RS Aggarwal 2019 2020 Solutions for Class 7 Math Chapter 5 - Exponents

Question 1:

Question 2:

(i) $\frac{5}{7}\times \frac{5}{7}\times \frac{5}{7}\times \frac{5}{7} = {\left(\frac{5}{7}\right)}^4$

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

(iii) $\left(\frac{-1}{6}\right)\times \left(\frac{-1}{6}\right)\times \left(\frac{-1}{6}\right)={\left(\frac{-1}{6}\right)}^3$

(iv) $\left(-8\right)\times \left(-8\right)\times \left(-8\right)\times \left(-8\right)\times \left(-8\right)={\left(-8\right)}^5$

Question 3:

(i) $\frac{25}{36}=\frac{5^2}{6^2}$                               [since 25 = 5² and 36 = 6²]
          $={\left(\frac{5}{6}\right)}^2$

(ii) $\frac{-27}{64}=\frac{{\left(-3\right)}^3}{4^3}$                     [since −27 = (−3)³ and 64 = 4³]

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

(iii) $\frac{-32}{243}=\frac{{\left(-2\right)}^5}{3^5}$                   [since −32 = (−2)⁵ and 243 = 3⁵]
               $={\left(\frac{-2}{3}\right)}^5$

(iv) $\frac{-1}{128}=\frac{{\left(-1\right)}^7}{2^7}$                    [since (−1)⁷ = −1 and 128 = 2⁷]
              $={\left(\frac{-1}{2}\right)}^7$

Question 4:

(i) ${\left(\frac{2}{3}\right)}^5=\frac{{\left(2\right)}^5}{{\left(3\right)}^5}=\frac{2\times 2\times 2\times 2\times 2}{3\times 3\times 3\times 3\times 3}=\frac{32}{243}$

(ii) ${\left(\frac{-8}{5}\right)}^3=\frac{{\left(-8\right)}^3}{{\left(5\right)}^3}=\frac{\left(-8\right)\times \left(-8\right)\times \left(-8\right)}{5\times 5\times 5}=\frac{-512}{125}$

(iii) ${\left(\frac{-13}{11}\right)}^2=\frac{{\left(-13\right)}^2}{{\left(11\right)}^2}=\frac{\left(-13\right)\times \left(-13\right)}{11\times 11}=\frac{169}{121}$

(iv) ${\left(\frac{1}{6}\right)}^3=\frac{{\left(1\right)}^3}{{\left(6\right)}^3}=\frac{1\times 1\times 1}{6\times 6\times 6}=\frac{1}{216}$

(v) ${\left(\frac{-1}{2}\right)}^5=\frac{{\left(-1\right)}^5}{{\left(2\right)}^5}=\frac{\left(-1\right)\times \left(-1\right)\times \left(-1\right)\times \left(-1\right)\times \left(-1\right)}{2\times 2\times 2\times 2\times 2}=\frac{-1}{32}$

(vi) ${\left(\frac{-3}{2}\right)}^4=\frac{{\left(-3\right)}^4}{{\left(2\right)}^4}=\frac{\left(-3\right)\times \left(-3\right)\times \left(-3\right)\times \left(-3\right)}{2\times 2\times 2\times 2}=\frac{81}{16}$

(vii) ${\left(\frac{-4}{7}\right)}^3=\frac{{\left(-4\right)}^3}{{\left(7\right)}^3}=\frac{\left(-4\right)\times \left(-4\right)\times \left(-4\right)}{7\times 7\times 7}=\frac{-64}{343}$

(viii) ${\left(-1\right)}^9=-1$       [Since (-1) ^{an odd natural number} = -1]

Question 5:

(i) ${\left(4\right)}^{-1}={\left(\frac{4}{1}\right)}^{-1}={\left(\frac{1}{4}\right)}^1=\frac{1}{4}$                                 [since ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$]

(ii) ${\left(-6\right)}^{-1}={\left(\frac{-6}{1}\right)}^{-1}={\left(\frac{1}{-6}\right)}^1=\frac{-1}{6}$                    [since ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$]

(iii) ${\left(\frac{1}{3}\right)}^{-1}={\left(\frac{3}{1}\right)}^1=\frac{3}{1}$                                           [since ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$]

(iv) ${\left(\frac{-2}{3}\right)}^{-1}={\left(\frac{3}{-2}\right)}^1=\frac{-3}{2}$                                  [since ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$]
                     

Question 6:

We know that the reciprocal of ${\left(\frac{a}{b}\right)}^m$ is ${\left(\frac{b}{a}\right)}^m$.

(i) Reciprocal of ${\left(\frac{3}{8}\right)}^4={\left(\frac{8}{3}\right)}^4$

(ii) Reciprocal of ${\left(\frac{-5}{6}\right)}^{11}={\left(\frac{-6}{5}\right)}^{11}$

(iii) Reciprocal of 6⁷ = Reciprocal of ${\left(\frac{6}{1}\right)}^7$= ${\left(\frac{1}{6}\right)}^7$

(iv) Reciprocal of (− 4)³ = Reciprocal of ${\left(\frac{-4}{1}\right)}^3$ = ${\left(\frac{-1}{4}\right)}^3$

Question 7:

(i) 8⁰ = 1
(ii) (−3)⁰ = 1
(iii) 4⁰ + 5⁰ = 1 + 1 = 2
(iv) 6⁰ × 7⁰ = 1 × 1 = 1

Note: a⁰ = 1

Question 8:

(i) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{1}{5}\right)}^2=\frac{3^4}{2^4}\times \frac{1^2}{5^2}=\frac{81\times 1}{16\times 25}=\frac{81}{400}$

(ii) ${\left(\frac{-2}{3}\right)}^5\times {\left(\frac{-3}{7}\right)}^3=\frac{{\left(-2\right)}^5}{{\left(3\right)}^5}\times \frac{{\left(-3\right)}^3}{{\left(7\right)}^3}$
                                  = $$\begin{gathered} =\frac{{\left(-2\right)}^5}{{\left(7\right)}^3}\times \frac{\left(-1\right){\left(3\right)}^3}{{\left(3\right)}^5} \\ =\frac{-32\times -1\times 3^{3-5}}{343} \\ =\frac{-32\times -1\times 3^{-2}}{343} \\ =\frac{-32\times -1\times 1}{343\times 9} \\ =\frac{32}{3087} \end{gathered}$$               $\left[s\mathrm{ince} 3^{-2}=\frac{1}{9}\right]$

                                
(iii) ${\left(\frac{-1}{2}\right)}^5\times 2^3\times {\left(\frac{3}{4}\right)}^2=\frac{{\left(-1\right)}^5}{2^5}\times 2^3\times \frac{3^2}{4^2}$
                                       $=\frac{{\left(-1\right)}^5}{2^5}\times 2^3\times \frac{3^2}{(2^2{)}^2}$
                                       $=\frac{-1\times 2^3\times 3^2}{2^5\times 2^4}$
                                       $=\frac{-1\times 2^3\times 3^2}{2^9}$$=-1\times 2^{3-9}\times 3^2$  = $-9\times 2^{-6}=\frac{-9}{2^6}=\frac{-9}{64}$                  $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$

(iv) ${\left(\frac{2}{3}\right)}^2\times {\left(\frac{-3}{5}\right)}^3\times {\left(\frac{7}{2}\right)}^2=\frac{2^2}{3^2}\times \frac{{\left(-3\right)}^3}{5^3}\times \frac{7^2}{2^2}$
                                             $\frac{-1\times 3^{3-2}\times 7^2}{5^3}=\frac{-1\times 3^1\times 7^2}{5^3}=\frac{-1\times 3\times 49}{125}=\frac{-147}{125}$

(v) $\left\{{\left(\frac{-3}{4}\right)}^3-{\left(\frac{-5}{2}\right)}^3\right\}\times 4^2=\left\{\left(\frac{-3^3}{4^3}\right)-\left(\frac{-5^3}{2^3}\right)\right\}\times 4^2$
                                             $=\left\{\left(\frac{-27}{64}\right)-\left(\frac{-125}{8}\right)\right\}\times 16$
                                             $=\left\{\frac{-27}{64}+\frac{125}{8}\right\}\times 16$
                                             $=\left(\frac{-27+1000}{64}\right)\times 16$
                                              $=\left(\frac{973}{64}\times 16\right)=\frac{973}{4}$

Question 9:

(i) ${\left(\frac{4}{9}\right)}^6\times {\left(\frac{4}{9}\right)}^{-4}={\left(\frac{4}{9}\right)}^{6+\left(-4\right)}$                                             $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{n}}\times {\mathrm{a}}^{\mathrm{m}}={\mathrm{a}}^{\mathrm{n}+\mathrm{m}}\right]$
                        =  ${\left(\frac{4}{9}\right)}^2$ = $\frac{{\left(4\right)}^2}{{\left(9\right)}^2}=\frac{4\times 4}{9\times 9}=\frac{16}{81}$

(ii) ${\left(\frac{-7}{8}\right)}^{-3}\times {\left(\frac{-7}{8}\right)}^2={\left(\frac{-7}{8}\right)}^{\left(-3\right)+2}$                                  $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{n}}\times {\mathrm{a}}^{\mathrm{m}}={\mathrm{a}}^{\mathrm{n}+\mathrm{m}}\right]$
                             = ${\left(\frac{-7}{8}\right)}^{-1}$            
                              =  ${\left(\frac{8}{-7}\right)}^1$                                          $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-1}= {\left(\frac{b}{a}\right)}^1\right]$
                              = $\left(\frac{8\times -1}{-7\times -1}\right)=\frac{-8}{7}$

(iii) ${\left(\frac{4}{3}\right)}^{-3}\times {\left(\frac{4}{3}\right)}^{-2}={\left(\frac{4}{3}\right)}^{\left(-3\right)+\left(-2\right)}$                                    $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{n}}\times {\mathrm{a}}^{\mathrm{m}}= {\mathrm{a}}^{\mathrm{n}+\mathrm{m}}\right]$
                           = ${\left(\frac{4}{3}\right)}^{-5}$
                           = ${\left(\frac{3}{4}\right)}^5$                                                $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$
                          = $\frac{{\left(3\right)}^5}{{\left(4\right)}^5}=\frac{3\times 3\times 3\times 3\times 3}{4\times 4\times 4\times 4\times 4}=\frac{243}{1024}$

Question 10:

Note: $\left[{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$


(i) 5⁻³ = ${\left(\frac{5}{1}\right)}^{-3}={\left(\frac{1}{5}\right)}^3=\frac{{\left(1\right)}^3}{{\left(5\right)}^3}=\frac{1}{125}$


(ii) (−2)⁻⁵ = ${\left(\frac{-2}{1}\right)}^{-5}={\left(\frac{1}{-2}\right)}^5=\frac{{\left(1\right)}^5}{{\left(-2\right)}^5}=\frac{1\times -1}{-32\times -1}=\frac{-1}{32}$


(iii) ${\left(\frac{1}{4}\right)}^{-4}={\left(\frac{4}{1}\right)}^4=\frac{{\left(4\right)}^4}{{\left(1\right)}^4}=\frac{256}{1}=256$


(iv) ${\left(\frac{-3}{4}\right)}^{-3}={\left(\frac{4}{-3}\right)}^3$ = $\frac{{\left(4\right)}^3}{{\left(-3\right)}^3}=\frac{64}{-27}=\frac{64\times -1}{-27\times -1}=\frac{-64}{27}$


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


(vi) ${\left(\frac{5}{7}\right)}^{-1}\times {\left(\frac{7}{4}\right)}^{-1}={\left(\frac{7}{5}\right)}^1\times {\left(\frac{4}{7}\right)}^1={\left(\frac{7\times 4}{5\times 7}\right)}^1=\frac{4}{5}$


(vii) ${\left(5^{-1}-7^{-1}\right)}^{-1}={\left(\frac{1}{5}-\frac{1}{7}\right)}^{-1}={\left(\frac{7-5}{35}\right)}^{-1}$
                                                     = ${\left(\frac{2}{35}\right)}^{-1}={\left(\frac{35}{2}\right)}^1=\frac{35}{2}$

(viii) ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$ = ${\left\{{\left(\frac{3}{4}\right)}^1-{\left(\frac{4}{1}\right)}^1\right\}}^{-1}={\left(\frac{3}{4}-\frac{4}{1}\right)}^{-1}$
                                                               $$\begin{gathered} = {\left(\frac{3-16}{4}\right)}^{-1}={\left(\frac{-13}{4}\right)}^{-1} \\ = {\left(\frac{4}{-13}\right)}^1=\left(\frac{4\times -1}{-13\times -1}\right) \\ = \frac{-4}{13} \end{gathered}$$

(ix) $\left\{{\left(\frac{3}{2}\right)}^{-1}÷{\left(\frac{-2}{5}\right)}^{-1}\right\}=\left\{{\left(\frac{2}{3}\right)}^1÷{\left(\frac{5}{-2}\right)}^1\right\}$
                                 $$\begin{gathered} =\left(\frac{2}{3}\times \frac{-2}{5}\right) \\ =\frac{-4}{15} \end{gathered}$$

(x) ${\left(\frac{23}{25}\right)}^0=1$          [since a⁰ = 1 for every integer a]

Question 11:

(i)

${\left[{\left\{{\left(-\frac{1}{4}\right)}^2\right\}}^{-2}\right]}^{-1}={\left[{\left(-\frac{1}{4}\right)}^{2\times -2}\right]}^{-1}$                 $\left[s\mathrm{ince} {\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$
                          $={\left[{\left(-\frac{1}{4}\right)}^{-4}\right]}^{-1}$
                           $$\begin{gathered} ={\left(-\frac{1}{4}\right)}^{\left(-4\right)\times \left(-1\right)} \\ ={\left(-\frac{1}{4}\right)}^4=\frac{{\left(-1\right)}^4}{{\left(4\right)}^4} \\ =\frac{1}{256} \end{gathered}$$

(ii)

${\left\{{\left(\frac{-2}{3}\right)}^2\right\}}^3={\left(\frac{-2}{3}\right)}^{2\times 3}$                               $\left[s\mathrm{ince} {\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$
                  = ${\left(\frac{-2}{3}\right)}^6$
                  = $\frac{{\left(-2\right)}^6}{{\left(3\right)}^6}=\frac{64}{729}$                      [since (−2)⁶ = 64 and (3)⁶ = 729]

(iii)

${\left(\frac{-3}{2}\right)}^3÷{\left(\frac{-3}{2}\right)}^6={\left(\frac{-3}{2}\right)}^{3-6}$                    $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$          
                             = ${\left(\frac{-3}{2}\right)}^{-3}$ 
                            $={\left(\frac{2}{-3}\right)}^3$                      $\left[s\mathrm{ince} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                            $$\begin{gathered} ={\left(\frac{2\times -1}{-3\times -1}\right)}^3={\left(\frac{-2}{3}\right)}^3 \\ =\frac{{\left(-2\right)}^3}{{\left(3\right)}^3}=\frac{-8}{27} \end{gathered}$$

(iv)

${\left(\frac{-2}{3}\right)}^7÷{\left(\frac{-2}{3}\right)}^4={\left(\frac{-2}{3}\right)}^{7-4}$                   $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$
                            $={\left(\frac{-2}{3}\right)}^3$
                             = $\frac{{\left(-2\right)}^3}{{\left(3\right)}^3}=\frac{-8}{27}$

Question 12:

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

Question 13:

Let the required number be x.
(3)^-3 x x = 4
⇒ $\frac{1}{3^3}\times x=4$
⇒ $\frac{1}{27}\times x=4$
∴ x = 4 x 27 = 108
Hence, the required number is 108.

Question 14:

Let the required number be x.
(-30)^-1 ÷ x = 6^-1
⇒ $\frac{1}{\left(-30\right)}\times \frac{1}{x}=\frac{1}{6}$
⇒ $\frac{1}{\left(-30x\right)}=\frac{1}{6}$
∴ x = $\frac{6}{\left(-30\right)}=\frac{1}{-5}$
                       =$\frac{-1}{5}$
Hence, the required number is $\frac{-1}{5}$.

Question 15:

${\left(\frac{3}{5}\right)}^3\times {\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$
⇒ ${\left(\frac{3}{5}\right)}^{3+(-6)}={\left(\frac{3}{5}\right)}^{2x-1}$      $\left[s\mathrm{ince} {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$
⇒ ${\left(\frac{3}{5}\right)}^{-3}={\left(\frac{3}{5}\right)}^{2x-1}$
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x   = −2
∴ x = $\left(\frac{-2}{2}\right)=-1$

Question 16:

$\frac{3^5\times {10}^5\times 25}{5^7\times 6^5}=\frac{3^5\times {\left(2\times 5\right)}^5\times 5^2}{5^7\times {\left(2\times 3\right)}^5}$
                        $=\frac{3^5\times 2^5\times 5^5\times 5^2}{5^7\times 2^5\times 3^5}$
                        $$\begin{gathered} =\frac{3^5\times 2^5\times 5^7}{3^5\times 2^5\times 5^7} \\ =3^{5-5}\times 2^{5-5}\times 5^{7-7} \\ =3^0\times 2^0\times 5^0 \\ =1\times 1\times 1=1 \end{gathered}$$

Question 17:

    $\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$
⇒ $\frac{2^4\times 2^{n+1}-2^2\times 2^n}{2^4\times 2^{n+2}-2^{n+1}\times 2^2}$
⇒ $\frac{2^2\times \left(2^{n+3}-2^n\right)}{2^2\times \left(2^{n+4}-2^{n+1}\right)}$
⇒ $\frac{2^n\times 2^3-2^n}{2^n\times 2^4-2^n\times 2}$
⇒ $\frac{2^n\left(2^3-1\right)}{2^n\left(2^4-2\right)}=\frac{8-1}{16-2}=\frac{7}{14}=\frac{1}{2}$

Question 18:

(i) 5²ⁿ × 5³ = 5⁹
     5²ⁿ⁺³ = 5⁹          [since aⁿ × a^m = a^m+n]

     On equating the coefficients:
     2n + 3 = 9
     ⇒ 2n = 9 − 3
     ⇒ 2n = 6
     ∴ n =$\frac{6}{2}=3$

(ii) 8 × 2ⁿ⁺² = 32
     ⇒ (2)³ × 2ⁿ⁺² = (2)⁵      [since 2³ = 8 and 2⁵ = 32]
     ⇒ (2)^{3+ (n+2)} = (2)⁵ 
     On equating the coefficients:
      3 + n + 2 = 5
      ⇒ n + 5 = 5
      ⇒ n = 5 − 5
      ∴ n = 0

(iii) 6²ⁿ⁺¹ ÷ 36 = 6³
       ⇒ 6²ⁿ⁺¹ ÷ 6² = 6³       [since 36 = 6²]
       ⇒ $\frac{6^{2n+1}}{6^2}=6^3$
       ⇒ $6^{2n+1-2}=6^3$          [since $\frac{a^m}{a^n}=a^{m-n}$ ]
       ⇒ 6^2n-1= 6³
        On equating the coefficients:
        2n - 1 = 3
        ⇒ 2n = 3 + 1
        ⇒ 2n = 4
         ∴ n = $\frac{4}{2}=2$

Question 1:

    $2^{n-7}\times 5^{n-4}=1250$
⇒ $\frac{2^n}{2^7}\times \frac{5^n}{5^4}=2\times 5^4$                       [since 1250 = 2 × 5⁴]
⇒ $\frac{2^{\mathrm{n}}\times 5^{\mathrm{n}}}{2^7\times 5^4}=2\times 5^4$
⇒  $2^n\times 5^n=2\times 5^4\times 2^7\times 5^4$           [using cross multiplication]
⇒  $2^n\times 5^n=2^{1+7}\times 5^{4+4}$               [since a^m × aⁿ= a^m+n ]
⇒ $2^n\times 5^n=2^8\times 5^8$
⇒ ${\left(2\times 5\right)}^{\mathrm{n}}={\left(2\times 5\right)}^8$                      [since aⁿ × bⁿ= (a × b)ⁿ ]
⇒ ${10}^{\mathrm{n}}={10}^8$
⇒ n = 8

Question 2:

(i) 538 = 5.38 $\times$ 10²                                        [since the decimal point is moved 2 places to the left]

(ii) 6428000 = 6.428 $\times$ 10⁶                             [since the decimal point is moved 6 places to the left]

(iii) 82934000000 = 8.2934 $\times$ 10¹⁰                [since the decimal point is moved 10 places to the left]

(iv) 940000000000 = 9.4 $\times$ 10¹¹                    [since the decimal point is moved 11 places to the left]

(v) 23000000 = 2.3 $\times$ 10⁷                               [since the decimal point is moved 7 places to the left]

Question 3:

(i) Diameter of the Earth = 1.2756 $\times$ 10⁷ m
     [since the decimal point is moved 7 places to the left]

(ii) Distance between the Earth and the Moon = 3.84 $\times$ 10⁸ m
      [since the decimal point is moved 8 places to the left]

(iii) Population of India in March 2001 = 1.027 $\times$ 10⁹
       [since the decimal point is moved 9 places to the left]

(iv) Number of stars in a galaxy = 1.0 $\times$ 10¹¹
       [since the decimal point is moved 11 places to the left]

(v) Present age of the universe = 1.2 $\times$ 10¹⁰ years
      [since the decimal point is moved 10 places to the left]

Question 4:

(i) 684502 = 6 x 10⁵ + 8 x 10⁴ + 4 x 10³ + 5 x 10² + 0 x 10¹ + 2 x 10⁰
(ii) 4007185 = 4 x 10⁶ + 0 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 1 x 10² + 8 x 10¹ + 5 x 10⁰
(iii) 5807294 = 5 x 10⁶ + 8 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 2 x 10² + 9 x 10¹ + 4 x 10⁰
(iv) 50074 = 5 x 10⁴ + 0 x 10³ + 0 x 10² + 7 x 10¹ + 4 x 10⁰

Note: a⁰ = 1

Question 1:

(i) 6 × 10⁴ + 3 × 10³ + 0 × 10² + 7 × 10¹ + 8 × 10⁰
      = 6 x 10000 + 3 x 1000 + 0 x 100 + 7 x 10 + 8 x 1= 63078
     
(ii) 9 × 10⁶ + 7 × 10⁵ + 0 × 10⁴ + 3 × 10³ + 4 × 10² + 6 × 10¹ + 2 × 10⁰
      = 9 x 1000000 + 7 x 100000 + 0 x 10000 + 3 x 1000 + 4 x 100 + 6 x 10 + 2 x 1 = 9703462

(iii) 8 × 10⁵ + 6 × 10⁴ + 4 × 10³ + 2 × 10² + 9 × 10¹ + 6 × 10⁰
       = 8 x 100000 + 6 x 10000 + 4 x 1000 + 2 x 100 + 9 x 10 + 6 x 1 = 864296