Number System - Aptitude MCQ Questions and Solutions with Explanations

61

One of the factors of (8^2k + 5^2k), where k is an odd number, is:

A) 88 B) 84 C) 89 D) 86

Discuss

Answer & Solution

Answer: Option C

Solution:

8^2k + 5^2k
If K is odd number
Then, K = 1, 3, 5 _ _ _ _
8^{2 × 1} + 5^{2 × 1}
= 64 + 25
= 89

62

The numerator of a fraction is multiple of two numbers. One of the numbers is greater than the other by 2. The greater number is smaller than the denominator by 4. If the denominator 7 + C (C > - 7) is a constant, then the minimum value of the fraction is:

A) 5 B) 1/5 C) -5 D) -1/5

Discuss

Answer & Solution

Answer: Option D

Solution:

According to the given condition
Denominator = 7 + C [C > -7]
\[\begin{array}{*{20}{c}} {{\text{Denominator}} = }&{{\text{Greator number}}}&{}&{{\text{Smaller number}}} \\ {}&{\left( {7 + C - 4} \right)}& \times &{\left( {7 + C - 4 - 2} \right)} \end{array}\]
Equation $$ = \frac{{\left( {3 + C} \right)\left( {1 + C} \right)}}{{7 + C}}$$
By putting value according to option
C = -2 for minimum value
= -1/5

63

A two-digit number, 9A, is a prime number. Find A.

A) 9 B) 7 C) 3 D) 1

Discuss

Answer & Solution

Answer: Option B

Solution:

9A is prime number then by option, 9A
Option A: A = 9, Number = 99 → Composite
Option B: A = 7, Number = 97 → Prime
Option C: A = 3, Number = 93 → Composite
Option D: A = 1, Number = 91 → Composite

64

What is the value of x so that the seven digit number 8439x53 is divisible by 99?

A) 9 B) 4 C) 3 D) 6

Discuss

Answer & Solution

Answer: Option B

Solution:

8 + 4 + 3 + 9 + x + 5 + 3 = 9
or multiple of x + 5 = 9
x = 4

65

The sum of the digits of a two digit number is $$\frac{1}{7}$$ of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:

A) 4 B) 6 C) 1 D) 5

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & x + y = \frac{1}{7}\left( {10x + y} \right) \cr & 7x + 7y = 10x + y \cr & 3x = 6y \cr & x:y = 2:1 \cr & 1{\text{ unit}} \to {\text{4}} \cr & {\text{Number will be}} = 84 \cr & \frac{{48}}{7}R \to 6 \cr} $$

66

Two positive whole numbers are such that the sum of the first and twice the second number is 8 and their difference is 2. The numbers are:

A) 7, 5 B) 3, 5 C) 6, 4 D) 4, 2

Discuss

Answer & Solution

Answer: Option D

Solution:

In such type of question always go through option to save your valuable time.
By option D: 4, 2
4 + 2 × 2 = 8
4 - 2 = 2 (Satisfied)

67

A and B are positive integers. If A + B + AB = 65, then what is the difference between A and B (A, B ≤ 15)?

A) 3 B) 4 C) 5 D) 6

Discuss

Answer & Solution

Answer: Option C

Solution:

A + B + AB = 65
(A, B ≤ 15)
Assume value of A & B
A = 10
B = 5
10 + 5 + 10 × 5 = 65
65 = 65 value satisfied
Difference between (A - B) = 10 - 5 = 5

68

How many times the keys of a typewriter have to be pressed in order to write number from 121 to 1346?

A) 3675 B) 4018 C) 4021 D) 4025

Discuss

Answer & Solution

Answer: Option D

Solution:

121 to 999 - 3 digit
Total number of terms = 999 - 121 + 1 = 878 + 1 = 879
Total keys from 121 to 999 = 879 × 3 = 2637
1000 to 1346 - 4 digit
Total number of terms = 1346 - 1000 + 1 = 347
Total keys from 1000 to 1346 = 347 × 4 = 1388
Total keys = 2637 + 1388 = 4025

69

Among the following statements, the statement which is not correct is:

A) Every natural number is a real number B) Every real number is a rational number C) Every integer is a rational number D) Every natural number is an integer

Discuss

Answer & Solution

Answer: Option B

Solution:

Every real number is a rational number.

70

If 1³ + 2³ . . . . . . + 10³ = 3025, then the value of 2³ + 4³ . . . . . . + 20³ is:

A) 7590 B) 5060 C) 24200 D) 12100

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \because {1^3} + {2^3} + {3^3}\,......\, + {10^3} = 3025 \cr & {\text{To find}} = {2^3} + {4^3} + {6^3}\,......\, + {20^3} = ? \cr & \Rightarrow {2^3}\left( {{1^3} + {2^3} + {3^3}\,......\, + {{10}^3}} \right) \cr & \left[ {{{\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)}^2} = {{\left( {\frac{{10\left( {11} \right)}}{2}} \right)}^2} = {{\left( {55} \right)}^2} = 3025} \right] \cr & \Rightarrow 8 \times 3025 \cr & \Rightarrow 24200 \cr} $$