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a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 7 | c | subtract(divide(add(add(add(2, 4), 7), 17), add(4, const_1)), 4) | for the positive integers x , x + 2 , x + 4 , x + 7 , and x + 17 , the mean is how much greater than the median ? | "mean = ( x + x + 2 + x + 4 + x + 7 + x + 17 ) / 5 = ( 5 x + 30 ) / 5 = x + 6 median = x + 4 thus mean - median = x + 6 - ( x + 4 ) = 2 answer = c" | a = 2 + 4
b = a + 7
c = b + 17
d = 4 + 1
e = c / d
f = e - 4
|
a ) 76 kmph , b ) 6 kmph , c ) 14 kmph , d ) 8 kmph , e ) 3 kmph | e | divide(subtract(divide(84, 4), divide(60, 4)), const_2) | a man rows his boat 84 km downstream and 60 km upstream , taking 4 hours each time . find the speed of the stream ? | "speed downstream = d / t = 84 / ( 4 ) = 21 kmph speed upstream = d / t = 60 / ( 4 ) = 15 kmph the speed of the stream = ( 21 - 15 ) / 2 = 3 kmph answer : e" | a = 84 / 4
b = 60 / 4
c = a - b
d = c / 2
|
['a ) 10 cm 20 cm', 'b ) 2 cm 10 cm', 'c ) 100 cm 20 cm', 'd ) 30 cm 100 cm', 'e ) none'] | a | sqrt(divide(divide(volume_cube(10), 0.5), 5)) | an iron cube of side 10 cm is hammered into a rectangular sheet of thickness 0.5 cm . if the sides of the sheet are in the ratio 1 : 5 , the sides are : | sol . let the sides of the sheet be x and 5 x . then , ⇒ x * 5 x * 1 / 2 = 10 * 10 * 10 ⇒ 5 x ² = 2000 ⇒ x ² = 400 ⇒ x = 20 . ∴ the sides are 20 cm and 100 cm . answer a | a = volume_cube / (
b = a / 0
c = math.sqrt(b)
|
a ) 7 second , b ) 7.5 second , c ) 8 second , d ) 8.5 second , e ) none of these | b | divide(add(110, 165), multiply(132, const_0_2778)) | a train is moving at a speed of 132 km / hour . if the length of the train is 110 meters , how long will it take to cross a railway platform 165 meters long . | explanation : as we need to calculate answer in seconds , so first convert speed into meter / sec . we know 1 km / hr = 1 * ( 5 / 18 ) m / sec so , speed = 132 * ( 5 / 18 ) = 110 / 3 m / sec distance need to be covered in passing the platform = length of train + length of platform = 110 + 165 = 275 meters time = distance / speed = > time = 275 ∗ 3 / 110 = 15 / 2 = 7.5 seconds option b | a = 110 + 165
b = 132 * const_0_2778
c = a / b
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 8 | d | subtract(divide(multiply(10, add(const_4, const_1)), 12), add(const_4, const_1)) | 10 machines , each working at the same constant rate , together can complete a certain job in 12 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 8 days ? | "another solution which is faster is since each machine works at a constant rate . the time needs to bought down from 12 to 8 . so the new time is 2 / 3 of the original time . thus to achieve this we need the rate to be 3 / 2 of original . so 3 / 2 * 10 = 15 so we need 15 - 10 = 5 more machines . answer : d" | a = 4 + 1
b = 10 * a
c = b / 12
d = 4 + 1
e = c - d
|
a ) 1 / 2 , b ) 1 / 5 , c ) 3 / 4 , d ) 1 / 3 , e ) 3 / 5 | a | divide(2, subtract(add(2, 3), const_1)) | given a bag with 2 red balls and 3 yellow balls , you randomly select one of them , in order . if the second ball you take is yellow , what is the probability that the first one was also yellow ? | this is a conditional probability question . basically we are told that out of 2 balls picked one is yellow and are asked to find the probability of the other one also to be yellow . so , possible options for this other ball would be : w , w , b , b ( as one yellow is already picked ) so the probability for that ball to be yellow too is 2 / 4 = 1 / 2 . option a | a = 2 + 3
b = a - 1
c = 2 / b
|
a ) 1520 , b ) 2520 , c ) 3520 , d ) 5140 , e ) 7660 | b | multiply(multiply(add(4, 1), add(4, 1)), multiply(add(4, 4), multiply(1, 1))) | how many 4 - digit positive integers are multiple of each integer from 1 to 10 | "least no which could be divisible by integers [ 1,2 , 3,4 , 5,6 , 7,8 , 9,10 ] is the lcm of the said integers . lcm [ 1,2 , 3,4 , 5,6 , 7,8 , 9,10 ] = 2520 no the nos which are divisible by 2520 will be divisible by [ 1 , 2,3 , 4,5 , 6,7 , 8,9 ] there are 3 values possible between 1000 to 9999 they are 2520 x 1 = 2520 2520 x 2 = 5040 2520 x 3 = 7560 answer : b" | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 + 4
e = 1 * 1
f = d * e
g = c * f
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | a | multiply(divide(const_1, add(const_1, const_4)), divide(const_1, subtract(divide(const_1, 6), divide(const_1, 10)))) | a water tank is one - fifth full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how many minutes will it take to empty or fill the tank completely ? | the combined rate of filling / emptying the tank = 1 / 10 - 1 / 6 = - 1 / 15 since the rate is negative , the tank will be emptied . a full tank would take 15 minutes to empty . since the tank is only one - fifth full , the time is ( 1 / 5 ) * 15 = 3 minutes the answer is a . | a = 1 + 4
b = 1 / a
c = 1 / 6
d = 1 / 10
e = c - d
f = 1 / e
g = b * f
|
a ) 4.54 % , b ) 9.09 % , c ) 10 % , d ) 11 % , e ) 12.25 % | a | multiply(divide(subtract(add(const_100, 10), divide(add(const_100, add(const_100, 10)), const_2)), add(const_100, 10)), const_100) | jar y has 10 % more marbles than jar x . what percent of marbles from jar y need to be moved into x so that both jars have equal marbles | no ; the choices will be approximated only when gmat question explicitly mentions that ; here the answer is very close 4.54 , not 5 . jar y has 10 % more marbles than x y = 1.1 x if k marbles are taken out of y . the remaining marbles in jar y = 1.1 x - k if k marbles are added to x . the new marbles ' count in jar x = x + k and we know they are equal ; x + k = 1.1 x - k 2 k = 0.1 x k = 0.05 x the marbles taken out of y = 0.05 x question is : 0.05 x is what percent of marbles that was originally there in y i . e . 1.1 x 0.05 x 1.1 x ∗ 1000.05 x 1.1 x ∗ 100 0.05 x 1.1 x ∗ 1000.05 x 1.1 x ∗ 100 5011 = 4.54 ¯ ≈ 4.545011 = 4.54 ¯ ≈ 4.54 answer : a | a = 100 + 10
b = 100 + 10
c = 100 + b
d = c / 2
e = a - d
f = 100 + 10
g = e / f
h = g * 100
|
a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability that out of the two , one is white and one is blue ? | "the required probability = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ." | a = 3 * 5
b = 3 + 5
c = 5 + 2
d = b * c
e = d / 2
f = a / e
|
a ) 334 km . , b ) 216 km . , c ) 314 km . , d ) 224 km . , e ) 544 km . | d | multiply(multiply(divide(multiply(10, 24), add(24, 21)), 21), const_2) | a woman complete a journey in 10 hours . she travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km . | "d 224 km . 0.5 x / 21 + 0.5 x / 24 = 10 - - > x / 21 + x / 24 = 20 - - > 15 x = 168 x 20 - - > x = ( 168 x 20 ) / 15 = 224 km ." | a = 10 * 24
b = 24 + 21
c = a / b
d = c * 21
e = d * 2
|
a ) 15 kmph , b ) 11 kmph , c ) 18 kmph , d ) 8 kmph , e ) 12 kmph | d | multiply(divide(6, const_60), 80) | the speed of a train is 80 kmph . what is the distance covered by it in 6 minutes ? | "80 * 6 / 60 = 8 kmph answer : d" | a = 6 / const_60
b = a * 80
|
a ) 560 , b ) 540 , c ) 520 , d ) 440 , e ) 500 | d | multiply(divide(880, const_100), subtract(const_100, 50)) | if 50 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | we know 50 % people study biology , therefore the no of people not studying = 100 - 50 = 50 % > therefore the people not studying biology out of a total 880 people are = 50 % of 880 > ( 50 / 100 ) * 880 = 440 people d | a = 880 / 100
b = 100 - 50
c = a * b
|
a ) 1360 , b ) 1450 , c ) 1600 , d ) 1800 , e ) none | e | divide(multiply(85, 70), subtract(85, 70)) | the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 85 and rs . 70 respectively . the sum is : | "sol . sum = s . i . * t . d . / ( s . i ) - ( t . d . ) = 85 * 70 / ( 85 - 70 ) = rs . 396.7 . answer e" | a = 85 * 70
b = 85 - 70
c = a / b
|
a ) a ) 7 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15 | a | subtract(10, divide(subtract(10, divide(5, const_2)), const_2)) | sum of two numbers is 10 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 10 2 x – 3 y = 5 x = 7 y = 3 a )" | a = 5 / 2
b = 10 - a
c = b / 2
d = 10 - c
|
a ) 3.2 hr , b ) 3.4 hr , c ) 3.6 hr , d ) 3.8 hr , e ) 4.2 hr | a | divide(subtract(multiply(8, 70), 400), subtract(70, 20)) | ram covers a part of the journey at 20 kmph and the balance at 70 kmph taking total of 8 hours to cover the distance of 400 km . how many hours has been driving at 20 kmph ? | let see d / t = v then x / 20 + ( 400 - x ) / 70 = 8 x = 64 kmh t = 64 / 20 = 3.2 hr answer : a | a = 8 * 70
b = a - 400
c = 70 - 20
d = b / c
|
a ) 20000 , b ) 16000 , c ) 18000 , d ) 17000 , e ) 15000 | b | divide(400, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 10), const_100)), multiply(divide(20, const_100), divide(10, const_100)))) | a shopkeeper sells 10 % of his stock at 20 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 400 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 10 * x * 6 / 5 = 6 x / 50 the sp of 80 % of the stock = 9 / 10 * x * 0.95 = 171 x / 200 total sp = 24 x / 200 + 171 / 200 = 49 x / 50 overall loss = x - 195 / 20050 = 5 x / 200 5 x / 200 = 400 = > x = 16000 answer : b" | a = 5 / 100
b = 100 - 10
c = b / 100
d = a * c
e = 20 / 100
f = 10 / 100
g = e * f
h = d - g
i = 400 / h
|
a ) 3 , b ) 5 , c ) 11 , d ) 9 , e ) 10 | c | divide(add(13, 9), const_2) | in one hour , a boat goes 13 km along the stream and 9 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 13 + 9 ) kmph . = 11 kmph . answer c" | a = 13 + 9
b = a / 2
|
a ) 10 , b ) 15 , c ) 30 , d ) 20 , e ) 16 | d | multiply(1, 5) | if 5 spiders make 2 webs in 8 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 5 : 1 ( webs ) 1 : 8 } : : x : 2 â ‡ ’ 5 ã — 1 ã — 8 = 1 ã — 2 ã — x â ‡ ’ x = 20 answer : option d" | a = 1 * 5
|
a ) 15 % , b ) 80 % , c ) 20 % , d ) 22 % , e ) 24 % | b | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | an employee ’ s annual salary was increased $ 5,000 . if her new annual salary now equals $ 25000 what was the percent increase ? | "new annual salary = $ 25000 salary increase = $ 5000 . original salary = $ 25000 - $ 5,000 . = $ 20,000 % increase = ( $ 15,000 / $ 75,000 ) * 100 = 80 % hence b ." | a = 100 - 10
b = a * 1000
c = 100 - 10
d = c * 1000
e = const_0_25 * 100
f = e * 1000
g = d - f
h = b / g
i = h - 1
j = i * 100
|
a ) 45 % , b ) 62.5 % , c ) 35 % , d ) 32.5 % , e ) 30 % | b | subtract(const_100, multiply(divide(subtract(const_100, 50), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 50 per cent . the two reductions together are equal to a single reduction of | "price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 50 percent is reduced which means the new price is merely 50 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 1 / 2 ) p = 3 / 8 p is the new price after both deduction which is 37.5 percent of the original value p . this implies this entire series of deduction is worth having discounted 62.5 % of p . so answer is b = 62.5 %" | a = 100 - 50
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
|
a ) 5 : 1 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(10, const_100), divide(1, 2)) | in a school 10 % of the boys are same in number as 1 / 2 th of the girls . what is the ratio of boys to the girls in the school ? | "10 % of b = 1 / 2 g 10 b / 100 = g / 2 b = 5 g / 1 b / g = 5 / 1 b : g = 5 : 1 answer is a" | a = 10 / 100
b = 1 / 2
c = a / b
|
a ) 2 , b ) 5 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(multiply(2, multiply(6, const_10)), multiply(divide(3, 2), multiply(6, const_10))), 6) | audrey 4 hours to complete a certain job . ferris can do the same job in 3 hours . audrey and ferris decided to collaborate on the job , working at their respective rates . while audrey worked continuously , ferris took 6 breaks of equal length . if the two completed the job together in 2 hours , how many minutes long was each of ferris ’ breaks ? | audery and ferris collective work rate : 1 / 4 + 1 / 3 = 7 / 12 collective work time = 12 / 7 = 1.7 hrs job was actually done in = 2 ( includes breaks ) breaks = actual time taken - collective work time = 2 - 1.7 = . 3 hrs = 1 / 2 so ferrais took 6 breaks = . 3 / 6 = . 06 hrs = 5 m so answer is b ) 10 mins | a = 6 * 10
b = 2 * a
c = 3 / 2
d = 6 * 10
e = c * d
f = b - e
g = f / 6
|
a ) s . 486 , b ) s . 455 , c ) s . 220 , d ) s . 480 , e ) s . 200 | e | divide(multiply(250, const_100), add(const_100, 25)) | by selling an article at rs . 250 , a profit of 25 % is made . find its cost price ? | "sp = 250 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 250 * [ 100 / ( 100 + 25 ) ] = 250 * [ 100 / 125 ] = rs . 200 answer : e" | a = 250 * 100
b = 100 + 25
c = a / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | add(divide(6, 6), divide(6, 8)) | a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 6 days . for how many number of days should a and c should work together to complete the remaining work ? | "b work 1 / 8 * 6 = 3 / 4 remaining work = 1 - 3 / 4 = 1 / 4 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 take reciprocial 4 * remaining work = 4 * 1 / 4 = 1 answer : a" | a = 6 / 6
b = 6 / 8
c = a + b
|
a ) 1000 , b ) 850 , c ) 875 , d ) 775 , e ) 750 | c | subtract(multiply(250, 4), const_100) | 4 guys buy same cellphones cost them 250 each . two guy have 25 % of for his first cellphone . what is the total cost all of them pay to seller . | solution : c . 250 * 2 = 500 for 2 guy without discount , for other 2 ( 250 * 2 ) with 25 % is 375 . total cost 500 + 375 = 875 | a = 250 * 4
b = a - 100
|
a ) - 10 , b ) - 18 , c ) - 28 , d ) - 30 , e ) - 32 | d | multiply(negate(add(const_2, const_3)), subtract(11, add(const_2, const_3))) | if x and y are integers and | x - y | = 11 , what is the minimum possible value of xy ? | sometimes the answer choices to a given question provide a big ' hint ' as to how you can go about solving it . this prompt can also be solved without any complex math ideas - you just need to do a bit of ' brute force ' math and you ' ll have the answer relatively quickly . we ' re told that x and y are integers and | x - y | = 11 . we ' re asked for the minimum possible value of ( x ) ( y ) . since all of the answer choices are negative , this tells us that one of the two variables must be negative ( and the other must be positive ) , so we should restrict our work to those options . if . . . x = 10 , y = - 1 , then xy = - 10 x = 9 , y = - 2 , then xy = - 18 x = 8 , y = - 3 , then xy = - 24 x = 7 , y = - 4 , then xy = - 28 x = 6 , y = - 5 , then xy = - 30 x = 5 , y = - 6 , then xy = - 30 x = 4 , y = - 7 , then xy = - 28 from this , we can conclude the xy will start to get bigger as x continues to decrease down to 1 , so there ' s no need to do any additional work . final answer : d | a = 2 + 3
b = negate * (
|
a ) 10 , b ) 15 , c ) 30 , d ) 25 , e ) 30 | c | multiply(10, divide(15, 5)) | 15 : 5 seconds : : ? : 10 minutes | "15 * 10 = 5 * x x = 30 answer : c" | a = 15 / 5
b = 10 * a
|
a ) 35 kmph , b ) 40 kmph , c ) 45 kmph , d ) 50 kmph , e ) 55 kmph | b | subtract(divide(divide(75, 6), const_0_2778), 5) | a train 75 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ? | "let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 75 / [ ( x + 5 ) * 5 / 18 ] = 6 30 ( x + 5 ) = 1350 x = 40 kmph answer is b" | a = 75 / 6
b = a / const_0_2778
c = b - 5
|
a ) 26 km / hr , b ) 17 km / hr , c ) 60 km / hr , d ) 42 km / hr , e ) 46 km / hr | d | divide(multiply(multiply(divide(add(100, 100), multiply(5, add(const_1, const_2))), const_2), const_3600), const_1000) | two trains , each 100 m long , moving in opposite directions , cross other in 5 sec . if one is moving twice as fast the other , then the speed of the faster train is ? | "let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 5 = 3 x = > x = 20 / 3 . so , speed of the faster train = 40 / 3 = 40 / 3 * 18 / 5 = 42 km / hr . answer : d" | a = 100 + 100
b = 1 + 2
c = 5 * b
d = a / c
e = d * 2
f = e * 3600
g = f / 1000
|
a ) 12 , b ) 7 , c ) 5 , d ) 15 , e ) 6 | c | divide(125, 25) | shreehari has 125 pencils . there are 25 students are studying in his class . he would like to give each student the same amount of pencils , how much should he give to each student ? | 125 / 25 = 5 the answer is c . | a = 125 / 25
|
a ) − 0.3 , b ) 0 , c ) 0.3 , d ) 1.08 , e ) 0.5 | e | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.8, 0.5)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.8, 0.5)), const_1000), const_1000)) | what is the value of 3 x ^ 2 − 1.8 x + 0.5 for x = 0.6 ? | "3 x ^ 2 - 1.8 x + 0.5 for x = 0.6 = 3 ( 0.6 * 0.6 ) - 3 * 0.6 * ( 0.6 ) + 0.5 = 0 + 0.5 = 0.5 answer : e" | a = 3 ** 2
b = 1 ** 8
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 1 ** 8
i = g - h
j = i / 1000
k = j / 1000
l = f - k
|
a ) 120 , b ) 1250 , c ) 1200 , d ) 1100 , e ) 1150 | c | add(multiply(divide(600, 60), 60), multiply(divide(600, 60), 60)) | the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 60 kmph its consumption is 1 litre per hour . if each litre costs $ 60 and other expenses per hous is $ 60 , then what would be the minimum expenditure required to cover a distance of 600 km ? | "60 kmph consumption is 1 lt / hr so 600 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 60 so 10 lt costs $ 600 extra expenses for 1 hr - $ 60 10 hrs - $ 600 total expense - $ 600 + $ 600 = $ 1200 answer : c" | a = 600 / 60
b = a * 60
c = 600 / 60
d = c * 60
e = b + d
|
a ) 456 , b ) 744 , c ) 912 , d ) 1200 , e ) 1400 | b | divide(multiply(divide(288, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 288 62 % - - - - - - - - ? = > 744 answer : b | a = 100 - 62
b = 62 - a
c = b / 100
d = 288 / c
e = d * 62
f = e / 100
|
a ) 980 , b ) 1150 , c ) 1360 , d ) 1540 , e ) 1720 | c | divide(multiply(multiply(17, subtract(17, const_1)), 10), const_2) | there are 17 teams in the hockey league , and each team faces all the other teams 10 times each . how many games are played in the season ? | "the number of ways to choose two teams is 17 c 2 = 17 * 16 / 2 = 136 the total number of games in the season is 10 * 136 = 1360 . the answer is c ." | a = 17 - 1
b = 17 * a
c = b * 10
d = c / 2
|
a ) 110 , b ) 230 , c ) 720 , d ) 420 , e ) 680 | c | subtract(divide(divide(7, const_1), const_3), const_3) | in how many ways can 7 boys be seated in a circular order ? | "number of arrangements possible = ( 7 − 1 ) ! = 6 ! = 6 x 5 x 4 x 3 x 2 x 1 = 720 answer : c" | a = 7 / 1
b = a / 3
c = b - 3
|
a ) 4 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | b | divide(log(8), log(power(8, 9))) | if n = 8 ^ 9 – 8 , what is the units digit of n ? | "8 ^ 9 - 8 = 8 ( 8 ^ 8 - 1 ) = = > 8 ( 2 ^ 24 - 1 ) last digit of 2 ^ 24 is 6 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 6 - 1 = 5 as the unit digit . now on multiply this with 8 , we get unit digit as 0 answer : b" | a = math.log(8)
b = 8 ** 9
c = math.log(b)
d = a / c
|
a ) 14700 , b ) 14500 , c ) 16800 , d ) 14300 , e ) 14000 | c | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 40,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 ⇒ 3 x + 14000 = 50000 ⇒ 3 x = 50000 – 14000 = 36000 ⇒ x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit × 21 / 50 = 40000 × 21 / 50 = 16,800 answer is c" | a = 5000 * 10
b = 4000 + 5000
c = b + 5000
d = a - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 3 * 12
i = h * 1000
j = 5000 * 10
k = i / j
l = g * k
|
a ) 165 , b ) 170 , c ) 175 , d ) 180 , e ) 185 | d | multiply(divide(subtract(5.94, divide(multiply(27, 200), const_1000)), 3), const_1000) | 30 carrots on a scale weigh 5.94 kg . when 3 carrots are removed from the scale , the average weight of the 27 carrots is 200 grams . what is the average weight ( in grams ) of the 3 carrots which were removed ? | 27 * 200 = 5400 . the other 3 carrots weigh a total of 540 grams . the average weight is 540 / 3 = 180 grams . the answer is d . | a = 27 * 200
b = a / 1000
c = 5 - 94
d = c / 3
e = d * 1000
|
a ) 540 km , b ) 675 km , c ) 276 km , d ) 178 km , e ) 176 km | b | multiply(add(20, 25), divide(75, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 75 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 75 km more than the other , 25 x â € “ 20 x = 75 5 x = 75 x = 15 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 15 = 675 km . answer : b" | a = 20 + 25
b = 25 - 20
c = 75 / b
d = a * c
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a ) 10 , b ) 6 , c ) 5 , d ) 0 , e ) 1 | c | add(const_3, const_4) | what is the smallest positive integer x such that 120 - x is the cube of a positive integer ? | "given 130 - x is a perfect cube so we will take 125 = 5 * 5 * 5 130 - x = 125 x = 130 - 125 = 5 correct option is c" | a = 3 + 4
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a ) x / 2 , b ) x , c ) 3 x / 2 , d ) 2 x , e ) 3 x | e | multiply(const_2, add(const_1, divide(const_1, const_2))) | two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of c crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of c ? | "lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( c ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e" | a = 1 / 2
b = 1 + a
c = 2 * b
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a ) 17.65 % , b ) 9 % , c ) 11.11 % , d ) 12 % , e ) none of these | a | multiply(divide(add(multiply(const_2, const_100), divide(const_100, const_2)), 850), const_100) | a shopkeeper forced to sell at cost price , uses a 850 grams weight for a kilogram . what is his gain percent ? | "shopkeeper sells 850 g instead of 1000 g . so , his gain = 1000 - 850 = 150 g . thus , % gain = ( 150 * 100 ) / 850 = 17.65 % . answer : option a" | a = 2 * 100
b = 100 / 2
c = a + b
d = c / 850
e = d * 100
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a ) 18000 , b ) 22000 , c ) 24000 , d ) 26000 , e ) 30000 | e | multiply(multiply(subtract(2, divide(3, 3)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(2, 3), 2)) | in a mayoral election , candidate x received 2 / 3 more votes than candidate y , and candidate y received 1 / 3 fewer votes than z . if z received 27,000 votes how many votes did candidate x received ? | "z = 27 - - > y received 2 / 3 fewer votes than z - - > y = z - 1 / 3 * z = 18 ; x received 2 / 3 more votes than y - - > x = y + 2 / 3 * y = 30 . answer : e ." | a = 3 / 3
b = 2 - a
c = const_0_25 * 100
d = c * 100
e = d * 10
f = b * e
g = 2 / 3
h = g + 2
i = f * h
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a ) 48 , b ) 32 , c ) 24 , d ) 18 , e ) 12 | b | multiply(4, multiply(const_2, const_4)) | a cubical block of metal weighs 4 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 4 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 4 pounds = 32 pounds so answer is b and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 4 * 8 = 32 so answer is b" | a = 2 * 4
b = 4 * a
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a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 40 % | e | multiply(divide(200, 500), const_100) | the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 200 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 200 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 200 ) * 100 / 500 = 40 % answer : option e" | a = 200 / 500
b = a * 100
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a ) 3 % , b ) 5 % , c ) 8 % , d ) 10 % , e ) 12 % | b | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(10, const_60), 8), add(multiply(10, const_60), 40)))) | bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 40 seconds and his sister ’ s time is currently 10 minutes 8 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | "bob ' s time = 640 secs . his sis ' time = 608 secs . percent increase needed = ( 640 - 608 / 640 ) * 100 = 32 / 640 * 100 = 5 % . ans ( b ) ." | a = 10 * 10
b = 10 * const_60
c = b + 8
d = 10 * const_60
e = d + 40
f = c / e
g = 1 - f
h = a * g
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a ) 54,000 , b ) 56,000 , c ) 58,000 , d ) 60,000 , e ) 62,000 | b | floor(multiply(divide(divide(divide(multiply(divide(multiply(24000, const_100), subtract(const_100, 40)), add(const_100, 40)), const_100), const_100), 40), const_2)) | when a mobile is sold for rs . 24000 , the owner loses 40 % . at what price must that mobile be sold in order to gain 40 % ? | "60 : 24000 = 140 : x x = ( 24000 x 140 ) / 60 = 56000 . hence , s . p . = rs . 56,000 . answer : option b" | a = 24000 * 100
b = 100 - 40
c = a / b
d = 100 + 40
e = c * d
f = e / 100
g = f / 100
h = g / 40
i = h * 2
j = math.floor(i)
|
a ) 10 hours , b ) 12.50 hours , c ) 15.80 hours , d ) 25 hours , e ) 1 day | c | divide(790, 50) | bella is taking a trip in her car from new york ( point a ) chicago , illinois ( point b ) . she is traveling at an average speed of 50 miles per hour , if the total distance from point a to point b is 790 miles , in approximately how long will bella reach her destination ? | answer is ( c ) . if she is traveling at a speed of 50 miles per hour and her trip is a total of 790 miles , 790 miles divided by 50 miles per hours would equal 15.80 hours , just shy of 16 hours . | a = 790 / 50
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a ) 32.5 % , b ) 17.5 % , c ) 20 % , d ) 36 % , e ) 37.5 % | b | multiply(const_100, divide(add(divide(10, const_100), multiply(3, divide(20, const_100))), add(const_1, 3))) | because he ’ s taxed by his home planet , mork pays a tax rate of 10 % on his income , while mindy pays a rate of 20 % on hers . if mindy earned 3 times as much as mork did , what was their combined tax rate ? | "say morks income is - 100 so tax paid will be 10 say mindys income is 3 * 100 = 300 so tax paid is 20 % * 300 = 60 total tax paid = 10 + 60 = 70 . combined tax % will be 70 / 100 + 300 = 15.5 %" | a = 10 / 100
b = 20 / 100
c = 3 * b
d = a + c
e = 1 + 3
f = d / e
g = 100 * f
|
a ) 28 , b ) 32 , c ) 43.2 , d ) 42 , e ) 45 | c | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | three numbers are in the ratio 4 : 5 : 6 and their average is 36 . the largest number is : | "explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 36 15 x = 108 x = 7.2 largest number = 6 x = 43.2 . answer c" | a = 4 * 6
b = a * 100
c = 5 * 6
d = b + c
|
a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.55 | c | subtract(add(0.6, 0.25), 0.4) | the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.6 , what is the probability that event b will occur ? | p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.6 = 0.4 + p ( b ) - 0.25 p ( b ) = 0.45 answer : c | a = 0 + 6
b = a - 0
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a ) 152 , b ) 253 , c ) 57 , d ) 171 , e ) 181 | b | multiply(multiply(8, subtract(11, 8)), 11) | there are 437 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | "given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 437 = 253 answer : b" | a = 11 - 8
b = 8 * a
c = b * 11
|
a ) 477 , b ) 420 , c ) 279 , d ) 276 , e ) 291 | b | add(multiply(divide(60, subtract(21, 14)), 14), multiply(divide(60, subtract(21, 14)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 14 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 14 + 21 = 35 t = 12 d = 35 * 12 = 420 . answer : b" | a = 21 - 14
b = 60 / a
c = b * 14
d = 21 - 14
e = 60 / d
f = e * 21
g = c + f
|
['a ) 4 : 1', 'b ) 6 : 1', 'c ) 8 : 1', 'd ) 16 : 1', 'e ) 64 : 1'] | d | multiply(4, 4) | if the length of the sides of two cubes are in the ratio 4 : 1 , what is the ratio of their total surface area ? | let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 4 x ) ^ 2 = 96 x ^ 2 . the ratio of surface areas is 16 : 1 . the answer is d . | a = 4 * 4
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(50, const_10) | how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 8 ? | "the numbers should be of the form 8 c + 3 . the minimum is 3 when c = 0 . the maximum is 43 when c = 5 . there are six such numbers . the answer is b ." | a = 50 / 10
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a ) 3 . , b ) 5 , c ) 1 / 2 . , d ) 1 / 3 . , e ) there is n ' t enough data to answer the question . | b | add(5, divide(multiply(5, 1), 5)) | two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores is 1 / 5 , what is the value of x / y ? | "answer is b : 5 x - y = ( x + y ) / 2 solving for x / y = 5" | a = 5 * 1
b = a / 5
c = 5 + b
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a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 550 , e ) $ 225 | e | multiply(multiply(power(const_100, const_2), divide(3, const_100)), divide(const_3, 3)) | how much interest will $ 10,000 earn in 9 months at an annual rate of 3 % ? | "soln : - 9 months = 3 / 4 of year ; 3 % = 3 / 100 = 3 / 100 ; $ 10,000 ( principal ) * 3 / 100 ( interest rate ) * 3 / 4 ( time ) = $ 225 . answer : e" | a = 100 ** 2
b = 3 / 100
c = a * b
d = 3 / 3
e = c * d
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a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | d | subtract(multiply(add(20, const_1), 5), 47) | the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 47 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "20 x + 47 = 21 ( x – 5 ) x = 58 answer : d" | a = 20 + 1
b = a * 5
c = b - 47
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a ) 8 , b ) 10 , c ) 14 , d ) 16 , e ) 19 | b | divide(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 9)), 3)), divide(const_1, 24)) | p can do a work in 24 days . q can do the samework in 9 days & r can do the same work in 12 days . q & r start the work and leave after 3 days . p finishes the remaining work in how many days . | work done by p in 1 day = 1 / 24 work done by q in 1 day = 1 / 9 work done by r in 1 day = 1 / 12 work done by q and r in 1 day = 1 / 9 + 1 / 12 = 7 / 36 work done by q and r in 3 days = 3 × 7 / 36 = 7 / 12 remaining work = 1 – 7 / 12 = 5 / 12 number of days in which p can finish the remaining work = ( 5 / 12 ) / ( 1 / 24 ) = 10 b | a = 1 / 12
b = 1 / 9
c = a + b
d = c * 3
e = 1 - d
f = 1 / 24
g = e / f
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a ) s . 3.26 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.83 , e ) s . 2.42 | a | subtract(multiply(8000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(8000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | what is the difference between the c . i . on rs . 8000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 8000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 8000 * 26 / 25 * 51 / 50 = rs . 8486.4 c . i . when interest is compounded half - yearly = [ 8000 * ( 1 + 2 / 100 ) 2 ] = ( 8000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 8489.66 difference = ( 8489.66 - 8486.4 = rs . 3.26 . answer : a" | a = 2 / 100
b = 1 + a
c = 2 / 100
d = 1 + c
e = b * d
f = 2 / 100
g = 1 + f
h = e * g
i = 8000 * h
j = 2 / 100
k = 1 + j
l = 4 / 100
m = 1 + l
n = k * m
o = 8000 * n
p = i - o
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a ) 60 , b ) 90 , c ) 120 , d ) 150 , e ) 240 | b | add(divide(multiply(210, 2), 5), multiply(2, 5)) | the number of timeshare condos available at sunset beach is 2 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 210 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 2 / 5 x = 210 hence , x = 150 . so , number of timeshare condos available at playa del mar = 150 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 2 / 5 x = 2 / 5 x = 3 / 5 ( 150 ) = 90 the correct answer is b" | a = 210 * 2
b = a / 5
c = 2 * 5
d = b + c
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a ) s . 380 , b ) s . 600 , c ) s . 420 , d ) s . 400 , e ) s . 480 | e | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3840) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3840 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 3840 × ( 1 / 8 ) = 480 answer : option e" | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 3840
|
a ) 743 , b ) 154 , c ) 852 , d ) 741 , e ) 785 | b | divide(subtract(13698, 14), 89) | on dividing 13698 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 13698 - 14 ) / 89 = 154 answer ( b ) | a = 13698 - 14
b = a / 89
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a ) $ 14,980 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 30 % next year and that the price of type b products will increase by 10 % next year . this year , the total amount paid for type a products was $ 4500 and the total price paid for type b products was $ 8300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.30 * 4500 = 5850 cost of type b products next year = 1.1 * 8300 = 9130 total 5850 + 9130 = 14980 option a" | a = 3 / 4
b = a * 1000
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a ) 14 , b ) 17 , c ) 11 , d ) 19 , e ) 99 | a | subtract(divide(38, const_2), 5) | a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 38 now , the son age 5 years back was | "let the son ' s present age be x years . then , ( 38 - x ) = x x = 19 . son ' s age 5 years back = ( 19 - 5 ) = 14 years answer : a" | a = 38 / 2
b = a - 5
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a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people x who purchased only book a ? | "this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b and this is 1 / 2 of those that purchased a . so , 1500 purchased a . less the 500 that purchased both , x = 1000 purchased a only . ( this is much simpler to solve using the venn diagram ) . correct answer is d . 1000" | a = 500 / 2
b = 500 + a
c = b * 2
d = c - 500
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a ) rs . 738 , b ) rs . 638 , c ) rs . 650 , d ) rs . 730 , e ) rs . 735 | a | subtract(825, divide(multiply(subtract(854, 825), 3), 4)) | a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 854 in 4 years . the sum is ? | "s . i . for 1 year = ( 854 - 825 ) = rs . 29 s . i . for 3 years = 29 * 3 = rs . 87 principal = ( 825 - 87 ) = rs . 738 . answer : a" | a = 854 - 825
b = a * 3
c = b / 4
d = 825 - c
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | divide(150, multiply(const_0_2778, 54)) | in how many seconds will a train 150 meters long pass an oak tree , if the speed of the train is 54 km / hr ? | "speed = 54 * 5 / 18 = 15 m / s time = 150 / 15 = 10 seconds the answer is c ." | a = const_0_2778 * 54
b = 150 / a
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a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | c | divide(subtract(4, 3), subtract(10, 4)) | cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 3 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 4 - 3 ) / ( 10 - 4 ) = 1 / 6 so they should be mixed in the ratio 1 : 6 answer - c" | a = 4 - 3
b = 10 - 4
c = a / b
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive e difference between n and m ? | we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result e will be 8 - 5 = 3 and answer b | a = 5 - 2
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a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | d | add(subtract(multiply(add(negate(subtract(4, multiply(4, const_2))), subtract(multiply(4, const_3.0), 4)), const_3), subtract(multiply(4, 4), 4)), add(negate(subtract(4, multiply(4, const_2))), subtract(multiply(4, 4), 4))) | 4 years ago , paula was 4 times as old as karl . in 4 years , paula will be twice as old as karl . what is the sum of their ages now ? | "p - 4 = 4 ( k - 4 ) and so p = 4 k - 12 p + 4 = 2 ( k + 4 ) ( 4 k - 12 ) + 4 = 2 k + 8 k = 8 p = 20 p + k = 28 the answer is d ." | a = 4 * 2
b = 4 - a
c = negate + (
d = 4 * 3
e = d - 4
f = c * e
g = f - 3
h = 4 * 4
i = h - 4
j = g + i
|
a ) 25 , b ) 75 , c ) 125 , d ) 225 , e ) 625 | c | power(5, const_3) | a palindrome is a word or a number that reads the same forward and backward . for example , 2442 and 111 are palindromes . if 5 - digit palindromes are formed using one or more of the digits 1 , 2 , 3 , 4 , and 5 , how many palindromes are possible ? | there are 5 choices for each of the first three digits . the number of possible palindromes is 5 ^ 3 = 125 . the answer is c . | a = 5 ** 3
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a ) $ 3500 , b ) $ 5000 , c ) $ 3150 , d ) $ 7000 , e ) $ 10000 | c | multiply(divide(45000, const_100), subtract(40, 33)) | if the personal income tax rate is lowered from 40 % to 33 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 45000 ? | "saving = ( 40 - 33 ) % of 45000 = 3150 . answer : c" | a = 45000 / 100
b = 40 - 33
c = a * b
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a ) 15.27 , b ) 16.27 , c ) 17.27 , d ) 18.27 , e ) 19.27 | a | multiply(divide(add(multiply(const_2, 10), const_2), add(const_3, const_4)), multiply(multiply(5.4, 5.4), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | the length of minute hand of a clock is 5.4 cm . what is the area covered by this in 10 minutes | "area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.4 ^ 2 ) * ( 60 / 360 ) = 15.27 cm ^ 2 answer : a" | a = 2 * 10
b = a + 2
c = 3 + 4
d = b / c
e = 5 * 4
f = 1 * const_60
g = 100 * const_3_6
h = f / g
i = e * h
j = d * i
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a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 70 min | d | multiply(const_10, multiply(const_1, 4)) | a pipe takes a hours to fill the tank . but because of a leakage it took 4 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 240 x = 80 min answer : d" | a = 1 * 4
b = 10 * a
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a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(30,000, 4347) | the compound interest on $ 30,000 at 7 % per annum is $ 4347 . the period ( in years ) is : | "amount = $ ( 30000 + 4347 ) = $ . 34347 . let the time be n years . then , 30000 1 + 7 n = 34347 100 107 n = 34347 = 11449 = 107 2 100 30000 10000 100 n = 2 years . answer a - 2" | a = 30 / 0
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a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 18 minutes , e ) 20 minutes | b | divide(50, add(divide(50, 24), add(divide(50, 40), divide(50, 30)))) | jonathan can type a 50 page document in 40 minutes , susan can type it in 30 minutes , and jack can type it in 24 minutes . working together , how much time will it take them to type the same document ? | "you may set up common equation like this : job / a + job / b + job / c = job / x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 50 / 40 + 50 / 30 + 50 / 24 = 50 / x if you solve this equation , you get the same answer b ( 10 )" | a = 50 / 24
b = 50 / 40
c = 50 / 30
d = b + c
e = a + d
f = 50 / e
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a ) 550 , b ) 1500 , c ) 250 , d ) 1300 , e ) 400 | b | divide(multiply(6800, 30), 136) | find the annual income derived by investing $ 6800 in 30 % stock at 136 . | by investing $ 136 , income obtained = $ 30 . by investing $ 6800 , income obtained = $ [ ( 30 / 136 ) * 6800 ] = $ 1500 . answer b . | a = 6800 * 30
b = a / 136
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a ) 20 , b ) 25 , c ) 55 , d ) 75 , e ) 80 | d | subtract(multiply(90, 4), multiply(85, const_3)) | joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 90 . he was allowed to drop his lowest score . after doing so , his average test score improved to 85 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 90 . so what we can assume is that we have 4 test scores , each 90 . he dropped his lowest score and the avg went to 95 . this means that the lowest score was not 90 and other three scores had given the lowest score 5 each to make it up to 90 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 90 . so the lowest score = 90 - 15 = 75 answer ( d )" | a = 90 * 4
b = 85 * 3
c = a - b
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a ) 87 , b ) 12 , c ) 30 , d ) 28 , e ) 17 | e | divide(add(add(10, 12), multiply(6, 2)), const_2) | the average age of 6 men increases by 2 years when two women are included in place of two men of ages 10 and 12 years . find the average age of the women ? | "10 + 12 + 6 * 2 = 34 / 2 = 17 answer : e" | a = 10 + 12
b = 6 * 2
c = a + b
d = c / 2
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a ) 18 , b ) 20 , c ) 21 , d ) 24 , e ) 25 | c | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 35), 15), 25)) | it takes 35 identical printing presses 15 hours to print 500,000 papers . how many hours would it take 25 of these printing presses to print 500,000 papers ? | "35 printing presses can do 1 / 15 of the job each hour . 25 printing presses can do 5 / 7 * 1 / 15 = 1 / 21 of the job each hour . the answer is c ." | a = 1000 / 2
b = a * 1000
c = 1000 / 2
d = c * 1000
e = d / 35
f = e / 15
g = f * 25
h = b / g
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a ) 407.09 , b ) 479.75 , c ) 523.93 , d ) 596.95 , e ) none of these | d | subtract(add(3639, 11.95), 3054) | 3639 + 11.95 - x = 3054 . find the value of x . | "explanation : let 3639 + 11.95 – x = 3054 then , x = ( 3639 + 11.95 ) – 3054 = 3650.95 – 3054 = 596.95 answer : d" | a = 3639 + 11
b = a - 3054
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a ) 8 % , b ) 15 % , c ) 45 % , d ) 42 % , e ) 56 % | d | multiply(divide(3, 20), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 104 million in sales ? | "( 9 / 104 ) / ( 3 / 20 ) = 30 / 54 = 57,6 % it means that 9 / 108 represents only 57,6 % . therefore a decrease of 42 % . answer d" | a = 3 / 20
b = a * 100
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a ) $ 4 , b ) $ 0.2 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | subtract(const_1, divide(80, const_100)) | the cost per pound of milk powder and coffee were the same in june . in july , the price of coffee shot up by 300 % and that of milk powder dropped by 80 % . if in july , a mixture containing equal quantities of milk powder and coffee costs $ 6.30 for 3 lbs , how much did a pound of milk powder cost in july ? | lets assume price of coffee in june = 100 x price of milk powder in june = 100 x price of coffee in july = 400 x ( because of 300 % increase in price ) price of milk powder in july = 20 x ( because of 80 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of milk powder in july will be = 600 x + 30 x = 630 x as per question 630 x = 6.30 $ x = 0.01 s so the price of milk powder in july = 20 x = 20 x 0.01 = 0.2 $ / pound answer b | a = 80 / 100
b = 1 - a
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a ) 6 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | e | lcm(18, 24) | a certain class of students is being divided into teams . the class can either be divided into 18 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | prime factorization of 18 = 2 * 3 ^ 2 prime factorization of 24 = 2 ^ 3 * 3 lcm of the given numbers = 2 ^ 3 * 3 ^ 2 = 72 answer e | a = math.lcm(18, 24)
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a ) 38 min , b ) 36 min , c ) 31 min , d ) 40 min , e ) 49 min | d | divide(multiply(50, divide(44, 50)), add(50, 5)) | a bus started its journey from mumbai and reached pune in 44 min with its average speed of 50 km / hr . if the average speed of the bus is increased by 5 km / hr , how much time will it take to cover the same distance ? | "sol . distance between ramgarh and devgarh = ( 50 * 44 ) / 60 = 110 / 3 average speed of the bus is increased by 5 km / hr then the speed of the bus = 55 km / hr required time = 110 / 3 * 60 / 55 = 40 min d" | a = 44 / 50
b = 50 * a
c = 50 + 5
d = b / c
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a ) a ) 19.7 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.7 , e ) e ) 22 | d | subtract(subtract(25, divide(30, const_100)), divide(30, 10)) | daniel went to a shop and bought things worth rs . 25 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 10 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 10 % ∴ ( 25 − 30 / 100 − x ) 10 / 100 = 30 / 100 ⇒ 10 ( 25 − 0.3 − x ) = 30 ⇒ ( 25 − 0.3 − x ) = 3 ⇒ x = 25 − 0.3 − 3 = 21.7 d" | a = 30 / 100
b = 25 - a
c = 30 / 10
d = b - c
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a ) 28 . , b ) 29 . , c ) 30 . , d ) 31 . , e ) 32 . | c | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), factorial(subtract(const_4, const_1)))) | how many positive integers less than 250 are there such that they are multiples of 15 or multiples of 16 ? | "250 / 15 = 16 ( plus remainder ) so there are 16 multiples of 15 250 / 16 = 15 ( plus remainder ) so there are 15 multiples of 16 we need to subtract 1 because 15 * 16 is a multiple of both so it was counted twice . the total is 16 + 15 - 1 = 30 the answer is c ." | a = 4 + 15
b = a - 1
c = math.factorial(b)
d = math.factorial(15)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) - 6 , b ) - 4 , c ) - 2 , d ) 0 , e ) 3 | d | subtract(multiply(5, 2), 5) | if ( 5 - x ) / ( 5 + x ) = x , what is the value of x ^ 2 + 6 x - 5 ? | "( 5 - x ) = x * ( 5 + x ) ( 5 - x ) = 5 x + x ^ 2 0 = x ^ 2 + 6 x - 5 the answer is d ." | a = 5 * 2
b = a - 5
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a ) 29 / 16 , b ) 19 / 16 , c ) 15 / 16 , d ) 9 / 13 , e ) 0 | e | divide(9, 16) | 1 / 2 - [ ( 2 / 7 * 7 / 32 ) + 1 ] + 9 / 16 = | 1 / 2 - [ ( 2 / 7 * 7 / 32 ) + 1 ] + 9 / 16 = 1 / 2 - [ ( 1 / 16 ) + 1 ] + 9 / 16 = 1 / 2 - [ 17 / 16 ] + 9 / 16 = 8 / 16 - 17 / 16 + 9 / 16 = 0 the answer is e . | a = 9 / 16
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a ) 162 , b ) 158 , c ) 159 , d ) 130 , e ) 160 | a | subtract(2468, divide(2468, add(const_1, divide(7, const_100)))) | if a person buy radio worth rs 2468 and pay 7 % sales . how much price of radio should reduce to pay only rs 2468 . | ( x + . 07 x ) = 2468 1.07 x = 2468 x = 2306 reduce = 2468 - 2306 = 162 answer : a | a = 7 / 100
b = 1 + a
c = 2468 / b
d = 2468 - c
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a ) 15 days , b ) 18 days , c ) 21 days , d ) 24 days , e ) 27 days | b | divide(multiply(48, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 30), multiply(divide(add(24, 60), 4), 3)), multiply(divide(8, 2), 3))) | 3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 48 days . if 60 more cows and 30 more buffaloes are brought in , how many days will the fodder last ? | 2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 48 days . 60 cows + 30 buffaloes = 30 oxen + 20 oxen = 50 oxen the new total is equal to 80 oxen instead of 30 oxen . 30 * 48 / 80 oxen = 18 days the answer is b . | a = 24 / 4
b = a * 3
c = 15 + b
d = 8 / 2
e = d * 3
f = c + e
g = 48 * f
h = 15 + 30
i = 24 + 60
j = i / 4
k = j * 3
l = h + k
m = 8 / 2
n = m * 3
o = l + n
p = g / o
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a ) 185 , b ) 190 , c ) 160 , d ) 120 , e ) 125 | e | power(add(50, const_4), const_4) | the difference between a number and its three - fifth is 50 . what is the number ? | "let the number be x . then , x - 3 / 5 x = 50 = > 2 / 5 x = 50 x = ( 50 * 5 ) / 2 = 125 . answer : e" | a = 50 + 4
b = a ** 4
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a ) 1 / 6 , b ) 1 / 4 , c ) 2 / 3 , d ) 3 / 4 , e ) 11 / 12 | e | subtract(const_1, multiply(subtract(const_1, divide(const_2, 3)), subtract(const_1, divide(3, 4)))) | two thirds of the roads from a to b are at least 5 miles long , and 3 / 4 of the roads from b to c are at least 5 miles long . if you randomly pick a road from a to b and then randomly pick a road from b to c , what is the probability that at least one of the roads you pick is at least 5 miles long ? | so please : find the probability of the event thatnoneof the roads you pick will be at least 5 miles long and subtract from 1 to get the probability thatat least oneof the roads you pick will be at least 5 miles long : p = 1 - 1 / 3 * 1 / 4 = 11 / 12 . answer : e . | a = 2 / 3
b = 1 - a
c = 3 / 4
d = 1 - c
e = b * d
f = 1 - e
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a ) − 100 , b ) 0.01 , c ) 0.25 , d ) 4 , e ) 8 | b | divide(divide(2, 2), const_1000) | if x = 6 and y = − 2 , what is the value of ( x − 2 y ) ^ y ? | "( x − 2 y ) ^ y = ( 6 − 2 ( − 2 ) ) ^ − 2 = 1 / ( 6 + 4 ) ^ 2 = 1 / 100 = 0.01 answer : b" | a = 2 / 2
b = a / 1000
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a ) 230 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 350 m | a | subtract(multiply(divide(add(120, 80), const_3_6), 9), 270) | a 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "speed of first train = ( 120 * 5 / 18 ) m / sec = 100 / 3 m / sec speed of other train = ( 80 * 5 / 18 ) m / sec = 200 / 9 m / sec time to cross each other = 9 sec let x be the length of 2 nd train therefore 9 = ( 270 + x ) / ( 100 / 3 + 200 / 9 ) 500 = 270 + x 230 m = x answer : a" | a = 120 + 80
b = a / const_3_6
c = b * 9
d = c - 270
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a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money t did bob receive ? | "notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ t = 20 answer : b" | a = 32 * 2
b = a / 3
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a ) 16 , b ) 17 , c ) 18 , d ) 15 , e ) 14 | d | multiply(3, divide(40, add(3, 5))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 40 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 5 km / h . what is the distance traveled by maxwell when they meet in the middle ? | "consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 3 xt = 40 - 5 xt ( i . e distance = speed x time ) = > 8 t = 40 hence t = 5 oa = 3 x 5 = 15 km answer : d" | a = 3 + 5
b = 40 / a
c = 3 * b
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a ) 5 % , b ) 7 % , c ) 3 % , d ) 9 % , e ) 15 % | a | subtract(const_100, add(add(add(subtract(const_100, 70), subtract(const_100, 75)), subtract(const_100, 65)), subtract(const_100, 95))) | in a urban village of india named ` ` owlna ' ' , 70 % people have refrigerator , 75 % people have television , 65 % people got computers and 95 % got air - conditionor . how many people ( minimum ) got all these luxury . | a 10 % 100 - [ ( 100 - 65 ) + ( 100 - 95 ) + ( 100 - 75 ) + ( 100 - 70 ) ] = 100 - ( 30 + 25 + 35 + 5 ) = 100 - 95 | a = 100 - 70
b = 100 - 75
c = a + b
d = 100 - 65
e = c + d
f = 100 - 95
g = e + f
h = 100 - g
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