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Edexcel GCSE Computer Science (1CP2) Exam Questions

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6 questions · 14 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).

2Multiple choice1 markStored Program Computers Swap Remove

Identify the component of the CPU that carries out arithmetic and logical operations.

[1 mark]
  • AArithmetic logic unit
  • BControl unit
  • CMain memory
  • DRegister
Mark scheme
  • A Arithmetic logic unit (1)
  • The only correct answer is A. The correct answer is arithmetic logic unit.
  • B is not correct because the control unit decodes instructions and controls the CPU
  • C is not correct because main memory is not part of the CPU and stores data and instructions
  • D is not correct because a register stores small amounts of data within the CPU
Example answers and how they were marked
Full marks1/1

A Arithmetic logic unit

Why: Correct option: the arithmetic logic unit carries out arithmetic and logical operations (1).

No marks0/1

B Control unit

Why: Control unit is a CPU component but its role is to decode and control, not perform arithmetic or logical operations, so no mark.

No marks0/1

D Register

Why: Register stores data but does not carry out operations; a common misconception that any CPU component performs calculations, so no mark.

3Short recall2 marksSecondary Memory Swap Remove

Computers have both primary storage and secondary storage.

Define the term secondary storage.

[2 marks]
Mark scheme

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

  • (Storage that is) non-volatile / keeps its data when the power is turned off (1)
  • (Storage that is) used to store programs and data/files (for later use) (1)
  • (Storage that is) not directly accessible by the CPU / not directly accessed by the CPU (1)

Allow both the non-volatile point and the stores programs and data point from one sentence if each idea is clearly stated, e.g. 'non-volatile storage that holds programs and data'.

Allow accessed via the CPU / must be copied into RAM before the CPU can use it for the third point.

Do not accept volatile / loses its data when the power is turned off.

Do not accept a description of primary storage, such as RAM or ROM.

Do not accept capacity.

Do not accept an example device on its own (e.g. hard disk drive).

Example answers and how they were marked
Full marks2/2

Secondary storage is non-volatile, so it does not lose data when the power is turned off. It is used to store programs and data files for later use.

Why: Non-volatile / keeps data when the power is off (1) and used to store programs and data files (1), each idea clearly stated.

Some marks1/2

Secondary storage is used to store programs and data files.

Why: Stores programs and data/files (1), but nothing is said about it being non-volatile or not directly accessible by the CPU.

No marks0/2

Secondary storage is volatile memory that holds the data the CPU is using at the moment.

Why: Describes primary storage (RAM) and says it is volatile, which the mark scheme says not to accept, so no marks are awarded.

4Describe2 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.

5Explain2 marksEmbedded Systems Swap Remove

An engine control unit (ECU) in a car is an embedded system.

Explain one benefit of using an embedded system for the engine control unit instead of a general purpose computer.

[2 marks]
Mark scheme

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

  • (The ECU) is more reliable/less likely to crash (1) because it only runs the one program/has a single purpose (1)
  • (The ECU) cannot be affected by other software/viruses (1) because it cannot run other software/it can only run the program it was made for (1)
  • (The ECU) is cheaper (1) because it has limited processing resources/less memory/simple components (1)
  • (The ECU) uses less power (1) because it has limited processing resources/a simpler processor (1)
  • (The ECU) is smaller (1) because it has fewer components/less memory than a general purpose computer (1)

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

A suitable benefit stated on its own earns one mark.

Do not accept a description or a location given as the reason, e.g. 'because it is part of the car', 'because it is inside the engine', 'because it is embedded'.

Do not accept a description of an embedded system on its own as the benefit, e.g. 'it does one specific job' or 'it is part of the car'.

Do not accept 'it is faster'/'it has a more powerful processor' as embedded systems have limited processing resources.

Example answers and how they were marked
Full marks2/2

It is more reliable than a general purpose computer (1) because it only runs the one program so it will not crash and stop the engine working (1).

Why: Benefit stated (1) with a linked reason (1); the reason shows why only running one program makes it reliable.

Some marks1/2

It is more reliable (1) because it is part of the car's engine (1).

Why: Only the benefit earns a mark (1); being part of the car is a location, not a linked reason why it is reliable.

No marks0/2

It is faster than a general purpose computer (0) because it is embedded in the engine (0).

Why: Misconception that an embedded system is faster; embedded systems have limited processing resources, so neither point nor reason is creditworthy.

6Discuss6 marksFetch Decode Execute Cycle Swap Remove

A small robotics company is designing a new robot. The robot's computer uses von Neumann architecture. The engineers are explaining how programs are stored and run.

Discuss how the von Neumann stored program concept and main memory (RAM) are used during the fetch-decode-execute cycle.

Your answer should include:

  • what is meant by the stored program concept
  • the role of RAM in the cycle
  • why the stored program concept is useful.
[6 marks]
Mark scheme

Indicative content

Stored program concept

  • Instructions and data are stored together in the same main memory (RAM).
  • Instructions are fetched one at a time from memory and executed in sequence.
  • The program can be changed by loading different instructions into RAM.
  • The CPU can only execute instructions that are in main memory.

Role of RAM

  • RAM holds the program instructions and the data being used.
  • The address of the next instruction is held in the program counter (PC).
  • The instruction is fetched from RAM using the address bus.
  • The data bus carries the instruction from RAM to the CPU.
  • RAM is volatile, so programs are lost when power is off.
  • RAM is directly accessible by the CPU, unlike secondary storage.

Benefits

  • The same hardware can run many different programs.
  • Programs can be loaded and changed quickly.
  • No need to rewire the computer for a new task.
  • Makes general-purpose computers possible.

Levels

  • Level 0 (0 marks): No rewardable content.
  • Level 1 (1–2 marks): Basic, independent points are made, showing elements of understanding of key concepts/principles of computer science. (AO1) The discussion will contain basic information with little linkage between points made or application to the context. (AO2)
  • Level 2 (3–4 marks): Demonstrates adequate understanding of key concepts/principles of computer science. (AO1) The discussion shows some linkages and lines of reasoning with some structure and application to the context. (AO2)
  • Level 3 (5–6 marks): Demonstrates comprehensive understanding of key concepts/principles of computer science to support the discussion being presented. (AO1) The discussion is well developed, with sustained lines of reasoning that are coherent and logically structured, and which clearly apply to the context. (AO2)
Example answers and how they were marked
Full marks6/6

The von Neumann stored program concept means that both the program instructions and the data are stored together in the same main memory, RAM. During the fetch-decode-execute cycle, the CPU fetches an instruction from RAM, decodes it in the control unit, and then executes it. The program counter holds the address of the next instruction, and this address is sent along the address bus to RAM. RAM then returns the instruction along the data bus to the CPU. RAM is volatile, so it only holds the program while the robot is powered on. The stored program concept is useful because the robot's computer can run different programs without changing any hardware; new instructions can simply be loaded into RAM. This makes the computer general-purpose and flexible.

Why: Comprehensive understanding: defines stored program concept, explains RAM role with PC, address bus and data bus, applies to robot, and gives benefits. Sustained lines of reasoning. Level 3.

Some marks3/6

The stored program concept is when instructions and data are stored in RAM. The CPU fetches instructions from RAM. The program counter keeps the address of the next instruction. RAM holds the program. The benefit is that you can change the program without rewiring the computer.

Why: Adequate understanding: correct basic points about stored program concept and RAM, with some linkage, but little development or application to the robot's fetch-decode-execute cycle. Level 2.

Some marks1/6

The von Neumann stored program concept is that the CPU uses RAM to store the program. The fetch-decode-execute cycle fetches, decodes and executes. RAM is random access memory. The benefit is that it is fast.

Why: Basic independent points only; no real linkage or application to the robot. Level 1.

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