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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 · 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 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.

3Describe2 marksCharacter Sets Swap Remove

Uppercase letters are represented in a computer using 7-bit ASCII codes.

Describe how the ASCII codes for the uppercase letters A to Z are arranged.

[2 marks]
Mark scheme

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

  • The codes run in alphabetical order/consecutively (1) so each letter has a code one more than the letter before it (1)
  • A has the code 65 and Z has the code 90 (1) so the codes for the letters in between increase by 1 each time (1)
  • The letters are stored in sequence (1) so B is A + 1, C is B + 1 and so on (1)

Accept the codes given in denary (65, 90) or as 7-bit binary (1000001, 1011010).

Do not accept 'the codes go up' or 'the codes are in order' on its own, as neither gives the step of 1 or links one letter to the next.

Do not accept 'each uppercase letter has a different code' on its own, as this does not describe the arrangement.

Example answers and how they were marked
Full marks2/2

The codes are in alphabetical order, so B is one more than A and C is one more than B.

Why: Codes are consecutive/in order (1) with the linked step of one more for each letter (1).

Some marks1/2

A is 65 and Z is 90.

Why: Gives the first half of the linked pair, the start and end codes (1), but does not give the linked development, the increase of 1 between letters (0).

No marks0/2

The ASCII codes for uppercase letters go up.

Why: 'Codes go up' on its own gives no step of 1 and no link to the next letter, so it is not creditworthy (0) under the 'Do not accept' line.

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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