OxfordAQA IGCSE Computer Science: Data Representation — Practice Questions (9210)

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: Data representation study guide | Data representation revision notes

These questions cover section 3.3 of the OxfordAQA International GCSE Computer Science (9210) specification, which is assessed on Paper 2. Use powers of 10 for units (1 kB = 1,000 bytes) and show your working.

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Section A: number bases and units

1. Explain why computers use binary to represent data and instructions. [2]

2. Explain why hexadecimal is often used in computer science. [2]

3. (a) Convert the binary number 1011 0101 to decimal and to hexadecimal. [2] (b) Convert the hexadecimal number 2F to decimal and to binary. [2] (c) Convert the decimal number 156 to binary and to hexadecimal. [2]

4. (a) State how many bytes are in 3 kB. [1] (b) A file is 2,500,000 bytes. State its size in megabytes. [1] (c) State how many megabytes are in 4 GB. [1]

Section B: binary arithmetic

5. Add the binary numbers 0110 0111 and 0001 1101. Give your answer in binary. [2]

6. (a) Apply a logical left shift of 3 places to 0000 1101 and state the effect on its value. [2] (b) Apply a logical right shift of 2 places to 0110 1000 and state the result in decimal. [2] (c) Describe one situation where a binary shift can be used. [1]

Section C: character encoding

7. In ASCII the character ‘A’ has code 65 and ‘a’ has code 97. (a) State the code for ‘H’ and for ‘d’. [2] (b) State the character with code 70. [1] (c) Explain why the codes for the other letters can be worked out once the code for ‘A’ is known. [1]

8. Describe the purpose of Unicode and give one advantage it has over ASCII. [3]

Section D: images and sound

9. (a) Explain what is meant by a pixel and by colour depth. [2] (b) State the colour depth needed for an image with 256 colours. [1] (c) Calculate the file size in megabytes of a 1200 × 800 pixel image with a colour depth of 24 bits. [2] (d) Explain the effect on file size of halving both the width and the height of the image. [2]

10. A 4 × 4 black-and-white bitmap uses 1 for black and 0 for white. Its rows are: 1001, 0110, 0110, 1001. (a) Draw (or describe) the image. [2] (b) Another 4 × 4 image has a black border one pixel wide around a white centre. Write the binary for each row. [2]

11. (a) Explain why sound must be sampled before it can be stored on a computer. [2] (b) Define sampling rate and sample resolution. [2] (c) Calculate the file size in megabytes of a 3-minute recording sampled at 22,050 Hz with an 8-bit sample resolution. [3]

Section E: compression

12. Give two reasons why data may be compressed. [2]

13. (a) Represent the row 1111000000001111 using run length encoding frequency/data pairs. [1] (b) Write the original bits for the RLE data 3 0 2 1 3 0. [1] (c) Explain why RLE would not compress the row 0101010101010101. [2]

14. The string COFFEE is to be compressed with Huffman coding. (a) Build a Huffman tree for the string. [3] (b) Give the Huffman code for each character, from your tree. [2] (c) Calculate the number of bits needed for COFFEE using your codes. [2] (d) Calculate the number of bits needed in 7-bit ASCII and the number of bits saved. [2]

15. In a given Huffman tree, a left branch is 0 and a right branch is 1. The character T is reached from the root by going right, then left, then left. State the code for T and explain why the most frequent characters have the shortest codes. [3]

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Answers

1. The circuits in a computer have two states (e.g. on and off) [1], which can be represented reliably by the two binary digits 0 and 1 [1].

2. It is a shorter way of representing binary – one hex digit represents 4 bits [1] – so it is easier for people to read and write, with fewer mistakes [1].

3. (a) 128 + 32 + 16 + 4 + 1 = 181 [1]; 1011 = B, 0101 = 5 -> B5 [1]. (b) 2 × 16 + 15 = 47 [1]; 2 = 0010, F = 1111 -> 0010 1111 [1]. (c) 156 = 128 + 16 + 8 + 4 -> 1001 1100 [1]; 1001 = 9, 1100 = C -> 9C [1].

4. (a) 3,000 bytes [1] (b) 2.5 MB [1] (c) 4,000 MB [1].

5.

  0110 0111   (103)
+ 0001 1101   (29)
  1000 0100   (132)

Correct method with carries [1]; answer 1000 0100 [1].

6. (a) 0110 1000 [1]; the value is multiplied by 8 (13 -> 104) [1]. (b) 0001 1010 [1] = 26 (104 ÷ 4) [1]. (c) Quick multiplication or division by a power of 2 [1].

7. (a) ‘H’ = 65 + 7 = 72 [1]; ‘d’ = 97 + 3 = 100 [1]. (b) 70 = 65 + 5 -> ‘F’ [1]. (c) Character codes are grouped and run in sequence, so each letter’s code is one more than the previous letter’s [1].

8. Unicode is a character set that can represent the characters of many different alphabets and special symbols [1], so text in any language can be stored and exchanged [1]. Advantage: a far greater range of characters than ASCII’s 128 [1].

9. (a) A pixel (picture element) is a single point in a graphical image [1]. Colour depth is the number of bits used to represent each pixel [1]. (b) 2⁸ = 256, so 8 bits [1]. (c) 1200 × 800 × 24 = 23,040,000 bits ÷ 8 = 2,880,000 bytes [1] = 2.88 MB [1]. (d) The number of pixels becomes ½ × ½ = ¼ [1], so the file size falls to a quarter (0.72 MB) [1].

10. (a) Black pixels in the four corners and the four centre squares, making an X-shaped pattern [2]. (b) 1111, 1001, 1001, 1111 [2] (1 mark for top and bottom rows, 1 for the middle rows).

11. (a) Sound is analogue – a continuously varying wave [1] – but a computer can only store digital (binary) values, so the amplitude is measured at regular intervals and each measurement is stored as a number [1]. (b) Sampling rate: the number of samples taken per second, measured in hertz [1]. Sample resolution: the number of bits used for each sample [1]. (c) 3 minutes = 180 s [1]; 22,050 × 8 × 180 = 31,752,000 bits [1]; ÷ 8 = 3,969,000 bytes = 3.969 MB [1].

12. Any two: files take up less storage space; files transfer faster across a network or the Internet; less bandwidth is used [2].

13. (a) 4 1 8 0 4 1 [1]. (b) 00011000 [1]. (c) There are no runs: each value appears only once before it changes [1], so every bit needs its own pair (1 0 1 1 1 0 …), which makes the data longer than the original [1].

14. (a) Frequencies: C = 1, O = 1, F = 2, E = 2 [1]. Join C and O (total 2) [1]; join F and E (total 4); then join (C, O) with (F, E) to make the root (total 6) [1]. (Other trees that always join the two lowest totals are also correct.) (b) From the tree described, with left = 0 and right = 1: C = 00, O = 01, F = 10, E = 11 [2]. (c) C 1 × 2 + O 1 × 2 + F 2 × 2 + E 2 × 2 = 12 bits [2]. (Any correct Huffman tree gives 12 bits for this string.) (d) 6 × 7 = 42 bits [1]; saving = 42 - 12 = 30 bits [1].

15. 100 [1]. Huffman coding places the most frequent characters closest to the root of the tree [1], so they need fewer branches – and therefore fewer bits – which minimises the total size of the compressed data [1].

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Where marks are usually lost

• Using 1,024 for kilo: 9210 assessments use powers of 10.
• Forgetting to convert minutes to seconds in sound calculations.
• In Huffman questions, adding up code lengths for each distinct character instead of multiplying by its frequency.
• Writing RLE pairs as value-then-frequency.
• Shifting the bits but not saying what happens to the value when the question asks for the effect.

Frequently asked questions

How large will numbers be in 9210 conversion questions? The specification uses decimal values from 0 to 255, so at most 8 bits or two hex digits (FF).

What if my Huffman tree is different from the mark scheme? Huffman trees can differ when frequencies tie. If you always join the two lowest totals, your codes will be valid and the total number of bits will be the same.