Number System - Aptitude MCQ Questions and Solutions with Explanations

31

Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is -

A) 24 B) 36 C) 72 D) 108

Discuss

Answer & Solution

Answer: Option C

Solution:

No. →	I	II	III
	3x	6x	2x	= 11x

$$\eqalign{ & \frac{{11x}}{3} = 44 \Rightarrow x = 12 \cr & {\text{Largest number = 6x = 72}} \cr} $$

32

A positive integer when divided by 425 gives a remainder 45. When the same number is divided by 17, the remainder will be :

A) 7 B) 8 C) 11 D) 10

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \frac{{{\text{Remainder of number }}}}{{17}} = \frac{{45}}{{17}} \cr & \Rightarrow {\text{Remainder = 11}} \cr} $$

33

The value of $${\text{0}}{\text{.}}\overline 2 + {\text{0}}{\text{.}}\overline 3 + {\text{0}}{\text{.}}\overline {32} $$ is :

A) $$0.\overline {87} $$ B) $$0.\overline {77} $$ C) $$0.\overline {82} $$ D) $$0.\overline {86} $$

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & 0.\overline 2 + 0.\overline 3 + 0.\overline {32} \cr & = \frac{2}{9} + \frac{3}{9} + \frac{{32}}{{99}} \cr & = \frac{{22 + 33 + 32}}{{99}} \cr & = \frac{{87}}{{99}} \cr & = 0.\overline {87} \cr} $$

34

$$0.\overline {142857} \div 0.\overline {285714} \,$$ is equal to :

A) 10 B) 2 C) $$\frac{1}{2}$$ D) $$\frac{1}{3}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & = 0.\overline {142857} \div 0.\overline {285714} \,{\text{ }} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \frac{{142857}}{{285714}} \cr & = \frac{1}{2} \cr} $$

35

A man read $$\frac{2}{5}$$ th of a book on the first day. He read $$\frac{1}{3}$$ rd more on second day than he read in the first day. 15 pages were left for the third day. The number of pages in the book is -

A) 100 B) 105 C) 225 D) 250

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Accroding to question}} \cr & ({\text{P are the pages)}} \cr & \Rightarrow {\text{P}} - \frac{2}{5}{\text{P}} - \frac{8}{{15}}{\text{P}} = 15, \cr & \because \left[ {\frac{8}{{15}} = \frac{2}{5} \times \frac{4}{3}} \right] \cr & \Rightarrow \frac{{15{\text{P}} - {\text{6P}} - 8{\text{P}}}}{{15}} = 15 \cr & \Rightarrow {\text{P}} = 225{\text{ pages}} \cr} $$

36

Unit digit of the number (22)²³ is :

A) 4 B) 6 C) 7 D) 8

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\left( {22} \right)^{23}} \cr & {\text{ Result unit digit}} \cr & {2^1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,2\, \leftarrow | \cr & {2^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,|\,\,{\text{Cycle}} \cr & {2^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,|\,\,\,{\text{Completes}} \cr & {2^4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,\,\,\,6 \leftarrow \,| \cr & {2^5}\,\,\,\,\,\,\,\,\,\,\,\,\,32\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\, \cr & {\text{So divided power of 22 by 4 }} \cr & \frac{{23}}{4} = {\text{ remainder 3}} \cr & {2^3}\,\,\,\, = \,\,\,\,\,8\, \cr & {\text{unit digit = 8}} \cr} $$

37

A number consists of two digits. If the number formed by interchange the digits is added to the original number, the resulting number (i.e., the sum) must be divisible by -

A) 11 B) 9 C) 5 D) 3

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the unit digit of the number be x
And ten's place number be y
Number = 10y + x
Interchange digit's = 10x + y
Adding result
= 10y + x + 10x + y
= 11x + 11y
= 11(x + y)
So number is divisible by 11

38

The unit digit in the sum of $${\left( {124} \right)^{372}}\,$$ $$ + \,{\left( {124} \right)^{373}}$$ = ?

A) 5 B) 4 C) 2 D) 0

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\left( {124} \right)^{372}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {124} \right)^{373}} \cr & \,\,\,\mathop {{4^{372}}}\limits^ \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {{4^{373}}}\limits^ \downarrow \cr} $$
When 4 has odd power then unit digit is 4, when 4 has even power then unit digit is 6

4¹	→	4	→	4
4²	→	16	→	6
4³	→	64	→	4
4⁴	→	256	→	6
4³⁷²				4³⁷³
↓				↓
6		+		4	= 10 = 0

Last (unit) digit = 0

39

The number of integers in between 100 and 600, which are divisible by 4 and 6 both, is :

A) 40 B) 42 C) 41 D) 50

Discuss

Answer & Solution

Answer: Option C

Solution:

Take L.C.M. of and 6 = 12 and numbers should be divisible by 12.
First number (after 100) = 108
Last of integers
$$\eqalign{ & = \frac{{{\text{ Last term }} - {\text{ First term}}}}{{{\text{Difference}}}} + 1 \cr & = \frac{{588 - 108}}{{12}} + 1 \cr & = \frac{{480}}{{12}} + 1 \cr & = 40 + 1 \cr & = 41 \cr} $$

40

Find the least number of five digits which is divisible by 333 ?

A) 10328 B) 10323 C) 10333 D) 10332

Discuss

Answer & Solution

Answer: Option B

Solution:

Five digit least value = 10000
10000 is divided by 333
therefore, remainder = 10
Number = 10000 + 333 - 10 = 10323