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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 · 7 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 recall2 marksSecondary Memory Swap Remove

A computer uses secondary storage to keep programs and data.

Give two examples of secondary storage devices that store data using magnetic media.

[2 marks]
Mark scheme

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

  • Hard disk drive/HDD (1)
  • External/portable hard disk drive (1)
  • Magnetic tape drive (1)
  • Magnetic tape (1)

Allow a named device or the storage media type, e.g. hard disk/hard disc for hard disk drive, tape for magnetic tape

Allow floppy disk/floppy disk drive

Do not accept solid state devices (SSD, USB flash drive/memory stick, SD card/memory card) or optical media (CD/DVD/Blu-ray) as these do not store data using magnetism

Example answers and how they were marked
Full marks2/2

Hard disk drive and magnetic tape

Why: Hard disk drive (1) and magnetic tape (1) are both magnetic secondary storage, so both marks are awarded.

Some marks1/2

Hard disk drive and SD card

Why: Hard disk drive is magnetic (1); an SD card is solid state, not magnetic, so the second mark is not awarded.

No marks0/2

CD and DVD

Why: Both are optical media, which the mark scheme says not to accept, so no marks are awarded.

3Describe2 marksStored Program Computers Swap Remove

The CPU contains an arithmetic logic unit (ALU).

Describe the purpose of the arithmetic logic unit (ALU).

[2 marks]
Mark scheme

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

  • Carries out arithmetic operations/calculations, such as addition or subtraction (1) so that the result can be stored in a register (1)
  • Carries out logical operations, such as comparing two data items (1) to see which one is greater/larger (1)
  • Compares two data items (1) to decide whether a condition is true / which path a program takes next (1)

Allow the result of an arithmetic or logical operation being held/stored in a register, but only where the operation is named.

Do not accept answers that say the ALU stores data or results without naming an operation (registers/main memory store data).

Do not accept answers about controlling components, decoding instructions or timing (that is the control unit or the clock).

Do not award credit for 'it processes data' on its own; an arithmetic or a logical operation must be named.

An example on its own, e.g. 'it adds two numbers', does not earn a second mark; the second mark needs a linked development.

Example answers and how they were marked
Full marks2/2

The ALU carries out arithmetic operations such as addition and subtraction, and the result of the calculation is stored in a register.

Why: Linked point (1) for arithmetic operations with an example, and development (1) for the result being stored in a register.

Some marks1/2

It does all the calculations in the CPU, for example adding two numbers together.

Why: Point (1) for arithmetic operations/calculations named with an example; no linked development, so the second mark is not earned.

No marks0/2

It stores the results of calculations so that they can be used by the CPU later.

Why: Storing results with no arithmetic or logical operation named, which the 'Do not accept' line rules out, so no mark is earned.

4Describe2 marksFetch Decode Execute Cycle Swap Remove

The control unit is one component of the CPU.

Describe the role of the control unit in the fetch-decode-execute cycle.

[2 marks]
Mark scheme

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

  • Decodes instructions (1) to determine what action/operation is required (1)
  • Sends control signals (1) along the control bus to coordinate/control the actions of other components (1)
  • Controls the fetch stage (1) by causing the next instruction to be fetched from main memory/RAM (1)
  • Controls the execute stage (1) by telling the ALU/other components when to carry out the required operation (1)

Do not accept responses relating to how the control unit affects performance, such as increasing clock speed.

Do not accept responses that describe the role of the ALU, registers or clock instead of the control unit.

Do not accept responses about the ALU performing calculations unless the control unit controlling the ALU is also stated.

Example answers and how they were marked
Full marks2/2

The control unit decodes instructions to determine what action is required.

Why: Linked point (1) and development (1): decodes instructions and determines the required action.

Some marks1/2

The control unit controls the fetch stage of the cycle.

Why: Point only (1): controls the fetch stage, but no linked development.

No marks0/2

The control unit carries out arithmetic and logical operations on data.

Why: No creditworthy point: describes the role of the ALU, which the mark scheme rejects.

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