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S3 ICT: Binary and Hexadecimal Data Representation (Paper 3: challenge)
S3 ICT: Binary and Hexadecimal Data Representation (Paper 3: challenge)
S3 ICT binary and hexadecimal paper 3, challenge: long hexadecimal strings, binary fractions, twos complement and overflow.
Direct answers Answers in order: F5C, FFF in three digits, 5.625, 0111 that is 7, 0x80 that is minus 128, 0101 0110 that is minus 86, 16,777,216, 1,000, 0100 0001 and 215.
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: S3 ICT: Binary and Hexadecimal Data Representation |
| Level | S3 |
02 Questions (10 items)
- FoundationWrite binary 1111 0101 1100 in hexadecimal.
- FoundationWrite denary 4,095 in hexadecimal and say how many digits are needed.
- FoundationWhat is the denary value of binary 101.101?
- IntermediateSubtract binary 0110 from 1101 and give the answer.
- IntermediateIn 8 bit twos complement, what does 7F become after adding 1?
- IntermediateWhat is the twos complement of 8 bit binary 1010 1010 and which value does it represent?
- ChallengeIf each pixel uses 24 bits, how many different colours can be shown?
- ChallengeWhat is the denary value of hexadecimal 3E8?
- ChallengeCharacter A has ASCII value 65. What is its 8 bit binary form?
- Error redoA student reads hexadecimal D7 digit by digit as 13 and 7 and writes 137. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | F5C | Group four bits from the right as 1111, 0101 and 1100, which are F, 5 and C. |
| 2 | FFF, three digits | 4,095 is 2 to the power 12 minus 1, so it is twelve ones, that is FFF. |
| 3 | 5.625 | The integer part is 4 plus 1, giving 5, and the fraction is 0.5 plus 0.125, giving 0.625. |
| 4 | 0111, that is 7 | 13 minus 6 gives 7, and 7 in four bits is 0111. |
| 5 | 0x80, that is minus 128, so overflow has occurred | 7F is 127, and adding 1 gives 1000 0000, whose leading 1 means minus 128. |
| 6 | 0101 0110, representing minus 86 | The ones complement is 0101 0101 and adding 1 gives 0101 0110, that is 86, so the original pattern is minus 86. |
| 7 | 16,777,216 | 2 to the power 24 is 16,777,216, roughly sixteen million colours. |
| 8 | 1,000 | 3 times 256 plus 14 times 16 plus 8 gives 1,000. |
| 9 | 0100 0001 | 64 plus 1 gives 65, which is 0100 0001. |
| 10 | It should be 13 times 16 plus 7, that is 215 | Hexadecimal expands in powers of 16 rather than writing the digits side by side; writing F5C as 1512 is the same mistake. |
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.
Expand binary by place value into denary first, then group four bits per hexadecimal digit; check for twos complement when negatives appear and confirm the answer stays within the bit width.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.