S4 ICT Data Representation, Paper 4: subtraction, capacity and image size
Paper 4 covers binary 11001 as 25, denary 100 as 01100100, hex 4D as 77, 1.5 KiB as 1536 bytes, 24 bits as 3 bytes, an ASCII gap of 17, 1011 minus 0110 giving 0101, hex A5 as 165, a 640 by 480 greyscale image of 307200 bytes and a hex 1A misread as 11.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 ICT: Data Representation |
| Level | S4 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 25 | Sixteen plus eight plus one is 25. |
| 2 | 01100100 | One hundred is 64 plus 32 plus 4, so those bits are 1 and the rest are 0. |
| 3 | 77 | Four times 16 is 64 and D stands for 13, giving 77 in total. |
| 4 | 1536 bytes | One KiB is 1024 bytes and half a KiB is 512, which add to 1536 bytes. |
| 5 | 3 bytes | Every 8 bits make one byte, and 24 divided by 8 is 3. |
| 6 | 17 | Sixty-five minus forty-eight is 17, since digit characters come before capital letters. |
| 7 | 0101 | Eleven minus six is 5, and 5 in four bits is 0101. |
| 8 | 165 | A stands for 10, and 10 times 16 is 160, plus 5 gives 165. |
| 9 | 307200 bytes, 300 KiB | The pixel count is 640 times 480, which is 307200, and one byte per pixel divides by 1024 to give 300 KiB. |
| 10 | 26 | The leading 1 means one sixteen and A means 10, so 16 plus 10 is 26 rather than a plain digit sum. |
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