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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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3 questions · 5 marks

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1Multiple choice1 markSound Swap Remove

A sound file is created by sampling an analogue sound wave.

Identify the term for the number of samples taken each second.

[1 mark]
  • ABit depth
  • BSample interval
  • CSample rate
  • DSample resolution
Mark scheme
  • C Sample rate (1)
  • The only correct answer is C. The correct answer is sample rate.
  • A is not correct because bit depth is the number of bits used to record each sample
  • B is not correct because the sample interval is the time between each sample
  • D is not correct because sample resolution is another name for the bit depth
Example answers and how they were marked
Full marks1/1

C Sample rate

Why: Selects the only correct option (1).

No marks0/1

B Sample interval

Why: Near miss: the sample interval is the time between samples, not the number of samples per second (0).

No marks0/1

D Sample resolution

Why: Misconception reported in examiners' reports: sample resolution means the bit depth, not the number of samples per second (0).

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 recall2 marksUnits & Conversion Swap Remove

A solid state drive has a capacity of 5 gibibytes.

Give an expression to calculate the number of bits in 5 gibibytes. You do not need to do the calculation.

[2 marks]
Mark scheme

Award one mark for each of:

  • 5 × 1024^3 / 5 × 1024 × 1024 × 1024 / 5 × 2^30, for the number of bytes (1)
  • A fully correct expression with × 8 to convert bytes to bits, in any order, e.g. 5 × 1024^3 × 8, 8 × 5 × 2^30 (1)

Accept 5 × 1 073 741 824 × 8.

Accept 5 × 2^33 or 42 949 672 960 as a fully correct expression (2 marks).

Do not accept 5 × 1000^3 × 8, which uses decimal multiples.

Do not accept ÷ 8, which converts bits to bytes.

Example answers and how they were marked
Full marks2/2

8 × 5 × 2^30

Why: 5 × 2^30 is the number of bytes (1) and × 8 converts bytes to bits (1).

Some marks1/2

5 × 2^30

Why: Correct number of bytes using the binary multiple 2^30 (1), but no conversion to bits.

No marks0/2

5 × 1000^3 × 8

Why: Uses decimal 1000^3 instead of binary 1024^3, so the expression is not fully correct; no creditworthy point.

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