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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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1Short recall2 marksTopologies Swap Remove

Mesh is a type of network topology.

State two characteristics of a mesh topology.

[2 marks]
Mark scheme

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

  • Each node/device has multiple direct connections to other nodes/devices (in a full mesh, every node is connected directly to every other node) (1)
  • There is no central device/switch (1)
  • Nodes act as routers/relay data for other nodes / data can take multiple/different routes to its destination / data can still be sent if a connection/node fails (1)
  • More connections/cables are needed as the number of nodes increases (1)
  • The network can be expanded without disrupting existing devices (1)

Allow answers that refer to full or partial mesh.

Allow 'more connections/cables are needed' and 'can be expanded' as characteristics (they may be described as advantages or disadvantages).

Do not accept 'more secure' or 'no collisions' without qualification.

Do not accept 'fast' or 'reliable' without qualification.

Do not accept all devices connect to one main cable (this describes a bus topology).

Do not accept it uses a central switch/hub or data is sent through a central hub (this describes a star topology).

Do not accept 'expensive' without qualification (e.g. because more cables/connections are needed).

Example answers and how they were marked
Full marks2/2

Each device is connected to several other devices, and data can be sent along more than one route.

Why: Multiple direct connections (1) and data can take multiple routes (1).

Some marks1/2

There is no central switch.

Why: No central device credited (1), but only one characteristic given.

No marks0/2

Data is sent through a central hub.

Why: Misconception: this describes a star topology, not a mesh topology.

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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