options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 28 : 15 , b ) 29 : 16 , c ) 25 : 23 , d ) 14 : 15 , e ) none of these | a | divide(multiply(35000, const_12), multiply(25000, add(const_4, const_3))) | x starts a business with rs . 35000 . y joins in the business after 3 months with rs . 25000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 35000 * 12 : 25000 * 9 = 35 * 12 : 25 * 9 = 7 * 4 : 5 * 3 = 28 : 15 . answer : option a" | a = 35000 * 12
b = 4 + 3
c = 25000 * b
d = a / c
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a ) 280 , b ) 216 , c ) 2240 , d ) 246 , e ) 2520 | a | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(3, 3)), multiply(const_4, const_2))), const_10) | how many 3 digit even numbers do not use any digit more than once | the way i solved it is a b c ( hundreds , tens , units ) c can be 0 2 4 6 8 ( any of the 5 digits ) a can be anything except ( d or 0 ) so 8 possibilities b can be anything execpt ( a d c ) so 7 possibilities total ways are 7 * 8 * 5 = 280 ans : a | a = 1000 - 10
b = 3 * 3
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
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a ) rs . 49.25 , b ) rs . 40.50 , c ) rs . 44.20 , d ) rs . 62.85 , e ) rs . 43.10 | d | multiply(25, subtract(circle_area(add(19, 2)), circle_area(19))) | a circular path of 19 m radius has marginal walk 2 m wide all round it . find the cost of leveling the walk at 25 p per m 2 ? | "explanation : Ο ( 21 ^ 2 - 19 ^ 2 ) = 22 / 7 * ( 441 - 361 ) = 251.43 251.43 * 1 / 4 = rs . 62.85 answer : option d" | a = 19 + 2
b = circle_area - (
c = 25 * b
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a ) 3.5 , b ) 3.75 , c ) 3.88 , d ) 4.25 , e ) 4.5 | c | divide(350, add(divide(multiply(3, 1000), const_100), divide(multiply(1200, 5), const_100))) | a money lender lent rs . 1000 at 3 % per year and rs . 1200 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years . | ( 1000 xtx 3 / 100 ) + ( 1200 xtx 5 / 100 ) = 350 Γ’ β β t = 3.88 answer c | a = 3 * 1000
b = a / 100
c = 1200 * 5
d = c / 100
e = b + d
f = 350 / e
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a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | subtract(60, divide(subtract(130, 60), 5)) | mother , her daughter and her grand child weighs 130 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | mother + daughter + child = 130 kg daughter + child = 60 kg mother = 130 - 60 = 70 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 70 = 14 kg so now daughter = 130 - ( mother + child ) = 130 - ( 70 + 14 ) = 46 kg answer : a | a = 130 - 60
b = a / 5
c = 60 - b
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a ) 2200 kmph , b ) 1800 kmph , c ) 1900 kmph , d ) 2100 kmph , e ) none | b | divide(divide(multiply(1500, 2), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 1500 kmph in 2 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "explanation : distance = ( 1500 x 2 ) = 3000 km . speed = distance / time speed = 3000 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = 3000 x 3 / 5 km / hr = 1800 km / hr . answer : option b" | a = 1500 * 2
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
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a ) 25650 , b ) 25750 , c ) 26550 , d ) 42750 , e ) 55725 | d | multiply(900, multiply(10, 4.75)) | the length of a room is 10 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre . | "area = 10 Γ 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 10 Γ 4.75 Γ 900 = 10 Γ 4275 = rs . 42750 answer is d ." | a = 10 * 4
b = 900 * a
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | d | add(8, divide(subtract(200, 20), add(25, 20))) | two stations a and b are 200 km apart on a straight line . one train starts from a at 7 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 7 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 200 45 x = 225 x = 5 . so , they meet at 12 a . m . answer : option d" | a = 200 - 20
b = 25 + 20
c = a / b
d = 8 + c
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a ) 1 / 7 , b ) 1 / 13 , c ) 1 / 15 , d ) 1 / 22 , e ) 1 / 26 | b | divide(const_4, divide(factorial(13), multiply(factorial(2), factorial(subtract(13, 2))))) | a certain box has 13 cards and each card has one of the integers from 1 to 13 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on the 13 cards ? | "the average of the numbers is 7 the total number of ways to choose 2 cards from 13 cards is 13 c 2 = 78 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } , { 1,4 } , { 1,5 } , { 2,3 } , { 2,4 } the probability is 6 / 78 = 1 / 13 the answer is b ." | a = math.factorial(13)
b = math.factorial(2)
c = 13 - 2
d = math.factorial(c)
e = b * d
f = a / e
g = 4 / f
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a ) 0.3825 , b ) 3.825 , c ) 38.25 , d ) 382.5 , e ) 0.0382 | d | divide(divide(multiply(multiply(0.625, 0.0729), 28.9), multiply(multiply(8.1, 0.025), 0.0017)), const_10) | the value of ( 0.625 * 0.0729 * 28.9 ) / ( 0.0017 * 0.025 * 8.1 ) is | ( 0.625 * 0.0729 * 28.9 ) / ( 0.0017 ) * 0.025 * 8.1 = { ( 625 / 1000 ) * ( 729 / 10000 ) * ( 289 / 10 ) } / { ( 17 / 10000 ) * ( 25 / 1000 ) * ( 81 / 10 ) = 382.5 answer : d | a = 0 * 625
b = a * 28
c = 8 * 1
d = c * 0
e = b / d
f = e / 10
|
['a ) 10000', 'b ) 10008', 'c ) 10005', 'd ) 10004', 'e ) 10001'] | a | multiply(add(25, const_100), subtract(const_100, 20)) | the length of a rectangle is increased by 25 % and its breadth is decreased by 20 % . what is the effect on its area ? | 100 * 100 = 10000 125 * 80 = 10000 no change answer : a | a = 25 + 100
b = 100 - 20
c = a * b
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a ) 10.06 % , b ) 10.25 % , c ) 10.35 % , d ) 16.09 % , e ) 16.19 % | b | add(add(divide(10, const_2), divide(10, const_2)), divide(multiply(divide(10, const_2), divide(10, const_2)), const_100)) | the effective annual rate of interest corresponding to a nominal rate of 10 % per annum payable half - yearly is ? | "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 5 / 100 ) 2 ] = rs . 110.25 effective rate = ( 110.25 - 100 ) = 10.25 % answer : b" | a = 10 / 2
b = 10 / 2
c = a + b
d = 10 / 2
e = 10 / 2
f = d * e
g = f / 100
h = c + g
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | divide(subtract(const_1, divide(80, const_100)), divide(5, const_100)) | in a certain parking lot , 5 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what percentage of cars in the parking lot are parked illegally ? | "let x be the number of cars and let y be the number of cars parked illegally . 5 % * x = 20 % * y y / x = 1 / 4 = 25 % the answer is b ." | a = 80 / 100
b = 1 - a
c = 5 / 100
d = b / c
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a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | b | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | in n is a positive integer less than 200 , and 14 n / 60 is an integer , then n has how many different positive prime factors ? | "14 n / 60 must be an integer . = > 7 n / 30 must be an integer . hence n must be a multiple of 2 * 3 * 5 . = > n has 3 different prime integers and can not have more than 3 different prime integers because 2 * 3 * 5 * 7 = 210 > 200 answer : b" | a = 1 + 4
b = 60 / 2
c = b / 2
d = c / 3
e = a / d
f = e + 2
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | a | divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(6, subtract(6, 1))) | if x is to be chosen at random from the integers between 1 to 6 , inclusive , and y is to be chosen at random from the integers between 7 and 11 , inclusive , what is the probability that x + y will be even ? | "x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 3 / 6 * 2 / 5 = 1 / 5 p ( x and y are both odd ) = 3 / 6 * 3 / 5 = 3 / 10 p ( x + y is even ) = 1 / 5 + 3 / 10 = 1 / 2 the answer is a ." | a = 2 * 2
b = 3 * 2
c = a + b
d = 6 - 1
e = 6 * d
f = c / e
|
a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 24.00 sec , e ) 72.0 sec | d | divide(add(110, 290), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 290 m in length ? | "d = 110 + 290 = 400 m s = 60 * 5 / 18 = 50 / 3 t = 400 * 3 / 50 = 24.00 sec answer : d" | a = 110 + 290
b = 60 * const_0_2778
c = a / b
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a ) 45 , b ) 40 , c ) 35 , d ) 30 , e ) 25 | e | add(divide(add(subtract(subtract(8, const_1), const_1), add(50, subtract(8, const_1))), const_3), const_3) | the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded down to the nearest integer ) . during its 8 th year , the plum tree grew 50 plums . if this trend continues , how many plums will it grow during its 6 th year ? | 1 st year : 0 - 1 ( age ) , we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear : 1 - 2 ( age ) , we take age = 1 3 rd year : 2 - 3 ( age ) , we take age = 2 4 th year : 3 - 4 ( age ) , we take age = 3 5 th year : 4 - 5 ( age ) , we take age = 4 6 th year : 5 - 6 ( age ) , we take age = 5 7 th year : 6 - 7 ( age ) , we take age = 6 8 th year : 7 - 8 ( age ) , we take age = 7 thus for the 3 rd year = 50 , 4 th year = 50 - 3 = 47 5 th year = 47 - 4 = 43 6 th year = 43 - 5 = 38 . 7 th year = 38 - 6 = 32 8 th year = 32 - 7 = 25 the correct answer is e . | a = 8 - 1
b = a - 1
c = 8 - 1
d = 50 + c
e = b + d
f = e / 3
g = f + 3
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a ) 77 kmph , b ) 77.15 kmph , c ) 77.25 kmph , d ) 79 kmph , e ) 80 kmph | c | divide(370, add(divide(180, 72), divide(190, 88))) | a man covers a distance of 180 km at 72 kmph and next 190 km at 88 kmph . what is his average speed for his whole journey of 370 km ? | formula = 2 * f . s * s . p / f . s + s . p = 2 * 72 * 88 / 88 + 72 = 79.2 kmph option ' c ' | a = 180 / 72
b = 190 / 88
c = a + b
d = 370 / c
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a ) 136 , b ) 468 , c ) 465 , d ) 463 , e ) 485 | a | add(divide(divide(16, divide(divide(divide(divide(divide(16, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(16, divide(divide(divide(divide(divide(16, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 16 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 16 ( 16 + 1 ) / 2 = 16 ( 17 ) / 2 = 136 answer : option a" | a = 16 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 16 / e
g = f / 2
h = 16 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 16 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 15 | d | multiply(lcm(1, 3), 2) | in a zoo , the ratio of the number of cheetahs to the number of pandas is 1 : 3 and was the same 5 years ago . if the increase in the number of cheetahs in the zoo since then is 2 , then what is the increase in the number of pandas ? | one short cut to solve the problem is c : p = 1 : 3 c increased to 3 = > 1 : 3 = 3 : x = > x = 9 = > p increased by 6 d is the answer | a = math.lcm(1, 3)
b = a * 2
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a ) 34778 , b ) 26888 , c ) 22899 , d ) 23100 , e ) 32778 | d | divide(multiply(add(const_100, 10), add(divide(multiply(16500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 16500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 10 % if no discount was offered ? | "price at which the tv set is bought = rs . 16,500 discount offered = 20 % marked price = 16500 * 100 / 80 = rs . 20625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 20625 + 375 = rs . 21000 the price at which the tv should be sold to get a profit of 10 % if no discount was offered = 21000 * 110 / 100 = rs . 23100 answer : d" | a = 100 + 10
b = 16500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 10 | e | multiply(divide(65, const_100), 16) | john bought a total of 16 mangoes and oranges . each mango costs 80 cents and each orange costs 60 cents . if the average price of the 16 mangoes and oranges that john originally purchased was 65 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | let number of mangoes be x , number of oranges be 16 - x 0.80 x + ( 16 - x ) 0.60 / 16 = 0.65 solving for x , we get x = 4 - - > mangoes 4 , oranges 12 now , number of oranges to be returned be y 0.80 * 4 + ( 12 - y ) * 0.60 / 16 - y = 0.72 solving for y , y = 10 ans : e | a = 65 / 100
b = a * 16
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a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | b | subtract(multiply(multiply(45, divide(8, 6)), divide(30, 30)), 45) | 45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 30 meters deep by working for 6 hours ? | "45 workers * 8 hours / 30 meters = x * 6 / 30 x = 60 total workers 60 - 45 = 15 new workers the answer is b ." | a = 8 / 6
b = 45 * a
c = 30 / 30
d = b * c
e = d - 45
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a ) rs . 32 , b ) rs . 36 , c ) rs . 42 , d ) rs . 26 , e ) rs . 104 | d | multiply(multiply(8, const_4), divide(200, add(8, 2))) | the cost of 8 pens and 4 pencils is rs . 200 and the cost of 2 pens and 2 pencils is rs . 48 . find the cost of each pen ? | "let the cost of each pen and pencil be ' p ' and ' q ' respectively . 8 p + 4 q = 200 - - - ( 1 ) 2 p + 2 q = 48 4 p + 4 q = 96 - - - ( 2 ) ( 1 ) - ( 2 ) = > 4 p = 104 = > p = 26 answer : d" | a = 8 * 4
b = 8 + 2
c = 200 / b
d = a * c
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | subtract(4, const_1) | if the product of 4 integers is negative , at most how many of the integers can be negative ? | the product of 4 integers is negative thus an odd number of integers need to be negative to have a negative product we are asked at most how many are required . so , the highest odd integer before 6 , i . e . 3 correct option : b | a = 4 - 1
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a ) 4.32 , b ) 4.42 , c ) 4.52 , d ) 4.62 , e ) 4.72 | b | multiply(divide(multiply(add(9, 1.2), subtract(9, 1.2)), add(add(9, 1.2), subtract(9, 1.2))), const_2) | a man can row 9 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 9 s = 1.2 ds = 9 + 1.2 = 10.2 us = 9 - 1.2 = 7.8 x / 10.2 + x / 7.8 = 1 x = 4.42 . answer : b" | a = 9 + 1
b = 9 - 1
c = a * b
d = 9 + 1
e = 9 - 1
f = d + e
g = c / f
h = g * 2
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a ) 820 acres , b ) 660 acres , c ) 620 acres , d ) 720 acres , e ) 920 acres | b | multiply(120, multiply(divide(44, 12), divide(54, 36))) | if 12 men can reap 120 acres of land in 36 days , how many acres of land can 44 men reap in 54 days ? | "12 men 120 acres 36 days 44 men ? 54 days 120 * 44 / 12 * 54 / 36 10 * 44 * 3 / 2 44 * 15 = 660 answer : b" | a = 44 / 12
b = 54 / 36
c = a * b
d = 120 * c
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a ) 3277 , b ) 2977 , c ) 16875 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 3)) | the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 3 years ? | "explanation : 40,000 ( 34 ) 3 = 1687540,000 ( 34 ) 3 = 16875 answer : c" | a = 100 * 100
b = 4 * a
c = 3 / 4
d = c ** 3
e = b * d
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a ) 18.75 % , b ) 23 % , c ) 30 % , d ) 50 % , e ) 7 % | e | multiply(divide(subtract(add(const_100, 60), add(const_100, 50)), add(const_100, 50)), const_100) | the wages earned by robin is 50 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? | "let wage of erica = 10 wage of robin = 1.5 * 10 = 15 wage of charles = 1.6 * 10 = 16 percentage by which wage earned by charles is more than that earned by robin = ( 16 - 15 ) / 15 * 100 % = 1 / 15 * 100 % = 7 % answer e" | a = 100 + 60
b = 100 + 50
c = a - b
d = 100 + 50
e = c / d
f = e * 100
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a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio r of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } Β = ( 1 / 2 ) ^ 2 Β = 1 / 4 therefore r , w _ r Β = { w _ o } / 4 = 8 / 4 Β = 2 litres . b" | a = 3 + 2
b = 3 / a
c = 20 - 10
d = b * c
e = d + 10
f = e / 20
g = 20 - 10
h = f * g
i = h + 10
j = 3 + 2
k = 2 / j
l = 20 - 10
m = k * l
n = m / 20
o = 20 - 10
p = n * o
q = i / p
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a ) 135 , b ) 105 , c ) 95 , d ) 75 , e ) 45 | d | subtract(multiply(divide(multiply(multiply(40, divide(45, const_60)), divide(subtract(const_100, 20), const_100)), 12), const_60), 45) | reb normally drives to work in 45 minutes at an average speed of 40 miles per hour . this week , however , she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20 % . if reb averages between 12 and 16 miles per hour when biking , how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives ? | reb normally drives to work in 45 minutes at an average speed of 40 miles per hour . use formula d = rt car : t 1 : 45 min r 1 : 40 mph d 1 : [ ( 40 * 45 ) / 60 ] = 30 miles bike : t 1 : ? r 2 : 12 - 16 mph d 2 : 08 * d 1 = 24 miles t 1 : [ ( 24 * 60 ) / 12 ] = 120 min ( only 12 mph speed yields an answer given in the choices ) therefore , deb has to leave 120 min - 45 min = 75 min early answer : d | a = 45 / const_60
b = 40 * a
c = 100 - 20
d = c / 100
e = b * d
f = e / 12
g = f * const_60
h = g - 45
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a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent p of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp p = = 50 % of cp thus , answer should bec ." | a = 100 - 20
b = a - 20
c = 100 - b
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a ) $ 14,305 , 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 15 % 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 = 1.15 * 4500 = 5175 cost of type b products next year = 1.1 * 8300 = 9130 total 5175 + 9130 = 14305 option a" | a = 3 / 4
b = a * 1000
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(72, subtract(multiply(4, divide(8, 2)), 8)), 8) | the area of one square is x ^ 2 + 8 x + 16 and the area of another square is 4 x ^ 2 β 28 x + 49 . if the sum of the perimeters of both squares is 72 , what is the value of x ? | "the areas are ( x + 4 ) ^ 2 and ( 2 x - 7 ) ^ 2 . the lengths of the sides are x + 4 and 2 x - 7 . if we add the two perimeters : 4 ( x + 4 ) + 4 ( 2 x - 7 ) = 72 12 x = 84 x = 7 the answer is e ." | a = 8 / 2
b = 4 * a
c = b - 8
d = 72 - c
e = d / 8
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a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 50 | e | divide(subtract(75, 65), subtract(const_1, divide(80, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 80 percent of books that were loaned out are returned and there are 65 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "there are 10 books less ( 75 - 65 ) which represents 20 % of the loaned books ( 100 - 80 ) so total loaned out books = 50 answer e" | a = 75 - 65
b = 80 / 100
c = 1 - b
d = a / c
|
a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 186 | e | multiply(subtract(5, divide(multiply(95, 5), add(95, 155))), const_60) | while driving from a - ville to b - town , harriet drove at a constant speed of 95 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 155 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ? | "5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = 186 min * leg a - b - - - > d = 95.186 / 60 = 17670 / 60 * leg b - a - - - - > d = 155 * 114 / 60 = 17670 / 60 so the correct ans . ise" | a = 95 * 5
b = 95 + 155
c = a / b
d = 5 - c
e = d * const_60
|
a ) 7297 , b ) 2977 , c ) 2871 , d ) 6725 , e ) 6525 | e | divide(8091, add(const_1, divide(24, const_100))) | the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8091 for a computer table , then what was the cost price of the computer table ? | "explanation : cp = sp * ( 100 / ( 100 + profit % ) ) = 8091 ( 100 / 124 ) = rs . 6525 . answer : e" | a = 24 / 100
b = 1 + a
c = 8091 / b
|
a ) 35 , b ) 34 , c ) 20 , d ) 35 , e ) 342 | b | add(lcm(lcm(2, 6), lcm(12, 24)), 10) | what is the smallest number which when diminished by 10 , is divisible 2 , 6 , 12 and 24 ? | "required number = ( lcm of 2 , 6 , 12 and 24 ) + 12 = 24 + 10 = 34 option b" | a = math.lcm(2, 6)
b = math.lcm(12, 24)
c = math.lcm(a, b)
d = c + 10
|
a ) 366 , b ) 106 , c ) 108 , d ) 192 , e ) 122 | c | subtract(108.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 108.25 , brokerage being 1 / 4 % is ? | "cash realised = rs . ( 108.25 - 0.25 ) = rs . 108 . answer : c" | a = 1 / 4
b = 108 - 25
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(divide(8, const_2), const_1) | if n is the greatest positive integer for which 4 ^ n is a factor of 8 ! , then n = ? | "8 ! = 40320 e . 4 ^ 10 = 1048576 ( 40320 / 1048576 ) - this is not a factor of 8 ! d . 4 ^ 8 = 65536 ( 40320 / 65536 ) - this is not a factor of 8 ! c . 4 ^ 6 = 4096 ( 40320 / 4096 ) - this is not a factor of 8 ! b . 4 ^ 4 = 256 ( 40320 / 256 ) - this is not a factor of 8 ! a . 4 ^ 2 = 16 ( 40320 / 16 ) - this is a factor of 8 !" | a = 8 / 2
b = a - 1
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a ) β 220 , b ) β 100 , c ) 100 , d ) 110 , e ) it can not be determined from the information given | d | subtract(multiply(const_60.0, const_2), multiply(215, const_2)) | if the average ( arithmetic mean ) of a and b is 215 , and the average of b and c is 160 , what is the value of a β c ? | "question : a - c = ? ( a + b ) / 2 = 215 = = = > a + b = 430 ( b + c ) / 2 = 160 = = = > b + c = 320 ( a + b ) - ( b + c ) = 430 - 320 = = = > a + b - b - c = 110 = = = > a - c = 110 answer : d" | a = const_60 * 0
b = 215 * 2
c = a - b
|
a ) 1 / 9 , b ) 1 / 6 , c ) 2 / 9 , d ) 1 / 4 , e ) 1 / 2 | a | subtract(add(multiply(divide(const_1, 3), divide(const_1, 3)), multiply(divide(const_2, 3), divide(const_2, 3))), add(multiply(divide(const_1, 3), divide(const_2, 3)), multiply(divide(const_1, 3), divide(const_2, 3)))) | 3 boys are ages 4 , 6 and 7 respectively . 3 girls are ages 5 , 8 and 9 , respectively . if two of the boys and two of the girls are randomly selected and the sum of the selected children ' s ages is w , what is the difference between the probability that w is even and the probability that w is odd ? | age of boys w : 4 , 6 , 7 sum of ages taken 2 at a time : 10 , 1311 ages of girls : 5 , 8 , 9 sum of ages taken 2 at a time : 13 , 1714 9 combinations of sum between sets ( 10 , 1211 ) ( 13 , 1714 ) = 23 , 2724 - 16 , 3017 - 24 , 2825 prob ( even ) = 5 / 9 prob ( odd ) = 4 / 9 answer = 5 / 9 - 4 / 9 = 1 / 9 | a = 1 / 3
b = 1 / 3
c = a * b
d = 2 / 3
e = 2 / 3
f = d * e
g = c + f
h = 1 / 3
i = 2 / 3
j = h * i
k = 1 / 3
l = 2 / 3
m = k * l
n = j + m
o = g - n
|
a ) 14 days , b ) 26 days , c ) 30 days , d ) 31 days , e ) 39 days | c | add(divide(const_1, 30), divide(const_1, 30)) | a can do a job in 30 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "c 1 / 30 + 1 / 30 = 1 / 15 15 * 2 = 30 days" | a = 1 / 30
b = 1 / 30
c = a + b
|
a ) 3.8 , b ) 10 , c ) 5 , d ) 5.5 , e ) 7.5 | d | divide(subtract(multiply(8, 2.25), multiply(5, 0.85)), 2.5) | 8 x 2.25 - 5 x 0.85 / 2.5 = ? | "given expression = ( 18 - 4.25 / 2.5 = 13.75 / 2.5 = 137.5 / 25 = 5.5 answer is d ." | a = 8 * 2
b = 5 * 0
c = a - b
d = c / 2
|
a ) 1978 , b ) 2707 , c ) 7728 , d ) 4800 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(25000, add(20000, 25000)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 20000 and b rs . 25000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | 20 : 25 = > 4 : 5 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 4 / 9 = 3840 + 960 = 4800 answer : d | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 20000 + 25000
e = 25000 / d
f = c * e
g = 9600 - f
|
a ) 80 , b ) 86 , c ) 92 , d ) 98 , e ) 104 | d | divide(subtract(multiply(170, divide(add(const_100, 50), const_100)), 10), add(divide(add(const_100, 50), const_100), const_1)) | if leo gains 10 pounds , he will weigh 50 % more than his sister kendra . currently their combined weight is 170 pounds . what is leo ' s current weight ? | "l + k = 170 and so k = 170 - l l + 10 = 1.5 k = 1.5 ( 170 - l ) 2.5 l = 245 l = 98 the answer is d ." | a = 100 + 50
b = a / 100
c = 170 * b
d = c - 10
e = 100 + 50
f = e / 100
g = f + 1
h = d / g
|
['a ) 48', 'b ) 49', 'c ) 50', 'd ) 51', 'e ) 52'] | a | divide(volume_rectangular_prism(12, 18, 6), volume_cube(3)) | 12 * 18 * 6 is the volume of some material . how many cubes of edge 3 can be insert into it ? | no . of such cubes = volume of material / volume of one cube ( 12 * 18 * 6 ) / ( 3 * 3 * 3 ) = 48 answer : a | a = volume_rectangular_prism / (
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a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84 | d | divide(0.63, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.63 more than jane , what was the original price of the dish ? | "the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.63 - - > 15 p - 13.5 p = 63 - - > p = 42 . answer : d ." | a = 15 / 100
b = 10 / 100
c = 1 - b
d = 15 / 100
e = c * d
f = a - e
g = 0 / 63
|
a ) 3623216 , b ) 3624216 , c ) 3624316 , d ) 3625116 , e ) 3625216 | e | multiply(divide(1904, 1904), const_100) | 1904 x 1904 = ? | "1904 x 1904 = ( 1904 ) 2 = ( 1900 + 4 ) 2 = ( 1900 ) 2 + ( 4 ) 2 + ( 2 x 1900 x 4 ) = 3610000 + 16 + 15200 . = 3625216 . e )" | a = 1904 / 1904
b = a * 100
|
a ) 1.2 % , b ) 2 % , c ) 4 % , d ) 5 % , e ) none of these | d | divide(const_100, 20) | at what rate percent per annum will sum of money double in 20 years ? | "explanation : hint : if sum of money becomes ( z times ) in ( t ) years at simple interest , then rate of interest ( r ) can be calculated using the formula : rate of interest ( r ) % = 100 ( z β 1 ) / t here , principal amount is not given . hence , we can directly use the trick to calculate the rate of interest . rate of interest ( r ) % = 100 ( 2 β 1 ) / 20 rate of interest ( r ) % = 5 % p . a . answer is d" | a = 100 / 20
|
a ) 1 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | divide(3, add(2, 3)) | n is a positive integer . when n + 1 is divided by 3 , the remainder is 2 . what is the remainder when n is divided by 3 ? | "n + 1 = 3 a + 2 i . e . n + 1 = 5 , 8 , 11 , 14 , . . . etc . i . e . n = 4 , 7 , 10 , 13 , . . . etc . when n is divided by 3 remainder is always 1 answer : a" | a = 2 + 3
b = 3 / a
|
a ) 337.83 , b ) 285.83 , c ) 284.83 , d ) 266.83 , e ) 299.83 | b | divide(multiply(490, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | i bought two books ; for rs . 490 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 490 - x ) 119 / 100 x = 285.83 answer : b" | a = 100 + 19
b = 490 * a
c = 100 - 15
d = 100 + 19
e = c + d
f = b / e
|
a ) 18 days , b ) 38 days , c ) 42 days , d ) 48 days , e ) 52 days | e | divide(multiply(13, 80), 20) | if 13 men do a work in 80 days , in how many days will 20 men do it ? | "13 * 80 = 20 * x x = 52 days answer : e" | a = 13 * 80
b = a / 20
|
a ) 3427 , b ) 3237 , c ) 3337 , d ) 2337 , e ) none of these | a | subtract(13200, 9773) | 9773 + x = 13200 , then x is ? | "answer x = 13200 - 9773 = 3427 option : a" | a = 13200 - 9773
|
a ) 705 , b ) 735 , c ) 685 , d ) 665 , e ) 725 | e | add(multiply(multiply(8, 9), const_10), 5) | when 5 + 6 = 305 , 6 + 7 = 425 , 7 + 8 = 565 , then 8 + 9 = ? | 5 + 6 = > 5 Γ£ β 6 = 30 = > 30 Γ£ β 10 + 5 = 305 6 + 7 = > 6 Γ£ β 7 = 42 = > 42 Γ£ β 10 + 5 = 425 7 + 8 = > 7 Γ£ β 8 = 56 = > 56 Γ£ β 10 + 5 = 565 then 8 + 9 = > 8 Γ£ β 9 = 72 = > 72 Γ£ β 10 + 5 = 725 answer : e | a = 8 * 9
b = a * 10
c = b + 5
|
a ) 9600 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000 | a | subtract(subtract(7800, multiply(7800, divide(10, const_100))), multiply(subtract(7800, multiply(7800, divide(10, const_100))), divide(10, const_100))) | the population of a town is 7800 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ? | "formula : ( after = 100 denominator ago = 100 numerator ) 7800 Γ£ β 100 / 90 Γ£ β 100 / 90 = 9629 a )" | a = 10 / 100
b = 7800 * a
c = 7800 - b
d = 10 / 100
e = 7800 * d
f = 7800 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | multiply(multiply(4, divide(8, 4)), divide(8, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 8 mat weavers in 8 days ? | "explanation : please always remember , that what exactly we need to calculate before applying chain formula . as in this question we need to find how many mats , right ! so we will compare everything with ` ` number of mats ' ' as , more men , more number of mats ( direct proportion ) more days , more number of mats ( direct proportion ) so , it can be solved as , [ weavers 4 4 days 8 8 ] : : 4 : x = > x β 4 β 4 = 8 β 8 β 4 = > x = 16 option c" | a = 8 / 4
b = 4 * a
c = 8 / 4
d = b * c
|
a ) 124 m 2 , b ) 128 m 2 , c ) 138 m 2 , d ) 148 m 2 , e ) none of these | d | subtract(rectangle_area(add(multiply(2, 2), 21), add(12, multiply(2, 2))), rectangle_area(21, 12)) | the floor of a rectangular room is 21 m long and 12 m wide . the room is surrounded by a veranda of width 2 m on all its sides . the area of the veranda is : | "area of the outer rectangle = 25 Γ£ β 16 = 400 m 2 area of the inner rectangle = 21 Γ£ β 12 = 252 m 2 required area = ( 400 Γ’ β¬ β 252 ) = 148 m 2 answer d" | a = 2 * 2
b = a + 21
c = 2 * 2
d = 12 + c
e = rectangle_area - (
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a ) 11 sec . , b ) 15 sec . , c ) 13.3 sec . , d ) 17 sec . , e ) 19 sec . | c | divide(add(120, 120), add(speed(120, 10), speed(120, 20))) | two bullet trains of equal lengths take 10 seconds and 20 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "speed of the first bullet train = 120 / 10 m / sec = 12 m / sec . speed of the second bullet train = 120 / 20 m / sec = 6 m / sec . relative speed = ( 12 + 6 ) = 18 m / sec . required time = ( 120 + 120 ) / 18 sec = 13.3 sec . c" | a = 120 + 120
b = speed + (
c = a / b
|
a ) 11 / 35 , b ) 14 / 35 , c ) 18 / 35 , d ) 31 / 70 , e ) 37 / 70 | c | add(multiply(divide(3, 7), divide(const_2.0, 5)), multiply(divide(3, 7), divide(1, 3))) | in a tree , 3 / 7 of the birds are robins while the rest are bluejays . if 1 / 3 of the robins are female and 3 / 5 of the bluejays are female , what fraction of the birds in the tree are male ? | "the fraction of birds that are male robins is ( 2 / 3 ) ( 3 / 7 ) = 2 / 7 . the fraction of birds that are male bluejays is ( 2 / 5 ) ( 4 / 7 ) = 8 / 35 . the total fraction of male birds is 2 / 7 + 8 / 35 = 18 / 35 . the answer is c ." | a = 3 / 7
b = 2 / 0
c = a * b
d = 3 / 7
e = 1 / 3
f = d * e
g = c + f
|
a ) 180 , b ) 270 , c ) 340 , d ) 140 , e ) 240 | e | multiply(multiply(const_4.0, 6), const_10) | the length of a rectangular landscape is 8 times its breadth . there is a playground in it whose area is 1200 square mtr & which is 1 / 6 rd of the total landscape . what is the length of the landscape ? | "sol . x * 8 x = 6 * 1200 x = 30 length = 8 * 30 = 240 e" | a = 4 * 0
b = a * 10
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a ) 26 , b ) 27 , c ) 25 , d ) 22 , e ) 23 | a | add(add(const_4, const_2), add(const_10, const_10)) | ratio and proportion 215 : 474 : : 537 : ? | 2 + 1 + 5 + 4 + 7 + 4 = 23 5 + 3 + 7 + x = 23 = > x = 8 combination that match 8 is 26 since 2 + 6 = 8 answer : a | a = 4 + 2
b = 10 + 10
c = a + b
|
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | d | divide(288, add(multiply(8, const_3), multiply(6, const_2))) | sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 288 per week . how much does she earn in dollars per hour ? | "let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 288 per week 36 x = 288 x = 8 correct option : d" | a = 8 * 3
b = 6 * 2
c = a + b
d = 288 / c
|
a ) 97.5 kg , b ) 103 kg , c ) 108 kg , d ) 125 kg , e ) 117 kg | a | multiply(subtract(const_1, divide(35, const_100)), multiply(subtract(const_1, divide(25, const_100)), multiply(200, subtract(const_1, divide(20, const_100))))) | a statue is being carved by a sculptor . the original piece of marble weighed 200 kg . in the first week 20 percent is cut away . in the second week 25 percent of the remainder is cut away . in the third week the statue is completed when 35 percent of the remainder is cut away . what is the weight of the final statue ? | "a 97.5 kg 250 Γ£ β 0.8 Γ£ β 0.75 Γ£ β 0.65 = 97.5 kg ." | a = 35 / 100
b = 1 - a
c = 25 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = 200 * f
h = d * g
i = b * h
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a ) 15 % , b ) 25 % , c ) 35 % , d ) 65 % , e ) 74 % | e | multiply(divide(multiply(add(const_3, const_4), const_10), add(multiply(subtract(const_100, multiply(add(const_3, const_4), const_10)), subtract(const_1, divide(25, const_100))), multiply(add(const_3, const_4), const_10))), const_100) | an offshore company sells two products , l and m . last year , seventy percent of the units sold were product l , and the price of product l was 25 percent greater than the price of product m . approximately what percent of the total revenue the company received last year was from the sale of product l ? | percentage of total units that is product l = 70 % price of product l = 25 percent greater than price of product m . now , we can consider this as a weighted average scenario . if the price of both the products l and m was equal , then contribution of product l to total sales would be = 70 % but since , price of product l is greater than m , then contribution of product l to total sales will be greater than 70 % the only option is 74 % answer e | a = 3 + 4
b = a * 10
c = 3 + 4
d = c * 10
e = 100 - d
f = 25 / 100
g = 1 - f
h = e * g
i = 3 + 4
j = i * 10
k = h + j
l = b / k
m = l * 100
|
a ) 15 , b ) 23 , c ) 37 , d ) 42 , e ) 60 | b | add(add(divide(divide(multiply(multiply(4, 5), 8), const_2), 4), divide(divide(multiply(multiply(4, 5), 8), const_2), 5)), divide(divide(multiply(multiply(4, 5), 8), const_2), 8)) | if x , y , and z are positive integers , and 4 x = 5 y = 8 z , then the least possible value of x + y + z is | "take lcm of 4,5 and 8 = 40 now 4 x = 40 = > x = 10 5 y = 40 = > y = 8 8 z = 40 = > z = 5 10 + 8 + 5 = 23 . option b ." | a = 4 * 5
b = a * 8
c = b / 2
d = c / 4
e = 4 * 5
f = e * 8
g = f / 2
h = g / 5
i = d + h
j = 4 * 5
k = j * 8
l = k / 2
m = l / 8
n = i + m
|
a ) s . 25 , b ) s . 48 , c ) s . 42 , d ) s . 20 , e ) s . 60 | a | divide(multiply(subtract(28, 20), const_100), const_2) | find the principal which yields a simple interest of rs . 20 and compound interest of rs . 28 in two years , at the same percent rate per annum ? | "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 28 % rate per annum = [ ( ci β si ) / ( si in 1 year ) ] * 100 = [ ( 28 β 20 ) / 20 ] * 100 = 40 % p . a . let the principal be rs . x time = t = 2 years % rate = 40 % p . a . si = ( prt / 100 ) 20 = ( x * 40 * 2 ) / 100 x = rs . 25 answer : a" | a = 28 - 20
b = a * 100
c = b / 2
|
a ) 1 / 15 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3 | d | divide(divide(factorial(6), multiply(factorial(2), factorial(2))), factorial(6)) | joshua and jose work at an auto repair center with 2 other workers . for a survey on health care insurance , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that joshua and jose will both be chosen ? | "two methods 1 ) probability of chosing josh first = 1 / 4 probability of chosing jose second = 1 / 3 total = 1 / 12 probability of chosing jose first = 1 / 4 probability of chosing josh second = 1 / 3 total = 1 / 12 final = 1 / 12 + 1 / 12 = 1 / 6 d" | a = math.factorial(6)
b = math.factorial(2)
c = math.factorial(2)
d = b * c
e = a / d
f = math.factorial(6)
g = e / f
|
a ) 65 % , b ) 67 % , c ) 69 % , d ) 72 % , e ) 75 % | c | divide(add(multiply(70, 60), multiply(90, subtract(100, 70))), 100) | in a certain accounting class of 100 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 60 % , and the students on the make - up date had an average score of 90 % , what was the average score for the entire class ? | "70 % of the class scored 60 % and 30 % of the class scored 90 % . the difference between 60 % and 90 % is 30 % . the average will be 60 % + 0.3 ( 30 % ) = 69 % . the answer is c ." | a = 70 * 60
b = 100 - 70
c = 90 * b
d = a + c
e = d / 100
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a ) 10 , b ) 8 , c ) 14 , d ) 16 , e ) 12 | c | add(subtract(add(42, 54), subtract(90, 8)), subtract(54, 42)) | of 90 applicants for a job , 42 had at least 4 years ' experience , 54 had degrees , and 8 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "90 - 8 = 82 82 - 42 - 54 = - 14 then 14 are in the intersection between 4 years experience and degree . answer : c" | a = 42 + 54
b = 90 - 8
c = a - b
d = 54 - 42
e = c + d
|
a ) $ 850 , b ) $ 875 , c ) $ 900 , d ) $ 925 , e ) $ 950 | d | multiply(const_100.0, divide(const_100, add(1110, 20))) | a shopkeeper sold an article at $ 1110 and gained a 20 % profit . what was the cost price ? | "let x be the cost price . 1.2 x = 1110 x = 1110 / 1.2 = 925 the answer is d ." | a = 1110 + 20
b = 100 / a
c = 100 * 0
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a ) 1 : 7 , b ) 8 : 7 , c ) 6 : 7 , d ) 6 : 5 , e ) 4 : 7 | a | divide(subtract(33, 31), subtract(47, 33)) | two trains running in opposite directions cross a pole placed on the platform in 47 seconds and 31 seconds respectively . if they cross each other in 33 seconds , what is the ratio of their speeds ? | explanation : let the speed of the trains be x and y respectively length of train 1 = 47 x length of train 2 = 31 y relative speed = x + y time taken to cross each other = 33 s = > = 33 = > ( 47 x + 31 y ) = 33 ( x + y ) = > 14 x = 2 y = > x / y = 2 / 14 = 1 / 7 = 1 : 7 answer : a | a = 33 - 31
b = 47 - 33
c = a / b
|
a ) 60 % , b ) 65 % , c ) 70 % , d ) 75 % , e ) 80 % | d | divide(add(add(multiply(25, 80), multiply(50, 65)), multiply(subtract(const_100, add(25, 50)), 90)), const_100) | if 25 % of a class averages 80 % on a test , 50 % of the class averages 65 % on the test , and the remainder of the class averages 90 % on the test , what is the overall class average ? | "this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 25 % - 50 % = 25 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.25 x 80 + 0.50 x 65 + 0.25 x 90 = 75 the class average is 75 % final answer d ) 75 %" | a = 25 * 80
b = 50 * 65
c = a + b
d = 25 + 50
e = 100 - d
f = e * 90
g = c + f
h = g / 100
|
a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord s , in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . s = pi * r = 30 . round pi = 3 , then r = 10 . chord is about 10 feet long . d" | a = 30 / 3
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a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 5 , d ) 5 , e ) 6 | a | divide(divide(6, 12), 12) | if xy > 0 , 1 / x + 1 / y = 6 , and 1 / xy = 12 , then ( x + y ) / 6 = ? | "( 1 / x + 1 / y ) = 6 canbe solved as { ( x + y ) / xy } = 12 . substituting for 1 / xy = 12 , we get x + y = 6 / 12 = = > ( x + y ) / 6 = 6 / ( 12 * 6 ) = 1 / 12 . a" | a = 6 / 12
b = a / 12
|
a ) 14 , b ) 24 , c ) 28 , d ) 48 , e ) 96 | d | divide(multiply(factorial(const_4), 4), const_2) | how many 4 digit numbers have no repeat digits , do not contain zero , and have a sum of digits d equal to 28 ? | first , look for all 4 digits without repeat that add up to 28 . to avoid repetition , start with the highest numbers first . start from the largest number possible 9874 . then the next largest number possible is 9865 . after this , you ' ll realize no other solution . clearly the solution needs to start with a 9 ( cuz otherwise 8765 is the largest possible , but only equals 26 ) . with a 9 , you also need an 8 ( cuz otherwise 9765 is the largest possible , but only equals 27 ) . with 98 __ only 74 and 65 work . so you have two solutions . each can be rearranged in 4 ! = 24 ways . so d = 24 + 24 = 48 . d | a = math.factorial(4)
b = a * 4
c = b / 2
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(choose(subtract(8, 1), 1), choose(8, 2)) | a basket contains 8 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "the total number of ways to choose 2 apples is 8 c 2 = 28 the number of ways that include the spoiled apple is 7 c 1 = 7 p ( the spoiled apple is included ) = 7 / 28 = 1 / 4 the answer is c ." | a = 8 - 1
b = math.comb(a, 1)
c = math.comb(8, 2)
d = b / c
|
a ) 20 seconds , b ) 30 seconds , c ) 40 seconds , d ) 48 seconds , e ) none of these | d | divide(add(460, 140), divide(multiply(45, const_1000), const_3600)) | a train is 460 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length | "explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 460 + 140 = 600 meter time = distance / speed = 600 β 2 / 25 = 48 seconds option d" | a = 460 + 140
b = 45 * 1000
c = b / 3600
d = a / c
|
a ) 192 , b ) 195 , c ) 200 , d ) 205 , e ) 208 | d | subtract(subtract(divide(multiply(divide(multiply(divide(subtract(645, 45), const_3), 3), 4), 20), 3), 645), divide(multiply(divide(subtract(645, 45), const_3), 3), 4)) | of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout , the number of males is 45 more than twice the number of females . if the ratio of female speckled trout to male rainbow trout is 4 : 3 and the ratio of male rainbow trout to all trout is 3 : 20 , how many female rainbow trout are there ? | it will be easier if you draw a table for this one . let t be the total trout . total speckled trout ( st ) = 645 if female speckled trout = fst then male speckled trout = 45 + 2 fst therefore 45 + 2 fst + fst = 645 we solve and get fst = 200 and mst = 445 ratio of fst to mrt = 4 : 3 we have fst = 200 solve ratio to get mrt = 150 now ratio of mrt to t = 3 : 20 we have mrt = 150 solve ratio to get t = 1000 now unless youve drawn the table it might get a little complex . from above we have mst and mrt mst + mrt = 445 + 150 = 595 so t - 595 = total female trouts 1000 - 595 = 405 female trouts we have from above that fst = 200 so ( whew ! ) frt = 405 - 200 = 205 answer is d | a = 645 - 45
b = a / 3
c = b * 3
d = c / 4
e = d * 20
f = e / 3
g = f - 645
h = 645 - 45
i = h / 3
j = i * 3
k = j / 4
l = g - k
|
a ) 3109 , b ) 3115 , c ) 3250 , d ) 3500 , e ) 4000 | a | subtract(add(add(8000, divide(multiply(8000, 15), const_100)), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 8000) | find compound interest on rs . 8000 at 15 % per annum for 2 years 4 months , compounded annually | "time = 2 yrs 4 mos = 2 ( 4 / 12 ) yrs = 2 ( 1 / 3 ) yrs . amount = rs . [ 8000 x ( 1 + ( 15 / 100 ) ) ^ 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = rs . 11109 . : . c . i . = rs . ( 11109 - 8000 ) = rs . 3109 . answer a" | a = 8000 * 15
b = a / 100
c = 8000 + b
d = 8000 * 15
e = d / 100
f = 8000 + e
g = f * 15
h = g / 100
i = c + h
j = i - 8000
|
a ) $ 30 . , b ) $ 31 , c ) $ 32 . , d ) $ 33 . , e ) $ 34 . | c | add(add(multiply(12, 2), 3), multiply(divide(12, const_60), divide(35, 3))) | it costs $ 3 for the first 1 / 3 hour to use the laundry machine at the laundromat . after the first ΒΌ hour it costs $ 12 per hour . if a certain customer uses the laundry machine for 2 hours and 35 minutes , how much will it cost him ? | "2 hrs 35 min = 155 min first 20 min - - - - - - > $ 3 time left is 135 min . . . now , 60 min costs $ 12 1 min costs $ 12 / 60 155 min costs $ 12 / 60 * 155 = > $ 31 so , total cost will be $ 31 + $ 3 = > $ 34 hence answer will be c" | a = 12 * 2
b = a + 3
c = 12 / const_60
d = 35 / 3
e = c * d
f = b + e
|
a ) 21 , b ) 42 , c ) 63 , d ) 72 , e ) 168 | b | sqrt(multiply(84, 21)) | which number should replace both the asterisks in ( * / 21 ) x ( * / 84 ) = 1 ? | "let ( y / 21 ) x ( y / 84 ) = 1 y ^ 2 = 21 x 84 = 21 x 21 x 4 y = ( 21 x 2 ) = 42 the answer is b ." | a = 84 * 21
b = math.sqrt(a)
|
a ) 30.5 kmph , b ) 31.5 kmph , c ) 32.5 kmph , d ) 33.5 kmph , e ) 21.5 kmph | b | multiply(const_3_6, divide(140, 16)) | a train 140 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 140 / 16 * 18 / 5 = 31.5 kmph answer : b" | a = 140 / 16
b = const_3_6 * a
|
a ) 10 litres , b ) 13 litres , c ) 15 litres , d ) 18 litres , e ) none of these | b | divide(20, add(1, divide(1, 2))) | how much water must be added to 52 litres of milk at 1 1 β 2 litres for 20 so as to have a mixture worth 10 2 β 3 a litre ? | c . p . of 1 litre of milk = ( 20 Γ 2 β 3 ) = 40 β 3 β΄ ratio of water and milk = 8 β 3 : 32 β 3 = 8 : 32 = 1 : 4 β΄ quantity of water to be added to 52 litres of milk = ( 1 β 4 Γ 52 ) litres = 13 litres . answer b | a = 1 / 2
b = 1 + a
c = 20 / b
|
a ) 175 kmph , b ) 150 kmph , c ) 162 kmph , d ) 145 kmph , e ) 100 kmph | a | subtract(divide(divide(500, 10), const_0_2778), 5) | a train 500 m long takes 10 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 500 / [ ( x + 5 ) * 5 / 18 ] = 10 10 ( x + 5 ) = 1800 x = 175 kmph answer is a" | a = 500 / 10
b = a / const_0_2778
c = b - 5
|
a ) 2 / 15 , b ) 2 / 13 , c ) 1 / 15 , d ) 4 / 13 , e ) 5 / 7 | d | divide(subtract(52, multiply(const_4, const_4)), 52) | a card is drawn from a pack of 52 cards . the probability of getting a face card ? | "clearly in the 52 cards out of which there are 16 face cards . probability of getting a face card = 16 / 52 = 4 / 13 correct option is d" | a = 4 * 4
b = 52 - a
c = b / 52
|
a ) 3 , b ) 5 , c ) 4.3 , d ) 6.5 , e ) 7.4 | e | divide(90, multiply(44, const_0_2778)) | in what time will a railway train 90 m long moving at the rate of 44 kmph pass a telegraph post on its way ? | "t = 90 / 44 * 18 / 5 = 7.4 sec answer : e" | a = 44 * const_0_2778
b = 90 / a
|
a ) 6 : 1 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8 | a | divide(divide(subtract(96, 72), subtract(96, 68)), subtract(const_1, divide(subtract(96, 72), subtract(96, 68)))) | in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 72 per kg ? | "( 96 - 72 ) / ( 72 - 68 ) = 24 / 4 = 6 / 1 answer : a" | a = 96 - 72
b = 96 - 68
c = a / b
d = 96 - 72
e = 96 - 68
f = d / e
g = 1 - f
h = c / g
|
a ) 24 , b ) 26 , c ) 42 , d ) 30 , e ) 32 | c | multiply(divide(7, 8), add(43, const_4)) | john was 43 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 43 john ' s age after 5 years = 48 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 48 ) = 42 therefore betty ' s current age = 42 option c" | a = 7 / 8
b = 43 + 4
c = a * b
|
a ) 372 , b ) 434 , c ) 468 , d ) 512 , e ) 564 | c | divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 120 % of 13 / 24 of 720 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 30 = 468 the answer is c ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 120 * n
p = o / 100
|
a ) 60 , b ) 100 , c ) 180 , d ) 240 , e ) 300 | b | multiply(multiply(subtract(6, const_4), const_10), const_10) | a β palindromic integer β is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 700,000 ? | "the first digit and last digit are the same so the only possibility is 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 1 * 10 * 10 = 100 the answer is b ." | a = 6 - 4
b = a * 10
c = b * 10
|
a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48 | b | divide(subtract(add(add(2.50, 5.00), multiply(0.25, 16)), 2.50), 0.25) | mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 16 miles , how many miles was mike ' s ride ? | "the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 16 ) = $ 11.50 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 11.5 0.25 * x = 9 x = 36 miles the answer is b ." | a = 2 + 50
b = 0 * 25
c = a + b
d = c - 2
e = d / 0
|
a ) 8 , b ) 4 , c ) 15 , d ) 6 , e ) 12 | d | multiply(add(add(multiply(divide(7, const_60), 5), multiply(divide(5, const_60), 6)), divide(7, const_60)), 5) | if a man walks at a rate of 5 kmph , he misses a train by 7 minutes . however , if he walks at the rate of 6 kmph , he reaches the station 5 minutes before the arrival of the train . find the distance covered by him to reach the station . | let the required distance x km difference in the times taken at two speeds = 12 min = 1 / 5 hr ( x / 5 ) - ( x / 6 ) = 1 / 5 x = 6 the required distance is 6 km answer is d | a = 7 / const_60
b = a * 5
c = 5 / const_60
d = c * 6
e = b + d
f = 7 / const_60
g = e + f
h = g * 5
|
a ) 100 , b ) 250 , c ) 750 , d ) 5812.5 , e ) 5835.5 | d | multiply(multiply(multiply(62, 25), divide(1, const_2)), 7.5) | the milk level in a rectangular box measuring 62 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 62 * 25 * 1 / 2 = 775 feet ^ 3 of milk must be removed , which equals to 775 * 7.5 = 5812.5 gallons . answer : d ." | a = 62 * 25
b = 1 / 2
c = a * b
d = c * 7
|
a ) 724 m , b ) 704 m , c ) 287 m , d ) 450.6 m , e ) 927 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 320), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 320 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 320 resolutions . = 320 * 2 * 22 / 7 * 22.4 = 45056 cm = 450.6 m answer : d" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 320
i = h / 100
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | subtract(6, 8) | what is 10 - 8 + 6 - 4 + . . . + ( - 12 ) ? | "the expression considers all even numbers between 10 and - 12 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , - 10 , and - 12 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) - ( - 12 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 = 42 - 30 = 12 hence the correct answer choice is c ." | a = 6 - 8
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 9 | c | multiply(2, divide(36, add(2, 4))) | 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 36 kilometers , maxwell ' s walking speed is 2 km / h , and brad ' s running speed is 4 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 = > 2 xt = 36 - 4 xt ( i . e distance = speed x time ) = > 6 t = 36 hence t = 6 oa = 2 x 6 = 12 km answer : c" | a = 2 + 4
b = 36 / a
c = 2 * b
|
a ) 50.5 % , b ) 44.4 % , c ) 22.2 % , d ) 14.3 % , e ) 25 % | d | multiply(divide(subtract(20, 15), add(20, 15)), const_100) | if 20 % of ( x - y ) = 15 % of ( x + y ) , then what percent of x is y ? | "20 % of ( x - y ) = 15 % of ( x + y ) 20 / 100 ( x - y ) = 15 / 100 ( x + y ) 5 x = 35 y required percentage = y / x * 100 = 5 y / 35 y * 100 = 14.28 % answer is d" | a = 20 - 15
b = 20 + 15
c = a / b
d = c * 100
|
a ) 35 hrs , b ) 36 hrs , c ) 37 hrs , d ) 38 hrs , e ) 39 hrs | b | inverse(subtract(divide(const_1, 12), divide(const_1, 18))) | if a tap could fill entire tank in 18 hrs due to leakage , then in how much time tank can be emptied by leakage if tap can fill entire tank in 12 hrs without leakage ? | if a tap could fill entire tank in 18 hrs due to leakage , tap can fill entire tank in 12 hrs without leakage . tank emptied in 0 ne hr by leakage = 1 / 12 - 1 / 18 = ( 6 - 4 ) / 72 = 2 / 72 = 1 / 36 . full tank can be emptied in 36 hrs . answer : b | a = 1 / 12
b = 1 / 18
c = a - b
d = 1/(c)
|
a ) 42 % , b ) 43 % , c ) 45 % , d ) 44 % , e ) 46 % | d | divide(subtract(multiply(add(const_100, 20), add(const_100, 20)), multiply(const_100, const_100)), const_100) | what is the percentage increase in the area of a rectangle , if each of its sides is increased by 20 % ? | let original length = l original breadth = b then original area = lb length is increased by 20 % β new length = l Γ 120100 = 1.2 l breadth is increased by 20 % β new breadth = b Γ 120100 = 1.2 b new area = 1.2 l Γ 1.2 b = 1.44 lb increase in area = new area - original area = 1.44 lb β lb = 0.44 lb percentage increase in area = increase in areaoriginal area Γ 100 = 0.44 lblb Γ 100 = 44 % answer is d . | a = 100 + 20
b = 100 + 20
c = a * b
d = 100 * 100
e = c - d
f = e / 100
|
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