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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).

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 marksEmbedded Systems Swap Remove

An embedded system is a computer system built into another device to perform a specific function.

Give two examples of devices that contain an embedded system.

[2 marks]
Mark scheme

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

  • Microwave (oven) (1)
  • Washing machine (1)
  • Dishwasher (1)
  • Digital camera (1)
  • Satellite navigation/satnav system (1)
  • Traffic light(s) (1)
  • Central heating controller/thermostat (1)
  • Digital watch/fitness tracker (1)
  • Vending machine (1)
  • Car (as it contains embedded systems) (1)

Allow any other device that contains an embedded system.

Do not accept general purpose devices such as a laptop, desktop computer, tablet or smartphone (these can run many different programs).

Example answers and how they were marked
Full marks2/2

A microwave oven and a washing machine.

Why: Two creditworthy devices, a microwave oven (1) and a washing machine (1).

Some marks1/2

A dishwasher and a laptop computer.

Why: Dishwasher credited (1); laptop is a general purpose device, so not accepted.

No marks0/2

A desktop computer and a smartphone.

Why: Both are general purpose devices, so neither earns a mark.

4Short 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.

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