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Edexcel GCSE Computer Science (1CP2) Exam Questions

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4 questions · 8 marks

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1Short recall3 marksFetch Decode Execute Cycle Swap Remove

The fetch-decode-execute cycle is used by the CPU.

State three things that happen during the fetch stage.

[3 marks]
Mark scheme

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

  • The address (of the next instruction) is sent from the CPU to main memory / placed on the address bus (1)
  • The (next) instruction is fetched/read from main memory (RAM) (1)
  • The instruction is brought/copied into the CPU / received into the MDR (1)
  • The program counter (PC) is incremented/updated to hold the address of the next instruction (1)

Allow 'the PC/MAR is placed on the address bus' for the address being sent to main memory.

Allow 'the MDR receives the instruction' for the instruction being brought into the CPU.

Do not award a second mark for the instruction being fetched from main memory and the instruction being brought into the CPU; these are the same point.

Do not award 'the instruction is transferred along the data bus' on its own; this is the same point as the instruction being fetched from main memory.

Do not accept answers about decoding the instruction, e.g. 'the instruction is fetched from RAM and decoded', as decoding happens in the decode stage.

Do not accept answers about the execute stage, e.g. 'the ALU carries out the calculation' or 'the result of a calculation is stored in a register', as these happen in the execute stage.

Example answers and how they were marked
Full marks3/3

The address of the next instruction is sent to main memory, the instruction is fetched from RAM, and the program counter is incremented.

Why: Address sent to main memory (1), instruction fetched from RAM (1) and program counter incremented (1).

Some marks1/3

The instruction is fetched from main memory and then decoded.

Why: Fetching the instruction from main memory (1); decoding happens in the decode stage, so it is not creditworthy in the fetch stage.

No marks0/3

The ALU carries out the calculation and the result is stored in a register.

Why: The ALU carrying out a calculation and storing a result happen in the execute stage, not the fetch stage, so no point is creditworthy.

2Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

3Describe2 marksLow and High Level Languages Swap Remove

A device driver is software that allows the operating system to control a piece of hardware. Device drivers are often written in a low-level programming language.

Describe one reason why a low-level programming language is suitable for writing a device driver.

[2 marks]
Mark scheme

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

  • Instructions can manipulate/control the hardware/memory/registers directly (1) so the driver can control the specific device it is written for (1)
  • The code is written for one specific processor/architecture (1) so it works with the exact hardware it will control (1)
  • The programmer controls exactly how memory/registers are used (1) so the code is efficient and needs little memory, which suits a device that has limited memory (1)
  • Each instruction maps to one machine code instruction (1) so the driver code is small enough to fit in the memory of the device (1)

Extra detail, such as naming the device the driver controls, is not needed to earn the marks.

Do not accept 'it is closer to machine code' on its own: it must be developed to show how this helps the driver, e.g. it can control the hardware directly.

Do not accept 'it is faster/quicker' with no link to the hardware/device, e.g. the device responds quickly.

Do not accept 'it is easier to learn', 'it is easier to debug' or 'it takes less time to write'.

Example answers and how they were marked
Full marks2/2

Low-level instructions can manipulate the hardware and memory directly (1) so the driver can control the specific device it was written for, such as a sound card (1).

Why: Point about direct control of hardware/memory (1) linked to controlling the specific device (1). The named device is extra detail and is not needed.

Some marks1/2

Low-level instructions can manipulate the hardware directly.

Why: Correct point (1) but no linked development saying how this helps the device driver.

No marks0/2

It is closer to machine code so it runs faster than a high-level language.

Why: 'Closer to machine code' is not developed into how this helps the driver, and 'runs faster' is not linked to the hardware/device, so no mark.

4Calculate2 marksAdding Binary Numbers Swap Remove

Computers manipulate binary patterns.

Complete the table to show the result of the 8-bit binary addition.

[2 marks]
1286432168421
10010101
01001111
Mark scheme

Award one mark for each correct nibble in the correct order:

  • 1110 (1)
  • 0100 (1)

0 marks if the response is not 8-bit binary.

Example answers and how they were marked
Full marks2/2

1110 0100

Why: 149 + 79 = 228, carrying where a column adds to 2: both nibbles are correct (2).

Some marks1/2

1110 0101

Why: A slip in the last bit. The first nibble is correct (1); the second is not.

No marks0/2

1101 1111

Why: Writes 1 in every column with a 1 and never carries. Neither nibble is correct.

Questions by topic

2.1 Binary

Build a paper on this topic 58
Adding Binary Numbers 6 questions Binary and Hexadecimal Basics 10 questions Binary Shifts 6 questions Binary to Denary Conversion 6 questions Binary to Hexadecimal Conversion 6 questions Denary to Binary Conversion 6 questions Denary to Hexadecimal Conversion 6 questions Denary to Negative Binary Conversion 6 questions Hexadecimal to Binary Conversion 6 questions

2.2 Data representation

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Character Sets 18 questions Images 19 questions Sound 19 questions

2.3 Data storage and compression

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Compression 18 questions Units & Conversion 6 questions

3.1 Hardware

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Embedded Systems 17 questions Fetch Decode Execute Cycle 15 questions Primary Memory 15 questions Secondary Memory 15 questions Stored Program Computers 15 questions

3.2 Software

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File Management 18 questions Memory Management 16 questions Peripheral Management 11 questions Process Management 12 questions Purpose & Function of Operating Systems 17 questions Software Basics 13 questions User Management 18 questions Utility Software 19 questions

3.3 Programming languages

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Low and High Level Languages 19 questions Translators 18 questions

4.1 Networks

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Addressing 16 questions LANs & WANs 16 questions Network Speed 5 questions Packet Switching 16 questions Protocols & Layers 17 questions The Internet 16 questions Topologies 17 questions Wired Connections 17 questions Wireless Connections 18 questions

4.2 Network security

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Network Security Risks 18 questions Preventing Network Security Risks 18 questions

5.1 Environmental

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E-waste 17 questions Manufacturing Technology 19 questions

5.2 Ethical and legal

Data Protection Act 11 questions

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