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AQA GCSE Computer Science (8525) · Fundamentals of data representation
Mini-Lesson

Fundamentals of data representation

This mini-lesson covers AQA 3.3 — Fundamentals of data representation: number bases (binary, denary, hexadecimal), converting between them, binary addition & shifts, units of data, characters (ASCII & Unicode), images and sound, and compression.

number bases units, images & sound compression computers store everything as binary — 1s and 0s

Work through each screen, answer the questions as you go (lots of quick binary and hex calculations) and collect ⭐ stars. Press Start when you're ready.

Number bases

Binary, denary & hexadecimal

Computers store all data as binary. AQA needs you to know three number bases:

  • Binarybase 2. Only two digits: 0 and 1. Each digit is a bit.
  • Denary (decimal) — base 10. The everyday numbers we count in, using digits 0–9.
  • Hexadecimalbase 16. Uses digits 0–9 then letters A–F, where A=10, B=11, C=12, D=13, E=14, F=15.

Why hex? One hex digit represents exactly 4 bits (a nibble), so hex is a short, human-friendly way to write long binary numbers — 2 hex digits = 1 byte (8 bits).

Number bases

Binary ↔ denary (8-bit place values)

Each bit in an 8-bit number has a place value. To convert binary to denary, add up the place values wherever there is a 1:

128 64 32 16 8 4 2 1 01101101 01101101 = 64+32+8+4+1 = 109
Place values 128 64 32 16 8 4 2 1 — add each column that holds a 1.

Denary → binary: work left to right, subtracting the biggest place value you can each time (put a 1), else put a 0. E.g. 45 = 32+8+4+1 → 00101101.

Calculate

Your turn — binary to denary

1Convert the binary number 00101101 to denary. (Place values: 128 64 32 16 8 4 2 1.)
Hint: the 1s sit under 32, 8, 4 and 1 → 32+8+4+1.
Number bases

Working with hexadecimal

Each hex digit is worth 4 bits (a nibble), so you can convert binary ↔ hex one nibble at a time, and hex ↔ denary using place values 16 and 1.

  • Hex → denary: multiply the left digit by 16, add the right digit. E.g. 2F = (2×16) + 15 = 32 + 15 = 47. (F = 15.)
  • Denary → hex: divide by 16. E.g. 156 = 9 remainder 12 → 9 and C → 9C.
  • Binary ↔ hex: split the 8 bits into two nibbles. E.g. 1001 1100 → 9 and C → 9C.

Remember the letters: A=10, B=11, C=12, D=13, E=14, F=15. One byte (8 bits) always fits in exactly 2 hex digits.

Calculate

Your turn — hex to denary

2Convert the hexadecimal number 2F to denary. (Left digit is worth 16s, right digit is worth 1s; remember F = 15.)
Hint: (2 × 16) + 15 = 32 + 15.
Binary arithmetic

Binary addition

Add binary the same way you add denary, but you carry whenever a column reaches 2. The rules are:

  • 0 + 0 = 0
  • 0 + 1 = 1
  • 1 + 1 = 0 carry 1
  • 1 + 1 + 1 = 1 carry 1
# 00110101 + 00010011 00110101 (= 53) + 00010011 (= 19) = 01001000 (= 72)

Overflow: if the answer needs more than 8 bits (a carry out of the leftmost column), an overflow error occurs — the byte can't hold the result.

Calculate

Your turn — add in binary

3Add the binary numbers 00110101 + 00010011. Give the denary value of the result. (The binary result is 01001000.)
Hint: 00110101 = 53 and 00010011 = 19, so 53 + 19.
Binary arithmetic

Binary shifts

A logical shift moves every bit left or right by a number of places, filling the gap with 0s:

  • Left shift by 1 place multiplies by 2. Shift by n places → × 2n.
  • Right shift by 1 place divides by 2 (whole-number division). Shift by n places → ÷ 2n.
00001010 = 10 00010100 = 20 10 20 left shift by 1 → × 2

Careful with right shifts: bits shifted off the right-hand end are lost, so information (odd amounts / precision) can be lost when dividing.

Calculate

Your turn — shift left

4The binary number 00001010 (= 10 in denary) is given a left shift of 1 place. What is the denary value of the result?
Hint: a left shift of 1 place multiplies by 2, so 10 × 2. (Result 00010100 = 16+4 = 20.)
Units of data

Bits, bytes and beyond

Data size is measured in units built up from the single bit:

  • Bit — a single 0 or 1.
  • Nibble4 bits.
  • Byte8 bits.

Bigger units go up in thousands (AQA uses decimal prefixes, powers of 1000):

1 kB = 1000 B · 1 MB = 1000 kB1 GB = 1000 MB · 1 TB = 1000 GB — each step ×1000

Watch out: AQA GCSE uses 1 kB = 1000 bytes (kilobyte, megabyte, gigabyte, terabyte), not 1024. Learn the order: bit → nibble → byte → kB → MB → GB → TB.

Quick check

How big is a byte?

?How many bits are there in one byte?
Characters

ASCII & Unicode

Text is stored by giving each character a number (a character code). Two character sets matter:

  • ASCII — a 7-bit code, so it can represent 27 = 128 characters (enough for English letters, digits and punctuation).
  • Unicode — uses more bits, so it can represent a far larger set of characters, covering the world's languages and emoji.

Character codes: in ASCII, capital 'A' = 65 and lowercase 'a' = 97. Letters are in order, so 'B' = 66, 'C' = 67, and so on. The gap between 'A' and 'a' is always 32.

Quick check

Which character set?

?Which character set can represent a much larger set of characters and languages (including emoji)?
Images

Bitmap images

A bitmap image is a grid of tiny dots called pixels. Two properties describe it:

  • Resolution — the number of pixels (width × height). More pixels = more detail.
  • Colour depth — the number of bits per pixel. Number of colours = 2colour depth (so 1 bit = 2 colours, 8 bits = 256 colours).
file size (bits) = W × H × colour depthe.g. 10 × 8 pixels at 3 bits = 240 bits

Trade-off: more pixels or more colours make a nicer image but a bigger file.

Calculate

Your turn — image file size

5A bitmap image is 10 pixels wide, 8 pixels high, with a colour depth of 3 bits. What is the file size in bits? (file size = width × height × colour depth.)
bits
Hint: 10 × 8 × 3.
Sound

Representing sound

Sound is analogue (a continuous wave). To store it, the computer takes samples — measuring the wave's height at regular intervals:

  • Sampling rate — how many samples per second are taken (measured in Hz). Higher rate = more accurate.
  • Sample resolution — the number of bits per sample. More bits = finer detail.
file size (bits) = rate × resolution × seconds(× number of channels for stereo)

Trade-off: a higher sampling rate or resolution gives better-quality sound but a larger file.

Compression

Lossy vs lossless

Compression makes a file smaller so it takes less storage and transfers faster. There are two types:

  • Lossy — permanently removes some data to shrink the file (you can't get it back). Quality drops, but files are much smaller. e.g. JPEG images, MP3 audio.
  • Losslessno data is lost; the original can be perfectly restored. e.g. run-length encoding (RLE) and ZIP.

RLE (a lossless method) replaces runs of the same value with a count + value, e.g. "AAAA" → "4A". Great for images with large areas of one colour.

Quick check

Which compression type?

?Which type of compression permanently loses some data to reduce the file size (used by JPEG and MP3)?
Sort it

Which number base?

Tap a description, then tap the number base it belongs to.

🔟 Binary (base 2)

🔢 Denary (base 10)

🅰️ Hexadecimal (base 16)

Match it

Match term to meaning

Tap a description on the left, then its matching term on the right.

Description
Term
Recap

The big ideas to know

Number bases: binary (base 2, 0/1) · denary (base 10, 0–9) · hex (base 16, 0–9 & A–F)

Convert: 8-bit place values 128 64 32 16 8 4 2 1; each hex digit = 4 bits (a nibble); 2 hex digits = 1 byte

Arithmetic: binary addition (carry; overflow if > 8 bits) · left shift ×2, right shift ÷2

Units: bit → nibble (4) → byte (8) → kB → MB → GB → TB (×1000 each)

Data: ASCII (7-bit, 128 chars) vs Unicode · image size = W×H×depth, colours = 2^depth · sound size = rate×resolution×seconds

Compression: lossy (removes data — JPEG/MP3) · lossless (no data lost — RLE/ZIP)

You've covered all of AQA 3.3 — number bases, conversion, binary arithmetic, units, characters, images, sound and compression. Press Finish to see your score.

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