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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 · 3 marks

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1Multiple choice1 markUnits & Conversion Swap Remove

Units of data storage can be binary or decimal.

Identify the unit of storage that is equivalent to 1024 gibibytes.

[1 mark]
  • AGibibyte
  • BMebibyte
  • CTebibyte
  • DTerabyte
Mark scheme
  • C Tebibyte (1)
  • The only correct answer is C. The correct answer is tebibyte.
  • A is not correct because a gibibyte is 1024 mebibytes
  • B is not correct because a mebibyte is 1024 kibibytes
  • D is not correct because a terabyte is 1000 gigabytes
Example answers and how they were marked
Full marks1/1

C Tebibyte

Why: Correct: 1024 gibibytes make one tebibyte (1).

No marks0/1

A Gibibyte

Why: Near miss: a gibibyte is 1024 mebibytes, not 1024 gibibytes (0).

No marks0/1

D Terabyte

Why: Typical misconception: uses the decimal terabyte (1000 gigabytes) instead of the binary tebibyte (0).

2Multiple choice1 markLANs & WANs Swap Remove

Networks can be classified by the area they cover.

Identify the type of network that the internet is an example of.

[1 mark]
  • ALAN
  • BMAN
  • CPAN
  • DWAN
Mark scheme
  • D WAN (1)
  • The only correct answer is D. The correct answer is WAN (wide area network).
  • A is not correct because a LAN covers a small geographical area, such as one building
  • B is not correct because a MAN covers a city or large town
  • C is not correct because a PAN connects devices around one person, such as with Bluetooth
Example answers and how they were marked
Full marks1/1

D WAN

Why: Correct option: the internet is the largest example of a WAN (1).

No marks0/1

B MAN

Why: Near-miss: a MAN covers a city or large town, not networks all over the world, so the mark scheme gives it 0.

No marks0/1

A LAN

Why: Misconception: the candidate treats the internet as a LAN because their home network is a LAN; the mark scheme says a LAN covers a small geographical area, so 0.

3Short 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.

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