MCQ Questions for Class 10 Maths Real Numbers Quiz 7

Choose the correct alternative answer for the question given below.
Decimal expansion of which of the following is non-terminating recurring?

0%
$\dfrac{2}{5}$
0%
$\dfrac{3}{16}$
0%
$\dfrac{3}{11}$
0%
$\dfrac{137}{25}$

Explanation

1) Since,

$5 = 2^0 \times 5^1$ , which is in the form of

$2^m \times 5^n$ , where m and n are no - negative integers.
So, the decimal expansion of

$\dfrac{2}{5}$ is terminating.
2) Since,

$16 = 2^4 \times 5^0$ , which in the form of

$2^m \times 5^n$ , where m and n are no - negative integers.
So, the decimal expansion of

$\dfrac{3}{16}$ is terminating.
3) Since,

$11 = 2^0 \times 5^0 \times 11^1$ , which is not in the form of

$2^m \times 5^n$ , where m and n are non - negative integers.
So, the decimal expansion of

$\dfrac{3}{11}$ is non - terminating recurring.
4) Since,

$25 = 2^0 x 5^2$ , which is in the form of

$2^m \times 5^n$ , where m and n are non - negative integers.
So, the decimal expansion of

$\dfrac{137}{25}$ is terminating.

If a rational number

$x = \dfrac{p}{q}$ such as a prime factor of

$q$ is not the form of

$2^{n} 5^{m}$ where

$n , m$ is a zero integer, then decimal expansion of

$x$ is

0%
terminating decimal expansion
0%
non-terminating repeating decimal
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choice (A) and (B) both correct
0%
none of these

Explanation

A rational number

$x$ expressed as

$x = \dfrac{p}{q}$ where the prime factorization of

$q$ in the form of

$2^n5^m$ is having the terminating decimal expansion.

So, if a rational number

$x = \dfrac{p}{q}$ such as a prime factor of

$q$ is not the form of

$2^{n} 5^{m}$ where

$n , m$ is a zero integer, then decimal expansion of

$x$ is non terminating repeating decimal

In the decimal expansion of rational number

$\dfrac{43}{{2}^{2} \times {5}^{3}}$ , after how many digits decimal will end ?

0%
$1$
0%
$2$
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$3$
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$4$

Explanation

Given,

$\dfrac{43}{2^2 \times 5^3}$ .
Multiply by

$2$ in numerator and denominator of number,

$\dfrac{43}{2^2 \times 5^3}$

$=\dfrac{43\times2}{2^3 \times 5^3} = \dfrac{86}{10^3} = 0.086$ .

Therefore, after three digits of decimal, the decimal expansion of the given rational number will terminate.

Decimal expansion of number

$\dfrac{441}{2^{2} \times 5^{7} \times 7^{2}}$ will be:

0%
Terminating
0%
Non-terminating repeating
0%
Terminating and non-terminating both
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Non-rational

Explanation

$\textbf{Step 1 : Simplify the given fraction. }$

$\text{The given fraction is :}$

$\dfrac{441}{2^2\times5^7\times7^2}$ .

$\text{It can be reduced to}$

$\dfrac{9}{2^2\times5^7}$

$\text{Since the new denominator is of the form}$

$2^{m} \times 5^{n}$ ,

$\text{the given number will have a terminating decimal expansion.}$

$\textbf{Hence, correct option is A.}$

HCF of three numbers

$17 , 23$ and

$29$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Given numbers are

$17 , 23$ and

$29$ which are all prime numbers.
It is known that HCF of all prime numbers is

$1$ cause the only common factor in prime numbers is

$1$
So, the correct choice is (A).

Which of the following is irrational?

0%
$0.54$
0%
$6.\overline{54}$
0%
$0.\overline{1416}$
0%
$0.4014001400014...$

Explanation

${\textbf{Step - 1: Checking option - A}}{\text{.}}$

${\text{We have 0}}{\text{.54 in the first option}}{\text{.}}$

${\text{0}}{\text{.54 is clearly a rational number as it can be written in the form }}\dfrac{p}{q}{\text{ as }}\dfrac{{54}}{{100.}}$

${\textbf{Step - 2: Checking option - B}}{\text{.}}$

${\text{We have }}6.\overline{54}{\text{ in the second option}}{\text{.}}$

${\text{As 6}}.\overline{54}{\text{ is a non - terminating recurring decimal, and as we know that }}$

${\text{non - terminating recurring decimal are rational, therefore, it is also a rational number}}{\text{.}}$

${\textbf{Step - 3: Checking option - C}}{\text{.}}$

${\text{Here we have }}0.\overline {1416}{\text{ , so this is also a non - terminating recurring decimal}}{\text{.}}$

${\text{Therefore, it also a rational number}}{\text{.}}$

${\textbf{Step - 4: Checking option - D}}{\text{.}}$

${\text{Here we have }}0.4014001400014...{\text{ which we can see is a non - terminating,}}$

${\text{non - recurring decimal}}{\text{. Therefore, it is an irrational number}}{\text{.}}$

${\textbf{Hennce, The decimal in option - D is an irrational number}}{\text{.}}$

$p$ is a positive prime number then

$\sqrt{p}$

0%
irrational number
0%
rational number
0%
integer number
0%
none of these

What is the greatest number which divides

$392,486$ and

$627$ so as to leave the same remainder in each case?

0%
$47$
0%
$43$
0%
$37$
0%
$34$

If P = 2

$^3 \times 3^{10} \times 5 and Q = 2^5 \times 3 \times 7,$ then HCF of P and Q is

0%
2 .3 .5. 7
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3 .2 $^3$
0%
2 $^2 . 3^7$
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2 $^5 . 3^{10} . 5 . 7$

The H.C.F of

$2$ and

$3$ by Euclid’s division algorithm

0%
$2$
0%
$3$
0%
$5$
0%
$1$

Explanation

Given numbers are

$a=3$ and

$b=2$

By Euclid's division lemma,

$a=bq+r$

where,

$a$ is dividend

$b$ is divisor

$q$ is a quotient

$r$ is remainder

Step 1-

$a\div b=3\div 2$

Quotient will be 1 and remainder will be 1.

$\therefore 3=\left( 2\times 1 \right) +1$

Step 2-

2 will be dividend and 1 will be divisor now.

$\therefore 2=\left( 1\times 2 \right) +0$

Thus, remainder is zero in this step. Thus, divisor in this step will be HCF.

$\therefore HCF=1$

CBSE Questions for Class 10 Maths Real Numbers Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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