## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

**NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1**

Ex 13.1 Class 7 Maths Question 1.

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

Ex 13.1 Class 7 Maths Question 2.

Exress 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 = 63

(ii) t × t = t2

(iii) b × b × b × b = b4

(iv) 5 × 5× 7 × 7 × 7 = 52 × 73 = 52 · 73

(v) 2 × 2 × a × a = 22 × a2 = 22 · a2

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

Ex 13.1 Class 7 Maths Question 3.

Express each of the following numbers using exponential notation:

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution:

Ex 13.1 Class 7 Maths Question 4.

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 or 34

43 = 4 × 4 × 4 = 64,

34 = 3 × 3 × 3 × 3 = 81

Since 81 > 64

∴ 34 is greater than 43.

(ii) 53 or 35

53 = 5 × 5 × 5 = 125

35 = 3 × 3 × 3 × 3 × 3 = 243

Since 243 > 125

∴ 35 is greater than 53.

(iii) 28 or 82

28 =2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

82 = 8 × 8 = 64

Since 256 > 64

∴ 28 is greater than 28.

(iv) 1002 or 2100

1002 = 100 × 100 = 10000

2100 = 2 × 2 × 2 × … 100 times

Here 2 × 2 × 2 ×2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 × 2 = 214 = 16384

Since 16384 > 10,000

∴ 2100 is greater than 1002.

(v) 210 or 102

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

102 = 10 × 10 = 100

Since 1024 > 100

∴ 210 is greater than 102.

Ex 13.1 Class 7 Maths Question 5.

Express each of the following as the product of powers of their prime

(i) 648

(ii) 405

(iii) 540

(iv) 3600

Solution:

Ex 13.1 Class 7 Maths Question 6.

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 = = 2000

(ii) 72 × 22 = = 7 × 7 × 2 × 2 = 196

(iii) 23 × 5 = 2 × 2 × 2 × 5 = 40

(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 768

(v) 0 × 102 = 0 × 10 × 10 = = 0

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

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

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

Ex 13.1 Class 7 Maths Question 7.

Simplify:

(i) (-4)3

(ii) (-3) × (-2)3

(iii) (-3)2 × (-5)2

(iv) (-2)3 × (-10)3

Solution:

(i) (-4)2 = (-4) × (-4) × (-4) = -64 [∵ (-a)odd number = -aodd number]

(ii) (-3) × (-2)3 = (-3) × (-2) × (-2) × (-2)

= (-3) × (-8) = 24

(iii) (-3)2 × (-5)2 = [(-3) × (-5)]2

= 152 = 225 [∵ am × bm = (ab)m)

(iv) (-2)3 × (-10)3 = [(-2) × (-10)]3

= 202 = 8000 [∵ am × bm = (ab)m]

Ex 13.1 Class 7 Maths Question 8.

Compare the following:

(i) 2.7 × 1012; 1.5 × 108

(ii) 4 × 1014; 3 × 1014

Solution:

(i) 2.7 × 1012; 1.5 × 108

Here, 1012 > 108

∴ 2.7 × 1012> 1.5 × 108

(ii) 4 × 1014; 3 × 1017

Here, 1017 > 1014

∴ 4 × 1014 < 3 × 1017

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