CBSE Class 11 Computer Science
Question 71 of 91
Data Representation — Question 2
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Question Question 2
Convert the following binary numbers to decimal :
(a) 1100
| Binary No | Power | Value | Result |
|---|---|---|---|
| 0 (LSB) | 20 | 1 | 0x1=0 |
| 0 | 21 | 2 | 0x2=0 |
| 1 | 22 | 4 | 1x4=4 |
| 1 (MSB) | 23 | 8 | 1x8=8 |
Equivalent decimal number = 4 + 8 = 12
Therefore, (1100)2 = (12)10
(b) 10010101
| Binary No | Power | Value | Result |
|---|---|---|---|
| 1 (LSB) | 20 | 1 | 1x1=1 |
| 0 | 21 | 2 | 0x2=0 |
| 1 | 22 | 4 | 1x4=4 |
| 0 | 23 | 8 | 0x8=0 |
| 1 | 24 | 16 | 1x16=16 |
| 0 | 25 | 32 | 0x32=0 |
| 0 | 26 | 64 | 0x64=0 |
| 1 (MSB) | 27 | 128 | 1x128=128 |
Equivalent decimal number = 1 + 4 + 16 + 128 = 149
Therefore, (10010101)2 = (149)10
(c) 11011100
| Binary No | Power | Value | Result |
|---|---|---|---|
| 0 (LSB) | 20 | 1 | 0x1=0 |
| 0 | 21 | 2 | 0x2=0 |
| 1 | 22 | 4 | 1x4=4 |
| 1 | 23 | 8 | 1x8=8 |
| 1 | 24 | 16 | 1x16=16 |
| 0 | 25 | 32 | 0x32=0 |
| 1 | 26 | 64 | 1x64=64 |
| 1 (MSB) | 27 | 128 | 1x128=128 |
Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220
Therefore, (11011100)2 = (220)10