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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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4 questions · 5 marks

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1Multiple choice1 markCharacter Sets Swap Remove

A new character set must be able to represent 70 different characters.

Identify the minimum number of bits needed to represent each character.

[1 mark]
  • A6
  • B7
  • C8
  • D70
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7.
  • A is not correct because 6 bits give only 64 combinations, which is not enough for 70 characters
  • C is not correct because 7 bits already give 128 combinations, so 8 bits is not the minimum needed
  • D is not correct because 70 is the number of characters needed, not a number of bits
Example answers and how they were marked
Full marks1/1

B 7

Why: The only correct answer: 7 bits give 128 combinations, enough for 70 characters, and it is the smallest number of bits that does (1).

No marks0/1

C 8

Why: Distractor, no credit: a single-tick item has no partial credit, and 8 bits is not the minimum because 7 bits already give 128 combinations, so the scheme's C line gives no mark.

No marks0/1

A 6

Why: Misconception, no credit: 6 bits give only 64 combinations, which cannot represent 70 different characters, so the scheme's A line gives no mark.

2Short recall1 markImages Swap Remove

Each pixel in a bitmap image is stored using a number of bits.

Define the term colour depth.

[1 mark]
Mark scheme

Award one mark for any of the following:

  • The number of bits per pixel (1)
  • The number of bits used to represent/store the colour of each pixel (1)

Allow bit depth

Do not accept the number of colours a pixel can have on its own

Do not accept the number of pixels in the image (this is the resolution/image size)

Example answers and how they were marked
Full marks1/1

The number of bits used to store the colour of one pixel.

Why: Gives the number of bits per pixel (1)

No marks0/1

The number of colours that each pixel can use.

Why: Gives the number of colours, not the number of bits; the mark scheme does not accept this on its own

No marks0/1

The number of pixels in the image.

Why: Confuses colour depth with resolution/image size, which the mark scheme does not accept

3Short recall1 markCharacter Sets Swap Remove

ASCII uses 7 bits to store each character. A system has only 6 bits available to store each character.

State the number of different characters the system can represent.

[1 mark]
Mark scheme

Award one mark for:

  • 64 (1)
  • 2^6 (1)
  • 2 × 2 × 2 × 2 × 2 × 2 (1)

Allow 64 characters / sixty-four.

Allow 2 to the power of 6 for 2^6.

Do not accept 128, which is the number of characters that can be represented using 7 bits.

Do not accept 63, the largest value that can be represented using 6 bits, rather than the number of different characters.

Do not accept 12 (2 × 6), ie the number of bits multiplied by 2.

Example answers and how they were marked
Full marks1/1

64

Why: Correct: 2^6 = 64 different characters (1).

No marks0/1

128

Why: Scores 0: 128 is the number of characters that 7 bits can represent, not the 6 bits the system has; the Do not accept line applies.

No marks0/1

63

Why: Scores 0: 63 is the largest value 6 bits can hold, not the number of different characters; the Do not accept line applies.

4Explain2 marksSound Swap Remove

Each sample of a sound wave is stored in binary using a fixed number of bits.

Explain one limitation of representing an analogue sound wave in binary when only a small number of bits is available for each sample.

[2 marks]
Mark scheme

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

  • Only a limited number of different amplitude values can be stored (1) so each sample is rounded to the nearest available value and the digital wave does not match the analogue wave exactly (1)
  • The wave is continuous but only a limited set of values can be represented (1) so amplitude detail/accuracy is lost (1)
  • Quiet and loud parts of the sound may be stored as the same value (1) because there are too few values to tell them apart (1)
  • The range of amplitudes that can be represented is limited (1) so the loudest/quietest parts of the sound cannot be stored accurately (1)

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

Do not award marks for answers that say the sound cannot be stored or heard at all.

Do not award answers about the number of samples taken/sample rate, as the number of bits limits the amplitude values, not how often the wave is sampled.

Do not accept 'lower quality'/'less accurate'/'less detail' on its own without a link to the amplitude values that can be stored/the number of bit combinations.

Do not accept 'the file size is smaller' without a link to the accuracy or the range of values that can be stored.

Example answers and how they were marked
Full marks2/2

Only a small number of amplitude values can be stored, so each sample is rounded to the nearest value and does not match the original wave exactly.

Why: Point (limited number of amplitude values) (1) with linked reason (samples are rounded and the wave is not exact) (1).

Some marks1/2

Only a small number of amplitude values can be stored.

Why: Creditworthy point (1) but no linked limitation such as rounding or lost accuracy, so only one mark.

No marks0/2

The sound cannot be stored at all because there are not enough bits to hold every sample.

Why: Misconception: the sound can still be stored, just less accurately, so no mark is awarded.

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