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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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3 questions · 6 marks

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1Short 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.

2Short 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.

3Describe2 marksMemory Management Swap Remove

An operating system keeps a record of the memory it has allocated to each program.

Describe how this record is used.

[2 marks]
Mark scheme

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

  • The start and end address/size of the block allocated to each program is recorded (1) so the operating system knows which areas of main memory are in use and which are free (1)
  • The record is checked before memory is allocated (1) so memory already in use is not given to another program/is not overwritten (1)
  • The record is searched for a free block that is large enough (1) so a newly loaded program is placed in a suitable area of main memory (1)
  • The record is updated when a program ends/closes (1) so the memory it was using is marked as free for other programs (1)

Allow the record to be called a memory map/memory table/allocation table.

A response that only says the operating system keeps a record/list of the memory each program is using, without saying what the record contains, earns 0 marks.

A response that states what the record contains, e.g. the addresses/the start and end address of each block, earns 1 mark even if it does not say how the record is used.

Do not accept responses about saving programs/files to secondary storage or about moving programs to virtual memory, as these are not uses of this record.

Example answers and how they were marked
Full marks2/2

The operating system records the start and end address of the block of memory given to each program so that it does not give the same memory to another program.

Why: The start and end address of each block is recorded (1) linked to memory already in use not being given to another program (1).

Some marks1/2

The operating system records the start and end address of the memory used by each program.

Why: States what the record contains (1), but does not develop how the record is used, so the second mark is not earned.

No marks0/2

The operating system saves the programs to the hard disk so there is enough memory for other programs.

Why: Describes moving programs to secondary storage rather than how the record of allocated memory is used; no creditworthy point is made.

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