Rs Aggarwal 2017 for Class 7 Math Chapter 5 - Exponents

Share with your friends

Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 5 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 7 students for Math Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 90:

Answer:

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

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

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

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

Page No 91:

Answer:

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

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

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

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

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

Page No 91:

Answer:

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

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

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

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

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

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

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

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

Page No 91:

Answer:

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

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

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

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

Page No 91:

Answer:

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

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

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

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

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

Page No 91:

Answer:

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

Note: a⁰ = 1

Page No 91:

Answer:

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

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


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

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

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

Page No 91:

Answer:

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

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

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

Page No 91:

Answer:

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

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

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

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

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

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

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

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

(viii) ${\{{(\frac{4}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}\}}^{- 1}$ = ${\{{(\frac{3}{4})}^{1} - {(\frac{4}{1})}^{1}\}}^{- 1} = {(\frac{3}{4} - \frac{4}{1})}^{- 1}$
                                                               $= {(\frac{3 - 16}{4})}^{- 1} = {(\frac{- 13}{4})}^{- 1} = {(\frac{4}{- 13})}^{1} = (\frac{4 \times - 1}{- 13 \times - 1}) = \frac{- 4}{13}$

(ix) $\{{(\frac{3}{2})}^{- 1} \div {(\frac{- 2}{5})}^{- 1}\} = \{{(\frac{2}{3})}^{1} \div {(\frac{5}{- 2})}^{1}\}$
                                 $= (\frac{2}{3} \times \frac{- 2}{5}) = \frac{- 4}{15}$

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

Page No 91:

Answer:

(i)

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

(ii)

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

(iii)

${(\frac{- 3}{2})}^{3} \div {(\frac{- 3}{2})}^{6} = {(\frac{- 3}{2})}^{3 - 6}$                 $[s ince a^{m} \div a^{n} = a^{m - n}]$
   = ${(\frac{- 3}{2})}^{- 3}$
                            $= {(\frac{2}{- 3})}^{3}$                   $[s ince {(\frac{a}{b})}^{- 1} = {(\frac{b}{a})}^{1}]$
                            $= {(\frac{2 \times - 1}{- 3 \times - 1})}^{3} = {(\frac{- 2}{3})}^{3} = \frac{{(- 2)}^{3}}{{(3)}^{3}} = \frac{- 8}{27}$

(iv)

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

Page No 91:

Answer:

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

Page No 91:

Answer:

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.

Page No 91:

Answer:

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

Page No 91:

Answer:

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

Page No 91:

Answer:

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

Page No 92:

Answer:

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

Page No 92:

Answer:

(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]
       ⇒ $\frac{6^{2 n + 1}}{6^{2}} = 6^{3}$
       ⇒ $6^{2 n + 1 - 2} = 6^{3}$          [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 = $\frac{4}{2} = 2$

Page No 92:

Answer:

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

Page No 92:

Answer:

(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]

Page No 92:

Answer:

(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 $\times$ 10¹⁰years
      [since the decimal point is moved 10 places to the left]

Page No 93:

Answer:

(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

Page No 93:

Answer:

(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

Page No 93:

Answer:

(d) 24

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

Page No 93:

Answer:

(c) 15

We have:

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

Page No 93:

Answer:

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

We have:

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

Page No 93:

Answer:

(b) 29

We have:

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

Page No 94:

Answer:

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

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

Page No 94:

Answer:

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

Page No 94:

Answer:

(b) $\frac{- 3}{8}$

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

Page No 94:

Answer:

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

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

Page No 94:

Answer:

Page No 94:

Answer:

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

${(\frac{2}{3})}^{- 5} = {(\frac{3}{2})}^{5}$     $[s ince {(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}]$

   = $\frac{3^{5}}{2^{5}} = \frac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} = \frac{243}{32}$

Page No 94:

Answer:

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

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

Page No 94:

Answer:

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

Page No 94:

Answer:

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

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

Page No 94:

Answer:

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

Page No 95:

Answer:

(c) (-5)⁵

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

Page No 95:

Answer:

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

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

Page No 95:

Answer:

Page No 95:

Answer:

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

Page No 95:

Answer:

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

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

Page No 95:

Answer:

(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 $\frac{- 4}{5}$ .

Page No 95:

Answer:

(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

Page No 96:

Answer:

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

(i) Reciprocal of ${(\frac{2}{3})}^{4}$ = ${(\frac{3}{2})}^{4}$

(ii) Reciprocal of ${(\frac{- 3}{5})}^{61} = {(\frac{- 5}{3})}^{61}$

(iii) Reciprocal of 2⁵= Reciprocal of ${(\frac{2}{1})}^{5} = {(\frac{1}{2})}^{5}$

(iv) Reciprocal of (−5)⁶ = Reciprocal of ${(\frac{- 5}{1})}^{6} = {(\frac{- 1}{5})}^{6}$

Page No 96:

Answer:

Let the required number be x.
(−6)^-1 × x = (9)^-1
⇒ $\frac{1}{- 6} \times x = \frac{1}{9}$
∴ x = $\frac{1}{9} \times (- 6) = \frac{(- 2)}{3}$
Hence, the required number is $\frac{- 2}{3}$ .

Page No 96:

Answer:

Let the required number be x.
(−20)^-1 ÷ x = (−10)^-1
⇒ $\frac{1}{(- 20)} \times \frac{1}{x} = \frac{1}{(- 10)}$
⇒ $\frac{1}{(- 20 x)} = \frac{1}{(- 10)}$
∴ x = $\frac{(- 10)}{(- 20)} = \frac{1}{2} = 2^{- 1}$
Hence, the required number is 2^-1.

Page No 96:

Answer:

(i) 2000000 = 2.000000 × 10⁶         [since the decimal point is moved 6 places to the left]
                    = 2 × 10⁶

(ii) 6.4 × 10⁵= 6.4 × 100000
                      = 640000

Page No 96:

Answer:

We have:
$\frac{16 \times 2^{n + 1} - 8 \times 2^{n}}{16 \times 2^{n + 2} - 4 \times 2^{n + 1}}$

⇒ $\frac{2^{4} \times 2^{n + 1} - 2^{3} \times 2^{n}}{2^{4} \times 2^{n + 2} - 2^{2} \times 2^{n + 1}}$

⇒ $\frac{2^{3} \times (2^{n + 2} - 2^{n})}{2^{3} \times (2^{n + 3} - 2^{n})}$

⇒ $\frac{2^{n} \times 2^{2} - 2^{n}}{2^{n} \times 2^{3} - 2^{n}}$

⇒ $\frac{2^{n} (2^{2} - 1)}{2^{n} (2^{3} - 1)} = \frac{4 - 1}{8 - 1} = \frac{3}{7}$

Page No 96:

Answer:

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

Page No 96:

Answer:

Page No 96:

Answer:

(d) $\frac{- 64}{27}$

${(\frac{- 3}{4})}^{- 3} = {(\frac{4}{- 3})}^{3}$            $[s ince {(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}]$

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

           = $\frac{4 \times 4 \times 4}{(- 3) \times (- 3) \times (- 3)} = \frac{64}{(- 27)}$

           = $\frac{64 \times - 1}{- 27 \times - 1} = \frac{- 64}{27}$

Page No 96:

Answer:

(b) $\frac{- 3}{5}$

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

Page No 96:

Answer:

(a) 19

$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} = \{{(\frac{3}{1})}^{3} - {(\frac{2}{1})}^{3}\}$     $[s ince {(\frac{a}{b})}^{- 1} = {(\frac{b}{a})}^{1}]$
                        = $\{{(3)}^{3} - {(2)}^{3}\}$
                        = (27 − 8) = 19

Page No 96:

Answer:

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

${(\frac{- 2}{3})}^{10} \div {(\frac{- 2}{3})}^{8}$ = ${(\frac{- 2}{3})}^{10 - 8}$             $[s ince a^{m} \div a^{n} = a^{m - n}]$
                              = ${(\frac{- 2}{3})}^{2}$
                              = $\frac{{(- 2)}^{2}}{3^{2}} = \frac{4}{9}$

Page No 96:

Answer:

(c) 3.263 x 10⁵

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: 3.263 x 10⁵

Page No 96:

Answer:

(i) If 9 × 3ⁿ = 3⁶, then n = 4.
Explanation:
If 9 × 3ⁿ = 3⁶
⇒ 3² × 3ⁿ = 3⁶
⇒ 3^{(2 + n)} = 3⁶
Equating the powers:
⇒ ( 2 + n) = 6
⇒ n = (6 - 2) = 4

(ii) (8)⁰ = 1
Explanation:
By definition, we have a⁰= 1 for every integer a.
∴ (8)⁰ = 1

(iii) ${(\frac{a}{b})}^{- 16}$ = ${(\frac{b}{a})}^{16}$
Explanation:
We know:
${(\frac{a}{b})}^{- 1} = {(\frac{b}{a})}^{1}$

(iv) (−2)⁻⁵ = $\frac{- 1}{32}$
Explanation:
(−2)⁻⁵= ${(\frac{- 2}{1})}^{- 5} = {(\frac{1}{- 2})}^{5}$ $[Since {(\frac{a}{b})}^{- 1} = {(\frac{b}{a})}^{1}]$
= $\frac{{(1)}^{5}}{{(- 2)}^{5}} = \frac{1 \times 1 \times 1 \times 1 \times 1}{(- 2) \times (- 2) \times (- 2) \times (- 2) \times (- 2)} = \frac{1}{- 32}$

Page No 96:

Answer:

(i) True
645 = 6.45 x 10²                  [since the decimal point is moved 2 places to the left]

(ii) False
27000 = 2.7 x 10⁴               [since the decimal point is moved 4 places to the left]

(iii) False
(3⁰ + 4⁰ + 5⁰) = 1               [since a⁰= 1 for every integer a]

(iv) False
Reciprocal of 5⁶= Reciprocal of ${(\frac{5}{1})}^{6} = {(\frac{1}{5})}^{6}$

(v) False
5⁻¹ × x = 8⁻¹
⇒ $\frac{1}{5} \times x = \frac{1}{8}$
⇒ x = $(\frac{1}{8} \times 5) = \frac{5}{8}$

View NCERT Solutions for all chapters of Class 7