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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 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 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 recall2 marksStored Program Computers Swap Remove

The central processing unit (CPU) contains an arithmetic logic unit (ALU).

State two types of operation carried out by the ALU.

[2 marks]
Mark scheme

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

  • An arithmetic operation, e.g. addition/subtraction/multiplication/division (1)
  • A logical operation, e.g. comparing two values (is A > B?) / using AND, OR, NOT (1)

Do not accept decoding instructions, fetching instructions or controlling the timing of the CPU, as these are carried out by the control unit.

Example answers and how they were marked
Full marks2/2

It carries out arithmetic operations such as addition and subtraction, and logical operations such as comparing two values to see if one is greater than the other.

Why: Arithmetic operation (1) and logical operation (1).

Some marks1/2

It adds numbers together and works out the result.

Why: Arithmetic operation, addition (1). No logical operation.

No marks0/2

It decodes the instructions that have been fetched from main memory.

Why: Decoding and fetching are the control unit's role; the mark scheme's do-not-accept line covers this.

3Short recall2 marksUnits & Conversion Swap Remove

A solid state drive has a capacity of 5 gibibytes.

Give an expression to calculate the number of bits in 5 gibibytes. You do not need to do the calculation.

[2 marks]
Mark scheme

Award one mark for each of:

  • 5 × 1024^3 / 5 × 1024 × 1024 × 1024 / 5 × 2^30, for the number of bytes (1)
  • A fully correct expression with × 8 to convert bytes to bits, in any order, e.g. 5 × 1024^3 × 8, 8 × 5 × 2^30 (1)

Accept 5 × 1 073 741 824 × 8.

Accept 5 × 2^33 or 42 949 672 960 as a fully correct expression (2 marks).

Do not accept 5 × 1000^3 × 8, which uses decimal multiples.

Do not accept ÷ 8, which converts bits to bytes.

Example answers and how they were marked
Full marks2/2

8 × 5 × 2^30

Why: 5 × 2^30 is the number of bytes (1) and × 8 converts bytes to bits (1).

Some marks1/2

5 × 2^30

Why: Correct number of bytes using the binary multiple 2^30 (1), but no conversion to bits.

No marks0/2

5 × 1000^3 × 8

Why: Uses decimal 1000^3 instead of binary 1024^3, so the expression is not fully correct; no creditworthy point.

4Describe2 marksMemory Management Swap Remove

A user closes a program that is running on a computer.

Describe how the operating system manages the main memory that the program was using.

[2 marks]
Mark scheme

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

  • The memory (that was) used by the program is freed/released/marked as free (1) so it can be allocated/reused by another program/process (1)
  • The operating system's memory allocation table/record (of memory in use) is updated (1) to show that the program's memory locations/blocks are now free/available (1)
  • The program's instructions/data are removed/overwritten in main memory (1) so the space they used is no longer reserved for that program (1)
  • The memory pages/blocks allocated to the program are returned to the list of free pages/blocks (1) that can be used when the next program is loaded (1)

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

Do not award two marks for the same idea described twice (e.g. the memory is freed and added to a list of free blocks); the two marks must come from two different linked descriptions.

Do not accept responses about closing the program window, stopping/terminating the program/process, saving the file or deleting the program from secondary storage/the hard disk.

Ignore statements that main memory is 'managed'/'cleaned up' with no development.

Example answers and how they were marked
Full marks2/2

The operating system updates its memory allocation table to show that the memory locations the program was using are now free, so another program can be loaded into them.

Why: The memory allocation table/record is updated (1) linked to showing the program's memory locations are now free/available (1).

Some marks1/2

The operating system frees up the memory the program was using.

Why: The memory used by the program is freed/released (1), but there is no linked development of what this allows.

No marks0/2

The operating system deletes the program from the hard disk to free up memory.

Why: This confuses main memory with secondary storage; deleting the program from secondary storage is named in the 'Do not accept' line, so no creditworthy point is made.

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