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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 · 6 marks

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1Multiple choice1 markEmbedded Systems Swap Remove

Identify the device that is a general purpose computer.

[1 mark]
  • ADesktop computer
  • BMicrowave oven
  • CWashing machine
  • DTraffic light
Mark scheme
  • A Desktop computer (1)
  • The only correct answer is A. The correct answer is Desktop computer.
  • B is not correct because a microwave oven uses an embedded system, built into it, to control a dedicated function
  • C is not correct because a washing machine uses an embedded system, built into it, to control a dedicated function
  • D is not correct because a traffic light uses an embedded system, built into it, to control a dedicated function
Example answers and how they were marked
Full marks1/1

A Desktop computer

Why: Correct option: the question asks for the general purpose computer, and a desktop computer is general purpose; it does not use an embedded system to control a dedicated function (1).

No marks0/1

B Microwave oven

Why: Near miss: a microwave oven is the kind of device that uses an embedded system, built into it, to control a dedicated function, so it is not a general purpose computer; mark scheme says B is not correct.

No marks0/1

C Washing machine

Why: Misconception: because a washing machine contains a computer, it must be a general purpose computer; it actually uses an embedded system built into it for a dedicated function, so no mark.

2Multiple choice1 markCharacter Sets Swap Remove

ASCII is a character set used for representing text.

Identify the number of bits used to encode each character in ASCII.

[1 mark]
  • A6
  • B7
  • C8
  • D16
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7 bits.
  • A is not correct because 6 bits give only 64 different characters
  • C is not correct because 8 bits are used by extended ASCII to give 256 characters
  • D is not correct because 16 bits give 65 536 different characters, which is far more than the 128 in ASCII
Example answers and how they were marked
Full marks1/1

B 7

Why: B is the only correct answer: each ASCII character is encoded using 7 bits (1).

No marks0/1

C 8

Why: Near miss scoring 0: 8 bits is extended ASCII, and the scheme's C line says extended ASCII, not ASCII, is not the answer.

No marks0/1

A 6

Why: Misconception scoring 0: 6 bits give only 64 different characters, so it cannot be ASCII.

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

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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