#### Page No 197:

#### Question 1:

Evaluate

(i) 3^{−2} (ii) (−4)^{−2} (iii)

#### Answer:

(i)

(ii)

(iii)

#### Question 2:

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

(i) (ii)

(iii) (iv)

(v)

#### Answer:

(i) (−4)^{5} ÷ (−4)^{8} = (−4)^{5 − 8} (*a*^{m} ÷ *a*^{n} = *a*^{m}^{ − }^{n})

= (− 4)^{−3}

(ii)

(iii)

(iv) (3^{−
7} ÷ 3^{−10}) × 3^{−5} = (3^{−7 − (−10)}) × 3^{−5} (*a*^{m} ÷ *a*^{n} = *a*^{m }^{− }^{n})

= 3^{3} × 3^{−5}

= 3^{3 + (− 5)} (*a*^{m} × *a*^{n} = *a*^{m}^{ + }^{n})

= 3^{−2}

(v) 2^{−3} × (−7)^{−3} =

#### Question 3:

Find the value of.

(i) (3^{0} + 4^{−1}) × 2^{2} (ii) (2^{−1} × 4^{−1}) ÷2^{−2}

(iii) (iv) (3^{−1} + 4^{−1} + 5^{−1})^{0}

(v)

#### Answer:

(i)

(ii) (2^{−1} × 4^{−1}) ÷ 2^{− 2 }= [2^{−1} × {(2)^{2}}^{− 1}] ÷ 2^{−
2}

= (2^{− 1} × 2^{− 2}) ÷ 2^{−
2}

= 2^{−1+ (−2)} ÷ 2^{−2} (*a*^{m} × *a*^{n} = *a*^{m}^{ + }^{n})

= 2^{−3} ÷ 2^{−2}

= 2^{−3} ^{− (−2)} (*a*^{m} ÷ *a*^{n} = *a*^{m}^{ − }^{n})

= 2^{−3 + 2} = 2^{ −1}

(iii)

(iv) (3^{−1} + 4^{−1} + 5^{−1})^{0}

= 1 (*a*^{0} = 1)

(v)

#### Page No 198:

#### Question 4:

Evaluate (i) (ii)

#### Answer:

(i)

(ii)

#### Question 5:

Find
the value of *m* for which 5^{m} ÷5^{−3} = 5^{5}.

#### Answer:

5^{m} ÷ 5^{−3} = 5^{5}

5^{m} ^{− (− 3)} = 5^{5} (*a*^{m} ÷ *a*^{n} = *a*^{m}^{ − }^{n})

5^{m}^{ + 3} = 5^{5}

Since the powers have same bases on both sides, their respective exponents must be equal.

*m* +
3 = 5

*m* =
5 − 3

*m* =
2

#### Question 6:

Evaluate (i) (ii)

#### Answer:

(i)

(ii)

#### Question 7:

Simplify. (i) (ii)

#### Answer:

(i)

(ii)

#### Page No 200:

#### Question 1:

Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

#### Answer:

(i) 0.0000000000085
= 8.5 × 10^{−12}

(ii) 0.00000000000942
= 9.42 × 10^{−12}

(iii) 6020000000000000
= 6.02 × 10^{15}

(iv) 0.00000000837
= 8.37 × 10^{−9}

(v) 31860000000
= 3.186 × 10^{10}

#### Question 2:

Express the following numbers in usual form.

(i) 3.02 ×
10^{−6} (ii) 4.5 ×
10^{4}

(iii) 3 ×
10^{−8} (iv) 1.0001 ×
10^{9}

(v) 5.8 ×
10^{12 }(vi) 3.61492 ×
10^{6}

#### Answer:

(i) 3.02 ×
10^{−6} = 0.00000302

(ii) 4.5 ×
10^{4} = 45000

(iii) 3 ×
10^{−8} = 0.00000003

(iv) 1.0001
× 10^{9} = 1000100000

(v) 5.8 ×
10^{12} = 5800000000000

(vi) 3.61492
× 10^{6} = 3614920

#### Question 3:

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

(i) 1 micron is equal to 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

#### Answer:

(i) = 1 × 10^{−6}

(ii) 0.000,
000, 000, 000, 000, 000, 16 = 1.6 × 10^{−19}

(iii) 0.0000005
= 5 × 10^{−7}

(iv) 0.00001275
= 1.275 × 10^{−5}

(v) 0.07 =
7 × 10^{−2}

#### Question 4:

In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

#### Answer:

Thickness of each book = 20 mm

Hence, thickness of 5 books = (5 × 20) mm = 100 mm

Thickness of each paper sheet = 0.016 mm

Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm

Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets

= (100 + 0.080) mm

= 100.08 mm

= 1.0008 × 10^{2} mm

**NCERT Solutions for Class 8 Math Chapters**

- Chapter 1 – Rational Numbers
- Chapter 2 – Linear Equations in One Variable
- Chapter 3 – Understanding Quadrilaterals
- Chapter 4 – Practical Geometry
- Chapter 5 – Data Handling
- Chapter 6 – Squares and Square Roots
- Chapter 7 – Cubes and Cube Roots
- Chapter 8 – Comparing Quantities
- Chapter 9 – Algebraic Expressions and Identities
- Chapter 10 – Visualising Solid Shapes
- Chapter 11 – Mensuration
- Chapter 12 – Exponents and Powers
- Chapter 13 – Direct and Inverse Proportions
- Chapter 14 – Factorisation
- Chapter 15 – Introduction to Graphs
- Chapter 16 – Playing with Numbers

**NCERT Solutions for Class 8:**