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 ) 320 m , b ) 350 m , c ) 650 m , d ) none of these , e ) can not be determined | b | divide(subtract(multiply(divide(300, divide(18, const_3)), 39), multiply(const_3, 300)), const_3) | a 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "solution speed = ( 300 / 18 ) m / sec = ( 50 / 3 ) m / sec . length of the platform be x metres . then = x + 300 / 39 = 50 / 3 m / sec = 3 ( x + 300 ) = 1950 . = 350 m answer b" | a = 18 / 3
b = 300 / a
c = b * 39
d = 3 * 300
e = c - d
f = e / 3
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a ) 53 % , b ) 57.7 % , c ) 62.5 % , d ) 64 % , e ) 75 % | b | multiply(divide(subtract(const_100, 25), add(30, const_100)), const_100) | marketing executives for a certain chewing gum company projected a 30 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ? | "last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 130 . actual / projected * 100 = 75 / 130 * 100 = 57.7 % . answer : b ." | a = 100 - 25
b = 30 + 100
c = a / b
d = c * 100
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a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 50 sec | c | multiply(divide(add(divide(190, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 190 m long ? | "explanation : speed = 45 km / hr = 45 Γ ( 10 / 36 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 190 = 550 meter time taken to cross the platform = 550 / ( 25 / 2 ) = 550 Γ 2 / 25 = 44 seconds answer : option c" | a = 190 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
|
a ) 3345 , b ) 1250 , c ) 1540 , d ) 2997 , e ) 1635 | d | multiply(divide(subtract(2500, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 2500 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 2500 = 6 x + 15 5 x = 2485 x = 497 large number = 497 + 1365 = 2997 d" | a = 2500 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) 7 , b ) 5 , c ) 4 , d ) 6 , e ) none of the above | c | multiply(328, divide(248, 3)) | 248 : 3 : : 328 : ? | "248 - - - - 24 / 8 . . . 3 328 - - - 32 / 8 . . . 4 answer c" | a = 248 / 3
b = 328 * a
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a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 48 | c | sqrt(multiply(72, 18)) | which number should replace both the asterisks in ( * / 18 ) x ( * / 72 ) = 1 ? | "let ( y / 18 ) x ( y / 72 ) = 1 y ^ 2 = 18 x 72 = 18 x 18 x 4 y = ( 18 x 2 ) = 36 the answer is c ." | a = 72 * 18
b = math.sqrt(a)
|
a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 125 | e | divide(power(1, 5), power(5, 5)) | the triplicate ratio of 1 : 5 is ? | "1 ^ 3 : 5 ^ 3 = 1 : 125 answer : e" | a = 1 ** 5
b = 5 ** 5
c = a / b
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a ) 60 m , b ) 100 m , c ) 43 m , d ) 20 m , e ) 23 m | b | subtract(500, divide(multiply(subtract(500, 200), 4), 3)) | in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 200 m . then , a wins by : | "to reach the winning post a will have to cover a distance of ( 500 - 200 ) m , i . e . , 300 m . while a covers 3 m , b covers 4 m . while a covers 300 m , b covers 4 x 300 / 3 m = 400 m . thus , when a reaches the winning post , b covers 400 m and therefore remains 100 m behind . a wins by 100 m . answer : b" | a = 500 - 200
b = a * 4
c = b / 3
d = 500 - c
|
a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount q of paint , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , lets further reduce 0.75 : 1.25 gives 1 + 1.5 = 2.5 gallons of mixture . this looks ok , but lets reduce further just to be sure 0.375 : 0.625 gives 0.5 + 1 = 1.5 gallons of mixture , thats less than 2 gallons of mixture q , so not acceptable . so correct ans is 2.5 gallons . b" | a = const_0_25 * 2
b = 2 * 5
c = a + b
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a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | d | subtract(multiply(9, const_2), const_1) | a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 9 multiples of 14 , how many multiples of 7 are there in q ? | "lets q ' s first number be 14 as per question , q contains 9 multiples of 14 , so let q ' s last number be 126 . ( 14 β 9 ) we have , a n = a + ( n β 1 ) d then 126 = 14 + ( n - 1 ) 7 solving for n , we get n = 17 answer : d" | a = 9 * 2
b = a - 1
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a ) 10.33 mps , b ) 17 mps , c ) 97 mps , d ) 17.33 mps , e ) 18.33 mps | e | multiply(const_0_2778, 66) | express a speed of 66 kmph in meters per second ? | "66 * 5 / 18 = 18.33 mps answer : e" | a = const_0_2778 * 66
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a ) s . 1090 , b ) s . 1360 , c ) s . 1190 , d ) s . 1202 , e ) s . 1256 | b | divide(multiply(subtract(const_100, 15), 1600), const_100) | a man buys a cycle for rs . 1600 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 1600 = rs . 85 / 100 x 1600 = rs . 1360 answer : b" | a = 100 - 15
b = a * 1600
c = b / 100
|
a ) 23.56 , b ) 23.59 , c ) 23.55 , d ) 23.53 , e ) 11.0 | e | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(8, 6)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 8 cm and breadth 6 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . | "let the side of the square be a cm . parameter of the rectangle = 2 ( 8 + 6 ) = 28 cm parameter of the square = 28 cm i . e . 4 a = 28 a = 7 diameter of the semicircle = 7 cm circimference of the semicircle = 1 / 2 ( Γ’ Λ Β ) 7 = 1 / 2 ( 22 / 7 ) ( 7 ) = 11 cm answer : e" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 12500 , b ) 13280 , c ) 17520 , d ) 14000 , e ) 12560 | d | divide(300, divide(multiply(divide(5, 7), 3), const_100)) | a tempo is insured to an extent of 5 / 7 of its original value . if the premium on it at the rate of 3 % amounts to $ 300 , the original value of the tempo is ? | "let the original value of the tempo is $ x 3 % of 5 / 7 of x = 300 ( 3 / 100 ) * ( 5 / 7 ) x = 300 x = $ 14000 answer is d" | a = 5 / 7
b = a * 3
c = b / 100
d = 300 / c
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a ) 1 / 4 , b ) 4 / 15 , c ) 3 / 14 , d ) 4 / 11 , e ) 4 / 5 | c | divide(3, add(subtract(const_12, const_1), 3)) | last year department store x had a sales total for december that was 3 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 = 30 year total = 11 * 10 + 30 = 140 answer = 30 / 140 = 3 / 14 = c | a = 12 - 1
b = a + 3
c = 3 / b
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a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | a | multiply(multiply(add(3, divide(subtract(sqrt(29), 3), 2)), divide(subtract(sqrt(29), 3), 2)), 2) | if a - b = 3 and a ^ 2 + b ^ 2 = 29 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 29 - 9 = 20 = > ab = 10 . answer : a" | a = math.sqrt(29)
b = a - 3
c = b / 2
d = 3 + c
e = math.sqrt(29)
f = e - 3
g = f / 2
h = d * g
i = h * 2
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 50 % , e ) 15 % | b | multiply(subtract(divide(5, 4), const_1), const_100) | the ratio of the cost price and selling price is 4 : 5 . the profit percent is ? | "let the c . p . = $ 4 x then s . p . = $ 5 x then , gain = 5 x - 4 x = $ x gain % = x / 4 x * 100 = 25 % answer is b" | a = 5 / 4
b = a - 1
c = b * 100
|
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) none | b | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 50) | farm tax is levied on the 50 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . willam . mr . willam paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | only trick n this question is to ignore 50 % information as farm tax is levied uniformly in the village and that includes mr william ' s land . what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 50 ) = 25 % and the answer is not there in the options . the correct answer is ( b ) . | a = 480 / 3840
b = a * 100
c = b * 100
d = c / 50
|
a ) 69990 , b ) 69993 , c ) 70000 , d ) 77000 , e ) 78000 | b | subtract(multiply(7, const_1000), 7) | the difference between the local value and the face value of 7 in the numeral 32675149 is | "explanation : ( local value of 7 ) - ( face value of 7 ) = ( 70000 - 7 ) = 69993 b )" | a = 7 * 1000
b = a - 7
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a ) 336 , b ) 784 , c ) 1120 , d ) 1804 , e ) 1936 | c | add(multiply(multiply(subtract(subtract(const_10, const_2), const_1), subtract(subtract(const_10, const_2), const_1)), subtract(subtract(divide(const_100, 4), const_3), multiply(const_2, const_3))), multiply(multiply(subtract(const_10, const_2), subtract(subtract(const_10, const_2), const_1)), multiply(const_2, const_3))) | how many even 4 - digit numbers can be formed , so that the numbers are divisible by 4 and no two digits are repeated ? | number is divisible by 4 if the last two digits form a number divisible by 4 . therefore last two digits can be : 00 ; 04 ; 08 ; 12 16 ; . . . 96 . basically multiples of 4 in the range 0 - 96 , inclusive . multiples of 4 in a range 0 - 96 , inclusive are last multiple in the range β first multiple in the range 4 + 1 = 25 last multiple in the range β first multiple in the range 4 + 1 = 25 but 3 numbers out of these 25 are not formed with distinct digits : 00 , 44 , and 88 . hence the numbers we are looking for can have only 22 endings . if there is 0 in the ending ( 04 , 08 , 20 , 40 , 60 , 80 - total 6 such numbers ) , then the first and second digit can take 8 and 7 choices each = 56 choices total . as there are 6 numbers with 0 in ending , hence total 6 * 56 = 336 . if there is no 0 in the ending ( total 22 - 6 with zero = 16 such numbers ) , then the first digit can take 7 choices ( 10 - 2 digits in the ending and zero , so total 3 digits = 7 , as 4 - digit number can not start with zero ) and the second digit can take 7 choices too ( 10 digits - 3 digits we ' ve already used ) = 7 * 7 = 49 choices total . as there are 16 numbers without zero in ending , hence total 16 * 49 = 784 . total : 336 + 784 = 1120 answer : c . | a = 10 - 2
b = a - 1
c = 10 - 2
d = c - 1
e = b * d
f = 100 / 4
g = f - 3
h = 2 * 3
i = g - h
j = e * i
k = 10 - 2
l = 10 - 2
m = l - 1
n = k * m
o = 2 * 3
p = n * o
q = j + p
|
a ) 79698 , b ) 80578 , c ) 80698 , d ) 180578 , e ) none of them | d | multiply(317, power(317, 283)) | 317 x 317 + 283 x 283 = ? | "= ( 317 ) ^ 2 + ( 283 ) ^ 2 = ( 300 + 17 ) ^ 2 + ( 300 - 17 ) ^ 2 = 2 [ ( 300 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 90000 + 289 ] = 2 x 90289 = 180578 answer is d" | a = 317 ** 283
b = 317 * a
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a ) 12 , b ) 36 , c ) 28 , d ) 20 , e ) 24 | a | multiply(const_3_6, divide(divide(add(120, 120), 36), const_2)) | two trains are running in opposite directions in the same speed . the length of each train is 120 meter . if they cross each other in 36 seconds , the speed of each train ( in km / hr ) is | explanation : distance covered = 120 + 120 = 240 m time = 36 s let the speed of each train = v . then relative speed = v + v = 2 v 2 v = distance / time = 240 / 36 = 6.667 m / s speed of each train = v = 6.667 / 2 = 3.33 m / s = 3.33 Γ 36 / 10 km / hr = 12 km / hr answer : option a | a = 120 + 120
b = a / 36
c = b / 2
d = const_3_6 * c
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a ) 12 , b ) 15 , c ) 17 , d ) r = 18 , e ) 20 | d | multiply(multiply(const_3, const_2), 3) | two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer r between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 1112 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 1113 ) , ( 717 ) , ( 5 , 19 ) 2 , 3 , 5 , 7 , 11 , 13 , 1517 , 19 , 23 , 29 , 31 , 37 - three pairs ( 13 , 17 ) , ( 11 , 19 ) , ( 7 , 23 ) 2 , 3 , 5 , 7 , 11 , 1317 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11 , 23 ) , ( 5 , 29 ) , ( 3 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 1819 , 23 , 29 , 31 , 37 - four pairs ( 17 , 19 ) , ( 13 , 23 ) , ( 7 , 29 ) , ( 5 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 2023 , 29 , 31 , 37 - definitely can not be more than 4 since there are only 4 primes more than 20 . so must be less than 4 pairs . ignore . answer ( d ) . | a = 3 * 2
b = a * 3
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a ) 1 / 3 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 , e ) 5 / 6 | c | divide(subtract(add(3, 3), add(divide(3, 2), add(2, 2))), 2) | pumps a , b , and c operate at their respective constant rates . pumps a and b , operating simultaneously , can fill a certain tank in 2 hours ; pumps a and c , operating simultaneously , can fill the tank in 3 / 2 hours ; and pumps b and c , operating simultaneously , can fill the tank in 2 hours . how many hours does it take pumps a , b , and c , operating simultaneously , to fill the tank . | a + b = 2 ; a + c = 3 / 2 , b + c = 2 ; add then 2 * ( a + b + c ) = 4 + 3 / 2 = 11 / 2 a + b + c = 11 / 4 hrs c | a = 3 + 3
b = 3 / 2
c = 2 + 2
d = b + c
e = a - d
f = e / 2
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a ) 15 , b ) 30 , c ) 32 , d ) 33 , e ) 46 | c | subtract(divide(subtract(subtract(100, 8), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 8 to 100 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 8,10 , 12,14 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 47 . for 6 , 12,18 , . . . 96 using ap formula , we can say 96 = 12 + ( n - 1 ) * 6 or n = 15 . hence , only divisible by 2 but not 3 = 47 - 15 = 32 . hence , answer c" | a = 100 - 8
b = a - 2
c = b / 2
d = 100 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 40 , b ) 50 , c ) 70 , d ) 60 , e ) none | c | add(add(20, subtract(50, 20)), 20) | in a group of 100 people , 50 like reading books , 20 like both reading books and listening songs . how many like listening songs ? | make a venn diagram , and enter your data . let the number of people who like listening songs be x x + 20 + 30 = 100 x = 50 so number who like listening songs = 20 + 50 = 70 answer c | a = 50 - 20
b = 20 + a
c = b + 20
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a ) 6 , b ) 5 , c ) 7 , d ) 3 , e ) 2 | d | divide(subtract(divide(90, 3), divide(72, 3)), const_2) | a man rows his boat 90 km downstream and 72 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / 3 = 30 kmph speed upstream = d / t = 72 / ( 3 ) = 24 kmph the speed of the stream = ( 30 - 24 ) / 2 = 3 kmph answer : d" | a = 90 / 3
b = 72 / 3
c = a - b
d = c / 2
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a ) 29 , b ) 49 , c ) 39 , d ) 59 , e ) 69 | c | subtract(add(floor(divide(97, const_3)), floor(divide(97, add(const_1, const_4)))), multiply(floor(divide(97, multiply(const_3, add(const_1, const_4)))), const_2)) | there are 97 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off position at the beggining , how many lights are switched on by the end of this operation ? | editing my solution : number of switches = 97 number of switches turned on by a : 3 , 6 , . . . 96 = 32 number of switches turned on by b : 5 , 10 , . . . . 95 = 19 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b as they are turned off . number of switches that are turned on = ( 32 - 6 ) + ( 19 - 6 ) = 39 answer : c | a = 97 / 3
b = math.floor(a)
c = 1 + 4
d = 97 / c
e = math.floor(d)
f = b + e
g = 1 + 4
h = 3 * g
i = 97 / h
j = math.floor(i)
k = j * 2
l = f - k
|
a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | multiply(power(const_2, 1), factorial(1)) | the product of two consecutive odd no is - 1 , then what is the value of the sum of them . | "two odd numbers are - 1 and + 1 . - 1 + 1 = 0 answer : c" | a = 2 ** 1
b = math.factorial(1)
c = a * b
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a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 125 | a | divide(multiply(10, add(const_4, const_1)), const_2) | to fill a tank , 10 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 10 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 10 x ) / ( 2 x / 5 ) = ( 10 x ) x 5 / 2 x = 50 / 2 = 25 answer is a ." | a = 4 + 1
b = 10 * a
c = b / 2
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a ) 30 days , b ) 20 days , c ) 15 days , d ) 18 days , e ) 21 days | b | divide(divide(30, 18), divide(divide(divide(1, 7), divide(subtract(7, const_1), 7)), 2)) | if 30 oxen can plough 1 / 7 th of a field in 2 days , how many days 18 oxen will take to do the remaining work ? | solution : we will use work equivalence method , 30 / 18 = ( 1 / 7 ) / ( 6 / 7 ) * x / 2 ; 5 / 3 = ( 1 / 6 ) * x / 2 ; or , x = 60 / 3 = 20 days . answer : option b | a = 30 / 18
b = 1 / 7
c = 7 - 1
d = c / 7
e = b / d
f = e / 2
g = a / f
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a ) 5 , b ) 4 , c ) 3 , d ) 6 , e ) 7 | a | subtract(9, const_4) | how many two - element subsets of { 6,7 , 8,9 } are there that do not contain the pair of elements 8 and 9 ? | required subsets are = { 6,7 } , { 6,8 } , { 6,9 } , { 7,8 } , { 7,9 } = 5 answer : a | a = 9 - 4
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a ) 74 % , b ) 64 % , c ) 42 % , d ) 28 % , e ) 26 % | e | multiply(divide(subtract(multiply(const_100, divide(14, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 14 percent sugar by weight . the second solution was what percent sugar by weight ? | say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.14 - - > x = 0.26 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.14 - - > x = 0.26 . answer : e . | a = 14 / 100
b = 100 * a
c = 1 / 4
d = c * 100
e = 100 - d
f = 10 / 100
g = e * f
h = b - g
i = 1 / 4
j = i * 100
k = h / j
l = k * 100
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a ) 72 , b ) 84 , c ) 136 , d ) 140 , e ) none of these | e | multiply(divide(15, multiply(multiply(divide(const_2, divide(15, 3)), divide(const_1, const_4)), divide(const_3, add(const_3, const_4)))), divide(40, const_100)) | two - fifth of one - third of 3 - seventh of a number is 15 . what is 40 % of that number ? | explanation : let the number be x . then , 2 / 5 of 1 / 3 of 3 / 7 of x = 15 = > x = ( 15 * 5 / 3 * 3 * 5 / 2 ) = 525 / 2 40 % of 525 / 2 = ( 40 / 100 * 525 / 2 ) = 105 answer is e | a = 15 / 3
b = 2 / a
c = 1 / 4
d = b * c
e = 3 + 4
f = 3 / e
g = d * f
h = 15 / g
i = 40 / 100
j = h * i
|
a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 7 | e | divide(522.416, 0.0168) | when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 522.416 is divided by 0.0168 ? | "for the 1 st no . there are 3 digits after decimal for the 2 nd no . there are 5 digits after decimal total no . of decimals = 8 req . no . of digits = ( n - 1 ) = ( 8 - 1 ) = 7 answer : e" | a = 522 / 416
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a ) 2 , b ) 4 , c ) 8 , d ) 39 , e ) 32 | d | add(subtract(69, add(add(add(15, 10), 5), 3)), 3) | there are 69 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | "lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 69 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 69 = 15 + 10 + 5 + 3 + x + y + s we need to know total snakes = x + y + s + 3 = 39 answer : d" | a = 15 + 10
b = a + 5
c = b + 3
d = 69 - c
e = d + 3
|
a ) 15 % , b ) 162 β 3 % , c ) 20 % , d ) 38.46 % , e ) 24 % | d | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | an employee β s annual salary was increased $ 25,000 . if her new annual salary now equals $ 90,000 , what was the percent increase ? | new annual salary = $ 90,000 salary increase = $ 25,000 . original salary = $ 90,000 - $ 25,000 . = $ 75,000 % increase = ( $ 25,000 / $ 65,000 ) * 100 = 38.46 % hence d . | a = 100 - 10
b = a * 1000
c = 100 - 10
d = c * 1000
e = const_0_25 * 100
f = e * 1000
g = d - f
h = b / g
i = h - 1
j = i * 100
|
a ) 87 days , b ) 10 days , c ) 15 days , d ) 44 days , e ) 22 days | c | divide(subtract(multiply(30, 40), multiply(40, 15)), 40) | 30 men can do a work in 40 days . when should 15 men leave the work so that the entire work is completed in 40 days after they leave the work ? | total work to be done = 30 * 40 = 1200 let 15 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 15 * 40 ) = 1200 40 p = 600 = > p = 15 days answer : c | a = 30 * 40
b = 40 * 15
c = a - b
d = c / 40
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | add(divide(subtract(multiply(65, 4), multiply(70, 4)), subtract(60, 65)), 4) | a car averages 70 mph for the first 4 hours of a trip and averages 60 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ? | "let the time for which car averages 60 mph = t 65 * ( t + 4 ) = 70 * 4 + 60 t = > 5 t = 20 = > t = 4 total duration of the trip = 4 + 4 = 8 answer a" | a = 65 * 4
b = 70 * 4
c = a - b
d = 60 - 65
e = c / d
f = e + 4
|
a ) 3.5 hours , b ) 6 hours , c ) 4 hours , d ) 5 hours , e ) 6 hours | a | divide(39, subtract(12, subtract(divide(39, 3), 12))) | the speed of the boat in still water in 12 kmph . it can travel downstream through 39 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 39 / 3 = 13 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 13 / 2 = 11 km / hr so time taken in upstream = 39 / 11 = 3.5 hrs answer : a" | a = 39 / 3
b = a - 12
c = 12 - b
d = 39 / c
|
a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | d | divide(log(7), log(power(7, 11))) | if n = 7 ^ 11 β 7 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 7 ^ 3 - 7 . now 7 ^ 3 has last digit 3 . since 7 ^ 11 ( or for that matter 7 ^ 3 ) is greater than 7 , we subtract 7 from 13 ( the 10 + 3 - - > 10 has come from carry over from the tenth place ) . thus 13 - 7 = 6 is the answer . option d | a = math.log(7)
b = 7 ** 11
c = math.log(b)
d = a / c
|
a ) 320 $ , b ) 380 $ , c ) 525 $ , d ) 456 $ , e ) 480 $ | c | multiply(multiply(0.65, 55), 14) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 14 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 14 ) + ( 0.65 * 14 * 55 ) = 24.5 + 500.5 = > 525 hence answer will be ( c ) 525" | a = 0 * 65
b = a * 14
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | add(divide(subtract(151, const_1), add(14, const_1)), const_1) | in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 14 feet between each tree . how many trees can be planted if the road is 151 feet long ? | "let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 14 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number of 14 feet spaces will be t - 1 . then , the length of sidewalk required for 14 feet spaces will be 14 * ( t - 1 ) it is given that total lenght of sidewalk is 166 feet . or 14 ( t - 1 ) + t = 151 or 14 t - 14 + t = 151 or 15 t = 165 or t = 11 answer : - d" | a = 151 - 1
b = 14 + 1
c = a / b
d = c + 1
|
a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability q that the next couple the maitre ' d seats will not order dessert ? | "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . q = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %" | a = 2 * 5
b = 2 * 5
c = a * b
d = 60 / c
e = 2 * 5
f = 2 * 5
g = e * f
h = 20 / g
i = 1 - h
j = d / i
k = 1 - j
l = 2 * 5
m = 2 * 5
n = l * m
o = k * n
|
a ) 23 , b ) 20 , c ) 12 , d ) 128 , e ) 171 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 and the sum of their squares is 289 . the sum of the number is : | "let the numbers be x and y . then , xy = 120 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 120 ) = 529 x + y = 529 = 23 . answer a" | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
|
a ) 44 % , b ) 125 % , c ) 105 % , d ) 225 % , e ) 400 % | b | multiply(divide(subtract(divide(3, 2), divide(2, 3)), divide(2, 3)), const_100) | when tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent greater than the quantifying chopped by tom ? . | tom chops 4 lbs in 6 minutes tammy chops 9 lbs in 6 minutes so in the same amount of time , tammy chops 125 % more than tom , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , and 9 x is still 125 % greater than 4 x . ans : b | a = 3 / 2
b = 2 / 3
c = a - b
d = 2 / 3
e = c / d
f = e * 100
|
a ) 5100 , b ) 5120 , c ) 5200 , d ) 5400 , e ) 5832 | e | subtract(subtract(8000, multiply(8000, divide(10, const_100))), multiply(subtract(8000, multiply(8000, divide(10, const_100))), divide(10, const_100))) | the population of a town is 8000 . it decreases annually at the rate of 10 % p . a . what will be its population after 3 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8000 Γ£ β 90 / 100 Γ£ β 90 / 100 x 90 / 100 = 5832 answer : e" | a = 10 / 100
b = 8000 * a
c = 8000 - b
d = 10 / 100
e = 8000 * d
f = 8000 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 2.8 , b ) 2.3 , c ) 2.5 , d ) 2.1 , e ) 4.3 | e | multiply(divide(divide(multiply(divide(30, const_100), 100), 10), multiply(divide(30, const_100), 100)), const_100) | a reduction of 30 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 30 / 100 ) = 30 - - - 10 ? - - - 1 = > rs . 3 100 - - - 70 ? - - - 3 = > rs . 4.3 answer : e" | a = 30 / 100
b = a * 100
c = b / 10
d = 30 / 100
e = d * 100
f = c / e
g = f * 100
|
a ) 3,000 , b ) 2,562 , c ) 2,500 , d ) 2,180 , e ) 2,766 | d | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 31), add(const_2, const_4)) | what is the sum of all the even numbers between 31 and 99 , inclusive ? | "to figure this problem out , you just have to add all the even numbers , starting with 32 . 32 + 34 + 36 + 38 and so on , all the way up to 98 . you ' ll get 2,180 , so final answer : d" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 31
t = 2 + 4
u = s + t
|
a ) $ 5,330 , b ) $ 3,360 , c ) $ 450 , d ) $ 360 , e ) $ 150 | c | multiply(30, divide(multiply(1080, 605), divide(multiply(1080, 605), const_10))) | a certain farmer pays $ 30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 1080 feet by 605 feet ? ( 43,560 square feet = 1 acre ) | basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 450 ( c ) | a = 1080 * 605
b = 1080 * 605
c = b / 10
d = a / c
e = 30 * d
|
a ) 9 , b ) 13.88 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(25, 5), 9) | according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 25 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 25 = 57 - - > c = 5 / 9 * 25 = 13.88 . increase = 13.88 degrees . answer : b ." | a = 25 * 5
b = a / 9
|
a ) 71 , b ) 69 , c ) 72 , d ) 75 , e ) none | a | divide(add(add(add(add(76, 60), 72), 65), 82), add(const_1, const_4)) | kamal obtained 76 , 60 , 72 , 65 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 60 + 72 + 65 + 82 / 5 ) = ( 355 / 5 ) = 71 . answer a" | a = 76 + 60
b = a + 72
c = b + 65
d = c + 82
e = 1 + 4
f = d / e
|
a ) 600 , b ) 715 , c ) 772 , d ) 662 , e ) 521 | b | add(550, multiply(550, divide(30, const_100))) | a person buys an article at rs . 550 . at what price should he sell the article so as to make a profit of 30 % ? | "cost price = rs . 550 profit = 30 % of 550 = rs . 165 selling price = cost price + profit = 550 + 165 = 715 answer : b" | a = 30 / 100
b = 550 * a
c = 550 + b
|
a ) 198 , b ) 364 , c ) 369 , d ) 207 , e ) 210 | b | add(280, multiply(280, divide(30, const_100))) | an article with cost price of 280 is sold at 30 % profit . what is the selling price ? | "sp = 1.30 * 280 = 364 answer : b" | a = 30 / 100
b = 280 * a
c = 280 + b
|
a ) 1 hrs , b ) 3 hrs , c ) 4 hrs , d ) 5 hrs , e ) 7 hrs | d | divide(multiply(const_1, divide(2, 5)), multiply(inverse(5), subtract(const_1, divide(3, 5)))) | john can type a set of pages in 5 hours at a constant rate . if john types for 3 hours and jack does the rest of the job , who works at 2 / 5 constant rate of john . how long will it take for jack alone to do the rest of the job ? | john will do 3 / 5 in 3 hrs , so john does 3 / 5 of the work . . therefore , jack will do the remaining 2 / 5 th work alone . . as the speed of jack is 2 / 5 rate of john , jack will do the 2 / 5 th work in same time that john takes to complete full job . . . ans 5 d | a = 2 / 5
b = 1 * a
c = 1/(5)
d = 3 / 5
e = 1 - d
f = c * e
g = b / f
|
a ) 23.75 , b ) 22 , c ) 20.8 , d ) 19.2 , e ) none of these | c | add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 30), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20))) | a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 30 % , what should be the sale price of the mixed tea ? | "c . p . of mixture = 80 Γ 15 + 20 Γ 20 / 80 + 20 = 16 β΄ s . p . = ( 100 + 30 ) / 100 Γ 16 = 20.8 answer c" | a = 80 * 15
b = 20 * 20
c = a + b
d = 80 + 20
e = c / d
f = e * 30
g = f / 100
h = 80 * 15
i = 20 * 20
j = h + i
k = 80 + 20
l = j / k
m = g + l
|
a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 5 / 3 | d | multiply(5, add(divide(const_1, 15), divide(const_1, 30))) | two persons a and b can complete a piece of work in 15 days and 30 days respectively . if they work together , what part of the work will be completed in 5 days ? | "a ' s one day ' s work = 1 / 15 b ' s one day ' s work = 1 / 30 ( a + b ) ' s one day ' s work = 1 / 15 + 1 / 30 = 1 / 10 the part of the work completed in 5 days = 5 ( 1 / 10 ) = 1 / 2 . answer : d" | a = 1 / 15
b = 1 / 30
c = a + b
d = 5 * c
|
a ) 63.75 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 50 minutes after it was poured was how many degrees fahrenheit ? | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 50 minutes after it was poured is : f = 120 ( 2 ^ - 50 / 10 ) + 60 f = 120 * 1 / 32 + 60 f = 15 / 4 + 60 f = 255 / 4 = 63.75 a" | a = 60 / 10
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
|
a ) 85 Β° f , b ) 92 Β° f , c ) 96 Β° f , d ) 99 Β° f , e ) 104 Β° f | e | add(multiply(divide(subtract(212, 32), 100), 40), 32) | water boils at 212 Β° f or 100 Β° c and ice melts at 32 Β° f or 0 Β° c . if the temperature of a pot of water is 40 Β° c , what is the temperature of the pot of water in Β° f ? | "let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) f = 9 c / 5 + 32 f = 9 ( 40 ) / 5 + 32 = 104 Β° f the answer is e ." | a = 212 - 32
b = a / 100
c = b * 40
d = c + 32
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | subtract(7, const_3) | the number of natural numbers n such that ( n + 1 ) ^ 2 / ( n + 7 ) is an integer , is ? | we can use ( n + 1 ) ^ 2 greater than ( n + 7 ) which gives n > 2 , - 3 . . . . . . . . now there are four integers - 2 . - 1 , 01 in between - 3 and 2 . so answer is 4 . answer : a | a = 7 - 3
|
a ) 11.84 , b ) 12.84 , c ) 23.84 , d ) 19.11 , e ) 10.12 | d | divide(multiply(139.00, add(divide(const_1, 10), const_1)), 8) | total dinning bill of 8 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared . | "dinner bill of 8 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) / 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 / 8 = $ 19.11 answer : d" | a = 1 / 10
b = a + 1
c = 139 * 0
d = c / 8
|
a ) 50 , b ) 100 , c ) 200 , d ) 1,000 , e ) 2,000 | c | multiply(multiply(divide(50, 100), divide(50, 100)), const_1000) | positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 100 . what is the value of x ? | "y = 50 % of 50 % 0 f x = x / 4 and y / 100 of x = 100 y / 100 * 4 y = 100 y = 50 and x = 200 answer - c" | a = 50 / 100
b = 50 / 100
c = a * b
d = c * 1000
|
a ) 392 , b ) 759 , c ) 753 , d ) 493 , e ) 540 | b | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5060) | 3 / 4 of 1 / 2 of 2 / 5 of 5060 = ? | "b 759 ? = 5060 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 759" | a = 3 / 4
b = 1 / 2
c = a * b
d = 2 / 5
e = c * d
f = e * 5060
|
a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | a | divide(subtract(add(multiply(40.2, 10), add(13, 18)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 18 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 18 = 402 hence , new average = 402 β 10 = 40.2 answer a | a = 40 * 2
b = 13 + 18
c = a + b
d = c - 31
e = d / 10
|
a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | d | subtract(190, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 100 and 190 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 190 , inclusive , is 91 . 3 * 34 = 102 and 3 * 63 = 189 so the number of multiples of 3 is 30 . 5 * 20 = 100 and 5 * 38 = 190 so the number of multiples of 5 is 19 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 12 = 180 so the number of multiples of 15 is 6 . thus the total number is 91 - 30 - 19 + 6 = 48 . the answer is d ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 190 - e
|
a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 120 , e ) $ 250 | d | multiply(divide(300, add(divide(2, 3), const_1)), divide(2, 3)) | $ 300 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 300 * 2 / 5 = $ 120 answer is d" | a = 2 / 3
b = a + 1
c = 300 / b
d = 2 / 3
e = c * d
|
a ) 96.9 , b ) 96.3 , c ) 96.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 4), divide(1, 5)) | what is the cp of rs 100 stock at 4 discount , with 1 / 5 % brokerage ? | explanation : use the formula , cp = 100 β discount + brokerage % cp = 100 - 4 + 1 / 5 96.2 thus the cp is rs 96.2 . answer : c | a = 100 - 4
b = 1 / 5
c = a + b
|
a ) $ 600 , b ) $ 900 , c ) $ 1200 , d ) $ 1500 , e ) $ 1800 | b | multiply(divide(63, 11), const_100) | if an article is sold at 18 % profit instead of 11 % profit , then the profit would be $ 63 more . what is the cost price ? | "7 % * cost price = $ 63 1 % * cost price = $ 63 / 7 = $ 9 the cost price is $ 900 . the answer is b ." | a = 63 / 11
b = a * 100
|
a ) 3 , b ) 16 , c ) 28 , d ) 32 , e ) 56 | d | multiply(power(const_2, 3), 4) | how many x ways can you group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen ? | ways to select 3 people from 8 people ( 4 twins x 2 ) = 8 c 3 = 56 ways to select 1 twin + 1 people = 4 c 1 * 6 c 1 = 24 ways to select a group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen x = 56 - 24 = 32 ans : d | a = 2 ** 3
b = a * 4
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(divide(16000, multiply(multiply(2, 4), 5)), power(const_10, const_2)) | in the game of dubblefud , yellow chips , blue chips and green chips are each worth 2 , 4 and 5 points respectively . in a certain selection of chips , the product of the point values of the chips is 16000 . if the number of blue chips in this selection equals the number of green chips , how many yellow chips are in the selection ? | this is equivalent to : - 2 x * 4 y * 5 z = 16000 y = z ( given ) 2 x * 4 y * 5 y = 16000 2 x * y ^ 2 = 16000 / 20 2 x * y ^ 2 = 800 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 4 as 2 * 4 * y ^ 2 = 800 y ^ 2 = 100 y = 10 number of yellow chips = 4 hence d | a = 2 * 4
b = a * 5
c = 16000 / b
d = 10 ** 2
e = c / d
|
a ) 152 , b ) 241 , c ) 345 , d ) 451 , e ) 542 | b | divide(multiply(divide(const_3600, const_4), const_3), 10) | if a light flashes every 10 seconds , how many times will it flash in 2 / 3 of an hour ? | "there are 60 minutes in an hour . in 2 / 3 of an hour there are ( 60 * 2 / 3 ) minutes = 40 minutes . in 2 / 3 of an hour there are ( 60 * 40 ) seconds = 2400 seconds . light flashed for every 10 seconds . in 2400 seconds 2400 / 10 = 240 times . the count start after the first flash , the light will flashes 241 times in 2 / 3 of an hour . answer : b" | a = 3600 / 4
b = a * 3
c = b / 10
|
a ) 95 / 8 , b ) 93 / 8 , c ) 90 / 8 , d ) 12 , e ) 10 | a | add(5, divide(subtract(multiply(20, 12), multiply(12, 5)), 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 5 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 ? | "in such questions , you need to start from the end . last 4 days john works alone and completes 5 * ( 1 / 20 ) = 1 / 4 of the work . so 3 / 4 of the work should have been completed by the two of them together before jane left . their combined rate of work is 1 / 20 + 1 / 12 = 8 / 60 time taken to complete 3 / 4 of the work = ( 3 / 4 ) / ( 8 / 60 ) = 45 / 8 days . so total number of days taken to complete the work = 45 / 8 + 5 = 95 / 8 days . a" | a = 20 * 12
b = 12 * 5
c = a - b
d = 12 + 20
e = c / d
f = 5 + e
|
a ) 18 , b ) 26 , c ) 14 , d ) 12 , e ) 9.3 | e | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 18)) | if the wheel is 18 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 18 * x = 1056 = > x = 9.3 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 18
h = 1056 / g
|
a ) 30 % , b ) 60 % , c ) 98 % , d ) 120 % , e ) 156 % | e | multiply(subtract(power(add(const_1, divide(60, const_100)), const_2), const_1), const_100) | if a large pizza has a radius that is 60 % larger that that of a medium pizza , what is the percent increase in area between a medium and a large pizza ? | "let the radius of medium pizza be r . then the radius of large pizza is 1.6 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.6 * r ) ^ 2 = 2.56 * pi * r ^ 2 , an increase of 156 % . the answer is e ." | a = 60 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 5 , b ) 8 , c ) 4 , d ) 10 , e ) 3 | b | divide(subtract(25, power(3, 2)), 2) | if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 25 , find the value of ab . | 2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 25 - 9 = 16 ab = 8 . answer is b . | a = 3 ** 2
b = 25 - a
c = b / 2
|
a ) $ 0.16 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | a | divide(multiply(add(const_1, const_1), 0.40), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with four times the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.40 per glass on the first day , what was the price per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 4 units of water was used to make 5 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 5 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 5 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 5 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.4 = 5 * x - - > x = $ 0.16 . answer : a ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
|
a ) 50 kmph , b ) 60 kmph , c ) 65 kmph , d ) 16 kmph , e ) none | d | divide(144, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 144 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 144 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 144 / 9 = 16 kmph d )" | a = 6 * 3
b = a / 2
c = 144 / b
|
a ) 350 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | a | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(5, 150))) | a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 140 . find the unknown rate of two blankets ? | "10 * 140 = 15 = 1400 3 * 100 + 5 * 150 = 1050 1400 β 1050 = 350 answer : a" | a = 10 * 150
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 2.07 , b ) 3.07 , c ) 5.07 , d ) 4.07 , e ) 2 | c | multiply(divide(divide(110, const_1000), 78), const_3600) | how long will it take a train travelling at 78 kmph to pass an electric pole if the train is 110 meters in lenght ? | sol . speed = [ 78 x 5 / 18 ] m / sec = 21.7 m / sec . time taken = ( 110 / 21.7 ) sec = 5.07 sec . answer c | a = 110 / 1000
b = a / 78
c = b * 3600
|
a ) 1394 , b ) 1450 , c ) 1245 , d ) 2564 , e ) 1256 | a | add(14,19, lcm(20,25, 35,40)) | find the least number which when divided by 20,25 , 35,40 leaves remainder 14,19 , 29,34 respectively . | "the difference of 20 - 14 = 6 , 25 - 19 = 6 , 35 - 29 = 6 , 40 - 34 = 6 lcm of 20,25 , 35,40 = 1400 required number = 1400 - 6 = 1394 answer is a" | a = math.lcm(20, 25)
b = 14 + 19
|
a ) 123 , b ) 106 , c ) 100 , d ) 156 , e ) 240 | b | subtract(106.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 106.25 , brokerage being 1 / 4 % is | explanation : cash realised = rs . ( 106.25 - 0.25 ) = rs . 106 . answer : b | a = 1 / 4
b = 106 - 25
|
a ) 36 kmh , b ) 40 kmh , c ) 41 kmh , d ) 44 kmh , e ) 50 kmh | c | divide(const_3, add(add(divide(const_1, 80), divide(const_1, 24)), divide(const_1, 54))) | if a car went the first third of the distance at 80 kmh , the second third at 24 kmh , and the last third at 54 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 2160 ( this number is convenient because it is divisible by 80 , 24 and 54 ) so : 2160 = 80 * t 1 = 27 hrs 2160 = 24 * t 2 = 90 hrs 2160 = 54 * t 3 = 40 hrs t = t 1 + t 2 + t 3 = 157 hrs d = rt ( 240 * 3 ) = r * 157 r = 41.27 answer : c" | a = 1 / 80
b = 1 / 24
c = a + b
d = 1 / 54
e = c + d
f = 3 / e
|
a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 12 and 9 | e | subtract(add(divide(22, 2), 1), 3) | one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 22 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x β 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x β 3 + 1 ) = ( x β 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) β
( x β 2 ) cm 2 , which is 22 cm 2 more than the first area . therefore a 1 + 22 = a 2 x ( x β 3 ) + 22 = ( x + 1 ) ( x β 2 ) x 2 β 3 x + 22 = x 2 + x β 2 x β 2 2 x = 24 x = 12 . so , the sides of the rectangle are 12 cm and ( 12 β 3 ) = 9 cm long . so answer is e ." | a = 22 / 2
b = a + 1
c = b - 3
|
a ) 22 , b ) 25 , c ) 28 , d ) 31 , e ) 34 | b | divide(multiply(divide(25, 4), 16), 4) | rice weighing 25 / 4 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces ) | "25 / 4 Γ· 4 = 25 / 16 pounds in each container 25 / 16 pounds * 16 ounces / pound = 25 ounces in each container the answer is b ." | a = 25 / 4
b = a * 16
c = b / 4
|
a ) 24 % , b ) 36 % , c ) 32 % , d ) 54 % , e ) 64 % | b | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 20), subtract(const_100, 20))), const_100) | a towel , when bleached , lost 20 % of its length and 20 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( β 20 β 20 + ( 20 Γ 20 ) / 100 ) % = β 36 % i . e . , area is decreased by 36 % answer : b" | a = 100 * 100
b = 100 - 20
c = 100 - 20
d = b * c
e = a - d
f = e / 100
|
a ) 7 , b ) 8 , c ) 10 , d ) 15 , e ) 36 / 3 | e | divide(36, divide(add(3.6, 3), const_2)) | sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 3.6 per pound , and german chocolate cost $ 3 per pound . if the total the she spent on chocolate was $ 36 and both types of chocolate were purchased in whole number of pounds , how many total pounds of chocolate she purchased ? | "if there were all the expensive ones , 3.6 . . . there would be 36 / 3.6 or 10 of them but since 3 $ ones are also there , answer has to be > 10 . . . . . if all were 3 $ ones , there will be 36 / 3 ans e" | a = 3 + 6
b = a / 2
c = 36 / b
|
a ) $ 90000 , b ) $ 108000 , c ) $ 119000 , d ) $ 126000 , e ) $ 140000 | b | multiply(divide(48000, subtract(9, 5)), 9) | a business finds that their ratio of spending to income is 5 to 9 . their total profit is $ 48000 . what is their total income ? | let x equal the amount of spending , and y equal the amount of income . we know that x = 5 / 9 y we also know that x = y - 48000 therefore we can simplify to : y - 48000 = 5 / 9 y we isolate y , and get : 4 / 9 y = 48000 and to solve for y , we get : y = 48000 answer : b | a = 9 - 5
b = 48000 / a
c = b * 9
|
a ) 219 , b ) 246 , c ) 252 , d ) 262 , e ) 270 | a | multiply(multiply(multiply(300, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 300 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "number of newborns that can die in first month = 1 / 10 * 300 = 30 survived = 270 number of newborns that can die in second month = 1 / 10 * 270 = 27 survived = 243 number of newborns that can die in third month = 1 / 10 * 243 = 24 survived = 219 answer : a" | a = 1 / 10
b = 1 - a
c = 300 * b
d = 1 / 10
e = 1 - d
f = c * e
g = 1 / 10
h = 1 - g
i = f * h
|
a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 5 | e | sqrt(6) | what is the range of all the roots of | x ^ 2 - 6 | = x ? | "we get 2 quadratic equations here . . 1 ) x ^ 2 - x - 2 = 0 . . . . . . . roots 2 , - 1 2 ) x ^ 2 + x - 2 = 0 . . . . . . . . roots - 2 , 1 inserting each root in given equation , it can be seen that - 1 and - 2 do not satisfy the equations . so value of x for given equation . . . . x = 6 or x = 1 i guess range is 6 - 1 = 5 e" | a = math.sqrt(6)
|
a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 16 | c | multiply(1, 6) | if 6 spiders make 3 webs in 7 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 6 : 1 ( webs ) 1 : 7 } : : x : 3 Γ’ β‘ β 6 Γ£ β 1 Γ£ β 7 = 1 Γ£ β 3 Γ£ β x Γ’ β‘ β x = 14 answer : option c" | a = 1 * 6
|
a ) 22 sec , b ) 27 sec , c ) 30 sec , d ) 24 sec , e ) 11 sec | c | divide(add(200, 100), multiply(36, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 200 meters long if the speed of the train is 36 kmph ? | "explanation : d = 100 + 200 = 300 s = 36 * 5 / 18 = 10 mps t = 300 / 10 = 10 sec answer : option c" | a = 200 + 100
b = 36 * const_0_2778
c = a / b
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | divide(divide(const_4, add(const_2, const_3)), subtract(inverse(6), inverse(15))) | a water tank is three - fifths full . pipe a can fill a tank in 15 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how many minutes will it take to empty or fill the tank completely ? | "the combined rate of filling / emptying the tank = 1 / 15 - 1 / 6 = - 1 / 10 since the rate is negative , the tank will be emptied . a full tank would take 10 minutes to empty . since the tank is only three - fifths full , the time is ( 3 / 5 ) * 10 = 6 minutes the answer is c ." | a = 2 + 3
b = 4 / a
c = 1/(6)
d = 1/(15)
e = c - d
f = b / e
|
a ) 5 / 18 , b ) 7 / 18 , c ) 1 / 3 , d ) 1 / 2 , e ) 1 / 5 | c | divide(6, add(add(5, 6), 7)) | in a bag , there are 5 green , 6 black and 7 red pens . one pen is picked up randomly . what is the probability that it is neither red nor green ? | explanation : neither red nor green means the pen drawn is black . total number of outcomes = ( 5 + 6 + 7 ) = 18 . number of favourable outcomes = 6 = number of black pens . hence , probability of the event = 6 / 18 = 1 / 3 . answer : c | a = 5 + 6
b = a + 7
c = 6 / b
|
a ) 10 , b ) 1 , c ) 1.5 , d ) 2 , e ) 2.5 | a | multiply(subtract(5, const_2.0), 5) | if ( c - a ) / ( c - b ) = 1 , then ( 5 b + 5 a ) / ( c - a ) = | "let ' s say c = 2 , b = 1 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 10 answer a ( hopefully ) ) )" | a = 5 - 2
b = a * 5
|
a ) 9 : 8 , b ) 11 : 10 , c ) 13 : 12 , d ) 15 : 14 , e ) 17 : 16 | d | divide(multiply(5, 9), multiply(6, 7)) | the marks obtained by polly and sandy are in the ratio 5 : 6 and those obtained by sandy and willy are in the ratio of 9 : 7 . the marks obtained by polly and willy are in the ratio of . . . ? | polly : sandy = 5 : 6 = 15 : 18 sandy : willy = 9 : 7 = 18 : 14 polly : sandy : willy = 15 : 18 : 14 polly : willy = 15 : 14 the answer is d . | a = 5 * 9
b = 6 * 7
c = a / b
|
a ) 12 , b ) 13 , c ) 16 , d ) 17 , e ) 18 | e | divide(subtract(const_1, multiply(12, divide(const_1, 24))), divide(const_1, 36)) | x can finish a work in 36 days . y can finish the same work in 24 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 36 work done by y in 1 day = 1 / 24 work done by y in 12 days = 12 / 24 = 1 / 2 remaining work = 1 β 1 / 2 = 1 / 2 number of days in which x can finish the remaining work = ( 1 / 2 ) / ( 1 / 36 ) = 18 e" | a = 1 / 24
b = 12 * a
c = 1 - b
d = 1 / 36
e = c / d
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 9 . | c | power(subtract(power(4, 2), 9), divide(9, 4)) | a ( 4 , w ^ 2 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 + 9 . what is the value of w ? | "y = x ^ 2 + 9 w ^ 2 = 4 ^ 2 + 9 w ^ 2 = 25 w = 5 answer c" | a = 4 ** 2
b = a - 9
c = 9 / 4
d = b ** c
|
a ) 1 hrs , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | c | divide(30, subtract(60, 50)) | two cars started at the same time , from the same point , driving along the same road . the rate of the first car is 50 mph and the rate of the second car is 60 mph . how long will it take for the distance between the two cars to be 30 miles ? | car 1 50 50 50 total = 150 car 2 60 60 60 total = 180 differece = 30 miles after 3 hrs . answer : c | a = 60 - 50
b = 30 / a
|
a ) 65 , b ) 80 , c ) 85 , d ) 82 , e ) 89 | e | divide(add(240, 650), speed(240, 24)) | a train 240 m long passes a pole in 24 seconds . how long will it take to pass a platform 650 m long ? | speed = ( 240 / 24 ) m / sec = 10 m / sec . therefore , required time = ( 240 + 650 ) / 10 sec = 89 sec . answer is e . | a = 240 + 650
b = a / speed
|
a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 55 % | d | multiply(divide(const_3, add(const_3, const_2)), const_100) | a feed store sells two varieties of birdseed : brand a , which is 40 % millet and 60 % sunflower , and brand b , which is 65 % millet and 35 % safflower . if a customer purchases a mix of the two types of birdseed that is 50 % millet , what percent of the mix is brand a ? | "another method to solve this question using equations : 1 kg of brand a will have 400 g millet and 600 g sunflower . 1 kg of brand b will gave 650 g millet and 350 g sunflower . suppose , x kg of brand a and y kg of brand b are in the mixture . then total weight of millet = 400 x + 650 y g total weight of mixture = 1000 x + 1000 y g given that millet is 500 % hence ( 400 x + 650 y ) / ( 1000 x + 1000 y ) = 1 / 2 or , 800 x + 1300 y = 1000 x + 1000 y or 300 y = 200 x or , x / y = 3 / 2 % of brand a = ( 3 / ( 3 + 2 ) ) * 100 = 60 % hence answer is d" | a = 3 + 2
b = 3 / a
c = b * 100
|
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