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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 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 markEmbedded Systems Swap Remove

Identify the device that is a general purpose computer.

[1 mark]
  • ADesktop computer
  • BMicrowave oven
  • CWashing machine
  • DTraffic light
Mark scheme
  • A Desktop computer (1)
  • The only correct answer is A. The correct answer is Desktop computer.
  • B is not correct because a microwave oven uses an embedded system, built into it, to control a dedicated function
  • C is not correct because a washing machine uses an embedded system, built into it, to control a dedicated function
  • D is not correct because a traffic light uses an embedded system, built into it, to control a dedicated function
Example answers and how they were marked
Full marks1/1

A Desktop computer

Why: Correct option: the question asks for the general purpose computer, and a desktop computer is general purpose; it does not use an embedded system to control a dedicated function (1).

No marks0/1

B Microwave oven

Why: Near miss: a microwave oven is the kind of device that uses an embedded system, built into it, to control a dedicated function, so it is not a general purpose computer; mark scheme says B is not correct.

No marks0/1

C Washing machine

Why: Misconception: because a washing machine contains a computer, it must be a general purpose computer; it actually uses an embedded system built into it for a dedicated function, so no mark.

2Multiple choice1 markCharacter Sets Swap Remove

A new character set must be able to represent 70 different characters.

Identify the minimum number of bits needed to represent each character.

[1 mark]
  • A6
  • B7
  • C8
  • D70
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7.
  • A is not correct because 6 bits give only 64 combinations, which is not enough for 70 characters
  • C is not correct because 7 bits already give 128 combinations, so 8 bits is not the minimum needed
  • D is not correct because 70 is the number of characters needed, not a number of bits
Example answers and how they were marked
Full marks1/1

B 7

Why: The only correct answer: 7 bits give 128 combinations, enough for 70 characters, and it is the smallest number of bits that does (1).

No marks0/1

C 8

Why: Distractor, no credit: a single-tick item has no partial credit, and 8 bits is not the minimum because 7 bits already give 128 combinations, so the scheme's C line gives no mark.

No marks0/1

A 6

Why: Misconception, no credit: 6 bits give only 64 combinations, which cannot represent 70 different characters, so the scheme's A line gives no mark.

3Short recall2 marksUnits & Conversion Swap Remove

A solid state drive has a capacity of 5 gibibytes.

Give an expression to calculate the number of bits in 5 gibibytes. You do not need to do the calculation.

[2 marks]
Mark scheme

Award one mark for each of:

  • 5 × 1024^3 / 5 × 1024 × 1024 × 1024 / 5 × 2^30, for the number of bytes (1)
  • A fully correct expression with × 8 to convert bytes to bits, in any order, e.g. 5 × 1024^3 × 8, 8 × 5 × 2^30 (1)

Accept 5 × 1 073 741 824 × 8.

Accept 5 × 2^33 or 42 949 672 960 as a fully correct expression (2 marks).

Do not accept 5 × 1000^3 × 8, which uses decimal multiples.

Do not accept ÷ 8, which converts bits to bytes.

Example answers and how they were marked
Full marks2/2

8 × 5 × 2^30

Why: 5 × 2^30 is the number of bytes (1) and × 8 converts bytes to bits (1).

Some marks1/2

5 × 2^30

Why: Correct number of bytes using the binary multiple 2^30 (1), but no conversion to bits.

No marks0/2

5 × 1000^3 × 8

Why: Uses decimal 1000^3 instead of binary 1024^3, so the expression is not fully correct; no creditworthy point.

4Explain2 marksThe Internet Swap Remove

Each data packet has a header that contains the destination IP address.

Explain one reason why a data packet contains the destination IP address.

[2 marks]
Mark scheme

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

  • Routers can forward the packet towards its destination (1) because they read the destination IP address in the packet header (1)
  • The packet is delivered to the correct device (1) because the receiving device compares the destination IP address with its own IP address (1)
  • Each packet can take a different route (1) because routers use the destination IP address to choose the next router (1)
  • The packet can be passed on by each router in turn (1) because the destination IP address shows which network the packet is for (1)

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

Statements such as 'so it reaches the right destination' or 'so routers know where to send it' earn the first mark only; do not award the second mark unless the reason says that the address is read from the packet header and used to forward the packet.

Do not accept 'so it knows where it came from' as this describes the source address, not the destination address.

Do not accept 'so it gets there faster' without a linked reason.

Example answers and how they were marked
Full marks2/2

Routers know where to send it because they use the address in the packet header to choose the next network.

Why: Linked point (1) and reason (1): routers read the destination IP address in the header to forward the packet.

Some marks1/2

So that it reaches the right destination because every packet has an address.

Why: First mark (1) for reaching the right destination, but the reason is circular and does not say the address is read and used to forward the packet.

No marks0/2

So the packet knows where it came from.

Why: Describes the source address, not the destination address; do not accept.

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

Build a paper on this topic 56
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

Build a paper on this topic 37
Low and High Level Languages 19 questions Translators 18 questions

4.1 Networks

Build a paper on this topic 138
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

Build a paper on this topic 36
Network Security Risks 18 questions Preventing Network Security Risks 18 questions

5.1 Environmental

Build a paper on this topic 36
E-waste 17 questions Manufacturing Technology 19 questions

5.2 Ethical and legal

Build a paper on this topic 18
Data Protection Act 14 questions Impact of AI 4 questions

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