AQA A-Level Computer Science: Fundamentals of Data 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: Fundamentals of data representation study guide | Fundamentals of data representation revision notes

These questions cover section 4.5 of the AQA A-Level Computer Science (7517) specification, which is assessed on Paper 2. Floating point questions use the AQA format: mantissa and exponent both in two’s complement. Show all working.

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Section A: number systems, bases and units

1. State the most appropriate number set for: (a) the number of people in a queue (b) the length of a table in metres (c) √2. [3]

2. (a) Convert 10110110₂ to decimal and to hexadecimal. [2] (b) Convert 3F₁₆ to binary and to decimal. [2] (c) Explain why hexadecimal is used as a shorthand for binary. [1]

3. (a) State how many different values can be represented with 10 bits, and the largest unsigned integer. [2] (b) State the number of bytes in 5 MiB and in 5 MB. [2]

Section B: binary arithmetic

4. (a) Add the unsigned binary numbers 01101011 and 00110110. [2] (b) Multiply the unsigned binary numbers 1011 and 101. [2]

5. Using 8-bit two’s complement: (a) Represent -77. [2] (b) Convert 10011100 to decimal. [1] (c) Calculate 50 - 75 by adding the two’s complement of 75, showing your working. [3] (d) State the range of integers that can be represented in 12-bit two’s complement. [1]

Section C: fixed and floating point

6. An 8-bit unsigned fixed point format has 3 bits after the binary point. (a) Convert 01011101 (i.e. 01011.101) to decimal. [1] (b) Represent 6.4 as closely as possible in this format. [2] (c) Calculate the absolute and relative errors in your answer to (b). [2]

7. A floating point format uses an 8-bit mantissa followed by a 4-bit exponent, both in two’s complement. (a) Convert mantissa 0.1011000, exponent 0100 to decimal. [2] (b) Convert mantissa 1.0100000, exponent 1110 to decimal. [2] (c) Represent 3.25 as a normalised floating point number. [2] (d) Represent -3.25 as a normalised floating point number. [3] (e) Normalise mantissa 1.1101000, exponent 0011. [2]

8. (a) Explain why floating point numbers are normalised. [2] (b) Compare fixed point and floating point in terms of range, precision and speed of calculation. [3] (c) Explain the difference between overflow and underflow. [2]

Section D: coding systems and error checking

9. Explain the difference between the character code of the digit ‘5’ and its pure binary representation, given that ASCII ‘0’ is 48. [2]

10. Explain why Unicode was introduced. [2]

11. (a) State the even parity bit for the 7-bit data 1011001. [1] (b) Using majority voting, each bit is sent three times. The receiver gets 111 010 001 110. State the data received and explain an advantage and a disadvantage of majority voting. [3] (c) Describe how a check digit is used. [2]

Section E: images, sound and other data

12. A bitmap is 800 × 600 pixels and can show 16 colours. (a) State the colour depth. [1] (b) Calculate the file size in kilobytes, ignoring metadata. [2] (c) State one item of metadata stored with the image. [1]

13. A mono recording lasts 2 minutes, sampled at 48 kHz with 24-bit sample resolution. (a) Calculate the file size in megabytes. [2] (b) Humans hear up to about 20 kHz. Use the Nyquist theorem to state the minimum sampling rate needed. [1]

14. (a) Give two properties stored for a circle in a vector graphic. [2] (b) Explain why a photograph is stored as a bitmap rather than a vector graphic. [1]

15. Describe what MIDI stores and give two advantages of MIDI files over sampled sound. [3]

16. Describe the role of an ADC in a digital thermometer, and the most common use of a DAC. [3]

Section F: compression and encryption

17. (a) Compress “AAAABBBCCD” using run length encoding. [1] (b) Explain why lossless compression must be used for a text file. [1] (c) Explain the principle of dictionary-based compression. [2]

18. (a) “LIPPS” was encrypted with a Caesar cipher with a shift of 4. Decrypt it. [1] (b) Explain why a Caesar cipher is easily cracked. [2] (c) Encrypt the character code 01000001 with the Vernam key 00110110 using XOR, and show that applying the key again recovers the plaintext. [2] (d) Explain why the Vernam cipher offers perfect security, and how this differs from ciphers that rely on computational security. [4]

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Answers

1. (a) Natural numbers ℕ – used for counting [1] (b) Real numbers ℝ – used for measurement [1] (c) Irrational (part of ℝ) – it cannot be written as a fraction [1].

2. (a) 128 + 32 + 16 + 4 + 2 = 182 [1]; 1011 0110 = B6₁₆ [1]. (b) 3 = 0011, F = 1111 -> 00111111₂ [1] = 63 [1]. (c) Each hex digit represents exactly 4 bits, so it is shorter and easier for people to read and write with fewer errors [1].

3. (a) 2¹⁰ = 1,024 values [1]; largest unsigned value 1,023 [1]. (b) 5 MiB = 5 × 2²⁰ = 5,242,880 bytes [1]; 5 MB = 5 × 10⁶ = 5,000,000 bytes [1].

4. (a) 01101011 + 00110110 = 10100001 (107 + 54 = 161) [2]. (b) 1011 × 101 = 1011 + 101100 = 110111 (11 × 5 = 55) [2].

5. (a) 77 = 01001101 [1]; invert and add 1 -> 10110011 [1]. (b) -128 + 16 + 8 + 4 = -100 [1]. (c) 75 = 01001011, so -75 = 10110101 [1]. 00110010 + 10110101 = 11100111 [1] = -128 + 64 + 32 + 4 + 2 + 1 = -25 [1]. (d) -2048 to 2047 [1].

6. (a) 8 + 2 + 1 + 0.5 + 0.125 = 11.625 [1]. (b) 6.4 lies between 110.011 (6.375) and 110.100 (6.5); 6.375 is closer [1], so 00110011 (00110.011) [1]. (c) Absolute error = 6.4 - 6.375 = 0.025 [1]. Relative error = 0.025 ÷ 6.4 = 0.0039 (about 0.39%) [1].

7. (a) 0.1011 = 0.6875 [1]; × 2⁴ = 11 [1]. (b) Mantissa = -1 + 0.25 = -0.75; exponent 1110 = -2 [1]; -0.75 × 2⁻² = -0.1875 [1]. (c) 3.25 = 11.01₂ = 0.1101 × 2² [1] -> mantissa 0.1101000, exponent 0010 [1]. (d) 3.25 = 011.01; two’s complement: invert -> 100.10, add 0.01 -> 100.11 (= -4 + 0.5 + 0.25) [1]. Normalised so it starts 1.0: 1.0011 × 2² [1] -> mantissa 1.0011000, exponent 0010 [1]. (e) The mantissa starts 1.1, so shift left until it starts 1.0: two places -> 1.0100000 [1]; reduce the exponent by 2 -> 0001 [1]. (Both forms equal -1.5.)

8. (a) To give the maximum precision for the number of bits in the mantissa [1] and a unique representation of each value [1]. (b) Floating point has a much greater range for the same number of bits [1]; fixed point has constant precision but floating point precision varies with the size of the number [1]; fixed point is faster to process because no exponent handling or normalising is needed [1]. (c) Overflow: the result is too large in magnitude to be represented [1]. Underflow: the result is too small (too close to zero) to be represented, e.g. a very small number divided by a very large one [1].

9. The character ‘5’ has ASCII code 48 + 5 = 53 = 0110101, which is used for text input and display [1]. Its pure binary value is 101, which is used for arithmetic [1].

10. ASCII’s 7 bits allow only 128 characters, enough for English but not other alphabets [1]. Unicode uses more bits per character so it can represent the characters of the world’s writing systems and many symbols [1].

11. (a) 1011001 has four 1s, so the even parity bit is 0 [1]. (b) Majority of each group: 111 -> 1, 010 -> 0, 001 -> 0, 110 -> 1, so 1001 [1]. Advantage: single-bit errors in a group can be corrected, not just detected [1]. Disadvantage: three times as much data is sent [1]. (c) A check digit is calculated from the other digits of a code using an agreed algorithm and added to it [1]; when the code is entered the calculation is repeated and, if the results differ, an error is reported [1].

12. (a) 16 = 2⁴, so 4 bits per pixel [1]. (b) 800 × 600 × 4 = 1,920,000 bits ÷ 8 = 240,000 bytes [1] = 240 kB [1]. (c) Any one: width, height, colour depth [1].

13. (a) 48,000 × 24 × 120 = 138,240,000 bits [1] ÷ 8 = 17,280,000 bytes = 17.28 MB [1]. (b) At least 40 kHz – twice the highest frequency [1].

14. (a) Any two: centre coordinates, radius, line colour, line thickness, fill colour [2]. (b) A photograph has continuous variation in colour that cannot be described efficiently as a list of geometric shapes [1].

15. MIDI stores event messages – instructions such as which note to play, when, and how loudly – rather than samples of the sound wave [1]. Advantages (any two): much smaller files; easy to edit the music, e.g. change tempo, key or instrument; the sound is generated at playback on the chosen instrument [2].

16. The temperature sensor produces an analogue signal [1]; the ADC samples it at intervals and converts each sample to a binary number the processor can use [1]. A DAC’s most common use is converting a digital audio signal to an analogue signal for a speaker or headphones [1].

17. (a) 4A3B2C1D [1]. (b) Every character must be restored exactly, or the text’s meaning changes [1]. (c) Repeated patterns (such as words) are stored once in a dictionary [1], and each occurrence in the data is replaced by a shorter reference to its dictionary entry [1].

18. (a) HELLO [1]. (b) There are only 25 possible shifts, so all can be tried quickly (brute force) [1]; letter frequencies are unchanged, so frequency analysis reveals the shift [1]. (c) 01000001 XOR 00110110 = 01110111 [1]; 01110111 XOR 00110110 = 01000001, the original plaintext [1]. (d) The key is truly random, at least as long as the message and used only once [1], so the ciphertext is uniformly distributed and gives no information about the plaintext [1]. Vernam is the only cipher mathematically proven to be completely secure [1]. Other ciphers are computationally secure: they could in theory be broken given enough ciphertext and time, but this would take an impractically long time with current computing power [1].

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Where marks are usually lost

• Writing a two’s complement mantissa’s first bit as +1 instead of -1.
• Adjusting the exponent in the wrong direction when normalising.
• Using 1,024 for “kB” or 1,000 for “KiB”.
• Forgetting to divide by 8 to convert bits to bytes, or to multiply minutes by 60.
• Stating that parity can correct an error.

Frequently asked questions

Will I be told the floating point format in the exam? Yes. AQA 7517 questions state the number of mantissa and exponent bits, and both are in two’s complement. You are not expected to know the IEEE standard.

Do I need to remember the ASCII table? No. Questions give you any codes you need, as question 9 does. What you must know is the difference between a digit’s character code and its pure binary value, and why Unicode was introduced.