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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 · 4 marks

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1Multiple choice1 markCompression Swap Remove

Identify the statement that describes lossless compression.

[1 mark]
  • AData is encrypted so that it cannot be read by others
  • BData is permanently removed to make the file smaller
  • CNo data is lost and the original file can be recreated
  • DThe file is split into packets that are sent separately
Mark scheme
  • C No data is lost and the original file can be recreated (1)
  • The only correct answer is C. The correct answer is no data is lost and the original file can be recreated.
  • A is not correct because encryption hides data and does not make the file smaller
  • B is not correct because permanently removing data is lossy compression
  • D is not correct because splitting a file into packets is part of transmission, not compression
Example answers and how they were marked
Full marks1/1

C No data is lost and the original file can be recreated

Why: Correct option (1).

No marks0/1

B Data is permanently removed to make the file smaller

Why: Near miss: this is lossy compression, not lossless, so no mark.

No marks0/1

D The file is split into packets that are sent separately

Why: Misconception that compression is packet switching; no mark.

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.

3Convert2 marksDenary to Binary Conversion Swap Remove

Binary patterns can represent numbers.

Convert the denary number 199 to 8-bit binary.

[2 marks]
Mark scheme

Award one mark for each correct nibble in the correct order:

  • 1100 (1)
  • 0111 (1)

0 marks if the response is not 8-bit binary.

Example answers and how they were marked
Full marks2/2

1100 0111

Why: 199 = 128 + 64 + 4 + 2 + 1: both nibbles are correct (2).

Some marks1/2

1100 1000

Why: Converts 200 instead (a slip in adding up the place values). The first nibble is correct (1); the second is not.

No marks0/2

1110 0011

Why: The bits are written in reverse, smallest place value first. Neither nibble is correct.

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