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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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3 questions · 4 marks

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1Multiple choice1 markCharacter Sets Swap Remove

A text file stores each character using 7-bit ASCII.

Identify the denary ASCII code for the space character.

[1 mark]
  • A32
  • B33
  • C48
  • D65
Mark scheme
  • A 32 (1)
  • The only correct answer is A. The correct answer is 32.
  • B is not correct because 33 is the ASCII code for the exclamation mark (!)
  • C is not correct because 48 is the ASCII code for the character 0
  • D is not correct because 65 is the ASCII code for the uppercase letter A
Example answers and how they were marked
Full marks1/1

A 32

Why: Correct option: the space character has the ASCII code 32 (1).

No marks0/1

C 48

Why: Near miss scoring 0: 48 is the ASCII code for the character 0, not the space character.

No marks0/1

D 65

Why: Misconception scoring 0: 65 is the ASCII code for the uppercase letter A, so a well-known code is given but not the code for a space.

2Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

3Describe2 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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