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Edexcel GCSE Computer Science (1CP2) Exam Questions

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4 questions · 7 marks

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1Multiple choice1 markStored Program Computers Swap Remove

Identify the function of the control unit in the fetch-decode-execute cycle.

[1 mark]
  • ADecodes instructions
  • BPerforms arithmetic operations
  • CStores data and instructions
  • DTemporarily stores data being processed
Mark scheme
  • A Decodes instructions (1)
  • The only correct answer is A. The correct answer is Decodes instructions.
  • B is not correct because the arithmetic logic unit performs arithmetic operations
  • C is not correct because main memory stores data and instructions
  • D is not correct because registers temporarily store data being processed
Example answers and how they were marked
Full marks1/1

A Decodes instructions

Why: Correct option: the control unit decodes instructions during the fetch-decode-execute cycle (1).

No marks0/1

C Stores data and instructions

Why: Main memory stores data and instructions, not the control unit; this is a common distractor, so no mark.

No marks0/1

B Performs arithmetic operations

Why: Arithmetic operations are performed by the ALU, not the control unit; a common confusion between CPU components, so no mark.

2Short 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.

3Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

4Describe2 marksTranslators Swap Remove

A high-level program must be translated into machine code before it can be run.

Describe how a compiler and an interpreter differ in when the translated code is executed.

[2 marks]
Mark scheme

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

  • The interpreter executes each line as soon as it has been translated (1) so translation and execution happen together/at the same time (1)
  • The compiler translates the whole program first (1) and the machine code is run later/separately (1)
  • The interpreter translates the code and runs it while the program is running (1) whereas the compiler does not run the code it produces (1)
  • The compiler produces translated code that is executed only after all the translation has finished (1) but the interpreter executes each line immediately (1)

Allow 'translates one line at a time' where it is linked to executing each line immediately.

Do not accept answers that only state that a compiler is faster/an interpreter is slower, without reference to when execution happens.

Do not accept answers that only give object code, error reporting or portability, without reference to when execution happens.

Do not accept the compiler executing the program as it translates it.

Example answers and how they were marked
Full marks2/2

The compiler translates the code first and the program is executed afterwards, but the interpreter executes each line as soon as it has been translated.

Why: Compiler translates the whole program first and it is executed afterwards (1) and interpreter executes each line as soon as it is translated (1).

Some marks1/2

The interpreter executes each line immediately after translating it.

Why: Correct interpreter timing (1); no comparison with when the compiler's translated code is executed.

No marks0/2

The compiler executes the program line by line as it translates it.

Why: Reverses the two translators; a compiler does not execute the program it translates, so no mark.

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 14 questions Impact of AI 4 questions

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