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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 · 6 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 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 recall2 marksPrimary Memory Swap Remove

A computer is designed using the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • The program instructions and data are stored together in main memory/RAM (1)
  • The same memory is used for both instructions and data (1)
  • The instructions are fetched from main memory and executed by the CPU (1)
  • The instructions are fetched and executed one at a time / in sequence (1)
  • The program can be changed/replaced by loading new instructions into memory, without changing the hardware (1)

Do not accept the program is stored in secondary storage / on the hard disk.

Do not accept the program is stored in the CPU / in registers.

Example answers and how they were marked
Full marks2/2

The program instructions and the data are stored together in main memory, and the CPU fetches the instructions one at a time and executes them.

Why: Creditworthy point that instructions and data are stored in the same main memory (1) and the linked point that the CPU fetches and executes them from memory (1).

Some marks1/2

The program is held in the main memory (RAM) so that the computer can run it.

Why: Earns the point that the program instructions are stored in main memory (1), but does not say that the CPU fetches and executes them, so the second mark is not earned.

No marks0/2

The program is stored on the hard disk and loaded into the CPU when it is run.

Why: Both parts are misconceptions named in the Do not accept lines: the stored program concept keeps the instructions in main memory, not on secondary storage or in the CPU.

4Explain2 marksFetch Decode Execute Cycle Swap Remove

The control unit is part of the CPU.

Explain the role of the control unit in the fetch-decode-execute cycle.

[2 marks]
Mark scheme

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

  • Decodes/interprets/translates the instruction (1) to work out what the instruction is/what operation is required (1)
  • Sends control signals to the other components (1) so that the correct component carries out the instruction (1), e.g. the ALU/main memory
  • Controls the fetch of the instruction from main memory (1) because the address of the next instruction is sent via the address bus (1)
  • Controls the fetch of the next instruction from main memory (1) by sending a control signal/read signal to main memory (1)

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

For the decode point, 'decodes the instruction' alone gains 1 mark; a second mark needs the purpose of decoding, i.e. working out what the instruction is/what operation is required.

Do not accept 'it controls the CPU' on its own, without naming a task in the cycle.

Do not accept 'it fetches the instruction from memory' on its own, as this describes the fetch step rather than the control unit's role; credit only if linked to controlling/signalling the fetch.

Do not accept carrying out arithmetic/logical operations, as these are carried out by the ALU, or holding the address of the next instruction, as this is the program counter.

Do not accept storing the instruction in a register (e.g. MAR/MDR), as this is not the role of the control unit.

Example answers and how they were marked
Full marks2/2

The control unit decodes each instruction, working out what it means, so that the CPU knows which component should carry it out.

Why: Linked point (1) and reason (1): decodes/interprets the instruction, and works out what it is/what operation is required.

Some marks1/2

It decodes the instruction that has been fetched from main memory.

Why: Decode point only (1); it does not explain the purpose of decoding, i.e. working out what the instruction is/what operation is required.

No marks0/2

It adds the two numbers together and stores the result in a register.

Why: Misconception: arithmetic is carried out by the ALU, which the mark scheme's Do not accept line covers.

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5.3 Cybersecurity

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