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 ) 52 , b ) 49 , c ) 95 , d ) 88 , e ) 89 | e | add(multiply(add(15, subtract(15, 3)), 3), 8) | in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 8 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 8 total people can seat = 81 + 8 = 89 answer : e" | a = 15 - 3
b = 15 + a
c = b * 3
d = c + 8
|
a ) a . 46000 , b ) b . 47000 , c ) c . 48000 , d ) d . 49000 , e ) e . 50000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | if a town of 45,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a = 3 + 4
b = 5 * a
c = b * 100
d = 3 + 4
e = d + 3
f = c * e
|
a ) 15 / 56 , b ) 3 / 28 , c ) 1 / 28 , d ) 1 / 15 , e ) 1 / 168 | d | multiply(3, multiply(divide(subtract(3, const_1), 10), divide(const_1, subtract(10, const_1)))) | a computer wholesaler sells 10 different computers and each is priced differently . if the wholesaler chooses 3 computers for display at a trade show , what is the probability ( all things being equal ) that the two most expensive computers will be among the 3 chosen for display ? | since , two of the choices are prefixed , we are free to choose 1 from the rest of the 8 avilable . so 8 c 1 is the numerator . total no of ways we can choose 3 from 10 is 10 c 3 , which is the denominator . so , the probability : 8 c 1 / 10 c 3 = 1 / 15 ans is d . | a = 3 - 1
b = a / 10
c = 10 - 1
d = 1 / c
e = b * d
f = 3 * e
|
['a ) 90', 'b ) 85', 'c ) 70', 'd ) 35', 'e ) 20'] | e | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | how many internal diagonals does a octagon ( eight sided polygon ) have ? | number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 8 no . of diagonals = 8 ( 8 - 3 ) / 2 = 20 ans e | a = 2 * 4
b = a - 3
c = 2 * 4
d = c / 2
e = b * d
|
a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | d | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 17) | on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 17 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 16 more transactions than cal . jade = 66 + 17 = 83 jade handled = 83 transactions . answer : d" | a = 90 / 100
b = 10 + 100
c = a * b
d = c * 2
e = d / 3
f = e + 17
|
a ) 30 , b ) 90 , c ) 120 , d ) 172 , e ) 180 | e | subtract(600, subtract(add(multiply(divide(600, const_100), 56), multiply(divide(600, const_100), 44)), multiply(divide(600, const_100), 30))) | among 600 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 180 using your analysis : we see that the total number of students who study either maths or sociology = 264 + 336 - 180 = 420 so , in the image we know that the number of students in the zone with the black boundary = 420 let ' s assume the number of students who studyonlybiology to beb ( this is the number that we have to maximize ) and , let ' s assume the number of students who study none of the three subjects , that is the number of students in the white space = w since the total number of students = 600 , we can write : 420 + b + w = 600 or , b + w = 600 - 420 = 180 that is , b = 180 - w so , the maximum value ofbwill happen forw = 0 this is how we get , the maximum value ofb = 180 e" | a = 600 / 100
b = a * 56
c = 600 / 100
d = c * 44
e = b + d
f = 600 / 100
g = f * 30
h = e - g
i = 600 - h
|
a ) 12 , b ) 24 , c ) 48 , d ) 60 , e ) 55 | d | multiply(subtract(multiply(3, 4), 20), 20) | if a farmer sells 15 of his goats , his stock of feed will last for 4 more days than planned , but if he buys 20 more goats , he will run out of feed 3 days earlier than planned . if no goats are sold or bought , the farmer will be exactly on schedule . how many goats does the farmer have ? | "say farmer has n goat and he is good for d days . : - we have 3 equations given in question : - ( n - 15 ) * d + 4 = ( n + 20 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 20 d - 3 n = 60 4 n - 15 d = 60 = > n = 60 ans d it is !" | a = 3 * 4
b = a - 20
c = b * 20
|
a ) 60 , b ) 20 , c ) 19 , d ) 29 , e ) 39 | a | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 3), divide(2, 3))) | 15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 3 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 3 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 3 * ( 2 / 3 ) ) days = > 60 days answer : a" | a = 15 * 21
b = a * 8
c = 21 * 3
d = 2 / 3
e = c * d
f = b / e
|
a ) 7 meters , b ) 9 meters , c ) 8 meters , d ) 5 meters , e ) 3 meters | e | multiply(divide(72, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | if a tire rotates at 400 revolutions per minute when the car is traveling 72 km / h , what is the circumference of the tire ? | "400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 72,000 m : c is the circumference c = 3 meters correct answer e" | a = 2 * 3
b = a * 10
c = b * 400
d = 72 / c
e = d * 1000
|
a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800 | d | divide(subtract(divide(multiply(divide(divide(subtract(50000, add(add(4000, 5000), 5000)), const_3), 50000), 35000), const_4), const_100), const_10) | a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 , c receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d | a = 4000 + 5000
b = a + 5000
c = 50000 - b
d = c / 3
e = d / 50000
f = e * 35000
g = f / 4
h = g - 100
i = h / 10
|
a ) 50 , b ) 40 , c ) 60 , d ) 90 , e ) 70 | d | divide(900, add(divide(900, const_100), const_1)) | the sum of number of boys and girls in a school is 900 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | we have x + x % of 900 = 900 x + x / 100 * 900 = 900 10 * x = 900 x = 90 answer is d | a = 900 / 100
b = a + 1
c = 900 / b
|
a ) 3 , b ) 9 , c ) 11 , d ) 12 , e ) 15 | a | add(reminder(multiply(3, const_4), const_10), const_1) | one half of a two digit number exceeds its one fourth by 3 . what is the sum of the digits of the number ? | explanation : x / 2 – x / 4 = 3 = > x = 12 1 + 2 = 3 a | a = 3 * 4
b = reminder + (
|
a ) 168 , b ) 100 , c ) 155 , d ) 160 , e ) 200 | c | divide(add(110, 200), 2) | a student chose a number , multiplied it by 2 , then subtracted 200 from the result and got 110 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 200 = 110 2 x = 310 x = 155 correct answer c" | a = 110 + 200
b = a / 2
|
a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 8 . | d | divide(multiply(3, 20), 12) | 20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ? | "total work = 20 * 3 = 60 beaver hours 12 beaver * x = 60 beaver hours x = 60 / 12 = 5 answer : d" | a = 3 * 20
b = a / 12
|
a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 81 | e | divide(add(add(multiply(90, 6), multiply(60, 4)), 30), add(6, 4)) | the average expenditure of a labourer for 6 months was 90 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | income of 6 months = ( 6 × 90 ) – debt = 540 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 810 average monthly income = 810 ÷ 10 = 81 answer e | a = 90 * 6
b = 60 * 4
c = a + b
d = c + 30
e = 6 + 4
f = d / e
|
a ) 300 kmph , b ) 360 kmph , c ) 600 kmph , d ) 720 kmph , e ) 750 kmph | d | divide(divide(multiply(240, 5), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 1 hours , it must travel at a speed of : | "explanation : distance = ( 240 x 5 ) = 1200 km . speed = distance / time speed = 1200 / ( 5 / 3 ) km / hr . required speed = ( 1200 x 3 / 5 ) km / hr = 720 km / hr . answer is d" | a = 240 * 5
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
|
a ) 432 , b ) 108 , c ) 252 , d ) 992 , e ) 212 | b | multiply(multiply(divide(48, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(48, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 48 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 48 l = 18 b = 6 lb = 18 * 6 = 108 answer : b" | a = 3 * 2
b = 1 * 2
c = a + b
d = 48 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 48 / h
j = e * i
|
a ) 195 / 63 , b ) 2.879 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | b | add(divide(sqrt(1.1), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | find the value of ( √ 1.1 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.1 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) = > 2.879 answer is b" | a = math.sqrt(1)
b = math.sqrt(0)
c = a / b
d = math.sqrt(1)
e = math.sqrt(0)
f = d / e
g = c + f
|
a ) 3377 , b ) 2688 , c ) 2688 , d ) 9490 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 6 years the amount to be paid is ? | "a = 7500 ( 26 / 25 ) ^ 6 = 9490 answer : d" | a = 100 + 4
b = a / 100
c = 100 + 4
d = c / 100
e = b * d
f = 7500 * e
|
a ) 8 , b ) 9 , c ) 14 , d ) 17 , e ) 20 | b | power(3, 2) | if x ^ 2 + y ^ 2 = 15 and xy = 3 , then ( x − y ) ^ 2 = | but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 15 - 2 * 3 = 9 . . answer b | a = 3 ** 2
|
a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | a | subtract(29, add(add(7, 1), 20)) | if h = { 1 , 7 , 18 , 20 , 29 , 33 } , how much less is the mean of the numbers in h than the median of the numbers in h ? | this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd , the middle term is the median . if the number of terms is even , the average of middle two terms is the median coming to this question , mean = ( 1 + 7 + 18 + 20 + 29 + 33 ) / 6 = 18 median = ( 18 + 20 ) / 2 = 19 difference = 1 option a | a = 7 + 1
b = a + 20
c = 29 - b
|
a ) 1800 , b ) 1760 , c ) 1400 , d ) 2600 , e ) 3600 | b | multiply(multiply(sqrt(484), const_4), 20) | what will be the cost of house to paint which area equal to 484 sq ft , if the price per foot of building is rs . 20 | let the side of the square plot be a ft . a 2 = 484 = > a = 22 length of the fence = perimeter of the plot = 4 a = 88 ft . cost of building the fence = 88 * 20 = rs . 1760 . answer : b | a = math.sqrt(484)
b = a * 4
c = b * 20
|
a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 0 , d ) 1 / 9 , e ) 1 / 5 | d | divide(multiply(1, power(2, const_2)), power(6, const_2)) | x varies inversely as square of y . given that y = 2 for x = 1 . the value of x for y = 6 will be equal to ? | "given x = k / y 2 , where k is a constant . now , y = 2 and x = 1 gives k = 4 . x = 4 / y 2 = > x = 4 / 62 , when y = 6 = > x = 4 / 36 = 1 / 9 . answer : d" | a = 2 ** 2
b = 1 * a
c = 6 ** 2
d = b / c
|
a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability r that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability r = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a = math.comb(4, 1)
b = a * 2
c = math.comb(6, 3)
d = b / c
e = 5 * 5
f = e * 4
g = d * f
|
a ) 6660 , b ) 3600 , c ) 2400 , d ) 4900 , e ) 5400 | d | multiply(multiply(subtract(50, const_1), const_4), divide(50, const_2)) | there are , in a certain league , 50 teams , and each team face another team for a total of 4 times . how many games are played in the season ? | by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 4900 option d . | a = 50 - 1
b = a * 4
c = 50 / 2
d = b * c
|
a ) 11 , b ) 20 , c ) 100 , d ) 77 , e ) 12 | c | multiply(divide(subtract(2000, 1000), 1000), const_100) | a cycle is bought for rs . 1000 and sold for rs . 2000 , find the gain percent ? | "1000 - - - - 1000 100 - - - - ? = > 100 % answer : c" | a = 2000 - 1000
b = a / 1000
c = b * 100
|
a ) 57.6 sec , b ) 20.2 sec , c ) 31.8 sec , d ) 50.4 sec , e ) none of the above | a | divide(add(360, 520), multiply(55, const_0_2778)) | a 360 m long train is running at a speed of 55 km / hr . it crossed a platform of length 520 m in ? | "speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 360 m + 520 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 880 m time = distance / speed = 880 x 18 / ( 5 x 55 ) = 57.6 sec ans is : a" | a = 360 + 520
b = 55 * const_0_2778
c = a / b
|
a ) 3 , b ) 0 , c ) 5 , d ) 4 , e ) 2 | d | divide(subtract(multiply(5, 1.6), multiply(2, 1.4)), 1.3) | 5 x 1.6 - 2 x 1.4 / 1.3 = ? | "given expression = ( 8 - 2.8 ) / 1.3 = 5.2 / 1.3 = 52 / 1.3 = 4 . answer is d ." | a = 5 * 1
b = 2 * 1
c = a - b
d = c / 1
|
a ) 76 kmph , b ) 75 kmph , c ) 87 kmph , d ) 56 kmph , e ) 84 kmph | e | divide(add(98, 70), const_2) | the speed of a car is 98 km in the first hour and 70 km in the second hour . what is the average speed of the car ? | "s = ( 98 + 70 ) / 2 = 84 kmph answer : e" | a = 98 + 70
b = a / 2
|
a ) 10 days , b ) 9 days , c ) 6 days , d ) 8 days , e ) 7 days | d | divide(multiply(multiply(12, 10), 8), multiply(10, 12)) | 12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 10 days , working 12 hours a day , the number of men required ? | "that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 12 × 10 ? ( i . e . no . of men required ) = 12 × 8 × 10 / 12 × 10 = 8 days d )" | a = 12 * 10
b = a * 8
c = 10 * 12
d = b / c
|
a ) 625 , b ) 367 , c ) 288 , d ) 209 , e ) 198 | a | divide(1301, add(const_1, divide(power(add(const_1, divide(4, const_100)), 9), power(add(const_1, divide(4, const_100)), 7)))) | divide rs . 1301 between a and b , so that the amount of a after 7 years is equal to the amount of b after 9 years , the interest being compounded at 4 % per annum . | explanation : let the two parts be rs . x and rs . ( 1301 - x ) . 625 x = 676 ( 1301 - x ) 1301 x = 676 x 1301 x = 676 . so , the parts are rs . 676 and rs . ( 1301 - 676 ) i . e rs . 676 and rs . 625 answer : a ) rs . 625 | a = 4 / 100
b = 1 + a
c = b ** 9
d = 4 / 100
e = 1 + d
f = e ** 7
g = c / f
h = 1 + g
i = 1301 / h
|
a ) 2.8 liters . , b ) 2.5 liters . , c ) 5 liters . , d ) 2.6 liters . , e ) 2.1 liters . | c | divide(multiply(divide(subtract(const_100, 90), const_100), 40), divide(subtract(const_100, 20), const_100)) | heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 40 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 40 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 5 liters . answer : c" | a = 100 - 90
b = a / 100
c = b * 40
d = 100 - 20
e = d / 100
f = c / e
|
a ) 3 / 5 , b ) 2 / 5 , c ) 4 / 5 , d ) 1 / 5 , e ) 6 / 5 | b | divide(4, add(subtract(const_10, 4), 4)) | david works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 4 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 4 x total tips for all months = 6 * x + 4 x = 10 x tips for august / total tips for all months = 4 x / 10 x = 2 / 5 answer : b | a = 10 - 4
b = a + 4
c = 4 / b
|
a ) 30 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 44 sec | e | multiply(const_3600, divide(divide(330, const_1000), add(25, 2))) | a train 330 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 × 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 330 × 2 / 15 ) sec = 44 sec answer : e" | a = 330 / 1000
b = 25 + 2
c = a / b
d = 3600 * c
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 65 | d | multiply(4, 15) | walking at 4 / 5 of her normal speed , a worker is 15 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is | "let v be her normal speed and let t be her normal time . d = ( 4 / 5 ) v * ( t + 15 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 4 / 5 ) v * ( t + 15 ) t / 5 = 12 t = 60 the answer is d ." | a = 4 * 15
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a ) 25 % , b ) 50 % , c ) 68 % , d ) 80 % , e ) 100 % | c | multiply(divide(divide(3, 5), divide(8, 9)), const_100) | a cylinder of height h is 8 / 9 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 8 / 9 new cylinder 3 / 5 so 3 / 5 / 8 / 9 = 3 / 5 * 9 / 8 = 12 / 15 = 4 / 5 = 0.680 or 68 % . answer c | a = 3 / 5
b = 8 / 9
c = a / b
d = c * 100
|
a ) 1 / 2 , b ) 2 , c ) 1 , d ) 4 , e ) 5 | a | multiply(const_60, divide(multiply(40, divide(15, const_60)), subtract(60, 40))) | mary passed a certain gas station on a highway while traveling west at a constant speed of 40 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least 2 hours , how long after he passed the gas station did paul catch up with mary ? | "d = rt m : r = 40 mph , t = t + 1 / 4 hr d = 40 ( t + 1 / 4 ) p : r = 60 , t = t d = 60 t since they went the same distance : 40 t + 40 / 4 = 60 t 20 t = 40 / 4 t = 1 / 2 or 1 / 2 hr a" | a = 15 / const_60
b = 40 * a
c = 60 - 40
d = b / c
e = const_60 * d
|
a ) 2 / 15 , b ) 3 / 5 , c ) 3 / 11 , d ) 1 / 12 , e ) 6 / 13 | b | subtract(const_1, multiply(4, add(divide(const_1, 15), divide(const_1, 30)))) | a can do a job in 15 days and b in 30 days . if they work on it together for 4 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 30 a + b 1 day work = 1 / 15 + 1 / 30 = 3 / 30 = 1 / 10 a + b 4 days work = 1 / 10 * 4 = 2 / 5 remaining work = 1 - 2 / 5 = 3 / 5 answer is b" | a = 1 / 15
b = 1 / 30
c = a + b
d = 4 * c
e = 1 - d
|
a ) 9 % , b ) 9.27 % , c ) 25.97 % , d ) 12 % , e ) none of these | c | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(9, 1)), subtract(9, 1)), const_100), subtract(9, 1))), subtract(9, 1)), const_100), add(divide(multiply(add(const_100, subtract(9, 1)), subtract(9, 1)), const_100), subtract(9, 1))) | the population of a city increases by 9 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 8 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 8 / 100 ) ^ 3 = 125.97 ∴ required percentage = 25.97 % answer : c" | a = 9 - 1
b = 100 + a
c = 9 - 1
d = b * c
e = d / 100
f = 9 - 1
g = e + f
h = 100 + g
i = 9 - 1
j = h * i
k = j / 100
l = 9 - 1
m = 100 + l
n = 9 - 1
o = m * n
p = o / 100
q = 9 - 1
r = p + q
s = k + r
|
a ) 90 cm , b ) 5 dm , c ) 1 m , d ) 1.1 cm , e ) none of these | b | subtract(multiply(multiply(3.3, 2.6), 1.5), divide(8000, const_1000)) | a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.5 m and its walls are 5 cm thick . the thickness of the bottom is : | "explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 150 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 150 - x ) = 8000 × 1000 = > ( 150 - x ) = 8000 × 1000 / 320 = 100 = > x = 50 cm = 5 dm . answer : b" | a = 3 * 3
b = a * 1
c = 8000 / 1000
d = b - c
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | add(const_1, divide(18, 3)) | if k is the greatest positive integer such that 3 ^ k is a divisor of 18 ! then k = | "18 / 3 = 6 18 / 9 = 2 6 + 2 = 8 k = 8 answer : e" | a = 18 / 3
b = 1 + a
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | gcd(18888, 11) | if 18888 – n is divisible by 11 , and 0 < n < 11 , what is n ? | 18,888 / 11 = 1717 with a remainder of 1 . we need to subtract the remainder to get a multiple of 11 . the answer is a . | a = math.gcd(18888, 11)
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a ) 12000 , b ) 15000 , c ) 1200 , d ) 1000 , e ) 10000 | e | divide(4400, subtract(const_1, divide(56, const_100))) | after decreasing 56 % in the price of an article costs rs . 4400 . find the actual cost of an article ? | "cp * ( 44 / 100 ) = 4400 cp = 100 * 100 = > cp = 10000 answer : e" | a = 56 / 100
b = 1 - a
c = 4400 / b
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a ) 22 % , b ) 24 % , c ) 25 % , d ) 29 % , e ) 28 % | d | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(45, const_100), subtract(800, 500))), 800), const_100) | for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 45 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amount of the sale ? | total sales = 800 comission = ( 20 / 100 ) * 500 + ( 45 / 100 ) * 300 = 100 + 135 = 235 % comission = ( 235 / 800 ) * 100 = 29.4 ~ 29 % answer is d | a = 20 / 100
b = a * 500
c = 45 / 100
d = 800 - 500
e = c * d
f = b + e
g = f / 800
h = g * 100
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a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | a | multiply(divide(10, subtract(50, 10)), const_100) | a shop owner sells 50 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 50 m cloth ) – ( c . p . of 50 m cloth ) selling price of 40 m cloth = selling price of 50 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 40 m cloth = rs . 4000 and s . p . of 40 m cloth = rs . rs . 5000 profit % = 10 / 40 × 100 = 25 % profit of 25 % was made by the merchant . a | a = 50 - 10
b = 10 / a
c = b * 100
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a ) rs . 512 , b ) rs . 552 , c ) rs . 612 , d ) rs . 622 , e ) none | c | subtract(add(add(7500, divide(multiply(7500, 4), const_100)), divide(multiply(add(7500, divide(multiply(7500, 4), const_100)), 4), const_100)), 7500) | find compound interest on rs . 7500 at 4 % per annum for 2 years , compounded annually . | "solution amount = rs [ 7500 x ( 1 + 4 / 100 ) ² ] = rs . ( 7500 x 26 / 25 x 26 / 25 ) = rs . 8112 . c . i = rs ( 8112 - 7500 ) = rs . 612 . answer c" | a = 7500 * 4
b = a / 100
c = 7500 + b
d = 7500 * 4
e = d / 100
f = 7500 + e
g = f * 4
h = g / 100
i = c + h
j = i - 7500
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a ) 632 , b ) 731 , c ) 533 , d ) 434 , e ) none of the above | a | add(multiply(subtract(38, subtract(38, 6)), const_100), subtract(38, 6)) | a number is 4 less than 6 times the other number . if the sum of both the number is 38 , what are the numbers ? | that numbers are x , 6 x - 4 so sum = x + 6 x - 4 = 38 7 x = 42 = = > x = 6 another no is 32 answer is 6 , 32 answer : a | a = 38 - 6
b = 38 - a
c = b * 100
d = 38 - 6
e = c + d
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a ) 15 / 10 , b ) 17 / 10 , c ) 16 / 10 , d ) 18 / 10 , e ) 16 / 10 | b | divide(50, multiply(add(64, 42), const_0_2778)) | two trains of length 250 m and 120 m are 50 m apart . they start moving towards each other on parallel tracks , at speeds 64 kmph and 42 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 64 + 42 ) * 5 / 18 = 7 * 5 = 29.4 mps . the time required = d / s = 50 / 29.4 = 17 / 10 sec . answer : b" | a = 64 + 42
b = a * const_0_2778
c = 50 / b
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a ) 30 m 2 , b ) 40 m 2 , c ) 47.7 m 2 , d ) 50 m 2 , e ) 55 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(38, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 38 ° is ? | "38 / 360 * 22 / 7 * 12 * 12 = 47.7 m 2 answer : c" | a = 12 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 38 / g
i = f * h
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a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | d | divide(18750, 750) | if the l . c . m of two numbers is 750 and their product is 18750 , find the h . c . f of the numbers . | "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18750 / 750 = 25 . answer : d" | a = 18750 / 750
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 2) | the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 200,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 200 ( 200,000 / 1,000 = 200 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 200 , where x represents the number of times the population doubles . a lot of people remember that 2 ^ 10 = 1,024 . hence , 2 ^ 8 = 256 , i . e . the population has to double 8 times . since it takes the population 2 minutes to double once it takes 8 * 2 minutes = 16 minutes to double 8 times . thus , solution d = 16 is correct ." | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 2
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a ) 3 : 4 , b ) 5 : 6 , c ) 7 : 8 , d ) 9 : 10 , e ) 11 : 12 | d | divide(multiply(3, 3), multiply(5, 2)) | the marks obtained by polly and sandy are in the ratio 3 : 5 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 3 : 5 = 9 : 15 sandy : willy = 3 : 2 = 15 : 10 polly : sandy : willy = 9 : 15 : 10 polly : willy = 9 : 10 the answer is d ." | a = 3 * 3
b = 5 * 2
c = a / b
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a ) 1 , b ) 9000 , c ) 9001 , d ) 90001 , e ) none | c | subtract(multiply(add(multiply(5, const_100), multiply(5, const_100)), const_10), subtract(add(multiply(5, const_100), multiply(5, const_100)), const_1)) | if the largest 3 digit number is subtracted from the smallest 5 digit number , then the remainder is | solution required remainder = ( 10000 - 999 ) = 9001 answer c | a = 5 * 100
b = 5 * 100
c = a + b
d = c * 10
e = 5 * 100
f = 5 * 100
g = e + f
h = g - 1
i = d - h
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a ) 3 : 2 , b ) 9 : 2 , c ) 5 : 4 , d ) 1 : 4 , e ) 18 : 4 | c | divide(50000, 40000) | x and y started a business investing rs . 50000 and rs 40000 respectively . in what ratio the profit earned after 2 years be divided between x and y respectively ? | x : y = 50000 : 40000 = 5 : 4 answer : c | a = 50000 / 40000
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a ) 2 , 9 , b ) 4 , 9 , c ) 8 , 9 , d ) 18 , 8 , e ) 7 , 9 | d | divide(divide(add(26, 10), const_2), const_2) | a man can row downstream at 26 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 26 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 8 . answer : d" | a = 26 + 10
b = a / 2
c = b / 2
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a ) 30 , b ) 28 , c ) 27 , d ) 50 , e ) 25 | d | add(multiply(const_4, 3.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 3.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 3.9421 + 1 ) = 10 ( 4.9421 ) = 49.421 x = antilog ( 49.421 ) therefore , number of digits in x = 50 . answer : d" | a = 4 * 3
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
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a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | c | multiply(12, const_1) | if $ 5,000 is invested in an account that earns 12 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.12 / 2 ) ^ 2 = 5412 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 12 / 100 ) = 5400 difference = 5412 - 5400 = 12 $ c" | a = 12 * 1
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a ) 90 , b ) 882 , c ) 277 , d ) 261 , e ) 281 | a | multiply(180, divide(180, add(add(5, 12), 13))) | the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 180 m , its area is ? | "5 x + 12 x + 13 x = 180 = > x = 6 a = 30 , b = 72 , c = 78 s = ( 30 + 72 + 78 ) / 2 = 90 answer : a" | a = 5 + 12
b = a + 13
c = 180 / b
d = 180 * c
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a ) 176 , b ) 288 , c ) 363 , d ) 396 , e ) 484 | a | multiply(multiply(multiply(power(2, 2), 2), divide(11, 2)), 2) | if 2 ^ 4 , 2 ^ 3 , and 11 ^ 3 are factors of the product of 1,452 and w , where w is a positive integer , what is the smallest possible value of w ? | i will go with a ( pending elements to match is 2 ^ 2 * 2 ^ 2 * 11 ^ 1 = 176 | a = 2 ** 2
b = a * 2
c = 11 / 2
d = b * c
e = d * 2
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a ) - 29 , b ) 39 , c ) - 30 , d ) 40 , e ) 60 | d | add(39, const_1) | what is the sum of 80 consecutive integers from - 39 inclusive , in a increasing order ? | "from - 39 to - 1 - - > 39 nos . zero - - > 1 number from + 1 to + 39 - - > 39 nos . when we add up nos . from - 39 to + 39 sum will be zero . total 79 nos will be added . 80 th number will be 40 . sum of these 80 nos . = 40 . d is the answer ." | a = 39 + 1
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a ) 277 , b ) 1088 , c ) 277 , d ) 266 , e ) 121 | b | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(70, const_100))) | the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 70 % profit ? | "let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 170 % of rs . 640 = 170 / 100 * 640 = rs . 1088 . answer : b" | a = 832 - 448
b = a / 2
c = 832 - b
d = 70 / 100
e = 1 + d
f = c * e
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a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 28 | b | divide(subtract(multiply(46, const_4), 136), const_2) | a man has some hens and cows . if the number of heads be 46 and the number of feet equals 136 , then the number of hens will be | "explanation : let number of hens = h and number of cows = c number of heads = 46 = > h + c = 46 - - - ( equation 1 ) number of feet = 136 = > 2 h + 4 c = 136 = > h + 2 c = 68 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 68 - 46 = > c = 22 substituting the value of c in equation 1 , we get h + 22 = 46 = > h = 46 - 22 = 24 i . e . , number of hens = 24 answer : b" | a = 46 * 4
b = a - 136
c = b / 2
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a ) 187 , b ) 211 , c ) 207 , d ) 219 , e ) 227 | b | subtract(multiply(13, 17), add(const_10, const_1)) | find the smallest number which when divided by 13 and 17 leaves respective remainders of 3 and 7 . | "let ' n ' is the smallest number which divided by 13 and 17 leaves respective remainders of 3 and 7 . required number = ( lcm of 13 and 17 ) - ( common difference of divisors and remainders ) = ( 221 ) - ( 10 ) = 211 . answer : b" | a = 13 * 17
b = 10 + 1
c = a - b
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a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 61 % | e | multiply(subtract(const_1, multiply(divide(subtract(const_100, 25), const_100), divide(13, 20))), const_100) | a ' s speed is 20 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 25 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 60 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 20 / 13 of b , therefore , b ' s speed is 13 / 20 of a distance covered by b = speed x time = ( 13 / 20 ) x ( 0.6 ) = 0.39 % ( which means b would have covered 0.39 of the race length during the time in which a has covered 0.6 of the race length . therefore to win , b needs a headstart of ( 1 - 0.39 = ) 0.61 of the race length . so the answer is e" | a = 100 - 25
b = a / 100
c = 13 / 20
d = b * c
e = 1 - d
f = e * 100
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a ) 11 , b ) 12 , c ) 17 , d ) 14 , e ) 15 | c | add(11, add(4, 2)) | find the constant k so that : - x 2 - ( k + 11 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "- x 2 - ( k + 11 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 11 ) x - 8 = - x 2 + 6 x - 8 - ( k + 11 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 17 : solve the above for k correct answer c" | a = 4 + 2
b = 11 + a
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a ) 4 , b ) 10 , c ) 16 , d ) 20 , e ) 25 | a | divide(multiply(20, 20), multiply(10, 10)) | what is the maximum number of pieces of birthday cake of size 10 ” by 10 ” that can be cut from a cake 20 ” by 20 ” ? | "the prompt is essentially asking for the maximum number of 10 x 10 squares that can be cut from a larger 20 by 20 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 2 ' pieces ' each , we have ( 2 ) ( 2 ) = 4 total smaller squares ( at maximum ) . a" | a = 20 * 20
b = 10 * 10
c = a / b
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a ) 120 cm 2 , b ) 750 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2 | b | divide(multiply(60, 25), const_2) | if the sides of a triangle are 65 cm , 60 cm and 25 cm , what is its area ? | "the triangle with sides 65 cm , 60 cm and 25 cm is right angled , where the hypotenuse is 65 cm . area of the triangle = 1 / 2 * 60 * 25 = 750 cm 2 answer : b" | a = 60 * 25
b = a / 2
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a ) 1250 m , b ) 1110 m , c ) 950 m , d ) 1500 m , e ) 1300 m | d | multiply(divide(multiply(15, const_1000), const_60), 30) | a man walking at a rate of 15 km / hr crosses a bridge in 30 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 30 minutes = 15 / 18 * 30 * 60 = 1500 m answer is d" | a = 15 * 1000
b = a / const_60
c = b * 30
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a ) 9 % , b ) 9.10 % , c ) 27 % , d ) 12 % , e ) none of these | b | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(7, 1)), subtract(7, 1)), const_100), subtract(7, 1))), subtract(7, 1)), const_100), add(divide(multiply(add(const_100, subtract(7, 1)), subtract(7, 1)), const_100), subtract(7, 1))) | the population of a city increases by 7 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 6 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 6 / 100 ) ^ 3 = 119.1016 ∴ required percentage = 9.1016 % answer : b" | a = 7 - 1
b = 100 + a
c = 7 - 1
d = b * c
e = d / 100
f = 7 - 1
g = e + f
h = 100 + g
i = 7 - 1
j = h * i
k = j / 100
l = 7 - 1
m = 100 + l
n = 7 - 1
o = m * n
p = o / 100
q = 7 - 1
r = p + q
s = k + r
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a ) 96 , b ) 240 , c ) r = 256 , d ) r = 384 , e ) r = 480 | e | multiply(multiply(16, const_2), 15) | at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses r at the restaurant ? ( assume that all boxes are filled to capacity . ) | most test takers would recognize thesystemof equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e | a = 16 * 2
b = a * 15
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a ) 400 , b ) 475 , c ) 480 , d ) 625 , e ) 700 | c | multiply(divide(600, 11), 8) | a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 360 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 360 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 660 a + b + c = 60 per hour so for eight hrs = 60 * 8 = 480 = answer = c" | a = 600 / 11
b = a * 8
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a ) 33.3 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 60 % | a | subtract(multiply(divide(240, subtract(240, 60)), const_100), const_100) | on increasing the number of lines in a page by 60 , they become 240 . what is the % of increase in the no . of lines in the page ? | "explanation : number of pages increased = 60 now , the number of pages of book = 240 number of pages of the books before increase = 240 – 60 = 180 % increase in the number of pages in the book = 60 / 180 x 100 % = 33.3 % a" | a = 240 - 60
b = 240 / a
c = b * 100
d = c - 100
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a ) 7 , b ) 8 , c ) 15 , d ) 16 , e ) 14 | e | power(divide(10, 5), const_4) | if ( 2 ^ 14 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ? | "given 2 ^ 14 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 14 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans e on comparing the power of 2 = > m = 14" | a = 10 / 5
b = a ** 4
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a ) 1 ⁄ 55 , b ) 54 ⁄ 55 , c ) 45 ⁄ 55 , d ) 3 ⁄ 55 , e ) none of these | b | divide(subtract(divide(multiply(multiply(add(const_10, const_1), const_12), const_10), multiply(3, const_2)), const_4), divide(multiply(multiply(add(const_10, const_1), const_12), const_10), multiply(3, const_2))) | in a box carrying one dozen of oranges , one - third have become bad . if 3 oranges are taken out from the box at random , what is the probability that at least one orange out of the 3 oranges picked up is good ? | n ( s ) = 12 c 3 = 12 × 11 × 10 / 3 × 2 = 2 × 11 × 10 = 220 no . of selection of 3 oranges out of the total 12 oranges = 12 c 3 = 2 × 11 × 10 = 220 no . of selection of 3 bad oranges out of the total 4 bad oranges = 4 c 3 = 4 \ n ( e ) = no . of desired selection of oranges = 220 – 4 = 216 \ p ( e ) = n ( e ) / n ( s ) = 216 / 220 = 54 / 55 answer b | a = 10 + 1
b = a * 12
c = b * 10
d = 3 * 2
e = c / d
f = e - 4
g = 10 + 1
h = g * 12
i = h * 10
j = 3 * 2
k = i / j
l = f / k
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a ) 382 , b ) 782 , c ) 300 , d ) 270 , e ) 881 | c | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 220) | a goods train runs at the speed of 72 km / hr and crosses a 220 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 220 ) / 26 = 20 x = 300 m . answer : c" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 220
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a ) 11 , b ) 13 , c ) 17 , d ) 49 / 5 , e ) 23 | d | divide(multiply(add(add(5, const_3), const_2), divide(5, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 5 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 3 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 5 ) / ( b + 5 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 3 ) / ( a + 3 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 49 / 5 ans : ( d )" | a = 5 + 3
b = a + 2
c = 5 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
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a ) 18 % , b ) 20 % , c ) 21 % , d ) 35 % , e ) can not be determined | d | add(divide(multiply(10, 4), 2), add(10, 5)) | two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 35 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of | answer : d . | a = 10 * 4
b = a / 2
c = 10 + 5
d = b + c
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(subtract(18, subtract(10, 4)), 6) | of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 18 boys and 42 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | "venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you will use logic : there are 6 happy boys . there are 4 sad girls but total 10 sad children . so rest 6 sad children must be sad boys . we have 6 happy boys and 6 sad boys . total we have 18 boys . so 18 - 6 - 6 = 6 boys must be neither happy nor sad . answer ( c )" | a = 10 - 4
b = 18 - a
c = b - 6
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a ) 0 , b ) − 2 , c ) − 25 , d ) − 49 , e ) − 51 | d | add(add(negate(26), const_1), add(add(negate(26), const_1), const_1)) | the sum of all the integers k such that − 26 < k < 24 is | "since k defines a range between − 26 < k < 24 we can set 0 as the reference point for the negative values and positive values . the negative values will range from - 25 to 0 whereas the positive values will range from 0 - 23 . we can conclude that for all but - 25 and - 24 the number pairs will add to 0 . so we have left - 25 - 24 = - 49 . answer d ." | a = negate + (
b = a + 1
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a ) 1400 , b ) 1500 , c ) 1505 , d ) none of these , e ) 1506 | b | divide(const_100.0, divide(05, 75)) | evaluate 75 / . 05 | "explanation : 75 / . 05 = 7500 / 5 = 1500 option b" | a = 5 / 75
b = 100 / 0
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['a ) 77 cm', 'b ) 88 cm', 'c ) 64 cm', 'd ) 11 cm', 'e ) 18 cm'] | c | multiply(add(sqrt(power(8, const_2)), multiply(sqrt(power(8, const_2)), const_3)), const_2) | the length of a rectangle is three times of its width . if the length of the diagonal is 8 √ 10 then find the perimeter of the rectangle . | let breadth = x cm , then , length = 3 x cm x 2 + ( 3 x ) 2 = ( 810 − − √ ) 2 = > 10 x 2 = 640 = > x = 8 x 2 + ( 3 x ) 2 = ( 810 ) 2 = > 10 x 2 = 640 = > x = 8 so , length = 24 cm and breadth = 8 cm perimeter = 2 ( l + b ) = 2 ( 24 + 8 ) = 64 cm answer : c | a = 8 ** 2
b = math.sqrt(a)
c = 8 ** 2
d = math.sqrt(c)
e = d * 3
f = b + e
g = f * 2
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | a | divide(225, add(subtract(26, 2), const_1)) | 225 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 225 / 25 ) = 10 a" | a = 26 - 2
b = a + 1
c = 225 / b
|
a ) 3 : 7 , b ) 4 : 9 , c ) 12 : 7 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(60, 8), multiply(70, 4)) | car a runs at the speed of 60 km / hr and reaches its destination in 8 hours . car b runs at the speed of 70 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | car a travels 60 × 8 = 480 km car b travels 70 × 4 = 280 km the ratio is 480 : 280 = 48 : 28 = 12 : 7 the answer is c . | a = 60 * 8
b = 70 * 4
c = a / b
|
a ) 3488 , b ) 3778 , c ) 12788 , d ) 1800 , e ) 2881 | d | multiply(multiply(const_3, const_60), const_60) | if an object travels at five feet per second , how many feet does it travel in one hour ? | "explanation : if an object travels at 5 feet per second it covers 5 x 60 feet in one minute , and 5 x 60 x 60 feet in one hour . answer = 1800 answer : d ) 1800" | a = 3 * const_60
b = a * const_60
|
a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 12 % | e | multiply(divide(4, add(29, 4)), const_100) | a house wife saved $ 4 in buying an item on sale . if she spent $ 29 for the item , approximately how much percent she saved in the transaction ? | "actual price = 29 + 4 = $ 33 saving = 4 / 33 * 100 = 12 % approximately answer is e" | a = 29 + 4
b = 4 / a
c = b * 100
|
a ) - 3 , b ) - 1 , c ) 0 , d ) - 5 , e ) - 2 | e | add(add(multiply(subtract(const_2, 3), multiply(divide(const_3, 2), const_3)), divide(const_3, 2)), const_1) | tough and tricky questions : exponents . if 2 ^ ( 3 x + 2 ) * 3 ^ ( 2 y - 3 ) = 32 ^ x * 9 ^ y , then x + y = | here is my solution . 2 ^ ( 3 x + 2 ) * 3 ^ ( y - 3 ) = 32 ^ x * 9 ^ y here rhs 32 ^ x * 9 ^ y = 2 ^ ( 5 x ) * 3 ^ ( 2 y ) equating powers on both sides - - > 3 x + 2 = 5 x , thus x = 1 and y - 3 = 2 y giving y = - 3 so , x + y = - 2 option : e | a = 2 - 3
b = 3 / 2
c = b * 3
d = a * c
e = 3 / 2
f = d + e
g = f + 1
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(add(const_4, const_3), const_2) | what is the units digit of ( 147 ^ 25 ) ^ 49 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 25 has the form 4 n + 1 so the units digit is 7 inside the bracket . the exponent 49 has the form 4 n + 1 , so the units digit is 7 . the answer is d . | a = 4 + 3
b = a + 2
|
a ) $ 1750 , b ) $ 2250 , c ) $ 1250 , d ) $ 1500 , e ) $ 2000 | a | divide(multiply(divide(divide(add(divide(multiply(4000, 10), const_100), divide(multiply(add(4000, divide(multiply(4000, 10), const_100)), 10), const_100)), 2), 3), const_100), 8) | simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on $ 4000 for 2 years at 10 % per annum . the sum placed on simple interest is : | "compound interest = ( 4000 x ( 1 + 10 / 100 ) ^ 2 - 4000 ) = 4000 x 11 / 10 x 11 / 10 - 4000 = 840 therefore the sum is 420 x 100 / ( 3 x 8 ) = 1750 . answer a ) $ 1750" | a = 4000 * 10
b = a / 100
c = 4000 * 10
d = c / 100
e = 4000 + d
f = e * 10
g = f / 100
h = b + g
i = h / 2
j = i / 3
k = j * 100
l = k / 8
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 10 | c | divide(add(11, 3), const_2) | in one hour , a boat goes 11 km along the stream and 3 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 3 ) km / hr = 7 kmph . answer c" | a = 11 + 3
b = a / 2
|
a ) 22 , b ) 24 , c ) 16 , d ) 18 , e ) 20 | c | subtract(power(6, const_2), multiply(const_2, 10)) | if a + b = − 6 , and a = 10 / b , what is the value of a ^ 2 + b ^ 2 ? | a ^ 2 + b ^ 2 should make you think of these formulas : ( a + b ) ( a + b ) = a ^ 2 + b ^ 2 + 2 ab we already know ( a + b ) = - 6 and a * b = 10 ( a + b ) ( a + b ) = ( - 6 ) ( - 6 ) = a ^ 2 + b ^ 2 + 2 * ( 10 ) a ^ 2 + b ^ 2 = 36 - 20 = 16 answer : c | a = 6 ** 2
b = 2 * 10
c = a - b
|
a ) 2010000 , b ) 2110000 , c ) 2150000 , d ) 2200000 , e ) 2210000 | e | divide(const_4, const_2) | a professional basketball player just signed a contract that has a 10 % raise in base salary from last year to this year . the contract also has provisions for a performance bonus of up to $ 10000 . if he earns the maximum performance bonus , his new effective salary will be a 10.5 % raise over last year ' s total salary , when he received no performance bonus . assuming he earns the full performance bonus , what will be his new salary ? | let x = old salary then ( 1 + . 10 ) * x + 10000 = ( 1 + . 105 ) * x i . e . new base salary + 10000 = effective new salary thus 1.1 * x + 10000 = 1.105 * x 10000 = . 005 * x 2000000 = x thus the new effective salary is 1.105 * 2000000 = 2210000 e | a = 4 / 2
|
a ) 12.9 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.25 sec | e | divide(add(110, 175), multiply(72, const_0_2778)) | how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 175 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 175 = 285 m . required time = 285 / 20 = 14.25 sec . answer : e" | a = 110 + 175
b = 72 * const_0_2778
c = a / b
|
a ) 15840 , b ) 15848 , c ) 15828 , d ) 27892 , e ) 22892 | a | multiply(square_perimeter(square_edge_by_area(69696)), 15) | find the length of the wire required to go 15 times round a square field containing 69696 m 2 . | "a 2 = 69696 = > a = 264 4 a = 1056 1056 * 15 = 15840 answer : a" | a = square_perimeter * (
|
a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 3 and 7 , e ) 2 and 9 | d | add(multiply(2, const_10), divide(add(44, 5), 7)) | 5 n + 2 > 17 and 7 n - 5 < 44 ; n must be between which numbers ? | "5 n > 15 - - > n > 3 7 n < 49 - - > n < 7 3 < n < 7 answer : d" | a = 2 * 10
b = 44 + 5
c = b / 7
d = a + c
|
a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 17.5 | e | divide(add(add(add(6, const_1), add(add(6, const_1), const_2)), add(subtract(5, 6), subtract(5, const_2))), 6) | find the average of first 6 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 = 17.5 answer is e" | a = 6 + 1
b = 6 + 1
c = b + 2
d = a + c
e = 5 - 6
f = 5 - 2
g = e + f
h = d + g
i = h / 6
|
a ) a ) 220 , b ) b ) 490 , c ) c ) 380 , d ) d ) 300 , e ) e ) 690 | d | multiply(15, 20) | a large box contains 15 small boxes and each small box contains 20 chocolate bars . how many chocolate bars are in the large box ? | "the only thing you have to do is a simple multiply we have : 15 * 20 = 300 correct answer is : d ) 300" | a = 15 * 20
|
a ) a ) 7 , b ) b ) 2 , c ) c ) 8.9 , d ) d ) 11.4 , e ) e ) 3 | d | divide(power(4, 2), 2) | the current of a stream runs at the rate of 4 kmph . a boat goes 10 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 4 m = x ds = x + 4 us = x - 4 10 / ( x + 4 ) + 10 / ( x - 4 ) = 2 x = 11.4 answer : d" | a = 4 ** 2
b = a / 2
|
a ) $ 198 , b ) $ 228 , c ) $ 220 , d ) $ 230 , e ) $ 240 | b | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 5)) | a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 5 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ? | "wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 5 emlpoyee paid = . 95 * 240 = 228 $ answer b" | a = 200 / 100
b = a * 20
c = 200 + b
d = 200 / 100
e = d * 20
f = 200 + e
g = f / 100
h = g * 5
i = c - h
|
a ) 2 , b ) 8 , c ) 15 , d ) 21 , e ) 48 | e | add(subtract(multiply(3, 6), 6), power(6, 2)) | if a # b = ab – b + b ^ 2 , then 3 # 6 = | "solution - simply substitute 3 and 6 in equation in the place of a and b respectively . 3 # 6 = 3 * 6 - 6 + 6 ^ 2 = 18 - 6 + 36 = 48 . ans e" | a = 3 * 6
b = a - 6
c = 6 ** 2
d = b + c
|
a ) rs 10 , b ) rs 9 , c ) rs 0 , d ) rs 55 , e ) rs 8 | a | subtract(multiply(divide(90, const_12), const_10), 65) | a man engaged a servant on a condition that he ' ll pay rs 90 and also give him a bag at the end of the yr . he served for 9 months and was given a turban and rs 65 . so the price of turban is | let the value of turban = rs x for 1 year = 90 + x for 9 months ( i , e 3 / 4 year ) = 65 + x i , e 1 : ( 90 + x ) : : 3 / 4 : ( 65 + x ) i , e ( 90 + x ) / ( 65 + x ) = 1 / ( 3 / 4 ) 270 + 3 x = 260 + 4 x thus x = 10 answer : a | a = 90 / 12
b = a * 10
c = b - 65
|
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