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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 · 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.

2Multiple choice1 markCharacter Sets Swap Remove

Characters are encoded using 7-bit ASCII.

Identify the number of different characters that can be represented using 7-bit ASCII.

[1 mark]
  • A64
  • B127
  • C128
  • D256
Mark scheme
  • C 128 (1)
  • The only correct answer is C. The correct answer is 128.
  • A is not correct because 6 bits give only 64 different combinations
  • B is not correct because 127 is the largest 7-bit value, not the number of characters that 7 bits can represent
  • D is not correct because 256 characters need 8 bits, as used by 8-bit extended ASCII
Example answers and how they were marked
Full marks1/1

C 128

Why: Selects the only correct option (1): 7 bits give 2 to the power of 7, which is 128 different characters.

No marks0/1

D 256

Why: Near miss scoring 0: 256 is the number of characters that 8 bits give in extended ASCII, not 7-bit ASCII.

No marks0/1

B 127

Why: Misconception scoring 0, the off-by-one error the examiners' report describes: 127 is the largest 7-bit value, but 7 bits give 128 different characters.

3Short recall1 markSecondary Memory Swap Remove

Data is stored on optical discs as pits and lands.

Name the component in an optical disc drive that reads this data.

[1 mark]
Mark scheme

Award one mark for:

  • Laser (1)

Allow laser beam

Do not accept read/write head

Do not accept pits

Do not accept lands

Do not accept reflective surface

Do not accept magnet/magnetic

Example answers and how they were marked
Full marks1/1

Laser

Why: Laser is the creditworthy answer (1).

No marks0/1

A read/write head

Why: Read/write head is not accepted by the mark scheme; it reads magnetic media, not optical discs, so no marks are awarded.

No marks0/1

Pits and lands

Why: Pits and lands are what the laser reads, not the component that reads the data, so the mark scheme does not accept them.

4Short recall2 marksUnits & Conversion Swap Remove

Data storage is measured in binary multiples.

Define the term bit.

[2 marks]
Mark scheme

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

  • A binary digit (1)
  • Has the value 0 or 1 (1)
  • The smallest unit of data storage (1)
  • A single 0 or 1 (1)

Allow binary digit/binary place value for binary digit

Allow a single binary digit for a single 0 or 1

Do not accept a group of 8 bits/half a nibble of 4 bits, which describe a byte or a nibble

Do not accept a binary number/binary on its own

Example answers and how they were marked
Full marks2/2

A binary digit that can have the value 0 or 1.

Why: A binary digit (1) and has the value 0 or 1 (1).

Some marks1/2

It is the smallest unit of data storage.

Why: Smallest unit of data storage (1), but does not state that it is a binary digit or that it has the value 0 or 1.

No marks0/2

A group of 8 bits.

Why: Describes a byte, not a bit; no creditworthy point and named in the Do not accept line.

5Describe2 marksTranslators Swap Remove

A compiler and an interpreter both translate source code.

Describe one difference between what a compiler produces and what an interpreter produces.

[2 marks]
Mark scheme

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

  • A compiler produces object code/machine code/an executable (file) (1) whereas an interpreter produces no object code/translated file (1)
  • A compiler produces a machine code file (1) that can be run on its own/without the compiler/translator being present (1)
  • An interpreter produces no object code/file (1) because it translates and executes each line instead of saving translated code (1)
  • The compiler's output can be saved/stored (1) and run again later, while interpreted code has to be translated again each time it is run (1)

Allow binary for machine code.

Allow executable/an executable for object code.

Do not accept responses about translating the whole program, reporting errors or the speed of execution.

Do not accept 'the interpreter translates/executes one line at a time' on its own: this is how it translates, not what it produces. It earns the development only when linked to no object code/translated file being produced.

Do not accept the interpreter producing object code/an executable.

Example answers and how they were marked
Full marks2/2

A compiler produces object code, which is saved as a file, whereas an interpreter does not produce a file because it translates and executes each line.

Why: Compiler produces object code (1) contrasted with the interpreter producing no object code/translated file (1).

Some marks1/2

The compiler produces machine code.

Why: Machine code/object code produced (1); there is no contrast with the interpreter, so the development is not earned.

No marks0/2

The interpreter produces object code for each line it translates.

Why: Reverses the translators' outputs: the interpreter produces no object code, so nothing is creditworthy.

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