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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 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.

2Short 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.

3Describe2 marksFile Management Swap Remove

A user saves a new file into a folder that already contains files.

Describe how the operating system prevents a new file overwriting an existing file.

[2 marks]
Mark scheme

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

  • Checks/compares the new file (name)/filename against the names of files already in the folder/directory (1) to see if the name is already in use (1)
  • Refuses to save/allow the new file when the (file) name already exists (1) so the existing file is not replaced/overwritten (1)
  • Prompts/asks the user to choose/enter a different (file) name (1) so both files can be saved without one replacing the other (1)

A point without a linked development can earn one mark.

For both marks, the development must follow/associate with the point.

Do not accept responses about defragmentation.

Do not accept the operating system making a backup copy, or warning the user that a file may be overwritten, without saying that the (file) name is checked or the new file is not saved.

Do not accept responses about process management or saving files to RAM.

Do not award a second mark for repeating the question, e.g. 'it stops the new file overwriting the existing file'.

Example answers and how they were marked
Full marks2/2

It checks the new file name against the names of files already in the folder (1) to see whether the name is already in use (1).

Why: Linked point (1) and development (1).

Some marks1/2

It checks the file name against the names already in the folder.

Why: Creditworthy point (1) on its own; no linked development, so one mark.

No marks0/2

It defragments the disk so there is room for the new file.

Why: Defragmentation is not accepted by the mark scheme.

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