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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 · 7 marks

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1Short 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.

2Describe2 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.

3Explain2 marksImages Swap Remove

Two versions of the same bitmap image are saved. The second version has a higher colour depth.

Explain why the second version may have a larger file size.

[2 marks]
Mark scheme

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

  • More binary data is stored for each pixel (1) because a higher colour depth uses more bits per pixel (1)
  • Each pixel has a longer binary code (1) so more storage is needed to hold the image (1)
  • More colours can be represented (1) so more bits are needed for each pixel (1)
  • The file contains more bits in total (1) because colour depth is the number of bits per pixel (1)

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

Allow 'more bits per pixel' for one mark.

Do not accept 'more pixels' or 'higher resolution' as the reason, because changing colour depth does not change the number of pixels.

Example answers and how they were marked
Full marks2/2

The second version stores more binary data for each pixel because a higher colour depth uses more bits per pixel.

Why: Linked point (1) and reason (1): more data per pixel because more bits per pixel.

Some marks1/2

The second version uses more bits for each pixel.

Why: One mark for identifying that more bits are stored per pixel; no linked statement about more data or file size.

No marks0/2

The second version has more pixels, so there is more data to store.

Why: A common misconception; colour depth does not change the number of pixels, so no creditworthy point 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

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Malware 18 questions Social Engineering 18 questions

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