Menu

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.

Build a paper

Topics (none ticked: all of them)
Question types (none ticked: all of them)

Practice paper

3 questions · 4 marks

Pick again
Print
1Multiple choice1 markImages Swap Remove

Identify the smallest identifiable area of a bitmap image.

[1 mark]
  • AColour depth
  • BFile size
  • CPixel
  • DResolution
Mark scheme
  • C Pixel (1)
  • The only correct answer is C. The correct answer is pixel.
  • A is not correct because colour depth is the number of bits used to store the colour of each pixel
  • B is not correct because file size is the amount of storage an image takes up, not an area of the image
  • D is not correct because resolution is a measure of the number of pixels in an image (such as its width × height), not the smallest identifiable area
Example answers and how they were marked
Full marks1/1

C Pixel

Why: Correct option; a pixel is the smallest identifiable area of a bitmap image (1).

No marks0/1

D Resolution

Why: Resolution is a measure of the number of pixels in an image, not the smallest identifiable area; no mark.

No marks0/1

A Colour depth

Why: Colour depth is the number of bits used to store each pixel's colour, not the smallest identifiable area; no mark.

2Short recall2 marksSound Swap Remove

A sound wave is sampled and the amplitude of each sample is stored using a small number of bits.

Give two limitations of storing the sound in binary when only a small number of bits is available for each sample.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • Only a limited/fixed number of amplitude values can be stored (1)
  • The amplitude of each sample is rounded/approximated to the nearest value that can be stored (1)
  • The recording is less accurate than the original analogue sound/does not match the analogue wave exactly (1)
  • Sounds louder than the highest amplitude that can be stored are clipped/distorted (1)
  • Sounds quieter than the lowest amplitude that can be stored cannot be represented/are lost (1)

Allow quantisation error/noise for the amplitude being rounded to the nearest value that can be stored, but do not award a second mark for the same idea

Allow a stated consequence as a separate mark only when it is a different point from the list, e.g. fewer amplitude values can be stored (1) so very quiet sounds are lost (1)

Do not accept 'lower quality'/'poorer sound' on its own; the answer must say the recording is less accurate than the original/does not match the analogue wave

Do not accept sample rate/sample interval/number of channels (stereo/mono), as these are not limited by the number of bits available for each sample

Do not accept remedies such as 'a larger bit depth would make it more accurate', as these do not describe a limitation

Do not accept answers about file size, compression or playback method

Do not accept a statement that repeats the stem, e.g. 'the sound is stored in binary using a fixed number of bits'

Example answers and how they were marked
Full marks2/2

The amplitude of each sample has to be rounded to the nearest value that can be stored, and very quiet sounds cannot be represented.

Why: Amplitude rounded to the nearest stored value (1) and quiet sounds cannot be represented (1): two different points from the list.

Some marks1/2

The recording is less accurate than the original sound.

Why: The recording is less accurate than the original analogue sound (1); a second, different limitation is not given.

No marks0/2

The sample rate is too low, so parts of the wave are missed.

Why: Sample rate is not limited by the number of bits per sample, so it is not accepted.

3Short 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.

Questions by topic

More ways to revise Edexcel GCSE Computer Science (1CP2)

Teachers: build a paper, print it with the mark schemes at the end, and sign your school up to set your classes activities on every topic.

Fetch Decode Explain isn't affiliated with or endorsed by Pearson Edexcel.