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 ) 1 / 15 , b ) 1 / 16 , c ) 1 / 11 , d ) 1 / 9 , e ) 1 / 12 | d | subtract(add(divide(const_1, 15), divide(const_1, 15)), divide(const_1, 45)) | two pipes can fill a tank in 15 minutes and 15 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 15 + 1 / 15 - 1 / 45 = 1 / 9 so , the tank becomes full in 9 minutes . answer : d" | a = 1 / 15
b = 1 / 15
c = a + b
d = 1 / 45
e = c - d
|
a ) 8 , b ) can not be determined , c ) 4 , d ) 6 , e ) 9 | b | subtract(subtract(9, 1), const_4) | set r is composed of 9 numbers , labeled r 1 through r 9 . set s is also composed of 9 numbers , labeled s 1 through s 9 . set s is defined as follows : s 1 = 1 + r 1 ; s 2 = 2 + r 2 ; and so on , including s 9 = 9 + r 9 . how much larger is the sum of set s ' s mean and range than the sum of set r ' s mean and range ? | yes , the correct answer is b | a = 9 - 1
b = a - 4
|
a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 75 | e | subtract(subtract(multiply(60, add(const_2, const_3)), multiply(45, const_4)), 45) | the average of temperatures at noontime from monday to friday is 60 ; the lowest one is 45 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 60 * 5 = 300 lowest is 45 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 45 = 180 . on the fifth day it is 300 - 180 = 120 range is therefore 120 - 45 = 75 answer e" | a = 2 + 3
b = 60 * a
c = 45 * 4
d = b - c
e = d - 45
|
a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 2 , e ) 4 : 7 | e | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(5, const_100)), const_1000)), subtract(multiply(add(const_1, divide(12, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | at a certain organisation , the number of male members went up by 12 % in the year 2001 from year 2000 , and the number of females members went down by 5 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ? | "men increase by 12 % = = > 1.12 m = males in 2001 women decrease by 5 % = = > 0.95 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.12 m + 0.95 f = 1.012 * ( m + f ) 1.12 m + 0.95 f = 1.012 m + 1.012 f 1.12 m - 1.012 m = 1.012 f - 0.95 f 0.108 m = 0.062 f m / f = ( 0.062 ) / ( 0.108 ) = 62 / 108 = 4 / 7 answer = ( e )" | a = 1 / 2
b = 1 + a
c = b * 1000
d = 5 / 100
e = 1 - d
f = e * 1000
g = c - f
h = 12 / 100
i = 1 + h
j = i * 1000
k = 1 / 2
l = 1 + k
m = l * 1000
n = j - m
o = g / n
|
a ) 21 , b ) 26 , c ) 27 , d ) 22 , e ) 29 | b | divide(subtract(multiply(9, 18), multiply(4, 8)), 5) | if the average of 9 digits is 18 and the average of 4 of them is 8 , calculate the average of the remaining 5 numbers ? | explanation : total of the 9 digits - 9 * 18 = 162 total of the 4 digits - 4 * 8 = 32 total of the remaining 5 digits - 162 - 32 = 130 average of the remaining 3 numbers = 130 / 5 = 26 answer : b | a = 9 * 18
b = 4 * 8
c = a - b
d = c / 5
|
a ) 2 , b ) 8 , c ) 6 , d ) 9 , e ) 0 | b | add(reminder(multiply(reminder(45, const_4), 3), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 3 ) ^ ( 45 ) is : | "any power of anything ending in 5 always has a units digit of 5 . so the first term has a units digit of 5 . 2 nd part 3 ^ 45 has a units digit of 3 therefore , the second term has a units digit of 3 . of course 5 + 3 = 8 , so something with a units digit of 5 plus something with a units digit of 3 answer = b" | a = reminder * (
b = reminder + (
|
a ) 75 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | a | multiply(divide(divide(multiply(7, 20), subtract(11, 7)), 7), 15) | in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 20 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ? | this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 20 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . algebraically , since the number of dogs is a multiple of 15 and the number of cats is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 20 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 20 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 20 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 20 ) ( 1 ) 11 x = 7 x + 20 4 x = 20 x = 5 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 5 ) = 75 final answer : a | a = 7 * 20
b = 11 - 7
c = a / b
d = c / 7
e = d * 15
|
a ) 59 , b ) 57 , c ) 59.5 , d ) 58 , e ) 55.5 | d | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 60 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "d 75 * t = 1 km = > t = 1 / 60 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 60 + 2 / 3600 ) = 1 = > v = 58 km / h" | a = 1 / 60
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
|
a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | e | subtract(subtract(multiply(3, 16), add(subtract(17, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 17 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 17 a ā d = 3 a + d = 17 2 d = 14 d = 7 anser e" | a = 3 * 16
b = 17 - 16
c = b + 3
d = a - c
e = d - 16
|
a ) 4038 , b ) 8076 , c ) 2052 , d ) 4845.6 , e ) none of these | c | multiply(divide(divide(24624, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 24624 , how much amount per month would he have spent on clothes and transport ? | "āµ amount , he have spent in 1 month on clothes transport = amount spent on saving per month āµ amount , spent on clothes and transport = 24624 ā 12 = 2052 answer c" | a = 100 - 60
b = a * 50
c = b / 100
d = c / 100
e = 24624 / d
f = 3 * 4
g = e / f
h = 100 - 60
i = h * 50
j = i / 100
k = j / 100
l = g * k
|
a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 18 sec , e ) 12 sec | d | multiply(divide(210, multiply(63, const_1000)), const_3600) | a train 210 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 210 * 2 / 35 = 18 sec answer : d" | a = 63 * 1000
b = 210 / a
c = b * 3600
|
a ) 4 / 5 , b ) 8 / 15 , c ) 11 / 15 , d ) 9 / 20 , e ) 13 / 20 | e | divide(add(multiply(divide(2, 5), const_12), multiply(divide(const_12, 2), divide(1, 2))), const_12) | drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 2 / 5 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ? | "( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 2 / 5 ) y = ( 13 / 20 ) y the answer is e ." | a = 2 / 5
b = a * 12
c = 12 / 2
d = 1 / 2
e = c * d
f = b + e
g = f / 12
|
a ) 42 , b ) 50 , c ) 50 , d ) 54 , e ) 56 | b | divide(add(180, 120), add(divide(180, 60), divide(120, 40))) | joe drives 180 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ? | "t 1 = 180 / 60 = 3 hours t 2 = 120 / 40 = 3 hours t = t 1 + t 2 = 6 hours avg speed = total distance / t = 300 / 6 = 50 mph = b" | a = 180 + 120
b = 180 / 60
c = 120 / 40
d = b + c
e = a / d
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | b | multiply(40, const_1) | at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 140 male associates , how many female managers are there ? | "250 employees : 90 male , 160 female 40 managers , 210 associates 140 male associates implies 70 female associates which means the remaining 20 females must be managers b . 20" | a = 40 * 1
|
a ) 80 kmph , b ) 60 kmph , c ) 70 kmph , d ) 100 kmph , e ) none of these | d | divide(add(250, 350), add(2, 4)) | a train travels 250 km in 2 hours and 350 km in 4 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 = 250 + 350 = 600 km thus , total speed = 6 hrs or , average speed = 600 / 6 or , 100 kmph . answer : d" | a = 250 + 350
b = 2 + 4
c = a / b
|
['a ) 9 cm', 'b ) 8 cm', 'c ) 7 cm', 'd ) 6 cm', 'e ) 5 cm'] | c | divide(subtract(32, multiply(2, const_2)), const_4) | the difference of the areas of two squares drawn on two line segments in 32 sq . cm . find the length of the greater line segment if one is longer than the other by 2 cm | explanation : let the lengths of the line segments be x and x + 2 cm then , ( x + 2 ) 2 ā x 2 = 32 x 2 + 4 x + 4 ā x 2 = 324 x = 28 x = 7 cm option c | a = 2 * 2
b = 32 - a
c = b / 4
|
a ) 20 % loss , b ) 50 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 24)), divide(multiply(const_100, 16), 24))) | if the cost price of 24 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ? | "explanation : let cost price of 1 article be re . 1 . therefore , cost price of 24 articles = rs . 24 . selling price of 16 articles = rs . 24 therefore , selling price of 24 articles is : - = > 24 / 16 Ć£ ā 24 = > 36 . therefore , profit = selling price - cost price . = > 36 Ć¢ Ė ā 24 = 12 . hence , the percentage of profit = profit x 100 / c . p . = > 12 / 24 Ć£ ā 100 . = > 50 % profit . answer : b" | a = 100 * 16
b = a / 24
c = 100 - b
d = 100 * 16
e = d / 24
f = c / e
g = 100 * f
|
a ) 6 days , b ) 10 days , c ) 15 days , d ) 16 days , e ) 20 days | b | add(4, divide(multiply(12, multiply(20, subtract(const_1, divide(4, 20)))), add(20, 12))) | x and y can do a piece of work in 20 days and 12 days respectively . x started the work alone and then after 4 days y joined him till the completion of the work . how long did the work last ? | work done by x in 4 days = ( 1 / 20 ) * 4 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 . ( x + y ) ' s 1 day ' s work = 1 / 20 + 1 / 12 = 8 / 60 = 2 / 15 . now , 2 / 15 work is done by x and y in 1 day . so , 4 / 5 work will be done by x and y in ( 15 / 2 x 4 / 5 ) = 6 days . hence , total time taken = ( 6 + 4 ) days = 10 days . option b | a = 4 / 20
b = 1 - a
c = 20 * b
d = 12 * c
e = 20 + 12
f = d / e
g = 4 + f
|
a ) 144 ft , b ) 88 ft , c ) 22 ft , d ) 112 ft , e ) 146 t | e | add(multiply(divide(680, 10), const_2), 10) | a rectangular field is to be fenced on three sides leaving a side of 10 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 10 x breadth = 680 breadth = 68 feet area to be fenced = 2 b + l = 2 ( 68 ) + 20 = 146 feet answer : e ) 146 ft" | a = 680 / 10
b = a * 2
c = b + 10
|
a ) 987 ways , b ) 346 ways , c ) 600 ways , d ) 200 ways , e ) 134 ways | d | multiply(divide(factorial(6), multiply(factorial(3), factorial(3))), divide(factorial(5), multiply(const_2, factorial(3)))) | a study group consists of 4 english majors , 6 biological sciences majors , and 5 engineers . in how many ways can 3 english majors , 3 biological science majors and 3 engineers selected from the given group ? | the number of ways of selecting three english majors , two biological science majors and three engineers is : = ā“ c ā * ā¶ c ā * āµ c ā = ( 4 * 3 * 2 ) / ( 3 * 2 * 1 ) * ( 6 * 5 ) / ( 3 * 2 * 1 ) * ( 5 * 4 * 3 ) / ( 3 * 2 * 1 ) = 4 * 5 * 10 = 200 ways . answer : d | a = math.factorial(6)
b = math.factorial(3)
c = math.factorial(3)
d = b * c
e = a / d
f = math.factorial(5)
g = math.factorial(3)
h = 2 * g
i = f / h
j = e * i
|
a ) 6.5 , b ) 5.5 , c ) 5.2 , d ) 4.6 , e ) 9.7 | b | divide(add(divide(20, 5), divide(35, 5)), const_2) | a man swims downstream 35 km and upstream 20 km taking 5 hours each time , what is the speed of the man in still water ? | "35 - - - 5 ds = 7 ? - - - - 1 20 - - - - 5 us = 4 ? - - - - 1 m = ? m = ( 7 + 4 ) / 2 = 5.5 answer : b" | a = 20 / 5
b = 35 / 5
c = a + b
d = c / 2
|
a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(90, const_3), divide(90, 6)) | in the third grade of windblown school there are 90 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "total = 90 failed in math = 90 / 3 = 30 failed in literature = 90 / 6 = 15 the least failed in both can be 0 while max can be 15 answer a" | a = 90 / 3
b = 90 / 6
c = min(a)
|
a ) 16.16 % , b ) 15.25 % , c ) 14.14 % , d ) 13.33 % , e ) 11.71 % | e | subtract(const_100, multiply(multiply(add(const_1, divide(9, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | a volunteer organization is recruiting new members . in the fall they manage to increase their number by 9 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 9 % ) * ( 100 % - 19 % ) = 1.09 * . 81 = 0.8829 . 1 - 0.8829 = 11.71 % lost = - 11.71 % the answer is e the organization has lost 11.71 % of its total volunteers from fall to spring ." | a = 9 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 2 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 15 : 19 , e ) 64 : 76 | c | divide(subtract(64, 60), subtract(76, 64)) | a theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 76 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ? | "deviation from the mean for the daytime staff = 76 - 64 = 12 . deviation from the mean for the evening staff = 64 - 60 = 4 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 4 : 12 = 1 : 3 . the answer is c ." | a = 64 - 60
b = 76 - 64
c = a / b
|
a ) 0 , b ) 1 , c ) 27 , d ) 54 , e ) can not be determined | a | divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4) | for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.367 ] = 1 and [ 1.988 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | "it can not be determined . as we do not know the actual value of k . however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer k = [ lcm of 80 and 45 ] * ( any integer ) for value of k ( 60 ) * any integer unit value will be always zero . a" | a = 3 * 2
b = 3 * 2
c = a * b
d = 3 * 2
e = c * d
f = e / 4
|
a ) 12 , b ) 27 , c ) 24 , d ) 29 , e ) 11 | c | add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is ? | "each row contains 12 plants . there are 11 gapes between the two corner trees ( 11 x 2 ) metres and 1 metre on each side is left . therefore length = ( 22 + 2 ) m = 24 m . answer : c" | a = 12 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
|
a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 2 and 5 , e ) 2 and 9 | d | add(multiply(2, const_10), divide(add(30, 5), 7)) | 5 n + 2 > 12 and 7 n - 5 < 30 ; n must be between which numbers ? | "5 n > 10 - - > n > 2 7 n < 35 - - > n < 5 2 < n < 5 answer : d" | a = 2 * 10
b = 30 + 5
c = b / 7
d = a + c
|
a ) 28 , b ) 16.0 , c ) 16.4 , d ) 16.8 , e ) 16.1 | a | divide(add(110, 170), multiply(36, const_0_2778)) | how long does a train 110 m long traveling at 36 kmph takes to cross a bridge of 170 m in length ? | "d = 110 + 170 = 280 m s = 36 * 5 / 18 = 10 t = 280 * 1 / 10 = 28 sec answer : a" | a = 110 + 170
b = 36 * const_0_2778
c = a / b
|
a ) 50 , b ) 33 , c ) 16 , d ) 35 , e ) 10 | e | divide(5, 2) | how many positive integer solutions does the equation 2 x + 5 y = 100 have ? | "formula : ( constant ) / ( lcm of two nos ) = 100 / ( 2 * 5 ) = 10 answer : e" | a = 5 / 2
|
a ) 35.67 % , b ) 66.90 % , c ) 68.57 % , d ) 69.57 % , e ) none of these | b | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(34, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 34 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | "girls = 900 , boys = 1100 passed = ( 34 % of 1100 ) + ( 32 % of 900 ) = 374 + 288 = 662 failed = 2000 - 662 = 1338 failed % = [ ( 1338 / 2000 ) x 100 ] % = 66.9 % . answer : b" | a = 100 * 10
b = 2 * a
c = 34 / 100
d = 100 * 10
e = 2 * d
f = e - 900
g = c * f
h = b - g
i = 32 / 100
j = i * 900
k = h - j
l = 100 * 10
m = 2 * l
n = k / m
o = n * 100
|
a ) 88 feet , b ) 86 feet , c ) 84 feet , d ) 82 feet , e ) 83 feet | a | add(multiply(divide(680, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 680 = > breadth = 34 feet area to be fenced = 2 b + l = 2 * 34 + 20 = 88 feet answer : option a" | a = 680 / 20
b = a * 2
c = b + 20
|
a ) 50 , b ) 78 , c ) 30 , d ) 29 , e ) 27 | c | subtract(add(200, 330), 500) | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 330 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 330 - - - - - - - - - - - - - - a + b + 2 c = 530 a + b + c = 500 - - - - - - - - - - - - - - - - c = 30 answer : c" | a = 200 + 330
b = a - 500
|
a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | d | divide(4, subtract(96.16, floor(96.16))) | when positive integer x is divided by positive integer y , the remainder is 4 . if x / y = 96.16 , what is the value of y ? | "by the definition of a remainder , the remainder here is equal to 4 / y . the remainder in decimal form is given as . 16 therefore , 4 / y = . 16 solve for y and get 25 . d" | a = math.floor(96, 16)
b = 96 - 16
c = 4 / b
|
a ) 20 , 28 , b ) 28 , 36 , c ) 35 , 45 , d ) 35 , 55 , e ) of these | c | subtract(multiply(divide(add(79, multiply(7, const_3)), add(add(4, 7), 9)), 7), 7) | the present ages of 3 persons are in proportions 4 : 7 : 9 . 7 years ago , the sum of their ages was 79 . find their present ages . | let their present ages be 4 x , 7 x and 9 x years respectively . then , ( 4 x - 7 ) + ( 7 x - 7 ) + ( 9 x - 7 ) = 79 20 x = 100 = > x = 5 their present ages are 20 , 35 and 45 years respectively . answer : c | a = 7 * 3
b = 79 + a
c = 4 + 7
d = c + 9
e = b / d
f = e * 7
g = f - 7
|
a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43 | d | divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), const_1)) | in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability e this college graduate has a graduate degree ? | in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability e of cgd / ( cg ) - > 3 / 19 answer d | a = 1 / 8
b = 2 / 3
c = a / b
d = 1 / 8
e = 2 / 3
f = d / e
g = f + 1
h = c / g
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | multiply(divide(add(multiply(40, divide(15, const_100)), multiply(80, divide(30, const_100))), add(40, 80)), const_100) | a car dealership has 40 cars on the lot , 15 % of which are silver . if the dealership receives a new shipment of 80 cars , 30 % of which are not silver , what percentage of total number of cars are silver ? | "the number of silver cars is 0.15 * 40 + 0.3 * 80 = 30 the percentage of cars which are silver is 30 / 120 = 25 % the answer is b ." | a = 15 / 100
b = 40 * a
c = 30 / 100
d = 80 * c
e = b + d
f = 40 + 80
g = e / f
h = g * 100
|
a ) 20 sec , b ) 25 sec , c ) 24 sec , d ) 30 sec , e ) 22 sec | c | multiply(500, inverse(multiply(const_0_2778, add(45, 30)))) | two train each 500 m long , and running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | speed = ( 45 + 30 ) = 75 * 5 / 18 = = > 125 / 6 req time = 500 * 6 / 125 = 24 sec answer c | a = 45 + 30
b = const_0_2778 * a
c = 1/(b)
d = 500 * c
|
a ) 106 seconds , b ) 130 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | inverse(add(divide(const_1, 210), divide(const_1, 214))) | geetha ' s home has one tap , working alone at its maximum constant rate , can fill a certain sink in 210 seconds . a second tap , working alone at its maximum constant rate , can fill the same sink in 214 seconds . if both taps work together at their respective maximum constant rates , the time it will take to fill the sink is closest to | tap 1 : 210 secs tap 2 : 214 secs considering the average of these 2 taps : 212 secs . so 1 tap can fill the tank in 212 secs , so 2 taps can fill the tank in 212 / 2 = 106 secs . closest answer is a . | a = 1 / 210
b = 1 / 214
c = a + b
d = 1/(c)
|
a ) 8 liters , b ) 10 liters , c ) 12 liters , d ) 15 liters , e ) 18 liters | a | multiply(divide(add(multiply(5, multiply(divide(3, add(3, 2)), 5)), multiply(3, subtract(5, multiply(divide(2, add(3, 2)), 5)))), subtract(multiply(5, 3), multiply(2, 3))), 3) | a bucket contains 2 mixtures a and b in the ratio 3 : 2 . if 5 liters of the mixture is replaced by 5 liters of liquid b , then the ratio of the two liquids becomes 3 : 5 . how much of the liquid a was there in the bucket ? | "1 st ratio = 3 : 2 2 nd ratio = 3 : 5 difference of cross products ratio = 3 * 5 - 3 * 2 = 9 common factor of the 1 st ratio = ( quantity replaced / sum of terms in 1 st ratio ) + ( quantity replaced * term a in 2 nd ratio / difference ) = ( 5 / 5 ) + ( 5 * 3 / 9 ) = 1 + 5 / 3 = 2.67 quantity of a = 3 * 2.67 = 8 liters approximately answer is a" | a = 3 + 2
b = 3 / a
c = b * 5
d = 5 * c
e = 3 + 2
f = 2 / e
g = f * 5
h = 5 - g
i = 3 * h
j = d + i
k = 5 * 3
l = 2 * 3
m = k - l
n = j / m
o = n * 3
|
a ) 49 , b ) 52 , c ) 36 , d ) 64 , e ) 56 | b | multiply(25, divide(34, 43)) | 34 : 43 : : 25 : ? | "ans 52 reverse of 25 answer : b" | a = 34 / 43
b = 25 * a
|
a ) 35 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 18 | e | subtract(50, multiply(divide(50, const_100), 2)) | how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 2 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 2 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 2 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 2 x = 3 since you need a 15 l solution , you must evaporate 18 of the original 50 l solution to get a 2 % solution . answer is e ." | a = 50 / 100
b = a * 2
c = 50 - b
|
a ) 482 , b ) 725 , c ) 964 , d ) 837 , e ) 923 | c | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 40000), multiply(power(add(divide(20, const_100), const_1), const_2), 40000)) | how much more would rs . 40000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | "40000 ( 11 / 10 ) 4 - 40000 ( 6 / 5 ) 2 = 964 answer : c" | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 4
e = d * 40000
f = 20 / 100
g = f + 1
h = g ** 2
i = h * 40000
j = e - i
|
a ) 75 kgs , b ) 64 kgs , c ) 72 kgs , d ) 65 kgs , e ) 70 kgs | a | add(divide(multiply(20, subtract(const_100, 25)), const_100), 80) | fresh grapes contain 80 % water by weight and raisins obtained by drying fresh grapes contain 25 % water by weight . how many kgs of fresh grapes are needed to get 20 kgs of raisins ? | "the weight of non - water in 20 kg of dried grapes ( which is 100 - 25 = 75 % of whole weight ) will be the same as the weight of non - water in x kg of fresh grapes ( which is 100 - 80 = 20 % of whole weight ) , so 20 * 0.75 = x * 0.2 - - > x = 75 . answer : a ." | a = 100 - 25
b = 20 * a
c = b / 100
d = c + 80
|
a ) 6 , b ) 5 , c ) 5.5 , d ) 2 , e ) 8 | d | divide(multiply(multiply(divide(const_1, 15), 10), 6), const_2) | a can finish a work in 6 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 x 10 ) = 2 . 15 3 remaining work = ( 1 - 2 ) = 1 . 3 3 now , 1 work is done by a in 1 day . 6 therefore 1 work is done by a in ( 6 x 1 ) = 2 days . d" | a = 1 / 15
b = a * 10
c = b * 6
d = c / 2
|
a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | e | divide(divide(500, const_2), divide(25, const_100)) | one night 25 percent of the female officers on a police force were on duty . if 500 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 500 number of female officers on duty on that night = 500 / 2 = 250 ( 25 / 100 ) * f = 250 = > f = 1000 answer e" | a = 500 / 2
b = 25 / 100
c = a / b
|
a ) 13 sec . , b ) 15 sec . , c ) 12 sec . , d ) 17 sec . , e ) 16.7 sec . | e | divide(add(120, 120), add(speed(120, 10), speed(120, 50))) | two bullet trains of equal lengths take 10 seconds and 50 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "speed of the first bullet train = 120 / 10 m / sec = 12 m / sec . speed of the second bullet train = 120 / 50 m / sec = 2.4 m / sec . relative speed = ( 12 + 2.4 ) = 14.4 m / sec . required time = ( 120 + 120 ) / 14.4 sec = 16.7 sec . e" | a = 120 + 120
b = speed + (
c = a / b
|
a ) 62.15 % , b ) 64.26 % , c ) 66.37 % , d ) 68.48 % , e ) 70.59 % | e | divide(935, add(5, 12)) | if $ 935 are divided between worker a and worker b in the ratio 5 : 12 , what is the share that worker b will get ? | "worker b will get 12 / 17 = 70.59 % the answer is e ." | a = 5 + 12
b = 935 / a
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 65 % , e ) 72 % | c | multiply(divide(10, subtract(30, 10)), const_100) | a shop owner sells 30 mtr of cloth and gains sp of 10 metres . find the gain % ? | "here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 30 m cloth ) ā ( c . p . of 30 m cloth ) selling price of 20 m cloth = selling price of 30 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 20 m cloth = rs . 2000 and s . p . of 20 m cloth = rs . rs . 3000 profit % = 10 Ć 100 = 50 % 20 profit of 50 % was made by the merchant . c" | a = 30 - 10
b = 10 / a
c = b * 100
|
a ) 5 , b ) 8 , c ) 10 , d ) 20 , e ) 25 | c | multiply(sqrt(divide(50, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 50 , then what is the largest positive integer that must divide n ? | "50 = 2 * 5 ^ 2 if 50 divides n ^ 2 , then n must be divisible by 2 * 5 = 10 the answer is c ." | a = 50 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | a | add(divide(add(29, 7), const_2), 7) | the sum of present age of abe and the age before 7 years is 29 . find the present age of abe . what will be his age after 7 years ? | "present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 29 2 x - 7 = 29 2 x = 29 + 7 x = 36 / 2 x = 18 z = x + 7 = 18 + 7 = 25 answer : a" | a = 29 + 7
b = a / 2
c = b + 7
|
a ) 49 , b ) 75 , c ) 150 , d ) 225 , e ) 343 | b | add(add(add(add(add(multiply(multiply(5, 2), 2), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | if both 5 ^ 2 and 3 ^ 3 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 * 5 ^ 2 = 75 the answer is b ." | a = 5 * 2
b = a * 2
c = 5 * 2
d = c * 2
e = b + d
f = 5 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 76 days , b ) 48 days , c ) 98 days , d ) 32 days , e ) 22 days | d | inverse(subtract(inverse(16), inverse(32))) | a and b can finish a work in 16 days while a alone can do the same work in 32 days . in how many days b alone will complete the work ? | b = 1 / 16 ā 1 / 32 = 1 / 32 = > 32 days answer : d | a = 1/(16)
b = 1/(32)
c = a - b
d = 1/(c)
|
a ) 10 , b ) 13 , c ) 9 , d ) 7 , e ) 5 | a | add(subtract(add(28, 36), subtract(60, 6)), subtract(36, 28)) | of 60 applicants for a job , 28 had at least 4 years ' experience , 36 had degrees , and 6 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "60 - 6 = 54 54 - 28 - 36 = - 10 then 10 are in the intersection between 4 years experience and degree . answer : a" | a = 28 + 36
b = 60 - 6
c = a - b
d = 36 - 28
e = c + d
|
a ) 0.3396 , b ) 0.3496 , c ) 0.3596 , d ) 0.3696 , e ) 0.3796 | a | multiply(0.3332, divide(0.3332, 0.3364)) | if log 0.317 = 0.3332 and log 0.318 = 0.3364 , then find log 0.319 | "log . 318 - log . 317 = . 0032 log . 318 + log . oo 1 = log . 319 . 3364 + . 0032 = . 3396 answer : a" | a = 0 / 3332
b = 0 * 3332
|
a ) 246 , b ) 448 , c ) 393 , d ) 710 , e ) 223 | c | divide(multiply(const_1000, 1), multiply(add(const_3, divide(add(multiply(const_3, const_4), const_2), power(add(multiply(const_4, const_2), const_2), const_2))), 0.81)) | a bicycle wheel has a diameter of 0.81 m . how many complete revolutions does it make in 1 km ? | "1 revolution = 3.14 * diameter . number of revolutions in 1 km = 1000 m / ( 3.14 * 0.81 m ) = 393.2 . hence 393 complete revolutions . answer c" | a = 1000 * 1
b = 3 * 4
c = b + 2
d = 4 * 2
e = d + 2
f = e ** 2
g = c / f
h = 3 + g
i = h * 0
j = a / i
|
a ) 3 ā 20 , b ) 31 ā 60 , c ) 9 ā 40 , d ) 1 ā 3 , e ) 11 ā 20 | b | divide(subtract(divide(subtract(const_100, 30), const_2), const_4), multiply(30, const_2)) | if there are half as many women as men in a group and an equal number of men and women do not own cars - a group that is 30 % of the total . what fraction of the total is men who own cars ? | consider a group of 100 women and 200 men , a total of 300 people . 30 % of them , which is 90 , form a group of people who do n ' t own a car . half of them are men , and the other half are women , more precisely 45 . it means that there are 200 - 45 = 155 men who own a car , and this represents 155 / 300 = 31 / 60 of the total . answer b | a = 100 - 30
b = a / 2
c = b - 4
d = 30 * 2
e = c / d
|
a ) 150 , b ) 131 , c ) 115 , d ) 90 , e ) 45 | a | add(divide(multiply(12, subtract(12, const_1)), const_2), multiply(12, 7)) | 12 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 12 business exec and in each handshake 2 business execs are involved . hence 12 c 2 = 66 also , each of 12 exec will shake hand with every 7 other chairmen for total of 84 handshake . total = 66 + 84 = 150 ans : a" | a = 12 - 1
b = 12 * a
c = b / 2
d = 12 * 7
e = c + d
|
a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(4, multiply(2, power(const_3, 2))) | if the function q is defined by the formula q = 5 w / ( 4 d ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , d is doubled , and z is tripled ? | "we just need to find the factor thats all , w - > quadrupled - > 4 w d - > doubled - > 2 d z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and d * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b" | a = 3 ** 2
b = 2 * a
c = 4 / b
|
a ) 5 min , b ) 2 min , c ) 8 min , d ) 1 min , e ) 10 min | c | multiply(88, divide(1, 11)) | a pipe can fill a cistern in 88 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | full cistern filled in = 88 minutes 1 / 11 part filled in = 88 * 1 / 11 = 8 minutes answer is c | a = 1 / 11
b = 88 * a
|
a ) 1250 m , b ) 1110 m , c ) 950 m , d ) 1000 m , e ) 1300 m | d | multiply(divide(multiply(15, const_1000), const_60), 20) | a man walking at a rate of 15 km / hr crosses a bridge in 20 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 20 minutes = 15 / 18 * 20 * 60 = 1000 m answer is d" | a = 15 * 1000
b = a / const_60
c = b * 20
|
a ) 32 , b ) 60 , c ) 27 , d ) 29 , e ) 21 | b | subtract(multiply(3, 120), divide(3000, const_10)) | a total of 3000 chocolates were distributed among 120 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates . find the respective number of boys and girls ? | let the number of boys be x . number of girls is 120 - x . total number of chocolates received by boys and girls = 2 x + 3 ( 120 - x ) = 300 = > 360 - x = 300 = > x = 60 . so , the number of boys or girls is 60 . answer : b | a = 3 * 120
b = 3000 / 10
c = a - b
|
a ) 987 , b ) 977 , c ) 967 , d ) 897 , e ) 997 | a | add(add(subtract(divide(divide(365, const_10), const_10), const_1), subtract(subtract(divide(365, const_10), const_1), subtract(divide(divide(365, const_10), const_10), const_1))), multiply(subtract(subtract(365, const_1), subtract(divide(365, const_10), const_1)), const_3)) | how many figures are required to number the pages the pages of a book containing 365 pages ? | "1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 365 = 266 * 3 = 798 - - - - - - - - - - - 987 answer : a" | a = 365 / 10
b = a / 10
c = b - 1
d = 365 / 10
e = d - 1
f = 365 / 10
g = f / 10
h = g - 1
i = e - h
j = c + i
k = 365 - 1
l = 365 / 10
m = l - 1
n = k - m
o = n * 3
p = j + o
|
a ) 30 , b ) 40 , c ) 20 , d ) 10 , e ) 25 | c | divide(multiply(10, divide(2, 3)), subtract(divide(3, add(3, 2)), multiply(divide(2, add(3, 2)), divide(2, 3)))) | a mixture contains milk and water in the ratio 3 : 2 . on adding 10 liters of water , the ratio of milk to water becomes 2 : 3 . total quantity of milk & water before adding water to it ? | milk : water = 3 : 2 after adding 10 liters of water milk : water = 2 : 3 olny water patrs increase when mixture of water milk : wate = 3 : 2 = 2 * ( 3 : 2 ) = 6 : 4 after adding 10 liters of water milk : water = 2 : 3 = 3 * ( 2 : 3 ) = 6 : 9 option ' c ' | a = 2 / 3
b = 10 * a
c = 3 + 2
d = 3 / c
e = 3 + 2
f = 2 / e
g = 2 / 3
h = f * g
i = d - h
j = b / i
|
a ) 55 days , b ) 77 days , c ) 20 days , d ) 24 days , e ) 44 days | d | divide(multiply(30, 16), 20) | 16 men can complete a piece of work in 30 days . in how many days can 20 men complete that piece of work ? | "16 * 30 = 20 * x = > x = 24 days answer : d" | a = 30 * 16
b = a / 20
|
a ) 5.0 , b ) 5.5 , c ) 6.0 , d ) 6.5 , e ) 7.0 | d | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 160 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 1.5 hours to travel 240 km , so c leaves 6.5 hours after a . the answer is d ." | a = 40 - 30
|
a ) 50 , b ) 60 , c ) 11 , d ) 80 , e ) 90 | d | divide(384, add(add(divide(7, 5), divide(7, 5)), const_2)) | the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 384 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 384 , 2 ( 7 / 5 ) w + 2 w = 384 solve the above equation to find : w = 80 m and l = 112 m . correct answer d ) 80" | a = 7 / 5
b = 7 / 5
c = a + b
d = c + 2
e = 384 / d
|
a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 982.9 hours , e ) 915 hours | d | add(divide(7740, add(16, 2)), divide(7740, subtract(16, 2))) | speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7740 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7740 / 18 + 7740 / 14 = 430 + 552.9 = 982.9 hours answer : option d" | a = 16 + 2
b = 7740 / a
c = 16 - 2
d = 7740 / c
e = b + d
|
a ) 52 , b ) 53 , c ) 54 , d ) 55 , e ) 56 | d | divide(multiply(add(72, divide(67, 66)), 61), const_100) | 72 67 66 61 60 . . . ? | "sequence = 72 67 66 61 60 55 there is 2 series : 1 st series = 72 66 60 ( decreasing by 6 ) 2 nd series = 67 61 55 ( decreasing by 6 ) answer : d" | a = 67 / 66
b = 72 + a
c = b * 61
d = c / 100
|
a ) 190 metres , b ) 160 metres , c ) 280 metres , d ) 120 metres , e ) 250 metres | c | multiply(25.2, multiply(40, const_0_2778)) | a train is running at a speed of 40 km / hr and it crosses a post in 25.2 seconds . what is the length of the train ? | "speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 25.2 s distance covered , d = vt = ( 400 / 36 ) Ć£ ā 25.2 = 280 m distance covered is equal to the length of the train = 280 m correct answer is 280 metres c" | a = 40 * const_0_2778
b = 25 * 2
|
a ) 120 hours , b ) 240 hours , c ) 160 hours , d ) 480 hours , e ) none | a | add(divide(210, add(8, 6)), divide(210, subtract(8, 6))) | speed of a boat in standing water is 8 kmph and the speed of the stream is 6 kmph . a man rows to place at a distance of 210 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 2 kmph ; speed downstream = 14 kmph . ā“ total time taken = [ 210 / 2 + 210 / 2 ] hours = 120 hours . answer a" | a = 8 + 6
b = 210 / a
c = 8 - 6
d = 210 / c
e = b + d
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 100 | e | multiply(divide(300, multiply(add(const_2, const_1), const_2)), const_2) | a rectangular garden is to be twice as long as it is wide . if 300 yards of fencing , including the gate , will completely enclose the garden , what will be the length of the garden , in yards ? | "alternate approach backsolving ( using answer options to reach the correct answer ) can work wonders here if one is fast in calculations . given perimeter is 300 so , 2 ( l + b ) = 300 or , l + b = 150 now use the answer options ( given length ; breath will be half the length ) ( a ) 40 l = 40 ; b = 20 l + b = 60 ( b ) 50 l = 50 ; b = 25 l + b = 75 ( c ) 60 l = 60 ; b = 30 l + b = 90 ( d ) 70 l = 70 ; b = 35 l + b = 105 ( e ) 100 l = 100 ; b = 50 l + b = 150 thus you see no , need of any calculations , u can reach the correct option only by checking options ; correct answer will be ( e )" | a = 2 + 1
b = a * 2
c = 300 / b
d = c * 2
|
a ) 14 years , b ) 17 years , c ) 19 years , d ) 21 years , e ) none | a | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 19 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ā“ required average = 266 / 19 = 14 years . answer a" | a = 35 * 16
b = a + 21
c = b + 35
d = 35 * 16
e = c - d
|
['a ) 26 %', 'b ) 21 %', 'c ) 15 %', 'd ) 18 %', 'e ) 12 %'] | b | multiply(subtract(power(divide(add(const_100, 10), const_100), const_2), const_1), const_100) | if each side of a right triangle is increased by 10 % and the base ( b ) is half of the height , find the percentage change in its area ? | a = 1 / 2 x b x h or 1 / 2 x 2 a x a ( for simplification ) new h = 220 a / 100 = 22 a / 10 new b = ( 110 a x 100 ) = = 11 a / 10 new area = 1 / 2 ( 22 a x 11 a ) / ( 10 x 10 ) = ( 121 a Ā² / 100 ) increased area = = ( 121 a Ā² / 100 ) - a Ā² increase % = [ ( 21 a Ā² / 100 ) x ( 1 / a Ā² ) x 100 ] % = 21 % answer : b | a = 100 + 10
b = a / 100
c = b ** 2
d = c - 1
e = d * 100
|
a ) 2060 , b ) 2050 , c ) 2040 , d ) 2033 , e ) 2034 | a | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2032) | the calendar of the year 2032 can be used again in the year ? | explanation : given year 2032 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2032 + 28 = 2060 answer : a | a = 4 * 4
b = a - 2
c = b * 2
d = c + 2032
|
a ) s . 738 , b ) s . 638 , c ) s . 777 , d ) s . 730 , e ) s . 735 | c | subtract(825, divide(multiply(subtract(840, 825), 3), 4)) | a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 840 in 4 years . the sum is ? | "s . i . for 1 year = ( 841 - 825 ) = rs . 16 s . i . for 3 years = 16 * 3 = rs . 48 principal = ( 825 - 48 ) = rs . 777 . answer : c" | a = 840 - 825
b = a * 3
c = b / 4
d = 825 - c
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | add(multiply(3, const_4), 3) | if p is the product of the integers from 1 to 35 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of p ? | "35 ! has 3 , 6 , 9 , . . . . 30 , 33 as factors , which are 11 multiples of 3 . we need to add 4 more to these 11 because of 9 , 18 , and 27 . the greatest integer of k is 15 . the answer is c ." | a = 3 * 4
b = a + 3
|
a ) 5 : 8 , b ) 4 : 9 , c ) 5 : 9 , d ) 5 : 3 , e ) 9 : 10 | e | divide(8, 9) | the simple form of the ratio 4 / 5 : 8 / 9 is ? | 4 / 5 : 8 / 9 = 9 : 10 answer : e | a = 8 / 9
|
a ) s . 2.04 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.83 , e ) s . 2.45 | e | subtract(multiply(6000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(6000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | what is the difference between the c . i . on rs . 6000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 6000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 6000 * 26 / 25 * 51 / 50 = rs . 6364.8 c . i . when interest is compounded half - yearly = [ 6000 * ( 1 + 2 / 100 ) 2 ] = ( 6000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 6367.25 difference = ( 6367.25 - 6364.8 ) = rs . 2.45 . answer : e" | a = 2 / 100
b = 1 + a
c = 2 / 100
d = 1 + c
e = b * d
f = 2 / 100
g = 1 + f
h = e * g
i = 6000 * h
j = 2 / 100
k = 1 + j
l = 4 / 100
m = 1 + l
n = k * m
o = 6000 * n
p = i - o
|
a ) 140 , b ) 141 , c ) 142 , d ) 143 , e ) 144 | d | power(add(power(multiply(subtract(1, divide(1, 8)), 150), 2), power(subtract(power(add(power(150, 2), power(90, 2)), divide(1, 2)), multiply(subtract(1, divide(1, 8)), 150)), 2)), divide(1, 2)) | in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 / 2 , and in every year men are reduces 8 % . what is population of after 2 year . | "next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 8 % so total man = 150 * . 92 = 138 women will be = 175 - 138 = 37 so population after two years = [ 135 ^ 2 + 37 ^ 2 ] ^ . 5 = 142.87 = 143 so population after two year = 143 answer : d" | a = 1 / 8
b = 1 - a
c = b * 150
d = c ** 2
e = 150 ** 2
f = 90 ** 2
g = e + f
h = 1 / 2
i = g ** h
j = 1 / 8
k = 1 - j
l = k * 150
m = i - l
n = m ** 2
o = d + n
p = 1 / 2
q = o ** p
|
a ) 600 , b ) 700 , c ) 990 , d ) 900 , e ) none | c | multiply(add(100, 10), 9) | i chose a number and divide it by 9 . then i subtracted 100 from the result and got 10 . what was the number i chose ? | "solution : let x be the number i chose , then x / 9 Ć¢ Ė ā 100 = 10 x / 9 = 110 x = 990 answer c" | a = 100 + 10
b = a * 9
|
a ) 333 , b ) 250 , c ) 288 , d ) 276 , e ) 1999 | b | subtract(divide(divide(6625, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 6625 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 6625 [ 4 l - 40 ] = 6625 / 26.50 = 250 4 l = 290 l = 240 / 4 = 72.5 meters . answer : b" | a = 6625 / 26
b = a / 2
c = 2 * 20
d = b - c
|
a ) 450 , b ) 500 , c ) 650 , d ) 600 , e ) 650 | e | divide(subtract(add(add(500, divide(500, divide(const_100, 20))), 50), 500), 500) | after giving a discount of rs . 50 the shopkeeper gets a profit of 20 % , if the cost price is rs . 500 . find the marked price % ? | "cost price = 500 s . p = 500 * 120 / 100 = 600 disc = 50 so . . . mark price = 600 + 50 = 650 . . . . . . answer : e" | a = 100 / 20
b = 500 / a
c = 500 + b
d = c + 50
e = d - 500
f = e / 500
|
a ) 2000 , b ) 500 , c ) 900 , d ) 1600 , e ) none | a | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 2.5) | find the value of 2.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 2.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 2.5 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 2.5 x 4 x 4 x 50 = 2000 correct option : a" | a = 3 * 6
b = a * 2
c = 0 * 12
d = c * 0
e = b / d
f = e * 2
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | subtract(const_100, multiply(divide(3, const_4), const_100)) | on an order of 3 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | clearly , the retailer gets 1 dozen out of 4 dozens free . equivalent discount = 1 / 4 * 100 = 25 % . answer b ) 25 % | a = 3 / 4
b = a * 100
c = 100 - b
|
a ) $ 9 , b ) $ 12 , c ) $ 13.20 , d ) $ 15 , e ) $ 16.80 | a | divide(subtract(multiply(61, 3), multiply(69, 2)), subtract(multiply(3, 3), multiply(2, 2))) | if bill can buy 3 pairs of jeans and 2 shirts for $ 69 or 2 pairs of jeans and 3 shirts for $ 61 , how much does one shirt cost ? | 3 j + 2 s = 69 2 j + 3 s = 61 - - - - - - - - - - - - - - - - 5 j + 5 s = 130 - - - - ( divide by 5 ) - - - > j + s = 26 3 j + 2 s = j + 2 ( j + s ) = j + 52 = 69 - - - > j = 17 3 * 17 + 2 s = 69 51 + 2 s = 69 2 s = 18 s = 9 answer : a | a = 61 * 3
b = 69 * 2
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
|
a ) 130 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(50.7, 30)), const_100) | 30 square stone slabs of equal size were needed to cover a floor area of 50.7 sq . m . find the length of each stone slab ? | "area of each slab = 50.7 / 30 m 2 = 1.69 m 2 length of each slab ā 1.69 = 1.3 m = 130 cm" | a = 50 / 7
b = math.sqrt(a)
c = b * 100
|
a ) rs . 849.40 , b ) rs . 877.40 , c ) rs . 901.60 , d ) rs . 815.20 , e ) none of these | b | add(multiply(divide(multiply(24, divide(multiply(6, 20.5), 2)), const_10), 4), add(multiply(divide(multiply(6, 20.5), 2), const_3), multiply(5, 20.5))) | the cost of 10 kg of apples is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is rs . 20.50 . find the total cost of 4 kg of apples , 3 kg of rice and 5 kg of flour ? | let the costs of each kg of apples and each kg of rice be rs . a and rs . r respectively . 10 a = 24 r and 6 * 20.50 = 2 r a = 12 / 5 r and r = 61.5 a = 147.6 required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5 = 590.4 + 184.5 + 102.5 = rs . 877.40 answer : b | a = 6 * 20
b = a / 2
c = 24 * b
d = c / 10
e = d * 4
f = 6 * 20
g = f / 2
h = g * 3
i = 5 * 20
j = h + i
k = e + j
|
a ) 12.5 % , b ) 13 % , c ) 13.15 % , d ) 14 % , e ) 15 % | a | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 70)), const_100), multiply(const_100, const_100)) | the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 70 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.7 x 1.25 = 0.875 decrease in price = ( 0.125 / 1 ) Ć 100 = 12.5 % a )" | a = 100 * 100
b = 100 + 25
c = b * 70
d = a - c
e = d * 100
f = 100 * 100
g = e / f
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 have a remainder of 5 when divided by 13 ? | "we have to include 5 also . as 13 * 0 + 5 = 5 if somebody says to divide 5 by 13 , we will be telling we have 0 quotient and remainder as 5 . answer is a" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 53 , b ) 66 , c ) 60 , d ) 85 , e ) 80 | e | multiply(multiply(80, divide(1, add(1, 2))), add(1, 2)) | a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 80 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ? | "yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 80 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 40 cents . fraction of spirit is 1 / 2 . 40 = k * ( 1 / 2 ) k = 80 cost per liter = 80 * ( 1 / 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 80 * ( 1 / 3 ) * 3 = 80 cents e . 80 cents" | a = 1 + 2
b = 1 / a
c = 80 * b
d = 1 + 2
e = c * d
|
a ) 30 minutes , b ) 8 minutes , c ) 15 minutes , d ) 10 minutes , e ) 12 minutes | b | subtract(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 45)) | two pipes can fill a tank in 10 minutes and 15 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 10 + 1 / 15 - 1 / 45 = ( 9 + 6 - 2 ) / 90 = 13 / 90 so , the tank becomes full in 8 minutes . answer : b" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / 45
e = c - d
|
a ) 296 minutes . , b ) 297 minutes . , c ) 295 minutes . , d ) 294 minutes . , e ) none | c | multiply(60, 5) | dean winchester has got a long wooden stock of size 60 feet , he need to cut small stock of size 1 feet long using his axe . he takes 5 minutes to cut one small stock ( 1 feet ) , how long will he take to make 60 such small stocks ? | solution : 295 minutes . when he cut 59 stocks , the 60 th stock will remain . 59 * 5 = 295 minutes . answer c | a = 60 * 5
|
a ) x = 2 , b ) x = 5 , c ) x = 8 , d ) 10 , e ) 11 | a | add(divide(multiply(divide(power(subtract(floor(add(sqrt(1000), const_1)), const_12), const_2), power(const_2, const_3)), const_2), const_100), const_1) | the difference between the squares of two consecutive odd integers is a square < 1000 whose root equals the combined sum of the digits of the two integers . what is the sum of the digits x of the larger integer ? | really hard . but let ' s start with few concepts : 1 . square of an odd integer will be odd . 2 . difference of odd and odd will be even . 3 . any odd integer can be expressed as difference of two squares . 4 . an even integer can be expressed as difference of two squares only if that even integer is a multiple of 4 . assume two odd integers to be ( 2 x - 1 ) and ( 2 x + 1 ) . difference of their squares = ( 2 x + 1 ) ^ 2 - ( 2 x - 1 ) ^ 2 = 4 x ^ 2 + 1 + 4 x - ( 4 x ^ 2 + 1 - 4 x ) = 4 x ^ 2 + 1 + 4 x - 4 x ^ 2 - 1 + 4 x = 8 x now root of 8 x needs to be an integer such that it is equal to the sum of the digits of the two odd integers . 8 = 2 ^ 3 , so x needs to be such that itcompletesa perfect square . if we find x , we can find the value of 2 x + 1 ( larger integer ) and then sum of its digits . let ' s check the options , starting with c . for sum to be 8 , few possibilities are : 17 , 35 , 53 , 71 if we take 17 , the pair is 15 and 17 , meaning x = 8 . 8 x = 64 root of 64 = 4 but 4 is not equal to 1 + 5 + 1 + 7 . reject . if we take 35 , the pair is 33 and 35 , meaning x = 17 . 8 x = 8 * 17 reject since perfect square is not possible . if we take 53 , the pair is 51 and 53 , meaning x = 26 . 8 x = 8 * 26 reject since perfect square is not possible . if we take 71 , the pair is 69 and 71 , meaning x = 35 . 8 x = 8 * 35 reject since perfect square is not possible . i tried each option and the possibilities and then got lost . then , it occurred to me that i was only checking 2 - digit integers . what about 3 - digits ? starting with option a , only 2 - digit odd integer whose sum is 2 is 101 . if we take 101 , the pair is 99 and 101 , meaning x = 50 . 8 x = 8 * 50 = 400 root of 400 = 20 which is indeed equal to 9 + 9 + 1 + 1 . answer ( a ) . | a = math.sqrt(1000)
b = a + 1
c = math.floor(b)
d = c - 12
e = d ** 2
f = 2 ** 3
g = e / f
h = g * 2
i = h / 100
j = i + 1
|
a ) 2.5 % , b ) 50 % , c ) 25 % , d ) 35 % , e ) 250 % | b | multiply(divide(divide(25, const_100), divide(50, const_100)), const_100) | if c is 25 % of a and 50 % of b , what percent of a is b ? | "answer = b 25 a / 100 = 50 b / 100 b = 25 a / 50 = 50 a / 100 = 50 %" | a = 25 / 100
b = 50 / 100
c = a / b
d = c * 100
|
a ) 55 , b ) 60 , c ) 73 , d ) 81 , e ) 87 | e | subtract(100, divide(subtract(100, 61), const_3)) | a teacher grades students ā tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 61 , how many questions did student a answer correctly ? | "let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 61 ( subtracting twice of incorrect from correct ) 3 x = 261 x = 87 answer : e" | a = 100 - 61
b = a / 3
c = 100 - b
|
a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | b | divide(power(divide(600, const_2), const_2), subtract(615, 600)) | if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 600 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 615 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 30,000 . now , put the value of x = 6000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 6000 + 5 % of 6000 = 6300 . 2 nd year , amount = 6300 + 5 % of 6300 = 6615 we can see after 2 years interest = 6615 - 6000 = 615 . hence , it satisfies the question . hence b is the correct answer" | a = 600 / 2
b = a ** 2
c = 615 - 600
d = b / c
|
a ) $ 14 , b ) $ 2 , c ) $ 10 , d ) $ 6 , e ) $ 8 | a | divide(divide(multiply(210, 20), const_100), const_3) | a store purchased a pair of pants for $ 210 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a store owner decided to decrease the selling price by 20 percent . what was the store ' s gross profit on this sale ? | sale price ( sp ) = 210 + markup ( mp ) - - > mp = sp - 210 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 210 3 sp / 4 = 210 sp = 280 now a discount of 20 % is given so new sp is . 8 * 280 = 224 profit = 224 - 210 = 14 answer is a | a = 210 * 20
b = a / 100
c = b / 3
|
a ) rs . 15000 , b ) rs . 8250 , c ) rs . 15600 , d ) rs . 16500 , e ) none of these | b | multiply(400, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 400 per sq . metre . | "solution area of the floor = ( 5.5 Ć 3.75 ) m 2 = 20.625 m 2 ā“ cost of paving = rs . ( 400 Ć 20.625 ) = 8250 . answer b" | a = 5 * 5
b = 400 * a
|
a ) 2.4658 , b ) 6.398 , c ) 1.488 , d ) 5.687 , e ) 9.652 | c | divide(125, 14) | find 125 / 14 * 6 | "answer = 125 / 14 * 6 = 125 / 84 = 1.488 option c is correct" | a = 125 / 14
|
a ) 50 , b ) 72 , c ) 25 , d ) 92 , e ) 33 | e | add(power(5, 2), multiply(power(2, 2), 2)) | if a 2 - b 2 = 5 and a * b = 2 , find a 4 + b 4 . | "a 2 - b 2 = 5 : given a 4 + b 4 - 2 a 2 b 2 = 52 : square both sides and expand . a * b = 2 : given a 2 b 2 = 22 : square both sides . a 4 + b 4 - 2 ( 4 ) = 25 : substitute a 4 + b 4 = 33 correct answer e" | a = 5 ** 2
b = 2 ** 2
c = b * 2
d = a + c
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.