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['a ) 22', 'b ) 16', 'c ) 11', 'd ) 12', 'e ) 10'] | b | divide(divide(64, const_2), const_2) | a cube is painted such that one pair of surfaces is painted brown and the other pair of surfaces is painted orange . the cube is cut in to 64 small cubes of equal size . find how many cubes have both the color brown and orange ? | there are 6 surfaces to the cube . in which opposite surfaces are painted brown and other 2 opposite surfaces are painted orange . then 4 edges of the cube have both the colors . since the big cube cut in to 64 smaller cubes , the 4 edges of the cube have both the colors . since the big cube cut into 64 smaller cubes , the 4 edges of the big cube has 4 × 4 = 16 smaller cubes have both brown and orange colors . answer = 16 answer : b | a = 64 / 2
b = a / 2
|
a ) 2 / 15 , b ) 2 / 21 , c ) 5 / 26 , d ) 3 / 29 , e ) 4 / 27 | a | divide(choose(4, 2), choose(add(add(4, 4), 2), 2)) | a bag contains 4 red , 4 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 4 c 210 c 2 = 4 c 210 c 2 = 6 / 45 = 2 / 15 a" | a = math.comb(4, 2)
b = 4 + 4
c = b + 2
d = math.comb(c, 2)
e = a / d
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 3 / 2 , e ) 4 | d | divide(subtract(add(5, 3), 5), const_2) | in the xy - coordinate system , if ( m , n ) and ( m + 3 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 3 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 2 - - - > 5 = 2 k + 2 - - - > k = 3 / 2 . the answer is , therefore , ( d ) . | a = 5 + 3
b = a - 5
c = b / 2
|
a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | b | divide(divide(choose(45, const_1), 45), power(const_3, const_2)) | each factor of 210 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 45 ? | "210 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 5 / 42 . answer : b ." | a = math.comb(45, 1)
b = a / 45
c = 3 ** 2
d = b / c
|
a ) 10 metres , b ) 20 metres , c ) 14 metres , d ) 12 metres , e ) 13 metres | d | divide(300, subtract(26, const_1)) | in a garden , 26 trees are planted at equal distances along a yard 300 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "between 26 trees , there are 25 gaps length of each gap = 300 / 25 = 12 i . e . , distance between two consecutive trees = 12 answer : d" | a = 26 - 1
b = 300 / a
|
a ) 384 , b ) 350 , c ) 400 , d ) 200 , e ) 250 | a | multiply(power(const_2, 4), factorial(const_4)) | the product of 4 consecutive even numbers is always divisible by : | the product of 4 consecutive numbers is always divisible by 4 ! . since , we have 4 even numbers , we have an additional 2 available with each number . now , using both the facts , we can say that the product of 4 consecutive even numbers is always divisible by , 2 ^ 4 * 4 ! 16 * 24 = 384 answer a | a = 2 ** 4
b = math.factorial(4)
c = a * b
|
a ) 300 , b ) 800 , c ) 1500 , d ) 1200 , e ) 1900 | c | subtract(subtract(2500, divide(2500, 5)), divide(subtract(2500, divide(2500, 5)), 4)) | workers at a campaign office have 2500 fliers to send out . if they send out 1 / 5 of them in the morning and 1 / 4 of the remaining ones out during the afternoon , how many are left for the next day ? | ( 1 / 5 ) * 2500 = 500 remaining = 2500 - 500 = 2000 ( 1 / 4 ) of remaining = ( 1 / 4 ) * 2000 = 500 remaining now = 2000 - 500 = 1500 answer : option c | a = 2500 / 5
b = 2500 - a
c = 2500 / 5
d = 2500 - c
e = d / 4
f = b - e
|
a ) 16 , b ) 56 , c ) 76 , d ) 87 , e ) 24 | a | divide(multiply(48, 8), 24) | two numbers n and 24 have lcm = 48 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 24 * n = 48 * 8 n = 48 * 8 / 24 = 16 correct answer a" | a = 48 * 8
b = a / 24
|
a ) 485 , b ) 72 , c ) 547 , d ) 104 , e ) 624 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 8), add(const_2, const_4)) | what is the sum of all the multiples of 8 between 30 and 100 ? | "you first have to know all the multiples of 8 between 30 and 100 . they are 8 , 16 , 24 , 32,40 , 48,56 , 64,72 , 80,88 and 96 . if you add all these numbers together , you get 624 . final answer : e" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 8
t = 2 + 4
u = s + t
|
a ) 1250 , b ) 6250 , c ) 600 , d ) 7500 , e ) 375 | b | multiply(5, 1250) | a trailer carries 3 , 4 and 5 crates on a trip . each crate weighs no less than 1250 kg . what is the maximum weight of the crates on a single trip ? | "max no . of crates = 5 . max weight = 1250 kg max . weight carried = 5 * 1250 = 6250 kg = b" | a = 5 * 1250
|
a ) $ 48.12 , b ) $ 52.64 , c ) $ 64.10 , d ) $ 72.85 , e ) $ 60.10 | c | subtract(subtract(multiply(1000, power(add(divide(10, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(10, const_100)), 4)) | what will be the difference between simple and compound interest @ 10 % per annum on a sum of $ 1000 after 4 years ? | "s . i . = 1000 * 10 * 4 / 100 = $ 400 c . i . = 1000 * ( 1 + 10 / 100 ) ^ 4 - 1000 = $ 464.10 difference = 464.10 - 400 = $ 64.10 answer is c" | a = 10 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 10 / 100
g = 1000 * f
h = g * 4
i = e - h
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | multiply(sqrt(divide(150, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 150 , then what is the largest positive integer that must divide n ? | "150 = 2 * 3 * 5 ^ 2 if 150 divides n ^ 2 , then n must be divisible by 2 * 3 * 5 = 30 the answer is c ." | a = 150 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 27 , b ) 64 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(196, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(196, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 196 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 49 x = 196 x = 4 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 4 = 64 answer b" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 196 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 196 / n
p = i * o
q = h - p
|
a ) $ 30.60 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60 | b | add(50.50, divide(multiply(50.50, 20), const_100)) | if tim had lunch at $ 50.50 and he gave 20 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.50 = ( 20 / 100 ) * 50.50 = 101 / 100 = $ 10.10 total spent 50.50 + 10.10 = $ 60.60 correct answer b" | a = 50 * 50
b = a / 100
c = 50 + 50
|
a ) 44 . , b ) 36 . , c ) 42 . , d ) 46 . , e ) 50 . | a | add(add(multiply(4, const_2), multiply(4, 4)), add(add(add(const_3, const_4), add(const_4, const_1)), 6)) | 20 . a certain church bell rings the bell twice at half past the hour and 4 times at the hour plus an additional number of rings equal to what ever time it is . how many rings will the clock make from 6 : 20 in the morning to 09 : 50 in the morning ? | unless i understood the problem correctly , i get 58 @ 6 : 30 - 2 @ 7 - 4 + 7 = 11 @ 7 : 30 - 2 @ 8 - 12 @ 8 : 30 - 2 @ 9 - 13 @ 9 : 30 - 2 totals to a = 44 | a = 4 * 2
b = 4 * 4
c = a + b
d = 3 + 4
e = 4 + 1
f = d + e
g = f + 6
h = c + g
|
a ) 75 , b ) 100 , c ) 60 , d ) 70 , e ) 80 | b | divide(multiply(const_100, divide(7, const_2)), 7) | if a book is sold at 7 % profit instead of 7 % loss , it would have brought rs 14 more . find out the cost price of the book | "let c . p . of the book be rs . ’ x ’ given , 1.07 x - 0.93 x = 14 = > 0.14 x = 14 = 14 / 0.14 = rs 100 answer : b" | a = 7 / 2
b = 100 * a
c = b / 7
|
a ) 18 % , b ) 21 % , c ) 25.9 % , d ) 19 % , e ) none of these | c | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 460), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 460 ? | cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 460 , discount = 621 - 460 = 161 discount % = 161 / 621 * 100 = 25.9 % answer : c | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 460
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
|
a ) 24 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | d | divide(multiply(subtract(add(multiply(divide(multiply(280, 50), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | a shopkeeper has 280 kg of apples . he sells 50 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total . | "if the total quantity was 100 then 50 x 20 % + 60 x 30 % = 28 this profit will remain same for any total quantity unless the % of products remains the same . hence ' d ' is the answer" | a = 280 * 50
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = 280 * 60
g = f / 100
h = 100 + 30
i = h / 100
j = g * i
k = e + j
l = k - 280
m = l * 100
n = m / 280
|
a ) y = 350 , b ) y = 353 , c ) y = 354 , d ) y = 356 , e ) y = 357 | b | add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits y are in s ( 99 ) ? | "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 . b" | a = const_60 + 4
b = a + 4
c = const_60 + 4
d = c + 4
e = d * 4
f = b + e
g = 3 * 1
h = 3 * 2
i = h * 2
j = g + i
k = f + j
|
a ) 15 , b ) 7.2 , c ) 7.7 , d ) 7.4 , e ) 8 | b | inverse(add(inverse(12), inverse(18))) | a can do a piece of work in 12 days and b can do a piece of work in 18 days how many days will the work be completed if both of them work together ? | "a ' s 1 day work = 1 / 12 , b ' s 1 day work = 1 / 18 ; ( a + b ) 1 day work = 1 / 12 + 1 / 18 = 5 / 36 ; they can finish the work in = 36 / 5 = 7.2 answer : b" | a = 1/(12)
b = 1/(18)
c = a + b
d = 1/(c)
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | subtract(add(2, multiply(16, 7)), multiply(16, 7)) | how many two - digit numbers yield a remainder of 2 when divided by both 7 and 16 ? | "easier to start with numbers that are of the form 16 p + 2 - - - > 18,34 , 50,66 , 82,98 . out of these no number is also of the form 7 q + 2 . thus 0 is the answer . a is the correct answer ." | a = 16 * 7
b = 2 + a
c = 16 * 7
d = b - c
|
a ) 9 miles , b ) 10 miles , c ) 11 miles , d ) 12 miles , e ) 13 miles | a | divide(add(34, 20), 6) | in track last week , the boys ran 34 laps . the girls ran 20 more laps . each lap is a 1 / 6 th of a mile . how many miles did the girls run ? | the girls ran 34 + 20 = 54 laps . 54 x 1 / 6 = 54 / 6 , which reduces to 9 . the girls ran 9 miles correct answer a | a = 34 + 20
b = a / 6
|
a ) 14 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | add(subtract(40, add(30, 1)), 1) | the average weight of a group of boys is 30 kg . after a boy of weight 40 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 40 kg joins the group , total weight of boys = 30 x + 40 so 30 x + 40 = 31 ( x + 1 ) = > x = 9 . answer : e" | a = 30 + 1
b = 40 - a
c = b + 1
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(divide(multiply(multiply(12, 8), 10), 20), 8) | in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 20 hours a day , the number of men required is : | explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 20 : 8 : : 12 : x = > 8 x 20 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 20 ) = > x = 6 answer : c | a = 12 * 8
b = a * 10
c = b / 20
d = c / 8
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | multiply(divide(50, const_100), const_2) | find the minimum value of n such that 50 ! is perfectly divisible by 2520 ^ n . | 50 ! / 2520 ^ n 2520 - > 2 ^ 3 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1 here 7 is the largest prime factor . . . so in order to find the minimum value of ` ` n ' ' , it is enough to find the minimum power of ` ` 7 ' ' . . . nd for maximum value of ` ` n ' ' , find max power of 7 . . . for max . value of n , find 50 / 7 ^ 1 + 50 / 7 ^ 2 = 7 + 1 = 8 [ quotient only ] max . value of n which is perfectly divisible by 2520 ^ n is ( 8 ) min . value is 1 max value : 8 min value : 1 answer : b | a = 50 / 100
b = a * 2
|
a ) 1 / 5 , b ) 1 / 6 , c ) 1 / 7 , d ) 3 / 10 , e ) none of these | d | add(divide(const_1, 10), multiply(divide(const_1, 10), const_2)) | a can finish a work in 10 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ? | "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 10 b ' s 1 day work = 1 / 5 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 / 10 + 1 / 5 ) = 3 / 10 so in one day 3 / 10 work will be done answer : d" | a = 1 / 10
b = 1 / 10
c = b * 2
d = a + c
|
a ) a ) 99 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 88 | a | subtract(divide(multiply(divide(multiply(81, 8), 30), 50), 6), 81) | 81 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 81 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 180 180 – 81 = 99 answer : a" | a = 81 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 81
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | add(add(3, multiply(const_3, const_2)), multiply(const_3, const_2)) | fog + fog + fog + fog = 1464 . if f , o and g are digits in a 3 - digit number in the preceding equation the f + o + g = ? | given , fog + fog + fog + fog = 1464 4 ( fog ) = 1464 = > fog = 366 face value of f is : 3 face value of o is : 6 face value of g is : 6 so f + o + g = 3 + 6 + 6 = 15 answer : e | a = 3 * 2
b = 3 + a
c = 3 * 2
d = b + c
|
a ) 1 / 4 , b ) 1 / 3 , c ) 3 / 4 , d ) 1 , e ) 5 / 4 | b | subtract(divide(subtract(51, const_1), add(49, const_1)), divide(add(39, const_1), subtract(61, const_1))) | a is an integer greater than 39 but less than 51 , b is an integer greater than 49 but less than 61 , what is the range of a / b ? | "min value of a / b will be when b is highest and a is lowest - - - > a = 40 and b = 60 so , a / b = 2 / 3 max value of a / b will be when b is lowest and a is highest - - - > a = 50 and b = 50 so , a / b = 1 range is 1 - ( 2 / 3 ) = 1 / 3 . answer should be b ." | a = 51 - 1
b = 49 + 1
c = a / b
d = 39 + 1
e = 61 - 1
f = d / e
g = c - f
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a ) 14 , b ) 42 , c ) 28 , d ) 12 , e ) it can not be determined from the information given . | b | subtract(multiply(6, const_10), multiply(const_4, 6)) | two sets of 6 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? | "a = ( 1,2 , 3,4 , 5,6 ) , sum of this = 21 b = ( 6 , 7,8 , 9,10 , 11,12 ) , sum of this = 63 , the differenct between 63 - 21 = 42 hence , 42 is the answer i . e . b" | a = 6 * 10
b = 4 * 6
c = a - b
|
a ) 99 , b ) 18 , c ) 165 , d ) 10 , e ) 15 | c | divide(multiply(50, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 50 m long and 16 1 / 2 m board ? | 50 * 33 / 2 = 5 / 2 * 2 * x = > x = 165 answer : c | a = 1 / 2
b = 16 + a
c = 50 * b
d = 1 / 2
e = 2 + d
f = e * 2
g = c / f
|
a ) 784 , b ) 672 , c ) 492 , d ) 372 , e ) 300 | b | add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(260, const_100), const_1), const_3)) | how many digits are required to number a book containing 260 pages ? | "9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 260 - ( 90 + 9 ) = 161 pages will require 161 * 3 = 483 digits . the total number of digits is 9 + 180 + 483 = 672 . the answer is b ." | a = 10 - 1
b = 10 - 1
c = b * 10
d = c * 2
e = a + d
f = 260 - 100
g = f + 1
h = g * 3
i = e + h
|
a ) 3375 , b ) 7292 , c ) 8291 , d ) 3929 , e ) 2727 | a | power(15, 3) | log 3 n + log 15 n what is 3 digit number n that will be whole number | "no of values n can take is 1 15 ^ 3 = 3375 answer : a" | a = 15 ** 3
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a ) 48 , b ) 42 , c ) 52 , d ) 32 , e ) 33 | b | subtract(subtract(divide(divide(500, 3), 3), const_10), const_3) | the sum of squares of 3 numbers is 764 and sum of their products taken two at a time is 500 . what is their sum ? | description : let the numbers be x , y , z . = > x 2 + y 2 + z 2 = 764 = > 2 ( xy + yz + xz ) = 2 * 500 = 1000 = > ( x + y + z ) 2 = ( x 2 + y 2 + z 2 ) + 2 ( xy + yz + xz ) = 1764 = > ( x + y + z ) = sqrt ( 1764 ) = 42 answer b | a = 500 / 3
b = a / 3
c = b - 10
d = c - 3
|
a ) 8 % , b ) - 37.5 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | b | multiply(divide(subtract(multiply(divide(2, 10), 120), 18), multiply(divide(2, 10), 120)), const_100) | a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 10 pounds of the body weight , by what percent was the prescribed dosage lesser than the typical dosage ? | typical dosage per 10 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 10 ) = 2 * 12 = 24 c . c dosage prescribed by doctor for 120 pound patient = 18 c . c % prescribed dosage greater than the typical dosage = ( 18 - 24 / 16 ) * 100 % = ( - 6 / 16 ) * 100 % = - 37.5 % answer b | a = 2 / 10
b = a * 120
c = b - 18
d = 2 / 10
e = d * 120
f = c / e
g = f * 100
|
a ) 25 , b ) 30 , c ) 20 , d ) 45 , e ) 90 | c | gcd(1340, 1280) | the maximum number of students among them 1340 pens and 1280 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 1340 number of pencils = 1280 required number of students = h . c . f . of 1340 and 1280 = 20 answer is c" | a = math.gcd(1340, 1280)
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(multiply(10, 2), 18) | if the remainder is 10 when the integer n is divided by 18 , what is the remainder when 2 n is divided by 9 ? | "n = 18 k + 10 2 n = 2 ( 18 k + 10 ) = 4 k * 9 + 20 = 4 k * 9 + 2 * 9 + 2 = 9 j + 2 the answer is c ." | a = 10 * 2
b = a - 18
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a ) 11,4 , b ) 12,3 , c ) 9,6 , d ) 3,2 , e ) 2,1 | b | divide(subtract(20, power(15, const_2)), const_2) | the sum of two numbers is 15 and their geometric mean is 20 % lower than arithmetic mean . find the numbers . | "let a and b are two numbers so a + b = 15 . . . . ( 1 ) now gm = am * 80 / 100 or , sqrt ( ab ) = [ ( a + b ) / 2 ] * 80 / 100 squaring both sides . . . and after that putting the value of ( a + b ) . . . we get . . . . . ab = 36 now go through from option then we get 12 * 3 = 36 answer : b" | a = 15 ** 2
b = 20 - a
c = b / 2
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a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.16 , e ) 2.35 | d | divide(divide(multiply(multiply(34.5, 0.473), 1.567), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | the value of ( 34.5 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.5 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 25.5710895 / 11.845 = 2.16 answer : d" | a = 34 * 5
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 134 | e | add(add(const_1, 67), 67) | in the land of oz only one or two - letter words are used . the local language has 67 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 67 letters is 67 ; the number of 2 letter words ( xx ) that can be made from 67 letters is 67 * 67 , since each x can take 67 values . total : 67 + 67 * 67 . similarly : the number of 1 letter words ( x ) that can be made from 66 letters is 66 ; the number of 2 letter words ( xx ) that can be made from 66 letters is 66 * 66 , since each x can take 66 values . total : 66 + 66 * 66 . the difference is ( 67 + 67 * 67 ) - ( 66 + 66 * 66 ) = 134 . answer : e ." | a = 1 + 67
b = a + 67
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a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 200 , e ) $ 250 | d | multiply(divide(500, add(divide(2, 3), const_1)), divide(2, 3)) | $ 500 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 500 * 2 / 5 = $ 200 answer is d" | a = 2 / 3
b = a + 1
c = 500 / b
d = 2 / 3
e = c * d
|
a ) 1859 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2991 | a | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 5720), divide(multiply(15, 5720), const_100)), const_2) | in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 5720 votes were polled , then how many valid votes did b get ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 5720 32.5 k = 32.5 k / 100 k * 85720 = 1859 the number of valid votes polled in favour of b is 1859 . answer : a" | a = 20 / 100
b = 1 - a
c = b * 5720
d = 15 * 5720
e = d / 100
f = c - e
g = f / 2
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a ) 30 , b ) 20 , c ) 15 , d ) 44 , e ) 50 | e | divide(subtract(multiply(10, 1.8), multiply(2, 1.5)), 0.3) | 10 x 1.8 - 2 x 1.5 / 0.3 = ? | given expression = ( 18 - 3.0 ) / 0.3 = 15 / 0.3 = 150 / 3 = 50 answer is e . | a = 10 * 1
b = 2 * 1
c = a - b
d = c / 0
|
a ) 1 , b ) 10 , c ) 100 , d ) 50 , e ) 20 | e | divide(multiply(divide(multiply(20, 1000), const_100), 10), const_100) | there are 1000 students in a school and among them 20 % of them attends chess class . 10 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "20 % of 1000 gives 200 . so 200 attends chess and 10 % of 200 gives 20 . so 20 enrolled for swimming answer : e" | a = 20 * 1000
b = a / 100
c = b * 10
d = c / 100
|
a ) 72 , b ) 90 , c ) 124 , d ) 132 , e ) 140 | b | add(multiply(negate(15), power(subtract(add(2, 2), 2), 2)), 150) | an object thrown directly upward is at a height of h feet after t seconds , where h = - 15 ( t - 2 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ? | "we see that h will be a maximum h = 150 when t - 2 = 0 , that is when t = 2 . at t = 4 , h = - 15 ( 4 - 2 ) ^ 2 + 150 = - 15 ( 4 ) + 150 = 90 the answer is b ." | a = negate * (
b = 2 + 2
c = b - 2
d = c ** 2
e = a + d
|
a ) 26 , b ) 88 , c ) 90 , d ) 42 , e ) 5 | e | subtract(divide(660, 66), 5) | a trader sells 66 meters of cloth for rs . 660 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 660 / 66 = rs . 10 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 10 - rs . 5 = rs . 5 answer : e" | a = 660 / 66
b = a - 5
|
a ) 56.25 , b ) 60 , c ) 52.5 , d ) 5.25 , e ) none of these | a | multiply(divide(15, const_2), divide(15, const_2)) | what is the sum of the first 15 terms of an a . p whose 11 th and 7 th terms are 5.25 and 3.25 respectively | a + 10 d = 5.25 , a + 6 d = 3.25 , 4 d = 2 , d = ½ a + 5 = 5.25 , a = 0.25 = ¼ , s 15 = 15 / 2 ( 2 * ¼ + 14 * ½ ) = 15 / 2 ( 1 / 2 + 14 / 2 ) = 15 / 2 * 15 / 2 = 225 / 4 = 56.25 answer : a | a = 15 / 2
b = 15 / 2
c = a * b
|
a ) 52.5 , b ) 52.9 , c ) 52.1 , d ) 58.75 , e ) 42.5 | d | divide(add(multiply(30, 40), multiply(50, 70)), add(30, 50)) | the average marks of a class of 30 students is 40 and that of another class of 50 students is 70 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 40 = 1200 sum of the marks for the class of 50 students = 50 * 70 = 3500 sum of the marks for the class of 80 students = 1200 + 3500 = 4700 average marks of all the students = 4700 / 80 = 58.75 answer : d" | a = 30 * 40
b = 50 * 70
c = a + b
d = 30 + 50
e = c / d
|
a ) 1 : 96 , b ) 1 : 99 , c ) 1 : 94 , d ) 1 : 92 , e ) 1 : 91 | d | divide(divide(sqrt(multiply(3, 6348)), const_2), 6348) | the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 6348 sq cm . what is the ratio of the width and the area of the rectangle ? | "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 6348 12 x ^ 2 = 6348 x ^ 2 = 529 x = 23 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 92 the answer is d ." | a = 3 * 6348
b = math.sqrt(a)
c = b / 2
d = c / 6348
|
a ) $ 430 , b ) $ 620 , c ) $ 650 , d ) $ 680 , e ) $ 710 | e | add(add(multiply(100, 5), multiply(30, 3)), multiply(multiply(20, 3), const_2)) | rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 3 per page each time a page is revised . if a certain manuscript has 100 pages , of which 30 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 100 - 30 - 20 = 50 pages only cost is 5 $ per page for the first time page is typed - 50 * 5 = 250 $ ; for 30 pages the cost is : first time 5 $ + 3 $ of the first revision - 30 * ( 5 + 3 ) = 240 $ ; for 20 pages the cost is : first time 5 $ + 3 $ of the first revision + 3 $ of the second revision - 20 ( 5 + 3 + 3 ) = 220 $ ; total : 250 + 240 + 220 = 710 $ . answer : e ." | a = 100 * 5
b = 30 * 3
c = a + b
d = 20 * 3
e = d * 2
f = c + e
|
a ) 23 years , b ) 22 years , c ) 25 years , d ) 26 years , e ) 27 years | b | divide(subtract(add(26, add(26, 2)), multiply(2, 2)), const_2) | the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 2 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x â € “ ( 26 + 28 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 45 2 x = 45 x = 22 . so , average age of the team is 22 years . answer b" | a = 26 + 2
b = 26 + a
c = 2 * 2
d = b - c
e = d / 2
|
a ) $ 10 , b ) $ 24 , c ) $ 26 , d ) $ 12 , e ) $ 36 | d | divide(68, add(add(const_1, const_2), divide(const_2, divide(3, const_4)))) | if josh , doug , and brad have a total of $ 68 between them , and josh has two times as much money as brad but only 3 - fourths as much as doug , how much money does brad have ? | josh + doug + brad = 68 ; josh = 2 brad , josh = 3 / 4 doug josh + 1 / 2 josh + 4 / 3 josh = 68 ( substituted the given values ) josh = 24 . 24 = 3 / 4 doug = > doug = 32 josh + doug + brad = 68 24 + 32 + brad = 68 brad = 12 answer is d . | a = 1 + 2
b = 3 / 4
c = 2 / b
d = a + c
e = 68 / d
|
a ) 54 , b ) 25 , c ) 37 , d ) 41 , e ) 30 | a | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 45))), 30)) | mahesh can do a piece of work in 45 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | work done by mahesh in 45 days = 20 * 1 / 45 = 4 / 9 remaining work = 1 - 4 / 9 = 5 / 9 5 / 9 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 9 / 5 = 54 days answer is a | a = 1 / 45
b = 20 * a
c = 1 - b
d = c / 30
e = 1 / d
|
a ) 50 kmph , b ) 65 kmph , c ) 75 kmph , d ) 85 kmph , e ) 65 kmph | e | divide(add(100, 30), const_2) | the speed of a car is 100 km in the first hour and 30 km in the second hour . what is the average speed of the car ? | "explanation : s = ( 100 + 30 ) / 2 = 65 kmph e )" | a = 100 + 30
b = a / 2
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | multiply(const_2, 6) | the hour hand of a watch rotates 30 degrees every hour . how many complete rotations does the hour hand make in 6 days ? | explanation : there are 360 degrees in a complete circle , so 360 / 30 = 12 hours to make one full circle . in 6 days there are 24 hours x 6 = 144 hours total . the total number of rotations will be 144 / 12 = 12 . answer : ( c ) | a = 2 * 6
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a ) 23.89 , b ) 72.9 , c ) 58.08 , d ) 78.3 , e ) 79.3 | c | subtract(subtract(75, multiply(const_3, const_3)), multiply(divide(subtract(75, multiply(const_3, const_3)), 75), multiply(const_3, const_3))) | a vessel of capacity 75 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem , t = 75 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 75 - 9 ) / 75 ] ^ 2 ( 75 ) = 58.08 litres . answer : option c" | a = 3 * 3
b = 75 - a
c = 3 * 3
d = 75 - c
e = d / 75
f = 3 * 3
g = e * f
h = b - g
|
a ) 18 kms , b ) 16 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | b | multiply(speed(add(25, 40), const_60), 15) | riya and priya set on a journey . riya moves eastward at a speed of 25 kmph and priya moves westward at a speed of 40 kmph . how far will be priya from riya after 15 minutes | "total eastward distance = 25 kmph * 1 / 4 hr = 6.25 km total westward distance = 40 kmph * 1 / 4 hr = 10 km total distn betn them = 6.25 + 10 = 16.25 km ans 16 km answer : b" | a = 25 + 40
b = speed * (
|
a ) 78 , b ) 68 , c ) 88 , d ) 97 , e ) 37 | d | sqrt(9506) | the product of two successive numbers is 9506 . which is the smaller of the two numbers ? | "d 97 from the given alternatives , 97 × 98 = 9506 ∴ smaller number = 97" | a = math.sqrt(9506)
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a ) 198 , b ) 288 , c ) 363 , d ) 360 , e ) 484 | d | multiply(multiply(multiply(power(const_2.0, 2), 3), divide(10, 2)), 2) | if 2 ^ 4 , 3 ^ 3 , and 10 ^ 3 are factors of the product of 1,452 and w , where w is a positive integer , what is the smallest possible value of w ? | i will go with d ( pending elements to match is 2 ^ 2 * 3 ^ 2 * 10 ^ 1 = 360 | a = 2 ** 0
b = a * 3
c = 10 / 2
d = b * c
e = d * 2
|
a ) 1200 , b ) 2000 , c ) 2500 , d ) 3200 , e ) 1250 | c | add(multiply(divide(2, multiply(100, 4)), 4), multiply(divide(2, multiply(100, 4)), 100)) | the l . c . m . of 2 numbers is 100 . the numbers are in the ratio 4 : 1 . find their product ? | "let the numbers be 4 x and x l . c . m . = 4 x 4 x = 100 x = 25 the numbers are = 100 and 25 required product = 100 * 25 = 2500 answer is c" | a = 100 * 4
b = 2 / a
c = b * 4
d = 100 * 4
e = 2 / d
f = e * 100
g = c + f
|
a ) s . 10,000 , b ) s . 10,200 , c ) s . 10,400 , d ) s . 10,700 , e ) s . 10,800 | b | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30,000 , b receives : | "let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 30000 x 17 / 50 ) = rs . 10,200 . b" | a = 10 * 5000
b = a - 5000
c = 4000 + 5000
d = b - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 10 * 5000
i = g / h
j = 3 + 4
k = j * 5000
l = i * k
m = l / 1000
n = math.floor(m)
o = n - 1
|
a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 5 , d ) 2 / 5 , e ) 3 / 4 | a | divide(const_4, subtract(9, const_4)) | the difference between a positive proper fraction and its reciprocal is 9 / 20 . the fraction is : | "let the required fraction be x . then 1 / x - x = 9 / 20 ( 1 - x ^ 2 ) / x = 9 / 20 = = > 20 - 20 x ^ 2 = 9 x 20 x ^ 2 + 9 x - 20 = 0 20 x ^ 2 + 25 x - 16 x - 20 = 0 5 x ( 4 x + 5 ) - 4 ( 4 x + 5 ) = 0 ( 4 x + 5 ) ( 5 x - 4 ) = 0 x = 4 / 5 ( neglecting - ve value ) answer a ) 4 / 5" | a = 9 - 4
b = 4 / a
|
a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 16 years | b | divide(subtract(20, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 20 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 20 ) + 2 = 2 ( x + 2 ) x + 22 = 2 x + 4 x = 18 . answer : b" | a = 2 * 2
b = a - 2
c = 20 - b
d = 2 - 1
e = c / d
|
a ) 90 , b ) 120 , c ) 80 , d ) 70 , e ) 60 | a | divide(multiply(30, divide(4, 5)), subtract(divide(5, add(5, 4)), multiply(divide(4, add(5, 4)), divide(4, 5)))) | a mixture contains milk and water in the ratio 5 : 4 . on adding 30 litters of water , the ratio of milk to water becomes 5 : 7 . total quantity of milk & water before adding water to it ? | "explanation : milk : water = 5 : 4 5 x : 4 x + 30 = 5 : 7 7 [ 5 x ] = 5 [ 4 x + 30 ] 35 x = 20 x + 150 35 x - 20 x = 150 15 x = 150 x = 10 the quantity of milk in the original mixture is = 5 : 4 = 5 + 4 = 9 9 x = 90 short cut method : milk : water = 5 : 4 after adding 30 liters of water milk : water = 5 : 7 milk is same but water increse 30 liters then the water ratio is increse 3 parts 3 part - - - - - > 30 liters the quantity of milk in the original mixture is = 5 : 4 = 5 + 4 = 9 9 parts - - - - - > 9 liters ( answer is = 90 ) short cut method - 2 : for only milk problems milk : water 5 : 4 5 : 7 milk ratio same but water ratio 3 parts incress per 30 liters 3 part of ratio - - - - - - - > 30 liters 1 part of ratio - - - - - - - > 10 liters 9 part of ratio - - - - - - > 90 liters answer : option a" | a = 4 / 5
b = 30 * a
c = 5 + 4
d = 5 / c
e = 5 + 4
f = 4 / e
g = 4 / 5
h = f * g
i = d - h
j = b / i
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) indeterminate | a | subtract(30, 20) | in a class , 30 students pass in english and 20 students in math , while some students among these pass in both . how many more students do only english as compared to those doing only maths ? | let us consider tht x student are tohse who passed in both english and maths . . . so first we remove x student from both of them therefore , english = 30 - x maths = 20 - x now , number of students more in english = ( 30 - x ) - ( 20 - x ) = 30 - x - 20 + x = 10 answer : a | a = 30 - 20
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a ) 36 , b ) 72 , c ) 120 , d ) 132 , e ) 180 | d | divide(550, add(divide(100, const_60), divide(150, const_60))) | a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 550 bolts ? | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 550 bolts 550 * 6 / 25 = 132 mins ans d" | a = 100 / const_60
b = 150 / const_60
c = a + b
d = 550 / c
|
a ) 7 / 8 , b ) 14 / 15 , c ) 29 / 30 , d ) 44 / 45 , e ) 131 / 135 | b | divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(1200, subtract(const_1, divide(40, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25))) | in a village of 2,700 people , 900 people are over 70 years old and 1200 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a female or younger than 70 years old ? | "the number of people younger than 70 years old is 2700 - 900 = 1800 the number of females older than 70 years old is 0.6 * 1200 = 720 the number of people who are either female or younger than 70 is 1800 + 720 = 2520 . p ( a person is younger than 70 or male ) = 2520 / 2700 = 14 / 15 the answer is b ." | a = 5 * 5
b = a / const_0_25
c = const_26 * b
d = 5 * 5
e = d / const_0_25
f = 40 / e
g = 1 - f
h = 1200 * g
i = c - h
j = 5 * 5
k = j / const_0_25
l = const_26 * k
m = i / l
|
a ) 4,888 , b ) 4,898 , c ) 4,889 , d ) 4,869 , e ) 4,896 | c | floor(divide(subtract(divide(subtract(852755, 362855), const_100), const_10), const_1000)) | how many integers between 362855 and 852755 have tens digit 1 and units digit 3 ? | there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 362,855 and 852,755 is 852,755 - 362,855 = 489,900 - one number per each hundred gives 133,900 / 100 = 4,889 numbers . answer : c . | a = 852755 - 362855
b = a / 100
c = b - 10
d = c / 1000
e = math.floor(d)
|
a ) 75 % , b ) 80 % , c ) 100 % , d ) 150 % , e ) 159.2 % | e | multiply(divide(subtract(subtract(square_area(add(add(const_1, divide(100, 100)), divide(multiply(add(const_1, divide(100, 100)), 80), 100))), const_1), const_4), add(const_1, square_area(add(const_1, divide(100, 100))))), 100) | the length of each side of square a is increased by 100 percent to make square b . if the length of the side of square b is increased by 80 percent to make square c , by what percent is the area of square c greater than the sum of the areas of squares a and b ? | "let length of each side of square a be 10 area of a = 10 ^ 2 = 100 since , length of each side of square a is increased by 100 percent to make square b length of each side of square b = 2 * 10 = 20 area of b = 20 ^ 2 = 400 since , length of the side of square b is increased by 80 percent to make square c length of each side of square c = 1.8 * 20 = 36 area of c = 36 ^ 2 = 1296 difference in areas of c and cummulative areas of a and b = 1296 - ( 400 + 100 ) = 796 percent is the area of square c greater than the sum of the areas of squares a and b = ( 796 / 500 ) * 100 % = 159.2 % answer e" | a = 100 / 100
b = 1 + a
c = 100 / 100
d = 1 + c
e = d * 80
f = e / 100
g = b + f
h = square_area - (
i = h - 1
j = i / 4
k = 100 / 100
l = 1 + k
m = 1 + square_area
n = j * m
|
a ) 96 , b ) 76 , c ) 56 , d ) 44 , e ) 16 | d | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 401 * 503 * 604 * 646 * 547 * 448 * 349 ) ^ 2 | "( ( 301 * 401 * 503 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,1 , 3,4 , 6,7 , 8,9 ; 1 & 5 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 01,03 , 04,46 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 03 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 52 * 44 ) ^ 2 = 88 ^ 2 = 44 hence answer is d" | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) 50 % , b ) 20 % , c ) 5 % , d ) 2 % , e ) 0.2 % | d | multiply(divide(multiply(200, power(10, add(const_3, const_3))), multiply(10, power(10, 9))), const_100) | a corporation that had $ 10 billion in profits for the year paid out $ 200 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 200 million ) / ( 10 billion ) ] * 100 = [ ( 200 * 10 ^ 6 ) / ( 10 * 10 ^ 9 ) ] * 100 = ( 20 / 1000 ) * 100 = 2 % so answer is ( d )" | a = 3 + 3
b = 10 ** a
c = 200 * b
d = 10 ** 9
e = 10 * d
f = c / e
g = f * 100
|
a ) $ 200,000 , b ) $ 210,000 , c ) $ 215,000 , d ) $ 220,000 , e ) $ 230,000 | c | divide(divide(subtract(subtract(multiply(200000, 15), multiply(5, 200000)), multiply(4, 170000)), subtract(15, add(5, 4))), const_1000) | the 15 homes in a new development are each to be sold for one of 3 different prices so that the developer receives an average ( arithmetic mean ) of $ 200000 per home . if 4 of the homes are to be sold for $ 170000 each and 5 are to be sold for $ 200000 each , what will be the selling price of each of the remaining 8 homes ? | imo the answer has to be d . 5 houses are being sold for 200,000 . 4 houses are being sold for $ 30,000 less , resulting in a loss of $ 120,000 . to make the average selling price intact i . e . $ 200,000 , the remaining 8 houses must be sold at such a profit that it compensates for the loss of 120,000 . hence 8 x = 120,000 . x = 15,000 . the 8 houses are sold at $ 15,000 profit or at $ 215,000 . ( answer c ) | a = 200000 * 15
b = 5 * 200000
c = a - b
d = 4 * 170000
e = c - d
f = 5 + 4
g = 15 - f
h = e / g
i = h / 1000
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a ) $ 11 , b ) $ 5 , c ) $ 45 , d ) $ 400 , e ) $ 4.4 | e | divide(subtract(200, multiply(subtract(const_1, divide(10, const_100)), 200)), subtract(5, divide(const_1, const_2))) | a reduction in the price of petrol by 10 % enables a motorist to buy 5 gallons more for $ 200 . find the original price of petrol ? | "price decreased by 10 % , so 9 / 10 times , which means that original gallons bought increased 10 / 9 times . since this increase equals to 5 gallons then 45 gallons were bought originally ( 45 * 10 / 9 = 50 - - > increase 5 gallons ) . hence original price was 200 / 45 = $ 4.4 answer : e ." | a = 10 / 100
b = 1 - a
c = b * 200
d = 200 - c
e = 1 / 2
f = 5 - e
g = d / f
|
a ) 62 , b ) 64 , c ) 65 , d ) 66 2 / 3 % , e ) 67 | d | add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(10, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(10, const_100))), const_100), 3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(10, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(10, const_100))), const_100), 3), const_10)) | at a summer camp with 1,500 participants , 1 / 2 of the campers are aged 10 to 12 . next year , the number of campers aged 10 to 12 will increase by 1 / 3 . after this change , what percentage of the total 1,500 campers will the 10 to 12 - year - olds represent ? | "total - 1,500 participants campers are aged 10 to 12 = ( 1 / 2 ) * 1500 = 750 next year , campers are aged 10 to 12 = ( 4 / 3 ) * 750 = 1000 percentage = ( 1000 / 1500 ) * 100 = 66 2 / 3 % answer : option d" | a = 10 * 100
b = 1000 + a
c = b / 2
d = 1 / 3
e = d + 1
f = c * e
g = 10 * 100
h = 1000 + g
i = f / h
j = i * 100
k = j * 3
l = 10 * 100
m = 1000 + l
n = m / 2
o = 1 / 3
p = o + 1
q = n * p
r = 10 * 100
s = 1000 + r
t = q / s
u = t * 100
v = u * 3
w = v / 10
x = k + w
|
a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | a | divide(20, subtract(divide(3, 4), divide(50, const_100))) | a big container is 50 % full with water . if 20 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container in liters ? | "20 liters is 25 % of the capacity c . 20 = 0.25 c c = 80 liters . the answer is a ." | a = 3 / 4
b = 50 / 100
c = a - b
d = 20 / c
|
a ) 70 , b ) 7 , c ) 0.7 , d ) 0.07 , e ) none of these | d | multiply(divide(0.0007, 0.01), const_100) | 0.0007 ? = 0.01 | "explanation : required answer = 0.0007 / 0.01 = 0.07 / 1 = 0.07 . answer : option d" | a = 0 / 7
b = a * 100
|
a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) 250 m | c | subtract(multiply(multiply(72, const_0_2778), 26), 260) | a goods train runs at a speed of 72 kmph and crosses a 260 m long platform in 26 seconds . what is the length of the goods train ? | s = 260 + x / t 72 * 5 / 18 = 260 + x / 26 x = 260 answer : c | a = 72 * const_0_2778
b = a * 26
c = b - 260
|
a ) 8.78 , b ) 8.67 , c ) 8.75 , d ) 10 , e ) 8.28 | d | multiply(divide(const_1, const_2), multiply(5, 4)) | the area of a sector of a circle of radius 5 cm formed by an arc of length 4 cm is ? | "( 5 * 4 ) / 2 = 10 answer : d" | a = 1 / 2
b = 5 * 4
c = a * b
|
a ) 68 , b ) 54 , c ) 21 , d ) 48 , e ) 22 | a | divide(divide(200, 5280), multiply(2, divide(1, const_3600))) | if an object travels 200 feet in 2 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 200 feet in 2 sec then it travels 200 / 2 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 100 feet in 1 hour = 360000 feet in 1 hr = 360000 / 5280 miles in 1 hour = 36000 / 528 miles / hr ~ 68 miles / hr answer - a" | a = 200 / 5280
b = 1 / 3600
c = 2 * b
d = a / c
|
a ) 12 , b ) 8 , c ) 28 , d ) 7 , e ) 27 | d | subtract(divide(14, divide(2, 3)), 14) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 14 white mice in the experiment , how many brown mice are in the experiment ? | let total number of mice = m number of white mice = 2 / 3 m = 14 m = 21 number of brown mice = 1 / 3 m = 1 / 3 * 21 = > brown mice = 7 answer d | a = 2 / 3
b = 14 / a
c = b - 14
|
a ) 1 / 48 , b ) 234 / 8 , c ) 1 / 24 , d ) 5 / 8 , e ) 3 / 4 | b | divide(add(divide(96, 2), divide(96, 12)), 96) | if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 12 ? | "i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 12 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 12 = not integer we can not forget multiples of 12 from 1 to 96 we have 8 numbers that are multiple of 12 therefore , 48 / 96 + 8 / 96 = 23 / 48 answer : b" | a = 96 / 2
b = 96 / 12
c = a + b
d = c / 96
|
a ) 2 : 3 , b ) 1 : 4 , c ) 5 : 6 , d ) 7 : 11 , e ) 3 : 7 | a | divide(subtract(68, 60), subtract(80, 68)) | a theater box office sold an average ( arithmetic mean ) of 68 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 80 , 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 = 80 - 68 = 12 . deviation from the mean for the evening staff = 68 - 60 = 8 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 8 : 12 = 2 : 3 . the answer is a ." | a = 68 - 60
b = 80 - 68
c = a / b
|
a ) 12 , b ) 11 , c ) 10 , d ) 14 , e ) 13 | d | power(multiply(4, power(5, 2)), 7) | if 7 log ( 4 * 5 ^ 2 ) = x , find x | "7 ( log 2 ^ 2 * 5 ^ 2 ) = x 7 log ( 5 * 2 ) ^ 2 = x 7 * 2 log ( 5 * 2 ) = x 14 log 10 = x log 10 base 10 = 1 so 14 * 1 = x x = 14 answer : d" | a = 5 ** 2
b = 4 * a
c = b ** 7
|
a ) 3 / 4 , b ) 4 / 7 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(8), factorial(subtract(8, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 8)), factorial(subtract(add(6, 8), const_2))), factorial(const_2))) | a bag contains 6 black and 8 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 8 ) = 14 . number of white balls = 8 . p ( drawing a white ball ) = 8 / 14 = 4 / 7 . option b ." | a = math.factorial(8)
b = 8 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 6 + 8
o = math.factorial(n)
p = 6 + 8
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t
v = m / u
|
a ) $ 3700 , b ) $ 4000 , c ) $ 4300 , d ) $ 4600 , e ) $ 4900 | b | divide(2800, subtract(const_1, divide(30, const_100))) | we had $ 2800 left after spending 30 % of the money that we took for shopping . how much money did we start with ? | "let x be the amount of money we started with . 0.7 x = 2800 x = 4000 the answer is b ." | a = 30 / 100
b = 1 - a
c = 2800 / b
|
a ) 545 , b ) 685 , c ) 865 , d ) 495 , e ) 534 | c | divide(multiply(173, 240), 48) | ? x 48 = 173 x 240 | "let y x 48 = 173 x 240 then y = ( 173 x 240 ) / 48 = 173 x 5 = 865 . answer : c" | a = 173 * 240
b = a / 48
|
['a ) 1 : 96', 'b ) 1 : 99', 'c ) 1 : 95', 'd ) 1 : 12', 'e ) 1 : 11'] | a | divide(multiply(divide(sqrt(multiply(4, divide(6912, 3))), 4), const_3), 6912) | the ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm . find the ratio of the breadth and the area of the rectangle ? | let the length and the breadth of the rectangle be 4 x cm and 3 x respectively . ( 4 x ) ( 3 x ) = 6912 12 x 2 = 6912 x 2 = 576 = 4 * 144 = 22 * 122 ( x > 0 ) = > x = 2 * 12 = 24 ratio of the breadth and the areas = 3 x : 12 x 2 = 1 : 4 x = 1 : 96 . answer : a | a = 6912 / 3
b = 4 * a
c = math.sqrt(b)
d = c / 4
e = d * 3
f = e / 6912
|
a ) 2.5 , b ) 3.0 , c ) 4.0 , d ) 6.5 , e ) 8.0 | c | divide(divide(0.25, const_0_25), const_0_25) | after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.25 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | "let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.25 x = 4 alternate solution : try to backsolve . initial volume = 4 after one hour - - > ( 1 / 4 ) 4 = 1 after two hours - - > ( 1 / 4 ) 1 = 0.25 answer : c" | a = 0 / 25
b = a / const_0_25
|
a ) 629 , b ) 729 , c ) 829 , d ) 929 , e ) 216 | e | power(6, 3) | log 3 n + log 6 n what is 3 digit number n that will be whole number | "no of values n can take is 1 6 ^ 3 = 216 answer : e" | a = 6 ** 3
|
a ) 2.8 % , b ) 4.4 % , c ) 4.6 % , d ) 5.2 % , e ) 6.0 % | b | add(multiply(divide(4, const_100), 50), multiply(divide(8, const_100), 30)) | of the total amount that rose spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if rose paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | assume she has $ 200 to spend . tax clothing = 50 % = $ 100 = $ 4.00 food = 20 % = $ 40 = $ 0.00 items = 30 % = $ 60 = $ 4.80 total tax = $ 8.80 % of total amount = 8.8 / 200 * 100 = 4.4 % b ) | a = 4 / 100
b = a * 50
c = 8 / 100
d = c * 30
e = b + d
|
a ) 84 m , b ) 88 m , c ) 42 m , d ) 137 m , e ) none | c | divide(33, multiply(power(divide(1, const_2), const_2), const_pi)) | 33 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be : | "sol . let the length of the wire b h . radius = 1 / 2 mm = 1 / 20 cm . then , 22 / 7 * 1 / 20 * 1 / 20 * h = 33 ⇔ = [ 33 * 20 * 20 * 7 / 22 ] = 4200 cm = 42 m . answer c" | a = 1 / 2
b = a ** 2
c = b * math.pi
d = 33 / c
|
a ) 41 / 50 , b ) 1 / 216 , c ) 1 / 221 , d ) 1 / 84 , e ) 1 / 42 | b | power(divide(const_1, multiply(3, const_2)), 3) | in a certain game of dice , the player ’ s score is determined as a sum of 3 throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if john plays this game against 22 other players , what is the probability of the minimum score that will guarantee john some monetary payoff ? | to guarantee that john will get some monetary payoff he must score the maximum score of 6 + 6 + 6 = 18 , because if he gets even one less than that so 17 , someone can get 18 and john will get nothing . p ( 18 ) = 1 / 6 ^ 3 = 1 / 216 . answer : b . | a = 3 * 2
b = 1 / a
c = b ** 3
|
a ) 10 , b ) 5 , c ) 16 , d ) 18 , e ) 21 | b | subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(divide(divide(68, const_2), const_2), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1), const_1) | if the product of the integers w , x , y and z is 68 , and if 0 < w < x < y < z , what is the value of w + z ? | 24 = 1 * 2 * 3 * 4 so w = 1 , x = 2 , y = 3 , z = 4 w + z = 1 + 4 = 5 answer - b | a = 68 / 2
b = a / 2
c = b - 1
d = c - 1
e = d - 1
f = e - 1
g = f - 1
h = g - 1
i = h - 1
j = i - 1
k = j - 1
l = k - 1
m = l - 1
n = m - 1
|
a ) 30 , b ) 44 , c ) 36 , d ) 56 , e ) 94 | e | add(lcm(lcm(5, 6), lcm(9, 18)), 4) | what is the least number which when divided by 5 , 6 , 9 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 5 , 6 , 9 and 18 is 90 required number = 90 + 4 = 94 answer : option e" | a = math.lcm(5, 6)
b = math.lcm(9, 18)
c = math.lcm(a, b)
d = c + 4
|
a ) 19 , b ) 18 , c ) 16 , d ) 17 , e ) 23 | e | power(divide(63, divide(subtract(63, 24), sqrt(9))), const_2) | the speed of a railway engine is 63 km per hour when no compartment is attached , and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached , the maximum number of compartments that can be carried by the engine is : | the reduction in speed is directly proportional to the square root of the number of compartments attached does reduction mean amount subtracted ? or percentage decrease ? there are at least two interpretations , and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words , if s = speed , and n = number of compartments , then s = 63 - k * sqrt ( n ) where k is a constant of the proportionality . in general , if a is directly proportional to b , we can write a = k * b and solve for k . if n = 9 , then s = 24 24 = 63 - k * sqrt ( 9 ) = 63 - 3 k k = 13 now , we need to know : what value of n makes s go to zero ? 0 = 63 - 13 * sqrt ( n ) 13 * sqrt ( n ) = 63 n = 23 with 24 compartments , the train does not budge . therefore , it would budge if there were one fewer cars . thus , 23 is the maximum number of cars the engine can pull and still move . e | a = 63 - 24
b = math.sqrt(9)
c = a / b
d = 63 / c
e = d ** 2
|
a ) 6 , b ) 15 , c ) 25 , d ) 33 , e ) 54 | c | subtract(79, subtract(add(41, 22), 9)) | in a class of 79 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 79 54 + not = 79 not = 79 - 54 = 25 so answer is c" | a = 41 + 22
b = a - 9
c = 79 - b
|
a ) 68 , b ) 84 , c ) 90 , d ) 120 , e ) 125 | e | divide(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), subtract(divide(multiply(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), const_3), 98), add(multiply(const_4, const_2), const_3))) | a train traveled the first d miles of its journey it an average speed of 60 miles per hour , the next d miles of its journey at an average speed of y miles per hour , and the final d miles of its journey at an average speed of 160 miles per hour . if the train ’ s average speed over the total distance was 98 miles per hour , what is the value of y ? | average speed = total distance traveled / total time taken 3 d / d / 60 + d / y + d / 160 = 98 solving for d and y , 15 y = 11 y + 480 4 y = 500 y = 125 answer e | a = 4 * 2
b = a * 10
c = 100 * 4
d = b + c
e = 4 * 2
f = e * 10
g = 100 * 4
h = f + g
i = h * 3
j = i / 98
k = 4 * 2
l = k + 3
m = j - l
n = d / m
|
a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 37 % . | e | multiply(divide(add(multiply(divide(35, const_100), 2), multiply(divide(50, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 35 % of alcohol and another vessel of capacity 6 litre had 50 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "35 % of 2 litres = 0.7 litres 50 % of 6 litres = 3 litres therefore , total quantity of alcohol is 3.7 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 37 % e" | a = 35 / 100
b = a * 2
c = 50 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
|
a ) 10,300 , b ) 10,030 , c ) 1,353 , d ) 1,352 , e ) 1,439 | e | subtract(468,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 468,600 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 324,700 and 468,600 is 468,600 - 324,700 = 133,900 - one number per each hundred gives 143,900 / 100 = 1,439 numbers . answer : e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 468 - 600
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | d | divide(add(62, const_2), 8) | find the unknown term 8 , x , 62 , - 4 , - 12 | 8 , x , 6,2 , - 4 , - 12 i guess each differ in d range of 2 8 - 8 = 0 8 - 6 = 2 6 - 2 = 4 2 - 6 = - 4 - 4 - 8 = - 12 answer : d | a = 62 + 2
b = a / 8
|
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