CBSE Class 11 Computer Science
Question 55 of 68
Data Representation and Boolean Logic — Question 26
Back to all questions(a) (1010100)10
| 2 | 1010100 | Remainder |
|---|---|---|
| 2 | 505050 | 0 (LSB) |
| 2 | 252525 | 0 |
| 2 | 126262 | 1 |
| 2 | 63131 | 0 |
| 2 | 31565 | 1 |
| 2 | 15782 | 1 |
| 2 | 7891 | 0 |
| 2 | 3945 | 1 |
| 2 | 1972 | 1 |
| 2 | 986 | 0 |
| 2 | 493 | 0 |
| 2 | 246 | 1 |
| 2 | 123 | 0 |
| 2 | 61 | 1 |
| 2 | 30 | 1 |
| 2 | 15 | 0 |
| 2 | 7 | 1 |
| 2 | 3 | 1 |
| 2 | 1 | 1 |
| 0 | 1 (MSB) |
Therefore, (1010100)10 = (11110110100110110100)2.
(b) (3674)8
| Octal Number | Binary Equivalent |
|---|---|
| 4 | 100 |
| 7 | 111 |
| 6 | 110 |
| 3 | 011 |
Therefore, (3674)8 = ()2.
(c) (266)10
| 8 | 266 | Remainder |
|---|---|---|
| 8 | 33 | 2 (LSB) |
| 8 | 4 | 1 |
| 0 | 4 (MSB) |
Therefore, (266)10 = (412)8.
(c) (9F2)16
| Hexadecimal Number | Binary Equivalent |
|---|---|
| 2 | 0010 |
| F | 1111 |
| 9 | 1001 |
Therefore, (9F2)16 = (100111110010)2.