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S6 ICT: Multimedia and Compression – Practice Paper 3 (Challenge, 10 Questions)
S6 ICT: Multimedia and Compression – Practice Paper 3 (Challenge, 10 Questions)
Paper three handles studio figures: 48-bit scans, 4:2:0 subsampling, 96 kHz masters and raw 4K at 746 MB per second, then works back from link bandwidth and disk capacity.
Quick answer A 48-bit 3 000 by 2 000 image takes 36 MB and 18 MB with 4:2:0; 96 kHz stereo for five minutes takes 172 800 000 bytes; raw 4K runs at 746.5 MB a second, so 20 Mbps encoding is about 299 times smaller.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 2 · Error redo 2 |
| Chapter | Chapter: S6 ICT | Multimedia and Compression: File Size, Sampling and Bitrate |
| Level | Challenge |
02 Questions (10 items)
- FoundationA 3 000 by 2 000 image at 16 bits per channel, so 48 bits per pixel: how many bytes uncompressed, and how many MB?
- FoundationStoring the same image with 4:2:0 chroma subsampling needs only 24 bits per pixel: what is the new size in MB?
- FoundationA five-minute master recording at 96 kHz, 24-bit, stereo: how many bytes uncompressed?
- IntermediateUncompressed 4K at 3 840 by 2 160, 24 bits, 30 fps: how many MB per second, and what bitrate in Gbps?
- IntermediateThe same 4K video encoded with H.264 at 20 Mbps for one minute: how many MB, and roughly what compression factor against raw?
- IntermediateA ten-second, 25 fps, 24-bit clip is 180 MB uncompressed: how many pixels per frame, and what frame size at a 3 to 2 ratio?
- ChallengeA 100 Mbps link carries 20 Mbps video streams, each with 320 kbps audio and ten per cent overhead: how many streams at once?
- ChallengeA 500 GB drive must hold four hours of video: what is the highest average bitrate in Mbps?
- Error redoError retry: a student divided by eight twice and got 2 700 000 bytes for a 96 kHz, 24-bit, five-minute stereo file. What is correct?
- Error redoError retry: a student thought 4:2:0 keeps luma only and gave 12 MB for a 3 000 by 2 000 image at 16-bit. What is correct and what was ignored?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 36 000 000 bytes, that is 36 MB | Six million pixels times six bytes give 36 000 000 bytes, which is 36 MB. |
| 2 | 18 MB | 4:2:0 halves the chroma data, so the depth falls from 48 to 24 bits and the file halves from 36 MB to 18 MB. |
| 3 | 172 800 000 bytes | 96 000 samples times three bytes times two channels times 300 seconds give 172 800 000 bytes. |
| 4 | 746.5 MB per second, about 5.97 Gbps | 24 883 200 bytes per frame times thirty gives 746 496 000 bytes a second, which is about 5.97 Gbps. |
| 5 | 150 MB, roughly one two-hundred-and-ninetieth to one three-hundredth | 20 Mbps for a minute is 1 200 Mb or 150 MB, while one raw minute is 44 789 760 000 bytes, so the factor is about 299. |
| 6 | 240 000 pixels per frame, about 600 by 400 | 250 frames share 180 000 000 bytes, so each frame is 720 000 bytes or 240 000 pixels, which at 3 to 2 is 600 by 400. |
| 7 | At most four streams | Each stream needs 20.32 Mbps, about 22.35 with overhead, and 100 over 22.35 is 4.47, so only four fit. |
| 8 | About 277.78 Mbps | 500 GB is 4 000 000 Mb, and dividing by 14 400 seconds gives 277.78 Mb per second. |
| 9 | 172 800 000 bytes | One division by eight converts bits to bytes; a second one shrinks the answer sixty-four times too far. |
| 10 | 18 MB, because the halved chroma data still has to be stored | Luma at sixteen bits takes 12 000 000 bytes and the halved chroma adds 6 000 000, making 18 MB in all. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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