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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 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.

3Describe2 marksMemory Management Swap Remove

An operating system keeps a record of the memory it has allocated to each program.

Describe how this record is used.

[2 marks]
Mark scheme

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

  • The start and end address/size of the block allocated to each program is recorded (1) so the operating system knows which areas of main memory are in use and which are free (1)
  • The record is checked before memory is allocated (1) so memory already in use is not given to another program/is not overwritten (1)
  • The record is searched for a free block that is large enough (1) so a newly loaded program is placed in a suitable area of main memory (1)
  • The record is updated when a program ends/closes (1) so the memory it was using is marked as free for other programs (1)

Allow the record to be called a memory map/memory table/allocation table.

A response that only says the operating system keeps a record/list of the memory each program is using, without saying what the record contains, earns 0 marks.

A response that states what the record contains, e.g. the addresses/the start and end address of each block, earns 1 mark even if it does not say how the record is used.

Do not accept responses about saving programs/files to secondary storage or about moving programs to virtual memory, as these are not uses of this record.

Example answers and how they were marked
Full marks2/2

The operating system records the start and end address of the block of memory given to each program so that it does not give the same memory to another program.

Why: The start and end address of each block is recorded (1) linked to memory already in use not being given to another program (1).

Some marks1/2

The operating system records the start and end address of the memory used by each program.

Why: States what the record contains (1), but does not develop how the record is used, so the second mark is not earned.

No marks0/2

The operating system saves the programs to the hard disk so there is enough memory for other programs.

Why: Describes moving programs to secondary storage rather than how the record of allocated memory is used; 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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