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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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3 questions · 5 marks

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1Multiple choice1 markPrimary Memory Swap Remove

A computer is running an application. The application's data and instructions are transferred from secondary storage into RAM before the CPU processes them.

Identify the reason for this transfer.

[1 mark]
  • ARAM is non-volatile
  • BRAM is faster to access than secondary storage
  • CRAM has a larger capacity than secondary storage
  • DRAM is cheaper than secondary storage
Mark scheme
  • B RAM is faster to access than secondary storage (1)
  • The only correct answer is B. The correct answer is RAM is faster to access than secondary storage.
  • A is not correct because RAM is volatile
  • C is not correct because RAM has a smaller capacity than secondary storage
  • D is not correct because RAM is more expensive than secondary storage
Example answers and how they were marked
Full marks1/1

B RAM is faster to access than secondary storage

Why: Selects the correct option: RAM is faster to access than secondary storage, so data and instructions can reach the CPU quickly (1).

No marks0/1

A RAM is non-volatile

Why: Near miss: RAM is volatile; this does not explain the transfer into RAM.

No marks0/1

C RAM has a larger capacity than secondary storage

Why: Misconception: assumes primary memory has a larger capacity, but RAM has a smaller capacity than secondary storage.

2Short recall2 marksStored Program Computers Swap Remove

The central processing unit (CPU) contains an arithmetic logic unit (ALU).

State two types of operation carried out by the ALU.

[2 marks]
Mark scheme

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

  • An arithmetic operation, e.g. addition/subtraction/multiplication/division (1)
  • A logical operation, e.g. comparing two values (is A > B?) / using AND, OR, NOT (1)

Do not accept decoding instructions, fetching instructions or controlling the timing of the CPU, as these are carried out by the control unit.

Example answers and how they were marked
Full marks2/2

It carries out arithmetic operations such as addition and subtraction, and logical operations such as comparing two values to see if one is greater than the other.

Why: Arithmetic operation (1) and logical operation (1).

Some marks1/2

It adds numbers together and works out the result.

Why: Arithmetic operation, addition (1). No logical operation.

No marks0/2

It decodes the instructions that have been fetched from main memory.

Why: Decoding and fetching are the control unit's role; the mark scheme's do-not-accept line covers this.

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.

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