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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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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.

4Short recall3 marksFetch Decode Execute Cycle Swap Remove

The fetch-decode-execute cycle is used by the CPU.

State three things that happen during the fetch stage.

[3 marks]
Mark scheme

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

  • The address (of the next instruction) is sent from the CPU to main memory / placed on the address bus (1)
  • The (next) instruction is fetched/read from main memory (RAM) (1)
  • The instruction is brought/copied into the CPU / received into the MDR (1)
  • The program counter (PC) is incremented/updated to hold the address of the next instruction (1)

Allow 'the PC/MAR is placed on the address bus' for the address being sent to main memory.

Allow 'the MDR receives the instruction' for the instruction being brought into the CPU.

Do not award a second mark for the instruction being fetched from main memory and the instruction being brought into the CPU; these are the same point.

Do not award 'the instruction is transferred along the data bus' on its own; this is the same point as the instruction being fetched from main memory.

Do not accept answers about decoding the instruction, e.g. 'the instruction is fetched from RAM and decoded', as decoding happens in the decode stage.

Do not accept answers about the execute stage, e.g. 'the ALU carries out the calculation' or 'the result of a calculation is stored in a register', as these happen in the execute stage.

Example answers and how they were marked
Full marks3/3

The address of the next instruction is sent to main memory, the instruction is fetched from RAM, and the program counter is incremented.

Why: Address sent to main memory (1), instruction fetched from RAM (1) and program counter incremented (1).

Some marks1/3

The instruction is fetched from main memory and then decoded.

Why: Fetching the instruction from main memory (1); decoding happens in the decode stage, so it is not creditworthy in the fetch stage.

No marks0/3

The ALU carries out the calculation and the result is stored in a register.

Why: The ALU carrying out a calculation and storing a result happen in the execute stage, not the fetch stage, so no point is creditworthy.

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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