Understand the bit and byte, the units of information capacity, binary and decimal prefixes (kibi versus kilo), and how the number of bits limits the range of values that can be represented.

A focused answer to AQA A-Level Computer Science 4.5.8, covering the bit and the byte, the units of information capacity, binary prefixes (kibi, mebi) versus decimal prefixes (kilo, mega), and how the number of bits limits the range of values.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

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

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What this dot point is asking
Bits and bytes
Units of capacity and prefixes
How many bits, how many values

What this dot point is asking

AQA wants you to define a bit and a byte, state the units of information capacity, distinguish binary prefixes (kibi, mebi) from decimal prefixes (kilo, mega), and explain how the number of bits limits the values that can be stored.

Bits and bytes

A byte became the standard unit because 8 bits is enough to hold a single character in the ASCII or extended character sets, and computer architectures address memory in whole bytes rather than individual bits. Word size (16, 32 or 64 bits) describes how many bits the processor handles at once, but storage is still counted in bytes and their multiples. Memory addresses, file sizes and data type widths are therefore all expressed in bytes, which keeps measurements consistent across very different machines.

It helps to keep the hierarchy straight: a single bit is the smallest unit and carries one yes or no decision; a nibble is 4 bits and maps to one hex digit; a byte is 8 bits and holds one character or a small integer; larger quantities are counted in multiples of bytes using prefixes. Bits are also the unit for network speed (megabits per second), which is why an 8 Mbps connection downloads at about 1 megabyte per second once you divide by 8.

Units of capacity and prefixes

This distinction explains why a "1 TB" drive (using $10^{12}$ bytes) appears smaller when an operating system reports capacity using powers of two: dividing the same byte count by $2^{40}$ gives roughly $0.91$ TiB. The mismatch is not faulty hardware; it is two valid but different counting conventions.

How many bits, how many values

The number of bits available sets the range of distinct values: $n$ bits give $2^n$ combinations. So 1 bit gives 2 values, 4 bits give 16, 8 bits give 256, and 16 bits give 65 536. To represent $k$ distinct values you need at least $\lceil \log_2 k \rceil$ bits. This is why widening a data type (more bits) increases the range it can hold, and why designers must size fields carefully: too few bits and values overflow; too many and storage is wasted.

Size a binary field for 1000 students and convert a capacity

Target: choose the field width, then convert a storage figure between units.

step 1: Find the minimum bits for 1000 unique IDs

We need $2^n \ge 1000$ . $2^9 = 512$ (too small), $2^{10} = 1024$ (enough). So 10 bits per student ID.

step 2: Total the storage for the IDs

$1000 \text{ students} \times 10 \text{ bits} = 10\,000$ bits. Divide by 8 for bytes: $10\,000 / 8 = 1250$ bytes.

step 3: Express the bytes in binary prefix form

$1250$ bytes is $1250 / 1024 \approx 1.22$ KiB. In decimal prefix form it would be $1.25$ kB, a small but real difference.

step 4: State the result

Each ID needs 10 bits; storing 1000 of them takes 1250 bytes, about 1.22 KiB. The two prefix conventions give slightly different KiB and kB figures for the same byte count.

Exam-style practice questions

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

AQA 20193 marksA file is 3 mebibytes (MiB) in size. Calculate the number of bytes this represents and explain why a file manager might report a different figure to the value printed on the storage device packaging.

Show worked answer →

A mebibyte is $2^{20} = 1\,048\,576$ bytes, so $3 \text{ MiB} = 3 \times 1\,048\,576 = 3\,145\,728$ bytes.

Device packaging usually quotes capacity using decimal (SI) prefixes, where 1 megabyte $= 10^6 = 1\,000\,000$ bytes, while many file managers compute and label sizes using the binary powers of two ( $2^{20}$ ). Because $2^{20}$ is larger than $10^6$ , the same number of bytes appears as a smaller figure when divided using binary prefixes, so a "1 TB" drive shows as roughly 0.91 TiB.

Markers reward the correct byte calculation ( $3\,145\,728$ ) and a clear explanation that the two prefix systems use different bases (powers of ten versus powers of two).

AQA 20212 marksA system needs to assign a unique binary code to each of 600 different products. State the minimum number of bits required and justify your answer.

Show worked answer →

With $n$ bits you can represent $2^n$ distinct codes. $2^9 = 512$ , which is fewer than 600, so 9 bits are insufficient. $2^{10} = 1024$ , which is at least 600, so 10 bits are required.

Equivalently, the minimum is $\lceil \log_2 600 \rceil = 10$ bits.

Markers award one mark for the answer (10 bits) and one for the justification that $2^9 < 600 \le 2^{10}$ .

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Sources & how we know this

AQA A-level Computer Science (7517) specification — AQA (2015)