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 ) 1 / 9 , b ) 2 / 9 , c ) 2 / 5 , d ) 3 / 5 , e ) 3 / 7 | b | multiply(divide(1, 3), subtract(1, divide(1, 3))) | every year taylor goes to the same carnival , and he attempts to shoot a basketball into a small hoop , hoping that he will win a 10 - foot tall teddy bear . if the probability that taylor does not make a basket is 1 / 3 , what is the probability that taylor makes exactly one basket in 3 tries ? | probability of basket = 2 / 3 probability of no basket = 1 / 3 required probability = 3 * 2 / 3 * 1 / 3 * 1 / 3 = 2 / 9 b is the answer . | a = 1 / 3
b = 1 / 3
c = 1 - b
d = a * c
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a ) 35 , b ) 37 , c ) 46 , d ) 90 , e ) 100 | e | subtract(add(multiply(8, 20), multiply(8, 40)), multiply(13, 20)) | the average of 13 result is 20 . average of the first 8 of them is 20 and that of the last 8 is 40 . find the 8 th result ? | "sum of all the 13 results = 13 * 20 = 260 sum of the first 7 of them = 8 * 20 = 160 sum of the last 7 of them = 8 * 40 = 320 so , the 8 th number = 260 + 160 - 320 = 100 . e" | a = 8 * 20
b = 8 * 40
c = a + b
d = 13 * 20
e = c - d
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a ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | d | divide(const_1, add(multiply(12, divide(divide(const_1, 6), 24)), multiply(4, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 4 boys and 12 girls will take to construct ? | explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 4 boys + 12 girls = 4 Γ£ β 6 / 4 + 12 = 6 + 12 = 18 girls 8 days to complete the work answer : option d | a = 1 / 6
b = a / 24
c = 12 * b
d = 1 / 6
e = d / 16
f = 4 * e
g = c + f
h = 1 / g
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a ) 5 kg , b ) 15 kg , c ) 25 kg , d ) 50 kg , e ) none | d | divide(const_100, divide(subtract(const_100, 20), 40)) | the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 40 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 40 kg . as the price falls by 20 % , new price for 40 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 40 = rs . 2 per kg . rice can bought now at = 100 / 2 = 50 kg . answer : option d" | a = 100 - 20
b = a / 40
c = 100 / b
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a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money j did bob receive ? | notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ j = 20 answer : b | a = 32 * 2
b = a / 3
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a ) 3 / 40,000 , b ) 1 / 3,600 , c ) 1 / 20,000 , d ) 1 / 60 , e ) 1 / 15 | c | divide(1, const_3) | a certain junior class has 1,000 students and a certain senior class has 800 students . among these students , there are 40 siblings pairs , each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected at will be a sibling pair ? | total number of ways of choosing one student from each group is = 800 * 1000 number of cases in which a sibling pair will be got is = 40 thus the probability that the 2 students selected will be a sibling pair is = 40 / ( 800 * 1000 ) = 1 / 20,000 c | a = 1 / 3
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a ) 4 , b ) 30 , c ) 8 , d ) 10 , e ) 12 | b | divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2) | running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 5 w widgets ? | "let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 5 * 6 = 30 days to produce 2 w widgets . answer : b" | a = 3 * 4
b = a * 2
c = 3 * 4
d = b - c
e = d / 2
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a ) - 5 , b ) 1 , c ) 4 , d ) 3 , e ) 2 | b | subtract(2, 1) | find value for x from below equation ? x + 1 = 2 | "1 . subtract 1 from both sides : x + 1 - 1 = 2 - 1 2 . simplify both sides : x = 1 b" | a = 2 - 1
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a ) 228 , b ) 278 , c ) 250 , d ) 500 , e ) 821 | c | divide(170, subtract(const_1, divide(multiply(4, 8), const_100))) | a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 170 less than the sum lent . what was the sum lent ? | "p - 170 = ( p * 4 * 8 ) / 100 p = 250 answer : c" | a = 4 * 8
b = a / 100
c = 1 - b
d = 170 / c
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a ) $ 1.85 , b ) $ 3.85 , c ) $ 4.85 , d ) $ 6.85 , e ) $ 5.85 | b | subtract(10.50, add(2.25, add(2.20, 2.20))) | little john had $ 10.50 . he spent $ 2.25 on sweets and gave to his two friends $ 2.20 each . how much money was left ? | "john spent and gave to his two friends a total of 2.25 + 2.20 + 2.20 = $ 6.65 money left 10.50 - 6.65 = $ 3.85 correct answer is b ) $ 3.85" | a = 2 + 20
b = 2 + 25
c = 10 - 50
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a ) 6 days , b ) 12 days , c ) 15 days , d ) 9 days , e ) 8 days | d | divide(subtract(3, add(divide(3, 16), divide(3, 12))), divide(3, 16)) | a and b can do a piece of work in 12 days and 16 days respectively . both work for 3 days and then a goes away . find how long will b take to complete the remaining work ? | "3 / 12 + ( 3 + x ) / 16 = 1 x = 9 days answer : d" | a = 3 / 16
b = 3 / 12
c = a + b
d = 3 - c
e = 3 / 16
f = d / e
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a ) 61 , b ) 62 , c ) 65 , d ) 63 , e ) 64 | d | multiply(add(8, const_1), subtract(8, const_1)) | a motorcycle importer is planning on increasing the price of a certain model by $ 1000 . at this new price 8 fewer motorcycles will be sold per month , but the total revenues will increase by $ 26000 to $ 594000 . what is the number of motorcycles the manufacturer will sell at this new price ? | responding to a pm : to solve this question , i will make an equation in x and then make educated guesses . here ' s how : assuming x motorcycles were sold every month initially . ( 568000 / x + 1000 ) ( x - 8 ) = 594000 ( 568 / x + 1 ) ( x - 8 ) = 594 now 568 = 8 * 71 assuming x = 71 ( we have all integers so it is obvious that 568 / x should be an integer . we get 9 * 66 = 594 ( matches ) so he will sell 71 - 8 = 63 bikes this month answer ( d ) | a = 8 + 1
b = 8 - 1
c = a * b
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a ) 50 , b ) 2.5 , c ) 25 , d ) . 25 , e ) none of these | a | divide(1, 0.02) | 1 / 0.02 is equal to | explanation : 1 / 0.02 = ( 1 * 100 ) / 2 = 100 / 2 = 50 option a | a = 1 / 0
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 14 | e | add(7, divide(multiply(7, subtract(10000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 10000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 10000 x 7 ) + 6000 ( x - 7 ) x = 14 . answer e" | a = 10000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(3, 3)) | what is the remainder when 3 ^ 29 is divided by 5 ? | "the units digit of powers of 3 follow a repeating cycle of four : { 3 , 9 , 7 , 1 } 29 has the form 4 k + 1 , so the units digit of 3 ^ 29 is 3 . the remainder when dividing by 5 is 3 . the answer is d ." | a = 5 / 2
b = 3 * 3
c = a - b
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a ) 6.67 % , b ) 12.5 % , c ) 41.66 % , d ) 25 % , e ) 33.3 % | c | subtract(subtract(multiply(multiply(divide(circumface(200), 200), const_3), const_2), const_4), const_2) | obra drove 200 Ο meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 Ο square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 Ο square meters so , Ο ( r ^ 2 ) = 57,600 Ο - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 Ο r = 480 Ο therefore , part of circumference covered = 200 Ο / 480 Ο = 41.66 % hence , answer is c ." | a = circumface / (
b = a * 200
c = b * 3
d = c - 2
e = d - 4
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a ) 540 , b ) 585 , c ) 619 , d ) 649 , e ) 700 | b | multiply(divide(351, divide(add(const_100, 20), const_100)), 2) | if the price of a certain computer increased 20 percent from d dollars to 351 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 20 / 100 ) * d = 1.2 d = 351 ( given ) i . e . d = 351 / 1.3 = $ 292,5 i . e . 2 d = 2 * 292.5 = 585 answer : option b" | a = 100 + 20
b = a / 100
c = 351 / b
d = c * 2
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a ) 7.5 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 20))) | two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 20 hr . if all pipes are opened simultaneously , then the cistern will be filled in | "solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 β 1 / 20 = 2 / 15 hence , tank will be filled in 15 / 2 = 7.5 hour option ( a )" | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / 20
e = c - d
f = 1/(e)
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a ) 2 , b ) 4 , c ) 6 , d ) 9 , e ) 8 | a | divide(subtract(multiply(4, 3), multiply(5, 2)), subtract(5, 4)) | what must be added to each term of the ratio 2 : 3 , so that it may become equal to 4 : 5 ? | let the number to be added be x , then ( 2 + x ) : ( 3 + x ) = 4 : 5 β ( 2 + x ) / ( 5 + x ) = 4 / 5 5 ( 2 + x ) = 4 ( 3 + x ) 10 + 5 x = 12 + 4 x 5 x - 4 x = 12 - 10 x = 2 answer : 2 | a = 4 * 3
b = 5 * 2
c = a - b
d = 5 - 4
e = c / d
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a ) 37.5 % , b ) 12.5 % , c ) 18.75 % , d ) 25 % , e ) 33.3 % | a | subtract(subtract(multiply(multiply(divide(circumface(180), 180), const_3), const_2), const_4), const_2) | obra drove 180 Ο meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 Ο square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 Ο square meters so , Ο ( r ^ 2 ) = 57,600 Ο - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 Ο r = 480 Ο therefore , part of circumference covered = 180 Ο / 480 Ο = 37.5 % hence , answer is a ." | a = circumface / (
b = a * 180
c = b * 3
d = c - 2
e = d - 4
|
a ) 168 , b ) 667 , c ) 288 , d ) 1991 , e ) 111 | a | multiply(21, 8) | the average of 8 numbers is 21 . if each of the number is multiplied by 8 , the average of the new set of numbers is : | "average of new numbers = ( 21 Γ 8 ) = 168 answer : a" | a = 21 * 8
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a ) 18 days , b ) 38 days , c ) 42 days , d ) 48 days , e ) 56 days | e | divide(multiply(14, 80), 20) | if 14 men do a work in 80 days , in how many days will 20 men do it ? | "14 * 80 = 20 * x x = 56 days answer : e" | a = 14 * 80
b = a / 20
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a ) 1 : 3 , b ) 1 : 2 , c ) 1 : 1 , d ) 2 : 1 , e ) 4 : 5 | e | divide(0.855, add(0.8, 0.9)) | in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.855 per pound ? | "using weighted average method : let x be the proportion in which the $ 0.8 per pound flour is mixed with $ 0.9 per pound flour . thus 0.8 * x + 0.9 * ( 1 - x ) = 0.855 0.9 β 0.1 x = 0.855 x = 0.45 thus ratio of both flours is 4 : 5 e" | a = 0 + 8
b = 0 / 855
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a ) 10 , b ) 30 , c ) 20 , d ) 40 , e ) 35 | d | multiply(subtract(multiply(5, 10), multiply(4, 10)), 4) | two terms are in 3 : 4 , if 10 is added to each of the terms , they will be 4 : 5 . find second term ? | solution : let the two numbers be 3 x and 4 x . then , ( 3 x + 10 ) / ( 4 x + 10 ) = 4 / 5 . = > 15 x + 50 = 16 x + 40 = > x = 10 . . ' . second term = 4 * 10 = 40 . answer is d | a = 5 * 10
b = 4 * 10
c = a - b
d = c * 4
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a ) 20 hours , b ) 15 hours , c ) 10 hours , d ) 10 2 / 7 hours , e ) 8 hours | d | divide(const_1, add(divide(const_1, 18), divide(const_1, 24))) | two pipes a and b can fill a tank in 18 hours and 24 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 18 part filled by b in 1 hour = 1 / 24 part filled by ( a + b ) in 1 hour = 1 / 18 + 1 / 24 = 7 / 72 both the pipes together fill the tank in 72 / 7 = 10 2 / 7 hours answer is d" | a = 1 / 18
b = 1 / 24
c = a + b
d = 1 / c
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a ) 18 cm , b ) 16 cm , c ) 40 cm , d ) 20 cm , e ) 28 cm | c | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 75 sq . cm . what is the length of the rectangle | "explanation : let breadth = x cm then , length = 2 x cm area = x Γ 2 x = 2 x ^ 2 sq . cm . new length = ( 2 x β 5 ) cm new breadth = ( x + 5 ) cm new area = ( 2 x β 5 ) ( x + 5 ) sq . cm . given that , new area = initial area + 75 sq . cm . β ( 2 x β 5 ) ( x + 5 ) = 2 x ^ 2 + 75 β 2 x ^ 2 + 10 x β 5 x β 25 = 2 x ^ 2 + 75 β 5 x β 25 = 75 β 5 x = 75 + 25 = 100 β x = 1005 = 20 cm length = 2 x = 2 Γ 20 = 40 cm answer : option c" | a = 100 / 3
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a ) 8 % , b ) 7 % , c ) 10 % , d ) 6 % , e ) 5 % | e | multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 59.85), subtract(70, multiply(70, divide(10, const_100)))), const_100) | the list price of an article is rs . 70 . a customer pays rs . 59.85 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "option e explanation : 70 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 59.85 x = 5 %" | a = 10 / 100
b = 70 * a
c = 70 - b
d = c - 59
e = 10 / 100
f = 70 * e
g = 70 - f
h = d / g
i = h * 100
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a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 . | a | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 9.8) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 9.8 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 9.8 = 6 * 0 + 2 * 7.2 + 8 - 9.8 = 0 + 14.4 - 1.8 12.6 answer a" | a = 2 * 7
b = 4 / 10
c = 8 - 4
d = a + c
e = d - 9
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a ) 144 , b ) 15 , c ) 18 , d ) 32 , e ) 38 | b | divide(250, multiply(add(20, 40), const_0_2778)) | two persons start running simultaneously around a circular track of length 250 m from the same point at speeds of 20 kmph and 40 kmph . when will they meet for the first time any where on the track if they are moving in the opposite direction ? | time taken to meet the first time = length of track / relative speed = 250 / ( 20 + 40 ) ( 5 / 18 ) = 250 / 60 * ( 18 / 5 ) = 15 sec . answer : b | a = 20 + 40
b = a * const_0_2778
c = 250 / b
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 63 % , e ) 75 % | d | multiply(divide(subtract(const_100, add(20, 5)), const_100), subtract(const_100, subtract(20, const_4))) | sixteen 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 84 % and 16 % 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 84 so , the required % is [ ( 3 / 4 ) β 84 / 100 ] β 100 = 63 % answer : d | a = 20 + 5
b = 100 - a
c = b / 100
d = 20 - 4
e = 100 - d
f = c * e
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a ) 3 hr , b ) 1 / 2 hr , c ) 1 hr , d ) 9 / 7 hr , e ) 5 / 3 hr | a | inverse(subtract(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 4))) | pipes a and b can fill a tank in 3 and 4 hours . pipe c can empty it in 4 hours . if all pipes are opened together , then the tank will be filled in ? | "net part filled in 1 hour = 1 / 3 + 1 / 4 - 1 / 4 = 1 / 3 the tank will be full in 3 hr answer is a" | a = 1 / 3
b = 1 / 4
c = a + b
d = 1 / 4
e = c - d
f = 1/(e)
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a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 20 : 9 , e ) 1 : 2 | d | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(5, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 5 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 5 - - > e : l = 3 : 5 = 12 : 20 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 12 : 9 . thus , l : s = 20 : 9 . answer : d ." | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 5 * 4
f = d / e
g = c / f
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a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 20 , e ) e ) 25 | c | multiply(divide(100000, 20000), const_3) | in a graduating class , the difference between the highest and lowest salaries is $ 100000 . the median salary is $ 50000 higher than the lowest salary and the average salary is $ 20000 higher than the median . what is the minimum number of students r in the class ? | the difference between the highest and lowest salaries is $ 100000 . so there are at least 2 people - say one with salary 0 and the other with 100 k . no salary will be outside this range . median = 50 k more than lowest . so median is right in the center of lowest and highest since lowest and highest differ by 100 k . in our example , median = 50 k . since there are more than 2 people , there would probably be a person at 50 k . mean = 20 k more than median so in our example , mean salary = 70 k on the number line , 0 . . . . . . . . 50 k ( median ) . . . . . . . . 100 k mean = 70 k so there must be people more toward 100 k to bring the mean up to 70 k . since we want to add minimum people , we will add people at 100 k to quickly make up the right side deficit . 0 and 50 k are ( 70 k + 20 k ) = 90 k away from 70 k . 100 k is 30 k away from 70 k . to bring the mean to 70 k , we will add two people at 100 k each to get : 0 . . . . 50 k . . . . . 100 k , 100 k , 100 k but when we add more people to the right of 70 k , the median will shift to the right . we need to keep the median at 50 k . so every time we add people to the right of 70 k , we need to add people at 50 k too to balance the median . 50 k is 20 k less than 70 k while 100 k is 30 k more than 70 k . to keep the mean same , we need to add 2 people at 100 k for every 3 people we add at 50 k . so if we add 3 people at 50 k and 2 people at 100 k , we get : 0 , . . . 50 k , 50 k , 50 k , 50 k , . . . 100 k , 100 k , 100 k , 100 k , 100 k the median is not at 50 k yet . add another 3 people at 50 k and another 2 at 100 k to get 0 , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k now the median is 50 k and mean is 70 k . total number of people is 15 . answer ( c ) | a = 100000 / 20000
b = a * 3
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a ) $ 11.00 , b ) $ 1.25 , c ) $ 5.00 , d ) $ 6.00 , e ) $ 20.00 | a | subtract(20.00, multiply(subtract(10, 1.00), 1.00)) | if 10 people contributed a total of $ 20.00 toward a gift and each of them contributed at least $ 1.00 , then the maximum possible amount any one person could have contributed is | "a for me 9 people with 1 $ each - > maximum = 11" | a = 10 - 1
b = a * 1
c = 20 - 0
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a ) 277 , b ) 665 , c ) 662 , d ) 552 , e ) 281 | b | floor(add(multiply(subtract(200, 125), divide(400, 200)), const_1)) | the average salary of workers in an industry is rs . 200 the average salary of technicians being rs . 400 and that of non - technicians being rs . 125 . what is the total number of workers ? | "8 * 84 = 672 β 7 = 665 answer : b" | a = 200 - 125
b = 400 / 200
c = a * b
d = c + 1
e = math.floor(d)
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a ) 22 , b ) 12 , c ) 9 , d ) 99 , e ) 21 | c | divide(multiply(120, const_2), add(speed(120, 8), speed(120, 10))) | two trains of equal lengths take 10 sec and 8 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 8 = 15 m / sec . relative speed = 12 + 15 = 27 m / sec . required time = ( 120 + 120 ) / 27 = 9 sec . answer : c | a = 120 * 2
b = speed + (
c = a / b
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a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | b | sqrt(divide(432, divide(1200, const_100))) | reema took a loan of rs 1200 with simple interest for as many years as the rate of interest . if she paid rs . 432 as interest at the end of the loan period , what was the rate of interest . | "explanation : let rate = r % then time = r years . = > 1200 β r β r / 100 = 432 = > r 2 = 36 = > r = 6 % option b" | a = 1200 / 100
b = 432 / a
c = math.sqrt(b)
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a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6 | c | subtract(add(26, 22), subtract(45, 8)) | each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 26 have long fur , 22 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | "no of dogs = 45 long fur = 26 brown = 22 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 26 + 22 - both both = 11 answer c" | a = 26 + 22
b = 45 - 8
c = a - b
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a ) 6819.59775 , b ) 6981.59775 , c ) 6918.59775 , d ) 6198.59775 , e ) 6891.59775 | e | subtract(6202.5, multiply(multiply(620.25, 62.025), 6.2025)) | evaluate : 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025 | "6202.5 620.25 62.025 6.2025 + 0.62025 - - - - - - - - - - - - - - - 6891.59775 answer is e ." | a = 620 * 25
b = a * 6
c = 6202 - 5
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a ) 56 , b ) 58 , c ) 60 , d ) 62 , e ) 64 | d | divide(465, multiply(divide(3, 2), 5)) | a van takes 5 hours to cover a distance of 465 km . what speed in kph should the van maintain to cover the same distance in 3 / 2 of the previous time ? | "( 3 / 2 ) * 5 = 7.5 hours 465 / 7.5 = 62 kph the answer is d ." | a = 3 / 2
b = a * 5
c = 465 / b
|
a ) 23 , b ) 21 , c ) 43 , d ) 56 , e ) 30 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 and the sum of their squares is 289 . the sum is : | "xy = 120 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 120 ) = 529 x + y = 529 = 23 . answer a" | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
|
a ) 4 sec , b ) 7 sec , c ) 2 sec , d ) 6 sec , e ) 9 sec | a | divide(60, multiply(54, const_0_2778)) | in what time will a railway train 60 m long moving at the rate of 54 kmph pass a telegraph post on its way ? | "t = 60 / 54 * 18 / 5 = 4 sec answer : a" | a = 54 * const_0_2778
b = 60 / a
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a ) 0.25 , b ) 0.5 , c ) 0.75 , d ) 1 , e ) 1.25 | c | add(divide(divide(4, const_2), 4), divide(divide(4, const_2), 8)) | a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 4 km , if the person completes half of the distance by walking and the other half by running ? | "time = 2 / 4 + 2 / 8 = 6 / 8 = 0.75 hours the answer is c ." | a = 4 / 2
b = a / 4
c = 4 / 2
d = c / 8
e = b + d
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a ) 2 , b ) 1.5 , c ) 1.6 , d ) 1.7 , e ) 1.25 | e | multiply(const_60, divide(multiply(50, divide(15, const_60)), subtract(60, 50))) | mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least 4 hours , how long after he passed the gas station did paul catch up with mary ? | "d = rt m : r = 50 mph , t = t + 1 / 4 hr d = 50 ( t + 1 / 4 ) p : r = 60 , t = t d = 60 t since they went the same distance : 50 t + 50 / 4 = 60 t 10 t = 50 / 4 t = 1.25 or 1 hr , 15 min e" | a = 15 / const_60
b = 50 * a
c = 60 - 50
d = b / c
e = const_60 * d
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a ) 14 , b ) 16 , c ) 18 , d ) 24 , e ) 36 | e | add(add(power(add(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 36 . find the middle number of the three ? | "three consecutive even numbers ( 2 p - 2 ) , 2 p , ( 2 p + 2 ) . ( 2 p - 2 ) + 2 p + ( 2 p + 2 ) = 36 6 p = 36 = > p = 6 . the middle number is : 6 p = 36 . answer : e" | a = 36 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 36 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 36 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 36 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 87 , b ) 16 , c ) 10 , d ) 76 , e ) 17 | d | subtract(multiply(add(10, const_1), add(4, 32)), multiply(10, 32)) | the average of runs of a cricket player of 10 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 36 required number of runs = ( 36 * 11 ) - ( 32 * 10 ) = 396 - 320 = 76 . answer : d" | a = 10 + 1
b = 4 + 32
c = a * b
d = 10 * 32
e = c - d
|
a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | c | add(floor(divide(323, 2)), const_1) | the guests at a football banquet consumed a total of 323 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 161 * 2 ( max possible amount of food a guest could consume ) = 322 pounds , so there must be more than 161 guests , next integer is 162 . answer : c ." | a = 323 / 2
b = math.floor(a)
c = b + 1
|
a ) 2 , b ) 7 , c ) 5 , d ) 9 , e ) 6 | a | divide(12, subtract(10, 4)) | a person can swim in still water at 10 km / h . if the speed of water 4 km / h , how many hours will the man take to swim back against the current for 12 km ? | "m = 10 s = 4 us = 10 - 4 = 6 d = 12 t = 12 / 6 = 2 answer : a" | a = 10 - 4
b = 12 / a
|
a ) $ 80,000 , b ) $ 150,000 , c ) $ 240,000 , d ) $ 290,000 , e ) $ 320,000 | b | divide(subtract(subtract(multiply(multiply(add(70, 20), const_1000), const_100), multiply(20, multiply(add(70, 20), const_1000))), multiply(multiply(multiply(divide(20, const_2), 20), const_1000), 70)), 20) | in plutarch enterprises , 70 % of the employees are marketers , 20 % are engineers , and the rest are managers . marketers make an average salary of $ 70,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | "for sake of ease , let ' s say there are 10 employees : 7 marketers , 2 engineers , and 1 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 7 * $ 70,000 ) and the engineers ( 2 * $ 80,000 ) > > > $ 800,000 - $ 490,000 - $ 160,000 = $ 150,000 . the correct answer is b ." | a = 70 + 20
b = a * 1000
c = b * 100
d = 70 + 20
e = d * 1000
f = 20 * e
g = c - f
h = 20 / 2
i = h * 20
j = i * 1000
k = j * 70
l = g - k
m = l / 20
|
a ) 22725 , b ) 25675 , c ) 22655 , d ) 27575 , e ) none of these | a | multiply(divide(2525, 9), const_100) | 2525 * 9 | "explanation : 2525 * ( 10 - 1 ) = 25250 - 2525 = 22725 option a" | a = 2525 / 9
b = a * 100
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a ) 120 , b ) 25 , c ) 614 , d ) 530 , e ) 210 | d | multiply(power(add(const_1, divide(2, const_100)), 2), 500) | rs . 500 amounts to rs . 620 in 2 years at simple interest . if the interest is increased by 2 % , it would amount to how much ? | "( 500 * 3 * 2 ) / 100 = 30 500 + 30 = 530 answer : d" | a = 2 / 100
b = 1 + a
c = b ** 2
d = c * 500
|
a ) rs . 150.40 , b ) rs . 44.80 , c ) rs . 140.40 , d ) rs . 50.48 , e ) none of these | b | subtract(multiply(add(divide(add(3, divide(1, 3)), const_100), 1), divide(6400, add(divide(add(divide(1, 2), 1), const_100), 1))), 6400) | if the sales tax be reduced from 3 ( 1 / 5 ) % to 2 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 6400 ? | "explanation : required difference = ( 3 ( 1 / 5 ) of rs . 6400 ) - ( 2 ( 1 / 2 ) of rs . 6400 ) = ( 16 / 5 β 5 / 2 ) % of rs . 6400 = ( 7 / 10 ) x ( 1 / 100 ) x 6400 = rs . 44.80 answer b" | a = 1 / 3
b = 3 + a
c = b / 100
d = c + 1
e = 1 / 2
f = e + 1
g = f / 100
h = g + 1
i = 6400 / h
j = d * i
k = j - 6400
|
a ) 13 , b ) pi / 3 , c ) pi / 2 , d ) 35 / 36 , e ) 8 / 9 | d | divide(subtract(power(6, const_2), const_1), power(6, const_2)) | circles x and y are concentric . if the radius of circle x is 6 times that of circle y , what is the probability that a point selected inside circle x at random will be outside circle y ? | small circle radius = a , so area = pi a ^ 2 large circle radius = 6 a , so area = 9 . pi a ^ 2 area of shaded region = 36 pi a ^ 2 - pi a ^ 2 = 35 pi a ^ 2 so , probability = ( 35 pi a ^ 2 ) / ( 36 pi a ^ 2 ) = 35 / 36 = answer = d | a = 6 ** 2
b = a - 1
c = 6 ** 2
d = b / c
|
a ) 1 / 7 , b ) 3 , c ) 15 , d ) 15 / 2 , e ) 60 / 7 | d | subtract(divide(add(multiply(8, 3), 7), add(const_3, const_1)), subtract(8, divide(add(multiply(8, 3), 7), add(const_3, const_1)))) | if x + y = 8 and y β 3 x = 7 , then what is the value of y - x ? | x + y = 8 . . . equation 1 - 3 x + y = 7 . . . equation 2 adding both the equations - 2 x + 2 y = 15 = > y - x = 15 / 2 correct answer option d | a = 8 * 3
b = a + 7
c = 3 + 1
d = b / c
e = 8 * 3
f = e + 7
g = 3 + 1
h = f / g
i = 8 - h
j = d - i
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 4 , e ) 8 | a | divide(6, add(6, 6)) | if 6 / w + 6 / x = 6 / y and wx = y , then the average ( arithmetic mean ) of w and x is | "given : 6 / w + 6 / x = 6 / ywx = y find : ( w + x ) / 2 = ? 6 ( 1 / w + 1 / x ) = 6 ( 1 / y ) - divide both sides by 6 ( 1 / w + 1 / x ) = 1 / y ( x + w ) / wx = 1 / wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) / 2 = 1 / 2 ans : a" | a = 6 + 6
b = 6 / a
|
a ) 39 % , b ) 20 % , c ) 23 % , d ) 50 % , e ) 83 % | d | multiply(divide(subtract(1350, 900), 900), const_100) | a cycle is bought for rs . 900 and sold for rs . 1350 , find the gain percent ? | "900 - - - - 450 100 - - - - ? = > 50 % answer : d" | a = 1350 - 900
b = a / 900
c = b * 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | d | multiply(multiply(24, divide(const_1, const_2)), divide(const_1, const_2)) | some persons can do a piece of work in 24 days . two times the number of these people will do half of that work in ? | "24 / ( 2 * 2 ) = 6 days answer : d" | a = 1 / 2
b = 24 * a
c = 1 / 2
d = b * c
|
a ) 120 , b ) 140 , c ) 150 , d ) 160 , e ) 180 | e | divide(50, 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 , 2 / 9 are from africa , 1 / 6 are from asia and the remaining 50 people are citizens of other continents , then how many passengers are on board the ship ? | "1 / 12 + 1 / 4 + 2 / 9 + 1 / 6 = ( 3 + 9 + 8 + 6 ) / 36 = 13 / 18 let x be the number of passengers on the ship . 50 = ( 5 / 18 ) x x = 180 the answer is e ." | 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 = 50 / h
|
a ) 897 , b ) 164,850 , c ) 164,749 , d ) 149,700 , e ) 156,720 | b | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 3), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 3 ? | "explanatory answer step 1 : identify the series the smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101 . the next couple of numbers that will leave a remainder of 2 when divided by 3 are 104 and 107 . the largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998 . it is evident that any number in the sequence will be a 3 digit positive integer of the form ( 3 n + 2 ) . so , the given numbers are in an arithmetic sequence with the first term being 101 and the last term being 998 and the common difference being 3 . step 2 : compute the sum sum of an arithmetic progression ( ap ) = [ first term + last term / 2 ] n we know the first term : 101 we know the last term : 998 . the only unknown is the number of terms , n . in an a . p . , the nth term an = a 1 + ( n - 1 ) * d in this case , therefore , 998 = 101 + ( n - 1 ) * 3 or 897 = ( n - 1 ) * 3 ( n - 1 ) = 299 or n = 300 . sum of the ap will therefore , be [ 101 + 998 / 2 ] β 300 = 164,850 choice b is the correct answer ." | a = 1000 - 3
b = 3 * 3
c = b * 10
d = 3 * 3
e = c + d
f = e + 3
g = a - f
h = g / 3
i = h + 1
j = i / 2
k = 1000 - 3
l = 3 * 3
m = l * 10
n = 3 * 3
o = m + n
p = o + 3
q = k + p
r = j * q
|
a ) 38 minutes , b ) 35 minutes , c ) 32 minutes , d ) 29 minutes , e ) 26 minutes | c | divide(multiply(40, 4), 5) | if it takes 4 movers 40 minutes to unpack their van , how long would 5 movers take ? | 4 * 40 = 5 * x x = 32 answer : c | a = 40 * 4
b = a / 5
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a ) 32 , b ) 28 , c ) 29 , d ) 24 , e ) 20 | d | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 5), add(add(9, 1), subtract(9, 1))) | a person can row at 9 kmph and still water . he takes 5 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 5.5 = > x / 10 + x / 8 = 11 / 2 = > ( 4 x + 5 x ) / 40 = 11 / 2 = > x = 24 km . answer : d" | a = 9 - 1
b = 9 + 1
c = a * b
d = c * 5
e = 9 + 1
f = 9 - 1
g = e + f
h = d / g
|
a ) rs . 5870 , b ) rs . 5991 , c ) rs . 6020 , d ) rs . 7670 , e ) none of these | d | subtract(multiply(add(5, const_1), 6300), add(add(add(add(5420, 5660), 6200), 6350), 6500)) | a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6300 ? | "explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6300 * 6 ) β 30,130 ] = rs . ( 37800 β 30,130 ) = rs . 7670 answer d" | a = 5 + 1
b = a * 6300
c = 5420 + 5660
d = c + 6200
e = d + 6350
f = e + 6500
g = b - f
|
a ) 60.8 , b ) 18 , c ) 30 , d ) 52.5 , e ) none | d | multiply(189, const_0_2778) | a train moves with a speed of 189 kmph . its speed in metres per second is | "solution speed = 189 kmph = ( 189 x 5 / 18 ) m / sec = 52.5 m / sec . answer d" | a = 189 * const_0_2778
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a ) 22 , b ) 23 , c ) 80 , d ) 90 , e ) 28 | d | divide(subtract(multiply(100, const_4), 180), const_2) | a man has some hens and cows . if the number of heads be 100 and the number of feet equals 180 , then the number of hens will be : | "explanation : let the number of hens be x and the number of cows be y . then , x + y = 100 . . . . ( i ) and 2 x + 4 y = 220 x + 2 y = 110 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = 90 , y = 10 . the required answer = 90 . answer : d" | a = 100 * 4
b = a - 180
c = b / 2
|
a ) 25 / 6 , b ) 36 / 25 , c ) 2 , d ) 5 / 6 , e ) 25 / 36 | c | multiply(divide(divide(1, 3), divide(1, 5)), divide(6, 5)) | if 5 x = 6 y and xy β 0 , what is the ratio of 1 / 3 * x to 1 / 5 * y ? | 5 x = 6 y = > x / y = 6 / 5 1 / 3 * x to 1 / 5 * y = x / y * 5 / 3 = ( 6 / 5 ) * ( 5 / 3 ) = 2 ans : c | a = 1 / 3
b = 1 / 5
c = a / b
d = 6 / 5
e = c * d
|
a ) 3 , b ) 2 1 / 3 , c ) 3 1 / 2 , d ) 4 , e ) 4 1 / 2 | b | divide(add(subtract(12, 9), subtract(9, subtract(multiply(divide(40, const_60), 12), multiply(divide(40, const_60), 9)))), subtract(multiply(divide(40, const_60), 12), multiply(divide(40, const_60), 9))) | john and jacob set out together on bicycle traveling at 12 and 9 miles per hour , respectively . after 40 minutes , john stops to fix a flat tire . if it takes john one hour to fix the flat tire and jacob continues to ride during this time , how many hours will it take john to catch up to jacob assuming he resumes his ride at 12 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible ) | "john ' s speed - 12 miles / hr jacob ' s speed - 9 miles / hr after 40 min ( i . e 2 / 3 hr ) , distance covered by john = 12 x 2 / 3 = 8 miles . jacob continues to ride for a total of 1 hour and 40 min ( until john ' s bike is repaired ) . distance covered in 1 hour 40 min ( i . e 5 / 3 hr ) = 9 x 5 / 3 = 15 miles . now , when john starts riding back , the distance between them is 7 miles . jacob and john are moving in the same direction . for john to catch jacob , the effective relative speed will be 12 - 9 = 3 miles / hr . thus , to cover 7 miles at 3 miles / hr , john will take 7 / 3 = 2 1 / 3 hours answer b" | a = 12 - 9
b = 40 / const_60
c = b * 12
d = 40 / const_60
e = d * 9
f = c - e
g = 9 - f
h = a + g
i = 40 / const_60
j = i * 12
k = 40 / const_60
l = k * 9
m = j - l
n = h / m
|
a ) 36.7 % , b ) 37.6 % , c ) 38.3 % , d ) 39.8 % , e ) 40.5 % | c | multiply(divide(add(1, 3), add(add(5, 2), add(1, 3))), const_100) | the proportion of water to alcohol in solution a is 5 : 1 and the proportion of water to alcohol in solution b is 2 : 3 . if an equal amount of each solution is mixed together , what is the concentration of alcohol in the new solution ? | "let v be the total volume of the new solution . then a volume of v / 2 was added from each solution a and b . the amount of alcohol added to the new solution was : ( 1 / 6 ) ( v / 2 ) + ( 3 / 5 ) ( v / 2 ) = v / 12 + 3 v / 10 = 23 v / 60 . the concentration of alcohol is 23 / 60 = 38.3 % the answer is c ." | a = 1 + 3
b = 5 + 2
c = 1 + 3
d = b + c
e = a / d
f = e * 100
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a ) 10 , b ) 13 , c ) 14 , d ) 15 , e ) 17 | e | divide(multiply(multiply(51, 12), 5), multiply(30, 6)) | 51 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "according to the chain rule , m 1 x t 1 = m 2 x t 2 therefore , 51 x 12 x 5 = 30 x 6 x x x = 17 hence , the number of days = 17 . answer : e" | a = 51 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 36 , b ) 96 , c ) 132 , d ) 216 , e ) 256 | d | power(5, 3) | a palindrome is a word or a number that reads the same forward and backward . for example , 2442 and 111 are palindromes . if 5 - digit palindromes are formed using one or more of the digits 1 , 2 , 3 , 4 , 5 , and 6 , how many palindromes are possible ? | "there are 6 choices for each of the first three digits . the number of possible palindromes is 6 ^ 3 = 216 . the answer is d ." | a = 5 ** 3
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a ) 277 , b ) 268 , c ) 270 , d ) 150 , e ) 282 | d | divide(multiply(const_100, divide(multiply(const_100, 225), add(25, const_100))), add(20, const_100)) | a sells a bicycle to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays rs . 225 for it , the cost price of the bicycle for a is ? | "125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d" | a = 100 * 225
b = 25 + 100
c = a / b
d = 100 * c
e = 20 + 100
f = d / e
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a ) 56.25 , b ) 52.9 , c ) 52.1 , d ) 52.3 , e ) 42.5 | a | divide(add(multiply(30, 50), multiply(50, 60)), add(30, 50)) | the average marks of a class of 30 students is 50 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 50 = 1500 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 80 students = 1500 + 3000 = 4500 average marks of all the students = 4500 / 80 = 56.25 answer : a" | a = 30 * 50
b = 50 * 60
c = a + b
d = 30 + 50
e = c / d
|
a ) 8 , b ) 50 , c ) 100 , d ) 200 , e ) 400 | a | add(power(multiply(2, 2), 2), power(multiply(2, 2), 2)) | if equation | x / 2 | + | y / 2 | = 1 encloses a certain region on the coordinate plane , what is the area of this region ? | the equation can be reduced to intercept form as | x / 2 | + | y / 2 | = 1 , so these are lines in four quadrants with x and y intercept as 2 , so it is a rhombus with diagonals of 4 each and hence area = 1 / 2 * d 1 * d 2 = 1 / 2 * 4 * 4 = 8 . answer a . | a = 2 * 2
b = a ** 2
c = 2 * 2
d = c ** 2
e = b + d
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a ) 1000 , b ) 1250 , c ) 1500 , d ) 1600 , e ) 2000 | c | add(divide(multiply(30, 40), multiply(2, 2)), divide(multiply(60, 80), multiply(2, 2))) | andrea wants to fill in two sections of her backyard with sod that must be purchased in 2 - x - 2 - foot squares . if the two sections measure 30 x 40 feet and 60 x 80 feet , how many squares of sod does she need to buy ? | to begin , find the area of each of the two sections by multiplying the lengths of the sides : 30 feet Γ 40 feet = 1200 square feet 60 feet Γ 80 feet = 4800 square feet the total area that needs sod has an area of 1200 + 4800 = 6000 square feet . each individual square is 2 feet by 2 feet , so each has an area of 4 square feet . because all the numbers are even , there will be no waste when the squares of sod are placed . therefore , you just have to find the number of squares needed by dividing : 6000 Γ· 4 = 1500 correct answer c ) 1500 | a = 30 * 40
b = 2 * 2
c = a / b
d = 60 * 80
e = 2 * 2
f = d / e
g = c + f
|
a ) 25 , b ) 23 , c ) 32 , d ) 30 , e ) 18 | d | divide(rectangle_area(5, 24), 4) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 4 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 4 w = 120 = > w = 30 answer d" | a = rectangle_area / (
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a ) 98.5 % , b ) 100.0 % , c ) 102.3 % , d ) 104.5 % , e ) 105.0 % | c | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 7), const_4)), multiply(const_100, 10)) | this year , mbb consulting fired 7 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ? | "100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 7 % reduction in employees lead to 93 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 93 employees is 93 * 1100 = 102300 now the new salary is more than previous salary by x % . x = ( 102300 / 100000 ) * 100 = 102.3 % so the answer is c" | a = 100 * 10
b = 100 + a
c = 100 - 7
d = c + 4
e = b * d
f = 100 * 10
g = e / f
|
a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | c | add(multiply(0.25, subtract(6, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 6 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s β 1 ) . an employee with a salary grade of 6 receives how many more dollars per hour than an employee with a salary grade of 1 ? | "salary grade of 6 is p ( 6 ) = 9.50 + 0.25 ( 6 β 1 ) = 9.50 + 0.25 * 5 ; salary grade of 1 is p ( 1 ) = 9.50 + 0.25 ( 1 β 1 ) = 9.50 ; p ( 6 ) - p ( 1 ) = 9.50 + 0.25 * 5 - 9.50 = 1.25 . answer : c ." | a = 6 - 1
b = 0 * 25
c = b + 0
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a ) 338 , b ) 277 , c ) 342 , d ) 334.4 , e ) 212 | d | subtract(subtract(400, divide(multiply(400, 12), const_100)), divide(multiply(subtract(400, divide(multiply(400, 12), const_100)), 5), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 12 % and 5 % is ? | "400 * ( 88 / 100 ) * ( 95 / 100 ) = 334.4 answer : d" | a = 400 * 12
b = a / 100
c = 400 - b
d = 400 * 12
e = d / 100
f = 400 - e
g = f * 5
h = g / 100
i = c - h
|
a ) 4 , b ) 9 / 2 , c ) 7 , d ) 41 / 4 , e ) 25 | b | divide(add(add(4, 4), const_1), const_2) | if 4 b ^ 4 β 41 b ^ 2 + 100 = 0 , then what is the sum of the two greatest possible values of b ? | just forget about the value of b ^ 4 for the moment and let b ^ 2 = x . ( its better to work with power of 2 than 4 ) now we have equation as 4 x ^ 2 - 41 x + 100 = 0 , factoring them gives us 4 x ^ 2 - 16 x - 25 x + 100 = 0 . which can be solved as ( 4 x - 25 ) ( x - 4 ) , so x = 4 and 25 / 4 . so b ^ 2 = 4 and 25 / 4 so b = + 2 and minus 2 and + 5 / 2 and minus 5 / 2 the two greatest values are + 2 and + 5 / 2 . so their sum 2 + 5 / 2 = 9 / 2 = b | a = 4 + 4
b = a + 1
c = b / 2
|
a ) 28 , b ) 27 , c ) 26 , d ) 22 , e ) 24 | b | add(subtract(99, multiply(19, const_4)), 4) | having scored 99 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | explanation : let the average score of the first 18 innings be n 18 n + 99 = 19 ( n + 4 ) = > n = 23 so , average score after 19 th innings = x + 4 = 27 . answer : b | a = 19 * 4
b = 99 - a
c = b + 4
|
a ) 4.3 , b ) 5.6 , c ) 6.2 , d ) 10.0 , e ) 12.0 | d | divide(const_100, 10) | a company decreased the price of its main product by 10 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | "for the total revenue to remain the same when the price is 1 / 10 , the number of products sold must 1 / 10 times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 10 % = 10.0 answer : d" | a = 100 / 10
|
a ) 1 / 4 , b ) 1 / 6 , c ) 1 / 8 , d ) 1 / 9 , e ) 1 / 10 | e | divide(add(1, const_2.0), multiply(3, 10)) | of all the homes on gotham street , 1 / 3 are termite - ridden , and 7 / 10 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | "the fraction of homes which are termite - infested but not collapsing is 3 / 10 * 1 / 3 = 1 / 10 the answer is e ." | a = 1 + 2
b = 3 * 10
c = a / b
|
a ) 32 , b ) 42 , c ) 54 , d ) 66 , e ) 09 | b | divide(divide(add(150, 320), const_1000), divide(40, const_3600)) | a train 150 meters long completely crosses a 320 meters long bridge in 40 seconds . what is the speed of the train is ? | "s = ( 150 + 320 ) / 40 = 470 / 40 * 18 / 5 = 42 answer : b" | a = 150 + 320
b = a / 1000
c = 40 / 3600
d = b / c
|
a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 5 / 12 , e ) 7 / 12 | b | add(divide(1, 2), multiply(divide(1, 2), divide(1, 2))) | jar x is 1 / 2 full of water . jar y , which has half the capacity of jar x , is 1 / 4 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | "let p be the capacity of jar x . the amount of water in jar y is 1 / 4 * p / 2 = p / 8 then the total amount in jar x is p / 2 + p / 8 = 5 p / 8 the answer is b ." | a = 1 / 2
b = 1 / 2
c = 1 / 2
d = b * c
e = a + d
|
a ) 12 , b ) 20 , c ) 18 , d ) 12 , e ) 81 | b | subtract(multiply(40, divide(80, const_100)), multiply(divide(4, 5), 15)) | how much is 80 % of 40 is greater than 4 / 5 of 15 ? | "( 80 / 100 ) * 40 Γ’ β¬ β ( 4 / 5 ) * 15 32 - 12 = 20 answer : b" | a = 80 / 100
b = 40 * a
c = 4 / 5
d = c * 15
e = b - d
|
a ) 240 , b ) 250 , c ) 260 , d ) 270 , e ) 300 | a | divide(divide(multiply(200, const_100), 50), const_2) | a rainstorm increased the amount of water stored in state j reservoirs from 160 billion gallons to 200 billion gallons . if the storm increased the amount of water in the reservoirs to 50 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ? | "let total capacity be x we know 200 = 0.50 x x = 200 / 0.50 = 400 prior to storm , we had 160 bn gallons 400 - 160 = 240 answer : a" | a = 200 * 100
b = a / 50
c = b / 2
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | b | multiply(1, 6) | if 6 spiders make 6 webs in 6 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 6 webs 6 : 1 1 x 6 x x = 6 x 1 x 6 = > x = 6 answer is b" | a = 1 * 6
|
a ) 250 / 3 , b ) 750 , c ) 400 , d ) 480 , e ) 600 | b | multiply(divide(subtract(25, 10), subtract(30, 25)), 250) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 250 milliliters of solution x to create a solution that is 25 percent alcohol by volume ? | "we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 250 then y = 750 answer : b" | a = 25 - 10
b = 30 - 25
c = a / b
d = c * 250
|
a ) 2 / 1 , b ) 3 / 8 , c ) 8 / 5 , d ) 5 / 8 , e ) 5 / 3 | a | divide(divide(600, 3), divide(300, 3)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 3 hours to complete the journey . if the distance between city a and city b is 600 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 600 km time taken by eddy = 3 hours average speed of eddy = 600 / 3 = 200 km / hour distance traveled by freddy = 300 km time taken by freddy = 3 hours average speed of freddy = 300 / 3 = 100 km / hour ratio of average speed of eddy to freddy = 200 / 100 = 2 / 1 answer a" | a = 600 / 3
b = 300 / 3
c = a / b
|
a ) 1 : 2 , b ) 1 : 7 , c ) 3 : 5 , d ) 1 : 5 , e ) 1 : 1 | c | divide(multiply(3, 100), 500) | a motorcyclist goes from nagpur to pune , a distance of 500 kms at an average of 100 kmph speed . another man starts from nagpur by car 3 Γ’ Β½ hours after the first , and reaches pune Γ’ Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | t = 500 / 100 = 10 h t = 10 - 4 = 6 time ratio = 10 : 6 = 5 : 3 speed ratio = 3 : 5 answer : c | a = 3 * 100
b = a / 500
|
a ) 178 , b ) 182 , c ) 186 , d ) 190 , e ) 194 | e | add(multiply(divide(subtract(287, 8), const_3), const_2), 8) | if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 287 pounds . what is jake β s present weight , in pounds ? | "j + k = 287 and so k = 287 - j j - 8 = 2 k j - 8 = 2 ( 287 - j ) 3 j = 582 j = 194 the answer is e ." | a = 287 - 8
b = a / 3
c = b * 2
d = c + 8
|
a ) q = 5 , b ) q = 9 , c ) q = 10 , d ) q = 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | what is the greatest positive integer q such that 3 ^ q is a factor of 9 ^ 10 ? | "what is the greatest positive integer q such that 3 ^ q is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a = 9 - 10
b = a * 10
|
a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum s of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum s = average * total count = 200 * 101 = 20,200 answer is b" | a = 301 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 0.391 days , b ) 0.491 days , c ) 0.331 days , d ) 1.391 days , e ) 0.301 days | a | divide(multiply(4, 15), divide(subtract(multiply(4, 15), multiply(add(divide(multiply(4, 15), 4), divide(multiply(4, 15), 15)), 24)), 24)) | a can do a piece of work in 4 days . b can do it in 15 days . with the assistance of c they completed the work in 24 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 24 - 1 / 15 = 0.391 days answer : a" | a = 4 * 15
b = 4 * 15
c = 4 * 15
d = c / 4
e = 4 * 15
f = e / 15
g = d + f
h = g * 24
i = b - h
j = i / 24
k = a / j
|
a ) 111 , b ) 1100 , c ) 121 , d ) 100 , e ) none of these | c | divide(10, 11) | find 10 / 11 = 110 / ? | "answer let 10 / 11 = 110 / n , then , 10 n = 11 x 110 β΄ n = ( 11 x 110 ) / 10 = 121 . option : c" | a = 10 / 11
|
a ) 1 , b ) 46 , c ) 20 , d ) 26 , e ) 30 | b | multiply(divide(46, 46), 46) | in a dairy farm , 46 cows eat 46 bags of husk in 46 days . in how many days one cow will eat one bag of husk ? | "explanation : one bag of husk = 46 cows per day β 46 Γ 1 Γ 46 = 1 Γ 46 Γ x for one cow = 46 days answer : b" | a = 46 / 46
b = a * 46
|
a ) 0.045388 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) none of these | a | divide(divide(multiply(3.242, 14), 100), const_10) | [ ( 3.242 x 14 ) / 100 ] = ? | "answer multiplying 3.242 x 14 = 4.5388 now divide 4.5388 by 100 so , 4.5388 Γ· 100 = 0.045388 β΄ shift the decimal two places to the left as 100 correct option : a" | a = 3 * 242
b = a / 100
c = b / 10
|
a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) none | c | multiply(add(154, 6), 5) | i chose a number and divide it by 5 . then i subtracted 154 from the result and got 6 . what was the number i chose ? | "solution : let xx be the number i chose , then x / 5 β 154 = 6 x / 5 = 160 x = 800 answer c" | a = 154 + 6
b = a * 5
|
a ) 6 , b ) 8 , c ) 5 , d ) 7 , e ) 9 | d | divide(subtract(divide(80, 2), divide(45, 2)), const_2) | a man rows his boat 80 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 80 / ( 2 1 / 2 ) = 32 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 32 - 18 ) / 2 = 7 kmph answer : d" | a = 80 / 2
b = 45 / 2
c = a - b
d = c / 2
|
a ) 367.25 , b ) 306.07 , c ) 306.04 , d ) 306.03 , e ) 306.01 | a | subtract(multiply(6000, power(add(const_1, divide(4, const_100)), divide(const_3, const_2))), 6000) | find out the c . i on rs . 6000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | a = 6000 ( 51 / 50 ) 3 = 6367.248 6000 - - - - - - - - - - - 367.25 answer : a | a = 4 / 100
b = 1 + a
c = 3 / 2
d = b ** c
e = 6000 * d
f = e - 6000
|
a ) 3 / 26 , b ) 1 / 26 , c ) 4 / 26 , d ) 6 / 26 , e ) 10 / 26 | b | divide(subtract(52, multiply(const_4, const_4)), 52) | a card is drawn from a pack of 52 cards . the probability of getting a queen club or a king of heart ? | "n ( s ) = 52 let e = event of getting a queen of club or a king of heart . n ( e ) = 2 p ( e ) = n ( e ) / n ( s ) = 2 / 52 = 1 / 26 option : b" | a = 4 * 4
b = 52 - a
c = b / 52
|
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