Practice MCQ Questions and Answer on Number System

If (-1)ⁿ + (-1)⁴ⁿ = 0, then n is

(A) Any positive integer
(B) Any negative integer
(C) Any odd natural number
(D) Any even natural number

What is the minimum number of four digits formed by using the digits 2, 4, 0, 7 ?

(A) 2047
(B) 2247
(C) 2407
(D) 2470

The decimal fraction $2.3 \overset{―}{49}$ ÃÂÃÂ is equal to :

(A) $\frac{2326}{999}$
(B) $\frac{2326}{990}$
(C) $\frac{2347}{999}$
(D) $\frac{2347}{990}$

Solution: $2.3 \overset{―}{49} = \frac{2349 - 23}{990} = \frac{2326}{990}$

If m and n are positive integers, then the digit in the unit's place of 5ⁿ + 6^m is always :

(A) 1
(B) 5
(C) 6
(D) n + m

If n is a whole number greater then 1, then n² (n² - 1) is always divisible by -

(A) 16
(B) 12
(C) 10
(D) 8

Solution: n2(n2 - 1) put n = 2 = 22(22 - 1) = 4(4 - 1) = 4 × 3 = 12 Check the option it is divisible by 12. $\begin{aligned} Take n = 3 \\ \Rightarrow 3^{3} (3^{3} - 1) \\ \Rightarrow 9 (9 - 1) \\ \Rightarrow 9 \times 8 = 72 \\ It is divisible by 12 \\ So 12 is answer \end{aligned}$

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

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

If 17²⁰⁰ is divided by 18, the remainder is -

(A) 17
(B) 16
(C) 1
(D) 2

Solution: $\begin{aligned} = 17^{200} \div 18 \\ = {(18 - 1)}^{200} \div 18 \end{aligned}$ Apply Binomial theorem $= {(18)}^{200} {(- 1)}^{0} + {(18)}^{199} {(- 1)}^{1} +$ $. . . . . + {(18)}^{1} {(- 1)}^{199}$ $+ {(18)}^{0} {(- 1)}^{200}$ $\Rightarrow$ Remainder always comes from last term is (18)0 (- 1)200 $\begin{aligned} = \frac{{(18)}^{0} {(- 1)}^{200}}{18} \\ = 1 \times 1^{200} \\ = 1 \\ So, remainder = 1 \end{aligned}$

What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?

(A) 1108
(B) 1683
(C) 2007
(D) 3363

The sum of all three digit numbers, each of which on divide by 5 leaves remainder 3, is -

(A) 180
(B) 1550
(C) 6995
(D) 99090

Solution: $\begin{aligned} S e r i e s : \\ 103 + 108 + . . . . . . . . + 998 \\ a = 103 \\ d = 5 \\ Last term = 998 \\ Number of term \\ = \frac{998 - 103}{5} + 1 \\ = \frac{895}{5} + 1 \\ = 180 \\ Sum of n terms \\ = \frac{n}{2} [2 a + (n - 1) d] \\ = \frac{180}{2} [2 \times 103 + (180 - 1) 5] \\ = 99090 \end{aligned}$

10.

The product of two positive numbers is 11520 and their quotient is $\frac{9}{5}$ . Find the difference between two numbers.

(A) 60
(B) 64
(C) 74
(D) 70

Solution: Let x and y are two numbers (x > y) According to the question, $\begin{aligned} x y = 11520 \\ \frac{x}{y} = \frac{9}{5} \Rightarrow x = \frac{9 y}{5} \\ put the value of x in eq . (i) \\ \frac{9 y}{5} \times y = 11520 \\ \Rightarrow y^{2} = 6400 \\ \Rightarrow y = 80 \\ 80 x = 11520 \\ \Rightarrow x = 144 \\ x - y = 144 - 80 = 64 Ans . \end{aligned}$

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Practice MCQ Questions and Answer on Number System

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