . m . , by which point the population had grown to 160,000 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ? | "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
|
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 to 8 percent of the cost of the tomatoes . | "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
|
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 pairs that still fulfill the requirements . answer : a" | 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 digits . so , there are 4 ways of filling it . required number of numbers = ( 1 x 5 x 4 ) = 20 . answer c | a = 5 * 4
|
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 ) / 8 = 105 : ( 11 × 12 ) = 35 : ( 11 × 4 ) = 35 : 44 answer is a . | a = 105 * 2
b = 88 * 3
c = a / b
|
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 1000 invested at 5 % ) 0.05 x + 0.06 ( 1 - x ) = 0.055 ( 0.05 ) x + 0.06 - ( 0.06 ) x = 0.055 - 0.01 x = - 0.005 x = - 0.005 / - 0.01 = 5 / 10 = 50 % so x = 50 % of the 1000 which is 500 answer : a" | 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
|
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
|
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 maximum number of different ages of the applicants ? | "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 by 3 ) . volume = multiple of ( 125 * 2 * 3 ) 750 = 125 * 2 * 3 answer : a | a = 5 * 2
b = 5 * 3
c = volume_rectangular_prism / (
|
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
|
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)
|
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
|
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
|
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
|
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
|
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
|
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 kilos of plankton , how many kilos did he consume on the sixth hour ? | "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
|
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
|
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
|
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
|
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
|
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
|
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 difference , in minutes , between the amount of time it takes tom to cover half of the distance that linda has covered and the amount of time it takes tom to cover twice the distance that linda has covered ? | 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
|
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
|
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
|
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 - 7 ways ; choosing the digit for s - 6 ways ; # of permutations of 3 digits in ppqrs - 5 ! / 2 ! total : 9 * 8 * 7 * 6 * 5 ! / 2 ! = 181440 . answer : d ." | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
|
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
|
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 2 , thus not a prime ; 7 ! + 3 = 3 ( 2 * 4 * 5 * 6 * 7 + 1 ) - - > a multiple of 3 , thus not a prime ; . . . 7 ! + 10 = 10 ( 3 * 4 * 6 * 7 + 1 ) - - > a multiple of 10 , thus not a prime . answer : a . | a = 1 + 1
b = a + 2
c = b + 10
d = 7 + 7
e = c - d
f = e + 2
|
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
|
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
|
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
|
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
|
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
|
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
|
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
|
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
|
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 / 525 = > d = 525 / 2 km therefore , t 1 = d / 75 = > t 1 = 525 / ( 2 x 75 ) = > t 1 = ( 7 x 60 ) / 2 - - in minutes = > t 1 = 210 minutes . 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 have are 4 * 3 * 2 = 24 l * w * h is 8 * 12 * 9 , then max . cube we can have are 2 * 4 * 3 = 24 l * w * h is 8 * 9 * 12 , then max . cube we can have are 2 * 3 * 4 = 24 in all cases we get q = 24 cubes . ans . c | 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
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.