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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 530 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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5 questions · 7 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.

2Multiple choice1 markCharacter Sets Swap Remove

A new character set must be able to represent 70 different characters.

Identify the minimum number of bits needed to represent each character.

[1 mark]
  • A6
  • B7
  • C8
  • D70
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7.
  • A is not correct because 6 bits give only 64 combinations, which is not enough for 70 characters
  • C is not correct because 7 bits already give 128 combinations, so 8 bits is not the minimum needed
  • D is not correct because 70 is the number of characters needed, not a number of bits
Example answers and how they were marked
Full marks1/1

B 7

Why: The only correct answer: 7 bits give 128 combinations, enough for 70 characters, and it is the smallest number of bits that does (1).

No marks0/1

C 8

Why: Distractor, no credit: a single-tick item has no partial credit, and 8 bits is not the minimum because 7 bits already give 128 combinations, so the scheme's C line gives no mark.

No marks0/1

A 6

Why: Misconception, no credit: 6 bits give only 64 combinations, which cannot represent 70 different characters, so the scheme's A line gives no mark.

3Short recall2 marksStored Program Computers Swap Remove

Computers use the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

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

  • Program instructions are stored in main memory/RAM (1)
  • The data used by the program is stored in main memory/RAM (1)
  • The CPU fetches/reads instructions from main memory to execute them (1)
  • The program can be run without the instructions being input again (1)

Allow 'the program is stored in memory' for one mark.

Do not accept answers about programs being stored on secondary storage, such as a hard disk or USB memory stick.

Do not accept answers that only describe the von Neumann architecture (e.g. having a CPU and memory) without stating that instructions are stored in main memory.

Example answers and how they were marked
Full marks2/2

Both the program instructions and the data the program uses are held in the same main memory, and the CPU fetches these instructions from memory to execute them.

Why: Instructions stored in main memory (1) and the CPU fetches instructions from memory to execute them (1).

Some marks1/2

Program instructions are stored in main memory.

Why: Instructions stored in main memory (1). No second point about data being stored in the same memory or about the CPU fetching instructions.

No marks0/2

The program is saved on the hard disk so that it can be loaded into RAM when the user runs it.

Why: A near-miss about secondary storage, which the mark scheme says not to accept; the stored program concept is about instructions being held in main memory.

4Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

5Calculate2 marksBinary Shifts Swap Remove

Computers manipulate binary patterns.

Complete the table with the result of applying the shift to the binary pattern.

[2 marks]
Binary patternShift8-bit binary result
1000 0001Logical shift right by 1
Mark scheme

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

  • 0100 (1)
  • 0000 (1)

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

Example answers and how they were marked
Full marks2/2

0100 0000

Why: Every bit moves 1 place right, with 0s in from the left: both nibbles are correct (2).

Some marks1/2

0100 0001

Why: A slip in the last bit. The first nibble is correct (1); the second is not.

No marks0/2

0000 0010

Why: Shifts the bits the wrong way. Neither nibble is correct.

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