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9618 · 1.2.1

Multimedia - Graphics — practice questions

Practice and worked examples for 9618 Multimedia - Graphics. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

An uncompressed bitmap image has dimensions of 1024 x 768 pixels and a colour depth of 16 bits. Calculate the file size in megabytes (MB). Show your working.

Show solution outline
  1. Calculate total pixels: 1024×768=786,4321024 \times 768 = 786,432 pixels
  2. Calculate total bits: Total pixels ×\times Colour Depth 786,432×16=12,582,912786,432 \times 16 = 12,582,912 bits
  3. Convert bits to bytes: Divide by 8 12,582,912÷8=1,572,86412,582,912 \div 8 = 1,572,864 bytes
  4. Convert bytes to kilobytes (KB): Divide by 1024 1,572,864÷1024=1,5361,572,864 \div 1024 = 1,536 KB
  5. Convert kilobytes to megabytes (MB): Divide by 1024 1,536÷1024=1.51,536 \div 1024 = 1.5 MB

Answer: 1.5 MB

Worked example 2

An image is stored with a colour depth that allows for 4096 different colours. What is the colour depth in bits?

Show solution outline
  1. Understand the relationship: The number of colours is determined by 2colour depth2^{\text{colour depth}}. We need to find the colour depth (let's call it dd).
  2. Set up the equation: 2d=40962^d = 4096
  3. Solve for d: We need to find the power to which 2 must be raised to get 4096. This is the logarithm base 2. d=log2(4096)d = \log_2(4096)
  4. Calculation: You can solve this by recognising powers of 2: 210=10242^{10} = 1024 211=20482^{11} = 2048 212=40962^{12} = 4096

Answer: The colour depth is 12 bits.