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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 bus that carries addresses in one direction only, from the CPU to main memory.

[1 mark]
  • AAddress bus
  • BControl bus
  • CData bus
  • DSystem bus
Mark scheme
  • A Address bus (1)
  • The only correct answer is A. The correct answer is address bus. Do not accept 'address' on its own.
  • B is not correct because the control bus carries control signals in both directions between the CPU and other components
  • C is not correct because the data bus carries data in both directions between the CPU and main memory
  • D is not correct because the system bus is the collection of all three buses
Example answers and how they were marked
Full marks1/1

A Address bus

Why: Correct option: the address bus carries addresses in one direction only, from the CPU to main memory (1).

No marks0/1

C Data bus

Why: The data bus carries data in both directions, so it is not the bus that carries addresses in one direction only from the CPU to main memory; no creditworthy point.

No marks0/1

B Control bus

Why: The control bus carries control signals in both directions, not addresses in one direction only; this is the common control/address bus confusion, 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).

3Describe2 marksFile Management Swap Remove

A user saves a new file into a folder that already contains files.

Describe how the operating system prevents a new file overwriting an existing file.

[2 marks]
Mark scheme

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

  • Checks/compares the new file (name)/filename against the names of files already in the folder/directory (1) to see if the name is already in use (1)
  • Refuses to save/allow the new file when the (file) name already exists (1) so the existing file is not replaced/overwritten (1)
  • Prompts/asks the user to choose/enter a different (file) name (1) so both files can be saved without one replacing the other (1)

A point without a linked development can earn one mark.

For both marks, the development must follow/associate with the point.

Do not accept responses about defragmentation.

Do not accept the operating system making a backup copy, or warning the user that a file may be overwritten, without saying that the (file) name is checked or the new file is not saved.

Do not accept responses about process management or saving files to RAM.

Do not award a second mark for repeating the question, e.g. 'it stops the new file overwriting the existing file'.

Example answers and how they were marked
Full marks2/2

It checks the new file name against the names of files already in the folder (1) to see whether the name is already in use (1).

Why: Linked point (1) and development (1).

Some marks1/2

It checks the file name against the names already in the folder.

Why: Creditworthy point (1) on its own; no linked development, so one mark.

No marks0/2

It defragments the disk so there is room for the new file.

Why: Defragmentation is not accepted by the mark scheme.

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.

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