NCERT Exemplar for Class 8 Maths Solutions Chapter 1 Rational Numbers

In examples 1 to 3 , there are four options out of which one is correct. Choose the correct answer.

Example 1. Which of the following is not true?

(a) $\frac{2}{3} + \frac{5}{4} = \frac{5}{4} + \frac{2}{3}$

(b) $\frac{2}{3} - \frac{5}{4} = \frac{5}{4} - \frac{2}{3}$

(c) $\frac{2}{3} \times \frac{5}{4} = \frac{5}{4} \times \frac{2}{3}$

(d) $\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5}$

Ans:

(b) is the correct answer.

Subtraction of rational numbers is not commutative.

Thus, $\frac{2}{3} - \frac{5}{4} = \frac{5}{4} - \frac{2}{3}$ is not true.

Therefore, (b) is the correct answer.

Example 2. Multiplicative inverse of $\frac{0}{1}$ is

(a) $1$

(b) $ - 1$

(c) $0$

(d) not defined

Ans:

(d) is the correct answer.

Reciprocal of $\frac{0}{1}$ is $\frac{1}{0}$, which is not defined.

Thus, Multiplicative inverse of $\frac{0}{1}$ is not defined.

Therefore, (d) is the correct answer.

Example 3. Three rational numbers lying between $\frac{{ - 3}}{4}$ and $\frac{1}{2}$ are

(a) $ - \frac{1}{2},0,\frac{3}{4}$

(b) $\frac{{ - 1}}{4},\frac{1}{4},\frac{3}{4}$

(c) $\frac{{ - 1}}{4},0,\frac{1}{4}$

(d) $\frac{{ - 5}}{4},0,\frac{1}{4}$

Ans:

(c) is the correct answer.

Three rational numbers lying between $\frac{{ - 3}}{4}$ and $\frac{1}{2}$ are,

$\frac{{ - 1}}{4},0,\frac{1}{4}$

Therefore, the correct answer is (c).

In examples 4 and 5 , fill in the blanks to make the statements true.

Example 4. The product of a non-zero rational number and its reciprocal is________.

Ans:

Let $x$ be the non-zero rational number.

Then, its reciprocal will be $\frac{1}{x}$.

Now, their product is $x \times \frac{1}{x} = 1$.

The product of a non-zero rational number and its reciprocal is one.

Example 5. If $x = \frac{1}{3}$ and $y = \frac{6}{7}$ then $xy - \frac{y}{x} =$_________.

Ans:

Given that, $x = \frac{1}{3}$ and $y = \frac{6}{7}$ 

Then, $xy - \frac{y}{x} = \left[ {\frac{1}{3} \times \frac{6}{7}} \right] - \frac{{\frac{6}{7}}}{{\frac{1}{3}}}$

Multiply and divide the numbers,

$xy - \frac{y}{x} = \frac{6}{{21}} - \frac{{18}}{7}$

Take LCM and simplify,

$xy - \frac{y}{x} = \frac{{ - 48}}{{21}} = \frac{{ - 16}}{7}$

Therefore, If $x = \frac{1}{3}$ and $y = \frac{6}{7}$ then $xy - \frac{y}{x} = \frac{{ - 16}}{7}$

In examples 6 and 7, state whether the given statements are true or false.

Example 6. Every rational number has a reciprocal.

Ans:

We know that The product of a non-zero rational number and its reciprocal is one.

But, there is no rational number which when multiplied with zero, gives one.

So, the rational number $0$ has no reciprocal or multiplicative inverse.

Therefore, the given statement is false.

Example 7. $\frac{{ - 4}}{5}$ is larger than $\frac{{ - 5}}{4}$.

Ans:

Make the denominators same:

Multiply and divide  $\frac{{ - 4}}{5}$ by $4,$

$\frac{{ - 4}}{5} \times \frac{4}{4} = \frac{{ - 16}}{{20}}$

Multiply and divide  $\frac{{ - 5}}{4}$ by $5,$

$\frac{{ - 5}}{4} \times \frac{5}{5} = \frac{{ - 25}}{{20}}$

Here, $\frac{{ - 16}}{{20}}$ is larger than $\frac{{ - 25}}{{20}}$.

Thus, $\frac{{ - 4}}{5}$ is larger than $\frac{{ - 5}}{4}$.

Therefore, the given statement is true.

Example 8. Find $\frac{4}{7} \times \frac{{14}}{3} \div \frac{2}{3}$.

Ans:

Rewrite the expression,

$\frac{4}{7} \times \frac{{14}}{3} \div \frac{2}{3} = \frac{4}{7} \times \left( {\frac{{14}}{3} \times \frac{3}{2}} \right)$ 

Simplify,

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

Example 9. Using appropriate properties, find $\frac{2}{3} \times \frac{{ - 5}}{7} + \frac{7}{3} + \frac{2}{3} \times \frac{{ - 2}}{7}$.

Ans:

$\frac{2}{3} \times \left( {\frac{{ - 5}}{7}} \right) + \frac{7}{3} + \frac{2}{3} \times \left( {\frac{{ - 2}}{7}} \right)$

Rewrite as,

$= \frac{{ - 5}}{7} \times \frac{2}{3} - \frac{2}{7} \times \frac{2}{3} + \frac{7}{3}$

$= \left( {\frac{{ - 5}}{7} - \frac{2}{7}} \right) \times \frac{2}{3} + \frac{7}{3}$

$=  - \frac{2}{3} + \frac{7}{3} = \frac{5}{3}$

Example 10. Let ${\text{O}},{\text{P}}$ and ${\text{Z}}$ represent the numbers \[0,3\] and $ - 5$ respectively on the number line. Points ${\text{Q,R}}$ and ${\text{S}}$ are between ${\text{O}}$and ${\text{P}}$ such that ${\text{OQ}} = {\text{QR}} = {\text{RS}} = {\text{SP}}$.

What are the rational numbers represented by the points ${\text{Q,R}}$and ${\text{S}}$. Next choose a point ${\text{T}}$ between ${\text{Z}}$ and ${\text{O}}$ so that ${\text{ZT = TO}}$ . Which rational number does ${\text{T}}$ represent?

Ans: As ${\text{OQ}} = {\text{QR}} = {\text{RS}} = {\text{SP}}$

and ${\text{OQ + QR + RS + SP = OP}}$

therefore ${\text{Q,R}}$and ${\text{S}}$ divide ${\text{OP}}$ into four equal parts.

So, ${\text{R}}$ is the midpoint of ${\text{OP}}$,  ${\text{R}} = \frac{{0 + 3}}{2} = \frac{3}{2}$

${\text{Q}}$ is the midpoint of ${\text{OR}}$, ${\text{Q}} = \frac{1}{2}\left( {0 + \frac{3}{2}} \right) = \frac{3}{4}$

And ${\text{S}}$ is the midpoint of ${\text{RP}}$, ${\text{S}} = \frac{1}{2}\left( {\frac{3}{2} + 3} \right) = \frac{9}{4}$

Therefore, ${\text{Q = }}\frac{{\text{3}}}{{\text{4}}}{\text{,R = }}\frac{{\text{3}}}{{\text{2}}}$ and ${\text{S = }}\frac{{\text{9}}}{{\text{4}}}$

Also ${\text{ZT = TO}}$

So, ${\text{T}}$ is the midpoint of ${\text{OZ}}$,${\text{T}} = \frac{{0 + ( - 5)}}{2} = \frac{{ - 5}}{2}$.

Example 11. A farmer has a field of area $49\frac{4}{5}$ ha. He wants to divide it equally among his one son and two daughters. Find the area of each one's share.(ha means hectare; $1$hectare${\text{ = 10,000}}{{\text{m}}^{\text{2}}}$ )

Ans:

$49\frac{4}{5}$ ha $= \frac{{249}}{5}$ ha

Each share $= \frac{1}{3} \times \frac{{249}}{5}{\text{ha}} = \frac{{83}}{5}{\text{ha}} = 16\frac{3}{5}$ ha.

Example 12. Let $a,{\text{ }}b,{\text{ }}c$be the three rational numbers where $a = \frac{2}{3},b = \frac{4}{5}$

and $c =  - \frac{5}{6}$

Verify:

(i) $a + (b + c) = (a + b) + c$ (Associative property of addition)

Ans: 

${\text{ L}}{\text{.H}}{\text{.S }} = a + (b + c) = \frac{2}{3} + \left[ {\frac{4}{5} + \left( {\frac{{ - 5}}{6}} \right)} \right]$

$= \frac{2}{3} + \left[ {\frac{{24 - 25}}{{30}}} \right]$ 

$= \frac{2}{3} + \left( {\frac{{ - 1}}{{30}}} \right) = \frac{{20 - 1}}{{30}} = \frac{{19}}{{30}}$

${\text{R}}{\text{.H}}{\text{.S}} = (a + b) + c$

$= \left( {\frac{2}{3} + \frac{4}{5}} \right) + \left( {\frac{{ - 5}}{6}} \right)$

$= \left( {\frac{{10 + 12}}{{15}}} \right) + \left( {\frac{{ - 5}}{6}} \right)$

$= \frac{{22}}{{15}} - \frac{5}{6} = \frac{{44 - 25}}{{30}} = \frac{{19}}{{30}}$

So, $\frac{2}{3} + \left[ {\frac{4}{5} + \left( {\frac{5}{2}} \right)} \right] = \left( {\frac{2}{3} + \frac{4}{5}} \right) + \left( {\frac{{ - 5}}{6}} \right)$.

Hence, verified.

(ii) $a \times (b \times c) = (a \times b) \times c$ (Associative property of multiplication)

Ans:

${\text{ L}}{\text{.H}}{\text{.S  }} = a \times (b \times c) = \frac{2}{3} \times \left[ {\frac{4}{5} \times \left( {\frac{{ - 5}}{6}} \right)} \right]$

$= \frac{2}{3} \times \left( {\frac{{ - 20}}{{30}}} \right) = \frac{2}{3} \times \left( {\frac{{ - 2}}{3}} \right)$

$= \frac{{2 \times ( - 2)}}{{3 \times 3}} = \frac{{ - 4}}{9}$

${\text{R}}{\text{.H}}{\text{.S}} = (a \times b) \times c$

$= \left( {\frac{2}{3} \times \frac{4}{5}} \right) \times \left( {\frac{{ - 5}}{6}} \right)$

$= \frac{{2 \times 4}}{{3 \times 5}} \times \frac{{ - 5}}{6}$

$= \frac{8}{{15}} \times \left( {\frac{{ - 5}}{6}} \right)$

$= \frac{{8 \times ( - 5)}}{{15 \times 6}} = \frac{{ - 40}}{{90}} = \frac{{ - 4}}{9}$

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

Hence, verified.

Example 13. Solve the following questions and write your observations.

(i) $\frac{5}{3} + 0 = ?$

Ans:

When we add zero to a rational number we get the same rational number.

$\frac{5}{3} + 0 = \frac{5}{3}$

(ii) $\frac{{ - 2}}{5} + 0 = ?$

Ans:

When we add zero to a rational number we get the same rational number.

$\frac{{ - 2}}{5} + 0 = \frac{{ - 2}}{5}$

(iii) $\frac{3}{7} + 0 = ?$

Ans:

When we add zero to a rational number we get the same rational number.

$\frac{3}{7} + 0 = \frac{3}{7}$

(iv) $\frac{2}{3} \times 1 = ?$

Ans:

When we multiply a rational number by one we get the same rational number.

$\frac{2}{3} \times 1 = \frac{2}{3}$

(v) $\frac{{ - 6}}{7} \times 1 = ?$

Ans:

When we multiply a rational number by one we get the same rational number.

$\frac{{ - 6}}{7} \times 1 = \frac{{ - 6}}{7}$

(vi) $\frac{9}{8} \times 1 = ?$

Ans:

When we multiply a rational number by one we get the same rational number.

$\frac{9}{8} \times 1 = \frac{9}{8}$

Example 14. Write any $5$ rational numbers between $\frac{{{\mathbf{ - 5}}}}{{\mathbf{6}}}$ and $\frac{{\mathbf{1}}}{{\mathbf{8}}}$.

Ans:

Rewrite the numbers,

$\frac{{ - 5}}{6} = \frac{{ - 5 \times 4}}{{6 \times 4}} = \frac{{ - 20}}{{24}}$

$\frac{7}{8} = \frac{{7 \times 3}}{{8 \times 3}} = \frac{{21}}{{24}}$

Thus, rational numbers $\frac{{ - 19}}{{24}},\frac{{ - 18}}{{24}},\frac{{ - 17}}{{24}}, \ldots  \cdots \frac{{20}}{{24}}$ lies between $\frac{{ - 5}}{6}$ and $\frac{7}{8}$.

Example 15. Identify the rational number which is different from the other three \[{\mathbf{:}}\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{,}}\frac{{{\mathbf{ -4}}}}{{\mathbf{5}}}{\mathbf{,}}\frac{{\mathbf{1}}}{{\mathbf{2}}}{\mathbf{,}}\frac{{\mathbf{1}}}{{\mathbf{3}}}\]. Explain your reasoning.

Ans:

$\frac{{ - 4}}{5}$ is the negative rational number which is different from the other three, as it lies on the left side of zero while others lie on the right side of zero on the number line.

Example 16. The product of two rational numbers is ${\mathbf{ - 7}}$. If one of the numbers is ${\mathbf{ - 10}}$, find the other.

Ans:

Let the other rational number be $x$

Then, $ - 10 \times x =  - 7$

$x = \frac{{ - 7}}{{ - 10}}$

Now, $x = \frac{7}{{10}}$

Therefore, the other number is $\frac{7}{{10}}$.

In questions 1 to 25, there are four options out of which one is correct. Choose the correct answer.

1. A number which can be expressed as $\frac{p}{q}$ where \[p\] and \[q\] are integers and $q \ne 0$ is

(a) Natural number.

(b) Whole number.

(c) Integer.

(d) Rational number.

Ans:

If $p$ and $q$ are integers and $q \ne 0$, a number expressed as $\frac{p}{q}$ will be a rational number.

Therefore, (d) is the correct answer.

2. A number of the form $\frac{p}{q}$ is said to be a rational number if

(a) \[p\] and \[q\] are integers.

(b) \[p\] and \[q\] are integers and $q \ne 0$

(c) \[p\] and \[q\] are integers and $p \ne 0$

(d) \[p\] and \[q\] are integers and $p \ne 0$ also $q \ne 0$.

Ans:

A number of the form $\frac{p}{q}$ is said to be a rational number if $p$ and $q$ are integers and $q \ne 0$.

Therefore, (b) is the correct answer.

3. The numerical expression \[\frac{{\mathbf{3}}}{{\mathbf{8}}}{\mathbf{ + }}\frac{{{\mathbf{( - 5)}}}}{{\mathbf{7}}}{\mathbf{ = }}\frac{{{\mathbf{ - 19}}}}{{{\mathbf{56}}}}\] shows that

(a) Rational numbers are closed under addition.

(b) Rational numbers are not closed under addition.

(c) Rational numbers are closed under multiplication.

(d) Addition of rational numbers is not commutative.

Ans:

The sum of any two rational numbers is always a rational number.

Here, $\frac{3}{8}$ and $\frac{{ - 5}}{7}$ are rational numbers and their sum $\frac{{ - 19}}{{56}}$ is also a rational number.

Thus, Rational numbers are closed under addition.

Therefore, (a) is the correct answer.

4. Which of the following is not true?

(a) Rational numbers are closed under addition.

(b) Rational numbers are closed under subtraction.

(c) Rational numbers are closed under multiplication.

(d) Rational numbers are closed under division.

Ans:

Rational numbers are not closed under division.

Here, $\frac{1}{0}$ is not defined but $1$ and $0$ are rational numbers.

Therefore, (d) is the correct answer.

5. $\frac{{{\mathbf{ - 3}}}}{{\mathbf{8}}}{\mathbf{ + }}\frac{{\mathbf{1}}}{{\mathbf{7}}}{\mathbf{ = }}\frac{{\mathbf{1}}}{{\mathbf{7}}}{\mathbf{ + }}\left( {\frac{{{\mathbf{ - 3}}}}{{\mathbf{8}}}} \right)$ is an example to show that

(a) Addition of rational numbers is commutative.

(b) Rational numbers are closed under addition.

(c) Addition of rational numbers is associative.

(d) Rational numbers are distributive under addition.

Ans:

Let, $a = \frac{{ - 3}}{8}$ and $b = \frac{1}{7}$

Given that, $a + b = b + a$, which is commutative.

So, Addition of rational numbers is commutative.

Therefore, (a) is the correct answer.

6. Which of the following expressions shows that rational numbers are associative under multiplication?

(a) $\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}} \right){\mathbf{ = }}\left( {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}} \right){\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}$

(b) $\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}} \right){\mathbf{ = }}\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\left( {\frac{{\mathbf{3}}}{{\mathbf{5}}}{\mathbf{ \times }}\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}} \right)$

(c) $\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}} \right){\mathbf{ = }}\left( {\frac{{\mathbf{3}}}{{\mathbf{5}}}{\mathbf{ \times }}\frac{{\mathbf{2}}}{{\mathbf{3}}}} \right){\mathbf{ \times }}\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}$

(d) $\left( {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}} \right){\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}{\mathbf{ = }}\left( {\frac{{{\mathbf{ - 6}}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{5}}}} \right){\mathbf{ \times }}\frac{{\mathbf{2}}}{{\mathbf{3}}}$

Ans:

For associative multiplication property-$a \times (b \times c) = (a \times b) \times c$

Let $a = \frac{2}{3},b = \frac{{ - 6}}{7},c = \frac{3}{5}$

Then, $\frac{2}{3} \times \frac{{ - 18}}{{35}} = \frac{{ - 12}}{{21}} \times \frac{3}{5}$

$\frac{{ - 12}}{{35}} = \frac{{ - 12}}{{35}}$

Thus, $\frac{2}{3} \times \left( {\frac{{ - 6}}{7} \times \frac{3}{5}} \right) = \left( {\frac{2}{3} \times \frac{{ - 6}}{7}} \right) \times \frac{3}{5}$ shows that rational numbers are associative under multiplication.

Therefore, (a) is the correct answer.

7. Zero ${\mathbf{(0)}}$ is

(a) The identity for addition of rational numbers.

(b) The identity for subtraction of rational numbers.

(c) The identity for multiplication of rational numbers.

(d) The identity for division of rational numbers.

Ans:

If $a$ is a rational number,

$a + 0 = 0 + a = a$

Thus, zero is the identity for addition of rational numbers.

Therefore, (a) is the correct answer.

8. One \[\left( {\mathbf{1}} \right)\] is

(a) The identity for addition of rational numbers.

(b) The identity for subtraction of rational numbers.

(c) The identity for multiplication of rational numbers.

(d) The identity for division of rational numbers.

Ans:

If $a$ is a rational number,

$a \cdot 1 = 1 \cdot a = a$

Thus, one is the identity for multiplication of rational numbers.

Therefore, (c) is the correct answer.

9. The additive inverse of $\frac{{{\mathbf{ - 7}}}}{{{\mathbf{19}}}}$ is

(a) $\frac{{{\mathbf{ - 7}}}}{{{\mathbf{19}}}}$

(b) $\frac{{\mathbf{7}}}{{{\mathbf{19}}}}$

(c) $\frac{{{\mathbf{19}}}}{{\mathbf{7}}}$

(d) $\frac{{{\mathbf{ - 19}}}}{{\mathbf{7}}}$

Ans:

For any number $a$ and $b$ additive inverse should be zero. 

That is, $a + b = 0$

let $a = \frac{{ - 7}}{{19}}$ and $b$ be the additive inverse of $a.$

Then,

$b - \frac{7}{{19}} = 0$

$b = \frac{7}{{19}}$

Thus, the additive inverse of $\frac{{ - 7}}{{19}}$ is $\frac{7}{{19}}$.

Therefore, (b) is the correct answer.

10. Multiplicative inverse of a negative rational number is

(a) A positive rational number.

(b) A negative rational number.

(c) ${\mathbf{0}}$

(d) ${\mathbf{1}}$

Ans:

Let the rational number be $\frac{{ - p}}{q}$.

Then, its multiplicative inverse $= \frac{{ - q}}{p}$

Now, $\frac{{ - p}}{q} \times \frac{{ - q}}{p} = 1$

It is observed that the product of the rational number and its multiplicative inverse will be $1.$

Thus, the multiplicative inverse of a negative rational number is a negative rational number.

Therefore, (b) is the correct answer.

11. If $x + 0 = 0 + x = x$, which is rational number, then ${\mathbf{0}}$ is called

(a) Identity for addition of rational numbers.

(b) Additive inverse of $x$.

(c) Multiplicative inverse of $x$.

(d) Reciprocal of $x$.

Ans:

Sum of any rational number and zero is always a rational number.

Therefore, zero is the identity for addition of rational numbers.

Therefore, (a) is the correct answer.

12. To get the product \[{\mathbf{1}}\], we should multiply $\frac{{\mathbf{8}}}{{{\mathbf{21}}}}$ by

(a) $\frac{{\mathbf{8}}}{{{\mathbf{21}}}}$

(b) $\frac{{{\mathbf{ - 8}}}}{{{\mathbf{21}}}}$

(c) $\frac{{{\mathbf{21}}}}{{\mathbf{8}}}$

(d) $\frac{{{\mathbf{ - 21}}}}{{\mathbf{8}}}$

Ans:

Let the number be $x$.

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

Divide each side with $\frac{8}{{21}}$, 

$x = \frac{1}{{\frac{8}{{21}}}} = \frac{{21}}{8}$

Thus, we should multiply $\frac{8}{{21}}$ by $\frac{{21}}{8}$ to get $1$.

Therefore, (c) is the correct answer.

13. $ - ( - x)$ is same as

(a) $ - x$

(b) $x$

(c) $\frac{1}{x}$

(d) $\frac{{ - 1}}{x}$

Ans:

 $ - ( - x)$ is the same as $x$.

Therefore, (b) is the correct answer.

14. The multiplicative inverse of ${\mathbf{ - 1}}\frac{{\mathbf{1}}}{{\mathbf{7}}}$ is

(a) $\frac{{\mathbf{8}}}{{\mathbf{7}}}$

(b) $\frac{{{\mathbf{ - 8}}}}{{\mathbf{7}}}$

(c) $\frac{{\mathbf{7}}}{{\mathbf{8}}}$

(d) $\frac{{\mathbf{7}}}{{{\mathbf{ - 8}}}}$

Ans: 

$ - 1\frac{1}{7}$ can be written as $\frac{{ - 8}}{7}$.

Let the multiplicative inverse is $x$.

The product of a rational number and its multiplicative inverse should be $1.$

Then, $\frac{{ - 8}}{7} \times x = 1$

Divide each side with $\frac{{ - 8}}{7}$, 

$x = \frac{1}{{\frac{{ - 8}}{7}}} = \frac{7}{{ - 8}}$

Thus, the multiplicative inverse of $ - 1\frac{1}{7}$ is $\frac{7}{{ - 8}}$.

Therefore, (d) is the correct answer.

15. If $x$ be any rational number then $x + 0$ is equal to

(a) $x$

(b) ${\mathbf{0}}$

(c) $ - x$

(d) Not defined

Ans:

It is known that, sum of any rational number and zero is always a rational number.

Then, if $x$ is any rational number then $x + 0$ is equal to $x$.

Therefore, (a) is the correct answer.

16. The reciprocal of ${\mathbf{1}}$ is

(a) ${\mathbf{1}}$

(b) $ - {\mathbf{1}}$

(c) ${\mathbf{0}}$

(d) Not defined

Ans:

The reciprocal of a number is a number (multiplicative inverse)  which when multiplied by the original number yields $1.$

Let the reciprocal be $x$.

Then, $1 \cdot x = 1$

That is, $x =1$

Thus, the reciprocal of $1$ is $1$.

Therefore, (a) is the correct answer.

17. The reciprocal of \[{\mathbf{ - 1}}\] is

(a) \[{\mathbf{1}}\]

(b) \[{\mathbf{ - 1}}\]

(c) ${\mathbf{0}}$ 

(d) Not defined

Ans:

The reciprocal of a number is a number (multiplicative inverse)  which when multiplied by the original number yields $1.$

Let the reciprocal be $x$.

Then, $-1 \cdot x = 1$

That is, $x =-1$

Thus, the reciprocal of $ - 1$ is $ - 1$.

Therefore, (b) is the correct answer.

18. The reciprocal of ${\mathbf{0}}$ is

(a) \[{\mathbf{1}}\]

(b) \[{\mathbf{ - 1}}\]

(c) ${\mathbf{0}}$

(d) Not defined

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Let the reciprocal be $x$.

Then, $0 \cdot x = 1$

That is, $x = \frac{1}{0}$, which is not defined.

Thus, the reciprocal of $0$ is not defined.

Therefore, (d) is the correct answer.

19. The reciprocal of any rational number $\frac{p}{q}$ where $p$ and $q$ are integers and $q \ne 0$

(a) $\frac{p}{q}$

(b) \[{\mathbf{1}}\]

(c) \[{\mathbf{0}}\]

(d) $\frac{q}{p}$

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Thus, the reciprocal of any rational number $\frac{p}{q}$ where $p$ and $q$ are integers and $q \ne 0$ is $\frac{q}{p}$.

Therefore, (d) is the correct answer.

20. If $y$ be the reciprocal of rational number $x$, then the reciprocal of $y$ will be

(a) $x$

(b) $y$

(c) $\frac{x}{y}$

(d) $\frac{y}{x}$

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

If $y$ be the reciprocal of rational number $x$, then the reciprocal of $y$ will be $x$.

Therefore, (a) is the correct answer.

21. The reciprocal of $\frac{{{\mathbf{ - 3}}}}{{\mathbf{8}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 7}}}}{{{\mathbf{13}}}}} \right)$ is

(a) $\frac{{{\mathbf{104}}}}{{{\mathbf{21}}}}$

(b) $\frac{{{\mathbf{ - 104}}}}{{{\mathbf{21}}}}$

(c) $\frac{{{\mathbf{21}}}}{{{\mathbf{104}}}}$

(d) $\frac{{{\mathbf{ - 21}}}}{{{\mathbf{104}}}}$

Ans:

Multiply the numbers,

$\frac{{ - 3}}{8} \times \left( {\frac{{ - 7}}{{13}}} \right) = \frac{{21}}{{104}}$

Let $x$ be the reciprocal.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{{21}}{{104}} \times x = 1$

Divide each side with $\frac{{21}}{{104}}$, 

$x = \frac{1}{{\frac{{21}}{{104}}}} = \frac{{104}}{{21}}$

Thus, the reciprocal of $\frac{{ - 3}}{8} \times \left( {\frac{{ - 7}}{{13}}} \right)$ is $\frac{{104}}{{21}}$.

Therefore, (a) is the correct answer.

22. Which of the following is an example of the distributive property of multiplication over addition for rational numbers?

(a) $\frac{{{\mathbf{ - 1}}}}{{\mathbf{4}}}{\mathbf{ \times }}\left\{ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)} \right\}{\mathbf{ = }}\left[ {{\mathbf{ - }}\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{ \times }}\frac{{\mathbf{2}}}{{\mathbf{3}}}} \right]{\mathbf{ + }}\left[ {{\mathbf{ - }}\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)} \right]$

(b) $\frac{{{\mathbf{ - 1}}}}{{\mathbf{4}}}{\mathbf{ \times }}\left\{ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)} \right\}{\mathbf{ = }}\left[ {\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{ \times }}\frac{{\mathbf{2}}}{{\mathbf{3}}}} \right]{\mathbf{ - }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)$

(c) $\frac{{{\mathbf{ - 1}}}}{{\mathbf{4}}}{\mathbf{ \times }}\left\{ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)} \right\}{\mathbf{ = }}\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {{\mathbf{ - }}\frac{{\mathbf{1}}}{{\mathbf{4}}}} \right){\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)$

(d) $\frac{{{\mathbf{ - 1}}}}{{\mathbf{4}}}{\mathbf{ \times }}\left\{ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{7}}}} \right)} \right\}{\mathbf{ = }}\left\{ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ + }}\left( {{\mathbf{ - }}\frac{{\mathbf{4}}}{{\mathbf{7}}}{\mathbf{ \times }}} \right)} \right\}{\mathbf{ - }}\frac{{\mathbf{1}}}{{\mathbf{4}}}$

Ans:

Distributive property is given as,

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

Thus, $\frac{{ - 1}}{4} \times \left\{ {\frac{2}{3} + \left( {\frac{{ - 4}}{7}} \right)} \right\} = \left[ { - \frac{1}{4} \times \frac{2}{3}} \right] + \left[ { - \frac{1}{4} \times \left( {\frac{{ - 4}}{7}} \right)} \right]$ is an example of distributive property of multiplication over addition for rational numbers.

Therefore, (a) is the correct answer.

23. Between two given rational numbers, we can find

(a) One and only one rational number.

(b) Only two rational numbers.

(c) Only ten rational numbers.

(d) Infinitely many rational numbers.

Ans:

Between two given rational numbers, we can find infinitely many rational numbers.

Therefore, (d) is the correct answer.

24. $\frac{{x + y}}{2}$ is a rational number

(a) Between $x$ and $y$

(b) Less than $x$ and $y$ both.

(c) Greater than $x$ and $y$ both.

(d) Less than $x$ but greater than $y$.

Ans:

$\frac{{x + y}}{2}$ is a rational number between $x$ and $y$.

Therefore, (a) is the correct answer.

25. Which of the following statements is always true?

(a) $\frac{{x - y}}{2}$ is a rational number between $x$ and $y$

(b) $\frac{{x + y}}{2}$ is a rational number between $x$ and $y$

(c) $\frac{{x \times y}}{2}$ is a rational number between $x$ and $y$

(d) $\frac{{x \div y}}{2}$ is a rational number between $x$ and $y$

Ans: 

$\frac{{x + y}}{2}$ is a rational number between $x$ and $y$.

Therefore, (b) is the correct answer.

In questions 26 to 47 , fill in the blanks to make the statements true.

26. The equivalent of $\frac{{\mathbf{5}}}{{\mathbf{7}}}$, whose numerator is ${\mathbf{45}}$ is ________.

Ans:

To make the numerator $45$multiply and divide the number with $9$,

$\frac{5}{7} \times \frac{9}{9} = \frac{{45}}{{63}}$

Therefore, the equivalent of $\frac{5}{7}$, whose numerator is $45$ is $\frac{{45}}{{63}}$.

27. The equivalent rational number of \[\frac{{\mathbf{7}}}{{\mathbf{9}}}\], whose denominator is ${\mathbf{45}}$ is _______.

Ans:

To make the denominator $45$multiply and divide the number with $5$,

$\frac{7}{9} \times \frac{5}{5} = \frac{{35}}{{45}}$

Therefore, the equivalent rational number of $\frac{7}{9}$, whose denominator is $45$ is $\frac{{35}}{{45}}$.

28. Between the numbers $\frac{{{\mathbf{15}}}}{{{\mathbf{20}}}}$ and $\frac{{{\mathbf{34}}}}{{{\mathbf{40}}}}$, the greater number is ________.

Ans:

Multiplying and dividing $\frac{{15}}{{20}}$ with $2$gives $\frac{{30}}{{40}}$

Now, we have to see which is greater in $\frac{{30}}{{40}}$ and $\frac{{34}}{{40}}$

As we see, the denominator is the same. 

So, the number which has a higher numerator value will be greater.

Therefore, $\frac{{34}}{{40}}$ will be greater

29. The reciprocal of a positive rational number is ________.

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Let $x$ be the reciprocal and $y$ be the original positive number.

Then, $x \cdot y = 1$ 

Thus, $x = \frac{1}{y}$, which is a positive number.

Therefore, the reciprocal of a positive rational number is always a positive rational number.

30. The reciprocal of a negative rational number is ________.

Ans: 

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Let $x$ be the reciprocal and $-y$ be the original negative number.

Then, $x \cdot ( - y) = 1$

Thus, $x = \frac{1}{-y}$, which is a negative number.

Therefore, the reciprocal of a negative rational number is always a negative rational number.

31. Zero has ________ reciprocal.

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Let $x$ be the reciprocal.

Then, $0 \cdot x = 1$

Thus, $x = \frac{1}{0}$, which is not defined.

Therefore, Zero has no reciprocal.

32. The numbers _______ and _______ are their own reciprocal.

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Let $x$ be the reciprocal of $1$.

Then, $1 \cdot x = 1$

Thus, $x = \frac{1}{1}$, which is one.

Let $y$ be the reciprocal of $-1$.

Then, $-1 \cdot y = 1$

Thus, $y = \frac{1}{-1}$, which is $-1$.

Therefore, the numbers $1$ and $ - 1$ are their own reciprocal.

33. If $y$ be the reciprocal of$x$, then the reciprocal of ${y^2}$ in terms of $x$ will be________.

Ans:

Given that, $\frac{1}{x} = y$

Now,  ${y^2} = \frac{1}{{{y^2}}} = \frac{1}{{{{\left( {\frac{1}{x}} \right)}^2}}} = {x^2}$

Therefore, if $y$ be the reciprocal of $x$, then the reciprocal of ${y^2}$ in terms of $x$ will be ${x^2}$.

34. The reciprocal of $\frac{{\mathbf{2}}}{{\mathbf{5}}}{\mathbf{ \times }}\left( {\frac{{{\mathbf{ - 4}}}}{{\mathbf{9}}}} \right)$ is _________.

Ans:

Given that the expression is $ - \frac{2}{5} \times \left( { - \frac{4}{9}} \right)$

Rewrite the expression,

$ - \frac{2}{5} \times \left( { - \frac{4}{9}} \right) = \frac{{ - 8}}{{45}}$

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Therefore, the reciprocal of $\frac{{ - 8}}{{45}}$ is $\frac{{ - 45}}{8}$

35. ${{\mathbf{(213 \times 657)}}^{{\mathbf{ - 1}}}}{\mathbf{ = 21}}{{\mathbf{3}}^{{\mathbf{ - 1}}}}{\mathbf{ \times }}$_________.

Ans:

Let $x$be the unknown number.

Then, $\frac{1}{{213 \times 657}} = \frac{1}{{213}} \times x$

Now, $x = \frac{1}{{657}}$

$x = {657^{ - 1}}$

Therefore, ${(213 \times 657)^{ - 1}} = {213^{ - 1}} \times {657^{ - 1}}$

36. The negative of ${\mathbf{1}}$ is ______.

Ans:

The negative of $1$ is $ - 1.$

37. For rational numbers $\frac{a}{b},\frac{c}{d}$, and $\frac{e}{f}$ we have $\frac{a}{b} \times \left( {\frac{c}{d} + \frac{e}{f}} \right) =$________.

Ans:

By distributive property of multiplication over addition,

$\frac{a}{b} \times \left( {\frac{c}{d} + \frac{e}{f}} \right) = \frac{a}{b} \times \frac{c}{d} + \frac{a}{b} \times \frac{e}{f}$

$\frac{a}{b} \times \left( {\frac{c}{d} + \frac{e}{f}} \right) = \frac{{ac}}{{bd}} + \frac{{ae}}{{bf}}$

Therefore, for rational numbers $\frac{a}{b},\frac{c}{d}$, and $\frac{e}{f}$ we have $\frac{a}{b} \times \left( {\frac{c}{d} + \frac{e}{f}} \right) = \frac{{ac}}{{bd}} + \frac{{ae}}{{bf}}$.

38. \[\frac{{{\mathbf{ - 5}}}}{{\mathbf{7}}}\] is _______ than ${\mathbf{ - 3}}$

Ans:

Multiply both the numbers with $7$,

$\frac{{ - 5}}{7} \times 7 =  - 5$ and $ - 3 \times 7 =  - 21$

Here, we can see that $ - 5$ is greater than $ - 21$.

That is, $\frac{{ - 5}}{7} >  - 3$

$\frac{{ - 5}}{7}$ is greater than $ - 3$.

39. There are ________ rational numbers between any two rational numbers.

Ans:

There are infinite rational numbers between any two rational numbers.

40. The rational numbers $\frac{{\mathbf{1}}}{{\mathbf{3}}}$ and $\frac{{{\mathbf{ - 1}}}}{{\mathbf{3}}}$ are on the ________ sides of zero on the number line.

Ans:

$\frac{1}{3}$ is greater than zero. So, it will be on the right side of zero on the number line.

$\frac{{ - 1}}{3}$ is less than zero. So, it will be on the left side of zero on the number line.

Therefore, the rational numbers $\frac{1}{3}$ and $\frac{{ - 1}}{3}$ are on the opposite sides of zero on the number line.

41. The negative of a negative rational number is always a _________ rational number.

Ans:

Let $ - x$be the negative rational number.

Then, the negative of this negative rational number is $ - ( - x) = x$, a positive rational number.

Therefore, the negative of a negative rational number is always a positive rational number.

42. Rational numbers can be added or multiplied in any ________.

Ans:

Rational numbers are commutative under addition and multiplication.

Therefore, rational numbers can be added or multiplied in any order.

43. The reciprocal of $\frac{{{\mathbf{ - 5}}}}{{\mathbf{7}}}$ is ________.

Ans:

Let $x$ be the reciprocal.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{{ - 5}}{7} \times x = 1$

Divide each side with $\frac{{ - 5}}{7}$, 

$x = \frac{1}{{\frac{{ - 5}}{7}}} = \frac{{ - 7}}{5}$

Therefore, the reciprocal of $\frac{{ - 5}}{7}$ is $\frac{{ - 7}}{5}$.

44. The multiplicative inverse of $\frac{{\mathbf{4}}}{{\mathbf{3}}}$ is __________.

Ans:

Let $x$ be the multiplicative inverse.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{4}{3} \times x = 1$

Divide each side with $\frac{4}{3}$, 

$x = \frac{1}{{\frac{4}{3}}} = \frac{3}{4}$

Therefore, the multiplicative inverse of $\frac{4}{3}$ is $\frac{3}{4}$.

45. The rational number ${\mathbf{10}}{\mathbf{.11}}$in the form $\frac{p}{q}$ is __________.

Ans:

Multiply and divide $10.11$by $100$,

 $\frac{{1011}}{{100}}$

Therefore, the rational number $10.11$in the form $\frac{p}{q}$ is $\frac{{1011}}{{100}}$.

46. $\frac{{\mathbf{1}}}{{\mathbf{5}}}{\mathbf{ \times }}\left[ {\frac{{\mathbf{2}}}{{\mathbf{7}}}{\mathbf{ + }}\frac{{\mathbf{3}}}{{\mathbf{8}}}} \right]{\mathbf{ = }}\left[ {\frac{{\mathbf{1}}}{{\mathbf{5}}}{\mathbf{ \times }}\frac{{\mathbf{2}}}{{\mathbf{7}}}} \right]{\mathbf{ + }}$__________.

Ans: 

By distributive law,

$\frac{1}{5} \times \left[ {\frac{2}{7} + \frac{3}{8}} \right] = \left[ {\frac{1}{5} \times \frac{2}{7}} \right] + \left[ {\frac{1}{5} \times \frac{3}{8}} \right]$

47. The two rational numbers lying between \[ - {\mathbf{2}}\] and $ - {\mathbf{5}}$ with denominator as ${\mathbf{1}}$ are _______ and _______.

Ans:

The two rational numbers lying between $ - 2$ and $ - 5$ with denominator as $1$ are $ - 3$ and $ - 4$.

In each of the following, state whether the statements are true (T) or false (F).

48. If $\frac{x}{y}$ is a rational number, then $y$ is always a whole number.

Ans: 

If $\frac{x}{y}$ is a rational number, then $y$ should always be a non-zero number.

Therefore, the given statement is false.

49. If $\frac{p}{q}$ is a rational number, then $p$ cannot be equal to zero.

Ans: 

If $\frac{p}{q}$ is a rational number, then $p$ can be any integer, including zero.

Therefore, the given statement is false.

50. If $\frac{r}{s}$ is a rational number, then $s$ cannot be equal to zero.

Ans:

It is known that rational numbers are not closed under division.

If $\frac{r}{s}$ is a rational number, then $s$ cannot be equal to zero.

Therefore, the given statement is true.

51. $\frac{5}{6}$ lies between $\frac{{\mathbf{2}}}{{\mathbf{3}}}$ and ${\mathbf{1}}$.

Ans:

Multiply each number with $6,$

$\frac{5}{6} \times 6,\frac{2}{3} \times 6,1 \times 6$ $= 5,4,6$ 

Here, $6 > 5 > 4$ 

Thus, $1 > \frac{5}{6} > \frac{2}{3}$

Therefore, the given statement is true.

52. $\frac{{\mathbf{5}}}{{{\mathbf{10}}}}$ lies between $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ and ${\mathbf{1}}$.

Ans:

Multiply each number with $10,$

$\frac{5}{{10}} \times 10,\frac{1}{2} \times 10,1 \times 10$ $= 5,5,10$ 

Here, $5 = 5 < 10$ 

Therefore, the given statement is false.

53. $\frac{{ - {\mathbf{7}}}}{{\mathbf{2}}}$ lies between $ - {\mathbf{3}}$ and $ - {\mathbf{4}}$

Ans:

Multiply each number with $2,$

$\frac{{ - 7}}{2} \times 2, - 3 \times 2, - 4 \times 2$ $=  - 7, - 6, - 8$ 

Here, $ - 6 >  - 7 >  - 8$ 

Thus, $ - 3 > \frac{{ - 7}}{2} >  - 4$

Therefore, the given statement is true.

54. $\frac{{\mathbf{9}}}{{\mathbf{6}}}$ lies between ${\mathbf{1}}$ and ${\mathbf{2}}$.

Ans:

Multiply each number with $6,$

$\frac{9}{6} \times 6,1 \times 6,2 \times 6$ $= 9,6,12$ 

Here, $12 > 9 > 6$ 

Thus, $2 > \frac{9}{6} > 1$

Therefore, the given statement is true.

55. If $a \ne {\mathbf{0}}$ the multiplicative inverse of $\frac{a}{b}$ is \[\frac{b}{a}\].

Ans:

The multiplicative inverse of a number is a number which when multiplied by the original number yields $1.$

Then, if $a \ne 0$ the multiplicative inverse of $\frac{a}{b}$ is $\frac{b}{a}$.

Therefore, the given statement is true.

56. The multiplicative inverse of $\frac{{ - {\mathbf{3}}}}{{\mathbf{5}}}$ is $\frac{{\mathbf{5}}}{{\mathbf{3}}}$.

Ans:

Let $x$ be the multiplicative inverse.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{{ - 3}}{5} \times x = 1$

Divide each side with $\frac{{ - 3}}{5}$, 

$x = \frac{1}{{\frac{{ - 3}}{5}}} = \frac{{ - 5}}{3}$

Thus, the multiplicative inverse of $\frac{{ - 3}}{5}$ is $\frac{{ - 5}}{3}$.

Therefore, the given statement is false.

57. The additive inverse of $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ is $ - {\mathbf{2}}$.

Ans:

The additive inverse of a number is the number that, when added to the original number, yields zero.

Let $a$be the additive inverse.

Then, $\frac{1}{2} + a = 0$

$a = \frac{{ - 1}}{2}$

That is, additive inverse of $\frac{1}{2}$ is $\frac{{ - 1}}{2}$.

Therefore, the given statement is false.

58. If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, then $\frac{x}{y} + \frac{c}{d} = 0$.

Ans:

The additive inverse of a number is the number that, when added to the original number, yields zero.

If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, then $\frac{x}{y} + \frac{c}{d} = 0$.

Therefore, the given statement is true.

59. For every rational number $x,x + 1 = x$.

Ans:

For every rational number $x,x + 1 \ne x$.

Therefore, the given statement is false.

60. 58. If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, then $\frac{x}{y} - \frac{c}{d} = 0$.

Ans:

The additive inverse of a number is the number that, when added to the original number, yields zero.

If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, then $\frac{x}{y} + \frac{c}{d} = 0$.

Therefore, the given statement is false.

61. The reciprocal of a non-zero rational number $\frac{q}{p}$ is the rational number $\frac{q}{p}$.

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

The reciprocal of a non-zero rational number $\frac{q}{p}$ is the rational number $\frac{p}{q}$.

Therefore, the given statement is true.

62. If $x + y = 0$, then $ - y$ is known as the negative of $x$, where $x$ and $y$ are rational numbers.

Ans:

If $x + y = 0$, then $y$ is known as the negative of $x$, where $x$ and $y$ are rational numbers.

Therefore, the given statement is false.

63. The negative of the negative of any rational number is the number itself.

Ans:

If $x$is the rational number, the negative of the negative of $x$ is $ - ( - x) = x$.

So, the negative of the negative of any rational number is the number itself.

Therefore, the given statement is true.

64. The negative of ${\mathbf{0}}$ does not exist.

Ans:

Zero has no sign, neither positive nor negative.

So, the negative of $0$ does not exist.

Therefore, the given statement is true.

65. The negative of \[{\mathbf{1}}\] is \[{\mathbf{1}}\] itself.

Ans:

The negative of $1$ is $ - 1.$

Therefore, the given statement is false.

66. For all rational numbers $x$ and $y,x - y = y - x$.

Ans:

Subtraction and division are not commutative for rational numbers.

That is, for all rational numbers $x$ and $y,x - y \ne y - x$.

Therefore, the given statement is false.

67. For all rational numbers $x$ and $y,x \times y = y \times x$.

Ans:

Multiplication is commutative for rational numbers.

That is, for all rational numbers $x$ and $y,x \times y = y \times x$.

Therefore, the given statement is true.

68. For every rational number $x,x \times 0 = x$.

Ans:

The product of any number and zero will be zero.

Therefore, the given statement is false.

69. For every rational numbers $x,y$ and $z,x + (y \times z) = (x + y) \times (x + z)$

Ans: 

Distributive property of addition over multiplication is not closed in rational numbers.

Therefore, the given statement is false.

70. For all rational numbers $a,b$ and $c,a(b + c) = ab + bc$.

Ans:

According to distributive property of multiplication over addition,

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

Therefore, the given statement is false,

71. ${\mathbf{1}}$ is the only number which is its own reciprocal.

Ans:

The numbers which are their own reciprocals are $1$and $ - 1.$

Therefore, the given statement is false.

72. \[ - {\mathbf{1}}\] is not the reciprocal of any rational number.

Ans:

The reciprocal of $ - 1$ is $ - 1$.

Therefore, the given statement is false.

73. For any rational number $x,x + ( - 1) =  - x$.

Ans:

If $x$ is a rational number, then $x + ( - 1) \ne  - x$.

Therefore, the given statement is false.

74. For rational numbers $x$ and $y$, if $x < y$ then $x - y$ is a positive rational number.

Ans:

For rational numbers $x$ and $y$, if $x < y$ then $x - y$ is a negative rational number.

Therefore, the given statement is false.

75. If $x$ and $y$ are negative rational numbers, then so is $x + y$ .

Ans:

If $x$ and $y$ are negative rational numbers, then their sum $x + y$ is also a negative number.

Therefore, the given statement is true.

76. Between any two rational numbers there are exactly ten rational numbers.

Ans:

Between any two rational numbers there are infinite rational numbers.

Therefore, the given statement is false.

77. Rational numbers are closed under addition and multiplication but not under subtraction.

Ans:

Rational numbers are closed under addition, multiplication and subtraction.

Therefore, the given statement is false.

78. Subtraction of rational numbers is commutative.

Ans:

Subtraction is not commutative for rational numbers because while performing subtraction, if the order of numbers is changed, then the result also changes.

Therefore, the given statement is false.

79.$\frac{{ - {\mathbf{3}}}}{{\mathbf{4}}}$ is smaller than $ - {\mathbf{2}}$.

Ans:

Multiply the numbers by $4,$

$\frac{{ - 3}}{4} \times 4, - 2 \times 4 =  - 3, - 8$

Here, $ - 3 >  - 8$

Then, $\frac{{ - 3}}{4} >  - 2$

Therefore, the given statement is false.

80. ${\mathbf{0}}$ is a rational number.

Ans: 

It is known that every integer is a rational number and Zero can be represented as the ratio of two integers.

Therefore, the given statement is true.

81. All positive rational numbers lie between ${\mathbf{0}}$ and ${\mathbf{1000}}{\mathbf{.}}$

Ans:

There are infinite positive rational numbers.

Therefore, the given statement is false.

82. The population of India in ${\mathbf{2004}} - {\mathbf{05}}$ is a rational number.

Ans:

it is known that every integer is a rational number.

So, the population of India in $2004 - 05$ is a rational number.

Therefore, the given statement is true.

83. There are countless rational numbers between $\frac{{\mathbf{5}}}{{\mathbf{6}}}$ and $\frac{{\mathbf{8}}}{{\mathbf{9}}}$.

Ans:

There are infinite positive rational numbers.

Therefore, the given statement is true.

84. The reciprocal of ${x^{ - 1}}$ is $\frac{1}{x}$.

Ans:

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Here, both are the same numbers.

Therefore, the given statement is false.

85. The rational number $\frac{{57}}{{23}}$ lies to the left of zero on the number line.

Ans:

$\frac{{57}}{{23}}$ is a positive rational number that lies  to the right of zero on the number line.

Therefore, the given statement is false.

86. The rational number $\frac{{\mathbf{7}}}{{ - {\mathbf{4}}}}$ lies to the right of zero on the number line.

Ans:

$\frac{7}{{ - 4}}$ is a negative rational number that lies  to the left of zero on the number line.

Therefore, the given statement is false.

87. The rational number $\frac{{ - {\mathbf{8}}}}{{ - {\mathbf{3}}}}$ lies neither to the right nor to the left of zero on the number line.

Ans:

$\frac{{ - 8}}{{ - 3}}$ is a positive rational number that  lies to the right of zero on the number line.

Therefore, the given statement is false.

88. The rational numbers $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ and $ - {\mathbf{1}}$ are on the opposite sides of zero on the number line.

Ans:

$\frac{1}{2}$ is a positive rational number that is to the right of zero on the number line.

$ - 1$ is a negative rational number that is to the left of zero on the number line.

That is, $\frac{1}{2}$ and $ - 1$ are on the opposite sides of zero on the number line.

Therefore, the given statement is true.

89. Every fraction is a rational number.

Ans:

It is known that the rational number is in the form of $\frac{p}{q}$ where $q$ is not equal to zero. 

Any fraction with non-zero denominators is a rational number.

Therefore, the given statement is true.

90. Every integer is a rational number.

Ans:

We can write integers in the form of $\frac{p}{q}$ where $q$ is not equal to zero. 

That means, every integer is a rational number.

Therefore, the given statement is true.

91. The rational numbers can be represented on the number line.

Ans:

Rational numbers can be represented on the number line.

Positive rational numbers are represented on the right side of zero and negative on left side of zero.

Therefore, the given statement is true.

92. The negative of a negative rational number is a positive rational number.

Ans:

If $x$is the rational number, the negative of the negative of $x$ is $ - ( - x) = x$.

So, the negative of the negative of a rational number is a positive rational number.

Therefore, the given statement is true.

93. If $x$ and $y$ are two rational numbers such that $x > y$, then $x - y$ is always a positive rational number.

Ans:

For rational numbers $x$ and $y$, if $x > y$ then $x - y$ is always a positive rational number.

Therefore, the given statement is true.

94. ${\mathbf{0}}$ is the smallest rational number.

Ans:

It is known that the rational number is in the form of $\frac{p}{q}$ where $q$ is not equal to zero. 

So, $0$ is not the smallest rational number.

Therefore, the given statement is false.

95. Every whole number is an integer.

Ans:

Whole numbers begin from $0$ to positive numbers.

That means, every whole number is an integer.

Therefore, the given statement is true.

96. Every whole number is a rational number.

Ans:

Whole numbers begin from $0$ to positive numbers.

They can be written in the form of $\frac{p}{q}$ where $q$ is not equal to zero. 

That means, every whole number is a rational number.

Therefore, the given statement is true.

97. ${\mathbf{0}}$ is the whole number but it is not a rational number.

Ans:

it is known that every integer is a rational number and Zero can be represented as the ratio of two integers.

It is also known that, whole numbers begin from $0$ to positive numbers.

That means, zero is a rational number as well as a whole number.

Therefore, the given statement is false.

98. The rational numbers $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ and $ - \frac{{\mathbf{5}}}{{\mathbf{2}}}$ are on the opposite sides of ${\mathbf{0}}$ on the number line.

Ans:

$\frac{1}{2}$ is a positive rational number which lies on the right side of zero.

$ - \frac{5}{2}$ is a negative rational number which lies on the left side of zero.

Therefore, the given statement is true.

99. Rational numbers can be added (or multiplied) in any order $\frac{{ - {\mathbf{4}}}}{{\mathbf{5}}}{\mathbf{ \times }}\frac{{ - {\mathbf{6}}}}{{\mathbf{5}}}{\mathbf{ = }}\frac{{ - {\mathbf{6}}}}{{\mathbf{5}}}{\mathbf{ \times }}\frac{{ - {\mathbf{4}}}}{{\mathbf{5}}}$.

Ans:

Rational numbers are closed under addition multiplication.

That means, they can be added or multiplied in any order.

Therefore, the given statement is true.

100. Solve the following: Select the rational numbers from the list which are also the integers.

$\frac{9}{4},\frac{8}{4},\frac{7}{6},\frac{6}{4},\frac{9}{3},\frac{8}{3},\frac{7}{3},\frac{6}{3},\frac{5}{2},\frac{4}{2},\frac{3}{2},\frac{1}{1},\frac{0}{1},\frac{{ - 1}}{1},\frac{{ - 2}}{1},\frac{{ - 3}}{2},\frac{{ - 4}}{2},\frac{{ - 5}}{2},\frac{{ - 6}}{2}$

Ans:

Integers in the form of $\frac{p}{q}$ where $q$ is not equal to zero are rational numbers.

So, the rational numbers which are also the integers are,

$\frac{8}{4},\frac{9}{3},\frac{6}{3},\frac{4}{2},\frac{3}{1},\frac{1}{1},\frac{{ - 1}}{1},\frac{{ - 2}}{2},\frac{{ - 4}}{2},\frac{{ - 6}}{2}$

101. Select those which can be written as a rational number with denominator ${\mathbf{4}}$ in their lowest form: $\frac{7}{8},\frac{{64}}{{16}},\frac{{36}}{{ - 12}},\frac{{ - 16}}{{17}},\frac{5}{{ - 4}},\frac{{140}}{{28}}$

Ans:

$\frac{5}{{ - 4}}$ can be written as a rational number with denominator $4$ in their lowest form.

102. Using suitable rearrangement and find the sum:

(a) $\frac{4}{7} + \left( {\frac{{ - 4}}{9}} \right) + \frac{3}{7} + \left( {\frac{{ - 13}}{9}} \right)$

Ans:

$\frac{4}{7} + \frac{3}{7} + \left( {\frac{{ - 4}}{9}} \right) + \left( {\frac{{ - 13}}{9}} \right)$

Add first two terms and add next two terms,

$= \frac{7}{7} + \left( {\frac{{ - 17}}{9}} \right)$

Take LCM,

$= \frac{{63 - 119}}{{63}}$

Subtract the numbers,

$= \frac{{ - 56}}{{63}}$

Divide numerator and denominator by $7,$

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

(b) $ - 5 + \frac{7}{{10}} + \frac{3}{7} + ( - 3) + \frac{5}{{14}} + \frac{{ - 4}}{5}$

Ans:

$ - 5 + \frac{7}{{10}} + \frac{3}{7} + ( - 3) + \frac{5}{{14}} + \frac{{ - 4}}{5}$

Rewrite the terms,

$ - 5 + ( - 3) + \left( { - \frac{{13}}{9}} \right) + \frac{7}{{10}} + \frac{3}{7} + \frac{5}{{14}}$

Add the first two numbers,

$=  - 8 + \frac{{7 - 8}}{{10}} + \frac{{6 + 5}}{{14}}$

Subtract the numbers,

$=  - 8 - \frac{1}{{10}} + \frac{{11}}{{14}}$

Take LCM,

$= \frac{{ - 560 - 7 - 55}}{{70}}$

Subtract the numbers,

$= \frac{{ - 512}}{{70}}$

Divide numerator and denominator by $2,$

$= \frac{{ - 256}}{{35}}$

103. Verify $ - ( - x) = x$ for

(i) $x = \frac{3}{5}$

Ans:

Here, $ - x =  - \frac{3}{5} \Rightarrow  - ( - x) =  - \left( { - \frac{3}{5}} \right) \Rightarrow  - ( - x) = \frac{3}{5} = x$

(ii) $x = \frac{{ - 7}}{9}$

Ans:

Here, $x = \frac{{ - 7}}{9} \Rightarrow  - x =  - \left( { - \frac{7}{9}} \right) \Rightarrow  - x = \frac{7}{9} \Rightarrow  - ( - x) = \frac{{ - 7}}{9} = x$

(iii) $x = \frac{{13}}{{ - 15}}$

Ans:

Here, $x = \frac{{13}}{{ - 15}} \Rightarrow  - x =  - \left( {\frac{{13}}{{ - 15}}} \right) \Rightarrow  - x = \frac{{13}}{{15}} \Rightarrow  - ( - x) = \frac{{ - 13}}{{15}} = x$

104. Give one example each to show that the rational numbers are closed under addition, subtraction, and multiplication. Are rational numbers closed under division? Give two examples in support of your answer.

Ans:

(i) Addition:

$\frac{1}{2} + \frac{2}{3} = \frac{{3 + 4}}{6} = \frac{7}{6}$

(ii) Subtraction:

$\frac{1}{2} - \frac{2}{3} = \frac{{3 - 4}}{6} = \frac{{ - 1}}{6}$

(iii) Multiplication:

$\frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Thus, we see that for addition, subtraction as well as multiplication, the result that we get is itself a rational number. This means that rational numbers are closed under addition, subtraction and multiplication. 

Rational Numbers are closed under division except for zero.

105. Verify the property $x + y = y + x$ of rational numbers by taking

(a) $x = \frac{1}{2},y = \frac{1}{2}$

Ans:

Substitute $x = \frac{1}{2},y = \frac{1}{2}$ in $x + y = y + x$,

$\frac{1}{2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2}$

LHS $= \frac{1}{2} + \frac{1}{2}$

$= \frac{2}{2}$

$= 1$

RHS $= \frac{1}{2} + \frac{1}{2}$

$= \frac{2}{2}$

$= 1$

Here, LHS $=$ RHS

Hence, verified.

(b) $x = \frac{{ - 2}}{3},y = \frac{{ - 5}}{6}$

Ans:

Substitute $x = \frac{{ - 2}}{3},y = \frac{{ - 5}}{6}$ in $x + y = y + x$,

$\frac{{ - 2}}{3} + \left( {\frac{{ - 5}}{6}} \right) = \frac{{ - 5}}{6} + \left( {\frac{{ - 2}}{3}} \right)$

LHS$= \frac{{ - 2}}{3} + \left( {\frac{{ - 5}}{6}} \right)$

$= \frac{{ - 4 + ( - 5)}}{6}$

$= \frac{{ - 9}}{6}$

RHS $= \frac{{ - 5}}{6} + \left( {\frac{{ - 2}}{3}} \right)$

$= \frac{{ - 5 + ( - 4)}}{6}$

$= \frac{{ - 9}}{6}$

Here, LHS$=$RHS

Hence, verified.

(c) $x = \frac{{ - 3}}{7},y = \frac{{ - 20}}{{21}}$

Ans:

Substitute $x = \frac{{ - 3}}{7},y = \frac{{ - 20}}{{21}}$ in $x + y = y + x$,

$\frac{{ - 3}}{7} + \left( {\frac{{ - 20}}{{21}}} \right) = \frac{{ - 20}}{{21}} + \left( {\frac{{ - 3}}{7}} \right)$

LHS $= \frac{{ - 3}}{7} + \left( {\frac{{ - 20}}{{21}}} \right)$

$= \frac{{ - 9 + ( - 20)}}{{21}}$

$= \frac{{ - 29}}{{21}}$

RHS $= \frac{{ - 20}}{{21}} + \left( {\frac{{ - 3}}{7}} \right)$

$= \frac{{ - 20 + ( - 9)}}{{21}}$

$= \frac{{ - 29}}{{21}}$

Here, LHS$=$RHS

Hence, verified.

(d) $x = \frac{{ - 2}}{5},y = \frac{{ - 9}}{{10}}$

Ans:

Substitute $x = \frac{{ - 2}}{5},y = \frac{{ - 9}}{{10}}$ in $x + y = y + x$,

$\frac{{ - 2}}{5} + \left( {\frac{{ - 9}}{{10}}} \right) = \frac{{ - 9}}{{10}} + \left( {\frac{{ - 2}}{5}} \right)$

LHS $= \frac{{ - 2}}{5} + \left( {\frac{{ - 9}}{{10}}} \right)$

$= \frac{{ - 4 + ( - 9)}}{{10}}$

$= \frac{{ - 13}}{{10}}$

RHS $= \frac{{ - 9}}{{10}} + \left( {\frac{{ - 2}}{5}} \right)$

$= \frac{{ - 9 + ( - 4)}}{{10}}$

$= \frac{{ - 13}}{{10}}$

Here, LHS$=$RHS

Hence, verified.

106. Simplify each of the following by using suitable property. Also name the property.

(a) $\left[ {\frac{1}{2} \times \frac{1}{4}} \right] + \left[ {\frac{1}{2} \times 6} \right]$

Ans:

Distributive property

$\left[ {\frac{1}{2} \times \frac{1}{4}} \right] + \left[ {\frac{1}{2} \times 6} \right]$

$ \Rightarrow \frac{1}{2}\left[ {\frac{{25}}{4}} \right]$

$ \Rightarrow \frac{{25}}{8}$

(b) $\left[ {\frac{1}{5} \times \frac{2}{{15}}} \right] - \left[ {\frac{1}{5} \times \frac{2}{5}} \right]$

Ans:

Distributive property

$\frac{1}{5}\left[ {\frac{2}{{15}} + \frac{2}{5}} \right]$

$ \Rightarrow \frac{1}{2}\left[ {\frac{7}{{15}}} \right]$

$ \Rightarrow \frac{7}{{30}}$

(c) $\frac{{ - 3}}{5} \times \left\{ {\frac{3}{7} + \left( {\frac{{ - 5}}{6}} \right)} \right\}$

Ans:

Distributive property

$\frac{{ - 3}}{5} \times \frac{3}{7} + \frac{{ - 5}}{6} \times \frac{{ - 3}}{5}$

$ \Rightarrow \frac{{ - 9}}{{35}} + \frac{1}{2}$

$ \Rightarrow \frac{{ - 18 + 35}}{{70}}$

$ \Rightarrow \frac{{17}}{{70}}$

107. Tell which property allows you to compute $\frac{1}{5} \times \left[ {\frac{5}{6} \times \frac{7}{9}} \right]{\text{as}}\left[ {\frac{1}{5} \times \frac{5}{6}} \right] \times \frac{7}{9}$.

Ans:

While multiplying three or more rational numbers, they can be grouped in any order.

Therefore, the given computation illustrates Associative property of multiplication.

108. Verify the property $x \times y = y \times x$ of rational numbers by using

(a) $x = 7$ and $y = \frac{1}{2}$

Ans:

Substitute $x = 7$ and $y = \frac{1}{2}$ in $x \times y = y \times x$,

$7 \times \frac{1}{2} = \frac{1}{2} \times 7$ 

LHS $= 7 \times \frac{1}{2}$

$= \frac{7}{2}$

RHS $= \frac{1}{2} \times 7$

$= \frac{7}{2}$

Here, LHS $=$ RHS

Hence, verified.

(b) $x = \frac{2}{3}$ and $y = \frac{9}{4}$

Ans:

Substitute $x = \frac{2}{3}$ and $y = \frac{9}{4}$ in $x \times y = y \times x$,

$\frac{2}{3} \times \frac{9}{4} = \frac{9}{4} \times \frac{2}{3}$

LHS $= \frac{2}{3} \times \frac{9}{4}$

$= \frac{{18}}{{12}}$

$= \frac{3}{2}$

RHS $= \frac{9}{4} \times \frac{2}{3}$

$= \frac{{18}}{{12}}$

$= \frac{3}{2}$

Here, LHS $=$ RHS

Hence, verified.

(c) $x = \frac{{ - 5}}{7}$ and $y = \frac{{14}}{{15}}$

Ans:

Substitute $x = \frac{{ - 5}}{7}$ and $y = \frac{{14}}{{15}}$ in $x \times y = y \times x$,

$\frac{{ - 5}}{7} \times \frac{{14}}{{15}} = \frac{{14}}{{15}} \times \frac{{ - 5}}{7}$

LHS $= \frac{{ - 5}}{7} \times \frac{{14}}{{15}}$

$= \frac{{ - 2}}{3}$

RHS $= \frac{{14}}{{15}} \times \frac{{ - 5}}{7}$

$= \frac{{ - 2}}{3}$

Here, LHS $=$ RHS

Hence, verified.

(d) $x = \frac{{ - 3}}{8}$ and $y = \frac{{ - 4}}{9}$

Ans:

Substitute$x = \frac{{ - 3}}{8}$ and $y = \frac{{ - 4}}{9}$ in $x \times y = y \times x$,

$\frac{{ - 3}}{8} \times \frac{{ - 4}}{9} = \frac{{ - 4}}{9} \times \frac{{ - 3}}{8}$

LHS $= \frac{{ - 3}}{8} \times \frac{{ - 4}}{9}$

$= \frac{{ - 1}}{6}$

RHS $= \frac{{ - 4}}{9} \times \frac{{ - 3}}{8}$

$= \frac{{ - 1}}{6}$

Here, LHS $=$ RHS

Hence, verified.

109. Verify the property $x \times (y \times z) = (x \times y) \times z$ of rational numbers by using

(a) $x = 1,y = \frac{{ - 1}}{2},z = \frac{1}{4}$

Ans:

Substitute $x = 1,y = \frac{{ - 1}}{2},{\text{z}} = \frac{1}{4}$ in $x \times (y \times z) = (x \times y) \times z$,

$1 \times \left( {\frac{{ - 1}}{2} \times \frac{1}{4}} \right) = \left( {1 \times \frac{{ - 1}}{2}} \right) \times \frac{1}{4}$

LHS $= 1 \times \left( {\frac{{ - 1}}{2} \times \frac{1}{4}} \right)$

$= \frac{{ - 1}}{8}$

RHS $= 1 \times \left( {\frac{{ - 1}}{2} \times \frac{1}{4}} \right)$

$= \frac{{ - 1}}{8}$

Here, LHS $=$ RHS

Hence, verified.

(b) $x = \frac{2}{3},y = \frac{{ - 3}}{7},z = \frac{1}{2}$

Ans:

Substitute $x = \frac{2}{3},y = \frac{{ - 3}}{7},z = \frac{1}{2}$ in $x \times (y \times z) = (x \times y) \times z$,

$\frac{2}{3} \times \left( {\frac{{ - 3}}{7} \times \frac{1}{4}} \right) = \left( {\frac{2}{3} \times \frac{{ - 3}}{7}} \right) \times \frac{1}{2}$

LHS $= \frac{2}{3} \times \left( {\frac{{ - 3}}{7} \times \frac{1}{4}} \right)$

$= \frac{{ - 1}}{7}$

RHS $= \left( {\frac{2}{3} \times \frac{{ - 3}}{7}} \right) \times \frac{1}{2}$

$= \frac{{ - 1}}{7}$

$ \Rightarrow \frac{{ - 1}}{7} = \frac{{ - 1}}{7}$

Here, LHS $=$ RHS

Hence, verified.

(c) $x = \frac{{ - 2}}{7},y = \frac{{ - 5}}{6},z = \frac{1}{4}$

Ans:

Substitute $x = \frac{{ - 2}}{7},y = \frac{{ - 5}}{6},z = \frac{1}{4}$ in $x \times (y \times z) = (x \times y) \times z$,

$\frac{{ - 2}}{7} \times \left( {\frac{{ - 5}}{6} \times \frac{1}{4}} \right) = \left( {\frac{{ - 2}}{7} \times \frac{{ - 5}}{6}} \right) \times \frac{1}{4}$ 

LHS $= \frac{{ - 2}}{7} \times \left( {\frac{{ - 5}}{6} \times \frac{1}{4}} \right)$

$= \frac{5}{{84}}$

RHS $= \left( {\frac{{ - 2}}{7} \times \frac{{ - 5}}{6}} \right) \times \frac{1}{4}$

$= \frac{5}{{84}}$

Here, LHS $=$ RHS

Hence, verified.

(d) $x = 0,y = \frac{1}{2}$

Ans:

Substitute $x = 0,y = \frac{1}{2}$ in $x \times (y \times z) = (x \times y) \times z$,

$0 \times \frac{1}{2} = \frac{1}{2} \times 0$

LHS $= 0 \times \frac{1}{2}$

$= 0$

RHS $= \frac{1}{2} \times 0$

$= 0$

Here, LHS $=$ RHS

Hence, verified.

110. Verify the property $x \times (y + z) = x \times y + x \times z$ of rational numbers by taking

(a) $x = \frac{{ - 1}}{2},y = \frac{3}{4},z = \frac{1}{4}$

Ans:

Substitute $x = \frac{{ - 1}}{2},y = \frac{3}{4},z = \frac{1}{4}$ in $x \times (y + z) = x \times y + x \times z$,

$ - \frac{1}{2} \times \left( {\frac{3}{4} \times \frac{1}{4}} \right) =  - \frac{1}{2} \times \frac{3}{4} + \left( {\frac{{ - 1}}{2} \times \frac{1}{4}} \right)$

LHS $=  - \frac{1}{2} \times \left( {\frac{3}{4} \times \frac{1}{4}} \right)$

$= \frac{{ - 1}}{2} \times \frac{{3 + 1}}{4}$

$= \frac{{ - 1}}{2}$

RHS $= - \frac{1}{2} \times \frac{3}{4} + \left( {\frac{{ - 1}}{2} \times \frac{1}{4}} \right)$

$= \frac{{ - 3}}{8} + \frac{{ - 1}}{8}$

$= \frac{{ - 1}}{2}$

Here, LHS $=$ RHS

Hence, verified.

(b) $x = \frac{{ - 1}}{2},y = \frac{2}{3},z = \frac{3}{4}$

Ans:

Substitute $x = \frac{{ - 1}}{2},y = \frac{2}{3},z = \frac{3}{4}$ in $x \times (y + z) = x \times y + x \times z$,

$ - \frac{1}{2} \times \left( {\frac{2}{3} \times \frac{3}{4}} \right) =  - \frac{1}{2} \times \frac{2}{3} + \left( {\frac{{ - 1}}{2} \times \frac{3}{4}} \right)$

LHS $=  - \frac{1}{2} \times \left( {\frac{2}{3} \times \frac{3}{4}} \right)$

$= \frac{{ - 1}}{2} \times \frac{{2 \times 4 + 3 \times 3}}{{12}}$

$= \frac{{ - 1}}{2} \times \frac{{8 + 9}}{{12}}$

$= \frac{{ - 17}}{{24}}$

RHS $=  - \frac{1}{2} \times \frac{2}{3} + \left( {\frac{{ - 1}}{2} \times \frac{3}{4}} \right)$

$= \frac{{ - 1}}{3} + \frac{{ - 3}}{8}$

$= \frac{{ - 8 - 9}}{{24}}$

$= \frac{{ - 17}}{{24}}$

Here, LHS $=$ RHS

Hence, verified.

(c) $x = \frac{{ - 2}}{3},y = \frac{{ - 4}}{6},z = \frac{{ - 7}}{9}$

Ans:

Substitute $x = \frac{{ - 2}}{3},y = \frac{{ - 4}}{6},z = \frac{{ - 7}}{9}$ in $x \times (y + z) = x \times y + x \times z$,

$\frac{{ - 2}}{3} \times \left( {\frac{{ - 4}}{6} + \frac{{ - 7}}{6}} \right) = \frac{{ - 2}}{3} \times \frac{{ - 4}}{6} + \left( {\frac{{ - 2}}{3} \times \frac{{ - 7}}{9}} \right)$ 

LHS $= \frac{{ - 2}}{3} \times \left( {\frac{{ - 4}}{6} + \frac{{ - 7}}{6}} \right)$

$= \frac{{ - 2}}{3} \times \frac{{ - 12 - 14}}{{18}}$

$= \frac{{ - 2}}{3} \times \frac{{ - 26}}{{18}}$

$= \frac{{26}}{{27}}$

RHS $= \frac{{ - 2}}{3} \times \frac{{ - 4}}{6} + \left( {\frac{{ - 2}}{3} \times \frac{{ - 7}}{9}} \right)$

$= \frac{4}{9} + \frac{{14}}{{27}}$

$= \frac{{12 + 14}}{{27}}$

$= \frac{{26}}{{27}}$

Here, LHS $=$ RHS

Hence, verified.

(d) $x = \frac{{ - 1}}{5},y = \frac{{ - 2}}{{15}},z = \frac{{ - 3}}{{10}}$

Ans:

Substitute $x = \frac{{ - 1}}{5},y = \frac{{ - 2}}{{15}},{\text{z}} = \frac{{ - 3}}{{10}}$ in $x \times (y + z) = x \times y + x \times z$,

$\frac{{ - 1}}{5} \times \left( {\frac{2}{{15}} + \frac{{ - 3}}{{10}}} \right) = \frac{{ - 1}}{5} \times \frac{2}{{15}} + \left( {\frac{{ - 1}}{5} \times \frac{{ - 3}}{{10}}} \right)$

LHS $= \frac{{ - 1}}{5} \times \left( {\frac{2}{{15}} + \frac{{ - 3}}{{10}}} \right)$

$= \frac{{ - 1}}{5} \times \frac{{4 - 9}}{{30}}$

$= \frac{{ - 1}}{5} \times \frac{{ - 5}}{{30}}$

$= \frac{1}{{30}}$

RHS $= \frac{{ - 1}}{5} \times \frac{2}{{15}} + \left( {\frac{{ - 1}}{5} \times \frac{{ - 3}}{{10}}} \right)$

$= \frac{{ - 2}}{{75}} + \frac{3}{{50}}$

$= \frac{{ - 4 + 9}}{{150}}$

$= \frac{5}{{150}}$

$= \frac{1}{{30}}$

Here, LHS $=$ RHS

Hence, verified.

111. Use the distributive of multiplication of rational numbers over addition to simplify

(a) $\frac{3}{5} \times \left[ {\frac{{35}}{{24}} + \frac{{10}}{1}} \right]$

Ans:

By the distributive of multiplication of rational numbers over addition, 

$\frac{3}{5} \times \frac{{35}}{{24}} + \frac{3}{5} \times \frac{{10}}{1}$

$ \Rightarrow \frac{7}{8} + \frac{6}{1}$

$ \Rightarrow \frac{{7 + 48}}{8}$

$ \Rightarrow \frac{{55}}{8}$

(b) $\frac{{ - 5}}{4} \times \left[ {\frac{8}{5} + \frac{{16}}{{15}}} \right]$

Ans:

By the distributive of multiplication of rational numbers over addition, 

$\frac{{ - 5}}{4} \times \frac{8}{5} - \frac{5}{4} \times \frac{{16}}{{15}}$

$=  - 2 - \frac{4}{3}$

$= \frac{{10}}{3}$

(c) $\frac{2}{7} \times \left[ {\frac{7}{{16}} - \frac{{21}}{4}} \right]$

Ans:

By the distributive of multiplication of rational numbers over addition, 

$\frac{2}{7} \times \frac{7}{{16}} - \frac{{ - 2}}{7} \times \frac{{21}}{4}$

$= \frac{1}{8} - \frac{3}{2}$

$= \frac{{ - 11}}{8}$

(d) $\frac{3}{4} \times \left[ {\frac{8}{9} - 40} \right]$

Ans:

By the distributive of multiplication of rational numbers over addition, 

$\frac{3}{4} \times \frac{8}{9} + \frac{3}{4} \times ( - 40)$

$= \frac{2}{3} - 30$

$=  - \frac{{88}}{3}$

112. Simplify-

(a) $\frac{{{\mathbf{32}}}}{{\mathbf{5}}}{\mathbf{ + }}\frac{{{\mathbf{23}}}}{{{\mathbf{11}}}}{\mathbf{ + }}\frac{{{\mathbf{22}}}}{{{\mathbf{15}}}}$

Ans:

$\frac{{32}}{5} + \frac{{23}}{{11}} + \frac{{22}}{{15}}$

$= \frac{{32}}{5} + \frac{{46}}{{15}}$

Take LCM,

$= \frac{{96 + 46}}{{15}}$

$= \frac{{142}}{{15}}$

(b) $\frac{{\mathbf{3}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{{\mathbf{28}}}}{{{\mathbf{15}}}}{\mathbf{ \div }}\frac{{{\mathbf{14}}}}{{\mathbf{5}}}$

Ans:

$\frac{3}{7} \times \frac{{28}}{{15}} \div \frac{{14}}{5}$

Rewrite the terms,

$= \frac{3}{7} \times \frac{{28}}{{15}} \times \frac{5}{{14}}$

$= \frac{2}{7}$

(c) $\frac{{\mathbf{3}}}{{\mathbf{7}}}{\mathbf{ + }}\frac{{ - {\mathbf{2}}}}{{{\mathbf{21}}}}{\mathbf{ \times }}\frac{{ - {\mathbf{5}}}}{{\mathbf{6}}}$

Ans:

$\frac{3}{7} + \frac{{ - 2}}{{21}} \times \frac{{ - 5}}{6}$ 

$= \frac{3}{7} + \frac{5}{{63}}$ 

Take LCM,

$= \frac{{27 + 5}}{{63}}$ 

$= \frac{{32}}{{63}}$ 

(d) $\frac{7}{8} + \frac{1}{{16}} - \frac{1}{{12}}$

Ans:

$\frac{7}{8} + \frac{1}{{16}} - \frac{1}{{12}}$ 

Take LCM,

$= \frac{{42 + 3 - 4}}{{48}}$ 

$= \frac{{41}}{{48}}$

113. Identify the rational number that does not belong with the other three. Explain your reasoning

$\frac{{ - 5}}{6},\frac{{ - 1}}{2},\frac{{ - 4}}{9},\frac{{ - 7}}{3}$.

Ans:

$\frac{{ - 1}}{2}$ does not belong to these because the value $0.5$ is non-recurring

Whereas $\frac{{ - 5}}{{11}} = 0.\overline {45} ,\frac{{ - 4}}{9} = 0.\bar 4,\frac{{ - 7}}{3} = 2.\bar 3$ are recurring.

114. The cost of $\frac{{{\mathbf{19}}}}{{\mathbf{4}}}$ meters of wire is Rs \[\frac{{{\mathbf{171}}}}{{\mathbf{2}}}\] Find the cost of one meter of the wire.

Ans:

$\frac{{19}}{4}$ meter wire costs $=$ Rs. $\frac{{171}}{2}$

$1$ meter wire will cost $=$ Rs. $\frac{{171}}{2} \times \frac{4}{{19}}$

Therefore, Cost of $1$ meter wire $=$ Rs.$38$

115. A train travels $\frac{{{\mathbf{1445}}}}{{\mathbf{2}}}{\mathbf{km}}$ in $\frac{{{\mathbf{17}}}}{{\mathbf{2}}}$ hours. Find the speed of the train in ${\mathbf{km/h}}$.

Ans:

Given that,

${\text{ Distance }} = \frac{{1445}}{2}{\text{km}}$

${\text{ Time }} = \frac{{17}}{2}{\text{ Hours }}$

It is known that,${\text{ Speed }} = \frac{{{\text{ Distance }}}}{{{\text{ Time }}}}$

$ \Rightarrow {\text{Speed }} = \frac{{\frac{{1445}}{2}}}{{\frac{{17}}{2}}}$

$ \Rightarrow {\text{Speed }} = \frac{{1445}}{2} \times \frac{2}{{17}}$

$ \Rightarrow {\text{Speed }} = 85{\text{kmph}}$

The speed of the train is $85{\text{kmph}}$.

116. If ${\mathbf{16}}$ shirts of equal size can be made out of ${\mathbf{24m}}$of cloth, how much cloth is needed for making one shirt?

Ans:

$16$ shirts can be made out of $= 24{\text{m}}$ of cloth

$1$ shirt can be made out of $= \frac{{24}}{{16}}{\text{m}}$ of cloth

Therefore, $1.5{\text{m}}$ cloth will be needed for one shirt.

117. $\frac{{\mathbf{7}}}{{{\mathbf{11}}}}$ of all the money in Hamid's bank account is Rs.\[{\mathbf{77,000}}\]. How much money does Hamid have in his bank account?

Ans:

Let all the Hamid's money is $x$.

${\text{ Total money }} = \frac{7}{{11}} \times x = 77000$

$ \Rightarrow x = \frac{{770000 \times 11}}{7}$

$ \Rightarrow x = 121000$

Therefore, Hamid have Rs. $121000$ in his bank account.

118. A ${\mathbf{117}}\frac{{\mathbf{1}}}{{\mathbf{3}}}{\mathbf{m}}$ long rope is cut into equal pieces measuring ${\mathbf{7}}\frac{{\mathbf{1}}}{{\mathbf{3}}}{\mathbf{m}}$ each. How many such small pieces are these?

Ans:

Given that,

A $117\frac{1}{3}{\mathbf{m}}$ long rope is cut into equal pieces measuring $7\frac{1}{3}{\mathbf{m}}$ each.

Now, Size of rope $= 117\frac{1}{3}{\text{m}} = \frac{{352}}{3}$

Size of small pieces $= 7\frac{1}{3}{\text{m}} = \frac{{22}}{3}$

Number of small pieces $= \frac{{\frac{{352}}{3}}}{{\frac{{22}}{3}}}$

Number of small pieces $= \frac{{352}}{2} \times \frac{3}{{22}}$

Number of small pieces $= 24$

119. $\frac{{\mathbf{1}}}{{\mathbf{6}}}$ of the class students are above average, $\frac{{\mathbf{1}}}{{\mathbf{4}}}$ are average and rest are below average. If there are ${\mathbf{48}}$ students in all, how many students are below average in the class?

Ans:

Here, $\frac{1}{6}$ of the class students are above average, $\frac{1}{4}$ are average and rest are below average and there are $48$ students

Then, Total number of below average students $= 1 - \frac{1}{6} - \frac{1}{4}$ $= 1 - \frac{1}{6} - \frac{1}{4} = \frac{{12 - 2 - 3}}{{12}} = \frac{7}{{12}}$

Number of students below average $= \frac{7}{{12}} \times 48 = 28$

120. $\frac{{\mathbf{2}}}{{\mathbf{5}}}$ of total number of students of a school come by car while $\frac{{\mathbf{1}}}{{\mathbf{4}}}$ of students come by bus to school. The other entire students walk to school of which \[\frac{{\mathbf{1}}}{{\mathbf{3}}}\] walk on their own and the rest are escorted by their parents. If \[{\mathbf{224}}\] students come to school walking on their own, how many students study in that school?

Ans:

Let the total number of students coming to school be $x$.

Students coming by car $= \frac{2}{5}x$

Students coming by bus $= \frac{1}{4}x$

Remaining students who walk to school$= x - \left( {\frac{2}{5}x + \frac{1}{4}x} \right) = x - \left( {\frac{{8x + 5x}}{{20}}} \right)$

$= x - \frac{{13x}}{{20}} = \frac{{20x - 13x}}{{20}} = \frac{{7x}}{{20}}$

Students walk on their own $= \frac{1}{3}$ of $\frac{{7x}}{{20}} = \frac{{7x}}{{60}}$

\[224\]students walk on their own

$\frac{{7x}}{{60}} = 224$

$x = \frac{{224 \times 60}}{7} = 32 \times 60 = 1920$

Therefore, Total number of students is \[1920{\text{ }}.\]

121. Huma, Hubna and Seema received a total of Rs.\[{\mathbf{2,016}}\] as monthly allowance from their mother such that Seema gets $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ of what Huma gets and Hubna gets ${\mathbf{2}}\frac{{\mathbf{1}}}{{\mathbf{3}}}$ times

Seema's share. How much money do the three sisters get individually?

Ans:

Allowances of Seema $= \frac{1}{2}$ of Huma

Allowances of Hubna $= 1\frac{2}{3}$ of Seema 's share

$= \frac{5}{3}$ of Seema 's share

$= \frac{5}{3}$ of $\frac{1}{2}$ of Huma's share

$= \frac{5}{3} \times \frac{1}{2}$ of Huma's share 

$= \frac{5}{6}$ of huma 's share

Total amount $= 2016$

$x + \frac{1}{2}x + \frac{5}{6}x =$ $\frac{{14}}{6}x = 2016$ $x = 2016 \times \frac{6}{{14}}$ $x = 864$

Huma's share $=$ Rs\[.{\text{ }}864\]

Seema's share $= \frac{1}{2} \times 864 = 432$

Hubna's share $= \frac{5}{3} \times \frac{1}{2} \times 864$$=$ Rs\[.{\text{ }}720\].

Therefore, Hubna's share is Rs\[.{\text{ }}720\].

122. A mother and her two daughters got a room constructed for Rs.\[{\mathbf{62}},{\mathbf{000}}\]. The elder daughter contributes $\frac{{\mathbf{3}}}{{\mathbf{8}}}$ of her mother's contribution while the younger daughter contributes $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ of her mother's share. How much do the three contribute individually?

Ans:

Let the share of the mother is $x$.

Then, Elder daughter's share $= \frac{3}{8}x$

Younger daughter's share $= \frac{1}{2}x$

Total amount $= 620000$

That is, $x + \frac{3}{8}x + \frac{1}{2}x = 62000$

$\frac{{8x + 3x + 4x}}{8} = 62000$

$\frac{{15x}}{8} = 62000$

$x = 62000 \times \frac{8}{{15}}$

$x = 33066$ 

So, Mother contributes Rs. $33066$

Elder daughter's share $= \frac{3}{8} \times 33066 \approx {\text{Rs}}.12400$

Younger daughter's share $= \frac{1}{2} \times 33066 = {\text{Rs}}{\text{.}}16533$

123. Tell which property allows you to compare

$\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\left[ {\frac{{\mathbf{3}}}{{\mathbf{4}}}{\mathbf{ \times }}\frac{{\mathbf{5}}}{{\mathbf{7}}}} \right]{\mathbf{ and }}\left[ {\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{ \times }}\frac{{\mathbf{5}}}{{\mathbf{7}}}} \right]{\mathbf{ \times }}\frac{{\mathbf{3}}}{{\mathbf{4}}}$.

Ans:

Use commutative property of multiplication in $\frac{3}{4} \times \frac{5}{7}$. 

$\frac{2}{3} \times \left[ {\frac{3}{4} \times \frac{5}{7}} \right] = \frac{2}{3} \times \left[ {\frac{5}{7} \times \frac{3}{4}} \right]$

Use associative property of multiplication in $\frac{2}{3} \times \left[ {\frac{5}{7} \times \frac{3}{4}} \right]$.

$\frac{2}{3} \times \left[ {\frac{3}{4} \times \frac{5}{7}} \right] = \frac{2}{3} \times \left[ {\frac{5}{7} \times \frac{3}{4}} \right]$

$= \left[ {\frac{2}{3} \times \frac{5}{7}} \right] \times \frac{3}{4}$

Hence, the statement $\frac{2}{3} \times \left[ {\frac{3}{4} \times \frac{5}{7}} \right] = \left[ {\frac{2}{3} \times \frac{5}{7}} \right] \times \frac{3}{4}$ shows the commutative and associative property of multiplication.

124. Name the property used in each of the following-

(i) $ - \frac{7}{{11}} \times \frac{{ - 3}}{5} = \frac{{ - 3}}{5} \times \frac{{ - 7}}{{11}}$

Ans:

Two rational numbers can be multiplied in any order. Multiplication is commutative for rational numbers.

In general, $a \times b = b \times a$ for any two rational numbers $a$ and $b$. 

Therefore, the statement $ - \frac{7}{{11}} \times \frac{{ - 3}}{5} = \frac{{ - 3}}{5} \times \frac{{ - 7}}{{11}}$ shows the commutative property of multiplication.

(ii) $\frac{{ - 2}}{3} \times \left[ {\frac{3}{4} + \frac{{ - 1}}{4}} \right] = \left[ {\frac{{ - 2}}{3} \times \frac{3}{4}} \right] + \left[ {\frac{{ - 2}}{3} \times \frac{{ - 1}}{2}} \right]$

Ans:

In general, $a \times (b + c) = (a \times b) + (a \times c)$ for any three rational numbers \[a,{\text{ }}b\]and $c$. 

Therefore, the statement $ - \frac{2}{3} \times \left[ {\frac{3}{4} + \frac{{ - 1}}{2}} \right] = \left[ {\frac{{ - 2}}{3} \times \frac{3}{4}} \right] + \left[ {\frac{{ - 2}}{3} \times \frac{{ - 1}}{2}} \right]$ shows the distributive property of multiplication over addition.

(iii) $\frac{{\mathbf{1}}}{{\mathbf{3}}}{\mathbf{ + }}\left[ {\frac{{\mathbf{4}}}{{\mathbf{9}}}{\mathbf{ + }}\left( {\frac{{ - {\mathbf{4}}}}{{\mathbf{3}}}} \right)} \right]{\mathbf{ = }}\left[ {\frac{{ - {\mathbf{1}}}}{{\mathbf{3}}}{\mathbf{ + }}\frac{{\mathbf{4}}}{{\mathbf{9}}}} \right]{\mathbf{ + }}\left[ {\frac{{ - {\mathbf{4}}}}{{\mathbf{3}}}} \right]$

Ans:

According to Associative Property of Addition $\frac{1}{3} + \left[ {\frac{4}{9} + \left( {\frac{{ - 4}}{3}} \right)} \right]$can be written as $\left[ {\frac{{ - 1}}{3} + \frac{4}{9}} \right] + \left[ {\frac{{ - 4}}{3}} \right]$

(iv) $\frac{{ - 2}}{7} + 0 = 0 + \frac{2}{7} = \frac{{ - 2}}{7}$

Ans:

If we add any rational number with 0, we get that rational number as the sum.

In general, $a + 0 = a = 0 + a$ for any rational number $a$.

Here, $\frac{{ - 2}}{7}$ is added with 0 , then the sum will be $\frac{{ - 2}}{7}$.

Therefore, the statement $\frac{{ - 2}}{7} + 0 = 0 + \frac{{ - 2}}{7} =  - \frac{2}{7}$ shows identity property of addition.

(v) \[\frac{{\mathbf{3}}}{{\mathbf{8}}}{\mathbf{ \times 1 = 1 \times }}\frac{{\mathbf{3}}}{{\mathbf{8}}}{\mathbf{ = }}\frac{{\mathbf{3}}}{{\mathbf{8}}}\]

Ans:

If we multiply any rational number with $1$, we get that rational number as the product. In general, $a \times 1 = a = 1 \times a$ for any rational number a.

Here, $\frac{3}{8}$ is multiplied with $1$, then the product will be $\frac{3}{8}$.

Therefore, the statement $\frac{3}{8} \times 1 = 1 \times \frac{3}{8} = \frac{3}{8}$ shows identity property of multiplication.

125. Find the multiplicative inverse of

(i) $ - 1\frac{1}{8}$

Ans:

Rewrite the fraction as $\frac{{ - 9}}{8}$.

Let $x$ be the multiplicative inverse.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{{ - 9}}{8} \times x = 1$

Divide each side with $\frac{{ - 9}}{8}$, 

$x = \frac{1}{{\frac{{ - 9}}{8}}} = \frac{{ - 8}}{9}$

Therefore, multiplicative inverse of $ - 1\frac{1}{8}$ is $\frac{{ - 8}}{9}$.

(ii) ${\mathbf{3}}\frac{{\mathbf{1}}}{{\mathbf{3}}}$

Ans:

Rewrite the fraction as $\frac{{10}}{9}$.

Let $x$ be the multiplicative inverse.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $\frac{{10}}{3} \times x = 1$

Divide each side with $\frac{{10}}{3}$, 

$x = \frac{1}{{\frac{{10}}{3}}} = \frac{3}{{10}}$

Therefore, multiplicative inverse of $3\frac{1}{3}$ is $\frac{3}{{10}}$ .

126. Arrange the numbers $\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{,}}\frac{{{\mathbf{13}}}}{{{\mathbf{16}}}}{\mathbf{,}}\frac{{\mathbf{5}}}{{\mathbf{8}}}$ in descending order.

Ans:

Let's make their denominator same:

$\frac{1}{4} \times \frac{4}{4} = \frac{4}{{16}}$

$\frac{{13}}{{16}}$

$\frac{5}{8} \times \frac{2}{2} = \frac{{10}}{{16}}$

Descending order:

$\frac{{20}}{{16}},\frac{{13}}{{16}},\frac{4}{{16}}$ or $\frac{5}{8},\frac{{13}}{{16}},\frac{1}{4}$.

127. The product of two rational numbers is $\frac{{ - {\mathbf{14}}}}{{{\mathbf{27}}}}$. If one of the numbers be $\frac{{\mathbf{7}}}{{\mathbf{9}}}$, find the other.

Ans:

Given that, one number is $\frac{7}{9}$.

Let the other number is $x$.

It is also given that, the product of two rational numbers is $\frac{{ - 14}}{{27}}$. 

Then,  $\frac{{ - 14}}{{27}} = \frac{7}{9} \times x$

$x = \frac{{ - 9 \times 14}}{{7 \times 27}}$

$x = \frac{{ - 2}}{3}$

Therefore, the other number is $\frac{{ - 2}}{3}$.

128. By what numbers should we multiply $\frac{{ - {\mathbf{15}}}}{{{\mathbf{20}}}}$ so that the product may be $\frac{{ - {\mathbf{5}}}}{{\mathbf{7}}}{\mathbf{?}}$

Ans:

Let the number is $x$.

Then, $x \times \frac{{ - 15}}{{20}} = \frac{{ - 5}}{7}$

$x = \frac{{ - 5 \times 20}}{{ - 15 \times 7}}$

$x = \frac{{22}}{{21}}$

Therefore, we should multiply $\frac{{ - 15}}{{20}}$ by $\frac{{22}}{{21}}$ so that the product may be $\frac{{ - 5}}{7}.$

129. By what number should we multiply $\frac{{ - {\mathbf{8}}}}{{{\mathbf{13}}}}$ so that the product may be ${\mathbf{24}}$?

Ans:

Let the number is $x$.

$x \times \frac{{ - 8}}{{13}} = 24$

Then, $x = \frac{{24 \times 13}}{{ - 8}}$

Multiply and divide the numbers,

$x =  - 39$

Therefore, we should multiply $\frac{{ - 8}}{{13}}$ by $ - 39$ so that the product may be $24$

130. The product of two rational numbers is $ - {\mathbf{7}}$. If one of the number is $ - {\mathbf{5}}$, find the other?

Ans:

Let the other number is $x$.

Then, $x \times ( - 5) =  - 7$

$x = \frac{{ - 7}}{{ - 5}}$

$x = \frac{7}{5}$

Therefore, the other number is $\frac{7}{5}$.

131.Can you find a rational number whose multiplicative inverse is \[ - {\mathbf{1}}\]?

Ans:

Let $x$ be the multiplicative inverse.

The reciprocal of a number is a number (multiplicative inverse) which when multiplied by the original number yields $1.$

Then, $ - 1 \times x = 1$

Divide each side with $ - 1$, 

$x = \frac{1}{{ - 1}} =  - 1$

Therefore, multiplicative inverse of $ - 1$ is $ - 1$ .

132. Find five rational numbers between ${\mathbf{0}}$ and ${\mathbf{1}}$.

Ans:

Rational numbers between $2$ numbers $= \frac{{a + b}}{2}$

(i) $\frac{{0 + 1}}{2} = \frac{1}{2}$

(ii) $\frac{{1 + \frac{1}{2}}}{2} = \frac{{\frac{3}{2}}}{2} = \frac{3}{4}$

(iii) $\frac{{\frac{1}{2} + \frac{3}{4}}}{2} = \frac{{\frac{{2 + 3}}{4}}}{2} = \frac{{\frac{5}{4}}}{2} = \frac{5}{8}$

133. Find two rational numbers whose absolute value is \[\frac{{\mathbf{1}}}{{\mathbf{5}}}\].

Ans:

Absolute numbers are the numbers which can be plotted on the number line.

For $\frac{1}{5} = 0.2$ the absolute numbers are $ - 0.1$ and $0.1$.

134. From a rope \[{\mathbf{40}}\] metres long, pieces of equal size are cut. If the length of one piece is $\frac{{{\mathbf{10}}}}{{\mathbf{3}}}$ metre, find the number of such pieces.

Ans:

Length of one piece $= \frac{{10}}{3}$

Now, Total number of pieces $= \frac{{40}}{{\frac{{10}}{3}}} = \frac{{40 \times 3}}{{10}} = 12$

135. ${\mathbf{5}}\frac{{\mathbf{1}}}{{\mathbf{2}}}$ metres long rope is cut into ${\mathbf{12}}$ equal pieces. What is the length of each piece?

Ans:

Length of rope $ - 5\frac{1}{2}$

Number of pieces $= 12$

Therefore, Length of each piece $= \frac{{5\frac{1}{2}}}{{12}} = \frac{{\frac{{11}}{2}}}{{12}} = \frac{{11}}{{24}}$

136. Write the following rational numbers in the descending order.

$\frac{{\mathbf{8}}}{{\mathbf{7}}}{\mathbf{,}}\frac{{ - {\mathbf{9}}}}{{\mathbf{8}}}{\mathbf{,}}\frac{{ - {\mathbf{3}}}}{{\mathbf{2}}}{\mathbf{,0,}}\frac{{\mathbf{2}}}{{\mathbf{5}}}$

Ans:

Let’s make the denominators same:

$\frac{8}{7} \times \frac{{40}}{{40}} = \frac{{320}}{{280}}$

$\frac{{ - 9}}{8} \times \frac{{35}}{{35}} = \frac{{ - 315}}{{280}}$

$\frac{{ - 3}}{2} \times \frac{{140}}{{140}} = \frac{{ - 420}}{{280}}$

$\frac{2}{5} \times \frac{{56}}{{56}} = \frac{{102}}{{280}}$

According to this,

The rational numbers in the descending order is,

$ - \frac{8}{7},\frac{{ - 9}}{8},\frac{{ - 3}}{2},0,\frac{2}{5}$

137. Find.

(i) ${\mathbf{0 \div }}\frac{{\mathbf{2}}}{{\mathbf{3}}}$

Ans:

Rewrite and simplify the expression,

$\frac{0}{{\frac{2}{3}}} = \frac{{0 \times 3}}{2} = 0$

(ii) $\frac{{\mathbf{1}}}{{\mathbf{3}}}{\mathbf{ \times }}\frac{{ - {\mathbf{5}}}}{{\mathbf{7}}}{\mathbf{ \times }}\frac{{ - {\mathbf{21}}}}{{{\mathbf{10}}}}$

Ans:

$\frac{1}{3} \times \frac{{ - 5}}{7} \times \frac{{ - 21}}{{10}}$

Cancel the numbers,

$\frac{1}{3} \times \frac{{ - 1}}{1} \times \frac{{ - 3}}{2}$

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

138. On a winter day the temperature at a place in Himachal Pradesh was $ - {\mathbf{1}}{{\mathbf{6}}^{\mathbf{^\circ }}}{\mathbf{C}}$ . Convert it in degree Fahrenheit ( $^{\mathbf{^\circ }}{\mathbf{F}}$ ) by using the formula.

$\frac{C}{5} = \frac{{F - 32}}{9}$

Ans:

Temperature in Himanchal Pradesh $=  - {16^\circ }{\text{C}}$

Formula to convert in Fahrenheit is,

$\frac{C}{5} = \frac{{F - 32}}{9}$

$\frac{{ - 16}}{5} = \frac{{F - 32}}{9}$

Cross multiply,

$9 \times \frac{{ - 16}}{5} = F - 32$

Multiply the numbers,

$\frac{{ - 144}}{5} = F - 32$

$F = \frac{{ - 144 + 160}}{5}$

$F = \frac{{16}}{5}$

$F = \frac{{16}}{5} = {3.2^\circ }{\text{F}}$

139. Find the sum of additive inverse and multiplicative inverse of \[{\mathbf{7}}\] .

Ans:

Additive inverse of $7$$=  - 7$

Multiplicative inverse of $7$$= \frac{1}{7}$

Now, the sum of additive inverse and multiplicative inverse of $7$is,

$ - 7 + \frac{1}{7} = \frac{{ - 48}}{7}$.

140. Find the product of additive inverse and multiplicative inverse of $ - \frac{{\mathbf{1}}}{{\mathbf{3}}}$.

Ans:

Additive inverse of $ - \frac{1}{3}$ $= \frac{1}{3}$

Multiplicative inverse of $ - \frac{1}{3}$$=  - 3$

Their product is $\frac{1}{3} \times ( - 3) =  - 1$.

141. The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:

(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull? 

Ans:

The length of Albatross wing $= 3\frac{3}{5}$

Length of Sea gull $= 1\frac{7}{{10}}$

Difference between the lengths of their wingspan is,

$\frac{3}{5} - 1\frac{7}{{10}}$

$\frac{{18}}{5} - \frac{{17}}{{10}}$

$\frac{{36 - 17}}{{10}} = \frac{{19}}{{10}} = 1\frac{9}{{10}}$

(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay?

Ans:

Length of Blue ray wings $= \frac{{41}}{{100}}{\text{m}}$

${\text{ Difference }} = \frac{5}{2} - \frac{{41}}{{100}}$

$\frac{{250 - 41}}{{100}} = \frac{{209}}{{100}} = 2\frac{9}{{100}}$

142. Shalini has to cut out circles of diameter ${\mathbf{1}}\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{cm}}$ from an Aluminium strip of dimensions ${\mathbf{8}}\frac{{\mathbf{3}}}{{\mathbf{4}}}{\mathbf{cm}}$ by ${\mathbf{1}}\frac{{\mathbf{1}}}{{\mathbf{4}}}{\mathbf{cm}}$. How many full circles can Shalini cut? Also calculate the wastage of the Aluminium strip.

Ans:

Diameter of the circle $= 1\frac{1}{4} = \frac{5}{4}{\text{cm}}$

Length of Aluminum strip $= 1\frac{1}{4} = \frac{5}{4}{\text{cm}}$

Width of Aluminium $= 8\frac{3}{4} = \frac{{35}}{4}{\text{cm}}$

Number of circles that can be cut $=$Length of aluminium

Strip/diameter of circle $= \frac{{\frac{{35}}{4}}}{{\frac{5}{4}}} = \frac{{35}}{4} \times \frac{4}{5} = 7$

Area of aluminium strip $=$ length $ \times $ width

$\frac{{35}}{4} \times \frac{5}{4} = \frac{{175}}{{16}}{\text{c}}{{\text{m}}^2}$

Area of circle $= \pi {{\text{r}}^2}$

$\frac{{22}}{7} \times \frac{5}{8} \times \frac{5}{8} = \frac{{550}}{{448}}$

Wastage of aluminium strip $= \frac{{175}}{{164}} - \frac{{550}}{{448}}$ $\frac{{4900 - 550}}{{448}} = \frac{{4350}}{{448}}$.

143. One fruit salad recipe requires $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ cup of sugar. Another recipe for the same fruit salad requires ${\mathbf{2}}$ tablespoons of sugar. If ${\mathbf{1}}$ tablespoon is equivalent to $\frac{{\mathbf{1}}}{{{\mathbf{16}}}}$ cup, how much more sugar does the first recipe require?

Ans:

$1$ tablespoon is equivalent $= \frac{1}{{16}}$ cup

Requirement for $1$ cup recipe $= \frac{1}{2}$ cup of sugar

Same recipe for same fruit $= 2$ tablespoon of sugar $=$$2 \times \frac{1}{{16}}$ cups of sugar $= \frac{1}{8}$ cups of sugar

First recipe requires $= \frac{1}{2} - \frac{1}{8}$$=$$\frac{{4 - 1}}{8} = \frac{3}{8}$ cups of sugar.

144. Four friends had a competition to see how far they could hop on one foot. The table given shows the distance covered by each-

(a) How farther did Soni hop than Nancy? 

Ans:

Distance covered by Soni $= \frac{1}{{20}}{\text{km}}$

Distance covered by Nancy $= \frac{1}{{32}}{\text{km}}$

The distance covered by Soni than Nancy $= \frac{1}{{20}} - \frac{1}{{32}} = \frac{{8 - 5}}{{160}} = \frac{3}{{160}}{\text{Km}}{\text{.}}$

(b) What is the total distance covered by Seema and Megha? 

Ans:

Distance covered by Seema $= \frac{1}{{25}}{\text{km}}$

Distance covered by Megha $= \frac{1}{{40}}{\text{km}}$

Total distance covered by Seema and Megha$= \frac{1}{{25}} + \frac{1}{{40}} = \frac{{8 + 5}}{{200}} = \frac{{13}}{{200}}$ ${\text{km}}$.

(c) Who walked farther, Nancy or Megha?

Ans:

Distance covered by Nancy $= \frac{1}{{32}}{\text{km}}$

Distance covered by Megha $= \frac{1}{{40}}{\text{km}}$

Therefore, Nancy walked further.

145. The table given below shows the distances, in kilometres, between four villages of a state. To find the distance between two villages, locate the square where the row for one village and the column for the other village intersect.

(a) Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh? 

Ans:

Distance between Himgton and Rawalpur $= 98\frac{3}{4}{\text{km}} = \frac{{395}}{4}{\text{km}}$

Distance between sonapur and ramgarh $= 40\frac{2}{3}{\text{km}} = \frac{{122}}{3}{\text{km}}$ 

On subtracting,

$\frac{{3954}}{4} - \frac{{122}}{3} = \frac{{1185}}{{12}} = \frac{{697}}{{12}} = \frac{{58}}{{58.08}}{\text{km}}$

Therefore, the distance between Sonapur and Ramgarh is more than from Himgton and Rawalpur

(b) If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive?

Ans:

Distance between Himgton to Sonpur $= 100\frac{5}{6}$

Distance between Sonapur to Rawalpur $= 16\frac{1}{2}$

Total distance $= 100\frac{5}{6} + 16\frac{1}{2}$

That is, $100\frac{5}{6} + 16\frac{1}{2} = \frac{{605}}{6} + \frac{{33}}{2} = \frac{{605 + 33}}{6} = \frac{{638}}{6}{\text{km}}$

146. The table shows the portion of some common materials that are recycled.

(a) Is the rational number expressing the amount of paper recycled more than $\frac{{\mathbf{1}}}{{\mathbf{2}}}$ or less than

$\frac{{\mathbf{1}}}{{\mathbf{2}}}{\mathbf{?}}$

Ans:

Paper recycled $= \frac{5}{{11}}$ , which is less than $\frac{1}{2}$.

(b) Which items have a recycled amount less than

$\frac{{\mathbf{1}}}{{\mathbf{2}}}{\mathbf{?}}$

Ans:

Paper and Glass have a recycled amount less than $\frac{1}{2}$.

(c) Is the quantity of aluminium cans recycled more (or less) than half of the quantity of aluminium cans?

Ans:

The quantity of aluminium cans recycled more than half of the quantity of aluminium

(d) Arrange the rate of recycling the materials from the greatest to the smallest.

Ans:

The rate of recycling the materials from the greatest to the smallest is,

$\frac{2}{5},\frac{5}{{11}},\frac{5}{8},\frac{1}{2}$

147. The overall width in cm of several wide-screen televisions are ${\mathbf{97}}{\mathbf{.28cm,98}}\frac{{\mathbf{4}}}{{\mathbf{9}}}{\mathbf{cm,98}}\frac{{\mathbf{1}}}{{{\mathbf{25}}}}{\mathbf{cm}}$ and ${\mathbf{97}}{\mathbf{.94cm}}$. Express these numbers as rational numbers in the form $\frac{p}{q}$ and arrange the widths in ascending order.

Ans:

Rewrite the numbers as improper fractions,

$97.28 = \frac{{9728}}{{100}}$

$98\frac{4}{9} = \frac{{895}}{9}$

$98\frac{1}{{25}} = \frac{{2451}}{{25}}$

$97.94 = \frac{{9794}}{{100}}$

The arrangement is,

$\frac{{9728}}{{100}},\frac{{895}}{9},\frac{{2451}}{{25}},\frac{{9794}}{{100}}$

Their ascending order is,

$\frac{{9728}}{{100}},\frac{{9795}}{{100}},\frac{{2451}}{{25}},\frac{{895}}{9}$

148. Roller Coaster at an amusement park is $\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{m}}$ high. If a new roller coaster is built that is \[\frac{{\mathbf{3}}}{{\mathbf{5}}}\] times the height of the existing coaster, what will be the height of the new roller coaster?

Ans:

Height of the roller coaster of the amusement park $= \frac{2}{3}{\text{m}}$

The height of new roller coaster is $\frac{3}{5}$ higher than the first roller coaster.

Then, the height of the new roller coaster $= \frac{3}{5} \times \frac{2}{3} = \frac{2}{5}{\text{m}}$.

149. Here is a table which gives the information about the total rainfall for several months compared to the average monthly rains of a town. Write each decimal in the form of rational number $\frac{p}{q}$.

Ans:

Consider \[2.6924\]:

Multiply and divide the decimal by $10000,$

 $2.6924 = \frac{{26924}}{{10000}}$

Consider \[0.6069\]:

Multiply and divide the decimal by $10000,$

$0.6096 = \frac{{6096}}{{10000}}$

Consider \[ - 6.9088\]:

Multiply and divide the decimal by $10000,$

$ - 6.9088 = \frac{{ - 69088}}{{10000}}$

Consider \[ - 8.636\]:

Multiply and divide the decimal by $1000,$

$ - 8.636 = \frac{{ - 8636}}{{1000}}$

150. The average life expectancies of males for several states are shown in the table. Express each decimal in the form $\frac{p}{q}$ and arrange the states from the least to the greatest male life expectancy.

State-wise data are included below; more indicators can be found in the "FACTFILE" section on the homepage for each state.

Source: Registrar General of India (\[{\mathbf{2003}}\]) SRS Based Abridged Lefe Tables. SRS Analytical Studies, Report No.${\mathbf{3}}$ of \[{\mathbf{2003}}\], New Delhi: Registrar General of India. The data are for the \[1995 - 99\] period; states subsequently divided are therefore included in their pre-partition states (Chhatisgarh in MP, Uttaranchal in UP and Jharkhand in Bihar)

Ans:

151. A skirt that is ${\mathbf{35}}\frac{{\mathbf{7}}}{{\mathbf{8}}}{\mathbf{cm}}$ long has a hem of ${\mathbf{3}}\frac{{\mathbf{1}}}{{\mathbf{8}}}{\mathbf{cm}}$. How long will the skirt be if the hem is let down?

Ans:

Length of skirt $= 35\frac{7}{8}{\text{cm}} = \frac{{297}}{8}{\text{cm}}$

Length of Hem $= \frac{{287}}{8} + \frac{{25}}{8}$

Total length of skirt $= \frac{{287 + 25}}{8}$ $= \frac{{312}}{8} = 39{\text{cm}}$

152. Manavi and Kuber each receives an equal allowance. The table shows the fraction of their allowance each deposits into his/her saving account and the fraction each spends at the mall. If the allowance of each is Rs.\[{\mathbf{1260}}\]. find the amount left with each.

Ans:

Let the total cost $=$ Rs $1$

For Manavi,

Left over $=$Total cost$ - $ All spends

That is, $1 - \left( {\frac{1}{2} + \frac{1}{4}} \right) = 1 - \frac{3}{4} = \frac{1}{4}$

Amount $= 1260 \times \frac{1}{4} =$ Rs \[315\]

For Kuber,

Left over $=$Total cost$ - $ All spends

$1 - \left( {\frac{1}{3} + \frac{3}{5}} \right) = 1 - \frac{{14}}{{15}} = \frac{1}{{15}}$

Amount $= 1260 \times \frac{1}{{15}} =$ Rs \[84\].

NCERT Solutions To Help You To Revise The Complete Syllabus And Score More Marks In Your Examinations

Numbers are often regarded as the most essential building component of mathematics. Students would have learnt about the different sorts of numbers in lower grades, such as natural numbers, whole numbers, integers, and so on. The rational numbers are mentioned in Chapter 1 of Class 8 Math to introduce students to a different set of numbers. This Chapter covers nearly all of the rational number topics that a student in Class 8 is required to master to understand the course. In addition, the method of expressing a rational number on a number line, as well as the method of finding rational numbers between two rational numbers, are described in this Chapter. Continue reading and studying Math Chapter 1 of the NCERT to understand better Rational Numbers and the ideas behind it.

NCERT Exemplar for Class 8 Math Chapter 1 is available online on Vedantu’s website. It is recommended that students complete the questions of each Chapter to achieve good results in the Class 8 Mathematics test. These NCERT Solutions for Class 8 assist the students in gaining a better knowledge of the subjects.

The Main Topics Covered In This Chapter Include

1.1 Introduction

1.2 Properties of Rational Numbers

1.2.1 Closure

1.2.2 Commutativity

1.2.3 Associativity

1.2.4 The role of zero

1.2.5 The role of 1

1.2.6 Negative of a number

1.2.7 Reciprocal

1.2.8 Distributivity of multiplication over addition for rational numbers.

1.3 Representation of Rational Numbers on the Number Line

1.4 Rational Numbers between Two Rational Numbers