Unit: 13: Exponents and Powers

## Exercise: 1 (Multiple Choice Questions and Answers 1-22)

In questions 1 to 22, there are four options, out of which one is correct. Write the correct one.

# Question: 1

${\left[{\left(-3\right)}^{2}\right]}^{3}$ is equal to

a.    ${\left(-3\right)}^{8}$

b.   ${\left(-3\right)}^{6}$

c.    ${\left(-3\right)}^{5}$

d.   ${\left(-3\right)}^{23}$

## Solution

(b)

We have studied that, if $‘a’$ is a rational number, $m$ and $n$  are natural numbers, then  $a=\left(-3\right),m=2,n=3$

# Question: 2

For a non-zero rational number $x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{8}÷{x}^{2}$ is equal to

a.    ${x}^{4}$

b.   ${x}^{6}$

c.    ${x}^{10}$

d.   ${x}^{16}$

## Solution

(b)

We have studied that, if $‘a’$ is a rational number,  are natural numbers such that $m>n,$ then ${a}^{m}÷{a}^{n}={a}^{\left(m-n\right)}$

Here, we have

So, $\begin{array}{c}{x}^{8}÷{x}^{2}=\frac{{x}^{8}}{{x}^{2}}\\ ={x}^{8-2}\\ ={x}^{6}\end{array}$

# Question: 3

$x$ is a non-zero rational number. Product of the square of $x$ with the cube of $x$ is equal to the

a.    second power of $x$

b.   third power of $x$

c.    fifth power of $x$

d.   sixth power of $x$

## Solution

(c)

Square of $x={x}^{2}$

Cube of $x={x}^{3}$

Product of square with the cube of $x$

i.e. fifth power of $x$.

# Question: 4

For any two non-zero rational numbers $x$ and $y$, ${x}^{5}÷{y}^{5}$ is equal to

a.    ${\left(x÷y\right)}^{1}$

b.   ${\left(x÷y\right)}^{0}$

c.    ${\left(x÷y\right)}^{5}$

d.   ${\left(x÷y\right)}^{10}$

## Solution

(c)

It is given that, ${x}^{5}÷{y}^{5}=\frac{{x}^{5}}{{y}^{5}}$

We have studied that, $\frac{{p}^{n}}{{q}^{n}}={\left(\frac{p}{q}\right)}^{n}$

Here,

Thus, $\begin{array}{c}\frac{{x}^{5}}{{y}^{5}}={\left(\frac{x}{y}\right)}^{5}\\ ={\left(}^{x}\end{array}$

# Question: 5

${a}^{m}×{a}^{n}$ is equal to

a.    ${\left({a}^{2}\right)}^{mn}$

b.   ${a}^{m-n}$

c.    ${a}^{m+n}$

d.   ${a}^{mn}$

## Solution

(c)

As we have studied that, if $‘a’$ is a rational number, $m$ and $n$ are natural numbers, then ${a}^{m}×{a}^{n}={a}^{m+n}$

# Question: 6

$\left({1}^{0}+{2}^{0}+{3}^{0}\right)$ is equal to

a.    $0$

b.   $1$

c.    $3$

d.   $6$

## Solution

(c)

As we have studied that, ${a}^{0}=1$

$\therefore$ $\begin{array}{l}{1}^{0}+{2}^{0}+{3}^{0}\\ =1+1+1\\ =3\end{array}$

# Question: 7

Value of $\frac{{10}^{22}+{10}^{20}}{{10}^{20}}$ is

a.    $10$

b.   ${10}^{42}$

c.    $101$

d.   ${10}^{22}$

## Solution

(c)

The given expression can be written as

$\begin{array}{l}\frac{{10}^{22}}{{10}^{20}}+\frac{{10}^{20}}{{10}^{20}}\\ ={10}^{22-20}+{10}^{20-20}\end{array}$

$\begin{array}{l}={10}^{2}+1\\ =10×10+1\\ =100+1\\ =101\end{array}$

# Question: 8

The standard form of the number $12345$ is

a.    $12345×{10}^{1}$

b.   $123.45×{10}^{2}$

c.    $12.345×{10}^{3}$

d.   $1.2345×{10}^{4}$

## Solution

(d)

A number in its standard form is written as ‘a $×\text{\hspace{0.17em}}{10}^{k},$ where a is a terminating decimal number such that $1\le a<10$ and $k$ is any integer.

Thus, the standard form of the number $12345=1.2345×{10}^{4}$

# Question: 9

If ${2}^{1998}-{2}^{1997}-{2}^{1996}-{2}^{1995}$ $=\text{K}\text{.}\text{\hspace{0.17em}}{2}^{1995}\text{\hspace{0.17em}}$, then the value of K is

a.    $1$

b.   $2$

c.    $3$

d.   $4$

## Solution

(c) It is given that

${2}^{1998}–{2}^{1997}–{2}^{1996}+{2}^{1995}=\text{K}{.2}^{1995}$

$⇒{2}^{1995}{}^{+3}-{2}^{1995+2}-{2}^{1995+1}+{2}^{1995}=\text{K}{.2}^{1995}$

$⇒{2}^{1995}\left[{2}^{3}-{2}^{2}-{2}^{1}+1\right]=\text{K}{.2}^{1995}$

$⇒{2}^{1995}\left[8-4-2+1\right]=\text{K}{.2}^{1995}$

$⇒3=\frac{\text{K}{.2}^{1995}}{{2}^{1995}}$

Thus, the value of $\text{K}$ is $3.$

# Question:10

Which of the following is equal to 1?

a.    ${2}^{0}+{3}^{0}+{4}^{0}$

b.   ${2}^{0}×{3}^{0}×{4}^{0}$

c.    $\left({3}^{0}-{2}^{0}\right)×{4}^{0}$

d.   $\left({3}^{0}-{2}^{0}\right)×\left({3}^{0}+{2}^{0}\right)$

## Solution

(b)

Let us solve all the expressions one by one,

Option (a),

Option (b),

Hence, option (b) is the answer.

# Question: 11

In standard form, the number $72105.4$ is written as $7.21054×{10}^{n}$ where $n$ is equal to

a.    $2$

b.   $3$

c.    $4$

d.   $5$

## Solution

(c)

We have studied that, if the given number is greater than or equal to $10,$ then the power of $10$ (i.e. $n$ ) is a positive integer and is equal to the number of places the decimal point has been shifted.

Hence, $72105.4=7.21054×{10}^{4}$

# Question: 12

Square of $\left(\frac{-2}{3}\right)$ is

a.    $\frac{-2}{3}$

b.   $\frac{2}{3}$

c.    $\frac{-4}{9}$

d.   $\frac{4}{9}$

## Solution

(d)

As per question, square of

Thus,

# Question: 13

Cube of $\left(\frac{-1}{4}\right)$ is

a.    $-\frac{1}{12}$

b.   $\frac{1}{16}$

c.    $-\frac{1}{64}$

d.   $\frac{1}{64}$

## Solution

(c)

As per question, cube of

Thus, ${\left(-\frac{\text{1}}{4}\right)}^{\text{3}}=\left(-\frac{\text{1}}{4}\right)×\left(-\frac{\text{1}}{4}\right)×\left(-\frac{\text{1}}{4}\right)$

# Question: 14

Which of the following is not equal to ${\left(\frac{-5}{4}\right)}^{4}$?

a.    $\frac{{\left(-5\right)}^{4}}{{4}^{4}}$

b.   $\frac{{5}^{4}}{{\left(-4\right)}^{4}}$

c.    $-\frac{{5}^{4}}{{4}^{4}}$

d.   $\left(-\frac{5}{4}\right)×\left(-\frac{5}{4}\right)×\left(-\frac{5}{4}\right)×\left(-\frac{5}{4}\right)$

## Solution

(c)

We have studied that, ${\left(\frac{p}{q}\right)}^{m}=\frac{{p}^{m}}{{q}^{m}}$

So,

Or  $\begin{array}{l}{\left(-\frac{5}{4}\right)}^{\text{4}}\\ =\left(-\frac{5}{4}\right)×\left(-\frac{5}{4}\right)×\left(-\frac{\text{5}}{4}\right)×\left(-\frac{5}{4}\right)\end{array}$

Hence, option (c) is not equal to $1.$

# Question: 15

Which of the following is not equal to $1$?

a.    $\frac{{2}^{3}×{3}^{2}}{4×18}$

b.   $\left[{\left(-2\right)}^{3}×{\left(-2\right)}^{4}\right]÷{\left(-2\right)}^{7}$

c.    $\frac{{3}^{0}×{5}^{3}}{5×25}$

d.   $\frac{{2}^{4}}{{\left({7}^{0}+{3}^{0}\right)}^{3}}$

## Solution

(d)

Let us solve each option one by one.

Option a,

$\begin{array}{l}\frac{{2}^{3}×{3}^{2}}{4×18}\\ =\frac{2×2×2×3×3}{4×18}\\ =\frac{4×18}{4×18}\\ =1\end{array}$

Option b,

$\begin{array}{l}\left[{\left(-2\right)}^{3}×{\left(-2\right)}^{4}\right]÷{\left(-2\right)}^{7}\\ =\frac{\left[{\left(-2\right)}^{3}×{\left(-2\right)}^{4}\right]}{{\left(-2\right)}^{7}}\end{array}$

$\begin{array}{l}=\frac{{\left(-2\right)}^{3+4}}{{\left(-2\right)}^{7}}\\ =\frac{{\left(-2\right)}^{7}}{{\left(-2\right)}^{7}}\\ =1\end{array}$

Option c,

$\begin{array}{l}=\frac{5×25}{5×25}\\ =1\end{array}$

Option d,

$\begin{array}{l}=-{2}^{1}\\ =2\end{array}$

Hence, option (d) is not equal to $1$.

# Question: 16

${\left(\frac{2}{3}\right)}^{3}×{\left(\frac{5}{7}\right)}^{3}$ is equal to

a.    ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{9}$

b.   ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{6}$

c.    ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{3}$

d.   ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{0}$

## Solution

(c)

We have studied that, if  are rational numbers, then ${a}^{m}×{b}^{m}={\left(ab\right)}^{m}$

Here,

$\begin{array}{l}a=\frac{2}{3},\\ b=\frac{5}{7},\\ m=3\end{array}$

Thus,

# Question: 17

In standard form, the number $829030000$ is written as $\text{K}\text{\hspace{0.17em}}×{10}^{8}$ where K is equal to

a.    $82903$

b.   $829.03$

c.    $82.903$

d.   $8.2903$

## Solution

(d)

We know that, a number in a standard form is written as K  where $\text{K}$ is a terminating decimal such that $1\le \text{K}<10.$

So, there is only one option, where $\text{K}=8.2903$ less then $10.$

# Question: 18

Which of the following has the largest value?

a.    $0.0001$

b.   $\frac{1}{10000}$

c.    $\frac{1}{{10}^{6}}$

d.   $\frac{1}{{10}^{6}}÷0.1$

## Solution

(a, b)

Let us solve each option one by one.

Option a,

$\begin{array}{l}0.0001\\ =\frac{1}{10000}\end{array}$

Option b,

$\frac{1}{\text{1}0000}$

Option c,

$\begin{array}{l}\frac{1}{{10}^{6}}\\ =\frac{1}{10×10×10×10×10×10}\\ =\frac{1}{1000000}\end{array}$

Option d,

$\begin{array}{l}\frac{1}{{10}^{6}}÷0.1\\ =\frac{1}{{10}^{6}}×\frac{1}{0.1}\\ =\frac{1}{{10}^{6}}×\frac{10}{1}\\ =\frac{10}{{10}^{6}}\\ =\frac{1}{{10}^{5}}\end{array}$

The fraction whose denominator is the smallest will be the largest fraction.

Hence,  are the largest.

# Question: 19

In standard form $72$ crore is written as

a.    $72×{10}^{7}$

b.   $72×{10}^{8}$

c.    $7.2×{10}^{8}$

d.   $7.2×{10}^{7}$

## Solution

(c)

We have studied that, a number in standard form is written as K $×\text{\hspace{0.17em}}{10}^{n},$ where K is the terminating decimal such that $1\le \text{K<}10$ and $n$ is any integer.

Thus, $72$ crore $\begin{array}{l}=720000000\\ =7.2×{10}^{8}\end{array}$

# Question: 20

For non-zero numbers a and b, ${\left(\frac{a}{b}\right)}^{m}÷{\left(\frac{a}{b}\right)}^{n}$, where $m>n$, is equal to

a.    ${\left(\frac{a}{b}\right)}^{mn}$

b.   ${\left(\frac{a}{b}\right)}^{m+n}$

c.    ${\left(\frac{a}{b}\right)}^{m-n}$

d.   ${\left[{\left(\frac{a}{b}\right)}^{m}\right]}^{n}$

## Solution

(c)

We have studied that,

Thus, $\begin{array}{l}{\left(\frac{a}{b}\right)}^{m}÷{\left(\frac{a}{b}\right)}^{n}\\ ={\left(\frac{a}{b}\right)}^{m-n}\end{array}$

# Question: 21

Which of the following is not true?

a.    ${3}^{2}>{2}^{3}$

b.   ${4}^{3}={2}^{6}$

c.    ${3}^{3}=9$

d.   ${2}^{5}>{5}^{2}$

## Solution

(c)

Let us solve each option one by one.

Option a,

$3×3>2×2×2$

$9>8$ (true)

Option b,

(true)

Option c,

$3×3×3=9$

$27\ne 9$ (false)

Option d,

${2}^{5}>{5}^{2}$

$2×2×2×2×2=5×5$

$32>25$ (true)

Hence, option (c) is not true.

# Question: 22

Which power of $8$ is equal to ${2}^{6}$?

a.    $3$

b.   $2$

c.    $1$

d.   $4$

## Solution

(b)

Let us suppose that the power of $8$ be $x$.

According to the question, we have

${8}^{x}={2}^{6}$

${\left({2}^{3}\right)}^{x}={2}^{6}$

${2}^{3x}={2}^{6}$

Since, bases are equal, by equating their exponents, we get

$3x=6$

$\frac{3x}{3}=\frac{6}{3}$

$x=2$

Hence, the power of $8$ is $2,$ which is equal to ${2}^{6}$.

In questions 23 to 39, fill in the blanks to make the statements true.

# Question: 23

${\left(-2\right)}^{31}×{\left(-2\right)}^{13}={\left(-2\right)}^{?}$

## Solution

Here,

$\begin{array}{l}{\left(-2\right)}^{31}×{\left(-2\right)}^{13}\\ \because {\text{(a)}}^{\text{m}}×{\text{(a)}}^{n}={\text{a}}^{\text{m+n}}\\ ={\left(-2\right)}^{31+13}\\ ={\left(-2\right)}^{44}\\ \therefore {\left(-2\right)}^{31}×{\left(-2\right)}^{13}\\ ={\left(-2\right)}^{44}\end{array}$

# Question: 24

${\left(-3\right)}^{8}÷{\left(-3\right)}^{5}={\left(-3\right)}^{?}$

## Solution

Here,

$\begin{array}{l}{\left(-3\right)}^{8}÷{\left(-3\right)}^{5}\\ \because {a}^{m}÷{a}^{n}={a}^{m-n}\\ ={\left(-3\right)}^{8-5}\\ ={\left(-3\right)}^{3}\\ \therefore {\left(-3\right)}^{8}÷{\left(-3\right)}^{5}\\ ={\left(-3\right)}^{3}\end{array}$

# Question: 25

${\left(\frac{11}{15}\right)}^{4}×{\left(?\right)}^{5}={\left(\frac{11}{15}\right)}^{9}$

## Solution

Let us suppose that

${\left(\frac{11}{15}\right)}^{4}×{\left(x\right)}^{5}={\left(\frac{11}{15}\right)}^{9}$

Since, in the above equation, the powers are same. Thus, $x=\frac{11}{15}$

Hence, ${\left(\frac{11}{15}\right)}^{4}×{\left(\frac{11}{15}\right)}^{5}$$={\left(\frac{11}{15}\right)}^{9}$

# Question: 26

${\left(\frac{-1}{4}\right)}^{3}×{\left(\frac{-1}{4}\right)}^{?}={\left(\frac{-1}{4}\right)}^{11}$

## Solution

Let us suppose that

${\left(-\frac{1}{4}\right)}^{3}×{\left(-\frac{1}{4}\right)}^{x}={\left(-\frac{1}{4}\right)}^{11}$

${\left(-\frac{1}{4}\right)}^{x}=\frac{{\left(-\frac{1}{4}\right)}^{11}}{{\left(-\frac{1}{4}\right)}^{3}}$

${\left(-\frac{1}{4}\right)}^{x}={\left(-\frac{1}{4}\right)}^{8}$

Since in the above equation the bases are equal. So, by equating the powers, we get, $x=8$

Hence, ${\left(-\frac{1}{4}\right)}^{3}×{\left(-\frac{1}{4}\right)}^{8}={\left(-\frac{1}{4}\right)}^{11}$

# Question: 27

${\left[{\left(\frac{7}{11}\right)}^{3}\right]}^{4}={\left(\frac{7}{11}\right)}^{?}$

Here,

# Question: 28

${\left(\frac{6}{13}\right)}^{10}÷{\left[{\left(\frac{6}{13}\right)}^{5}\right]}^{2}={\left(\frac{6}{13}\right)}^{?}$

## Solution

$={\left(\frac{6}{13}\right)}^{10}÷{\left(\frac{6}{13}\right)}^{5×2}$

$={\left(\frac{6}{13}\right)}^{0}$

# Question: 29

${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^{2}={\left(\frac{-1}{4}\right)}^{?}$

## Solution

Here,

$={\left(-\frac{1}{4}\right)}^{16×2}$

$={\left(-\frac{1}{4}\right)}^{32}$

$\therefore {\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^{2}={\left(-\frac{1}{4}\right)}^{32}$

# Question: 30

${\left(\frac{13}{14}\right)}^{5}÷{\left(?\right)}^{2}={\left(\frac{13}{14}\right)}^{3}$

## Solution

Let us suppose that

${\left(\frac{13}{14}\right)}^{5}÷{\left(x\right)}^{2}={\left(\frac{13}{14}\right)}^{3}$

${\left(\frac{13}{14}\right)}^{5}×\frac{1}{{\left(x\right)}^{2}}={\left(\frac{13}{14}\right)}^{3}$

${\left(x\right)}^{2}=\frac{{\left(\frac{13}{14}\right)}^{5}}{{\left(\frac{13}{14}\right)}^{3}}$

${\left(x\right)}^{2}={\left(\frac{13}{14}\right)}^{5-3}$

${\left(x\right)}^{2}={\left(\frac{13}{14}\right)}^{2}$

Since in the above equation, the powers are same.

Thus, $x=\frac{13}{14}$

Hence, ${\left(\frac{13}{14}\right)}^{5}÷{\left(\frac{13}{14}\right)}^{2}={\left(\frac{13}{14}\right)}^{3}$

# Question: 31

${a}^{6}×{a}^{5}×{a}^{0}={a}^{?}$

## Solution

The expression can be written as ${a}^{6}×{a}^{5}×{a}^{0}={a}^{6+5+0}={a}^{11}$

# Question: 32

$1$ lakh $={10}^{?}$

We know that,

# Question: 33

1 million $={10}^{?}$

We know that,

# Question: 34

$729={3}^{?}$

## Solution

Here, we have to find out the factors of the given expression.

So, $729=3×3×3×3×3×3={3}^{6}$

 3 729 3 243 3 81 3 27 3 9 3 3 1

# Question: 35

$432={2}^{4}×{3}^{?}$

## Solution

Here, we are suppose to find out the factors of given expression.

So, $432=2×2×2×2×3×3×3={2}^{4}×{3}^{3}$

# Question: 36

$53700000=\text{\hspace{0.17em}}?×1{0}^{7}$

## Solution

The given number =   $53700000$

In standard form, it can be written as $53700000=537×{10}^{5}$

Also, $537=5.37×{10}^{2}×{10}^{5}=5.37×{10}^{7}$

$53700000=5.37×{10}^{7}$

# Question: 37

$88880000000=\text{\hspace{0.17em}}?\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{10}$

## Solution

The given number =  $88880000000$

In standard form, it can be written as $88880000000=8888×{10}^{7}$

Also, $8888=8.888×{10}^{3}$

So, $8.888×{10}^{3}×{10}^{7}=8.888×{10}^{10}$

$88880000000=8.888×{10}^{10}$

# Question: 38

$27500000=2.75×{10}^{?}$

## Solution

The given number =  $27500000$

This number can be expressed in standard form as

$\begin{array}{l}27500000\\ =275×{10}^{5}\end{array}$

Also,

$\begin{array}{l}275\\ =2.75×{10}^{2}\end{array}$

So,

$\begin{array}{l}2.75×{10}^{2}×{10}^{5}\\ =2.75×{10}^{7}\end{array}$                    $\left[\because {a}^{m}×{a}^{n}={a}^{m+n}\right]$

# Question: 39

$340900000=3.409×{10}^{?}$

## Solution

The given number can be written in standard form as,

$\begin{array}{l}340900000\\ =3409×{10}^{5}\end{array}$

Also,

$\begin{array}{l}3409\\ =3.409×{10}^{3}\end{array}$

So,

$3.409×{10}^{3}×{10}^{5}$

$=3.409×{10}^{8}$

$\begin{array}{l}340900000\\ =3.409×{10}^{8}\end{array}$

# Question: 40

Fill in the blanks with <, > or = sign.

a.    ${3}^{2}______\text{\hspace{0.17em}}15$

b.   ${2}^{3}______{3}^{2}$

c.    ${7}^{4}______\text{\hspace{0.17em}}{5}^{4}$

d.   $10,000______{10}^{5}$

e.

## Solution

a.    ${3}^{2}____15$
We can write,

So, $9<15$
Therefore, ${3}^{2}<15$

b.   ${2}^{3}____{3}^{2}$
We can write,

And ${3}^{2}=3×3=9$
So,
$8<9$
Therefore, ${2}^{3}<{3}^{2}$

c.    ${7}^{4}____{5}^{4}$
In the above expression, as base 7 is greater than base 5 and power is same, Therefore,
${7}^{4}>{5}^{4}$

d.   $10000__{10}^{5}$
We can write,
$10000={10}^{4}$
So, ${10}^{4}<{10}^{5}$
Therefore, $10000<{10}^{5}$

e.    ${6}^{3}___{4}^{4}$
We can write,
${6}^{3}=6×6×6=216$
And
So,
$216<256$
Therefore, ${6}^{3}<{4}^{4}$

In questions 41 to 65, state whether the given statements are True or False.

# Question: 41

One million $={10}^{7}$

False

We know that,

Hence,

# Question: 42

One hour $={60}^{2}$ seconds

True

We know that,

False

We know that,

False

We know that,

# Question: 45

${3}^{4}>{4}^{3}$

## Solution

True

We know that, $\because {3}^{4}=3×3×3×3=81$ and

$81>64$

Therefore,

# Question: 46

${\left(-\frac{3}{5}\right)}^{100}=\frac{-{3}^{100}}{-{5}^{100}}$

## Solution

True

Considering left hand side of the equation, we have

$=\frac{{\left(-1\right)}^{100}×{3}^{100}}{{5}^{100}}$

$=\frac{1×{3}^{100}}{{5}^{100}}$

$=\frac{{3}^{100}}{{5}^{100}}$

Now, taking RHS, we have

LHS $=$ RHS

Hence, ${\left(-\frac{3}{5}\right)}^{100}=\frac{-{3}^{100}}{-{5}^{100}}$

# Question: 47

${\left(10+10\right)}^{10}={10}^{10}+{10}^{10}$

## Solution

False

Let us solve LHS,

${\left(10+10\right)}^{10}={\left(20\right)}^{10}={\left(2×10\right)}^{10}={2}^{10}×{10}^{10}$

Let us solve RHS,

${10}^{10}+{10}^{10}=2×{10}^{10}$

Now,

${2}^{10}×{10}^{10}\ne 2×{10}^{10}$

Hence, LHS is not equal to RHS

# Question: 48

is true for all non-zero values of $x$.

## Solution

True

As we know that

${x}^{0}×{x}^{0}=1×1=1$                  $\left[\because {a}^{0}=1\right]$ and

${x}^{0}÷{x}^{0}=1÷1=1$                 $\left[\because {a}^{0}=1\right]$

Hence,

# Question: 49

In the standard form, a large number can be expressed as a decimal number between $0$ and $1$, multiplied by a power of $10$.

## Solution

False

We know that, a number in standard form is written as  where  is any integer.

# Question: 50

${4}^{2}$ is greater than ${2}^{4}.$

## Solution

False

${4}^{2}=4×4=16$          and

${2}^{4}=2×2×2×2=16$

Therefore,

# Question: 51

${x}^{m}+{x}^{m}={x}^{2}{}^{m}$ where $x$ is a non-zero rational number and $m$ is a positive integer.

## Solution

False

We have studied that

${a}^{m}×{a}^{n}={a}^{m+n}$

${x}^{m}×{x}^{m}={x}^{m+m}={x}^{2m}$

Also,

So, ${x}^{m}+{x}^{m}=2{x}^{m}\ne {x}^{2m}$

# Question: 52

${x}^{m}×{y}^{m}={\left(x×y\right)}^{2}{}^{m}$, where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.

## Solution

False

We have studied that if  are rational numbers, then ${x}^{m}×{y}^{m}={\left(xy\right)}^{m}$

${x}^{m}×{y}^{m}={\left(xy\right)}^{m}={\left(x×y\right)}^{m}$

Hence, ${x}^{m}×{y}^{m}\ne {\left(x×y\right)}^{2m}$

# Question: 53

${x}^{m}÷{y}^{m}={\left(x÷y\right)}^{m}$ where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.

## Solution

True

We have studied that, if  are rational numbers, then

# Question: 54

${x}^{m}×{x}^{n}={x}^{m}{}^{+n}$ where $x$ is a non-zero rational number and $m,n$ are positive integers.

## Solution

True

We have studied that, if $a$ is a rational number and $m$ and $n$ are positive integers, then

${a}^{m}×{a}^{n}={a}^{m+n}$

$\therefore {x}^{m}×{x}^{n}={x}^{m+n}$

# Question: 55

${4}^{9}$ is greater than ${16}^{3}.$

## Solution

True

Now, in ${4}^{9}>{4}^{6}$ as powers $9>6$

# Question: 56

${\left(\frac{2}{5}\right)}^{3}÷{\left(\frac{5}{2}\right)}^{3}=1$

## Solution

False

Taking LHS we get,      ${\left(\frac{2}{5}\right)}^{3}÷{\left(\frac{5}{2}\right)}^{3}={\left(\frac{2}{5}\right)}^{3}×{\left(\frac{2}{5}\right)}^{3}$

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

Hence, ${\left(\frac{2}{5}\right)}^{3}÷{\left(\frac{5}{2}\right)}^{3}\ne 1$

# Question: 57

${\left(\frac{4}{3}\right)}^{5}×{\left(\frac{5}{7}\right)}^{5}={\left(\frac{4}{3}+\frac{5}{7}\right)}^{5}$

## Solution

False

Taking LHS we get,      ${\left(\frac{4}{3}\right)}^{5}×{\left(\frac{5}{7}\right)}^{5}={\left(\frac{4}{3}×\frac{5}{7}\right)}^{5}={\left(\frac{20}{21}\right)}^{5}$

And      ${\left[\left(\frac{4}{3}\right)+\left(\frac{5}{7}\right)\right]}^{5}={\left(\frac{4}{3}+\frac{5}{7}\right)}^{5}={\left(\frac{43}{21}\right)}^{5}$

As the base is not same for LHS value,

So, LHS $\ne$ RHS

Hence, ${\left(\frac{4}{3}\right)}^{5}×{\left(\frac{5}{7}\right)}^{5}\ne {\left(\frac{4}{3}+\frac{5}{7}\right)}^{5}$

# Question: 58

${\left(\frac{5}{8}\right)}^{9}÷{\left(\frac{5}{8}\right)}^{4}={\left(\frac{5}{8}\right)}^{4}$

## Solution

False

Taking LHS we get, ${\left(\frac{5}{8}\right)}^{9}÷{\left(\frac{5}{8}\right)}^{4}={\left(\frac{5}{8}\right)}^{9-4}={\left(\frac{5}{8}\right)}^{5}\ne {\left(\frac{5}{8}\right)}^{4}$

Hence, ${\left(\frac{5}{8}\right)}^{9}÷{\left(\frac{5}{8}\right)}^{4}\ne {\left(\frac{5}{8}\right)}^{4}$

# Question: 59

${\left(\frac{7}{3}\right)}^{2}×{\left(\frac{7}{3}\right)}^{5}={\left(\frac{7}{3}\right)}^{10}$

## Solution

False

Taking LHS we get, ${\left(\frac{7}{3}\right)}^{2}×{\left(\frac{7}{3}\right)}^{5}={\left(\frac{7}{3}\right)}^{2+5}$

$={\left(\frac{7}{3}\right)}^{7}\ne {\left(\frac{7}{3}\right)}^{10}$

Hence, ${\left(\frac{7}{3}\right)}^{2}×{\left(\frac{7}{3}\right)}^{5}\ne {\left(\frac{7}{3}\right)}^{10}$

# Question: 60

${5}^{0}×{25}^{0}×{125}^{0}={\left({5}^{0}\right)}^{6}$

## Solution

True

Taking LHS we get,

$\begin{array}{l}{5}^{0}×{25}^{0}×{125}^{0}\\ =1×1×1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\because {a}^{0}=1\right)\\ =1\end{array}$

Now, taking RHS we get,

$\begin{array}{l}{\left({5}^{0}\right)}^{6}=\text{\hspace{0.17em}}{\left({1}^{}\right)}^{6}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\because {a}^{0}=1\right)\\ =1\end{array}$

Hence, ${5}^{0}×{25}^{0}×{125}^{0}={\left({5}^{0}\right)}^{6}$

# Question: 61

$\begin{array}{l}876543=8×{10}^{5}+7×{10}^{4}+6×{10}^{3}+5×{10}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+4×{10}^{1}+3×{10}^{0}\end{array}$

## Solution

True

Considering the right hand side of the given equation, we get $8×{10}^{5}+7×{10}^{4}+6×{10}^{3}+5×{10}^{2}$ $+4×{10}^{1}+3×{10}^{0}$

$=8×100000+7×10000+6×1000+5×100$

$=800000+70000+6000+500+40+3$

$=876543=\text{LHS}$

Hence, RHS $=$ LHS

# Question: 62

$600060=6×{10}^{5}+6×{10}^{2}$

## Solution

False

Considering the right hand side of the given equation, we get  $=6×{10}^{5}+6×{10}^{2}=6×100000+6×100$

$=600000+600=600600\ne 600060$

Hence, $\text{RHS}\ne \text{LHS}$

# Question: 63

$4×{10}^{5}+3×{10}^{4}+2×{10}^{3}+1×{10}^{0}=432010$

## Solution

False

Considering the left hand side of the given equation, we get $=\text{4}×\text{1}{0}^{\text{5}}+\text{3}×\text{1}{0}^{\text{4}}+\text{2}×\text{1}{0}^{\text{3}}+\text{1}×\text{1}{0}^{0}$

$=4×100000+3×10000+2×1000+1×1$

$=400000+30000+2000+1=432001\ne 432010$

Hence, $\text{LHS}\ne \text{RHS}$

# Question: 64

$8×{10}^{6}+2×{10}^{4}+5×{10}^{2}+9×{10}^{0}=8020509$

## Solution 

True

Considering the left hand side of the given equation, we get $=8×{10}^{6}+2×{10}^{4}+5×{10}^{2}+9×{10}^{0}$

$=8×1000000+2×10000+5×100+9×1$   $\left[\because a°=1\right]$

$=8000000+20000+500+9=8020509=\text{RHS}$

Hence,

# Question: 65

${4}^{0}+{5}^{0}+{6}^{0}={\left(4+5+6\right)}^{0}$

## Solution

False

Considering the left hand side of the given equation, we get, ${4}^{0}+{5}^{0}+{6}^{0}=1+1+1=3$    $\left[\because {a}^{0}=1\right]$
and ${\left(4+5+6\right)}^{0}={\left(15\right)}^{0}=1$

Hence, ${4}^{0}+{5}^{0}+{6}^{0}\ne {\left(4+5+6\right)}^{0}$

# Question: 66

Arrange in ascending order:

## Solution

Ascending order means arranging the numbers from least to greatest.

We have, ${2}^{5}=2×2×2×2×2=32.$

${3}^{3}=3×3×3=27.$

${2}^{3}×2=2×2×2×2=16.$

${3}^{6}=3×3×3×3×3×3=729.$

${3}^{5}=3×3×3×3×3=243.$

and ${2}^{3}×3=2×2×2×3=24$

Thus, the required ascending order will be

# Question: 67

Arrange in descending order:

## Solution

Descending order means arranging the numbers from greatest to least.

We have, ${2}^{2+3}={2}^{5}=2×2×2×2×2=32.$

${\left({2}^{2}\right)}^{3}={2}^{6}=2×2×2×2×2×2=64.$

${3}^{2}×{3}^{0}={3}^{2+0}={3}^{2}=9.$

${2}^{3}×{5}^{2}=2×2×2×5×5$

$=8×25=200.$

Thus, the required descending order will be

$\left({2}^{3}×{5}^{2}\right)>{\left({2}^{2}\right)}^{3}>{2}^{2+3}>\frac{{3}^{5}}{{3}^{2}}>\left({3}^{2}×{3}^{0}\right)>\left(2×{2}^{2}\right)$

# Question: 68

By what number should ${\left(-4\right)}^{5}$ be divided so that the quotient may be equal to ${\left(-4\right)}^{3}$?

## Solution

In order to find the number, that should divide  to get the quotient  we will divide  by

Hence, the required number is

# Question: 69

Find $m$ so that ${\left(\frac{2}{9}\right)}^{3}×{\left(\frac{2}{9}\right)}^{6}={\left(\frac{2}{9}\right)}^{2m-1}$

## Solution

We have,

$⇒{\left(\frac{2}{9}\right)}^{3+6}={\left(\frac{2}{9}\right)}^{2m-1}$

When the bases are same, we can equate the powers.

Therefore, $9=2m-1$

$9+1=2m$

$10=2m$

$\frac{10}{2}=\frac{2m}{2}$

$5=m$

Hence, $m=5$

# Question: 70

If $\frac{p}{q}={\left(\frac{3}{2}\right)}^{2}÷{\left(\frac{9}{4}\right)}^{0}$ find the value of ${\left(\frac{p}{q}\right)}^{3}.$

## Solution

Considering the given equation, $\frac{p}{q}={\left(\frac{3}{2}\right)}^{2}÷{\left(\frac{9}{4}\right)}^{0}$

$\frac{p}{q}={\left(\frac{3}{2}\right)}^{2}÷\frac{1}{1}$

$\frac{p}{q}={\left(\frac{3}{2}\right)}^{2}$

$\frac{p}{q}=\frac{{3}^{2}}{{2}^{2}}$

$\frac{p}{q}=\frac{9}{4}$

Cubing both sides, we get

${\left(\frac{p}{q}\right)}^{3}={\left(\frac{9}{4}\right)}^{3}$

$\begin{array}{l}{\left(\frac{p}{q}\right)}^{3}=\frac{9×9×9}{4×4×4}\\ =\frac{729}{64}\end{array}$

# Question: 71

Find the reciprocal of the rational number ${\left(\frac{1}{2}\right)}^{2}÷{\left(\frac{2}{3}\right)}^{3}$

## Solution

Considering the given expression

$\begin{array}{l}{\left(\frac{1}{2}\right)}^{2}÷{\left(\frac{2}{3}\right)}^{3}\\ =\frac{{\left(\frac{1}{2}\right)}^{2}}{{\left(\frac{2}{3}\right)}^{3}}\end{array}$

$\begin{array}{l}=\frac{\frac{{\left(1\right)}^{2}}{{\left(2\right)}^{2}}}{\frac{{\left(2\right)}^{3}}{{\left(3\right)}^{3}}}\\ =\frac{\frac{1}{4}}{\frac{8}{27}}\end{array}$

$\begin{array}{l}=\frac{1}{4}×\frac{27}{8}\\ =\frac{27}{4×8}\\ =\frac{27}{32}\end{array}$

The reciprocal is $\frac{32}{27}$

# Question: 72

Find the value of:

a.    ${7}^{0}$

b.   ${7}^{7}÷{7}^{7}$

c.    ${\left(-7\right)}^{2×7-6-8}$

d.

e.    $2×3×4÷{2}^{0}×{3}^{0}×{4}^{0}$

f.$\left({8}^{0}-{2}^{0}\right)×\left({8}^{0}+{2}^{0}\right)$

## Solution

a.    ${7}^{0}=1$, as we know that any number to the power of zero is one.

b.

$\begin{array}{l}={7}^{0}\\ =1\end{array}$

c.    $\begin{array}{l}{\left(-7\right)}^{2×7-6-8}\\ ={\left(}^{-}\\ ={\left(-7\right)}^{0}\\ =1\end{array}$

d.   $\begin{array}{l}\left({2}^{0}+{3}^{0}+{4}^{0}\right)\left({4}^{0}-{3}^{0}-{2}^{0}\right)\\ =\left(1+1+1\right)\left(1-1-1\right)\end{array}$

$\begin{array}{l}=\left(3\right)\left(-1\right)\\ =-3\end{array}$

e.    $\begin{array}{l}2×3×4÷{2}^{0}×{3}^{0}×{4}^{0}\\ =2×3×4÷1×1×1\end{array}$

$\begin{array}{l}=\frac{2×3×4}{1×1×1}\\ =2×3×4\\ =24\end{array}$

f.     $\begin{array}{l}\left({8}^{0}-{2}^{0}\right)×\left({8}^{0}+{2}^{0}\right)\\ =\left(1-1\right)×\left(1+1\right)\\ =0×2=0\end{array}$

# Question: 73

Find the value of $n$, where n is an integer and

## Solution

Considering the given equation,

${2}^{n-5}×{6}^{2}{}^{n-4}=\frac{1}{{12}^{4}×2}$

$\frac{{2}^{n}}{{2}^{5}}×\frac{{6}^{2n}}{{6}^{4}}=\frac{1}{{12}^{4}×2}$

$\frac{{2}^{n}×{6}^{2n}}{{2}^{5}×{6}^{4}}=\frac{1}{{\left(2×6\right)}^{4}×2}$

${2}^{n}×{\left({6}^{2}\right)}^{n}=\frac{\left({2}^{5}×{6}^{4}\right)}{{2}^{4}×{6}^{4}×2}$

${2}^{n}×{36}^{n}=\frac{{2}^{5}×{6}^{4}}{{2}^{5}×{6}^{4}}$

${2}^{n}×{36}^{n}=1$

${\left(2×36\right)}^{n}=1$

${72}^{n}={72}^{0}$

$n=0$

# Question: 74

Express the following in usual form:

a.    $8.01×{10}^{7}$

b.   $1.75×{10}^{-3}$

## Solution

a.    The given number $8.01×{10}^{7}$ can be written as

$\begin{array}{l}=\frac{801}{100}×100000000\\ =80100000\end{array}$

b.   The given number $1.75×{10}^{-3}$ can be written as

$\begin{array}{l}=\frac{175}{100}×\frac{1}{{10}^{3}}\\ =\frac{175}{100000}\\ =0.00175\end{array}$

# Question: 75

Find the value of

a.    ${2}^{5}$

b.   $\left(-{3}^{5}\right)$

c.    $-{\left(-4\right)}^{4}$

## Solution

We have studied that,

a.     $\begin{array}{l}{2}^{5}\\ =2×2×2×2×2\\ =32\end{array}$

b.   $\begin{array}{l}{\left(}^{-}\\ ={\left(}^{-}×{3}^{5}\\ =-1×3×3×3×3×3\\ =-243\end{array}$

c.    $\begin{array}{l}-{\left(}^{-}\\ =-\left[\left(-4\right)×\left(-4\right)×\left(-4\right)×\left(-4\right)\right]\end{array}$

$\begin{array}{l}=-\left[{\left(}^{-}\left(4×4×4×4\right)\right]\\ =-\left(256\right)\\ =-256\end{array}$

# Question: 76

Express the following in exponential form:

a.    $3×3×3×a×a×a×a$

b.   $a×a×b×b×b×c×c×c×c$

c.    $s×s×t×t×s×s×t$

## Solution

We have studied that,

a.    $\begin{array}{l}3×3×3×a×a×a×a\\ ={3}^{3}·\text{\hspace{0.17em}}{a}^{4}\end{array}$

b.   $\begin{array}{l}a×a×b×b×b×c×c×c×c\\ ={a}^{2}·\text{\hspace{0.17em}}{b}^{3}·\text{\hspace{0.17em}}{c}^{4}\end{array}$

c.    $\begin{array}{l}s×s×t×t×s×s×t\\ =s×s×s×s×t×t×t\\ ={s}^{4}·\text{\hspace{0.17em}}{t}^{3}\end{array}$

# Question: 77

How many times of $30$ must be added together to get a sum equal to ${30}^{7}$?

## Solution

Let $n$ be the number of times that $30$ must be added together to get a sum equal to ${30}^{7}.$

Therefore, we can write the above statement as -

$30+30+\dots +30={30}^{7}$

$30×n={30}^{7}$

$\frac{30×n}{30}=\frac{{30}^{7}}{30}$

$n={30}^{7-1}$

$n={30}^{6}$

Hence, if $30$ is added ${30}^{6}$ times, then we get ${30}^{7}.$

# Question: 78

Express each of the following numbers using exponential notations:

a.    $1024$

b.   $1029$

c.    $\frac{144}{875}$

## Solution

a.    The number 1024 using prime factorisation can be written as -
$1024=2×2×2×2×2×2×2×2×2×2$
$={2}^{10}$

b.   The number 1029 using prime factorisation can be written as -
$\begin{array}{l}=3×7×7×7\\ =3×{7}^{3}\end{array}$

c.    The number  $\frac{144}{875}$ using prime factorisation can be written as -

$\begin{array}{l}=\frac{2×2×2×2×3×3}{5×5×5×7}\\ =\frac{{2}^{4}×{3}^{2}}{{5}^{3}×{7}^{1}}\end{array}$

# Question: 79

Identify the greater number, in each of the following:

a.

b.

c.    $7.9×{10}^{4}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5.28×{10}^{5}$

## Solution

a.    We know that, ${2}^{6}=2×2×2×2×2×2=64$ and ${6}^{2}=6×6=36$
S
o, ${\text{2}}^{\text{6}}>{\text{6}}^{\text{6}}$

b.   We know that, ${2}^{9}=2×2×2×2×2×2×2×2×2$

$=512$ and ${9}^{2}=9×9=81$

So,

c.    We know that,

$5.28=5.28×100000=528000$

So, $5.28×{10}^{5}>7.9×{10}^{4}$

# Question: 80

Express each of the following as a product of powers of their prime factors:

a.    $9000$

b.   $2025$

c.    $800$

## Solution

The number 9000 using prime factorisation can be written as - $9000={2}^{3}×{3}^{2}×{5}^{3}$

The number 2025 using prime factorisation can be written as -  The number 800 using prime factorisation can be written as -  # Question: 81

Express each of the following in single exponential form:

a.    ${2}^{3}×{3}^{3}$

b.   ${2}^{4}×{4}^{2}$

c.    ${5}^{2}×{7}^{2}$

d.   ${\left(-5\right)}^{5}×\left(-5\right)$

e.    ${\left(-3\right)}^{3}×{\left(-10\right)}^{3}$

f.${\left(-11\right)}^{2}×{\left(-2\right)}^{2}$

## Solution

a.    We have,
$\begin{array}{l}{2}^{3}×{3}^{3}\\ ={\left(}^{2}\end{array}$ b.   We have,
$\begin{array}{l}{2}^{4}×{4}^{2}\\ ={2}^{4}×{\left(}^{{2}^{2}}\end{array}$   c.    We have,

$\begin{array}{l}{5}^{2}×{7}^{2}\\ ={\left(}^{5}.\end{array}$

$={35}^{2}$

d.   We have,

$\begin{array}{l}{\left(}^{-}×\left(-5\right)\\ ={\left(}^{-}\\ ={\left(}^{-}\end{array}$

$\begin{array}{l}={\left(}^{-}\\ ={\left(}^{-}×{\left(}^{5}\\ =1×{5}^{6}\\ ={5}^{6}\end{array}$

e.    We have,
$\begin{array}{l}{\left(}^{-}×{\left(}^{-}\\ ={\left[}^{\left(}\end{array}$

$={\left(30\right)}^{3}$

f.     We have,
$\begin{array}{l}{\left(}^{-}×{\left(}^{-}\\ ={\left[}^{\left(}\end{array}$ # Question: 82

Express the following numbers in standard form:

a.    $76,47,000$

b.   $8,19,00,000$

c.    $5,83,00,00,00,000$

d.   $24$ billion

## Solution

a.    We have, $76,47,000=7647000.00$

We have studied that, a number in standard form is written as a

where $a$ is the terminating decimal such that  is any integer.

So, $7647000=7647×{10}^{3}=7.647×{10}^{3}×{10}^{3}$

$=7.647×{10}^{6}$

Similarly,

b.   $8,19,00,000=81900000.00=819×{10}^{5}$

$=8.19×{10}^{2}×{10}^{5}=8.19×{10}^{7}$

c.    $5,83,00,00,00,000=583000000000.00$

$=583×{10}^{9}=5.83×{10}^{2}×{10}^{9}=5.83×{10}^{11}$

d.   $24$ $\text{billion}$ $=24,00,00,00,000=24×{10}^{9}$

$=2.4×{10}^{1}×{10}^{9}=2.4×{10}^{10}$

# Question: 83

The speed of light in vacuum is . Sunlight takes about $8$ minutes to reach the earth. Express distance of Sun from Earth in standard form.

## Solution

It is given that, speed of light $=3×{10}^{8}$ m/s

Since the speed is given in m/s, we need to convert the time taken into seconds.

Time taken by light to reach the Earth $=8\mathrm{min}=8$

We have studied that, Distance $=$ Speed $×$ Time $=3×{10}^{8}×480=1440×{10}^{8}$

$=1.440×{10}^{3}×{10}^{8}=1.44×{10}^{11}$

Hence, the distance of Sun from the Earth is

# Question: 84

Simplify and express each of the following in exponential form:

a.    $\left[{\left(\frac{3}{7}\right)}^{4}×{\left(\frac{3}{7}\right)}^{5}\right]÷{\left(\frac{3}{7}\right)}^{7}$

b.   $\left[{\left(\frac{7}{11}\right)}^{5}÷{\left(\frac{7}{11}\right)}^{2}\right]×{\left(\frac{7}{11}\right)}^{2}$

c.    ${\left({3}^{7}÷{3}^{5}\right)}^{4}$

d.   $\left(\frac{{a}^{6}}{{a}^{4}}\right)×{a}^{5}×{a}^{0}$

e.    $\left[{\left(\frac{3}{5}\right)}^{3}×{\left(\frac{3}{5}\right)}^{8}\right]÷\left[{\left(\frac{3}{5}\right)}^{2}×{\left(\frac{3}{5}\right)}^{4}\right]$

f.$\left({5}^{15}÷{5}^{10}\right)×{5}^{5}$

## Solution

a.    We have, $\left[{\left(\frac{3}{7}\right)}^{4}×{\left(\frac{3}{7}\right)}^{5}\right]÷{\left(\frac{3}{7}\right)}^{7}$
$={\left(\frac{3}{7}\right)}^{4+5}÷{\left(\frac{3}{7}\right)}^{7}$
$\begin{array}{l}={\left(\frac{3}{7}\right)}^{9}÷{\left(\frac{3}{7}\right)}^{7}\\ =\frac{{\left(\frac{3}{7}\right)}^{9}}{{\left(\frac{3}{7}\right)}^{7}}\\ ={\left(\frac{3}{7}\right)}^{9-7}\end{array}$
$={\left(\frac{3}{7}\right)}^{2}$

b.   We have, $\left[{\left(\frac{7}{11}\right)}^{5}÷{\left(\frac{7}{11}\right)}^{2}\right]×{\left(\frac{7}{11}\right)}^{2}$
$\begin{array}{l}=\left[\frac{{\left(\frac{7}{11}\right)}^{5}}{{\left(\frac{7}{11}\right)}^{2}}\right]×{\left(\frac{7}{11}\right)}^{2}\\ ={\left(\frac{7}{11}\right)}^{5-2}×{\left(\frac{7}{11}\right)}^{2}\end{array}$
$\begin{array}{l}={\left(\frac{7}{11}\right)}^{3}×{\left(\frac{7}{11}\right)}^{2}\\ ={\left(\frac{7}{11}\right)}^{3+2}\end{array}$
$={\left(\frac{7}{11}\right)}^{5}$

c.    We have, ${\left({3}^{7}÷{3}^{5}\right)}^{4}$
$={\left({3}^{7-5}\right)}^{4}$
$\begin{array}{l}={\left({3}^{2}\right)}^{4}\\ ={3}^{2×4}\end{array}$

$={3}^{8}$

d.   We have, $\left(\frac{{a}^{6}}{{a}^{4}}\right)×{a}^{5}×{a}^{0}$
$=\left({a}^{6-4}×{a}^{5}×1\right)$

$\begin{array}{l}={a}^{2}×{a}^{5}\\ ={a}^{2+5}\end{array}$
$={a}^{7}$

e.    We have, $\left[{\left(\frac{3}{5}\right)}^{3}×{\left(\frac{3}{5}\right)}^{8}\right]÷\left[{\left(\frac{3}{5}\right)}^{2}×{\left(\frac{3}{5}\right)}^{4}\right]$
$={\left(\frac{3}{5}\right)}^{3+8}÷{\left(\frac{3}{5}\right)}^{2+4}$
$\begin{array}{l}={\left(\frac{3}{5}\right)}^{11}÷{\left(\frac{3}{5}\right)}^{6}\\ =\frac{{\left(\frac{3}{5}\right)}^{11}}{{\left(\frac{3}{5}\right)}^{6}}\\ ={\left(\frac{3}{5}\right)}^{11-6}\end{array}$
$={\left(\frac{3}{5}\right)}^{5}$

f.We have, $\left({5}^{15}÷{5}^{10}\right)×{5}^{5}$
$=\left(\frac{{5}^{15}}{{5}^{10}}\right)×{5}^{5}$ $\begin{array}{l}={5}^{5}×{5}^{5}\\ ={5}^{5+5}\\ ={5}^{10}\end{array}$

# Question: 85

Evaluate

a.    $\frac{{7}^{8}×{a}^{10}{b}^{7}{c}^{12}}{{7}^{6}×{a}^{8}{b}^{4}{c}^{12}}$

b.   $\frac{{5}^{4}×{7}^{4}×{2}^{7}}{8×49×{5}^{3}}$

c.    $\frac{125×{5}^{2}×{a}^{7}}{{10}^{3}×{a}^{4}}$

d.   $\frac{{3}^{4}×{12}^{3}×36}{{2}^{5}×{6}^{3}}$

e.    ${\left(\frac{6×10}{{2}^{2}×{5}^{3}}\right)}^{2}×\frac{25}{27}$

f.$\frac{{15}^{4}×{18}^{3}}{{3}^{3}×{5}^{2}×{12}^{2}}$

g.   $\frac{{6}^{4}×{9}^{2}×{25}^{3}}{{3}^{2}×{4}^{2}×{15}^{6}}$

## Solution

a.    Considering the expression, $\frac{{7}^{8}×{a}^{10}{b}^{7}{c}^{12}}{{7}^{6}×{a}^{8}{b}^{4}{c}^{12}}$

$=\left(\frac{{7}^{8}}{{7}^{6}}\right)×\left(\frac{{a}^{10}}{{a}^{8}}\right)×\left(\frac{{b}^{7}}{{b}^{4}}\right)×\left(\frac{{c}^{12}}{{c}^{12}}\right)$

$={7}^{8-6}×{a}^{10-8}×{b}^{7-4}×{c}^{12-12}$

$\begin{array}{l}={7}^{2}×{a}^{2}×{b}^{3}×{c}^{0}\\ =49{a}^{2}{b}^{3}\end{array}$

b.      Considering the expression, $\frac{{5}^{4}×{7}^{4}×{2}^{7}}{8×49×{5}^{3}}$

$=\frac{{5}^{4}×{7}^{4}×{2}^{7}}{{2}^{3}×{7}^{2}×{5}^{3}}$

$\begin{array}{l}=\left(\frac{{5}^{4}}{{5}^{3}}\right)×\left(\frac{{2}^{7}}{{2}^{3}}\right)×\left(\frac{{7}^{4}}{{7}^{2}}\right)\\ ={5}^{4-3}×{2}^{7-3}×{7}^{4-2}\end{array}$

$\begin{array}{l}=5×{2}^{4}×{7}^{2}\\ =5×16×49\end{array}$

$=3920$

c.    Considering the expression, $\frac{125×{5}^{2}×{a}^{7}}{{10}^{3}×{a}^{4}}$

$=\frac{{5}^{3}×{5}^{2}×{a}^{7}}{{\left(2×5\right)}^{3}×{a}^{4}}$

$=\frac{{5}^{3+2}×{a}^{7}}{{2}^{3}×{5}^{3}×{a}^{4}}$

$=\frac{{5}^{5}×{a}^{7}}{{2}^{3}×{5}^{3}×{a}^{4}}$

$=\left(\frac{{5}^{5}}{{5}^{3}}\right)×\left(\frac{{a}^{7}}{{a}^{4}}\right)×\left(\frac{1}{{2}^{3}}\right)$

$=\left(\frac{{5}^{5-3}×{a}^{7-4}}{{2}^{3}}\right)$

$\begin{array}{l}=\frac{{5}^{2}×{a}^{3}}{{2}^{3}}\\ =\frac{25{a}^{3}}{8}\end{array}$

d.   Considering the expression, $\frac{{3}^{4}×{12}^{3}×36}{{2}^{5}×{6}^{3}}$

$=\frac{{3}^{4}×\left({2}^{6}×{3}^{3}\right)×\left({2}^{2}×{3}^{2}\right)}{{2}^{5}×{\left(2×3\right)}^{3}}$

$=\frac{{3}^{4}×{2}^{6}×{3}^{3}×{2}^{2}×{3}^{2}}{{2}^{5}×{2}^{3}×{3}^{3}}$

$\frac{\left({3}^{4}×{3}^{3}×{3}^{2}\right)×\left({2}^{6}×{2}^{2}\right)}{{2}^{5}×{2}^{3}×{3}^{3}}$

$=\frac{{3}^{4+3+2}×{2}^{6+2}}{{2}^{5+3}×{3}^{3}}$

$=\frac{{3}^{9}×{2}^{8}}{{3}^{3}×{2}^{8}}={3}^{9-3}×{2}^{8-8}$

$\begin{array}{l}={3}^{6}×{2}^{0}={3}^{6}×1\\ =729\end{array}$

e.    Considering the expression, ${\left(\frac{6×10}{{2}^{2}×{5}^{3}}\right)}^{2}×\frac{25}{27}$

$\begin{array}{l}={\left(\frac{2×3×2×5}{{2}^{2}×{5}^{3}}\right)}^{2}×\frac{{5}^{2}}{{3}^{3}}\\ ={\left(\frac{3}{{5}^{2}}\right)}^{2}×\frac{{5}^{2}}{{3}^{3}}\end{array}$

$\begin{array}{l}=\frac{{3}^{2}}{{5}^{4}}×\frac{{5}^{2}}{{3}^{3}}\\ =\frac{1}{{5}^{2}×3}\\ =\frac{1}{25×3}\\ =\frac{1}{75}\end{array}$

f.     Considering the expression,, $\frac{{15}^{4}×{18}^{3}}{{3}^{3}×{5}^{2}×{12}^{2}}$

$=\frac{{\left(3×5\right)}^{4}×{\left(2×{3}^{2}\right)}^{3}}{{3}^{3}×{5}^{2}×{\left({2}^{2}×3\right)}^{2}}$

$=\frac{{3}^{4+6-3-2}×{5}^{4}{}^{-2}}{{2}^{4-3}}$

$\begin{array}{l}=\frac{{3}^{5}×{5}^{2}}{2}\\ =\frac{\left(243×25\right)}{2}\end{array}$

$=\frac{6075}{2}$

g.   Considering the expression,, $=\frac{{6}^{4}×{9}^{2}×{25}^{3}}{{3}^{2}×{4}^{2}×{15}^{6}}$

$=\frac{{\left(2×3\right)}^{4}×{\left({3}^{2}\right)}^{2}×{\left({5}^{2}\right)}^{3}}{{3}^{2}×{2}^{4}×{3}^{6}×{5}^{6}}$

$={2}^{4-4}×{3}^{4+4-2-6}×{5}^{6-6}$

$={2}^{0}×{3}^{0}×{5}^{0}$

$\begin{array}{l}=1×1×1\\ =1\end{array}$

# Question: 86

Express the given information in Scientific notation (standard form) and then arrange them in ascending order of their size.

 Sl.No. Deserts of the World Area (Sq. Kilometres) (1) Kalahari, South Africa $932,400$ (2) Thar, India $199,430$ (3) Gibson, Australia $155,400$ (4) Great Victoria, Australia $647,500$ (5) Sahara, North Africa $8,598,800$

## Solution

1.   Area of Kalahari, South Africa

$=932,400=932400.00$

[Since we know that the standard form is written as $a×{10}^{k}$ ]

$\begin{array}{l}=9324×{10}^{2}\\ =9.324×{10}^{3}×{10}^{2}\end{array}$

$=9.324×{10}^{5}$

2.   Area of Thar, India $=199,430=199430.00$

$\begin{array}{l}=19943×{10}^{1}\\ =1.9943×{10}^{4}×{10}^{1}\\ =1.9943×{10}^{5}\end{array}$

3.   Area of Gibson, Australia $=155,400$

$=155400.00$

$\begin{array}{l}=1554×{10}^{2}\\ =1.554×{10}^{3}×{10}^{2}\\ =1.554×{10}^{5}\end{array}$

4.   Area of Great Victoria, Australia $=647,500$

$\begin{array}{l}=6475×{10}^{2}\\ =6.475×{10}^{3}×{10}^{5}\\ =6.475×{10}^{5}\end{array}$

5.   Area of Sahara, North-Africa $=8,598,800$

$\begin{array}{l}=85988×{10}^{2}\\ =8.5988×{10}^{4}×{10}^{2}\end{array}$

$=8.5988×{10}^{6}$

We have studied that, to compare two numbers written in scientific notation: The number with the larger power of $10$ is greater than the number with the smaller power of $10$. If the powers of ten are the same, then the number with larger factor is the larger number. Hence, the ascending order of the sizes of the deserts will be Gibson, Australia $<$ Thar, India $<$ Great Victoria, Australia $<$ Kalahari, South-Africa $<$ Sahara, North-Africa.

# Question: 87

Express the given information in Scientific notation and then arrange them in descending order of their size.

 Sl.No. Name of the Planet Mass (in kg) (1) Mercury $330000000000000000000000$ (2) Venus $4870000000000000000000000$ (3) Earth $5980000000000000000000000$ (4) Mars $642000000000000000000000$ (5) Jupiter $\begin{array}{l}190000000000000000000000\\ 0000\end{array}$ (6) Saturn $\begin{array}{l}56900000000000000000000\\ 0000\end{array}$ (7) Uranus $86900000000000000000000000$ (8) Neptune $\begin{array}{l}10200000000000000000000\\ 0000\end{array}$ (9) Pluto $13100000000000000000000$

## Solution

We have studied that, a number is written in standard form as a $×\text{\hspace{0.17em}}\text{\hspace{0.17em}}{10}^{k}$, where a is terminating decimal and k is an integer.

 Sl.No. Name of the Planet Mass (in kg)