AS · Practice questions · Two's complement and binary arithmetic

Invert and add one.

Six original Cambridge-style questions on signed binary integers and binary arithmetic, with worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

The sign bit is negative.

Two's complement: leftmost place value is −128 in 8 bits; range −128 to +127.
Negate: invert every bit, then add 1 (inverting alone gives one's complement).
Subtract: A − B = A + (−B); ignore a carry out of the leftmost bit.
Overflow: two positives give a negative, or two negatives give a positive.

Convert

[2 marks]

Convert the denary number −77 to 8-bit two's complement. Show your working.

+77 = 01001101, inverted 10110010. ✓
Add 1: 10110011. ✓

Convert

[1 marks]

The 8-bit two's complement integer 10101010 is stored. Give its denary value.

−128 + 32 + 8 + 2 = −86. ✓

Calculate

[3 marks]

Calculate 38 − 91 using 8-bit two's complement arithmetic. Show your working and give the answer in binary and denary.

38 = 00100110; −91 = 10100101 (01011011 inverted is 10100100, plus 1). ✓
00100110 + 10100101 = 11001011. ✓
11001011 = −128 + 64 + 8 + 2 + 1 = −53. ✓

Explain

[3 marks]

The 8-bit two's complement integers 01011010 and 01000110 are added. Give the 8-bit result and explain what has happened.

Result 10100000. ✓
The true answer is 90 + 70 = 160, which is above the maximum of +127. ✓
Overflow: two positive numbers have given a negative result (−96). ✓

Explain

[3 marks]

State the range of integers that can be stored in 8 bits using (a) one's complement and (b) two's complement, and explain why the ranges differ.

(a) −127 to +127. ✓
(b) −128 to +127. ✓
One's complement has two representations of zero (00000000 and 11111111), wasting one pattern; two's complement has only one zero, so it can represent one extra negative number. ✓

Analysis

[2 marks]

A 16-bit register holds the hexadecimal value FF9C, which is a two's complement integer. Give its denary value and show how you found it.

FF9C = 1111 1111 1001 1100; it is negative because the leftmost bit is 1. ✓
Invert to 0000 0000 0110 0011 and add 1: 0000 0000 0110 0100 = 100, so the value is −100. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.