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 ) 105 / 2 , b ) 45 , c ) 44 , d ) 52 , e ) 50 | a | divide(subtract(multiply(60, 2), multiply(60, divide(60, const_100))), add(divide(60, const_100), 1)) | a store sells chairs and tables . if the price of 2 chairs and 1 table is 60 % of the price of 1 chair and 2 tables , and the price of 1 table and 1 chair is $ 60 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . ) | "let c = chair ; t = table 2 c + 1 t = 0.6 ( 1 c + 2 t ) or c ( 2 - 0.6 ) = 1.2 t - 1 t or 1.4 c = 0.2 t therefore c = 0.2 / 1.4 t = 1 / 7 t ic + 1 t = 60 or 1 / 7 t + 1 t = 60 therefore t = 60 * 7 / 8 = 105 / 2 a" | a = 60 * 2
b = 60 / 100
c = 60 * b
d = a - c
e = 60 / 100
f = e + 1
g = d / f
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a ) 45 % , b ) 46 % , c ) 50 % , d ) 55 % , e ) 44 % | e | subtract(60, divide(60, 10)) | you hold some gold in a vault as an investment . over the past year the price of gold increases by 60 % . in order to keep your gold in the vault , you must pay 10 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 60 % ) * ( 100 % - 10 % ) = 160 * 0.90 = 144 % an increase of 44 % your gold holdings have increased in value by 44 % . the answer is e" | a = 60 / 10
b = 60 - a
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['a ) 11 / 14', 'b ) 1 / 14', 'c ) 13 / 14', 'd ) 12 / 14', 'e ) 5 / 14'] | a | divide(circle_area(divide(6, const_2)), square_area(6)) | there is a square of side 6 cm . a circle is inscribed inside the square . find the ratio of the area of circle to square . | let the side of square = a unit so area of square = a ^ 2 diameter of inscribed circle = side of square = a radius of circle = a / 2 area of circle = ( 22 / 7 ) ( a / 2 ) ^ 2 ratio of area of circle to square = { ( 22 / 7 ) ( a / 2 ) ^ 2 } / a ^ 2 = 11 / 14 answer : a | a = 6 / 2
b = circle_area / (
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a ) 30 , b ) 35 , c ) 40 , d ) 50 , e ) 60 | d | multiply(add(4, const_1), add(9, const_1)) | if p and q are prime numbers , how many divisors does the product p ^ 4 * q ^ 9 have ? | "when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is d . 5 * 10 = 50" | a = 4 + 1
b = 9 + 1
c = a * b
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a ) 200 , b ) 997 , c ) 266 , d ) 288 , e ) 271 | a | divide(add(30, 50), divide(40, const_100)) | 30 is subtracted from 40 % of a number , the result is 50 . find the number ? | "( 40 / 100 ) * x β 30 = 50 4 x = 800 x = 200 answer : a" | a = 30 + 50
b = 40 / 100
c = a / b
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a ) 2.6 days , b ) 3.6 days , c ) 5.6 days , d ) 6 days , e ) 2 days | a | divide(multiply(multiply(2, 4), 65), multiply(40, 5)) | calculate how many days it will take for 5 boys to paint a 65 m long wall if 2 boys can paint a 40 m long wall in 4 days , | "the length of wall painted by one boy in one day = 40 / 2 * 1 / 4 = 5 m no . of days required to paint 50 m cloth by 8 boys = 65 / 5 * 1 / 5 = 2.6 days . a" | a = 2 * 4
b = a * 65
c = 40 * 5
d = b / c
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a ) 10 , b ) 12 , c ) 14 , d ) 17 , e ) 18 | a | divide(subtract(29, power(3, 2)), 2) | if a - b = 3 and a 2 + b 2 = 29 , find the value of ab ? | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 29 - 9 = 20 ab = 10 a )" | a = 3 ** 2
b = 29 - a
c = b / 2
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a ) $ 200 , b ) $ 220 , c ) $ 230 , d ) $ 240 , e ) $ 250 | a | subtract(260, multiply(divide(subtract(380, 260), 4), 2)) | mary invested a certain sum of money in a bank that paid simple interest . the amount grew to $ 260 at the end of 2 years . she waited for another 4 years and got a final amount of $ 380 . what was the principal amount that she invested at the beginning ? | "what shall be the rate of interest . ? does that is not required for the calculation ? not really ! keep in mind that the interest earned each year will be the same in simple interest . at the end of 2 years , amount = $ 260 at the end of 6 years , amount = $ 380 this means she earned an interest of $ 120 in 4 years . or $ 30 in each year . we know that the interest earned each year will be the same . therefore she must have earned $ 60 in 2 years . hence principal amount = $ 260 - $ 60 = $ 200 option a" | a = 380 - 260
b = a / 4
c = b * 2
d = 260 - c
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a ) 4 minutes , b ) 2 / 7 minutes , c ) 32 / 7 minutes , d ) 224 / 7 minutes , e ) 28 minutes | c | inverse(add(divide(const_1, 32), add(divide(const_1, 32), divide(5, 32)))) | pipe a can fill a tank in 32 minutes . pipe b can fill the same tank 5 times faster than pipe a . if both the pipes are connected to the tank so that they fill the tank simultaneously , how long will it take for the empty tank to overflow ? | "pipe a can fill a tank in 32 minutes - - > the rate of a = 1 / 32 tank / minute ; pipe b can fill the same tank 5 times fasterthan pipe a - - > the rate of b = 1 / 32 + 5 / 32 = 6 / 32 tank / minute . their combined rate = 1 / 32 + 6 / 32 = 7 / 32 tank / minute ; therefore , a and b can fill the tank in 32 / 7 minutes . answer : c" | a = 1 / 32
b = 1 / 32
c = 5 / 32
d = b + c
e = a + d
f = 1/(e)
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a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 3 miles , e ) 6 1 / 4 miles | d | divide(divide(multiply(subtract(4.90, 1.85), const_100), const_3), const_4) | a taxi company charges $ 1.85 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ? | "if we start out with $ 4.90 and have to spend $ 1.85 for the first quarter - mile , we will have $ 3.05 left to spend on quarter - mile intervals . since $ 3.05 / $ 0.15 = 20.33 , we can buy 20.33 more quarter - miles , and will travel 21.33 quarter miles in all : 21.33 Γ 1 / 4 = 5 1 / 3 miles . the correct answer is choice ( d ) ." | a = 4 - 90
b = a * 100
c = b / 3
d = c / 4
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a ) 10 kmph , b ) 11 kmph , c ) 88 kmph , d ) 18 kmph , e ) 12 kmph | a | multiply(divide(9, const_60), 64) | the speed of a train is 64 kmph . what is the distance covered by it in 9 minutes ? | "64 * 9 / 60 = 9.6 kmph answer : a" | a = 9 / const_60
b = a * 64
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a ) 1012 , b ) 1011 , c ) 1212 , d ) 1122 , e ) 1021 | d | add(subtract(add(269, subtract(add(269, 123), 50)), 123), subtract(add(subtract(add(269, 123), 50), subtract(add(196, 269), 50)), 123)) | if it is a 3 x 3 matrix , for which elements in first row are ( v , 50 , w ) , second row ( 196 , x , y ) and in third row are ( 269 , z , 123 ) . it is given that the sum of the numbers in each row , column , diagonal are same . then find the value of ( y + z ) . | v 50 w 196 x y 269 z 123 in 1 st row and 1 st column v + 196 + 269 = v + 50 + w w = 415 again , in 2 nd column and 3 rd row 50 + x + z = 269 + z + 123 x = 342 as sum of each row column and diagonal are same . so , 269 + 342 + 415 = 1026 415 + y + 123 = 1026 y = 488 269 + z + 123 = 1026 z = 634 y + z = 488 + 634 = 1122 answer : d | a = 269 + 123
b = a - 50
c = 269 + b
d = c - 123
e = 269 + 123
f = e - 50
g = 196 + 269
h = g - 50
i = f + h
j = i - 123
k = d + j
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a ) 540 , b ) 400 , c ) 700 , d ) 650 , e ) 840 | c | divide(multiply(455, const_100), subtract(const_100, 35)) | in an examination 35 % of the students passed and 455 failed . how many students appeared for the examination ? | "let the number of students appeared be x then , 65 % of x = 455 65 x / 100 = 455 x = 455 * 100 / 65 = 700 answer is c" | a = 455 * 100
b = 100 - 35
c = a / b
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a ) 20 coins , b ) 30 coins , c ) 40 coins , d ) 80 coins , e ) none of these | d | divide(140, add(add(inverse(const_4), inverse(const_2)), const_1)) | a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 140 , how many coins of each type are there ? | let number of each type of coin = x . then , 1 Γ x + . 50 Γ x + . 25 x = 140 β 1.75 x = 140 β x = 80 coins answer d | a = 1/(4)
b = 1/(2)
c = a + b
d = c + 1
e = 140 / d
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a ) 20 , b ) 12 , c ) 13 , d ) 14 , e ) 16 | a | divide(multiply(80, subtract(10, 8)), 8) | a 80 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ? | "amount of salt = 8.0 assume x gallons of water are added . 8.0 / 80 + x = 8 / 100 800 = 8 x + 640 8 x = 160 x = 20 correct option : a" | a = 10 - 8
b = 80 * a
c = b / 8
|
a ) 0 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(5, subtract(5, 5)) | what least value should be replaced by * in 223431 * so the number become divisible by 5 | explanation : trick : number is divisible by 5 , if last digit is 0 or 5 so that least number is 0 . answer : option a | a = 5 - 5
b = 5 - a
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a ) 187.89 cm , b ) 182.29 cm , c ) 123.98 cm , d ) 149.98 cm , e ) 146.89 cm | b | floor(divide(add(subtract(multiply(35, 184), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 184 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ? | "calculated average height of 35 boys = 184 cm . wrong total height of 35 boys = 184 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 183 cm - ( 166 cm + 106 cm ) / 35 = 184 cm - 60 / 35 cm = 184 cm - 1.71 cm = 182.29 cm . answer : b" | a = 35 * 184
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
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a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 5,000 | b | divide(multiply(2.4, power(10, 11)), multiply(240, power(10, add(const_4, const_2)))) | a certain country had a total annual expenditure of $ 2.4 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ? | "total expenditure / population = per capita expenditure hence , ( 2,4 x 10 ^ 11 ) / 240 000 000 = ( 2,4 x 10 ^ 11 ) / ( 2,4 x 10 ^ 8 ) = 1 x 10 ^ ( 11 - 8 ) = 1 x 10 ^ 3 = 1000 . answer is b ." | a = 10 ** 11
b = 2 * 4
c = 4 + 2
d = 10 ** c
e = 240 * d
f = b / e
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a ) 288 , b ) 744 , c ) 788 , d ) 868 , e ) 177 | d | divide(multiply(divide(336, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 336 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 336 62 % - - - - - - - - ? = > 868 answer : d" | a = 100 - 62
b = 62 - a
c = b / 100
d = 336 / c
e = d * 62
f = e / 100
|
a ) 120 metres , b ) 180 metres , c ) 324 metres , d ) 50 metres , e ) 180 metres | d | multiply(divide(multiply(60, const_1000), const_3600), 3) | a train running at the speed of 60 km / hr crosses a pole in 3 seconds . what is the length of the train ? | "speed = ( 60 x 5 / 18 ) / m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 x 3 / 3 m = 50 m . answer : option d" | a = 60 * 1000
b = a / 3600
c = b * 3
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a ) a ) 3 , b ) b ) 1 , c ) c ) 5 , d ) d ) 21 , e ) 4 | e | add(5, 5) | two positive integers differ by 4 , and sum of their reciprocals is 5 then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 5 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 6 or ( n ^ 2 + 4 n ) * 5 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 4 . hence , e" | a = 5 + 5
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a ) $ 12 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 0 | a | subtract(90, add(add(multiply(90, divide(1, 5)), multiply(90, divide(1, 6))), multiply(90, divide(1, 2)))) | jennifer had $ 90 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | "1 / 5 x $ 90 = $ 15 for sandwich 1 / 6 x $ 90 = $ 45 for museum 1 / 2 x $ 90 = $ 78 for book $ 15 + $ 45 + $ 18 = $ 78 spent $ 90 - $ 78 = $ 12 left over correct answer a" | a = 1 / 5
b = 90 * a
c = 1 / 6
d = 90 * c
e = b + d
f = 1 / 2
g = 90 * f
h = e + g
i = 90 - h
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(add(11, 5), const_2) | in one hour , a boat goes 11 km along the stream and 5 km against it . find the speed of the boat in still water | "explanation : we know we can calculate it by 1 / 2 ( a + b ) = > 1 / 2 ( 11 + 5 ) = 1 / 2 ( 16 ) = 8 km / hr answer is c" | a = 11 + 5
b = a / 2
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a ) 52.66 , b ) 53.66 , c ) 54.66 , d ) 58.66 , e ) 59.66 | b | divide(add(145, 660), multiply(54, const_0_2778)) | how long does a train 145 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 145 ) / 54 * 18 / 5 t = 53.66 answer : b" | a = 145 + 660
b = 54 * const_0_2778
c = a / b
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a ) 10 , b ) 12 , c ) 13 , d ) 15 , e ) 18 | e | divide(36, const_2) | in a group of ducks and cows , the total number of legs are 36 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 36 = > 2 x = 36 = > x = 18 e" | a = 36 / 2
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a ) 16 % , b ) 20 % , c ) 31 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(60, 48), 60), const_100) | in town x , 60 percent of the population are employed , and 48 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 60 % , out of which 48 are employed males , hence 12 % are employed females . ( employed females ) / ( total employed people ) = 12 / 60 = 1 / 5 = 20 % answer : b ." | a = 60 - 48
b = a / 60
c = b * 100
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a ) 1254 , b ) 1376 , c ) 1456 , d ) 1599 , e ) 1635 | d | multiply(divide(subtract(1335, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1335 . 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 + 1335 ) . x + 1335 = 6 x + 15 5 x = 1320 x = 264 large number = 264 + 1335 = 1599 d" | a = 1335 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) 1 / 4 , b ) 2 / 3 , c ) 4 / 7 , d ) 1 / 2 , e ) 3 / 7 | a | divide(subtract(subtract(const_100, 20), multiply(const_3, const_10)), multiply(const_3, const_10)) | in a sample of college students , 20 percent are third year students and 80 percent are not second - year students . what fraction of those students who are not third - year students are second - year students ? | "= 20 / 80 . = 1 / 4 my answer is a too" | a = 100 - 20
b = 3 * 10
c = a - b
d = 3 * 10
e = c / d
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a ) β 29 , b ) β 12 , c ) 12 , d ) 20 , e ) 168 | d | divide(20, const_1) | what is the product of all the possible values of x if x ^ 2 - 9 x + 20 ? | "explanation : = > y = x ^ 2 - 9 x + 20 = > y = ( x - 4 ) ( x - 5 ) = > x = 4 , y = 5 product y = ( 4 ) ( 5 ) = 20 answer option 20 answer : d" | a = 20 / 1
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a ) 14 , b ) 12 , c ) 16 , d ) 8 , e ) 6 | a | divide(28, const_2) | in a group of ducks and cows , the total number of legs are 28 more than twice the number of heads . find the total number of cows . | "explanation : let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 28 more than twice the number of heads = > 2 ( d + 2 c ) = 28 + 2 ( c + d ) = > d + 2 c = 14 + c + d = > 2 c = 14 + c = > c = 14 i . e . , total number of cows = 14 answer : a" | a = 28 / 2
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a ) 9 : 8 , b ) 9 : 9 , c ) 9 : 6 , d ) 9 : 1 , e ) 9 : 2 | a | divide(multiply(3, 3), 8) | a dog takes 3 leaps for every 8 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 8 leaps of hare = 9 : 5 . answer : a ) 9 : 8" | a = 3 * 3
b = a / 8
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a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 9 | c | subtract(subtract(subtract(50, 25), const_4), const_2) | a certain no . when divided by 50 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 50 + 25 = 75 / 15 = 5 ( remainder ) c" | a = 50 - 25
b = a - 4
c = b - 2
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(multiply(divide(76, add(divide(90, 100), const_1)), const_2), 76) | a retailer bought a coat at wholesale and marked it up 90 % to its initial price of $ 76 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | let x be the wholesale price . then 1.9 x = 76 and x = 76 / 1.9 = 40 . to achieve a 100 % markup , the price needs to be $ 80 . the retailer needs to increase the price by $ 4 more . the answer is b . | a = 90 / 100
b = a + 1
c = 76 / b
d = c * 2
e = d - 76
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a ) 42 , b ) 45 , c ) 34 , d ) 76 , e ) 25 | a | subtract(add(57, const_3), multiply(multiply(const_3, const_3), const_2)) | if bat = 58 hat = 52 cat = 57 then whats rat = ? | a 42 | a = 57 + 3
b = 3 * 3
c = b * 2
d = a - c
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a ) 24500 , b ) 21600 , c ) 36600 , d ) 22400 , e ) 12500 | b | divide(add(700, 200), 4) | ( 700 + 200 ) Γ£ β 4 Γ£ β ( 4 + 2 ) = ? | "( 700 + 200 ) Γ£ β 4 Γ£ β ( 4 + 2 ) = ? or , ? = 900 Γ£ β 4 Γ£ β 6 = 21600 answer b" | a = 700 + 200
b = a / 4
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a ) 10 , b ) 20 , c ) 40 , d ) 60 , e ) 120 | c | divide(multiply(5, 16), const_2) | there are 5 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 16 other amateurs , what is the total number of chess games possible to be played in the tournament ? | "each person is one participant of 4 games . so there are in all 16 * 5 = 80 instances of one participant games . but each game has 2 participants so total number of games = 80 / 2 = 40 c" | a = 5 * 16
b = a / 2
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a ) 12 , b ) 10 , c ) 16 , d ) 19 , e ) 18 | a | gcd(36, 84) | find the highest common factor of 36 and 84 . | "36 = 2 ^ 2 x 3 ^ 2 84 = 2 ^ 2 x 3 x 7 h . c . f . = 2 ^ 2 x 3 = 12 . answer : option a" | a = math.gcd(36, 84)
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a ) 1 / 5 , b ) 2 / 7 , c ) 2 / 9 , d ) 3 / 5 , e ) 4 / 11 | b | divide(choose(subtract(7, 1), 1), choose(7, 2)) | a basket contains 7 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "the total number of ways to choose 2 apples is 7 c 2 = 21 the number of ways that include the spoiled apple is 6 c 1 = 6 p ( the spoiled apple is included ) = 6 / 21 = 2 / 7 the answer is b ." | a = 7 - 1
b = math.comb(a, 1)
c = math.comb(7, 2)
d = b / c
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a ) 15 % , b ) 16 % , c ) 17 % , d ) 78 % , e ) 28 % | a | multiply(divide(subtract(1500, 1275), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1275 , find the loss % ? | "1500 - - - - 225 100 - - - - ? = > 15 % answer : a" | a = 1500 - 1275
b = a / 1500
c = b * 100
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a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) none of these | d | divide(54, add(22, 5)) | a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 54 km downstream | "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 54 km time taken = distance / speed = 54 / 27 = 2 hours . answer : option d" | a = 22 + 5
b = 54 / a
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a ) 4 year , b ) 18 year , c ) 10 year , d ) none of these , e ) 9 year | b | divide(subtract(divide(60, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | the sum of the ages of 5 children born at the intervals of 3 year each is 60 year . what is the age of the eldest child ? | "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 60 Γ’ β‘ β 5 x = 30 Γ’ β‘ β x = 6 . Γ’ Λ Β΄ age of the eldest child = x + 12 = 6 + 12 = 18 years . answer b" | a = 5 / 3
b = 60 / a
c = 5 - 1
d = c * 3
e = b - d
f = e / 3
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a ) 22 , b ) 37 , c ) 99 , d ) 38 , e ) 27 | b | add(divide(add(21, 23), const_2), multiply(const_1, 15)) | the average age of 15 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 15 * 2 ) years = 30 years . sum of ages of two new men = ( 21 + 23 + 30 ) years = 74 years average age of two new men = ( 74 / 2 ) years = 37 years . answer : b" | a = 21 + 23
b = a / 2
c = 1 * 15
d = b + c
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a ) 12.9 sec , b ) 9.68 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.9 sec | b | divide(add(110, 132), multiply(90, const_0_2778)) | how long does a train 110 m long running at the speed of 90 km / hr takes to cross a bridge 132 m length ? | "speed = 90 * 5 / 18 = 25 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 25 = 9.68 sec . answer : b" | a = 110 + 132
b = 90 * const_0_2778
c = a / b
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 75 % | d | add(divide(multiply(200, subtract(20, 5)), 100), divide(multiply(200, subtract(20, 5)), 100)) | 20 percent of the programmers in a startup company weigh 200 pounds or more . 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less . what percent of the programmers in the startup company weigh between 100 and 200 pounds ? | initially 80 % and 20 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 % let total be ' 100 ' then that 75 % is ( 3 / 4 ) β 80 so , the required % is [ ( 3 / 4 ) β 80 / 100 ] β 100 = 60 % answer : d | a = 20 - 5
b = 200 * a
c = b / 100
d = 20 - 5
e = 200 * d
f = e / 100
g = c + f
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | divide(12, subtract(3, const_1)) | lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 12 more photos than claire . how many photos has claire taken ? | "explanation : l = r l = 3 c r = c + 12 we can substitute r for l in the second equation : r = 3 c . if r is equal to both 3 c and c + 12 , we can say 3 c = c + 12 , and solve for c . 3 c = c + 12 2 c = 12 c = 6 answer : ( a )" | a = 3 - 1
b = 12 / a
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a ) 15 , b ) 21 , c ) 17 , d ) 37 , e ) 57 | e | divide(subtract(multiply(500, 72), 10350), subtract(500, 50)) | a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 72 . find the number of notes of rs . 50 denomination . | "total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 72 - x now , 50 * x + 500 * ( 72 - x ) = 10350 50 x + 36000 - 500 x = 10350 - 450 x = - 25650 x = 57 . no . of 50 rupees note = 57 . answer : option e" | a = 500 * 72
b = a - 10350
c = 500 - 50
d = b / c
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a ) 25 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | e | multiply(divide(10.5, 17.5), const_100) | mike earns $ 17.5 per hour and phil earns $ 10.5 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "what % less of 17.5 is 10.5 let it be x % less , then = 17.5 ( 1 - x / 100 ) = 10.5 1 - x / 100 = 10.5 / 17.5 x = 200 / 5 x = 40 % ans e" | a = 10 / 5
b = a * 100
|
a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | e | multiply(divide(subtract(const_1, add(divide(const_1, 3), divide(const_1, 5))), divide(const_1, 5)), const_60) | it takes avery 3 hours to build a brick wall while tom can do it in 5 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery takes 3 hours tom takes 2 hours efficiency of avery is 1 / 3 units / hr efficiency of tom is 1 / 5 units / hr combined efficiency of tom and avery is 1 / 3 + 1 / 5 = 8 / 15 units / hr since they worked for 1 hour they completed 8 / 15 units of work and 7 / 15 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 7 / 15 / 1 / 5 hours = > 7 / 3 * 60 = 140 minutes . . . answer will be ( e )" | a = 1 / 3
b = 1 / 5
c = a + b
d = 1 - c
e = 1 / 5
f = d / e
g = f * const_60
|
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | c | multiply(65, divide(18, 45)) | a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 65 meters , what is the height of the building ( in meters ) ? | "the height : length ratio will be equal in both cases . 18 / 45 = x / 65 x = 26 the answer is c ." | a = 18 / 45
b = 65 * a
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a ) 1360 , b ) 637.5 , c ) 1600 , d ) 1800 , e ) none | b | divide(multiply(85, 75), subtract(85, 75)) | the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 85 and rs . 75 respectively . the sum is : | "sol . sum = s . i . * t . d . / ( s . i ) - ( t . d . ) = 85 * 75 / ( 85 - 75 ) = rs . 637.5 . answer b" | a = 85 * 75
b = 85 - 75
c = a / b
|
a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,700 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,700 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.7 so , answer is d . | a = 100 * 100
b = a * 4
|
a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024 | c | power(5, 4) | a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many w ways can the test be completed if every question is unanswered ? | "5 choices for each of the 4 questions , thus total w of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c ." | a = 5 ** 4
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a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 75 | e | divide(multiply(36, 25), 12) | if 36 men can do a piece of work in 25 hours , in how many hours will 12 men do it ? | "explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 12 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 12 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 12 } = 75 hence , 12 men can do it in 75 hours . answer : e ) 75" | a = 36 * 25
b = a / 12
|
a ) 20 , b ) 25 , c ) 10 , d ) 30 , e ) 35 | c | divide(subtract(80, 10), 6) | in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 10 and the other one , that is cheaper , costs $ 6 . the factory holds a $ 80 worth stock of 12 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 6 modules must be 10 so that the leftover 2 modules are of $ 10 which will give a total value $ 80 . 10 * 6 + 2 * 10 = 60 + 20 = 80 answer : c" | a = 80 - 10
b = a / 6
|
a ) 1 / 3125 , b ) 1 / 3000 , c ) 1 / 658 , d ) 1 / 500 , e ) 1 / 75 | a | divide(multiply(add(add(const_100, const_60), const_1), 5), const_100) | what is the value of ( 5 ) ^ - 5 ? | "5 ^ - 5 = 1 / ( 5 ) ^ 5 = 1 / 3125 answer : a" | a = 100 + const_60
b = a + 1
c = b * 5
d = c / 100
|
a ) s . 14 , b ) s . 18 , c ) s . 12 , d ) s . 11 , e ) s . 10 | a | divide(add(600, 240), add(40, 20)) | rahim bought 40 books for rs . 600 from one shop and 20 books for rs . 240 from another . what is the average price he paid per book ? | "average price per book = ( 600 + 240 ) / ( 40 + 20 ) = 840 / 60 = rs . 14 answer : a" | a = 600 + 240
b = 40 + 20
c = a / b
|
a ) rs . 7000 , b ) rs . 8200 , c ) rs . 8500 , d ) rs . 9000 , e ) none | b | divide(add(add(add(add(9000, 5000), 11000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 9000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 9000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8200 correct option : b" | a = 9000 + 5000
b = a + 11000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
|
a ) 83 % , b ) 28 % , c ) 20 % , d ) 17 % , e ) 12 % | b | subtract(40, const_1) | at a local appliance manufacturing facility , the workers received a 40 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | "let ' s say he works usually 40 hours and earns 100 per hour . 10 * 100 = 1000 10 * 140 = 1400 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 140 = 7.14 so he can reduce his work by 2.86 hours . which is > 28 % . answer b" | a = 40 - 1
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a ) 500 , b ) 2000 , c ) 5000 , d ) 6000 , e ) 7000 | d | subtract(multiply(add(5, const_1), 7500), add(add(add(add(7435, 7920), 7855), 8230), 7560)) | a grocer has a sale of rs . 7435 , rs . 7920 , rs . 7855 , rs . 8230 and rs . 7560 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 7500 ? | "total sale for 5 months = rs . ( 7435 + 7920 + 7855 + 8230 + 7560 ) = rs . 39000 . required sale = rs . [ ( 7500 x 6 ) - 39000 ] = rs . ( 45000 - 39000 ) = rs . 6000 answer : option d" | a = 5 + 1
b = a * 7500
c = 7435 + 7920
d = c + 7855
e = d + 8230
f = e + 7560
g = b - f
|
a ) 75 , b ) 95 , c ) 93 , d ) 165 , e ) 11 | c | divide(add(add(add(add(96, 95), 82), 97), 95), add(const_2, const_3)) | david obtained 96 , 95 , 82 , 97 and 95 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | "explanation : average = ( 96 + 95 + 82 + 97 + 95 ) / 5 = 465 / 5 = 93 . answer : c" | a = 96 + 95
b = a + 82
c = b + 97
d = c + 95
e = 2 + 3
f = d / e
|
a ) β 100 , b ) 0.2 , c ) 0.01 , d ) 4 , e ) 8 | c | divide(divide(2, 2), const_1000) | if x = 5 and y = β 2 , what is the value of ( x β 2 y ) ^ y ? | "quickly we can spot that answer is neither integer nor negative . eliminate a , de 6 + 4 = 10 by inversing and squaring 0.01 answer : c" | a = 2 / 2
b = a / 1000
|
a ) 2177 , b ) 2876 , c ) 4500 , d ) 2981 , e ) 6100 | e | subtract(multiply(8000, const_4), subtract(multiply(8100, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8100 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8100 = 32400 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 400 salary of may is rs . 6500 salary of january = rs . 6100 . answer : e" | a = 8000 * 4
b = 8100 * 4
c = b - 6500
d = a - c
|
a ) rs . 80 , b ) rs . 115.20 , c ) rs . 120 , d ) rs . 125.40 , e ) none of these | b | multiply(divide(96, 10), 12) | a invested some money in 10 % stock at 96 . if b wants to invest in an equally good 12 % stock , he must purchase a stock worth of : | explanation : for an income of rs . 10 , investment = rs . 96 . for an income of rs . 12 , investment = ( 96 / 10 x 12 ) = 115.20 answer : b | a = 96 / 10
b = a * 12
|
a ) a ) 1 , b ) b ) 2 , c ) c ) 7 , d ) d ) 18 / 5 , e ) e ) 4 | c | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 7) | if 2 x + y = 7 and x + 2 y = 5 , then 7 xy / 3 = ? | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 7 * 3 * 1 ) / 3 = 21 / 3 = 7 c" | a = 7 * 2
b = a - 5
c = 2 * 2
d = c - 1
e = b / d
f = 7 * 2
g = f - 5
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 7 - k
m = e + l
n = m / 7
|
a ) 1000 , b ) 2876 , c ) 1977 , d ) 1167 , e ) 2711 | d | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 8))) | a watch was sold at a loss of 8 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "92 % 104 % - - - - - - - - 12 % - - - - 140 100 % - - - - ? = > rs . 1167 answer : d" | a = 140 * 100
b = 100 + 4
c = 100 - 8
d = b - c
e = a / d
|
a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | b | multiply(4400, divide(const_2, add(add(multiply(const_2, const_3), multiply(divide(const_2, const_3), const_3)), const_3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 4400 . what is the share of b ? | let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 4400 x ( 2 / 11 ) ) = rs . 800 . answer is b | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 4400 * f
|
a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | j is 25 % less than p and 20 % less than t . t is e % less than p . what is the value of e ? | "usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( e / 100 ) ] * p i . e . 100 - e = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . e = 6.25 answer : option d" | a = 25 * 25
b = a / 100
|
a ) 25 m , b ) 45 m , c ) 32 m , d ) 50 m , e ) 60 m | e | multiply(5, 12) | walking at 5 / 6 th of its usual speed a cab is 12 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 12 m ut / 5 = 12 m ut = 60 m answer is e" | a = 5 * 12
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a ) 4 , b ) 6 , c ) 5 , d ) 15 , e ) 20 | c | subtract(divide(factorial(6), multiply(factorial(const_3), factorial(const_3))), divide(factorial(6), multiply(factorial(const_4), factorial(const_2)))) | 6 points lie on the circumference of a ellipse . what is the positive difference between the number of triangles and the number of quadrilaterals that can be created by connecting these points ? | number of triangles can be formed out of 6 points = 6 c 3 = 20 number of quadrilaterals can be formed out of 6 points = 6 c 4 = 15 20 - 15 = 5 . answer = c | a = math.factorial(6)
b = math.factorial(3)
c = math.factorial(3)
d = b * c
e = a / d
f = math.factorial(6)
g = math.factorial(4)
h = math.factorial(2)
i = g * h
j = f / i
k = e - j
|
a ) 2 / 3 , b ) 3 / 5 , c ) 4 / 7 , d ) 5 / 9 , e ) none of these | d | divide(subtract(multiply(3, const_2), const_1), subtract(multiply(5, const_2), const_1)) | the denominator of a fraction is 1 less than twice the numerator . if the numerator and denominator are both increased by 1 , the fraction becomes 3 / 5 . find the fraction ? | let the numerator and denominator of the fraction be ' n ' and ' d ' respectively . d = 2 n - 1 ( n + 1 ) / ( d + 1 ) = 3 / 5 5 n + 5 = 3 d + 3 5 n + 5 = 3 ( 2 n - 1 ) + 3 = > n = 5 d = 2 n - 1 = > d = 9 hence the fraction is : 5 / 9 answer : d | a = 3 * 2
b = a - 1
c = 5 * 2
d = c - 1
e = b / d
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) 2 / 3 | b | divide(const_2, divide(divide(factorial(const_4), 2), 2)) | set s consists of integers { 13 , 57 } . if two numbers are chosen from set s at random , what is the probability that the product of 2 numbers is more than 15 ? | number of ways you can choose 2 from 4 = 4 c 2 = 6 e = event of getting the 2 numbers product is more than 15 = ( 3,7 ) , ( 5,7 ) = 2 probability = 2 / 6 = 1 / 3 answer is b | a = math.factorial(4)
b = a / 2
c = b / 2
d = 2 / c
|
a ) 50 % , b ) 55 % , c ) 60 % , d ) 66 + ( 2 / 3 ) % , e ) 80 % | d | multiply(subtract(divide(divide(divide(add(const_100, 20), const_100), subtract(const_1, divide(subtract(10, 9), 10))), divide(subtract(const_100, 20), const_100)), const_1), const_100) | a dealer offers a cash discount of 20 % and still makes a profit of 20 % when he further allows 10 articles to be sold at the cost price of 9 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | "given cash discount - 20 % profit - 20 % items sold - 10 price sold at = list price of 9 assume list price = $ 10 total invoice = $ 90 - 20 % cash discount = $ 72 let cost price of 10 items be x so total cost = 10 * x given the shopkeeper had a profit of 20 % 10 * x * 120 / 100 = 72 or x = $ 6 which means his products were listed at $ 10 which is a 66 + ( 2 / 3 ) % markup over $ 6 answer d" | a = 100 + 20
b = a / 100
c = 10 - 9
d = c / 10
e = 1 - d
f = b / e
g = 100 - 20
h = g / 100
i = f / h
j = i - 1
k = j * 100
|
a ) 8 , b ) 64 , c ) 48 , d ) 102 , e ) 144 | c | subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 40), const_100), 48) | in a certain warehouse , 40 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ? | "if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 40 % of the packages weigh less than 75 pounds so , 40 % oft = number of packages that weigh less than 75 pounds 40 % of 240 = 96 , so 96 packages weigh less than 75 pounds of those 96 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 96 - 48 = 48 = c" | a = 100 - 80
b = 48 / a
c = b * 100
d = c * 40
e = d / 100
f = e - 48
|
a ) rs . 190 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | a | divide(95, 0.5) | if 0.5 % of a = 95 paise , then the value of a is ? | "answer β΅ 0.5 / 100 of a = 95 / 100 β΄ a = rs . ( 95 / 0.5 ) = rs . 190 correct option : a" | a = 95 / 0
|
a ) 12000 , b ) 4500 , c ) 5000 , d ) 8000 , e ) 9000 | a | divide(60000, 5) | a company produces 60000 bottles of water everyday . if a case can hold 5 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 5 number of cases required to hold 60000 bottles = 60000 / 5 = 12000 cases . so the answer is a = 12000" | a = 60000 / 5
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a ) e = 200 , b ) e = 600 , c ) e = 800 , d ) e = 1600 , e ) 50 | d | multiply(power(const_2, 4), const_100) | cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost e is what percent of the original cost ? | "original cost e , c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600" | a = 2 ** 4
b = a * 100
|
a ) 96 , b ) 100 , c ) 104 , d ) 108 , e ) 112 | d | divide(42, subtract(1, add(add(add(divide(1, 12), divide(1, 4)), divide(1, 9)), divide(1, 6)))) | if 1 / 12 of the passengers on a ship are from north america , 1 / 4 are europeans , 1 / 9 are from africa , 1 / 6 are from asia and the remaining 42 people are citizens of other continents , then how many passengers are on board the ship ? | "1 / 12 + 1 / 4 + 1 / 9 + 1 / 6 = ( 3 + 9 + 4 + 6 ) / 36 = 11 / 18 let x be the number of passengers on the ship . 42 = ( 7 / 18 ) x x = 108 the answer is d ." | a = 1 / 12
b = 1 / 4
c = a + b
d = 1 / 9
e = c + d
f = 1 / 6
g = e + f
h = 1 - g
i = 42 / h
|
a ) s . 600 , b ) s . 480 , c ) s . 300 , d ) s . 450 , e ) s . 550 | c | subtract(divide(multiply(500, const_100), add(25, const_100)), divide(multiply(divide(multiply(500, const_100), add(25, const_100)), 25), const_100)) | by selling an article at rs . 500 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ? | "sp = 500 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 500 * [ 100 / 125 ] = 400 loss = 25 % = 25 % of 400 = rs . 100 sp = cp - loss = 400 - 100 = rs . 300 answer : c" | a = 500 * 100
b = 25 + 100
c = a / b
d = 500 * 100
e = 25 + 100
f = d / e
g = f * 25
h = g / 100
i = c - h
|
a ) 420 / 177 , b ) 420 / 179 , c ) 410 / 87 , d ) 90 / 14 , e ) 77 / 56 | b | divide(multiply(multiply(7, 12), 5), multiply(add(add(7, 12), 5), 7)) | a can do a work in 7 days b in 12 days and c in 5 days . if they work on it together then in how many days required to complete the work ? | "person ( a ) ( b ) ( c ) ( a + b + c ) time - ( 7 ) ( 12 ) ( 5 ) rate - ( 60 ) ( 35 ) ( 84 ) ( 179 ) work - ( 420 ) ( 420 ) ( 420 ) ( 420 ) therefore a + b + c requires ( 420 / 179 ) days to complete entire work = 420 / 179 answer is b" | a = 7 * 12
b = a * 5
c = 7 + 12
d = c + 5
e = d * 7
f = b / e
|
a ) 25 , b ) 28 , c ) 30 , d ) 34 , e ) 36 | b | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(5, const_2)), add(add(5, const_2), const_2)), add(add(add(5, const_2), const_2), const_4)), add(add(add(add(5, const_2), const_2), const_4), const_2)), add(add(add(add(add(5, const_2), const_2), const_4), const_2), const_4)) | find a sum for 1 st 5 prime number ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 ) = 28 note : 1 is not a prime number option b" | a = 2 + 3
b = 2 + 3
c = a + b
d = 2 + 3
e = d + 2
f = c + e
g = 5 + 2
h = f + g
i = 5 + 2
j = i + 2
k = h + j
l = 5 + 2
m = l + 2
n = m + 4
o = k + n
p = 5 + 2
q = p + 2
r = q + 4
s = r + 2
t = o + s
u = 5 + 2
v = u + 2
w = v + 4
x = w + 2
y = x + 4
z = t + y
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | multiply(divide(subtract(multiply(25, divide(5, const_60)), multiply(5, divide(5, const_60))), 5), const_60) | a hiker walking at a constant rate of 5 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 25 kilometers per hour . the cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate . how many minutes must the cyclist wait until the hiker catches up ? | "in 5 minutes , the cyclist travels a distance of ( 5 / 60 ) * 25 = 25 / 12 km . the time it takes the hiker to complete this distance is ( 25 / 12 ) / 5 = 5 / 12 hours = 25 minutes the cyclist needs to wait 25 - 5 = 20 minutes the answer is c ." | a = 5 / const_60
b = 25 * a
c = 5 / const_60
d = 5 * c
e = b - d
f = e / 5
g = f * const_60
|
a ) 20 % , b ) 24 % , c ) 36 % , d ) 37.5 % , e ) 70 % | b | multiply(divide(add(add(multiply(divide(20, const_100), 60), multiply(divide(40, const_100), 50)), multiply(divide(10, const_100), 40)), add(add(60, 50), 40)), const_100) | each employee of company x is a member of precisely 1 of 3 shifts of employees . among the 60 members of the first shift , 20 percent participate in the pension program ; among the 50 members of the second shift , 40 percent participate in the pension program ; and among the 40 members of the third shift , 10 percent participate in the pension program . what percent of the workers at company x participate in the pension program ? | 60 members & 20 percent = 12 50 members & 40 percent = 20 40 members & 10 percent = 4 = > 36 / 150 β 100 = 24 % answer b | a = 20 / 100
b = a * 60
c = 40 / 100
d = c * 50
e = b + d
f = 10 / 100
g = f * 40
h = e + g
i = 60 + 50
j = i + 40
k = h / j
l = k * 100
|
a ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 51 | e | add(multiply(35, divide(40, const_100)), 35) | a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 60 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.4 = $ 14 - - > x = $ 35 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 35 + 21 = $ 51 . answer : e ." | a = 40 / 100
b = 35 * a
c = b + 35
|
a ) $ 60 , b ) $ 80 , c ) $ 100 , d ) $ 120 , e ) $ 140 | d | divide(30, subtract(divide(3, 2), divide(5, 4))) | the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 5 : 4 . if the oil bill for february had been $ 30 more , the corresponding ratio would have been 3 : 2 . how much was the oil bill for january ? | "5 : 4 = 15 : 12 and 3 : 2 = 18 : 12 . an increase in $ 30 increases the ratio by 3 : 12 . therefore an increase in $ 10 increases the ratio by 1 : 12 . therefore , january ' s bill was 12 ( $ 10 ) = $ 120 . the answer is d ." | a = 3 / 2
b = 5 / 4
c = a - b
d = 30 / c
|
a ) 38 , b ) 40 , c ) 44 , d ) 45 , e ) 48 | c | divide(50, add(divide(25, 66), divide(subtract(50, 25), 33))) | a driver goes on a trip of 50 kilometers , the first 25 kilometers at 66 kilometers per hour and the remaining distance at 33 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | the time for the first part of the trip was 25 / 66 hours . the time for the second part of the trip was 25 / 33 hours . the total time fro the trip was 25 / 66 + 25 / 33 = 75 / 66 = 25 / 22 hours . the average speed for the trip was 50 / ( 25 / 22 ) = 44 kph the answer is c . | a = 25 / 66
b = 50 - 25
c = b / 33
d = a + c
e = 50 / d
|
['a ) 47 1 / 5 cm 3', 'b ) 47 3 / 5 cm 3', 'c ) 47 7 / 5 cm 3', 'd ) 47 9 / 5 cm 3', 'e ) none of these'] | b | subtract(volume_sphere(divide(6, const_2)), volume_sphere(multiply(divide(6, const_2), divide(1, 2)))) | a hollow spherical metallic ball has an external diameter 6 cm and is 1 / 2 cm thick . the volume of metal used in the metal is : | explanation : please note we are talking about ` ` hollow ' ' ball . do not ignore this word in this type of question in a hurry to solve this question . if we are given with external radius and thickness , we can get the internal radius by subtracting them . then the volume of metal can be obtained by its formula as , external radius = 3 cm , internal radius = ( 3 - 0.5 ) cm = 2.5 cm volume of sphere = 4 / 3 Ο r 3 = 4 / 3 β 22 / 7 β [ 3 ( 2 ) β 2.52 ] cm 3 = 4 / 3 β 22 / 7 β 91 / 8 cm 3 = 143 / 3 cm 3 = 47 2 / 3 cm 3 option b | a = 6 / 2
b = volume_sphere - (
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a ) 75 , b ) 72 , c ) 60 , d ) 70 , e ) 125 | e | divide(multiply(const_100, divide(6, const_2)), 6) | if a book is sold at 6 % profit instead of 6 % loss , it would have brought rs 15 more . find out the cost price of the book | "let c . p . of the book be rs . β x β given , 1.06 x - 0.94 x = 15 = > 0.12 x = 15 = 15 / 0.12 = rs 125 answer : e" | a = 6 / 2
b = 100 * a
c = b / 6
|
a ) 10 % , b ) 15 % , c ) 18 % , d ) 20 % , e ) 25 % | a | divide(const_100, 10) | at what rate percent per annum will a sum of money double in 10 years ? | "let principle = p s . i . = p t = 10 yrs rate = 100 * p / p * 10 = 10 % answer is a" | a = 100 / 10
|
a ) 45 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(5, 5)), multiply(5, 9)) | the average of 9 observations was 5 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 5 = 45 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 β 45 = 45 option a" | a = 10 * 5
b = 5 * 5
c = a + b
d = 5 * 9
e = c - d
|
a ) $ 92 , b ) $ 312 , c ) $ 104 , d ) $ 120 , e ) $ 240 | b | multiply(add(divide(0.8, const_2), multiply(multiply(subtract(1.1, 0.8), const_10), 0.04)), 600) | a certain company expects quarterly earnings of $ 0.80 per share of stock , half of which will be distributed as dividends to shareholders while the rest will be used for research and development . if earnings are greater than expected , shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings . if quarterly earnings are $ 1.10 per share , what will be the dividend paid to a person who owns 600 shares of the company ' s stock ? | eps actual > eps expected . each gets and additional . 12 per share . thus . 52 * 600 - - > $ 312 answer is b | a = 0 / 8
b = 1 - 1
c = b * 10
d = c * 0
e = a + d
f = e * 600
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | a | multiply(divide(multiply(10, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 10 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,10 = 10 / 100 x = 80 * 10 / 100 x = 8 8 / 10 = 0.8 = 800 % , answer a" | a = 20 / 100
b = 1 - a
c = 10 * b
d = c / 10
e = d * 100
|
a ) 1 , b ) 3 , c ) 4 , d ) 12 , e ) 8 | d | divide(60, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 60 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 60 = 6 k + 4 k = > k = 6 total red ' s in bag a will be 2 k = 12 d" | a = 3 * 2
b = a + 4
c = 60 / b
|
a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | divide(divide(80, 20), divide(80, 20)) | solution a is 20 % barium and solution b is 80 % barium . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % barium solution ? | forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 - 20 ) = 1 / 1 so the volume of the two solutions will be equal . answer has to be 4 : 4 = c | a = 80 / 20
b = 80 / 20
c = a / b
|
a ) 8.1 , b ) 8.3 , c ) 9.6 , d ) 8.9 , e ) 9.0 | c | divide(add(12, 8), const_2) | a cyclist bikes x distance at 12 miles per hour and returns over the same path at 8 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ? | "distance = d 1 = x miles speed = s 1 = 12 miles per hour time = t 1 = distance / speed = x / 12 2 . going from b to a distance = d 2 = x miles speed = s 2 = 8 miles per hour time = t 2 = distance / speed = x / 8 3 . average speed = total distance / total time total distance = x + x = 2 x total time = x / 12 + x / 8 = x ( 1 / 12 + 1 / 8 ) = = 5 x / 24 speed = 2 x / ( 5 x / 24 ) = 48 / 5 = 9.6 answer : c" | a = 12 + 8
b = a / 2
|
a ) 50 % , b ) 59 % , c ) 54 % , d ) 55 % , e ) 57 % | b | add(multiply(60, divide(85, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 85 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ? | "say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.85 = 51 democrats are expected to vote for candidate a ; 40 * 0.20 = 8 republicans are expected to vote for candidate a . thus total of 51 + 8 = 59 registered voters are expected to vote for candidate a , which is 59 % of the total number of registered voters . answer : b ." | a = 85 / 100
b = 60 * a
c = 100 - 60
d = 20 / 100
e = c * d
f = b + e
|
a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 180 | e | multiply(subtract(5, divide(multiply(100, 5), add(100, 150))), const_60) | while driving from a - ville to b - town , harriet drove at a constant speed of 100 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 150 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ? | "5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = 180 min * leg a - b - - - > d = 100.180 / 60 = 18000 / 60 * leg b - a - - - - > d = 150 * 120 / 60 = 18000 / 60 so the correct ans . ise" | a = 100 * 5
b = 100 + 150
c = a / b
d = 5 - c
e = d * const_60
|
a ) - 6 , b ) - 21 , c ) - 2 , d ) - 3 , e ) 4 | b | multiply(subtract(7, const_4), 3) | find the value for x from below equation : x / 3 = - 7 ? | "1 . multiply both sides by 3 : x * 3 / 3 = - 7 / 3 simplify both sides : x = - 21 b" | a = 7 - 4
b = a * 3
|
a ) 3 / 25 , b ) 11 / 36 , c ) 5 / 9 , d ) 2 / 3 , e ) 25 / 36 | c | divide(divide(multiply(50, 5), 30), 15) | at a speed of 50 miles per hour , a certain car uses 1 gallon of gasoline every 30 miles . if the car starts with a full 15 gallon tank of gasoline and travels for 5 hours at 50 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | gas used = ( 5 hours ) * ( 50 miles / hour ) * ( 1 gallon / 30 miles ) = 8 + 1 / 3 gallons portion used = ( 8 + 1 / 3 ) / 15 = 5 / 9 ans c | a = 50 * 5
b = a / 30
c = b / 15
|
a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2 | c | subtract(13, 4) | robert ate 13 chocolates , nickel ate 4 chocolates . how many more chocolates did robert ate than nickel ? | 13 - 4 = 9 . answer is c | a = 13 - 4
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a ) s . 486 , b ) s . 455 , c ) s . 487 , d ) s . 120 , e ) s . 489 | d | divide(multiply(150, const_100), add(const_100, 25)) | by selling an article at rs . 150 , a profit of 25 % is made . find its cost price ? | "sp = 150 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 150 * [ 100 / ( 100 + 25 ) ] = 150 * [ 100 / 125 ] = rs . 120 answer : d" | a = 150 * 100
b = 100 + 25
c = a / b
|
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