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 ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(multiply(3, 6), add(subtract(13, 6), 3)), 6) | the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 5 . if the sum of the first and the last number is 13 . what is the last numbers ? | "a + b + c = 18 b + c + d = 15 a + d = 13 a – d = 3 a + d = 13 2 d = 10 d = 5 answer : b" | a = 3 * 6
b = 13 - 6
c = b + 3
d = a - c
e = d - 6
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a ) 22 , b ) 28 , c ) 17 , d ) 12 , e ) 47 | e | divide(4136, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | if the wheel is 14 cm then the number of revolutions to cover a distance of 4136 cm is ? | "2 * 22 / 7 * 14 * x = 4136 = > x = 47 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 14
h = 4136 / g
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a ) 23.56 , b ) 23.59 , c ) 44.0 , d ) 23.53 , e ) 23.57 | c | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(36, 20)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 36 cm and breadth 20 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 36 + 20 ) = 112 cm parameter of the square = 112 cm i . e . 4 a = 112 a = 28 diameter of the semicircle = 28 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 28 ) = 1 / 2 ( 22 / 7 ) ( 28 ) = 44 cm to two decimal places answer : c" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) s . 45 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 72 | e | divide(720, subtract(15, 5)) | on selling 15 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 15 balls ) - ( s . p . of 15 balls ) = ( c . p . of 5 balls ) c . p . of 10 balls = s . p . of 15 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 10 = rs . 72 . answer : option e" | a = 15 - 5
b = 720 / a
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a ) 179 : 2 , b ) 178 : 3 , c ) 179 : 1 , d ) 179 : 54 , e ) 179 : 45 | d | divide(add(divide(multiply(62.5, 3), const_100), divide(multiply(87.5, 4), const_100)), add(subtract(3, divide(multiply(62.5, 3), const_100)), subtract(4, divide(multiply(87.5, 4), const_100)))) | two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 3 litres from vessel p is mixed with 4 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 3 = 15 / 8 litres quantity of alcohol in vessel q = 87.5 / 100 * 4 = 7 / 2 litres quantity of alcohol in the mixture formed = 15 / 8 + 7 / 2 = 43 / 8 = 5.375 litres as 7 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 5.375 : 1.625 = 179 : 54 . answer : d" | a = 62 * 5
b = a / 100
c = 87 * 5
d = c / 100
e = b + d
f = 62 * 5
g = f / 100
h = 3 - g
i = 87 * 5
j = i / 100
k = 4 - j
l = h + k
m = e / l
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a ) 2.5 , b ) 3 , c ) 2 , d ) 1.3 , e ) 1.5 | e | subtract(add(multiply(9, const_2), 5), add(3, 6)) | x and y are positive integers . when x is divided by 6 , the remainder is 3 , and when x is divided by 16 , the remainder is 5 . when y is divided by 9 , the remainder is 5 , and when y is divided by 7 , the remainder is 4 . what is the least possible value of x / y ? | "when x is divided by 6 , the remainder is 3 : so , the possible values of x are : 3 , 9 , 15 , 21 , etc . when x is divided by 16 , the remainder is 5 : so , the possible values of x are : 5,21 . . . stop . since both lists include 21 , the smallest possible value of x is 21 . when y is divided by 9 , the remainder is 5 : so , the possible values of y are : 5 , 14 , 23,32 etc . when y is divided by 7 , the remainder is 4 : so , the possible values of y are : 6 , 14 , . . . stop . since both lists include 14 , the smallest possible value of y is 14 since the smallest possible values of x and y are 21 and 14 respectively , the smallest possible value of x / y is 1.5 . so , e is the correct answer to the original question ." | a = 9 * 2
b = a + 5
c = 3 + 6
d = b - c
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a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 15 seconds , e ) none of these | d | divide(600, multiply(144, const_0_2778)) | in what time will a train 600 meters long cross an electric pole , if its speed is 144 km / hr | "explanation : first convert speed into m / sec speed = 144 * ( 5 / 18 ) = 40 m / sec time = distance / speed = 600 / 40 = 15 seconds answer : d" | a = 144 * const_0_2778
b = 600 / a
|
a ) 6 , b ) 36 , c ) 120 , d ) 60 , e ) 72 | c | lcm(10, 24) | a certain class of students is being divided into teams . the class can either be divided into 10 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | "let total no of students in the class be n so , we are told that n is divisible by both 10 24 so , lets find the least common multiple of 10 24 , ie 120 so our answer is ( c ) 120" | a = math.lcm(10, 24)
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a ) rs . 2000 , b ) rs . 10,000 , c ) rs . 15,000 , d ) rs . 20,000 , e ) none | c | divide(divide(multiply(5400, const_100), multiply(3, 12)), multiply(divide(const_60, const_4), const_100)) | a man took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was : | "solution principal = rs . ( 100 x 5400 / 12 x 3 ) = rs . 15000 . answer c" | a = 5400 * 100
b = 3 * 12
c = a / b
d = const_60 / 4
e = d * 100
f = c / e
|
a ) 39 , b ) 45 , c ) 51 , d ) 57 , e ) 63 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 7), add(2, 2)), 6) | the ratio of the present age of sandy to that of molly is 7 : 2 . twelve years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | "let the present age of sandy be 7 x years and that of molly be 2 x years . ( 7 x + 12 ) / ( 2 x + 12 ) = 5 / 2 4 x = 36 x = 9 six years ago , sandy ' s age was 7 ( 9 ) - 6 = 57 the answer is d ." | a = 10 + 6
b = 5 * a
c = 10 + 6
d = 2 * c
e = b - d
f = e * 7
g = 2 + 2
h = f / g
i = h - 6
|
a ) 7 , b ) 30 , c ) 28 , d ) 25 , e ) 16 | c | multiply(60, divide(60, add(add(3, 5), 7))) | the sides of a triangle are in the ratio 3 : 5 : 7 . if the pereimeter is 60 cms , the longest side is | "3 x + 5 x + 7 x = 60 15 x = 60 x = 4 longest side = 7 * 4 = 28 answer : c" | a = 3 + 5
b = a + 7
c = 60 / b
d = 60 * c
|
a ) 1500 , b ) 1000 , c ) 999 , d ) 1001 , e ) 1005 | b | divide(subtract(5000, 5), 5) | how many positive integers between 5 and 5000 ( both are inclusive ) are there such that they are multiples of 5 ? | "multiples of 5 = 5 , 10,15 - - - - - , 5000 number of multiples of 5 = > 5000 - 5 / 5 + 1 = 1000 answer is b" | a = 5000 - 5
b = a / 5
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a ) 288 , b ) 277 , c ) 380 , d ) 290 , e ) 262 | c | subtract(subtract(528, divide(multiply(528, 20), const_100)), divide(multiply(subtract(528, divide(multiply(528, 20), const_100)), 10), const_100)) | the sale price sarees listed for rs . 528 after successive discount is 20 % and 10 % is ? | "528 * ( 80 / 100 ) * ( 90 / 100 ) = 380 answer : c" | a = 528 * 20
b = a / 100
c = 528 - b
d = 528 * 20
e = d / 100
f = 528 - e
g = f * 10
h = g / 100
i = c - h
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a ) 100 , b ) 105 , c ) 110 , d ) 120 , e ) 125 | e | divide(9, subtract(divide(9, 6), 5)) | a train covers a distance of 9 km in 6 min . if it takes 5 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 9 / 6 * 60 ) km / hr = ( 90 * 5 / 18 ) m / sec = 25 m / sec . length of the train = 25 * 5 = 125 m . answer : option e" | a = 9 / 6
b = a - 5
c = 9 / b
|
a ) 11 , b ) 20 , c ) 24 , d ) 27 , e ) 32 | b | floor(multiply(divide(3, 8), 52)) | the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 52 black cars in the lot , how many red cars are there in the lot ? | "b is correct r / b = 3 / 8 and b = 52 r = 52 * 3 / 8 = 20" | a = 3 / 8
b = a * 52
c = math.floor(b)
|
a ) 51 , b ) 66 , c ) 98 , d ) 112 , e ) 154 | a | multiply(divide(204, add(3, 9)), 3) | in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 3 : 5 : 9 . if the shop carries 204 dogs and bunnies total in stock , how many dogs are there ? | let us assume the number of dogs , cats and bunnies to be 3 x , 5 x and 9 x total dogs and bunnies = 12 x . and we are given that 12 x = 204 . hence x = 17 . dogs = 3 x = 3 * 17 = 51 ( option a ) | a = 3 + 9
b = 204 / a
c = b * 3
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a ) a ) 6 , b ) b ) 10 , c ) c ) 11 , d ) d ) 15 , e ) e ) 34 | e | add(20, 5) | a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 900 . how many workers were part of the latter production run ? | "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 900 . . thus for the current period = 900 - 500 = 400 ( 5 n + 45 ) k = 400 k = 400 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 400 / 5 n + 45 500 ( 5 n + 45 ) = 400 ( 5 n + 20 ) 25 n + 225 = 20 n + 80 5 n = 145 n = 29 thus n + 5 = 34 hence e" | a = 20 + 5
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 25 | e | multiply(divide(multiply(4, 5), subtract(5, const_1)), 5) | there are 5 guys who have equal number of bullets . all of them shoot 4 bullets . the remaining bullets are equal to the total number of bullets remaining is equal to the bullets each had after division . what was the original number of bullets each one had ? | let initially each was having x bullets they shot 4 bullets so 5 x - 20 = x x = 5 so each was having 5 bullets . total was 25 . answer : e | a = 4 * 5
b = 5 - 1
c = a / b
d = c * 5
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a ) 40.5 % , b ) 48.7 % , c ) 54.5 % , d ) 55.6 % % , e ) none | c | multiply(subtract(divide(subtract(const_100, 15), 55), const_1), const_100) | the cost price of an article is 55 % of the marked price . calculate the gain percent after allowing a discount of 15 % . | sol . let marked price = rs . 100 . then , c . p . = rs . 55 . s . p = rs . 85 . â ˆ ´ gain % = [ 30 / 55 * 100 ] % = 54.5 % . answer c | a = 100 - 15
b = a / 55
c = b - 1
d = c * 100
|
a ) 1.6 , b ) 1 , c ) 0.4 , d ) 0.16 , e ) 0.016 | e | divide(volume_cylinder(4, 10), circle_area(100)) | a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 4 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ? | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 16 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 160 / 10,000 = 16 / 1000 = 0.016 ft the answer is e ." | a = volume_cylinder / (
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a ) 1 : 5 , b ) 2 : 5 , c ) 1 : 2 , d ) 3 : 2 , e ) 2 : 3 | d | divide(subtract(const_1, divide(2, 5)), divide(2, 5)) | peter and tom shared the driving on a certain trip . if peter and tom both drove for the same amount of time , but peter only drove 2 / 5 of the total distance , what was the ratio of tom ' s average speed to peter ' s average speed ? | the answer is suppose to be d . 3 : 2 . it ' s from the gmatprep option d is answer | a = 2 / 5
b = 1 - a
c = 2 / 5
d = b / c
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a ) 4857 , b ) 4184 , c ) 4012 , d ) 4691 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6735, 6927)), 7230), 6562)) | a grocery shop has a sale of rs . 6735 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6735 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6735 + 6927 + 6855 + 7230 + 6562 + x = 6 × 6500 = > 34309 + x = 39000 = > x = 39000 − 34309 = 4691 answer : d" | a = 5 + 1
b = 6500 * a
c = 6735 + 6927
d = 6855 + c
e = d + 7230
f = e + 6562
g = b - f
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a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 6 | e | divide(12, subtract(4, 2)) | a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 12 km ? | "m = 4 s = 2 us = 4 - 2 = 2 d = 12 t = 12 / 2 = 6 answer : e" | a = 4 - 2
b = 12 / a
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a ) 4 / 9 , b ) 1 / 2 , c ) 2 / 3 , d ) 1 / 3 , e ) 5 / 6 | b | divide(add(multiply(divide(2, 3), multiply(4, 3)), multiply(multiply(divide(2, 3), multiply(4, 3)), divide(1, 4))), add(multiply(4, 3), multiply(divide(2, 3), multiply(4, 3)))) | 1 / 4 of all the juniors and 2 / 3 of all the seniors are going on a trip . if there are 2 / 3 as many juniors as seniors , what fraction of the students are not going on the trip ? | j be the number of juniors and s be the number of seniors . there are 2 / 3 as many juniors as seniors 23 ∗ s = j j / s = 23 going on trip = j 4 + 23 ∗ s fraction = ( j 4 + 23 s ) / ( j + s ) = ( 2 / 3 * 1 / 4 + 2 / 3 ) / ( 2 / 3 + 1 ) = 1 / 2 fraction not on trip = 1 / 2 answer : b | a = 2 / 3
b = 4 * 3
c = a * b
d = 2 / 3
e = 4 * 3
f = d * e
g = 1 / 4
h = f * g
i = c + h
j = 4 * 3
k = 2 / 3
l = 4 * 3
m = k * l
n = j + m
o = i / n
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a ) 9680 , b ) 8790 , c ) 8760 , d ) 9020 , e ) 5000 | c | add(multiply(multiply(add(divide(4, const_100), divide(divide(subtract(8400, 4500), 2), 4500)), 4500), 2), 4500) | sonika deposited rs . 4500 which amounted to rs . 8400 after 2 years at simple interest . had the interest been 4 % more . she would get how much ? | "( 4500 * 2 * 4 ) / 100 = 360 8400 - - - - - - - - 8760 answer : c" | a = 4 / 100
b = 8400 - 4500
c = b / 2
d = c / 4500
e = a + d
f = e * 4500
g = f * 2
h = g + 4500
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a ) s . 12557 , b ) s . 14993 , c ) s . 15560 , d ) s . 12589 , e ) s . 12500 | a | subtract(multiply(add(5, const_1), 7000), add(add(add(add(4000, 6524), 5689), 7230), 6000)) | a grocer has a sale of rs . 4000 , rs . 6524 , rs . 5689 , rs . 7230 and rs . 6000 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 7000 ? | "total sale for 5 months = rs . ( 4000 + 6524 + 5689 + 7230 + 6000 ) = rs . 29443 required sale = rs . [ ( 7000 x 6 ) - 29443 ] = rs . ( 42000 - 29443 ) = rs . 12557 . option a" | a = 5 + 1
b = a * 7000
c = 4000 + 6524
d = c + 5689
e = d + 7230
f = e + 6000
g = b - f
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a ) 62.50 % , b ) 60 % , c ) 70 % , d ) 65 % , e ) 66 / 3 % | a | multiply(divide(multiply(divide(2000, const_2), add(const_1, divide(1, 4))), 2000), const_100) | at a contest with 2000 participants , 1 / 2 of the people are aged 16 to 18 . next year , the number of people aged 16 to 18 will increase by 1 / 4 . after this change , what percentage of the total 2000 people will the 16 - to 18 - year - olds represent ? | i just wanted to mention a couple of things here : * this is a pure ratio question ; the number 2000 is completely irrelevant , and you can ignore it if you like . when we increase something by 1 / 4 , we are multiplying it by 1 + 1 / 4 = 5 / 4 , so the answer here must be ( 1 / 2 ) * ( 5 / 4 ) = 5 / 8 = 62.50 % . answer : a | a = 2000 / 2
b = 1 / 4
c = 1 + b
d = a * c
e = d / 2000
f = e * 100
|
a ) 26 sec , b ) 28 sec , c ) 12 sec , d ) 77 sec , e ) 66 sec | c | divide(250, multiply(80, const_0_2778)) | two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively . in what time will they cross each other completely ? | "d = 250 m + 250 m = 500 m rs = 80 + 70 = 150 * 5 / 18 = 125 / 3 t = 500 * 3 / 125 = 12 sec answer : c" | a = 80 * const_0_2778
b = 250 / a
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a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | the closest approximation of f ( 69.28 × 0.004 ) / 0.03 is | f ( 69.28 × 0.004 ) / 0.03 1 . 0.004 = 4 × 10 ^ ( - 3 ) 2 . 0.03 = 3 × 10 ^ ( - 2 ) 3 . ( a × b ) / c = a × ( b / c ) 4 . 0.004 / 0.03 = 4 × 10 ^ ( - 3 ) / ( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 × 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ) therefore , ( 69.28 × 0.004 ) / 0.03 = 69.28 × ( 0.004 / 0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33 / 10 = 6.928 * 1.33 now , 7 × 2 = 14 7 × 1 = 7 or better : 6.9 × 1 = 6.9 6.9 × 2 = 13.8 hence , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c | a = 0 / 4
b = a * 69
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a ) 97 , b ) 99 , c ) 101 , d ) 103 , e ) 105 | e | subtract(factorial(subtract(9, 4)), add(multiply(4, const_3), const_3)) | there are 9 executives , including the ceo and cfo , that are asked to form a small team of 4 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | the total number of ways to form a team of 4 is 9 c 4 = 126 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 7 c 2 = 21 . the number of ways to form an acceptable team is 126 - 21 = 105 . the answer is e . | a = 9 - 4
b = math.factorial(a)
c = 4 * 3
d = c + 3
e = b - d
|
a ) 14.55 , b ) 11.11 , c ) 12.11 , d ) 13.11 , e ) 14.11 | a | add(11, const_1) | the average of first 11 prime numbers is ? | "sum of 11 prime no . = 160 average = 160 / 11 = 14.55 answer : a" | a = 11 + 1
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a ) 10 days , b ) 30 days , c ) 20 days , d ) 80 days , e ) 40 days | d | inverse(add(divide(9, multiply(12, 108)), divide(12, multiply(20, 108)))) | if 12 men or 20 women can do a piece of work in 108 days , then in how many days can 9 men and 12 women together do the work ? | "d 80 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 108 days . so , 27 women can do it in ( 20 * 108 ) / 27 = 80 days ." | a = 12 * 108
b = 9 / a
c = 20 * 108
d = 12 / c
e = b + d
f = 1/(e)
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a ) 103.92 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | a | subtract(multiply(divide(7500, 756), 10.5), divide(1, 4)) | the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 7500 , brokerage being 1 / 4 % , is : | "face value = rs . 7500 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 7500 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs . 100 . = market value of rs . 100 stock + brokerage per rs . 100 = rs . 104.17 = market value of rs . 100 stock = rs . 104.17 - re . 0.25 . = market value of rs . 100 stock = rs . 103.92 . answer : a" | a = 7500 / 756
b = a * 10
c = 1 / 4
d = b - c
|
a ) 3 / 4 , b ) 5 / 11 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(5), factorial(subtract(5, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 5)), factorial(subtract(add(6, 5), const_2))), factorial(const_2))) | a bag contains 6 black and 5 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 5 ) = 11 . number of white balls = 5 p ( drawing a white ball ) = 5 / 11 option b ." | a = math.factorial(5)
b = 5 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 6 + 5
o = math.factorial(n)
p = 6 + 5
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t
v = m / u
|
a ) 2 : 7 , b ) 1 : 2 , c ) 2 : 9 , d ) 2 : 2 , e ) 2 : 6 | b | divide(subtract(const_1, divide(8, 12)), divide(8, 12)) | in what ratio water be mixed with milk costing rs . 12 per liter to obtain a mixture worth of rs . 8 per litre ? | answer with explanation : by the rule of allegation , i . c . p of 1 liter of water = 0 ii . c . p of 1 liter of milk = 12 iii . mean price ( p ) = 8 iv . d – m = 12 - 8 = 4 v . m – c = 8 - 0 = 8 ratio of water and milk = 4 : 8 = 1 : 2 answer : b | a = 8 / 12
b = 1 - a
c = 8 / 12
d = b / c
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a ) 30 min , b ) 45 min , c ) 1 hr , d ) 3 min , e ) 1 min | d | divide(divide(subtract(60, 50), multiply(50, divide(2, const_60))), const_2) | a reindeer chased by a cheetah passed a certain tree at a constant speed of 50 miles per hour while being chased by a cheetah . then , 2 minutes later , the cheetah passed the same tree at a constant speed of 60 miles per hour . if both animals maintained their speeds , how long after the cheetah passed the tree did the cheetah catch up with the reindeer ? | when the cheetah is at the tree , the reindeer is 50 / 30 miles ahead on the highway . ( the distance covered in 2 min ) every hour , cheetah runs 10 miles more than the deer . how many hours will it takes it to cover 50 / 30 miles more ? the answer is ( 50 / 30 ) / 10 = 1 / 60 = 1 min answer e | a = 60 - 50
b = 2 / const_60
c = 50 * b
d = a / c
e = d / 2
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a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 100 , d ) 1 , e ) 2 | c | divide(1, 100) | if the numbers 1 to 100 are written on 100 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | "there are 25 primes , 74 composite numbers from 1 to 100 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 100 . answer : c" | a = 1 / 100
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a ) 44 , b ) 42 , c ) 40 , d ) 38 , e ) 36 | b | divide(multiply(150, add(25, 10)), subtract(150, 25)) | a fort had provision of food for 150 men for 45 days . after 10 days , 25 men left the fort . find out the number of days for which the remaining food will last | "explanation : given that fort had provision of food for 150 men for 45 days hence , after 10 days , the remaining food is sufficient for 150 men for 35 days remaining men after 10 days = 150 - 25 = 125 assume that after 10 days , the remaining food is sufficient for 125 men for x days more men , less days ( indirect proportion ) ⇒ men 150 : 125 } : : x : 35 ⇒ 150 × 35 = 125 x ⇒ 6 × 35 = 5 x ⇒ x = 6 × 7 = 42 ⇒ the remaining food is sufficient for 125 men for 42 days . answer : option b" | a = 25 + 10
b = 150 * a
c = 150 - 25
d = b / c
|
a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 3 , d ) 4 / 3 , e ) 5 / 2 | c | add(divide(2, 6), divide(2, 8)) | a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 2 days . for how many number of days should a and c should work together to complete the remaining work ? | "b work 1 / 8 * 2 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 take reciprocal 3 / 4 * remaining work = 4 / 3 * 1 / 4 = 1 / 3 answer : c" | a = 2 / 6
b = 2 / 8
c = a + b
|
a ) 1 / 8 , b ) 3 / 8 , c ) 3 / 28 , d ) 5 / 28 , e ) 1 / 56 | e | multiply(multiply(divide(1, 7), divide(1, 8)), subtract(1, divide(1, 6))) | jack , jill , and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are 1 / 6 , 1 / 7 , and 1 / 8 respectively , what is the probability that jack and sandy will make a basket but jill will miss ? | "the probability that jack and sandy will make a basket but jill will miss is 1 / 6 * 6 / 7 * 1 / 8 = 1 / 56 . the answer is e ." | a = 1 / 7
b = 1 / 8
c = a * b
d = 1 / 6
e = 1 - d
f = c * e
|
a ) 90 , b ) 100 , c ) 110 , d ) 120 , e ) 140 | b | divide(multiply(20, divide(3, 7)), subtract(divide(7, add(7, 3)), multiply(divide(3, add(7, 3)), divide(3, 7)))) | a mixture contains milk and water in the ratio 7 : 3 . on adding 20 liters of water , the ratio of milk to water becomes 7 : 5 . total quantity of milk & water before adding water to it ? | "milk : water = 7 : 3 7 x : 3 x + 20 = 7 : 5 5 [ 7 x ] = 7 [ 3 x + 20 ] 35 x = 21 x + 140 35 x - 21 x = 140 14 x = 140 x = 10 the quantity of milk in the original mixture is = 7 : 3 = 7 + 3 = 10 10 x = 100 option ' b '" | a = 3 / 7
b = 20 * a
c = 7 + 3
d = 7 / c
e = 7 + 3
f = 3 / e
g = 3 / 7
h = f * g
i = d - h
j = b / i
|
a ) 40 years , b ) 41 years , c ) 42 years , d ) 43 years , e ) 36 years | e | subtract(multiply(add(20, 2), add(14, 2)), multiply(20, 14)) | the average age of a class of 20 students is 14 years . the average increased by 2 when the teacher ' s age also included . what is the age of the teacher ? | "if age of the teacher was 14 , average would not have changed . since average increased by 2 , age of the teacher = 14 + 22 × 1 = 36 answer : e" | a = 20 + 2
b = 14 + 2
c = a * b
d = 20 * 14
e = c - d
|
a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23 | e | add(subtract(100, multiply(12, 7)), 7) | a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points w that an individual player might have scored ? | "general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points w for 12 th player is 100 - 77 = 23 . answer : e ." | a = 12 * 7
b = 100 - a
c = b + 7
|
a ) 75 . , b ) 85 . , c ) 90 . , d ) 94 . , e ) 100 . | b | divide(subtract(multiply(80, 6), multiply(divide(6, const_3), 70)), subtract(6, divide(6, const_3))) | the average length of 6 strings is 80 cm . if the average length of one third of the strings is 70 cm , what is the average of the other strings ? | 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 strings is 70 . that means out 6 / 3 = 2 strings . let the avg length of two strings 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 strings . imo correct option is b . | a = 80 * 6
b = 6 / 3
c = b * 70
d = a - c
e = 6 / 3
f = 6 - e
g = d / f
|
a ) 10 , b ) 12 , c ) 13 , d ) 11 , e ) 15 | c | add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(24), inverse(15))), 2))), inverse(15)), 4) | a can do a piece of work in 24 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ? | "2 / 24 + 4 / 15 + x / 20 = 1 x = 13 answer : c" | a = 4 - 2
b = 1/(15)
c = 1/(20)
d = b + c
e = a * d
f = 1/(20)
g = 1/(24)
h = 1/(15)
i = g + h
j = f + i
k = j * 2
l = e + k
m = 1 - l
n = 1/(15)
o = m / n
p = o + 4
|
a ) 40 , b ) 40000 , c ) 500 , d ) 5,000 , e ) 100,000 | b | multiply(const_4, power(const_10, floor(divide(9, const_2)))) | a palindrome is a number that reads the same forward and backward , such as 242 . how many even 9 - digit numbers are palindromes ? | if the palindrome must be even , it must begin and end in 2,4 6 and 8 ( 4 ways ) e . g 222212222,444414444 . the middle number in the nine - digit could be 0 - 9 ( 10 ways ) , e . g 22522 , 22822 the 2 nd and fourth numbers , the 3 rd and 5 th and 4 th and 6 th could must also be same i . e 0 - 9 ( 10 ways ) e . g 242343242 , 648565846 we thus have , 4 * 10 * 10 * 10 * 10 = 40000 b | a = 9 / 2
b = math.floor(a)
c = 10 ** b
d = 4 * c
|
a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 30 | d | divide(subtract(64, multiply(multiply(const_4, const_2), const_3)), const_2) | a cube is divided into 64 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ? | "each side of the cube has 4 x 4 = 16 cubelets . only the interior cubelets are painted one colour . on each side , 2 x 2 = 4 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 4 = 24 the answer is d ." | a = 4 * 2
b = a * 3
c = 64 - b
d = c / 2
|
a ) 10 / 216 , b ) 12 / 216 , c ) 21 / 216 , d ) 23 / 216 , e ) 25 / 216 | e | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting a sum of 12 when rolling 3 dice simultaneously ? | "sum of 12 can be achieved in following ways 6 , 5,1 - - - total cases = 3 ! = 6 6 , 4,2 - - - total cases = 3 ! = 6 6 , 3,3 - - - total cases = 3 ! / 2 ! = 3 5 , 5,2 - - - total cases = 3 ! / 2 ! = 3 5 , 4,3 - - - total cases = 3 ! = 6 4 , 4,4 - - - total cases = 3 ! / 3 ! = 1 total cases = 25 probability = 25 * ( 1 / 6 * 1 / 6 * 1 / 6 ) = 25 / 216 answer : e" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) $ 1 , b ) $ 27 , c ) $ 31 , d ) $ 41 , e ) it can not be determined | a | subtract(add(110, subtract(164.5, 110)), multiply(subtract(164.5, 110), const_3)) | at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 110 . at the same place it would cost $ 164.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 110 4 b + 10 s + c = 164.5 subtracting the first equation from the second gives us b + 3 s = 54.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 11 . in any case , there is no necessity to know each item ' s price , just the sum . answer : a | a = 164 - 5
b = 110 + a
c = 164 - 5
d = c * 3
e = b - d
|
a ) 100.5 , b ) 188.5 , c ) 190.5 , d ) 168.5 , e ) 175.5 | b | multiply(circumface(divide(30, const_2)), 2) | find the cost of fencing around a circular field of diameter 30 m at the rate of rs . 2 a meter ? | "2 * 22 / 7 * 15 = 94.25 94.25 * 2 = rs . 188.5 answer : b" | a = 30 / 2
b = circumface * (
|
a ) 32 : 23 , b ) 16 : 56 , c ) 71 : 85 , d ) 32 : 39 , e ) 41 : 52 | b | divide(divide(multiply(40, 2), subtract(7, 2)), add(40, divide(multiply(40, 2), subtract(7, 2)))) | what is the ratio whose term differ by 40 and the measure of which is 2 / 7 ? | let the ratio be x : ( x + 40 ) then , x / ( x + 40 ) = 2 / 7 x = 16 required ratio = 16 : 56 answer is b | a = 40 * 2
b = 7 - 2
c = a / b
d = 40 * 2
e = 7 - 2
f = d / e
g = 40 + f
h = c / g
|
a ) 48 , b ) 288 , c ) 12 , d ) 256 , e ) none | a | multiply(factorial(4), factorial(2)) | in how many ways 4 boys and 2 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , b g b g b b 4 boys can be seated in 4 ! ways . girl can be seated in 2 ! ways . required number of ways , = 4 ! * 2 ! = 48 answer : option a" | a = math.factorial(4)
b = math.factorial(2)
c = a * b
|
a ) 1 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | a | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(352, multiply(floor(divide(352, const_4)), const_4))), multiply(const_2, const_10)) | find the ones digit of 73 ^ 352 | "the units digit of 73 ^ 352 will be the same as the units digit of 3 ^ 352 . 3 ^ 1 = 3 - - > the units digit is 3 ; 3 ^ 2 = 9 - - > the units digit is 9 ; 3 ^ 3 = 27 - - > the units digit is 7 ; 3 ^ 4 = 81 - - > the units digit is 1 ; 3 ^ 5 = 243 - - > the units digit is 3 again ; . . . so , as you can see the units digit repeats in blocks of 4 : { 3 , 9 , 7 , 1 } , { 3 , 9 , 7 , 1 } , . . . now , since 352 = ( multiple of 4 ) + 0 , since 3 is an odd number , the units digit of 3 ^ 352 will be 1 . answer : a ." | a = 3 + 4
b = a * 10
c = 73 - b
d = 352 / 4
e = math.floor(d)
f = e * 4
g = 352 - f
h = c ** g
i = 2 * 10
j = h - i
|
a ) rs . 16500 / - , b ) rs . 17000 / - , c ) rs . 18000 / - , d ) rs . 19000 / - , e ) rs . 20000 / - | c | divide(multiply(multiply(const_100, const_100), 45), const_100) | the difference between the compound interest and simple interest on a certain sum of money at 5 % per annum for 2 years is 45 . then the original sum is ? | "for 2 years = ( 1002 d ) / r 2 = ( 1002 × 45 ) / ( 5 × 5 ) = ( 10000 × 45 ) / 25 = rs . 18000 / - c" | a = 100 * 100
b = a * 45
c = b / 100
|
a ) 448 , b ) 553 , c ) 542 , d ) 548 , e ) 560 | b | multiply(54, const_10) | the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 13 is : | "required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 13 = 540 + 13 = 553 . answer : b" | a = 54 * 10
|
a ) 7 / 12 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15 | a | subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(4))) | sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 4 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card , what portion of her limit on the platinum card will remain unspent ? | "let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 4 x = 3 / 4 x so if gold card balance is transferred then the rest unspent will be 3 / 4 x - x / 6 = 7 / 12 x so the ans is a" | a = 1/(3)
b = 1/(2)
c = a * b
d = 1/(4)
e = c + d
f = 1 - e
|
a ) 20 % , b ) 25 % , c ) 50 % , d ) 75 % , e ) 80 % | e | multiply(divide(divide(multiply(subtract(120, divide(120, const_3)), subtract(subtract(divide(120, const_3), 30), 5)), add(5, subtract(subtract(divide(120, const_3), 30), 5))), subtract(120, divide(120, const_3))), const_100) | at the end of the day , february 14 th , a florist had 120 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 30 of which were white and 5 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ? | "r + w + p = 120 s + l = 120 1 / 3 * 120 = 40 short - stemmed white = 30 short - stemmed pink = 5 = > short - stemmed red = 5 5 / p = 5 / r = > r = p so total long stemmed = 80 and long stemmed red + long stemmed pink = 80 so long stemmed red / long stemmed = ? total white = 30 ( as no long stemmed white ) = > r + r + 30 = 120 = > 2 r = 90 and r = 45 long stemmed r = 45 - 5 = 40 so long stemmed red / r = 40 / 80 = 50 % answer - e" | a = 120 / 3
b = 120 - a
c = 120 / 3
d = c - 30
e = d - 5
f = b * e
g = 120 / 3
h = g - 30
i = h - 5
j = 5 + i
k = f / j
l = 120 / 3
m = 120 - l
n = k / m
o = n * 100
|
a ) 300 m , b ) 150 m , c ) 200 m , d ) 450 m , e ) 500 m | d | multiply(divide(108, const_3_6), 15) | a car is running at a speed of 108 kmph . what distance will it cover in 15 sec ? | "speed = 108 kmph = 108 * 5 / 18 = 30 m / s distance covered in 15 sec = 30 * 15 = 450 m answer is d" | a = 108 / const_3_6
b = a * 15
|
a ) 15 sec , b ) 19 sec , c ) 12 sec , d ) 12.9 sec , e ) 11 sec | d | divide(250, multiply(80, const_0_2778)) | two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 60 kmph respectively . in what time will they cross each other completely ? | "explanation : d = 250 m + 250 m = 500 m rs = 80 + 60 = 140 * 5 / 18 = 350 / 9 t = 500 * 9 / 350 = 12.9 sec answer : option d" | a = 80 * const_0_2778
b = 250 / a
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(52, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(52, 10)), subtract(60, 40))), 60)), multiply(44, 10)), subtract(60, 44)) | at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 52 ¢ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 44 ¢ ? | if the average price is 52 , then mary has 6 oranges and 4 apples ( a ratio of 3 : 2 ) . for the average price to be 44 , the ratio should be 1 : 4 . mary must put back 5 oranges . the answer is e . | a = 60 * 10
b = 52 * 10
c = a - b
d = 60 - 40
e = c / d
f = e * 40
g = 60 * 10
h = 52 * 10
i = g - h
j = 60 - 40
k = i / j
l = 10 - k
m = l * 60
n = f + m
o = 44 * 10
p = n - o
q = 60 - 44
r = p / q
|
a ) rs . 450 , b ) rs . 320 , c ) rs . 550 , d ) rs . 420 , e ) none | d | multiply(divide(add(42, add(35, 7)), 7), 35) | there were 35 students in a hostel . due to the admission of 7 new students the expenses of the mess were increased by rs . 42 per day while the average expenditure per head diminished by re 1 . what was the original expenditure of the mess ? | "solution : let the original average expenditure be rs . x then , 42 ( x - 1 ) - 35 x = 42 . = > 7 x = 84 = > x = 12 therefore original expenditure , = rs . ( 35 × 12 ) = rs . 420 . answer : option d" | a = 35 + 7
b = 42 + a
c = b / 7
d = c * 35
|
a ) 22 , b ) 10 , c ) 99 , d ) 77 , e ) 29 | b | multiply(const_60, divide(subtract(54, 45), 54)) | excluding stoppages , the speed of a bus is 54 km / hr and including stoppages , it is 45 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 9 km less . time taken to cover 9 km = 9 / 54 * 60 = 10 min . answer : b" | a = 54 - 45
b = a / 54
c = const_60 * b
|
a ) 2 % , b ) 2.25 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | b | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 15)), 675958), multiply(divide(const_100, subtract(const_100, 15)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 15)), 675958), multiply(divide(const_100, subtract(const_100, 15)), 675958))), const_100) | a man two flats for $ 675958 each . on one he gains 15 % while on the other he loses 15 % . how much does he gain or lose in the whole transaction ? | "in such a case there is always a loss loss % = ( 15 / 10 ) ^ 2 = 9 / 4 = 2.25 % answer is b" | a = 100 + 15
b = 100 / a
c = b * 675958
d = 100 - 15
e = 100 / d
f = e * 675958
g = c + f
h = 675958 + 675958
i = g - h
j = 100 + 15
k = 100 / j
l = k * 675958
m = 100 - 15
n = 100 / m
o = n * 675958
p = l + o
q = i / p
r = q * 100
|
a ) 11 , b ) 15 , c ) 19 , d ) 38 , e ) 56 | c | multiply(power(const_2, const_4.0), factorial(990)) | the product of three consecutive numbers is 990 . then the sum of the smallest two numbers is ? | "product of three numbers = 990 990 = 9 * 10 * 11 . so , the three numbers are 9 , 10 and 11 . and sum of smallest of these two = 9 + 10 = 19 . answer : option c" | a = 2 ** 4
b = math.factorial(990)
c = a * b
|
a ) 22 , b ) 77 , c ) 29 , d ) 62 , e ) 21 | d | add(61, divide(multiply(5, 12), divide(180, 3))) | 61 + 5 * 12 / ( 180 / 3 ) = ? | "61 + 5 * 12 / ( 180 / 3 ) = 61 + 5 * 12 / ( 60 ) = 61 + ( 5 * 12 ) / 60 = 61 + 1 = 62 . answer : d" | a = 5 * 12
b = 180 / 3
c = a / b
d = 61 + c
|
a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 10 | e | divide(add(add(5, 7), 8), 7) | the area of a triangle will be when a = 5 m , b = 7 m , c = 8 m , a , b , c being lengths of respective sides ? | s = ( 5 + 7 + 8 ) / 2 = e answer : e | a = 5 + 7
b = a + 8
c = b / 7
|
a ) 10 days , b ) 50 days , c ) 100 days , d ) 150 days , e ) 200 days | d | divide(multiply(600, 25), 100) | there is enough provisions for 600 men in an army camp for 25 days . if there were 100 men less , how long will the provision last ? | "exp : we have , m 1 d 1 = m 2 d 2 600 * 25 = 100 * d 2 d 2 = 600 * 25 / 100 = 150 days . answer : d" | a = 600 * 25
b = a / 100
|
a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | b | subtract(1000, subtract(add(divide(1000, 11), divide(1000, 34)), divide(1000, multiply(11, 34)))) | what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 34 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 890 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? b" | a = 1000 / 11
b = 1000 / 34
c = a + b
d = 11 * 34
e = 1000 / d
f = c - e
g = 1000 - f
|
a ) 88 , b ) 108 , c ) 82 , d ) 90 , e ) 93 | c | subtract(subtract(subtract(182, 22), subtract(79, const_1)), const_4) | veena ranks 79 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 79 + 1 = 82 answer : c | a = 182 - 22
b = 79 - 1
c = a - b
d = c - 4
|
a ) 76 , b ) 65 , c ) 75 , d ) 89 , e ) 90 | a | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(4, subtract(multiply(80, const_4), multiply(84, 4))))) | the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 4 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ? | "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 4 = 72 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 72 + 4 = 76 a" | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 4
e = c - d
f = 4 + e
g = b - f
h = a - g
|
a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 50 % | e | multiply(divide(subtract(150, 100), 100), const_100) | a certain telescope increases the visual range at a particular location from 100 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | "original visual range = 100 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 100 ) / 100 * 100 % = 1 / 2 * 100 % = 50 % answer e" | a = 150 - 100
b = a / 100
c = b * 100
|
a ) 50 , b ) 70 , c ) 60 , d ) 20 , e ) 25 | a | divide(square_area(10), const_2) | what is the area of a square field whose diagonal of length 10 m ? | "d 2 / 2 = ( 10 * 10 ) / 2 = 50 answer : a" | a = square_area / (
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a ) 945 , b ) 745 , c ) 845 , d ) 145 , e ) 345 | c | add(add(multiply(const_100, const_4), const_100), add(multiply(15, const_4), add(const_4, const_3))) | if a wholesaler distributes items to several outlets ( a , b , c and d ) in the ratio of 1 / 10 : 1 / 11 : 1 / 15 : 1 / 16 , then find the total number of items the wholesaler distributes ? | "here , a : b : c : d = 1 / 10 : 1 / 11 : 1 / 15 : 1 / 16 1 ) l . c . m of 10 , 11 , 15,16 is 2640 2 ) find the number of books each friend received - - - - - - - - - ( to find no . of books each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 10 ) x 2640 = 264 b = ( 1 / 11 ) x 2640 = 240 c = ( 1 / 15 ) x 2640 = 176 d = ( 1 / 16 ) x 2640 = 165 3 ) total number of toys = ( 264 x + 240 x + 176 x + 165 x ) = 845 x minimum number of pens ( x ) = 1 therefore , total number of items = 845 items . correct option : c" | a = 100 * 4
b = a + 100
c = 15 * 4
d = 4 + 3
e = c + d
f = b + e
|
a ) 420 , b ) 840 , c ) 1260 , d ) 2520 , e ) 5020 | d | lcm(1, 9) | what is the lowest positive integer that is divisible by each of the integers 1 through 9 inclusive ? | "we have to find the lcm of 1 , 2 , 3 , 2 ^ 2 , 5 , 2 * 3 , 7 , 2 ^ 3 , and 3 ^ 2 . the lcm is 1 * 2 ^ 3 * 3 ^ 2 * 5 * 7 = 2520 the answer is d ." | a = math.lcm(1, 9)
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a ) 12 , b ) 15 , c ) 18 , d ) 24 , e ) 10 | b | add(subtract(36, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 36 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 36 kg joins the group , total weight of boys = 20 x + 36 so 20 x + 36 = 21 ( x + 1 ) = > x = 15 answer : b" | a = 20 + 1
b = 36 - a
c = b + 1
|
a ) 17.1 , b ) 17.3 , c ) 17.5 , d ) 17.7 , e ) 17.2 | c | divide(175, divide(14.5, 1.45)) | if 2994 ã · 14.5 = 175 , then 29.94 ã · 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 175 in the place of 2994 / 14.5 ] = 175 / 10 = 17.5 answer is c ." | a = 14 / 5
b = 175 / a
|
a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | what is the remainder when the number y = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus y = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a = 5 / 5
|
a ) 43 , b ) 36 , c ) 28 , d ) 49 , e ) 11 | d | multiply(subtract(const_1, divide(2, 50)), 50) | there are 50 students in a class . if 2 % are absent on a particular day , find the number of students present in the class . | "number of students absent on a particular day = 1 % of 50 i . e . , 2 / 100 × 50 = 1 therefore , the number of students present = 50 - 1 = 49 students . answer : d" | a = 2 / 50
b = 1 - a
c = b * 50
|
a ) two , b ) three , c ) four , d ) five , e ) six | c | multiply(const_3, const_1) | if k and w are the dimensions of a rectangle that has area 40 , and if k and w are integers such that k > w , what is the total number of possible values of k ? | "kw = 40 = 40 * 1 = 20 * 2 = 10 * 4 = 8 * 5 = - - > k can take 4 values , namely : 8 , 10,20 and 40 answer : c ." | a = 3 * 1
|
a ) 113 miles , b ) 432 miles , c ) 456 miles , d ) 887 miles , e ) 985 miles | e | subtract(1200, 215) | jim drove 215 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ? | "the number of miles to drive to finish his journey is given by 1200 - 215 = 985 miles correct answer e" | a = 1200 - 215
|
a ) 30240 , b ) 60480 , c ) 91440 , d ) 98240 , e ) 101040 | c | add(add(divide(factorial(const_10), factorial(subtract(const_10, 5))), multiply(multiply(const_10, divide(factorial(subtract(const_10, const_1)), multiply(factorial(const_3), factorial(subtract(subtract(const_10, const_1), const_3))))), divide(factorial(5), const_2))), multiply(multiply(divide(factorial(const_10), multiply(factorial(subtract(const_10, const_2)), factorial(const_2))), subtract(const_10, const_2)), divide(factorial(5), const_4))) | for a certain alarm system , each code is comprised of 5 digits , but no digit can be used more than twice . how many codes can be made ? | case 1 : all digits distinct select 5 digits out of 10 in 10 c 5 ways and arrange them in 5 ! ways which gives us 10 c 5 x 5 ! = 30240 case 2 : two digits are same , other 3 are distinct e . g . 45722 select a digit which is to be repeated in 10 c 1 ways , select other 3 digits in 9 c 3 ways and arrange them in 5 ! / 2 ! ways to get 10 c 1 x 9 c 3 x 5 ! / 2 ! = 50400 ways case 3 : two digits repeated and one other digit e . g . 33448 select 2 digits to be repeated in 10 c 2 ways , select the single digit in 8 c 1 ways and arrange them all in 5 ! / 2 ! . 2 ! ways to get 10 c 2 x 8 c 1 x 5 ! / 2 ! . 2 ! = 10800 sum of all = 30240 + 50400 + 10800 = 91440 ways ans : c | a = math.factorial(10)
b = 10 - 5
c = math.factorial(b)
d = a / c
e = 10 - 1
f = math.factorial(e)
g = math.factorial(3)
h = 10 - 1
i = h - 3
j = math.factorial(i)
k = g * j
l = f / k
m = 10 * l
n = math.factorial(5)
o = n / 2
p = m * o
q = d + p
r = math.factorial(10)
s = 10 - 2
t = math.factorial(s)
u = math.factorial(2)
v = t * u
w = r / v
x = 10 - 2
y = w * x
z = math.factorial(5)
A = z / 4
B = y * A
C = q + B
|
a ) - 2 , b ) 0 , c ) 1 / 2 , d ) 1 , e ) 2 / 3 | e | divide(2, const_3) | in the rectangular coordinate system , if the line x = 3 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ? | x = 3 y + 5 , and thus y = x / 3 - 5 / 3 the slope is 1 / 3 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) / ( m + 2 - m ) = p / 2 p / 2 = 1 / 3 and thus p = 2 / 3 the answer is e . | a = 2 / 3
|
a ) . 3 , b ) . 0003 , c ) . 003 , d ) 0.03 , e ) none of these | d | divide(30, const_1000) | what decimal fraction is 30 ml of a litre ? | "answer required fraction = 30 / 1000 = 3 / 100 = . 03 correct option : d" | a = 30 / 1000
|
a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 9 days | b | divide(const_1, add(divide(const_1, 12), divide(const_1, 10))) | a can do a piece of work in 12 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 ∗ 5 = 1 / 2 remaining work = 1 − 1 / 2 = 1 / 2 a can finish work = 12 ∗ 1 / 2 = 6 days option b" | a = 1 / 12
b = 1 / 10
c = a + b
d = 1 / c
|
a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | e | divide(add(multiply(10, 20), subtract(86, 26)), 10) | the average of 10 numbers was calculated as 20 . it is discovered later on that while calculating the average , one number , namely 86 , was incorrectly read as 26 . what is the correct average ? | "10 * 20 - 26 + 86 = 260 260 / 10 = 26 the answer is e ." | a = 10 * 20
b = 86 - 26
c = a + b
d = c / 10
|
a ) $ 120 , b ) $ 150 , c ) $ 874 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 3)), 1200) | find the compound interest on $ 1200 for 3 years at 20 % p . a . if ci is component yearly ? | "a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 3 = $ 2074 ci = $ 874 answer is c" | a = 20 / 100
b = 1 + a
c = b ** 3
d = 1200 * c
e = d - 1200
|
a ) 39 , b ) 28 , c ) 27 , d ) 69 , e ) 71 | d | divide(divide(divide(690, divide(add(12, 8), const_2)), 8), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 12 m wide at the top and 8 m wide at the bottom and the area of cross - section is 690 sq m , the depth of cannel is ? | "1 / 2 * d ( 12 + 8 ) = 690 d = 69 answer : d" | a = 12 + 8
b = a / 2
c = 690 / b
d = c / 8
e = d / 2
|
a ) 76 hrs , b ) 70 / 3 hrs , c ) 55 hrs , d ) 90 hrs , e ) 11 hrs | b | inverse(subtract(divide(const_1, 7), divide(const_1, const_10))) | a cistern which could be filled in 7 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? | 1 / 7 - 1 / x = 1 / 10 = > 70 / 3 hrs answer : b | a = 1 / 7
b = 1 / 10
c = a - b
d = 1/(c)
|
a ) 123.07 mph , b ) 100 mph , c ) 150.07 mph , d ) 200 mph , e ) 300 mph | a | divide(add(divide(800, 10), divide(800, 5)), const_2) | pooja travels from a to b a distance of 800 miles in 10 hours . he returns to a in 5 hours . find his average speed ? | speed from a to b = 800 / 8 = 100 mph speed from b to a = 800 / 5 = 160 mph average speed = 2 * 100 * 160 / 260 = 123.07 mph answer is a | a = 800 / 10
b = 800 / 5
c = a + b
d = c / 2
|
a ) 7500 , b ) 6500 , c ) 8500 , d ) 9500 , e ) none of them | c | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 68,000 at 16 2 / 3 % per year for 9 months . | "p = rs . 68000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . simple interest = ( p * r * t ) / 100 = rs . ( 68,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8500 answer is c ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 11,025 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 4 * 11025 = 44 , 100 , which is a square of 210 b" | a = 2 + 3
b = a * 2
|
a ) 300 , b ) 375 , c ) 400 , d ) 500 , e ) 750 | e | multiply(subtract(divide(400, subtract(const_1, divide(const_3, 5))), 400), add(1, divide(1, 4))) | when jessica withdrew $ 400 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 4 of the remaining balance , what will be the final balance in her bank account ? | as per the question 400 = 2 a / 5 thus - a which is the total amount = 1000 the amount thus left = 600 she then deposited 1 / 4 of 600 = 150 total amount in her account = 750 answer e | a = 3 / 5
b = 1 - a
c = 400 / b
d = c - 400
e = 1 / 4
f = 1 + e
g = d * f
|
a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 52 sec | e | multiply(divide(add(divide(290, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 290 m long ? | "explanation : speed = 45 km / hr = 45 × ( 10 / 36 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 290 = 650 meter time taken to cross the platform = 650 / ( 25 / 2 ) = 650 × 2 / 25 = 52 seconds answer : option e" | a = 290 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
|
a ) - 16 , b ) - 20 , c ) 0 , d ) 14 , e ) 16 | b | add(sqrt(81), sqrt(64)) | if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 81 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | "( x + 1 ) ^ 2 < = 81 x < = 8 x > = - 10 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 10 × - 6 = 60 minimum possible value of xy is - 10 × 8 = - 80 - 80 + 60 = - 20 answer : b" | a = math.sqrt(81)
b = math.sqrt(64)
c = a + b
|
a ) 1800 , b ) 2888 , c ) 1200 , d ) 2699 , e ) 2771 | a | add(multiply(multiply(divide(900, 10), 5), const_3), multiply(divide(900, 10), 5)) | the simple interest on a sum of money will be rs . 900 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 900 p - - - 5 - - - - - 450 3 p - - - 5 - - - - - 1350 - - - - - - = > 1800 answer : a" | a = 900 / 10
b = a * 5
c = b * 3
d = 900 / 10
e = d * 5
f = c + e
|
['a ) 0', 'b ) 1', 'c ) 2', 'd ) 3', 'e ) 4'] | d | divide(multiply(6, 3), 6) | a number when divided by 6 leaves a remainder 3 . when the square of the same number is divided by 6 , the remainder is : | let x = 6 q + 3 . then , x 2 = ( 6 q + 3 ) 2 = 36 q 2 + 36 q + 9 = 6 ( 6 q 2 + 6 q + 1 ) + 3 . so , when 2 n is divided by 4 , remainder = 3 . answer : d | a = 6 * 3
b = a / 6
|
a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(40, 25), 25) | the present ratio of students to teachers at a certain school is 40 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 40 to 1 . we can rewrite this using variable multipliers . students : teachers = 40 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers 25 / 1 = ( 40 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 40 x + 50 25 x + 125 = 40 x + 50 5 = x since x is the present number of teachers , currently there are 5 teachers . answer a ." | a = 40 + 25
b = a / 25
|
a ) - 1 , b ) 6 , c ) 7 , d ) 12 , e ) 14 | d | add(add(const_4, 6), subtract(6, const_4)) | what is the sum of all possible solutions to | x - 6 | ^ 2 + | x - 6 | = 20 ? | denote | x - 6 | as y : y ^ 2 + y = 20 - - > y = - 5 or y = 4 . discard the first solution since y = | x - 6 | , so it ' s an absolute value and thus can not be negative . y = | x - 6 | = 4 - - > x = 10 or x = - 2 . the sum = 12 . answer : d . | a = 4 + 6
b = 6 - 4
c = a + b
|
['a ) 80', 'b ) 90', 'c ) 77', 'd ) 88', 'e ) 99'] | c | divide(multiply(subtract(15, const_1), subtract(subtract(15, const_1), const_3)), const_2) | how many diagonals does a polygon with 15 sides have , if one of its vertices does not connect to any diagonal ? | if i calculate it using the formulae , # diagonals = n ( n - 3 ) / 2 each vertex sends of n - 3 diagonals n = 15 - 1 then 14 * ( 14 - 3 ) / 2 = 77 correct option : c | a = 15 - 1
b = 15 - 1
c = b - 3
d = a * c
e = d / 2
|
a ) 300 , b ) 450 , c ) 750 , d ) 800 , e ) none of them | a | multiply(divide(const_1, add(add(const_4, 3), const_1)), 600) | a and b undertake to do a piece of work for rs . 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of a | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a ’ s share = rs . ( 600 * 4 / 8 ) = rs . 300 , b ' s share = rs . ( 600 * 3 / 8 ) = rs . 225 . c ' s share = rs . [ 600 - ( 300 + 225 » ) = rs . 75 . answer is a" | a = 4 + 3
b = a + 1
c = 1 / b
d = c * 600
|
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