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 ) 2 / 125 , b ) 4 / 25 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 5 | b | divide(subtract(divide(multiply(800, 1), 5), 32), 800) | in a group of 800 people , 1 / 5 play at least one instrument , 32 play two or more . what is the probability that one student play exactly one instrument ? | "p ( playing 2 or more instruments ) = 32 / 800 = 1 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 1 / 5 - 1 / 25 = 4 / 25 . answer b ." | a = 800 * 1
b = a / 5
c = b - 32
d = c / 800
|
a ) 60 , b ) 70 , c ) 65 , d ) 75 , e ) 80 | d | divide(subtract(subtract(80, multiply(divide(10, const_100), 95)), multiply(divide(20, const_100), 90)), subtract(const_1, add(divide(10, const_100), divide(20, const_100)))) | teacher took exam for english , average for the entire class was 80 marks . if we say that 10 % of the students scored 95 marks and 20 % scored 90 marks then calcualte average marks of the remaining students of the class . | lets assume that total number of students in class is 100 and required average be x . then from the given statement we can calculate : ( 10 * 95 ) + ( 20 * 90 ) + ( 70 * x ) = ( 100 * 80 ) 70 x = 8000 - ( 950 + 1800 ) = 5250 answer : d x = 75 . | a = 10 / 100
b = a * 95
c = 80 - b
d = 20 / 100
e = d * 90
f = c - e
g = 10 / 100
h = 20 / 100
i = g + h
j = 1 - i
k = f / j
|
a ) a ) 4000 , b ) b ) 2900 , c ) c ) 2000 , d ) d ) 2393 , e ) e ) 2009 | a | multiply(multiply(subtract(4, 3), 2000), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 2000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 2000 = > x = 2000 . b ' s share = rs . 2 x = 2 * 2000 = rs . 4000 . answer : a" | a = 4 - 3
b = a * 2000
c = b * 3
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a ) 484 , b ) 460 , c ) 550 , d ) 664 , e ) none of them | a | multiply(divide(const_2, add(const_3, const_2)), 1210) | a and b together have rs . 1210 . if of a ' s amount is equal to of b ' s amount , how much amount does b have ? | 4 / 15 a = 2 / 5 b a = ( 2 / 5 x 15 / 4 ) b a = 3 / 2 b a / b = 3 / 2 a : b = 3 : 2 . therefore , b ' s share = rs . ( 1210 x 2 / 5 ) = rs . 484 . answer is a . | a = 3 + 2
b = 2 / a
c = b * 1210
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | d | multiply(10, multiply(10, divide(4, multiply(4, 4)))) | 4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 10 weavers in 10 days ? | "1 weaver can weave 1 mat in 4 days . 10 weavers can weave 10 mats in 4 days . 10 weavers can weave 25 mats in 10 days . the answer is d ." | a = 4 * 4
b = 4 / a
c = 10 * b
d = 10 * c
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a ) 1.44 % , b ) 2.02 % , c ) 1.04 % , d ) 2.25 % , e ) 3.40 % | d | divide(multiply(15, 15), const_100) | if a trader sold two cars each at rs . 404415 and gains 15 % on the first and loses 15 % on the second , then his profit or loss percent on the whole is ? | "explanation : sp of each car is rs . 404415 , he gains 15 % on first car and losses 15 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 15 ) ( 15 ) / 100 % = 2.25 % answer is d" | a = 15 * 15
b = a / 100
|
a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2 | d | subtract(10, 5) | robert ate 10 chocolates , nickel ate 5 chocolates . how many more chocolates did robert ate than nickel ? | 10 - 5 = 5 . answer is d | a = 10 - 5
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a ) 31 , b ) 21 , c ) 41 , d ) 61 , e ) 51 | b | subtract(power(9, 2), multiply(2, 30)) | if a + b = β 9 , and a = 30 / b , what is the value of a ^ 2 + b ^ 2 ? | "a ^ 2 + b ^ 2 should make you think of these formulas : ( a + b ) ( a + b ) = a ^ 2 + b ^ 2 + 2 ab we already know ( a + b ) = - 9 and a * b = 30 ( a + b ) ( a + b ) = ( - 9 ) ( - 9 ) = a ^ 2 + b ^ 2 + 2 * ( 30 ) a ^ 2 + b ^ 2 = 81 - 60 = 21 answer : b" | a = 9 ** 2
b = 2 * 30
c = a - b
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(choose(subtract(6, 1), 1), choose(6, 2)) | a basket contains 6 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "the total number of ways to choose 2 apples is 6 c 2 = 15 the number of ways that include the spoiled apple is 5 c 1 = 5 p ( the spoiled apple is included ) = 5 / 15 = 1 / 3 the answer is a ." | a = 6 - 1
b = math.comb(a, 1)
c = math.comb(6, 2)
d = b / c
|
a ) 104 , b ) 60 , c ) 52 , d ) 50 , e ) 102 | e | add(subtract(60, multiply(power(2, 2), 2)), multiply(2, power(5, 2))) | if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 60 , what is the value of f ( 5 ) ? | "f ( x ) = 2 x ^ 2 + y f ( 2 ) = 60 = > 2 * ( 2 ) ^ 2 + y = 60 = > 8 + y = 60 = > y = 52 f ( 5 ) = 2 * ( 5 ) ^ 2 + 52 = 102 answer e" | a = 2 ** 2
b = a * 2
c = 60 - b
d = 5 ** 2
e = 2 * d
f = c + e
|
a ) s . 608 , b ) s . 690 , c ) s . 698 , d ) s . 700 , e ) s . 760 | a | subtract(815, divide(multiply(subtract(884, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 884 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 884 - 815 ) = rs . 69 . s . i . for 3 years = rs . ( 69 x 3 ) = rs . 207 . principal = rs . ( 815 - 207 ) = rs . 608 . answer : option a" | a = 884 - 815
b = a * 3
c = b / 4
d = 815 - c
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6 | d | multiply(2, 2) | if n divided by 7 has a remainder of 2 , what is the remainder when 2 times n is divided by 7 ? | "as per question = > n = 7 p + 2 for some integer p hence 2 n = > 14 q + 4 = > remainder = > 4 for some integer q alternatively = > n = 2 > 2 n = > 4 = > 4 divided by 7 will leave a remainder 4 hence d" | a = 2 * 2
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a ) 4966 , b ) 6477 , c ) 7391 , d ) 2676 , e ) 1881 | c | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6900), 5) | sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6900 ? | "total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6900 x 6 ) - 34009 ] = rs . ( 41400 - 34009 ) = rs . 7391 answer : c" | a = 6835 + 9927
b = a + 6855
c = b + 7230
d = c + 6562
e = d / 5
f = e - 6900
g = f * 5
|
a ) 600 , b ) 720 , c ) 820 , d ) 800 , e ) 920 | b | divide(multiply(60, 600), 50) | if 60 percent of 600 is 50 percent of x , then x = ? | "0.6 * 600 = 0.5 * x x = 6 / 5 * 600 = 720" | a = 60 * 600
b = a / 50
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 33 % | e | multiply(divide(subtract(89.95, 59.95), 89.95), const_100) | a $ 89.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 89.95 $ discounted selling price of chair = 59.95 $ discount = 89.95 - 59.95 = 30 $ % decrease in price of chair = ( 30 / 89.95 ) * 100 % = 33 % approx answer e" | a = 89 - 95
b = a / 89
c = b * 100
|
a ) 3.5 , b ) 3.75 , c ) 4 , d ) 3.9 , e ) 4.5 | d | divide(390, add(divide(multiply(3, 1000), const_100), divide(multiply(1400, 5), const_100))) | a money lender lent rs . 1000 at 3 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 390 . find the number of years . | ( 1000 xtx 3 / 100 ) + ( 1400 xtx 5 / 100 ) = 390 Γ’ β β t = 3.9 answer d | a = 3 * 1000
b = a / 100
c = 1400 * 5
d = c / 100
e = b + d
f = 390 / e
|
a ) $ 2,250 , b ) $ 2,500 , c ) $ 4,100 , d ) $ 3,250 , e ) $ 4,500 | c | floor(divide(subtract(subtract(multiply(500, 12.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000)) | company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 12.00 each , what was company c β s gross profit ? | "cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 12 = $ 6000 profit : 6000 - 1900 = $ 4100 option c is correct" | a = 500 * 12
b = 5 * 0
c = a - b
d = 500 - 100
e = d * 3
f = c - e
g = f / 1000
h = math.floor(g)
|
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 4 | a | add(divide(subtract(multiply(65, const_4), multiply(50, 4)), subtract(80, 65)), 4) | a car averages 50 mph for the first 4 hours of a trip and averages 80 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ? | let the time for which car averages 80 mph = t 65 * ( t + 4 ) = 50 * 4 + 80 t = > 15 t = 60 = > t = 4 total duration of the trip = 4 + 4 = 8 answer a | a = 65 * 4
b = 50 * 4
c = a - b
d = 80 - 65
e = c / d
f = e + 4
|
a ) rs . 45 , b ) rs . 70 , c ) rs . 39 , d ) rs . 72 , e ) rs . 48 | e | multiply(divide(subtract(divide(1192, const_10), const_100), 4), const_10) | a sum of rs . 1000 at simple interest amounts to rs . 1192 in 4 years . the s . i . for 1 year is : | s . i . for 4 years = rs . ( 1192 - 1000 ) = rs . 192 . s . i . for 1 year = rs . 192 / 4 = rs . 48 . answer : option e | a = 1192 / 10
b = a - 100
c = b / 4
d = c * 10
|
a ) 12.11 , b ) 11.11 , c ) 13.11 , d ) 14.11 , e ) 15.11 | b | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 900), 900), const_100) | a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 900 grams . find the profit percent . | "explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 900 100 + g / 100 + 0 = 1000 / 900 100 + g = 10 / 9 * 100 g = 11.11 answer : b" | a = 4 + 1
b = 4 + 1
c = a + b
d = c * 100
e = d - 900
f = e / 900
g = f * 100
|
a ) rs . 150 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | a | divide(75, 0.5) | if 0.5 % of a = 75 paise , then the value of a is ? | "answer β΅ 0.5 / 100 of a = 75 / 100 β΄ a = rs . ( 75 / 0.5 ) = rs . 150 correct option : a" | a = 75 / 0
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a ) 2 , b ) - 2 , c ) 4 , d ) - 5 , e ) 6 | b | divide(subtract(46, 6), 4) | if | 4 x + 6 | = 46 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 6 = 46 or 4 x + 6 = - 46 = > x = 10 or x = - 12 sum of both the values will be - 12 + 10 = - 2 answer is b" | a = 46 - 6
b = a / 4
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a ) 2 β 2 , b ) 2 β 3 , c ) 3 β 2 , d ) 11 / ( 3 * β 2 ) , e ) β 2 | d | divide(add(sqrt(27), sqrt(192)), sqrt(54)) | ( β 27 + β 192 ) / β 54 = ? | "( β 27 + β 192 ) / β 54 = ( 3 β 3 + 8 β 3 ) / 3 β 3 * 2 = 11 β 3 / 3 β 3 * 2 = 11 / ( 3 * β 2 ) = 11 / ( 3 * β 2 ) . hence , the correct answer is d ." | a = math.sqrt(27)
b = math.sqrt(192)
c = a + b
d = math.sqrt(54)
e = c / d
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a ) 100.5 , b ) 200.5 , c ) 300.5 , d ) 350.5 , e ) 412.5 | e | divide(110, subtract(divide(60, const_100), divide(1, 3))) | 60 % of x is greater than 1 / 3 rd of x by 110 . what is x ? | "6 x / 10 - x / 3 = 110 4 x / 15 = 110 x = 412.5 answer : e" | a = 60 / 100
b = 1 / 3
c = a - b
d = 110 / c
|
a ) 280 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | a | subtract(multiply(divide(multiply(72, const_1000), const_3600), 32), multiply(divide(multiply(72, const_1000), const_3600), 18)) | a train traveling at 72 kmph crosses a platform in 32 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ? | "speed of train = 72 * ( 5 / 18 ) = 20 m / s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 32 = 640 m . so , length of platform = 640 - 360 = 280 m imo , answer a" | a = 72 * 1000
b = a / 3600
c = b * 32
d = 72 * 1000
e = d / 3600
f = e * 18
g = c - f
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a ) s . 1090 , b ) s . 1160 , c ) s . 1190 , d ) s . 1202 , e ) s . 1330 | e | divide(multiply(subtract(const_100, 5), 1400), const_100) | a man buys a cycle for rs . 1400 and sells it at a loss of 5 % . what is the selling price of the cycle ? | "since , c . p = 1400 loss % = ( c . p - s . p ) / c . p * 100 5 = ( 1400 - s . p ) / 1400 * 100 so , after solving answer = 1330 . answer : e" | a = 100 - 5
b = a * 1400
c = b / 100
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a ) 512 , b ) 768 , c ) 64 , d ) 2048 , e ) 4096 | c | subtract(power(2, add(5, const_1)), const_1) | the population of a bacteria colony doubles every day . if it was started 5 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ? | "2 ^ 5 ( 2 ) = 2 ^ 6 = 64 the answer is c" | a = 5 + 1
b = 2 ** a
c = b - 1
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a ) 470 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) none of these | a | multiply(subtract(36, divide(250, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | a goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 36 seconds . what is the length of the goods train ? | "explanation : speed = [ 72 x ( 5 / 18 ) ] m / sec = 20 m / sec . time = 36 sec . let the length of the train be x metres . then , [ ( x + 250 ) / 36 ] = 20 = > x + 250 = 720 = > x = 470 . answer : a" | a = const_0_2778 * 72
b = 250 / a
c = 36 - b
d = const_0_2778 * 72
e = c * d
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a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 9 , e ) 2 / 3 | d | divide(const_2, 9) | if an integer n is to be selected at random from 1 to 900 , inclusive , what is probability n ( n + 1 ) will be divisible by 9 ? | for n ( n + 1 ) to be a multiple of 9 , either n or n + 1 has to be a multiple of 9 . thus n must be of the form 9 k or 9 k - 1 . the probability is 2 / 9 . the answer is d . | a = 2 / 9
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a ) 1500 , b ) 6000 , c ) 2500 , d ) 1400 , e ) 2000 | e | multiply(multiply(divide(100, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 100 of money ? | "p = 100 ( 100 / 10 ) 2 = > p = 2000 answer : e" | a = 10 * 2
b = 100 / a
c = b * 100
d = 10 * 2
e = c * d
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a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 65 | c | subtract(multiply(multiply(45, divide(8, 6)), divide(45, 30)), 45) | 45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 45 meters deep by working for 6 hours ? | "45 workers * 8 hours / 30 meters = x * 6 / 45 x = 90 total workers 90 - 45 = 45 new workers the answer is c ." | a = 8 / 6
b = 45 * a
c = 45 / 30
d = b * c
e = d - 45
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a ) 3 / 7 , b ) 34 , c ) 1 , d ) 2 , e ) 3 | c | add(divide(4, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5)))) | if p / q = 4 / 5 , then the value of 4 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ? | "answer given exp . = 4 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 4 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 4 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 4 / 7 + 6 / 14 = 4 / 7 + 3 / 7 = 7 / 7 = 1 . correct option : c" | a = 4 / 7
b = 4 / 5
c = 2 - b
d = 4 / 5
e = 2 + d
f = c / e
g = a + f
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a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.28 , e ) 1.35 | d | inverse(divide(70, add(70, 20))) | patrick purchased 70 pencils and sold them at a loss equal to the selling price of 20 pencils . the cost of 70 pencils is how many times the selling price of 70 pencils ? | "say the cost price of 70 pencils was $ 70 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 70 - 70 p = 20 p - - > p = 7 / 9 . ( cost price ) / ( selling price ) = 1 / ( 7 / 9 ) = 9 / 7 = 1.28 . answer : d ." | a = 70 + 20
b = 70 / a
c = 1/(b)
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a ) 13 / 14 , b ) 13 / 10 , c ) 13 / 18 , d ) 13 / 16 , e ) 13 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 4), 5), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | 3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 4 men and 5 women will do that task . | 3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 4 m + 5 w = 13 w answer is 13 / 14 answer : a | a = 8 - 2
b = 6 - 3
c = a / b
d = c * 4
e = d + 5
f = 8 - 2
g = 6 - 3
h = f / g
i = 6 * h
j = i + 2
k = e / j
|
a ) 1100 , b ) 2100 , c ) 1300 , d ) 1650 , e ) 1400 | d | divide(110, subtract(divide(40, const_100), divide(1, 3))) | 40 % of x is greater than 1 / 3 rd of x by 110 . what is x ? | "4 x / 10 - x / 3 = 110 1 x / 15 = 110 x = 1650 answer : d" | a = 40 / 100
b = 1 / 3
c = a - b
d = 110 / c
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a ) 48 , b ) 66 , c ) 72 , d ) 80 , e ) 84 | b | subtract(divide(multiply(divide(330, 5), 12), 2), 330) | the ratio of boarders to day students at a school was originally 5 to 12 . however , after a number of new boarders join the initial 330 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ? | let x be the number of new boarders . the ratio changed from 5 : 12 up to 1 : 2 = 6 : 12 . 330 / ( 330 + x ) = 5 / 6 x = 66 the answer is b . | a = 330 / 5
b = a * 12
c = b / 2
d = c - 330
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a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | c | multiply(70, divide(18, 45)) | a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 70 meters , what is the height of the building ( in meters ) ? | "the height : length ratio will be equal in both cases . 18 / 45 = x / 70 x = 28 the answer is c ." | a = 18 / 45
b = 70 * a
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a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | a | add(subtract(33, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 33 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 33 kg joins the group , total weight of boys = 20 x + 33 so 20 x + 33 = 21 ( x + 1 ) = > x = 12 . answer : a" | a = 20 + 1
b = 33 - a
c = b + 1
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a ) 20 , b ) 19 , c ) 21 , d ) 18 , e ) 22 | a | subtract(28, multiply(const_4, const_2)) | presently the ratio between the ages of dan and james is 6 : 5 . after 4 years dan will be 28 . what is the present age of james ? | let the present ages dan and james be 6 x years and 5 x years respectively 6 x + 4 = 28 6 x = 24 x = 4 kim ' s age = 5 x = 20 years answer is a | a = 4 * 2
b = 28 - a
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a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 42 , e ) 25 - 37 | e | divide(subtract(62, 12), const_2) | the sum of two numbers is 62 , and one of them is 12 more than the other . what are the two numbers ? | "in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 62 : word problem = 62 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 62 Γ’ Λ β 12 = 50 . x = 50 / 2 = 25 . this is the first number . therefore the other number is x + 12 = 25 + 12 = 37 . the sum of 25 + 37 is 62 . e" | a = 62 - 12
b = a / 2
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a ) 25 hr , b ) 35 hr , c ) 40 hr , d ) 20 hr , e ) 50 hr | b | multiply(add(add(multiply(const_2, const_2), const_2), const_1), 5) | a tank is filled by 3 pipes a , b , c in 5 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? | "suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / 2 and x / 4 hours respectively to fill the tank . 1 / x + 2 / x + 4 / x = 1 / 5 7 / x = 1 / 5 x = 35 hours answer is b" | a = 2 * 2
b = a + 2
c = b + 1
d = c * 5
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a ) 2 : 1 , b ) 2 : 8 , c ) 2 : 9 , d ) 2 : 5 , e ) 2 : 2 | a | divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13))) | if 12 men and 16 boys can do a piece of work in 5 days ; 13 men and 24 boys can do it in 4 days , then the ratio of the daily work done by a man to that of a boy is ? | "let 1 man ' s 1 day work = x and 1 boy ' s 1 day work = y . then , 12 x + 16 y = 1 / 5 and 13 x + 24 y = 1 / 4 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 required ratio = x : y = 1 / 100 : 1 / 200 = 2 : 1 . answer : a" | a = 4 * 24
b = 5 * 16
c = a - b
d = 5 * 12
e = 4 * 13
f = d - e
g = c / f
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a ) 5 % , b ) 18 % , c ) 33 % , d ) 44 % , e ) 38 % | d | multiply(subtract(multiply(add(divide(20, const_100), const_1), add(const_1, divide(20, const_100))), const_1), const_100) | a certain company β s profit in 1996 was 20 percent greater than its profit in 1995 , and its profit in 1997 was 20 percent greater than its profit in 1996 . the company β s profit in 1997 was what percent greater than its profit in 1995 ? | "profit in 1995 - 100 profit in 1996 - 120 % increae profit in 1997 in comparison to 1995 = 20 + 120 * 20 % = 44 correct option : d" | a = 20 / 100
b = a + 1
c = 20 / 100
d = 1 + c
e = b * d
f = e - 1
g = f * 100
|
a ) 31 % , b ) 35 % , c ) 54 % , d ) 38 % , e ) 80 % | a | multiply(divide(subtract(880, 670), 670), const_100) | a cycle is bought for rs . 670 and sold for rs . 880 , find the gain percent ? | "explanation : 670 - - - - 210 100 - - - - ? = > 31 % answer : a" | a = 880 - 670
b = a / 670
c = b * 100
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a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | a | subtract(subtract(subtract(20, 25), const_4), const_2) | a certain no . when divided by 20 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 20 + 25 = 45 / 15 = 3 ( remainder ) a" | a = 20 - 25
b = a - 4
c = b - 2
|
a ) 603.75 , b ) 555.75 , c ) 569.55 , d ) 256.25 , e ) 563.23 | a | divide(divide(subtract(multiply(7000, power(add(const_1, divide(7, const_100)), 2)), 7000), 2), multiply(6, divide(14, const_100))) | the s . i . on a certain sum of money for 6 years at 14 % per annum is half the c . i . on rs . 7000 for 2 years at 7 % per annum . the sum placed on s . i . is ? | c . i . = [ 7000 * ( 1 + 7 / 100 ) 2 - 7000 ] = ( 7000 * 11 / 10 * 11 / 10 - 7000 ) = rs . 1014.3 . sum = ( 507.15 * 100 ) / ( 6 * 14 ) = rs . 603.75 answer : a | a = 7 / 100
b = 1 + a
c = b ** 2
d = 7000 * c
e = d - 7000
f = e / 2
g = 14 / 100
h = 6 * g
i = f / h
|
a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 20 % | a | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(20, const_100))), divide(multiply(30, const_100), add(20, const_100))), const_100) | selling an kite for rs . 30 , a shop keeper gains 20 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | explanation : marked price = rs . 30 c . p . = 100 / 120 * 30 = rs . 25 sale price = 90 % of rs . 30 = rs . 27 required gain % = 0.2 / 25 * 100 = 8 % . answer : a | a = 30 / 100
b = 100 - 10
c = a * b
d = 30 * 100
e = 20 + 100
f = d / e
g = c - f
h = 30 * 100
i = 20 + 100
j = h / i
k = g / j
l = k * 100
|
a ) rs 8.81 , b ) rs 9.35 , c ) rs 10.35 , d ) rs 12.35 , e ) none of these | d | divide(multiply(10, add(const_100, 5)), subtract(const_100, 15)) | a fruit seller sells mangoes at the rate of rs . 10 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 85 : 10 = 105 : x x = ( 10 Γ 105 / 85 ) = rs 12.35 option d" | a = 100 + 5
b = 10 * a
c = 100 - 15
d = b / c
|
a ) 800 , b ) 6,000 , c ) 8,000 , d ) 12,000 , e ) 80,000 | b | multiply(divide(multiply(divide(multiply(3, 10), 50), power(10, const_4)), const_1000), 3) | a certain galaxy is known to comprise approximately 3 x 10 ^ 11 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | "total no . of stars on galaxy = 3 * 10 ^ 11 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 3 * 10 ^ 11 / 50 * 10 ^ 6 = 30 * 10 ^ 3 / 5 = 6000 therefore answer is b" | a = 3 * 10
b = a / 50
c = 10 ** 4
d = b * c
e = d / 1000
f = e * 3
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | b | divide(multiply(250, 32), subtract(250, 50)) | a hostel had provisions for 250 men for 32 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 32 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 32 250 Γ 32 = 200 x 5 Γ 32 = 4 x x = 5 Γ 8 = 40 answer b" | a = 250 * 32
b = 250 - 50
c = a / b
|
a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16 | e | add(5, 5) | how many different values of positive integer x , for which | x + 5 | < x , are there ? | "answer e i opted to put the random value option . i used 0 , 5 , - 5 and the the extreme of 40 and - 40 . . i was able to solve it in 1 : 09 e" | a = 5 + 5
|
a ) 0.49 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.411 | b | multiply(divide(80, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 80 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time | explanation : probability that a speaks truth is 80 / 100 = 0.8 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) Γ p ( b ) = 0.8 Γ 0.6 = 0.48 answer : b | a = 4 * 5
b = a * 5
c = 80 / b
d = 4 * 5
e = d * 5
f = 60 / e
g = c * f
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | a | divide(divide(27, 3), 27) | tickets numbered from 1 to 27 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , β¦ . , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 ] p ( e ) = n ( e ) / n ( s ) = 9 / 27 = 1 / 3 the answer is a ." | a = 27 / 3
b = a / 27
|
a ) 9,6 , b ) 6,3 , c ) 9,3 , d ) 6,6 , e ) none of these | c | divide(subtract(12, 6), const_2) | a man can row downstream at the rate of 12 km / hr and upstream at 6 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 12 + 6 ) = 9 km / hr rate of current = 1 / 2 ( 12 - 6 ) = 3 km / hr answer : option c" | a = 12 - 6
b = a / 2
|
a ) 21 / 34 , b ) 1 / 12 , c ) 15 / 64 , d ) 17 / 35 , e ) 7 / 6 | d | subtract(divide(subtract(25, const_1), add(34, const_1)), divide(add(7, const_1), subtract(44, const_1))) | if a is an integer greater than 7 but less than 25 and b is an integer greater than 34 but less than 44 , what is the range of a / b ? | "the way to approach this problem is 7 < a < 25 and 34 < b < 41 minimum possible value of a is 8 and maximum is 24 minimum possible value of b is 35 and maximum is 40 range = max a / min b - min a / max b ( highest - lowest ) 24 / 35 - 8 / 40 = 17 / 35 hence d" | a = 25 - 1
b = 34 + 1
c = a / b
d = 7 + 1
e = 44 - 1
f = d / e
g = c - f
|
a ) s . 56.37 , b ) s . 53.22 , c ) s . 56.219 , d ) s . 51.18 , e ) s . 51.11 | a | subtract(add(add(divide(multiply(divide(55, multiply(divide(5, const_100), 2)), 5), const_100), divide(55, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(55, multiply(divide(5, const_100), 2)), 5), const_100), divide(55, multiply(divide(5, const_100), 2))), 5), const_100)), divide(55, multiply(divide(5, const_100), 2))) | if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 55 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 55 * 100 ) / ( 2 * 5 ) = rs . 550 amount = [ 550 * ( 1 + 5 / 100 ) 2 ] = rs . 606.375 c . i . = ( 606.375 - 550 ) = rs . 56.37 answer : a" | a = 5 / 100
b = a * 2
c = 55 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 55 / g
i = e + h
j = 5 / 100
k = j * 2
l = 55 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 55 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 55 / w
y = u - x
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | divide(multiply(const_12, log(2)), log(2)) | if 2 ^ ( 2 w ) = 8 ^ ( w β 4 ) , what is the value of w ? | 2 ^ ( 2 w ) = 8 ^ ( w β 4 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w β 4 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 12 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 12 w = 12 the answer is d . | a = math.log(2)
b = 12 * a
c = math.log(2)
d = b / c
|
a ) 25 , b ) 27 , c ) 15 , d ) 18 , e ) 21 | b | inverse(multiply(divide(inverse(18), add(const_2, const_1)), const_2)) | a is twice as good as b . and together they finish a piece of work in 18 days . in how many days will a alone finish the work | ( a ' s 1 day work ) : ( b ' s 1 day work ) = 2 : 1 a + b 1 day work = 1 / 18 a ' s 1 day work = ( 1 / 18 ) * ( 2 / 3 ) = 1 / 27 a alone can finish the work in 27 days answer is b | a = 1/(18)
b = 2 + 1
c = a / b
d = c * 2
e = 1/(d)
|
a ) 14 , b ) 12 , c ) 11 , d ) 15 , e ) 16 | c | divide(divide(multiply(55, const_2), const_3), const_4) | everyone shakes hands with everyone else in a room . total number of handshakes is 55 . number of persons = ? | "in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 55 or n ( n - 1 ) = 110 or n = 11 answer is ( c )" | a = 55 * 2
b = a / 3
c = b / 4
|
a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | d | divide(circle_area(divide(4, const_2)), const_2) | what will be the fraction of 4 % | "explanation : 4 * 1 / 100 = 1 / 25 . friends i know it is quite simple , but trust me while solving percentage questions in hurry we use to do these types of mistake only . so i recommend you to have a bit practise of this option d" | a = 4 / 2
b = circle_area / (
|
a ) a ) 2 , b ) b ) 3 , c ) c ) 4 , d ) d ) 5 , e ) e ) 6 | c | subtract(670, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | the least number which must be subtracted from 670 to make it exactly divisible by 9 is : | "on dividing 670 by 9 , we get remainder = 4 therefore , required number to be subtracted = 4 answer : c" | a = 4 + 1
b = a * 10
c = 4 + 2
d = b + c
e = d * 9
f = 670 - e
|
a ) 630 , b ) 700 , c ) 535 , d ) 450 , e ) 815 | b | divide(476, divide(subtract(84, subtract(const_100, 84)), const_100)) | in an election a candidate who gets 84 % of the votes is elected by a majority of 476 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 84 ) % of x = 16 % of x 84 % of x - 16 % of x = 476 68 x / 100 = 476 x = 476 * 100 / 68 = 700 answer is b" | a = 100 - 84
b = 84 - a
c = b / 100
d = 476 / c
|
a ) a ) 54 , b ) b ) 75 , c ) c ) 48 , d ) d ) 42 , e ) e ) 63 | d | multiply(divide(49, subtract(add(5, 8), 6)), 6) | 3 numbers are in the ratio 5 : 6 : 8 . the sum of its longest and smallest numbers equals the sum of the third number and 49 . find the third number ? | let the numbers be 5 x , 6 x , 8 x . largest number = 8 x . smallest number = 5 x . third number = 6 x . 8 x + 5 x = 6 x + 49 7 x = 49 = > x = 7 6 x = 42 = > third number is 42 . answer : d | a = 5 + 8
b = a - 6
c = 49 / b
d = c * 6
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(divide(80, 100), const_100)) | the price of an item changed from $ 120 to $ 100 . then later the price decreased again from $ 100 to $ 80 . which of the two decreases was larger in percentage term ? | first decrease in percent part / whole = ( 120 - 100 ) / 120 = 0.17 = 17 % second decrease in percent part / whole = ( 100 - 80 ) / 100 = 0.20 = 20 % the second decrease was larger in percent term . the part were the same in both cases but the whole was smaller in the second decrease . answer b | a = 80 / 100
b = a * 100
c = 100 - b
|
a ) 631 , b ) 731 , c ) 831 , d ) 849 , e ) 901 | b | subtract(subtract(subtract(multiply(83, const_10), const_100), 3), 3) | what is the next number : 3 , 11 , 83 , __ | "3 ^ 0 + 2 = 3 3 ^ 2 + 2 = 11 3 ^ 4 + 2 = 83 3 ^ 6 + 2 = 731 the answer is b ." | a = 83 * 10
b = a - 100
c = b - 3
d = c - 3
|
a ) rs . 22325.58 , b ) rs . 32325.58 , c ) rs . 52325.58 , d ) rs . 62325.58 , e ) none of these | c | divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 14)) | a man sells a car to his friend at 14 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was : | "explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 14 % . now s . p = rs . 45000 and loss = 14 % c . p . rs . [ 100 / 86 Γ£ β 45000 ] = rs . 52325.58 . correct option : c" | a = 54000 * 100
b = 100 + 20
c = a / b
d = c * 100
e = 100 - 14
f = d / e
|
a ) 4.16 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(960, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 960 gm . for a kg . his gain is β¦ % . | "his percentage gain is 100 * 40 / 960 as he is gaining 40 units for his purchase of 960 units . so 4.16 % . answer : a" | a = 4 + 1
b = a * 2
c = b * 100
d = 960 / c
e = 1/(d)
f = e - 1
g = f * 100
|
a ) 20 % , b ) 20.5 % , c ) 20.8 % , d ) 21 % , e ) 21.8 % | d | multiply(const_100, subtract(multiply(add(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), const_1)) | increasing the original price of an article by 10 percent and then increasing the new price by 10 percent is equivalent to increasing the original price by | "1.1 * 1.1 * x = 1.21 * x the answer is d ." | a = 10 / 100
b = 1 + a
c = 10 / 100
d = 1 + c
e = b * d
f = e - 1
g = 100 * f
|
a ) 90 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) 190 m | b | subtract(multiply(1.5, const_1000), multiply(divide(multiply(1.5, const_1000), add(multiply(2, const_60), 30)), add(multiply(1.5, const_60), 20))) | a can run 1.5 km distance in 2 min 20 seconds , while b can run this distance in 2 min 30 sec . by how much distance can a beat b ? | "a takes time 2.20 minutes = 140 sec b takes time 2.30 minutes = 150 sec diffrence = 150 - 140 = 10 sec now we are to find distance covered in 10 sec by b 150 sec = 1500 m 1 sec = 10 m 10 sec = 10 x 10 = 100 m answer : b" | a = 1 * 5
b = 1 * 5
c = 2 * const_60
d = c + 30
e = b / d
f = 1 * 5
g = f + 20
h = e * g
i = a - h
|
a ) rs . 11 , b ) rs . 75 , c ) rs . 17 , d ) rs . 16 , e ) rs . 22 | b | multiply(divide(const_100, 8), 6) | a 6 % stock yields 8 % . the market value of the stock is ? | "for an income of rs . 8 , investment = rs . 100 . for an income of rs 6 , investment = rs . \ inline \ left ( \ frac { 100 } { 8 } \ times 6 \ right ) = rs . 75 market value of rs . 100 stock = rs . 75 . answer : b" | a = 100 / 8
b = a * 6
|
a ) 5 , b ) 7 , c ) 15 , d ) 35 , e ) 45 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 6300 multiplied by y is the square of an integer , then y must be | "6300 = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 * 7 to be perfect square , we need to multiply by at least 7 . the answer is b ." | a = 2 + 3
b = a * 2
|
a ) 60 , b ) 120 , c ) 240 , d ) 275 , e ) 300 | e | multiply(divide(multiply(20, divide(75, divide(subtract(100, 75), subtract(20, 15)))), subtract(20, divide(75, divide(subtract(100, 75), subtract(20, 15))))), divide(subtract(100, 75), subtract(20, 15))) | if the farmer sells 75 of his chickens , his stock of feed will last for 20 more days than planned , but if he buys 100 more chickens , he will run out of feed 15 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | let v denote the volume of feed one chicken consumes per day . then the total volume of feed in stock will be v β d β c where d is the number of days the feed will last if the number of chickens does not change and cc is the current number of chickens . from the question it follows that v ( d + 20 ) ( c β 75 ) = vdc v ( d β 15 ) ( c + 100 ) = vdc the first equation simplifies to 20 c β 75 d = 1500 . the second equation simplifies to ( β 15 ) c + 100 d = 1500 after dividing everything by 5 we get the linear system : 4 c β 15 d = 300 ( β 3 ) c + 20 d = 300 solving it we get c = 300 d = 60 answer : e | a = 100 - 75
b = 20 - 15
c = a / b
d = 75 / c
e = 20 * d
f = 100 - 75
g = 20 - 15
h = f / g
i = 75 / h
j = 20 - i
k = e / j
l = 100 - 75
m = 20 - 15
n = l / m
o = k * n
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(multiply(12, add(const_2, const_3)), const_4) | if the average of t , b , c , 14 and 15 is 12 . what is the average value of t , b , c and 29 | t + b + c + 14 + 15 = 12 * 5 = 60 = > t + b + c = 60 - 29 = 31 t + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d | a = 2 + 3
b = 12 * a
c = b / 4
|
['a ) 0.76', 'b ) 0.86', 'c ) 0.96', 'd ) 1.16', 'e ) 1.26'] | b | subtract(square_area(2), multiply(circle_area(divide(divide(2, const_2), const_2)), const_4)) | a square of 2 unit is given and four circles are made on its corners having equal radius of each . and another circle of equal radius is made on the center of the square . find out the area of square which is not covered by circles ? | let the radius of circle is ' r ' area of the circle which is at corner = ( 22 / 7 * r ^ 2 ) / 4 hence the area of all four circles are = 22 / 7 * r ^ 2 areaof circle which is made on center of suare is = 22 / 7 * r ^ 2 total area of all circles are = 2 * 22 / 7 * r ^ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) if we construct the diagram we get the relation 4 r = 2 sqrt 2 r = 1 / sqrt 2 put the value of r in eqn 1 and we get 2 * 22 / 7 * 1 / 2 = 22 / 7 = 3.14 area of suare = 2 ^ 2 = 4 area of remaining part = 4 - 3.14 = 0.86 answer : b | a = square_area - (
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a ) 24.58 , b ) 14 , c ) 15 , d ) 24.9 , e ) 24.1 | c | multiply(divide(6, subtract(9, 7)), 7) | sachin is younger than rahul by 6 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 6 , so ( x - 6 ) / x = 7 / 9 = > 9 x - 42 = 7 x = > 2 x = 42 = > x = 21 so sachin age is 21 - 6 = 15 answer : c" | a = 9 - 7
b = 6 / a
c = b * 7
|
a ) a ) 11 , b ) b ) 18 , c ) c ) 24 , d ) d ) 25 , e ) d ) 36 | d | add(add(3, divide(subtract(subtract(51, 31), 3), const_2)), 13) | 31 of the scientists that attended a certain workshop were wolf prize laureates , and 13 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 51 of the scientists attended that workshop , how many of them were nobel prize laureates ? | "lets solve by creating equation . . w = 31 . . total = 52 . . not w = 52 - 31 = 21 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 21 or x = 9 . . so who won nobel but not wolf = x + 3 = 12 . . but people who won both w and n = 13 . . so total who won n = 12 + 13 = 25 . . d" | a = 51 - 31
b = a - 3
c = b / 2
d = 3 + c
e = d + 13
|
['a ) 400', 'b ) 1550', 'c ) 1575', 'd ) 1600', 'e ) 1625'] | c | add(rectangle_area(200, 5), rectangle_area(120, 5)) | a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area covered by the two roads ? | area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads = 1000 + 600 - 25 = 1575 answer : option c | a = rectangle_area + (
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a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | if p # q denotes the least common multiple of p and q , then q = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , q = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c ." | a = 18 * 24
b = 12 * 16
c = a - b
d = c / 2
e = d + 24
|
a ) 8 , b ) 2 , c ) 7 , d ) can not be determined , e ) none of these | e | divide(9, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the digits is 9 . what is the difference between the two digits of the number ? | "suppose the two - digit number be 10 x + y . then we have been given l 0 x + y β ( 10 y + x ) = 9 β 9 x β 9 y = 9 β x β y = 1 hence , the required difference = 1 note that if the difference between a two - digit number and the number obtained by interchanging the digits is d , then the difference between the two digits of the number = d β 9 answer e" | a = 10 - 1
b = 9 / a
|
a ) 12 , b ) 4 , c ) 9 , d ) 3 , e ) 1 | c | subtract(negate(16), multiply(subtract(64, 27), divide(subtract(64, 27), subtract(81, 64)))) | 81 , 64 , 27 , 16 . . . | "81 / 3 = 27 64 / 4 = 16 27 / 3 = 9 answer : c" | a = negate - (
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a ) 84 , b ) 89 , c ) 90 , d ) 92 , e ) 95 | c | subtract(subtract(100, const_3), 7) | set x consists of all two - digit primes and set y consists of all positive odd multiples of 7 less than 100 . if the two sets are combined into one , what will be the range of the new set ? | set x = { 11 , 13 , 17 , . . . . . . . . . . . . . , 83 , 89 , 97 } set y = { 7 , 21 , 35 , . . . . . . . . . . . . . . . , 77 , 91 , } combining two sets , say set z set z = { 7 , 11 , 13 , 17,21 , . . . . . . . . . . . . . . . . . . . , 77 , 83 , 89 , 91 , 97 } range = max value - min value range ( z ) = 97 - 7 = 90 oa c is the answer . | a = 100 - 3
b = a - 7
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a ) 39 % , b ) 52 % , c ) 15 % , d ) 21 % , e ) 28 % | a | subtract(divide(multiply(add(const_100, 25), const_100), subtract(const_100, 10)), const_100) | a shopkeeper sold an article offering a discount of 10 % and earned a profit of 25 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be $ 100 . then , s . p . = $ 125 let marked price be $ x . then , 90 / 100 * x = 125 x = 12500 / 90 = $ 139 now , s . p . = $ 139 , c . p . = rs . 100 profit % = 39 % . answer : a" | a = 100 + 25
b = a * 100
c = 100 - 10
d = b / c
e = d - 100
|
a ) 430 , b ) 440 , c ) 450 , d ) 460 , e ) 470 | d | subtract(500, subtract(add(30, 20), 10)) | a certain high school has 500 students . of these students , 30 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ? | "we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 30 y = 20 z = 10 so , we have 30 students taking music , 20 taking art and 10 taking both music and art . 20 student taking just music 10 student taking just art 10 student taking both music and art total = 40 students we ' re asked for the total number of students who are taking neither course . that is 500 - 40 = 460 . d" | a = 30 + 20
b = a - 10
c = 500 - b
|
a ) 130 , b ) 141 , c ) 152 , d ) 163 , e ) 174 | d | add(add(lcm(10, multiply(divide(10, 5), 8)), 3), lcm(10, multiply(divide(10, 5), 8))) | a ranch has both horses and ponies . exactly 3 / 10 of the ponies have horseshoes , and exactly 5 / 8 of the ponies with horseshoes are from iceland . if there are 3 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ? | "3 / 10 * p are ponies with horseshoes , so p is a multiple of 10 . 5 / 8 * 3 / 10 * p = 3 / 16 * p are icelandic ponies with horseshoes , so p is a multiple of 16 . the minimum value of p is 80 . then h = p + 3 = 83 . the minimum number of horses and ponies is 163 . the answer is d ." | a = 10 / 5
b = a * 8
c = math.lcm(10, b)
d = c + 3
e = 10 / 5
f = e * 8
g = math.lcm(10, f)
h = d + g
|
a ) 13 sec . , b ) 15 sec . , c ) 12 sec . , d ) 17 sec . , e ) 19 sec . | c | divide(add(120, 120), add(speed(120, 10), speed(120, 15))) | two bullet trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "c 12 sec . speed of the first bullet train = 120 / 10 m / sec = 12 m / sec . speed of the second bullet train = 120 / 15 m / sec = 8 m / sec . relative speed = ( 12 + 8 ) = 20 m / sec . required time = ( 120 + 120 ) / 20 sec = 12 sec ." | a = 120 + 120
b = speed + (
c = a / b
|
a ) 1122 , b ) 1140 , c ) 1199 , d ) 1188 , e ) 1166 | b | subtract(divide(multiply(760, subtract(22, const_2)), 8), 760) | there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 8 days more ? | 760 - - - - 22 760 - - - - 20 x - - - - - 8 x * 8 = 760 * 20 x = 1900 760 - - - - - - - 1140 answer : b | a = 22 - 2
b = 760 * a
c = b / 8
d = c - 760
|
a ) 4 % increase , b ) 24 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 55)), const_100), const_100) | if the price of a tv is first decreased by 20 % and then increased by 55 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 155 % of ( 80 % of 100 ) = rs . 155 / 100 * 80 / 100 * 100 = rs . 124 . . ' . increase = 24 % answer : b" | a = 100 - 20
b = 100 + 55
c = a * b
d = c / 100
e = d - 100
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | what is the smallest integer c for which 27 ^ c > 3 ^ 24 ? | "27 ^ c > 3 ^ 24 converting into the same bases : 27 ^ c > 27 ^ 8 therefore for the equation to hold true , c > 8 or c = 9 option c" | a = 24 / 3
b = math.floor(a)
c = b + 1
|
a ) 5363 , b ) 6240 , c ) 2368 , d ) 7632 , e ) 7732 | d | multiply(divide(106, 12), 600) | in order to obtain an income of rs . 600 from 12 % stock at rs . 106 , one must make an investment of | "to obtain rs . 10 , investment = rs . 106 . to obtain rs . 600 , investment = = rs . 7632 answer : d" | a = 106 / 12
b = a * 600
|
a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 905 hours , e ) 915 hours | a | add(divide(7200, add(16, 2)), divide(7200, subtract(16, 2))) | speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7200 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7200 / 18 + 7200 / 14 = 400 + 514.2 = 914.2 hours answer : option a" | a = 16 + 2
b = 7200 / a
c = 16 - 2
d = 7200 / c
e = b + d
|
a ) s . 375 , b ) s . 400 , c ) s . 600 , d ) s . 800 , e ) s . 875 | e | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 7000), 3) | a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 7000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 7000 ) = rs . 875 answer = e" | a = 1/(3)
b = 1/(8)
c = 1/(6)
d = b + c
e = a - d
f = e * 7000
g = f * 3
|
a ) 190 , b ) 195 , c ) 200 , d ) 205 , e ) 210 | d | divide(divide(multiply(add(10, 400), add(divide(subtract(400, 10), 10), const_1)), const_2), add(divide(subtract(400, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 400 inclusive ? | "we ' re asked for the average of all of the multiples of 10 from 10 to 400 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 40 ( 10 ) = 400 so we know that there are 40 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 400 = 410 20 + 390 = 410 30 + 380 = 410 etc . since there are 40 total terms , this pattern will create 20 ' pairs ' of 410 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 20 ) ( 410 ) / ( 40 ) = 410 / 2 = 205 final answer : d" | a = 10 + 400
b = 400 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 400 - 10
h = g / 10
i = h + 1
j = f / i
|
a ) 10 , b ) 6.5 , c ) 7 , d ) 7.13 , e ) 10.2 | d | divide(130, multiply(add(60, 4), const_0_2778)) | a train 130 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 4 = 64 km / hr . = 64 * 5 / 18 = 160 / 9 m / sec . time taken to pass the men = 130 * 9 / 160 = 7.13 sec . answer : option d" | a = 60 + 4
b = a * const_0_2778
c = 130 / b
|
a ) 31 % , b ) 33 % , c ) 28 % , d ) 30 % , e ) 32 % | a | divide(subtract(add(add(200, divide(200, divide(const_100, 10))), 80), 200), 200) | after giving a discount of rs . 80 the shopkeeper still gets a profit of 10 % , if the cost price is rs . 200 . find the markup % ? | "cost price = 180 s . p = 200 * 110 / 100 = 220 disc = 80 so . . . mark price = 216 + 80 = 296 . . . . . . mark up % = 261 - 200 / 200 = 61 / 200 = 0.305 or 31 % answer : a" | a = 100 / 10
b = 200 / a
c = 200 + b
d = c + 80
e = d - 200
f = e / 200
|
a ) 0.27 , b ) 0.027 , c ) 27 , d ) 0.0027 , e ) 2.7 e - 06 | c | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(27, subtract(multiply(10, 10), const_1))), const_1000), 10) | if the digits 27 in the decimal 0.00027 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00027 ) ? | "99 * 0.27 = 26.73 approx . 27 answer : c" | a = 10 * 10
b = 10 * 10
c = b - 1
d = a * c
e = 10 * 10
f = e - 1
g = 27 / f
h = d * g
i = h / 1000
j = i * 10
|
a ) 80 cm , b ) 90 cm , c ) 100 cm , d ) 120 cm , e ) 130 cm | a | divide(200, add(const_2, divide(50, const_100))) | one ball will drop from a certain height . the height it will reach after rebounding from the floor is 50 percent of the previous height . the total travel is 200 cm when it touches the floor on third time . what is the value of the original height ? | "when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = h ( 1 + 1 + 0.5 ) = 80 2.5 h = 80 h = 80 . a is the answer ." | a = 50 / 100
b = 2 + a
c = 200 / b
|
a ) 100 , b ) 169 , c ) 225 , d ) 324 , e ) 256 | e | multiply(multiply(3, 5), multiply(3, 5)) | in the coordinate plane , one of the vertices of a square is the point ( - 3 , - 4 ) . if the diagonals of that square intersect at point ( 5 , 3 ) , what is the area of that square ? | "one point ( - 3 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 3 - 5 = - 8 is the midpoint of the square - - > whole side 16 , 16 * 16 = 256 e" | a = 3 * 5
b = 3 * 5
c = a * b
|
a ) 150 , b ) 200 , c ) 250 , d ) 163 , e ) none | d | subtract(multiply(12, 652), multiply(add(11, divide(9, 12)), 652)) | in a school with 652 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is | "sol . let the number of grils be x . then , number of boys = ( 652 - x ) . then , ( 11 3 / 4 Γ 652 ) β 11 x + 12 ( 652 - x ) β x = 7824 - 7661 β 163 . answer d" | a = 12 * 652
b = 9 / 12
c = 11 + b
d = c * 652
e = a - d
|
a ) 28213 , b ) 84307 , c ) 50694 , d ) 54094 , e ) none of them | d | divide(add(35423, 7164), 41720) | ( 35423 + 7164 + 41720 ) - ( 317 x 89 ) = ? | "= ( 84307 ) - 317 x 89 = 84307 - ( 317 x ( 90 - 1 ) ) = 84307 - ( 317 x 90 - 317 ) = 84307 - 28213 = 54094 answer is d" | a = 35423 + 7164
b = a / 41720
|
a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8 | a | multiply(subtract(1, divide(2, 3)), divide(1, 2)) | in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 2 / 3 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 2 / 3 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 3 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 1 / 3 * 1 / 2 = 1 / 6 , or a ." | a = 2 / 3
b = 1 - a
c = 1 / 2
d = b * c
|
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