CBSE Class 11 Computer Science
Question 34 of 68
Data Representation and Boolean Logic — Question 5
Back to all questions(a) 75
| Octal No | Power | Value | Result |
|---|---|---|---|
| 5 (LSB) | 80 | 1 | 5x1=5 |
| 7 (MSB) | 81 | 8 | 7x8=56 |
Equivalent decimal number = 5 + 56 = 61
Therefore, (75)8 = (61)10.
(b) 321
| Octal No | Power | Value | Result |
|---|---|---|---|
| 1 (LSB) | 80 | 1 | 1x1=1 |
| 2 | 81 | 8 | 2x8=16 |
| 3 (MSB) | 82 | 64 | 3x64=192 |
Equivalent decimal number = 1 + 16 + 192 = 209
Therefore, (321)8 = (209)10.
(c) 142
| Octal No | Power | Value | Result |
|---|---|---|---|
| 2 (LSB) | 80 | 1 | 2x1=2 |
| 4 | 81 | 8 | 4x8=32 |
| 1 (MSB) | 82 | 64 | 1x64=64 |
Equivalent decimal number = 2 + 32 + 64 = 98
Therefore, (142)8 = (98)10.
(d) 205
| Octal No | Power | Value | Result |
|---|---|---|---|
| 5 (LSB) | 80 | 1 | 5x1=5 |
| 0 | 81 | 8 | 0x8=56 |
| 2 (MSB) | 82 | 64 | 2x64=128 |
Equivalent decimal number = 5 + 128 = 133
Therefore, (205)8 = (133)10.