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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 markFetch Decode Execute Cycle Swap Remove

Identify the component that stores instructions and data during the fetch-decode-execute cycle.

[1 mark]
  • AArithmetic logic unit
  • BClock
  • CMain memory (RAM)
  • DProgram counter
Mark scheme
  • C Main memory (RAM) (1)
  • The only correct answer is C. The correct answer is main memory (RAM); accept 'main memory', 'RAM' or 'random access memory'.
  • A is not correct because the arithmetic logic unit performs arithmetic and logical operations
  • B is not correct because the clock sends regular pulses to synchronise the components of the CPU
  • D is not correct because the program counter holds the address of the next instruction, not the instructions and data
Example answers and how they were marked
Full marks1/1

C Main memory (RAM)

Why: Correct option: main memory (RAM) stores instructions and data during the fetch-decode-execute cycle (1).

No marks0/1

D Program counter

Why: Near-miss distractor: the program counter holds the address of the next instruction, not the instructions and data (0).

No marks0/1

A Arithmetic logic unit

Why: Misconception: the ALU performs arithmetic and logical operations, so it does not store instructions and data (0).

2Short recall1 markSecondary Memory Swap Remove

Data is stored on optical discs as pits and lands.

Name the component in an optical disc drive that reads this data.

[1 mark]
Mark scheme

Award one mark for:

  • Laser (1)

Allow laser beam

Do not accept read/write head

Do not accept pits

Do not accept lands

Do not accept reflective surface

Do not accept magnet/magnetic

Example answers and how they were marked
Full marks1/1

Laser

Why: Laser is the creditworthy answer (1).

No marks0/1

A read/write head

Why: Read/write head is not accepted by the mark scheme; it reads magnetic media, not optical discs, so no marks are awarded.

No marks0/1

Pits and lands

Why: Pits and lands are what the laser reads, not the component that reads the data, so the mark scheme does not accept them.

3Describe2 marksSecondary Memory Swap Remove

A computer uses a magnetic hard disk.

Describe how data is stored on a magnetic hard disk.

[2 marks]
Mark scheme

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

  • A read/write head moves over the surface of the disk (1) to read/write data by detecting or changing the magnetic alignment of areas of the surface (1)
  • The head magnetises areas of the disk surface in different directions (1) so each area represents a binary 1 or 0 (1)
  • Data is stored on concentric circles called tracks (1) which are divided into sectors (1)

Allow one mark for tracks/concentric circles on their own.

For the second mark in the first two bullets, the response must refer to the magnetic alignment/charge being detected or changed, not just that the head reads/writes data.

Do not award a mark for 'it is magnetic' or 'it stores data in binary' with no description of how the data is stored.

Do not accept descriptions of optical storage (e.g. a laser, pits and lands) or solid state storage (e.g. electric charges in chips).

Example answers and how they were marked
Full marks2/2

The read/write head moves over the magnetic surface and magnetises areas of the disk to store the 1s and 0s.

Why: Head moves over the surface (1) and magnetises areas to represent 1s and 0s (1).

Some marks1/2

The data is stored on tracks on the disk.

Why: Tracks/concentric circles (1), but no development such as tracks divided into sectors.

No marks0/2

It is magnetic and saves the data in 1s and 0s.

Why: Near-miss: 'magnetic' alone and 'data in binary/1s and 0s' with no description of how data is stored, which the mark scheme does not award.

4Explain2 marksSecondary Memory Swap Remove

A Blu-ray disc and a CD are the same physical size.

Explain why a Blu-ray disc can store more data than a CD.

[2 marks]
Mark scheme

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

  • The pits and lands on a Blu-ray disc are smaller/closer together (1) because the blue laser has a shorter wavelength than the infrared laser used by a CD (1)
  • More pits and lands can be stored in the same length of track (1) because the blue laser has a shorter wavelength, so it can focus on a smaller area/read smaller details (1)
  • A blue laser can focus on a smaller area (1) so smaller pits and lands can be used, allowing more data to be stored on the disc (1)

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

Do not accept 'a Blu-ray is newer/better' or 'a Blu-ray has more space/bigger storage' without a linked reason about the laser/pits and lands.

Do not accept answers about read/write speed, for example 'a Blu-ray spins faster', because the question asks about the amount of data stored, not how quickly it is read.

Do not award 'a Blu-ray disc can have more layers' on its own as the reason; it must be linked to the smaller pits and lands/shorter wavelength of the laser.

Do not accept 'red laser' for a CD; a CD uses an infrared laser (red is used by DVD).

Example answers and how they were marked
Full marks2/2

A Blu-ray is read with a blue laser which has a shorter wavelength so the pits and lands on the disc can be smaller and more of them fit on the disc.

Why: Linked point (1) shorter wavelength of the blue laser with reason (1) smaller pits and lands so more data fits on the same disc.

Some marks1/2

The pits and lands on a Blu-ray are smaller than the ones on a CD so it can store more data.

Why: Point (1) that the pits and lands are smaller, but the second part repeats the question and does not give the linked reason (shorter wavelength of the blue laser/infrared CD laser), so only one mark.

No marks0/2

A Blu-ray spins faster than a CD so more data can be read from it in the same time.

Why: Do not accept answers about read/write speed; the question asks about the amount of data stored, not how quickly it is read.

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