Unit 9: Rational Numbers

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

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.   $q\ne 0$ 

(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.

Which of the following rational numbers is positive?

a.    $\frac{-8}{7}$ 

b.   $\frac{19}{-13}$ 

c.    $\frac{-3}{-4}$ 

d.   $\frac{-21}{13}$ 

(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.

Which of the following rational numbers is negative?

a.    $-\left(\frac{-3}{7}\right)$ 

b.   $\frac{-5}{-8}$ 

c.    $\frac{9}{8}$ 

d.   $\frac{3}{-7}$ 

(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 $\frac{3}{-7}=-\frac{3}{7}$.

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

a.    $0$ 

b.   $1$ 

c.    $-2$ 

d.   $2$ 

(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.

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

a.    $1$ 

b.   $2$ 

c.    $\frac{1}{2}$ 

d.   $0$ 

(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.

The reciprocal of $\frac{1}{2}$ is

a.    $3$ 

b.   $2$ 

c.    $-1$ 

d.   $0$ 

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

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

a.    $\frac{48}{60}$ 

b.   $\frac{-60}{48}$

c.    $\frac{-4}{5}$ 

d.   $\frac{-4}{-5}$ 

(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 $-\frac{4}{5}.$  

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

a.    $\frac{5}{4}$ 

b.   $\frac{16}{25}$ 

c.    $\frac{16}{20}$ 

d.   $\frac{15}{25}$ 

(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

 

$\frac{16}{20}=\frac{16÷4}{20÷4}=\frac{4}{5}$

$\frac{15}{25}=\frac{15÷5}{25÷5}=\frac{3}{5}$

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

How many rational numbers are there between two rational numbers?

a.    $1$ 

b.   $0$ 

c.    Unlimited

d.   $100$ 

(c)

There are unlimited rational numbers between two rational numbers.

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

a.    $0$ 

b.   negative integer

c.    positive integer

d.   $1$ 

(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.

 

 

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

(b)

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

Which is greater number in the following:

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

(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.

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

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.

$1$ is a _____ rational number.

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

Hence, $1$ is a positive rational number

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

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 $\frac{2}{9}$ 

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

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 $-\frac{3}{4}$ 

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

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.

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

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.

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

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 $\frac{1}{5}$ 

$-\frac{3}{5}$ is ______ than 0.

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

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

 

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

 

And $\frac{20}{-16}=-\frac{5}{4}$ 

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

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

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

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.

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

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 $\frac{-2}{3}$ 

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

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

$=-\frac{1}{5}$ 

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

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

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

$=\frac{-5-1}{6}=-\frac{6}{6}=-1$ 

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

$\frac{3}{4}\times \left(\frac{-2}{3}\right)=\_\_\_\_\_\_.$ 

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

Product of rational numbers $=$ $\frac{\text{product numerators}}{\text{product of denominators}}$ $=\frac{3\times (-2)}{4\times 3}=\frac{-6}{12}$ 

$=-\frac{1}{2}$ 

$\frac{-5}{3}\times \left(\frac{-3}{5}\right)=\_\_\_\_\_\_\_.$ 

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

Product of rational numbers $=$$\frac{\text{product numerators}}{\text{product of denominators}}$ $=\frac{(-5)\times (-3)}{3\times 5}=\frac{15}{15}=1$ 

$-\frac{6}{7}=\frac{\_\_}{42}$ 

Let given expression is written as:

$-\frac{6}{7}=\frac{x}{42}$ 

$x=\frac{42\times (-6)}{7}=6\times (-6)$ 

$x=-36$ 

Hence, $-\frac{6}{7}=-\frac{36}{42}$ 

$\frac{1}{2}=\frac{6}{\_\_}$ 

Let the denominator will be $x$

So, we can write the expression as:

$\frac{1}{2}=\frac{6}{x}$ 

$x=12$ 

Hence, $\frac{1}{2}=\frac{6}{12}$ 

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

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 =.

$\frac{7}{-8}\boxed{}\frac{8}{9}$

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, $\frac{7}{-8}<\frac{8}{9}.$

$\frac{3}{7}\boxed{}\frac{-5}{6}$

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, $\frac{3}{7}>-\frac{5}{6}.$ 

$\frac{5}{6}\boxed{}\frac{8}{4}$

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

 $\frac{5\times 2}{6\times 2}=\frac{10}{12}$ and

 $\frac{8\times 3}{4\times 3}=\frac{24}{12}$ 

 i.e., $24>10$ 

$\frac{24}{12}>\frac{10}{12}$ 

Hence, $\frac{5}{6}<\frac{8}{4}$ 

$\frac{-9}{7}\boxed{}\frac{4}{-7}$

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

Hence, $-\frac{9}{7}<\frac{4}{-7}$  

$\frac{8}{8}\boxed{}\frac{2}{2}$

Given $\frac{8}{8}=1$ and

          $\frac{2}{2}=1$ 

Hence, $\frac{8}{8}=\frac{2}{2}$ 

The reciprocal of _______ does not exist.

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

The reciprocal of $1$  is _______.

The reciprocal of $1=\frac{1}{1}$ 

Hence, the reciprocal of $1$ is $1.$

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

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

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

Product of rational numbers $=$ $\frac{\text{product numerators}}{\text{product of denominators}}$

$=\frac{(-3\times 3)}{7\times (-7)}=\frac{-9}{-49}=\frac{9}{49}$ 

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

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

Here, $0÷\left(-\frac{5}{6}\right)=0$ 

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

$0\times \left(-\frac{5}{6}\right)=\_\_\_\_\_\_\_.$

Here, $0\times \left(-\frac{5}{6}\right)=0$ 

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

$\_\_\_\_\_\_\_\times \left(\frac{-2}{5}\right)=1$.

Let $x\times \left(-\frac{2}{5}\right)=1$

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

$-2x=5$

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

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

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

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

For standard form $=-1÷1=-1$

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

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

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

$\frac{a}{b}=\frac{a÷m}{b÷m}$

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

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.

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

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

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

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.

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

True

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

Zero is a rational number.

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.

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

True

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

Every negative integer is not a negative rational number.

False

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

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

True

e.g. Let m $=1,2,3,\mathrm{...}$

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

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

Hence, $\frac{p}{q}=\frac{p\times m}{q\times m}$

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

True

e.g. Let $m=1,2,3,\mathrm{...}$

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

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

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

Hence, $\frac{p}{q}=\frac{p÷m}{q÷m}$

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

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.

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

False

Let $m=1,2,3,\mathrm{...}$

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

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

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

Sum of two rational numbers is always a rational number.

True

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

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

All decimal numbers are also rational numbers.

True

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

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

The quotient of two rationals is always a rational number.

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

Every fraction is a rational number.

True

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

Two rationals with different numerators can never be equal.

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.

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

False

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

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

True

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

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

False

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

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

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.

Every rational number is a whole number.

False

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

Zero is the smallest rational number.

False

A rational number can be negative for example,

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

which are less than zero.

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)	$\frac{a}{b}÷(-1)$	(c)	$1$
(iv)	$\frac{a}{b}÷\frac{-a}{b}$	(d)	$\frac{bc}{ad}$
(v)	$\frac{b}{a}÷\left(\frac{d}{c}\right)$	(e)	$\frac{ad}{bc}$

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

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

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

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

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

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

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

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

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

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

a.    $36$

b.   $-80$

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

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

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

(i)        $\frac{-60}{72}$

(ii)    $\frac{91}{-364}$ 

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

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

$91=7\times 13$

$-364=2\times 2\times 7\times 13\times -1$

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

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

$\Rightarrow \frac{91÷91}{-364÷91}$

$\Rightarrow \frac{1}{-4}$

$\Rightarrow -\frac{1}{4}$

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

(i)        $\frac{-12}{-30}$

(ii)    $\frac{14}{-49}$

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

(iv)  $\frac{299}{-161}$

(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 $\frac{299÷23}{-161÷23}$
$\frac{13}{-7}=\frac{13÷(-1)}{13÷(-1)}=\frac{-13}{7}$
Hence, the standard form of $\frac{299}{-161}$ is $\frac{-13}{7}$

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

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.

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 $10,8,3,4\&5$ which is $120$

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

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

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

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

 

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

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

$7\times x=-5\times 28$

$x=\frac{-5\times 28}{7}=-5\times 4$

$x=-20$

Hence, the value of $x$ is $-20.$

Give three rational numbers equivalent to:

(i)        $\frac{-3}{4}$

(ii)    $\frac{7}{11}$

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

 $\frac{-3\times 4}{4\times 4}=-\frac{12}{16}$
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
$\begin{array}{l}\frac{7\times 2}{11\times 2}=\frac{14}{22},\\ \frac{7\times 3}{11\times 3}=\frac{21}{33}\end{array}$ and

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

Write the next three rational numbers to complete the pattern:

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

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

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

$\frac{4\times 7}{-5\times 7}=\frac{28}{-35}$
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
$\begin{array}{l}\frac{-8\times 5}{7\times 5}=-\frac{40}{35},\\ \frac{-8\times 6}{7\times 6}=-\frac{48}{42}\end{array}$ and

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

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

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

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

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

Find the sum of

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

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

(i)             Given $\frac{8}{13}$ and $\frac{3}{11}$
Sum $\begin{array}{l}=\frac{8}{13}+\frac{3}{11}=\frac{8\times 11}{13\times 11}+\frac{3\times 13}{11\times 13}\\ =\frac{88}{143}+\frac{39}{143}\end{array}$
        $=\frac{88\times 39}{143}=\frac{127}{143}$
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.

Solve:

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

(ii)    $\frac{5}{13}-\frac{-8}{26}$

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

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

Find the product of:

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

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

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

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

Simplify:

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

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

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

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

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

Simplify:

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

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

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

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

Which is greater in the following?

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

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

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

and $\frac{7}{8}=\frac{7\times 1}{8\times 1}=\frac{7}{8}$ 
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 $3\frac{1}{9}$ 

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

Let $-7\times 11=p=-77$ 

And $12+4=q=16$ 

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

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

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

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

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

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

$=-\frac{11}{40}$ 

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

$=\frac{1\times 4}{10\times 4}+\frac{3\times 5}{8\times 5}$ 

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

$=\frac{19}{40}$ 

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

And $x÷y=\frac{1}{10}\times \frac{8}{-3}$ 

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

So, $x÷y=\frac{1}{10}\times \frac{8}{-3}$ 

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

$=-\frac{4}{15}$ 

Find the reciprocal of the following:

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

(ii)    $\frac{20}{51}\times \frac{4}{91}$ 

(iii) $\frac{3}{13}÷\frac{-4}{65}$

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

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

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

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 $\frac{65}{-4}$ 

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

$\frac{65\times 3}{13\times (-4)}=-\frac{15}{4}$

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

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

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

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

Complete the following table by finding the sums:

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

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, $a=-\frac{1}{9}+\frac{2}{3}$ 

= $\frac{2}{3}-\frac{1}{9}$ 

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

$b=\frac{2}{3}+\frac{4}{11}$ 

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

$=\frac{34}{33}$ 

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

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

$=\frac{4}{6}-\frac{5}{6}$ 

$=\frac{4-5}{6}=-\frac{1}{6}$ 

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

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

$=-\frac{45}{36}-\frac{4}{36}$ 

$=\frac{-45-4}{36}=-\frac{49}{36}$ 

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

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

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

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

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

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

$g=-\frac{1}{3}+\frac{4}{11}$ 

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

$=-\frac{11}{33}+\frac{12}{33}$ 

$=\frac{-11+12}{33}=\frac{1}{33}$ 

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

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

$=-\frac{2}{6}-\frac{5}{6}$ 

$=\frac{-2-5}{6}=-\frac{7}{6}$ 

Hence, the complete table is

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

 

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

a.    six-eights $=\frac{6}{8}$ 

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

c.    opposite of  $1=-1$

d.   one-fourth $=\frac{1}{4}$ 

e.    zero $=\frac{0}{1}$ 

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

If $p=m\times t,$ and $q=n\times t$  then $\frac{p}{q}=\frac{\boxed{}}{\boxed{}}$ 

Given, $p=m\times t$ and $q=n\times t$ 

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

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.    $\frac{\boxed{}}{\boxed{}}<\frac{\boxed{}}{\boxed{}},$ If $p\times s<r\times q$ 

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

c.    $\frac{\boxed{}}{\boxed{}}>\frac{\boxed{}}{\boxed{}},$ If  $p\times s>r\times q$

a.    Given, $p\times s<r\times q$ 

By cross multiplication, we get

$\frac{p}{q}<\frac{r}{s}$ 

b.   Given, $\frac{p}{q}=\frac{r}{s}$ 

$p\times s=r\times q$ 

c.    Given $p\times s>r\times q$

$\frac{p}{q}>\frac{r}{s}$ 

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.   $\left(–4\right)\times 6$ and $8÷2$ 

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

d.   $25+15$ and $81÷40$ 

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, $p=(-4)\times 6=-24$ 

& denominator $q=8÷2=\frac{8}{2}=4$ 

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

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

& 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 $=\frac{p}{q}=\frac{1600}{81}$

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.    $115÷207$

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, $=-6\frac{3}{7}=-\left(\frac{6\times 7+3}{7}\right)=-\frac{45}{7}$

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

HCF of $240\text{\&}840=120$ 

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 $115=5\times 23$ and $207=3\times 23\times 3$ 

HCF of $115$ and $207=23$ 

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

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 $\frac{1}{12}$

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

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

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

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 $b=-\frac{7}{9}$

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

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

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

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

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 $b=\frac{1}{12}$ 

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

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

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

$=\frac{9}{60\times 2}=\frac{9}{120}$ 

$=\frac{3}{40}$ 

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.   $\left(x÷y\right)\times z$ 

h.   $\left(x–y\right)+z$ 

i.  $x+\left(y+z\right)$ 

j.  $x÷\left(y÷z\right)$ 

k.   $x–\left(y+z\right)$ 

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

a.    Let we add P to $x$ get $y$ 

$P+x=y$ 

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

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

$=\frac{5\times 3+4\times 4}{36}$ 

$=\frac{15+16}{36}=\frac{31}{36}$ 

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

$y-A=z$ 

$\frac{5}{12}-A=\frac{7}{18}$ 

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

$=\frac{14-15}{5}=-\frac{1}{36}$ 

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

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

$P+z=x$ 

$P+\frac{7}{18}=-\frac{4}{9}$ 

$P=-\frac{4}{9}-\frac{7}{18}$ 

$=\frac{(-4\times 2-7\times 1)}{18}$

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

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

$A\times \frac{5}{12}=-\frac{4}{9}$ 

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

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

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

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, $(x÷y)\times z$ 

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

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

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

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

$=\frac{-4\times 4-5\times 3}{36}+\frac{7}{18}$ 

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

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

$=-\frac{17}{36}$ 

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

$=-\frac{4}{9}+\frac{5\times 3+7\times 2}{36}$ 

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

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

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

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

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

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

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

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

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.

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

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.

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

Let number be $p$

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

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

$p=-1\times \frac{8}{-5}=\frac{8}{5}$

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

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

Let the number be $p$

Greatest negative integer is $-1$

According to the question,

$p÷\frac{1}{2}=-1$

$\begin{array}{l}\Rightarrow p\times 2=-1\\ \Rightarrow p=\frac{-1}{2}\end{array}$ 

 

 

 

 

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

Given length of the rope $=68\text{ m}$

& length of small pieces $=4\frac{1}{4}\text{ m}=\frac{(4\times 4)+1}{4}\text{ m}=\frac{17}{4}\text{ m}$

Number of pieces $=\frac{\text{Total length of the rope}}{\text{Length of small piece}}=\frac{68}{\frac{17}{4}}$

$=\frac{68}{1}\times \frac{4}{17}=4\times 4=16$

Hence, the number of pieces is $16$. 

If $12$ shirts of equal size can be prepared from $27\text{ m}$ cloth, what is length of cloth required for each shirt?

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

Total size of available cloth $=27\text{ m}$

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

Length of cloth required for each shirt $=\frac{\text{Total available cloth}}{\text{Number of shirts}}$

$=\frac{27}{12}=\frac{9}{4}=2.25\text{ m}$

Hence, $2.25\text{ m}$ cloth required for each shirt.

Insert $3$ equivalent rational numbers between

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

(ii)         $0$ and $-10$ 

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

$\frac{-1\times 5}{2\times 5}=-\frac{5}{10}$ and

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

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 = $\frac{-15}{3}$

 

 

 

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.     $0$

 

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.     $0$		√	√	√	√

 

‘ $a$ ’ 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?

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. 

$150$ 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?

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

Student studying English $=62\%$ 

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

Student studying Maths $=68\%$ 

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

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

$=93+102-150=45$

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.

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

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

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

 

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

a.    $\frac{4}{3}\times \frac{3}{4}$ 

b.   $\frac{-3}{2}\times \frac{-2}{3}$

c.    $2\times \frac{1}{2}$

d.   $\frac{-1}{3}\times \frac{3}{1}$

a.    Given, $\frac{4}{3}\times \frac{3}{4}$

Product of denominators $=\frac{4\times 3}{3\times 4}=\frac{12}{12}=1$

2. Similarly, $-\frac{3}{2}\times -\frac{2}{3}$

$=\frac{(-3)\times (-2)}{2\times 3}=\frac{6}{6}=1$

3. $\frac{2}{1}\times \frac{1}{2}=\frac{2\times 1}{1\times 2}=\frac{2}{2}=1$
4. $-\frac{1}{3}\times \frac{3}{1}=\frac{(-1)\times 3}{3\times 1}=-\frac{3}{3}=-1$

Since, the value of options (a), (b), (c) are $1$ and option (d) is $-1.$

Hence, option (d) is odd one out.

 

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

 

a.    $\frac{4}{-9}$ 

b.   $\frac{-16}{36}$

c.    $\frac{-20}{-45}$

d.   $\frac{28}{-63}$

(c)

From the above given rational numbers, $\frac{-20}{-45}$ is odd one out among others, because all are standard form of $\frac{4}{-9}$ except $\frac{-20}{-45}$ which is $\frac{4}{9}$.

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

 

a.    $\frac{-4}{3}$ 

b.   $\frac{-7}{6}$

c.    $\frac{-10}{3}$

d.   $\frac{-8}{7}$

 

(b)

From the above given rational numbers, $-\frac{7}{6}$ is odd one out because all the three rational number except $-\frac{7}{6}$ has even numerator .

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

 

a.    $\frac{-3}{7}$ 

b.   $\frac{-9}{15}$

c.    $\frac{+24}{20}$

d.   $\frac{+35}{25}$

Standard form of rational numbers are:

$\frac{-3}{7}$, $\frac{-3}{5},\frac{6}{5},\frac{7}{5}$ 

All rational numbers have 5 as denominator, except $\frac{-3}{7}$.

What’s the Error? Chhaya simplified a rational number in this manner $\frac{-25}{-30}=\frac{-5}{6}.$ What error did the student make?

If a negative $(-)$ sign comes in both numerator & denominator, then it will be cancelled and become positive number. $\frac{-25}{-30}=\frac{25}{30}=\frac{5}{6}$

Here, Chhaya divided numerator by $5$ but denominator by $-5.$