RS Aggarwal 2019,2020 solution class 7 chapter 5 Exponents Exercise 5C

Exercise 5C

Page-93

Question 1:

Mark (✓) against the correct answer
(6⁻¹ − 8⁻¹)⁻¹=?

(a) $-\frac{1}{2}$
(b) −2
(c) $\frac{1}{24}$
(d) 24

Answer 1:

(d) 24

${\left(6^{-1}-8^{-1}\right)}^{-1}={\left(\frac{1}{6}-\frac{1}{8}\right)}^{-1}$
                  = ${\left(\frac{4-3}{24}\right)}^{-1}$        [since L.C.M. of 6 and 8 is 24]
                  = ${\left(\frac{1}{24}\right)}^{-1}$
                  = ${\left(\frac{24}{1}\right)}^1=24$      $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$

Question 2:

Mark (✓) against the correct answer
(5⁻¹ × 3⁻¹)⁻¹=?

(a) $\frac{1}{15}$
(b) $\frac{-1}{15}$
(c) 15
(d) −15

Answer 2:

(c) 15

We have:

${\left(5^{-1}\times 3^{-1}\right)}^{-1}={\left(\frac{1}{5}\times \frac{1}{3}\right)}^{-1}$
                      = ${\left(\frac{1}{15}\right)}^{-1}$      
                      = ${\left(\frac{15}{1}\right)}^1=15$      $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$

Question 3:

Mark (✓) against the correct answer

(2⁻¹ − 4⁻¹)² = ?

(a) 4
(b) −4
(c) $\frac{1}{16}$
(d) $\frac{-1}{16}$

Answer 3:

(c) $\frac{1}{16}$

We have:

${\left(2^{-1}-4^{-1}\right)}^2={\left(\frac{1}{2}-\frac{1}{4}\right)}^2$
                   = ${\left(\frac{2-1}{4}\right)}^2$        [since L.C.M. of 2 and 4 is 4]
                   = ${\left(\frac{1}{4}\right)}^2$
                   = $\left(\frac{1}{4}\times \frac{1}{4}\right)=\frac{1}{16}$     

Question 4:

Mark (✓) against the correct answer

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

(a) $\frac{61}{144}$
(b) 29
(c) $\frac{144}{61}$
(d) none of these

Answer 4:

(b) 29

We have:

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}={\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2$         $\left[s\mathrm{ince}{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$
                                 = (2² + 3² + 4²)
                                 = (4 + 9 + 16)
                                 = 29

Page-94

Question 5:

Mark (✓) against the correct answer

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

(a) $\frac{2}{3}$
(b) $\frac{5}{6}$
(c) $\frac{6}{5}$
(d) none of these

Answer 5:

(c) $\frac{6}{5}$

We have:
${\left\{6^{-1}+{\left(\frac{3}{2}\right)}^{-1}\right\}}^{-1}={\left(\frac{1}{6}+\frac{2}{3}\right)}^{-1}$
                             = ${\left(\frac{1+4}{6}\right)}^{-1}$    [since L.C.M. of 3 and 6 is 6]
                             = ${\left(\frac{5}{6}\right)}^{-1}$        
                             = ${\left(\frac{6}{5}\right)}^1=\left(\frac{6}{5}\right)$             $\left[s\mathrm{ince}{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$

Question 6:

Mark (✓) against the correct answer

${\left(\frac{-1}{2}\right)}^{-6}=?$

(a) −64
(b) 64
(c) $\frac{1}{64}$
(d) $\frac{-1}{64}$

Answer 6:

(b) 64
We have:
${\left(\frac{-1}{2}\right)}^{-6}={\left(\frac{2}{-1}\right)}^6$                            $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n\right]$
                $$\begin{gathered} ={\left(-2\right)}^6 \\ =\left(-2\right)\times \left(-2\right)\times \left(-2\right)\times \left(-2\right)\times \left(-2\right)\times \left(-2\right) \\ = 64 \end{gathered}$$

Question 7:

Mark (✓) against the correct answer

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

(a) $\frac{3}{8}$
(b) $\frac{-3}{8}$
(c) $\frac{8}{3}$
(d) $\frac{-8}{3}$

Answer 7:

(b)  $\frac{-3}{8}$

${\left\{{\left(\frac{3}{4}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}={\left(\frac{4}{3}-\frac{4}{1}\right)}^{-1}$
                                  = ${\left(\frac{4-12}{3}\right)}^{-1}$     [ since L.C.M. of 1 and 3 is 3]
                                  = ${\left(\frac{-8}{3}\right)}^{-1}$
                                  = ${\left(\frac{3}{-8}\right)}^1$                      $\left[s\mathrm{ince} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                                  = $\left(\frac{3\times -1}{-8\times -1}\right)=\frac{-3}{8}$

Question 8:

Mark (✓) against the correct answer

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

(a) $\frac{1}{16}$
(b) 16
(c) $\frac{-1}{16}$
(d) −16

Answer 8:

(a) $\frac{1}{16}$

${\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1}={\left[{\left(-\frac{1}{2}\right)}^{2\times -2}\right]}^{-1}$                           $\left[s\mathrm{ince}{\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$
                          $={\left[{\left(-\frac{1}{2}\right)}^{-4}\right]}^{-1}$
                           $$\begin{gathered} ={\left(-\frac{1}{2}\right)}^{\left(-4\right)\times \left(-1\right)} \\ ={\left(-\frac{1}{2}\right)}^4=\frac{{\left(-1\right)}^4}{{\left(2\right)}^4} \\ =\frac{1}{16} \end{gathered}$$

Question 9:

Mark (✓) against the correct answer
${\left(\frac{5}{6}\right)}^0=?$

(a) $\frac{6}{5}$
(b) 0
(c) 1
(d) none of these

Answer 9:

(c) 1

(a)⁰ = 1
∴ ${\left(\frac{5}{6}\right)}^0=1$

Question 10:

Mark (✓) against the correct answer
${\left(\frac{2}{3}\right)}^{-5}=?$

(a) $\frac{32}{243}$
(b) $\frac{243}{32}$
(c) $\frac{-32}{243}$
(d) $\frac{-243}{32}$

Answer 10:

(b) $\frac{243}{32}$

${\left(\frac{2}{3}\right)}^{-5}={\left(\frac{3}{2}\right)}^5$                                                    $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n\right]$
     
         = $\frac{3^5}{2^5}=\frac{3\times 3\times 3\times 3\times 3}{2\times 2\times 2\times 2\times 2}=\frac{243}{32}$

Question 11:

Mark (✓) against the correct answer
${\left\{{\left(\frac{1}{3}\right)}^2\right\}}^4=?$

(a) ${\left(\frac{1}{3}\right)}^6$
(b) ${\left(\frac{1}{3}\right)}^8$
(c) ${\left(\frac{1}{3}\right)}^{16}$
(d) ${\left(\frac{1}{3}\right)}^{24}$

Answer 11:

(b) ${\left(\frac{1}{3}\right)}^8$

${\left\{{\left(\frac{1}{3}\right)}^2\right\}}^4={\left(\frac{1}{3}\right)}^{2\times 4}={\left(\frac{1}{3}\right)}^8$                  $\left[s\mathrm{ince}{\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$

Question 12:

Mark (✓) against the correct answer
${\left(\frac{-3}{2}\right)}^{-1}=?$

(a) $\frac{2}{3}$
(b) $\frac{-2}{3}$
(c) $\frac{3}{2}$
(d) none of these

Answer 12:

(b) $\frac{-2}{3}$
We have:
${\left(\frac{-3}{2}\right)}^{-1}={\left(\frac{2}{-3}\right)}^1$       $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-n}= {\left(\frac{b}{a}\right)}^n\right]$
               = $\frac{-2}{3}$

Question 13:

Mark (✓) against the correct answer
$\left(3^2-2^2\right)\times {\left(\frac{2}{3}\right)}^{-3}=?$

(a) $\frac{45}{8}$
(b) $\frac{8}{45}$
(c) $\frac{8}{135}$
(d) $\frac{135}{8}$

Answer 13:

(d) $\frac{135}{8}$

$\left(3^2-2^2\right)\times {\left(\frac{2}{3}\right)}^{-3}=\left(9-4\right)\times {\left(\frac{3}{2}\right)}^3$         $\left[s\mathrm{ince} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                             $=5\times \frac{3^3}{2^3}$ =  $5\times \frac{27}{8}$ = $\frac{135}{8}$

Question 14:

Mark (✓) against the correct answer
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$

(a) $\frac{19}{64}$
(b) $\frac{64}{19}$
(c) $\frac{27}{16}$
(d) none of these

Answer 14:

(a) $\frac{19}{64}$
We have:
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$ = $\left\{{\left(\frac{3}{1}\right)}^3- {\left(\frac{2}{1}\right)}^3\right\} ÷ {\left(\frac{4}{1}\right)}^3$                               $\left[s\mathrm{ince} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                                             = $\left\{\left(3^3\right)-{\left(2\right)}^3\right\}÷{\left(4\right)}^3$
                                             = $\left(27 - 8\right) ÷ 64$
                                             = $19 ÷ 64$
                                             = $19\times \frac{1}{64}=\frac{19}{64}$

Page-95

Question 15:

Mark (✓) against the correct answer
${\left(\frac{-1}{5}\right)}^3÷{\left(\frac{-1}{5}\right)}^8=?$

(a) ${\left(-\frac{1}{5}\right)}^5$
(b) ${\left(\frac{-1}{5}\right)}^{11}$
(c) (−5)⁵
(d) ${\left(\frac{1}{5}\right)}^5$

Answer 15:

(c) (-5)⁵

We have:
${\left(\frac{-1}{5}\right)}^3÷{\left(\frac{-1}{5}\right)}^8={\left(\frac{-1}{5}\right)}^{3-8}$               $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$          
                             = ${\left(\frac{-1}{5}\right)}^{-5}$ 
                            $={\left(\frac{5}{-1}\right)}^5$                  $\left[\mathrm{Since} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                            $={\left(\frac{5\times -1}{-1\times -1}\right)}^5={\left(\frac{-5}{1}\right)}^5={\left(-5\right)}^5$
                                

Question 16:

Mark (✓) against the correct answer
${\left(\frac{-2}{5}\right)}^7÷{\left(\frac{-2}{5}\right)}^5=?$

(a) $\frac{4}{25}$
(b) $\frac{-4}{25}$
(c) ${\left(\frac{-2}{5}\right)}^{12}$
(d) $\frac{25}{4}$

Answer 16:

(a) $\frac{4}{25}$

${\left(\frac{-2}{5}\right)}^7÷{\left(\frac{-2}{5}\right)}^5={\left(\frac{-2}{5}\right)}^{7-5}$            $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$
                            $={\left(\frac{-2}{5}\right)}^2$
                             = $\frac{{\left(-2\right)}^2}{{\left(5\right)}^2}=\frac{4}{25}$

Question 17:

Mark (✓) against the correct answer
${\left(\frac{-2}{3}\right)}^2=?$

(a) $\frac{4}{3}$
(b) $\frac{-2}{9}$
(c) $\frac{4}{9}$
(d) $\frac{-4}{9}$

Answer 17:

(c) $\frac{4}{9}$

${\left(\frac{-2}{3}\right)}^2=\frac{-2}{3}\times \frac{-2}{3}=\frac{4}{9}$

Question 18:

Mark (✓) against the correct answer
${\left(\frac{-1}{2}\right)}^3=?$

(a) $\frac{-3}{2}$
(b) $\frac{-1}{8}$
(c) $\frac{-1}{6}$
(d) none of these

Answer 18:

(b) $\frac{-1}{8}$
We have:
${\left(\frac{-1}{2}\right)}^3=\frac{-1}{2}\times \frac{-1}{2}\times \frac{-1}{2}=\frac{-1}{8}$

Question 19:

Mark (✓) against the correct answer
If ${\left(\frac{5}{3}\right)}^{-5}\times {\left(\frac{5}{3}\right)}^{11}={\left(\frac{5}{3}\right)}^{8x}$, then x = ?

(a) $\frac{-1}{2}$
(b) $\frac{-3}{4}$
(c) $\frac{3}{4}$
(d) $\frac{4}{3}$

Answer 19:

(c) $\frac{3}{4}$

${\left(\frac{5}{3}\right)}^{-5}\times {\left(\frac{5}{3}\right)}^{11}={\left(\frac{5}{3}\right)}^{8x}$
⇒ ${\left(\frac{5}{3}\right)}^{-5+11}={\left(\frac{5}{3}\right)}^{8x}$        [ since $a^m\times a^n=a^{m+n}$]
⇒ ${\left(\frac{5}{3}\right)}^6={\left(\frac{5}{3}\right)}^{8x}$
 On equating the coefficients:
  6 = 8x
  ∴ x = $\frac{6}{8}=\frac{3}{4}$

Question 20:

Mark (✓) against the correct answer
By what number should (−8)⁻¹ be multiplied to get 10⁻¹?

(a) $\frac{4}{5}$
(b) $\frac{-5}{4}$
(c) $\frac{-4}{5}$
(d) none of these

Answer 20:

(c) $\frac{-4}{5}$
Let the required number be x.
(−8)^-1 x x = (10)^-1
⇒ $\frac{1}{-8}\times x=\frac{1}{10}$
∴ x = $\frac{1}{10}\times (-8)$ = $\frac{-4}{5}$
Hence, the required number is $\frac{-4}{5}$.

Question 21:

Mark (✓) against the correct answer
Which of the following numbers is in standard form?

(a) 21.56 × 10⁵
(b) 215.6 × 10⁴
(c) 2.156 × 10⁶
(d) none of these

Answer 21:

(c) 2.156 × 10⁶
A given number is said to be in standard form if it can be expressed as k x 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 2.156 × 10⁶