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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 · 4 marks

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1Multiple choice1 markUnits & Conversion Swap Remove

A computer stores data in binary.

Identify the number of bits in one kibibyte.

[1 mark]
  • A1024
  • B8000
  • C8192
  • D8 388 608
Mark scheme
  • C 8192 (1)
  • The only correct answer is C. The correct answer is 8192.
  • A is not correct because 1024 is the number of bytes in a kibibyte
  • B is not correct because 8000 is the number of bits in 1000 bytes
  • D is not correct because 8 388 608 is the number of bits in a mebibyte
Example answers and how they were marked
Full marks1/1

C 8192

Why: Correct: one kibibyte is 1024 bytes and 1024 × 8 = 8192 bits (1).

No marks0/1

A 1024

Why: Near miss: 1024 is the number of bytes in a kibibyte, not the number of bits (0).

No marks0/1

B 8000

Why: Typical misconception: uses 1000 bytes for a kibibyte, giving 1000 × 8 = 8000 bits (0).

2Short recall2 marksTopologies Swap Remove

Mesh is a type of network topology.

State two characteristics of a mesh topology.

[2 marks]
Mark scheme

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

  • Each node/device has multiple direct connections to other nodes/devices (in a full mesh, every node is connected directly to every other node) (1)
  • There is no central device/switch (1)
  • Nodes act as routers/relay data for other nodes / data can take multiple/different routes to its destination / data can still be sent if a connection/node fails (1)
  • More connections/cables are needed as the number of nodes increases (1)
  • The network can be expanded without disrupting existing devices (1)

Allow answers that refer to full or partial mesh.

Allow 'more connections/cables are needed' and 'can be expanded' as characteristics (they may be described as advantages or disadvantages).

Do not accept 'more secure' or 'no collisions' without qualification.

Do not accept 'fast' or 'reliable' without qualification.

Do not accept all devices connect to one main cable (this describes a bus topology).

Do not accept it uses a central switch/hub or data is sent through a central hub (this describes a star topology).

Do not accept 'expensive' without qualification (e.g. because more cables/connections are needed).

Example answers and how they were marked
Full marks2/2

Each device is connected to several other devices, and data can be sent along more than one route.

Why: Multiple direct connections (1) and data can take multiple routes (1).

Some marks1/2

There is no central switch.

Why: No central device credited (1), but only one characteristic given.

No marks0/2

Data is sent through a central hub.

Why: Misconception: this describes a star topology, not a mesh topology.

3Short recall1 markCharacter Sets Swap Remove

A computer can store a maximum of 80 bits for a text message. Each character is stored using 7-bit ASCII.

State the maximum number of whole characters that can be stored in the message.

[1 mark]
Mark scheme

Award one mark for:

  • 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Example answers and how they were marked
Full marks1/1

11

Why: Correct: 80 ÷ 7 = 11 remainder 3, so 11 whole characters can be stored (1).

No marks0/1

11.4

Why: Near miss: gives a decimal result rather than the number of whole characters; the mark scheme does not accept 11.4.

No marks0/1

12

Why: Rounds up to 12, but 12 characters would need 84 bits, more than the 80 available; the mark scheme does not accept 12.

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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