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Edexcel GCSE Computer Science (1CP2) Exam Questions

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5 questions · 13 marks

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1Multiple choice1 markStored Program Computers Swap Remove

Identify the function of the control unit in the fetch-decode-execute cycle.

[1 mark]
  • ADecodes instructions
  • BPerforms arithmetic operations
  • CStores data and instructions
  • DTemporarily stores data being processed
Mark scheme
  • A Decodes instructions (1)
  • The only correct answer is A. The correct answer is Decodes instructions.
  • B is not correct because the arithmetic logic unit performs arithmetic operations
  • C is not correct because main memory stores data and instructions
  • D is not correct because registers temporarily store data being processed
Example answers and how they were marked
Full marks1/1

A Decodes instructions

Why: Correct option: the control unit decodes instructions during the fetch-decode-execute cycle (1).

No marks0/1

C Stores data and instructions

Why: Main memory stores data and instructions, not the control unit; this is a common distractor, so no mark.

No marks0/1

B Performs arithmetic operations

Why: Arithmetic operations are performed by the ALU, not the control unit; a common confusion between CPU components, so no mark.

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.

3Describe2 marksUtility Software Swap Remove

A hard disk can become fragmented as files are saved and deleted.

Describe how disk defragmentation software improves the performance of a hard disk.

[2 marks]
Mark scheme

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

  • Parts of a file are moved so they are stored together/in consecutive locations (1) so the read/write head has less distance to travel to read the file (1)
  • Files are rearranged so that each file is stored in one block (1) which reduces the time taken to open/load the file (1)
  • Free space on the disk is collected into one area (1) so new files can be saved without becoming fragmented (1)
  • The parts of each fragmented file are put in order (1) so fewer movements of the read/write head are needed to read it (1)

Do not accept responses about compressing files, deleting files or increasing the amount of storage, as these are not what defragmentation software does.

Do not accept 'it makes the disk faster' without saying how the files are changed.

Example answers and how they were marked
Full marks2/2

It moves the parts of a file so they are stored together in consecutive locations (1), so the read/write head does not have to move as far and the file opens more quickly (1).

Why: Linked point (1) and development (1): the file parts are stored together, which reduces read/write head movement.

Some marks1/2

It reorganises the files on the hard disk so that they are stored together.

Why: The point is made (1) but there is no development saying how this improves performance, so the second mark is not awarded.

No marks0/2

It compresses the files so that they take up less space on the disk.

Why: This describes data compression software, which the mark scheme says is not accepted; no creditworthy point is made.

4Calculate3 marksAdding Binary Numbers Swap Remove

Binary patterns can represent numbers.

(a)

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

[2 marks]
1286432168421
11011011
11011100
Mark scheme

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

  • 1011 (1)
  • 0111 (1)

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

Ignore a carry shown beyond the eighth bit.

Example answers and how they were marked
Full marks2/2

1011 0111

Why: 219 + 220 = 439, which needs 9 bits; the 8 bits that are kept are both correct (2).

Some marks1/2

0000 0111

Why: Forgets to carry the 1s. The second nibble is correct (1); the first is not.

No marks0/2

1101 1111

Why: Never carries. Neither nibble is correct.

(b)

State the name of the error that occurs when this addition is carried out.

[1 mark]
Mark scheme

Award one mark for:

  • overflow (error) (1)
Example answers and how they were marked
Full marks1/1

Overflow

Why: The result needs more bits than are available (1).

No marks0/1

Underflow

Why: A near miss: the result is too big for 8 bits, not too small.

No marks0/1

A syntax error

Why: A syntax error is a mistake in a program's code, not in arithmetic.

5Discuss6 marksLow and High Level Languages Swap Remove

A company is developing the firmware for a robotic vacuum cleaner. The firmware must read the distance sensors and control the motors. The vacuum cleaner has a small amount of memory.

Discuss the characteristics of a low-level programming language that make it suitable for writing the firmware.

Your answer should include:

  • control of the hardware
  • the efficiency of the code
  • the link between the language and the processor.
[6 marks]
Mark scheme

Indicative content

Control of the hardware

  • Low-level languages can read and write the contents of registers and memory addresses directly, so the firmware can take the values from the distance sensors without calling a library routine.
  • Assembly language instructions can read from and write to specific input/output ports on the vacuum cleaner's processor, so the motors can be switched on and off directly.
  • The programmer decides exactly which processor operations are carried out and in which order, which matters when the vacuum must reverse away from a step or an obstacle.
  • A low-level language puts nothing between the program and the hardware, so the programmer can control the hardware of the vacuum cleaner precisely.

Efficiency of the code

  • Assembly language code can be hand-optimised instruction by instruction, so the code that checks a sensor is as short as possible.
  • The programmer can choose which registers hold values that are used again and again, which cuts down the number of memory accesses the firmware needs.
  • A hand-written low-level program can be made compact, so it is more likely to fit in the small amount of memory the vacuum cleaner has.
  • Because each instruction does one defined job, the time taken to react to a sensor reading is predictable, which suits firmware that must not overrun.

Link between the language and the processor

  • Low-level languages are processor-specific, so the instructions used are those of the exact processor in the vacuum cleaner rather than general statements that need translating.
  • Each assembly language instruction maps to one machine code instruction, so the programmer knows precisely which processor operation will be carried out by each line of code.
  • An assembler turns each mnemonic into its machine code equivalent, so the machine code running on the vacuum cleaner's processor is exactly what the programmer wrote.
  • Code written for a different processor would have to be rewritten, which is acceptable here because the vacuum cleaner always uses the same processor.

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)

Within Level 1, award 1 mark for a single basic point and 2 marks for two or more independent basic points; Level 2 requires the points to be linked to the vacuum cleaner; Level 3 requires both control of the hardware and efficiency of the code, and a few concise, linked sentences can reach Level 3, so extra length is not required.

Accept registers, memory addresses, input/output ports, machine code and mnemonics as ways of describing access to the hardware, and accept 'a low-level language is faster' as a basic point on its own. Do not award 'binary is a low-level language' (machine code is written in binary, but binary is not a language), or characteristics of high-level languages such as being easy to read or portable, as characteristics of a low-level language.

Example answers and how they were marked
Full marks6/6

A low-level language can read and write to registers and to input/output ports directly, so the firmware can take the distance sensor readings and switch the motors without needing library routines. It can be hand-optimised instruction by instruction, so the code that checks a sensor is short and the time it takes is predictable, and a compact program is more likely to fit in the vacuum cleaner's small memory. The language is processor-specific and each assembly instruction maps to one machine code instruction, so the programmer knows exactly which operation the vacuum's processor will carry out, and the assembled machine code is exactly what they wrote.

Why: Covers all three bullets: direct control of registers and ports applied to the sensors and motors, hand-optimised compact code with predictable timing applied to the small memory, and the one-to-one link to the processor's own instruction set. Linked and applied throughout, so Level 3.

Some marks2/6

A low-level language can control the hardware directly. It can read the sensors and control the motors. It is also faster than a high-level language, so the vacuum responds quickly. This is useful because it has little memory.

Why: Three basic, independent points: direct control of the hardware applied to the sensors and motors, faster execution, and suitability for limited memory. They are asserted without development (no registers, ports or predictable timing) and the link between the language and the processor is missing, so Level 1, two or more basic points (2).

Some marks1/6

A low-level language is written in binary, so the computer can understand it directly. It is faster than a high-level language.

Why: 'Faster than a high-level language' is a creditworthy basic point (1). 'Written in binary' is not awarded, as binary itself is not a language and the do-not-award rule applies. Only one undeveloped point with no application to the vacuum cleaner, so Level 1, single basic point (1).

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