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EnglandComputer ScienceSyllabus dot point

How are text, images and sound represented as binary, and what determines the file size and quality?

Representing text, images and sound: character sets (ASCII and Unicode), bitmap images with resolution, colour depth and the file-size calculation, and sampled sound with sample rate, bit depth and the file-size calculation.

An Eduqas Component 2 answer on representing text, images and sound: the ASCII and Unicode character sets, bitmap images with resolution and colour depth and the file-size calculation, and sampled sound with sample rate and bit depth and the file-size calculation.

Generated by Claude Opus 4.814 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to explain how text, images and sound are represented in binary: the character sets (ASCII and Unicode), bitmap images (resolution, colour depth and the file-size calculation), and sampled sound (sample rate, bit depth and the file-size calculation). The image and sound file-size calculations are examined frequently.

The answer

Representing text: ASCII and Unicode

Representing images: bitmaps and file size

Representing sound: sampling and file size

Examples in context

These representations underlie every file you use: a text document (character set), a photo (bitmap resolution and colour depth) and a music track (sample rate and bit depth). The file-size calculations explain why a high-resolution photo or a high-fidelity audio file is large, and why compression (the next dot point) is so important for storage and transmission. Unicode is why modern software can display any language and emoji. The binary skills from the number-systems dot point are exactly what is being counted in these file-size sums.

Try this

Q1. How many colours can a colour depth of 55 bits represent? [1 mark]

  • Cue. 25=322^5 = 32 colours.

Q2. Calculate the file size in bytes of a 100×100100 \times 100 pixel image at 88 bits per pixel. [2 marks]

  • Cue. 100×100×8=80000100 \times 100 \times 8 = 80\,000 bits; ÷8=10000\div 8 = 10\,000 bytes.

Q3. Define sample rate. [1 mark]

  • Cue. The number of samples (amplitude measurements) taken per second, measured in hertz (Hz).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20195 marksA bitmap image is 200200 pixels wide and 150150 pixels high, using a colour depth of 44 bits per pixel. Calculate the file size in bytes (ignoring any header), showing your working, and state what is meant by colour depth.
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Colour depth (up to 1 mark): the number of bits used to represent the colour of each pixel; nn bits gives 2n2^n possible colours.

File-size calculation (up to 4 marks): number of pixels =200×150=30000= 200 \times 150 = 30\,000. Bits =30000×4=120000= 30\,000 \times 4 = 120\,000 bits. Bytes =120000÷8=15000= 120\,000 \div 8 = 15\,000 bytes.

Markers reward the bits-per-pixel definition of colour depth, multiplying width by height by colour depth for the total bits, and dividing by 88 to get 1500015\,000 bytes. A common error is forgetting to divide by 88.

Eduqas 20216 marksExplain how sound is represented digitally by sampling, define sample rate and bit depth, and calculate the file size of a 1010 second mono recording sampled at 4410044\,100 Hz with a bit depth of 1616 bits.
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Sampling (up to 2 marks): the analogue sound wave's amplitude is measured (sampled) at regular intervals and each measurement is stored as a binary number; more frequent, finer samples reproduce the wave more accurately.

Definitions (up to 1 mark): sample rate is the number of samples taken per second (in Hz); bit depth is the number of bits used to store each sample.

File-size calculation (up to 3 marks): samples =44100×10=441000= 44\,100 \times 10 = 441\,000. Bits =441000×16=7056000= 441\,000 \times 16 = 7\,056\,000 bits. Bytes =7056000÷8=882000= 7\,056\,000 \div 8 = 882\,000 bytes (mono, so one channel).

Markers reward the sample-the-amplitude-at-intervals description, correct definitions, and sample rate ×\times duration ×\times bit depth, divided by 88, giving 882000882\,000 bytes.

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