NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers | class 8th maths

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NCERT Solutions Class 8 Maths chapter 12 Exponents and Powers
class 8th maths

NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers
class 8th maths
Ex 12.1

Exercise 12.1

1. Evaluate.

(i) 3−2 (ii) (− 4)2 (iii) \(\displaystyle {{\left( {\frac{1}{2}} \right)}^{{-5}}}\)

Solutions :
(i) 32

= \(\displaystyle \frac{1}{{{{3}^{2}}}}\)  (∵ \(\displaystyle {{a}^{{-m}}}=\frac{1}{{{{a}^{m}}}}\) )
= \(\displaystyle \frac{1}{9}\)

(ii) (− 4)2
= \(\displaystyle \frac{1}{{{{{(-4)}}^{{2}}}}}\)  (∵ \(\displaystyle {{a}^{{-m}}}=\frac{1}{{{{a}^{m}}}}\) )
= \(\displaystyle \frac{1}{{16}}\)

(iii) \(\displaystyle {{\left( {\frac{1}{2}} \right)}^{{-5}}}\)
= (2)5 (∵ \(\displaystyle \frac{1}{{{{a}^{{-m}}}}}={{a}^{m}}\) )
= 32

2. Simplify and express the result in power notation with positive exponent.

(i) (− 4)5 ÷ (− 4)8 (ii) \(\displaystyle {{\left( {\frac{1}{{{{2}^{3}}}}} \right)}^{2}}\)
(iii) \(\displaystyle {{(-3)}^{4}}\times {{\left( {\frac{5}{3}} \right)}^{4}}\) (iv) (37 ÷ 3-10) × 35
(v) 23 × (− 7)3  

Solutions :
(i) (− 4)5 ÷ (− 4)8
= \(\displaystyle \left( {\frac{{-{{4}^{5}}}}{{-{{4}^{8}}}}} \right)\)  (∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) )
= \(\displaystyle \frac{1}{{{{{(-4)}}^{3}}}}\)

(ii) \(\displaystyle {{\left( {\frac{1}{{{{2}^{3}}}}} \right)}^{2}}\)
= \(\displaystyle \frac{1}{{{{{(2)}}^{{3\times 2}}}}}\)  (∵ (am)n = amn)
= \(\displaystyle \frac{1}{{{{2}^{6}}}}\)

(iii) \(\displaystyle {{(-3)}^{4}}\times {{\left( {\frac{5}{3}} \right)}^{4}}\)
= \(\displaystyle {{\left\{ {\frac{{(-3)\times 5}}{3}} \right\}}^{4}}\)  {∵ am × bm = (ab)m}
= 54

(iv) (37 ÷ 310) × 35
= \(\displaystyle \frac{{{{3}^{{-7}}}}}{{{{3}^{{-10}}}}}\times {{3}^{{-5}}}\)
= \(\displaystyle \frac{{{{3}^{{-7+(-5)}}}}}{{{{3}^{{-10}}}}}\) {∵ am × bm = (ab)m}
= \(\displaystyle \frac{{{{3}^{{-12}}}}}{{{{3}^{{-10}}}}}\)
= \(\displaystyle \frac{1}{{{{3}^{{-10-(-12)}}}}}\)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) )
= \(\displaystyle \frac{1}{{{{{(3)}}^{2}}}}\)

(v) 23 × (− 7)3
= {2 × (− 7)}3 {∵ am × bm = (ab)m}
= (− 14)3
= \(\displaystyle \frac{1}{{{{{(-14)}}^{3}}}}\)

3. Find the value of.

(i) (30 + 4-1) × 22 (ii) (21 × 41) ÷ 22
(iii) \(\displaystyle {{\left( {\frac{1}{2}} \right)}^{{-2}}}+{{\left( {\frac{1}{3}} \right)}^{{-2}}}+{{\left( {\frac{1}{4}} \right)}^{{-2}}}\) (iv) (31 + 41 +51)0
(v) \(\displaystyle {{\left\{ {{{{\left( {\frac{{-2}}{3}} \right)}}^{{-2}}}} \right\}}^{2}}\)  

Solutions :
(i) (30 + 41) × 22
= \(\displaystyle \left( {1+\frac{1}{4}} \right)\times {{2}^{2}}\)   {∵ a0 = 1}
= \(\displaystyle \left( {\frac{{4+1}}{4}} \right)\times {{2}^{2}}\)
= \(\displaystyle \frac{5}{{{{2}^{2}}}}\times {{2}^{2}}\)
= 5

(ii) (21 × 41) ÷ 22
= \(\displaystyle \left( {\frac{1}{2}\times \frac{1}{4}} \right)\div \frac{1}{{{{2}^{2}}}}\)  (∵ \(\displaystyle {{a}^{{-m}}}=\frac{1}{{{{a}^{m}}}}\) )
= \(\displaystyle \frac{1}{8}\div \frac{1}{{{{2}^{2}}}}\)
= \(\displaystyle \frac{1}{{{{2}^{3}}}}\div \frac{1}{{{{2}^{2}}}}\)
= \(\displaystyle \frac{1}{{{{2}^{3}}}}\times {{2}^{2}}\)
= \(\displaystyle \frac{1}{{{{2}^{{3-2}}}}}\)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= \(\displaystyle \frac{1}{2}\)

(iii) \(\displaystyle {{\left( {\frac{1}{2}} \right)}^{{-2}}}+{{\left( {\frac{1}{3}} \right)}^{{-2}}}+{{\left( {\frac{1}{4}} \right)}^{{-2}}}\)
= 22 + 32 + 42  {∵ \(\displaystyle {{a}^{{-m}}}=\frac{1}{{{{a}^{m}}}}\) }
= 4 + 9 + 16
= 29

(iv) (31 + 41 +51)0
= 1   {∵ a0 = 1}

(v) \(\displaystyle {{\left\{ {{{{\left( {\frac{{-2}}{3}} \right)}}^{{-2}}}} \right\}}^{2}}\)
= \(\displaystyle {{\left( {\frac{{-2}}{3}} \right)}^{{-2\times 2}}}\)  (∵ (am)n = amn)
= \(\displaystyle {{\left( {\frac{{-2}}{3}} \right)}^{{-4}}}\)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= \(\displaystyle {{\left( {\frac{3}{{-2}}} \right)}^{4}}\)
= \(\displaystyle \frac{{81}}{{16}}\)

4. Evaluate
(i) \(\displaystyle \frac{{{{8}^{{-1}}}\times {{5}^{3}}}}{{{{2}^{{-4}}}}}\)
(ii) (5−1 × 2−1) × 6−1
Solutions :
(i) \(\displaystyle \frac{{{{8}^{{-1}}}\times {{5}^{3}}}}{{{{2}^{{-4}}}}}\)
= \(\displaystyle \frac{{{{{\left( {{{2}^{3}}} \right)}}^{{-1}}}\times {{5}^{3}}}}{{{{2}^{{-4}}}}}\)  (Taking the prime factors of all the numbers)
= \(\displaystyle \frac{{{{2}^{{3\times (-1)}}}\times {{5}^{3}}}}{{{{2}^{{-4}}}}}\)  (∵ (am)n = amn)
= \(\displaystyle \frac{{{{2}^{{-3}}}\times {{5}^{3}}}}{{{{2}^{{-4}}}}}\)
= 2−3−(−4) × 53  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= 2−3+4 × 53
= 2 × 53
= 250

(ii) (5−1 × 2−1) × 6−1
= \(\displaystyle \left( {\frac{1}{5}\times \frac{1}{2}} \right)\times \frac{1}{6}\)  
= \(\displaystyle \frac{1}{{10}}\times \frac{1}{6}\)
= \(\displaystyle \frac{1}{{60}}\)

5. Find the value of m for which 5m ÷ 5−3 = 55
Solutions :
\(\displaystyle \frac{{{{5}^{m}}}}{{{{5}^{{-3}}}}}={{5}^{5}}\)
5m−(−3) = 55  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
5m+3 = 55  (because base is same)
m + 3 = 5
m = 5 − 3
m = 2

6. Evaluate
(i) \(\displaystyle {{\left\{ {{{{\left( {\frac{1}{3}} \right)}}^{{-1}}}-{{{\left( {\frac{1}{4}} \right)}}^{{-1}}}} \right\}}^{{-1}}}\)
(ii) \(\displaystyle {{\left( {\frac{5}{8}} \right)}^{{-7}}}\times {{\left( {\frac{8}{5}} \right)}^{{-4}}}\)
Solutions :
(i) \(\displaystyle {{\left\{ {{{{\left( {\frac{1}{3}} \right)}}^{{-1}}}-{{{\left( {\frac{1}{4}} \right)}}^{{-1}}}} \right\}}^{{-1}}}\)
= (3 − 4)−1  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= (− 1)−1
= − 1

(ii) \(\displaystyle {{\left( {\frac{5}{8}} \right)}^{{-7}}}\times {{\left( {\frac{8}{5}} \right)}^{{-4}}}\)
= \(\displaystyle {{\left( {\frac{8}{5}} \right)}^{7}}\times {{\left( {\frac{8}{5}} \right)}^{{-4}}}\)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= \(\displaystyle {{\left( {\frac{8}{5}} \right)}^{{7+(-4)}}}\)
= \(\displaystyle {{\left( {\frac{8}{5}} \right)}^{{7-4}}}\)
= \(\displaystyle {{\left( {\frac{8}{5}} \right)}^{3}}\)
= \(\displaystyle \frac{{512}}{{125}}\)

7. Simplify.

(i) \(\displaystyle \frac{{25\times {{t}^{{-4}}}}}{{{{5}^{{-3}}}\times 10\times {{t}^{{-8}}}}}\,\,\,(t\ne 0)\) (ii) \(\displaystyle \frac{{{{3}^{{-5}}}\times {{{10}}^{{-5}}}\times 125}}{{{{5}^{{-7}}}\times {{6}^{{-5}}}}}\)

Solutions :
(i) \(\displaystyle \frac{{25\times {{t}^{{-4}}}}}{{{{5}^{{-3}}}\times 10\times {{t}^{{-8}}}}}\,\,\,(t\ne 0)\)
= \(\displaystyle \frac{{{{5}^{2}}\times {{t}^{{-4}}}}}{{{{5}^{{-3}}}\times 2\times 5\times {{t}^{{-8}}}}}\)  (Taking the prime factors of all the numbers)
= \(\displaystyle \frac{{{{5}^{{2-(-3)-1}}}{{t}^{{-4-(-8)}}}}}{2}\)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= \(\displaystyle \frac{{{{5}^{{2+3-1}}}\times {{t}^{{-4+8}}}}}{2}\)
= \(\displaystyle \frac{{{{5}^{4}}\times {{t}^{4}}}}{2}\)
= \(\displaystyle \frac{{625\times {{t}^{4}}}}{2}\)

(ii) \(\displaystyle \frac{{{{3}^{{-5}}}\times {{{10}}^{{-5}}}\times 125}}{{{{5}^{{-7}}}\times {{6}^{{-5}}}}}\) = \(\displaystyle \frac{{{{3}^{{-5}}}\times {{2}^{{-5}}}\times {{5}^{{-5}}}\times {{5}^{3}}}}{{{{5}^{{-7}}}\times {{2}^{{-5}}}\times {{3}^{{-5}}}}}\)  (सभी संख्याओं के अभाज्य गुणनखंड लेने पर)
= 3−5−(−5) × 25−(−5) × 53−5−(−7)  {∵ \(\displaystyle \frac{{{{a}^{m}}}}{{{{a}^{n}}}}={{a}^{{m-n}}}\) }
= 35+5 × 25+5 × 510−5
= 30 × 20 × 55  
= 1 × 1 × 55  {∵ a0 = 1}
= 55

NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers
class 8th maths
Ex 12.2

Excrcise 12.2

1. Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942
(iii) 6020000000000000 (iv) 0.00000000837
(v) 31860000000  

Solutions :
(i) 0.0000000000085
= 8.5 × 10−12

(ii) 0.00000000000942
= 9.42 × 10−12

(iii) 6020000000000000
= 6.02 × 1015

(iv) 0.00000000837
= 8.37 × 10−9

(v) 31860000000
= 3.186 × 1010

2. Express the following numbers in usual form.

(i) 3.02 × 10−6 (ii) 4.5 × 104 (iii) 3 × 10−8
(iv) 1.0001 × 109 (v) 5.8 × 1012 (vi) 3.61492 × 106

Solutions :
(i) 3.02 × 10−6
= 0.00000302

(ii) 4.5 × 104
= 45000

(iii) 3 × 10−8
= 0.00000003

(iv) 1.0001 × 109
= 1000100000

(v) 5.8 × 1012
= 5800000000000

(vi) 3.61492 × 106
= 3614920

3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to \(\displaystyle \frac{1}{{1000000}}\) m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm

Solutions :
(i) 1 micron is equal to \(\displaystyle \frac{1}{{1000000}}\) m.
= 1 × 10−6

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
= 1.6 × 10−19

(iii) Size of a bacteria is 0.0000005 m
= 5 × 10−7

(iv) Size of a plant cell is 0.00001275 m
= 1.275 × 10−5

(v) Thickness of a thick paper is 0.07 mm
= 7 × 10−2

4. In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.
Solutions :
∵ Thickness of a book = 20 mm
∴ Thickness of five books = 20 × 5 = 100 mm

∵ Thickness of a paper sheet = 0.0016 mm
∴ Thickness of five paper sheets = 0.0016 × 5 = 0.0080 mm

The total thickness of the stack = Thickness of five books + Thickness of five paper sheets
= 100 + 0.0080
= 100.0080 mm
= 1.0008 × 102

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