Number System - Aptitude MCQ Questions and Solutions with Explanations

31

The number of factors of 196 which are divisible by 4 is:

A) 228 B) 4 C) 57 D) 3

Discuss

Answer & Solution

Answer: Option D

Solution:

196 = 7² × 2²
Factors ⇒ (1, 7, 49) × (1, 2, 4)
Divisible by 4 = 4, 28, 196 = 3 factors

32

Let x be the least number which when divided by 15,18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to make it a perfect square?

A) 39 B) 37 C) 43 D) 36

Discuss

Answer & Solution

Answer: Option A

Solution:

L.C.M. of 15,18, 20, 27 = 540
x = 540K + 10
Number System mcq question image

x = 540 × 4 + 10 = 2170
46² < 2170 > 47²
47² = 2209
To make perfect square = 2209 - 2170 = 39

33

If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:

A) $$\frac{3}{{52}}$$ B) $$\frac{1}{{52}}$$ C) $$\frac{5}{{52}}$$ D) $$\frac{7}{{52}}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Lets first number}} = a \cr & {\text{Second number}} = b \cr & a + b = 65 \cr & \sqrt {a \times b} = 26 \cr & ab = 676 \cr & {\text{Sum of reciprocal}} = \frac{1}{a} + \frac{1}{b} \cr & = \frac{{b + a}}{{ab}} \cr & = \frac{{65}}{{676}} \cr & = \frac{5}{{52}} \cr} $$

34

The sum of three consecutive natural numbers is always divisible by . . . . . . . .?

A) 3 B) 9 C) 15 D) 21

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the number be x, x + 1, x + 2
Sum = x + x + 1 + x + 2
= 3x + 3
= 3(x + 1)
It is always divisible by 3

35

Three numbers are in Arithmetic progression (AP) whose sum is 30 and the product is 910. Then the greatest number in the AP is:

A) 17 B) 15 C) 13 D) 10

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the three number is a - d, a, a + d
a is first term, d is common difference
a + d + a + a - d = 30 (Given)
3a = 30
a = 10
(a + d)(a)(a - d) = 910
(10 + d)(10)(10 - d) = 91 × 10
(10 + d)(10 - d) = 91
Put d = 3
$$\boxed{\left( {10 + 3} \right)\left( {10 - 3} \right) = 91}$$
So, d = 3
So, greater number is = a + b = 10 + 3 = 13

36

What is the value of $$99\frac{{11}}{{99}} + 99\frac{{13}}{{99}} + 99\frac{{15}}{{99}} + ........ + 99\frac{{67}}{{99}}?$$

A) $$\frac{{95120}}{{33}}$$ B) $$\frac{{96220}}{{33}}$$ C) $$\frac{{94220}}{{33}}$$ D) $$\frac{{97120}}{{33}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & 99\frac{{11}}{{99}} + 99\frac{{13}}{{99}} + 99\frac{{15}}{{99}} + ........ + 99\frac{{67}}{{99}} \cr & {\text{Number of term}} = \frac{{67 - 11}}{2} + 1 = 29 \cr & = 99 \times 29 + \frac{1}{{99}}\left( {11 + 13 + 15 + ........ + 67} \right) \cr & = 2871 + \frac{1}{{99}}\left[ {\frac{{29}}{2}\left( {2 \times 11 + \left( {29 - 1} \right)2} \right)} \right] \cr & = 2871 + \frac{1}{{99}}\left[ {\frac{{99}}{2}\left( {22 + 56} \right)} \right] \cr & = 2871 + \frac{1}{{99}}\left( {29 \times 39} \right) \cr & = 2871 + \frac{{1131}}{{99}} \cr & = 2871 + \frac{{377}}{{33}} \cr & = \frac{{95120}}{{33}} \cr} $$

37

The digit at Hundred's place value of 17! is:

A) 1 B) 0 C) 2 D) 3

Discuss

Answer & Solution

Answer: Option B

Solution:

We know that after 5! we get one zero at the end of the number.
We can say that in 17! we get minimum three zeros.
The hundred's place value od 17! = 0

38

Ifa number K = 42 × 25 × 54 × 135 is divisible by 3^a, then find the maximum value of a

A) 6 B) 7 C) 4 D) 5

Discuss

Answer & Solution

Answer: Option B

Solution:

K = 42 × 25 × 54 × 135
K = 3 × 14 × 25 × 3³ × 2 × 3³ × 5
= 3⁷ × 14 × 25 × 2 × 5
= It is divided by 3⁷, which is equal to 3^a (given)
⇒ So, 3⁷ = 3^a
a = 7

39

Select the correct option:
Convert decimal 99 to binary.

A) 1100101 B) 1101001 C) 11100011 D) 1100011

Discuss

Answer & Solution

Answer: Option D

Solution:

99 can be written as 64 + 32 + 2 + 1 = 2⁶ + 2⁵ + 2¹ + 2⁰
⇒ 99 in binary is 1100011
∴ 99 in binary is 1100011

40

What is the largest 4 digit number that is exactly divisible by 93?

A) 9961 B) 9971 C) 9981 D) 9951

Discuss

Answer & Solution

Answer: Option D

Solution:

Largest 4 digit number which is exactly divisible by 93
= 9999 - Remainder
= 9999 - 48
= 9951