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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 514 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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5 questions · 8 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 markSound Swap Remove

A sound wave is sampled and the amplitude of each sample is stored using 2 bits.

Identify the number of different amplitude values that can be stored for each sample.

[1 mark]
  • A2
  • B4
  • C8
  • D16
Mark scheme
  • B 4 (1)
  • The only correct answer is B. The correct answer is 4 (2 bits give 2^2 = 4 values).
  • A is not correct because 2 is the number of bits used to store each sample
  • C is not correct because 8 different amplitude values would need 3 bits
  • D is not correct because 16 different amplitude values would need 4 bits
Example answers and how they were marked
Full marks1/1

B 4

Why: Selects the only correct option (1).

No marks0/1

C 8

Why: Near miss: 8 values would need 3 bits, one more than this recording uses (0).

No marks0/1

A 2

Why: Typical misconception: 2 is the number of bits, not the number of amplitude values those bits can store (0).

3Short recall2 marksStored Program Computers Swap Remove

Computers use the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

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

  • Program instructions are stored in main memory/RAM (1)
  • The data used by the program is stored in main memory/RAM (1)
  • The CPU fetches/reads instructions from main memory to execute them (1)
  • The program can be run without the instructions being input again (1)

Allow 'the program is stored in memory' for one mark.

Do not accept answers about programs being stored on secondary storage, such as a hard disk or USB memory stick.

Do not accept answers that only describe the von Neumann architecture (e.g. having a CPU and memory) without stating that instructions are stored in main memory.

Example answers and how they were marked
Full marks2/2

Both the program instructions and the data the program uses are held in the same main memory, and the CPU fetches these instructions from memory to execute them.

Why: Instructions stored in main memory (1) and the CPU fetches instructions from memory to execute them (1).

Some marks1/2

Program instructions are stored in main memory.

Why: Instructions stored in main memory (1). No second point about data being stored in the same memory or about the CPU fetching instructions.

No marks0/2

The program is saved on the hard disk so that it can be loaded into RAM when the user runs it.

Why: A near-miss about secondary storage, which the mark scheme says not to accept; the stored program concept is about instructions being held in main memory.

4Describe2 marksUtility Software Swap Remove

A hard disk can become fragmented as files are saved and deleted.

Describe how disk defragmentation software improves the performance of a hard disk.

[2 marks]
Mark scheme

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

  • Parts of a file are moved so they are stored together/in consecutive locations (1) so the read/write head has less distance to travel to read the file (1)
  • Files are rearranged so that each file is stored in one block (1) which reduces the time taken to open/load the file (1)
  • Free space on the disk is collected into one area (1) so new files can be saved without becoming fragmented (1)
  • The parts of each fragmented file are put in order (1) so fewer movements of the read/write head are needed to read it (1)

Do not accept responses about compressing files, deleting files or increasing the amount of storage, as these are not what defragmentation software does.

Do not accept 'it makes the disk faster' without saying how the files are changed.

Example answers and how they were marked
Full marks2/2

It moves the parts of a file so they are stored together in consecutive locations (1), so the read/write head does not have to move as far and the file opens more quickly (1).

Why: Linked point (1) and development (1): the file parts are stored together, which reduces read/write head movement.

Some marks1/2

It reorganises the files on the hard disk so that they are stored together.

Why: The point is made (1) but there is no development saying how this improves performance, so the second mark is not awarded.

No marks0/2

It compresses the files so that they take up less space on the disk.

Why: This describes data compression software, which the mark scheme says is not accepted; no creditworthy point is made.

5Describe2 marksTranslators Swap Remove

Source code can contain syntax errors.

Describe how a compiler and an interpreter differ in the way they deal with syntax errors.

[2 marks]
Mark scheme

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

  • The compiler reports all syntax errors after translation has finished (1) whereas the interpreter stops at the first error it finds (1)
  • The interpreter reports a syntax error as soon as it finds one (1) and does not translate/execute any more of the program (1)
  • The compiler continues translating after an error (1) and reports the errors at the end (1)
  • The interpreter stops translating when it reaches an error (1) but the compiler finishes translating before reporting errors (1)

Do not accept answers about fixing/correcting errors or about object code.

Do not award 'the compiler finds more errors' without saying errors are reported at the end.

Example answers and how they were marked
Full marks2/2

A compiler reports all the syntax errors at the end of translation, whereas an interpreter stops and reports the error as soon as it finds one.

Why: Compiler reports errors at the end (1) and interpreter stops/reports at the first error (1).

Some marks1/2

An interpreter stops translating when it finds a syntax error.

Why: Interpreter stops at the first error (1); no comparison with when the compiler reports errors.

No marks0/2

A compiler reports a syntax error immediately, before translating the rest of the program.

Why: Reverses the compiler's behaviour; it finishes translating before reporting errors.

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