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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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3 questions · 9 marks

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1Multiple choice1 markCompression Swap Remove

Identify the definition of lossy compression.

[1 mark]
  • AData is removed only while the file is being transmitted
  • BData is permanently removed to reduce the file size
  • CData is restored exactly when it is decompressed
  • DData is stored using a dictionary of repeated patterns
Mark scheme
  • B Data is permanently removed to reduce the file size (1)
  • The only correct answer is B. The correct answer is data is permanently removed to reduce the file size.
  • A is not correct because the data removed by a lossy algorithm is not restored after transmission
  • C is not correct because that describes lossless compression, where the original file is rebuilt exactly
  • D is not correct because dictionary encoding is a lossless method that records repeated patterns once
Example answers and how they were marked
Full marks1/1

B Data is permanently removed to reduce the file size

Why: Correct option (1).

No marks0/1

C Data is restored exactly when it is decompressed

Why: Near miss: this describes lossless compression, so no mark.

No marks0/1

A Data is removed only while the file is being transmitted

Why: Misconception that compression only applies during transmission; the data is removed permanently, so no mark.

2Describe2 marksFile Management Swap Remove

A user saves a new file into a folder that already contains files.

Describe how the operating system prevents a new file overwriting an existing file.

[2 marks]
Mark scheme

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

  • Checks/compares the new file (name)/filename against the names of files already in the folder/directory (1) to see if the name is already in use (1)
  • Refuses to save/allow the new file when the (file) name already exists (1) so the existing file is not replaced/overwritten (1)
  • Prompts/asks the user to choose/enter a different (file) name (1) so both files can be saved without one replacing the other (1)

A point without a linked development can earn one mark.

For both marks, the development must follow/associate with the point.

Do not accept responses about defragmentation.

Do not accept the operating system making a backup copy, or warning the user that a file may be overwritten, without saying that the (file) name is checked or the new file is not saved.

Do not accept responses about process management or saving files to RAM.

Do not award a second mark for repeating the question, e.g. 'it stops the new file overwriting the existing file'.

Example answers and how they were marked
Full marks2/2

It checks the new file name against the names of files already in the folder to see whether the name is already in use.

Why: Linked point (1) and development (1).

Some marks1/2

It checks the file name against the names already in the folder.

Why: Creditworthy point (1) on its own; no linked development, so one mark.

No marks0/2

It defragments the disk so there is room for the new file.

Why: Defragmentation is not accepted by the mark scheme.

3Discuss6 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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