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 ) 73 , b ) 32 , c ) 34 , d ) 43 , e ) 96 | e | sqrt(divide(multiply(square_area(8), 72), inverse(const_2))) | the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 72 of the area of the field . what is the length of the field ? | "explanation : a / 72 = 8 * 8 = > a = 8 * 8 * 72 x * 2 x = 8 * 8 * 72 x = 48 = > 2 x = 96 answer : option e" | a = square_area * (
b = a / 72
c = 1/(2)
d = math.sqrt(b)
|
a ) 0 , b ) 1 , c ) 27 , d ) 54 , e ) 0 | e | divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4) | for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.367 ] = 1 and [ 1.899 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | "k = [ lcm of 80 and 45 ] * ( any integer ) however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer for value of k ( 60 ) * any integer unit value will be always zero . e" | a = 3 * 2
b = 3 * 2
c = a * b
d = 3 * 2
e = c * d
f = e / 4
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a ) 648 , b ) 300 , c ) 252 , d ) 225 , e ) 26 | d | add(add(divide(subtract(1000, 80), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | how many times digit 6 is used while writing numbers from 80 to 1000 ? | there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' so far we have calculated 190 . the total now comes to 280 . the nearest to which is 225 . hence d . | a = 1000 - 80
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 6 * 2
h = f + g
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a ) rs . 65,000 , b ) rs . 70,000 , c ) rs . 80,000 , d ) rs . 90,000 , e ) rs . 60,000 | d | multiply(multiply(add(add(3, 2), 2), add(3, 2)), 2) | p and q invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if p invested rs . 60 , 000 , the amount invested by q is : | "suppose q invested rs . y . then , y = 60,000 / y = 2 / 3 or y = 60000 / 2 * 3 = 90,000 . answer : d" | a = 3 + 2
b = a + 2
c = 3 + 2
d = b * c
e = d * 2
|
a ) 5 , b ) 4 , c ) 2 , d ) 3 , e ) 1 | c | subtract(add(5, 2), 5) | if m = | | n – 3 | – 2 | , for how many values of n is m = 5 ? | m = | | n – 3 | – 2 | can be 4 only and only when n - 3 = + / - 7 . so there are 2 values of n answer : c | a = 5 + 2
b = a - 5
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a ) 24 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(30, 20), multiply(20, 30)), add(30, 20)) | the average of 30 results is 20 and the average of other 20 results is 30 . what is the average of all the results ? | "answer sum of 50 result = sum of 30 result + sum of 20 result . = 30 x 20 + 20 x 30 = 1200 correct option : a" | a = 30 * 20
b = 20 * 30
c = a + b
d = 30 + 20
e = c / d
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a ) 5050 , b ) 5020 , c ) 5040 , d ) 5030 , e ) 5075 | c | divide(80, subtract(divide(const_1, 14), divide(const_1, 18))) | when an amount was distributed among 14 boys , each of them got $ 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys . what was the amount ? | "let the total amount be rs . x the , x / 14 - x / 18 = 80 = = > 2 x / 126 = 80 = = > x / 63 = 63 x 80 = 5040 . hence the total amount is 5040 . answer c ." | a = 1 / 14
b = 1 / 18
c = a - b
d = 80 / c
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a ) 2 : 5 , b ) 20 : 3 , c ) 3 : 7 , d ) 6 : 11 , e ) 2 : 3 | b | divide(divide(const_1, const_4), divide(5, const_100)) | if 5 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ? | "let 5 % of a = 1 / 3 b then 5 a / 100 = 1 b / 3 a / 20 = b / 3 a / b = 20 / 3 a : b = 20 : 3 answer is b" | a = 1 / 4
b = 5 / 100
c = a / b
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a ) 28 % , b ) 55 % , c ) 32 % , d ) 36 % , e ) 72 % | b | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(50, const_100))), const_100)) | a baseball card decreased in value 50 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 50 / 100 ) * 100 = 50 after second year , value of baseball card = ( 1 - 10 / 100 ) * 50 = 45 total percent decrease of the card ' s value over the two years = ( 100 - 45 ) / 100 * 100 % = 55 % answer b" | a = 10 / 100
b = 1 - a
c = 50 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 2 , b ) 3 , c ) 1 , d ) 10 , e ) 25 | e | subtract(26, 1) | 2 ^ 35 ^ 26 ^ 1 find largest value ? | explanation : 2 ^ 3 = 2 * 2 * 2 = 8 5 ^ 2 = 5 * 5 = 25 6 ^ 1 = 6 * 1 = 6 hence 5 ^ 2 is the largest one answer : e | a = 26 - 1
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a ) 52 , b ) 63 , c ) 74 , d ) 85 , e ) 96 | b | 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 585 ? | "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 585 , or 1 / 9 of 585 . 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 585 is between 540 and 630 , so ( 1 / 9 ) * 585 will be about 65 . 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 closest answer is 63 , so this is the answer to choose . the answer is b ." | 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 ) 15 , b ) 45 , c ) 75 , d ) 125 , e ) 150 | a | add(add(add(add(add(multiply(multiply(5, 2), 2), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) has two appearances of 3 ( in 12 ^ 2 ) and one appearance of 5 ( in the 10 ) . thus n must include at least 3 * 5 = 15 the answer is a ." | a = 5 * 2
b = a * 2
c = 5 * 2
d = c * 2
e = b + d
f = 5 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 5 / 16 , b ) 3 / 4 , c ) 9 / 20 , d ) 7 / 10 , e ) 5 / 7 | b | divide(subtract(divide(60, const_100), multiply(subtract(const_1, divide(64, const_100)), subtract(const_1, divide(2, 3)))), divide(64, const_100)) | at a small company , 64 percent of the employees are women , and 60 percent of the employees are married . if 2 / 3 of the men are single , what fraction of the women are married ? | "lets take total employees are 100 . given that , total women = 64 and total married = 60 . total men = 100 - 64 = 36 and single men = 2 / 3 * 36 = 24 . married men = total men - single men = 36 - 24 = 12 . married women = total married - married men = 60 - 12 = 48 . fraction of women are married = married women / total women = 48 / 64 = 3 / 4 . ans b" | a = 60 / 100
b = 64 / 100
c = 1 - b
d = 2 / 3
e = 1 - d
f = c * e
g = a - f
h = 64 / 100
i = g / h
|
a ) 450 , b ) 310 , c ) 250 , d ) 410 , e ) 390 | a | multiply(power(divide(subtract(const_100, 10), const_100), 1), 500) | if annual decrease in the population of a town is 10 % and the present number of people is 500 what will the population be in 1 year ? | "population in 1 year = 500 ( 1 - 10 / 100 ) = 500 * 90 / 100 = 450 answer is a" | a = 100 - 10
b = a / 100
c = b ** 1
d = c * 500
|
a ) 1 , b ) 1 / 2 , c ) 1 and 1 / 2 , d ) 2 , e ) - 1 | d | divide(5, subtract(7, const_1)) | if | x | = 7 x - 5 , then x = ? | "answer : a approach : substituted option a i . e x = 1 . inequality satisfied . d" | a = 7 - 1
b = 5 / a
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a ) 36 - 48 , b ) 22 - 34 , c ) 60 - 24 , d ) 42 - 42 , e ) 21 - 63 | b | divide(subtract(56, 12), const_2) | the sum of two numbers is 56 , 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 56 : word problem = 56 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 56 â ˆ ’ 12 = 44 . x = 44 / 2 = 22 . this is the first number . therefore the other number is x + 12 = 22 + 12 = 34 . the sum of 22 + 34 is 56 . b" | a = 56 - 12
b = a / 2
|
a ) 154 cm 2 , b ) 308 m 2 , c ) 227 m 2 , d ) 407 m 2 , e ) none of these | c | divide(multiply(power(17, const_2), const_pi), const_4) | a horse is tethered to one corner of a rectangular grassy field 46 m by 20 m with a rope 17 m long . over how much area of the field can it graze ? | "area of the shaded portion = 1 ⁄ 4 × π × ( 17 ) 2 = 227 m 2 answer c" | a = 17 ** 2
b = a * math.pi
c = b / 4
|
a ) $ 1260 , b ) $ 1150 , c ) $ 1542 , d ) $ 1000 , e ) $ 1292 | a | divide(1140, subtract(const_1, divide(5, const_100))) | a person incurs 5 % loss by selling a watch for $ 1140 . at what price should the watch be sold to earn 5 % profit ? | let the new selling price be $ x ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) ( 100 - 5 ) / 1140 = ( 100 + 5 ) / x x = 105 * 1140 / 95 = 1260 answer is a | a = 5 / 100
b = 1 - a
c = 1140 / b
|
a ) 15 , b ) 20 , c ) 23 , d ) 27 , e ) 25 | d | add(multiply(10, const_2), multiply(subtract(22.5, multiply(10, const_2)), 10)) | if the average of 10 consecutive integers is 22.5 then the 10 th integer is : - | "the average falls between the 5 th and 6 th integers , integer 5 = 22 , integer 6 = 23 . counting up to the tenth integer we get 27 . answer : d" | a = 10 * 2
b = 10 * 2
c = 22 - 5
d = c * 10
e = a + d
|
a ) 281 , b ) 284 , c ) 704 , d ) 640 , e ) 920 | c | multiply(32, 22) | find the area of a parallelogram with base 32 cm and height 22 cm ? | "area of a parallelogram = base * height = 32 * 22 = 704 cm 2 answer : c" | a = 32 * 22
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a ) $ 1.50 , b ) $ 1.88 , c ) $ 2.09 , d ) $ 2.50 , e ) $ 3.25 | c | multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, 1), const_2)), 350), add(multiply(subtract(multiply(add(const_4, 1), const_2), 4), 32), multiply(multiply(divide(add(const_100, 25), const_100), 4), 32))), divide(add(const_100, 25), const_100)) | sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 350 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 4 of the days were hot , then what was the price of 1 cup on a hot day ? | "6 regular days - - > sales = 6 * 32 * x = 192 x ; 4 hot days - - > sales = 4 * 32 * ( 1.25 x ) = 160 x ; total sales = 192 x + 160 x = 352 x . total cost = 10 * 32 * 0.75 = 240 . profit = 352 x - 240 = 350 - - > x = 1.676 . 1.25 x = ~ 2.09 . answer : c ." | a = 75 / 100
b = 32 * a
c = 4 + 1
d = c * 2
e = b * d
f = e + 350
g = 4 + 1
h = g * 2
i = h - 4
j = i * 32
k = 100 + 25
l = k / 100
m = l * 4
n = m * 32
o = j + n
p = f / o
q = 100 + 25
r = q / 100
s = p * r
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | min(12, 2) | if 1 = 6 2 = 12 3 = 18 4 = 24 5 = 30 6 = 36 then 12 = ? hint : its a logic riddle not a mathematical riddle | b 2 as stated 2 = 12 = > 12 = 2 answer is b | a = min(12)
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a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | c | subtract(900, subtract(add(divide(900, 11), divide(900, 35)), divide(900, multiply(11, 35)))) | what is the number of integers from 1 to 900 ( inclusive ) that are divisible by neither 11 nor by 35 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 892 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? c" | a = 900 / 11
b = 900 / 35
c = a + b
d = 11 * 35
e = 900 / d
f = c - e
g = 900 - f
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a ) 229.3657 , b ) 229.7563 , c ) 229.6357 , d ) 229.5637 , e ) 221.2357 | e | add(add(add(217, 2.017), 0.217), 2.0017) | 217 + 2.017 + 0.217 + 2.0017 = ? | 217 2.017 0.217 + 2.0017 - - - - - - - - 221.2357 - - - - - - - - - answer is e . | a = 217 + 2
b = a + 0
c = b + 2
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a ) 3 : 9 , b ) 3 : 10 , c ) 3 : 21 , d ) 3 : 12 , e ) 3 : 14 | b | divide(divide(divide(3, 2), const_4), divide(5, const_4)) | two men amar and bhuvan have the ratio of their monthly incomes as 6 : 5 . the ratio of their monthly expenditures is 3 : 2 . if bhuvan saves one - fourth of his income , find the ratio of their monthly savings ? a . 3 : 5 | let the monthly incomes of amar and bhuvan be 6 x and 5 x respectively . let the monthly expenditure of amar and bhuvan be 3 y and 2 y respectively . savings of bhuvan every month = 1 / 4 ( 5 x ) = ( his income ) - ( his expenditure ) = 5 x - 2 y . = > 5 x = 20 x - 8 y = > y = 15 x / 8 . ratio of savings of amar and bhuvan = 6 x - 3 y : 1 / 4 ( 5 x ) = 6 x - 3 ( 15 x / 8 ) : 5 x / 4 = 3 x / 8 : 5 x / 4 = 3 : 10 . answer : option b | a = 3 / 2
b = a / 4
c = 5 / 4
d = b / c
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a ) 6 , b ) 3 , c ) 1 , d ) 4 , e ) 2 | a | add(add(3, 2), 1) | if 3 cm of a pencil is purple , 2 cm of the remaining is black and the remaining 1 cm is blue , what is the total length of the pencil ? | answer is a ) 6 | a = 3 + 2
b = a + 1
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a ) 104 kmph , b ) 194 kmph , c ) 109 kmph , d ) 260 kmph , e ) 271 kmph | d | divide(624, add(2, divide(2, 5))) | a car covers a distance of 624 km in 2 2 / 5 hours . find its speed ? | "624 / 2 2 / 5 = 260 kmph answer : d" | a = 2 / 5
b = 2 + a
c = 624 / b
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['a ) 27', 'b ) 26', 'c ) 28', 'd ) 25', 'e ) 24'] | a | add(add(add(4, add(3, 3)), sqrt(add(power(subtract(multiply(4, 3), add(add(3, 3), sqrt(subtract(power(4, const_2), power(3, const_2))))), const_2), power(3, const_2)))), multiply(4, 3)) | find the perimetri of trapezium with sides 4 , 6,12 and distance between two parallel side is 3 . | 1 / 2 * ( 6 + 12 ) * 3 = 27 answer : a | a = 3 + 3
b = 4 + a
c = 4 * 3
d = 3 + 3
e = 4 ** 2
f = 3 ** 2
g = e - f
h = math.sqrt(g)
i = d + h
j = c - i
k = j ** 2
l = 3 ** 2
m = k + l
n = math.sqrt(m)
o = b + n
p = 4 * 3
q = o + p
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a ) 32.2 , b ) 32.98 , c ) 32.3 , d ) 32.8 , e ) 32.4 | a | add(20, const_1) | the average of first five prime numbers greater than 20 is ? | "23 + 29 + 31 + 37 + 41 = 161 / 5 = 32.2 answer : a" | a = 20 + 1
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a ) 12 , b ) 9 , c ) 15 , d ) 21 , e ) 30 | a | divide(subtract(0.60, multiply(0.06, 6)), subtract(0.08, 0.06)) | a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 6 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company charges the same amount under plan a and under plan b ? | "let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 6 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 6 ) = 0.08 t - - > 60 + 6 ( t - 6 ) = 8 t - - > t = 12 . answer : a ." | a = 0 * 6
b = 0 - 60
c = 0 - 8
d = b / c
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a ) 80 % , b ) 75 % , c ) 70 % , d ) 65 % , e ) 60 % | a | multiply(divide(divide(subtract(20, 12), subtract(12, 10)), add(divide(subtract(20, 12), subtract(12, 10)), const_1)), const_100) | solution x is 10 % chemical a and 90 % chemical b by volume . solution y is 20 % chemical a and 80 % chemical b by volume . if a mixture of x and y is 12 % chemical a , what percent of the mixture is solution x ? | "the volume of the mixture be x + y . 0.1 x + 0.2 y = 0.12 ( x + y ) x = 4 y x / ( x + y ) = 4 / 5 = 80 % . the answer is a ." | a = 20 - 12
b = 12 - 10
c = a / b
d = 20 - 12
e = 12 - 10
f = d / e
g = f + 1
h = c / g
i = h * 100
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a ) 38 , b ) 34 , c ) 58 , d ) 63 , e ) 73 | c | subtract(add(42, 17), const_1) | the flowerman sale , all of the prices of the flowers sold were different . if the price of a radio sold at the flowerman sale was both the 17 th highest price and the 42 th lowest price among the prices of the slowers sold , how many flowers were sold at the flowerman sale ? | 16 + 41 + 1 = 58 answer : c | a = 42 + 17
b = a - 1
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a ) 15 litres , b ) 16 litres , c ) 17 litres , d ) 25 litres , e ) 35 litres | a | subtract(divide(multiply(divide(5, const_100), 10), divide(2, const_100)), 10) | milk contains 5 % water . what quantity of pure milk should be added to 10 litres of milk to reduce this 2 % ? | present ration of milk to water is 95 : 5 amount of milk is 10 litres hence , pure milk content is 9.5 litre and water content is 0.5 litres now , we should not disturb the water content so , 0.5 litres of water is equivalent to 2 % so 100 % mixture is : ( 0.5 / 2 ) * 100 = 25 litre thus , milk should be added = 25 - 10 = 15 litres of pure milk answer : a | a = 5 / 100
b = a * 10
c = 2 / 100
d = b / c
e = d - 10
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a ) 180 , b ) 245 , c ) 320 , d ) 260 , e ) 270 | c | divide(multiply(80, multiply(const_3, 5)), 5) | in the set of positive integers from 1 to 80 , what is the sum of all the odd multiples of 5 ? | "reduce to 1 - 80 5 - 15 - 25 - 35 - 45 - - 55 - - 65 - - 75 are valid multiples . add them - - > 320 c" | a = 3 * 5
b = 80 * a
c = b / 5
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | b | divide(add(18, multiply(3, 8)), subtract(8, const_1)) | dan ' s age after 18 years will be 8 times his age 3 years ago . what is the present age of dan ? | "let dan ' s present age be x . x + 18 = 8 ( x - 3 ) 7 x = 42 x = 6 the answer is b ." | a = 3 * 8
b = 18 + a
c = 8 - 1
d = b / c
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a ) 23 % , b ) 18.75 % , c ) 30 % , d ) 30.55 % , e ) 100 % | d | multiply(divide(subtract(add(const_100, 60), add(const_100, 44)), add(const_100, 44)), const_100) | the wages earned by robin is 44 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ? | "explanatory answer let the wages earned by erica be $ 100 then , wages earned by robin and charles will be $ 144 and $ 160 respectively . charles earns $ 44 more than robin who earns $ 144 . therefore , charles ' wage is 44 / 144 * 100 = 30.55 % . the correct choice is ( d )" | a = 100 + 60
b = 100 + 44
c = a - b
d = 100 + 44
e = c / d
f = e * 100
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a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | multiply(divide(subtract(divide(67.5, const_100), divide(55, const_100)), add(subtract(divide(67.5, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(67.5, const_100)))), const_100) | solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 67.5 % percent carbonated water , what percent of the volume of the mixture is p ? | "67.5 % is 12.5 % - points below 80 % and 12.5 % - points above 55 % . so the ratio of solution p to solution q is 1 : 1 . mixture p is 1 / 2 = 50 % of the volume of mixture pq . the answer is c ." | a = 67 / 5
b = 55 / 100
c = a - b
d = 67 / 5
e = 55 / 100
f = d - e
g = 80 / 100
h = 67 / 5
i = g - h
j = f + i
k = c / j
l = k * 100
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a ) 33 , b ) 18 , c ) 21 , d ) 12 , e ) 19 | a | subtract(divide(subtract(multiply(6, 40), 7), const_4), 7) | the average age of a family of 6 members is 40 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 40 = 240 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 240 - 42 = 198 therefore , avg age at the birth of youngest member = 198 / 6 = 33 . answer : a" | a = 6 * 40
b = a - 7
c = b / 4
d = c - 7
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a ) 6 , b ) 9 , c ) 7 , d ) 8 , e ) 2 | b | divide(subtract(divide(80, 3), divide(38, 3)), const_2) | a man rows his boat 80 km downstream and 38 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 80 / ( 3 ) = 27 kmph speed upstream = d / t = 38 / ( 3 ) = 9 kmph the speed of the stream = ( 27 - 9 ) / 2 = 9 kmph answer : b" | a = 80 / 3
b = 38 / 3
c = a - b
d = c / 2
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a ) – 2 , b ) – 1 , c ) 0 , d ) 1 , e ) 2 | c | divide(5, 4) | if x ⁄ 2 + 5 ⁄ 4 = 5 ⁄ 4 , what is the value of x ? | x ⁄ 2 + 5 ⁄ 4 = 5 ⁄ 4 let ' s multiply both sides by 4 . 2 x + 5 = 5 2 x = 0 x = 0 the answer is c . | a = 5 / 4
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a ) 12.5 % , b ) 13.5 % , c ) 14.2 % , d ) 14.5 % , e ) 21.5 % | c | divide(const_100, 7) | at what rate percent per annum will a sum of money double in 7 years . | "let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 8 ) ] % = 14.2 % per annum . answer : c" | a = 100 / 7
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a ) 20 , b ) 24 , c ) 19 , d ) 28 , e ) 30 | c | sqrt(add(multiply(131, const_2), 99)) | sum of the squares of 3 no . ' s is 99 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 99 + 2 * 131 a + b + c = √ 361 = 19 c" | a = 131 * 2
b = a + 99
c = math.sqrt(b)
|
a ) 2288 , b ) 2779 , c ) 2779 , d ) 3900 , e ) 2600 | e | multiply(multiply(subtract(add(80, 60), 10), 10), 2) | a rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 2 per sq m ? | "area = ( l + b â € “ d ) d ( 80 + 60 â € “ 10 ) 10 = > 1300 m 2 1300 * 2 = rs . 2600 answer : e" | a = 80 + 60
b = a - 10
c = b * 10
d = c * 2
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a ) 148 , b ) 154 , c ) 156 , d ) 144 , e ) none | b | multiply(39, divide(150, divide(add(add(37, 39), const_2), const_2))) | the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 150 . what is the number of frogs in the pond ? | "solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 150 . so , total number of ducks and frogs in the pond , = 2 * 150 = 300 . therefore , number of frogs , = ( 300 * 39 ) / 76 = 154 . answer : option b" | a = 37 + 39
b = a + 2
c = b / 2
d = 150 / c
e = 39 * d
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a ) rs . 420000 , b ) rs . 180000 , c ) rs . 201600 , d ) rs . 470400 , e ) none of these | d | multiply(multiply(multiply(14000, add(const_1, divide(12, const_100))), divide(5, 2)), 12) | the monthly incomes of a and b are in the ratio 5 : 2 . b ' s monthly income is 12 % more than c ' s monthly income . if c ' s monthly income is rs . 14000 , then find the annual income of a ? | "b ' s monthly income = 14000 * 112 / 100 = rs . 15680 b ' s monthly income = 2 parts - - - - > rs . 15680 a ' s monthly income = 5 parts = 5 / 2 * 15680 = rs . 39200 a ' s annual income = rs . 39200 * 12 = rs . 470400 answer : d" | a = 12 / 100
b = 1 + a
c = 14000 * b
d = 5 / 2
e = c * d
f = e * 12
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above | a | sqrt(power(2, const_2)) | in the rectangular coordinate system , points ( 2 , 0 ) and ( – 2 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ? | "the answer is a it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 2" | a = 2 ** 2
b = math.sqrt(a)
|
a ) 10 , b ) 5 , c ) 8 , d ) 6 , e ) 13 | d | inverse(add(divide(const_1, 10), divide(const_1, 15))) | if ram and gohul can do a job in 10 days and 15 days independently , how many days would they take to complete the same job working simultaneously ? | if total work is x . ram working rate = x / 10 per day . working rate of gohul = x / 15 per day . rate of work = ( x / 10 ) + ( x + 15 ) = 30 x / 5 x = 6 days the answer is option d | a = 1 / 10
b = 1 / 15
c = a + b
d = 1/(c)
|
a ) 34 , b ) 36 , c ) 38 , d ) 40 , e ) ( f ) 42 | b | add(add(add(const_10, add(2, 4)), add(2, 6)), add(8, 4)) | trapezoid jklm in the x - y plane has coordinates j = ( – 2 , – 4 ) , k = ( – 2 , 2 ) , l = ( 6 , 8 ) , and m = ( 6 , – 4 ) . what is its perimeter ? | "jk = 6 lm = 12 kl = using distance formula 10 jm = using distance formula 8 sum of all is 36 b" | a = 2 + 4
b = 10 + a
c = 2 + 6
d = b + c
e = 8 + 4
f = d + e
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | d | add(add(7, 11), const_1) | a box contains 7 purple , 5 blue and 11 yellow balls . what is the minimum number of tries required to get one blue and one yellow ball ? | 11 yellow + 7 purple + after that we have only one option that is to be blue so answer is 19 answer : d | a = 7 + 11
b = a + 1
|
a ) 18 , b ) 19 , c ) 20 , d ) 22 , e ) 24 | b | multiply(sqrt(divide(divide(361, 3.00001), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 361 is required to paint the floor at the rate of rs . 3.00001 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 361 / 3 = 120.33 sq m l b = 120.33 i . e . , l * l / 3 = 120.33 l ^ 2 = 361 = > l = 19 . b" | a = 361 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) 12 , b ) 15 , c ) 18 , d ) 36 , e ) 24 | d | multiply(factorial(3), factorial(3)) | 3 men and 3 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | the list should be wmwmw . hence , from women 3 ! and men 3 ! , we get ( 3 ! ) ( 3 ! ) = 36 . therefore , the correct answer is d . | a = math.factorial(3)
b = math.factorial(3)
c = a * b
|
['a ) 24', 'b ) 26', 'c ) 28', 'd ) 30', 'e ) 32'] | d | divide(rectangle_area(12, 15), 6) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 12 inches by 15 inches and jordan ' s rectangle is 6 inches long , how wide is jordan ' s rectangle , in inches ? | area of first rectangle is 12 * 15 = 180 hence area of second would be 6 x = 180 x x = 30 answer is d | a = rectangle_area / (
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a ) 2105 , b ) 1955 , c ) 1945 , d ) 1935 , e ) 2108 | e | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2080) | the calendar of the year 2080 can be used again in the year ? | "explanation : given year 2080 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2080 + 28 = 2108 answer : e" | a = 4 * 4
b = a - 2
c = b * 2
d = c + 2080
|
a ) 1764 , b ) 1774 , c ) 1784 , d ) 1794 , e ) 1754 | a | multiply(multiply(42, const_2), divide(42, const_2)) | sum of 42 odd numbers is ? | "sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 42 odd numbers = 42 ^ 2 = 1764 answer : a" | a = 42 * 2
b = 42 / 2
c = a * b
|
a ) 1.25 % , b ) 3.75 % , c ) 6.25 % , d ) 6.67 % , e ) 11.11 % | e | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 5,500 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | the remaining solution will be approximately what percent sodium chloride ? means : what percent of the remaining solution is sodium chloride . now , since the remaining solution is 10,000 - 5,500 = 4,500 gallons and sodium chloride is 500 gallons ( 5 % of initial solution of 10,000 gallons ) then sodium chloride is 500 / 4,500 * 100 = ~ 11.11 % of the remaining solution of 4,500 gallons . answer : e . | a = 100 * 100
b = 5 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 149 , b ) 169 , c ) 189 , d ) 529 , e ) 219 | d | subtract(negate(289,361), multiply(subtract(25,49, 121,169), divide(subtract(25,49, 121,169), subtract(4, 25,49)))) | 4 , 25,49 , 121,169 , 289,361 , | "23 ^ 2 = 529 because follow sequence of square of the prime numbers answer : d" | a = negate - (
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a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | b | multiply(divide(add(20, divide(20, const_2)), 6), const_2) | the distance from steve ' s house to work is 20 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 20 km = > speed = 10 kmph b" | a = 20 / 2
b = 20 + a
c = b / 6
d = c * 2
|
a ) 4 days , b ) 21 days , c ) 13 days , d ) 12 days , e ) 3 days | d | inverse(subtract(3, divide(3, 4))) | a and b can do a piece of work in 4 days . with the help of c they finish the work in 3 days . c alone can do that piece of work in ? | "c = 1 / 3 – 1 / 4 = 1 / 12 = > 12 days answer : d" | a = 3 / 4
b = 3 - a
c = 1/(b)
|
a ) 32 m , b ) 22 m , c ) 24 m , d ) 26 m , e ) 28 m | a | add(add(multiply(subtract(16, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 16 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is | "explanation : each row contains 16 plants . there are 15 gapes between the two corner trees ( 15 x 2 ) metres and 1 metre on each side is left . therefore length = ( 30 + 2 ) m = 32 m . answer : a" | a = 16 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
|
a ) 15 , b ) 16 , c ) 18 , d ) 20 , e ) 24 | a | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 10)))) | on a certain date , pat invested $ 9,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 10 years will be $ 36,000 , in how many years total will the total value of the investment plus interest increase to $ 72,000 ? | "36,000 = 9,000 ( 1 + x ) ^ 10 4 = ( 1 + x ) ^ 10 = 2 ^ 2 ( 1 + x ) ^ 10 = ( ( 1 + x ) ^ 5 ) ^ 2 = 2 ^ 2 therefore , ( 1 + x ) ^ 5 = 2 72,000 = 9,000 ( 1 + x ) ^ n 8 = ( 1 + x ) ^ n 2 ^ 3 = ( 1 + x ) ^ n ( 1 + x ) ^ n = ( ( 1 + x ) ^ 5 ) ^ 3 = ( 1 + x ) ^ 15 therefore , n = 15 . the answer is a ." | a = 3 * 10
b = 4 + 1
c = a / b
d = math.log(c)
e = 2 * 10
f = 4 + 1
g = e / f
h = 1 / 10
i = g ** h
j = math.log(i)
k = d / j
|
a ) 10 , b ) 11 , c ) 12 , d ) can not be determined , e ) none of the above | c | divide(subtract(510, multiply(75, 2.00)), 30) | 30 pens and 75 pencils were purchased for 510 . if the average price of a pencil was 2.00 , find the average price of a pen . | "since average price of a pencil = 2 ∴ price of 75 pencils = 150 ∴ price of 30 pens = ( 510 – 150 ) = 360 ∴ average price of a pen = 360 ⁄ 60 = 12 answer c" | a = 75 * 2
b = 510 - a
c = b / 30
|
['a ) 5 m', 'b ) 12 m', 'c ) 13 m', 'd ) 14.5 m', 'e ) 15.5 m'] | b | divide(60, sqrt(divide(multiply(60, const_10), subtract(power(5, const_2), const_1)))) | a rectangular carpet has an area of 60 sq . m . if its diagonal and longer side together equal 5 times the shorter side , the length of the carpet is : | let , length = x meters and breadth = y meters then xy = 60 and ( x 2 + y 2 ) + x = 5 therefore , x = 60 and ( x 2 + y 2 ) = ( 5 y - x ) 2 or xy = 60 and 24 y 2 - 10 xy = 0 . therefore , 24 y 2 - 10 * 60 = 0 or y 2 = 25 or = 5 . therefore , x = ( 60 / 5 ) m = 12 m . so , length of the carpet = 12 m answer : b | a = 60 * 10
b = 5 ** 2
c = b - 1
d = a / c
e = math.sqrt(d)
f = 60 / e
|
a ) 44 , b ) 42.343 , c ) 50 , d ) 39.252 , e ) 27.851 | b | divide(add(357, 137), divide(multiply(42, const_1000), const_3600)) | a train is 357 meter long is running at a speed of 42 km / hour . in what time will it pass a bridge of 137 meter length ? | "speed = 42 km / hr = 42 * ( 5 / 18 ) m / sec = 35 / 3 m / sec total distance = 357 + 137 = 494 meter time = distance / speed = 494 * ( 3 / 35 ) = 42.343 seconds . answer : b" | a = 357 + 137
b = 42 * 1000
c = b / 3600
d = a / c
|
a ) 9 , b ) 16 , c ) 18 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(34, divide(multiply(3, 2), multiply(4, 3))) | a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 34 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 34 pints will require 9 / 17 * 34 = 18 points . . second way . . get both red and white paints in same denominator . . . 3 / 4 + 2 / 3 = ( 9 / 12 ) + ( 8 / 12 ) pints . . from here we get the ratio of red : white = 9 : 8 . . so in 34 pints red will be 9 / ( 9 + 8 ) * 34 = 18 . . answer : c" | a = 3 * 2
b = 4 * 3
c = a / b
d = 34 * c
|
a ) 0.039 , b ) 0.235 , c ) 0.25 , d ) 4 , e ) 5 | d | divide(subtract(power(add(0.137, 0.098), const_2), power(subtract(0.137, 0.098), const_2)), multiply(0.137, 0.098)) | the value of ( 0.137 + 0.098 ) ² - ( 0.137 - 0.098 ) ² / 0.137 × 0.098 is | solution given expression = ( a + b ) ² - ( a - b ) ² / ab = 4 ab / ab = 4 . answer d | a = 0 + 137
b = a ** 2
c = 0 - 137
d = c ** 2
e = b - d
f = 0 * 137
g = e / f
|
a ) s . 4991 , b ) s . 5991 , c ) s . 6001 , d ) s . 6991 , e ) s . 7391 | e | subtract(multiply(add(5, const_1), 6900), add(add(add(add(6435, 6927), 6855), 7230), 6562)) | a grocer has a sale of rs . 6435 , rs . 6927 , 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 fr 5 mnths = ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs 34009 . reqd . sale = rs . [ ( 6900 * 6 ) - 34009 ] = rs . ( 41400 - 34009 ) = rs . 7391 . answer : e" | a = 5 + 1
b = a * 6900
c = 6435 + 6927
d = c + 6855
e = d + 7230
f = e + 6562
g = b - f
|
a ) 26.4 % , b ) 21 % , c ) 20 % , d ) 19 % , e ) none of these | a | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 457), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 457 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 457 , discount = 621 - 457 = 164 discount % = 164 / 621 * 100 = 26.4 % answer : a" | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 457
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
|
a ) 1 , b ) 5 , c ) 4 , d ) 6 , e ) 2 | a | multiply(multiply(multiply(const_2, const_2), const_2), 8) | a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of 8 cubic feet . if a small block has half the width , half the depth , and half the length of a normal block , what is the volume of cheese in a small block in cubic feet ? | "volume of cube = lbh = 8 new cube l , b , h are decreases of . 5 l , . 5 b , . 5 h new volume of cube = . 5 l * . 5 b * . 5 h = . 125 * lbh = . 125 * 8 = 1 answer : a" | a = 2 * 2
b = a * 2
c = b * 8
|
a ) rs . 400 , b ) rs . 450 , c ) rs . 550 , d ) rs . 600 , e ) rs . 650 | c | add(divide(450, subtract(const_1, divide(10, const_100))), multiply(divide(450, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | boy sells a book for rs . 450 he gets a loss of 10 % , to gain 10 % , at price he should sell ? | "find selling price to gain 10 % . now , we are asked to find selling price to gain 10 % profit . hint : selling price = ( 100 + gain % ) × c . p . 100 selling price = ( 100 + 10 ) × 500 100 selling price = ( 110 ) × 500 100 therefore , selling price = rs . 550 c" | a = 10 / 100
b = 1 - a
c = 450 / b
d = 10 / 100
e = 1 - d
f = 450 / e
g = 10 / 100
h = f * g
i = c + h
|
a ) 100 , b ) 120 , c ) 140 , d ) 160 , e ) 180 | c | divide(700, multiply(18, const_0_2778)) | how many seconds does sandy take to cover a distance of 700 meters , if sandy runs at a speed of 18 km / hr ? | "18 km / hr = 18000 m / 3600 s = 5 m / s time = 700 / 5 = 140 seconds the answer is c ." | a = 18 * const_0_2778
b = 700 / a
|
a ) 272258 , b ) 272358 , c ) 273258 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 51), const_100) | 5358 x 51 = ? | "5358 x 51 = 5358 x ( 50 + 1 ) = 5358 x 50 + 5358 x 1 = 267900 + 5358 = 273258 . c )" | a = 5358 / 51
b = a * 100
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a ) 1 / 5 , b ) 2 / 8 , c ) 5 / 9 , d ) 27 / 40 , e ) 15 / 30 | d | subtract(subtract(1, divide(const_1, 4)), divide(subtract(1, divide(const_1, 4)), 10)) | a company purchased raw material worth 1 / 4 of it capital , 1 / 10 of the remaining is spent on machinery . how much fraction of the capital was left with the company after its expenditures on raw material and machinery ? | the amount spent by the company on raw material is 1 / 4 , so remaining is 3 / 4 . 1 / 10 of the remaining is spent on machinery , 1 / 10 * 3 / 4 = 3 / 40 total amount spent on raw material and machinery is 1 / 4 + 3 / 40 to make the denominator common we use 40 , so 1 / 4 is 10 / 40 10 / 40 + 3 / 40 = 13 / 40 is spent on raw material and machinery remaining 27 / 40 is left with the company answer is d | a = 1 / 4
b = 1 - a
c = 1 / 4
d = 1 - c
e = d / 10
f = b - e
|
a ) 12 , b ) 8 , c ) 16 , d ) 6 , e ) 2 | b | power(multiply(256, 2), divide(const_1, const_3)) | if x = y = 2 z and xyz = 256 then what is the value of x ? | given that x = y = z . so , x = y , x = 2 z , y = 2 z . given xyz = 256 = > ( 2 z ) * ( 2 z ) * z = 256 = > 4 z ^ 3 = 256 = > z = 4 then x = 2 z = 2 ( 4 ) = 8 answer : b | a = 256 * 2
b = 1 / 3
c = a ** b
|
a ) 123 , b ) 167 , c ) 178 , d ) 199 , e ) 288 | e | divide(power(multiply(const_3, power(5184, const_2)), divide(const_1, const_3)), 1.5) | danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 5184 , what is the area of the side face of the box ? | "lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 / 2 w * d ( l = 1 / 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 5184 l * 1.5 l * 2 l = 5184 l = 12 side face area = l * d = l * 2 l = 12 * 2 * 12 = 288 e is the answer" | a = 5184 ** 2
b = 3 * a
c = 1 / 3
d = b ** c
e = d / 1
|
a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed r of the train . | "let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y / x * 1 / 3 = 7 / 12 y / x = 7 / 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) / ( 3 x / 4 ) - ( y - 24 ) / x = 25 / 60 4 ( y - 24 ) / 3 x - ( y - 24 ) / x = 5 / 12 ( y - 24 ) / x ( 4 / 3 - 1 ) = 5 / 12 ( y - 24 ) / x * 1 / 3 = 5 / 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = r ans = c" | a = 24 * 4
b = a / 3
c = b - 24
d = 35 - 25
e = d + 50
f = 35 - 25
g = e / f
h = c * g
|
a ) 31 , b ) 32 , c ) 43 , d ) 45 , e ) 26 | a | add(add(subtract(multiply(5, 4), multiply(3, const_2.0)), multiply(4, const_2.0)), add(subtract(multiply(2, 2), multiply(3, 4)), multiply(2, 4))) | if a ã — b = 2 a - 3 b + ab , then 4 ã — 5 + 5 ã — 4 is equal to : | "explanation : 4 ã — 5 + 5 ã — 4 = ( 2 ã — 4 - 3 ã — 5 + 4 ã — 5 ) + ( 2 ã — 5 - 3 ã — 4 + 5 ã — 4 ) = ( 8 - 15 + 20 + 10 - 12 + 20 ) = 31 . answer : a" | a = 5 * 4
b = 3 * 2
c = a - b
d = 4 * 2
e = c + d
f = 2 * 2
g = 3 * 4
h = f - g
i = 2 * 4
j = h + i
k = e + j
|
a ) 50 years , b ) 20 years , c ) 5 years , d ) 10 years , e ) 90 years | c | multiply(divide(100, add(const_10, const_10)), const_12) | my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 100 years . can you tell my grandson age in years ? | let m be my age in years . if s is my son ' s age in years , then my son is 52 s weeks old . if g is my grandson ' s age in years , then my grandson is 365 g days old . thus , 365 g = 52 s . since my grandson is 12 g months old , 12 g = m . since my grandson , my son and i together are 120 years , g + s + m = 100 . the above system of 3 equations in 3 unknowns ( g , s and m ) can be solved as follows : g + 365 g / 52 + 12 g = 100 or 52 g + 365 g + 624 g = 5,200 or g = 5,200 / 1,041 = 5 years answer : c | a = 10 + 10
b = 100 / a
c = b * 12
|
a ) 699990 , b ) 99990 , c ) 99980 , d ) 69300 , e ) none of these | d | subtract(multiply(multiply(add(const_3, const_4), const_10), const_1000), multiply(add(const_3, const_4), const_10)) | what is the difference between the place values of two sevens in the numeral 54179759 ? | explanation : required difference = 70000 - 700 = 69300 answer is d | a = 3 + 4
b = a * 10
c = b * 1000
d = 3 + 4
e = d * 10
f = c - e
|
a ) 17 litres , b ) 22.67 litres , c ) 11 litres , d ) 07 litres , e ) 38.67 litres | b | divide(subtract(multiply(34, 1399.45), multiply(34, 262.85)), subtract(3104.35, 1399.45)) | the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 34 litres of the mixed fruit juice . how many litres of the a ç ai berry juice does he need to add ? | "262.85 ( 34 ) + 3 , 104.35 x = 1 , 399.45 ( 34 + x ) solve the equation . 262.85 ( 34 ) + 3 , 104.35 x = 1 , 399.45 ( 34 + x ) 8936.9 + 3 , 104.35 x = 47 , 581.3 + 1 , 399.45 x 8936.9 + 1 , 704.9 x = 47 , 581.3 1 , 704.9 x = 38 , 644.40 x ≈ 22.67 answer is b ." | a = 34 * 1399
b = 34 * 262
c = a - b
d = 3104 - 35
e = c / d
|
a ) 48 , b ) 50 , c ) 60 , d ) 75 , e ) 100 | c | divide(multiply(100, const_3), sqrt(add(power(4, const_2), power(3, const_2)))) | the equation of line s is y = 4 / 3 * x - 100 . what is the smallest possible distance in the xy - plane from the point with coordinates ( 0 , 0 ) to any point on line s ? | this can be solve in two steps and without any complex calculation . given : equation of line s as y = ( 4 / 3 ) x - 100 . so the line intercept the axes at ( 0 , - 100 ) and ( 75,0 ) . this can be considered a right angle triangle with right angle at ( 0,0 ) . so base = 100 , height = 75 and hypotenuse = 125 ( by pythagoras triplet ) so a perpendicular from the ( 0,0 ) to hypotenuse will be the answer . area of triangle = 0.5 * 100 * 75 = 0.5 * 125 * x = > x = 60 ; so answer is 60 = c | a = 100 * 3
b = 4 ** 2
c = 3 ** 2
d = b + c
e = math.sqrt(d)
f = a / e
|
a ) $ 5.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.40 | a | add(multiply(divide(3.6, divide(2, 5)), 0.35), 2.0) | jim â € ™ s taxi service charges an initial fee of $ 2.0 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim â € ™ s taxi service = 2 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2 + ( 3.6 / . 4 ) * . 35 = 2 + 9 * . 35 = 5.15 $ answer a" | a = 2 / 5
b = 3 / 6
c = b * 0
d = c + 2
|
a ) 25 % , b ) 30 % , c ) 15 % , d ) 20 % , e ) 18 % | a | multiply(divide(multiply(divide(50, const_100), 10), add(10, const_1)), const_100) | to a sugar solution of 10 liters containing 50 % sugar , 10 liter of water is added . the percentage of sugar in the new solution is ? | "quantity of sugar = 50 * 10 / 100 = 5 kg new percentage = 5 / 20 * 100 = 25 % answer is a" | a = 50 / 100
b = a * 10
c = 10 + 1
d = b / c
e = d * 100
|
a ) 18 , b ) 144 , c ) 175 , d ) 216 , e ) 264 | e | add(multiply(divide(360, 3), 2), divide(subtract(360, multiply(divide(360, 3), 2)), 5)) | joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 2 / 3 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | total paint initially = 360 gallons paint used in the first week = ( 2 / 3 ) * 360 = 240 gallons . remaning paint = 120 gallons paint used in the second week = ( 1 / 5 ) * 120 = 24 gallons total paint used = 264 gallons . option e | a = 360 / 3
b = a * 2
c = 360 / 3
d = c * 2
e = 360 - d
f = e / 5
g = b + f
|
a ) 1 , b ) 2 , c ) 6 , d ) 8 , e ) 9 | b | divide(18, subtract(const_10, const_1)) | a two - digit number is such that the product of the digits is 8 . when 18 is added to the number , then the digits are reversed . the number is ? | let the ten ' s and unit digit be x and 8 respectively . x then , 10 x + 8 + 18 = 10 x 8 + x x x 10 x 2 + 8 + 18 x = 80 + x 2 9 x 2 + 18 x - 72 = 0 x 2 + 2 x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0 x = 2 option b | a = 10 - 1
b = 18 / a
|
a ) 12 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | add(divide(subtract(multiply(floor(divide(47, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1) | how many numbers from 10 to 47 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , . 12 numbers . 10 / 3 = 3 and 47 / 3 = 15 = = > 15 - 3 = 12 . therefore 12 digits a )" | a = 47 / 3
b = math.floor(a)
c = b * 3
d = 10 / 3
e = math.floor(d)
f = e + 1
g = f * 3
h = c - g
i = h / 3
j = i + 1
|
a ) rs 60 , b ) rs 70 , c ) rs 90 , d ) rs 75 , e ) rs 50 | b | multiply(divide(const_100, 10), 7) | a 7 % stock yields 10 % . the market value of the stock is : | "explanation : for an income of rs . 10 , investment = rs . 100 . for an income of rs 7 , investment = rs . 100 / 10 x 7 = rs 70 market value of rs . 100 stock = rs . 70 answer is b" | a = 100 / 10
b = a * 7
|
a ) 288 , b ) 744 , c ) 788 , d ) 298 , e ) 930 | e | divide(multiply(divide(360, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 360 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 360 62 % - - - - - - - - ? = > 930 answer : e" | a = 100 - 62
b = 62 - a
c = b / 100
d = 360 / c
e = d * 62
f = e / 100
|
a ) 550 , b ) 600 , c ) 650 , d ) 700 , e ) 750 | b | divide(const_3600, 6) | a light flashes every 6 seconds , how many times will it flash in ? of an hour ? | "1 flash = 6 sec for 1 min = 10 flashes so for 1 hour = 10 * 60 = 600 flashes . answer : b" | a = 3600 / 6
|
a ) 66 % , b ) 64 % , c ) 68 % , d ) 52 % , e ) 72 % | d | subtract(const_100, add(multiply(6, 4), multiply(4, 6))) | uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought more of toyota than honda at the ratio of 6 : 4 . if 60 % of the toyota bought and 40 % of the honda bought were suv ã ¢ â ‚ ¬ â „ ¢ s . how many suv ã ¢ â ‚ ¬ â „ ¢ s did uba capital buy in the aforementioned purchase ? | "let total no of vehicles bought be 100 , toyota 60 and honda 40 , so total number of suv ' s bought for toyota and honda respectively 60 * 60 / 100 = 36 and 40 * 40 / 100 = 16 so total 52 suv ' s were bought out of 100 vehicles bought . . so required % is 52 % answer : d" | a = 6 * 4
b = 4 * 6
c = a + b
d = 100 - c
|
a ) $ 14 , b ) $ 5.4 , c ) $ 4.4 , d ) $ 1.2 , e ) $ 5.0 | b | subtract(divide(multiply(subtract(const_100, 20), add(divide(81, const_3), 81)), const_100), 81) | a clothing store purchased a pair of pants for $ 81 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a clothing store owner decided to decrease the selling price by 20 percent . what was the clothing store ' s gross profit on this sale ? | "sale price ( sp ) = 81 + markup ( mp ) - - > mp = sp - 81 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 81 3 sp / 4 = 81 sp = 108 now a discount of 20 % is given so new sp is . 8 * 108 = 86.4 profit = 86.4 - 81 = 5.4 $ answer is b" | a = 100 - 20
b = 81 / 3
c = b + 81
d = a * c
e = d / 100
f = e - 81
|
a ) 88 , b ) 60 , c ) 80 , d ) 89.6 , e ) 92 | d | divide(add(add(add(80, const_1), add(add(80, const_1), const_2)), add(subtract(100, 80), subtract(100, const_2))), 80) | find the average of all prime numbers between 80 and 100 | "prime numbers between 80 and 100 are 83 , 89 , 97 required average = ( 83 + 89 + 97 ) / 3 = 269 / 3 = 89.6 answer is d" | a = 80 + 1
b = 80 + 1
c = b + 2
d = a + c
e = 100 - 80
f = 100 - 2
g = e + f
h = d + g
i = h / 80
|
a ) 4 , b ) 6 , c ) 12 , d ) 9 , e ) 13 | e | multiply(const_3600, divide(divide(240, const_1000), add(60, 6))) | a train 240 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 240 * 3 / 55 ] sec = 13 sec answer : e" | a = 240 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 18.75 % , b ) 23 % , c ) 30 % , d ) 31 % , e ) 100 % | d | multiply(divide(subtract(add(const_100, 70), add(const_100, 30)), add(const_100, 30)), const_100) | the wages earned by robin is 30 % more than that earned by erica . the wages earned by charles is 70 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? | "let wage of erica = 10 wage of robin = 1.3 * 10 = 13 wage of charles = 1.7 * 10 = 17 percentage by which wage earned by charles is more than that earned by robin = ( 17 - 13 ) / 13 * 100 % = 4 / 13 * 100 % = 31 % answer d" | a = 100 + 70
b = 100 + 30
c = a - b
d = 100 + 30
e = c / d
f = e * 100
|
a ) 3 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 5 / 7 , e ) 7 / 5 | a | multiply(divide(3, 7), divide(1, 5)) | two brothers ram and ravi appeared for an exam . the probability of selection of ram is 3 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 3 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 3 / 7 x 1 / 5 = 3 / 35 answer : a" | a = 3 / 7
b = 1 / 5
c = a * b
|
a ) 540 km , b ) 767 km , c ) 276 km , d ) 178 km , e ) 176 km | a | multiply(add(20, 25), divide(60, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 60 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 60 km more than the other , 25 x – 20 x = 60 5 x = 60 x = 12 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 12 = 540 km . answer : a" | a = 20 + 25
b = 25 - 20
c = 60 / b
d = a * c
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | a | subtract(18, 12) | a set consists of 18 numbers , all are even or multiple of 5 . if 6 numbers are even and 12 numbers are multiple of 5 , how many numbers is multiple of 10 ? | { total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 18 = 6 + 12 - { both } + 0 ; { both } = 0 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : a . | a = 18 - 12
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(add(multiply(factorial(17), factorial(83)), multiply(factorial(17), factorial(52))), 17) | what is the units digit of 17 ^ 83 × 13 ^ 52 × 11 ^ 87 ? | "to find : the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 87 let ' s reduce the clutter and simplify the product ( 7 ^ 83 ) ( 3 ^ 82 ) ( 1 ^ 87 ) 7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4 th power so 7 ^ 5 has the same last digit as 7 ^ 1 , 7 ^ 9 , 7 ^ 13 thus , 7 ^ 83 has the same last digit as 7 ^ 3 , 7 ^ 7 , 7 ^ 11 i . e . 3 3 has a cyclicity of 4 : exactly the same routine as above thus , 3 ^ 82 has the same last digit as 3 ^ 2 , 3 ^ 6 , 3 ^ 10 i . e . 9 any power of 1 will result in 1 as the last digit so , product of our last digits = 3 x 9 x 1 = 27 . . . . last digit is 5 correct option : b" | a = math.factorial(17)
b = math.factorial(83)
c = a * b
d = math.factorial(17)
e = math.factorial(52)
f = d * e
g = c + f
h = g / 17
|
a ) 5 a , b ) 10 a , c ) 20 a , d ) 25 a , e ) 28 a | b | floor(divide(50, add(9, const_1))) | during a certain two - week period , 50 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 9 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | "movies : 50 % comedies . 50 % remaining genre . now in this 50 % , there are only 2 categories . action movies and drama movies . if action = x ; drama movies = 9 x . total 10 x . 10 x = 50 ; x = 5 action movies : 5 % drama movies : 45 % we can say that out of 100 z , : comedies : 50 z action : 5 z drama : 45 z now action movies were a this means : a = 5 z . z = ( a / 5 ) comedies : 50 z = 50 * ( a / 5 ) 10 a b is the answer ." | a = 9 + 1
b = 50 / a
c = math.floor(b)
|
a ) 200 , b ) 276 , c ) 230 , d ) 240 , e ) 250 | b | subtract(320, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(320, 20.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | a snooker tournament charges $ 45.00 for vip seats and $ 20.00 for general admission ( “ regular ” seats ) . on a certain night , a total of 320 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | "let no of sits in vip enclosure is x then x * 45 + 20 ( 320 - x ) = 7500 or 25 x = 7500 - 6400 , x = 1100 / 25 = 44 vip = 44 general 320 - 44 = 276 b" | a = 3 + 4
b = a * 1000
c = 2 + 3
d = c * 100
e = b + d
f = 320 * 20
g = e - f
h = 2 + 3
i = 2 + 3
j = h * i
k = g / j
l = 320 - k
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 78 | d | multiply(inverse(add(inverse(4), inverse(12))), const_2) | a and b can do a work in 4 hours and 12 hours respectively . a starts the work at 6 am and they work alternately for one hour each . when will the work be completed ? | "work done by a and b in the first two hours , working alternately = first hour a + second hour b = 1 / 4 + 1 / 12 = 1 / 3 . total time required to complete the work = 2 * 3 = 6 days . answer : d" | a = 1/(4)
b = 1/(12)
c = a + b
d = 1/(c)
e = d * 2
|
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