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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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3 questions · 5 marks

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1Short recall2 marksTopologies Swap Remove

Networks can use different topologies.

Define the term network topology.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • The layout/arrangement/structure of a network (1)
  • How devices/nodes are connected together (1)

Allow 'the physical or logical layout/arrangement of a network' for the first point.

Allow 'the way parts of a system/network are connected' for the second point.

Do not award the first mark for 'how devices are connected' on its own.

Do not accept a named topology (e.g. star, bus, mesh) on its own.

Do not accept a network device or how data is transmitted on its own.

Example answers and how they were marked
Full marks2/2

The layout of a network and how the devices in it are connected together.

Why: Layout/arrangement of a network (1) and how devices/nodes are connected together (1).

Some marks1/2

The way devices are connected in a network.

Why: How devices/nodes are connected together (1), but no reference to the layout/arrangement/structure of the network.

No marks0/2

A star network.

Why: Near miss: names a topology but does not define the term.

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.

Questions by topic

5.3 Cybersecurity

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