DAV Class 7 Maths Chapter 4 Worksheet 5

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DAV Class 7 Maths Chapter 4 Worksheet 5

Exponents and Powers Worksheet 5


1. Simplify

(i) \( \displaystyle (2^{-1} - 3^{-1})^{-1} + (6^{-1} - 8^{-1})^{-1} \)

Answer

\begin{align*} & = (2^{-1} - 3^{-1})^{-1} + (6^{-1} - 8^{-1})^{-1} \\ \\ &= \left(\frac{1}{2 } - \frac{1 }{3 }\right)^{-1} + \left(\frac{1}{6} - \frac{1}{8}\right)^{-1} \\ \\ &= \left(\frac{1 {\color{green} \times 3}}{2 {\color{green} \times 3}} - \frac{1 {\color{green} \times 2}}{3 {\color{green} \times 2}}\right)^{-1} + \left(\frac{1 {\color{green} \times 4}}{6 {\color{green} \times 4}} - \frac{1 {\color{green} \times 3}}{8 {\color{green} \times 3}}\right)^{-1} \\ \\ &= \left(\frac{3-2}{6}\right)^{-1} + \left(\frac{4-3}{24} \right)^{-1} \\ \\ &= \left(\frac{1}{6}\right)^{-1} + \left(\frac{1}{24}\right)^{-1} \\ \\ &= 6 + 24 \\ &= \color{red} 30 \end{align*}

(ii) \( \displaystyle (2^{-1} \times 3^{-1})^{2} \times \left(-\frac{3}{8}\right)^{-1} \)

Answer

\begin{align*} &= (2^{-1} \times 3^{-1})^{2} \times \left(-\frac{3}{8}\right)^{-1} \\\\ &= \left(\frac{1}{2} \times \frac{1}{3}\right)^{2} \times \left(-\frac{8}{3}\right) \\ \\ &= \left(\frac{1}{6}\right)^{2} \times \left(-\frac{8}{3}\right) \\ \\ &= \frac{1}{\cancel{36}_9} \times \left(-\frac{\cancel8^2}{3}\right) \\ \\ &= \color{red} -\frac{2}{27} \end{align*}

(iii) \( \displaystyle (4^{-1} \times 3^{-1}) \div 12^{-1} \)

Answer

\begin{align*} &= (4^{-1} \times 3^{-1}) \div 12^{-1} \\ \\ &= \left(\frac{1}{4} \times \frac{1}{3}\right) \div \frac{1}{12} \\ \\ &= \left(\frac{1}{12}\right) \div \frac{1}{12} \\ \\ &= \frac{1}{\cancel{12}} \times \cancel{12} \\ \\ &= \color{red} 1 \end{align*}

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

Answer

\begin{align*} &= \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 \\ &= \color{red}29 \end{align*}

(v) \( \displaystyle \left(2^{-1} \div 5^{-1}\right)^2 \times \left(-\frac{5}{8}\right)^{-1}\)

Answer

\begin{align*} &= \left(2^{-1} \div 5^{-1}\right)^2 \times \left(-\frac{5}{8}\right)^{-1} \\ \\ &= \left(\frac{1}{2} \div \frac{1}{5}\right)^2 \times \left(-\frac{8}{5}\right) \\ \\ &= \left(\frac{1}{2} \times \frac{5}{1}\right)^2 \times \left(-\frac{8}{5}\right) \\ \\ &= \left(\frac{5}{2}\right)^2 \times \left(-\frac{8}{5}\right) \\ \\ &= \frac{\cancel{25}^5}{\cancel4_1} \times \left(-\frac{\cancel8^2}{\cancel5_1}\right) \\ \\ &= 5 \times (-2) \\ &= \color{red} -10 \end{align*}

(vi) \( \displaystyle {\left(\frac{3}{7}\right)^{-2}} \div {\left(\frac{4}{7}\right)^{-2}} \)

\begin{align*} \end{align*}

Answer

\begin{align*} &={\left(\frac{3}{7}\right)^{-2}} \div {\left(\frac{4}{7}\right)^{-2}} \\ \\ &= {\left(\frac{7}{3}\right)^{2}} \div {\left(\frac{7}{4}\right)^{2}} \\ \\ &= \frac{49}{9} \div \frac{49}{16} \\ \\ &= \frac{\cancel{49}^1}{9} \times \frac{16}{\cancel{49}_1} \\ \\ &= \color{red} \frac{16}{9} \end{align*}

(vii) \( \displaystyle \left[\left(\frac{2}{5}\right)^{-1} \times \left(\frac{3}{4}\right)^{-1}\right]^{-1} \)

Answer

\begin{align*} & = \left[\left(\frac{2}{5}\right)^{-1} \times \left(\frac{3}{4}\right)^{-1}\right]^{-1} \\ \\ &= \left[\frac{5}{\cancel2_1} \times \frac{\cancel4^2}{3}\right]^{-1} \\ \\ &= \left[\frac{5 \times 2}{1 \times 3}\right]^{-1} \\ \\ &= \left[\frac{10}{3}\right]^{-1} \\ \\ &= \color{red} \frac{3}{10} \\ \end{align*}

(viii) \( \displaystyle ({4^5} \div {4^8}) \times 64 \)

Answer

\begin{align*} & = ({4^5} \div {4^8}) \times 64 \\ \\ &= \left(\frac{1}{4^{8-5}}\right) \times 64 \\ \\ &= \frac{1}{4^3} \times 64 \\ \\ &= \frac{1}{\cancel{64}_1} \times \cancel{64}^1 \\ \\ &= \color{red} 1 \end{align*}

(ix) \( \displaystyle \left(\frac{4}{5}\right)^{-3} \times \left(\frac{4}{5}\right)^{2} \times \left(\frac{4}{5}\right)^{3} \)

Answer

\begin{align*} & =\left(\frac{4}{5}\right)^{-3} \times \left(\frac{4}{5}\right)^{2} \times \left(\frac{4}{5}\right)^{3} \\ \\ & = \left(\frac{4}{5}\right)^{-3 + 2 + 3} \\ \\ & = \left(\frac{4}{5}\right)^{-3 + 5} \\ \\ & = \left(\frac{4}{5}\right)^{2} \\ \\ & = \frac{4}{5} \times \frac{4}{5} \\ \\ & = \color{red} \frac{16}{25} \end{align*}

(x) \( \displaystyle \frac{\left(-\frac{1}{2}\right)^{-3}}{\left(-\frac{1}{2}\right)^{-4}} - \frac{\left(-\frac{1}{2}\right)^{-5}}{\left(-\frac{1}{2}\right)^{-6}} \)

Answer

\begin{align*} &= {\color{magenta} \frac{\left(-\frac{1}{2}\right)^{-3}}{\left(-\frac{1}{2}\right)^{-4}}} - {\color{green} \frac{\left(-\frac{1}{2}\right)^{-5}}{\left(-\frac{1}{2}\right)^{-6}}} \\ \\ & = {\color{magenta} \left(-\frac{1}{2}\right)^{-3} \div \left(-\frac{1}{2}\right)^{-4}} - {\color{green} \left(-\frac{1}{2}\right)^{-5} \div \left(-\frac{1}{2}\right)^{-6}} \\ \\ &= {\color{magenta} \left(-\frac{1}{2}\right)^{(-3) - (-4)}} - {\color{green} \left(-\frac{1}{2}\right)^{(-5) - (-6)}} \\ \\ &= {\color{magenta} \left(-\frac{1}{2}\right)^{-3 +4}} - {\color{green} \left(-\frac{1}{2}\right)^{-5+6}} \\ \\ &= {\color{magenta} \left(-\frac{1}{2}\right)^1} - {\color{green} \left(-\frac{1}{2}\right)^1 }\\ \\ &= -\frac{1}{2} + \frac{1}{2} \\ \\ &= \color{red} 0 \end{align*}

2. By what number should we multiply \( (2^{-5}) \) so that the product may be equal to \( (2^{-1}) \)?

Solution

\begin{align*} \text{Let } x & \text{ be multiplied to } (2^{-5}) \\ x \times 2^{-5} &= 2^{-1} \\ x &= 2^{-1} \div 2^{-5} \\ x &= 2^{(-1) - (-5)} \\ x &= 2^{-1 + 5} \\ x &= 2^{4} \\ x &= 16 \end{align*}

Answer The number is \( \color{red} 16 \)

3. By what number should \( \displaystyle \left( -\frac{1}{5} \right)^{-1} \) be multiplied so that the product may be equal to \( \displaystyle \left( \frac{1}{5} \right)^{-3} \)?

Solution

\begin{align*} \text{Let } x & \text{ be multiplied to } \left(-\frac{1}{5}\right)^{-1} \\ \\ x \times \left(-\frac{1}{5}\right)^{-1} &= \left(\frac{1}{5}\right)^{-3} \\ \\ x &= \left(\frac{1}{5}\right)^{-3} \div \left(-\frac{1}{5}\right)^{-1} \\ \\ x &= 5^{3} \div (-5) \\ \\ x &= \cancel{125}^{25} \times \left(-\frac{1}{\cancel5_1}\right) \\ x &= 25 \times (-1) \\ x &= -25 \\ \end{align*}

Answer The number is \( \color{red} -25 \)

4. By what number should \( (24)^{-1} \) be divided so that the quotient may be equal to \( (4)^{-1} \)?

Solution

\begin{align*} \text{Let } (24)^{-1} & \text{ be divided by } x \\ \\ (24)^{-1} \div x &= (4)^{-1} \\ \\ (24)^{-1} &= (4)^{-1} \times x \\ \\ (24)^{-1} \div (4)^{-1} &= x \\ \\ \frac{1}{24} \div \frac{1}{4} &= x \\ \\ \frac{1}{\cancel{24}_6} \times \cancel4^1 &= x \\ \frac{1}{6} &= x \\ x & = \frac{1}{6} \end{align*}

Answer The number is \( \displaystyle \color{red} \frac{1}{6} \)

5. By what number should \( \displaystyle \left( -\frac{2}{3} \right)^3 \) be divided so that the quotient may be equal to \( \displaystyle \left( \frac{9}{4} \right)^{-2} \)?

Solution

\begin{align*} \text{Let } \left(-\frac{2}{3}\right)^{3} & \text{ be divided by } x \\ \\ \left(-\frac{2}{3}\right)^{3} \div x &= \left(\frac{9}{4}\right)^{-2} \\ \\ \left(-\frac{2}{3}\right)^{3} &= \left(\frac{9}{4}\right)^{-2} \times x \\ \\ \left(-\frac{2}{3}\right)^{3} \div \left(\frac{9}{4}\right)^{-2} &= x \\ \\ \left(-\frac{2}{3}\right)^{3} \div \left(\frac{4}{9}\right)^{2} &= x \\ \\ \left(-\frac{2}{3}\right)^{3} \times \left(\frac{9}{4}\right)^{2} &= x \\ \\ -\frac{\cancel8^{\color{magenta}1}}{\cancel{27}_{\color{green}1}} \times \frac{\cancel{81}^{\color{green}3}}{\cancel{16}_{\color{magenta}2}} &= x \\ \\ -\frac{3}{2} &= x \\ x &= -\frac{3}{2} \end{align*}

Answer The number is \( \color{red} \displaystyle -\frac{3}{2} \)

6. Find the value of x so that -

(i) \( \displaystyle \left(\frac{3}{4}\right)^{-9} \times \left(\frac{3}{4}\right)^{-7} = \left(\frac{3}{4}\right)^{4x} \)

Solution

\begin{align*} \left(\frac{3}{4}\right)^{-9} \times \left(\frac{3}{4}\right)^{-7} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \left(\frac{3}{4}\right)^{-9 + (-7)} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \left(\frac{3}{4}\right)^{-9 -7} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \left(\frac{3}{4}\right)^{-16} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \\ 4x &= -16 \\ x &= \frac{-\cancel{16}^4}{\cancel4_1} \\ \\ \color{red} x &= \color{red} -4 \end{align*}

(ii) \( \displaystyle \left(\frac{2}{9}\right)^{-6} \times \left(\frac{2}{9}\right)^{3} = \left(\frac{2}{9}\right)^{2x - 1} \)

Solution

\begin{align*} \left(\frac{2}{9}\right)^{-6} \times \left(\frac{2}{9}\right)^{3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \\ \left(\frac{2}{9}\right)^{-6 + 3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \\ \left(\frac{2}{9}\right)^{-3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 2x - 1 &= -3 \\ 2x &= -3 + 1 \\ 2x &= -2 \\ x &= \frac{-\cancel2^1}{\cancel2_1} \\ \\ \color{red} x &= \color{red} -1 \end{align*}

(iii) \( \displaystyle \left(\frac{5}{7}\right)^2 \div \left(\frac{5}{7}\right)^{3x + 1} = \left(\frac{5}{7}\right)^{4} \)

Solution

\begin{align*} \left(\frac{5}{7}\right)^2 \div \left(\frac{5}{7}\right)^{3x + 1} &= \left(\frac{5}{7}\right)^{4} \\ \\ \left(\frac{5}{7}\right)^{2 - (3x + 1)} & = \left(\frac{5}{7}\right)^{4} \\ \\ \left(\frac{5}{7}\right)^{2 - 3x - 1} & = \left(\frac{5}{7}\right)^{4} \\ \\ \left(\frac{5}{7}\right)^{1 - 3x} & = \left(\frac{5}{7}\right)^{4} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 1 - 3x &= 4 \\ 3x &= 4 - 1 \\ -3x &= 3 \\ -x &= \frac{\cancel3^1}{\cancel3_1} \\ -x &= 1 \\ \color{red} x &= \color{red} -1 \end{align*}

(iv) \( \displaystyle \left(-\frac{6}{11}\right)^x \div \left[\left(-\frac{6}{11}\right)^{-2}\right]^{-1} = \left[\left(-\frac{6}{11}\right)^2\right]^{-3} \)

Solution

\begin{align*} \left(-\frac{6}{11}\right)^x \div \left[\left(-\frac{6}{11}\right)^{-2}\right]^{-1} &= \left[\left(-\frac{6}{11}\right)^2\right]^{-3} \\ \\ \left(-\frac{6}{11}\right)^x \div \left(-\frac{6}{11}\right)^2 &= \left(-\frac{6}{11}\right)^{-6} \\ \\ \left(-\frac{6}{11}\right)^{x-2} &= \left(-\frac{6}{11}\right)^{-6} \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ x - 2 &= -6 \\ x &= -6 + 2 \\ \color{red} x &= \color{red} -4 \end{align*}

7. If \( \ \displaystyle \frac{p}{q} = \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{3}\right)^{-4} \quad \text{Find the value of } \left(\frac{p}{q}\right)^{-2}\)

Solution

\begin{align*} \frac{p}{q} &= \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{3}\right)^{-4} \\ \\ \frac{p}{q} &= \left(\frac{2}{3}\right)^2 \times (3)^{4} \\ \\ &= \frac{4}{\cancel9_1} \times \cancel{81}^9 \\ \\ &= 4 \times 9 \\ \\ \color{green} \frac{p}{q} &= \color{green} 36 \\ \\ \left(\frac{p}{q}\right)^{-2} &= \left(36\right)^{-2} \\ \\ &= \left(\frac{1}{36}\right)^2 \\ \\ &= \left(\frac{1}{36}\right) \times \left(\frac{1}{36}\right) \\ \\ \color{red} \left(\frac{p}{q}\right)^{-2} &= \color{red} \frac{1}{1296} \end{align*}

8. If \( \displaystyle a = \left(\frac{3}{5}\right)^{-2} \div \left(\frac{7}{5}\right)^0, \quad \text{Find the value of } a^{-3} \)

Solution

\begin{align*} a &= \left(\frac{3}{5}\right)^{-2} \div \left(\frac{7}{5}\right)^0 \\ \\ a &= \left(\frac{5}{3}\right)^{2} \div 1 \\ \\ &= \left(\frac{5}{3}\right)^2 \\ \\ \color{green} a &= \color{green} \frac{25}{9} \\ \\ a^{-3} &= \left(\frac{25}{9}\right)^{-3} \\ \\ &= \left(\frac{9}{25}\right)^3 \\ \\ &= \frac{9}{25} \times \frac{9}{25} \times \frac{9}{25} \\ \\ \color{red} a^{-3} &= \color{red} \frac{729}{15625} \end{align*}

9. Simplify \( \displaystyle \left[\left(\frac{2}{3}\right)^2\right]^3 \times \left(\frac{2}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \)

Solution

\begin{align*} & = \left[\left(\frac{2}{3}\right)^2\right]^3 \times \left(\frac{2}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \\ \\ &= \left(\frac{2}{3}\right)^{6} \times \left(\frac{2}{3}\right)^{-4} \times \frac{1}{3} \times \frac{1}{6} \\ \\ &= \left(\frac{2}{3}\right)^{6 + (-4)} \times \frac{1}{3} \times \frac{1}{6} \\ \\ &= \left(\frac{2}{3}\right)^{6 -4} \times \frac{1}{18} \\ \\ &= \left(\frac{2}{3}\right)^{2} \times \frac{1}{18} \\ \\ &= \frac{\cancel4^2}{9} \times \frac{1}{\cancel{18}_9} \\ \\ &= \color{red} \frac{2}{81} \end{align*}

10. Find the reciprocals of

(i) \( \displaystyle {\left(\frac{1}{2}\right)^{-2}} \div {\left(\frac{2}{3}\right)^{-3}} \)

Solution

\begin{align*} & = {\left(\frac{1}{2}\right)^{-2}} \div {\left(\frac{2}{3}\right)^{-3}} \\ \\ &= 2^2 \div {\left(\frac{3}{2}\right)^3} \\ \\ &= 4 \div \frac{27}{8} \\ \\ &= 4 \times \frac{8}{27} \\ \\ &= \frac{32}{27} \\ \\ \color{red} \text{Reciprocal } &= \color{red} \frac{27}{32} \end{align*}

(ii) \( \displaystyle \left(\frac{2}{5}\right)^3 \times \left(\frac{5}{4}\right)^2 \)

Solution

\begin{align*} & =\left(\frac{2}{5}\right)^3 \times \left(\frac{5}{4}\right)^2 \\ \\ & = \frac{\cancel8^1}{\cancel{125}_5} \times \frac{\cancel{25}^1}{\cancel{16}_2} \\ \\ &= \frac{1}{5 \times 2} \\ \\ &= \frac{1}{10} \\ \\ \color{red} \text{Reciprocal } &= \color{red} 10 \end{align*}

11. Express the following as a rational number with positive exponent.

(i) \( \displaystyle \left(\frac{3}{2}\right)^{-4} \)

Solution

\begin{align*} & = \left(\frac{3}{2}\right)^{-4} \\ \\ & = \color{red} \left(\frac{2}{3}\right)^{4} \end{align*}

(ii) \( \displaystyle (3^{-3})^2 \)

Solution

\begin{align*} & = (3^{-3})^2 \\ &= 3^{-3 \times 2} \\ &= 3^{-6} \\ &= \ \color{red}\left(\frac{1}{3}\right)^{6} \end{align*}

(iii) \( \displaystyle 7^2 \times 7^{-3} \)

Solution

\begin{align*} & = 7^2 \times 7^{-3} \\ &= 7^{2+ (- 3)} \\ &= 7^{2 - 3} \\ &= 7^{-1} \\ &= \ \color{red} \frac{1}{7} \end{align*}

(iv) \( \displaystyle \left[\left(\frac{5}{8}\right)^{-2}\right]^3 \)

Solution

\begin{align*} & = \left[\left(\frac{5}{8}\right)^{-2}\right]^3 \\ \\ &= \left(\frac{5}{8}\right)^{-2 \times 3} \\ \\ &= \left(\frac{5}{8}\right)^{-6} \\ \\ &= \color{red} \left(\frac{8}{5}\right)^{6} \end{align*}

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

Solution

\begin{align*} & = \left(\frac{2}{5}\right)^{-3} \times \left(\frac{2}{5}\right)^{5} \\ \\ &= \left(\frac{2}{5}\right)^{-3 + 5} \\ \\ &= \color{red} \left(\frac{2}{5}\right)^{2} \end{align*}

(vi) \( \displaystyle (8^{3} \div 8^{5}) \times 8^{-4} \)

Solution

\begin{align*} & = (8^{3} \div 8^{5}) \times 8^{-4} \\ &= 8^{3 - 5} \times 8^{-4} \\ &= 8^{-2} \times 8^{-4} \\ &= 8^{-2 + (-4)} \\ &= 8^{-2 -4} \\ &= 8^{-6} \\ &= \color{red} \left(\frac{1}{8}\right)^{6} \end{align*}

12. Express the following as a rational number with negative exponent.

(i) \( \displaystyle \left(\frac{1}{7}\right)^{5} \)

Solution

\begin{align*} &= \left(\frac{1}{7}\right)^{5} \\ \\ &= \color{red} (7)^{-5} \end{align*}

(ii) \( \displaystyle (3^2)^9 \)

Solution

\begin{align*} & = (3^2)^9 \\ & = (3)^{2 \times 9} \\ & = (3)^{18} \\ &= \color{red} \left(\frac{1}{3}\right)^{-18} \end{align*}

(iii) \( \displaystyle 5^3 \times 5^2 \)

Solution

\begin{align*} & = 5^{3 + 2} \\ & = 5^{5} \\ &= \color{red} \left(\frac{1}{5}\right)^{-5} \end{align*}

(iv) \( \displaystyle \left[\left(-\frac{8}{9}\right)^3\right]^2 \)

Solution

\begin{align*} & = \left[\left(-\frac{8}{9}\right)^3\right]^2 \\ \\ &= \left(-\frac{8}{9}\right)^{3 \times 2} \\ \\ &= \left(-\frac{8}{9}\right)^{6} \\ \\ &= \color{red} \left(-\frac{9}{8}\right)^{-6} \end{align*}

(v) \( \displaystyle \left(\frac{5}{7}\right)^3 \div \left(\frac{5}{7}\right)^2 \)

Solution

\begin{align*} & = \left(\frac{5}{7}\right)^3 \div \left(\frac{5}{7}\right)^2 \\ \\ &= \left(\frac{5}{7}\right)^{3 - 2} \\ \\ &= \left(\frac{5}{7}\right)^{1} \\ \\ &= \color{red} \left(\frac{7}{5}\right)^{-1} \end{align*}

(vi) \( \displaystyle \left(2^6 \div 2^5\right) \times 2^2 \)

Solution

\begin{align*} &= \left(2^6 \div 2^5\right) \times 2^2 \\ &= 2^{6-5} \times 2^2 \\ &= 2^1 \times 2^2 \\ &= 2^{1+2} \\ &= 2^3 \\ &= \color{red} \left(\frac{1}{2}\right)^{-3} \end{align*}