Number Systems
Chapter 1
Exercise 1.6
EX 1.6 QUESTION 1.
Find:
(i)641/2
(ii)321/5
(iii)1251/3
Solution:
(i)641/2
641/2
= (8×8)1/2
= (82)½
= 81 [⸪2×1/2 = 2/2 =1]
= 8
(ii)321/5
321/5
= (25)1/5
= (25)⅕
= 21 [⸪5×1/5 = 1]
= 2
(iii)1251/3
(125)1/3
= (5×5×5)1/3
= (53)⅓
= 51 (3×1/3 = 3/3 = 1)
= 5
EX 1.6 QUESTION 2.
Find:
(i) 93/2
(ii) 322/5
(iii)163/4
(iv) 125-1/3
Solution:
(i) 93/2
93/2
= (3×3)3/2
= (32)3/2
= 33 [⸪2×3/2 = 3]
=27
(ii) 322/5
322/5 = (2×2×2×2×2)2/5
= (25)2⁄5
= 22 [⸪5×2/5= 2]
= 4
(iii)163/4
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
EX 1.6 QUESTION 3.
Simplify:
(i) 22/3×21/5
(ii) (1/33)7
(iii) 111/2/111/4
(iv) 71/2×81/
Solution:
(i) 22/3×21/5
22/3×21/5
= 2(2/3)+(1/5) [ am×an=am+n]
= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]
(ii) (1/33)7
(1/33)7 = (3-3)7 [(am)n = am x n]
= 3-21
(iii) 111/2/111/4
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
71/2×81/2 = (7×8)1/2 [am×bm = (a×b)m ]
= 561/2