MCQ Questions for Class 8 Maths Rational Numbers Quiz 3

Which of the following statements is incorrect, if

$a, b, c$ and

$d$ are any rational numbers?

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$\dfrac {a}{b} + \dfrac {c}{d} = \dfrac {c}{d} + \dfrac {a}{b}$
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$\dfrac {a}{b}\times \dfrac {c}{d} = \dfrac {c}{d} \times \dfrac {a}{b}$
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$\dfrac {a}{b} - \dfrac {c}{d} = \dfrac {c}{d} - \dfrac {a}{b}$
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none of these

Explanation

For rational number, subtraction is not commutative.

i.e.

$p - q \neq q - p$ .

$\therefore \dfrac {a}{b} - \dfrac {c}{d} = \dfrac {c}{d} - \dfrac {a}{b}$ is not true.

Thus, it is an incorrect statement.

Fill the blank spaces:

$\dfrac {2}{8} + ......= \dfrac {-1}{6} + .......$

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$\dfrac {-1}{6}, \dfrac {2}{8}$
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$\dfrac {1}{6}, \dfrac {-2}{8}$
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$\dfrac {2}{8}, \dfrac {-1}{6}$
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$\dfrac {1}{6}, \dfrac {2}{8}$

Explanation

Addition is commutative for rational numbers i.e.

$a + b = b + a$

$\therefore \dfrac {2}{8} + \dfrac {-1}{6} = \dfrac {-1}{6} + \dfrac {2}{8}$

$\therefore$ The missing numbers are

$\dfrac {-1}{6}, \dfrac {2}{8}$

Find the missing value:

$\dfrac {-4}{7} + \dfrac {-7}{11} = \dfrac {-7}{11} + .....$

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$\dfrac {-4}{7}$
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$\dfrac {-7}{4}$
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$\dfrac {-7}{11}$
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$\dfrac {7}{11}$

Explanation

Addition is commutative for rational numbers. i.e.

$a + b = b + a.$

$\therefore \dfrac {-4}{7} + \dfrac {-7}{11} = \dfrac {-7}{11} + \dfrac {-4}{7}$

Fill the blank spaces:

$\dfrac {4}{-9} \times ..... = \dfrac {7}{6} \times ....$

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$\dfrac {-4}{9}, \dfrac {7}{6}$
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$\dfrac {4}{-9}, \dfrac {7}{6}$
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$\dfrac {7}{6}, \dfrac {4}{-9}$
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$\dfrac {7}{6}, \dfrac {7}{6}$

Explanation

Multiplication is commutative for rational numbers. i.e.

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

$\therefore \dfrac {4}{-9} \times \dfrac {7}{6} = \dfrac {7}{6} \times \dfrac {4}{-9}$

$\therefore$ The missing terms are

$\dfrac {7}{6}, \dfrac {4}{-9}.$

Fill the blank spaces:

$....... + \dfrac {3}{7} = ...... + \dfrac {3}{8}$

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$\dfrac {3}{7}, \dfrac {3}{8}$
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$\dfrac {3}{8}, \dfrac {3}{7}$
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$\dfrac {8}{3}, \dfrac {8}{7}$
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$\dfrac {8}{3}, \dfrac {7}{3}$

Explanation

Addition is commutative for rational numbers. i.e.

$a + b = b + a$

$\therefore \dfrac {3}{8} + \dfrac {3}{7} = \dfrac {3}{7} + \dfrac {3}{8}$

$\therefore$ The missing terms are

$\dfrac {3}{8}, \dfrac {3}{7}$

The missing value in

$........ + \dfrac {2}{7} = \dfrac {2}{7} + \dfrac {-11}{13}$ is

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$\dfrac {2}{7}$
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$\dfrac {-11}{13}$
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$\dfrac {-2}{7}$
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$\dfrac {11}{13}$

Explanation

Addition is commutative for rational numbers. i.e.

$a + b = b + a$

$\therefore \dfrac {-11}{13} + \dfrac {2}{7} = \dfrac {2}{7} + \dfrac {-11}{13}$

......... are associative for rational numbers.

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Multiplication and addition
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Multiplication and subtraction
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Division and subtraction
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Division and addition

Explanation

Multiplication and addition are associative for rational numbers because when any three rational numbers are added or multiplied, the result so obtained will remain the same irrespective of the change in their order. i.e.

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

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

Find the missing value:

$-61 + ....... = \dfrac {2}{3} + (-61)$

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$61$
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$-6$
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$\dfrac {2}{3}$
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$\dfrac {-2}{3}$

Explanation

Addition is commutative for rational numbers. i.e.

$a + b = b + a.$

$\therefore -61 + \dfrac {2}{3} = \dfrac {2}{3} + (-61)$

Find the missing value:

$\dfrac {-5}{6} + \dfrac {19}{11} = ..... + \dfrac {-5}{6}$

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$\dfrac {19}{11}$
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$-\dfrac {19}{11}$
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$\dfrac {5}{6}$
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$\dfrac {-5}{6}$

Explanation

Addition is commutative for rational numbers.
i.e.

$a + b = b + a$

$\therefore \dfrac {-5}{6} + \dfrac {19}{11} = \dfrac {19}{11} + \dfrac {-5}{6}$

Simplify using commutative and associative property :

$\dfrac {2}{9} + \dfrac {-3}{5} + \dfrac {1}{3}$

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$\dfrac {10}{18}$
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$\dfrac {-2}{45}$
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$\dfrac {2}{45}$
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$\dfrac {2}{27}$

Explanation

$\dfrac {2}{9} + \dfrac {-3}{5} + \dfrac {1}{3} = \dfrac {2}{9} + \left (\dfrac {-3}{5} + \dfrac {1}{3}\right )$

$= \dfrac {2}{9} + \left (\dfrac {1}{3} + \dfrac {-3}{5}\right )$ (commutative property)

$=\left (\dfrac {2}{9} + \dfrac {1}{3}\right ) + \dfrac {-3}{5}$ (associative property)

$= \dfrac {2 + 3}{9} + \dfrac {-3}{5}$

$= \dfrac {5}{9} + \dfrac {-3}{5}$

$= \dfrac {25 - 27}{45}$

$= \dfrac {-2}{45}$

Fill in the blank:

$(-12) + \left (4 + \dfrac {1}{8}\right ) = [(-12) + .....] + \dfrac {1}{8}$

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$(-12)$
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$4$
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$1$
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$\dfrac {1}{8}$

Explanation

Addition is associative for rational numbers
i.e.

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

$\therefore (-12) + \left (4 + \dfrac {1}{8}\right ) = [(-12) + 4] + \dfrac {1}{8}$

Fill in the blank:

$\left (-\dfrac {1}{3}\right ) + \left [\left (\dfrac {-4}{3}\right ) + \dfrac {3}{7}\right ] = \left [\left (\dfrac {-1}{3}\right ) + ..... \right ] + \dfrac {3}{7}$

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$\dfrac {-1}{3}$
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$\dfrac {-4}{3}$
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$\dfrac {3}{7}$
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$\dfrac {-3}{7}$

Explanation

Addition is associative for rational numbers
i.e.

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

$\therefore \left (-\dfrac {1}{3}\right ) + \left [\left (\dfrac {-4}{3}\right ) + \dfrac {3}{7}\right ] = \left [\left (\dfrac {-1}{3}\right ) + \left (\dfrac {-4}{3}\right ) \right ] + \dfrac {3}{7}$

Simplify using commutative and associative property :

$\left [\dfrac {2}{5} + \dfrac {5}{7} + \dfrac {-12}{5}\right ]$

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$-2\dfrac {5}{7}$
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$\dfrac {9}{7}$
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$\dfrac {-7}{9}$
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$\dfrac {-9}{7}$

Explanation

$\dfrac {2}{5} + \dfrac {5}{7} + \dfrac {-12}{5} = \dfrac {2}{5} + \left (\dfrac {5}{7} + \dfrac {-12}{5}\right )$

$= \dfrac {2}{5} + \left (\dfrac {-12}{5} + \dfrac {5}{7}\right )$ (commutative property)

$= \left (\dfrac {2}{5} + \dfrac {-12}{5}\right ) + \dfrac {5}{7}$ (associative property)

$= \dfrac {2 - 12}{5} + \dfrac {5}{7}$

$= \dfrac {-10}{5} + \dfrac {5}{7}$

$= \dfrac {-70 + 25}{35}$

$= -\dfrac {45}{35}$

$= \dfrac {-9}{7}$
Or

$= -2 + \dfrac {5}{7}$

$= \dfrac {-14 + 5}{7}$

$= \dfrac {-9}{7}$

Which property is depicted by

$\dfrac {1}{2} \times \left (6\times \dfrac {4}{3}\right ) = \left (\dfrac {1}{2} \times 6 \right )\times \dfrac {4}{3}$ ?

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Commutative
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Closure
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Associative
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Distributive

Explanation

For any three rational numbers, their product remains the same irrespective of their orders. i.e.

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

$\therefore$ Here, associative property is depicted.

Fill in the blank:

$\dfrac {2}{3} \times \left (-6 \times \dfrac {4}{5}\right ) = [ ..... \times -6] \times ....$

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$\dfrac {2}{3}, \dfrac {4}{5}$
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$\dfrac {-4}{5}, \dfrac {2}{3}$
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$\dfrac {-2}{3}, \dfrac {-4}{5}$
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$-6, -6$

Explanation

Multiplication is associative for rational numbers.
i.e.

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

$\therefore \dfrac {2}{3} \times \left (-6\times \dfrac {4}{5}\right ) = \left [\dfrac {2}{3} \times (-6)\right ] \times \dfrac {4}{5}$

$\therefore$ The missing terms are

$\dfrac {2}{3}, \dfrac {4}{5}$ .

The value of

$\dfrac {4}{5} + \dfrac {2}{3} + \dfrac {2}{5}$ is

(Use associative and commutative properties)

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$\dfrac {28}{15}$
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$\dfrac {15}{26}$
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$\dfrac {10}{15}$
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$\dfrac {2}{3}$

Explanation

$\dfrac {4}{5} + \dfrac {2}{3} + \dfrac {2}{5} = \dfrac {4}{5} + \left (\dfrac {2}{3} + \dfrac {2}{5}\right )$

$= \dfrac {4}{5} + \left (\dfrac {2}{5} + \dfrac {2}{3}\right )$ (commutative property)

$= \left (\dfrac {4}{5} + \dfrac {2}{5} \right )+ \dfrac {2}{3}$ (associative property)

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

$= \dfrac {28}{15}$

Simplify using associative property :

$\dfrac {-11}{7}\times \dfrac {4}{14}\times \dfrac {21}{33}$

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$1$
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$-1$
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$\dfrac {-2}{7}$
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$\dfrac {2}{7}$

Explanation

$\dfrac {-11}{7}\times \dfrac {4}{14}\times \dfrac {21}{33} = \dfrac {-11}{7} \times \left (\dfrac {4}{14} \times \dfrac {21}{33}\right ) = \dfrac {-11}{7} \times \left (\dfrac {21}{33} \times \dfrac {4}{14}\right )$ (

$\because$ commutative property)

$= \left (\dfrac {-11}{7}\times \dfrac {21}{33}\right ) \times \dfrac {4}{14} (\because$ associative property)

$= -1\times \dfrac {4}{14} = -\dfrac {2}{7}$

................. are not associative for rational numbers.

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Addition and multiplication
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Subtraction and multiplication
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Subtraction and division
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Addition and division

Explanation

Let us add three rational numbers

$\dfrac{-7}{5},\ \dfrac{2}{-11}$ and

$\dfrac{-13}{25}$

Then we know

$\displaystyle \frac{-7}{5} + \left(\displaystyle \frac{2}{-11} + \frac{-13}{25} \right) = \left(\displaystyle \dfrac{-7}{5} + \frac{2}{-11} \right) + \frac{-13}{25}$

So we can say that addition is associative for rational numbers.

Now let us multiply three rational numbers

$\dfrac 16,\ 6, \ \dfrac 43$

Then we get

$\displaystyle \frac{1}{3}\times \left ( 6 \times\frac{4}{3} \right )= \left ( \frac{1}{3} \times 6\right )\times \frac{4}{3}$

So we can say that multiplication is associative for rational numbers.

But when we divide three rational numbers in an order, the result so obtained will change if the order is changed.

For example:

$(81\div 9)\div 3\ne$

$81\div(9\div 3)$

And when we subtract three rational numbers, the result changes when we change the order of subtraction

For example:

$5-(3-9)\ne (5-3)-2$

Therefore, subtraction and division are not associative for rational numbers.

........... is the only rational number which is equal to its additive inverse.

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

Explanation

$0$ is the only number whose negative is

$0$ itself. Thus,

$0$ is the only rational number which is equal to its additive inverse.

................. is the additive inverse of

$\dfrac {-p}{q}$ , where

$\dfrac {-p}{q}$ is a rational number.

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$\dfrac {p}{q}$
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$\dfrac {p}{-q}$
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$\dfrac {-p}{q}$
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$\dfrac {q}{p}$

Explanation

$\left (\dfrac {-p}{q}\right )$ is a rational number, then its negative

$\left [ - \left (\dfrac {-p}{q}\right ) = \dfrac {p}{q}\right ]$ is called the additive inverse of

$\dfrac {-p}{q}$ .
Also addition of

$\left (\dfrac {-p}{q}\right )$ and

$\left (\dfrac {p}{q}\right )$ results zero.

For any integer

$a$ , its additive inverse is ..........

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$-a$
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$a$
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$\dfrac {1}{a}$
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$\dfrac {-1}{a}$

Explanation

The additive inverse of a number

$a$ is the number that, when added to

$a$ yields zero.

This number is also known as the negative, or negation.

$a$ is an integer, then its negative or additive inverse is

$(-a)$

The multiplicative identity for integers is .....

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$0$
0%
$-1$
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$1$
0%
None of these

Explanation

${\textbf{Step-1: Identity of any integer a is a number.}}$

${\text{The multiplicative identity of any integer a is a number b which when multiplied with}}\\ \text{a, leaves it unchanged.}$

${\text{ i.e. b is called as the multiplicative identity of any integer a if}}$

$a \times b = a.$

${\textbf{Step-2: When we multiply 1 with any integers.}}$

${\text{Now, when we multiply 1 with any of the integers a we get}}$

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

${\text{So, 1 is the multiplicative identity for integers.}}$

${\textbf{Hence ,The correct option is (C).The multiplicative identity for integers is 1.}}$

The sum of an integer and its additive inverse is always .........

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

Explanation

The sum of an integer and its additive inverse is always

$0$ , because for any integer

$a$ and its additive inverse

$(-a)$ , the sum is

$a + (-a)$

$=a - a$

$= 0$ .

............ is the additive inverse of

$\dfrac {-3}{11}.$

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$\dfrac {-3}{11}$
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$\dfrac {11}{3}$
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$-1$
0%
$\dfrac {3}{11}$

Explanation

In mathematics, the additive inverse of a number is the number that, when added to a, yields zero.

This number is also known as the opposite (number), sign change, and negation.

For a real number, it reverses its sign: the opposite to a positive number is negative, and the opposite to a negative number is positive.

The given rational number is

$\dfrac {-3}{11}$ and its negative is

$-\left (\dfrac {-3}{11}\right ) = \dfrac {3}{11}$ . Thus, the additive inverse of

$\dfrac {-3}{11}$ is

$\dfrac {3}{11}$ .

The additive inverse of

$\dfrac {2}{7}$ is ..........

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$\dfrac {2}{7}$
0%
$\dfrac {-2}{7}$
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$\dfrac {7}{2}$
0%
$\dfrac {-7}{2}$

Explanation

The additive inverse of a number

$a$ is the number that, when added to

$a$ , yields zero.

$a$ is an integer, then its negative

$(-a)$ is called the additive inverse of

$a$ .

The given rational number is

$\dfrac {2}{7}$ and its negative is

$\dfrac {-2}{7}$ .

Thus, the additive inverse of

$\dfrac {2}{7}$ is

$\dfrac {-2}{7}$ .

The additive inverse of

$6$ is .........

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$6$
0%
$-6$
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$0$
0%
$1$

Explanation

For the given integer

$6$ , its negative is

$-6$ .

Addition of both results to zero.

$6+(-6)=0$

Hence,

$-6$ is the additive inverse of

$6$ .

Find using additive inverse property, what should be added to

$\dfrac {-1}{3}$ , so that the sum is zero ?

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$\dfrac {-1}{3}$
0%
$\dfrac {1}{3}$
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$1$
0%
$0$

Explanation

We know that the sum of a rational number and its additive inverse is zero.

Since the given rational number is

$\dfrac {-1}{3}$ , its negative is

$-\left (\dfrac {-1}{3}\right ) = \dfrac {1}{3}$

Now,

$\dfrac {-1}{3} + \dfrac {1}{3} = 0$ . Hence,

$\dfrac {1}{3}$ should be added to

$\dfrac {-1}{3}$ to get the sum as

$0$ .

The product of ........... and any rational number is the rational number itself.

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$1$
0%
$-1$
0%
$0$
0%
the given rational number

Explanation

The product of

$1$ and any rational number is the rational number itself because

$1$ is the multiplicative identity for rational numbers.

Ex:

$\dfrac 3 2$ when multiplied by

$1$ gives

$\dfrac 3 2$ itself, which is a rational number.

Which of the following number lines, represents rational numbers?

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0%
0%
0%
All of the above

Which of the following rational number does not have a reciprocal?

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$1$
0%
$-1$
0%
$0$
0%
none

Explanation

If we try to find the reciprocal of 0 then, we get the result which is not defined.

As,

$\dfrac{1}{0}=$ Not a number

So, we can say that

$0$ does not have a reciprocal.

CBSE Questions for Class 8 Maths Rational Numbers Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Rational Numbers Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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