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#### Question 1:

Find the value of each of the following:
(i) 132
(ii) 73
(iii) 34

We have
(i) 132 = 13 × 13 = 169
(ii) 73 = 7 × 7 × 7 = 343
(iii) 34 = 3 × 3 × 3 × 3 = 81

#### Question 2:

Find the value of each of the following:
(i) (−7)2
(ii) (−3)4
(iii) (−5)5

We know that if 'a' is natural number, then
(−a)even number = Positive number
(−a)odd number  = Negative number

We have
(i) (−7)2 = −7 × −7 = 49
(ii) (−3)4 = −3 × −3 × −3 × −3 = 81
(iii) (−5)5 = −5 × −5 × −5 × −5 × −5 = −3125

#### Question 3:

Simplify:
(i) 3 × 102
(ii) 22 × 53
(ii) 33 × 52

We have
(i) 3 × 102 = 3 × 100 = 300             [since 102 = 10 × 10 = 100]
(ii) 22 × 53 = 4 × 125 = 500            [since 22 = 2 × 2 = 4 and 53 = 5 × 5 × 5 = 125]
(iii) 33 × 52 = 27 × 25 = 675           [ since 33 = 3 × 3 × 3 = 27 and 52 = 5 × 5 = 25]

#### Question 4:

Simplify:
(i) 32 × 104
(ii) 24 × 32
(ii) 52 × 34

We have
(i) 32 × 104 = 9 × 10000 = 90000       [since 32 = 3 × 3 = 9 and 104 = 10 × 10 × 10 × 10 = 10000]
(ii) 24 × 32 = 16 × 9 = 144                  [since 24 = 2 × 2 × 2 × 2 = 16 and 32 = 3 × 3 = 9]
(iii) 52 × 34 = 25 × 81 = 2025              [since 52 = 5 × 5 = 25 and 34 = 3 × 3 × 3 × 3 = 81]

#### Question 5:

Simplify:
(i) (−2) × (−3)3
(ii) (−3)2 × (−5)3
(iii) (−2)5 × (−10)2

We know that if 'a' is natural number, then
(−a)even number = Positive number
(−a)odd number = Negative number

We have

(i) (−2) × (−3)3 = ( −2 )(−27) = 54               [since (−3)3 = −3 ×−3 × − 3 = −27]
(ii) (−3)2 × ( −5)3 = 9 (−125) =  −1125        [ since (−3)2 = −3 ×− 3 = 9 and (−5 )3 = −5 ×−5 × − 5 = −125]
(iii) ( −2)5 × (−10)2 = −32 × 100 = −3200    [ since (−2)5= −2 ×−2 × −2 ×−2 ×−2 = −32 and (−10)2 = −10 ×− 10 = 100]

#### Question 6:

Simplify:
(i) ${\left(\frac{3}{4}\right)}^{2}$
(ii) ${\left(\frac{-2}{3}\right)}^{4}$
(iii) ${\left(\frac{-4}{5}\right)}^{5}$

We have

(i)
(ii)
(iii)

#### Question 7:

Identify the greater number in each of the following:
(i) 25 or 52
(ii) 34 or 43
(iii) 35 or 53

We have
(i) 25 = 2 × 2 × 2 × 2 × 2 = 32  and 52 = 5 × 5 = 25
Therefore, 32 > 25.
Thus, 25 > 52.

(ii) 34 = 3 × 3 × 3 × 3 = 81 and 43= 4 × 4 × 4 = 64
Therefore, 81 > 64.
Thus, 34 > 43.

(iii) 35 = 3 × 3 × 3 × 3 × 3 = 243 and 53 = 5 × 5 × 5 = 125
Therefore, 243 > 125.
Thus, 35 > 53.

#### Question 8:

Express each of the following in exponential form:
(i) (−5) × (−5) × (−5)
(ii) $\frac{-5}{7}×\frac{-5}{7}×\frac{-5}{7}×\frac{-5}{7}$
(iii)

We have
(i) (−5) × (−5) × (−5) = ( −5)3

(ii)

(iii)

#### Question 9:

Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (−2) × (−2) × (−2) × (−2) × a × a × a
(iii) $\left(\frac{-2}{3}\right)×\left(\frac{-2}{3}\right)×x×x×x$

We have
(i)
(ii)
(iii)

#### Question 10:

Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729

We have
(i) Prime factorisation of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29

(ii) Prime factorisation of 625 = 5 x 5 x 5 x 5 = 54

(iii) Prime factorisation of 729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

#### Question 11:

Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392

We have
(i) Prime factorisation of 36 = 2 x 2 x 3 x 3 = 22 x 32

(ii) Prime factorisation of 675 = 3 x 3 x 3 x 5 x 5 = 33 x 52

(iii) Prime factorisation of 392 = 2 x 2 x 2 x 7 x 7 = 23 x 72

#### Question 12:

Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000

We have
(i) Prime factorisation of 450 = 2 x 3 x 3 x 5 x 5 = 2 x 32 x 52

(ii) Prime factorisation of 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7 = 24 x 52 x 7

(iii) Prime factorisation of 24000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 = 26 x 3 x 53

#### Question 13:

Express each of the following as a rational number of the form $\frac{p}{q}$:
(i) ${\left(\frac{3}{7}\right)}^{2}$
(ii) ${\left(\frac{7}{9}\right)}^{3}$
(iii) ${\left(\frac{-2}{3}\right)}^{4}$

We have

(i)
(ii)

(iii)

#### Question 14:

Express each of the following rational numbers in power notation:
(i) $\frac{49}{64}$
(ii) $-\frac{64}{125}$
(iii) $-\frac{1}{216}$

We have
(i)
(ii)
(iii) $-\frac{1}{216}=\frac{-1}{6}×\frac{-1}{6}×\frac{-1}{6}=\frac{\left(-1{\right)}^{3}}{\left(6{\right)}^{3}}={\left(\frac{-1}{6}\right)}^{3}$

#### Question 15:

Find the value of each of the following:
(i) ${\left(\frac{-1}{2}\right)}^{2}×{2}^{3}×{\left(\frac{3}{4}\right)}^{2}$
(ii) ${\left(\frac{-3}{5}\right)}^{4}×{\left(\frac{4}{9}\right)}^{4}×{\left(\frac{-15}{18}\right)}^{2}$

We have

(i)

(ii)

#### Question 16:

If a = 2 and b = 3, then find the values of each of the following:
(i) (a + b)a
(ii) (ab)b
(iii) ${\left(\frac{b}{a}\right)}^{b}$
(iv) ${\left(\frac{a}{b}+\frac{b}{a}\right)}^{a}$

We have a = 2 and b = 3.
Thus,
(i) (a + b)a = (2 + 3)2 = (5)2 = 25

(ii) (ab)b = (2 x 3 )3 = (6)3 = 216

(iii)

(iv)

#### Question 1:

Using laws of exponents, simplify and write the answer in exponential form:
(i) 23 × 24 × 25
(ii) 512 ​÷ 53
(iii) (72)3
(iv) (32)5 ​÷ 34
(v) 37 ​× 27
(vi) (521 ÷ 513) × 57

We have

(i)  23 x 24 x 25 = 2(3 + 4 + 5) = 212             [since am + an + ap = a(m+n+p)]

(ii) 512 ÷ 53 = $\frac{{5}^{12}}{{5}^{3}}$ = 512 - 3 = 59                [ since am ÷ an = am-n ]
(iii) (72)3 = 76                                                       [since (am)n = amn ]

(iv)(32)5 ÷ 34 = 310 ÷ 34                             [since (am)n = amn ]
= 3(10 - 4) = 36                      [since am ÷ an = am-n ]

(v) 37 ​× 27 = (3 x 2)7 = 67                          [since am x bm = (a x b)m ]

(vi) (521 ÷  513) x 57 = 5(21 -13) x 57         [since am ÷ an = am-n ]
= 58 x 57                  [since am x bn =a(m +n)]
= 5(8+7)
= 515

#### Question 2:

Simplify and express each of the following in exponential form:
(i) $\left\{\left({2}^{3}{\right)}^{4}×{2}^{8}\right\}÷{2}^{12}$
(ii) (82 × 84) ÷ 83
(iii) $\left(\frac{{5}^{7}}{{5}^{2}}\right)×{5}^{3}$
(iv)

We have
(i)  {(23)4 x 28} ÷ 212
= {212 x 28} ÷ 212
= 2(12 + 8) ÷ 212

= 220 ÷ 212

= 2 (20 - 12) =  28

(ii) (82 x 84)  ÷ 83
= 8(2 + 4) ÷ 83
= 86 ÷ 83
= 8(6-3) = 83 = (23)3 = 29

(iii) $\left(\frac{{5}^{7}}{{5}^{2}}\right)$ x 53 = 5(7-2) x 53
= 55 x 53
= 5(5 + 3 ) = 58

(iv) = ${5}^{\left(4-4\right)}×{x}^{\left(10-7\right)}×{y}^{\left(5-4\right)}$
= ${5}^{0}×{x}^{3}×y$               [since 50 = 1]
=

#### Question 3:

Simplify and express each of the following in exponential form:
(i) $\left\{\left({3}^{2}{\right)}^{3}×{2}^{6}\right\}×{5}^{6}$
(ii) ${\left(\frac{x}{y}\right)}^{12}×{y}^{24}×\left({2}^{3}{\right)}^{4}$
(iii) ${\left(\frac{5}{2}\right)}^{6}×{\left(\frac{5}{2}\right)}^{2}$
(iv) ${\left(\frac{2}{3}\right)}^{5}×{\left(\frac{3}{5}\right)}^{5}$

We have
(i)   {(32)3 x 26} x 56
= {36 x 26} x 56             [since (am)n = amn]
= 66 x 56                        [since am x bm = (a x b)m ]
= 306

(ii)
${\left(\frac{x}{y}\right)}^{12}×{y}^{24}×\left({2}^{3}{\right)}^{4}$
$=\frac{{x}^{12}}{{y}^{12}}×{y}^{24}×{2}^{12}$
$={x}^{12}×\frac{{y}^{24}}{{y}^{12}}×{2}^{12}$
$={x}^{12}×{y}^{24-12}×{2}^{12}$
$={x}^{12}×{y}^{12}×{2}^{12}$

(iii)
${\left(\frac{5}{2}\right)}^{6}×{\left(\frac{5}{2}\right)}^{2}$

(iv)
${\left(\frac{2}{3}\right)}^{5}×{\left(\frac{3}{5}\right)}^{5}$

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

#### Question 4:

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

We have
9 x 9 x 9 x 9 x 9  = (9)5 =(32)5 = 310

#### Question 5:

Simplify and write each of the following in exponential form:
(i) (25)3 ÷ 53
(ii) (81)5 ÷ (32)5
(iii)
(iv)

We have
(i) (25)3 ÷ 53
= (52)3÷ 53
= 56 ÷ 53
= $\frac{{5}^{6}}{{5}^{3}}={5}^{6-3}={5}^{3}$

(ii) (81)5 ÷ (32)5
= (34)5 ÷ (32)5
= (3)20 ÷ (3)10
= $\frac{{3}^{20}}{{3}^{10}}={3}^{20-10}={3}^{10}$

(iii)

$\frac{{9}^{8}×\left({x}^{2}{\right)}^{5}}{\left(27{\right)}^{4}×\left({x}^{3}{\right)}^{2}}$
$=\frac{\left({3}^{2}{\right)}^{8}×\left({x}^{2}{\right)}^{5}}{\left({3}^{3}{\right)}^{4}×\left({x}^{3}{\right)}^{2}}$
$=\frac{{3}^{16}×\left(x{\right)}^{10}}{{3}^{12}×\left(x{\right)}^{6}}$
$=3$16-12× (x)10-6 = 34× x4= (3x)4

(iv)

$\frac{{3}^{2}×{7}^{8}×{13}^{6}}{{21}^{2}×{91}^{3}}$
$=\frac{{3}^{2}×{7}^{2}×{7}^{6}×{13}^{6}}{{21}^{2}×\left(13×7{\right)}^{3}}$
$=\frac{\left(21{\right)}^{2}×{7}^{6}×{13}^{6}}{{21}^{2}×{13}^{3}×{7}^{3}}$
$=\frac{{7}^{6}×{13}^{6}}{{13}^{3}×{7}^{3}}$
$=\frac{{91}^{6}}{{91}^{3}}={91}^{6-3}={91}^{3}$

#### Question 6:

Simplify:
(i) $\left({3}^{5}{\right)}^{11}×\left({3}^{15}{\right)}^{4}-\left({3}^{5}{\right)}^{18}×\left({3}^{5}{\right)}^{5}$
(ii) $\frac{16×{2}^{n+1}-4×{2}^{n}}{16×{2}^{n+2}-2×{2}^{n+2}}$
(iii) $\frac{10×{5}^{n+1}+25×{5}^{n}}{3×{5}^{n+2}+10×{5}^{n+1}}$
(iv)

We have
(i) (35)11× (315)4- (35)18×(35)5
= 355 x 360 - 390 x 325
= 3(55 + 60) - 3(90 + 25)
= 3115 - 3115
= 0

(ii) $\frac{16×{2}^{n+1}-4×{2}^{n}}{16×{2}^{n+2}-2×{2}^{n+2}}$
=
$=\frac{{2}^{2}×\left({2}^{n+3}-{2}^{n}\right)}{{2}^{2}×\left({2}^{n+4}-{2}^{n+1}\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{{2}^{n}×{2}^{3}-{2}^{n}}{{2}^{n}×{2}^{4}-{2}^{n}×2}\phantom{\rule{0ex}{0ex}}$
$=\frac{{2}^{n}\left({2}^{3}-1\right)}{{2}^{n}\left({2}^{4}-2\right)}=\frac{8-1}{16-2}=\frac{7}{14}=\frac{1}{2}$

(iii)

$\frac{10×{5}^{n+1}+25×{5}^{n}}{3×{5}^{n+2}+10×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}$
$=\frac{10×{5}^{n+1}+\left(5{\right)}^{2}×{5}^{n}}{3×{5}^{n+2}+2×5×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}$
$=\frac{10×{5}^{n+1}+5×{5}^{n+1}}{3×{5}^{n+2}+2×5×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}$
$=\frac{{5}^{n+1}\left(10+5\right)}{3×5×{5}^{n+1}+10×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}$
$=\frac{{5}^{n+1}\left(15\right)}{{5}^{n+1}\left(15+10\right)}=\frac{{5}^{n+1}×15}{{5}^{n+1}×25}=\frac{15}{25}=\frac{3}{5}$

(iv)
$\frac{\left(16{\right)}^{7}×\left(25{\right)}^{5}×\left(81{\right)}^{3}}{\left(15{\right)}^{7}×\left(24{\right)}^{5}×\left(80{\right)}^{3}}$
$=\frac{\left(16{\right)}^{7}×\left({5}^{2}{\right)}^{5}×\left({3}^{4}{\right)}^{3}}{\left(3×5{\right)}^{7}×\left(3×8{\right)}^{5}×{\left(16×5\right)}^{3}}$
$=\frac{\left(16{\right)}^{7}×\left(5{\right)}^{10}×\left(3{\right)}^{12}}{{3}^{7}×{5}^{7}×{3}^{5}×{8}^{5}×{16}^{3}×{5}^{3}}$
$=\frac{\left(16{\right)}^{7}×\left(5{\right)}^{10}×\left(3{\right)}^{12}}{{3}^{7}×{3}^{5}×{5}^{7}×{5}^{3}×{8}^{5}×{16}^{3}}$
$=\frac{\left(16{\right)}^{7}×\left(5{\right)}^{10}×\left(3{\right)}^{12}}{{3}^{12}×{5}^{10}×{8}^{5}×{16}^{3}}$
$=\frac{\left(16{\right)}^{7}}{{8}^{5}×{16}^{3}}$
$=\frac{\left(16{\right)}^{7-3}}{{8}^{5}}=\frac{\left(16{\right)}^{4}}{{8}^{5}}=\frac{\left(2×8{\right)}^{4}}{{8}^{5}}=\frac{{2}^{4}×{8}^{4}}{{8}^{5}}=\frac{{2}^{4}}{8}=\frac{16}{8}=2$

#### Question 7:

Find the values of n in each of the following:
(i) ${5}^{2n}×{5}^{3}={5}^{11}$
(ii) $9×{3}^{n}={3}^{7}$
(iii)
(iv) ${7}^{2n+1}÷49={7}^{3}$
(v) ${\left(\frac{3}{2}\right)}^{4}×{\left(\frac{3}{2}\right)}^{5}={\left(\frac{3}{2}\right)}^{2n+1}$
(vi) ${\left(\frac{2}{3}\right)}^{10}+{\left\{{\left(\frac{3}{2}\right)}^{2}\right\}}^{5}={\left(\frac{2}{5}\right)}^{2n-2}$

We have

(i) 52n x 53 = 511
= 52n+3 = 511
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n =$\frac{8}{2}=4$
(ii) 9 x 3n = 37
= (3)2 x 3n = 37
= (3)2+n = 37
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2  = 5

(iii) 8 x 2n+2 = 32
= (2)3 x 2n+2 = (2)5      [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0

(iv) 72n+1 ÷ 49 = 73
= 72n+1 ÷ 72 = 73  [since 49 = 72]
$=\frac{{7}^{2n+1}}{{7}^{2}}={7}^{3}$

= 72n-1 =73
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = $\frac{4}{2}=2$
(v) ${\left(\frac{3}{2}\right)}^{4}×{\left(\frac{3}{2}\right)}^{5}={\left(\frac{3}{2}\right)}^{2n+1}$
$={\left(\frac{3}{2}\right)}^{\left(4+5\right)}={\left(\frac{3}{2}\right)}^{\left(2n+1\right)}$
$={\left(\frac{3}{2}\right)}^{9}={\left(\frac{3}{2}\right)}^{2n+1}$
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n =$\frac{8}{2}=4$
(vi) ${\left(\frac{2}{3}\right)}^{10}×{\left\{{\left(\frac{3}{2}\right)}^{2}\right\}}^{5}={\left(\frac{2}{5}\right)}^{2n-2}$
$={\left(\frac{2}{3}\right)}^{10}×{\left(\frac{3}{2}\right)}^{10}={\left(\frac{2}{5}\right)}^{2n-2}$
$=\frac{{2}^{10}×{3}^{10}}{{3}^{10}×{2}^{10}}={\left(\frac{2}{5}\right)}^{2n-2}\phantom{\rule{0ex}{0ex}}$
$=1={\left(\frac{2}{5}\right)}^{2n-2}$

On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = $\frac{2}{2}=1$

#### Question 8:

If , find the value of n.

We have

$\frac{{9}^{n}×{3}^{2}×{3}^{n}-\left(27{\right)}^{n}}{\left({3}^{3}{\right)}^{5}×{2}^{3}}=\frac{1}{27}$
$=\frac{\left({3}^{2}{\right)}^{n}×{3}^{2}×{3}^{n}-\left({3}^{3}{\right)}^{n}}{\left(3{\right)}^{15}×{2}^{3}}=\frac{1}{27}$
$=\frac{\left(3{\right)}^{2n+2+n}-\left(3{\right)}^{3n}}{\left(3{\right)}^{15}×{2}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(3{\right)}^{3n+2}-\left(3{\right)}^{3n}}{\left(3{\right)}^{15}×{2}^{3}}=\frac{1}{27}$
$=\frac{\left(3{\right)}^{3n}×\left(3{\right)}^{2}-\left(3{\right)}^{3n}}{\left(3{\right)}^{15}×{2}^{3}}=\frac{1}{27}$
$=\frac{\left(3{\right)}^{3n}\left({3}^{2}-1\right)}{\left(3{\right)}^{15}×{2}^{3}}=\frac{1}{27}$

$=\frac{{3}^{3n}}{{3}^{15}}=\frac{1}{27}$
$={3}^{3n-15}=\frac{1}{{3}^{3}}$
$={3}^{3n-15}={3}^{-3}$
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n =$\frac{12}{3}=4$

#### Question 1:

Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 107
(v) 723 × 109

We have

(i) 3908.78 = 3.90878 x 103                                       [since the decimal point is moved 3 places to the left]

(ii) 5,00,00,000 = 5,00,00,000.00 = 5 x 107               [since the decimal point is moved 7 places to the left]

(iii) 3,18,65,00,000 = 3,18,65,00,000.00
= 3.1865 x 109                            [since the decimal point is moved 9 places to the left]

(iv) 846 × 107 = 8.46 x 102 x 107                               [since the decimal point is moved 2 places to the left]
= 8.46 x 109                                        [since am x an = am+n]

(v) 723 × 109 = 7.23 x 102 x 109                               [since the decimal point is moved 2 places to the left]
= 7.23 x 1011                                        [ since am x an = am+n]

#### Question 2:

Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.21 × 105
(iii) 3.5 × 103

We have
(i) 4.83 × 107 = 483 × 107-2                        [since the decimal point is moved two places to the right]
= 483 × 105 = 4,83,00,000

(ii) 3.21 × 105 = 321 x 105-2                       [since the decimal point is moved two places to the right]
= 321 x 103 = 3,21,000

(iii) 3.5 × 103 = 35 x 103-1                          [since the decimal point is moved one place to the right]
= 35 x 102 = 3,500

#### Question 3:

Express the numbers appearing in the following statements in the standard form:
(i) The distance between the Earth and the Moon is 384,000,000 metres.
(ii) Diameter of the Earth is 1,27,56,000 metres.
(iii) Diameter of the Sun is 1,400,000,000 metres.
(iv) The universe is estimated to be about 12,000,000,000 years old.

We have
(i) The distance between the Earth and the Moon is 3.84 x 108 metres.
[Since the decimal point is moved 8 places to the left.]

(ii) The diameter of the Earth is 1.2756 x 107  metres.
[Since the decimal point is moved 7 places to the left.]

(iii) The diameter of the Sun is 1.4 x 109 metres.
[Since the decimal point is moved 9 places to the left.]

(iv) The universe is estimated to be about 1.2x 1010 years old.
[Since the decimal point is moved 10 places to the left.]

#### Question 1:

Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303

We have
(i) 20068 = 2 x 104 + 0 x 103 + 0 x 102 + 6 x 101 + 8 x 100
(ii) 420719 = 4 x 105 + 2 x 104 + 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100
(iii) 7805192 = 7 x 106 + 8 x 105 + 0 x 104 + 5 x 103 + 1 x 102 + 9 x 101 + 2 x 100
(iv) 5004132 = 5 x 106 + 0 x 105 + 0 x 104  4 x 103 + 1 x 102 + 3 x 101 + 2 x 100
(v) 927303 = 9 x 105 + 2 x 104  + 7 x 103 + 3 x 102 +  0 x 101 + 3 x 100

Note: a0 = 1

#### Question 2:

Find the number from each of the following expanded forms:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
(iii) 9 × 105 + 5 × 102 + 3 × 101
(iv) 3 × 104 + 4 × 102 + 5 × 100

We have

(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 7 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1 = 76045

(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
=
5 x 100000 + 4 x 10000 + 2 x 1000 + 3 x 1 = 542003

(iii) 9 × 105 + 5 × 102 + 3 × 101

= 9 x 100000 + 5 x 100 + 3 x 10 = 900530

(iv)
3 × 104 + 4 × 102 + 5 × 100
= 3 x 10000 + 4 x 100 + 5 x 1 = 30405

#### Question 1:

Mark the correct alternative in the following question:

Hence, the correct alternative is option (b).

#### Question 2:

Mark the correct alternative in the following question:

Since, ${2}^{{3}^{2}}={2}^{9}=512$

Hence, the correct alternative is option (d).

#### Question 3:

Mark the correct alternative in the following question:

Hence, the correct alternative is option (c).

#### Question 4:

Mark the correct alternative in the folowing question:

Since,

Hence, the correct alternative is option (c).

#### Question 5:

Mark the correct alternative in the folowing question:

Since,

Hence, the correct alternative is option is (d).

#### Question 6:

Mark the correct alternative in the folowing question:

Since,

Hence, the correct alternative is option (b).

#### Question 7:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (a).

#### Question 8:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (b).

#### Question 9:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (c).

#### Question 10:

Mark the correct alternative in the following question:

Hence, the correct alternative is option (a).

#### Question 11:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (b).

#### Question 12:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (b).

#### Question 13:

Mark the correct alternative in the following question:

#### Question 14:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (c).

#### Question 15:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (c).

#### Question 16:

Mark the correct alternative in the following question:

Since,

Hence, the correct alternative is option (a).

#### Question 17:

Mark the correct alternative in the following question:

Hence, the correct alternative is option (a).

#### Question 18:

Mark the correct alternative in the following question:

So, 65 should be multiplied.

Hence, the correct alternative is option (b).

#### Question 19:

Mark the correct alternative in the following question:

Since,

Hence, the correct option is (a).

#### Question 20:

Choose the correct alternative in the following question:

Hence, the correct alternative is option (a).

#### Question 21:

Choose the correct alternative in the following question:

The number 4,70,394 in standard form is written as

(a) 4.70394 $×$ 105                   (b) 4.70394 $×$ 104                   (c) 47.0394 $×$ 104                   (d) 4703.94 $×$ 102

Since, 4,70,394 = 4.70394 $×$ 100000 = 4.70394 $×$ 105.

So, the number 4,70,394 in standard form is written as 4.70394 $×$ 105.

Hence, the correct alternative is option (a).

#### Question 22:

Choose the correct alternative in the following question:

The number 2.35 $×$ 104 in the usual form is written as

(a) 2.35 $×$ 103                            (b) 23500                            (c) 2350000                            (d) 235 $×$ 104

Since, 2.35 $×$ 104 =  2.35 $×$ 10000 = 23500

So, the number 2.35 $×$ 104 in the usual form is written as 23500.

Hence, the correct alternative is option (b).

#### Question 23:

Choose the correct alternative in the following question:

Hence, the correct alternative is option (b).

#### Question 24:

Choose the correct alternative in the following question:

Hence, the correct alternative is option (b).

#### Question 25:

Choose the correct alternative in the following question: