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Page No 197:

Question 1:

Evaluate

(i) 3−2 (ii) (−4)−2 (iii)

Answer:

(i)

(ii)

(iii)

Page No 197:

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 (am ÷ an = amn)

= (− 4)−3

(ii)

(iii)

(iv) (3− 7 ÷ 3−10) × 3−5 = (3−7 − (−10)) × 3−5 (am ÷ an = am n)

= 33 × 3−5

= 33 + (− 5) (am × an = am + n)

= 3−2

(v) 2−3 × (−7)−3 =

Page No 197:

Question 3:

Find the value of.

(i) (30 + 4−1) × 22 (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 (am × an = am + n)

= 2−3 ÷ 2−2

= 2−3− (−2) (am ÷ an = amn)

= 2−3 + 2 = 2 −1

(iii)

(iv) (3−1 + 4−1 + 5−1)0

= 1 (a0 = 1)

(v)



Page No 198:

Question 4:

Evaluate (i) (ii)

Answer:

(i)

(ii)

Page No 198:

Question 5:

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

Answer:

5m ÷ 5−3 = 55

5m− (− 3) = 55 (am ÷ an = amn)

5m + 3 = 55

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

m + 3 = 5

m = 5 − 3

m = 2

Page No 198:

Question 6:

Evaluate (i) (ii)

Answer:

(i)

(ii)

Page No 198:

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 × 1015

(iv) 0.00000000837 = 8.37 × 10−9

(v) 31860000000 = 3.186 × 1010

Page No 200:

Question 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

Answer:

(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

Page No 200:

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

Page No 200:

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 × 102 mm
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Video Solution for Exponents and Powers (Page: 200 , Q.No.: 4)

NCERT Solution for Class 8 math - Exponents and Powers 200 , Question 4



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