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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 component that stores instructions and data during the fetch-decode-execute cycle.

[1 mark]
  • AArithmetic logic unit
  • BClock
  • CMain memory (RAM)
  • DProgram counter
Mark scheme
  • C Main memory (RAM) (1)
  • The only correct answer is C. The correct answer is main memory (RAM); accept 'main memory', 'RAM' or 'random access memory'.
  • A is not correct because the arithmetic logic unit performs arithmetic and logical operations
  • B is not correct because the clock sends regular pulses to synchronise the components of the CPU
  • D is not correct because the program counter holds the address of the next instruction, not the instructions and data
Example answers and how they were marked
Full marks1/1

C Main memory (RAM)

Why: Correct option: main memory (RAM) stores instructions and data during the fetch-decode-execute cycle (1).

No marks0/1

D Program counter

Why: Near-miss distractor: the program counter holds the address of the next instruction, not the instructions and data (0).

No marks0/1

A Arithmetic logic unit

Why: Misconception: the ALU performs arithmetic and logical operations, so it does not store instructions and data (0).

2Multiple choice1 markCharacter Sets Swap Remove

Characters are encoded using 7-bit ASCII.

Identify the number of different characters that can be represented using 7-bit ASCII.

[1 mark]
  • A64
  • B127
  • C128
  • D256
Mark scheme
  • C 128 (1)
  • The only correct answer is C. The correct answer is 128.
  • A is not correct because 6 bits give only 64 different combinations
  • B is not correct because 127 is the largest 7-bit value, not the number of characters that 7 bits can represent
  • D is not correct because 256 characters need 8 bits, as used by 8-bit extended ASCII
Example answers and how they were marked
Full marks1/1

C 128

Why: Selects the only correct option (1): 7 bits give 2 to the power of 7, which is 128 different characters.

No marks0/1

D 256

Why: Near miss scoring 0: 256 is the number of characters that 8 bits give in extended ASCII, not 7-bit ASCII.

No marks0/1

B 127

Why: Misconception scoring 0, the off-by-one error the examiners' report describes: 127 is the largest 7-bit value, but 7 bits give 128 different characters.

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.

4Describe2 marksUtility Software Swap Remove

A hard disk can become fragmented as files are saved and deleted.

Describe how disk defragmentation software improves the performance of a hard disk.

[2 marks]
Mark scheme

Award up to two marks for a linked description such as:

  • Parts of a file are moved so they are stored together/in consecutive locations (1) so the read/write head has less distance to travel to read the file (1)
  • Files are rearranged so that each file is stored in one block (1) which reduces the time taken to open/load the file (1)
  • Free space on the disk is collected into one area (1) so new files can be saved without becoming fragmented (1)
  • The parts of each fragmented file are put in order (1) so fewer movements of the read/write head are needed to read it (1)

Do not accept responses about compressing files, deleting files or increasing the amount of storage, as these are not what defragmentation software does.

Do not accept 'it makes the disk faster' without saying how the files are changed.

Example answers and how they were marked
Full marks2/2

It moves the parts of a file so they are stored together in consecutive locations (1), so the read/write head does not have to move as far and the file opens more quickly (1).

Why: Linked point (1) and development (1): the file parts are stored together, which reduces read/write head movement.

Some marks1/2

It reorganises the files on the hard disk so that they are stored together.

Why: The point is made (1) but there is no development saying how this improves performance, so the second mark is not awarded.

No marks0/2

It compresses the files so that they take up less space on the disk.

Why: This describes data compression software, which the mark scheme says is not accepted; no creditworthy point is made.

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