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Class 9 NCERT Solutions – Chapter 1 Number System – Exercise 1.6

Last Updated : 03 Apr, 2024
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Note: This exercise has been renumbered as Exercise 1.5 in the updated NCERT syllabus.

Question 1. Find the Value of:

(i) 641/2

641/2 

= (8 × 8)1/2

= (82)½

= 81 

= 8

(ii) 321/5

321/5

= (2 x 2 x 2 x 2 x 2)1/5

= (25)

= 21 

= 2

(iii) 1251/3

(125)1/3

= (5 × 5 × 5)1/3

= (53)

= 51 

= 5

Question 2. Find the value of:

(i) 93/2

93/2 

= (3 × 3)3/2

= (32)3/2

= 33 

= 27

(ii) 322/5

322/5 

= (2 × 2 × 2 × 2 × 2)2/5

= (25)2⁄5

= 22 

= 4

(iii) 163/4

163/4 

= (2 × 2 × 2 × 2)3/4

= (24)3⁄4

= 23 

= 8

(iv) 125-1/3

125-1/3 

= (5 × 5 × 5)-1/3

= (53)-1⁄3

= 5-1 

= 1/5

Question 3. Simplify the followings:

(i) 22/3 × 21/5

22/3 × 21/5

= 2(2/3) + (1/5)   ⸪ As, am × an = am + n

= 213/15

(ii) (1/33)7

(1/33)7 

= (3-3)7   ⸪ As, (am)n = am x n

= 3-21

(iii) 111/2/111/4

111/2/111/4 

= 11(1/2) – (1/4)    ⸪ As, am × a-n = am – n

= 111/4

(iv) 71/2 × 81/2

71/2 × 81/2 

= (7 × 8)1/2  ⸪ As, am × bm = (a × b)m

= 561/2


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