OxfordAQA A-Level Computer Science: Representing data (9645) – 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 from your board or school.

These questions cover section 3.5, Representing data (3.5.1 to 3.5.8), of the OxfordAQA International AS and A-level Computer Science (9645) specification, Version 1.1, for International AS exams from May/June 2025 and International A-level exams from May/June 2026. All of it is AS content, assessed in the Unit 2 written exam; none is International A-level only.

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Questions

1. Number bases and units.

(a) State the number of bytes in 4 KiB and in 4 kB. [2] (b) Convert hexadecimal 9C to binary and to decimal. [2] (c) Convert 214 to hexadecimal. [1]

2. Unsigned binary.

(a) State the largest value that 7-bit unsigned binary can represent. [1] (b) Add the unsigned binary numbers 01011110 and 01101101. Show your working. [2] (c) Multiply the unsigned binary numbers 1011 and 110. Give your answer in binary. [2] (d) Explain what would happen if the result of (b) had to be stored in 7 bits. [1]

3. Two’s complement, 8 bits unless stated.

(a) Represent −100. [2] (b) Convert 11100011 to decimal. [1] (c) Calculate 58 − 91 using two’s complement. Show your working. [3] (d) State the range of 12-bit two’s complement. [1]

4. A system uses 10-bit unsigned fixed point: four bits before the binary point and six after.

(a) Represent 9.375. [2] (b) 3.7 cannot be stored exactly. Give the closest value it can store, then calculate the absolute error and the relative error. [4] (c) Explain why relative error is a more useful measure than absolute error. [2]

5. A floating point format has an 8-bit mantissa and a 4-bit exponent, both in two’s complement, with the binary point after the first mantissa bit.

(a) Convert mantissa 0.1010110, exponent 0101 to decimal. [2] (b) Convert mantissa 1.0011000, exponent 0100 to decimal. [2] (c) Represent −0.21875 as a normalised number in this format. [3] (d) The format is changed to a 9-bit mantissa and 3-bit exponent. State the effect on precision and on range. [2]

6. Using the format from question 5:

(a) Normalise mantissa 0.0001101, exponent 0010. [2] (b) Normalise mantissa 1.1110010, exponent 0100. [2] (c) Give two reasons why floating point numbers are normalised. [2]

7. A weather station stores readings from 0.002 to 85,000.

(a) Compare fixed point and floating point for this data in terms of range, precision and speed of calculation. [6] (b) Explain what underflow is and when it occurs. [2]

8. Characters.

(a) The ASCII code for A is 65. Give the code for P in decimal and in 7-bit binary. [2] (b) Explain why Unicode was introduced. [2] (c) Explain how UTF-8 was designed to be backwards compatible with ASCII. [2] (d) Give the pure binary representation of the digit 5 and its 7-bit ASCII code in binary. [2]

9. Graphics.

(a) Calculate the storage, in MiB, of a 2048 × 1536 bitmap with a colour depth of 4 bits, excluding metadata. [3] (b) State how many colours this colour depth allows. [1] (c) A club logo will be printed on badges and on large banners. Explain why a vector graphic suits this, and give two properties stored for each object. [3]

10. Sound.

(a) Describe how an ADC is used when recording analogue audio. [3] (b) A mono clip lasts 3 minutes, sampled at 32,000 Hz with a sample resolution of 8 bits. Calculate its size in MB. [3] (c) The highest frequency in the clip is 12 kHz. Use Nyquist’s theorem to decide whether 32,000 Hz is enough. [2]

11. Encryption.

(a) Encrypt BRIDGE with a Caesar cipher, shift 9. [2] (b) Give two reasons why the Caesar cipher is easily cracked. [2] (c) In a Vernam cipher the character K (01001011) is XORed with the key bits 01110010. Give the ciphertext. [2] (d) Compare the Vernam cipher with ciphers that rely on computational security. [4]

12. Error detection.

(a) Give the 8 bits sent for 0110010 with an even parity bit added at the start. [1] (b) Explain why a parity check may not detect corruption. [1] (c) With majority voting, each bit is sent three times. Decode 101 000 011 110. [2] (d) A checksum is the sum of the data bytes modulo 256. Calculate it, in binary, for 11001010 and 01110001. [2] (e) Compare the effectiveness of parity bits and majority voting. [2]

Answers

1. (a) 4 × 2¹⁰ = 4096 bytes [1]; 4 × 10³ = 4000 bytes [1]. (b) 9 = 1001, C = 1100, so 10011100 [1]; 9 × 16 + 12 = 156 [1]. (c) 214 = 1101 0110 = D6 [1]. Examiner insight: Show the nibble grouping; a bare hex answer leaves nothing to credit if one digit is wrong.

2. (a) 2⁷ − 1 = 127 [1]. (b) Carries shown column by column [1]; 01011110 + 01101101 = 11001011 (94 + 109 = 203) [1]. (c) Partial products 10110 and 101100 [1]; sum 1000010 (11 × 6 = 66) [1]. (d) 203 is more than 127, so overflow: the result cannot be represented in 7 bits [1]. Examiner insight: Write carries above the columns so a correct method is visible.

3. (a) +100 = 01100100, flip to 10011011 [1], add 1: 10011100 [1]. (b) −128 + 64 + 32 + 2 + 1 = −29 [1]. (c) 58 = 00111010 [1]; −91 = 10100101 [1]; sum 11011111 = −33 [1]. (d) −2048 to +2047 [1]. Examiner insight: “Using two’s complement” means showing the negative number being added; a decimal subtraction does not show the method asked for.

4. (a) 9.375 = 8 + 1 + 0.25 + 0.125 [1] = 1001.011000 [1]. (b) 3.7 × 64 = 236.8, so use 237/64 [1]: 0011.101101 = 3.703125 [1]. Absolute error = 3.703125 − 3.7 = 0.003125 [1]. Relative error = 0.003125 ÷ 3.7 = 0.000845 (about 0.0845%) [1]. (c) Relative error compares the error with the size of the value [1], so the same absolute error, such as 0.5, is tiny on 1000 (0.05%) but large on 2 (25%) [1]. Examiner insight: Divide by the true value, not the stored one, and show the closest value before the errors.

5. (a) Exponent +5 [1]; 010101.10 = 21.5 [1]. (b) Mantissa −1 + 0.125 + 0.0625 = −0.8125, exponent +4 [1]; −0.8125 × 16 = −13 [1]. (c) 0.21875 = 0.00111 = 0.111 × 2⁻² [1]; exponent −2 = 1110 [1]; two’s complement of 0.1110000 is 1.0010000 [1]. (d) One more mantissa bit gives greater precision [1]; one fewer exponent bit gives a smaller range [1]. Examiner insight: For a negative value, give the final mantissa in two’s complement; a positive mantissa with a minus sign written in front is not this format.

6. (a) Shift left 3: mantissa 0.1101000 [1]; exponent 2 − 3 = −1 = 1111 [1]. (b) Shift left 3: mantissa 1.0010000 [1]; exponent 4 − 3 = 1 = 0001 [1]. (c) Each number has one unique representation [1], and precision is maximised because no mantissa bits are wasted on redundant leading bits [1]. Examiner insight: Check the value is unchanged (0.40625 in (a), −1.75 in (b)); a changed value means the exponent moved the wrong way.

7. (a) Range: floating point covers both ends because the exponent scales the value [1]; fixed point would need many bits [1]. Precision: fixed point has the same absolute precision everywhere, so small readings lose significant figures [1]; floating point precision is set by the mantissa and stays similar relative to each value [1]. Speed: fixed point is faster, like integer arithmetic [1]; floating point is slower as exponents must be handled [1]. (b) A result so close to zero [1] that its closest representation is zero, for example when two very small numbers are multiplied [1]. Examiner insight: Cover both representations under each criterion; one-sided points are not a comparison.

8. (a) 65 + 15 = 80 [1]; 1010000 [1]. (b) ASCII has only 128 characters, too few for other alphabets and symbols [1]; Unicode gives characters from the world’s writing systems unique codes [1]. (c) UTF-8 uses 8 bits for codes 0 to 127 [1], and these codes represent the same characters as in ASCII, so ASCII text is valid UTF-8 [1]. (d) Pure binary 101 [1]; ASCII code 53 = 0110101 [1]. Examiner insight: Link Unicode to ASCII’s 128-character limit; “more characters” alone is thin.

9. (a) 2048 × 1536 = 3,145,728 pixels [1]; × 4 = 12,582,912 bits [1]; ÷ 8 = 1,572,864 bytes = 1.5 MiB [1]. (b) 2⁴ = 16 [1]. (c) Vectors scale to any size without losing quality, so one file serves badges and banners [1]. Any two properties, one mark each [1] [1]: x and y coordinates of the top left corner; outline colour; outline line width; fill colour. Examiner insight: MiB means dividing by 2²⁰; dividing by 1,000,000 answers a different question.

10. (a) The microphone gives an analogue voltage [1]; the ADC samples it at regular intervals [1] and converts each sample to the nearest binary value using the sample resolution [1]. (b) 3 minutes = 180 s; 32,000 × 8 × 180 = 46,080,000 bits [1]; ÷ 8 = 5,760,000 bytes [1] = 5.76 MB [1]. (c) Minimum rate = 2 × 12 kHz = 24 kHz [1]; 32 kHz is more than this, so yes [1]. Examiner insight: Convert minutes to seconds before using the formula, and say which unit each line of working is in.

11. (a) Each letter moved 9 places, wrapping Z to A [1]: KARMPN [1]. (b) Any two, one mark each [1] [1]: only 25 possible keys; letter frequencies survive, allowing frequency analysis; finding one letter’s shift reveals all the others. (c) XOR bit by bit, 1 where the bits differ [1]: 00111001 [1]. (d) Vernam can be perfectly secure: nothing about the plaintext can be learned from the ciphertext [1], if the key is random, at least as long as the plaintext, used once and known only to sender and receiver [1]. Computationally secure ciphers leak some information, but cracking takes unfeasible time [1]. Vernam needs a new message-length key shared securely each time, so the key exchange problem recurs [1]. Examiner insight: When perfect security is part of a comparison, state all four key conditions; a vague “the key is secret” does not cover them.

12. (a) Three 1s, so the parity bit is 1: 10110010 [1]. (b) If an even number of bits flip, the count of 1s stays even [1]. (c) Majority of each group [1]: 1 0 1 1 [1]. (d) 202 + 113 = 315; 315 mod 256 = 59 [1] = 00111011 [1]. (e) Parity adds one bit but only detects odd numbers of errors [1]; majority voting can correct isolated errors but sends three times the data [1]. Examiner insight: Compare cost and capability; describing each method separately is not a comparison.

Where marks are usually lost

• Flipping bits for a negative number but not adding 1.
• Reading a mantissa starting 1 as positive.
• Moving the point the wrong way for a negative exponent.
• Adding, not subtracting, the shift when normalising.
• Forgetting to convert minutes to seconds in sound calculations.
• Describing parity as able to correct errors.

Next steps

• Representing data revision notes
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• Exam preparation
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Official syllabus

OxfordAQA International AS and A-level Computer Science (9645) specification, Version 1.1, for International AS exams May/June 2025 onwards and International A-level exams May/June 2026 onwards (OxfordAQA), section 3.5 Representing data.