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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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1Multiple choice1 markImages Swap Remove

Identify the smallest identifiable area of a bitmap image.

[1 mark]
  • AColour depth
  • BFile size
  • CPixel
  • DResolution
Mark scheme
  • C Pixel (1)
  • The only correct answer is C. The correct answer is pixel.
  • A is not correct because colour depth is the number of bits used to store the colour of each pixel
  • B is not correct because file size is the amount of storage an image takes up, not an area of the image
  • D is not correct because resolution is a measure of the number of pixels in an image (such as its width × height), not the smallest identifiable area
Example answers and how they were marked
Full marks1/1

C Pixel

Why: Correct option; a pixel is the smallest identifiable area of a bitmap image (1).

No marks0/1

D Resolution

Why: Resolution is a measure of the number of pixels in an image, not the smallest identifiable area; no mark.

No marks0/1

A Colour depth

Why: Colour depth is the number of bits used to store each pixel's colour, not the smallest identifiable area; no mark.

2Short 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.

3Describe2 marksSound Swap Remove

Sound is converted into binary using a limited bit depth.

Describe one limitation of representing a sound sample with a limited bit depth.

[2 marks]
Mark scheme

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

  • Only a fixed number of amplitude values/levels can be represented (1) so the amplitude is rounded to the nearest available value/level (1)
  • Fewer amplitude values/levels are available to choose from (1) so similar/close amplitudes cannot be stored as different values (1)
  • The exact amplitude often falls between available values/levels (1) so the stored amplitude is not exact/less accurate (1)

Answers must relate to the number of bits per sample (bit depth) and the amplitude values/levels it can represent.

Do not accept answers about the sample rate, sample interval or the number of samples taken, or that confuse bit depth with the number of samples.

Do not award 'lower quality'/'more detail'/'less detail' unless it is linked to the number of amplitude values/levels.

Do not accept 'the sound cannot be played/stored' without a linked limitation.

Example answers and how they were marked
Full marks2/2

Only a fixed number of amplitude values can be represented, so the amplitude is rounded to the nearest available value and some detail is lost.

Why: Linked points: fixed number of amplitude values (1) and rounding to the nearest available value (1).

Some marks1/2

Only a fixed number of amplitude values can be represented.

Why: Gives the first linked point only (1); it does not develop what this limitation leads to.

No marks0/2

A lower bit depth means fewer samples are taken, so parts of the sound are missed.

Why: Confuses bit depth with the number of samples/sample rate; no creditworthy point about amplitude values is given.

4Explain2 marksSound Swap Remove

Analogue sound is converted to binary by sampling at regular intervals.

Explain one reason why increasing the sample rate improves the quality of a digitised sound.

[2 marks]
Mark scheme

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

  • More samples are taken every second / the sample rate is higher (1) so the digital wave follows the shape of the analogue wave more closely (1)
  • The amplitude is measured more often / at shorter intervals (1) so fewer changes in the sound are missed between measurements (1)
  • The sample interval is shorter (1) so less of the sound wave is lost/goes unrecorded between samples (1)
  • More data points are recorded for the same length of sound (1) so the digital representation is a closer copy of the original sound (1)

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

Do not accept 'the quality is better/higher' or 'the sound is clearer' as a point on its own without a reason such as more samples or more detail.

Do not accept 'the samples are more accurate' or 'more data is collected' without saying what becomes more accurate or what the extra data records.

Do not accept 'the file size is larger' as a reason for improved quality.

Do not accept answers about using more bits per sample, as these describe bit depth, not the sample rate.

Example answers and how they were marked
Full marks2/2

More samples are taken every second, so the digital representation follows the shape of the analogue wave more closely.

Why: Point (more samples each second) (1) with linked reason (follows the analogue wave more closely) (1).

Some marks1/2

More samples are taken every second.

Why: Creditworthy point (1) but no linked reason about what the extra samples achieve, so stays on one mark.

No marks0/2

Each sample is stored using more bits, so every measurement is more accurate.

Why: This describes bit depth, not the sample rate, so neither mark is awarded.

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

Build a paper on this topic 36
Malware 18 questions Social Engineering 18 questions

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