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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 term for a computer that forms part of another device and performs a specific task.

[1 mark]
  • AEmbedded system
  • BGeneral purpose computer
  • CNetworked computer
  • DServer computer
Mark scheme
  • A Embedded system (1)
  • The only correct answer is A. The correct answer is embedded system.
  • B is not correct because a general purpose computer can run many different application programs
  • C is not correct because a networked computer is connected to other computers to share resources
  • D is not correct because a server computer provides services or resources to other computers on a network
Example answers and how they were marked
Full marks1/1

A Embedded system

Why: Correct option: an embedded system forms part of another device and performs a specific task (1).

No marks0/1

B General purpose computer

Why: Near miss: a general purpose computer can run many different application programs; it is not built into another device to perform a specific task, so no mark.

No marks0/1

D Server computer

Why: Misconception: a server provides services or resources to other computers on a network and is not an embedded system, 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.

3Short recall2 marksUnits & Conversion Swap Remove

Data storage is measured in binary multiples.

Define the term bit.

[2 marks]
Mark scheme

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

  • A binary digit (1)
  • Has the value 0 or 1 (1)
  • The smallest unit of data storage (1)
  • A single 0 or 1 (1)

Allow binary digit/binary place value for binary digit

Allow a single binary digit for a single 0 or 1

Do not accept a group of 8 bits/half a nibble of 4 bits, which describe a byte or a nibble

Do not accept a binary number/binary on its own

Example answers and how they were marked
Full marks2/2

A binary digit that can have the value 0 or 1.

Why: A binary digit (1) and has the value 0 or 1 (1).

Some marks1/2

It is the smallest unit of data storage.

Why: Smallest unit of data storage (1), but does not state that it is a binary digit or that it has the value 0 or 1.

No marks0/2

A group of 8 bits.

Why: Describes a byte, not a bit; no creditworthy point and named in the Do not accept line.

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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