CBSE Class 11 Computer Science
Question 30 of 68
Data Representation and Boolean Logic — Question 1
Back to all questions(a) 52
| 2 | 52 | Remainder |
|---|---|---|
| 2 | 26 | 0 (LSB) |
| 2 | 13 | 0 |
| 2 | 6 | 1 |
| 2 | 3 | 0 |
| 2 | 1 | 1 |
| 0 | 1 (MSB) |
Therefore, (52)10 = (110100)2.
(b) 44
| 2 | 44 | Remainder |
|---|---|---|
| 2 | 22 | 0 (LSB) |
| 2 | 11 | 0 |
| 2 | 5 | 1 |
| 2 | 2 | 1 |
| 2 | 1 | 0 |
| 0 | 1 (MSB) |
Therefore, (44)10 = (101100)2.
(c) 25.80
Let us first convert 25 into binary as shown below:
| 2 | 25 | Remainder |
|---|---|---|
| 2 | 12 | 1 (LSB) |
| 2 | 6 | 0 |
| 2 | 3 | 0 |
| 2 | 1 | 1 |
| 0 | 1 (MSB) |
The binary equivalent of integer part (25)10 = (11001)2.
Now let us convert (0.80)10 into binary as shown below:
| Multiply | = | Integer | Fraction |
|---|---|---|---|
| 0.80 x 2 | = | 1 | 0.6 (MSB) |
| 0.6 x 2 | = | 1 | 0.2 |
| 0.2 x 2 | = | 0 | 0.4 |
| 0.4 x 2 | = | 0 | 0.8 |
| 0.8 x 2 | = | 1 | 0.6 (LSB) |
The binary equivalent of fractional part (0.8)10 = (11001)2.
Therefore, (25.80)10 = (11001.11001)2.
(d) 62.325
Let us first convert 62 into binary as shown below:
| 2 | 62 | Remainder |
|---|---|---|
| 2 | 31 | 0 (LSB) |
| 2 | 15 | 1 |
| 2 | 7 | 1 |
| 2 | 3 | 1 |
| 2 | 1 | 1 |
| 0 | 1 (MSB) |
The binary equivalent of integer part (62)10 = (111110)2.
Now let us convert (0.325)10 into binary as shown below:
| Multiply | = | Integer | Fraction |
|---|---|---|---|
| 0.325 x 2 | = | 0 | 0.65 (MSB) |
| 0.65 x 2 | = | 1 | 0.3 |
| 0.3 x 2 | = | 0 | 0.6 |
| 0.6 x 2 | = | 1 | 0.2 |
| 0.2 x 2 | = | 0 | 0.4 |
| 0.4 x 2 | = | 0 | 0.8 |
| 0.8 x 2 | = | 1 | 0.6 (LSB) |
The binary equivalent of fractional part (0.325)10 = (0101001)2.
Therefore, (62.325)10 = (111110.0101001)2.