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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 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.

2Multiple choice1 markImages Swap Remove

Identify the smallest identifiable area of a bitmap image.

[1 mark]
  • AColour depth
  • BFile size
  • CPixel
  • DResolution
Mark scheme
  • C Pixel (1)
  • The only correct answer is C. The correct answer is pixel.
  • A is not correct because colour depth is the number of bits used to store the colour of each pixel
  • B is not correct because file size is the amount of storage an image takes up, not an area of the image
  • D is not correct because resolution is a measure of the number of pixels in an image (such as its width × height), not the smallest identifiable area
Example answers and how they were marked
Full marks1/1

C Pixel

Why: Correct option; a pixel is the smallest identifiable area of a bitmap image (1).

No marks0/1

D Resolution

Why: Resolution is a measure of the number of pixels in an image, not the smallest identifiable area; no mark.

No marks0/1

A Colour depth

Why: Colour depth is the number of bits used to store each pixel's colour, not the smallest identifiable area; no mark.

3Describe2 marksMemory Management Swap Remove

A program is stored on secondary storage before it is run.

Describe how the operating system manages memory when a program is started.

[2 marks]
Mark scheme

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

  • The program (and its data) is copied/loaded from secondary storage into main memory/RAM (1) so that the CPU can fetch/execute its instructions (1)
  • An area of main memory is allocated/reserved for the program (1) that is large enough to hold the program and its data (1)
  • The operating system checks that there is enough free main memory (1) before the program can be started/loaded, otherwise it cannot run/an error is reported (1)
  • The operating system keeps a record of the memory addresses used by the program (1) so that its instructions and data can be found when it runs (1)

Allow 'loaded into RAM', 'stored in memory' and 'put into main memory' as equivalent to copied into main memory.

A single statement that the program is loaded/stored in main memory, with no linked development, can be given one mark only.

Do not accept descriptions of the CPU fetching/executing instructions that do not refer to the operating system loading or allocating memory.

Do not accept responses about virtual memory (moving processes to secondary storage) or about which process runs next on the CPU.

Do not accept the program being run from secondary storage/hard disk.

Example answers and how they were marked
Full marks2/2

The operating system copies the program and its data from secondary storage into main memory so that the CPU can fetch and execute its instructions.

Why: Copying the program and data into main memory (1) linked to the CPU fetching/executing instructions (1).

Some marks1/2

The operating system loads the program into main memory.

Why: The program is copied/loaded into main memory (1), but there is no linked development of how or what this leads to, so only one mark is given.

No marks0/2

The CPU fetches the program's instructions from main memory and executes them one at a time.

Why: This describes the fetch-decode-execute cycle, not the operating system loading or allocating memory, which is the Do not accept line; no creditworthy point is made.

4Describe2 marksTranslators Swap Remove

High-level code can be translated by a compiler or by an interpreter.

Describe how a compiler translates high-level code differently from an interpreter.

[2 marks]
Mark scheme

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

  • The compiler translates the whole program/complete source code in one go (1) whereas the interpreter translates one line/statement at a time (1)
  • The compiler translates all of the source code before any of it is run/executed (1) but the interpreter translates each line/statement as it reads it (1)
  • The interpreter translates one line/statement at a time (1) and the compiler translates the whole program/source code together (1)

Accept the compiler point and the interpreter point in either order.

Do not accept responses about object code/executable files, when the translated code is executed, how errors are reported, or speed/memory use.

Do not accept 'the interpreter executes/runs each line' on its own, without saying that it translates one line/statement at a time.

Example answers and how they were marked
Full marks2/2

A compiler translates the whole program in one go, while an interpreter translates one line at a time.

Why: Whole program translated in one go (1) and interpreter translates one line at a time (1).

Some marks1/2

A compiler translates the whole program in one go.

Why: Whole program translated in one go (1); no linked point about the interpreter translating a line at a time.

No marks0/2

A compiler produces an executable file, but an interpreter does not.

Why: Describes the output, not the amount translated at a time; object code/executable files are not accepted.

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