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
|
|---|---|---|---|---|---|
a ) 941,1009 , b ) 991,1001 , c ) 991,1009 , d ) 992,1008 , e ) 931,1009 | d | divide(999936, add(multiply(const_100, const_10), add(const_3, const_2))) | there are cats got together and decided to kill the mice of 999936 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999936 can be written as 1000000 Γ’ β¬ β 64 = 10002 Γ’ β¬ β 82 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 8 ) * ( 1000 - 8 ) = ( 1008 ) * ( 992 ) given that number of cats is less than number if mice . so number of cats is 992 and number of mice were 1008 answer d" | a = 100 * 10
b = 3 + 2
c = a + b
d = 999936 / c
|
a ) 4 km , b ) 2 km , c ) 5 km , d ) 3 km , e ) 1 km | d | subtract(add(add(7, divide(const_1, const_2)), 1.5), subtract(add(7, divide(const_1, const_2)), 1.5)) | a man can row 7 Γ’ Β½ kmph in still water . if in a river running at 1.5 km / hr an hour , it takes him 50 minutes to row to a place and back , how far off is the place ? | speed downstream = ( 7.5 + 1.5 ) km / hr = 9 km / hr ; speed upstream = ( 7.5 - 1.5 ) kmph = 6 kmph . let the required distance be x km . then , x / 9 + x / 6 = 50 / 60 . 2 x + 3 x = ( 5 / 6 * 18 ) 5 x = 15 x = 3 . hence , the required distance is 3 km . answer d | a = 1 / 2
b = 7 + a
c = b + 1
d = 1 / 2
e = 7 + d
f = e - 1
g = c - f
|
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | d | multiply(subtract(divide(60, const_2), divide(multiply(60, 30), const_100)), const_2) | a chemical supply company has 60 liters of a 30 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | 60 liters of a 30 % hno 3 solution means hno 3 = 18 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 18 + x = 50 % of ( 60 + x ) or x = 24 . hence , d | a = 60 / 2
b = 60 * 30
c = b / 100
d = a - c
e = d * 2
|
a ) 233 , b ) 245 , c ) 257 , d ) 270 , e ) 285 | b | multiply(add(divide(subtract(subtract(28, 10), const_2), const_2), 10), divide(add(subtract(28, 10), const_2), const_2)) | what is the sum of all digits for the number 10 ^ 28 - 44 ? | "10 ^ 28 is a 29 - digit number : 1 followed by 28 zeros . 10 ^ 28 - 44 is a 28 - digit number : 26 9 ' s and 56 at the end . the sum of the digits is 26 * 9 + 5 + 6 = 245 . the answer is b ." | a = 28 - 10
b = a - 2
c = b / 2
d = c + 10
e = 28 - 10
f = e + 2
g = f / 2
h = d * g
|
a ) 160 km , b ) 170 km , c ) 180 km , d ) 190 km , e ) 130 km | c | multiply(divide(120, add(30, 10)), 60) | mr and mrs a are opposite to each other . the distance between mr a and mrs a are 120 km . the speed of mr a and mrs a are 30 kmph , 10 kmph respectively . one bee is running between mr a nose to mrs a nose and returning back to mr a nose . the speed of bee is 60 kmph . then how far bee traveled ? | 30 x + 10 x = 120 x = 3 hrs speed of bee = 60 kmph distance traveled by bee = speed * time taken = 60 * 3 = 180 km answer : c | a = 30 + 10
b = 120 / a
c = b * 60
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | add(divide(32, add(5, 3)), divide(40, add(2, 3))) | each machine of type a has 2 steel parts and 3 chrome parts . each machine of type b has 3 steel parts and 5 chrome parts . if a certain group of type a and type b machines has a total of 40 steel parts and 32 chrome parts , how many machines are in the group | look at the below representation of the problem : steel chrome total a 2 3 40 > > no . of type a machines = 40 / 5 = 8 b 3 5 32 > > no . of type b machines = 32 / 8 = 4 so the answer is 12 i . e a . hope its clear . | a = 5 + 3
b = 32 / a
c = 2 + 3
d = 40 / c
e = b + d
|
a ) 32 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 38 % | a | multiply(const_100, divide(add(divide(40, const_100), multiply(4, divide(30, const_100))), add(const_1, 4))) | because he β s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ? | "let x be mork ' s income , then mindy ' s income is 4 x . the total tax paid is 0.4 x + 1.2 x = 1.6 x 1.6 x / 5 x = 0.32 the answer is a ." | a = 40 / 100
b = 30 / 100
c = 4 * b
d = a + c
e = 1 + 4
f = d / e
g = 100 * f
|
a ) 32 , b ) 20 , c ) 28 , d ) 11 , e ) 18 | b | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 6 and 34 which are divisible by 5 ? | "average = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 100 / 5 = 20 . answer : b" | a = 6 + 4
b = 34 - 4
c = a + b
d = c / 2
|
a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | c | multiply(divide(const_3, add(const_3, 2)), divide(add(const_3, 2), add(add(const_3, 2), 1))) | set # 1 = { a , b , c , d , e } set # 2 = { k , l , m , n , o , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? | so not a vowel in set - 1 : 3 / 5 and not a vowel in ser - 2 : 5 / 6 now , 3 / 5 β 5 / 6 = 12 this is for not a vowel . then for at least one vowel will be = 1 β 1 / 2 = 1 / 2 answer will be c . | a = 3 + 2
b = 3 / a
c = 3 + 2
d = 3 + 2
e = d + 1
f = c / e
g = b * f
|
a ) 23 / 30 , b ) 11 / 15 , c ) 7 / 10 , d ) 4 / 5 , e ) 2 / 15 | d | divide(add(15, 5), multiply(15, const_2)) | in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 15 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 15 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 5 ) = 1 / 15 therefore only video cassette recorder = 1 / 15 - 1 / 15 = 0 total = 1 / 5 + 0 + neither cable tv nor videocassette recorders 1 = 1 / 5 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = 1 - 1 / 5 = 4 / 5 hence d ." | a = 15 + 5
b = 15 * 2
c = a / b
|
a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these | d | multiply(divide(multiply(5, const_1000), const_60), 15) | a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in metres ) is | "explanation : speed = ( 5 Γ 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 Γ 15 Γ 60 ) m = 1250 m . answer : d" | a = 5 * 1000
b = a / const_60
c = b * 15
|
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | d | subtract(add(17, 5), const_1) | a student is ranked 17 th from right and 5 th from left . how many students are there in totality ? | "from right 17 , from left 5 total = 17 + 5 - 1 = 21 answer : d" | a = 17 + 5
b = a - 1
|
a ) 2.45 , b ) 8.45 , c ) 7.45 , d ) 3.45 , e ) 1.45 | d | multiply(const_12, divide(multiply(46, divide(46, const_100)), 64)) | a reduction of 46 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "explanation : 40 * ( 46 / 100 ) = 18.4 - - - 64 ? - - - 12 = > rs . 3.45 answer : d" | a = 46 / 100
b = 46 * a
c = b / 64
d = 12 * c
|
a ) 12 , b ) 9 , c ) 18 , d ) 77 , e ) 26 | b | subtract(11, reminder(11002, 11)) | what should be the least number to be added to the 11002 number to make it divisible by 11 ? | "answer : 9 option : b" | a = 11 - reminder
|
a ) 8 / 5 , b ) 5 / 8 , c ) 1 , d ) 10 , e ) it can not be determined | e | divide(multiply(8, 5), multiply(5, 8)) | the ratio between x and y is 8 / 5 ; x is increased by 10 and y is multiplied by 10 , what is the ratio between the new values of x and y ? | "ratio = 8 k / 5 k = 8 / 5 , 16 / 10 , etc . x and y are decreased by 5 - - > ( 8 k + 10 ) / ( 5 k * 10 ) new ratio can be 18 / 50 , 26 / 100 , etc . answer : e" | a = 8 * 5
b = 5 * 8
c = a / b
|
a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) 39 | c | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive multiples of 3 is 72 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 72 9 x = 63 x = 7 largest number = 3 x + 6 = 27 . answer : c" | a = 3 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 3 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 3 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 3 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 103 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 22) | cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 22 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 27 shows a cube of a number could be right . here 64 + 22 = 86 72 + 22 = 94 86 + 22 = 108 98 + 22 = 120 103 + 22 = 125 - - - ( 5 * 5 * 5 ) so we have 103 as the answer ! e" | a = 1 + 3
b = 1 + a
c = b ** 3
d = c - 22
|
a ) 6 , b ) 12 , c ) 16 , d ) 20 , e ) 22 | a | divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 34)), 4) | the average age of an adult class is 40 years . 12 new students with an avg age of 34 years join the class . therefore decreasing the average by 4 years . find what was the original average age of the class ? | "let original strength = y then , 40 y + 12 x 34 = ( y + 12 ) x 36 Γ’ β‘ β 40 y + 408 = 36 y + 432 Γ’ β‘ β 4 y = 48 Γ’ Λ Β΄ y = 6 a" | a = 40 - 4
b = 12 * a
c = 12 * 34
d = b - c
e = d / 4
|
a ) 169 : 121 , b ) 169 : 127 , c ) 169 : 191 , d ) 121 : 169 , e ) 121 : 182 | a | power(divide(2197, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 2197 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = Β³ β 2197 : Β³ β 1331 = 13 : 11 ratio of surface areas = 13 ^ 2 : 11 ^ 2 = 169 : 121 answer : option a" | a = 2197 / 1331
b = 1 / 3
c = a ** b
|
a ) 87 kmph , b ) 65 kmph , c ) 97 kmph , d ) 16 kmph , e ) 18 kmph | c | subtract(multiply(10, multiply(120, const_0_2778)), 150) | a train 150 m long crosses a platform 120 m long in 10 sec ; find the speed of the train ? | "d = 150 + 120 = 270 t = 10 s = 270 / 10 * 18 / 5 = 97 kmph answer : c" | a = 120 * const_0_2778
b = 10 * a
c = b - 150
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | subtract(100, add(add(add(subtract(100, 85), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 85 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 85 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 15 - 25 - 15 - 30 = 100 - 85 = 15 answer : e" | a = 100 - 85
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
|
a ) 1 : 5 , b ) 9 : 25 , c ) 5 : 1 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(10, const_2))) | two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 10 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 5 / speed b since the times are same , 50 pi / speed a = 10 pi / speed b speeda / speed b = 50 pi / 30 pi = 5 / 1 correct option : c" | a = 50 / 2
b = circumface / (
|
a ) rs . 80 , b ) rs . 86 , c ) rs . 90 , d ) rs . 95 , e ) none of these | b | subtract(divide(4500, 45), 14) | a trader sells 45 meters of cloth for rs . 4500 at the profit of rs . 14 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4500 / 45 = rs . 100 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 14 = rs . 86 . answer : b" | a = 4500 / 45
b = a - 14
|
a ) 2.75 , b ) 3.25 , c ) 3.75 , d ) 4.25 , e ) 4.75 | c | add(divide(divide(20, const_2), 4), divide(divide(20, const_2), 8)) | a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 20 km , if the person completes half of the distance by walking and the other half by running ? | "time = 10 / 4 + 10 / 8 = 30 / 8 = 3.75 hours the answer is c ." | a = 20 / 2
b = a / 4
c = 20 / 2
d = c / 8
e = b + d
|
a ) 11 , b ) 12 , c ) 13 , d ) none , e ) can not be determined | c | divide(156, divide(156, 12)) | 12 times a number gives 156 . the number is | "explanation : let the number be ' n ' 12 Γ n = 156 β n = 13 correct option : c" | a = 156 / 12
b = 156 / a
|
a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150 | d | divide(multiply(28, 10), const_2) | if the sides of a triangle are 30 cm , 28 cm and 10 cm , what is its area ? | "the triangle with sides 30 cm , 28 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 28 * 10 = 140 cm 2 answer : option d" | a = 28 * 10
b = a / 2
|
['a ) 9', 'b ) 6', 'c ) 4', 'd ) 3', 'e ) 2'] | d | power(sqrt(3), const_2) | if the side length of square b is sqrt ( 3 ) times that of square a , the area of square b is how many times the area of square a ? | let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( sqrt ( 3 ) x ) ^ 2 = 3 x ^ 2 . the answer is d . | a = math.sqrt(3)
b = a ** 2
|
a ) 28 % , b ) 40 % , c ) 68 % , d ) 70 % , e ) 72 % | c | add(multiply(divide(divide(20, const_100), subtract(1, divide(1, 10))), const_100), 2) | the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 15 percent , and on day 3 , it is discounted an additional 20 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 15 % discount = 765 price in day 3 after 20 % discount = 612 so , price in day 3 as percentage of the sale price on day 1 will be = 612 / 900 * 100 = > 68 % answer will definitely be ( c )" | a = 20 / 100
b = 1 / 10
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 700,000 , b ) 300,000 , c ) 380,000 , d ) 400,000 , e ) 420,000 | a | divide(add(divide(subtract(260000, multiply(0.8, 100000)), 0.3), 100000), const_1000) | an author received $ 0.80 in royalties for each of the first 100000 copies of her book sold , and $ 0.30 in royalties for each additional copy sold . if she received a total of $ 260000 in royalties , how many copies of her book were sold ? | total royalties for first 100.000 books = . 8 * 100,000 = 80,000 total royalties for the rest of the books = 260,000 - 80,000 = 180,000 remaining books = 180,000 / 0.3 = 600,000 total books = 600,000 + 100,000 = 700,000 answer a | a = 0 * 8
b = 260000 - a
c = b / 0
d = c + 100000
e = d / 1000
|
a ) 1 / 7 , b ) 1 / 8 , c ) 1 / 4 , d ) 3 / 7 , e ) 7 / 8 | d | multiply(subtract(1, divide(1, 6)), divide(1, 2)) | in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 / 6 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 / 6 of the garden is green , so , since all the flowers must be either green or yellow , we know that 5 / 6 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 5 / 6 * 1 / 2 = 3 / 7 , or d ." | a = 1 / 6
b = 1 - a
c = 1 / 2
d = b * c
|
a ) 23 , b ) 32 , c ) 35 , d ) 43 , e ) 44 | d | add(divide(subtract(102, add(add(9, 9), 9)), const_3), add(9, 9)) | virginia , adrienne , and dennis have taught history for a combined total of 102 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 102 = > 3 d = 102 + 27 = 129 = > d = 43 answer d" | a = 9 + 9
b = a + 9
c = 102 - b
d = c / 3
e = 9 + 9
f = d + e
|
a ) 5 : 2 , b ) 7 : 3 , c ) 9 : 2 , d ) 13 : 4 , e ) none of these | b | divide(add(multiply(subtract(add(multiply(10, const_2), multiply(10, const_2)), multiply(10, const_2)), const_3), 10), add(subtract(add(multiply(10, const_2), multiply(10, const_2)), multiply(10, const_2)), 10)) | the age of father 10 years ago was thirce the age of his son . 10 years hence , father β s age will be twice that of his son . the ration of their present ages is : | solution let the ages of father and son 10 year ago be 3 x and x years respectively . then , ( 3 x + 10 ) + 10 = 2 [ ( x + 10 ) + 10 β 3 x + 20 = 2 x + 40 β x = 20 . β΄ required ratio = ( 3 x + 10 ) : ( x + 10 ) = 70 : 30 : 7 : 3 . answer b | a = 10 * 2
b = 10 * 2
c = a + b
d = 10 * 2
e = c - d
f = e * 3
g = f + 10
h = 10 * 2
i = 10 * 2
j = h + i
k = 10 * 2
l = j - k
m = l + 10
n = g / m
|
a ) 8 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(2, 4) | a one - foot stick is marked in 1 / 2 and 1 / 4 portion . how many total markings will there be , including the end points ? | "lcm of 8 = 4 1 / 2 marking are ( table of 2 ) 0 . . . . . . 2 . . . . . . . . . . . 4 ( total = 3 ) 1 / 4 marking are ( table of 1 ) 0 . . . . . . . 1 . . . . . . 2 . . . . . . 3 . . . . . . . . 4 ( total = 5 ) overlapping markings are 0 . . . . . . . . 2 . . . . . . . . . 4 ( total = 3 ) total markings = 3 + 5 - 3 = 5 answer = c" | a = 2 + 4
|
a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | d | divide(add(divide(subtract(multiply(floor(divide(600, 25)), 25), multiply(floor(divide(200, 25)), 25)), 25), const_1), const_2) | if integer k is equal to the sum of all even multiples of 25 between 200 and 600 , what is the greatest prime factor of k ? | "if we break down what the stem is asking what is the sum of all mult of 50 between 200 and 600 . using arithmetic progression to find n : 600 = 200 + ( n - 1 ) 50 400 + 50 = 50 n 450 = 50 n = > n = 9 the sum would be : 9 * mean mean = [ 600 + 200 ] / 2 = 400 9 * 400 = 3600 d" | a = 600 / 25
b = math.floor(a)
c = b * 25
d = 200 / 25
e = math.floor(d)
f = e * 25
g = c - f
h = g / 25
i = h + 1
j = i / 2
|
a ) 10,300 , b ) 8,030 , c ) 1,253 , d ) 1,252 , e ) 1,239 | e | subtract(448,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 448,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 448,513 the number of integers is 4486 - 3247 = 1239 the answer is e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 448 - 600
|
a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44 | b | multiply(2, 12) | g ( x ) is defined as the product of all even integers k such that 0 < k β€ x . for example , g ( 14 ) = 2 Γ 4 Γ 6 Γ 8 Γ 10 Γ 12 Γ 14 . if g ( r ) is divisible by 4 ^ 11 , what is the smallest possible value for r ? | "g ( r ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b" | a = 2 * 12
|
a ) 110 , b ) 115 , c ) 120 , d ) 125 , e ) 130 | c | divide(35, subtract(const_1, add(add(add(divide(1, 12), divide(1, 8)), divide(1, 3)), divide(1, 6)))) | if 1 / 12 of the passengers on a ship are from north america , 1 / 8 are europeans , 1 / 3 are from africa , 1 / 6 are from asia and the remaining 35 people are citizens of other continents , then how many passengers are on board the ship ? | 1 / 12 + 1 / 8 + 1 / 3 + 1 / 6 = ( 2 + 3 + 8 + 4 ) / 24 = 17 / 24 let x be the number of passengers on the ship . 35 = ( 7 / 24 ) x x = 120 the answer is c . | a = 1 / 12
b = 1 / 8
c = a + b
d = 1 / 3
e = c + d
f = 1 / 6
g = e + f
h = 1 - g
i = 35 / h
|
a ) 2449 , b ) 5449 , c ) 6749 , d ) 6449 , e ) 6468 | c | subtract(multiply(divide(54671, const_100), 14456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 14456))) | 54671 - 14456 - 33466 = ? | "c if we calculate we will get 6749" | a = 54671 / 100
b = a * 14456
c = 1 / 3
d = 54671 / 100
e = d * 14456
f = c * e
g = b - f
|
a ) $ 2,040 , b ) $ 2,120 , c ) $ 2,240 , d ) $ 1,920 , e ) $ 1,400 | c | subtract(multiply(const_3, const_1000), add(multiply(divide(add(13, 3), 3), 20), multiply(subtract(divide(add(13, const_2.0), 3), 3), 100))) | a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 13 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 3 more or 3 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 13 : x + x - 3 = 13 - - > x = 8 : 8 $ 20 chips were lost and 8 - 2 = 6 $ 100 chips were lost . total worth of chips lost is 8 * 20 + 6 * 100 = $ 760 , so the gambler kept $ 3,000 - $ 760 = $ 2,240 . answer : c ." | a = 3 * 1000
b = 13 + 3
c = b / 3
d = c * 20
e = 13 + 2
f = e / 3
g = f - 3
h = g * 100
i = d + h
j = a - i
|
a ) 30 , 10 , b ) 32 , 12 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | b | add(divide(add(multiply(5, 7), subtract(20, 7)), subtract(5, const_1)), 20) | the ages of two person differ by 20 years . if 7 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 7 ) = ( x + 20 - 7 ) = > 4 x = 48 = > x = 12 their present ages are 32 years and 12 year . answer : b" | a = 5 * 7
b = 20 - 7
c = a + b
d = 5 - 1
e = c / d
f = e + 20
|
a ) 161 , b ) 171 , c ) 181 , d ) 191 , e ) 201 | b | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 1 and 10 ^ 18 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 17 and there are 18 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 17 thus , the total number of integers is 1 + 2 + . . . + 17 + 18 = 18 * 19 / 2 = 171 the answer is b ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 10 - e
|
a ) 27 , b ) 25 , c ) 24 , d ) 20 , e ) 22 | e | subtract(subtract(subtract(27, 2), const_1), const_1) | how many positive integers less than 27 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 3 odd multiples of 5 : { 5 , 15 , 25 } 11 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 } notice , that 5 is in two sets , thus total # of integers satisfying the given conditions is 9 + 3 + 11 - 1 = 22 . answer : e ." | a = 27 - 2
b = a - 1
c = b - 1
|
a ) 3 / 5 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | d | divide(subtract(50, 42), subtract(50, 40)) | in a corporation , 50 percent of the male employees and 40 percent of the female employees are at least 35 years old . if 42 percent of all the employees are at least 35 years old , what fraction of the employees in the corporation are females ? | you can use the weighted averages formula for a 10 sec solution . no of females / no of males = ( 50 - 42 ) / ( 42 - 40 ) = 4 / 1 no of females as a fraction of total employees = 4 / ( 4 + 1 ) = 4 / 5 ; answer : d | a = 50 - 42
b = 50 - 40
c = a / b
|
a ) rs 840 , b ) rs 1320 , c ) rs 1620 , d ) rs 1890 , e ) none of these | d | multiply(divide(multiply(9, 6), multiply(4, 10)), 1400) | if 4 men working 10 hours a day earn rs . 1400 per week , then 9 men working 6 hours a day will earn how much per week ? | "explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1400 : x hence 4 * 10 * x = 9 * 6 * 1400 or x = 9 * 6 * 1400 / 4 * 10 = 1890 answer : d" | a = 9 * 6
b = 4 * 10
c = a / b
d = c * 1400
|
a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | c | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 120)))), const_2) | if a rectangular billboard has an area of 120 square feet and a perimeter of 46 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 120 and a perimeter of 46 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 120 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 120 , let ' s test answer c : 8 if . . . the shorter side = 8 . . . the area = 120 . . . . 120 / 8 = 15 = the longer side perimeter = 8 + 8 + 15 + 15 = 46 c" | a = 46 / 2
b = 46 / 2
c = b ** 2
d = 4 * 120
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
|
a ) 168 , b ) 172 , c ) 176 , d ) 180 , e ) 184 | c | subtract(multiply(multiply(divide(33, 75), const_100), 16), multiply(33, 16)) | a car gets 33 miles to the gallon . if it is modified to use a solar panel , it will use only 75 percent as much fuel as it does now . if the fuel tank holds 16 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ? | "originally , the distance the car could go on a full tank was 16 * 33 = 528 miles . after it has been modified , the car can go 33 / 0.75 = 44 miles per gallon . on a full tank , the car can go 16 * 44 = 704 miles , thus 176 miles more . the answer is c ." | a = 33 / 75
b = a * 100
c = b * 16
d = 33 * 16
e = c - d
|
a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 2 , d ) 4 / 9 , e ) 2 / 3 | c | multiply(subtract(1, divide(2, 7)), subtract(1, divide(subtract(1, divide(2, 7)), 2))) | for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 2 / 7 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ? | "p ( of not burning out in a six mnth period ) = 1 / 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 2 / 7 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 / 7 = 5 / 7 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 5 / 7 = 1 / 3 - - - > p ( of burning out in a six mnth period ) = 1 - 1 / 3 = 2 / 3 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 5 / 7 * 2 / 3 = 1 / 2 ans c" | a = 2 / 7
b = 1 - a
c = 2 / 7
d = 1 - c
e = d / 2
f = 1 - e
g = b * f
|
a ) 65 , b ) 69 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 55), add(const_1, const_4)) | shekar scored 76 , 65 , 82 , 67 and 55 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 67 + 55 ) / 5 = 375 / 5 = 69 hence average = 69 answer : b" | a = 76 + 65
b = a + 82
c = b + 67
d = c + 55
e = 1 + 4
f = d / e
|
a ) 1187 m , b ) 1704 m , c ) 2179 m , d ) 3520 m , e ) 4297 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 2500), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 2500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 2500 resolutions . = 2500 * 2 * 22 / 7 * 22.4 = 352000 cm = 3520 m answer : d" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 2500
i = h / 100
|
a ) 20 , b ) 27 , c ) 72 , d ) 28 , e ) 82 | b | multiply(23, 4) | the average of 10 numbers is 23 . if each number is increased by 4 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 4 , the total increase = 4 * 10 = 40 the new sum = 230 + 40 = 270 the new average = 270 / 10 = 27 . answer : b" | a = 23 * 4
|
a ) 69 minutes , b ) 68 minutes , c ) 58 minutes , d ) 48 minutes , e ) 67 minutes | b | add(add(add(multiply(20, multiply(4, divide(50, const_100))), 20), 4), 4) | if a person has two rectangular fields . the larger field has thrice the length and 4 times the width of the smaller field . if the smaller field has a length 50 % more than the width . if a person takes 20 minutes to complete one round of a smaller field then what is the time required to complete one round of a larger field ? | let the width of the smaller rectangle is 4 units then , the length of the smaller rectangle is 6 units ( that is 50 % more than the width ) now the perimeter of the rectangle is 2 ( 6 + 4 ) = 20 units so , 20 units is covered in 20 units , implies covers one unit in one minute so now coming to the larger rectangle the width = 16 units , length = 18 units , perimeter = 2 ( 16 + 18 ) 68 units thus , to cover the larger field , 68 minutes required to cover the rectangle . answer : b | a = 50 / 100
b = 4 * a
c = 20 * b
d = c + 20
e = d + 4
f = e + 4
|
a ) 2 , b ) 4 , c ) 5 , d ) 40 , e ) 160 | c | subtract(divide(480, 40), 7) | the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 7 miles per gallon , will not get him there . by how many miles per gallon must the aircraft β s fuel efficiency be improved to make the flight to cleveland possible ? | "actual miles / gallon is = 480 / 40 = 12 miles / gallon . current engine miles / gallon is 7 miles / gallon . additional 5 miles / gallon is required to match the actual mileage . imo option c ." | a = 480 / 40
b = a - 7
|
a ) 2 / 5 , b ) 2 / 3 , c ) 1 / 3 , d ) 1 / 2 , e ) 1 / 1 | e | divide(subtract(divide(20, const_100), divide(40, const_100)), subtract(divide(20, const_100), divide(40, const_100))) | some of 40 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 20 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 40 ( 1 - x ) + 20 x = 20 x = 1 / 1 answer : e" | a = 20 / 100
b = 40 / 100
c = a - b
d = 20 / 100
e = 40 / 100
f = d - e
g = c / f
|
a ) 11 , b ) 111 , c ) 211 , d ) 311 , e ) 411 | b | multiply(divide(subtract(subtract(power(multiply(10, divide(add(const_100, 20), const_100)), const_2), power(multiply(6, divide(subtract(const_100, 50), const_100)), const_2)), subtract(power(10, const_2), power(6, const_2))), subtract(power(10, const_2), power(6, const_2))), const_100) | there are two concentric circles with radii 10 and 6 . if the radius of the outer circle is increased by 20 % and the radius of the inner circle decreased by 50 % , by what percent does the area between the circles increase ? | "the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 100 pi . the area of the small circle is 36 pi . the area a 1 between the circles is 64 pi . when the big circle ' s radius increases , the new area is 144 pi . when the small circle ' s radius decreases , the new area is 9 pi . the area a 2 between the circles is 135 pi . the ratio of a 2 / a 1 is 135 / 64 = 2.11 which is an increase of 111 % . the answer is b ." | a = 100 + 20
b = a / 100
c = 10 * b
d = c ** 2
e = 100 - 50
f = e / 100
g = 6 * f
h = g ** 2
i = d - h
j = 10 ** 2
k = 6 ** 2
l = j - k
m = i - l
n = 10 ** 2
o = 6 ** 2
p = n - o
q = m / p
r = q * 100
|
a ) 10 , b ) 20 , c ) 50 , d ) 8.6 , e ) 40 | d | subtract(subtract(15, 3.2), 3.2) | a man ' s speed with the current is 15 km / hr and the speed of the current is 3.2 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 3.2 km / hr hence , speed of the man = 15 - 3.2 = 11.8 km / hr man ' s speed against the current = speed of the man - speed of the current = 11.8 - 3.2 = 8.6 km / hr answer is d ." | a = 15 - 3
b = a - 3
|
a ) $ 55,000 , b ) $ 75,000 , c ) $ 95,000 , d ) $ 115,000 , e ) $ 125,000 | b | divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2))) | the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 46,875 , how much profit did the business make ? | "let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = 46875 p = $ 75,000 the answer is b ." | a = 1 + 4
b = 100 * a
c = 100 * b
d = 25 / 100
e = 25 / 100
f = 25 / 100
g = 1 - f
h = e * g
i = h * 2
j = d + i
k = c / j
|
a ) 30 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | a | multiply(24, inverse(subtract(const_1, divide(8, 40)))) | x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 24 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 24 days whole work will be done by y in 24 * 5 / 4 = 30 days answer is a" | a = 8 / 40
b = 1 - a
c = 1/(b)
d = 24 * c
|
a ) $ 6.80 , b ) $ 8.40 , c ) $ 7.70 , d ) $ 4.70 , e ) $ 3.90 | a | add(add(multiply(3, 1.2), 1.5), 1.7) | linda bought 3 notebooks at $ 1.20 each ; a box of pencils at $ 1.50 and a box of pens at $ 1.70 . how much did linda spend ? | linda spent 1.20 ? 3 = $ 3.60 on notebooks the total amount of money that linda spent is equal to 3.60 + 1.50 + 1.70 = $ 6.80 correct answer a | a = 3 * 1
b = a + 1
c = b + 1
|
['a ) 576 metres', 'b ) 589 metres', 'c ) 600 metres', 'd ) 700 metres', 'e ) none'] | c | multiply(divide(add(1764, power(multiply(3, const_2), const_2)), multiply(const_2, multiply(3, const_2))), const_4) | a park square in shape has a 3 metre wide road inside it running along its sides . the area occupied by the road is 1764 square metres . what is the perimeter along the outer edge of the road ? | solution let the length of the outer edges be x metres . then , length of the inner edge = ( x - 6 ) m . β΄ x 2 - ( x - 6 ) 2 = 1764 βΉ = βΊ x 2 - ( x 2 - 12 x + 36 ) = 1764 βΉ = βΊ 12 x = 1800 βΉ = βΊ x = 150 . β΄ required perimeter = ( 4 x ) m βΉ = βΊ ( 4 x 150 ) m = 600 m . answer c | a = 3 * 2
b = a ** 2
c = 1764 + b
d = 3 * 2
e = 2 * d
f = c / e
g = f * 4
|
a ) 45 % , b ) 80 % , c ) 78 % , d ) 94 % , e ) 68 % | b | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | two numbers are respectively 20 % and 50 % more than a third number . the percentage that is first of the second is ? | "i ii iii 120 150 100 150 - - - - - - - - - - 120 100 - - - - - - - - - - - ? = > 80 % answer : b" | a = 20 + 100
b = 50 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 30 % | e | multiply(divide(subtract(22932, 17640), 17640), const_100) | an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 22932 / - in 3 years at the same rate of interest . find the rate percentage ? | "explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 22932 / - - 17640 / - = rs . 5292 / - rate of interest = ( 5292 / 22932 ) Γ ( 100 / 1 ) = > 30 % answer : option e" | a = 22932 - 17640
b = a / 17640
c = b * 100
|
a ) 10 , b ) 15 , c ) 8 , d ) 12 , e ) 20 | b | subtract(floor(divide(subtract(60, 7), 4)), floor(divide(subtract(1, 7), 4))) | for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 60 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 60 4 n < 53 n < 13 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 13 ) combining the two - 1.5 < n < 13.25 n = 1 to 13 ( 48 integers ) and n = - 1 and 0 so 15 integers . b" | a = 60 - 7
b = a / 4
c = math.floor(b)
d = 1 - 7
e = d / 4
f = math.floor(e)
g = c - f
|
a ) 14 , b ) 13 , c ) 15 , d ) 18 , e ) 12 | b | add(25, const_1) | the average of first 25 natural numbers is ? | "sum of 25 natural no . = 650 / 2 = 325 average = 325 / 25 = 13 answer : b" | a = 25 + 1
|
a ) 0.15 , b ) 0.20 , c ) 0.25 , d ) 0.30 , e ) 0.33 | b | multiply(divide(4, 4), divide(const_1, 5)) | r = { 2 , 3 , 4 , 5 } b = { 4 , 5 , 6 , 7 , 8 } two integers will be randomly selected from the sets above , one integer from set r and one integer from set b . what is the probability that the sum of the two integers will equal 9 ? | the total number of pairs r , b possible is 4 * 5 = 20 . out of these 20 pairs only 4 sum up to 9 : ( 2 , 7 ) ; ( 3 , 6 ) , ( 4 , 5 ) and ( 5 , 4 ) . the probability thus is 4 / 20 = 0.2 . answer : b . | a = 4 / 4
b = 1 / 5
c = a * b
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 80 % | c | multiply(divide(subtract(79.95, 59.95), 79.95), const_100) | a $ 79.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 79.95 $ discounted selling price of chair = 59.95 $ discount = 79.95 - 59.95 = 20 $ % decrease in price of chair = ( 20 / 79.95 ) * 100 % = 25 % approx answer c" | a = 79 - 95
b = a / 79
c = b * 100
|
a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | e | subtract(multiply(4, 5), 5) | 5 drainage pipes , each draining water from a pool at the same constant rate , together can drain a certain pool in 16 days . how many additional pipes , each draining water at the same constant rate , will be needed to drain the pool in 4 days ? | this is an inverse proportional problem . . . . . . 5 pipes in 16 days ; so for 4 days , it will be = 16 x 5 / 4 = 20 so , 20 - 5 = 15 answer e | a = 4 * 5
b = a - 5
|
a ) a ) 31 , b ) b ) 41 , c ) c ) 106 , d ) d ) 61 , e ) e ) 71 | c | lcm(lcm(3, 5), 7) | a heap of grapes is divided into groups of 3 , 5 and 7 and each time one coconut is left over . the least number of grapes in the heap is ? a . 31 b . 41 c . 51 d . 61 | lcm = 105 = > 105 + 1 = 106 answer : c | a = math.lcm(3, 5)
b = math.lcm(a, 7)
|
a ) 5 / 0 , b ) 5 / 9 , c ) 5 / 1 , d ) 5 / 3 , e ) 5 / 6 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | three 6 faced dice are thrown together . the probability that no two dice show the same number on them is | "explanation : no two dice show same number would mean all the three faces should show different numbers . the first can fall in any one of the six ways . the second die can show a different number in five ways . the third should show a number that is different from the first and second . this can happen in four ways . thus 6 * 5 * 4 = 120 favourable cases . the total cases are 6 * 6 * 6 = 216 . the probability = 120 / 216 = 5 / 9 . answer : b" | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 121 , b ) 124 , c ) 127 , d ) 122 , e ) 129 | c | gcd(subtract(1657, 6), subtract(2037, 5)) | find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively . | "required number = h . c . f . of ( 1657 - 6 ) and ( 2037 - 5 ) = h . c . f . of 1651 and 2032 _______ 1651 ) 2032 ( 1 1651 1651 _______ 381 ) 1651 ( 4 1524 _________ 127 ) 381 ( 3 381 0 required number = 127 . answer is c ." | a = 1657 - 6
b = 2037 - 5
c = math.gcd(a, b)
|
a ) 1478 , b ) 1578 , c ) 1678 , d ) 1778 , e ) 1798 | a | add(divide(multiply(add(divide(100, 2), 3), 52), 2), 100) | ram - leela has $ 100 in her piggy bank . how much will she have in her bank 52 weeks from now if she puts $ 1 in the bank next week , $ 2 two weeks from now , $ 3 3 weeks from now , and continues to increase the amount that she puts in by $ 1 each week ? | the dollar deposits are in an a . p . 1,2 , 3,4 . . . 52 with common difference 1 sum of the terms is n ( n + 1 ) / 2 i . e 52 * ( 52 + 1 ) / 2 = 52 * 53 / 2 = 1378 total deposit therefore with chiu - lihas is 100 + 1378 = 1478 $ | a = 100 / 2
b = a + 3
c = b * 52
d = c / 2
e = d + 100
|
a ) 250 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 350), add(5, 2)) | a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 350 pounds of the mixture ? | almonds : walnuts = 5 : 2 total mixture has 7 parts in a 350 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 350 = 250 pounds answer ( a ) | a = 5 * 350
b = 5 + 2
c = a / b
|
a ) 1 / 4 , b ) - 1 , c ) 5 / 4 , d ) 3 / 2 , e ) 3 / 4 | a | subtract(multiply(2, divide(5, 4)), power(divide(3, 2), 2)) | if x = - 5 / 4 and y = - 3 / 2 , what is the value of the expression - 2 x β y ^ 2 ? | x = - 5 / 4 and y = - 3 / 2 = = > - 2 ( - 5 / 4 ) - ( 3 / 2 ) ^ 2 = 10 / 4 - 9 / 4 = 1 / 4 ans : a | a = 5 / 4
b = 2 * a
c = 3 / 2
d = c ** 2
e = b - d
|
a ) 3252 / 3257 , b ) 3456 / 3461 , c ) 3591 / 3596 , d ) 3641 / 3656 , e ) 3453 / 3458 | d | divide(subtract(const_3600, multiply(const_3, const_3)), subtract(const_3600, const_4)) | which expression is the greatest | "options can be re - written as ( x - 5 ) x = > 1 - ( 5 / x ) a ) 1 - ( 5 / 3257 ) b ) 1 - ( 5 / 3461 ) c ) 1 - ( 5 / 3596 ) d ) 1 - ( 5 / 3656 ) e ) 1 - ( 5 / 3458 ) to get the largest among these second half should be the least and so denominator to be largest . hence ' d ' ." | a = 3 * 3
b = 3600 - a
c = 3600 - 4
d = b / c
|
a ) 87 , b ) 89 , c ) 86 , d ) 93 , e ) 95 | c | subtract(multiply(4, add(78, 2)), multiply(78, 3)) | jerry β s average ( arithmetic mean ) score on the first 3 of 4 tests is 78 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 78 * 3 = 234 jerry wants the average to be = 80 hence total score on 4 tests should be = 80 * 4 = 320 score required on the fourth test = 320 - 234 = 86 option c" | a = 78 + 2
b = 4 * a
c = 78 * 3
d = b - c
|
a ) 12 , b ) 14 , c ) 16 , d ) 24 , e ) 40 | d | divide(divide(multiply(16, 30), const_2), divide(60, multiply(const_3, const_2))) | guy drives 60 miles to attend a meeting . halfway through , he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half . his average speed for the entire trip is 30 miles per hour . guy drives on average how many miles per hour during the first half of the way ? | let x be the average speed for 1 st half of the distance . then the average speed for 2 nd half of the distance will be x + 16 avg speed = total distance / total time 30 = 60 / { ( 30 / x ) + ( 30 / ( x + 16 ) ) } solving we get x ^ 2 - 14 x - 240 = 0 x = - 10 or 24 x cant be negative hence x = 24 answer : d | a = 16 * 30
b = a / 2
c = 3 * 2
d = 60 / c
e = b / d
|
a ) 5 days , b ) 15 days , c ) 28 days , d ) 4 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 18 men , 4 women and 10 children together complete the piece of work ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 18 men β s 1 day β s work = 1 Γ 18 / 96 = 3 / 16 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 18 men + 4 women + 10 children ) β s 1 day β s work = 3 / 16 + 1 / 48 + 1 / 24 = 1 / 4 the required no . of days = 4 days answer : option d" | a = 24 * 10
b = 1/(a)
c = 10 * b
d = 12 * 8
e = 1/(d)
f = e * 10
g = 48 * 4
h = 1/(g)
i = h * 4
j = f + i
k = c + j
l = 1/(k)
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | divide(add(multiply(4, 15), 15), subtract(4, const_1)) | 15 years hence , rohan will be just 4 times as old as he was 15 years ago . how old is rohan at present ? | let the present age of rohan be x years then , given : x + 15 = 4 ( x - 15 ) x = 25 answer : b | a = 4 * 15
b = a + 15
c = 4 - 1
d = b / c
|
a ) 236 , b ) 354 , c ) 432 , d ) 512 , e ) 670 | c | divide(multiply(54, 32), 4) | dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 54 of gas ? | 54 / 4 = 13.5 gallons 13.5 * 32 = 432 miles the answer is c . | a = 54 * 32
b = a / 4
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | c | divide(power(2, divide(power(2, 3), power(2, 2))), power(2, 1)) | if the operation ΓΈ is defined for all positive integers x and w by x ΓΈ w = ( 2 ^ x ) / ( 2 ^ w ) then ( 3 ΓΈ 1 ) ΓΈ 1 = ? | "3 ΓΈ 1 = 2 ^ 3 / 2 ^ 1 = 4 4 ΓΈ 1 = 2 ^ 4 / 2 = 8 the answer is c ." | a = 2 ** 3
b = 2 ** 2
c = a / b
d = 2 ** c
e = 2 ** 1
f = d / e
|
a ) 52 , b ) 62 , c ) 66 , d ) 68 , e ) 72 | c | power(const_2.0, const_2) | what will be the remainder when ( 67 ^ 67 ) + 67 is divided by 68 ? | "x ^ n + 1 will be divisible by x + 1 only when n is odd 67 ^ 67 + 1 will be divisible by 67 + 1 ( 67 ^ 67 + 1 ) + 66 , when divided by 68 will give 66 as remainder answer is c" | a = 2 ** 0
|
a ) 12 min , b ) 36 min , c ) 40 min , d ) 48 min , e ) 60 min | b | multiply(divide(15, multiply(5, 5)), const_60) | walking at the rate of 5 kmph a man cover certain distance in 5 hr . running at a speed of 15 kmph the man will cover the same distance in . | distance = speed * time 5 * 5 = 25 km new speed = 15 kmph therefore time = 25 / 15 = 5 / 3 = 36 min answer : b | a = 5 * 5
b = 15 / a
c = b * const_60
|
a ) 8 , b ) 10 , c ) 2 , d ) 14 , e ) 16 | c | floor(subtract(divide(300, 40), divide(50, 10))) | subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 18 pages . | subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 5 / 2.5 ) = 2 hrs answer : c | a = 300 / 40
b = 50 / 10
c = a - b
d = math.floor(c)
|
a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600 | d | divide(multiply(subtract(const_100, 10), divide(720, const_2)), const_100) | a pair of articles was bought for $ 720 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 720 / 2 = $ 360 let m . p = $ x 90 % of x = 360 x = 360 * 100 / 90 = $ 400 answer is d" | a = 100 - 10
b = 720 / 2
c = a * b
d = c / 100
|
a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | b | add(330, divide(multiply(4, subtract(420, 330)), add(3, 4))) | average expenditure of a person for the first 3 days of a week is rs . 330 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 330 total excess than assumed mean = 4 Γ ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 330 + rs . 40 = rs . 370 correct option : b" | a = 420 - 330
b = 4 * a
c = 3 + 4
d = b / c
e = 330 + d
|
a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50 | e | add(30, 30) | set x consists of 10 integers and has median of 30 and a range of 30 . what is the value of the greatest possible integer that can be present in the set ? | "note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 40 , 50 this is possible . hence it is the greatest such number . answer ( e )" | a = 30 + 30
|
a ) 132 , b ) 26 , c ) 17 , d ) 11 , e ) 12 | c | subtract(subtract(subtract(55, 18), 10), 10) | in a class there are 55 pupil , out of them 10 are in debate only and 18 in singing only . then how many in both ? | explanation : total pupil = 55 debate + singing = 10 + 18 = 28 the intersection for two = 55 Γ’ β¬ β 10 Γ’ β¬ β 28 = 17 play both games . answer : c | a = 55 - 18
b = a - 10
c = b - 10
|
a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 9.4 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 8), add(const_100, 20)))), const_100) | in a hostel , the number of students decreased by 8 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ? | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 8 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.92 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.92 * 1.2 ) = > f ( new ) = 0.906 f therefore the new cost of food must be 90.6 % of the old cost , or the cost of food must decrease by 9.4 % answer : e" | a = 100 * 100
b = 100 - 8
c = 100 + 20
d = b * c
e = a / d
f = 1 - e
g = f * 100
|
a ) 5.7 , b ) 6.0 , c ) 6.7 , d ) 9.7 , e ) 14.0 | e | divide(add(divide(divide(88, const_3), const_3), divide(multiply(divide(88, const_3), const_2), const_3)), const_2) | the total circumference of two circles is 88 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 Ο r + 2 Ο r = 88 where 2 r = r thus : 2 Ο r + 4 Ο r = 88 6 Ο r = 88 r = approx 4.7 Ο r + 2 r Ο = 88 3 Ο r = 88 r = approx 9.3 r + r = approx 14.0 answer : e" | a = 88 / 3
b = a / 3
c = 88 / 3
d = c * 2
e = d / 3
f = b + e
g = f / 2
|
a ) 250 , b ) 440 , c ) 510 , d ) 575 , e ) 449 | e | add(divide(subtract(999, 101), 2), const_1) | how many multiples of 2 are there between 101 and 999 ? | "2 multiples are 102 , 104,106 , - - - - - - - - - , 994 , 996,998 it should be mentioned whether 1 and 89 are inclusive . the answer is ( 998 - 102 ) / 2 + 1 = 449 answer is e" | a = 999 - 101
b = a / 2
c = b + 1
|
a ) 128 , b ) 197 , c ) 127 , d ) 182 , e ) 091 | b | subtract(multiply(13, 16), add(const_10, const_1)) | find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5 . | "explanation : let ' n ' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5 . required number = ( lcm of 13 and 16 ) - ( common difference of divisors and remainders ) = ( 208 ) - ( 11 ) = 197 . answer : b" | a = 13 * 16
b = 10 + 1
c = a - b
|
a ) 12 , b ) 27 , c ) 18 , d ) 16 , e ) 81 | d | subtract(multiply(40, divide(80, const_100)), multiply(divide(4, 5), 20)) | how much is 80 % of 40 is greater than 4 / 5 of 20 ? | "( 80 / 100 ) * 40 Γ’ β¬ β ( 4 / 5 ) * 20 32 - 16 = 16 answer : d" | a = 80 / 100
b = 40 * a
c = 4 / 5
d = c * 20
e = b - d
|
a ) 15 % , b ) 20 % , c ) 23 % , d ) 30 % , e ) none | c | multiply(subtract(const_1, divide(multiply(const_1, 10), 13)), const_100) | if the price of sugar rises from rs . 10 per kg to rs . 13 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x Γ 13 = x = 77 kg . β΄ reduction in consumption = 23 % . answer c" | a = 1 * 10
b = a / 13
c = 1 - b
d = c * 100
|
a ) 375 m , b ) 750 m , c ) 400 m , d ) 800 m , e ) 300 m | c | multiply(divide(800, subtract(45, 15)), 15) | a train crosses a bridge of length 800 m in 45 seconds and a lamp post on the bridge in 15 seconds . what is the length of the train in metres ? | "let length of train = l case - 1 : distance = 800 + l ( while crossing the bridge ) time = 45 seconds i . e . speed = distance / time = ( 800 + l ) / 45 case - 2 : distance = l ( while passing the lamp post ) time = 15 seconds i . e . speed = distance / time = ( l ) / 15 but since speed has to be same in both cases so ( 800 + l ) / 45 = ( l ) / 15 i . e . 800 + l = 3 l i . e . 2 l = 800 i . e . l = 400 answer : option c" | a = 45 - 15
b = 800 / a
c = b * 15
|
a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 70 % , e ) 72 % | e | add(multiply(divide(divide(20, const_100), subtract(1, divide(1, 10))), const_100), 2) | the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 20 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 10 % discount = 810 price in day 3 after 20 % discount = 648 so , price in day 3 as percentage of the sale price on day 1 will be = 648 / 900 * 100 = > 72 % answer will definitely be ( e )" | a = 20 / 100
b = 1 / 10
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 26 , b ) 28 , c ) 29 , d ) 30 , e ) 31 | d | divide(add(87, const_1), const_2) | the sum of three consecutive numbers is 87 . the greatest among these three number is : | "let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 87 3 x = 84 x = 28 greatest number , ( x + 2 ) = 30 . answer : d" | a = 87 + 1
b = a / 2
|
a ) 80 kmph , b ) 69 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | b | divide(add(290, 400), add(4.5, 5.5)) | a train travels 290 km in 4.5 hours and 400 km in 5.5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 290 + 400 = 690 km thus , total speed = 10 hrs or , average speed = 690 / 10 or , 69 kmph . answer : b" | a = 290 + 400
b = 4 + 5
c = a / b
|
a ) 419 , b ) 551 , c ) 601 , d ) 620 , e ) 858.54 | e | divide(add(add(add(add(add(add(12, 13), 14), 510), 520), 530), 12), add(const_3, const_4)) | what is the average of 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1 , 125,2140 , 2345 ? | "add 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1,125 , 2140 , 2345 grouping numbers together may quicken the addition sum = 9444 4959 / 11 = 858.54 . e" | a = 12 + 13
b = a + 14
c = b + 510
d = c + 520
e = d + 530
f = e + 12
g = 3 + 4
h = f / g
|
a ) 3 / 10 , b ) 2 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | b | divide(4, subtract(multiply(subtract(6, 4), 3), 4)) | when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 3 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 3 ) ( 4 + 4 x ) or , x = 2 answer b" | a = 6 - 4
b = a * 3
c = b - 4
d = 4 / c
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | how many integerskgreater than 100 and less than 800 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be six pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 6 * 10 = 60 hence b ." | a = 100 / 10
b = 2 * 4
c = a * b
d = 100 / 10
e = c + d
|
a ) 80 , b ) 70 , c ) 60 , d ) 50 , e ) 45 | a | multiply(40, const_3.0) | local kennel has cats and dogs in the ratio of 5 : 10 . if there are 40 fewer cats than dogs , how many dogs are in the kennel ? | "lets work with the data given to us . we know that there ratio of cats to dogs is 5 : 10 or cats 5 dogs 10 we can write number of cats as 5 x and number of dogs as 10 x and we know that 10 x - 5 x = 40 ( therefore 5 x = 40 = > x = 8 ) then # of dogs = 10 x 8 = 80 answer is a" | a = 40 * 3
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.