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Practise 34 exam-style questions on Data storage units and file size calculations for Cambridge IGCSE Computer Science (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Write the number of bytes in one tebibyte as a power of 2.
2 to the power 40 bytes.
1024 is 2 to the power 10, and a tebibyte is four steps up from a byte (KiB, MiB, GiB, TiB). So it is 1024 × 1024 × 1024 × 1024 = 2 to the power (10 × 4) = 2 to the power 40 bytes.
2 to the power 40
accept 2^40
Identify which is larger, 3000 KiB or 3 MiB. Show your working.
3 MiB = 3 × 1024 = 3072 KiB.
3072 KiB is more than 3000 KiB, so 3 MiB is larger.
This question tests whether you use 1024. If you think 1 MiB is 1000 KiB, the two look equal. Converting to the same unit with 1024 shows 3 MiB is 72 KiB larger.
3 MiB = 3 × 1024 = 3072 KiB
working
3 MiB is larger (by 72 KiB)
An image has a resolution of 256 × 256 pixels. It uses 16 different colours, with the smallest possible colour depth.
Calculate the file size of the image in kibibytes (KiB). Show your working.
16 colours need 4 bits per pixel, because 2 to the power 4 = 16.
File size = 256 × 256 × 4 = 262 144 bits.
262 144 ÷ 8 = 32 768 bytes.
32 768 ÷ 1024 = 32 KiB.
First find the colour depth from the number of colours, then use width × height × colour depth. 4 bits is half a byte, so the image needs 65 536 ÷ 2 = 32 768 bytes, which is 32 KiB.
Colour depth = 4 bits (2 to the power 4 = 16)
256 × 256 × 4 = 262 144 bits / 32 768 bytes
follow through their colour depth
32 KiB
must use 1024, not 1000
A digital camera takes photographs with a resolution of 4096 × 2048 pixels and a colour depth of 24 bits. The camera's memory card has 6 GiB of free space.
Calculate how many photographs can be stored in the free space. Show your working.
24 bits = 3 bytes per pixel.
One photograph = 4096 × 2048 × 3 = 25 165 824 bytes.
25 165 824 ÷ 1024 ÷ 1024 = 24 MiB (or by cancelling: 4 × 2 × 3 = 24 MiB).
Free space = 6 × 1024 = 6144 MiB.
6144 ÷ 24 = 256 photographs.
Work out the size of one photograph, put it in the same unit as the free space, then divide. Cancelling the 1024s (4096 = 4 × 1024 and 2048 = 2 × 1024) means you never need to handle the 8-digit number. Using 1000 anywhere gives a wrong answer.
4096 × 2048 × 3 (bytes) / 4096 × 2048 × 24 (bits)
working for one photograph
One photograph = 24 MiB
accept 25 165 824 bytes or 24 576 KiB
6 GiB = 6144 MiB
accept the free space in the same unit as their photograph size
6144 ÷ 24 = 256 photographs
follow through their values
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