ncert solutions class 8 maths chapter 12

 

ncert solutions class 8 maths chapter 12 deals with Exponents and Powers.

 

ncert solutions for class 8 maths chapter 12 exercise 12.1

class 8 maths chapter 12 exercise 12.1 (1)

1. Evaluate.

(i) (3)⁻²      (ii) (-4)⁻²       (iii) (½)⁻⁵

Solution to class 8 maths chapter 12 exercise 12.1 (1)

(i) (3)⁻²

= (¹/₃)²= ¹/₉     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

(ii) (-4)⁻²

= (- ¼)²= ¹/₁₆     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

(iii) (½)⁻⁵

= (²/₁)⁵= 32     [∵ a⁻ᵐ= (¹/ₐ)ᵐ]

class 8 maths chapter 12 exercise 12.1 (2)

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

(i) (-4)⁵ ÷ (-4)⁸  (ii) [(¹/₂)³]²   (iii) (-3)⁴ ×(⁵/₃)⁴ 

(iv) (3⁻⁷ ÷ 3⁻¹⁰)×3⁻⁵  (v) 2⁻³ × (-7)⁻³

Solution to class 8 maths chapter 12 exercise 12.1 (2)

(i) (-4)⁵ ÷ (-4)⁸ 

$={(–4)}^{5–8}$ $[\because a^m÷a^n=a^{m–n}]$

$={(–4)}^{–3}=\frac{1}{{(–4)}^3}$ $[\because a^{–m}=\frac{1}{a^m}]$

$\left(ii\right) {\left(\frac{1}{2^3}\right)}^2=\frac{1^2}{{\left(2^3\right)}^2}$ $\left[\because {\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}\right]$

$=\frac{1}{2^{3\times 2}}=\frac{1}{2^6}$

${\left(iii\right)(–3)}^4\times {\left(\frac{5}{3}\right)}^4={\left(–3\right)}^4\times \frac{5^4}{3^4}$

$=\left\{{\left(–1\right)}^4\times 3^4\right\}\times \frac{5^4}{3^4} [\because {\left(ab\right)}^m=a^m{b}^m]$

$=3^{4–4}\times 5^4 [\because a^m÷a^n=a^{m–n}]$

$=3^0\times 5^4=5^4 \left[\because a^0=1\right]$

$\left(iv\right) (3^{–7}÷ 3^{–10})\times 3^{–5}$

$=3^{–7–(–10)}\times 3^{–5}$

$=3^{–7+10}\times 3^{–5}=3^3\times 3^{–5}$

$=3^{3+(–5) } [\because a^m\times a^n=a^{m+n}]$

$=3^{–2}=\frac{1}{3^2} \left[\because a^{–m}=\frac{1}{a^m}\right]$

$\left(v\right) 2^{–3}\times {\left(–7\right)}^{–3}=\frac{1}{2^3}\times \frac{1}{{\left(–7\right)}^3} \left[\because a^{–m}=\frac{1}{a^m}\right]$

$=\frac{1}{{\left\{2\times \left(–7\right)\right\}}^3}=\frac{1}{{\left(–14\right)}^3} \left[\because {\left(ab\right)}^m=a^m{b}^m\right]$

class 8 maths chapter 12 exercise 12.1 (3)

3  Find the value of 

(i) (3⁰+4⁻¹) ×2² (ii) (2⁻¹ ×4⁻¹)÷2⁻²

(iii) (½)⁻²+(⅓)⁻²+(¼)⁻² (iv) (3⁻¹+4⁻¹+5⁻¹)⁰

(v){(– ⅔)⁻²}²

Solution to class 8 maths chapter 12 exercise 12.1 (3)

(i) (3⁰+4⁻¹) ×2²=(1+¼) ×2²  (∵a⁻¹=¹/ₐ)

=(⁵/₄) ×4 =5

(ii) (2⁻¹ ×4⁻¹)÷2⁻²

=(½ ×¼) ÷ ¹/₍₂₎²

=⅛ ÷ ¼ =⅛ × 4=½

(iii) (½)⁻²+(⅓)⁻²+(¼)⁻²

=(2⁻¹)⁻² + (3⁻¹)⁻² +(4⁻¹)⁻²

=2⁻¹ˣ⁽⁻²⁾+3⁻¹ˣ⁽⁻²⁾+4⁻¹ˣ⁽⁻²⁾ {∵ (aᵐ)ⁿ =aᵐⁿ}

=2² + 3² + 4²

=4+9+16=29

(iv) (3⁻¹+4⁻¹+5⁻¹)⁰

=(⅓ + ¼ + ⅕)⁰

=[⁽²⁰⁺¹⁵⁺¹²⁾/₆₀]⁰

=(⁴⁷/₆₀)⁰

=1

(v){(– ⅔)⁻²}²

=(– ⅔)⁻⁴   {∵ (aᵐ)ⁿ =aᵐⁿ}

=( –  ³/₂)⁴      {∵a⁻ᵐ =¹/₍ₐ₎ᵐ}

= ⁸¹/₁₆

class 8 maths ch÷apter 12 exercise 12.1 (4)

4  Evaluate :

$\left(i\right) \frac{8^{–1}\times 5^3}{2^{–4}} \left(ii\right) \left(5^{–1}\times 2^{–1}\right)\times 6^{–1}$

Solution to class 8 maths chapter 12 exercise 12.1 (4)

$\left(i\right) \frac{8^{–1}\times 5^3}{2^{–4}}=\frac{{\left(2^3\right)}^{–1}\times 5^3}{2^{–4}} \left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$

$=2^{–3–\left(–4\right)}\times 5^3$

$=2^{–3+4}\times 5^3 \left[\because a^m÷a^n=a^{m–n}\right]$

=2×125=250

$\left(ii\right) \left(5^{–1}\times 2^{–1}\right)\times 6^{–1}$

$=\left(\frac{1}{5}\times \frac{1}{2}\right)\times \frac{1}{6} \left[\because a^{–m}=\frac{1}{a^m}\right]$

$=\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$

class 8 maths ch÷apter 12 exercise 12.1 (5)

5. Find the value of m for which 5ᵐ÷5⁻³=5⁵

Solution to class 8 maths chapter 12 exercise 12.1 (5)

5ᵐ÷5⁻³=5⁵

⇒ 5ᵐ ⁻ ⁽⁻³⁾=5⁵

Comparing exponents both sides, we get 

⇒m+3=5

⇒m=5 –3 

⇒m=2

class 8 maths ch÷apter 12 exercise 12.1 (6)

6  Evaluate:

$\left(i\right) {\left\{{\left(\frac{1}{3}\right)}^{–1}–{\left(\frac{1}{4}\right)}^{–1}\right\}}^{–1} \left(ii\right) {\left(\frac{5}{8}\right)}^{–7}\times {\left(\frac{8}{5}\right)}^{–4}$

Solution to class 8 maths chapter 12 exercise 12.1 (6)

(i)

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

$\left[\because a^{–m}=\frac{1}{a^m}\right]$

=[3-4]=-1

(ii)

${\left(\frac{5}{8}\right)}^{–7}\times {\left(\frac{8}{5}\right)}^{–4}=\frac{5^{–7}}{8^{–7}}\times \frac{8^{–4}}{5^{–4}}$

$\left[\because {\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}\right]$

$=5^{–7–\left(–4\right)}\times 8^{–4–\left(–7\right)} \left[\because a^m÷a^n=a^{m–n}\right]$

$=5^{–7+4}\times 8^{–4+7}$

$=5^{–3}\times 8^3=\frac{8^3}{5^3} \left[\because a^{–m}=\frac{1}{a^m}\right]$

$=\frac{512}{125}$

class 8 maths ch÷apter 12 exercise 12.1 (7)

7  Simplify:

$\left(i\right) \frac{25\times t^{–4}}{5^{–3}\times 10\times t^{–8}} \left(t\ne 0\right) \left(ii\right)\frac{3^{–5}\times {10}^{–5}\times 125}{5^{–7}\times 6^{–5}}$

Solution to class 8 maths chapter 12 exercise 12.1 (7)

(i)

$\frac{25\times t^{–4}}{5^{–3}\times 10\times t^{–8}}=\frac{5^2\times t^{–4}}{5^{–3}\times 5\times 2\times t^{–8}}$

$=\frac{5^{2–\left(–3\right)–1}\times t^{–4–\left(–8\right)}}{2} \left[\because a^m÷a^n=a^{m–n}\right]$

$=\frac{5^{2+3–1}\times t^{–4+8}}{2}$

$=\frac{5^4\times t^4}{2}=\frac{625}{2} t^4$

(ii)

$\frac{3^{–5}\times {10}^{–5}\times 125}{5^{–7}\times 6^{–5}}=\frac{3^{–5}\times {\left(2\times 5\right)}^{–5}\times 5^3}{5^{–7}\times {\left(2\times 3\right)}^{–5}}$

$=\frac{3^{–5}\times 2^{–5}\times 5^{–5}\times 5^3}{5^{–7}\times 2^{–5}\times 3^{–5}} \left[\because {\left(ab\right)}^m=a^m{b}^m\right]$

$=\frac{3^{–5}\times 2^{–5}\times 5^{–5+3}}{5^{–7}\times 2^{–5}\times 3^{–5}}$

$=\frac{3^{–5}\times 2^{–5}\times 5^{–2}}{5^{–7}\times 2^{–5}\times 3^{–5}} \left[\because a^m\times a^n=a^{m+n}\right]$

$=3^{–5–\left(–5\right)}\times 2^{–5–\left(–5\right)}\times 5^{–2–\left(–7\right)}$

$\left[\because a^m÷a^n=a^{m–n}\right]$

$=3^{–5+5}\times 2^{–5+5}\times 5^{–2+7}=3^0\times 2^0\times 5^5$

=1×1×3125 

$\left[\because a^0=1\right]$

=3125

ncert solutions for class 8 maths chapter 12 exercise 12.2

Scientific notation: 

Very large or very small numbers can be expressed in the simplest form by using scientific notation. In this notation any number can be written as the product of a number ‘x’  such that 1≤ x < 10 and a power of 10.

Example: 

 (i) 100.08 mm =1.0008 ×10² mm

(ii) 0.00000000000942=9.42 ×10⁻¹²

Note: 1) when we move the decimal point towards right side then multiply with negative power (power=number of places we moved) of 10.

Example: 0.00000000000942=9.42 ×10⁻¹²

2) when we move the decimal point towards left side then multiply with positive power (power=number of places we moved) of 10.

Example: 100.08 mm =1.0008 ×10² mm

class 8 maths chapter 12 exercise 12.2 (1)

1. Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

Solution to class 8 maths chapter 12 exercise 12.2 (1)

(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

class 8 maths chapter 12 exercise 12.2 (2)

2. Express the following numbers in usual form.

(i) 3.06 ×10⁻⁶  (ii) 4.5 ×10⁴  (iii) 3 ×10⁻⁸

(iv) 1.0001 ×10⁹  (v) 5.8 ×10¹²

(vi) 3.61492 ×10⁶

Solution to class 8 maths chapter 12 exercise 12.2 (2)

(i) 3.06 ×10⁻⁶ = ⁽³·⁰⁶⁾/₍₁₀₀₀₀₀₀₎ =0.00000302

(ii) 4.5 ×10⁴ =4.5 ×10000 =45000

(iii) 3 ×10⁻⁸ =³/₍₁₀₀₀₀₀₀₀₀₎ =0.00000003

(iv) 1.0001 ×10⁹ =1.0001 ×1000000000

=1000100000

(v) 5.8 ×10¹² =5.8 ×1000000000000

=5800000000000

(vi) 3.61492 ×10⁶ =3.61492 ×1000000

=3614920

class 8 maths chapter 12 exercise 12.2 (3)

3. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 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 

Solution to class 8 maths chapter 12 exercise 12.2 (3)

(i)1 micron is equal to ¹/₍₁₀₀₀₀₀₀₎ m.=1 ×10⁻⁶ m 

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb

=1.6 ×10⁻¹⁹ coulomb.

(iii) Size of a bacteria is 0.0000005 m

=5.0 ×10⁻⁷ m 

(iv) Size of a plant cell is 0.00001275 m

= 1.275 ×10⁻⁵ m 

(v) Thickness of a thick paper is 0.07 mm 

= 7.0 ×10⁻² mm

class 8 maths chapter 12 exercise 12.2 (4)

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.

Solution to class 8 maths chapter 12 exercise 12.2 (4)

Thickness of one book = 20 mm

Thickness of 5 books = 20 x 5 = 100 mm

Thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016 x 5 = 0.08 mm

Total thickness of a stack = 100 + 0.08 

=100.08 mm =1.0008 ×10² mm