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 ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 19 | b | subtract(30, divide(add(multiply(7.50, 30), 360), add(7.50, 25))) | a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 360 in all . for how many days was he absent ? | "30 * 25 = 750 360 - - - - - - - - - - - 380 25 + 7.50 = 32.5 390 / 32.5 = 12 b" | a = 7 * 50
b = a + 360
c = 7 + 50
d = b / c
e = 30 - d
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a ) 349 , b ) 300 , c ) 500 , d ) 450 , e ) 250 | a | add(multiply(13, add(add(const_10, const_12), const_4)), 11) | when a number is divided by 13 , the remainder is 11 . when the same number is divided by 17 , then remainder is 9 . what is the number ? | x = 13 p + 11 and x = 17 q + 9 13 p + 11 = 17 q + 9 17 q - 13 p = 2 q = 2 + 13 p / 17 least value of p which q = 2 + 13 p / 17 = > 26 x = ( 13 * 26 + 11 ) = 349 answer a | a = 10 + 12
b = a + 4
c = 13 * b
d = c + 11
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a ) 4 , b ) 6 , c ) 8 , d ) 11 , e ) 12 | d | add(subtract(multiply(const_4, 3), multiply(multiply(const_4, 5), 0.2)), 3) | a football player scores 3 goals in his fifth match thus increasing his average goals score by 0.2 . 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 go higher or lower with your next test . here , i ' ll start with answer d = 11 goals if . . . . total goals = 11 goals 5 th game = 3 goals 1 st 4 games = 8 goals avg . for 4 games = 8 / 4 = 2 goals / game avg . for all 5 games = 6 / 5 = 2.2 goals / game 2.2 * 5 = 11 so answer d | a = 4 * 3
b = 4 * 5
c = b * 0
d = a - c
e = d + 3
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | c | divide(350, add(subtract(26, 2), const_1)) | 350 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 350 / 25 ) = 14 c" | a = 26 - 2
b = a + 1
c = 350 / b
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a ) 4300 , b ) 4500 , c ) 5120 , d ) 5230 , e ) 5366 | c | subtract(subtract(8000, multiply(8000, divide(20, const_100))), multiply(subtract(8000, multiply(8000, divide(20, const_100))), divide(20, const_100))) | the population of a town is 8000 . it decreases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8000 Γ 80 / 100 Γ 80 / 100 = 5120 c" | a = 20 / 100
b = 8000 * a
c = 8000 - b
d = 20 / 100
e = 8000 * d
f = 8000 - e
g = 20 / 100
h = f * g
i = c - h
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a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600 | b | divide(multiply(subtract(const_100, 10), divide(900, const_2)), const_100) | a pair of articles was bought for $ 900 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 900 / 2 = $ 450 let m . p = $ x 90 % of x = 450 x = 450 * 100 / 90 = $ 500 answer is b" | a = 100 - 10
b = 900 / 2
c = a * b
d = c / 100
|
a ) 1 / 10 , b ) 1 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | a | divide(circle_area(divide(10, const_2)), const_2) | what will be the fraction of 10 % | "explanation : 10 * 1 / 100 = 1 / 10 option a" | a = 10 / 2
b = circle_area / (
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a ) rs . 480 , b ) rs . 760 , c ) rs . 860 , d ) rs . 960 , e ) none of these | a | multiply(600, subtract(const_1, divide(20, const_100))) | a man buys an item at rs . 600 and sells it at the loss of 20 percent . then what is the selling price of that item | "explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 600 = 80 / 100 * 600 = 480 option a" | a = 20 / 100
b = 1 - a
c = 600 * b
|
a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | d | divide(subtract(0.75, 0.5), subtract(0.8, 0.5)) | last month , john rejected 0.5 % of the products that he inspected and jane rejected 0.8 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.8 ( x ) + 0.5 ( 1 - x ) = 0.75 0.3 x = 0.75 - 0.5 x = 0.25 / 0.3 x = 5 / 6 therefore the answer is d : 5 / 6 ." | a = 0 - 75
b = 0 - 8
c = a / b
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(log(divide(10, 0.00010101)), log(10)) | if k is an integer and 0.00010101 x 10 ^ k is greater than 10 , what is the least possible value of k ? | "0.00010101 * 10 ^ k > 10 we need to move the decimal point to the right 5 places to get 10.101 this is equivalent to multiplying by 10 ^ 5 . the answer is c ." | a = 10 / 0
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | d | subtract(42398, multiply(floor(divide(42398, 15)), 15)) | what least number must be subtracted from 42398 so that remaining no . is divisible by 15 | "explanation : on dividing 42398 by 15 we get the remainder 8 , so 8 should be subtracted option d" | a = 42398 / 15
b = math.floor(a)
c = b * 15
d = 42398 - c
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(multiply(35, 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 35 % of 4 / 13 of 650 ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 650 , or 1 / 9 of 650 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate , the errors will partially cancel each other out . our estimate will be relatively accurate . the number 650 is a bit more than 630 , so ( 1 / 9 ) * 630 will be about 70 . keeping track not only of your current estimate , but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently . the answer is c ." | 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 = 35 * n
p = o / 100
|
a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 17 years | d | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 x = 22 . answer d" | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
|
a ) 10 ^ 7 , b ) 10 ^ 9 , c ) 10 ^ 8 , d ) 10 ^ 10 , e ) 10 ^ 11 | c | multiply(multiply(79, 48), 25073) | 79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25073 bacteria each . how many bacteria are there approximately ? | 79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,073 bacteria each . how many bacteria are there approximately ? a . 10 ^ 7 b . 10 ^ 8 c . 10 ^ 9 d . 10 ^ 10 e . 10 ^ 11 - > due to approximately , 79 = 80 , 48 = 50 , 25,073 = 25,000 are derived , which makes ( 79 ) ( 48 ) ( 25,075 ) = ( 80 ) ( 50 ) ( 25,000 ) = 10 ^ 8 . the answer is c . | a = 79 * 48
b = a * 25073
|
a ) 10 : 6 , b ) 10 : 3 , c ) 25 : 6 , d ) 25 : 3 , e ) 30 : 3 | c | divide(2, 5) | the simple form of the ratio 5 / 3 : 2 / 5 is ? | "5 / 3 : 2 / 5 = 25 : 6 answer : c" | a = 2 / 5
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a ) 18 , b ) 36 , c ) 72 , d ) 90 , e ) 100 | e | multiply(divide(180, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1)) | pat , kate and mark charged a total of 180 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate . | let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 180 - total hours charged for , x = 20 . mat charged 6 x - x or 5 x for more hours than kate , or for 100 hours . e is correct | a = 1 + 2
b = 2 * 3
c = a + b
d = 180 / c
e = 2 * 3
f = e - 1
g = d * f
|
a ) 142 seconds , b ) 150 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | divide(const_1, add(divide(const_1, 287), divide(const_1, 283))) | one tap , working alone at its maximum constant rate , can fill a certain sink in 287 seconds . a second tap , working alone at its maximum constant rate , can fill the same sink in 283 seconds . if both taps work together at their respective maximum constant rates , the time it will take to fill the sink is closest to | tap 1 : 283 secs tap 2 : 287 secs considering the average of these 2 taps : 285 secs . so 1 tap can fill the tank in 285 secs , so 2 taps can fill the tank in 285 / 2 = 142.5 secs . closest answer is a . | a = 1 / 287
b = 1 / 283
c = a + b
d = 1 / c
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | e | subtract(85, add(add(subtract(50, 35), subtract(40, 35)), 35)) | of the 85 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | "neither car nor garage = total - garage - ( swim - common ) = 85 - 50 - ( 40 - 35 ) = 85 - 55 = 30 answer e" | a = 50 - 35
b = 40 - 35
c = a + b
d = c + 35
e = 85 - d
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a ) 1300 m , b ) 1350 m , c ) 1400 m , d ) 1450 m , e ) 1500 m | c | multiply(divide(multiply(240, const_1000), const_3600), 21) | a train running at the speed of 240 km / hr crosses a pole in 21 seconds . what is the length of the train ? | "speed = ( 240 * 5 / 18 ) m / sec = ( 200 / 3 ) m / sec length of the train = ( speed x time ) = ( 200 / 3 * 21 ) m = 1400 m . answer : c" | a = 240 * 1000
b = a / 3600
c = b * 21
|
a ) m = 2 , b ) m = 5 , c ) m = 8 , d ) 10 , e ) 11 | a | divide(sqrt(subtract(power(add(divide(1000, const_10), const_1), const_2), power(subtract(add(divide(1000, const_10), const_1), const_2), const_2))), const_10) | the difference between the squares of two consecutive odd integers is a square < 1000 whose root equals the combined sum of the digits of the two integers . what is the sum of the digits m of the larger integer ? | really hard . but let ' s start with few concepts : 1 . square of an odd integer will be odd . 2 . difference of odd and odd will be even . 3 . any odd integer can be expressed as difference of two squares . 4 . an even integer can be expressed as difference of two squares only if that even integer is a multiple of 4 . assume two odd integers to be ( 2 x - 1 ) and ( 2 x + 1 ) . difference of their squares = ( 2 x + 1 ) ^ 2 - ( 2 x - 1 ) ^ 2 = 4 x ^ 2 + 1 + 4 x - ( 4 x ^ 2 + 1 - 4 x ) = 4 x ^ 2 + 1 + 4 x - 4 x ^ 2 - 1 + 4 x = 8 x now root of 8 x needs to be an integer such that it is equal to the sum of the digits of the two odd integers . 8 = 2 ^ 3 , so x needs to be such that itcompletesa perfect square . if we find x , we can find the value of 2 x + 1 ( larger integer ) and then sum of its digits . let ' s check the options , starting with c . for sum to be 8 , few possibilities are : 17 , 35 , 53 , 71 if we take 17 , the pair is 15 and 17 , meaning x = 8 . 8 x = 64 root of 64 = 4 but 4 is not equal to 1 + 5 + 1 + 7 . reject . if we take 35 , the pair is 33 and 35 , meaning x = 17 . 8 x = 8 * 17 reject since perfect square is not possible . if we take 53 , the pair is 51 and 53 , meaning x = 26 . 8 x = 8 * 26 reject since perfect square is not possible . if we take 71 , the pair is 69 and 71 , meaning x = 35 . 8 x = 8 * 35 reject since perfect square is not possible . i tried each option and the possibilities and then got lost . then , it occurred to me that i was only checking 2 - digit integers . what about 3 - digits ? starting with option a , only 2 - digit odd integer whose sum is 2 is 101 . if we take 101 , the pair is 99 and 101 , meaning x = 50 . 8 x = 8 * 50 = 400 root of 400 = 20 which is indeed equal to 9 + 9 + 1 + 1 . answer ( a ) . | a = 1000 / 10
b = a + 1
c = b ** 2
d = 1000 / 10
e = d + 1
f = e - 2
g = f ** 2
h = c - g
i = math.sqrt(h)
j = i / 10
|
a ) 3 , b ) 2 , c ) 9 , d ) 7 , e ) 1 | d | divide(add(multiply(const_4, 3), 2), 2) | if f ( x ) = 2 x + 3 , & g ( x ) = ( x - 3 ) / 2 , then what is the value of , fo ( fo ( go ( go ( fo ( fo ( go ( go . . . . . . . . . . . . . . . . ( fo ( fo ( go ( gof ( x ) ) ) ) ) . . . . . ) ) ) ) ) ) ? | f ( x ) = 2 x + 3 g ( x ) = ( x - 3 ) / 2 gof ( x ) = 2 x + 3 β 32 = x 2 x + 3 β 32 = x gogof ( x ) = x β 32 x β 32 fogogof ( x ) = 2 ( x β 32 ) + 32 ( x β 32 ) + 3 = 2 x β 6 + 622 x β 6 + 62 = x fofogogof ( x ) = 2 x + 3 this means , when we apply two times g ( x ) and two time f ( x ) on f ( x ) we get f ( x ) . this pattern continues , hence the answer is option ( d ) . | a = 4 * 3
b = a + 2
c = b / 2
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a ) 1208 , b ) 3108 , c ) 241 , d ) 217 , e ) 3147 | b | add(3105, subtract(21, reminder(3105, 21))) | find the number which is nearest to 3105 and is exactly divisible by 21 | "on dividing 3105 by 21 , we get 18 as remainder . ο number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . answer b 3108" | a = 21 - reminder
b = 3105 + a
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a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 48 | b | add(multiply(5, 5), 5) | the ratio of pens to pencils is 5 to 6 . there are 5 more pencils than pens . how many pencils are there ? | "let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 5 x = 5 the number of pencils is 30 . the answer is b ." | a = 5 * 5
b = a + 5
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a ) 21 , b ) 24 , c ) 27 , d ) 30 , e ) 33 | d | subtract(multiply(3, 11), 3) | p is 3 times as old as q . in 11 years , p will be twice as old as q . how old was p 3 years ago ? | p = 3 q so q = p / 3 p + 11 = 2 ( q + 11 ) = 2 ( p / 3 + 11 ) p / 3 = 11 p = 33 three years ago , p was 30 . the answer is d . | a = 3 * 11
b = a - 3
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a ) 32 , b ) 38 , c ) 29 , d ) 54 , e ) 20 | b | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 8), add(add(9, 1), subtract(9, 1))) | a person can row at 9 kmph and still water . he takes 8 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 8.5 = > x / 10 + x / 8 = 17 / 2 = > ( 4 x + 5 x ) / 40 = 17 / 2 = > x = 38 km . answer : b" | a = 9 - 1
b = 9 + 1
c = a * b
d = c * 8
e = 9 + 1
f = 9 - 1
g = e + f
h = d / g
|
a ) 1200 km , b ) 1500 km , c ) 2000 km , d ) 2250 km , e ) 3600 km | d | multiply(30, 50) | a walks at 30 kmph and 30 hours after his start , b cycles after him at 50 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 50 = 30 x = 2250 km answer is d" | a = 30 * 50
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a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3 | c | divide(add(1, 5), 5) | a box contains 1 red balls and 5 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 5 black balls be bbbbb and 1 red balls be r they can be arranged in 6 slots _ _ _ _ _ _ _ _ in ( 6 ! ) / ( 5 ! x 1 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 5 slots and 4 black ( bbbb ) and 1 red ( r ) they can be arranged in ( 5 ! ) / ( 4 ! x 1 ! ) hence required probability = [ ( 6 ! ) / ( 5 ! x 1 ! ) ] / [ ( 5 ! ) / ( 4 ! x 1 ! ) ] = 6 / 5 c" | a = 1 + 5
b = a / 5
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 13), 2)), 1), 3), add(multiply(12, 1), multiply(13, 2)))) | in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 13 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the class borrowed a total of 40 * 2 = 80 books . the 27 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 13 * 2 = 38 . to maximize the number of books borrowed by 1 student , let ' s assume that 12 students borrowed 3 books and 1 student borrowed the rest . 80 - 38 - 3 * 12 = 6 the maximum number of books borrowed by any student is 6 . the answer is d ." | a = 40 * 2
b = 12 * 1
c = b + 13
d = c + 2
e = 40 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 13 * 2
j = h + i
k = g + j
l = a - k
|
a ) 11 , b ) 18 , c ) 55 , d ) 25 , e ) 33 | c | multiply(4, power(const_3, 4)) | if 5400 mn = k ^ 4 , where m , n , and k are positive integers , what is the least possible value of 4 m + n ? | "k ^ 4 = 5400 β mn k ^ 4 = 3 ^ 3.2 ^ 3.5 ^ 2 . mn in order to make rhs a perfect power of 4 , we require it to be multiplied by 3 , 2 & 5 ^ 2 mn = 3 . 2.5 ^ 2 mn = 150 = 10 * 15 ( least possible ) answer = 4 * 10 + 15 = 55 answer : c" | a = 3 ** 4
b = 4 * a
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a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | b | subtract(add(const_4, 3), divide(divide(add(26, 10), const_2), 3)) | on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 3 cups of tea . last week mo drank a total of 26 cups of tea and hot chocolate together . if during that week mo drank 10 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 26 t - c = 10 - > t = 18 . c = 8 . mo drinks 3 cups of tea a day then number of days that are not rainy = 18 / 3 = 6 so number of rainy days = 7 - 6 = 1 b is the answer ." | a = 4 + 3
b = 26 + 10
c = b / 2
d = c / 3
e = a - d
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a ) 36.7 , b ) 36.4 , c ) 36.5 , d ) 36.9 , e ) 36.3 | b | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 45 was wrongly taken as 23 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 45 - 23 ) = 1822 . correct mean = 1822 / 50 = 36.4 answer : b" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 44 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23 | a | divide(multiply(add(add(4, const_3), const_2), divide(4, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 4 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 4 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let bucket a be a and bucket b be b scenario 1 a - 4 = 1 / 3 ( b + 4 ) - - - - > 3 a - 12 = b + 4 scenario 2 b - 4 = 1 / 2 ( a + 4 ) - - - - - > 2 b - 8 = a + 4 from scenario 1 , b = 3 a - 16 substitute b with this information in stmt 2 2 ( 3 a - 16 ) - 8 = a + 4 - - - - - - > 6 a - 32 - 8 = a + 4 - - - - - - > 6 a - a = 40 + 4 - - - > 5 a = 44 a = 44 / 5 , answer choice a" | a = 4 + 3
b = a + 2
c = 4 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
|
a ) 12 , b ) 67 , c ) 95 , d ) 36 , e ) 66 | a | divide(multiply(6, 4), subtract(6, 4)) | pipe a can fill a tank in 4 hours . due to a leak at the bottom , it takes 6 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 4 - 1 / x = 1 / 6 = > 1 / x = 1 / 4 - 1 / 6 = ( 3 - 2 ) / 12 = 1 / 12 = > x = 12 . answer : a" | a = 6 * 4
b = 6 - 4
c = a / b
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a ) rs . 500000 , b ) rs . 50000 , c ) rs . 5000 , d ) rs . 15000 , e ) rs . 6000 | b | divide(750, multiply(divide(30, const_100), multiply(divide(20, const_100), divide(25, const_100)))) | an individual deposited 20 % of 25 % of 30 % of what he had into his account with a bank . if he deposited rs . 750 , how much was he having initially with him before depositing ? | let x amount he has then 20 % of 25 % of 30 % of x = 750 so x = 50000 answer : b | a = 30 / 100
b = 20 / 100
c = 25 / 100
d = b * c
e = a * d
f = 750 / e
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a ) 0.33 , b ) 0.44 , c ) 0.55 , d ) 44 , e ) 55 | d | multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many r cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer r = 44 . d" | a = 1 / 2
b = 8 / a
c = b / 12
d = 0 * 33
e = c * d
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a ) 79 , b ) 99 , c ) 88 , d ) 88 , e ) 98 | e | divide(add(add(add(add(96, 98), 99), 100), 98), divide(const_10, const_2)) | dacid obtained 96 , 98 , 99 , 100 and 98 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 96 + 98 + 99 + 100 + 98 ) / 5 = 491 / 5 = 98 . answer : e" | a = 96 + 98
b = a + 99
c = b + 100
d = c + 98
e = 10 / 2
f = d / e
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a ) 10 , b ) 3 , c ) 15 , d ) 52 / 7 , e ) 60 / 7 | a | multiply(divide(add(8, 4), add(3, 4)), divide(subtract(multiply(3, divide(add(8, 4), add(3, 4))), 4), 3)) | if 4 x + 3 y = 8 and y β 3 x = 2 , then what is the value of x + 4 y ? | "4 x + 3 y = 8 β 3 x + y = 2 adding both the equations - - > x + 4 y = 10 answer : a" | a = 8 + 4
b = 3 + 4
c = a / b
d = 8 + 4
e = 3 + 4
f = d / e
g = 3 * f
h = g - 4
i = h / 3
j = c * i
|
a ) 7 / 16 , b ) 7 / 15 , c ) 10 / 21 , d ) 17 / 35 , e ) 1 / 10 | e | divide(multiply(subtract(const_10, multiply(divide(3, 5), const_10)), multiply(divide(1, 6), multiply(divide(3, 5), const_10))), add(multiply(multiply(divide(3, 5), const_10), multiply(divide(3, 5), const_10)), multiply(subtract(const_10, multiply(divide(3, 5), const_10)), multiply(divide(1, 6), multiply(divide(3, 5), const_10))))) | a lemonade stand sold only small and large cups of lemonade on tuesday . 3 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 1 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | "let the total # of cups be 10 . # of small cups sold 3 / 5 * 10 = 6 ; # of large cups sold 10 - 6 = 4 ; let the price of small cup be $ 6 , then the price of larges cup would be 1 / 6 * 6 = $ 1 ; revenue from small cups : 6 * $ 6 = $ 36 ; revenue from large cups cups : 4 * $ 1 = $ 4 ; fraction of total revenue from large cups : 4 / ( 36 + 4 ) = 1 / 10 . answer : e ." | a = 3 / 5
b = a * 10
c = 10 - b
d = 1 / 6
e = 3 / 5
f = e * 10
g = d * f
h = c * g
i = 3 / 5
j = i * 10
k = 3 / 5
l = k * 10
m = j * l
n = 3 / 5
o = n * 10
p = 10 - o
q = 1 / 6
r = 3 / 5
s = r * 10
t = q * s
u = p * t
v = m + u
w = h / v
|
a ) 330 , b ) 390 , c ) 450 , d ) 510 , e ) 570 | c | subtract(1000, multiply(add(add(multiply(add(1, 10), divide(10, const_2)), divide(10, const_2)), 1), 10)) | there are 10 baskets numbered from 1 to 10 and filled with apples . 10 children are asked to pick apples one by one from each basket such that the number of apples picked by each child from each basket is equal to the number marked on the basket . if there were 1000 apples in total and the baskets were filled in such a way that none of the basket became empty in the process , how many apples were left in the end ? | "each child takes a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 10 * 11 / 2 = 55 apples . the total number of apples taken by 10 children is 10 * 55 = 550 the number of apples left in the end is 1000 - 550 = 450 apples . the answer is c ." | a = 1 + 10
b = 10 / 2
c = a * b
d = 10 / 2
e = c + d
f = e + 1
g = f * 10
h = 1000 - g
|
a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | c | multiply(const_100, subtract(multiply(multiply(divide(add(30, const_100), const_100), divide(add(30, const_100), const_100)), divide(subtract(const_100, 30), const_100)), const_1)) | the positive numbers w , x , y , and z are such that x is 30 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 20 - answer c" | a = 30 + 100
b = a / 100
c = 30 + 100
d = c / 100
e = b * d
f = 100 - 30
g = f / 100
h = e * g
i = h - 1
j = 100 * i
|
a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 219 / 256 , e ) 15 / 16 | d | add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 7), multiply(7, power(divide(const_1, const_2), 7))), multiply(multiply(7, const_3), power(divide(const_1, const_2), 7)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 7), multiply(7, power(divide(const_1, const_2), 7))), multiply(multiply(7, const_3), power(divide(const_1, const_2), 7))))) | an exam consists of 7 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 70 % is a passing grade , what is the probability that brian passes ? | if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure out the number of ways that brian can get 6 out of 8 , 7 out of 8 , and 8 out of 8 questions correct . which is 8 choose 6 , equals is 28 , 8 choose 7 , equals 8 , and 8 choose 8 , equals 1 . this sums to 37 . the number of possible questions outcomes - the sum of 8 choose 8 , 7 choose 8 , 6 choose 8 β¦ . 2 choose 8 , 1 choose 8 , and 0 choose 8 is 256 , so the chance of him passing is 219 / 256 . d | a = 1 / 2
b = a ** 7
c = 1 / 2
d = c ** 7
e = 7 * d
f = b + e
g = 7 * 3
h = 1 / 2
i = h ** 7
j = g * i
k = f + j
l = 1 - k
m = l / 10
n = 1 / 2
o = n ** 7
p = 1 / 2
q = p ** 7
r = 7 * q
s = o + r
t = 7 * 3
u = 1 / 2
v = u ** 7
w = t * v
x = s + w
y = 1 - x
z = m + y
|
a ) 2 / 15 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) none of these | a | subtract(divide(4, 5), divide(2, 3)) | what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 7 | explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 7 = . 71 so biggest is 4 / 5 and smallest is 2 / 3 their difference is 4 / 5 - 2 / 3 = 2 / 15 option a | a = 4 / 5
b = 2 / 3
c = a - b
|
a ) 1 / 4 , b ) 4 / 15 , c ) 1 / 3 , d ) 5 / 16 , e ) 4 / 5 | d | divide(5, add(subtract(const_12, const_1), 5)) | last year department store x had a sales total for december that was 5 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ? | "let avg for 11 mos . = 10 therefore , dec = 50 year total = 11 * 10 + 50 = 160 answer = 50 / 160 = 5 / 16 = d" | a = 12 - 1
b = a + 5
c = 5 / b
|
a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | a | subtract(add(3.4, 12.7), 6.2) | if [ x ] is the greatest integer less than or equal to x , what is the value of [ - 6.2 ] + [ - 3.4 ] + [ 12.7 ] ? | "you are asked what the closest lesser integer value to [ x ] is . [ - 6.2 ] = - 7.0 [ - 3.4 ] = - 4.0 [ 12.7 ] = 12.0 therefore , answer is : - 7.0 - 4.0 + 12.0 = - 1.0 option a ." | a = 3 + 4
b = a - 6
|
a ) 13 % , b ) 14 % , c ) 18 % , d ) 20 % , e ) 28 % | e | multiply(subtract(const_1, multiply(divide(subtract(const_100, 20), const_100), divide(subtract(const_100, 10), const_100))), const_100) | a towel , when bleached , was found to have lost 20 % of its length and 10 % of its breadth . the percentage of decrease in area is : | let original length = x and original breadth = y . decrease in area = xy - ( 80 / 100 * x * 90 / 100 * y ) = ( xy - 18 / 25 * xy ) = 7 / 25 * xy . decrease % = ( 7 / 25 * xy * 1 / xy * 100 ) % = 28 % . answer is e | a = 100 - 20
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = 1 - e
g = f * 100
|
a ) 4 % , b ) 74 % , c ) 64 % , d ) 74 % , e ) 94 % | a | divide(multiply(divide(2, 5), const_100), 10) | at what rate percent per annum will the simple interest on a sum of money be 2 / 5 of the amount in 10 years ? | let sum = x . then , s . i . = 2 x / 5 , time = 10 years . rate = ( 100 * 2 x ) / ( x * 5 * 10 ) = 4 % answer : a | a = 2 / 5
b = a * 100
c = b / 10
|
a ) 27 % , b ) 20 % , c ) 80 % , d ) 30 % , e ) 25 % | e | multiply(divide(subtract(1125, 900), 900), const_100) | a cycle is bought for rs . 900 and sold for rs . 1125 , find the gain percent ? | "900 - - - - 225 100 - - - - ? = > 25 % answer : e" | a = 1125 - 900
b = a / 900
c = b * 100
|
a ) 3 , b ) 6 , c ) 1 / 3 , d ) 1 / 2 , e ) 1 / 6 | c | divide(1, divide(add(subtract(5, 1), 2), 2)) | the vertex of a rectangle are ( 1 , 0 ) , ( 5 , 0 ) , ( 1 , 2 ) and ( 5 , 2 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | "if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 3 , 1 ) . the slope of a line passing through ( 0,0 ) and ( 3 , 1 ) is 1 / 3 . the answer is c ." | a = 5 - 1
b = a + 2
c = b / 2
d = 1 / c
|
a ) 41.1 sec , b ) 20.2 sec , c ) 31.8 sec , d ) 43.2 sec , e ) none of the above | d | divide(add(140, 520), multiply(55, const_0_2778)) | a 140 m long train is running at a speed of 55 km / hr . it crossed a platform of length 520 m in ? | speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 140 m + 520 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 660 m time = distance / speed = 660 x 18 / ( 5 x 55 ) = 43.2 sec ans is : d | a = 140 + 520
b = 55 * const_0_2778
c = a / b
|
a ) 80 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(50, const_4), multiply(9, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 50 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 50 * 4 = 200 m time taken = 200 * 2 / 5 = 80 sec answer is a" | a = 50 * 4
b = 1000 / 3600
c = 9 * b
d = a / c
|
a ) 3 , b ) 1 / 3 , c ) 2 , d ) 9 , e ) 1 / 9 | a | power(subtract(28, const_1), divide(const_1, 3)) | if the ratio of the sum of the first 6 terms of a g . p . to the sum of the first 3 terms of the g . p . is 28 , what is the common ratio of the g . p ? | "28 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) factorize the same terms 28 = 1 + ( a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) write every term with respect to r a 1 = a 1 a 2 = a 1 * r ^ 1 a 3 = a 1 * r ^ 2 . . . . . . . . . 28 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) / ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 27 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) / ( ( 1 + r ^ 1 + r ^ 2 ) ) 27 = r ^ 3 r = 3 a" | a = 28 - 1
b = 1 / 3
c = a ** b
|
a ) 32 , b ) 87 , c ) 38 , d ) 99 , e ) 77 | c | add(floor(divide(multiply(multiply(21, 8), multiply(19, 3)), multiply(multiply(21, 2), 6))), const_1) | 19 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | "3 w = 2 m 19 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 19 * 21 * 8 = 14 * x * 6 x = 38 answer : c" | a = 21 * 8
b = 19 * 3
c = a * b
d = 21 * 2
e = d * 6
f = c / e
g = math.floor(f)
h = g + 1
|
a ) $ 500 , b ) $ 400 , c ) $ 700 , d ) $ 900 , e ) $ 950 | b | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,056 . how much did she invest at 5 % ? | "et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1056 / 1000 ) - 1 = 0.053 = 5.3 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of 1000 invested at 5 % ) 0.05 x + 0.06 ( 1 - x ) = 0.056 ( 0.05 ) x + 0.06 - ( 0.06 ) x = 0.056 - 0.01 x = - 0.004 x = - 0.004 / - 0.01 = 4 / 10 = 40 % so x = 40 % of the 1000 which is 400 answer : b" | a = 5 * 2
b = a * 100
c = 5 * 2
d = c * 100
e = 5 * 2
f = e * 5
g = f + 4
h = d + g
i = 5 * 2
j = i * 100
k = h - j
l = k * 100
m = 5 * 2
n = m * 100
o = 5 * n
p = l - o
q = b - p
|
a ) 2 hours , b ) 2.5 hours , c ) 3 hours , d ) 3.5 hours , e ) 8 hours | e | subtract(10, subtract(8, subtract(const_2.0, 1))) | there are two inlets and one outlet to a cistern . one of the inlets takes 4 hours to fill up the cistern and the other inlet takes twice as much time to fill up the same cistern . both of the inlets are turned on at 8 : 30 am with the cistern completely empty , and at 10 : 30 am , the outlet is turned on and it takes 1 more hour to fill the cistern completely . how much time does the outlet working alone takes to empty the cistern when the cistern is full ? | "the combined inflow rate of the two inlets is 1 / 4 + 1 / 8 = 3 / 8 cistern / hour . thus , working together , it takes 8 / 3 hours ( time is reciprocal of rate ) to fill the cistern . from 8 : 30 am to 10 : 30 am , so in 2 hours , the inlet pipes will fill ( time ) * ( rate ) = 2 * 3 / 8 = 3 / 4 th of the cistern . then the outlet is turned on and the remaining 1 / 4 th of the cistern is filled in 1 hour . letting x to be the rate of the outlet , we would have : 3 / 8 - x = 1 / 4 - - > x = 1 / 8 cistern / hour , which means that it takes 8 hours the outlet working alone to empty the cistern . answer : e ." | a = 2 - 0
b = 8 - a
c = 10 - b
|
a ) $ 640 , b ) $ 680 , c ) $ 720 , d ) $ 760 , e ) $ 800 | c | subtract(multiply(80, divide(const_100, 10)), 80) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 80 , how much more remains to be paid ? | "90 % remains to be paid so the remaining amount is 9 * 80 = $ 720 . the answer is c ." | a = 100 / 10
b = 80 * a
c = b - 80
|
a ) 343 , b ) 777 , c ) 780 , d ) 867 , e ) 232 | c | divide(add(212, 22), divide(30, const_100)) | mike needs 30 % to pass . if he scored 212 marks and falls short by 22 marks , what was the maximum marks he could have got ? | "if mike had scored 22 marks more , he could have scored 30 % therefore , mike required 212 + 22 = 234 marks let the maximum marks be m . then 30 % of m = 234 ( 30 / 100 ) Γ m = 234 m = ( 234 Γ 100 ) / 30 m = 23400 / 30 m = 780 answer : c" | a = 212 + 22
b = 30 / 100
c = a / b
|
a ) 3 : 4 , b ) 4 : 5 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 2 | b | divide(400, 500) | if shares of two persons in profits are rs . 400 and rs . 500 then ratio of their capitals is | "profit = time * capital so 400 : 500 = 4 : 5 answer : b" | a = 400 / 500
|
a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | b | add(add(2, 2), 1) | in the xy - plane , the point ( 2 , 1 ) is the center of a circle . the point ( - 2 , 1 ) lies inside the circle and the point ( 2 , - 5 ) lies outside the circle . if the radius r of the circle is an integer , then r = | an easy way to solve this question will be just to mark the points on the coordinate plane . you ' ll see that the distance between the center ( 2 , 1 ) and the point inside the circle ( - 2 , 1 ) is 4 units ( both points are on y = 1 line so the distance will simply be 2 - ( - 2 ) = 4 ) so the radius must be more than 4 units . the distance between the center ( 2,1 ) and the point outside the circle ( 2 , - 5 ) is 6 units ( both points are on x = 2 line so the distance will simply be 1 - ( - 5 ) = 6 ) so the radius must be less than 6 units , which implies 4 < r < 6 , thus as r is an integer then r = 5 . answer : b . | a = 2 + 2
b = a + 1
|
a ) 15 , b ) 33 , c ) 56 , d ) 96 , e ) 88 | b | add(add(power(add(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive integers is 102 . find the lowest of the three ? | "explanation : three consecutive numbers can be taken as ( p - 1 ) , p , ( p + 1 ) . so , ( p - 1 ) + p + ( p + 1 ) = 102 3 p = 102 = > p = 34 . the lowest of the three = ( p - 1 ) = 34 - 1 = 33 . answer b" | a = 102 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 102 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 102 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 102 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 45 % , b ) 125 % , c ) 147 % , d ) 150 % , e ) 225 % | c | divide(subtract(21,947, 8,902), 8,902) | in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 21,947 . approximately what was the percent increase ? | "the percent increase is ( 21947 - 8902 ) / 8902 = 13045 / 8902 = 1.47 so the approximate answer is c" | a = 21 - 947
b = a / 8
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(const_1, divide(16, 4)) | if k is the greatest positive integer such that 4 ^ k is a divisor of 16 ! then k = | "16 / 4 = 4 16 / 16 = 1 4 + 1 = 5 k = 5 answer : c" | a = 16 / 4
b = 1 + a
|
a ) 565 , b ) 444 , c ) 676 , d ) 767 , e ) 432 | e | add(multiply(divide(60, subtract(20, 16)), 16), multiply(divide(60, subtract(20, 16)), 20)) | 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 20 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 4 ? - - - - - - 60 12 h rs = 16 + 20 = 36 t = 12 d = 36 * 12 = 432 answer : e" | a = 20 - 16
b = 60 / a
c = b * 16
d = 20 - 16
e = 60 / d
f = e * 20
g = c + f
|
a ) $ 30.50 , b ) $ 16.50 , c ) $ 28.50 , d ) $ 12 , e ) $ 9.60 | b | multiply(8.25, 2) | johnny makes $ 8.25 per hour at his work . if he works 2 hours , how much money will he earn ? | 8.25 * 2 = 16.50 . answer is b . | a = 8 * 25
|
a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number t ? | "( 1 + 1 / 8 ) x - ( 1 - 1 / 4 ) x = 30 ( 9 / 8 ) x - ( 3 / 4 ) x = 30 x = 80 = t answer : b" | a = 12 / 5
b = 1 + a
c = 25 / 100
d = 1 - c
e = b - d
f = 30 / e
|
a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | d | divide(factorial(subtract(add(const_4, 18), const_1)), multiply(factorial(18), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 18 nor 21 ? | "integers less than 5000 divisible by 18 5000 / 18 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 18 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 18 and 21 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4428 answer d ." | a = 4 + 18
b = a - 1
c = math.factorial(b)
d = math.factorial(18)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | d | subtract(subtract(4, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | evaluate : | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = ? | "according to order of operations , inner brackets first . hence | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 4 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 4 + 72 | - | 5 - 11 | = | 76 | - | - 6 | = 76 - 6 = 70 correct answer d ) 70" | a = 3 - 12
b = 8 * a
c = 4 - b
d = 5 - 11
e = c - negate
|
a ) 154 , b ) 677 , c ) 784 , d ) 148 , e ) 132 | e | subtract(subtract(add(subtract(subtract(reminder(reminder(19714181, power(const_100, const_2)), const_100), const_10), const_10), reminder(floor(divide(19413844, power(const_100, const_2))), const_100)), const_10), const_10) | if xy denotes x is raised to the power y , find the last two digits of 19413844 + 19714181 | "1941 ^ 3844 unit digit = 1 ^ 4 = 1 10 th digit = 4 * 4 = 16 = 2 means last two digit = 61 and 1971 ^ 4181 unit digit = 1 10 th digit = 7 * 1 = 7 mean last two digit = 71 hence , 71 + 61 = 132" | a = 100 ** 2
b = reminder - (
c = b - 100
d = c + 10
e = d - 10
f = 100 ** 2
g = 19413844 / f
h = math.floor(g)
i = e - reminder
|
a ) 430 , b ) 438 , c ) 436 , d ) 434 , e ) 422 | a | divide(172, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 172 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 172 = > 4 x / 10 = 172 = > 4 x = 1720 = > x = 430 . answer : a" | a = 100 - 70
b = 70 - a
c = b / 100
d = 172 / c
|
a ) 142 , b ) 130 , c ) 132 , d ) 123 , e ) 120 | c | divide(multiply(55, 12), 5) | 12 buckets of water fill a tank when the capacity of each bucket is 55 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 5 litres ? | "capacity of the tank = ( 12 Γ£ β 55 ) litre number of buckets required of capacity of each bucket is 17 litre = 12 Γ£ β 55 / 5 = 12 Γ£ β 11 = 132 answer is c" | a = 55 * 12
b = a / 5
|
a ) 10 : 00 , b ) 11 : 00 , c ) 09 : 00 , d ) 12 : 00 , e ) 07 : 00 | d | add(multiply(1, const_10), 2) | yesterday i noticed that in my clock , there is time when the hour needle and the minute needle are exactly between 1 and 2 . also both the needles lie on top of each other . what time i and talking about ? | d 12 : 00 both minute and hour clock lie exactly between the number one and two in the number . | a = 1 * 10
b = a + 2
|
a ) 280 , b ) 288 , c ) 279 , d ) 277 , e ) 290 | a | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 46))) | a watch was sold at a loss of 46 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "54 % 104 % - - - - - - - - 50 % - - - - 140 100 % - - - - ? = > rs . 280 answer : a" | a = 140 * 100
b = 100 + 4
c = 100 - 46
d = b - c
e = a / d
|
a ) a ) 354 , b ) b ) 175 , c ) c ) 225 , d ) d ) 375 , e ) e ) 524 | a | divide(multiply(divide(168, const_100), 1265), 6) | ( 168 % of 1265 ) Γ· 6 = ? | "explanation : ? = ( 168 x 1265 / 100 ) Γ· 6 = 212520 / 600 = 354 answer : option a" | a = 168 / 100
b = a * 1265
c = b / 6
|
a ) 650 , b ) 750 , c ) 850 , d ) 950 , e ) 1050 | e | divide(multiply(multiply(15, subtract(15, const_1)), 10), const_2) | there are 15 teams in the hockey league , and each team faces all the other teams 10 times each . how many games are played in the season ? | "the number of ways to choose two teams is 15 c 2 = 15 * 14 / 2 = 105 the total number of games in the season is 10 * 105 = 1050 . the answer is e ." | a = 15 - 1
b = 15 * a
c = b * 10
d = c / 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(add(multiply(12, 3), 12), 16) | if 3 : 12 : : x : 16 , then find the value of x | explanation : treat 3 : 12 as 3 / 12 and x : 16 as x / 16 , treat : : as = so we get 3 / 12 = x / 16 = > 12 x = 48 = > x = 4 option b | a = 12 * 3
b = a + 12
c = b / 16
|
a ) $ 32 , b ) $ 36 , c ) $ 52 , d ) $ 60 , e ) $ 68 | a | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(8, const_2), 10)), const_2) | bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 8 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 10 left in his pocket . what was the value of n ? | "started to test answer b if he had 36 , then he spent 9 at hardware store now he was left with 27 $ he spent 8 dollars on cleaning , thus he remained with 19 $ he then spent 1 / 2 of 19 , or 9.50 , hence , the only option that can be right is a . ans a" | a = 1 / 4
b = 1 - a
c = 1/(b)
d = 8 / 2
e = d + 10
f = c * e
g = f * 2
|
a ) 6 , b ) 10 , c ) 8 , d ) 7.5 , e ) 3.5 | b | add(4, divide(subtract(multiply(20, 12), multiply(12, 4)), add(12, 20))) | john can complete a given task in 20 days . jane will take only 12 days to complete the same task . john and jane set out to complete the task by beginning to work together . however , jane was indisposed 4 days before the work got over . in how many days did the work get over from the time john and jane started to work on it together ? | "john 6 * 20 = 120 jane 10 * 12 = 120 in 4 days working alone john did : 4 days * rate = 4 * 6 = 24 , so working together they did 120 - 24 = 96 - - > rate ( a + b ) * t = 96 - > 16 * t = 96 , t = 6 . hence the answer is b ( john worked 4 days + 6 days they worked together = 10 days )" | a = 20 * 12
b = 12 * 4
c = a - b
d = 12 + 20
e = c / d
f = 4 + e
|
a ) 30 , b ) 45 , c ) 270 , d ) 90 , e ) 135 | c | divide(multiply(divide(multiply(multiply(1, 3), 30), 1), 3), 1) | in cliff β s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 1 / 3 are shiny and the rest are matte , while 1 / 5 of the sedimentary rocks are shiny . if there are 30 shiny igneous rocks , how many total rocks does cliff have ? | "we can start with the known quantity and then go on to find the others . shiny igneous ricks are 30 . these are ( 1 / 3 ) of total igneous rocks . ( 1 / 3 ) * total igneous rocks = 30 total igneous rocks = 30 * ( 3 ) = 90 total sedimentary rocks = 2 * total igneous rocks = 2 * 90 = 180 total number of rocks = 90 + 180 = 270 answer ( c )" | a = 1 * 3
b = a * 30
c = b / 1
d = c * 3
e = d / 1
|
a ) 4 / 7 , b ) 5 / 9 , c ) 6 / 11 , d ) 7 / 12 , e ) 8 / 13 | e | divide(multiply(6, 4), add(multiply(6, 4), multiply(3, 5))) | at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 5 / 6 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 5 x / 6 ) ( 3 y / 4 ) = 5 xy / 8 the total number of boxes is xy + 5 xy / 8 = 13 xy / 8 the fraction loaded by the day crew is xy / ( 13 xy / 8 ) = 8 / 13 the answer is e ." | a = 6 * 4
b = 6 * 4
c = 3 * 5
d = b + c
e = a / d
|
a ) 22 , b ) 23 , c ) 24 , d ) 30 , e ) 28 | d | divide(subtract(multiply(50, const_4), 140), const_2) | a man has some hens and cows . if the number of heads be 50 and the number of feet equals 140 , then the number of hens will be : | "let hens be x and cows be y now , feet : x * 2 + y * 4 = 140 heads : x * 1 + y * 1 = 50 implies , 2 x + 4 y = 140 and x + y = 50 solving these two equations , we get x = 30 and y = 20 therefore , hens are 26 . answer : d" | a = 50 * 4
b = a - 140
c = b / 2
|
a ) 45 , b ) 48 , c ) 51 , d ) 54 , e ) 57 | a | add(add(add(add(divide(4.4, divide(8, const_60)), 3), 3), 3), 3) | the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 4.4 miles in the fifth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | "( 4.4 miles / 8 minutes ) * 60 minutes / hour = 33 mph let x be the original speed . x - 4 ( 3 ) = 33 x = 45 mph the answer is a ." | a = 8 / const_60
b = 4 / 4
c = b + 3
d = c + 3
e = d + 3
f = e + 3
|
a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 36.5 % , e ) 40 % | d | subtract(multiply(divide(subtract(const_100, 9), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 9 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.91 * 150 x = 136.5 x hence total profit = 36.5 % option d" | a = 100 - 9
b = a / 100
c = 100 + 20
d = 100 + 25
e = d / 100
f = c * e
g = b * f
h = g - 100
|
a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 19 | e | add(multiply(6, 2), 7) | if p / q = 6 / 7 , then 2 p + q = ? | "let p = 6 , q = 7 then 2 * 6 + 7 = 19 so 2 p + q = 19 . answer : e" | a = 6 * 2
b = a + 7
|
a ) $ 0.62 , b ) $ 0.70 , c ) $ 0.80 , d ) $ 0.90 , e ) $ 1.00 | a | add(multiply(multiply(multiply(0.08, const_2), const_2), const_2), 0.30) | a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.08 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the fourth day it is overdue ? | "1 st day fine - 0.08 2 nd day fine - 0.08 * 2 = 0.16 ( as doubling gives lower value ) 3 rd day fine - 0.16 * 2 = 0.32 ( as doubling gives lower value ) 4 th day fine - 0.32 + 0.3 = 0.62 ( as doubling gives higher value we add 0.3 this time ) answer : a ." | a = 0 * 8
b = a * 2
c = b * 2
d = c + 0
|
a ) $ 3000 , b ) $ 3500 , c ) $ 4000 , d ) $ 4200 , e ) $ 4500 | a | divide(1200, multiply(subtract(divide(800, const_1000), divide(200, const_1000)), subtract(const_1, divide(const_1, const_3)))) | mary and mike enter into a partnership by investing $ 800 and $ 200 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if mary received $ 1200 more than mike did , what was the profit made by their business in that year ? | "explanatory answer let the profit made during the year be $ 3 x therefore , $ x would have been shared equally and the remaining $ 2 x would have been shared in the ratio 8 : 2 . i . e . , 80 % of 2 x would go to mary and 20 % of 2 x would go to mike . hence , mary would get ( 80 - 20 ) % of 2 x more than mike or 60 % of 2 x = $ 1200 i . e . , ( 60 / 100 ) * 2 x = 1200 or 2 x = 2000 . hence , the profit made by the company during the year $ 3 x = $ 3000 . answer : a" | a = 800 / 1000
b = 200 / 1000
c = a - b
d = 1 / 3
e = 1 - d
f = c * e
g = 1200 / f
|
a ) 1 : 72 , b ) 1 : 76 , c ) 1 : 80 , d ) 1 : 84 , e ) 1 : 88 | b | divide(divide(sqrt(multiply(3, 4332)), const_2), 4332) | the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 4332 sq cm . what is the ratio of the width and the area of the rectangle ? | "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 4332 12 x ^ 2 = 4332 x ^ 2 = 361 x = 19 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 76 the answer is b ." | a = 3 * 4332
b = math.sqrt(a)
c = b / 2
d = c / 4332
|
a ) 0 and 3 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 7 , e ) 7 and 9 | b | add(multiply(floor(power(30, inverse(3))), const_10), add(floor(power(30, inverse(3))), const_1)) | if a and b are positive numbers , and a ^ 3 + b ^ 3 = 30 , then the greatest possible value of a is between : | "if a = 3.1 and b is very small , then a ^ 3 + b ^ 3 can equal 30 . if a > 4 , then a ^ 3 + b ^ 3 > 30 . the answer is b ." | a = 1/(3)
b = 30 ** a
c = math.floor(b)
d = c * 10
e = 1/(3)
f = 30 ** e
g = math.floor(f)
h = g + 1
i = d + h
|
a ) 3000 , b ) 2000 , c ) 4000 , d ) 2500 , e ) 3500 | b | subtract(5000, multiply(const_3, const_1000)) | an owner of a pizza stand sold small slices of pizza for rs . 150 each and large slices for rs . 250 each . one night he sold 5000 slices , for a total of rs . 10.50 lakh . how many small slices were sold ? | explanatory answer this question is a word problem in linear equations . the problem is solved by framing two linear equations from the information given in the question and solving them . let ' s ' be the number of small slices and ' b ' the number of large slices sold on that night . therefore , s + b = 5000 . . . eqn ( 1 ) each small slice was sold for rs . 150 . therefore , ' s ' small slices would have fetched rs . 150 s . each large slice was sold for rs . 250 . therefore , ' b ' large slices would have fetched rs . 250 b . total value of sale = 150 s + 250 b = 10 , 50000 ( note the left hand side is in rupees and hence we have to convert right hand also to rupees from lakhs of rupees ) or 150 s + 250 b = 10 , 50000 . . . eqn ( 2 ) multiplying equation ( 1 ) by 150 , we get 150 s + 150 b = 7 , 50000 . . . eqn ( 3 ) subtracting eqn ( 3 ) from eqn ( 2 ) , we get 100 b = 300000 or b = 3000 we know that s + b = 5000 so , s = 5000 - b = 5000 - 3000 = 2000 . 2000 small slices were sold . answer b | a = 3 * 1000
b = 5000 - a
|
a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) 7 hours | e | divide(189, add(22, 5)) | a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 189 km downstream | "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 189 km time taken = distance / speed = 189 / 27 = 7 hours . answer : option e" | a = 22 + 5
b = 189 / a
|
a ) 380 times , b ) 236 times , c ) 382 times , d ) 392 times , e ) 482 times | b | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 21) | light glows for every 21 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 21 ) = 236 times . answer : b" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 21
|
a ) 40 sec , b ) 42 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec | d | divide(360, multiply(subtract(45, 240), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 240 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 240 = 600 m required time = 600 * 2 / 25 = 48 sec answer : d" | a = 45 - 240
b = a * const_0_2778
c = 360 / b
|
a ) s . 1090 , b ) s . 1160 , c ) s . 1232 , d ) s . 1202 , e ) s . 1204 | c | divide(multiply(subtract(const_100, 12), 1400), const_100) | a man buys a cycle for rs . 1400 and sells it at a loss of 12 % . what is the selling price of the cycle ? | "since , c . p = 1400 loss % = ( c . p - s . p ) / c . p * 100 12 = ( 1400 - s . p ) / 1400 * 100 so , after solving answer = 1232 . answer : c" | a = 100 - 12
b = a * 1400
c = b / 100
|
a ) 4 , b ) 8 , c ) 16 , d ) 24 , e ) 2 | d | divide(divide(64, const_4), const_2) | all the faces of cubes are painted with red colour . it is then cut into 64 equal small cubes . find how many small cubes have only one face coloured ? | "there are 64 small cubes , hence one side side of the big cube is 3 β 64 = 4 cm number of small cubes having only one face coloured = ( x - 2 ) 2 x no . of faces = ( 4 - 2 ) ^ 2 x 6 = 24 answer : d" | a = 64 / 4
b = a / 2
|
a ) 12.45 , b ) 12.5 , c ) 12.65 , d ) 12.35 , e ) 12.25 | e | multiply(divide(add(multiply(20, divide(10, const_100)), multiply(30, divide(15, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 20 percent on clothing , 50 percent on food , and 30 percent on other items . if jill paid a 10 percent tax on the clothing , 15 percent tax on the food , and no tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "assume she has $ 200 to spend . tax clothing = 20 % = $ 40 = $ 10.00 food = 50 % = $ 100 = $ 15.00 items = 30 % = $ 60 = $ 0.00 total tax = $ 25.00 % of total amount = 25 / 200 * 100 = 12.25 % answer e" | a = 10 / 100
b = 20 * a
c = 15 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 39.5 % , b ) 38.5 % , c ) 36.5 % , d ) 37.5 % , e ) 30.5 % | b | divide(multiply(10, const_100), 26) | if there are 26 laborers in a crew , and on a certain day , 10 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | "10 / 26 * 100 = 38.46 38.5 % correct answer b" | a = 10 * 100
b = a / 26
|
a ) 20 , b ) 5 , c ) 10 , d ) 36 , e ) 25 | d | multiply(5, const_4) | how many 3 - digit numbers can be formed from the digits 2 , 3 , 5 , 6 , 7 and 9 , which are divisible by 5 and if the repetition is allowed ? | "since , each desired number is divisible by 5 , so we must have 5 at the unit place . so , the tens place can now be filled by any of the remaining 6 digits ( 2 , 3 , 6 , 7 , 9 , 5 ) therefore the 3 - digit no can be formed by ( 6 * 6 * 1 ) = 36 ways - - - - - - ( since repetition is allowed ) ans - d" | a = 5 * 4
|
a ) rs . 40 , b ) rs . 50 , c ) rs . 49 , d ) rs . 59 , e ) none of these | d | divide(add(66, 52), const_2) | the profit obtained by selling an article for rs . 66 is the same as the loss obtained by selling it for rs . 52 . what is the cost price of the article ? | "s . p 1 - c . p = c . p β s . p 2 66 - c . p = c . p - 52 2 c . p = 66 + 52 ; c . p = 118 / 2 = 59 answer : d" | a = 66 + 52
b = a / 2
|
a ) 2 , b ) 3 , c ) 7 , d ) 14 , e ) 21 | e | lcm(7, 9) | if 7 and 9 are factors of 60 n , what is the minimum value of n ? | "60 n / 7 * 9 should be integer = > 2 * 2 * 3 * 5 * n / 7 * 3 * 3 = 2 * 2 * 5 * n / 21 must be an integer for this to be true n must multiple of 21 , thus min of n = 21 hence e" | a = math.lcm(7, 9)
|
a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 33 , d ) 4 / 27 , e ) 3 / 23 | c | divide(choose(5, 2), choose(add(add(5, 4), 3), 2)) | a bag contains 5 red , 4 blue and 3 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 5 c 212 c 2 = 5 c 212 c 2 = 10 / 66 = 5 / 33 c" | a = math.comb(5, 2)
b = 5 + 4
c = b + 3
d = math.comb(c, 2)
e = a / d
|
a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | a | divide(multiply(240, const_100), subtract(const_100, 70)) | in an election between two candidates , the first candidate got 70 % of the votes and the second candidate got 240 votes . what was the total number of votes ? | "let v be the total number of votes . 0.3 v = 240 v = 800 the answer is a ." | a = 240 * 100
b = 100 - 70
c = a / b
|
a ) 10 % , b ) 33.33 % , c ) 40 % , d ) 86.66 % , e ) 66.66 % | d | divide(subtract(38, 25), subtract(divide(40, const_100), divide(25, const_100))) | seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 38 percent ryegrass , what percent of the weight of this mixture is x ? | - - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 38 % 0.4 x + ( m - x ) 0.25 = 0.38 m 0.15 x = 0.13 m x = 0.8666 m x = 86.66 % of m d | a = 38 - 25
b = 40 / 100
c = 25 / 100
d = b - c
e = a / d
|
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