| OCR |
GCSE (9-1) Computer Science
Mark Scheme
J277/01: Unit 1.2 Number Conversions
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| Question | Answer | Marks | Guidance |
|---|---|---|---|
| 1a | 21 | 1 |
AO2
Reject if answer is in binary or hex.
|
| 1b | 131 | 1 |
AO2
|
| 2 |
Working: Correct method shown (e.g., subtraction from 170 or table with correct headers 128, 32, 8, 2) (1) Answer: 1010 1010 (1) |
2 |
Examiner Note: If the answer is correct (10101010), award full marks (2) even if no working is shown.
The answer must be 8 bits. If they write 10101010, accept it. If they miss leading zeros e.g. 101010 (if applicable for smaller numbers), generally penalise in J277 if "8-bit" was specified.
|
| 3a | E6 | 1 |
Examiner Note: Mark final answer. Do not need to see splitting of nibbles.
Accept lower case 'e'.
|
| 3b | 0010 1011 | 1 |
Must be 8 bits.
0010 = 2, 1011 = B.
|
| 3c |
|
2 |
AO1
Do not accept: "It takes up less storage space". This is a critical misconception. Hex is just a representation; the computer still stores it as binary.
Do not accept: "It is faster for the computer".
|
| 4 |
Working: Showing 3 x 16 and 14 x 1 (or E = 14) (1) Answer: 62 (1) |
2 |
AO2
Allow follow-through if they identify E=14 correctly but multiply wrongly.
Answer only = 2 marks.
|
| 5 |
Working: Method shown (e.g., 214 ÷ 16 = 13 remainder 6 OR converting 214 to binary 1101 0110 first) (1) Answer: D6 (1) |
2 |
Most students convert to Binary first, then Hex. This is a valid method.
13 = D.
Answer only = 2 marks.
|
| 6a | 15 | 1 |
AO2
|
| 6b | F | 1 |
AO2
|
| 6c |
0000 (1) With an overflow error indicated / carry out (1) Or simply: (1)0000 |
2 |
AO2
The question asks for the result in 4-bit binary. 1111 + 0001 = 10000.
In a 4-bit system, this results in 0000 (1 mark) and an overflow (1 mark).
Examiner Note: This is a "stretch" question testing understanding of binary limits (Topic 1.2.4).
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