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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 · 4 marks

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1Short recall1 markSound Swap Remove

To store sound in binary, an analogue sound wave is sampled at regular intervals.

Define the term 'sample rate'.

[1 mark]
Mark scheme

Award one mark for:

  • The number of samples taken per second (1)

Allow captured/recorded for taken.

Allow 'samples per second'.

Allow 'each second' for 'per second'.

Allow Hz/hertz as the unit.

Do not accept the time between samples/sample interval.

Do not accept 'the number of samples' without a per second/time reference.

Do not accept the number of bits used per sample/bit depth.

Example answers and how they were marked
Full marks1/1

The number of samples taken per second.

Why: States samples per second (1); 'per second' is accepted.

No marks0/1

The time between each sample.

Why: Defines sample interval, not sample rate; do not accept.

No marks0/1

The number of bits used for each sample.

Why: Describes bit depth, not sample rate; do not accept.

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.

3Describe2 marksSound Swap Remove

Sound is converted into binary using a limited bit depth.

Describe one limitation of representing a sound sample with a limited bit depth.

[2 marks]
Mark scheme

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

  • Only a fixed number of amplitude values/levels can be represented (1) so the amplitude is rounded to the nearest available value/level (1)
  • Fewer amplitude values/levels are available to choose from (1) so similar/close amplitudes cannot be stored as different values (1)
  • The exact amplitude often falls between available values/levels (1) so the stored amplitude is not exact/less accurate (1)

Answers must relate to the number of bits per sample (bit depth) and the amplitude values/levels it can represent.

Do not accept answers about the sample rate, sample interval or the number of samples taken, or that confuse bit depth with the number of samples.

Do not award 'lower quality'/'more detail'/'less detail' unless it is linked to the number of amplitude values/levels.

Do not accept 'the sound cannot be played/stored' without a linked limitation.

Example answers and how they were marked
Full marks2/2

Only a fixed number of amplitude values can be represented, so the amplitude is rounded to the nearest available value and some detail is lost.

Why: Linked points: fixed number of amplitude values (1) and rounding to the nearest available value (1).

Some marks1/2

Only a fixed number of amplitude values can be represented.

Why: Gives the first linked point only (1); it does not develop what this limitation leads to.

No marks0/2

A lower bit depth means fewer samples are taken, so parts of the sound are missed.

Why: Confuses bit depth with the number of samples/sample rate; no creditworthy point about amplitude values is given.

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5.3 Cybersecurity

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