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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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4 questions · 6 marks

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1Short 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.

2Short recall1 markImages Swap Remove

A bitmap image has a resolution.

State what is meant by the resolution of an image.

[1 mark]
Mark scheme

Award one mark for any of the following:

  • The concentration/density of pixels in an area (1)
  • The number of pixels per inch (PPI) / pixels per unit area (1)
  • The width and height/dimensions of the image in pixels (1)

Do not accept the number of pixels in the image on its own, without dimensions or an area, e.g. 'the total number of pixels'

Do not accept colour depth / the number of bits used per pixel / the number of colours

Do not accept file size in bytes / the amount of storage the image uses / screen size / image quality

Example answers and how they were marked
Full marks1/1

The concentration of pixels in an area, such as the number of pixels per inch.

Why: Gives the density/concentration of pixels in an area, and pixels per inch (1)

No marks0/1

The number of pixels in the image.

Why: A bare total pixel count is not accepted without dimensions or an area

No marks0/1

The number of bits used to store each pixel's colour.

Why: Describes colour depth, not resolution, and colour depth is not accepted

3Describe2 marksCharacter Sets Swap Remove

A computer uses 7-bit ASCII to represent characters.

The character M has the ASCII code 77. Describe how the character M is represented in the computer.

[2 marks]
Mark scheme

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

  • The denary code 77 is converted to binary (1) to give the 7-bit pattern 1001101 (1)
  • The character M is represented by a unique ASCII code/number (1) that is stored as a 7-bit binary pattern (1)
  • 7 bits are used to store the code for M (1) so 128 different characters can be represented (1)
  • The binary pattern 1001101 is stored in memory (1) using one 7-bit code for each character (1)

Do not accept answers that only state the ASCII code 77, as this is given in the question.

Do not accept 01001101, as this is an 8-bit pattern.

Do not accept ASCII described as a programming language or a method of encryption.

Example answers and how they were marked
Full marks2/2

77 is converted to binary, giving 1001101. This is stored using 7 bits for the character M.

Why: Linked description: converts the code to binary (1) to give the 7-bit pattern 1001101 (1).

Some marks1/2

77 is changed into binary so it can be stored.

Why: Gives the conversion of 77 to binary (1) but does not give the 7-bit pattern 1001101 (0).

No marks0/2

ASCII is a programming language that stores letters as numbers.

Why: Describes ASCII as a programming language, which the mark scheme says is not creditworthy; no linked description of how M is represented.

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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5.3 Cybersecurity

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