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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 631 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 markTopologies Swap Remove

A bus topology uses terminators at each end of the main cable.

State the purpose of terminators in a bus topology.

[1 mark]
Mark scheme

Award one mark for:

  • Absorb signals (at the ends of the bus cable) / prevent signals reflecting (back along the cable) (1)

Allow 'stop data/signals bouncing back' for prevent signals reflecting

Do not accept 'terminate the cable' / 'close the end of the cable' on its own

Do not accept boost/repeat/strengthen the signal

Example answers and how they were marked
Full marks1/1

To absorb signals so that they do not reflect back along the bus cable.

Why: Absorbing signals and preventing reflection are alternatives within the one creditworthy point (1).

No marks0/1

To terminate the cable at each end.

Why: Near miss: this is rejected by the 'terminate the cable' guidance because it does not say signals are absorbed or prevented from reflecting.

No marks0/1

To make the signal stronger at the end of the cable.

Why: Misconception: terminators absorb signals; they do not boost, repeat or strengthen them.

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 marksUser Management Swap Remove

A computer can be used by more than one person.

Describe how an operating system uses access rights to protect each user's files.

[2 marks]
Mark scheme

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

  • Each user/file is given permissions/access rights (1) so other users cannot read/write/open/change that user's files (1)
  • A user may be given read-only/read access to a file (1) so they cannot edit/delete/change it (1)
  • Each file is given permissions/access rights for different users (1) so the operating system allows/refuses access depending on which user is logged in (1)
  • The operating system checks a user's permissions/access rights when a file is opened (1) so a user with read-only/read access cannot save/delete changes (1)

Do not accept responses about asking for a password/username, as this checks who the user is, not what they are allowed to do.

Do not accept 'files are protected/safe' or 'access rights protect files' without saying how access is limited (e.g. read-only access, permissions for different users).

Example answers and how they were marked
Full marks2/2

A user can be given read-only access to a file, so they can open it but cannot edit or delete it.

Why: Read-only/read access point (1) and linked development that the user cannot edit/delete the file (1).

Some marks1/2

A user can be given read-only access to a file.

Why: Read-only/read access point (1); no development saying what this prevents, such as being unable to edit/delete the file (0).

No marks0/2

The operating system asks for the password again before the file opens.

Why: Describes asking for a password, which checks who the user is, not access rights to the file; no creditworthy point.

4Describe2 marksLow and High Level Languages Swap Remove

A programmer has written a program in assembly language.

Describe why it is necessary to translate the program into machine code before it can be run.

[2 marks]
Mark scheme

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

  • The processor/CPU can only execute machine code/binary instructions (1) so the assembly instructions must be converted into machine code before the program can run (1)
  • Each assembly instruction/mnemonic matches/corresponds to one machine code instruction (1) so each line must be replaced with the machine code the processor executes (1)
  • The program is written using mnemonics (1) that must be translated into machine code for the processor to carry out the instructions (1)
  • The hardware/processor does not understand assembly language directly (1) so it must be translated into machine code that the processor can execute (1)

Do not accept 'assembly language is a low-level language'.

Do not accept 'it is hard to understand' or 'it is not English'.

Do not accept benefits, such as 'machine code runs faster' or 'the program will run more quickly'.

Do not accept 'an assembler/a translator converts it' on its own, with no reference to the processor only being able to execute machine code/binary.

Do not accept responses about high-level languages or about compilers/interpreters.

Example answers and how they were marked
Full marks2/2

The processor can only execute machine code, so the assembly instructions must be converted into machine code before the program can run.

Why: States what the processor can execute (1) and links it to converting the program's instructions into machine code (1).

Some marks1/2

Each assembly instruction matches one machine code instruction.

Why: Correct point about the one-to-one match (1) but no development saying that this is because the processor only executes machine code.

No marks0/2

Because an assembler translates assembly language into machine code.

Why: Names the translator but gives no reason tied to the processor only being able to execute machine code, so no mark; this is in the 'Do not accept' guidance.

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 16 questions Intellectual Property 12 questions

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