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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 631 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 · 7 marks

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1Short recall1 markCharacter Sets Swap Remove

A computer can store a maximum of 80 bits for a text message. Each character is stored using 7-bit ASCII.

State the maximum number of whole characters that can be stored in the message.

[1 mark]
Mark scheme

Award one mark for:

  • 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Example answers and how they were marked
Full marks1/1

11

Why: Correct: 80 ÷ 7 = 11 remainder 3, so 11 whole characters can be stored (1).

No marks0/1

11.4

Why: Near miss: gives a decimal result rather than the number of whole characters; the mark scheme does not accept 11.4.

No marks0/1

12

Why: Rounds up to 12, but 12 characters would need 84 bits, more than the 80 available; the mark scheme does not accept 12.

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 marksUser Management Swap Remove

A computer can be used by more than one person.

Describe how an operating system uses access rights to protect each user's files.

[2 marks]
Mark scheme

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

  • Each user/file is given permissions/access rights (1) so other users cannot read/write/open/change that user's files (1)
  • A user may be given read-only/read access to a file (1) so they cannot edit/delete/change it (1)
  • Each file is given permissions/access rights for different users (1) so the operating system allows/refuses access depending on which user is logged in (1)
  • The operating system checks a user's permissions/access rights when a file is opened (1) so a user with read-only/read access cannot save/delete changes (1)

Do not accept responses about asking for a password/username, as this checks who the user is, not what they are allowed to do.

Do not accept 'files are protected/safe' or 'access rights protect files' without saying how access is limited (e.g. read-only access, permissions for different users).

Example answers and how they were marked
Full marks2/2

A user can be given read-only access to a file, so they can open it but cannot edit or delete it.

Why: Read-only/read access point (1) and linked development that the user cannot edit/delete the file (1).

Some marks1/2

A user can be given read-only access to a file.

Why: Read-only/read access point (1); no development saying what this prevents, such as being unable to edit/delete the file (0).

No marks0/2

The operating system asks for the password again before the file opens.

Why: Describes asking for a password, which checks who the user is, not access rights to the file; no creditworthy point.

4Describe2 marksTranslators Swap Remove

High-level code can be translated by a compiler or by an interpreter.

Describe how a compiler translates high-level code differently from an interpreter.

[2 marks]
Mark scheme

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

  • The compiler translates the whole program/complete source code in one go (1) whereas the interpreter translates one line/statement at a time (1)
  • The compiler translates all of the source code before any of it is run/executed (1) but the interpreter translates each line/statement as it reads it (1)
  • The interpreter translates one line/statement at a time (1) and the compiler translates the whole program/source code together (1)

Accept the compiler point and the interpreter point in either order.

Do not accept responses about object code/executable files, when the translated code is executed, how errors are reported, or speed/memory use.

Do not accept 'the interpreter executes/runs each line' on its own, without saying that it translates one line/statement at a time.

Example answers and how they were marked
Full marks2/2

A compiler translates the whole program in one go, while an interpreter translates one line at a time.

Why: Whole program translated in one go (1) and interpreter translates one line at a time (1).

Some marks1/2

A compiler translates the whole program in one go.

Why: Whole program translated in one go (1); no linked point about the interpreter translating a line at a time.

No marks0/2

A compiler produces an executable file, but an interpreter does not.

Why: Describes the output, not the amount translated at a time; object code/executable files are not accepted.

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