The identity for addition of rational numbers.

The identity for subtraction of rational numbers.

The identity for division of rational numbers.

MCQ Questions for Class 8 Maths Rational Numbers Quiz 8

Study the following statements.
Statement - 1 : Rational numbers are always closed under division.
Statement - 2 : Division by zero is not defined.
Which of the following options hold?

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Both statement - 1 and statement - 2 are true.
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Statement - 1 is true but statement -2 is false.
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Statement - 1 is false but statement -2 is true.
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Both statement - 1 and statement - 2 are false.

Explanation

Statement

$1:$ Rational number can even be simply integers which can be further represented as

$\cfrac {p}{q}$ form. So, statement

$1$ is false.

Statement

$2:$ Any number divided by

$0$ is not defined. So, statement

$2$ is true.

If a, b, c are rational numbers, then associativity of rational numbers under addition is given by

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$a + b = b + a$
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$a + (b + c) = (a + b) + c$
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$a \times (b \times c) = (a \times b) \times c$
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$a + (b - c) = (a + b) - c$

Explanation

Associative property of rational number under addition is given by following:

$a+(b+c)=(a+b)+c$

where,

$a,b$ and

$c$ are rational numbers.

Hence, B is the correct option.

Find two rational numbers lying between

$\dfrac{-1}{3}$ and

$\dfrac{1}{2}$ .

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-1/6,1 /6
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-2/3, 2/3
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none
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both

$x$ and

$y$ are rational numbers then, then the following numbers:

$x^{2}- y^{2}$ is

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rational number
0%
irrational number
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natural number
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whole number

Explanation

Given:

$x$ and

$y$ are rational numbers.

To Prove:

$x^2-y^2$ are rational numbers.

Proof: Consider

$x^2-y^2$

$=(x-y)(x+y)$ which is a rational number because we know addition, subtraction and product of two rational numbers is also a rational number.

So,

$A$ is correct option.

The rational number that lies between

$\dfrac{3}{7}$ and

$\dfrac{2}{3}$ is

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$\dfrac{2}{5}$
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$\dfrac{4}{7}$
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$\dfrac{3}{7}$
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$\dfrac{2}{3}$

Explanation

$\dfrac{3}{7}\times \dfrac{3}{3} = \dfrac{9}{21},$

$\dfrac{2}{3}\times \dfrac{7}{7} = \dfrac{14}{21}$

Rational numbers between

$\dfrac{3}{7}$ &

$\dfrac{2}{3}$ are

$\dfrac{10}{21},\dfrac{11}{21},\dfrac{12}{21},\dfrac{13}{21}$

$\dfrac{12}{21}= \dfrac{4}{7}$

Hence,

$\dfrac{4}{7}$ lies between

$\dfrac37$ and

$\dfrac23$ .

Which of the following statements are false about

$\dfrac{-2}{3}?$

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It lies to the left side of zero on the number line.
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It lies to the right side of zero on the number line.
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It is not possible to represent on the number line.
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Negative rational numbers lie on the left of $0$ on the number line.

Explanation

The given number is

$-\dfrac{2}{3}$

So, on the number line it will lie to the left of

$0$

$\therefore$ (A) is true and (B) is false

It is possible to represent the given number on the number line.

$\therefore$ (C) is false.

We know, any negative number lies on the number line to the left of

$0$

$\therefore$ (D) is true

Hence, the correct option is (B) and (C)

The rational number lies between

$\cfrac{3}{7}$ and

$\cfrac{2}{3}$ is

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$\cfrac{2}{5}$
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$\cfrac{4}{7}$
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$\cfrac{3}{7}$
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$\cfrac{2}{3}$

What should be subtracted from

$\dfrac {a}{b}$ so that the resulting fraction will be multiplicative inverse of the fraction

$\dfrac {a}{b}$ ?

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$\dfrac {a^{2} - b^{2}}{ab}$
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$a^{2} - b^{2}$
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$\dfrac {ab}{a^{2} - b^{2}}$
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$a^{2} + b^{2}$
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None of these

What is the additive inverse of

$-\dfrac { 3 }{ 13 }$ ?

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$\dfrac{-13}{3}$
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$\dfrac{3}{13}$
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$1\dfrac{1}{3}$
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None of the above

Express the following decimal in the form

$\dfrac{p}{q}$ :

$0.\bar{4}$

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$\dfrac{4}{9}$
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$\dfrac{40}{9}$
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$\dfrac{4}{90}$
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None of the above

Explanation

Given,

$0.\bar{4}$

Let

$x=0.4444$ .....(1)

$10x=4.4444$ .........(2)

(2)-(1) gives,

$9x=4$

$\therefore x=\dfrac{4}{9}$

Express the following decimal in the form

$\dfrac{p}{q}$ :

$2.15$

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$\dfrac{215}{1000}$
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$\dfrac{43}{20}$
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$\dfrac{75}{100}$
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None of the above

Explanation

Given,

$2.15$

multiply and divide by

$100$

$=2.15 \times \dfrac{100}{100}$

$=\dfrac{215}{100}$

$=\dfrac{43}{20}$

Express the following decimal in the form

$\dfrac{p}{q}$ :

$7.010$

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$\dfrac{7.5}{100}$
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$\dfrac{710}{100}$
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$\dfrac{701}{100}$
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None of the above

Explanation

Given,

$7.010$

multiply and divide by

$1000$

$=7.010 \times \dfrac{1000}{1000}$

$=\dfrac{7010}{1000}$

$=\dfrac{701}{100}$

State whether the statement is true/false.

The rational numbers

$\dfrac{-12}{-5}$ and

$\dfrac{-7}{7}$ are on the opposite side of zero on the number line.

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True
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False

Explanation

Since the given rational number

$\dfrac {-12}{-17}$ can be multiplied by

$-1$ on both numerator and denominator, it becomes positive number

That is,

$\dfrac {-12\times -1}{-17\times -1} = \dfrac {12}{17}$

Since

$\dfrac {12}{17}$ Is a positive number, all the positive numbers are

$> 0$ and so all the positive number lie to the right of

$0$ on the number line.

Since

$\dfrac {-7}{7}$ is a negative number, all the negative numbers are

$< 0$ and so all the negative numbers lie to the left of

$0$ on the number line.

$\therefore\$ Both the rational numbers lie on opposite sides of zero on the number line.

Hence, the given statement is true.

State whether the statement is true/false.

The rational number

$\dfrac{3}{4}$ lies to the right of zero on the number line.

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True
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False

Explanation

The given rational number

$\dfrac{3}{4}$ is a positive rational number that lies on the right side of zero in the number line.

Hence the given statement is true.

State whether the statement is true/false.

The rational number

$\dfrac{-12}{-17}$ lies to the left of zero on the number line.

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True
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False

Explanation

The given rational number

$\dfrac {-12}{-17}$ can be multiplied by

$-1$ on both numerator and denominator so it becomes positive number

$\dfrac {-12\times -1}{-17\times -1} = \dfrac {12}{17}$

Since

$\dfrac {12}{17}$ Is a positive number, all the positive numbers are

$> 0$ and so all the positive number lie to the right of

$0$ on the number line.

Hence the given statement is false.

State whether the statement is true/false.

The rational number

$\dfrac{29}{23}$ lies to the left of zero on the number line.

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True
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False

Explanation

The given rational number

$\dfrac{29}{23}$ is a positive rational number that lies on the right side of zero in the number line.

Hence the given statement is false.

State whether the statement is true/false.

The rational numbers

$\dfrac{-21}{5}$ and

$\dfrac{7}{-31}$ are on the opposite side of zero on the number line.

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True
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False

Explanation

Since

$\dfrac {-21}{5}$ is a negative number, all the negative numbers are

$< 0$ and so all the negative numbers lie to the left of

$0$ on the number line.

Since

$\dfrac {7}{-31}$ is a negative number, all the negative numbers are

$< 0$ and so all the negative numbers lie to the left of

$0$ on the number line

$\therefore$ The above statement is false both the rational numbers lie on opposite sides.

State whether the statement is true/false.

The rational number

$\dfrac{-3}{-5}$ is on the right of

$\dfrac{-4}{7}$ on the number line.

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True
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False

State true or false:

Between any two distinct rational numbers there are infinitely many rational numbers.

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True
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False

Explanation

Consider

$a\,\&\,b$ as rational numbers
Then a rational number

$n_1$ can be find in between

$a\,\&\,b$ such that

$a<n_1<b$
using the formula

$n_1=\dfrac{a+b}{2}$

similarly, again we can find a rational number between

$a\,\&\,n_1$ and

$n_1\,\&\,b$
and so on...

Hence, between any two distinct rational numbers there are infinitely many rational numbers.

State true or false.

The rational numbers

$\dfrac {1}{3}$ and

$\dfrac {-5}{2}$ are on opposite sides of

$0$ on the number line.

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True
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False

Explanation

Since

$\dfrac {1}{3}$ is a positive number, all the positive numbers lie to the right of

$0$ on the number line.
Since

$\dfrac {-5}{2}$ is a negative number, all the negative numbers lie to the left of

$0$ on the number line.

$\therefore$ The above statement is true both the rational numbers lie on opposite sides of

$0$ .

State whether the statements are true (T) or false (F).
Rational numbers can be added (or multiplied) in any order

$\dfrac{-4}{5} \times \dfrac{-6}{5} = \dfrac{-6}{5} \times \dfrac{-4}{5}$

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True
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False

State true or false.

The rational number

$\dfrac {-18}{-13}$ lies to the left of

$0$ on the number line.

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True
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False

Explanation

The given rational number

$\dfrac {-18}{-13}$ can be multiplied by

$-1$ on both numerator and denominator so it becomes positive number

$\dfrac {-18\times (-1)}{-13\times (-1)} = \dfrac {18}{13}$
Since

$\dfrac {18}{13}$ Is a positive number, all the positive numbers are

$> 0$ and so all the positive number lie to the right of

$0$ on the number line.

$\therefore$ The above statement is False.

Multiplicative inverse of

$\dfrac{2}{-3}$ is

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$\dfrac{2}{3}$
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$\dfrac{-3}{2}$
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$\dfrac{3}{2}$
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None of these

Explanation

${\textbf{Step -1: Write the number whose inverse is to be done}}{\textbf{.}}$

$\dfrac{2}{{ - 3}}$

${\textbf{Step -2: Get it's reciprocal to get the multiplicative inverse}}{\textbf{.}}$

$\text{Multiplicative inverse }=\dfrac{1}{\dfrac{2}{-3}}=\dfrac{{ - 3}}{2}$

${\textbf{Hence, multiplicative inverse of}}$

${\mathbf{\dfrac{2}{{ - 3}}}}$

${\textbf{is}}$

${\mathbf{\dfrac{{ - 3}}{2.}}}$

State whether the statements given are True or False

The rational number $\dfrac{-12}{-5}$ and $\dfrac{-7}{17}$ are on the opposite sides of zero on the number line.

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True
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False

Which of the following rational numbers is equal to its reciprocal?

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$1$
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$2$
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$\dfrac{1}{2}$
0%
$0$

Explanation

(a) Reciprocal of

$1=\dfrac{1}{1}=1$

(b) Reciprocal of

$2=\dfrac{1}{2}$

(c) Reciprocal of

$\dfrac{1}{2}=\dfrac{1}{\dfrac{1}{2}}=2$

(d) Reciprocal of

$0=\dfrac{1}{0}$

Therefore, only

$1$ is the only number that is equal to its reciprocal.

To get the product 1, we should multiply

$\dfrac{8}{21}$ by

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$\dfrac{8}{21}$
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$\dfrac{-8}{21}$
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$\dfrac{21}{8}$
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$\dfrac{-21}{8}$

Explanation

Let the number be

$x$

Then

$x \times \dfrac{8}{21} = 1$

$\Rightarrow x = \dfrac{21}{8}$

Therefore, we should multiply with

$\dfrac{21}{8}$ to get

$1.$

$\therefore$ Option C is the correct answer.

$x$ be any rational number, then

$x + 0$ is equal to

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$x$
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$0$
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$-x$
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Not defined

Explanation

Given

$x$ is a rational number.

To find the value of

$(x+0)$ .

We know that, the sum of any rational number and '

$0$ ' is the rational number itself.

$\therefore x+0=x$

Hence,

$Op-A$ is correct.

Multiplication inverse of a negative rational number is

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A positive rational number.
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A negative rational number
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Both positive and negative rational numbers
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None of the above

Explanation

We know that

Product of any rational number and its multiplicative inverse is

$1$ .

Hence the multiplicative inverse of a negative rational number must be

$\text {negative}$ so that their product is a positive rational number

$1$ .

Hence option B is the correct answer.

Which of the following is true for the number

$1$ ?

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It is identity for addition of rational numbers.
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It is identity for subtraction of rational numbers.
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It is identity for multiplication of rational numbers.
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It is identity for division of rational numbers.

Explanation

${\textbf{Step-1: Identify for multiplication.}}$

${\text{If a is a rational number.}}$

$\Rightarrow a \times 1 =1 \times a =a$

$\therefore$

${\text{1 is the identity for multiplication of rational numbers.}}$

${\textbf{Hence, option C is the correct answer.}}$

Zero (0) is

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The identity for addition of rational numbers.
0%
The identity for subtraction of rational numbers.
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The identity for multiplication of rational numbers.
0%
The identity for division of rational numbers.

CBSE Questions for Class 8 Maths Rational Numbers Quiz 8 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 8 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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