WJEC Eduqas Eduqas 2021-style practice: Convert the hexadecimal number 3C to denary and to binary, and convert the denary number 214 to hexadecimal, showing your method.

Hex to denary (up to 2): 3C has digits 3 and C (=12). Value = 3 × 16 + 12 = 48 + 12 = 60. Hex to binary (up to 1): each hex digit maps to four bits, so 3 = 0011 and C = 1100, giving 0011\,1100. Denary to hex (up to 2): 214 16 = 13 remainder 6. 13 = D and 6 = 6, so reading quotient then remainder gives D6. Check: 13 × 16 + 6 = 208 + 6 = 214. Markers reward the place-value method, the four-bit grouping for binary, and the divide-by-16 method with correct letter digits.

EnglandComputer ScienceSyllabus dot point

How are numbers represented in binary and hexadecimal, including negatives, and how is binary arithmetic performed?

Data representation of numbers: converting between binary, denary and hexadecimal, representing negative numbers with sign and magnitude and two's complement, binary addition and subtraction, and detecting overflow.

An Eduqas Component 2 answer on number representation: converting between binary, denary and hexadecimal, representing negative numbers with sign and magnitude and two's complement, binary addition and subtraction, and detecting overflow.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

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

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Quick answer

A binary digit (bit) has place values that are powers of two. To convert binary to denary, add the place values where there is a 1; to convert denary to binary, repeatedly subtract the largest fitting power of two. Hexadecimal is base 16 (digits $0$ to $9$ then $\text{A}$ to $\text{F}$ ), and one hex digit equals exactly four bits, so it is a compact way to write binary. Negative numbers use sign and magnitude (the most significant bit is a sign, giving two zeros and an 8-bit range $-127$ to $+127$ ) or two's complement (invert all bits and add 1, giving one zero and an 8-bit range $-128$ to $+127$ ), which is preferred because addition and subtraction work directly. Binary addition carries when a column sums to $2$ or more; subtraction is done by adding the two's complement. Overflow occurs when the result is too large for the available bits, detected when the carry into the sign bit differs from the carry out of it.

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What this dot point is asking
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What this dot point is asking

Eduqas wants fluent conversion between binary, denary and hexadecimal, representation of negative numbers in sign and magnitude and two's complement, binary addition and subtraction, and detection of overflow. This is a calculation-heavy dot point assessed without a calculator, so shown working is essential.

The answer

Binary, denary and hexadecimal

Key fact

In an 8-bit number the place values are $128, 64, 32, 16, 8, 4, 2, 1$ . Binary to denary: sum the place values where the bit is $1$ , so $0110\,1010 = 64 + 32 + 8 + 2 = 106$ . Denary to binary: repeatedly take the largest power of two that fits and set that bit. Hexadecimal is base 16 with digits $0$ to $9$ then $\text{A}=10$ through $\text{F}=15$ ; each hex digit maps to exactly four bits, so $1011\,0110$ groups into $1011 = \text{B}$ and $0110 = 6$ , giving $\text{B6}$ . Hex to denary uses powers of $16$ : $\text{B6} = 11 \times 16 + 6 = 182$ . Hexadecimal is used because it is far shorter to read and write than binary while mapping cleanly onto it.

Representing negative numbers

Key fact

Sign and magnitude uses the most significant bit as a sign ( $0$ positive, $1$ negative) and the rest as the magnitude, so in 8 bits $1000\,0101 = -5$ ; it is simple but has two zeros ( $+0$ and $-0$ ), an 8-bit range of $-127$ to $+127$ , and awkward arithmetic. Two's complement makes the most significant bit a negative weight ( $-128$ in 8 bits): to negate a number, invert every bit and add 1. So $+5 = 0000\,0101 \to 1111\,1010 + 1 = 1111\,1011 = -5$ . Two's complement has a single zero, an 8-bit range of $-128$ to $+127$ , and, crucially, ordinary binary addition and subtraction work directly on it, which is why processors use it.

Binary arithmetic and overflow

Convert and add in two's complement

Work out $0101\,1010$ in denary, then compute $90 - 34$ in 8-bit two's complement.

step 1: Binary to denary

$0101\,1010 = 64 + 16 + 8 + 2 = 90$ .

step 2: Represent $-34$ in two's complement

$+34 = 0010\,0010$ . Invert: $1101\,1101$ . Add 1: $1101\,1110$ . So $-34 = 1101\,1110$ .

step 3: Add $90 + (-34)$

  0101 1010   (+90)
+ 1101 1110   (-34)
-----------
  0011 1000   (final carry out of bit 8 discarded)

step 4: Interpret

$0011\,1000 = 32 + 16 + 8 = 56$ , and indeed $90 - 34 = 56$ . Subtraction was performed as addition of the two's complement, and the carry out of the top bit was discarded. The carry into and out of the sign bit match, so there is no overflow.

Examples in context

Memory addresses and colour values are written in hexadecimal because it is compact (the colour #FF8000 is three bytes). Two's complement is used inside every processor so one adder circuit can do both addition and subtraction, linking this directly to the logic-gate adders from Component 1. Overflow detection matters in real systems: an unchecked overflow can crash software or corrupt data. This dot point is the foundation for floating-point representation (next), which uses two's complement for both its mantissa and exponent.

Try this

Q1. Convert $1100\,0110$ to denary. [1 mark]

Cue. $128 + 64 + 4 + 2 = 198$ .

Q2. Represent $-12$ in 8-bit two's complement. [2 marks]

Cue. $+12 = 0000\,1100$ ; invert to $1111\,0011$ ; add 1 to get $1111\,0100$ .

Q3. Convert the denary number $174$ to hexadecimal. [1 mark]

Cue. $174 \div 16 = 10$ remainder $14$ , giving $\text{A}$ then $\text{E}$ , so $\text{AE}$ (or $1010\,1110 \to \text{AE}$ ).

Exam-style practice questions

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

Eduqas 20195 marksUsing 8-bit two's complement, represent

-18

in binary, and add it to the 8-bit two's complement representation of

+45

, showing your working and interpreting the result.

Show worked answer →

Represent $-18$ (up to 3): start with $+18 = 0001\,0010$ . Invert all bits: $1110\,1101$ . Add 1: $1110\,1110$ . So $-18 = 1110\,1110$ .

$+45 = 0010\,1101$ .

Add (up to 2):

  0010 1101   (+45)
+ 1110 1110   (-18)
-----------
  0001 1011   (carry out of the 8th bit is discarded)

The 8-bit result is $0001\,1011 = +27$ , correct since $45 + (-18) = 27$ . Markers reward the correct two's complement of $-18$ (invert and add 1), the binary addition, and discarding the final carry to read $+27$ .

Eduqas 20215 marksConvert the hexadecimal number

\text{3C}

to denary and to binary, and convert the denary number

214

to hexadecimal, showing your method.

Show worked answer →

Hex to denary (up to 2): $\text{3C}$ has digits $3$ and $\text{C}$ ( $=12$ ). Value $= 3 \times 16 + 12 = 48 + 12 = 60$ .

Hex to binary (up to 1): each hex digit maps to four bits, so $3 = 0011$ and $\text{C} = 1100$ , giving $0011\,1100$ .

Denary to hex (up to 2): $214 \div 16 = 13$ remainder $6$ . $13 = \text{D}$ and $6 = 6$ , so reading quotient then remainder gives $\text{D6}$ . Check: $13 \times 16 + 6 = 208 + 6 = 214$ . Markers reward the place-value method, the four-bit grouping for binary, and the divide-by-16 method with correct letter digits.

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Sources & how we know this

WJEC Eduqas GCE AS/A Level Computer Science specification (from 2015) — Eduqas (2015)