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 ) 48 kmph , b ) 50 kmph , c ) 52 kmph , d ) 56 kmph , e ) 60 kmph | a | divide(432, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 432 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 432 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 432 / 9 = 48 kmph a" | a = 6 * 3
b = a / 2
c = 432 / b
|
a ) 901 , b ) 989 , c ) 990 , d ) 991 , e ) 1,001 | a | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), const_0.5)) | in a recent election , james received 5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 50 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 5 / 100 ) * 2000 = 100 votes to win = ( 50 / 100 ) * total votes + 1 = ( 50 / 100 ) * 2000 + 1 = 1001 remaining voted needed to win election = 1001 - 100 = 901 answer : option a" | a = 1000 + 1000
b = 1000 + 1000
c = b * 0
d = a - c
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | add(floor(divide(subtract(subtract(40, 14), subtract(20, 5)), const_2)), 1) | the scoring system in a certain football competition goes as follows : 3 points for victory , 1 point for a draw , and 0 points for defeat . each team plays 20 matches . if a team scored 14 points after 5 games , what is the least number of the remaining matches it has to win to reach the 40 - point mark by the end of the tournament ? | "to get 40 points as end of season we need another 26 points or more from remaining 15 matches : option a = 6 * 3 + 9 * 1 = 27 hence option a - 6" | a = 40 - 14
b = 20 - 5
c = a - b
d = c / 2
e = math.floor(d)
f = e + 1
|
a ) 1245 , b ) 1345 , c ) 1455 , d ) 1577 , e ) 1665 | e | multiply(divide(subtract(1390, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1390 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1390 ) . x + 1390 = 6 x + 15 5 x = 1375 x = 275 large number = 275 + 1390 = 1665 e" | a = 1390 - 15
b = 6 - 1
c = a / b
d = c * 6
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | b | subtract(divide(subtract(105, 50), subtract(55, 50)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 105 units raises the average to 55 units per day , what is the value of n ? | "the daily production was raised by 5 units for n days , which has a weighting of 5 n . 5 n = 105 - 55 = 50 n = 10 the answer is b ." | a = 105 - 50
b = 55 - 50
c = a / b
d = c - 1
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(subtract(add(5, const_100), subtract(const_100, 1)), subtract(subtract(const_100, 1), subtract(const_100, 13))) | a company sells pens and pencils . the revenue from pens in 2007 was up 5 % from 2006 . the revenue from pencils declined 13 % over the same period . overall revenue was down 1 % from 06 to 07 . what was the ratio of pencil revenue to pen revenue in 2006 | let the revenue of pens be ' x ' let the revenue of pencils be ' y ' the revenue of pens up from 2006 by 5 % = > 105 % the revenue of pencils down from 2006 by 13 % = > 87 % overall revenue down by 1 % = > 99 % hence the ratio of pencils to pens = 6 / 12 = 1 / 2 answer : a | a = 5 + 100
b = 100 - 1
c = a - b
d = 100 - 1
e = 100 - 13
f = d - e
g = c / f
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a ) 6 % , b ) 12 % , c ) 14 % , d ) 18.66 % , e ) 20 % | d | multiply(divide(subtract(add(multiply(9, 57), 21), subtract(750, 300)), subtract(750, 300)), const_100) | a family made a down payment of $ 300 and borrowed the balance on a set of encyclopedias that cost $ 750 . the balance with interest was paid in 9 monthly payments of $ 57 each and a final payment of $ 21 . the amount of interest paid was what percent of the amount borrowed ? | interest paid = 57 * 9 + 21 - 450 = 84 % off the balance = ( 84 / 450 ) * 100 = 18.66 % answer is d . | a = 9 * 57
b = a + 21
c = 750 - 300
d = b - c
e = 750 - 300
f = d / e
g = f * 100
|
a ) 30 , b ) 36 , c ) 42 , d ) 45 , e ) 50 | d | divide(multiply(10, subtract(10, const_1)), const_2) | there are 10 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "10 c 2 = 45 the answer is d ." | a = 10 - 1
b = 10 * a
c = b / 2
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['a ) 20 * ( sqr 2 - 1 )', 'b ) 8 * ( sqr 3 - 1 )', 'c ) 4 * ( sqr 7 - 1 )', 'd ) 12 * ( sqr 7 - 1 )', 'e ) none of these'] | a | multiply(20, subtract(sqrt(const_2), const_1)) | consider a quarter of a circle of radius 20 . let r be the radius of the circle inscribed in this quarter of a circle . find r . | i got 20 / ( sqr 2 + 1 ) and just forgot to multiply by ( sqr 2 - 1 ) . answer a | a = math.sqrt(2)
b = a - 1
c = 20 * b
|
a ) 95 , b ) 99 , c ) 26 , d ) 73 , e ) none of the above | b | add(multiply(const_10, add(subtract(12, 5), const_1)), 5) | what two - digit number is less than the sum of the square of its digits by 12 and exceeds their doubled product by 5 ? | "let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 12 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 12 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . . thus the two numbers can be 15 or 99 thus the answer is b" | a = 12 - 5
b = a + 1
c = 10 * b
d = c + 5
|
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(5, subtract(5, const_1))) | if x is to be chosen at random from the integers between 1 to 5 , inclusive , and y is to be chosen at random from the integers between 7 and 10 , 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 ) = 2 / 5 * 2 / 4 = 1 / 5 p ( x and y are both odd ) = 3 / 5 * 2 / 4 = 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 = 5 - 1
e = 5 * d
f = c / e
|
a ) 10 , b ) 11 , c ) 62 , d ) 14 , e ) 15 | c | subtract(power(2, 2), 2) | if x ^ 2 + ( 1 / x ^ 2 ) = 8 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 8 ^ 2 - 2 = 62 . thus , the answer is c ." | a = 2 ** 2
b = a - 2
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['a ) 5', 'b ) 10', 'c ) 15', 'd ) 20', 'e ) 25'] | b | multiply(sqrt(subtract(power(divide(52, const_4), const_2), power(divide(24, const_2), const_2))), const_2) | the perimeter of a rhombus is 52 units . one of its diagonal is 24 units . what is its second diagonals length ? | when the perimeter of a triangle is 52 units each side is 13 units next , the diagonal of a rhombus bisect each other at 90 degree thus the half of the diagonal is 12 units now in a right angle triangle so formed , the perpendicular is 12 and the hypptenuse is 13 ( that is the side of the triangle ) finding the base ( which is half of the other diagonal ) : 13 ^ 2 - 12 ^ 2 = 25 thus half of the diagonal is 5 and the diagonal is 10 units answer : b | a = 52 / 4
b = a ** 2
c = 24 / 2
d = c ** 2
e = b - d
f = math.sqrt(e)
g = f * 2
|
a ) 4 , b ) 6 , c ) 10 , d ) 8 , e ) 13 | a | subtract(multiply(40, divide(60, const_100)), multiply(divide(4, 5), 25)) | how much is 60 % of 40 is greater than 4 / 5 of 25 ? | "60 / 100 ) * 40 Γ’ β¬ β ( 4 / 5 ) * 25 24 - 20 = 4 answer : a" | a = 60 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | b | subtract(24, 18) | if x and y are integers , what is the least positive number of 24 x + 18 y ? | "24 x + 21 y = 6 ( 4 x + 3 y ) which will be a minimum positive number when 4 x + 3 y = 1 . 4 ( 1 ) + 3 ( - 1 ) = 1 then 6 ( 4 x + 3 y ) can have a minimum positive value of 6 . the answer is b ." | a = 24 - 18
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a ) 5.2 minutes , b ) 5.21 minutes , c ) 5.25 minutes , d ) 5.28 minutes , e ) 6 minutes | d | divide(726, divide(add(multiply(4.5, const_1000), multiply(3.75, const_1000)), const_60)) | the jogging track in a sports complex is 726 m in circumference . suresh and his wife start from the same point and walk in opposite direction at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | let both of them meet after tt min 4500 m are covered by suresh in 60 m . in tt min he will cover 4500 t 604500 t 60 likewise , in tt min suresh ' s wife will cover 3750 t 603750 t 60 given , 4500 t 60 + 3750 t 60 = 7264500 t 60 + 3750 t 60 = 726 tt = 5.28 minutes option ( d ) is correct | a = 4 * 5
b = 3 * 75
c = a + b
d = c / const_60
e = 726 / d
|
a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) 1 hours | d | divide(3, divide(add(multiply(divide(1, 30), const_60), divide(3, 3)), const_2)) | a boatman goes 3 km against the current of the stream in 3 hour and goes 1 km along the current in 30 minutes . how long will it take to go 3 km in stationary water ? | "explanation : speed upstream = 3 / 3 = 1 km / hr speed downstream = 1 / ( 30 / 60 ) = 2 km / hr speed in still water = 1 / 2 ( 2 + 1 ) = 3 / 2 km / hr time taken to travel 3 km in still water = 3 / ( 3 / 2 ) = 6 / 3 = 2 hours answer : option d" | a = 1 / 30
b = a * const_60
c = 3 / 3
d = b + c
e = d / 2
f = 3 / e
|
a ) 39 , b ) 27 , c ) 30 , d ) 26 , e ) 29 | c | divide(add(add(20, 24), multiply(8, 2)), const_2) | the average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 24 years . find the average age of the women ? | "20 + 24 + 8 * 2 = 60 / 2 = 30 answer : c" | a = 20 + 24
b = 8 * 2
c = a + b
d = c / 2
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a ) 66 , b ) 18 , c ) 16 , d ) 10 , e ) 15 | a | divide(multiply(20, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 20 m long and 16 1 / 2 m board ? | "20 * 33 / 2 = 5 / 2 * 2 * x = > x = 66 answer : a" | a = 1 / 2
b = 16 + a
c = 20 * b
d = 1 / 2
e = 2 + d
f = e * 2
g = c / f
|
a ) 6 % , b ) 14 % , c ) 20 % , d ) 140 % , e ) 43 % | d | multiply(divide(subtract(divide(add(30, 5), 150), divide(5, 50)), divide(5, 50)), const_100) | a corporation paid $ 5 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 million in profits ? | difference in ratios = ( 30 / 150 ) - ( 5 / 50 ) = ( 5 / 50 ) % change = ( change ( 5 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 140 % answer - d | a = 30 + 5
b = a / 150
c = 5 / 50
d = b - c
e = 5 / 50
f = d / e
g = f * 100
|
a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | d | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(70, 1), 70), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(70, 1), 70))), 75000)) | a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 69 / 70 ) Γ’ Λ β 150000 Γ£ β ( 69 / 70 ) Γ’ Λ β 75000 . = > rs 2 , 70,000 . decrease in profit = > 2 , 75,000 Γ’ Λ β 2 , 70,000 = > rs . 5,000 . answer : d | a = 5 * 1000
b = a * 100
c = 75000 * 2
d = c + 75000
e = b - d
f = 70 - 1
g = f / 70
h = 5 * 1000
i = h * 100
j = g * i
k = 75000 * 2
l = 70 - 1
m = l / 70
n = k * m
o = j - n
p = o - 75000
q = e - p
|
a ) 124 , b ) 125 , c ) 126 , d ) 127 , e ) 128 | d | add(add(add(multiply(5, multiply(5, 4)), multiply(5, 4)), 5), 2) | there are 5 thieves , each loot a bakery one after the other such that the first one takes 1 / 2 of the total no . of the breads plus 1 / 2 of a bread . similarly 2 nd , 3 rd , 4 th and 5 fth also did the same . after the fifth one no . of breads remained are 3 . initially how many breads were there ? | let the number of breads initial be x you should go in reverse order as left bread is 3 i . e x = ? after 1 st person ate a = x - ( x / 2 ) - ( 1 / 2 ) i . e a = ( 2 x - x ) / 2 - ( 1 / 2 ) i . e a = ( x / 2 ) - ( 1 / 2 ) i . e a = ( 2 x - 2 ) / 4 i . e a = ( x - 1 ) / 2 like wise - - - - after 2 nd person ate b = ( a - 1 ) / 2 like wise - - - - after 3 rd person ate c = ( b - 1 ) / 2 like wise - - - - after 4 th person ate d = ( c - 1 ) / 2 like wise - - - - after 5 th person ate e = ( d - 1 ) / 2 now as after 5 th person ate the value of remaining bread is - - - 3 therefore the value of e = 3 ; now go reverse from botom to top to get the value of x after calcuating value of d = 7 value of c = 15 value of b = 31 value of a = 63 value of x = 127 so the final answer is 127 breads were there answer : d | a = 5 * 4
b = 5 * a
c = 5 * 4
d = b + c
e = d + 5
f = e + 2
|
a ) a ) 540 , b ) b ) 400 , c ) c ) 700 , d ) d ) 650 , e ) e ) 840 | b | divide(multiply(260, const_100), subtract(const_100, 35)) | in an examination 35 % of the students passed and 260 failed . how many students appeared for the examination ? | "let the number of students appeared be x then , 65 % of x = 260 65 x / 100 = 260 x = 260 * 100 / 65 = 400 answer is b" | a = 260 * 100
b = 100 - 35
c = a / b
|
a ) 6 days , b ) 4 days , c ) 5 days , d ) 15 days , e ) 10 days | c | divide(500, multiply(10, divide(400, multiply(8, 5)))) | if 8 women can color 400 m long cloth in 5 days , then 10 women can color 500 m long cloth in ? | "the length of cloth painted by one woman in one day = 400 / 8 Γ 5 = 10 m no . of days required to paint 500 m cloth by 10 women = 500 / 10 Γ 10 = 5 days answer : c" | a = 8 * 5
b = 400 / a
c = 10 * b
d = 500 / c
|
a ) 4.7 , b ) 4.8 , c ) 4.3 , d ) 4.6 , e ) 4.9 | d | divide(subtract(3.95, add(multiply(3.4, divide(const_2, 6)), multiply(3.85, divide(const_2, 6)))), divide(const_2, 6)) | average of 6 numbers comes out to be 3.95 . average of two of the numbers is 3.4 while the average of other two is 3.85 . determine the average of the two numbers left . | explanation : in the given question we have taken average of 6 numbers taken 2 at a time which makes 3 numbers . = > 3.95 ( average of 1 st and 2 nd number ) = > 3.85 ( average of 3 rd and 4 th number ) = > 3.4 ( average of 5 th and 6 th number ) 3.95 = ( 3.4 + 3.85 + x ) / 3 x = 4.6 answer d | a = 2 / 6
b = 3 * 4
c = 2 / 6
d = 3 * 85
e = b + d
f = 3 - 95
g = 2 / 6
h = f / g
|
a ) 80 kg , b ) 87 kg , c ) 90 kg , d ) 100 kg , e ) 110 kg | b | add(multiply(8, 4), 55) | the average weight of 8 person ' s increases by 4 kg when a new person comes in place of one of them weighing 55 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 4 ) kg = 32 kg . weight of new person = ( 55 + 32 ) kg = 87 kg . b )" | a = 8 * 4
b = a + 55
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a ) 1000 , b ) 1050 , c ) 1060 , d ) 370 , e ) 380 | b | subtract(multiply(speed(900, 18), 39), 900) | a 900 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 900 / 18 = 50 / 1 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 50 / 1 = > x = 1950 m . l = 1950 - 900 = 1050 answer : option b" | a = speed * (
b = a - 39
|
['a ) 154 cm 2', 'b ) 308 m 2', 'c ) 150 m 2', 'd ) 407 m 2', 'e ) none of these'] | a | divide(multiply(power(14, const_2), const_pi), const_4) | a horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long . over how much area of the field can it graze ? | area of the shaded portion = 1 β 4 Γ Ο Γ ( 14 ) 2 = 154 m 2 answer a | a = 14 ** 2
b = a * math.pi
c = b / 4
|
a ) 70 , b ) 75 , c ) 67 , d ) 85 , e ) 90 | c | divide(add(add(add(add(55, 67), 76), 82), 55), add(const_4, const_1)) | reeya obtained 55 , 67 , 76 , 82 and 55 out of 100 in different subjects , what will be the average | "explanation : ( 55 + 67 + 76 + 82 + 55 / 5 ) = 67 option c" | a = 55 + 67
b = a + 76
c = b + 82
d = c + 55
e = 4 + 1
f = d / e
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['a ) 1', 'b ) 2', 'c ) 4', 'd ) 8', 'e ) 16'] | e | multiply(const_2, sqrt(power(8, const_2))) | a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 8 of th area of the garden . what is the radius of the circular garden ? | as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 8 ( pir ^ 2 ) pir ^ 2 = 16 pir r = 16 answer : e | a = 8 ** 2
b = math.sqrt(a)
c = 2 * b
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 33.33 % , e ) 40 % | d | multiply(subtract(divide(add(const_100, 60), add(const_100, 20)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 20 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 60 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | "another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 60 = 20 + b + 20 b / 100 b = 33.33 answer ( d )" | a = 100 + 60
b = 100 + 20
c = a / b
d = c - 1
e = d * 100
|
a ) 20 , b ) 31 , c ) 45 , d ) 90 , e ) 89 | a | multiply(2, 15) | the class mean score on a test was 50 , and the standard deviation was 15 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ? | "1 sd from the mean ranges from 35 to 65 , where 65 is within sd above the mean and 35 within 1 sd below the mean 2 sd = 15 twice = 30 from the the mean , which is 80 to 20 , where 80 is within 2 sd above the mean and 20 is within 2 sd below the mean . answer = a" | a = 2 * 15
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['a ) 120 cm cube', 'b ) 110 cm cube', 'c ) 220 cm cube', 'd ) 430 cm cube', 'e ) 480 cm cube'] | a | multiply(multiply(4, 6), 5) | find the area of a cuboid of length 4 cm , breadth 6 cm . and height 5 cm . | area of a cuboid = lxbxh = 4 cm x 6 cm x 5 cm = 120 cm cube answer : a | a = 4 * 6
b = a * 5
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a ) 660 , b ) 680 , c ) 720 , d ) 760 , e ) 800 | c | divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3)) | a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is two - thirds of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ? | "total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = 2 x / 3 x + 2 x / 3 + 600 = 1800 5 x / 3 = 1200 x = 720 km the answer is c ." | a = 1800 * 2
b = a / 3
c = b * 3
d = 2 + 3
e = c / d
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a ) rs . 12250 , b ) rs . 13375 , c ) rs . 16750 , d ) rs . 15000 , e ) none of these | d | multiply(55000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 55000 . what is the share of c ? | explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 Γ ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) Γ 55000 ] = rs . 15000 answer : option d | a = 2 / 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 1
f = 1/(e)
g = 55000 * f
|
a ) 12 , b ) 42 , c ) 44 , d ) 76 , e ) 84 | c | subtract(multiply(add(5, const_1), add(20, 3)), multiply(20, 5)) | the average runs of a cricket player of 5 innings was 20 . how many runs must he make in his next innings so as to increase his average of runs by 3 ? | "explanation : average after 6 innings = 23 required number of runs = ( 24 * 6 ) β ( 20 * 5 ) = 144 β 100 = 44 answer c" | a = 5 + 1
b = 20 + 3
c = a * b
d = 20 * 5
e = c - d
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a ) 7 , b ) 9 , c ) 11 , d ) can not be determined , e ) none of these | b | add(add(power(add(add(divide(subtract(subtract(20, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(20, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(20, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(20, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 20 more than the first of these numbers . what is the middle number ? | "solution let the numbers be x , x + 2 and x + 4 . then , x + ( x + 2 ) + ( x + 4 ) = x + 20 β 2 x = 14 β x = 7 . β΄ middle number = x + 2 = 9 . answer b" | a = 20 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 20 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 20 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 20 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 7000 , b ) 7029 , c ) 2778 , d ) 2800 , e ) 5000 | e | divide(600, divide(multiply(subtract(18, 12), const_2), const_100)) | a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 600 . find the sum ? | "let the sum be rs . x . ( x * 18 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 600 = > 36 x / 100 - 24 x / 100 = 600 = > 12 x / 100 = 600 = > x = 5000 . answer : e" | a = 18 - 12
b = a * 2
c = b / 100
d = 600 / c
|
a ) 25 , b ) 32 , c ) 36 , d ) 48 , e ) 62 | e | add(add(divide(multiply(subtract(100, add(add(add(17, 10), 9), 20)), 3), const_4), 20), 9) | there are 100 freshmen at a particular college , all of whom must take at least one of the 3 core classes : art , biology , and calculus . of these freshmen , 17 take only biology , 10 take only calculus , 9 take all 3 classes , and 20 take art and exactly one of the other two core classes . if the number of freshmen who take only art is 3 times the number of freshmen who take every core class except art , how many freshmen take art ? | make a venn diagram to get a clear picture . look at the diagram : each letter represents only one color . b represents the people who take only art . d represents people who take only art and bio etc . d + f = 20 ( people who take art and one other class ) b = 3 e ( people who take only art is 3 times the people who take bio and calculus ) 17 + 10 + 9 + b + d + e + f = 100 ( total people ) b + b / 3 = 44 b = 33 number of freshmen who take art = 33 + 20 + 9 = 62 answer e | a = 17 + 10
b = a + 9
c = b + 20
d = 100 - c
e = d * 3
f = e / 4
g = f + 20
h = g + 9
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a ) 512 , b ) 656 , c ) 740 , d ) 896 , e ) 972 | d | multiply(subtract(power(3, 4), 3), multiply(4, 4)) | in how many ways can an answer key for a quiz be written if the quiz contains 4 true - false questions followed by 3 multiple - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | "there are 2 ^ 4 = 16 possibilities for the true - false answers . however we need to remove two cases for tttt and ffff . there are 4 * 4 * 4 = 64 possibilities for the multiple choice questions . the total number of possibilities is 14 * 64 = 896 . the answer is d ." | a = 3 ** 4
b = a - 3
c = 4 * 4
d = b * c
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a ) 228 , b ) 2000 , c ) 267 , d ) 270 , e ) 274 | d | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 250) | a goods train runs at the speed of 72 km / hr and crosses a 250 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 + 250 ) / 26 = 20 x = 270 m . answer : d" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 250
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a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 8 | b | divide(52.416, 0.0168) | when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 52.416 is divided by 0.0168 ? | "for the 1 st no . there are 2 digits after decimal for the 2 nd no . there are 5 digits after decimal total no . of decimals = 7 req . no . of digits = ( n - 1 ) = ( 7 - 1 ) = 6 answer : b" | a = 52 / 416
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a ) 299 , b ) 277 , c ) 276 , d ) 370 , e ) 281 | d | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 150) | a goods train runs at the speed of 72 km / hr and crosses a 150 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 + 150 ) / 26 = 20 x = 370 m . answer : d" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 150
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a ) 19 , b ) 91 , c ) 41 , d ) 18 , e ) 14 | a | subtract(add(floor(divide(subtract(76, 49), 3)), divide(subtract(76, 49), 2)), floor(divide(subtract(76, 49), multiply(2, 3)))) | if w is the set of all the integers between 49 and 76 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 75 , the smallest is 51 ) : 75 - 51 = 24 step 2 . divide by 3 : 24 / 3 = 8 step 3 . add 1 : 8 + 1 = 9 . so there are 9 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 76 , the smallest is 50 ) : 76 - 50 = 26 step 2 . divide by 2 : 26 / 2 = 13 step 3 . add 1 : 13 + 1 = 14 . so there are 14 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 9 multiples of 3 and the 14 multiples of 2 : 9 + 14 = 23 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 9 + 14 = 23 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 23 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 72 - 54 = 18 step 2 . divide by 6 : 18 / 6 = 3 step 3 . add 1 : 3 + 1 = 4 . so there are 4 multiples of 6 within the range : we counted 4 numbers twice . subtract the 4 multiples of 6 from the sum of the multiples of 2 and 3 : = 9 + 14 - 4 = 23 - 4 = 19 therefore , the final number of multiples of 2 , 3 or 6 is 19 . hence , this is the correct answer . ( a )" | a = 76 - 49
b = a / 3
c = math.floor(b)
d = 76 - 49
e = d / 2
f = c + e
g = 76 - 49
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
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a ) 2 : 9 , b ) 2 : 5 , c ) 3 : 7 , d ) 2 : 0 , e ) 2 : 1 | c | divide(subtract(10, 7), 7) | the ratio between the sale price and the cost price of an article is 10 : 7 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 7 x and s . p . = rs . 10 x . then , gain = rs . 3 x required ratio = 3 x : 7 x = 3 : 7 answer : c" | a = 10 - 7
b = a / 7
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a ) 20 sec , b ) 15 sec , c ) 30 sec , d ) 50 sec , e ) 1 min | a | divide(300, add(8, 7)) | two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 300 meters , after what time will they meet at the starting point ? | they meet every 300 / 7 + 8 = 20 sec answer is a | a = 8 + 7
b = 300 / a
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a ) 315 , b ) 345 , c ) 325 , d ) 335 , e ) 400 | e | multiply(multiply(const_100.0, divide(12, 1344)), 3) | what annual installment will discharge a debt of rs . 1344 due in 3 years at 12 % simple interest ? | "let each installment be rs . x then , ( x + ( ( x * 12 * 1 ) / 100 ) ) + ( x + ( ( x * 12 * 2 ) / 100 ) ) + x = 1344 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1344 Γ― Ζ βΊ ( 28 x + 31 x + 25 x ) = ( 1344 * 25 ) x = ( 1344 * 25 ) / 84 = rs . 400 . therefore , each installment = rs . 400 . answer is e ." | a = 12 / 1344
b = 100 * 0
c = b * 3
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a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 4 : 1 , e ) 10 : 1 | e | multiply(5, const_2) | city x has a population 5 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | "x = 5 y , y = 2 * z x : y , y : z 5 : 1 , 2 : 1 10 : 2 , 2 : 1 so , x : z = 10 : 1 ( e )" | a = 5 * 2
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a ) 60 , b ) 50 , c ) 70 , d ) 55 , e ) 40 | b | multiply(divide(8, add(subtract(multiply(const_3, const_4), 7), 11)), const_100) | a working mom wakes up every day at 7 am and goes to sleep at 11 pm . she works 8 hours a day . spends 2 hours working out at the gym . she spends 1.5 hours cooking dinner and doing dishes . she spends . 5 of an hour giving her daughter a bath . she spends 1 hour helping with homework and getting her daughter ready for bed . she spends another . 5 of an hour packing lunches for the family for the next day . she spends . 5 of an hour cleaning the house and 2 hours taking a shower and reading or watching t . v . before she goes to sleep . what percent of her day does she spend at work ? | 8 + 2 + 1.5 + . 5 + 1 + . 5 + . 5 + 2 = 16 = total number of hours 8 / 16 = . 05 = 50 percemt the answer is b | a = 3 * 4
b = a - 7
c = b + 11
d = 8 / c
e = d * 100
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a ) 588 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1400) | in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1400 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1400 female employees , 60 % of 1400 . = ( 60 * 1400 ) / 100 = 840 . then male employees , = 560 50 % of male are computer literate , = 280 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1400 ) / 100 = 868 computer literate . thus , female computer literate = 868 - 280 = 588 . answer : option a" | a = 62 / 100
b = 60 / 100
c = 1 - b
d = 50 / 100
e = c * d
f = a - e
g = f * 1400
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a ) 32 , b ) 33 , c ) 39 , d ) 40 , e ) 51 | e | subtract(81, divide(81, add(const_1, divide(170, const_100)))) | sales price is $ 81 , gross profit is 170 % of cost , what is the value of gross profit ? | "cost + profit = sales cost + ( 170 / 100 ) cost = 81 cost = 30 profit = 81 - 30 = 51 answer ( e )" | a = 170 / 100
b = 1 + a
c = 81 / b
d = 81 - c
|
a ) 2574 , b ) 2500 , c ) 1485 , d ) 1574 , e ) none of these | c | divide(17820, 12) | if the product of two numbers is 17820 and their h . c . f . is 12 , find their l . c . m . | "explanation : hcf * lcm = 17820 , because we know product of two numbers = product of hcf and lcm lcm = 17820 / 12 = 1485 option c" | a = 17820 / 12
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a ) 10 , b ) 25 , c ) 35 , d ) 46 , e ) 50 | b | subtract(divide(add(160, 20), const_4), 20) | thabo owns exactly 160 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books does thabo own ? | i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fiction books is 0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f 2 x + 40 0 nf x + 20 x total 3 x + 60 x 160 3 x + 60 + x = 160 x = 25 answer ( b . ) | a = 160 + 20
b = a / 4
c = b - 20
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a ) - 4.5 . , b ) - 2 . , c ) - 1.7 . , d ) 3 . , e ) 2.5 . | c | multiply(21, 2) | if x + y = 2 x - 2 z , x - 2 y = 4 z and x + y + z = 21 , what is the value of y / z ? | given : x + y = 2 x - 2 z , - - > eq 1 x - 2 y = 4 z - - > eq 2 x + y + z = 21 - - > eq 3 . now we are asked to find the value of y / z . now let choose some option randomly , take c , which is 3 , i randomly chose this value as it is positive : ) y / z = 3 , this is the value we need to get . then y = 3 z - - eq 4 sub this in eq 1 . x + 3 z = 2 x - 2 z = > x = 5 z - - > eq 5 . sub x and y value in eq 3 . 5 z + 3 z + z = 21 = > z = 21 / 9 . now sub this value to get y , we get y = 21 / 3 . then y / z = 21 / 3 / 21 / 9 = we get 3 again . so this option is correct . answer : option c is correct . | a = 21 * 2
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a ) 400 , b ) 510 , c ) 500 , d ) none of these , e ) 506 | a | divide(100, divide(25, const_100)) | evaluate 100 / . 25 | explanation : 100 / . 25 = 10000 / 25 = 400 option a | a = 25 / 100
b = 100 / a
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a ) 5 kg , b ) 12,5 kg , c ) 25 kg , d ) 30 kg , e ) none | b | divide(const_100, divide(subtract(const_100, 20), 10)) | the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 10 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 10 kg . as the price falls by 20 % , new price for 10 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 10 = rs . 8 per kg . rice can bought now at = 100 / 8 = 12,5 kg . answer : option b" | a = 100 - 20
b = a / 10
c = 100 / b
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a ) 147 , b ) 142 , c ) 173 , d ) 144 , e ) 186 | d | add(add(70, const_1), add(70, const_3)) | what is the smallest integer that is the sum of 2 distinct prime integers , each greater than 70 ? | a prime number ( or a prime ) is a natural number greater than 1 that has no positive divisors other than 1 and itself . here are the first few prime numbers : 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131 , 137 , 139 , 149 , 151 , 157 , 163 , 167 , 173 , 179 , 181 , 191 , 193 , 197 , 199 , etc . immediate two prime numbers after 70 are - 71 and 73 sum = 71 + 73 = 144 answer - d | a = 70 + 1
b = 70 + 3
c = a + b
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a ) 24 , b ) 22 , c ) 16 , d ) 18 , e ) 20 | c | add(add(4, 3), add(add(4, 4), 1)) | for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 940 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 940 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( opps / / it exceeds 1000 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which haslength of integeris 9 . now , since x = 512 , x + 3 y < 940 so , 3 y < 428 = = > y < 428 / 3 so , y can take any value which is less than 428 / 3 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . c" | a = 4 + 3
b = 4 + 4
c = b + 1
d = a + c
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a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | b | divide(100, multiply(7, const_3.0)) | how many 7 in between 1 to 100 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78,79 , 87,97 20 7 ' s between 1 to 100 answer : b" | a = 7 * 3
b = 100 / a
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a ) 31.5 , b ) 28 , c ) 98 , d ) 37 , e ) 13 | a | divide(162, add(const_2, const_pi)) | the perimeter of a semi circle is 162 cm then the radius is ? | "36 / 7 r = 162 = > r = 31.5 answer : a" | a = 2 + math.pi
b = 162 / a
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a ) 22 hours , b ) 30 hours , c ) 44 hours , d ) 60 hours , e ) it can not be determined from the information above . | c | divide(const_1, divide(add(divide(const_1, 4), divide(const_1, 11)), 15)) | together , 15 type a machines and 7 type b machines can complete a certain job in 4 hours . together 8 type b machines and 15 type c machines can complete the same job in 11 hours . how many hours r would it take one type a machine , one type b machine , and one type c machine working together to complete the job ( assuming constant rates for each machine ) ? | "say the rates of machines a , b and c are a , b , and c , respectively . together 15 type a machines and 7 type b machines can complete a certain job in 4 hours - - > 15 a + 7 b = 1 / 4 ; together 8 type b machines and 15 type c machines can complete the same job in 11 hours - - > 8 b + 15 c = 1 / 11 . sum the above : 15 a + 15 b + 15 c = 1 / 4 + 1 / 11 = 15 / 44 - - > reduce by 15 : a + b + c = 1 / 44 - - > so , the combined rate of the three machines is 1 / 44 job / hour - - > time is reciprocal of the rate , thus machines a , b and c can do the job r in 44 hours . answer : c ." | a = 1 / 4
b = 1 / 11
c = a + b
d = c / 15
e = 1 / d
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a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | d | divide(subtract(add(add(28, 3), 28), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 28 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 28 + 31 ) = 9 ( x - 1 ) = > 11 x - 9 x = 50 = > 2 x = 50 = > x = 25 . so , average age of the team is 25 years . d" | a = 28 + 3
b = a + 28
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
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a ) 3999 , b ) 7799 , c ) 2500 , d ) 5000 , e ) 2912 | d | divide(multiply(400, const_100), 2) | a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 400 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 400 b capital = 30000 * 1 / 6 = 5000 . answer : d" | a = 400 * 100
b = a / 2
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a ) 1 , b ) 2 , c ) 6 , d ) 7 , e ) 10 | b | multiply(divide(divide(multiply(30, 30), const_100), 54), const_12) | a reduction of 30 % in the price of apples would enable a man to obtain 54 more for rs . 30 , what is reduced price per dozen ? | "b 30 * ( 30 / 100 ) = 9 - - - 54 ? - - - 12 = > rs . 2" | a = 30 * 30
b = a / 100
c = b / 54
d = c * 12
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a ) $ 120 , b ) $ 135 , c ) $ 161 , d ) $ 165 , e ) $ 192 | c | add(multiply(divide(subtract(350, 35), 5), 2), 35) | a certain psychologist charges $ 35 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 350 , what is the total charge to a patient who receives only 2 hours of therapy ? | "let the charge for first hour = x + 35 then charge for each additional hour = x x + 35 + 4 x = 350 = > 5 x = 315 = > x = 63 total charge for patient for 3 hours of therapy = x + 35 + x = 2 x + 35 = 161 $ answer c" | a = 350 - 35
b = a / 5
c = b * 2
d = c + 35
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a ) 75 . , b ) 85 . , c ) 90 . , d ) 94 . , e ) 100 . | b | multiply(subtract(80, divide(multiply(70, const_2), 6)), divide(6, const_4)) | the average length of 6 ropes is 80 cm . if the average length of one third of the ropes measure to 70 cm , what is the average of the other ropes ? | edit : given ( x 1 + x 2 . . . + x 6 ) / 6 = 80 ( x 1 + x 2 . . . + x 6 ) = 480 - - > eq 1 . now given avg length of one third ropes is 70 . that means out 6 / 3 = 2 ropes . let the avg length of two ropes be ( x 1 + x 2 ) / 2 = 70 . ( x 1 + x 2 ) = 140 . - - > eq 2 . now we are asked to find the average of the remaining i . e . ( x 3 + x 4 + x 5 + x 6 ) substitute eq 2 in eq 1 then we get 140 + x 3 + x 4 + x 5 + x 6 = 480 = > x 3 + x 4 + x 5 + x 6 = 340 now divide 340 by 4 we get 85 . = > ( x 3 + x 4 + x 5 + x 6 ) / 4 = 85 = avg length of remaining ropes . imo correct option is b . | a = 70 * 2
b = a / 6
c = 80 - b
d = 6 / 4
e = c * d
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a ) 39.6 , b ) 43 , c ) 40 , d ) 38 , e ) 29 | a | divide(add(add(subtract(44, const_10), const_100), add(subtract(44, const_10), const_100)), add(divide(add(subtract(44, const_10), const_100), 44), divide(add(subtract(44, const_10), const_100), 36))) | x and y are two towns . ganesh covers the distance from x to y at an average speed of 44 km / hr . however , he covers the distance from y to x at an average speed of 36 km / hr . his average speed during the whole journey in km / hr . is : | "solution : average speed = 2 xy / x + y = 2 * 44 * 36 / 44 + 36 = 39.6 answer : a" | a = 44 - 10
b = a + 100
c = 44 - 10
d = c + 100
e = b + d
f = 44 - 10
g = f + 100
h = g / 44
i = 44 - 10
j = i + 100
k = j / 36
l = h + k
m = e / l
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a ) 5.26 , b ) 5.16 , c ) 5.61 , d ) 5.52 , e ) 5 | a | multiply(divide(subtract(const_100, 95), 95), const_100) | if the cost price is 95 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 95 : profit = rs 5 . profit = { ( 5 / 95 ) * 100 } % = 5.26 % answer is a ." | a = 100 - 95
b = a / 95
c = b * 100
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a ) 65 % , b ) 50 % , c ) 22 % , d ) 18 % , e ) 8.5 % | b | multiply(divide(subtract(multiply(const_100, divide(20, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 20 percent sugar by weight . the second solution was what percent sugar by weight ? | "say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.20 - - > x = 0.5 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.20 - - > x = 0.5 . answer : b ." | a = 20 / 100
b = 100 * a
c = 1 / 4
d = c * 100
e = 100 - d
f = 10 / 100
g = e * f
h = b - g
i = 1 / 4
j = i * 100
k = h / j
l = k * 100
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a ) 50 , b ) 80 , c ) 100 , d ) 44 , e ) 250 | d | multiply(divide(subtract(power(add(divide(multiply(20, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | a circular logo is enlarged to fit the lid of a jar . the new diameter is 20 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 4.8 , so radius is 2.4 new area = 5.76 Ο increase in area is 1.76 Ο % increase in area = 1.76 / 4 * 100 so , % increase is 44 % answer will be ( d )" | a = 20 * 2
b = a / 1000
c = b + 3
d = c ** 2
e = d - 4
f = e / 4
g = f * 100
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a ) 82 m , b ) 32 m , c ) 27 m , d ) 26 m , e ) 76 m | b | divide(add(add(sqrt(subtract(power(8, const_2), power(2, const_2))), 2), add(sqrt(subtract(power(8, const_2), power(2, const_2))), 2)), 2) | what is the perimeter of a square field whose diagonal is 8 β 2 ? | "4 a = 32 m answer : b" | a = 8 ** 2
b = 2 ** 2
c = a - b
d = math.sqrt(c)
e = d + 2
f = 8 ** 2
g = 2 ** 2
h = f - g
i = math.sqrt(h)
j = i + 2
k = e + j
l = k / 2
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a ) 55287 , b ) 29887 , c ) 27768 , d ) 55053 , e ) 17191 | d | add(multiply(120, 456), 333) | in a division sum , the quotient is 120 , the divisor 456 and the remainder 333 , find the dividend ? | "explanation : 120 * 456 + 333 = 55053 answer : d" | a = 120 * 456
b = a + 333
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a ) 10 / 30 , b ) 20 / 30 , c ) 1 , d ) 5 / 30 , e ) 15 / 30 | a | divide(multiply(const_5, const_2), 30) | what is the probability that a number selected from number 1 to 30 is prime number , when each of the given numbers is equally likely to be selected ? | x = { 2,3 , 5,7 , 11,13 , 17,19 , 23,29 } n ( x ) = 10 n ( s ) = 30 hence required probability = n ( x ) / n ( s ) = 10 / 30 option a is answer | a = 5 * 2
b = a / 30
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a ) 40 , b ) 60 , c ) 96 , d ) 120 , e ) 240 | c | divide(8.00, subtract(divide(1.00, 4), divide(0.50, 3))) | a grocer purchased a quantity of bananas at 3 pounds for $ 0.50 and sold the entire quantity at 4 pounds for $ 1.00 . how many pounds did the grocer purchase if the profit from selling the bananas was $ 8.00 ? | "cost price of 1 pound of bananas = 0.5 / 3 = 1 / 6 selling price of 1 pound of bananas = 1 / 4 profit per pound = ( 1 / 4 - 1 / 6 ) = ( 1 / 12 ) total profit is given as 8 ( 1 / 12 ) * x = 8 x = 96 answer : c" | a = 1 / 0
b = 0 / 50
c = a - b
d = 8 / 0
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a ) 27 , b ) 36 , c ) 42 , d ) 22 , e ) 147 | d | multiply(divide(48, 112), 53) | the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 53 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 53 ounces of a substance has a volume of ( 48 / 112 ) * 53 = 22 cubic inches answer d" | a = 48 / 112
b = a * 53
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a ) 1 : 1 , b ) 3 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | a | divide(subtract(27, 22), subtract(22, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 22 seconds , what is the ratio of their speeds ? | let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 22 s = ( 27 x + 17 y ) / ( x + y ) = 22 = ( 27 x + 17 y ) / = 22 ( x + y ) = 5 x = 5 y = x / y = 5 / 5 = 1 / 1 i . e 1 : 1 answer : a | a = 27 - 22
b = 22 - 17
c = a / b
|
a ) 156 , b ) 162 , c ) 317 , d ) 324 , e ) 325 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 5.30))) | roses can be purchased individually for $ 5.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | "buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 5.3 = ~ 5 roses for that amount . total = 312 + 5 = 317 . answer : c ." | a = 680 / 50
b = math.floor(a)
c = 12 * 2
d = b * c
e = 680 / 50
f = math.floor(e)
g = f * 50
h = 680 - g
i = h / 5
j = math.floor(i)
k = d + j
|
a ) rs . 6000 , b ) rs . 8000 , c ) rs . 7500 , d ) rs . 6500 , e ) rs . 8500 | b | divide(subtract(14000, multiply(14000, divide(80, const_100))), subtract(subtract(add(const_1, const_1), divide(80, const_100)), divide(85, const_100))) | the salaries of a and b together is rs . 14000 . a spend 80 % of his salary and b spends 85 % of his salary . what is the salary of b if their savings are equal ? | let the salaries of a and b are x and y respectively x + y = 14000 savings of a = 20 x / 100 = savings of b = 15 y / 100 x = ΒΎ y 3 / 4 y + y = 14000 , 7 y / 4 = 14000 , y = 8000 answer : b | a = 80 / 100
b = 14000 * a
c = 14000 - b
d = 1 + 1
e = 80 / 100
f = d - e
g = 85 / 100
h = f - g
i = c / h
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | how many of the positive divisors q of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . q = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a = 120 / 4
b = a / 2
c = b / 3
|
a ) 10,300 , b ) 10,030 , c ) 1,353 , d ) 1,352 , e ) 1,349 | e | subtract(458,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 323,700 and 458,600 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 323,700 and 458,600 is 458,600 - 323,700 = 134,900 - one number per each hundred gives 134,900 / 100 = 1,349 numbers . answer : e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 458 - 600
|
a ) 0 , b ) 1 / 12 , c ) 2 / 15 , d ) 1 / 10 , e ) 2 / 10 | b | multiply(divide(1, 6), divide(subtract(1, divide(2, 3)), divide(2, 3))) | two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 6 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | "suppose the water in each jug is l liters cx x ( 1 / 6 ) = l cx = 6 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 6 l = 1 / 12 answer will be b" | a = 1 / 6
b = 2 / 3
c = 1 - b
d = 2 / 3
e = c / d
f = a * e
|
a ) 1361 , b ) 1376 , c ) 1363 , d ) 1364 , e ) 1365 | b | divide(multiply(51, add(51, 11)), 12) | 11 + 12 + 13 + . . . 51 + 52 + 53 = ? | "sum = 11 + 12 + 13 + . . . 51 + 52 + 53 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 53 positive integers = 53 * 54 / 2 sum of first 10 positive integers = 11 * 10 / 2 sum = 11 + 12 + 13 + . . . 51 + 52 + 53 = 53 * 54 / 2 - 11 * 10 / 2 = 1376 answer : b" | a = 51 + 11
b = 51 * a
c = b / 12
|
a ) 7.6 % , b ) 7.7 % , c ) 11.32 % , d ) 11.91 % , e ) 7.8 % | d | multiply(const_100, divide(subtract(multiply(52, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | a man buys 52 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 52 pens = rs . 46 s . p . of 52 pens = 99 % of rs . 52 = rs . 51.48 profit % = ( profit / c . p . ) x 100 profit % = ( 5.48 / 46 ) x 100 = 11.91 % answer : d" | a = 100 - 1
b = 52 * a
c = 46 * 100
d = b - c
e = 46 * 100
f = d / e
g = 100 * f
|
a ) 172 , b ) 160 , c ) 150 , d ) 180 , e ) 120 | d | multiply(divide(50, const_1000), const_3600) | express 50 mps in kmph ? | "25 * 18 / 5 = 180 kmph answer : d" | a = 50 / 1000
b = a * 3600
|
a ) 250 , b ) 125 , c ) 500 , d ) 100 , e ) 80 | a | multiply(250, const_1) | the total price of a kilogram each of shimla apples and red delicious apples is 250 rupees more than the total price of a kilogram each of red delicious apples and fuji apples . fuji apples is how much cheaper than shimla apples ? | ( shimla + red delicious ) - ( red delicious + fuji ) = 250 shimla - fuji = 250 answer : a | a = 250 * 1
|
a ) 82.1 sec . , b ) 12.4 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec . | b | divide(add(110, 138), 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 138 m length ? | speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 138 = 248 m . required time = 248 / 20 = 12.4 sec . answer : b | a = 110 + 138
b = 72 * const_0_2778
c = a / b
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(6, 11) | what is the 25 th digit to the right of the decimal point in the decimal form of 6 / 11 ? | "explanation : to determine the greatest possible number of members , first recognize that each member had to contribute the lowest amount given . build an inequality for the individual contributions and the total amount collected , with n = the number of members in the club , and solve for n . 12 n β€ 599 n β€ 49 11 / 12 since n represents individual people it must be a whole number . thus the greatest number of people is 49 . answer : option c" | a = 6 / 11
|
a ) $ 490 , b ) $ 500 , c ) $ 510 , d ) $ 520 , e ) $ 530 | b | subtract(590, multiply(subtract(815, 590), divide(2, 5))) | a sum of money lent out at s . i . amounts to a total of $ 590 after 2 years and to $ 815 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 815 - $ 590 = $ 225 the s . i . is $ 45 / year s . i . for 2 years = $ 90 principal = $ 590 - $ 90 = $ 500 the answer is b ." | a = 815 - 590
b = 2 / 5
c = a * b
d = 590 - c
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(add(7, const_10), const_10) | if 5 < x < 9 and y = x + 7 , what is the greatest possible integer value of x + y ? | "x + y = x + x + 7 = 2 x + 7 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 8.5 then x + y = 8.5 + 15.5 = 24 . the answer is d ." | a = 7 + 10
b = a + 10
|
a ) 18 / 35 , b ) 16 / 35 , c ) 14 / 35 , d ) 13 / 35 , e ) 12 / 35 | a | divide(multiply(choose(4, divide(4, const_2)), choose(4, divide(4, const_2))), choose(8, 4)) | there are 8 students . 4 of them are men and 4 of them are women . if 4 students are selected from the 8 students . what is the probability that the number of men is equal to that of women ? | men = women means 2 men 2 women total such combinations = 4 c 2 * 4 c 2 = 4 ! / 2 ! . 2 ! * 4 ! / 2 ! . 2 ! = 6 * 6 total combinations = 8 c 4 = 8 * 7 * 6 * 5 / 4 * 3 * 2 * 1 = 70 so probability = 36 / 70 = 18 / 35 hence a | a = 4 / 2
b = math.comb(4, a)
c = 4 / 2
d = math.comb(4, c)
e = b * d
f = math.comb(8, 4)
g = e / f
|
a ) 10 sec , b ) 20 sec , c ) 25 sec , d ) 30 sec , e ) 35 sec | b | divide(200, multiply(const_0_2778, 36)) | how many seconds will a train 200 meters long running at the rate of 36 kmph take to pass a certain telegraph post ? | distance = length of train = 200 meters speed = 36 kmph = 36 * 5 / 18 = 10 m / s required time = 200 / 10 = 20 sec answer is b | a = const_0_2778 * 36
b = 200 / a
|
a ) 50 , b ) 96 , c ) 100 , d ) 125 , e ) 250 | b | multiply(divide(subtract(power(add(divide(multiply(40, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | a circular logo is enlarged to fit the lid of a jar . the new diameter is 40 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 5.6 , so radius is 2.8 new area = 7.84 Ο increase in area is 3.84 Ο % increase in area = 3.84 / 4 * 100 so , % increase is 96 % answer will be ( b )" | a = 40 * 2
b = a / 1000
c = b + 3
d = c ** 2
e = d - 4
f = e / 4
g = f * 100
|
a ) $ 1865 , b ) $ 2087 , c ) $ 2217 , d ) $ 2431 , e ) $ 2662 | d | divide(multiply(multiply(11000, divide(3315, divide(multiply(15000, 8), const_100))), 8), const_100) | a , b and c enter into a partnership by investing $ 11000 , $ 15000 and $ 23000 respectively . at the end of 8 months , b receives $ 3315 as his share . find the share of a . | the ratio of capital of a , b and c = 11000 : 15000 : 23000 = 11 : 15 : 23 a ' s share = ( 11 / 15 ) * 3315 = $ 2431 the answer is d . | a = 15000 * 8
b = a / 100
c = 3315 / b
d = 11000 * c
e = d * 8
f = e / 100
|
a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | c | divide(divide(multiply(800, 35), const_100), 5) | a reduction of 35 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | 800 * ( 35 / 100 ) = 280 - - - - 5 ? - - - - 1 = > rs . 56 answer : c | a = 800 * 35
b = a / 100
c = b / 5
|
a ) 16 , b ) 90 , c ) 112 , d ) 128 , e ) 142 | b | multiply(2, divide(90, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 90 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is in 90 grams of water . 2 x + 16 x = 90 18 x = 90 x = 5 since x is 5 , we know that there are 16 x 5 = 80 grams of oxygen in 90 grams of water . answer b ." | a = 2 + 16
b = 90 / a
c = 2 * b
|
a ) 9 , b ) 10 , c ) 11 , d ) 3 , e ) 13 | d | divide(6, divide(6, 3)) | suppose 3 monkeys take 3 minutes to eat 3 bananas . how many monkeys would it take to eat 6 bananas in 6 minutes ? | "one monkey takes 3 min to eat 1 banana , so in 6 mins 1 monkey will eat 2 bananas , so for 6 bananas in 6 min we need 6 / 2 = 3 monkeys answer : d" | a = 6 / 3
b = 6 / a
|
a ) 1.5 , b ) 2.5 , c ) 3 , d ) 4.5 , e ) 5 | b | divide(divide(subtract(add(50, 50), 50), subtract(50, 40)), const_2) | a dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce . a chef accidentally makes 15 cups of the sauce with 40 % melted chocolate and 60 % raspberry puree instead . how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50 % of each ? | we have 15 cups of sauce with 4040 choc and 6060 rasb cups of choc = 0.4 β 15 = 6 cups of rasb = 0.6 β 15 = 9 now let say we removed x cup of original mix and replaced with x cups of choc . therefore final number of cups of choc = 6 β 0.4 x + x now this number of cup should be 50 % of total = 15 / 2 = 7.5 therefore 6 β 0.4 x + x = 7.5 on solving x = 2.5 answer : b | a = 50 + 50
b = a - 50
c = 50 - 40
d = b / c
e = d / 2
|
a ) 3 , b ) 4 , c ) 6 , d ) 12 , e ) 8 | e | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | if x and y are both odd prime numbers andx < y , then how many different positive integer factors does 2 xyhave ? | "x and y are both odd prime numbers - it means either x or y is not 2 and since prime numbers have only two factors - 1 and the number itself x and y each will have ( 1 + 1 ) = 2 factors hence 2 xy will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors e is the answer" | a = 1 + 1
b = 2 * a
c = 1 + 1
d = b * c
|
a ) 336 , b ) 366 , c ) 330 , d ) 660 , e ) 770 | a | subtract(negate(60), multiply(subtract(6, 24), divide(subtract(6, 24), subtract(1, 6)))) | 1 , 6 , 24 , 60 , 120 , 210 , ___ ? | "it follow the pattern like this . . 2 ^ 3 - 2 = 6 3 ^ 3 - 3 = 24 4 ^ 3 - 4 = 60 5 ^ 3 - 5 = 120 6 ^ 3 - 6 = 210 so . . . 7 ^ 3 - 7 = 336 answer : a" | a = negate - (
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | multiply(divide(2, add(1, 2)), 30) | there are 30 students in a classroom . the ratio of the # of girls to boys was 1 : 2 , how many boys are there ? | # of girls = x # of boys = 2 x x + 2 x = 30 3 x = 30 x = 30 / 3 = 10 boys = 2 x = 2 * 10 = 20 answer : c | a = 1 + 2
b = 2 / a
c = b * 30
|
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