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OCR H446 1.4.1 Binary, hexadecimal and denary

Part 2 of 11 · H446 1.4.1 · Data types

A number base changes how a value is written, not the quantity stored, and H446 1.4.1 expects fluent conversion in both directions. Built around colour channels and device identifiers, this worksheet keeps place values and nibble grouping visible so students convert between 8-bit binary, two-digit hexadecimal and denary by method rather than by memory.

Students will:

  • read an 8-bit pattern using the place values 128 down to 1
  • convert positive denary values to 8-bit binary and back
  • group bits into nibbles to move between binary and two-digit hexadecimal
  • explain why hexadecimal is easier for a person to read without changing what is stored
  • complete the conversion definitions from memory in a closed-book check

Inside: 7 explanation cells, 1 multiple-choice question, 1 fill-in-the-blanks cell, 1 written answer and 6 number answers. 24 marks, about 35 to 45 minutes.

Series: H446 1.4.1 · Data types, part 2 of 11.

Shared by Coding PathwayVerified teacher

  • 16 cells
  • About 45 minutes
  • CC BY-SA 4.0
  • Shared 31 Aug 2026
  • Updated 3 Sept 2026

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The whole resource, exactly as a class sees it. Answers and marking are held back.

Binary, hexadecimal and denary

Digital artwork stores colour channels and device identifiers as bit patterns. The number base changes how a person writes a value, not the underlying quantity.

By the end, you will be able to

  • read an 8-bit value using place values
  • convert positive integers between denary, binary and hexadecimal
  • use one hexadecimal digit for each four-bit nibble
  • explain why hexadecimal is useful without claiming it increases storage

Reactivate: powers of two double from right to left.

Read the representation

Digital electronic systems can reliably distinguish two physical states, so binary uses two symbols, 0 and 1, to represent data. Binary place values are powers of two. An 8-bit pattern has place values 128, 64, 32, 16, 8, 4, 2 and 1. Hexadecimal is base 16 and uses sixteen symbols: 0–9 then A–F. One hexadecimal digit corresponds exactly to four binary bits (a nibble).

Worked conversion: 10110110 has set place values 128 + 32 + 16 + 4 + 2, so it is 182 in denary. Regroup the same bits as 1011 0110: 1011 is B and 0110 is 6, so the hexadecimal notation is B6. The quantity has not changed; only its written representation has.

One value written in binary, denary and hexadecimal Binary place-value model 1286432168421 10110110 1011₂ = B₁₆0110₂ = 6₁₆ 128 + 32 + 16 + 4 + 2 = 182₁₀ = B6₁₆

The same bit pattern can mean different things under different interpretations; the representation needs a stated purpose.

Worked model: binary to denary

Eight-bit place-value modelPlace valueBit128643216842110110110128 + 32 + 16 + 4 + 2 = 182

Only place values with a 1 contribute. For 10110110₂, add 128 + 32 + 16 + 4 + 2 = 182₁₀.

Sense check: the leftmost bit is 1, so an unsigned 8-bit value must be at least 128. The answer 182 is plausible.

Guided method

For each practice value: write place values → circle the 1 columns → add them → check the range 0–255.

Number systems3 marks

Convert each 8-bit unsigned binary value to denary.

a)What is 01000110 in denary?

Binary
01000110
DenaryNot answered

b)What is 01011110 in denary?

Binary
01011110
DenaryNot answered

c)What is 11001111 in denary?

Binary
11001111
DenaryNot answered
Number systems3 marks

Convert each positive denary value to 8-bit binary.

a)Convert 210 to 8-bit binary.

Binary

b)Convert 126 to 8-bit binary.

Binary

c)Convert 53 to 8-bit binary.

Binary

Use nibbles for hexadecimal

Split an 8-bit value into two groups of four. Convert each nibble independently. For example, 1010 0111 becomes A7. This is a regrouping of the same bits, not a change to the stored value.

Worked model: preserve the nibbles

Convert 1101 0110₂ one group at a time:

  • 1101₂ = 13₁₀ = D₁₆;
  • 0110₂ = 6₁₀ = 6₁₆;
  • therefore 11010110₂ = D6₁₆.

Reverse the check: D → 1101 and 6 → 0110, which reconstructs the original eight bits. The hexadecimal form is shorter for a person to read; it does not store extra information.

Number systems2 marks

Convert each positive denary value to two-digit hexadecimal.

a)Convert 16 to hexadecimal.

Hexadecimal

b)Convert 106 to hexadecimal.

Hexadecimal
Number systems2 marks

Convert each two-digit hexadecimal value to denary.

a)What is 5C in denary?

Hexadecimal
5C
DenaryNot answered

b)What is F4 in denary?

Hexadecimal
F4
DenaryNot answered
Number systems3 marks

Convert each 8-bit binary pattern to two hexadecimal digits.

a)Convert 11001000 to hexadecimal, one nibble at a time.

Binary
1100
1000
Hexadecimal

b)Convert 00011011 to hexadecimal, one nibble at a time.

Binary
0001
1011
Hexadecimal

c)Convert 11101000 to hexadecimal, one nibble at a time.

Binary
1110
1000
Hexadecimal
Number systems3 marks

Convert each two-digit hexadecimal value to 8-bit binary without place-value prompts.

a)Convert CD to binary, one nibble at a time.

Hexadecimal
C
D
Binary

b)Convert 9F to binary, one nibble at a time.

Hexadecimal
9
F
Binary

c)Convert D2 to binary, one nibble at a time.

Hexadecimal
D
2
Binary
Multiple choice1 mark

Why is hexadecimal often used when people read long binary values?

  • AIt stores more information in the same bits.
  • BIt makes arithmetic exact when binary would not.
  • CIt is a compact notation with one digit per four bits.
  • DProcessors execute hexadecimal instead of binary.
Written answer3 marks

A colour channel is stored as 11010110. Give its hexadecimal value and denary value, then explain why the hexadecimal form is convenient for a developer.

Show the two nibbles and use 8-bit place values.

Students type their answer here.

Closed-book checkpoint

Complete each sentence from memory. There is no answer bank and correctness is held for teacher review.

Fill in the blanks4 marks
Binary place values are completion 1. Four bits form a completion 2. One hexadecimal digit corresponds to completion 3 binary bits. Rewriting a bit pattern in hexadecimal leaves its stored quantity completion 4.

Closed-book retrieval

Explain: (1) why 10110110₂ is at least 128; (2) why D6₁₆ maps to exactly eight bits; (3) why hexadecimal is convenient but does not change capacity.