RS Aggarwal solution class 8 chapter 2 Exponents Exercise 2A

Exercise 2A

Page-33

Q1 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 1:

Evaluate:
(i) 4⁻³
(ii) ${\left(\frac{1}{2}\right)}^{-5}$
(iii) ${\left(\frac{4}{3}\right)}^{-3}$
(iv) ${\left(-3\right)}^{-4}$
(v) ${\left(\frac{-2}{3}\right)}^{-5}$

Answer 1:

(i) $4^{-3}$

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

$=\frac{1}{64}$

(ii) ${\left(\frac{1}{2}\right)}^{-5}$

$=2^5$

$=32$

(iii) ${\left(\frac{4}{3}\right)}^{-3}$

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

$=\frac{3^3}{4^3}$

$=\frac{27}{64}$

(iv) ${\left(-3\right)}^{-4}$

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

$=\frac{(-1{)}^4}{3^4}$

$=\frac{1}{81}$

(v) ${\left(\frac{-2}{3}\right)}^{-5}$

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

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

$=\frac{-243}{32}$

Q2 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 2:

Evaluate:
(i) ${\left(\frac{5}{3}\right)}^2\times {\left(\frac{5}{3}\right)}^2$
(ii) ${\left(\frac{5}{6}\right)}^6\times {\left(\frac{5}{6}\right)}^{-4}$
(iii) ${\left(\frac{2}{3}\right)}^{-3}\times {\left(\frac{2}{3}\right)}^{-2}$
(iv) ${\left(\frac{9}{8}\right)}^{-3}\times {\left(\frac{9}{8}\right)}^2$

Answer 2:

(i) ${\left(\frac{5}{3}\right)}^2\times {\left(\frac{5}{3}\right)}^2$

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

$=\frac{5^4}{3^4}$

$=\frac{625}{81}$

(ii) ${\left(\frac{5}{6}\right)}^6\times {\left(\frac{5}{6}\right)}^{-4}$

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

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

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

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

$=\frac{25}{36}$

(iii)${\left(\frac{2}{3}\right)}^{-3}\times {\left(\frac{2}{3}\right)}^{-2}$

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

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

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

$=\frac{3^5}{2^5}$

$=\frac{243}{32}$

(iv) ${\left(\frac{9}{8}\right)}^{-3}\times {\left(\frac{9}{8}\right)}^2$

$={\left(\frac{9}{8}\right)}^{(-3+2)}$

$={\left(\frac{9}{8}\right)}^{-1}=\frac{8}{9}$

Q3 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 3:

Evaluate:
(i) ${\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}\times {\left(\frac{3}{5}\right)}^0$
(ii) ${\left(\frac{-3}{5}\right)}^{-4}\times {\left(\frac{-2}{5}\right)}^2$
(iii) ${\left(\frac{-2}{3}\right)}^{-3}\times {\left(\frac{-2}{3}\right)}^{-2}$

Answer 3:

(i)

​${\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}\times {\left(\frac{3}{5}\right)}^0$

$={\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3+0}$

$={\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}$

$={\left(\frac{9}{5}\right)}^2\times {\left(\frac{5}{3}\right)}^3$

$=\frac{9^2}{5^2}\times \frac{5^3}{3^3}$

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

$=\frac{3^4}{5^2}\times \frac{5^3}{3^3}$

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

$=3\times 5=15$


(ii)

​${\left(\frac{-3}{5}\right)}^{-4}\times {\left(\frac{-2}{5}\right)}^2$

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

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

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

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

$=25\times \frac{4}{81}$

$=\frac{100}{81}$


(iii)

${\left(\frac{-2}{3}\right)}^{-3}\times {\left(\frac{-2}{3}\right)}^{-2}$

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

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

$=\frac{3^{(3+2)}}{-2^{(3+2)}}$

$=\frac{3^5}{-2^5}$

$=\frac{-243}{32}$
​

Q4 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 4:

Evaluate:
(i) ${\left\{{\left(\frac{-2}{3}\right)}^2\right\}}^{-2}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^2$

Answer 4:

(i) ${\left\{{\left(\frac{-2}{3}\right)}^2\right\}}^{-2}$

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

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

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

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

$=\frac{3^4}{2^4}$

$=\frac{81}{16}$

(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}$

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

$={\left[{\left(\frac{-1}{3}\right)}^{-4}\right]}^{-1}$

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

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

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

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

$=\frac{1}{81}$

(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^2$

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

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

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

$=\frac{2^4}{3^4}$

$=\frac{16}{81}$

Q5 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 5:

Evaluate $\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}.$

Answer 5:

$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$

$=\left\{3^3-2^3\right\}÷4^3$

$=\left\{27-8\right\}÷64$

$=\frac{19}{64}$

Q6 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 6:

Evaluate ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}.$

Answer 6:

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

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

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

$\mathrm{The} \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 4 \mathrm{and} 1 \mathrm{is} 4.$

$\therefore {\left\{\left(\frac{3\times 1}{4\times 1}\right)-\left(\frac{4\times 4}{1\times 4}\right)\right\}}^{-1}$

$= {\left\{\frac{3}{4}-\frac{16}{4}\right\}}^{-1}$

$={\left\{\frac{3-16}{4}\right\}}^{-1}$

$={\left\{\frac{-13}{4}\right\}}^{-1}$

$={\left\{\frac{4}{-13}\right\}}^1$

$=\frac{4}{-13}$

$=\frac{4\times -1}{-13\times -1}$

$=\frac{-4}{13}$

Q7 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 7:

Evaluate $\left[{\left(5^{-1}\times 3^{-1}\right)}^{-1}÷6^{-1}\right].$

Answer 7:

$\left[{\left(5^{-1}\times 3^{-1}\right)}^{-1}÷6^{-1}\right]$

$=\left[{\left(\frac{1}{5}\times \frac{1}{3}\right)}^{- 1}÷\frac{1}{6}\right]$

$=\left[{\left(\frac{1}{15}\right)}^{- 1}÷\frac{1}{6}\right]$

$=\left[15\times 6\right]$

$=90$

Q8 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 8:

Find the value of:
(i) (2⁰ + 3⁻¹) × 3²
(ii) (2⁻¹ × 3⁻¹) ÷ 2⁻³
(iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

Answer 8:

(i)
​$\left(2^0+3^{-1}\right)\times 3^2$

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

$(\mathrm{because} 2^0=1 \mathrm{and} 3^{-1}=\frac{1}{3})$

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

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

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

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

$=4\times 3=12$

(ii)

$\left(2^{-1}\times 3^{-1}\right)÷2^{-3}$

$$\begin{gathered} =\left(\frac{1}{2}\times \frac{1}{3}\right)÷{\left(\frac{1}{2}\right)}^3 \\ \left(\frac{1}{6}\right)÷\frac{1^3 }{2^3} \end{gathered}$$

$=\left(\frac{1}{6}\right)÷\left(\frac{1}{8}\right)$

$=\frac{1}{6}\times 8$

$=\frac{8}{6}$

$=\frac{4}{3}$


(iii)

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

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

$=2^2+3^2+4^2$

$=4+9+16$

$=29$

Page-34

Q9 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 9:

Find the value of x for which ${\left(\frac{5}{3}\right)}^{-4}\times {\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{3x}.$

Answer 9:

Consider the left side:
 ${\left(\frac{5}{3}\right)}^{-4}\times {\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{(-4+(-5\left)\right)}={\left(\frac{5}{3}\right)}^{-9}$

Given:
${\left(\frac{5}{3}\right)}^{-9}$

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

Comparing the powers:
$-9=3x$

$\Rightarrow x=-3$

Q10 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 10:

Find the value of x for which ${\left(\frac{4}{9}\right)}^4\times {\left(\frac{4}{9}\right)}^{-7}={\left(\frac{4}{9}\right)}^{2x-1}.$

Answer 10:

Given:
 ${\left(\frac{4}{9}\right)}^4\times {\left(\frac{4}{9}\right)}^{-7}$

$={\left(\frac{4}{9}\right)}^{2x-1}$

$\therefore$ ${\left(\frac{4}{9}\right)}^{(4-7)}={\left(\frac{4}{9}\right)}^{-3}={\left(\frac{4}{9}\right)}^{2x-1}$

$$\begin{gathered} \Rightarrow 2x-1=-3 \\ 2x=-3+1=-2 \\ \Rightarrow x=-1 \end{gathered}$$

Q11 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 11:

By what number should (−6)⁻¹ be multiplied so that the product becomes 9⁻¹?

Answer 11:

Let the required number be $x$.

$\therefore$ $x\times (-6{)}^{-1}=9^{-1}$

$x\times \frac{1}{-6}=\frac{1}{9}$

$$\begin{gathered} \Rightarrow \frac{x}{-6}=\frac{1}{9} \\ \mathrm{or} x=\frac{-6}{9} \end{gathered}$$
The greatest common divisor for the numerator and the denominator is 3.

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

$=\frac{(-6)÷3}{9÷3}=\frac{-2}{3}$

Q12 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 12:

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient may be ${\left(\frac{4}{27}\right)}^{-2}$?

Answer 12:

Let the number be $x$.

$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{27}\right)}^{-2}$
$\Rightarrow {\left(\frac{3}{-2}\right)}^3÷x={\left(\frac{27}{4}\right)}^2$

$\Rightarrow {\left(\frac{-3}{2}\right)}^3÷x={\left(\frac{27}{4}\right)}^2$

$\Rightarrow {\left(\frac{-3}{2}\right)}^3\times \frac{1}{x}={\left(\frac{27}{4}\right)}^2$

$\Rightarrow \frac{-3^3}{2^3}\times \frac{1}{x}=\frac{{27}^2}{4^2}$

$\Rightarrow \frac{-27}{8}\times \frac{1}{x}=\frac{{27}^2}{4^2}=\frac{27\times 27}{4\times 4}$

$=\frac{27 \times 27}{4 \times 2 \times 2}=\frac{27 \times 27}{8 \times 2}$

∴ $\frac{1}{x}=\left(\frac{27 \times 27}{8 \times 2}\right)/\left(\frac{-27}{8}\right)$

$\Rightarrow x=\frac{\left({\displaystyle \frac{-27}{8}}\right)}{\left({\displaystyle \frac{27\times 27}{8\times 2}}\right)}$

$=\left({\displaystyle \frac{-27}{8}}\right)\times \left(\frac{8\times 2}{27\times 27}\right)=\frac{-2}{27}$

Q13 | Ex-2A | Class 8 | RS AGGARWAL | chapter 2 | Exponents Solution  | myhelper

OPEN IN YOUTUBE

Question 13:

If 5^2x+1÷25=125, find the value of x.

Answer 13:

Given :

5^2x+1÷25=125

We know:

25=5×5=5²

125=5×5=5=5³

$\therefore$$\frac{5^{2 x+1}}{5^{2}}=5^{3}$

$\Rightarrow$5^[(2x+1)-2]=5³ or 5^[(2x+1)-2]=5^[2x-1]=5³

$\Rightarrow 2x-1=3$

$\Rightarrow 2x=3+1$

$x=\frac{4}{2}=2$

∴ x=2