MCQ Questions for Class 8 Maths Rational Numbers Quiz 2

Which of the following statements is correct?

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0 is called the additive identity for rational numbers.
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1 is called the multiplicative identity for rational numbers.
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The additive inverse of 0 is zero itself.
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All the above

Explanation

Option A.

For any rational number, additive element is an element which when added to the rational number, gives the same number as its sum. Like if

$\dfrac{p}{q}$ is a rational number, where

$p$ and

$q$ are integers,

$q$ not equal to zero. The above mentioned equation is additive identity and

$0$ is an additive element.

Option B.

The multiplicative identity property states that any time you multiply a number by

$1$ , the result, or product, is that original number. This is why

$1$ is called the multiplicative identity for rational numbers.

Option C.

$0$ being neutral, has no sense of being written as

$+0$ or

$-0$ , hence, it is additive inverse of itself.

All the statements are correct.

Choose the correct option for following statement.

Subtraction of rational numbers is not commutative.

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True
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False
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Cannot be determined
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None of these

Explanation

Subtraction of two rational numbers is not commutative.

$\dfrac{a}{b}$ and

$\dfrac{c}{d}$ are any two rational numbers,

Then,

$\dfrac{a}{b}-\dfrac{c}{d}\ne \dfrac{c}{d}-\dfrac{a}{b}$

For example:

$\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}$

$\dfrac{3}{4}-\dfrac{5}{4}=\dfrac{-2}{4}=\dfrac{-1}{2}$

$\therefore$

$\dfrac{5}{4}-\dfrac{3}{4}\ne$

$\dfrac{3}{4}-\dfrac{5}{4}$

$\therefore$ The given statement is true.

$x$ and

$\displaystyle\frac{8}{9}$ are the additive inverse of each other. Then the value of

$x$ is

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$\displaystyle\frac{-8}{9}$
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$\displaystyle\frac{-4}{9}$
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$\displaystyle\frac{-5}{9}$
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$\displaystyle\frac{-7}{9}$

Explanation

$\displaystyle\frac{-8}{9}+\displaystyle\frac{8}{9}=0$
Thus

$\displaystyle\frac{-8}{9}$ and

$\displaystyle\frac{8}{9}$ are additive inverse of each other.

$\therefore$ the value of

$x$ is

$\displaystyle\frac{-8}{9}$

The statement ''Rational numbers are closed under division'' is

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True
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False
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Cannot be determined
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None

$x$ is the multiplicative identity. The value of

$x$ is

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1
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2
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0
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None

Explanation

Multiplying a number by 1 leaves it unchanged

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

$\therefore$

$1$ is the multiplicative identity.

The rational number between

$\cfrac{1}{2}$ and

$\cfrac{6}{10}$ is

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$\cfrac{1}{4}$
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$\cfrac{3}{4}$
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$\cfrac{21}{40}$
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$\cfrac{33}{100}$

Explanation

$\dfrac{1}{2}$ =

$\dfrac{5}{10}$

$\dfrac{5}{10}$ and

$\dfrac{6}{10}$

Multiplying the denominator and numerator by

$4$ we get,

$\dfrac{5}{10}\times 4=\dfrac{20}{40}$ and

$\dfrac{6}{10}\times 4=\dfrac{24}{40}$

Number between

$\dfrac{20}{40}$ and

$\dfrac{24}{40}$ is

$\dfrac{21}{40}$

Find the value of

$x$ in

$\displaystyle\frac{13}{10}\times x=1$

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$\dfrac{10}{13}$
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$\dfrac{13}{10}$
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$\dfrac{23}{10}$
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$\dfrac{10}{23}$

Explanation

$x$ is a reciprocal of

$\dfrac{13}{10}$ . Hence,

$x = \dfrac{10}{13}$

Find the additive inverse

$\displaystyle\frac{24}{25}$ is

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$\displaystyle\frac{-24}{-25}$
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$\displaystyle\frac{-24}{25}$
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0
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None

Explanation

Let the additive inverse of

$\displaystyle\frac{24}{25}$ be

$x$
Thus,

$\displaystyle\frac{24}{25}+x=0$

$=>x=-\displaystyle\frac{24}{25}$

$\therefore$ The additive inverse of

$\displaystyle\frac{24}{25}=-\displaystyle\frac{24}{25}$

How many rational numbers are there between

$-1$ and

$0$ ?

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Infinite
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1000
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4990
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None

The number of rational numbers between two given rational numbers is

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Infinite
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Finite
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Two
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One

Explanation

A rational number between two rational numbers

$a$ and

$b = \frac {(a + b)}{2}$

Like this, using this rational number

$\frac {(a + b)}{2}$ and

$b$ , we can find another rational number.
Hence, if we continue this, we get infinite rational numbers between two given rational numbers

$4\div 5= a$ and

$5\div 4= b$ , then which of the following is true?

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$a=b$
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$a\neq b$
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This is closure property
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None of the above

Explanation

$4\div 5 =0.8 = a$

$5\div 4=1.25 = b$

$\therefore a \neq b$

So, option

$B$ is correct.

Out of the following, identify which option states commutative property.

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$-2 \times (1+8) = (-2 \times 1) + (-2 \times 8)$
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$-16 \times 19 = 19 \times -16$
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$1+ (2+3) = (1+2) + 3$
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All of the above

Explanation

Commutative property means

$A \times B = B \times A$
Hence option

$B$ is correct.

Which property is being used in the following

$\displaystyle \frac{2}{3}$ +

$\displaystyle \frac{4}{7}$ =

$\displaystyle \frac{4}{7}$ +

$\displaystyle \frac{2}{3}$

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Additive identity
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Commutative property of addition
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Commutative property of multipication
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Associative property of addition

Explanation

commutative property of addition states that

(a+b) = (b+a)

Here we have

$\displaystyle \frac{2}{3}$ +

$\displaystyle \frac{4}{7}$ =

$\displaystyle \frac{4}{7}$ +

$\displaystyle \frac{2}{3}$ which is of the form

$a+b=b+a$

The number with which when

$82$ is multiplied product remains the same.

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$82$
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$0$
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$\displaystyle \frac{1}{82}$
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$1$

Explanation

Any number when multiplied by

$1$ , results in that number itself.

Hence,

$82 \displaystyle \times 1 =82$

So, option D is correct.

Commutative property in rational numbers is not satisfied with respect to which one of the following operations?

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addition
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subtraction
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multiplication
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division

Explanation

Subtraction and division is not commutative with respect to rational numbers.
For example,

$\dfrac 12 - \dfrac 13 \neq \dfrac 13 - \dfrac 12$ .

and

$5\div 3 \neq 3\div 5$

Hence, options

$B$ and

$D$ are correct.

Given that,

$\dfrac {-15}{2} \times \dfrac {12}{5} = \dfrac {12}{5} \times \dfrac {-15}{2}$
This is the best example for

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associative property of multiplication
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distributive property of multiplication
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closure property of multiplication
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commutative property of multiplication

Explanation

Commutativity is satisfied with respect to multiplication in rational numbers.

$a \times b = b \times a$

Hence, option

$D$ is correct.

Which of the following is/are correct?

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$6\div 3=2$
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$3\div 6=\dfrac{1}{2}$
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$6\div 3\neq 3\div 6$
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None of the above

Explanation

$\dfrac{6}{3}=2$

$...(1)$

$\dfrac{3}{6}=\dfrac{1}{2}$

$...(2)$

$(1)\neq (2)$

So, options

$A, B$ and

$C$ are correct.

Rational Numbers are

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Closed under addition
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Closed under substraction
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Closed under multiplication
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None of the above

Explanation

Lets take Rational numbers

$\dfrac{1}{5}$ and

$5$

Addition

$=5+\dfrac{1}{5}=\dfrac{26}{5}$ (rational number)

Subtraction

$=5-\dfrac{1}{5}=\dfrac{24}{5}$ (rational number)

Multiplication

$\dfrac{1}{5} \times 5=1$ (rational number)

Rational numbers are closed under addition ,subtraction and multiplication.

For any rational number, multiplicative identity is

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$1$
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$0$
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$100$
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$11$

Explanation

Option

$A$ is correct.
eg.

$1 \times 0 = 0$ and

$508 \times 1 = 508$
Here, the identity of a number maintained.

Select the appropriate property:

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

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commutative property of multiplication
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identity property of multiplication
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distributive property
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associative property of multiplication

Rational numbers are closed under ..........

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addition
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subtraction
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multiplication
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all of the above

Explanation

For any two rational numbers

$a$ and

$b,$

$(a + b), (a - b), (b - a)$ &

$(a\times b)$ are all rational numbers.

Thus, rational numbers are closed under addition, subtraction and multiplication.

Rational numbers are commutative under .........

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subtraction
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multiplication
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division
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none of these

Explanation

For any two rational numbers

$a$ and

$b, a\times b = b\times a$ . Thus, multiplication is commutative for rational numbers.

If zero is excluded from the set of rational numbers, then closure property holds true for the operations of

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multiplication
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division
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subtraction
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addition

Explanation

We know that rational numbers are closed under

$\text{addition, subtraction}$ and

$\text{multiplication}$ .

In case of division, if

$0$ is excluded from the set of rational number then they are closed under division too.

This is because, for any rational number

$"a", a\div 0$ is not a rational number.

Thus, in general, if

$0$ is excluded from the set of rational numbers, then closure property holds true for all the four basic operations.

The missing value in

$\dfrac {2}{3} \times \dfrac {5}{6} =..... \times \dfrac {2}{3}$ is

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$\dfrac {2}{3}$
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$\dfrac {3}{2}$
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$\dfrac {5}{6}$
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$\dfrac {6}{5}$

Explanation

Multiplication is commutative for rational numbers. i.e.

$a\times b = b\times a.$

$\therefore \dfrac {2}{3} \times \dfrac {5}{6} = \dfrac {5}{6} \times \dfrac {2}{3}$

If .............. is excluded from the collection of rational numbers, then they are closed under division.

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$2$
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$1$
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$0$
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$-1$

Explanation

For any rational number

$"a", a\div 0$ is not a rational number. So, if

$0$ is excluded from the set of rational numbers, then they are closed under division operation.

Which of the following statements is incorrect?

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Rational numbers are not closed under addition and division.
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Rational numbers are not closed under subtraction.
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Rational numbers are not closed under division and addition.
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Rational numbers are not closed under division.

Explanation

For any rational number

$a, a\div 0$ is not a rational number. Therefore, it is true that rational numbers are not closed under division.

Operation of ....... is not commutative for rational numbers.

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addition
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multiplication
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division
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none of these

Explanation

For any two rational numbers

$a$ and

$b, a\div b$ is not necessarily equal to

$b\div a$ . Thus, division is not commutative for rational numbers.

Operation of ........... are commutative for rational numbers.

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subtraction and division
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multiplication and division
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addition and subtraction
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addition and multiplication

Explanation

Two rational numbers can be added/ multiplied in any order and the result remains same.
For any two rational numbers

$a$ and

$b$ ,

$a + b = b + a$

$a\times b = b\times a$

$\therefore$ Addition and multiplication are commutative for rational numbers.

.............. are not commutative for rational numbers.

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Subtraction and division
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Multiplication and division
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Multiplication and addition
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Addition and subtraction

Find the missing value:

$\dfrac {-5}{9} \times \dfrac {6}{-1} = ..... \times \dfrac {-5}{9}$

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$6$
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$-6$
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$\dfrac {5}{9}$
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$\dfrac {-5}{9}$

Explanation

Multiplication is commutative for rational numbers. i.e.

$a\times b = b\times a$

$\therefore \dfrac {-5}{9} \times \dfrac {6}{-1} = \dfrac {6}{-1} \times \dfrac {-5}{9}$

$= -6\times \dfrac {-5}{9}$ .

CBSE Questions for Class 8 Maths Rational Numbers Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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