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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 · 5 marks

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1Short recall1 markTopologies Swap Remove

A bus topology uses terminators at each end of the main cable.

State the purpose of terminators in a bus topology.

[1 mark]
Mark scheme

Award one mark for:

  • Absorb signals (at the ends of the bus cable) / prevent signals reflecting (back along the cable) (1)

Allow 'stop data/signals bouncing back' for prevent signals reflecting

Do not accept 'terminate the cable' / 'close the end of the cable' on its own

Do not accept boost/repeat/strengthen the signal

Example answers and how they were marked
Full marks1/1

To absorb signals so that they do not reflect back along the bus cable.

Why: Absorbing signals and preventing reflection are alternatives within the one creditworthy point (1).

No marks0/1

To terminate the cable at each end.

Why: Near miss: this is rejected by the 'terminate the cable' guidance because it does not say signals are absorbed or prevented from reflecting.

No marks0/1

To make the signal stronger at the end of the cable.

Why: Misconception: terminators absorb signals; they do not boost, repeat or strengthen them.

2Convert2 marksBinary to Hexadecimal Conversion Swap Remove

Binary patterns can represent numbers.

Convert the binary pattern 1110 0100 to hexadecimal.

[2 marks]
Mark scheme

Award one mark for each correct digit in the correct position:

  • E (1)
  • 4 (1)

Answer: E4

Example answers and how they were marked
Full marks2/2

E4

Why: Each nibble is one hexadecimal digit: 1110 = E, 0100 = 4: both digits are correct (2).

Some marks1/2

144

Why: One digit is correct (1), but a digit over 9 is written in denary instead of as a letter.

No marks0/2

4E

Why: The digits are the wrong way round, so neither is in the correct position.

3Describe2 marksLow and High Level Languages Swap Remove

A programmer has written a program in assembly language.

Describe why it is necessary to translate the program into machine code before it can be run.

[2 marks]
Mark scheme

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

  • The processor/CPU can only execute machine code/binary instructions (1) so the assembly instructions must be converted into machine code before the program can run (1)
  • Each assembly instruction/mnemonic matches/corresponds to one machine code instruction (1) so each line must be replaced with the machine code the processor executes (1)
  • The program is written using mnemonics (1) that must be translated into machine code for the processor to carry out the instructions (1)
  • The hardware/processor does not understand assembly language directly (1) so it must be translated into machine code that the processor can execute (1)

Do not accept 'assembly language is a low-level language'.

Do not accept 'it is hard to understand' or 'it is not English'.

Do not accept benefits, such as 'machine code runs faster' or 'the program will run more quickly'.

Do not accept 'an assembler/a translator converts it' on its own, with no reference to the processor only being able to execute machine code/binary.

Do not accept responses about high-level languages or about compilers/interpreters.

Example answers and how they were marked
Full marks2/2

The processor can only execute machine code, so the assembly instructions must be converted into machine code before the program can run.

Why: States what the processor can execute (1) and links it to converting the program's instructions into machine code (1).

Some marks1/2

Each assembly instruction matches one machine code instruction.

Why: Correct point about the one-to-one match (1) but no development saying that this is because the processor only executes machine code.

No marks0/2

Because an assembler translates assembly language into machine code.

Why: Names the translator but gives no reason tied to the processor only being able to execute machine code, so no mark; this is in the 'Do not accept' guidance.

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