Understand how bitmap images are represented using pixels, colour depth and resolution, how analogue sound is sampled, and the effect of sample rate, resolution and metadata on quality and file size.

Number of pixels $= 800 \times 600 = 480\,000$. Bits per pixel $= 24$, so total bits $= 480\,000 \times 24 = 11\,520\,000$ bits.

Convert to bytes: $11\,520\,000 / 8 = 1\,440\,000$ bytes.

Convert to mebibytes: $1\,440\,000 / 1\,048\,576 \approx 1.37$ MiB.

Markers award marks for the pixel count, multiplying by the colour depth, dividing by 8 for bytes, and dividing by $2^{20}$ for MiB. Carrying an early slip correctly still earns the method marks.

File size in bits $= \text{sample rate} \times \text{sample resolution} \times \text{duration} = 44\,100 \times 16 \times 30 = 21\,168\,000$ bits.

Stereo records two channels, so the file size is multiplied by 2, giving $42\,336\,000$ bits (twice the size). Quality, in the sense of frequency range and amplitude precision per channel, is unchanged because sample rate and resolution are the same; the benefit of stereo is spatial separation of the sound, not higher fidelity per channel.

Markers reward the correct mono calculation, doubling for stereo, and recognising that the per-channel sample rate and resolution (so the captured quality) are unchanged.