Q.22 Convert the following :
(i) (478A)16 = (X)10 (ii) (975.55)10 = (Y)2 (iii) (11001100.10)2 = (Z)16
(iv) (678)10 = (W)3 (v) (752)8 = (V)16 (AKTU. 2008 - 09)
Ans. (i) (478A)16 
= (4 ´ 163 + 7 ´ 162 + 8 ´ 161 + A ´ 160)10 = (4 ´ 4096 + 7 ´ 256 + 128 + 10 ´ 1)10
= (16384 + 1792 + 128 + 10)10 = (18314)10
= (478A)16 = (18314)10
(ii) (975.55)10
= (975)10 + (0.55)10
Converting (975)10 to base 2 number
= (1111 00 1111)2
Converting (0.55)10 to base 2 number
= (.10001100011- -)2
Thus (975)10 = (1111001111)2
and (0.55)10 = (0.10001100011- -)2
\ (975.55)10 = (1111001111.10001100011- -)2
= (1111001111.100011)2
(975.55)10 = (1111001111.100011)2
(iii) (11001100.10)2 = (Z)16
Now (11001100.10)2
= (11001100)2 + (0.10)2
= (1 ´ 27 + 1 ´ 26 + 0 ´ 25 + 0 ´ 24 + 1 ´ 23 + 1 ´ 22 + 0 ´ 21 + 0 ´ 20)
+ (1 ´ 2-1 + 0 ´ 2-2)
+ (1 ´ 2-1 + 0 ´ 2-2)
= (1 ´ 128 + 1 ´ 64 + 0 + 0 + 1 ´ 8 + 1 ´ 4 + 0 + 0) + (1 ´ 0.5 + 0)
= (128 + 64 + 8 + 4) + (0.5) = (204.5)10
Now we have that
(1100 1100.10)2 = (204.5)10
We now convert (204.5)10 to (Z)16 as follows
(204.5)10 = (204)10 + (0.5)10
= (CC)16
Þ (204)10 = (CC)16 \ (0.5)10 = (0.8)16
\ (204)10 + (0.5)10 = (CC)16 + (0.8)16
Þ (204.5)10 = (CC.8)16
Thus the have (11001100.10)2 = (204.5)10 = (CC.8)16
(iv) (678)10 = (w)3 Þ (678)10 = (221010)3
(v) (752)8 = (v)16
Step I : Converting (752)8 to a base 10 number 
= (7 ´ 82 + 5 ´ 81 + 2 ´ 80)10 = (7 ´ 64 + 5 ´ 8 + 2 ´ 1)10
= (448 + 40 + 2)10 = (490)10
Thus (752)8 = (490)10
Step II : Converting (490)10 to (v)16 as follows
\ (752)8 = (490)10 = (IEA)16
Alternative Method: -
(7 5 2)8
¯ ¯ ¯
111 101 010
¯ ¯ ¯
1 14 10
¯ ¯ ¯
Þ (752)8 = (I E A)16
We convert each digit to binary equivalent.
Now we make a group of 4 digits and replace each group by its equivalent hexadecimal number.
