Unit 9: Rational Numbers

## Exercise 1: (Multiple Choice Questions and Answers 1-12)

In each of the following questions 1 to 12, there are four options, out of which, only one is correct. Write the correct one.

# Question: 1

A rational number is defined as a number that can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are integers and

a.    $q=0$

b.   $q=1$

c.    $q\ne 1$

d.   $q\ne 0$

## Solution

(d)

From definition, a number that can be expressed in the form $\frac{p}{q},$ where p and q are integers and $q\ne 0,$  is called a rational number.

# Question: 2

Which of the following rational numbers is positive?

a.    $\frac{-8}{7}$

b.   $\frac{19}{-13}$

c.    $\frac{-3}{-4}$

d.   $\frac{-21}{13}$

## Solution

(c)

A rational number is said to be positive only if it fulfils any of the following conditions:

(1) Both numerator and denominator are greater than 0, i.e. positive.

(2) Both numerator and denominator are less than 0, i.e. negative. Here, both numerator and denominator of the rational number are negative. Therefore, it is a positive rational number.

Hence, among the given rational numbers $\left(\frac{-3}{-4}\right)$ is positive.

# Question: 3

Which of the following rational numbers is negative?

a.    $-\left(\frac{-3}{7}\right)$

b.   $\frac{-5}{-8}$

c.    $\frac{9}{8}$

d.   $\frac{3}{-7}$

## Solution

(d)

a.    $-\left(-\frac{3}{7}\right)=\frac{3}{7}$

b.   $\frac{-5}{-8}=\frac{5}{8}$

c.    $\frac{9}{8}=\frac{9}{8}$

d.   $\frac{3}{-7}=-\frac{3}{7}$

In option d, numerator is positive but denominator is negative. Hence, rational number $\frac{3}{-7}=-\frac{3}{7}$.

# Question: 4

In the standard form of a rational number, the common factor of numerator and denominator is always:

a.    $0$

b.   $1$

c.    $-2$

d.   $2$

## Solution

(b)

Common factor means, a number which divides both the given two numbers. Standard form is the simplest form.

By definition, in the standard form of a rational number, the common factor of numerator and denominator is always 1.

# Question: 5

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

a.    $1$

b.   $2$

c.    $\frac{1}{2}$

d.   $0$

## Solution

(a)

To find the reciprocal of fraction, interchange its numerator and denominator.

a.    Reciprocal of $1=\frac{1}{1}=1$

b.   Reciprocal of $2=\frac{1}{2}$

c.    Reciprocal of $\frac{1}{2}=\frac{2}{1}=2$

d.   Reciprocal of $0=\frac{1}{0}$

Hence, option (a) $1=\frac{1}{1}=1$ rational numbers is equal to its reciprocal.

# Question: 6

The reciprocal of $\frac{1}{2}$ is

a.    $3$

b.   $2$

c.    $-1$

d.   $0$

## Solution

(b) Reciprocal of $\frac{1}{2}=\frac{1}{\frac{1}{2}}=2$

# Question: 7

The standard form of $\frac{-48}{60}$ is

a.    $\frac{48}{60}$

b.   $\frac{-60}{48}$

c.    $\frac{-4}{5}$

d.   $\frac{-4}{-5}$

## Solution

(c)

Given rational number is $-\frac{48}{60}$

For standard form, divide numerator & denominator by their HCF

i.e., $\frac{-48÷12}{60÷12}=-\frac{4}{5}$

Hence, the standard form of $-\frac{48}{60}$ is $-\frac{4}{5}.$

# Question: 8

Which of the following is equivalent to $\frac{4}{5}?$

a.    $\frac{5}{4}$

b.   $\frac{16}{25}$

c.    $\frac{16}{20}$

d.   $\frac{15}{25}$

## Solution

(c)

Common factor of 4 and 5 is only 1

$\therefore \frac{4}{5}$ is in standard form

Common factor of 5 and 4 is only 1

$\therefore \frac{5}{4}$ is in standard form

Common factor of 16 and 25 is only 1

$\therefore \frac{16}{25}$ is in standard form

$\frac{16}{20}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{16÷4}{20÷4}=\frac{4}{5}$

$\frac{15}{25}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{15÷5}{25÷5}\text{\hspace{0.17em}}=\frac{3}{5}$

Hence, $\frac{16}{25}$ is equivalent to $\frac{4}{5}$.

# Question: 9

How many rational numbers are there between two rational numbers?

a.    $1$

b.   $0$

c.    Unlimited

d.   $100$

## Solution

(c)

There are unlimited rational numbers between two rational numbers.

# Question: 10

In the standard form of a rational number, the denominator is always a

a.    $0$

b.   negative integer

c.    positive integer

d.   $1$

## Solution

(c)

Standard form is such that 1 is the only common factor of numerator and denominator.

By definition, a rational number is said to be in the standard form, if it is denominator is a positive integer.

# Question: 11

To reduce a rational number to its standard form, we divide its numerator and denominator by their

a.    LCM

b.   HCF

c.    Product

d.   Multiple

## Solution

(b)

To reduce a rational number to its standard form, we divide its numerator and denominator by their HCF.

# Question: 12

Which is greater number in the following:

1. $\frac{-1}{2}$
2. $0$
3. $\frac{1}{2}$
4. $-2$

## Solution

(c)

Positive number >zero > negative number.

Obviously, $\frac{1}{2}$ is greater, since this is the only number which is on the rightmost side of the number line among others.

In Questions 13 to 46, fill in the blanks to make the statements true.

# Question: 13

$-\frac{3}{8}$ is a _____ rational number.

## Solution

The given rational number $-\frac{3}{8}$ is a negative number, because its numerator is negative integer. A rational number is negative if either of numerator and denominator is negative but not the both.

Hence, $-\frac{3}{8}$ is a negative rational number.

# Question: 14

$1$ is a _____ rational number.

## Solution

The given rational number $1$ is positive number, because its numerator and denominator are positive integer.

Hence, $1$ is a positive rational number

# Question: 15

The standard form of $\frac{-8}{-36}$ is ______.

## Solution

Given rational no. is $\frac{-8}{-36}$

For standard form $\frac{-8÷4}{-36÷4}=\frac{-2}{-9}=\frac{2}{9}$

Hence, the standard form of $\frac{-8}{-36}$  is $\frac{2}{9}$

# Question: 16

The standard form of $\frac{18}{-24}$ is ______.

## Solution

Rational no. is $\frac{18}{-24}$

For standard form $\frac{18÷6}{-24÷6}=\frac{3}{-4}$

Hence, the standard form of $\frac{18}{-24}$ is $-\frac{3}{4}$

# Question: 17

On a number line, $\frac{-1}{2}$ is to the ______ of zero (0).

## Solution

On a no. line $-\frac{1}{2}$ is to the left of the $0.$

All the negative numbers lie on the left side of zero on the number line.

# Question: 18

On a number line, $\frac{4}{3}$ is to the ______ of zero (0).

## Solution

On a number line $\frac{4}{3}$ is to the right of the $0.$

All the positive numbers lie on the right side of zero on the number line.

# Question: 19

$-\frac{1}{2}$ is ______ than $\frac{1}{5}.$

## Solution

Given rational numbers are $-\frac{1}{2}$ & $\frac{1}{5}$

$-\frac{1}{2}$ is negative rational number.

$\frac{1}{5}$ is positive rational number

Negative number < zero (0) < positive number.

$\frac{1}{5}>-\frac{1}{2}$

Hence, $-\frac{1}{2}$  is smaller than $\frac{1}{5}$

# Question: 20

$-\frac{3}{5}$ is ______ than 0.

## Solution

Since, $-\frac{3}{5}$ is smaller than 0. Hence, $-\frac{3}{5}$  lies on the left side of the number line, i.e., $-\frac{3}{5}<0.$

# Question: 21

$\frac{-16}{24}$ and $\frac{20}{-16}$ represent ______ rational numbers.

## Solution

Given numbers are $\begin{array}{l}-\frac{16}{24}\\ =-\frac{4}{6}\\ =-\frac{2}{3}\end{array}$

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

$\therefore -\frac{16}{24}\ne \frac{20}{-16}$

$\therefore -\frac{16}{24}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{20}{-16}$ represent different rational numbers.

# Question: 22

$\frac{-27}{45}$ and $\frac{-3}{5}$ represent ______ rational numbers.

## Solution

Given numbers are $-\frac{27}{45}$ and $-\frac{3}{5}$

Now, $-\frac{27}{45}=-\frac{9}{15}=-\frac{3}{5}$

Hence, $-\frac{27}{45}$  and $-\frac{3}{5}$ represent same rational numbers.

# Question: 23

Additive inverse of $\frac{2}{3}$ is ______.

## Solution

Additive inverse of a number is a number, which when added to the given number, gives the result as zero.

Since, additive inverse is the negative of a number.

Hence, additive inverse of $\frac{2}{3}$  is $\frac{-2}{3}$

# Question: 24

$\frac{-3}{5}+\frac{2}{5}=_____.$

## Solution

Given, $-\frac{3}{5}+\frac{2}{5}=\frac{-3+2}{5}$

$\text{\hspace{0.17em}}=-\frac{1}{5}$

Hence, $-\frac{3}{5}+\frac{2}{5}=-\frac{1}{5}$

# Question: 25

$\frac{-5}{6}+\frac{-1}{6}=______.$

## Solution

Given, $-\frac{5}{6}+\frac{-1}{6}=-\frac{5}{6}-\frac{1}{6}$

$=\frac{-5-1}{6}=-\frac{6}{6}=-1$

Hence, $-\frac{5}{6}+\frac{-1}{6}=-1$

# Question: 26

$\frac{3}{4}×\left(\frac{-2}{3}\right)=______.$

## Solution

Given, $\frac{3}{4}×\left(-\frac{2}{3}\right)$

Product of rational numbers $=$  $=\frac{3×\left(-2\right)}{4×3}=\frac{-6}{12}$

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

# Question: 27

$\frac{-5}{3}×\left(\frac{-3}{5}\right)=_______.$

## Solution

Given, $\frac{-5}{3}×\left(\frac{-3}{5}\right)$

Product of rational numbers $=$ $=\frac{\left(-5\right)×\left(-3\right)}{3×5}=\frac{15}{15}=1$

# Question: 28

$-\frac{6}{7}=\frac{__}{42}$

## Solution

Let given expression is written as:

$-\frac{6}{7}=\frac{x}{42}$

$x=\frac{42×\left(-6\right)}{7}=6×\left(-6\right)$

$x=-36$

Hence, $-\frac{6}{7}=-\frac{36}{42}$

# Question: 29

$\frac{1}{2}=\frac{6}{__}$

## Solution

Let the denominator will be $x$

So, we can write the expression as:

$\frac{1}{2}=\frac{6}{x}$

$x=12$

Hence, $\frac{1}{2}=\frac{6}{12}$

# Question: 30

$\frac{-2}{9}-\frac{7}{9}=_______.$

## Solution

Given, $-\frac{2}{9}-\frac{7}{9}=\frac{-2-7}{9}$                   [taking LCM]

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

Hence, $-\frac{2}{9}-\frac{7}{9}=-1$

In questions 31 to 35, fill in the boxes with the correct symbol >,< or =.

# Question: 31

$\frac{7}{-8}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\frac{8}{9}$

## Solution

Every positive rational number is greater than negative rational number.

Since, $\frac{7}{-8}=-\frac{7}{8}$  is a negative rational number & $\frac{8}{9}$ is a positive rational number. Hence, $\frac{7}{-8}<\frac{8}{9}.$

# Question: 32

$\frac{3}{7}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\frac{-5}{6}$

## Solution

Every positive rational number is greater than negative rational number.

Since, $-\frac{5}{6}$ is a negative rational number & $\frac{3}{7}$ is a positive rational number. Hence, $\frac{3}{7}>-\frac{5}{6}.$

# Question: 33

$\frac{5}{6}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\frac{8}{4}$

## Solution

Convert the given rational numbers to the rational numbers with the same denominators.

$\frac{5×2}{6×2}=\frac{10}{12}$ and

$\frac{8×3}{4×3}=\frac{24}{12}$

i.e., $24>10$

$\frac{24}{12}>\frac{10}{12}$

Hence, $\frac{5}{6}<\frac{8}{4}$

# Question: 34

$\frac{-9}{7}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\frac{4}{-7}$

## Solution

Since, both fractions have same denominator, the fraction which has greater nominator is greater.

Hence, $-\frac{9}{7}<\frac{4}{-7}$

# Question: 35

$\frac{8}{8}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\frac{2}{2}$

## Solution

Given $\frac{8}{8}=1$ and

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

Hence, $\frac{8}{8}=\frac{2}{2}$

# Question: 36

The reciprocal of _______ does not exist.

## Solution

The reciprocal of zero does not exist, as reciprocal of $0$ is $\frac{1}{0}$ , which is not defined.

# Question: 37

The reciprocal of $1$  is _______.

## Solution

The reciprocal of $1=\frac{1}{1}$

Hence, the reciprocal of $1$ is $1.$

# Question: 38

$\frac{-3}{7}÷\left(\frac{-7}{3}\right)=_______.$

## Solution

Reciprocal of $-\frac{7}{3}$  is $\frac{3}{-7}$

$\frac{-3}{7}×\left(\frac{3}{-7}\right)$

Product of rational numbers $=$

$=\frac{\left(-3×3\right)}{7×\left(-7\right)}=\frac{-9}{-49}=\frac{9}{49}$

Hence, $-\frac{3}{7}÷\left(-\frac{7}{3}\right)=\frac{9}{49}$

# Question: 39

$0÷\left(\frac{-5}{6}\right)=_______.$

## Solution

Here, $0÷\left(-\frac{5}{6}\right)=0$

Because, $0$ divided by any numbers is 0.

# Question: 40

$0×\left(-\frac{5}{6}\right)=_______.$

## Solution

Here, $0×\left(-\frac{5}{6}\right)=0$

Because, $0$ multiplied by any number is $0$

# Question: 41

$_______×\left(\frac{-2}{5}\right)=1$.

## Solution

Let $x×\left(-\frac{2}{5}\right)=1$

$-\frac{2x}{5}=1$

$-2x=5$

$x=-\frac{5}{2}$

Hence, $-\frac{5}{2}×\left(-\frac{2}{5}\right)=1$

# Question: 42

The standard form of rational number $-1$ is _______.

## Solution

HCF of given rational number $-1$ is $1$.

For standard form $=-1÷1=-1$

Hence, the standard form of rational number $-1$ is $-1.$

# Question: 43

If $m$ is a common divisor of $a$ and $b$, then $\frac{a}{b}=\frac{a÷m}{____}$

## Solution

If $m$ is a common divisor of $a$ and $b$ then

$\frac{a}{b}=\frac{a÷m}{b÷m}$

# Question: 44

If $p$ and $q$ are positive integers, then $\frac{p}{q}$ is a _______ rational number and $\frac{p}{-q}$ is a _______ rational number.

## Solution

If $p$ and $q$ are positive integers, then $\frac{p}{q}$ is a positive rational number, because both numerator and denominator are positive and $\frac{p}{-q}$ is a negative rational number, because denominator is negative.

# Question: 45

Two rational numbers are said to be equivalent or equal, if they have the same _______ form.

## Solution

Two rational numbers are said to be equivalent or equal, if they have the same simplest form.

# Question: 46

If $\frac{p}{q}$ is a rational number, then $q$ cannot be ______.

## Solution

By definition, if $\frac{p}{q}$ is a rational number, then $q$ cannot be zero.

State whether the statements given in question 47 to 65 are True or False.

# Question: 47

Every natural number is a rational number but every rational number need not be a natural number.

## Solution

True

e.g. $\frac{1}{5}$ is a rational number, but not a natural number.

# Question: 48

Zero is a rational number.

## Solution

True

e.g. Zero can be written $0=\frac{0}{1}.$ We know that, a number of the form $\frac{p}{q}$, where $p$, $q$ are integers and $q\ne 0$ is a rational number. So, zero is a rational number.

# Question: 49

Every integer is a rational number but every rational number need not be an integer.

## Solution

True

Every integer is rational number, but every rational number is not an integer.

# Question: 50

Every negative integer is not a negative rational number.

## Solution

False

Because all the integers are rational number, whether it is negative or positive but vice-versa is not true.

# Question: 51

If $\frac{p}{q}$ is a rational number and $m$ is a non-zero integer, then $\frac{p}{q}=\frac{p×m}{q×m}$

## Solution

True

e.g. Let m $=1,2,3,...$

When $m=1,$ then $\frac{p}{q}=\frac{p×1}{1×q}=\frac{p}{q}$

When $m=2,$ then $\frac{p}{q}=\frac{p×2}{q×2}=\frac{p}{q}$

Hence, $\frac{p}{q}=\frac{p×m}{q×m}$

# Question: 52

If $\frac{p}{q}$ is a rational number and $m$ is a non-zero common divisor of $p$ and $q$, then $\frac{p}{q}=\frac{p÷m}{q÷m}$.

## Solution

True

e.g. Let $m=1,2,3,...$

When $m=1,$ then $\frac{p}{q}=\frac{p÷1}{q÷1}=\frac{p}{1}$

$=\frac{p}{1}÷\frac{q}{1}=\frac{p}{1}×\frac{1}{q}=\frac{p}{q}$

When $m=2,$ then $\begin{array}{l}\frac{p}{q}=\frac{p÷2}{q÷2}=\frac{p}{2}÷\frac{q}{2}=\frac{p}{2}×\frac{2}{q}\\ =\frac{p}{q}\end{array}$

Hence, $\frac{p}{q}=\frac{p÷m}{q÷m}$

# Question: 53

In a rational number, denominator always has to be a non-zero integer.

## Solution

True

From definition, a rational number should be in the form of $\frac{p}{q}$, where $q\ne 0.$ This is because any number divided by zero cannot be defined.

# Question: 54

If $\frac{p}{q}$ is a rational number and $m$ is a non-zero integer, then $\frac{p×m}{q×m}$ is a rational number not equivalent to $\frac{p}{q}$.

## Solution

False

Let $m=1,2,3,...$

When $m=1,$ then $\frac{p×m}{q×m}=\frac{p×1}{q×1}=\frac{p}{q}$

When $m=2,$ then $\frac{p×m}{q×m}=\frac{p×2}{q×2}=\frac{p}{q}$

For any non-zero value of $m,\frac{p×m}{q×m}$ is always equivalent to $\frac{p}{q}$

# Question: 55

Sum of two rational numbers is always a rational number.

## Solution

True

Sum of two rational numbers is always a rational number, it is true.

$\frac{1}{3}+\frac{1}{4}=\frac{4+3}{12}=\frac{7}{12}$

# Question: 56

All decimal numbers are also rational numbers.

## Solution

True

All decimal numbers are also rational numbers, it is true.

$0.4=\frac{4}{10}=\frac{2}{5}$

# Question: 57

The quotient of two rationals is always a rational number.

## Solution

False

Let’s take 1 and 0 as two numbers,

Both of these are rational numbers but their division is not defined.

Also, $\frac{1}{0}$ is not a rational number as q = 0.

Hence, the quotient of two rational is not always a rational number.

e.g. $\frac{1}{0}.$

# Question: 58

Every fraction is a rational number.

## Solution

True

Every fraction is a rational number but vice-versa is not true.

# Question: 59

Two rationals with different numerators can never be equal.

## Solution

False

Let $\frac{2}{3}$ and $\frac{10}{15}$ be two rational numbers, then $\frac{10}{15}$ can be written as $\frac{2}{3}$ in its lowest form.

$\frac{10}{15}=\frac{10÷5}{15÷5}=\frac{2}{3}$

Hence, two rational numbers with different numerators can be equal.

# Question: 60

$8$ can be written as a rational number with any integer as denominator.

## Solution

False

$8$ can be written as a rational number with only $1$ as denominator i.e. $\frac{8}{1}$.

# Question: 61

$\frac{4}{6}$ is equivalent to $\frac{2}{3}$.

## Solution

True

Given, $\frac{4}{6}=\frac{4÷2}{6÷2}=\frac{2}{3}$

# Question: 62

The rational number $\frac{-3}{4}$ lies to the right of zero on the number line.

## Solution

False

Because every negative rational number lies to the left of $0$ on the number line.

# Question: 63

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

## Solution

True

Given rational number are $\frac{-12}{-5}$ i.e., $\frac{12}{5}$ and $\frac{-7}{17}$

Hence, it is true, that the rational number $\frac{-12}{-15}$ and $-\frac{7}{17}$ are on the opposite sides of $0$ on the number line as one is negative & one is positive.

# Question: 64

Every rational number is a whole number.

## Solution

False

e.g., $-\frac{7}{8}$ is a rational number, but it is not a whole number, because whole numbers are $0,1,2,...$

# Question: 65

Zero is the smallest rational number.

## Solution

False

A rational number can be negative for example,

$-1,-8,-\frac{8}{5}......$

which are less than zero.

# Question: 66

Match the following:

 Column I Column II (i) $\frac{a}{b}÷\frac{a}{b}$ (a) $\frac{-a}{b}$ (ii) $\frac{a}{b}÷\frac{c}{d}$ (b) $-1$ (iii) $\frac{a}{b}÷\left(-1\right)$ (c) $1$ (iv) $\frac{a}{b}÷\frac{-a}{b}$ (d) $\frac{bc}{ad}$ (v) $\frac{b}{a}÷\left(\frac{d}{c}\right)$ (e) $\frac{ad}{bc}$

## Solution

(i)             (c) Given, $\frac{a}{b}÷\frac{a}{b}=\frac{a}{b}×\frac{b}{a}=1$

(ii)         – (e) Given, $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}×\frac{d}{c}=\frac{ad}{bc}$

(iii)      –(a) Given, $\frac{a}{b}÷\left(-1\right)=\frac{a}{b}×\left(-1\right)=\frac{-a}{b}$

(iv)       –(b) Given, $\frac{a}{b}÷\frac{-a}{b}=\frac{a}{b}×\left(\frac{-b}{a}\right)=-1$

(v)          –(d) Given, $\frac{b}{a}÷\left(\frac{d}{c}\right)=\frac{b}{a}×\frac{c}{d}=\frac{bc}{ad}$

# Question: 67

Write each of the following rational numbers with positive denominators: $\frac{5}{-8},\frac{15}{-28},\frac{-17}{-13}.$

## Solution

We can write, $\frac{5}{-8}=\frac{5×\left(-1\right)}{-8×\left(-1\right)}=\frac{-5}{8}$ [multiplying numerators & denominators by $\left(-1\right)$ ]

$\frac{15}{-28}$ can be written as $\frac{15×\left(-1\right)}{-28×\left(-1\right)}=\frac{-15}{28}$

And $\frac{-17}{-13}$ can be written as $\frac{-17×\left(-1\right)}{-13×\left(-1\right)}=\frac{17}{13},$ as both negative signs are cancelled.

# Question: 68

Express $\frac{3}{4}$ as a rational number with denominator:

a.    $36$

b.   $-80$

## Solution

a.    To make the denominator $36$, we have to multiply numerator & denominator by $9$.
$\frac{3×9}{4×9}=\frac{27}{36}$

b.   To make the denominator $-80,$ we have to multiply numerator & denominator by $-20$.
$\frac{3×\left(-20\right)}{4×\left(-20\right)}=\frac{-60}{-80}$

# Question: 69

Reduce each of the following rational numbers in its lowest form:

(i)        $\frac{-60}{72}$

(ii)    $\frac{91}{-364}$

## Solution

(i)        $-\frac{60}{72}$ can be written as
$\frac{-60÷12}{72÷12}$
$\frac{-60×\frac{1}{12}}{72×\frac{1}{12}}=-\frac{5}{6}$ which is the lowest form.

(ii)    considering the rational number $=\frac{91}{-364}$

$91=7×13$

$-364=2×2×7×13×-1$

$\text{H}\text{.C}\text{.F}\text{.}=91$

To convert $\frac{91}{-364}$ into its lowest form, divide the numerator and denominator by H.C.F.

# Question: 70

Express each of the following rational numbers in its standard form:

(i)        $\frac{-12}{-30}$

(ii)    $\frac{14}{-49}$

(iii) $\frac{-15}{35}$

(iv)  $\frac{299}{-161}$

## Solution

(i)             Rational number is $\frac{-12}{-30}$
In standard form $=\frac{-12÷6}{-30÷6}=\frac{-2}{-5}=\frac{2}{5}$
Hence, the standard form is $\frac{-12}{-30}$ is $\frac{2}{5}$

(ii)         Rational number is $\frac{14}{-49}$
In, standard form $\frac{14÷7}{-49÷7}=\frac{2}{-7}=\frac{-2}{7}$
Hence, the standard form of $\frac{14}{-49}$ is $\frac{-2}{7}$

(iii)      Rational number $\frac{-15}{53}$
In, standard form $\frac{-15÷5}{35÷5}=\frac{-3}{7}$
Hence, standard form of $\frac{-15}{35}$ is $\frac{-3}{7}$

(iv)       Given rational number is $\frac{299}{-161}$
For standard form $\frac{299÷23}{-161÷23}$
$\frac{13}{-7}=\frac{13÷\left(-1\right)}{13÷\left(-1\right)}=\frac{-13}{7}$
Hence, the standard form of $\frac{299}{-161}$ is $\frac{-13}{7}$

# Question: 71

Are the rational numbers $\frac{-8}{28}$ and $\frac{32}{-112}$ equivalent? Give reason.

## Solution

Given rational numbers are $-\frac{8}{28}$ and $\frac{32}{-112}$

For standard form $\begin{array}{l}-\frac{8}{28}=\frac{-8÷4}{28÷4}\\ =-\frac{2}{7}\end{array}$

And, $\begin{array}{l}\frac{32}{-112}=\frac{32÷16}{-112÷16}\\ =\frac{2}{-7}\\ =\frac{-2}{7}\end{array}$

The standard form of $-\frac{8}{28}$ and $\frac{32}{-112}$ are equal.

Hence, they are equivalent.

# Question: 72

## Solution

Rational numbers are $-\frac{7}{10},\frac{5}{-8},\frac{2}{-3},-\frac{1}{4},-\frac{3}{5}$   [given]

To arrange in any order, we make denominators of all rational numbers same.

Take LCM of $10,\text{\hspace{0.17em}}8,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}4\text{\hspace{0.17em}}&\text{\hspace{0.17em}}5$ which is $120$

$\begin{array}{l}\frac{-7×12}{10×12},\\ \frac{5×15}{-8×15},\\ \frac{2×40}{-3×40},\\ \frac{-1×30}{4×30},\\ \frac{-3×24}{5×24},\end{array}$
$=\frac{-84}{120},\frac{75}{120},\frac{80}{-120},-\frac{30}{120},\frac{72}{120}$

$=-\frac{84}{120},-\frac{75}{120},-\frac{80}{120},-\frac{30}{120},-\frac{72}{120}$

Since, denominators are same, as ascending order of numerators are $-84,-80,-75,-72,-30$

Hence, $-\frac{84}{120}<-\frac{80}{12}<-\frac{75}{120}<-\frac{72}{120}<-\frac{30}{120}$

i.e., $-\frac{7}{10}<\frac{2}{-3}<\frac{5}{-8}<-\frac{3}{5}<-\frac{1}{4}$

# Question: 73

Represent the following rational numbers on a number line:
$\frac{3}{8},\frac{-7}{3},\frac{22}{-6}.$

## Solution # Question: 74

If $\frac{-5}{7}=\frac{x}{28},$ find the value of $x.$

## Solution

Give, $-\frac{5}{7}=\frac{x}{28}$

$7×x=-5×28$

$x=\frac{-5×28}{7}=-5×4$

$x=-20$

Hence, the value of $x$ is $-20.$

# Question: 75

Give three rational numbers equivalent to:

(i)        $\frac{-3}{4}$

(ii)    $\frac{7}{11}$

## Solution

(i)             Given rational number is $-\frac{3}{4}$
so, the equivalent rational numbers are
$\begin{array}{l}\frac{-3×2}{4×2}=-\frac{6}{8},\\ \frac{-3×3}{4×3}=-\frac{9}{12}\end{array}$ and

$\frac{-3×4}{4×4}=-\frac{12}{16}$
Hence, three equivalent rational numbers are $-\frac{6}{8},-\frac{9}{12}$ and $-\frac{12}{16}$

(ii)         Given rational number is $\frac{7}{11}$
so, the equivalent rational numbers are
$\begin{array}{l}\frac{7×2}{11×2}=\frac{14}{22},\\ \frac{7×3}{11×3}=\frac{21}{33}\end{array}$ and

$\frac{7×4}{11×4}=\frac{28}{44}$
Hence, three equivalent rational numbers are $\frac{14}{22},\frac{21}{33}$ and $\frac{28}{44}$

# Question: 76

Write the next three rational numbers to complete the pattern:

(i)        $\frac{4}{-5},\frac{8}{-10},\frac{12}{-15},\frac{16}{-20},______,______,_______.$

(ii)    $\frac{-8}{7},\frac{-16}{14},\frac{-24}{21},\frac{-32}{28},______,______,_______.$

## Solution

(i)             Given rational numbers are $\frac{4}{-5}$
so, the next three equivalent rational numbers are
$\begin{array}{l}\frac{4×5}{-5×5}=\frac{20}{-25},\\ \frac{4×6}{-5×6}=\frac{24}{-30}\end{array}$ and

$\frac{4×7}{-5×7}=\frac{28}{-35}$
Hence, the next equivalent numbers are $\frac{20}{-25},\frac{24}{-30},\frac{28}{-35}$

(ii)         Given rational numbers are $-\frac{8}{7}$
so, the next three equivalent rational
numbers are
$\begin{array}{l}\frac{-8×5}{7×5}=-\frac{40}{35},\\ \frac{-8×6}{7×6}=-\frac{48}{42}\end{array}$ and

$\frac{-8×7}{7×7}=-\frac{56}{49}$
Hence, the next equivalent numbers are $-\frac{40}{35},-\frac{48}{42},-\frac{56}{49}$

# Question: 77

List four rational numbers between $\frac{5}{7}$ and $\frac{7}{8}.$

## Solution

Given rational numbers are $\frac{5}{7}$ and $\frac{7}{8}$

For making the same denominators: LCM of $7$ and $8=56$
i.e., $\frac{5×8}{7×8}=\frac{40}{56}$ and $\frac{7×7}{8×7}=\frac{49}{56}$

So, the four rational numbers between $\frac{40}{56}$ and $\frac{49}{56}$ are $\frac{42}{56},\frac{44}{56},\frac{46}{56},\frac{48}{56}$

# Question: 78

Find the sum of

(i)        $\frac{8}{13}$ and $\frac{3}{11}$

(ii)    $\frac{7}{3}$ and $\frac{-4}{3}$

## Solution

(i)             Given $\frac{8}{13}$ and $\frac{3}{11}$
Sum $\begin{array}{l}=\frac{8}{13}+\frac{3}{11}=\frac{8×11}{13×11}+\frac{3×13}{11×13}\\ =\frac{88}{143}+\frac{39}{143}\end{array}$
$=\frac{88×39}{143}=\frac{127}{143}$
Hence, the sum of $\frac{8}{13}$ and $\frac{3}{11}$ is $\frac{127}{143}$

(ii)         Given $\frac{7}{3}$ and $-\frac{4}{3}$
Sum $=\frac{7}{3}+\left(-\frac{4}{3}\right)$
$=\frac{7}{3}-\frac{4}{3}$
$=\frac{7-4}{3}=\frac{3}{3}=1$
Hence, the sum of $\frac{7}{3}$ and $-\frac{4}{3}$ is 1.

# Question: 79

Solve:

(i)        $\frac{29}{4}-\frac{30}{7}$

(ii)    $\frac{5}{13}-\frac{-8}{26}$

## Solution

(i)             Given $\frac{29}{4}-\frac{30}{7}=\frac{29×7}{4×7}-\frac{30×4}{7×4}$
$=\frac{203}{28}-\frac{120}{28}$
$=\frac{203-120}{28}=\frac{83}{28}$

(ii)         Given $\frac{5}{13}-\frac{-8}{26}$
$=\frac{5}{13}+\frac{8}{26}=\frac{5×2}{13×2}+\frac{8×1}{26×1}$
$=\frac{10}{26}+\frac{8}{26}$
$=\frac{10+8}{26}=\frac{18}{26}$
$=\frac{18÷2}{26÷2}=\frac{9}{13}$

# Question: 80

Find the product of:

(i)             $\frac{-4}{5}$ and $\frac{-5}{12}$

(ii)         $\frac{-22}{11}$ and $\frac{-21}{11}$

## Solution

(i)             Given $-\frac{4}{5}$  and  $-\frac{5}{12}$
$=\frac{\left(-4\right)×\left(-5\right)}{5×12}=\frac{20}{60}$
$=\frac{20÷20}{60÷20}$    [dividing numerator & denominator by $20$ ]
$=\frac{1}{3}$

(ii)         Given $-\frac{22}{11}$ and $-\frac{21}{11}$
$=\frac{\left(-22\right)×\left(-21\right)}{11×11}=\frac{462}{121}$
$=\frac{462÷11}{121÷11}$   [dividing numerator & denominator by $20$ ]
$=\frac{42}{11}$

# Question: 81

Simplify:

(i)             $\frac{13}{11}×\frac{-14}{5}+\frac{13}{11}×\frac{-7}{5}+\frac{-13}{11}×\frac{34}{5}$

(ii)         $\frac{6}{5}×\frac{3}{7}-\frac{1}{5}×\frac{3}{7}$

## Solution

(i)             Given $\frac{13}{11}×-\frac{14}{5}+\frac{13}{11}×-\frac{7}{5}+\frac{-13}{11}×\frac{34}{5}$
$=\frac{13×\left(-14\right)}{11×5}+\frac{13×\left(-7\right)}{11×5}+\frac{\left(-13\right)×34}{11×5}$

$=-\frac{182}{55}+\frac{-91}{55}+\frac{-442}{55}$
$=\frac{-182-91-442}{55}=-\frac{715}{55}=-13$

(ii)         Given $\frac{6}{5}×\frac{3}{7}-\frac{1}{5}×\frac{3}{7}$
$\begin{array}{l}=\frac{6×3}{5×7}-\frac{1×3}{5×7}\\ =\frac{18}{35}-\frac{3}{35}\end{array}$
$=\frac{18-3}{35}=\frac{15}{35}$
$=\frac{15÷5}{35÷5}=\frac{3}{7}$

# Question: 82

Simplify:

(i)             $\frac{3}{7}÷\left(\frac{21}{-55}\right)$

(ii)         $1÷\left(-\frac{1}{2}\right)$

## Solution

(i)             Given $\frac{3}{7}÷\left(\frac{21}{-55}\right)$
The reciprocal of $\left(\frac{21}{-55}\right)$ is $-\frac{55}{21}$
So, $\frac{3}{7}÷\left(\frac{21}{-55}\right)=\frac{3}{7}×\frac{-55}{21}$
$=\frac{\left(-55\right)×3}{7×21}=\frac{-55}{49}$

(ii)         Given $1÷\left(-\frac{1}{2}\right)$
The reciprocal of $\left(-\frac{1}{2}\right)$ is $\frac{2}{-1}$
So, $\begin{array}{l}1÷\left(-\frac{1}{2}\right)\\ =\frac{1}{1}×\frac{2}{-1}\end{array}$
$\begin{array}{l}=\frac{1×2}{1×\left(-1\right)}\\ =\frac{2}{-1}=-2\end{array}$

# Question: 83

Which is greater in the following?

(i)             $\frac{3}{4},\frac{7}{8}$

(ii)         $-3\frac{5}{7},3\frac{1}{9}$

## Solution

(i)             To compare the rational numbers, we need to make their denominators same.
Here, $\frac{3}{4}=\frac{3×2}{4×2}=\frac{6}{8}$ ,

and $\frac{7}{8}=\frac{7×1}{8×1}=\frac{7}{8}$
From the above expression,

$7>6$
So, $\frac{7}{8}>\frac{3}{4}$
Hence, the greater number is $\frac{7}{8}.$

(ii)         Given rational number are  $-3\frac{5}{7},3\frac{1}{9}$

Positive numbers are always greater than negative numbers.
Hence, the greater number is $3\frac{1}{9}$

# Question: 84

Write a rational number in which the numerator is less than ‘ $-7×11$ ’ and the denominator is greater than ‘ $12+4$ ’.

## Solution

Let $-7×11=p=-77$

And $12+4=q=16$

Rational number $=\frac{p}{q}=-\frac{77}{16}$

Hence, it has more than one answers like $-\frac{78}{17},-\frac{79}{18},-\frac{80}{19}$

# Question: 85

If $x=\frac{1}{10}$ and $y=\frac{-3}{8},$ then evaluate $x+y,x-y,x×y$ and $x÷y.$

## Solution

Given, $x=\frac{1}{10}$ and $y=-\frac{3}{8}$

Now, $\begin{array}{l}x+y=\frac{1}{10}+\frac{\left(-3\right)}{8}\\ \frac{1}{10}-\frac{3}{8}\end{array}$

$=\frac{1×4}{10×4}-\frac{3×5}{8×5}$
$\begin{array}{l}=\frac{4}{40}-\frac{15}{40}\\ =\frac{4-15}{40}\end{array}$

$=-\frac{11}{40}$

And $x-y=\frac{1}{10}-\left(-\frac{3}{8}\right)=\frac{1}{10}+\frac{3}{8}$

$=\frac{1×4}{10×4}+\frac{3×5}{8×5}$

$=\frac{4}{40}+\frac{15}{40}=\frac{4+15}{40}$

$=\frac{19}{40}$

$\begin{array}{l}x×y=\frac{1}{10}×\frac{-3}{8}=\frac{1×\left(-3\right)}{10×8}\\ =-\frac{3}{80}\end{array}$

And $x÷y=\frac{1}{10}×\frac{8}{-3}$

The reciprocal of $\left(-\frac{3}{8}\right)$ is $\frac{8}{-3}$

So, $x÷y=\frac{1}{10}×\frac{8}{-3}$

$=\frac{1×8}{10×-3}=-\frac{8}{30}=\frac{-8÷2}{30÷2}$

$=-\frac{4}{15}$

# Question: 86

Find the reciprocal of the following:

(i)        $\left(\frac{1}{2}×\frac{1}{4}\right)+\left(\frac{1}{2}×6\right)$

(ii)    $\frac{20}{51}×\frac{4}{91}$

(iii) $\frac{3}{13}÷\frac{-4}{65}$

(iv)  $\left(-5×\frac{12}{15}\right)-\left(-3×\frac{2}{9}\right)$

## Solution

(i)             Given, $\left(\frac{1}{2}×\frac{1}{4}\right)+\left(\frac{1}{2}×6\right)$
$=\frac{1×1}{2×4}+\frac{1×6}{2×1}=\frac{1}{8}+\frac{6}{2}$
$=\frac{1×1}{8×1}+\frac{6×4}{2×4}$
$=\frac{1}{8}+\frac{24}{8}=\frac{1+24}{8}$
$=\frac{25}{8}$
Hence, the reciprocal of $\frac{25}{8}$ is $\frac{8}{25}$

(ii)         Given, $\frac{20}{51}×\frac{4}{91}$
$=\frac{20×4}{51×91}=\frac{80}{4641}$

Hence, the reciprocal of $\frac{80}{4641}$ is $\frac{4641}{80}$

(iii)      Given, $\frac{3}{13}÷-\frac{4}{65}$

The reciprocal of $-\frac{4}{65}$ is $\frac{65}{-4}$

$\frac{3}{13}÷-\frac{4}{65}=\frac{3}{13}×\frac{65}{-4}$

$\frac{65×3}{13×\left(-4\right)}=-\frac{15}{4}$

Hence, the reciprocal of $\frac{15}{-4}$ is $-\frac{4}{15}$

(iv)       Given, $\left(-5×\frac{12}{15}\right)-\left(-3×\frac{2}{9}\right)$

$=\left(-\frac{-5×12}{15}\right)-\left(-\frac{-3×2}{9}\right)$
$\begin{array}{l}\left(-\frac{12}{3}\right)-\left(\frac{-2}{3}\right)=\frac{-12+2}{3}\\ ⇒-\frac{10}{3}\end{array}$

Hence, the reciprocal of $-\frac{10}{3}$ is $-\frac{3}{10}$

# Question: 87

Complete the following table by finding the sums:

 $+$ $-\frac{1}{9}$ $\frac{4}{11}$ $\frac{-5}{6}$ $\frac{2}{3}$ $-\frac{5}{4}$ $\frac{-39}{44}$ $-\frac{1}{3}$

## Solution

Let

 $+$ $-\frac{1}{9}$ $\frac{4}{11}$ $\frac{-5}{6}$ $\frac{2}{3}$ $a$ $b$ $c$ $-\frac{5}{4}$ $d$ $\frac{-39}{44}$ $e$ $-\frac{1}{3}$ $f$ $g$ $h$

Here, $a=-\frac{1}{9}+\frac{2}{3}$

= $\frac{2}{3}-\frac{1}{9}$

$=\frac{6}{9}-\frac{1}{9}=\frac{6-1}{9}=\frac{5}{6}$

$b=\frac{2}{3}+\frac{4}{11}$

$\begin{array}{l}=\frac{2×11}{3×11}+\frac{4×3}{11×3}=\frac{22}{33}+\frac{12}{33}\\ =\frac{22+12}{33}\end{array}$

$=\frac{34}{33}$

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

$=\frac{2}{3}-\frac{5}{6}=\frac{2×2}{3×2}-\frac{5×1}{6×1}$

$=\frac{4}{6}-\frac{5}{6}$

$=\frac{4-5}{6}=-\frac{1}{6}$

And $d=-\frac{5}{4}+\left(-\frac{1}{9}\right)=\frac{-5}{4}-\frac{1}{9}$

$=\frac{-5×9}{4×9}-\frac{1×4}{9×4}$

$=-\frac{45}{36}-\frac{4}{36}$

$=\frac{-45-4}{36}=-\frac{49}{36}$

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

$=\frac{-5×3}{4×3}-\frac{5×2}{6×2}=-\frac{15}{12}-\frac{10}{12}$

$=\frac{-15-10}{12}=-\frac{25}{12}$

And $f=-\frac{1}{3}+\left(-\frac{1}{9}\right)=-\frac{1}{3}-\frac{1}{9}$

$=\frac{-1×3}{3×3}-\frac{1×1}{9×1}$

$=-\frac{3}{9}-\frac{1}{9}=\frac{-3-1}{9}=-\frac{4}{9}$

$g=-\frac{1}{3}+\frac{4}{11}$

$=\frac{-1×11}{3×11}+\frac{4×3}{11×3}$

$=-\frac{11}{33}+\frac{12}{33}$

$=\frac{-11+12}{33}=\frac{1}{33}$

$h=-\frac{1}{3}+\left(-\frac{5}{6}\right)=-\frac{1}{3}-\frac{5}{6}$

$=\frac{-1×2}{3×2}-\frac{5×1}{6×1}$

$=-\frac{2}{6}-\frac{5}{6}$

$=\frac{-2-5}{6}=-\frac{7}{6}$

Hence, the complete table is

 $+$ $-\frac{1}{9}$ $\frac{4}{11}$ $-\frac{5}{6}$ $\frac{2}{3}$ $\frac{5}{6}$ $\frac{34}{33}$ $-\frac{1}{6}$ $-\frac{5}{4}$ $-\frac{49}{36}$ $\frac{-39}{44}$ $-\frac{25}{12}$ $-\frac{1}{3}$ $-\frac{4}{9}$ $\frac{1}{33}$ $-\frac{7}{6}$

# Question: 88

Write each of the following numbers in the form $\frac{p}{q}$, where $p$ and $q$ are integers:

1. six-eighths
2. three and half
3. opposite of 1
4. one-fourth
5. zero
6. opposite of three-fifths

## Solution

a.    six-eights $=\frac{6}{8}$

b.   three and half $=3\frac{1}{2}=\frac{3×2+1}{2}=\frac{7}{2}$

c.    opposite of  $1=-1$

d.   one-fourth $=\frac{1}{4}$

e.    zero $=\frac{0}{1}$

f.     opposite of three $–$ fifths $=-\frac{3}{5}$

# Question: 89

If $p=m×t,$ and $q=n×t$  then $\frac{p}{q}=\frac{\overline{)}}{\overline{)}}$

## Solution

Given, $p=m×t$ and $q=n×t$

$\frac{p}{q}=\frac{m×t}{n×t}=\frac{m}{n}$

# Question: 90

Given that $\frac{p}{q}$ and $\frac{r}{s}$ are two rational numbers with different denominators and both of them are in standard form. To compare these rational numbers we say that:

a.    $\frac{\overline{)}}{\overline{)}}<\frac{\overline{)}}{\overline{)}},$ If $p×s

b.   $\frac{p}{q}=\frac{r}{s,}$ If $______=_____$

c.    $\frac{\overline{)}}{\overline{)}}>\frac{\overline{)}}{\overline{)}},$ If  $p×s>r×q$

## Solution

a.    Given, $p×s

By cross multiplication, we get

$\frac{p}{q}<\frac{r}{s}$

b.   Given, $\frac{p}{q}=\frac{r}{s}$

$p×s=r×q$

c.    Given $p×s>r×q$

$\frac{p}{q}>\frac{r}{s}$

# Question: 91

In each of the following cases, write the rational number whose numerator and denominator are respectively as under:

a.    $5–39$ and $54–6$

b.    and $8÷2$

c.    $35÷\left(–7\right)$ and $35–18$

d.   $25+15$ and $81÷40$

## Solution

a.    Let numerator $p=5-39=-34$

And denominator $q=54-6=48$

Hence, rational number $=\frac{p}{q}=-\frac{34}{48}$

b.   Let numerator, $p=\left(-4\right)×6=-24$

& denominator $q=8÷2=\frac{8}{2}=4$

Hence, rational number $=\frac{p}{q}=-\frac{24}{4}$

c.    Let numerator, $p=34÷\left(-7\right)=\frac{35}{-7}=-5$

& denominator, $q=35-18=17$

Hence, rational number $=\frac{p}{q}=-\frac{5}{17}$

d.   Let numerator, $p=25+15=40$

and denominator $q=81÷40=\frac{81}{40}$

Rational number = $\frac{40}{\frac{81}{40}}=\frac{1600}{81}$

Hence, rational number $=\frac{p}{q}=\frac{1600}{81}$

# Question: 92

Write the following as rational numbers in their standard forms:

a.    $35%$

b.   $1.2$

c.    $–6\frac{3}{7}$

d.   $240÷\left(–840\right)$

e.    $115÷207$

## Solution

a.    Given, $35%=\frac{35}{100}$

On dividing numerator & denominator by their HCF, we get  $\frac{35÷5}{100÷5}=\frac{7}{20}$

b.   Here, $1.2=\frac{12}{10}=\frac{12÷2}{10÷2}=\frac{6}{5}$

c.    Here, $=-6\frac{3}{7}=-\left(\frac{6×7+3}{7}\right)=-\frac{45}{7}$

d.   Here, $\begin{array}{l}240÷\left(-840\right)=240×\frac{1}{-840}\\ =\frac{240}{-840}\end{array}$

HCF of $240\text{\hspace{0.17em}}\text{&}\text{\hspace{0.17em}}840=120$

On dividing numerator & denominator by their HCF, we get

$\frac{240÷120}{-840÷120}=\frac{2}{-7}=\frac{-2}{7}$

e.    Given, $115÷207=\frac{115}{207}$

By using prime factorisation, we get $115=5×23$ and $207=3×23×3$

HCF of $115$ and $207=23$

On dividing numerator & denominator by their HCF, we get $\frac{115÷23}{207÷23}=\frac{5}{9}$

# Question: 93

Find a rational number exactly halfway between:

a.    $\frac{-1}{3}$ and $\frac{1}{3}$

b.   $\frac{1}{6}$ and $\frac{1}{9}$

c.    $\frac{5}{-13}$ and $\frac{-7}{9}$

d.   $\frac{1}{15}$ and $\frac{1}{12}$

## Solution

a.    Given rational numbers are $-\frac{1}{3}$ and $\frac{1}{3}$

here, $a=-\frac{1}{3}$ and $b=\frac{1}{3}$

$\frac{a+b}{2}=\frac{-\frac{1}{3}+\frac{1}{3}}{2}=\frac{0}{2}=0$

Here, exactly halfway between $-\frac{1}{3}$ and $\frac{1}{3}$ is $0$

b.   Given rational number are $\frac{1}{6}$ and $\frac{1}{9}$

Here, $a=\frac{1}{6}$  and $b=\frac{1}{9}$

$\begin{array}{l}\frac{a+b}{2}=\frac{\frac{1}{6}+\frac{1}{9}}{2}\\ =\frac{\frac{1×3}{6×3}+\frac{1×2}{9×2}}{2}\end{array}$
$=\frac{\frac{3}{18}+\frac{2}{18}}{2}=\frac{\frac{3+2}{18}}{2}$

$=\frac{\frac{5}{18}}{2}=\frac{5}{18×2}=\frac{5}{36}$

Hence, the exactly halfway between $\frac{1}{6}$ and $\frac{1}{9}$ is $\frac{5}{36}.$

c.    $\frac{5}{-13}$ and $-\frac{7}{9}$   [given]

Here, $a=-\frac{5}{13}$ and $b=-\frac{7}{9}$

$=\frac{a+b}{2}=\frac{-\frac{5}{13}+\left(-\frac{7}{9}\right)}{2}$

$\begin{array}{l}=\frac{\left(-\frac{5}{13}\right)-\frac{7}{9}}{2}\\ =\frac{\frac{-5×9}{13×9}-\frac{7×13}{9×13}}{2}\end{array}$

$=\frac{-\frac{45}{117}-\frac{91}{117}}{2}=\frac{\frac{-45-91}{117}}{2}$

$=-\frac{136}{117×2}=-\frac{136}{234}$

Hence, the exactly the halfway between $\frac{5}{-13}$ and $-\frac{7}{9}$ is $-\frac{136}{234}$

d.   Given rational numbers are $\frac{1}{15}$ and $\frac{1}{12}$

hence, $a=\frac{1}{15}$ and $b=\frac{1}{12}$

$\frac{a+b}{2}=\frac{\frac{1}{15}+\frac{1}{12}}{2}$

$=\frac{\frac{1×4}{15×4}+\frac{1×5}{12×5}}{2}$

$=\frac{\frac{4}{60}+\frac{5}{60}}{2}=\frac{\frac{4+5}{60}}{2}$

$=\frac{9}{60×2}=\frac{9}{120}$

$=\frac{3}{40}$

# Question: 94

Taking $x=\frac{-4}{9},y=\frac{5}{12}$ and $z=\frac{7}{18},$ find:

a.    the rational number which when added to $x$ gives $y$.

b.   the rational number which subtracted from $y$ gives $z$.

c.    the rational number which when added to $z$ gives us $x$.

d.   the rational number which when multiplied by $y$ to get $x$.

e.    the reciprocal of $x+y$.

f.     the sum of reciprocals of $x$ and $y$.

g.   $\left(x÷y\right)×z$

h.   $\left(x–y\right)+z$

i.  $x+\left(y+z\right)$

j.  $x÷\left(y÷z\right)$

k.   $x–\left(y+z\right)$

## Solution

Given, $x=-\frac{4}{9},y=\frac{5}{12}$ and $z=\frac{7}{18}$

a.    Let we add P to $x$ get $y$

$P+x=y$

$P+\left(-\frac{4}{9}\right)=\frac{5}{12}$

$P=\frac{5}{12}-\left(-\frac{4}{9}\right)=\frac{5}{12}+\frac{4}{9}$

$=\frac{5×3+4×4}{36}$

$=\frac{15+16}{36}=\frac{31}{36}$

b.   Let we subtract A from $y$ get $z$

$y-A=z$

$\frac{5}{12}-A=\frac{7}{18}$

$-A=\frac{7}{18}-\frac{5}{12}=\frac{\left(7×2-5×3\right)}{36}$

$=\frac{14-15}{5}=-\frac{1}{36}$

$A=\frac{1}{36}$

c.    Let P be added to $z$ to get $x$

$P+z=x$

$P+\frac{7}{18}=-\frac{4}{9}$

$P=-\frac{4}{9}-\frac{7}{18}$

$=\frac{\left(-4×2-7×1\right)}{18}$

$=\frac{-8-7}{18}=-\frac{15}{18}=-\frac{5}{6}$

d.   Suppose, if $A$ is multiply by $y$, then we get $x$ i.e., $A×y=x$

$A×\frac{5}{12}=-\frac{4}{9}$

$A=-\frac{4}{9}×\frac{12}{5}=-\frac{48}{45}$

e.    Here, $x+y=-\frac{4}{9}+\frac{5}{12}=\frac{-4×4+5×3}{36}$

$x+y=\frac{-16+15}{36}=-\frac{1}{36}$

Reciprocal of $x+y=\frac{1}{-\frac{1}{36}}=-36$

f.     Reciprocal of $x$ and $y$ is $\frac{1}{x}$ and $\frac{1}{y}$ sum of reciprocal $=\frac{1}{-\frac{4}{9}}+\frac{1}{\frac{5}{12}}$

$=-\frac{9}{4}+\frac{12}{5}=\frac{-45+48}{20}$

$=\frac{3}{20}$

g.   We have, $\left(x÷y\right)×z$

$\left(-\frac{4}{9}÷\frac{5}{12}\right)×\frac{7}{18}$

$=\left(-\frac{4}{9}×\frac{12}{5}\right)×\frac{7}{18}$

$=\frac{-4×12×7}{9×5×18}=-\frac{56}{135}$

h.   We have, $\left(x-y\right)+z=\left(-\frac{4}{9}-\frac{5}{12}\right)+\frac{7}{18}$

$=\frac{-4×4-5×3}{36}+\frac{7}{18}$

$=\frac{-16-15}{36}+\frac{7}{18}=\left(-\frac{31}{36}+\frac{7}{18}\right)$

$=\frac{-31+7×2}{36}=\frac{-31+14}{36}$

$=-\frac{17}{36}$

i.  Here, $x+\left(y+z\right)=-\frac{4}{9}+\left(\frac{5}{12}+\frac{7}{18}\right)$

$=-\frac{4}{9}+\frac{5×3+7×2}{36}$

$=-\frac{4}{9}+\left(\frac{15+14}{36}\right)$

$=-\frac{4}{9}+\frac{29}{36}=\frac{-4×4+29}{36}=\frac{13}{36}$

j.  Here, $x÷\left(y÷z\right)=-\frac{4}{9}÷\left(\frac{5}{12}÷\frac{7}{18}\right)=-\frac{4}{9}÷\left(\frac{5}{12}×\frac{18}{7}\right)$$=-\frac{4}{9}÷\frac{15}{14}=-\frac{4}{9}×\frac{14}{15}=-\frac{56}{135}$

k.   Here, $x-\left(y+z\right)=-\frac{4}{9}-\left(\frac{5}{12}+\frac{7}{18}\right)$

$=-\frac{4}{9}-\left(\frac{5×3+7×2}{36}\right)=-\frac{4}{9}-\left(\frac{15+14}{36}\right)$

$=-\frac{4}{9}-\frac{29}{36}=\frac{-4×4-29}{36}$

$=\frac{-16-29}{36}=-\frac{45}{36}=-\frac{5}{4}$

# Question: 95

What should be added to $\frac{-1}{2}$ to obtain the nearest natural number?

## Solution

Given rational numbers is $-\frac{1}{2}$

We know that, nearest natural number of $-\frac{1}{2}$ is $1$

Then, $-\frac{1}{2}+x=1$

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

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

Hence, $\frac{3}{2}$ should be added to $-\frac{1}{2}$ to obtain nearest natural number.

# Question: 96

What should be subtracted from $\frac{-2}{3}$ to obtain the nearest integer?

## Solution

Given rational number is $-\frac{2}{3}$

We know that, nearest natural number of $-\frac{2}{3}$ is $-1$

Let $x$ be subtracted to $-\frac{2}{3}$ to obtain $-1$

Then, $-\frac{2}{3}-x=-1$

$x=-\frac{2}{3}+1=\frac{1}{3}$

So, we subtract $\frac{1}{3}$ from $-\frac{2}{3}$ to get the nearest integer.

# Question: 97

What should be multiplied with $\frac{-5}{8}$ to obtain the nearest integer?

## Solution

Let number be $p$

Nearest integer of $-\frac{5}{8}$ is $-1$

According to the question, $-\frac{5}{8}×p=-1$

$p=-1×\frac{8}{-5}=\frac{8}{5}$

Hence, the required number is $\frac{8}{5}.$

# Question: 98

What should be divided by $\frac{1}{2}$ to obtain the greatest negative integer?

## Solution

Let the number be $p$

Greatest negative integer is $-1$

According to the question,

$p÷\frac{1}{2}=-1$

$\begin{array}{l}⇒p×2=-1\\ ⇒p=\frac{-1}{2}\end{array}$

# Question: 99

From a rope $68$ m long, pieces of equal size are cut. If length of one piece is $4\frac{1}{4}\text{\hspace{0.17em}}\text{m}$, find the number of such pieces.

## Solution

Given length of the rope

& length of small pieces

Number of pieces

$=\frac{68}{1}×\frac{4}{17}=4×4=16$

Hence, the number of pieces is $16$

# Question: 100

If $12$ shirts of equal size can be prepared from  cloth, what is length of cloth required for each shirt?

## Solution

Let total length of cloth required for each shirt = $x$ m

Total size of available cloth

Since, $12$ shirts can be made from $27$ m long cloth.

Length of cloth required for each shirt

Hence,  cloth required for each shirt.

# Question: 101

Insert $3$ equivalent rational numbers between

(i)             $\frac{-1}{2}$ and $\frac{1}{5}$

(ii)         $0$ and $-10$

## Solution

(i)             For common denominator, take LCM of $2$ & $5=10$

$\frac{-1×5}{2×5}=-\frac{5}{10}$ and

$\frac{1×2}{5×2}=\frac{2}{10}$

Hence, three equivalent rational numbers between $-\frac{5}{10}$ and $\frac{2}{10}$ are $-\frac{3}{10},-\frac{6}{20},-\frac{9}{30}$

(ii)         Three equivalent rational numbers between $0$ & $-10$ are

First rational number = $\frac{1}{2}\left[0-\left(10\right)\right]=-5$

Similarly, second rational number = $\frac{-10}{2}$

And third rational number = $\frac{-15}{3}$

# Question: 102

Put the ( $\surd$ ), wherever applicable

 Number Natural Number Whole Number Integer Fraction Rational Number a.    $-114$ b.   $\frac{19}{27}$ c.    $\frac{623}{1}$ d.   $-19\frac{3}{4}$ e.    $\frac{73}{71}$ f.     $0$

## Solution

We know that,

Natural number are 1, 2, 3, 4 .

Whole number are 0, 1, 2, 3,

Integers are $-2,-1,0,1,2$

Fractions are $\frac{1}{2},\frac{2}{5},\frac{3}{7}$

Rational numbers are $\frac{1}{2},-\frac{3}{2},-\frac{7}{9}$

1. $-114$ integer & rational number.
2. $\frac{19}{27}$ Fraction & rational number
3. $\frac{623}{1}$ Natural number, whole number, integer, fraction, & rational number.
4. $-19\frac{3}{4}=-\frac{79}{4}$ rational number
5. $\frac{73}{71}$ Fraction & rational number
6. $0$ Whole number, integer, fraction & rational number hence the table is
 Number Natural Number Whole Number Integer Fraction Rational Number a.    $-114$ √ √ b.   $\frac{19}{27}$ √ √ c.    $\frac{623}{1}$ √ √ √ √ √ d.   $-19\frac{3}{4}$ √ e.    $\frac{73}{71}$ √ √ f.     $0$ √ √ √ √

# Question: 103

$a$ ’ and ‘ $b$ ’ are two different numbers taken from the numbers $1-50.$ What is the largest value that $\frac{a-b}{a+b}$ can have? What is the largest value that $\frac{a+b}{a-b}$ can have?

## Solution

Given, a and b are two different numbers between $1$ to $50.$

For largest value of $\frac{a-b}{a+b}$

Let $a$$=$$50$ and $b$$=$$1$

$\frac{a-b}{a+b}=\frac{50-1}{50+1}=\frac{49}{51},$ which is the largest value

Similarly,

For largest value of $\frac{a+b}{a-b}$

Let $a=50$ and $b=49$

$\frac{a+b}{a-b}=\frac{50+49}{50-49}=\frac{99}{1}=99,$ which is the largest value.

# Question: 104

$150$ students are studying English, Maths or both. $62$ per cent of the students are studying English and $68$ per cent are studying Maths. How many students are studying both?

## Solution

Total number of student in the class studying English, Maths or both $=150$

Student studying English $=62%$

i.e $62%$ of $150=\frac{62}{100}×150=93$

Student studying Maths $=68%$

i.e  $68%$ of $\begin{array}{l}150\\ =\frac{68}{100}×150\\ =102\end{array}$

Total students studying both $=$ Students studying English $+$ Students studying Maths $-$ Students studying English, Maths or Both

$=93+102-150=45$

# Question: 105

A body floats $\frac{2}{9}$ of its volume above the surface. What is the ratio of the body submerged volume to its exposed volume? Re-write it as a rational number.

## Solution

Let the volume of the body = V

Exposed volume = $\frac{2}{9}\text{V}$

Volume of body submerged = $\left(\text{V}-\frac{2}{9}\text{V}\right)$

= $\frac{7}{9}\text{V}$

Required ratio $=\frac{7}{9}V:\frac{2}{9}V$

$=\frac{7}{9}×\frac{9}{2}=\frac{7}{2}=7:2$

As a rational number $=\frac{7}{2}.$

# Question: 106

Find the odd one out of the following and give reason.

a.    $\frac{4}{3}×\frac{3}{4}$

b.   $\frac{-3}{2}×\frac{-2}{3}$

c.    $2×\frac{1}{2}$

d.   $\frac{-1}{3}×\frac{3}{1}$

a.    Given,