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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 markFetch Decode Execute Cycle Swap Remove

Identify the CPU component that decodes instructions during the fetch-decode-execute cycle.

[1 mark]
  • AArithmetic logic unit
  • BControl unit
  • CMain memory
  • DRegister
Mark scheme
  • B Control unit (1)
  • The only correct answer is B. The correct answer is control unit.
  • A is not correct because the arithmetic logic unit performs arithmetic and logical operations, not decoding
  • C is not correct because main memory stores instructions and data, not decoding
  • D is not correct because a register holds data or addresses temporarily, not decoding
Example answers and how they were marked
Full marks1/1

B Control unit

Why: Correct option: the control unit decodes instructions (1).

No marks0/1

D Register

Why: Near-miss: a register holds data or addresses temporarily, it does not decode instructions (0).

No marks0/1

A Arithmetic logic unit

Why: Misconception: the arithmetic logic unit performs arithmetic and logical operations; it does not decode (0).

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.

4Calculate2 marksAdding Binary Numbers Swap Remove

Computer systems manipulate bit patterns.

Complete the table to show the result of the 8-bit binary addition.

[2 marks]
1286432168421
10111000
00011111
Mark scheme

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

  • 1101 (1)
  • 0111 (1)

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

Example answers and how they were marked
Full marks2/2

1101 0111

Why: 184 + 31 = 215, carrying where a column adds to 2: both nibbles are correct (2).

Some marks1/2

1010 0111

Why: Writes 0 for 1 + 1 but forgets to carry the 1. The second nibble is correct (1); the first is not.

No marks0/2

1011 1111

Why: Writes 1 in every column with a 1 and never carries. Neither nibble is correct.

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