A Level Computer Science: Information Representation — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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. A floating point number uses a 6-bit mantissa and a 4-bit exponent, both in two’s complement.

(a) Convert 0.1101 × 2³ into this format. [3] (b) Explain what normalisation means and why it is used. [4] (c) Explain the trade-off between allocating more bits to the mantissa and more to the exponent. [4]

5. Explain two causes of error in floating point representation. [4]

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]

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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) Mantissa 011010 (the leading 0 is the sign bit, then .11010) [1]; exponent 3 = 0011 [1]; full representation 011010 0011 [1]. (b) A normalised number has its mantissa adjusted so that the first bit after the sign bit is different from the sign bit — 01… for a positive number, 10… for a negative one [1] [1]. It is used so that each value has exactly one representation, which avoids ambiguity [1], and so that the maximum number of significant bits is retained, giving the greatest possible precision for the bits available [1]. (c) More bits in the mantissa give greater precision — more significant figures, so less rounding error [1] [1]. More bits in the exponent give greater range — much larger and much smaller magnitudes can be represented [1]. Since the total number of bits is fixed, increasing one necessarily reduces the other, so the designer must decide whether the application needs accuracy or reach [1].

5. Any two, 2 marks each: many values, such as 0.1 in denary, cannot be represented exactly in binary, so they are stored as the nearest available value — a rounding error [1] [1]. Truncation — when a result needs more bits than the mantissa provides, the surplus bits are discarded [1] [1]. Accumulated error — small errors compound over repeated calculations, so a long loop can produce a significantly wrong result [1] [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].

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Where marks are usually lost

• Forgetting to add 1 after inverting the bits.
• Saying overflow has occurred whenever a carry is generated — check the sign bit.
• Normalising by making the first two bits the same rather than different.
• Not stating the precision/range trade-off explicitly.
• 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.