What is not stating overflow?

If the answer needs more bits than are available, you must say overflow has occurred; do not just drop the extra bit silently.

EnglandComputer ScienceSyllabus dot point

How do you add two binary numbers, and what does shifting the bits left or right do?

Binary addition of two 8-bit numbers including carrying and overflow, and binary shifts (left and right) and their effect of multiplying or dividing by powers of two.

An Eduqas GCSE Computer Science answer on adding two 8-bit binary numbers (with carrying and overflow) and performing left and right binary shifts, including how a shift multiplies or divides by powers of two.

Generated by Claude Opus 4.811 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
Binary addition
Overflow
Binary shifts
Try this

What this dot point is asking

Eduqas wants you to add two 8-bit binary numbers (carrying correctly and spotting overflow) and to carry out binary shifts left and right, explaining that a shift multiplies or divides the number by a power of two. Both are short, method-marked questions on Component 1, so showing the working is essential.

Binary addition

Add two 8-bit binary numbers

Add $01011101$ and $00101110$ as 8-bit numbers.

Step 1: Line them up and add from the right

01011101

+ \ 00101110

Working right to left:

1+0=1

;

0+1=1

;

1+1=10

(write

0

, carry

1

);

1+1+1=11

(write

1

, carry

1

);

1+0+1=10

(write

0

, carry

1

);

0+1+1=10

(write

0

, carry

1

);

1+0+1=10

(write

0

, carry

1

);

0+0+1=1

Step 2: Read off the result

01011101 + 00101110 = 10001011

Step 3: Check

$01011101 = 93$ and $00101110 = 46$ , and $93 + 46 = 139$ . The result $10001011 = 128 + 8 + 2 + 1 = 139$ . Correct, and it fits in 8 bits, so there is no overflow.

Overflow

Binary shifts

Shift a binary number both ways

Start with $00010100$ (denary $20$ ).

Step 1: Left shift by 1

Move every bit one place left, fill the right with a zero: $00101000$ . In denary this is $40$ , which is $20 \times 2^1 = 40$ . Correct.

Step 2: Right shift by 2 (from the original)

Move every bit two places right, fill the left with zeros, and drop the bits that fall off the right: $00010100$ becomes $00000101$ . In denary this is $5$ , which is $20 \div 2^2 = 5$ . Correct.

Step 3: Note the rounding

If a $1$ is shifted off the right-hand end, it is lost, so a right shift effectively rounds the result down (for example shifting $00000011 = 3$ right by one gives $00000001 = 1$ , that is $3 \div 2$ rounded down).

Try this

Q1. Add the 8-bit numbers $00111100$ and $00000110$ . [2 marks]

Cue. $01000010$ (that is $60 + 6 = 66$ ).

Q2. Perform a left shift of one place on $00001001$ and give the denary value before and after. [2 marks]

Cue. Before $9$ , after $00010010 = 18$ (multiplied by 2).

Q3. State the effect of a right shift of three places on a binary number. [1 mark]

Cue. It divides the number by $2^3 = 8$ (rounding down).

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 Component 1, 20223 marksAdd the two 8-bit binary numbers 01101011 and 00110110. Give your answer as an 8-bit number and state whether overflow has occurred.

Show worked answer →

Add the columns from the right, carrying a 1 whenever the column total reaches two.

01101011 + 00110110 = 10100001, with the carries propagating through the middle columns.

The result fits in 8 bits (the leading bit is part of the answer, not a carry out of bit 8), so there is no overflow.

Markers award method marks for correct carrying. If the sum had needed a 9th bit, you would state that overflow has occurred because the result is too large for 8 bits.

Eduqas Component 1, 20234 marksA program performs a left shift of two places on the 8-bit binary number 00000110. State the result, give its denary value before and after, and explain the effect of the shift.

Show worked answer →

Before: 00000110 is denary 6.

A left shift of two places moves every bit two columns to the left, filling the empty right-hand positions with zeros: 00011000.

After: 00011000 is denary 24.

Effect: a left shift of n places multiplies the number by 2 to the power n. Here a shift of two multiplies by 4, and 6 times 4 = 24, which matches (1 mark each for before, result, after, and the multiply-by-4 explanation).

Related dot points

Sources & how we know this

WJEC Eduqas GCSE Computer Science specification (from 2016) — Eduqas (2020)