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S4 ICT: Data Representation and Binary Encoding (Paper 2 (intermediate))
Tutor Prep Guide · data representation: binary and hexadecimal conversion, bits and bytes, two’s complement and sign-magnitude, character encoding (ASCII and Unicode), bitmap image and sampled sound size, lossy versus lossless compression
S4 ICT: Data Representation and Binary Encoding (Paper 2 (intermediate))
Paper 2 (intermediate): ten data representation questions with answers and one-line notes.
Direct answers Level: S4 ICT; format: number base conversion, bits and bytes, two’s complement, character encoding, image and sound size, compression.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 ICT: Data Representation and Binary Encoding |
| Level | S4 |
02 Questions (10 items)
- FoundationRepresent minus five in 8-bit two’s complement.
- FoundationWhat denary value does the two’s complement pattern 11111011 represent?
- FoundationWhat range of integers can 8-bit two’s complement hold?
- IntermediateRepresent minus five in 8-bit sign-magnitude form.
- IntermediateWhat are the ASCII codes of A and a, and how far apart are they?
- IntermediateHow many distinct characters can 7-bit ASCII represent?
- ChallengeConvert hexadecimal 3F into 8-bit binary and denary.
- ChallengeA 640 by 480 black-and-white bitmap uses one bit per pixel. What is its uncompressed size in bytes and KiB?
- ChallengeA mono recording uses an 8 kHz sample rate at 16 bits for 10 seconds. How many bytes is that?
- Error redoA student adds two negative numbers written in sign-magnitude form and gets a wrong answer. Why?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 11111011 | Five is 00000101, which inverts to 11111010 and becomes 11111011 after adding one. |
| 2 | Minus five | The leading one marks a negative number, and subtracting 256 from the pattern gives minus five. |
| 3 | Minus 128 to 127 | The top bit carries a negative weight, so there is one more negative value than positive ones. |
| 4 | 10000101 | The leading bit is set to one for the sign and the remaining seven bits hold 0000101 for five. |
| 5 | A is 65 and a is 97, a difference of 32 | Upper and lower case differ by a single bit, so the gap is always 32. |
| 6 | 128 | Two to the seventh power is 128, enough for English letters, digits and symbols. |
| 7 | 0011 1111, which is 63 | Each hex digit maps to four bits, so 3 gives 0011 and F gives 1111. |
| 8 | 38 400 bytes, which is 37.5 KiB | 640 times 480 times 1 divided by 8 gives 38 400, and dividing by 1 024 gives 37.5. |
| 9 | 160 000 bytes | 8 000 samples per second times two bytes times ten seconds gives 160 000 bytes. |
| 10 | The sign bits are added as ordinary values, so the sign becomes meaningless; two’s complement should be used | Two’s complement lets one adder handle both addition and subtraction without special sign handling. |
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.
A computer only knows 0 and 1, so text, images and sound are all encoded into bits first. Keep bits apart from bytes, remember the top bit of a two’s complement number is a negative weight, and compute image or sound size as count times bits per item divided by eight, then check the unit in bytes, KiB or MiB.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
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