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#### Question 1:

Write each of the following in power notation:

(i) $\frac{5}{7}×\frac{5}{7}×\frac{5}{7}×\frac{5}{7}$
(ii) $\left(\frac{-4}{3}\right)×\left(\frac{-4}{3}\right)×\left(\frac{-4}{3}\right)×\left(\frac{-4}{3}\right)×\left(\frac{-4}{3}\right)$
(iii) $\left(\frac{-1}{6}\right)×\left(\frac{-1}{6}\right)×\left(\frac{-1}{6}\right)$
(iv) (−8) × (−8) × (−8) × (−8) × (−8)

(i)

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

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

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

#### Question 2:

Express each of the following in power notation:

(i) $\frac{25}{36}$
(ii) $\frac{-27}{64}$
(iii) $\frac{-32}{243}$
(iv) $\frac{-1}{128}$

(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 3:

Express each of the following as a rational number:

(i) ${\left(\frac{2}{3}\right)}^{5}$
(ii) ${\left(\frac{-8}{5}\right)}^{3}$
(iii) ${\left(\frac{-13}{11}\right)}^{2}$
(iv) ${\left(\frac{1}{6}\right)}^{3}$
(v) ${\left(\frac{-1}{2}\right)}^{5}$
(vi) ${\left(\frac{-3}{2}\right)}^{4}$
(vii) ${\left(\frac{-4}{7}\right)}^{3}$
(viii) (−1)9

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

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

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

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

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

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

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

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

#### Question 4:

Express each of the following as a rational number:

(i) (4)−1
(ii) (−6)−1
(iii) ${\left(\frac{1}{3}\right)}^{-1}$
(iv) ${\left(\frac{-2}{3}\right)}^{-1}$

(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 5:

Find the reciprocal of each of the following:

(i) ${\left(\frac{3}{8}\right)}^{4}$
(ii) ${\left(\frac{-5}{6}\right)}^{11}$
(iii) 67
(iv) (−4)3

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 6:

Find the value of each of the following:

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

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

Note: a0 = 1

#### Question 7:

Simplify each of the following and express each as a rational number:

(i) ${\left(\frac{3}{2}\right)}^{4}×{\left(\frac{1}{5}\right)}^{2}$
(ii) ${\left(\frac{-2}{3}\right)}^{5}×{\left(\frac{-3}{7}\right)}^{3}$
(iii) ${\left(\frac{-1}{2}\right)}^{5}×{2}^{3}×{\left(\frac{3}{5}\right)}^{2}$
(iv) ${\left(\frac{2}{3}\right)}^{2}×{\left(\frac{-3}{5}\right)}^{3}×{\left(\frac{7}{2}\right)}^{2}$
(v) $\left\{{\left(\frac{-3}{4}\right)}^{3}-{\left(\frac{-5}{2}\right)}^{3}\right\}×{4}^{2}$

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

(ii) ${\left(\frac{-2}{3}\right)}^{5}×{\left(\frac{-3}{7}\right)}^{3}=\frac{{\left(-2\right)}^{5}}{{\left(3\right)}^{5}}×\frac{{\left(-3\right)}^{3}}{{\left(7\right)}^{3}}$
= $=\frac{{\left(-2\right)}^{5}}{{\left(7\right)}^{3}}×\frac{\left(-1\right){\left(3\right)}^{3}}{{\left(3\right)}^{5}}\phantom{\rule{0ex}{0ex}}=\frac{-32×-1×{3}^{3-5}}{343}\phantom{\rule{0ex}{0ex}}=\frac{-32×-1×{3}^{-2}}{343}\phantom{\rule{0ex}{0ex}}=\frac{-32×-1×1}{343×9}\phantom{\rule{0ex}{0ex}}=\frac{32}{3087}$

(iii) ${\left(\frac{-1}{2}\right)}^{5}×{2}^{3}×{\left(\frac{3}{4}\right)}^{2}=\frac{{\left(-1\right)}^{5}}{{2}^{5}}×{2}^{3}×\frac{{3}^{2}}{{4}^{2}}$
$=\frac{{\left(-1\right)}^{5}}{{2}^{5}}×{2}^{3}×\frac{{3}^{2}}{\left({2}^{2}{\right)}^{2}}$
$=\frac{-1×{2}^{3}×{3}^{2}}{{2}^{5}×{2}^{4}}$
$=\frac{-1×{2}^{3}×{3}^{2}}{{2}^{9}}$ $=-1×{2}^{3-9}×{3}^{2}$  = $-9×{2}^{-6}=\frac{-9}{{2}^{6}}=\frac{-9}{64}$

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

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

#### Question 8:

Simplify and express each as a rational number:

(i) ${\left(\frac{4}{9}\right)}^{6}×{\left(\frac{4}{9}\right)}^{-4}$
(ii) ${\left(\frac{-7}{8}\right)}^{-3}×{\left(\frac{-7}{8}\right)}^{2}$
(iii) ${\left(\frac{4}{3}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-2}$

(i) ${\left(\frac{4}{9}\right)}^{6}×{\left(\frac{4}{9}\right)}^{-4}={\left(\frac{4}{9}\right)}^{6+\left(-4\right)}$
=  ${\left(\frac{4}{9}\right)}^{2}$$\frac{{\left(4\right)}^{2}}{{\left(9\right)}^{2}}=\frac{4×4}{9×9}=\frac{16}{81}$

(ii) ${\left(\frac{-7}{8}\right)}^{-3}×{\left(\frac{-7}{8}\right)}^{2}={\left(\frac{-7}{8}\right)}^{\left(-3\right)+2}$
= ${\left(\frac{-7}{8}\right)}^{-1}$
=  ${\left(\frac{8}{-7}\right)}^{1}$
= $\left(\frac{8×-1}{-7×-1}\right)=\frac{-8}{7}$

(iii) ${\left(\frac{4}{3}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-2}={\left(\frac{4}{3}\right)}^{\left(-3\right)+\left(-2\right)}$
= ${\left(\frac{4}{3}\right)}^{-5}$
= ${\left(\frac{3}{4}\right)}^{5}$
= $\frac{{\left(3\right)}^{5}}{{\left(4\right)}^{5}}=\frac{3×3×3×3×3}{4×4×4×4×4}=\frac{243}{1024}$

#### Question 9:

Express each of the following as a rational number:

(i) 5−3
(ii) (−2)−5
(iii) ${\left(\frac{1}{4}\right)}^{-4}$
(iv) ${\left(\frac{-3}{4}\right)}^{-3}$
(v) ${\left(-3\right)}^{-1}×{\left(\frac{1}{3}\right)}^{-1}$
(vi) ${\left(\frac{5}{7}\right)}^{-1}×{\left(\frac{7}{4}\right)}^{-1}$
(vii) (5−1−7−1)−1
(viii) ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$
(ix) $\left\{{\left(\frac{3}{2}\right)}^{-1}÷{\left(\frac{-2}{5}\right)}^{-1}\right\}$
(x) ${\left(\frac{23}{25}\right)}^{0}$

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

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

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

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

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

(ix) $\left\{{\left(\frac{3}{2}\right)}^{-1}÷{\left(\frac{-2}{5}\right)}^{-1}\right\}=\left\{{\left(\frac{2}{3}\right)}^{1}÷{\left(\frac{5}{-2}\right)}^{1}\right\}$
$=\left(\frac{2}{3}×\frac{-2}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{-4}{15}$

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

#### Question 10:

Simplify:

(i) ${\left[{\left\{{\left(\frac{-1}{4}\right)}^{2}\right\}}^{-2}\right]}^{-1}$
(ii) ${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{3}$
(iii) ${\left(\frac{-3}{2}\right)}^{3}÷{\left(\frac{-3}{2}\right)}^{6}$
(iv) ${\left(\frac{-2}{3}\right)}^{7}÷{\left(\frac{-2}{3}\right)}^{4}$

(i)

${\left[{\left\{{\left(-\frac{1}{4}\right)}^{2}\right\}}^{-2}\right]}^{-1}={\left[{\left(-\frac{1}{4}\right)}^{2×-2}\right]}^{-1}$
$={\left[{\left(-\frac{1}{4}\right)}^{-4}\right]}^{-1}$
$={\left(-\frac{1}{4}\right)}^{\left(-4\right)×\left(-1\right)}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{4}\right)}^{4}=\frac{{\left(-1\right)}^{4}}{{\left(4\right)}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{1}{256}$

(ii)

${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{3}={\left(\frac{-2}{3}\right)}^{2×3}$
= ${\left(\frac{-2}{3}\right)}^{6}$
= $\frac{{\left(-2\right)}^{6}}{{\left(3\right)}^{6}}=\frac{64}{729}$                      [since (−2)6 = 64 and (3)6 = 729]

(iii)

${\left(\frac{-3}{2}\right)}^{3}÷{\left(\frac{-3}{2}\right)}^{6}={\left(\frac{-3}{2}\right)}^{3-6}$
= ${\left(\frac{-3}{2}\right)}^{-3}$
$={\left(\frac{2}{-3}\right)}^{3}$
$={\left(\frac{2×-1}{-3×-1}\right)}^{3}={\left(\frac{-2}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{\left(-2\right)}^{3}}{{\left(3\right)}^{3}}=\frac{-8}{27}$

(iv)

${\left(\frac{-2}{3}\right)}^{7}÷{\left(\frac{-2}{3}\right)}^{4}={\left(\frac{-2}{3}\right)}^{7-4}$
$={\left(\frac{-2}{3}\right)}^{3}$
= $\frac{{\left(-2\right)}^{3}}{{\left(3\right)}^{3}}=\frac{-8}{27}$

#### Question 11:

By what number should (−5)−1 be multiplied so that the product is (8)−1?

Let the required number be x.
(−5)-1 $×$ x = (8)-1

x = $\frac{1}{8}×\left(-5\right)$ = $\frac{-5}{8}$
Hence, the required number is $\frac{-5}{8}$.

#### Question 12:

By what number should 3−3 be multiplied to obtain 4?

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

#### Question 13:

By what number should (−30)−1 be divided to get 6−1?

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

#### Question 14:

Find x such that ${\left(\frac{3}{5}\right)}^{3}×{\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$.

${\left(\frac{3}{5}\right)}^{3}×{\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$
${\left(\frac{3}{5}\right)}^{3+\left(-6\right)}={\left(\frac{3}{5}\right)}^{2x-1}$
${\left(\frac{3}{5}\right)}^{-3}={\left(\frac{3}{5}\right)}^{2x-1}$
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x   = −2
x = $\left(\frac{-2}{2}\right)=-1$

#### Question 15:

Simplify: $\frac{{3}^{5}×{10}^{5}×25}{{5}^{7}×{6}^{5}}$.

$\frac{{3}^{5}×{10}^{5}×25}{{5}^{7}×{6}^{5}}=\frac{{3}^{5}×{\left(2×5\right)}^{5}×{5}^{2}}{{5}^{7}×{\left(2×3\right)}^{5}}$
$=\frac{{3}^{5}×{2}^{5}×{5}^{5}×{5}^{2}}{{5}^{7}×{2}^{5}×{3}^{5}}$
$=\frac{{3}^{5}×{2}^{5}×{5}^{7}}{{3}^{5}×{2}^{5}×{5}^{7}}\phantom{\rule{0ex}{0ex}}={3}^{5-5}×{2}^{5-5}×{5}^{7-7}\phantom{\rule{0ex}{0ex}}={3}^{0}×{2}^{0}×{5}^{0}\phantom{\rule{0ex}{0ex}}=1×1×1=1$

#### Question 16:

Simplify: $\frac{16×{2}^{\mathrm{n}+1}-4×{2}^{\mathrm{n}}}{16×{2}^{\mathrm{n}+2}-2×{2}^{\mathrm{n}+2}}$.

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

#### Question 17:

Find the value of n when:

(i) 52n × 53 = 59
(ii) 8 × 2n+2 = 32
(iii) 62n+1 ÷ 36 = 63

(i) 52n × 53 = 59
52n+3 = 59          [since an × am = am+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 $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ ]
⇒ 62n-1 = 63
On equating the coefficients:
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
∴ n = $\frac{4}{2}=2$

#### Question 18:

If 2n−7 × 5n−4 = 1250, find the value on n.

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

#### Question 1:

Express each of the following numbers in standard form:

(i) 538
(ii) 6428000
(iii) 82934000000
(iv) 940000000000
(v) 23000000

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

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

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

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

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

#### Question 2:

Express each of the following numbers in standard form:

(i) Diameter of Earth = 12756000 m.
(ii) Distance between Earth and Moon = 384000000 m.
(iii) Population of India in March 2001 = 1027000000.
(iv) Number of stars in a galaxy = 100000000000.
(v) The present age of universe = 12000000000 years

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

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

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

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

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

#### Question 3:

Write the following numbers in expanded form:

(i) 684502
(ii) 4007185
(iii) 5807294
(iv) 50074

(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 4:

Write the numeral whose expanded form is given below:

(i) 6 × 104 + 3 × 103 + 0 × 102 + 7 × 101 + 8 × 100
(ii) 9 × 106 + 7 × 105 + 0 × 104 + 3 × 103 + 4 × 102 + 6 × 101 + 2 × 100
(iii) 8 × 105 + 6 × 104 + 4 × 103 + 2 × 102 + 9 × 101 + 6 × 100

(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

#### Question 1:

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

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

(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]
= ${\left(\frac{1}{24}\right)}^{-1}$
= ${\left(\frac{24}{1}\right)}^{1}=24$

#### Question 2:

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

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

(c) 15

We have:

${\left({5}^{-1}×{3}^{-1}\right)}^{-1}={\left(\frac{1}{5}×\frac{1}{3}\right)}^{-1}$
= ${\left(\frac{1}{15}\right)}^{-1}$
= ${\left(\frac{15}{1}\right)}^{1}=15$

#### Question 3:

Mark (✓) against the correct answer

(2−1 − 4−1)2 = ?

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

(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]
= ${\left(\frac{1}{4}\right)}^{2}$
= $\left(\frac{1}{4}×\frac{1}{4}\right)=\frac{1}{16}$

#### 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

(b) 29

We have:

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

#### 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

(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) $\frac{-1}{64}$

(b) 64
We have:
${\left(\frac{-1}{2}\right)}^{-6}={\left(\frac{2}{-1}\right)}^{6}$

#### 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) $\frac{-8}{3}$

(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]
= ${\left(\frac{-8}{3}\right)}^{-1}$
= ${\left(\frac{3}{-8}\right)}^{1}$
= $\left(\frac{3×-1}{-8×-1}\right)=\frac{-3}{8}$

#### 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

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

${\left[{\left\{{\left(-\frac{1}{2}\right)}^{2}\right\}}^{-2}\right]}^{-1}={\left[{\left(-\frac{1}{2}\right)}^{2×-2}\right]}^{-1}$                           $\left[s\mathrm{ince}{\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$
$={\left[{\left(-\frac{1}{2}\right)}^{-4}\right]}^{-1}$
$={\left(-\frac{1}{2}\right)}^{\left(-4\right)×\left(-1\right)}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{2}\right)}^{4}=\frac{{\left(-1\right)}^{4}}{{\left(2\right)}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}$

#### 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

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

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

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

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

#### 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) ${\left(\frac{1}{3}\right)}^{24}$

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

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

#### 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

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

#### Question 13:

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

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

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

$\left({3}^{2}-{2}^{2}\right)×{\left(\frac{2}{3}\right)}^{-3}=\left(9-4\right)×{\left(\frac{3}{2}\right)}^{3}$
$=5×\frac{{3}^{3}}{{2}^{3}}$$5×\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

(a) $\frac{19}{64}$
We have:
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$ =
= $\left\{\left({3}^{3}\right)-{\left(2\right)}^{3}\right\}÷{\left(4\right)}^{3}$
=
=
= $19×\frac{1}{64}=\frac{19}{64}$

#### 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) ${\left(\frac{1}{5}\right)}^{5}$

(c) (-5)5

We have:
${\left(\frac{-1}{5}\right)}^{3}÷{\left(\frac{-1}{5}\right)}^{8}={\left(\frac{-1}{5}\right)}^{3-8}$
= ${\left(\frac{-1}{5}\right)}^{-5}$
$={\left(\frac{5}{-1}\right)}^{5}$
$={\left(\frac{5×-1}{-1×-1}\right)}^{5}={\left(\frac{-5}{1}\right)}^{5}={\left(-5\right)}^{5}$

#### 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) $\frac{25}{4}$

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

${\left(\frac{-2}{5}\right)}^{7}÷{\left(\frac{-2}{5}\right)}^{5}={\left(\frac{-2}{5}\right)}^{7-5}$
$={\left(\frac{-2}{5}\right)}^{2}$
= $\frac{{\left(-2\right)}^{2}}{{\left(5\right)}^{2}}=\frac{4}{25}$

#### 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) $\frac{-4}{9}$

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

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

#### 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

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

#### Question 19:

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

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

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

${\left(\frac{5}{3}\right)}^{-5}×{\left(\frac{5}{3}\right)}^{11}={\left(\frac{5}{3}\right)}^{8x}$
${\left(\frac{5}{3}\right)}^{-5+11}={\left(\frac{5}{3}\right)}^{8x}$        [ since ${a}^{m}×{a}^{n}={a}^{m+n}$]
${\left(\frac{5}{3}\right)}^{6}={\left(\frac{5}{3}\right)}^{8x}$
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

(c) $\frac{-4}{5}$
Let the required number be x.
(−8)-1 x x = (10)-1
⇒ $\frac{1}{-8}×x=\frac{1}{10}$
x = $\frac{1}{10}×\left(-8\right)$ = $\frac{-4}{5}$
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

(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

#### Question 1:

Write the reciprocal of:

(i) ${\left(\frac{2}{3}\right)}^{4}$
(ii) ${\left(\frac{-3}{5}\right)}^{61}$
(iii) 25
(iv) (−5)6

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

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

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

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

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

#### Question 2:

By what number should we multiply (−6)−1 to obtain a product equal to 9−1?

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

#### Question 3:

By what number should (−20)−1 be divided to obtain (−10)−1?

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

#### Question 4:

(i) Express 2000000 in standard form.
(ii) Express 6.4 × 105 in usual form.

(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

#### Question 5:

Simplify: $\frac{16×{2}^{n+1}-8×{2}^{n}}{16×{2}^{n+2}-4×{2}^{n+1}}$

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

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

$\frac{{2}^{3}×\left({2}^{n+2}-{2}^{n}\right)}{{2}^{3}×\left({2}^{n+3}-{2}^{n}\right)}$

$\frac{{2}^{n}×{2}^{2}-{2}^{n}}{{2}^{n}×{2}^{3}-{2}^{n}}$

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

#### Question 6:

If 2n-7 × 5n-4 = 1250, find the value of n.

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

#### Question 7:

Mark (✓) against the correct answer
${\left(\frac{3}{4}\right)}^{0}=?$

(a) 0
(b) $\frac{4}{3}$
(c) 1
(d) none of these

(c)  1
We know:
(a)0 = 1
${\left(\frac{3}{4}\right)}^{0}=1$

#### Question 8:

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

(a) $\frac{27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{-27}{64}$
(d) $\frac{-64}{27}$

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

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

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

= $\frac{4×4×4}{\left(-3\right)×\left(-3\right)×\left(-3\right)}=\frac{64}{\left(-27\right)}$

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

#### Question 9:

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

(a) $\frac{3}{5}$
(b) $\frac{-3}{5}$
(c) $\frac{5}{3}$
(d) $\frac{-5}{3}$

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

${\left(\frac{-5}{3}\right)}^{-1}={\left(\frac{3}{-5}\right)}^{1}$
= $\left(\frac{3}{-5}×\frac{-1}{-1}\right)=\frac{-3}{5}$

#### Question 10:

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

(a) 19
(b) $\frac{1}{19}$
(c) −19
(d) $\frac{-1}{19}$

(a) 19

= $\left\{{\left(3\right)}^{3}-{\left(2\right)}^{3}\right\}$
= (27 − 8) = 19

#### Question 11:

Mark (✓) against the correct answer
$\left({\frac{-2}{3}}^{10}\right)÷{\left(\frac{-2}{3}\right)}^{8}=?$

(a) $\frac{4}{9}$
(b) $\frac{-4}{9}$
(c) ${\left(\frac{-2}{3}\right)}^{18}$
(d) none of these

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

${\left(\frac{-2}{3}\right)}^{10}÷{\left(\frac{-2}{3}\right)}^{8}$ = ${\left(\frac{-2}{3}\right)}^{10-8}$
= ${\left(\frac{-2}{3}\right)}^{2}$
= $\frac{{\left(-2\right)}^{2}}{{3}^{2}}=\frac{4}{9}$

#### Question 12:

Which of the following numbers is in standard form?

(a) 32.63 × 104
(b) 326.3 × 103
(c) 3.263 × 105
(d) none of these

(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

#### Question 13:

Fill in the blanks.

(i) If 9 × 3n = 36, then n = ...... .
(ii) (8)0 = ?
(iii) ${\left(\frac{a}{b}\right)}^{-16}=......$
(iv) (−2)−5 = ......

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

(iv) (−2)−5 = $\frac{-1}{32}$
Explanation:
(−2)−5 = ${\left(\frac{-2}{1}\right)}^{-5}={\left(\frac{1}{-2}\right)}^{5}$
= $\frac{{\left(1\right)}^{5}}{{\left(-2\right)}^{5}}=\frac{1×1×1×1×1}{\left(-2\right)×\left(-2\right)×\left(-2\right)×\left(-2\right)×\left(-2\right)}=\frac{1}{-32}$

#### Question 14:

Write 'T' for true and 'F' for false for each of the following.

(i) 645 in standard form is 6.45 × 102.
(ii) 27000 in standard form is 27 × 103.
(iii) (30 + 40 + 50) = 12.
(iv) Reciprocal of 56 is 65.
(v) If 5−1 × x = 8−1, then x = $\frac{8}{5}$.

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

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

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