Aptitude test - Problems on Numbers

1) 36 is divided in 2 parts such that 8 times the first part added to 3 times the second part makes 203. What is the first part?

a) 15
b) 19
c) 23
d) None of these

ANSWER:  19
SOLUTION
Let the first part be x.
8x + 3 (36 – x) = 203
8x + 108 – 3x = 203
5x + 108 = 203
5x = 95
x = 19

2) 6/7 of a certain number is 96. Find quarter of that number.

a) 112
b) 32
c) 56
d) 28

ANSWER:  28
SOLUTION
6/7x = 96
x = 112
y = x/4 = 112/4 = 28

3) 3/4th of a number exceeds its 2/3rd by 8. What is the number?

a) 96
b) 36
c) 72
d) 144

ANSWER:  72
SOLUTION
Let the number be x.
3/4x = 2/3x + 8
Multiply by 12
9x = 8x + 72
x = 72

4) How many numbers between 100 and 600 are divisible by 2, 3, and 7 together?

a) 11
b) 12
c) 14
d) cannot be determined

ANSWER:  12
SOLUTION
As the division is by 2, 3, 7 together, the numbers are to be divisible by: 2*3*7 = 42
The limits are 100 and 600
The first number divisible is 42*3 = 126
To find out the last number divisible by 42 within 600:
600/42 = 14.28
Hence, 42*14 = 588 is the last number divisible by 42 within 600
Hence, total numbers divisible by 2, 3, 7 together are (14 – 2) = 12

5) The sum of two umbers is 73 and their difference is 28. Find the difference between their squares.

a) 101
b) 45
c) 2044
d) cannot be determined

ANSWER:  2044
SOLUTION
Let x and y be the two given numbers
(x + y) = 73; (x – y) = 28
(x² – y²) = (x + y) (x – y) = 73*28 = 2044

6) What is the remainder when (1! + 2! + 3! + … + 50!) is divided by 7?

a) 1
b) 3
c) 5
d) cannot be determined

ANSWER:  5
SOLUTION
As the division is by 7, numbers 7! onwards are eliminated as they are completely divisible by 7.
Therefore, we need to find the remainder when (1!+2!+3!+4!+5!+6!) is divided by 7.
(1!+2!+3!+4!+5!+6!) = (1+2+6+24+120+720) = 873
On division of 873 by 7, the remainder is 5.

7) What is the remainder when 570 is divided by 6?

a) 1
b) 5
c) 4
d) 0

ANSWER:  1
SOLUTION
5² = 25, on division by 6, the remainder is 1.
5³ = 125, on division by 6, the remainder is 5.
5⁴ = 625, on division by 6, the remainder is 1.
5⁵ = 3125, on division by 6, the remainder is 5.
Hence, for 5 ^(2x) on division by 6, the remainder is 1, and for 5 ^(2x-1) on division by 6, the remainder is 5.
Therefore, for 5⁷⁰, on division by 6, the remainder is 1.

8) 68n – 58n, where n > 0 is an integer, is divisible by

a) 6
b) 30
c) 671
d) cannot be determined

ANSWER:  671
SOLUTION
For n = 1,
6⁸ – 5⁸ = (6⁴) ² – (5⁴) ² = (6⁴ + 5⁴) (6⁴ - 5⁴)
= (1921) (671)

9) What is the value of (22 + 42 + 62 + … + 402) – (12 + 32 + 52 + … + 392)?

a) 410
b) 820
c) 800
d) 780

ANSWER:  820
SOLUTION
(2n) ² – (2n-1) ² …. (n = 1 to 20)
= (2n – 2n + 1) (2n + 2n - 1)
= (1) (2n + 2n - 1)
= {x + (x – 1)} …. (put 2n = x)
For n = 1, x = 2
For n = 2, x = 4
= (2 + 1) + (4 + 3) + … + (40 + 39)
= (1 + 2 + 3 + 4 +…+ 39 + 40)
= {n( n+1)}/2
= 820

10) If x, x+2 and x+4 are prime numbers, then find the number of possible solutions for x.

a) 1
b) 2
c) 3
d) 4

ANSWER:  1
SOLUTION
For x=2, the numbers are: 2, 4, 6 … (only first number is a prime number)
For x=3, the numbers are: 3, 5, 7 … (all numbers are prime numbers)
For x=5, the numbers are 5, 7, 9 … (first two numbers are prime numbers)
Hence for all other possibilities of x as a prime number, x, x+2 and x+4 all are not prime numbers.
Therefore there is only one possibility of x.

11) If an operation @ is defined by the equation p@q = 3p +q, find the value of z if 7@z = z@5?

a) 0
b) 6
c) 8
d) 7

ANSWER:  8
SOLUTION
p@q = 3p + q
Using the above analogy,
7@z = 3(7) + z = 21 + z
z@5 = 3(z) + 5 = 3z + 5
21 + z = 3z + 5
2z = 16
z = 8

12) A positive integer which when added to 100, gives a sum which is greater than when it is multiplied by 100. What is the number?

a) 5
b) 2
c) 3
d) 1

ANSWER:  1
SOLUTION
The number is 1

13) a, b, and c are even numbers and x, y, and z are odd numbers. Which of the following relationships cannot be true at any cost?

I. ab/c = xy II. ac/x = yz III. xy/z = ab
a) only II
b) only III
c) only II and III
d) all of these

ANSWER:  only II and III

14) Find the least value of x and y so that the number 5x423y is divisible by 88.

a) 8, 2
b) 9, 4
c) 7, 3
d) 8, 5

ANSWER:  8, 2
SOLUTION
As 88 is an even number, option (c) and (d) are eliminated. Substituting the values of (a) and (b) and solving the answer is (a).

15) Sum of two positive numbers are multiplied by each number separately. The products thus obtained are 510 and 390. What is the difference of the two numbers?

a) 2
b) 5
c) 3
d) 4

ANSWER:  4
SOLUTION

Let the numbers be x, y.
x(x+y) = 510, y(x+y) = 390
x² + xy = 510 …… (1)
y² + xy = 390 …… (2)
Adding (1) and (2)
x² + xy + xy + y² = 510 + 390 = 900
(x+y) ² = 900 = (30) ²
(x+y) = 30
Subtracting (2) from (1)
x² + xy – xy - y² = 510 - 390 = 120
x² - y² = 120
(x+y) (x-y) = 120
(x-y) = 4.