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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 659 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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Practice paper

2 questions · 2 marks

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1Multiple choice1 markCompression Swap Remove

A student downloads a compressed music file and then decompresses it.

Identify what is meant by decompression.

[1 mark]
  • AData is encrypted so that it cannot be read
  • BData is permanently removed to reduce the file size
  • CData is restored to its original format
  • DData is stored using fewer bits
Mark scheme
  • C Data is restored to its original format (1)
  • The only correct answer is C. The correct answer is data is restored to its original format.
  • A is not correct because encryption changes data so that it cannot be read without a key; it does not restore the data
  • B is not correct because permanently removing data to reduce the file size is lossy compression, not decompression
  • D is not correct because storing data using fewer bits is compression, not decompression
Example answers and how they were marked
Full marks1/1

C Data is restored to its original format

Why: Correct option (1).

No marks0/1

D Data is stored using fewer bits

Why: Near miss: this describes compression, not decompression.

No marks0/1

B Data is permanently removed to reduce the file size

Why: Misconception: this describes lossy compression, not decompression.

2Short recall1 markCharacter Sets Swap Remove

A computer can store a maximum of 80 bits for a text message. Each character is stored using 7-bit ASCII.

State the maximum number of whole characters that can be stored in the message.

[1 mark]
Mark scheme

Award one mark for:

  • 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Example answers and how they were marked
Full marks1/1

11

Why: Correct: 80 ÷ 7 = 11 remainder 3, so 11 whole characters can be stored (1).

No marks0/1

11.4

Why: Near miss: gives a decimal result rather than the number of whole characters; the mark scheme does not accept 11.4.

No marks0/1

12

Why: Rounds up to 12, but 12 characters would need 84 bits, more than the 80 available; the mark scheme does not accept 12.

Questions by topic

5.3 Cybersecurity

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