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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 659 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 · 5 marks

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1Multiple choice1 markPrimary Memory Swap Remove

RAM is the main memory of a computer.

Identify the effect on the contents of RAM when the computer is switched off.

[1 mark]
  • AThey are lost
  • BThey are stored permanently
  • CThey are saved to secondary storage
  • DThey are written to ROM
Mark scheme
  • A They are lost (1)
  • The only correct answer is A. The correct answer is They are lost
  • B is not correct because RAM is volatile and its contents are lost when the power is switched off, so they are not stored permanently
  • C is not correct because RAM is volatile and its contents are lost when the power is switched off, so they are not saved to secondary storage
  • D is not correct because ROM is read-only memory that holds the start-up instructions, so the contents of RAM are not written to it
Example answers and how they were marked
Full marks1/1

A They are lost

Why: Selected the correct option. RAM is volatile, so its contents are lost when the power is switched off (1).

No marks0/1

C They are saved to secondary storage

Why: Near miss: thinks RAM contents are automatically saved, but RAM is volatile and its contents are not saved to secondary storage (0).

No marks0/1

B They are stored permanently

Why: Misconception: thinks RAM is non-volatile and keeps data permanently, but RAM is volatile and loses its contents when the power is switched off (0).

2Multiple choice1 markSecondary Memory Swap Remove

A company stores films on optical media.

Identify the optical storage medium that can store the most data.

[1 mark]
  • ABlu-ray Disc
  • BCD
  • CDVD
  • DMagnetic tape
Mark scheme
  • A Blu-ray Disc (1)
  • The only correct answer is A. The correct answer is Blu-ray Disc.
  • B is not correct because a CD stores less data than a Blu-ray Disc, up to about 700 MB
  • C is not correct because a DVD stores less data than a Blu-ray Disc, up to about 4.7 GB (single-layer) or 8.4 GB (dual-layer)
  • D is not correct because magnetic tape is not an optical storage medium
Example answers and how they were marked
Full marks1/1

A Blu-ray Disc

Why: Correct option (1): a Blu-ray Disc has the greatest capacity of the optical media.

No marks0/1

C DVD

Why: Near miss scoring 0: a DVD is optical but stores less than a Blu-ray Disc (mark scheme line C). Only A is credited.

No marks0/1

D Magnetic tape

Why: Misconception scoring 0: candidates pick tape for its large capacity but magnetic tape is not an optical storage medium (mark scheme line D).

3Short recall1 markSecondary Memory Swap Remove

Data is stored on optical discs as pits and lands.

Name the component in an optical disc drive that reads this data.

[1 mark]
Mark scheme

Award one mark for:

  • Laser (1)

Allow laser beam

Do not accept read/write head

Do not accept pits

Do not accept lands

Do not accept reflective surface

Do not accept magnet/magnetic

Example answers and how they were marked
Full marks1/1

Laser

Why: Laser is the creditworthy answer (1).

No marks0/1

A read/write head

Why: Read/write head is not accepted by the mark scheme; it reads magnetic media, not optical discs, so no marks are awarded.

No marks0/1

Pits and lands

Why: Pits and lands are what the laser reads, not the component that reads the data, so the mark scheme does not accept them.

4Explain2 marksPrimary Memory Swap Remove

A computer follows the von Neumann stored program concept. A program's instructions are stored on secondary storage before the program is run.

Explain one reason why the instructions of a program must be held in RAM before they can be executed.

[2 marks]
Mark scheme

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

  • The CPU can only fetch/read instructions from main memory/RAM (1) so the instructions of a running program must be copied/loaded into RAM before they can be executed (1)
  • RAM is faster to access than secondary storage (1) so the instructions can be fetched quickly enough for the CPU to execute them (1)
  • The CPU needs the instructions in RAM to fetch and execute them (1) because the fetch-decode-execute cycle reads each instruction from main memory/RAM (1)
  • The address of the next instruction is sent to RAM on the address bus (1) so the instruction can be transferred to the CPU/registers for decoding/execution (1)

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

Allow the point and reason in either order if they are linked.

Do not accept answers that only say the CPU needs instructions, or that RAM is faster, without linking this to fetching/executing the instructions.

Do not accept answers that only say RAM holds data and instructions without linking this to the fetch-decode-execute cycle.

Do not accept 'RAM is permanent/non-volatile storage, so the instructions are kept safe' - RAM is volatile.

The response does not need to name the address bus or the fetch stage for full marks; a linked reason that explains why instructions must be in RAM (for example, the CPU can only fetch from RAM, or RAM is faster than secondary storage) is enough.

Example answers and how they were marked
Full marks2/2

The CPU can only fetch instructions from main memory, so the instructions of a running program must be copied into RAM before they can be executed.

Why: Linked point (1) for the CPU only being able to fetch instructions from main memory/RAM, and reason (1) for instructions being copied into RAM before execution.

Some marks1/2

The CPU can only fetch instructions from main memory.

Why: Point (1) only; there is no linked reason that explains why this means the instructions must be held in RAM before execution, such as RAM being faster than secondary storage.

No marks0/2

Instructions are kept in RAM because RAM is permanent storage and does not lose the instructions when the power is turned off.

Why: Misconception: RAM is described as permanent/non-volatile, which the mark scheme's 'Do not accept' line rules out; no creditworthy linked point.

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 12 questions

5.3 Cybersecurity

Malware 8 questions

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