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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 530 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

ASCII is a character set used for representing text.

Identify the number of bits used to encode each character in ASCII.

[1 mark]
  • A6
  • B7
  • C8
  • D16
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7 bits.
  • A is not correct because 6 bits give only 64 different characters
  • C is not correct because 8 bits are used by extended ASCII to give 256 characters
  • D is not correct because 16 bits give 65 536 different characters, which is far more than the 128 in ASCII
Example answers and how they were marked
Full marks1/1

B 7

Why: B is the only correct answer: each ASCII character is encoded using 7 bits (1).

No marks0/1

C 8

Why: Near miss scoring 0: 8 bits is extended ASCII, and the scheme's C line says extended ASCII, not ASCII, is not the answer.

No marks0/1

A 6

Why: Misconception scoring 0: 6 bits give only 64 different characters, so it cannot be ASCII.

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

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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