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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 markPrimary Memory Swap Remove

Identify what is fetched from RAM during the fetch stage.

[1 mark]
  • AThe next instruction
  • BThe data at a memory address
  • CThe decoded instruction
  • DThe address of the next instruction
Mark scheme
  • A The next instruction (1)
  • The only correct answer is A. The correct answer is the next instruction.
  • B is not correct because data is fetched during the execute stage
  • C is not correct because an instruction is decoded after it is fetched
  • D is not correct because the program counter holds the address used to fetch the instruction
Example answers and how they were marked
Full marks1/1

A The next instruction

Why: The next instruction is fetched from RAM during the fetch stage (1).

No marks0/1

B The data at a memory address

Why: Near miss: data is not fetched during the fetch stage; data is used during the execute stage.

No marks0/1

C The decoded instruction

Why: Misconception: an instruction is decoded after it is fetched, in the decode stage.

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 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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