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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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5 questions · 8 marks

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1Multiple choice1 markPrimary Memory Swap Remove

Identify what is fetched from RAM during the fetch stage.

[1 mark]
  • AThe next instruction
  • BThe data at a memory address
  • CThe decoded instruction
  • DThe address of the next instruction
Mark scheme
  • A The next instruction (1)
  • The only correct answer is A. The correct answer is the next instruction.
  • B is not correct because data is fetched during the execute stage
  • C is not correct because an instruction is decoded after it is fetched
  • D is not correct because the program counter holds the address used to fetch the instruction
Example answers and how they were marked
Full marks1/1

A The next instruction

Why: The next instruction is fetched from RAM during the fetch stage (1).

No marks0/1

B The data at a memory address

Why: Near miss: data is not fetched during the fetch stage; data is used during the execute stage.

No marks0/1

C The decoded instruction

Why: Misconception: an instruction is decoded after it is fetched, in the decode stage.

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.

4Describe2 marksSecondary Memory Swap Remove

A laptop uses a solid state drive (SSD) for secondary storage.

Describe how data is stored on a solid state drive.

[2 marks]
Mark scheme

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

  • Data is stored on (flash) memory chips/cells (1) as a pattern of electrical charges (1)
  • A current/voltage forces electrons through a barrier (1) so that they are trapped (or not trapped) to hold/represent a value (1)
  • Cells with a trapped charge and cells with no trapped charge (1) represent 1s and 0s/binary (1)
  • The trapped charge remains when the power is turned off (1) so the data is not lost/it is non-volatile (1)

Do not accept responses that describe magnetic storage, e.g. magnetised platters, spinning disks or read/write heads, or optical storage, e.g. lasers, pits and lands.

Do not award a mark for 'it is solid state' or 'it has no moving parts' without a description of how the data is stored.

Do not award a mark for restating the stem, e.g. 'a laptop uses an SSD' or 'an SSD uses flash memory'.

'Data is stored as electrical charges' on its own, with no reference to memory chips/cells or to how the charge is held, earns a maximum of one mark (1).

A response that only describes electrons being trapped, with no link to how the value/data is represented, earns one mark (1). The second mark needs a link such as trapped and not trapped charge representing 1s and 0s, or the trapped charge being kept so the data is not lost.

Example answers and how they were marked
Full marks2/2

A current is used to force electrons through a barrier so that they become trapped, and a cell with a trapped charge or a cell with no trapped charge represents a 1 or a 0.

Why: Linked description: a current forces electrons through a barrier (1) and trapped/not trapped charge represents 1s and 0s (1).

Some marks1/2

The data is stored in flash memory chips.

Why: Gives the creditworthy point that data is stored on flash memory chips (1) but does not develop it, e.g. by saying the data is held as electrical charges or by describing trapped electrons.

No marks0/2

The data is stored as magnetised areas on a spinning platter.

Why: This describes magnetic storage, which the mark scheme says not to accept for a solid state drive.

5Explain2 marksEmbedded Systems Swap Remove

An engine control unit (ECU) in a car is an embedded system.

Explain one benefit of using an embedded system for the engine control unit instead of a general purpose computer.

[2 marks]
Mark scheme

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

  • (The ECU) is more reliable/less likely to crash (1) because it only runs the one program/has a single purpose (1)
  • (The ECU) cannot be affected by other software/viruses (1) because it cannot run other software/it can only run the program it was made for (1)
  • (The ECU) is cheaper (1) because it has limited processing resources/less memory/simple components (1)
  • (The ECU) uses less power (1) because it has limited processing resources/a simpler processor (1)
  • (The ECU) is smaller (1) because it has fewer components/less memory than a general purpose computer (1)

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

A suitable benefit stated on its own earns one mark.

Do not accept a description or a location given as the reason, e.g. 'because it is part of the car', 'because it is inside the engine', 'because it is embedded'.

Do not accept a description of an embedded system on its own as the benefit, e.g. 'it does one specific job' or 'it is part of the car'.

Do not accept 'it is faster'/'it has a more powerful processor' as embedded systems have limited processing resources.

Example answers and how they were marked
Full marks2/2

It is more reliable than a general purpose computer because it only runs the one program so it will not crash and stop the engine working.

Why: Benefit stated (1) with a linked reason (1); the reason shows why only running one program makes it reliable.

Some marks1/2

It is more reliable because it is part of the car's engine.

Why: Only the benefit earns a mark (1); being part of the car is a location, not a linked reason why it is reliable.

No marks0/2

It is faster than a general purpose computer because it is embedded in the engine.

Why: Misconception that an embedded system is faster; embedded systems have limited processing resources, so neither point nor reason 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

Build a paper on this topic 77
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

Build a paper on this topic 124
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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