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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 693 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 · 5 marks

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1Multiple choice1 markCompression Swap Remove

Identify the definition of lossy compression.

[1 mark]
  • AData is removed only while the file is being transmitted
  • BData is permanently removed to reduce the file size
  • CData is restored exactly when it is decompressed
  • DData is stored using a dictionary of repeated patterns
Mark scheme
  • B Data is permanently removed to reduce the file size (1)
  • The only correct answer is B. The correct answer is data is permanently removed to reduce the file size.
  • A is not correct because the data removed by a lossy algorithm is not restored after transmission
  • C is not correct because that describes lossless compression, where the original file is rebuilt exactly
  • D is not correct because dictionary encoding is a lossless method that records repeated patterns once
Example answers and how they were marked
Full marks1/1

B Data is permanently removed to reduce the file size

Why: Correct option (1).

No marks0/1

C Data is restored exactly when it is decompressed

Why: Near miss: this describes lossless compression, so no mark.

No marks0/1

A Data is removed only while the file is being transmitted

Why: Misconception that compression only applies during transmission; the data is removed permanently, so no mark.

2Convert2 marksBinary to Hexadecimal Conversion Swap Remove

Computer systems manipulate bit patterns.

Convert the binary pattern 1000 1010 to hexadecimal.

[2 marks]
Mark scheme

Award one mark for each correct digit in the correct position:

  • 8 (1)
  • A (1)

Answer: 8A

Example answers and how they were marked
Full marks2/2

8A

Why: Each nibble is one hexadecimal digit: 1000 = 8, 1010 = A: both digits are correct (2).

Some marks1/2

810

Why: One digit is correct (1), but a digit over 9 is written in denary instead of as a letter.

No marks0/2

A8

Why: The digits are the wrong way round, so neither is in the correct position.

3Describe2 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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