Solutions of Ncert (Cbse) Mathematics Textbook Exercise 13.1 | Class 7, Chapter-13, Exponents and Powers

Class VII, NCERT (CBSE) Mathematics

EXPONENTS AND POWERS

Solutions of NCERT Math Exercise 13.1

Q1: Find the value of:

(i) 2⁶ (ii) 9³ (iii) 11² (iv)5⁴

Solution:

(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³ = 9 × 9 × 9 = 729

(iii) 11² = 11 × 11 = 121

(iv) 5⁴ = 5 × 5 × 5 × 5 = 625

Q2: Express the following in exponential form:

(i) 6 × 6 × 6 × 6 (ii) t × t (iii) b × b × b × b (iv) 5 × 5 × 7 ×7 × 7

(v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d

Solution:

(i) 6 × 6 × 6 × 6 = 6⁴

(ii) t × t= t²

(iii) b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d = a³ c⁴ d

Q3: Express the following numbers using exponential notation:

(i) 512 (ii) 343 (iii) 729 (iv) 3125

Solution:

(i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

(ii) 343 = 7 × 7 × 7 = 7³

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Q4: Identify the greater number, wherever possible, in each of the following?

(i) 4³ or 3⁴ (ii) 5³ or 3⁵ (iii) 2⁸ or 8² (iv) 100² or 2¹⁰⁰ (v) 2¹⁰ or 10²

Solution:

 (i) 4³ = 4 × 4 × 4 = 64

3⁴ = 3 × 3 × 3 × 3 = 81

So, 3⁴ > 4³

(ii) 5³ = 5 × 5 × 5 =125

3⁵ = 3 × 3 × 3 × 3 × 3 = 243

So, 3⁵ > 5³

(iii) 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8² = 8 × 8 = 64

Therefore, 2⁸ > 8²

(iv)100² or 2¹⁰⁰

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

2¹⁰⁰ = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024

100² = 100 × 100 = 10000

Therefore, 2¹⁰⁰ > 100²

(v) 2¹⁰ and 10²

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

10² = 10 × 10 = 100

Therefore, 2¹⁰ > 10²

Q5: Express each of the following as product of powers of their prime factors:

(i) 648 (ii) 405 (iii) 540 (iv) 3,600

Solution:

(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³. 3⁴

(ii) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ . 5

(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2². 3³. 5

(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴. 3². 5²

Q6: Simplify:

(i) 2 × 10³ (ii) 7² × 2² (iii) 2³ × 5 (iv) 3 × 4⁴

(v) 0 × 10² _­­­­­ (vi) 5² × 3³ (vii) 2⁴ × 3² (viii) 3² × 10⁴

Solution:

(i) 2 × 10³ = 2 × 10 × 10 × 10 = 2 × 1000 = 2000

(ii) 7² × 2² = 7 × 7 × 2 × 2 = 49 × 4 = 196

(iii) 2³ × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40

(iv) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768

(v) 0 × 10² = 0 × 10 × 10 = 0

(vi) 5² × 3³ = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Q7: Simplify:

(i) (− 4)³ (ii) (− 3) × (− 2)³ (iii) (− 3)² × (− 5)² (iv)(− 2)³ × (−10)³

Solution:

(i) (−4)³ = (−4) × (−4) × (−4) = −64

(ii) (−3) × (−2)³

= (−3) × (−2) × (−2) × (−2) = 24

(iii) (−3)² × (−5)² = (−3) × (−3) × (−5) × (−5)

= 9 × 25 = 225

(iv) (−2)³ × (−10)³

= (−2) × (−2) × (−2) × (−10) × (−10) × (−10)

= (−8) × (−1000)

= 8000

Q8: Compare the following numbers:

(i) 2.7 × 10¹²; 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷

Solution:

(i) 2.7 × 10¹² > 1.5 × 10⁸

(ii) 3 × 10¹⁷ > 4 × 10¹⁴