NCERT Solutions for class 9 Maths Chapter 1- Number Systems Exercise 1.6

NCERT Solutions Class 9 Maths Chapter 1 Number Systems Exercise 1.6 given here provides easy reference while solving the exercise problems. The sixth exercise in Number Systems Exercise 1.6 explains the laws of exponents for Real Numbers. They provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook. These NCERT solutions are always prepared by following NCERT guidelines so that it should cover the whole syllabus accordingly. These are very helpful in scoring well in first and second term examinations.

Download PDF of NCERT Solutions for Class 9 Maths Chapter 1- Number Systems Exercise 1.6

ncert sol nov11 for class 9 maths chapter 1 18
ncert sol nov11 for class 9 maths chapter 1 19

Access other exercise solutions of Class 9 Maths Chapter 1- Number Systems

Exercise 1.1 Solutions 4 Questions ( 2 long, 2 short)
Exercise 1.2 Solutions 4 Questions ( 3 long, 1 short)
Exercise 1.3 Solutions 9 Questions ( 9 long)
Exercise 1.4 Solutions 2 Questions ( 2 long)
Exercise 1.5 Solutions 5 Questions ( 4 long 1 short)

Access Answers of Maths NCERT class 9 Chapter 1 – Number Systems Exercise 1.6

1. Find:

(i)641/2

Solution:

641/2 = (8×8)1/2

= (82)½

= 81 [⸪2×1/2 = 2/2 =1]

= 8

(ii)321/5

Solution:

321/5 = (25)1/5

= (25)

= 21 [⸪5×1/5 = 1]

= 2

(iii)1251/3

Solution:

(125)1/3 = (5×5×5)1/3

= (53)

= 51 (3×1/3 = 3/3 = 1)

= 5

2. Find:

(i) 93/2

Solution:

93/2 = (3×3)3/2

= (32)3/2

= 33 [⸪2×3/2 = 3]

=27

(ii) 322/5

Solution:

322/5 = (2×2×2×2×2)2/5

= (25)2⁄5

= 22 [⸪5×2/5= 2]

= 4

(iii)163/4

Solution:

163/4 = (2×2×2×2)3/4

= (24)3⁄4

= 23 [⸪4×3/4 = 3]

= 8

(iv) 125-1/3

125-1/3 = (5×5×5)-1/3

= (53)-1⁄3

= 5-1 [⸪3×-1/3 = -1]

= 1/5

3. Simplify:

(i) 22/3×21/5

Solution:

22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]

= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/33)7

Solution:

(1/33)7 = (3-3)7 [⸪Since,(am)n = am x n____ Laws of exponents]

= 3-21

(iii) 111/2/111/4

Solution:

111/2/111/4 = 11(1/2)-(1/4)

= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 71/2×81/2

Solution:

71/2×81/2 = (7×8)1/2 [⸪Since, (am×bm = (a×b)m ____ Laws of exponents]

= 561/2

2 Comments

  1. Very good teacher
    Outstanding

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*