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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 693 exam-style questions, each with its mark scheme and example answers. Print it, or answer it on paper and then open the mark schemes. Or browse the questions by topic.

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4 questions · 6 marks

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1Short recall1 markCharacter Sets Swap Remove

A computer can store a maximum of 80 bits for a text message. Each character is stored using 7-bit ASCII.

State the maximum number of whole characters that can be stored in the message.

[1 mark]
Mark scheme

Award one mark for:

  • 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Example answers and how they were marked
Full marks1/1

11

Why: Correct: 80 ÷ 7 = 11 remainder 3, so 11 whole characters can be stored (1).

No marks0/1

11.4

Why: Near miss: gives a decimal result rather than the number of whole characters; the mark scheme does not accept 11.4.

No marks0/1

12

Why: Rounds up to 12, but 12 characters would need 84 bits, more than the 80 available; the mark scheme does not accept 12.

2Short recall1 markCharacter Sets Swap Remove

ASCII uses 7 bits to store each character. A system has only 6 bits available to store each character.

State the number of different characters the system can represent.

[1 mark]
Mark scheme

Award one mark for:

  • 64 (1)
  • 2^6 (1)
  • 2 × 2 × 2 × 2 × 2 × 2 (1)

Allow 64 characters / sixty-four.

Allow 2 to the power of 6 for 2^6.

Do not accept 128, which is the number of characters that can be represented using 7 bits.

Do not accept 63, the largest value that can be represented using 6 bits, rather than the number of different characters.

Do not accept 12 (2 × 6), ie the number of bits multiplied by 2.

Example answers and how they were marked
Full marks1/1

64

Why: Correct: 2^6 = 64 different characters (1).

No marks0/1

128

Why: Scores 0: 128 is the number of characters that 7 bits can represent, not the 6 bits the system has; the Do not accept line applies.

No marks0/1

63

Why: Scores 0: 63 is the largest value 6 bits can hold, not the number of different characters; the Do not accept line applies.

3Describe2 marksCharacter Sets Swap Remove

A computer uses 7-bit ASCII to represent characters.

The character M has the ASCII code 77. Describe how the character M is represented in the computer.

[2 marks]
Mark scheme

Award up to two marks for a linked description such as:

  • The denary code 77 is converted to binary (1) to give the 7-bit pattern 1001101 (1)
  • The character M is represented by a unique ASCII code/number (1) that is stored as a 7-bit binary pattern (1)
  • 7 bits are used to store the code for M (1) so 128 different characters can be represented (1)
  • The binary pattern 1001101 is stored in memory (1) using one 7-bit code for each character (1)

Do not accept answers that only state the ASCII code 77, as this is given in the question.

Do not accept 01001101, as this is an 8-bit pattern.

Do not accept ASCII described as a programming language or a method of encryption.

Example answers and how they were marked
Full marks2/2

77 is converted to binary, giving 1001101. This is stored using 7 bits for the character M.

Why: Linked description: converts the code to binary (1) to give the 7-bit pattern 1001101 (1).

Some marks1/2

77 is changed into binary so it can be stored.

Why: Gives the conversion of 77 to binary (1) but does not give the 7-bit pattern 1001101 (0).

No marks0/2

ASCII is a programming language that stores letters as numbers.

Why: Describes ASCII as a programming language, which the mark scheme says is not creditworthy; no linked description of how M is represented.

4Explain2 marksImages Swap Remove

Two versions of the same photograph are displayed at the same size on a screen. Version A is 800 × 600 pixels. Version B is 1600 × 1200 pixels.

Explain why Version B can show more detail.

[2 marks]
Mark scheme

Award up to two marks for a linked explanation such as:

  • Version B has more pixels in the same area (1) so each pixel represents a smaller part of the image (1)
  • Version B has a higher pixel density (1) so finer detail can be represented (1)
  • The pixels in Version B are smaller (1) so more detail can be captured in the same space (1)
  • Version B has a higher resolution (1) because there are more pixels per unit of area (1)

For both marks, the expansion must follow/associate with the statement.

Do not accept 'more pixels' on its own without saying same area/pixel density/smaller pixels.

Do not accept higher colour depth or more colours.

Example answers and how they were marked
Full marks2/2

Version B has more pixels in the same area so each pixel represents a smaller part of the image.

Why: Linked point (1) and reason (1): more pixels in the same area means smaller pixels and more detail.

Some marks1/2

Version B has more pixels in the same area.

Why: One mark for the point about pixel density; the linked reason about smaller pixels or finer detail is missing.

No marks0/2

Version B has a higher colour depth, so it can display more colours.

Why: Colour depth is not the reason for more detail here; this is the common misconception that colour depth increases resolution.

Questions by topic

2.1 Binary

Build a paper on this topic 58
Adding Binary Numbers 6 questions Binary and Hexadecimal Basics 10 questions Binary Shifts 6 questions Binary to Denary Conversion 6 questions Binary to Hexadecimal Conversion 6 questions Denary to Binary Conversion 6 questions Denary to Hexadecimal Conversion 6 questions Denary to Negative Binary Conversion 6 questions Hexadecimal to Binary Conversion 6 questions

2.2 Data representation

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Character Sets 18 questions Images 19 questions Sound 19 questions

2.3 Data storage and compression

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Compression 18 questions Units & Conversion 6 questions

3.1 Hardware

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Embedded Systems 17 questions Fetch Decode Execute Cycle 15 questions Primary Memory 15 questions Secondary Memory 15 questions Stored Program Computers 15 questions

3.2 Software

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File Management 18 questions Memory Management 16 questions Peripheral Management 11 questions Process Management 12 questions Purpose & Function of Operating Systems 17 questions Software Basics 13 questions User Management 18 questions Utility Software 19 questions

3.3 Programming languages

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Low and High Level Languages 19 questions Translators 18 questions

4.1 Networks

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Addressing 16 questions LANs & WANs 16 questions Network Speed 5 questions Packet Switching 16 questions Protocols & Layers 17 questions The Internet 16 questions Topologies 17 questions Wired Connections 17 questions Wireless Connections 18 questions

4.2 Network security

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Network Security Risks 18 questions Preventing Network Security Risks 18 questions

5.1 Environmental

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E-waste 17 questions Manufacturing Technology 19 questions

5.2 Ethical and legal

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Data Protection Act 17 questions Impact of AI 18 questions Intellectual Property 18 questions Software Licences 18 questions

5.3 Cybersecurity

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Malware 18 questions Social Engineering 18 questions

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