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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 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 markPrimary Memory Swap Remove

A computer is running an application. The application's data and instructions are transferred from secondary storage into RAM before the CPU processes them.

Identify the reason for this transfer.

[1 mark]
  • ARAM is non-volatile
  • BRAM is faster to access than secondary storage
  • CRAM has a larger capacity than secondary storage
  • DRAM is cheaper than secondary storage
Mark scheme
  • B RAM is faster to access than secondary storage (1)
  • The only correct answer is B. The correct answer is RAM is faster to access than secondary storage.
  • A is not correct because RAM is volatile
  • C is not correct because RAM has a smaller capacity than secondary storage
  • D is not correct because RAM is more expensive than secondary storage
Example answers and how they were marked
Full marks1/1

B RAM is faster to access than secondary storage

Why: Selects the correct option: RAM is faster to access than secondary storage, so data and instructions can reach the CPU quickly (1).

No marks0/1

A RAM is non-volatile

Why: Near miss: RAM is volatile; this does not explain the transfer into RAM.

No marks0/1

C RAM has a larger capacity than secondary storage

Why: Misconception: assumes primary memory has a larger capacity, but RAM has a smaller capacity than secondary storage.

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.

3Short recall2 marksUnits & Conversion Swap Remove

Data storage is measured in binary multiples.

Define the term bit.

[2 marks]
Mark scheme

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

  • A binary digit (1)
  • Has the value 0 or 1 (1)
  • The smallest unit of data storage (1)
  • A single 0 or 1 (1)

Allow binary digit/binary place value for binary digit

Allow a single binary digit for a single 0 or 1

Do not accept a group of 8 bits/half a nibble of 4 bits, which describe a byte or a nibble

Do not accept a binary number/binary on its own

Example answers and how they were marked
Full marks2/2

A binary digit that can have the value 0 or 1.

Why: A binary digit (1) and has the value 0 or 1 (1).

Some marks1/2

It is the smallest unit of data storage.

Why: Smallest unit of data storage (1), but does not state that it is a binary digit or that it has the value 0 or 1.

No marks0/2

A group of 8 bits.

Why: Describes a byte, not a bit; no creditworthy point and named in the Do not accept line.

4Describe2 marksTranslators Swap Remove

A high-level program must be translated into machine code before it can be run.

Describe how a compiler and an interpreter differ in when the translated code is executed.

[2 marks]
Mark scheme

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

  • The interpreter executes each line as soon as it has been translated (1) so translation and execution happen together/at the same time (1)
  • The compiler translates the whole program first (1) and the machine code is run later/separately (1)
  • The interpreter translates the code and runs it while the program is running (1) whereas the compiler does not run the code it produces (1)
  • The compiler produces translated code that is executed only after all the translation has finished (1) but the interpreter executes each line immediately (1)

Allow 'translates one line at a time' where it is linked to executing each line immediately.

Do not accept answers that only state that a compiler is faster/an interpreter is slower, without reference to when execution happens.

Do not accept answers that only give object code, error reporting or portability, without reference to when execution happens.

Do not accept the compiler executing the program as it translates it.

Example answers and how they were marked
Full marks2/2

The compiler translates the code first and the program is executed afterwards, but the interpreter executes each line as soon as it has been translated.

Why: Compiler translates the whole program first and it is executed afterwards (1) and interpreter executes each line as soon as it is translated (1).

Some marks1/2

The interpreter executes each line immediately after translating it.

Why: Correct interpreter timing (1); no comparison with when the compiler's translated code is executed.

No marks0/2

The compiler executes the program line by line as it translates it.

Why: Reverses the two translators; a compiler does not execute the program it translates, so no mark.

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