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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 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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4 questions · 6 marks

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1Multiple choice1 markCharacter Sets Swap Remove

Characters are encoded using 7-bit ASCII.

Identify the number of different characters that can be represented using 7-bit ASCII.

[1 mark]
  • A64
  • B127
  • C128
  • D256
Mark scheme
  • C 128 (1)
  • The only correct answer is C. The correct answer is 128.
  • A is not correct because 6 bits give only 64 different combinations
  • B is not correct because 127 is the largest 7-bit value, not the number of characters that 7 bits can represent
  • D is not correct because 256 characters need 8 bits, as used by 8-bit extended ASCII
Example answers and how they were marked
Full marks1/1

C 128

Why: Selects the only correct option (1): 7 bits give 2 to the power of 7, which is 128 different characters.

No marks0/1

D 256

Why: Near miss scoring 0: 256 is the number of characters that 8 bits give in extended ASCII, not 7-bit ASCII.

No marks0/1

B 127

Why: Misconception scoring 0, the off-by-one error the examiners' report describes: 127 is the largest 7-bit value, but 7 bits give 128 different characters.

2Short recall2 marksStored Program Computers Swap Remove

Computers use the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

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

  • Program instructions are stored in main memory/RAM (1)
  • The data used by the program is stored in main memory/RAM (1)
  • The CPU fetches/reads instructions from main memory to execute them (1)
  • The program can be run without the instructions being input again (1)

Allow 'the program is stored in memory' for one mark.

Do not accept answers about programs being stored on secondary storage, such as a hard disk or USB memory stick.

Do not accept answers that only describe the von Neumann architecture (e.g. having a CPU and memory) without stating that instructions are stored in main memory.

Example answers and how they were marked
Full marks2/2

Both the program instructions and the data the program uses are held in the same main memory, and the CPU fetches these instructions from memory to execute them.

Why: Instructions stored in main memory (1) and the CPU fetches instructions from memory to execute them (1).

Some marks1/2

Program instructions are stored in main memory.

Why: Instructions stored in main memory (1). No second point about data being stored in the same memory or about the CPU fetching instructions.

No marks0/2

The program is saved on the hard disk so that it can be loaded into RAM when the user runs it.

Why: A near-miss about secondary storage, which the mark scheme says not to accept; the stored program concept is about instructions being held in main memory.

3Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

4Explain2 marksTopologies Swap Remove

A bus topology has a terminator at each end of its cable.

Explain why terminators are needed in a bus topology.

[2 marks]
Mark scheme

Award up to two marks for a linked explanation such as:

  • Terminators absorb the signal at the ends of the cable (1) so it is not reflected back along the bus (1)
  • Terminators prevent signals reflecting/bouncing back (1) so the reflected signal does not corrupt data on the cable (1)
  • Terminators absorb the signal (1) so it does not collide/interfere with data being sent by other devices (1)
  • Terminators stop the signal reflecting back down the cable (1) so data can be transmitted and received without errors (1)

For both marks, the expansion must follow/associate with the statement.

Allow 'interfere'/'corrupt' only where the answer says the reflected signal mixes with/collides with data on the cable.

Do not accept 'to stop the signal' on its own, as stopping the signal is not the same as stopping it reflecting back along the cable.

Do not accept 'to terminate the cable/network', 'to end the cable' or 'to stop the cable breaking'.

Do not award responses that only repeat the stem, e.g. 'because there is a terminator at each end'.

Example answers and how they were marked
Full marks2/2

Terminators absorb the signal at each end of the cable (1) so that it does not reflect back along the bus (1).

Why: Linked point (1) and reason (1): absorbing the signal is linked to it not being reflected back along the bus.

Some marks1/2

Terminators absorb the signal at the ends of the cable.

Why: Awarded 1 mark for the point that the signal is absorbed, but no linked consequence is given.

No marks0/2

They terminate the network so the cable does not break.

Why: No credit: it does not say the signal is absorbed or that reflections are prevented, and the mark scheme rejects 'to terminate the network' and 'to stop the cable breaking'.

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