A Level Computer Science: Information Representation — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Information Representation revision notes

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Section A

1. Represent −45 in 8-bit two’s complement, showing your working. [3]

2. State the range of values representable in 8-bit two’s complement. [2]

Section B

3. Perform the following in 8-bit two’s complement, stating whether overflow occurs:

(a) 00110101 + 01011010 [3] (b) 01000000 − 11000000 [3]

4. An 8-bit register performs the unsigned addition 11111111 + 00000001.

(a) State the 8-bit result stored in the register, and whether the carry flag is set. [2] (b) Explain why the overflow flag is not the correct flag to check for this case, and describe the situation in which the overflow flag would be set instead. [3]

5. State two reasons why hexadecimal, rather than binary, is used to represent values such as memory addresses and colour codes. [2]

6. Explain how run-length encoding works, apply it to the string AAAABBBCCCCCCD, and state one case where it would increase the file size. [5]

7. Convert the binary number 10110110 into hexadecimal, showing your working. [2]

8. State the effect of a logical left shift by 2 places on the 8-bit value 00000011, and explain why an arithmetic (not logical) right shift is needed for a negative two’s complement number. [3]

9. Calculate the file size, in bits, of an uncompressed image measuring 200 × 150 pixels with a colour depth of 8 bits per pixel. [2]

10. State one advantage of Unicode over ASCII, and one cost of this advantage. [2]

11. Convert the denary number 45 into (a) 8-bit binary and (b) hexadecimal. [2]

12. The 8-bit binary number 11101100 is a two’s complement integer. Convert it to denary. [1]

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Answers

1. 45 in binary = 00101101 [1]; invert all bits = 11010010 [1]; add 1 = 11010011 [1].

2. −128 to +127 [1] [1].

3. (a) 00110101 = 53; 01011010 = 90 [1]; sum = 10001111 [1]. Adding two positives has produced a negative sign bit, so overflow occurs — 143 exceeds +127 [1]. (b) 11000000 = −64, so this is 64 − (−64) = 64 + 64 [1]; result = 10000000 [1]; the sign bit is negative although both operands imply a positive result, so overflow occurs — 128 exceeds +127 [1].

4. (a) The true sum is 100000000 (9 bits), but the 8-bit register can only store the lower 8 bits, so the stored result is 00000000 [1]; because an unsigned result exceeded the register’s range, the carry flag is set [1]. (b) The overflow flag detects invalid signed (two’s complement) results — for example, adding two positive numbers and getting a result with a negative sign bit, or adding two negatives and getting a positive sign bit [1] [1]. This addition is being evaluated as unsigned, so the relevant boundary is the register’s unsigned range (255), not the sign bit, which is why the carry flag — not the overflow flag — is the one that applies here [1].

5. Any two, 1 mark each: hexadecimal is far more compact than binary, so long binary values such as memory addresses are shorter and easier for people to read, write and compare [1]; each hex digit maps to exactly four binary digits, making conversion between the two trivial [1]; using hexadecimal reduces the chance of human transcription error compared with copying out long strings of 1s and 0s [1].

6. Run-length encoding replaces a run of identical consecutive values with the value and a count of how many times it repeats [1] [1]. AAAABBBCCCCCCD becomes 4A3B6C1D [1] — from 14 characters to 8. It would increase the file size where there are few or no repeated runs [1], for example ABCDEFG, which would become 1A1B1C1D1E1F1G — twice the length [1].

7. Split into 4-bit groups from the right: 1011 0110 [1] → B6 [1]. Hexadecimal is used in preference to binary because it is far more compact and each hex digit maps to exactly four binary digits, making conversion straightforward.

8. A logical left shift by 2 gives 00001100, equivalent to multiplying by 4 (2² = 4) [1]. A logical right shift would discard the sign bit and turn a negative number positive [1]; an arithmetic right shift preserves the sign bit, so it works correctly on negative two’s complement values [1].

9. File size = width × height × colour depth = 200 × 150 × 8 [1] = 240,000 bits [1].

10. Advantage: Unicode can represent a very wide range of the world’s writing systems, not just Latin script [1]. Cost: Unicode text can require more storage per character than ASCII — though UTF-8, the most common Unicode encoding, keeps plain ASCII characters at one byte, so the extra storage cost applies mainly to non-ASCII characters [1].

11. (a) 00101101 [1] (32 + 8 + 4 + 1). (b) 2D [1].

Examiner insight (June 2025): the most common mistake was leaving out the leading zeros; an 8-bit answer must have 8 bits. Try the real question next: Cambridge International AS & A Level Computer Science 9618, June 2025, Paper 12, Question 2.

12. −128 + 64 + 32 + 8 + 4 = −20 [1].

Examiner insight (June 2025): some candidates worked out the magnitude correctly and then forgot the minus sign. A leading 1 in two’s complement means the number is negative. Try the real question next: Cambridge International AS & A Level Computer Science 9618, June 2025, Paper 12, Question 2.

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Where marks are usually lost

• Forgetting to add 1 after inverting the bits.
• Naming the overflow flag for an unsigned range overflow — that case sets the carry flag; the overflow flag is for signed (two’s complement) overflow only.
• Defining resolution as a pixel density rather than a pixel count.
• Splitting a binary number into 4-bit groups from the left rather than the right when converting to hexadecimal.
• Using a logical right shift on a negative two’s complement value instead of an arithmetic one.
• Forgetting to convert file size from bits to bytes (divide by 8) when a question asks for bytes specifically.