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 Ex 13.1

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 Ex 13.1 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 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.

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 = (2a)2
(vi) a × a × a × c × c × c × c × 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 > 43.

(ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Since 243 > 125
35 > 53.

(iii) 28 or 82
28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Since 256 > 64
28 > 82.

(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 > 10000
2100 > 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 > 102.

 

Ex 13.1 Class 7 Maths Question 5.

Express each of the following as the product of powers of their prime factors.
(i) 648
(ii) 405
(iii) 540
(iv) 3,600

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)3 = (-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
= 203 = 8000          [
am × bm = (ab)m]

 

Ex 13.1 Class 7 Maths Question 8.

Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017

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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