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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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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 marksSecondary Memory Swap Remove

A computer uses secondary storage to keep programs and data.

Give two examples of secondary storage devices that store data using magnetic media.

[2 marks]
Mark scheme

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

  • Hard disk drive/HDD (1)
  • External/portable hard disk drive (1)
  • Magnetic tape drive (1)
  • Magnetic tape (1)

Allow a named device or the storage media type, e.g. hard disk/hard disc for hard disk drive, tape for magnetic tape

Allow floppy disk/floppy disk drive

Do not accept solid state devices (SSD, USB flash drive/memory stick, SD card/memory card) or optical media (CD/DVD/Blu-ray) as these do not store data using magnetism

Example answers and how they were marked
Full marks2/2

Hard disk drive and magnetic tape

Why: Hard disk drive (1) and magnetic tape (1) are both magnetic secondary storage, so both marks are awarded.

Some marks1/2

Hard disk drive and SD card

Why: Hard disk drive is magnetic (1); an SD card is solid state, not magnetic, so the second mark is not awarded.

No marks0/2

CD and DVD

Why: Both are optical media, which the mark scheme says not to accept, so no marks are awarded.

4Explain2 marksFetch Decode Execute Cycle Swap Remove

The von Neumann architecture is used in most computers.

Explain one feature of the von Neumann stored program concept.

[2 marks]
Mark scheme

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

  • The instructions/program instructions and data are stored together in main memory/RAM (1) so the CPU can fetch each instruction (from its address) and execute it (1)
  • The instructions/program instructions are held in main memory/RAM (1) rather than being built into/part of the hardware, so the computer does not need to be rewired/rebuilt to run a different program (1)
  • The next instruction to be executed/instruction is held in main memory/RAM (1) so the CPU can fetch it using its address before it is decoded and executed (1)

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

A point/statement on its own can only earn one mark.

Do not accept a response that only refers to data being stored in main memory/RAM without referring to the instructions/program instructions.

Do not accept instructions/programs being stored on the hard disk/backing store.

Do not accept 'the program is stored in memory' without saying that it is the instructions/program instructions that are stored there.

Example answers and how they were marked
Full marks2/2

Both the instructions of the program and the data it uses are stored together in the same main memory, so the CPU can fetch each instruction in turn and execute it.

Why: Linked point (1) and reason (1).

Some marks1/2

The instructions of the program are stored in main memory.

Why: Point only (1); no linked reason about fetching/executing the instructions.

No marks0/2

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

Why: Misconception covered by the 'Do not accept' line: the instructions must be stored in main memory/RAM, not on the hard disk/backing store, for the CPU to fetch them.

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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