Page 74 - Computer Science V2.0 Class 11

iii. (76F) = (?)
10
16
iv. (4D9) = (?) 10
16
v. (11001010) = (?) 10
2
vi. (1010111) = (?) 10
2
Ans. i.
Digits 5 1 4
Position 2 1 0
Weight 8 2 8 1 8 0
Therefore,
1
2
Decimal number = 5×8 + 1×8 + 4×8 0
= 5×64 + 1×8 + 4× 1
= 320 + 8 + 4
= (332) 10
ii.

Octal Digits 2 2 0
Binary value 010 010 000
(3 bits)
Therefore,
Binary number = (10010000) 2
iii.

Digits 7 6 F(15)
Position 2 1 0
Weight 16 2 16 1 16 0
Therefore,
1
2
Decimal number = 7×16 + 6×16 + F×16
= 7×256 + 6×16 + F× 1
= 1792 + 96 + 15
= (1903) 10
iv.

Digits 4 D 9
Position 2 1 0
Weight 16 2 16 1 16 0
Therefore,
2
1
Decimal number = 4×16 + 13×16 + 9×16 0
= 4×256 + 13×16 + 9× 1
= 1024 + 208 + 9
= (1241) 10
v.
Digits 1 1 0 0 1 0 1 0
Position 7 6 5 4 3 2 1 0
Weight 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0
Therefore,
3
5
4
2
7
1
6
Decimal number = 1×2 + 1×2 +0×2 +0×2 +1×2 +0×2 +1×2 +0×2 0
= 128+64 +8 + 2
= (202) 10
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