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a ) 52 , b ) 54 , c ) 55 , d ) 56 , e ) 58
b
multiply(reminder(60, subtract(const_10, const_1)), subtract(const_10, const_1))
q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 60 , what is the greatest possible value of q minus r ?
"a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 60 . the greatest multiple of 9 which is less than 60 is 54 . the answer is b ."
a = 10 - 1 b = reminder * (
a ) rs 450 , b ) rs 920 , c ) rs 550 , d ) rs . 630 , e ) none of these
d
multiply(divide(add(84, add(35, 7)), 7), 35)
there were 35 students in a hostel . due to the admission of 7 new students the expenses of the mess were increased by rs . 84 per day while the average expenditure per head diminished by re 1 . what was the original expenditure of the mess ?
explanation : let the original average expenditure be rs . x then , 42 ( x - 1 ) - 35 x = 84 ? 7 x = 126 ? x = 18 therefore original expenditure = rs . ( 35 18 ) = rs . 630 . answer : d
a = 35 + 7 b = 84 + a c = b / 7 d = c * 35
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 11
c
divide(subtract(27, const_3), const_3)
27 is a perfect cube . when x is added to the prime factor of 27 , the result is also a perfect cube . what is x ?
"27 is 3 * 3 * 3 a 3 + 1 = 4 , 4 = 2 * 2 b 3 + 3 = 6 , 6 = 2 * 3 c 3 + 5 = 8 , 8 = 2 * 2 * 2 d 3 + 7 = 10 , 10 = 2 * 5 e 3 + 11 = 14 , 14 = 2 * 7 c is the only addition that gives a perfect cube ."
a = 27 - 3 b = a / 3
a ) 41.1 sec , b ) 72 sec , c ) 31.8 sec , d ) 50.4 sec , e ) none of the above
b
divide(add(480, 620), multiply(55, const_0_2778))
a 480 m long train is running at a speed of 55 km / hr . it crossed a platform of length 620 m in ?
"speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 480 m + 620 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 1100 m time = distance / speed = 1100 x 18 / ( 5 x 55 ) = 72 sec ans is : b"
a = 480 + 620 b = 55 * const_0_2778 c = a / b
a ) 12 , b ) 150 , c ) 160 , d ) 180 , e ) 18
b
subtract(power(divide(add(15, 10), const_2), const_2), power(subtract(15, divide(add(15, 10), const_2)), const_2))
if the sum and difference of two numbers are 15 and 10 respectively , then the difference of their square is :
"let the numbers be x and y . then , x + y = 15 and x - y = 10 x 2 - y 2 = ( x + y ) ( x - y ) = 15 * 10 = 150 . answer : b"
a = 15 + 10 b = a / 2 c = b ** 2 d = 15 + 10 e = d / 2 f = 15 - e g = f ** 2 h = c - g
a ) 45 ( 4 / 11 ) % , b ) 40 % , c ) 45 ( 5 / 11 ) % , d ) 44 ( 5 / 11 ) % , e ) none of these
b
multiply(divide(subtract(100, add(multiply(3, 8), multiply(8, 3))), 100), const_100)
a batsman scored 100 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ?
"explanation : total runs scored = 100 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 100 - 60 = 40 required % = ( 40 / 100 ) × 100 = 40 % answer : option b"
a = 3 * 8 b = 8 * 3 c = a + b d = 100 - c e = d / 100 f = e * 100
a ) $ 120 , b ) $ 100 , c ) $ 91 , d ) $ 72 , e ) $ 69
e
multiply(100, divide(100, add(100, 45)))
a shopkeeper sold an article at $ 100 with 45 % profit . then find its cost price ?
"cost price = selling price * 100 / ( 100 + profit ) c . p . = 100 * 100 / 145 = $ 69 ( approximately ) answer is e"
a = 100 + 45 b = 100 / a c = 100 * b
a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28
d
multiply(3, add(divide(subtract(48, multiply(3, 3)), add(4, 3)), 3))
lionel left his house and walked towards walt ' s house , 48 miles away . two hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 3 miles per hour and walt ' s 4 miles per hour , how many miles had lionel walked when he met walt ?
"in the first 2 hours lionel at the rate of 3 miles per hour covered distance = rate * time = 3 * 2 = 6 miles . so , the distance between him and walt was 48 - 6 = 42 miles when walt left his house . now , their combined rate to cover this distance was 3 + 4 = 7 miles per hour , hence they will meet ( they will cover that distance ) in time = distance / rate = 42 / 7 = 6 hours . total time that lionel was walking is 2 + 6 = 8 hours , which means that he covered in that time interval distance = rate * time = 3 * 8 = 24 miles . answer : d ."
a = 3 * 3 b = 48 - a c = 4 + 3 d = b / c e = d + 3 f = 3 * e
a ) 459 , b ) 954 , c ) 594 , d ) 584 , e ) 458
c
subtract(multiply(6, const_1000), 6)
the difference between the local value and the face value of 6 in the numeral 85475641 is
"explanation : ( local value of 6 ) - ( face value of 6 ) = ( 600 - 6 ) = 594 c )"
a = 6 * 1000 b = a - 6
a ) 500 , b ) 334 , c ) 555 , d ) 664 , e ) 526
e
divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 17))
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 17 % , then the cost price of the article is :
110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 117 = rs . 526 answer : option e
a = 616 * 100 b = 100 + 10 c = a / b d = 100 + 10 e = c * d f = 100 + 17 g = e / f
a ) 13.25 , b ) 38.25 , c ) 33.25 , d ) 12.25 , e ) 31.25
d
divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 28), const_2)
28 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ?
"x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 28 = 12.25 answer : d"
a = 4 + 3 b = 4 + 3 c = b + 2 d = a / c e = d * 28 f = e / 2
a ) 10 , b ) 20 , c ) 88 , d ) 77 , e ) 14
a
subtract(35, divide(35, add(divide(2, 5), const_1)))
a 35 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ?
"1 : 2 / 5 = 5 : 2 2 / 7 * 35 = 10 answer : a"
a = 2 / 5 b = a + 1 c = 35 / b d = 35 - c
a ) 1.5 , b ) 4.5 , c ) 4.75 , d ) 3.75 , e ) 3.5
c
divide(add(30, 8), subtract(58, 50))
car a is 30 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ?
"relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 30 miles and drive 8 miles ahead so to drive 38 miles it ' ll need 38 / 8 = 3.5 hours . answer : c"
a = 30 + 8 b = 58 - 50 c = a / b
a ) 34 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh
a
divide(const_3, add(add(divide(const_1, 80), divide(const_1, 18)), divide(const_1, 48)))
if a car went the first third of the distance at 80 kmh , the second third at 18 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ?
"assume d / 3 = 720 ( this number is convenient because it is divisible by 80 , 18 and 48 ) so : 720 = 80 * t 1 = 9 hrs 720 = 18 * t 2 = 40 hrs 720 = 48 * t 3 = 15 hrs t = t 1 + t 2 + t 3 = 64 hrs d = rt ( 720 * 3 ) = r * 64 r = 33.75 answer : a"
a = 1 / 80 b = 1 / 18 c = a + b d = 1 / 48 e = c + d f = 3 / e
a ) 11 , b ) 12 , c ) 13 , d ) 15 , e ) 16
a
divide(22, const_2)
in a group of ducks and cows , the total number of legs are 22 more than twice the no . of heads . find the total no . of buffaloes .
"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 22 = > 2 x = 22 = > x = 11 a"
a = 22 / 2
a ) 24.2 sec , b ) 30.6 sec , c ) 32.6 sec , d ) 28.1 sec , e ) 20.4 sec
a
divide(add(110, 132), multiply(36, const_0_2778))
how long does a train 110 meters long running at the rate of 36 km / hr take to cross a bridge 132 meters in length ?
"distance = length of train + length of bridge = 110 + 132 = 242 speed = 36 km / hr = 36 * 5 / 18 = 10 m / s required time = 242 / 10 = 24.2 seconds answer is a"
a = 110 + 132 b = 36 * const_0_2778 c = a / b
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18
b
divide(100, add(3, 4))
a certain airline ' s fleet consisted of 100 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ?
"let x be the number of years . 4 x > 100 - 3 x 7 x > 100 x > 14 + 2 / 7 the answer is b ."
a = 3 + 4 b = 100 / a
a ) 10 % , b ) 9 1 / 11 % , c ) 11 1 / 9 % , d ) 11 % , e ) none
c
add(subtract(const_100, subtract(const_100, 10)), const_2)
the price of an article is cut by 10 % . to restore it to the former value , the new price must be increased by
"explanation : let original price = rs . 100 then , new price = rs . 90 increase on rs . 90 = rs . 100 increase % = ( 10 / 90 × 100 ) % = 11 1 / 9 % correct option : c"
a = 100 - 10 b = 100 - a c = b + 2
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 7
c
floor(divide(23, 5))
on dividing 23 by a number , the quotient is 5 and the remainder is 3 . find the divisor .
d = ( d - r ) / q = ( 23 - 3 ) / 5 = 20 / 5 = 4 c
a = 23 / 5 b = math.floor(a)
a ) 88 , b ) 560 , c ) 99 , d ) 550 , e ) 540
b
divide(multiply(multiply(add(add(6, const_1), const_1), multiply(2, add(6, const_1))), add(14, const_1)), const_3)
( 2 ^ 2 + 4 ^ 2 + 6 ^ 2 + . . . . . + 14 ^ 2 ) = ?
= ( 1 x 2 ) ^ 2 + ( 2 x 2 ) ^ 2 , + ( 2 x 3 ) ^ 3 + . . . . . . ( 2 x 7 ) ^ 2 = 2 ^ 2 x ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . . . . . + 7 ^ 2 ) formula is = 1 / 6 n ( n + 1 ) ( 2 n + 1 ) = ( 4 x 1 / 6 x 7 x 8 x 15 ) = ( 4 x 140 ) = 560 answer is b
a = 6 + 1 b = a + 1 c = 6 + 1 d = 2 * c e = b * d f = 14 + 1 g = e * f h = g / 3
a ) 33 , b ) 46 , c ) 49 , d ) 54 , e ) 86
d
add(add(multiply(5, const_2), 2), add(multiply(5, multiply(const_2, 2)), 2))
a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 2 students left over . what is the sum of the two smallest possible values of n ?
"n = 4 k + 1 = 5 j + 2 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 5 j + 2 . 1 , 5 , 9 , 13 , 17 = 5 ( 3 ) + 2 the next such number is 17 + 4 * 5 = 37 . 17 + 37 = 54 the answer is d ."
a = 5 * 2 b = a + 2 c = 2 * 2 d = 5 * c e = d + 2 f = b + e
a ) 120 , b ) 100 , c ) 112 , d ) 150 , e ) 105
c
multiply(divide(subtract(100, 4), subtract(9, const_1)), 9)
find large number from below question the difference of two numbers is 100 . on dividing the larger number by the smaller , we get 9 as quotient and the 4 as remainder
"let the smaller number be x . then larger number = ( x + 100 ) . x + 100 = 9 x + 4 8 x = 96 x = 12 large number = 12 + 100 = 112 c"
a = 100 - 4 b = 9 - 1 c = a / b d = c * 9
a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 5.20 , e ) $ 2.80
d
add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.30))
if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.30 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was
"in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.3 ) = $ 4.2 ( as the charge is 0.3 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 4.2 = 5.2 answer is d ."
a = 1 / 5 b = 1 / 0 c = b - 1 d = c * 0 e = 1 + 0
a ) 30 , b ) 66 , c ) 77 , d ) 88 , e ) 44
a
divide(multiply(18, 5), const_4)
there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 18 , then find the smaller number ?
"let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 18 3 x = 18 = > x = 6 = > smaller number = 5 x = 30 . answer : a"
a = 18 * 5 b = a / 4
a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) none of these
c
divide(subtract(add(multiply(3, subtract(4, 2)), multiply(4, 4)), multiply(4, add(1, 4))), subtract(multiply(4, 2), 3))
the function f ( x ) = x + 4 is given . solve the equation : [ 3 . f ( x - 2 ) ] / f ( 0 ) + 4 = f ( 2 x + 1 )
solution : we calculate f ( 0 ) , f ( x - 2 ) , f ( 2 x + 1 ) , namely f ( 0 ) = 0 + 4 = 4 ; f ( x - 2 ) = x - 2 + 4 = x + 2 ; f ( 2 x + 1 ) = 2 x + 1 + 4 = 2 x + 5 the equation gets this look [ 3 ( x + 2 ) ] / 4 + 4 = 2 x + 5 < = > 3 ( x + 2 ) + 16 = 4 ( 2 x + 5 ) < = > 3 x + 6 + 16 = 8 x + 20 < = > 22 - 20 = 8 x - 3 x < = > 2 = 5 x < = > x = 0.4 answer c
a = 4 - 2 b = 3 * a c = 4 * 4 d = b + c e = 1 + 4 f = 4 * e g = d - f h = 4 * 2 i = h - 3 j = g / i
a ) 6 % . , b ) 7.5 % . , c ) 9.2 % . , d ) 10.5 % . , e ) 14 % .
e
multiply(divide(add(subtract(subtract(340, multiply(340, divide(80, const_100))), multiply(340, divide(6, const_100))), 3.2), add(add(add(340, 3.2), 10), 6.8)), const_100)
a 340 - liter solution of kola is made from 80 % water , 6 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 10 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ?
denominator : 340 + 10 + 3.2 + 6.8 = 360 numerator : 340 ( 1 - . 80 - . 06 ) + 3.2 340 ( 0.14 ) + 3.2 47.6 + 3.2 50.8 ratio : 50.8 / 360 = 0.14 answer : e
a = 80 / 100 b = 340 * a c = 340 - b d = 6 / 100 e = 340 * d f = c - e g = f + 3 h = 340 + 3 i = h + 10 j = i + 6 k = g / j l = k * 100
a ) 998765 , b ) 998907 , c ) 999944 , d ) 999954 , e ) 999990
c
multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(6, const_10)), const_3), 88)
find the largest 6 digit number which is exactly divisible by 88 ?
largest 6 digit number is 999999 after doing 999999 ÷ 88 we get remainder 55 hence largest 6 digit number exactly divisible by 88 = 999999 - 55 = 999944 c
a = 100 * 100 b = 100 * 10 c = a + b d = 100 * 3 e = c + d f = 6 * 10 g = e + f h = g + 3 i = h * 88
a ) 0.68 , b ) 0.086 , c ) 0.86 , d ) 0.068 , e ) none of them
e
divide(subtract(power(0.12, 3), power(0.1, 3)), add(add(power(0.12, 2), 0.012), power(0.1, 2)))
( 0.12 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.12 ) ( power 2 ) + 0.012 + ( 0.1 ) ( power 2 ) is :
"given expression = ( 0.12 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.12 ) ( power 2 ) + ( 0.12 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.12 - 0.1 ) = 0.02 answer is e ."
a = 0 ** 12 b = 0 ** 1 c = a - b d = 0 ** 12 e = d + 0 f = 0 ** 1 g = e + f h = c / g
a ) 40 , b ) 44 , c ) 52 , d ) 88 , e ) 90
c
divide(subtract(power(18, const_2), 220), const_2)
if the sum of two numbers is 18 and the sum of their squares is 220 , then the product of the numbers is
"sol . let the numbers be x and y . then , ( x + y ) = 18 and x 2 + y 2 = 220 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 18 ) 2 - 220 = 324 - 220 = 104 xy = 52 . answer c"
a = 18 ** 2 b = a - 220 c = b / 2
a ) 31 , b ) 51 , c ) 56 , d ) 62 , e ) 93
c
divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1))))
how many positive integers less than 10,000 are there in which the sum of the digits equals 5 ?
basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a separator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit separators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . answer : c
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 ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42
e
multiply(divide(const_100, add(const_100, 20)), 50)
from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 20 percent . the enrollment on january 1 , 1993 , was 50 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ?
1.20 x = 50 - - > 6 / 5 * x = 50 - - > x = 50 * 5 / 6 = 250 / 6 = ~ 42 . answer : e .
a = 100 + 20 b = 100 / a c = b * 50
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 2
e
divide(10, 5)
if the remainder is 7 when positive integer n is divided by 10 , what is the remainder when n is divided by 5 ?
"assume x is quotient here , n = 10 x + 7 - - - - - - - - - - ( 1 ) and n = 5 x + ? we can also write equation ( 1 ) as : n = ( 10 x + 5 ) + 2 . ie 5 ( 2 x + 1 ) + 1 ie the first term is perfectly divisible by 5 . so , the remainder left is 2 so , answer ( e ) is right choice ."
a = 10 / 5
a ) 15 % , b ) 14.25 % , c ) 6.06 % , d ) 10.5 % , e ) 11.5 %
c
divide(multiply(const_100, subtract(subtract(340, divide(330, 2)), divide(330, 2))), divide(330, 2))
on a sum of money , the simple interest for 2 years is rs . 330 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is
"explanation : the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r ã — si ) / ( 2 ã — 100 ) difference between the compound interest and simple interest = 340 - 330 = 10 ( r ã — si ) / ( 2 ã — 100 ) = 10 ( r ã — 330 ) / ( 2 ã — 100 ) = 10 r = 6.06 % answer : option c"
a = 330 / 2 b = 340 - a c = 330 / 2 d = b - c e = 100 * d f = 330 / 2 g = e / f
a ) 59 , b ) 55 , c ) 61 , d ) 41 , e ) 36
d
subtract(add(add(multiply(30, 10), 1), 30), multiply(30, 10))
the average age of 30 students in a class is 10 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ?
"total age of 50 students = 30 * 10 = 300 total age of 51 persons = 31 * 11 = 341 age of teacher = 341 - 300 = 41 years answer is d"
a = 30 * 10 b = a + 1 c = b + 30 d = 30 * 10 e = c - d
a ) 8 , b ) 7.5 , c ) 6.5 , d ) 5 , e ) 6
d
divide(subtract(multiply(3, 11), multiply(4, 7)), subtract(4, 3))
what must be added to each term of the ratio 7 : 11 , so as to make it equal to 3 : 4 ?
"solution : let x be added to each term . according to question , ( 7 + x ) / ( 11 + x ) = 3 / 4 ; or , 33 + 3 x = 28 + 4 x ; or , x = 5 . answer : option d"
a = 3 * 11 b = 4 * 7 c = a - b d = 4 - 3 e = c / d
a ) 5 , b ) 11 , c ) 17 , d ) 25 , e ) 30
a
add(divide(lcm(22, 55), 22), const_10)
jaime earned enough money by selling seashells at 22 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ?
"let ' s test answer a : 5 seashells . . . with 5 seashells , jamie would have 5 ( 22 ) = 110 cents . this would allow him to buy 2 books for 110 cents total , so this is the correct answer . a"
a = math.lcm(22, 55) b = a / 22 c = b + 10
a ) 100 , b ) 250 , c ) 4500 , d ) 1200 , e ) 5625
c
multiply(volume_rectangular_prism(60, 20, divide(6, add(const_10, const_2))), 7.5)
the water level in a rectangular swimming pool measuring 60 feet by 20 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons )
"6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 60 * 20 * 1 / 2 = 600 feet ^ 3 of water must be removed , which equals to 600 * 7.5 = 4500 gallons . answer : c ."
a = 10 + 2 b = 6 / a c = volume_rectangular_prism * (
a ) 3944 , b ) 920 , c ) 7290 , d ) 3876 , e ) 3729
d
multiply(square_perimeter(sqrt(289)), 57)
what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 57 ?
"explanation : let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 57 = rs . 3876 . answer : option d"
a = math.sqrt(289) b = square_perimeter * (
a ) $ 495 , b ) $ 510 , c ) $ 480 , d ) $ 375 , e ) $ 360
a
add(multiply(divide(subtract(1105, add(add(10, 20), 15)), add(add(11, 18), 24)), 24), 15)
3 friends alice , bond and charlie divide $ 1105 amongs them in such a way that if $ 10 , $ 20 and $ 15 are removed from the sums that alice , bond and charlie received respectively , then the share of the sums that they got will be in the ratio of 11 : 18 : 24 . how much did charlie receive ?
explanatory answer let the sums of money received by a , b and c be x , y and z respectively . then x - 10 : y - 20 : z - 15 is 11 a : 18 a : 24 a when $ 10 , $ 20 and $ 15 are removed , we are removing a total of $ 45 from $ 1105 . therefore , 11 a + 18 a + 24 a = 1105 - 45 = 1060 i . e . , 53 a = 1060 or a = 20 . we know that z - 15 = 24 a = 24 * 20 = 480 therefore , z = 480 + 15 = $ 495 correct choice is ( a )
a = 10 + 20 b = a + 15 c = 1105 - b d = 11 + 18 e = d + 24 f = c / e g = f * 24 h = g + 15
a ) 46 , b ) 47 , c ) 42 , d ) 49 , e ) 50
c
subtract(60, divide(subtract(150, 60), 5))
mother , her daughter and her grand child weighs 150 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ?
"mother + daughter + child = 150 kg daughter + child = 60 kg mother = 150 - 60 = 90 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 90 = 18 kg so now daughter = 120 - ( mother + child ) = 150 - ( 90 + 18 ) = 42 kg answer : c"
a = 150 - 60 b = a / 5 c = 60 - b
a ) 225 , b ) 275 , c ) 325 , d ) 350 , e ) 400
e
divide(add(900, 300), const_3)
in a weight - lifting competition , the total weight of joe ' s two lifts was 900 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ?
"this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 900 pounds . we sum the two variables to obtain : f + s = 900 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f – 300 = s we can now plug in ( 2 f – 300 ) for s into the first equation , so we have : f + 2 f – 300 = 900 3 f = 1,200 f = 400 answer is e ."
a = 900 + 300 b = a / 3
a ) 276 , b ) 299 , c ) 312 , d ) 400 , e ) none
d
multiply(25, 16)
the h . c . f . of two numbers is 25 and the other two factors of their l . c . m . are 14 and 16 . the larger of the two numbers is
"solution clearly , the numbers are ( 25 x 14 ) and ( 25 x 16 ) . larger number = ( 25 x 16 ) = 400 . answer d"
a = 25 * 16
a ) 5 seconds , b ) 4.3 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these
b
divide(180, multiply(154, const_0_2778))
in what time will a train 180 meters long cross an electric pole , if its speed is 154 km / hr
"explanation : first convert speed into m / sec speed = 154 * ( 5 / 18 ) = 42 m / sec time = distance / speed = 180 / 42 = 4.3 seconds option b"
a = 154 * const_0_2778 b = 180 / a
a ) 50 % , b ) 67 % , c ) 80 % , d ) 90 % , e ) 95 %
d
multiply(subtract(1, power(divide(divide(9, const_2), 9), const_2)), const_100)
a miniature roulette wheel is divided into 9 equal sectors , each bearing a distinct integer from 1 to 9 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun four times , approximately what is the probability that the product of the four winning sectors ’ integers will be even ?
"the only way to have an odd product is if all 4 integers are odd . p ( odd product ) = 5 / 9 * 5 / 9 * 5 / 9 * 5 / 9 = 625 / 6561 p ( even product ) = 1 - 625 / 6561 = 5936 / 6561 which is about 90 % the answer is d ."
a = 9 / 2 b = a / 9 c = b ** 2 d = 1 - c e = d * 100
a ) 17.39 % , b ) 33 1 / 6 % , c ) 33 1 / 3 % , d ) 38 1 / 3 % , e ) 33 2 / 3 %
a
subtract(const_100, divide(multiply(675, const_100), 575))
an article is bought for rs . 575 and sold for rs . 675 , find the gain percent ?
"575 - - - - 100 100 - - - - ? = > 17.39 % answer : a"
a = 675 * 100 b = a / 575 c = 100 - b
a ) 15 , b ) 20 , c ) 27.77 , d ) 40 , e ) 12
c
multiply(divide(subtract(1150, 900), 900), const_100)
a cycle is bought for rs . 900 and sold for rs . 1150 , find the gain percent ?
"900 - - - - 250 100 - - - - ? = > 27.77 % answer : c"
a = 1150 - 900 b = a / 900 c = b * 100
a ) 10 , b ) 40 , c ) 30 , d ) 20 , e ) 50
b
divide(subtract(multiply(20, const_60), 280), negate(subtract(negate(3), 20)))
an employer pays rs . 20 for each day a worker works and for felts rs . 3 for each day is ideal at the end of sixty days a worker gets rs . 280 . for how many days did the worker remain ideal ?
suppose a worker remained ideal for x days then he worked for 60 - x days 20 * ( 60 - x ) - 3 x = 280 1200 - 23 x = 280 23 x = 920 x = 40 answer is b .
a = 20 * const_60 b = a - 280 c = negate - ( d = b / negate
a ) 140 m , b ) 142 m , c ) 146 m , d ) 152 m , e ) 158 m
a
multiply(divide(multiply(const_1000, 72), const_3600), 7)
a train usually travels at a speed of 72 kmph crossing a pole in 7 sec , what would be the length of the train ?
explanation : d = 72 * 5 / 18 * 7 = 140 m answer : option a
a = 1000 * 72 b = a / 3600 c = b * 7
a ) rs 5.30 , b ) rs 6.44 , c ) rs 60.44 , d ) rs 16.44 , e ) none of these
b
add(subtract(multiply(divide(divide(42, 5), 6), 5), const_1), const_0_25)
a man received a cheque . the amount in rs . has been transposed for paise and vice versa . after spending rs . 5 and 42 paise , he discovered he now had exactly 6 times the value of the correct cheque amount . what amount he should have received ?
explanation : option 4 : - on reversal will be 44 rs 16 ps , after spending 5 rs 42 ps , remaining will be around 40 rs which is not 6 times 16 rs 44 ps . option 3 : - on reversal will lead a smaller value . hence we can eliminate that as well . option 1 : - on reversal is 30 rs 5 paise , after spending 5 rs 42 ps , remaining will be 24 rs 63 paise , 24 rs 63 ps is not 6 times 5 rs 30 paise . option 2 on reversal is 44 rs 06 paise , after spending 5 rs 42 ps , remaining will be 38 rs 64 ps , which is 6 times 6 rs 44 paise . hence rs 6.44 is the answer . we need not even consider option 4 and option 3 as reversed number itself is not 6 times the original number . option 1 when reversed is odd , and when even ( 5 rs 42 ps ) is subtracted from odd , we get a odd number , and an odd number can not be 6 times another number answer : b
a = 42 / 5 b = a / 6 c = b * 5 d = c - 1 e = d + const_0_25
a ) 50.5 % , b ) 44.4 % , c ) 22.2 % , d ) 33.3 % , e ) 20 %
e
multiply(divide(subtract(30, 20), add(30, 20)), const_100)
if 30 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ?
"30 % of ( x - y ) = 20 % of ( x + y ) 30 / 100 ( x - y ) = 20 / 100 ( x + y ) x = 5 y required percentage = y / x * 100 = y / 5 y * 100 = 20 % answer is e"
a = 30 - 20 b = 30 + 20 c = a / b d = c * 100
a ) 5.45 % , b ) 6.23 % , c ) 7 % , d ) 8.12 % , e ) 10 %
a
multiply(divide(subtract(5800, add(4700, 800)), add(4700, 800)), const_100)
alfred buys an old scooter for $ 4700 and spends $ 800 on its repairs . if he sells the scooter for $ 5800 , his gain percent is ?
"c . p . = 4700 + 800 = $ 5500 s . p . = $ 5800 gain = 5800 - 5500 = $ 300 gain % = 300 / 5500 * 100 = 5.45 % answer is a"
a = 4700 + 800 b = 5800 - a c = 4700 + 800 d = b / c e = d * 100
a ) 50 , b ) 27 , c ) 29 , d ) 47 , e ) 28
a
divide(add(380, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36))
a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 380 m long platform ?
"let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 380 ) / 36 * 5 / 18 = 50 min . answer : a"
a = const_0_2778 * 36 b = a * 12 c = 380 + b d = const_0_2778 * 36 e = c / d
a ) ( a ) 25 , b ) ( b ) 13 , c ) ( c ) 17 , d ) ( d ) 24 , e ) ( e ) 15
d
multiply(add(divide(2, 8), 3), 8)
find the value of 3 + 2 • ( 8 – 3 )
"3 + 2 • ( 8 – 3 ) = 3 + 2 ( 5 ) = 3 + 2 × 5 = 3 + 10 = 13 answer : ( d )"
a = 2 / 8 b = a + 3 c = b * 8
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3
d
floor(divide(log(divide(210,000, 2.134)), log(10)))
if x is an integer and 2.134 × 10 ^ x is less than 210,000 , what is the greatest possible value for x ?
"21340 * 10 ^ - 4 * 10 ^ x < 210,000 , x = 4 because the next zero on the right side will make 21,340 > 210,000 answer d ,"
a = 210 / 0 b = math.log(a) c = math.log(10) d = b / c e = math.floor(d)
a ) 43,100 , b ) 43,200 , c ) 43,300 , d ) 43,400 , e ) 43,450
b
multiply(multiply(subtract(4, 3), const_3600), const_12)
in a renowned city , the average birth rate is 4 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day .
"this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 4 - 3 = 1 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 1 = 43,200 . hence b ."
a = 4 - 3 b = a * 3600 c = b * 12
a ) 42 , b ) 70 , c ) 540 , d ) 565 , e ) 560
e
multiply(multiply(8, 3), 5)
a certain university will select 2 of 5 candidates eligible to fill a position in the mathematics department and 3 of 8 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ?
"5 c 2 * 8 c 3 = 10 * 56 = 560 the answer is ( e )"
a = 8 * 3 b = a * 5
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) none of them
a
subtract(multiply(multiply(multiply(784, 618), 917), 463), subtract(multiply(multiply(multiply(784, 618), 917), 463), add(const_4, const_4)))
the unit digit in the product ( 784 x 618 x 917 x 463 ) is
"unit digit in the given product = unit digit in ( 4 x 8 x 7 x 3 ) = 2 answer is a"
a = 784 * 618 b = a * 917 c = b * 463 d = 784 * 618 e = d * 917 f = e * 463 g = 4 + 4 h = f - g i = c - h
a ) $ 2000 , b ) $ 2500 , c ) $ 3000 , d ) $ 3600 , e ) $ 4200
c
multiply(multiply(subtract(4, 3), 500), 3)
a sum of money is to be distributed among a , b , c , d in the proportion of 6 : 2 : 4 : 3 . if c gets $ 500 more than d , what is d ' s share ?
"let the shares of a , b , c and d be 6 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 a ' s share = 6 x = 6 * $ 500 = $ 3000 the answer is c ."
a = 4 - 3 b = a * 500 c = b * 3
a ) 63 , b ) 37 , c ) 54 , d ) 99 , e ) 01
a
divide(multiply(300, subtract(const_100, add(25, 54))), const_100)
in an examination , 300 students appeared . out of these students ; 25 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed .
"the number of students with first division = 25 % of 300 = 25 / 100 × 300 = 8400 / 100 = 75 and , the number of students with second division = 54 % of 300 = 54 / 100 × 300 = 16200 / 100 = 162 therefore , the number of students who just passed = 300 – ( 75 + 162 ) = 63 answer : a"
a = 25 + 54 b = 100 - a c = 300 * b d = c / 100
a ) 8 min . , b ) 10 min . , c ) 12 min . , d ) 15 min . , e ) 17 min .
c
divide(9, subtract(inverse(divide(4, 7)), const_1))
walking at 4 / 7 of its usual speed , a train is 9 minutes late . find its usual time to cover the journey .
explanation : solution : new speed = 4 / 7 of the usual speed . let usual time be x . . ' . new time taken = 7 / 4 of usual time = 7 x / 4 so , ( 7 x / 4 ) - x = 9 min . = > 3 x = 36 = > x = 12 min . answer : c
a = 4 / 7 b = 1/(a) c = b - 1 d = 9 / c
a ) 145646 , b ) 236578 , c ) 645353 , d ) 456546 , e ) 220025
e
add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 25)
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 25 as remainder . find the no . is ?
"( 555 + 445 ) * 2 * 110 + 25 = 220000 + 25 = 220025 e"
a = 555 + 445 b = a * 2 c = 555 - 445 d = b * c e = d + 25
a ) 3999 , b ) 10000 , c ) 2500 , d ) 2772 , e ) 2912
b
divide(multiply(800, const_100), 2)
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 800 when the rate to profit rises from 5 to 7 percent . find the capital of b ?
"a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 800 b capital = 60000 * 1 / 6 = 10000 . answer : b"
a = 800 * 100 b = a / 2
a ) 11 , b ) 17 , c ) 11 , d ) 19 , e ) 99
a
subtract(divide(32, const_2), 5)
a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 32 now , the son age 5 years back was
let the son ' s present age be x years . then , ( 32 - x ) = x x = 16 . son ' s age 5 years back = ( 16 - 5 ) = 11 years answer : a
a = 32 / 2 b = a - 5
a ) 69 miles , b ) 84 miles , c ) 96 miles , d ) 100 miles , e ) 112 miles
b
divide(multiply(multiply(subtract(20, 8), add(20, 8)), 10), add(add(20, 8), subtract(20, 8)))
the current in a river is 8 mph . a boat can travel 20 mph in still water . how far up the river can the boat travel if the round trip is to take 10 hours ?
"upstream speed = 20 - 8 = 12 mph downstream speed = 20 + 8 = 28 mph d / 12 + d / 28 = 10 hours solving for d we get d = 84 answer : b"
a = 20 - 8 b = 20 + 8 c = a * b d = c * 10 e = 20 + 8 f = 20 - 8 g = e + f h = d / g
a ) 34 , b ) 36 , c ) 32 , d ) 40 , e ) 44
c
divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 9))
to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 9 hours a day ?
"man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 9 = 180 therefore number of days = 5760 / 180 = 32 days ans = c"
a = 30 * 12 b = a * 8 c = b * 2 d = 20 * 9 e = c / d
a ) 0 , b ) 12 , c ) 32 , d ) 20 , e ) 25
c
divide(multiply(12, 56), 21)
in a division sum , the remainder is 0 . as student mistook the divisor by 12 instead of 21 and obtained 56 as quotient . what is the correct quotient ?
"12 * 56 = 672 672 % 21 = 32 answer : c"
a = 12 * 56 b = a / 21
a ) 320 , b ) 350 , c ) 360 , d ) 330 , e ) 320
c
multiply(36, 160)
the h . c . f of two numbers is 36 and their l . c . m is 1600 . if one of the numbers is 160 , find the other ?
"explanation : the other number is = 36 * 1600 / 160 = 360 answer : option c"
a = 36 * 160
a ) 33 , b ) 54 , c ) 63 , d ) 17 , e ) 01
c
multiply(subtract(70, 21), divide(90, 70))
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 21 adults have their meal , find the total number of children that can be catered with the remaining food .
"explanation : as there is meal for 70 adults and 21 have their meal , the meal left can be catered to 49 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 49 adults = 63 children hence , the meal can be catered to 63 children . answer : c"
a = 70 - 21 b = 90 / 70 c = a * b
a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 0
d
divide(30, 51)
how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 30 / 51 ?
"there is no certain way to solve 2 unknown with 1 equation . the best way is to look at the question and retrospect the most efficient way . in this question , a and b are only positive integers . so that is a big relief . now , we can start with putting a = 1,2 , . . and so on till the time we are confident about one of the options . so , we start with a = 1 , we get b as - ve . out a = 2 , we get b as 6 . yes ( now ( a , b ) = ( 2,6 ) . we can directly see that ( a , b ) = ( 6,2 ) will also satisfy . so we have 2 possible solutions ) a = 3 , we get b as 3 . yes ( now we have 3 possible solutions ) a = 4 , we get b as fraction . out a = 5 , we get b again as some fraction . out a = 6 already taken . we have a , b options left . c , d , e are out . a is 6 . to have 6 as the answer , we will need one more pair like 2,6 and one more solution where a = b . when a = b , we have only 1 solution = 1 . so , one more solution , where a = b is not possible . so , answer will be d ."
a = 30 / 51
a ) s . 17 , b ) s . 18 , c ) s . 12 , d ) s . 11 , e ) s . 10
c
divide(add(520, 248), add(42, 22))
rahim bought 42 books for rs . 520 from one shop and 22 books for rs . 248 from another . what is the average price he paid per book ?
"average price per book = ( 520 + 248 ) / ( 42 + 22 ) = 768 / 64 = rs . 12 answer : c"
a = 520 + 248 b = 42 + 22 c = a / b
a ) 120 metres , b ) 180 metres , c ) 260 metres , d ) 150 metres , e ) 100 meters
c
multiply(divide(multiply(52, const_1000), const_3600), 18)
a train running at the speed of 52 km / hr crosses a pole in 18 seconds . what is the length of the train ?
"speed = 52 x 5 / 18 m / sec = 130 / 9 m / sec . length of the train = ( speed x time ) . length of the train = 130 / 9 x 18 m = 260 m . answer : c"
a = 52 * 1000 b = a / 3600 c = b * 18
a ) 20000 , b ) 10000 , c ) 25000 , d ) 34000 , e ) 30000
c
multiply(45000, subtract(const_1, divide(multiply(30000, multiply(2, multiply(2, const_3))), add(multiply(45000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(30000, multiply(2, multiply(2, const_3)))))))
tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 45000 after completion of one year . what will be jose ' s share of profit ?
"sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 45000 = 25000 c"
a = 2 * 3 b = 2 * a c = 30000 * b d = 2 * 3 e = 2 * d f = e - 2 g = 45000 * f h = 2 * 3 i = 2 * h j = 30000 * i k = g + j l = c / k m = 1 - l n = 45000 * m
a ) 2750 , b ) 2850 , c ) 2950 , d ) 3050 , e ) 3150
a
multiply(divide(multiply(300, const_1000), const_3600), 33)
a train running at the speed of 300 km / hr crosses a pole in 33 seconds . what is the length of the train ?
"speed = ( 300 x ( 5 / 18 ) m / sec = ( 250 / 3 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 250 / 3 ) x 33 ) m = 2750 m a"
a = 300 * 1000 b = a / 3600 c = b * 33
a ) 25 % , b ) 19.23 % , c ) 40 % , d ) 75 % , e ) none of these
b
divide(multiply(105, 40), multiply(105, divide(add(multiply(const_2, const_100), 5), const_100)))
a rectangle having length 105 cm and width 40 cm . if the length of the rectangle is increased by 20 5 percent then how much percent the breadth should be decreased so as to maintain the same area .
explanation : solution : ( 25 / ( 105 + 25 ) * 100 ) % = 19.23 % answer : b
a = 105 * 40 b = 2 * 100 c = b + 5 d = c / 100 e = 105 * d f = a / e
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6
d
multiply(6, 2)
if n divided by 7 has a remainder of 2 , what is the remainder when 6 times n is divided by 7 ?
"as per question = > n = 7 p + 2 for some integer p hence 6 n = > 42 q + 12 = > remainder = > 12 for some integer q alternatively = > n = 2 > 6 n = > 12 = > 12 divided by 7 will leave a remainder 5 hence d"
a = 6 * 2
a ) 540 , b ) 528 , c ) 530 , d ) 520 , e ) 510
b
multiply(divide(880, const_100), subtract(const_100, 40))
if 40 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ?
"we know 37.5 % people study biology , therefore the no of people not studying = 100 - 40 = 60 % > therefore the people not studying biology out of a total 880 people are = 60 % of 880 > ( 60 / 100 ) * 880 = 528 people b"
a = 880 / 100 b = 100 - 40 c = a * b
a ) 1008 , b ) 1011 , c ) 1022 , d ) 1032 , e ) 1043
b
add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 3)
the smallest number which when diminished by 3 , is divisible by 12 , 16 , 18 , 21 and 28 is
"required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 3 = 1008 + 3 = 1011 answer : b"
a = 3 ** 2 b = 2 ** 0 c = a * b d = 3 + 4 e = c * d f = e + 3
a ) a ) 3820 , b ) b ) 930 , c ) c ) 9309 , d ) d ) 3600 , e ) e ) 8302
d
multiply(multiply(subtract(add(80, 50), 10), 10), 3)
a rectangular lawn of dimensions 80 m * 50 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ?
"explanation : area = ( l + b – d ) d ( 80 + 50 – 10 ) 10 = > 1200 m 2 1200 * 3 = rs . 3600 answer : option d"
a = 80 + 50 b = a - 10 c = b * 10 d = c * 3
a ) 7 : 3 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 1 : 1
e
divide(subtract(15.8, 15.2), subtract(16.4, 15.8))
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.2 years , the ratio of the number of boys to the number of girls in the class is
"explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.2 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.2 ) < = > k = 0.6 / 0.6 = 1 / 1 . required ratio = 1 / 1 : 1 = 1 : 1 . answer : e"
a = 15 - 8 b = 16 - 4 c = a / b
a ) 45 % , b ) 50 % , c ) 57.14 % , d ) 60 % , e ) 65.12 %
c
multiply(divide(4000, add(add(1000, 2000), 4000)), const_100)
3 candidates in an election and received 1000 , 2000 and 4000 votes respectively . what % of the total votes did the winning candidate got in that election ?
"total number of votes polled = ( 1000 + 2000 + 4000 ) = 7000 so , required percentage = 4000 / 7000 * 100 = 57.14 % c"
a = 1000 + 2000 b = a + 4000 c = 4000 / b d = c * 100
a ) 5 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 12 %
a
subtract(multiply(divide(subtract(37, 20), subtract(const_100, 20)), const_100), const_10)
two numbers are less than third number by 20 % and 37 % respectively . how much percent is the second number less than by the first
"let the third number is x . then first number = ( 100 - 20 ) % of x = 80 % of x = 80 x / 100 second number is ( 63 x / 100 ) difference = 80 x / 100 - 63 x / 100 = 17 x / 100 so required percentage is , difference is what percent of first number ( 17 x / 100 * 100 / 80 x * 100 ) % = 5 % answer : a"
a = 37 - 20 b = 100 - 20 c = a / b d = c * 100 e = d - 10
a ) 600 , b ) 887 , c ) 256 , d ) 654 , e ) 725
e
add(500, multiply(500, divide(45, const_100)))
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 45 % ?
"cost price = rs . 500 profit = 45 % of 500 = rs . 225 selling price = cost price + profit = 500 + 225 = 725 answer : e"
a = 45 / 100 b = 500 * a c = 500 + b
a ) $ 100 , b ) $ 120 , c ) $ 140 , d ) $ 160 , e ) $ 180
b
divide(20, subtract(divide(5, 3), divide(3, 2)))
the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 3 : 2 . if the oil bill for february had been $ 20 more , the corresponding ratio would have been 5 : 3 . how much was the oil bill for january ?
"3 : 2 = 9 : 6 and 5 : 3 = 10 : 6 . an increase in $ 20 increases the ratio by 1 : 6 . therefore , january ' s bill was 6 ( $ 20 ) = $ 120 . the answer is b ."
a = 5 / 3 b = 3 / 2 c = a - b d = 20 / c
a ) $ 7000 , b ) $ 9000 , c ) $ 11,000 , d ) $ 13,000 , e ) $ 15,000
b
divide(add(divide(subtract(240, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(240, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000)
salesperson a ' s compensation for any week is $ 240 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ?
"240 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 180 x = $ 9,000 the answer is b ."
a = 6 / 100 b = a * 1 c = 240 - b d = 8 / 100 e = 6 / 100 f = d - e g = c / f h = 6 / 100 i = h * 1 j = 240 - i k = 8 / 100 l = 6 / 100 m = k - l n = j / m o = g + n p = o / 1
a ) a ) 6 , b ) b ) 10 , c ) c ) 9 , d ) d ) 15 , e ) e ) 23.5
c
add(20, 5)
a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 1300 . how many workers were part of the latter production run ?
"the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1300 . . thus for the current period = 1300 - 500 = 800 ( 5 n + 45 ) k = 800 k = 800 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 800 / 5 n + 45 500 ( 5 n + 45 ) = 800 ( 5 n + 20 ) 25 n + 225 = 40 n + 160 - 15 n = - 65 n = 4 thus n + 5 = 9 hence c"
a = 20 + 5
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60
d
divide(rectangle_area(15, 20), 6)
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 15 inches by 20 inches and jordan ' s rectangle is 6 inches long , how wide is jordan ' s rectangle , in inches ?
"area of first rectangle is 15 * 20 = 300 hence area of second would be 6 x = 300 x x = 50 answer is d"
a = rectangle_area / (
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
c
subtract(multiply(10, 2.00), multiply(1.60, 10))
martin bought 10 concert tickets , some at the full price of $ 2.00 per ticket , and some at a discounted price of $ 1.60 per ticket . if he spent a total of $ 18.00 , how many discounted tickets did he buy ?
"let x be the number of tickets he bought at $ 2 per ticket . then 2 x + ( 10 - x ) 1.6 = 18 0.4 x = 2 = > x = 5 discounted tickets = 10 - x = 5 ans : c"
a = 10 * 2 b = 1 * 60 c = a - b
a ) 15 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 %
d
add(multiply(divide(12, 48), const_100), 15)
the purchase price of an article is $ 48 . in order to include 15 % of cost for overhead and to provide $ 12 of net profit , the markup should be
"cost price of article = 48 $ % of overhead cost = 15 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 15 = 40 % answer d"
a = 12 / 48 b = a * 100 c = b + 15
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9
e
divide(divide(multiply(multiply(12, 8), 10), 13.33), 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 13.33 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 13.33 : 8 : : 12 : x = > 8 x 13.33 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 13.33 ) = > x = 9 answer : e"
a = 12 * 8 b = a * 10 c = b / 13 d = c / 8
a ) 7 / 30 , b ) 7 / 21 , c ) 2 / 5 , d ) 12 / 29 , e ) 4 / 7
c
divide(multiply(divide(4, 7), multiply(7, 3)), add(add(add(multiply(const_2, 7), multiply(divide(4, 7), multiply(7, 3))), multiply(3, 1)), 1))
papaya , a popular soft drink , contains only 4 ingredients . soda water comprise 4 / 7 of papaya and natural lemon juice makes 1 / 3 of papaya . the amounts of sugar and papaya puree in papaya are equal . due to a malfunction , the mixing machine mixes double the regular amount of lemon juice and 3 times the regular amount of papaya puree . if no other changes were made to the relative quantities of the ingredients , what is the fractional portion of soda water in the drink that comes out of the malfunctioning machine ?
soda water = 12 / 21 natural lemon juice = 7 / 21 the quantities of sugar and papaya puree are equal . total units are 21 of which 12 + 7 = 19 are used up by above two . remaining is 2 which is equally divided between sugar and papaya puree : sugar = 1 / 21 papaya puree = 1 / 21 thus ratio of soda : lemon : sugar : puree = 12 : 7 : 1 : 1 now , lemon is doubled , puree is three times and remaining are the same . resultant ratio is = 12 : 14 : 1 : 3 ratio of soda water in the resultant mixture = 12 / ( 12 + 14 + 1 + 3 ) = 12 / 30 = 2 / 5 answer : c
a = 4 / 7 b = 7 * 3 c = a * b d = 2 * 7 e = 4 / 7 f = 7 * 3 g = e * f h = d + g i = 3 * 1 j = h + i k = j + 1 l = c / k
a ) 13.9 sec , b ) 12.3 sec , c ) 17.9 sec , d ) 61.9 sec , e ) 47.98 sec
b
divide(add(130, 140), multiply(80, const_0_2778))
how long does a train 130 m long running at the speed of 80 km / hr takes to cross a bridge 140 m length ?
"speed = 80 * 5 / 18 = 22 m / sec total distance covered = 130 + 140 = 270 m . required time = 270 / 22 = 12.3 sec . answer : b"
a = 130 + 140 b = 80 * const_0_2778 c = a / b
a ) 14000 , b ) 15000 , c ) 16000 , d ) 12000 , e ) 17000
d
multiply(36000, divide(9, add(add(divide(multiply(5, 8), 4), 8), 9)))
ramesh , xyz and rajeev put a partnership . profit is 36000 , if ramesh and xyz ratio is 5 : 4 and xyz and rajeev 8 : 9 . find rajeev ' s share .
ramesh : xyz = 5 : 4 xyz : rajeev = 8 : 9 then ramesh : xyz / rajeev : xyz = 5 : 4 / 9 : 8 ramesh / rajeev = 10 / 9 therefore ramesh : xyz : rajeev = 10 : 8 : 9 then rajeev share = ( 9 / 27 ) * 36000 = 12000 answer : d
a = 5 * 8 b = a / 4 c = b + 8 d = c + 9 e = 9 / d f = 36000 * e
a ) 20 % , b ) 43 % , c ) 99.33 % , d ) 75.109 % , e ) 47 %
b
subtract(multiply(divide(add(const_100, 30), const_100), add(const_100, 10)), const_100)
a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 10 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 30 % of that is added to the weight , the weight claimed by the trader is obtained . if he charges the cost price of the weight that he claims , find his profit percentage .
anyways , one can infer that he ' steals ' 10 % from suppliers and then charges 30 % extra to customers so basically 1.1 * 1.3 = 1.43 given that 1 is start point , we get 43 % more hence answer is b
a = 100 + 30 b = a / 100 c = 100 + 10 d = b * c e = d - 100
a ) 3.5 % , b ) 2.4 % , c ) 8.3 % , d ) 5 % , e ) 2.6 %
c
subtract(subtract(14, 5), divide(multiply(14, 5), const_100))
in measuring the sides of a rectangle , one side is taken 14 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements .
"say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . now , the area obtained with wrong measurements would be 114 * 95 = 10,830 , which is 8.3 % greater than the actual area . answer : c ."
a = 14 - 5 b = 14 * 5 c = b / 100 d = a - c
a ) 1040 , b ) 1170 , c ) 1150 , d ) 4160 , e ) none of these
b
multiply(divide(multiply(5200, 45), const_100), divide(subtract(const_100, 50), const_100))
in an office , totally there are 5200 employees and 45 % of the total employees are males . 50 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ?
"number of male employees = 5200 * 45 / 100 = 2340 required number of male employees who are less than 50 years old = 4160 * ( 100 - 50 ) % = 2340 * 50 / 100 = 1170 . answer : b"
a = 5200 * 45 b = a / 100 c = 100 - 50 d = c / 100 e = b * d
a ) 345 , b ) 384 , c ) 378 , d ) 398 , e ) 300
b
divide(multiply(multiply(12, 16), 6), 3)
a lady builds 12 cm length , 16 cm width , and 6 cm height box using 3 cubic cm cubes . what is the minimum number of cubes required to build the box ?
"number of cubes required = volume of box / volume of cube = 12 * 16 * 6 / 3 = 384 cubes answer : b"
a = 12 * 16 b = a * 6 c = b / 3
a ) $ 400 , b ) $ 450 , c ) $ 500 , d ) $ 550 , e ) $ 600
c
multiply(power(10, 1), 10)
oil cans x and y are right circular cylinders and the height and radius of y are each 10 times those of x . if the oil in can x filled to capacity sells for $ 1 , how much does the oil in y sell for if y is only half filled ?
"formula for vol of a cyl is pi * r ^ 2 * h so vy = 1000 * vy y when half filled will cost 500 times x so ans is c"
a = 10 ** 1 b = a * 10
a ) 17 , b ) 28 , c ) 24 , d ) 26 , e ) 12
c
divide(121, add(const_2, const_pi))
the perimeter of a semi circle is 121 cm then the radius is ?
"36 / 7 r = 121 = > r = 24 answer : c"
a = 2 + math.pi b = 121 / a
a ) a ) 32 , b ) b ) 38 , c ) c ) 30 , d ) d ) 31 , e ) e ) 33
e
divide(subtract(100, const_1), 3)
if you write down all the numbers from 1 to 100 , then how many multiples of 3 ?
division rule for 3 - sum of all digits is multiple of 3 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99 ans - e
a = 100 - 1 b = a / 3
a ) q = 21 , b ) q = 22 , c ) q = 23 , d ) 24 , e ) 27
c
subtract(power(5, 2), 2)
if x + ( 1 / x ) = 5 , what is the value q of x ^ 2 + ( 1 / x ) ^ 2 ?
squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c
a = 5 ** 2 b = a - 2