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 ) 20 , b ) 30 , c ) 40 , d ) 24 , e ) 60 | d | divide(400, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 400 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 400 * 3 / 50 = 24 sec . answer : option d" | a = 63 - 3
b = a * const_0_2778
c = 400 / b
|
a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 1997), add(const_100, 1997)), subtract(add(const_100, 1997), subtract(const_100, 11))) | a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996 ? | "let c = revenue from car sales in 1996 let t = revenue from truck sales in 1996 equating the total revenue in 1997 with the individual revenue from selling cars and trucks in 1997 , c ( 0.89 ) + t ( 1.07 ) = ( c + t ) ( 1.01 ) dividing throughout by t , ( c / t ) ( 0.89 ) + 1.07 = ( c / t + 1 ) ( 1.01 ) = > ( c / t ) ( 0.12 ) = 0.06 = > ( c / t ) = 1 / 2 option ( a ) ." | a = 100 + 1997
b = 100 + 1997
c = a - b
d = 100 + 1997
e = 100 - 11
f = d - e
g = c / f
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a ) s . 120 , b ) s . 130 , c ) s . 140 , d ) s . 150 , e ) s . 300 | e | divide(150, multiply(divide(5, const_100), 10)) | a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 150 more . what was the sum ? | at 5 % more rate , the increase in s . i for 10 years = rs . 150 ( given ) so , at 5 % more rate , the increase in si for 1 year = 150 / 10 = rs . 15 / - i . e . rs . 15 is 5 % of the invested sum so , 1 % of the invested sum = 15 / 5 therefore , the invested sum = 15 Γ 100 / 5 = rs . 300 answer : e | a = 5 / 100
b = a * 10
c = 150 / b
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a ) 15 women , b ) 85 women , c ) 75 women , d ) 65 women , e ) 35 women | a | multiply(multiply(divide(const_1, divide(divide(const_1, 4), 15)), subtract(const_1, multiply(multiply(6, 2), divide(divide(const_1, 4), 12)))), divide(const_1, 3)) | 12 men can complete a piece of work in 4 days , while 15 women can complete the same work in 4 days . 6 men start working on the job and after working for 2 days , all of them stopped working . how many women should be put on the job to complete the remaining work , if it so to be completed in 3 days ? | "1 man ' s 1 day work = 1 / 48 ; 1 woman ' s 1 day work = 1 / 60 . 6 men ' s 2 day ' s work = 6 / 48 * 2 = 1 / 4 . remaining work = ( 1 - 1 / 4 ) = 3 / 4 now , 1 / 60 work is done in 1 day by 1 woman . so , 3 / 4 work will be done in 3 days by ( 60 * 3 / 4 * 1 / 3 ) = 15 women . answer : a" | a = 1 / 4
b = a / 15
c = 1 / b
d = 6 * 2
e = 1 / 4
f = e / 12
g = d * f
h = 1 - g
i = c * h
j = 1 / 3
k = i * j
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a ) 11 , b ) 15 , c ) 22 , d ) 18 , e ) 25 | c | divide(add(10, 10), divide(55, const_60)) | in a stream running at 2 kmph , a motor boat goes 10 km upstream and back again to the starting point in 55 minutes . find the speed of motor boat in still water ? | "let the speed of motor boat instill water be x kmph then , speed in downstream = ( x + 2 ) km and . speed in upstream = ( x - 2 ) kmph time taken to row 10 km & back = ( 10 / x + 2,10 / x - 2 ) 10 / x + 2 + 10 / x - 2 = 55 / 60 11 x 2 - 240 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 or x = - 2 / 11 then x = 22 kmph answer is c ." | a = 10 + 10
b = 55 / const_60
c = a / b
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a ) 19 kmph , b ) 18 kmph , c ) 72 kmph , d ) 17 kmph , e ) 91 kmph | b | divide(divide(100, const_1000), divide(20, const_3600)) | a train 100 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 100 / 20 s = 5 m / sec speed = 5 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 18 kmph answer : b" | a = 100 / 1000
b = 20 / 3600
c = a / b
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a ) rs . 100.2 , b ) rs . 50.2 , c ) rs . 20.5 , d ) rs . 100 , e ) none of these | e | subtract(multiply(add(divide(add(4, divide(1, 4)), const_100), 1), divide(5400, add(divide(add(divide(1, 4), 1), const_100), 1))), 5400) | if the sales tax be reduced from 4 ( 1 / 5 ) % to 3 ( 1 / 4 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 5400 ? | "explanation : required difference = ( 4 ( 1 / 5 ) of rs . 5400 ) - ( 3 ( 1 / 4 ) of rs . 5400 ) = ( 21 / 5 β 13 / 4 ) % of rs . 5400 = ( 19 / 20 ) x ( 1 / 100 ) x 5400 = rs . 51.30 answer e" | a = 1 / 4
b = 4 + a
c = b / 100
d = c + 1
e = 1 / 4
f = e + 1
g = f / 100
h = g + 1
i = 5400 / h
j = d * i
k = j - 5400
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a ) 35 kg , b ) 37 kg , c ) 42 kg , d ) 38 kg , e ) 29.78 kg | d | divide(add(multiply(30, 40), multiply(20, 35)), add(30, 20)) | there are 2 sections a and b in a class , consisting of 30 and 20 students respectively . if the average weight of section a is 40 kg and that of section b is 35 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 30 * 40 + 20 * 35 = 1900 average weight of the class is = 1900 / 50 = 38 kg answer is d" | a = 30 * 40
b = 20 * 35
c = a + b
d = 30 + 20
e = c / d
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a ) 3.5 , b ) 4.5 , c ) 5 , d ) 9 , e ) 6 | a | divide(14, subtract(6, 2)) | a person can swim in still water at 6 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 14 km ? | "m = 6 s = 2 us = 6 - 2 = 4 d = 42 t = 14 / 4 = 3.5 answer : a" | a = 6 - 2
b = 14 / a
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a ) 50 % , b ) 60 % , c ) 70 % , d ) 80 % , e ) 90 % | c | subtract(const_100, multiply(divide(add(12, const_100), add(60, const_100)), const_100)) | two numbers are respectively 12 % and 60 % more than a third number . the percentage that is first of the second is ? | "i ii iii 112 160 100 160 - - - - - - - - - - 112 100 - - - - - - - - - - - ? = > 70 % answer : c" | a = 12 + 100
b = 60 + 100
c = a / b
d = c * 100
e = 100 - d
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a ) 36 , b ) 72 , c ) 132 , d ) 144 , e ) 180 | b | multiply(subtract(divide(multiply(const_2, 40), subtract(80, multiply(const_2, 40))), divide(40, subtract(80, 40))), const_60) | buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of 40 miles per hour . one hour later , bus b begins to travel in a straight line in the exact opposite direction at a constant rate of 80 miles per hour . if both buses travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ? | "1 st part : - in 1 hr , bus a covers 30 miles . relative speed of bus abus b is ( 80 - 30 ) = 50 mph . so time required for bus b to cover the exact distance as a is 50 * t = 30 t = 3 / 5 = 36 min 2 nd part 80 * t = 2 d - b has to cover twice the distance 30 * ( t + 1 ) = d - a traveled 1 hr more and has to travel only only d so d / 30 - 2 d / 80 = 1 d = 120 t = 3 hrs = 180 min question asks for + ve difference between part 1 and part 2 in minutes = 180 - 36 = 72 min b" | a = 2 * 40
b = 2 * 40
c = 80 - b
d = a / c
e = 80 - 40
f = 40 / e
g = d - f
h = g * const_60
|
a ) 53 , b ) 50 , c ) 99 , d ) 288 , e ) 12 | a | divide(divide(subtract(125, multiply(multiply(8, const_0_2778), 8)), 8), const_0_2778) | a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 8 ) km / hr . x - 8 = 45 = = > x = 53 km / hr . answer : a | a = 8 * const_0_2778
b = a * 8
c = 125 - b
d = c / 8
e = d / const_0_2778
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a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 32 | d | add(power(divide(const_1, const_2), 5), power(divide(const_1, const_2), 5)) | when tossed , a certain coin has an equal probability of landing on either side . if the coin is tossed 5 times , what is the probability that it will land on the same side each time ? | "on the first toss , the coin will land on one side or the other . on the second toss , the probability of landing on the same side is 1 / 2 . on the third toss , the probability of landing on the same side is 1 / 2 . on the fourth toss , the probability of landing on the same side is 1 / 2 . on the fifth toss , the probability of landing on the same side is 1 / 2 . p ( same side all five times ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 . the answer is d ." | a = 1 / 2
b = a ** 5
c = 1 / 2
d = c ** 5
e = b + d
|
a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | the closest approximation of d ( 69.28 Γ 0.004 ) / 0.03 is | "d ( 69.28 Γ 0.004 ) / 0.03 1 . 0.004 = 4 Γ 10 ^ ( - 3 ) 2 . 0.03 = 3 Γ 10 ^ ( - 2 ) 3 . ( a Γ b ) / c = a Γ ( b / c ) 4 . 0.004 / 0.03 = 4 Γ 10 ^ ( - 3 ) / ( 3 Γ 10 ^ ( - 2 ) ) = 4 Γ 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 Γ 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) Γ 10 ^ ( - 1 ) = 1.333 Γ 10 ^ ( - 1 ) therefore , ( 69.28 Γ 0.004 ) / 0.03 = 69.28 Γ ( 0.004 / 0.03 ) = 69.28 Γ 1.33 Γ 10 ^ ( - 1 ) = 69.28 Γ 1.33 / 10 = 6.928 * 1.33 now , 7 Γ 2 = 14 7 Γ 1 = 7 or better : 6.9 Γ 1 = 6.9 6.9 Γ 2 = 13.8 hence , 6.9 < 6.928 Γ 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c" | a = 0 / 4
b = a * 69
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a ) 5 m / s , b ) 8 m / s , c ) 9 m / s , d ) 4 m / s , e ) 2 m / s | b | divide(200, 25) | in a 1000 m race , a beats b by 200 meters or 25 seconds . find the speed of b ? | b 8 m / s since a beats b by 200 m or 25 seconds , i t implies that b covers 200 m in 25 seconds . hence speed of b = 200 / 25 = 8 m / s . | a = 200 / 25
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a ) 10 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 50 % | c | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 50), add(const_100, 50))), multiply(const_100, const_100)), const_100) | robert ' s salary was decreased by 50 % and subsequently increased by 50 % . how much percentage does he lose ? | "let original salary be $ 100 new final salary = 150 % of ( 50 % of $ 100 ) = ( 150 / 100 ) * ( 50 / 100 ) * 100 = $ 75 decrease = 25 % correct option is c" | a = 100 * 100
b = 100 - 50
c = 100 + 50
d = b * c
e = a - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) 35 , b ) 15 , c ) 40 , d ) 52 , e ) 14 | c | multiply(10, 4) | the average age of 10 members of a committee is the same as it was 4 years ago , because an old member has been replaced by a young member . find how much younger is the new member ? | c let the sum of nine member ( total ) = x and the age of old one = z so its average 4 yrs before = ( x + z ) / 10 . after 4 yrs let z be replaced by y . so now avg = ( x + 4 * 10 + y ) / 10 now ( x + z ) / 10 = ( x + 40 + y ) / 10 so after solving it found z = y + 40 . so old person is 40 yrs older than young one . | a = 10 * 4
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a ) rs . 179.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none | a | divide(subtract(multiply(157, add(add(1, 1), 2)), add(126, 126)), 2) | tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 157 per kg , the price of the third variety per kg will be | solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 157 / 22.50 = 1 = Γ’ β¬ ΒΊ x - 157 = 22.50 = Γ’ β¬ ΒΊ x = 179.50 . hence , price of the third variety = rs . 179.50 per kg . answer a | a = 1 + 1
b = a + 2
c = 157 * b
d = 126 + 126
e = c - d
f = e / 2
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a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | b | divide(multiply(75, 600), 50) | if 75 percent of 600 is 50 percent of x , then x = ? | "0.75 * 600 = 0.5 * x x = 7.5 / 5 * 600 = 900" | a = 75 * 600
b = a / 50
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a ) 400 , b ) 400 , c ) 445 , d ) 465 , e ) none of the above | b | subtract(multiply(divide(add(42, 42), subtract(42, 35)), 35), add(subtract(subtract(42, divide(add(42, 42), subtract(42, 35))), divide(add(42, 42), subtract(42, 35))), const_2)) | there were 35 students in a hostel . due to the admission of 7 new students , the expenses of mess were increased by 42 per day while the average expenditure per head diminished by 1 . what was the original expenditure of the mess ? | let the original average expenditure be ` x . then , 42 ( x β 1 ) β 35 x = 42 β 7 x = 84 β x = 12 . β΄ original expenditure = ( 35 Γ 12 ) = 420 . answer b | a = 42 + 42
b = 42 - 35
c = a / b
d = c * 35
e = 42 + 42
f = 42 - 35
g = e / f
h = 42 - g
i = 42 + 42
j = 42 - 35
k = i / j
l = h - k
m = l + 2
n = d - m
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a ) 32 km , b ) 25 km , c ) 28 km , d ) 24 km , e ) 20 km | d | divide(multiply(multiply(subtract(10, 2), add(10, const_2)), 5), add(add(10, const_2), subtract(10, 2))) | a person can row at 10 kmph and still water . he takes 5 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 2 kmph ? | let the distance between a and b be x km . total time = x / ( 10 + 2 ) + x / ( 10 - 2 ) = 5 = > x / 12 + x / 8 = 5 = > ( 2 x + 3 x ) / 24 = 5 = > x = 24 km . answer : d | a = 10 - 2
b = 10 + 2
c = a * b
d = c * 5
e = 10 + 2
f = 10 - 2
g = e + f
h = d / g
|
a ) $ 22 , b ) $ 23.94 , c ) $ 36 , d ) $ 25.20 , e ) $ 30 | c | divide(multiply(divide(945, 25), subtract(const_100, 5)), const_100) | a tour group of 25 people paid a total of $ 945 for entrance to a museum . if this price included a 5 % sales tax , and all the tickets cost the same amount , what was the face value of each ticket price without the sales tax ? choices | soln : - 945 / 25 = x + 0.05 x 945 / 25 = 1.05 x x = 36 answer : c | a = 945 / 25
b = 100 - 5
c = a * b
d = c / 100
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a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 171.6 , e ) $ 73.1025 | e | floor(multiply(9, 8.55)) | carrie likes to buy t - shirts at the local clothing store . they cost $ 8.55 each . one day , she bought 9 t - shirts . how much money did she spend ? | $ 8.55 * 9 = $ 73.1025 . answer is e . | a = 9 * 8
b = math.floor(a)
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a ) 55.5 , b ) 61.5 , c ) 53.5 , d ) 78.5 , e ) 60.5 | e | multiply(divide(divide(multiply(10, add(10, const_1)), const_2), 10), 11) | what is the average of the first 10 multiples of 11 ? | "( 11 + 22 + . . . + 110 ) / 10 = 11 ( 1 + 2 + . . . + 10 ) / 10 = 11 ( 10 ) ( 11 ) / ( 10 * 2 ) = 121 / 2 = 60.5 option e" | a = 10 + 1
b = 10 * a
c = b / 2
d = c / 10
e = d * 11
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['a ) 38 and 19 .', 'b ) 28 and 19 .', 'c ) 18 and 19 .', 'd ) 18 and 29 .', 'e ) none'] | c | divide(add(37, const_1), const_2) | the difference between the squares of two consecutive numbers is 37 . find the numbers . | solution : if the difference between the squares of two consecutive numbers is x , then the numbers are ( x - 1 ) / 2 and ( x + 1 ) / 2 ( 37 - 1 ) / 2 and ( 37 + 1 ) / 2 36 / 2 and 38 / 2 therefore the required answer = 18 and 19 . answer c | a = 37 + 1
b = a / 2
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a ) 50 % , b ) 67 % , c ) 88 % , d ) 94 % , e ) 98 % | d | multiply(subtract(1, power(divide(divide(10, const_2), 10), const_2)), const_100) | a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun four times , approximately what is the probability that the product of the four winning sectors β integers will be even ? | "the only way to have an odd product is if all 4 integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 p ( even product ) = 1 - 1 / 16 = 15 / 16 which is about 94 % the answer is d ." | a = 10 / 2
b = a / 10
c = b ** 2
d = 1 - c
e = d * 100
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a ) 436 , b ) 478 , c ) 512 , d ) 546 , e ) 620 | d | divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 120 % of 13 / 24 of 840 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 35 = 546 the answer is d ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 120 * n
p = o / 100
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a ) 45 , b ) 65 , c ) 77 , d ) 89 , e ) 90 | c | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(const_3.0, subtract(multiply(80, const_4), multiply(84, 5))))) | the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 5 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ? | "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 5 = 73 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 β 316 = 4 a = e + 4 = 73 + 4 = 77 c" | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 5
e = c - d
f = 3 + 0
g = b - f
h = a - g
|
a ) 20 , b ) 24 , c ) 36 , d ) 48 , e ) 96 | b | lcm(8, 12) | at golds gym class can be divided into 8 teams with an equal number of players on each team or into 12 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | we are given that golds gym class can be divided into 8 teams or 12 teams , with an equal number of players on each team . translating this into two mathematical expressions we can say , where g is the total number of students in the gym class , that : g / 8 = integer and g / 12 = integer this means that g is a multiple of both 8 and 12 . we are asked to determine the lowest number of students in the class , or the lowest value for variable β g β . because we know that g is a multiple of 8 and of 12 , we need to find the least common multiple of 8 and 12 . although there are technical ways for determining the least common multiple , the easiest method is to analyze the multiples of 8 and 12 until we find one in common . starting with 8 , we have : 8 , 16 , 24 , 32 for 12 , we have : 12 , 24 for the multiples of 12 , we stopped at 24 , because we see that 24 is also a multiple of 8 . thus , 24 is the least common multiple of 8 and 12 , and therefore we know that the lowest possible number of students in the gym class is 24 . answer b . | a = math.lcm(8, 12)
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a ) 5 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | d | divide(multiply(6, subtract(6, const_3)), const_2) | how many internal diagonals does a hexagon ( 6 sided polygon ) have ? | number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 6 no . of diagonals = 6 ( 6 - 3 ) / 2 = 9 ans d | a = 6 - 3
b = 6 * a
c = b / 2
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a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | e | divide(multiply(14.70, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 14.70 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 14.70 when c . p = ( 100 * 14.70 ) / 21 = 70 answer : e" | a = 14 * 70
b = 25 + 100
c = 30 + 100
d = 100 - 20
e = c * d
f = e / 100
g = b - f
h = a / g
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a ) 1000 , b ) 2777 , c ) 2889 , d ) 1400 , e ) 2771 | d | divide(1568, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1568 ? | "1568 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1400 . answer : d" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1568 / f
|
a ) 10 , b ) 25 , c ) 50 , d ) 100 , e ) 200 | c | subtract(divide(add(100, 0), const_2), divide(add(200, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 200 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 100 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 200 is 200 / 2 = 100 the average of the even numbers from 0 to 100 is 100 / 2 = 50 the answer is c ." | a = 100 + 0
b = a / 2
c = 200 + 0
d = c / 2
e = b - d
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a ) 83 , b ) 63 , c ) 53 , d ) 33 , e ) 13 | d | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 22) | by selling 22 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 22 120 % - - - ? 80 / 120 * 22 = 33 answer : d" | a = 100 + 20
b = 100 - 20
c = 1 / b
d = a * c
e = 1 / d
f = e * 22
|
a ) 1 , b ) 2 , c ) - 1 , d ) - 3 , e ) - 4 | c | divide(add(17, 21), add(19, 19)) | solve the equation for x : 19 ( x + y ) + 17 = 19 ( - x + y ) - 21 | "c - 1 19 x + 19 y + 17 = - 19 x + 19 y - 21 38 x = - 38 = > x = - 1" | a = 17 + 21
b = 19 + 19
c = a / b
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a ) 23 sec , b ) 30 sec , c ) 27 sec , d ) 36 sec , e ) 42 sec | e | divide(700, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 700 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 700 * 3 / 50 = 42 sec . answer : e" | a = 63 - 3
b = a * const_0_2778
c = 700 / b
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) none of them | b | subtract(divide(divide(divide(518, 7), const_2), const_pi), divide(divide(divide(352, 7), const_2), const_pi)) | two concentric circles form a ring . the inner and outer circumferences of ring are ( 352 / 7 ) m and ( 518 / 7 ) m respectively . find the width of the ring . | "let the inner and outer radii be r and r meters . then 2 ( 22 / 7 ) r = ( 352 / 7 ) = r = ( ( 352 / 7 ) x ( 7 / 22 ) x ( 1 / 2 ) ) = 8 m . 2 ( 22 / 7 ) r = ( 528 / 7 ) = r = ( ( 528 / 7 ) x ( 7 / 22 ) x ( 1 / 2 ) ) = 12 m . therefore , width of the ring = ( r - r ) = ( 12 - 8 ) m = 4 m . answer is b ." | a = 518 / 7
b = a / 2
c = b / math.pi
d = 352 / 7
e = d / 2
f = e / math.pi
g = c - f
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a ) 7 , b ) 5 , c ) 2 , d ) 3 , e ) 1 | b | subtract(100, 25) | a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 25 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ? | "200 men do the rest of the work in 100 - 25 = 75 days 100 men can do the rest of the work in 75 * 200 / 100 = 75 days required number of days = 75 - 80 = 5 days answer is b" | a = 100 - 25
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a ) rs . 2928 , b ) rs . 5928 , c ) rs . 3928 , d ) rs . r 4928 , e ) rs . 6928 | d | add(subtract(4000, divide(multiply(4000, 20), const_100)), divide(multiply(subtract(4000, divide(multiply(4000, 20), const_100)), 20), const_100)) | the initial price of an article is rs . 4000 which increases 40 % increse in its price in the first year , a 20 % decrease in the second year and a 20 % increase in the next year . what is the final price of the article ? | "the initial price of the article is rs . 4000 . in the 1 st year , price of the article = 4000 + 1600 = rs . 5600 . in the 2 nd year , price = 5600 - 20 % of 5600 = 5600 - 1120 = rs . 4480 . in the 3 rd year , price = 4480 + 10 % of 4480 = 4480 + 448 = rs . 4928 required price = = rs . 4928 . answer : d" | a = 4000 * 20
b = a / 100
c = 4000 - b
d = 4000 * 20
e = d / 100
f = 4000 - e
g = f * 20
h = g / 100
i = c + h
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a ) 19 / 60 , b ) 9 / 60 , c ) 2 / 3 , d ) 5 / 18 , e ) 6 / 24 | a | divide(10, add(multiply(5, divide(10, 3)), multiply(6, divide(10, 4)))) | if 3 men or 4 women can reap a field in 10 days how long will 5 men and 6 women take to reap it ? | explanation : 3 men reap 1 / 10 field in 1 day 1 man reap 1 / ( 3 x 10 ) 4 women reap 1 / 10 field in 1 day 1 woman reap 1 / ( 10 x 4 ) 5 men and 6 women reap ( 5 / ( 3 x 10 ) + 6 / ( 4 x 10 ) ) = 19 / 60 in 1 day 5 men and 6 women will reap the field in 19 / 60 days answer : option a | a = 10 / 3
b = 5 * a
c = 10 / 4
d = 6 * c
e = b + d
f = 10 / e
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a ) 40 minutes . , b ) 1 hour and 30 minutes . , c ) 1 hour . , d ) 1 hour and 40 minutes . , e ) 2 hours and 15 minutes . | c | divide(const_60, divide(divide(150, add(divide(25, const_60), divide(25, 20))), divide(const_3, const_2))) | qames can eat 25 marshmallows is 20 minutes . dylan can eat 25 in one hour . in how much time will the two eat 150 marshmallows ? | rate = output / time qames rate = 25 / 20 = 5 / 4 dylan rate = 25 / 60 = 5 / 12 combined rate = 5 / 4 + 5 / 12 = 20 / 12 combinedrate * combinedtime = combinedoutput 20 / 12 * t = 150 t = 90 mins = > 1 hr 30 min | a = 25 / const_60
b = 25 / 20
c = a + b
d = 150 / c
e = 3 / 2
f = d / e
g = const_60 / f
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a ) 16 days , b ) 24 days , c ) 36 days , d ) 48 days , e ) 58 days | b | multiply(inverse(subtract(subtract(const_1, multiply(5, inverse(12))), multiply(subtract(7, 5), inverse(16)))), subtract(13, const_2)) | ajay and balu together can do a piece of work in 12 days . balu and charan together can do the same work in 16 days . after ajay has been working at it for 5 days and balu for 7 days , charan finishes it in 13 days . in how many days will charan alone be able to do the work ? | let the total work be 48 units . ajay and balu β s one day work = 48 / 12 = 4 units . balu and charan β s one day work = 48 / 16 = 3 units . ajay β s one day work = 4 - balu β s one day work . balu β s one day work = 3 - charan β s one day work . therefore , ajay β s one day work = 1 + charan β s one day work . also given , 5 x ajay β s one day work + 7 x balu β s one day work + 13 x charan β s one day work = 48 equating all the equations formed , we get charan β s one day work is 2 units . hence , charan can finish the whole work in 48 / 2 = 24 days . answer : b | a = 1/(12)
b = 5 * a
c = 1 - b
d = 7 - 5
e = 1/(16)
f = d * e
g = c - f
h = 1/(g)
i = 13 - 2
j = h * i
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a ) 57 , b ) 57.5 , c ) 58.2 , d ) 59 , e ) 55 | e | divide(add(multiply(1, 40), multiply(3, 60)), add(1, 3)) | city a to city b , john drove for 1 hour at 40 mph and for 3 hours at 60 mph . what was the average speed for the whole trip ? | the total distance is 1 Γ 40 + 3 Γ 60 = 220 and the total time is 4 hours . hence , average speed = ( total distance / total time ) = 220 / 4 = 55 answer : e | a = 1 * 40
b = 3 * 60
c = a + b
d = 1 + 3
e = c / d
|
a ) 720 , b ) 830 , c ) 940 , d ) 1050 , e ) 1160 | a | factorial(6) | if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 } is used exactly once , in how many ways can the digits be arranged ? | 6 ! = 720 the answer is a . | a = math.factorial(6)
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a ) 30 , b ) 33.6 , c ) 18 , d ) 20 , e ) none | c | divide(add(add(add(12, const_1), add(add(12, const_1), const_2)), add(subtract(24, 12), subtract(24, const_2))), 12) | find the average of all prime numbers between 12 and 24 . | "sol . there are five prime numbers between 12 and 24 . they are 13 , 17 , 19 , 23 Γ’ Λ Β΄ required average = [ 13 + 17 + 19 + 23 / 4 ] = 72 / 4 = 18 answer c" | a = 12 + 1
b = 12 + 1
c = b + 2
d = a + c
e = 24 - 12
f = 24 - 2
g = e + f
h = d + g
i = h / 12
|
a ) 1 / 19 , b ) 3 / 19 , c ) 1 / 10 , d ) 1 / 20 , e ) 3 / 10 | d | divide(subtract(20, 19), 20) | a β s speed is 20 / 19 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "let d be the full distance . let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 20 v / 19 ) = xd / v ( 19 / 20 ) * d / v = x * d / v x = 19 / 20 b should have a head start of 1 / 20 of the full distance . the answer is d ." | a = 20 - 19
b = a / 20
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a ) 2 , b ) 8 , c ) 9 , d ) 14 , e ) 24 | d | divide(504, add(multiply(8, const_3), multiply(6, const_2))) | sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 504 per week . how much does she earn in dollars per hour ? | explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 504 . hourly wage = 504 / 36 = 14 answer : d | a = 8 * 3
b = 6 * 2
c = a + b
d = 504 / c
|
a ) 5 , b ) 3 , c ) 1 , d ) 7 , e ) 9 | c | subtract(2, 1) | if 1 = 6 , 2 = 12 , 3 = 18 , 4 = 24 , 5 = 30 , then 6 = ? | solution : 1 as stated 1 = 6 = > 6 = 1 answer c | a = 2 - 1
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a ) 2000 , b ) 4000 , c ) 5000 , d ) 6000 , e ) none | d | divide(16500, add(add(divide(const_3, const_2), const_1), divide(const_1, 4))) | 3 partners a , b , c start a business . twice a β s capital is equal to thrice b β s capital and b β s capital is 4 times c β s capital . out of a total profit of rs . 16500 at the end of the year , b β share is : | sol . let c = x . then , b = 4 x and 2 a = 3 * 4 x = 12 x or a = 6 x . β΄ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so , b β s capital = rs . [ 16500 * 4 / 11 ] = rs . 6000 . answer d | a = 3 / 2
b = a + 1
c = 1 / 4
d = b + c
e = 16500 / d
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a ) 47.5 % , b ) 44 % , c ) 50 % , d ) 35 % , e ) 38 % | c | subtract(multiply(add(const_100, 20), divide(add(const_100, 25), const_100)), const_100) | a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 25 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.25 = 150 increase in 1993 from 1991 = 150 - 100 = 50 answer : c | a = 100 + 20
b = 100 + 25
c = b / 100
d = a * c
e = d - 100
|
a ) 5 , b ) 8 , c ) 3 , d ) 2 , e ) 12 | c | inverse(add(inverse(divide(12, const_4)), inverse(12))) | a is three times as fast as b . if b alone can do a piece of work in 12 days , in what time can a and b together complete the work ? | "a can do the work in 12 / 3 i . e . , 4 days . a and b ' s one day ' s work = 1 / 4 + 1 / 12 = ( 3 + 1 ) / 12 = 1 / 3 so a and b together can do the work in 3 days . answer : c" | a = 12 / 4
b = 1/(a)
c = 1/(12)
d = b + c
e = 1/(d)
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a ) 8.39 , b ) 6.39 , c ) 7.39 , d ) 9.39 , e ) 2.39 | c | multiply(divide(const_1, multiply(add(const_100, 15), divide(const_1, subtract(const_100, 15)))), 10) | by selling 10 pencils for a rupee a man loses 15 % . how many for a rupee should he sell in order to gain 15 % ? | 85 % - - - 10 115 % - - - ? 85 / 115 * 10 = 7.39 answer : c | a = 100 + 15
b = 100 - 15
c = 1 / b
d = a * c
e = 1 / d
f = e * 10
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a ) 18 % , b ) 13 % , c ) 60 % , d ) 40 % , e ) 15 % | c | multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 100)), const_100), const_100), const_1), const_100) | a trader bought a car at 20 % discount on its original price . he sold it at a 100 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 80 s = 80 * ( 200 / 100 ) = 160 100 - 160 = 60 % answer : c" | a = 100 - 20
b = 100 + 100
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
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a ) 5 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 4.8 | a | divide(multiply(7, 40), const_100) | a can complete a certain job in 7 days . b is 40 % more efficient than a . in how many days can b complete the same job ? | "let , total work unit = 70 units a can finish in 7 days = 70 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 40 / 100 ) * 10 = 14 unit work days in which b will complete the work alone = 70 / 14 = 5 days answer : option a" | a = 7 * 40
b = a / 100
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['a ) 5 Β½ ft', 'b ) 7 Β½ ft', 'c ) 8 Β½ ft', 'd ) 9 Β½ ft', 'e ) 10 ft'] | b | subtract(divide(12, const_2), divide(const_1, const_2)) | the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least Β½ foot when they are inside the traffic lane , what should be the limit t on the height of vehicles that are allowed to use the tunnel ? | let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer t is 8 - . 5 = 7.5 . . . choose ( b ) ! from a strategic guessing point of view , as soon as we realize that the height of the tunnel is 10 in the middle , we should quickly eliminate ( d ) and ( e ) as too big ; worse case you have a 1 / 3 shot at picking up the points . b | a = 12 / 2
b = 1 / 2
c = a - b
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a ) 100 , b ) 120 , c ) 140 , d ) 150 miles , e ) 160 | d | add(multiply(subtract(70, 5), 2), multiply(2, 10)) | if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 70 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day ? | let v and t be initial parameter vt = d from the stem ( t + 1 ) ( v + 5 ) = 70 + d = > v + 5 t = 65 - - - - - - - 1 ) we need the additional distance traveled second time at v + 10 and t + 2 ( t + 2 ) ( v + 10 ) = vt + 2 v + 10 t + 20 but vt = d hence d + 2 ( v + 5 t ) + 20 but v + 5 t = 65 hence the additional distance traveled is 2 ( v + 5 t ) + 20 = 2 * 65 + 20 = 130 + 20 = 150 d | a = 70 - 5
b = a * 2
c = 2 * 10
d = b + c
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a ) 0.005 % , b ) 0.02 % , c ) 0.5 % , d ) 5 % , e ) 20 % | c | multiply(divide(multiply(0.01, 20), 40), const_100) | a glass was filled with 40 ounces of water , and 0.01 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ? | "we are given that 0.01 ounces of water evaporated each day . furthermore , we know that this process happened over a 20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply 0.01 by 20 . this gives us : 0.01 x 20 = 0.2 ounces finally , we are asked for β what percent β of the original amount of water evaporated during this period . to determine this percentage , we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated / original amount ) x 100 % ( 0.2 / 40 ) x 100 % = 0.5 % answer c" | a = 0 * 1
b = a / 40
c = b * 100
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a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,400 , d ) $ 1,600 , e ) $ 2,000 | d | subtract(const_1000, 800) | a family pays $ 800 per year for an insurance plan that pays 60 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ? | assuming the medical expenses are $ 1000 or more , the family pays $ 800 + $ 400 = $ 1200 . the total amount paid by insurance plan for the first $ 1000 of expenses is $ 600 . the insurance will pay another $ 600 when the medical expenses are $ 1600 . the answer is d . | a = 1000 - 800
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a ) 60 , b ) 30 , c ) 300 , d ) 600 , e ) none of the above | d | divide(multiply(6, const_60), 0.6) | mike and his friend are going on a road trip and hoping to miss traffic . if they are driving at a rate of 6 miles per minute , what is their speed in kilometers per hour ? [ 1 km = 0.6 miles ] | to calculate the equivalent of kilometres in a mile 6 miles = 6 * ( 0.6 ) ^ - 1 = 10 kilometres in 1 hour there are 60 minutes speed in kmph = 10 * 60 = 600 correct answer - d | a = 6 * const_60
b = a / 0
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a ) $ 240,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 540,000 | a | divide(const_3600, const_10) | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 30,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 30,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 42 years , the investment will get doubled 42 / 14 = 3 times . so the amount invested will get doubled thrice . so , 30000 * 2 ^ 3 = 240000 hence , the answer is a ." | a = 3600 / 10
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a ) 62 kg , b ) 60 kg , c ) 70 kg , d ) 72 kg , e ) none of these | b | subtract(120, multiply(10, 6)) | the average weight of 10 students decreases by 6 kg when one of them weighing 120 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 6 kg = > ( 120 - x ) / 10 = 6 = > x = 60 answer : b" | a = 10 * 6
b = 120 - a
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['a ) 90', 'b ) 100', 'c ) 120', 'd ) 130', 'e ) 140'] | c | multiply(multiply(4, 3), const_10) | the length of a rectangular landscape is 4 times its breadth . there is a playground in it whose area is 1200 square mtr & which is 1 / 3 rd of the total landscape . what is the length of the landscape ? | sol . x * 4 x = 3 * 1200 x = 30 length = 4 * 30 = 120 c | a = 4 * 3
b = a * 10
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a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | b | add(divide(subtract(add(40, 2), 30), 2), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 2 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "james worked for 41 hours hence he earned 40 * x + 1 * 2 x = 42 x dollars ; we know that harry also earned the same 42 x dollars , out of which he earned 30 x dollars for thefirst 30 hoursplus 12 x additional dollars . since for each additional hour he gets 2 x dollars then he worked for 12 x / 2 x = 6 additional hours , so harry worked for total of 30 + 6 = 36 hours . answer : b ." | a = 40 + 2
b = a - 30
c = b / 2
d = c + 30
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a ) 20 , b ) 30 , c ) 60 , d ) 70 , e ) 10 | b | divide(multiply(98, multiply(3, 5)), add(add(multiply(3, 5), multiply(2, 5)), multiply(8, 3))) | the sum of 3 numbers is 98 . the ratio of the first to the second is 2 / 3 , and the ratio of the second to the third is 5 / 8 . the second number is : | b 30 let the three numbers be x , y and z . sum of the numbers is 98 . x + y + z = 98 β¦ β¦ β¦ β¦ β¦ β¦ ( i ) the ratio of the first to the second is 2 / 3 . x / y = 2 / 3 . x = 2 / 3 Γ y . x = 2 y / 3 . the ratio of the second to the third is 5 / 8 . y / z = 5 / 8 . z / y = 8 / 5 . z = 8 / 5 Γ y . z = 8 y / 5 . put the value of x = 2 y / 3 and z = 8 y / 5 in ( i ) . 2 y / 3 + y + 8 y / 5 = 98 49 y / 15 = 98 . 49 y = 98 Γ 15 . 49 y = 1470 . y = 1470 / 49 . y = 30 . therefore , the second number is 30 . | a = 3 * 5
b = 98 * a
c = 3 * 5
d = 2 * 5
e = c + d
f = 8 * 3
g = e + f
h = b / g
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a ) 140 % , b ) 120 % , c ) 100 % , d ) 90 % , e ) 80 % | a | subtract(multiply(const_2, add(const_100, 20)), const_100) | a trader sold an article on a certain price with 20 % profit . if he sold double of previous selling price then find its profit % | let cost price = 100 % selling price = 120 % new s . p . = 240 % p % = 240 - 100 = 140 % answer is a | a = 100 + 20
b = 2 * a
c = b - 100
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a ) 1 : 4 , b ) 3 : 2 , c ) 2 : 3 , d ) 2 : 5 , e ) none of these | b | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 6000 and rs . 4000 respectively at the end of year the ratio of their investments are ? | "profit = investment * time so , 6000 = i 1 * 1 = 6000 4000 = i 2 * 1 = 4000 i 1 / i 2 = 6000 / 4000 = 3 / 2 answer : b" | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
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a ) 15 , b ) 17 , c ) 19 , d ) 18 , e ) 16 | c | subtract(57, divide(multiply(57, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 57 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 57 / 3 = 19 kmph . answer : c" | a = 57 * 2
b = a / 3
c = 57 - b
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a ) 15 % , b ) 25 % , c ) 65 % , d ) 53.8 % , e ) 35 % | d | subtract(multiply(divide(const_100, 650), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 650 grams per kg , what is his percent ? | "explanation : 650 - - - 350 100 - - - ? = > 53.8 % answer : d" | a = 100 / 650
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
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a ) 33 , b ) 88 , c ) 48 , d ) 77 , e ) 27 | c | multiply(multiply(const_pi, 4), 12) | the slant height of a cone is 12 cm and radius of the base is 4 cm , find the curved surface of the cone . | "Ο * 12 * 4 = 48 answer : c" | a = math.pi * 4
b = a * 12
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a ) 13 , b ) 9 , c ) 16 , d ) 17 , e ) 18 | a | divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 40), 4) | the sum of ages of 4 children born 2 years different each is 40 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) = 40 4 x + 12 = 40 = > 4 x = 28 x = 7 x + 6 = 7 + 6 = 13 answer : a" | a = 2 + 4
b = 2 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 40
g = f / 4
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a ) 60 , b ) 55 , c ) 50 , d ) 45 , e ) 40 | e | multiply(divide(200, 100), divide(40, const_2)) | the ratio , by volume , of bleach to detergent to water in a certain solution is 2 : 40 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 200 liters of water , how many liters of detergent will it contain ? | "b : d : w = 2 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 40 ) = ( 1 / 60 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 200 dnew = wnew / 5 = 200 / 5 = 40 so , answer will be e" | a = 200 / 100
b = 40 / 2
c = a * b
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a ) rs . 258 , b ) rs . 358 , c ) rs . 458 , d ) rs . 558 , e ) none of these | d | multiply(divide(75, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 75 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) Γ h ] + ( l Γ b ) = [ 2 ( 25 + 12 ) Γ 6 ] + ( 25 Γ 12 ) = 744 sq m cost of plastering = 744 Γ ( 75 / 100 ) = rs . 558 answer : d" | a = 75 / 100
b = 25 * 12
c = 25 * 6
d = 2 * c
e = 12 * 6
f = e * 2
g = d + f
h = b + g
i = a * h
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a ) 12 , b ) 18 , c ) 20 , d ) 42 , e ) 30 | d | divide(multiply(140, add(multiply(4, const_2), 4)), 40) | it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 140 pages ? | "if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 140 / 50 = 42 minutes to print 140 pages ( answer d )" | a = 4 * 2
b = a + 4
c = 140 * b
d = c / 40
|
a ) 2 / 3 , b ) 3 / 5 , c ) 7 / 10 , d ) 9 / 20 , e ) 11 / 20 | e | subtract(1, add(divide(1, 4), divide(1, 5))) | there is a total of 100 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 5 . what is the probability that the marble will be either red or blue ? | "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 20 / 20 - 5 / 20 - 4 / 20 = 11 / 20 the answer is e ." | a = 1 / 4
b = 1 / 5
c = a + b
d = 1 - c
|
a ) $ 9 , b ) $ 3 , c ) $ 7 , d ) $ 6 , e ) $ 5 | c | add(divide(subtract(11, 3), const_2), 3) | you and your friend spent a total of $ 11 for lunch . your friend spent $ 3 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 1 ( l ) + ( l + 3 ) = 11 l + l + 3 - 3 = 11 - 3 2 l = 8 l = 4 my friends lunch l + 3 = 4 + 3 = $ 7 , the answer is c" | a = 11 - 3
b = a / 2
c = b + 3
|
a ) $ 11,000 , b ) $ 50,000 , c ) $ 16,500 , d ) $ 6,600 , e ) $ 3,600 | c | divide(reminder(multiply(add(3, const_2), divide(multiply(2200, 3), const_2)), const_1000), const_10) | 3 business people wish to invest in a new company . each person is willing to pay one third of the total investment . after careful calculations , they realize that each of them would pay $ 2200 less if they could find two more equal investors . how much is the total investment in the new business ? | total investment between 5 : ( x / 5 ) total investment including 2200 less between 3 people ( x - ( 2200 * 3 ) ) / 3 set both eq . equal to each other . 16,500 answer c ) | a = 3 + 2
b = 2200 * 3
c = b / 2
d = a * c
e = reminder / (
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 7 | b | divide(add(21, 51), add(1, 7)) | haresh went to buy fruits . she bought 1 apple , 3 bananas , 5 oranges and paid 21 rupees . another person bought 7 apple , 5 bananas , 3 oranges and paid 51 rupees . if i want to buy 1 apple , 1 banana , 1 orange then how much i have to pay ? | there can be multiple possible rates of apple , banana and orange to satisfy given conditions , but cost of 1 apple , 1 banana , 1 orange will be rs 9 . 1 x + 3 y + 5 z = 21 . . . . . . . . ( 1 ) 7 x + 5 y + 3 z = 51 . . . . . . . . ( 2 ) adding ( 1 ) and ( 2 ) , we get 8 x + 8 y + 8 z = 72 8 * ( x + y + z ) = 72 x + y + z = 9 answer : b | a = 21 + 51
b = 1 + 7
c = a / b
|
a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 30 | a | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 60))), 30)) | mahesh can do a piece of work in 60 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 60 days = 20 * 1 / 60 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 2 / 3 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 3 / 2 = 45 days answer is a" | a = 1 / 60
b = 20 * a
c = 1 - b
d = c / 30
e = 1 / d
|
a ) 22 , b ) 33 , c ) 36 , d ) 44 , e ) 51 | b | subtract(add(multiply(7, divide(add(add(const_2, const_10), const_10), 7)), multiply(const_2, 7)), 3) | a land owner needs to fence his semicircular land leaving an opening of length of 3 m for access . the radius of the semicircular plot is 7 m . how long would be the fence in meters ? | perimeter of the semicircle = Γ― β¬ xd / 2 + d = Γ― β¬ x 7 + 14 = 22 + 14 = 36 m length of the fence = 36 - 3 = 33 m answer : b | a = 2 + 10
b = a + 10
c = b / 7
d = 7 * c
e = 2 * 7
f = d + e
g = f - 3
|
a ) 222 km , b ) 333 km , c ) 555 km , d ) 444 km , e ) 666 km | d | add(multiply(divide(60, subtract(21, 16)), 16), multiply(divide(60, subtract(21, 16)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively . when they meet , it is found that one train has travelled 60 km more than the other one . the distance between the two stations is ? | d 444 km 1 h - - - - - 5 ? - - - - - - 60 12 h rs = 16 + 21 = 37 t = 12 d = 37 * 12 = 444 | a = 21 - 16
b = 60 / a
c = b * 16
d = 21 - 16
e = 60 / d
f = e * 21
g = c + f
|
a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000 | d | multiply(const_1000, divide(1, 10000)) | a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 10000 that a person who randomly guesses the answer to each question will be a winner ? | "a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . 5 b . 10 c . 50 d . 100 e . 1000 soln : ans is b probability that one question is answered right is 1 / 2 . now for minimum number of questions needed to take probability less than 1 / 1000 is = > ( 1 / 2 ) ^ n < 1 / 10000 n = 100 satisfies this . d" | a = 1 / 10000
b = 1000 * a
|
a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | if a = 105 and a ^ 3 = 21 Γ 25 Γ 45 Γ q , what is the value of q ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 Γ 25 Γ 45 Γ q = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x q = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x q = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x q q = 7 ^ 2 = 49 answer d" | a = 105 ** 3
b = 21 * 25
c = b * 45
d = a / c
|
a ) 20 , b ) 23 , c ) 22 , d ) 21 , e ) 26 | b | add(add(add(5, 5), 5), 5) | what number should replace the question mark ? 5 , 24 , 11 , 20 , 17 , 16 , - - - ? | "answer : b 5 , 24 , 11 , 20 , 17 , 16 , 23 ? there are two alternate sequences : + 6 and - 4 ." | a = 5 + 5
b = a + 5
c = b + 5
|
a ) 1 , b ) 19 , c ) 21 , d ) 30 , e ) 33 | c | subtract(subtract(subtract(multiply(72, 4), 90), subtract(90, const_1)), subtract(90, const_2)) | the average ( arithmetic mean ) of 4 different integers is 72 . if the largest integer is 90 , what is the least possible value of the smallest integer ? | total of integers = 72 * 4 = 288 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 288 - ( 90 + 89 + 88 ) = 288 - 267 = 21 answer : c | a = 72 * 4
b = a - 90
c = 90 - 1
d = b - c
e = 90 - 2
f = d - e
|
a ) 9 , b ) 8 , c ) 11 , d ) 8.5 , e ) 6 | c | divide(66, 6) | stacy has a 66 page history paper due in 6 days . how many pages per day would she have to write to finish on time ? | 66 / 6 = 11 answer : c | a = 66 / 6
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(const_10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent w ? | c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . w = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
|
a ) $ 79.25 , b ) $ 79.50 , c ) $ 79.75 , d ) $ 80.00 , e ) $ 80.25 | e | add(75, multiply(divide(7, const_100), 75)) | diana took out a charge account at the general store and agreed to pay 7 % simple annual interest . if she charges $ 75 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | 1.07 * $ 75 = $ 80.25 the answer is e . | a = 7 / 100
b = a * 75
c = 75 + b
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(add(subtract(9, const_0.5), add(9, 3)), const_2) | a rower whose speed is 9 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 3 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? | "time upstream = d / 6 time downstream = d / 12 total time = d / 6 + d / 12 = d / 4 average speed = 2 d / ( d / 4 ) = 8 km / hr the answer is b ." | a = 9 - 0
b = 9 + 3
c = a + b
d = c / 2
|
a ) 6 , b ) 12 , c ) 7 , d ) 15 , e ) 5 | a | divide(multiply(10, subtract(const_1, divide(add(const_2, const_3), multiply(const_2, const_4)))), divide(add(const_2, const_3), multiply(const_2, const_4))) | a man completes of a job in 10 days . at this rate , how many more days will it takes him to finish the job ? | work done = 5 / 8 balance work = ( 1 - 5 / 8 ) = 3 / 8 let the required number of days be x . then , ( 5 / 8 ) : ( 3 / 8 ) = : : 10 : x ( 5 / 8 ) * x = ( 3 / 8 ) * 10 x = ( 3 / 8 ) * 10 * ( 8 / 5 ) x = 6 . answer is a . | a = 2 + 3
b = 2 * 4
c = a / b
d = 1 - c
e = 10 * d
f = 2 + 3
g = 2 * 4
h = f / g
i = e / h
|
a ) 42 , b ) 48 , c ) 56 , d ) 208 , e ) 256 | b | subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 22), subtract(const_100, 74)), 800), const_100), divide(multiply(subtract(const_100, 74), 800), const_100)) | in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 22 percent are juniors and 74 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ? | "sophomores = 26 % juniors = 22 % seniors = 160 / 800 = 20 % total : 68 % 1 - 68 % = 32 % ( freshmen ) 0,32 * 800 - 0,26 * 800 = 256 - 208 = 48 . answer choice b" | a = 160 / 800
b = a * 100
c = 100 - b
d = c - 22
e = 100 - 74
f = d - e
g = f * 800
h = g / 100
i = 100 - 74
j = i * 800
k = j / 100
l = h - k
|
a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 450 | b | divide(add(80, 100), divide(60, const_100)) | a student needs 60 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 100 marks , find the maximum marks set for the test . | "60 % = 180 marks 1 % = 3 marks 100 % = 300 marks the answer is b ." | a = 80 + 100
b = 60 / 100
c = a / b
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | add(subtract(multiply(const_4, 2), multiply(multiply(const_4, 5), 0.1)), 2) | a football player scores 2 goals in his fifth match thus increasing his average goals score by 0.1 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the answer is not correct , then you would have a gauge for whether you should gohigherorlowerwith your next test . here , i ' ll start with answer c = 8 goals if . . . . total goals = 8 goals 5 th game = 2 goals 1 st 4 games = 6 goals avg . for 1 st 4 games = 6 / 4 = 1.5 goal / game avg . for all 5 games = 8 / 5 = 1.6 goals / game this is an exact match for what we ' re told in the prompt , so answer c must be the answer ." | a = 4 * 2
b = 4 * 5
c = b * 0
d = a - c
e = d + 2
|
a ) 8 kg . , b ) 10.8 kg . , c ) 7 kg . , d ) 18.0 kg , e ) none | c | divide(multiply(6, 14), 12) | if the weight of 12 meters long rod is 14 kg . what is the weight of 6 meters long rod ? | "answer β΅ weight of 12 m long rod = 14 kg β΄ weight of 1 m long rod = 14 / 12 kg β΄ weight of 6 m long rod = 14 x 6 / 12 = 7 kg option : c" | a = 6 * 14
b = a / 12
|
a ) 13 , 3 , b ) 12 , 6 , c ) 15 , 3 , d ) 14 , 4 , e ) 17 , 1 | e | divide(divide(add(18, 16), const_2), const_2) | a man can row downstream at 18 kmph and upstream at 16 kmph . find the speed of the man in still water and the speed of stream respectively ? | "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 16 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 34 = > x = 17 , y = 1 . answer : option e" | a = 18 + 16
b = a / 2
c = b / 2
|
a ) 0.4 , b ) 0.55 , c ) 0.45 , d ) 0.48 , e ) 0.58 | c | divide(subtract(add(add(180, divide(180, divide(const_100, 20))), 45), 180), 180) | after giving a discount of rs . 45 the shopkeeper still gets a profit of 20 % , if the cost price is rs . 180 . find the markup % ? | cost price = 180 s . p = 180 * 120 / 100 = 216 disc = 45 so . . . mark price = 216 + 45 = 261 . . . . . . mark up % = 261 - 180 / 180 = 81 / 180 = . 45 or 45 % answer : c | a = 100 / 20
b = 180 / a
c = 180 + b
d = c + 45
e = d - 180
f = e / 180
|
a ) 15 , b ) 25 , c ) 30 , d ) 42 , e ) 13 | e | add(subtract(divide(multiply(multiply(3, 2), 4), 3), divide(multiply(multiply(3, 2), 4), 2)), divide(multiply(multiply(3, 2), 4), 4)) | if x , y , and z are positive integers and 3 x = 2 y = 4 z , then the least possible value of x + y + z is | "given 3 x = 2 y = 4 z x + y + z in terms of x = x + ( 3 x / 2 ) + ( 3 x / 4 ) = 26 x / 8 = 13 x / 4 now checking with each of the answers and see which value gives a minimum integer value . a x = 4 / 13 * 15 , not an integer b , c , e can be ruled out similarly . d is minimum value as x = 13 * 4 / 13 = 4 answer is e" | a = 3 * 2
b = a * 4
c = b / 3
d = 3 * 2
e = d * 4
f = e / 2
g = c - f
h = 3 * 2
i = h * 4
j = i / 4
k = g + j
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(multiply(divide(subtract(18, 12), const_60), 80), divide(70, const_10)) | a train travelling 80 kmph to pass through a tunnel of 70 km at 5 : 12 am . the train leaves the tunnel at 5 : 18 am . find the length of train ? | distance = speed * time let x is length of train 5 : 18 - 5 : 12 = 6 minutes = 6 / 60 hr = 1 / 10 hr now d = ( x + 7 ) = 80 kmph * 1 / 10 x = 1 km answer : a | a = 18 - 12
b = a / const_60
c = b * 80
d = 70 / 10
e = c - d
|
a ) 810 , b ) 811 , c ) 830 , d ) 850 , e ) 851 | e | add(multiply(divide(add(30, 50), const_2), add(subtract(50, 30), const_1)), add(divide(subtract(50, 30), const_2), const_1)) | if x is equal to the sum of the integers from 30 to 50 , inclusive , and y is the number of even integers from 30 to 50 , inclusive , what is the value of x + y ? | since the integers from 30 - 50 are consecutive we can use the ` ` median x # of terms formula ' ' to determine x . there are 21 terms and the median is 40 . this gives a total of 21 x 40 = 840 . there are 11 even integers from 30 - 50 = y . therefore , x + y = 851 . answer : e | a = 30 + 50
b = a / 2
c = 50 - 30
d = c + 1
e = b * d
f = 50 - 30
g = f / 2
h = g + 1
i = e + h
|
a ) 32 , b ) 34 , c ) 36 , d ) 38 , e ) 40 | e | multiply(100, divide(2, 5)) | of the 100 people in a room , 2 / 5 are women . if 1 / 4 of the people are married , what is the maximum number of women in the room who could be unmarried ? | women = 2 / 5 * 100 = 40 married = 1 / 4 * 100 = 25 unmarried = 75 max ( un - married women ) = 40 answer e | a = 2 / 5
b = 100 * a
|
a ) 960 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,694 | a | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | yesterday ' s closing prices of 1,760 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ? | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 1760 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 960 . a is my answer ." | a = 20 * 100
b = a - 10
c = b - 10
d = 20 / 100
e = 1 + d
f = e + 1
g = c / f
h = 20 / 100
i = 1 + h
j = g * i
|
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