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Cambridge IGCSE Computer Science (Paper 1) Smart practice A mixed round aimed at your weak spots
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Binary, denary and hexadecimal conversions

Practise 36 exam-style questions on Binary, denary and hexadecimal conversions for Cambridge IGCSE Computer Science (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

  • 36 questions
  • 1–6 marks each
  • Instant marking
  • No sign-up needed
20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

Convert the hexadecimal number 2F into denary.

Model answer

2 × 16 + 15 × 1 = 32 + 15 = 47

Explanation

The left digit is worth 16s and the right digit 1s. F stands for 15, so 2F = 32 + 15 = 47.

Mark scheme
  1. 1 mark

    47

Question 2

2 marks

Convert the denary number 200 into hexadecimal.

Show your working.

Model answer

200 ÷ 16 = 12 remainder 8
12 = C
So 200 = C8

Explanation

12 sixteens make 192 and 8 is left over, so the digits are 12 and 8. The 12 must be written as C. An alternative is to convert to binary first (1100 1000) and turn each nibble into one hexadecimal digit.

Mark scheme
  1. 1 mark

    Working, e.g. 200 ÷ 16 = 12 remainder 8 or 200 = 1100 1000 in binary

  2. 1 mark

    C8

    accept lower case (c8)

Question 3

3 marks

A microcontroller in a weather station stores a 12-bit reading. A technician's screen shows the reading as the hexadecimal number 3B7.

Convert 3B7 into 12-bit binary.

Model answer

3 = 0011
B = 11 = 1011
7 = 0111
3B7 = 0011 1011 0111

Explanation

Three hexadecimal digits give three nibbles, which is 12 bits. Convert each digit using place values 8, 4, 2, 1 and keep the leading zeros so that every nibble has four bits.

Mark scheme
  1. 1 mark

    0011 (for 3)

    one mark for each correct nibble in the correct position; accept with or without spaces; correct answer with no working scores full marks

  2. 1 mark

    1011 (for B)

  3. 1 mark

    0111 (for 7)

    final answer 0011 1011 0111

Question 4

4 marks

Four values are written in different number systems.

P: binary 1001 1010
Q: hexadecimal 9C
R: denary 155
S: hexadecimal 97

Convert P, Q and S into denary. Then write the letters P, Q, R and S in order from smallest to largest.

Model answer

P: 128 + 16 + 8 + 2 = 154
Q: 9 × 16 + 12 = 144 + 12 = 156
S: 9 × 16 + 7 = 144 + 7 = 151
Order from smallest to largest: S (151), P (154), R (155), Q (156)

Explanation

Values in different bases can only be compared once they are all in the same base, and denary is the easiest. All four values are close together, so each conversion must be exact - an error of one or two changes the order.

Mark scheme
  1. 1 mark

    P = 154

    correct answer with no working scores full marks

  2. 1 mark

    Q = 156

  3. 1 mark

    S = 151

  4. 1 mark

    Order: S, P, R, Q

    allow follow-through from incorrect conversions

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

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