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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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3 questions · 3 marks

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1Multiple choice1 markPrimary Memory Swap Remove

A computer is running an application. The application's data and instructions are transferred from secondary storage into RAM before the CPU processes them.

Identify the reason for this transfer.

[1 mark]
  • ARAM is non-volatile
  • BRAM is faster to access than secondary storage
  • CRAM has a larger capacity than secondary storage
  • DRAM is cheaper than secondary storage
Mark scheme
  • B RAM is faster to access than secondary storage (1)
  • The only correct answer is B. The correct answer is RAM is faster to access than secondary storage.
  • A is not correct because RAM is volatile
  • C is not correct because RAM has a smaller capacity than secondary storage
  • D is not correct because RAM is more expensive than secondary storage
Example answers and how they were marked
Full marks1/1

B RAM is faster to access than secondary storage

Why: Selects the correct option: RAM is faster to access than secondary storage, so data and instructions can reach the CPU quickly (1).

No marks0/1

A RAM is non-volatile

Why: Near miss: RAM is volatile; this does not explain the transfer into RAM.

No marks0/1

C RAM has a larger capacity than secondary storage

Why: Misconception: assumes primary memory has a larger capacity, but RAM has a smaller capacity than secondary storage.

2Multiple choice1 markFetch Decode Execute Cycle Swap Remove

Identify the CPU component that decodes instructions during the fetch-decode-execute cycle.

[1 mark]
  • AArithmetic logic unit
  • BControl unit
  • CMain memory
  • DRegister
Mark scheme
  • B Control unit (1)
  • The only correct answer is B. The correct answer is control unit.
  • A is not correct because the arithmetic logic unit performs arithmetic and logical operations, not decoding
  • C is not correct because main memory stores instructions and data, not decoding
  • D is not correct because a register holds data or addresses temporarily, not decoding
Example answers and how they were marked
Full marks1/1

B Control unit

Why: Correct option: the control unit decodes instructions (1).

No marks0/1

D Register

Why: Near-miss: a register holds data or addresses temporarily, it does not decode instructions (0).

No marks0/1

A Arithmetic logic unit

Why: Misconception: the arithmetic logic unit performs arithmetic and logical operations; it does not decode (0).

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.

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