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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 · 7 marks

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1Multiple 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.

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 marksMemory Management Swap Remove

An operating system keeps a record of the memory it has allocated to each program.

Describe how this record is used.

[2 marks]
Mark scheme

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

  • The start and end address/size of the block allocated to each program is recorded (1) so the operating system knows which areas of main memory are in use and which are free (1)
  • The record is checked before memory is allocated (1) so memory already in use is not given to another program/is not overwritten (1)
  • The record is searched for a free block that is large enough (1) so a newly loaded program is placed in a suitable area of main memory (1)
  • The record is updated when a program ends/closes (1) so the memory it was using is marked as free for other programs (1)

Allow the record to be called a memory map/memory table/allocation table.

A response that only says the operating system keeps a record/list of the memory each program is using, without saying what the record contains, earns 0 marks.

A response that states what the record contains, e.g. the addresses/the start and end address of each block, earns 1 mark even if it does not say how the record is used.

Do not accept responses about saving programs/files to secondary storage or about moving programs to virtual memory, as these are not uses of this record.

Example answers and how they were marked
Full marks2/2

The operating system records the start and end address of the block of memory given to each program so that it does not give the same memory to another program.

Why: The start and end address of each block is recorded (1) linked to memory already in use not being given to another program (1).

Some marks1/2

The operating system records the start and end address of the memory used by each program.

Why: States what the record contains (1), but does not develop how the record is used, so the second mark is not earned.

No marks0/2

The operating system saves the programs to the hard disk so there is enough memory for other programs.

Why: Describes moving programs to secondary storage rather than how the record of allocated memory is used; no creditworthy point is made.

4Describe2 marksLow and High Level Languages Swap Remove

A programmer has written a program in assembly language.

Describe why it is necessary to translate the program into machine code before it can be run.

[2 marks]
Mark scheme

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

  • The processor/CPU can only execute machine code/binary instructions (1) so the assembly instructions must be converted into machine code before the program can run (1)
  • Each assembly instruction/mnemonic matches/corresponds to one machine code instruction (1) so each line must be replaced with the machine code the processor executes (1)
  • The program is written using mnemonics (1) that must be translated into machine code for the processor to carry out the instructions (1)
  • The hardware/processor does not understand assembly language directly (1) so it must be translated into machine code that the processor can execute (1)

Do not accept 'assembly language is a low-level language'.

Do not accept 'it is hard to understand' or 'it is not English'.

Do not accept benefits, such as 'machine code runs faster' or 'the program will run more quickly'.

Do not accept 'an assembler/a translator converts it' on its own, with no reference to the processor only being able to execute machine code/binary.

Do not accept responses about high-level languages or about compilers/interpreters.

Example answers and how they were marked
Full marks2/2

The processor can only execute machine code, so the assembly instructions must be converted into machine code before the program can run.

Why: States what the processor can execute (1) and links it to converting the program's instructions into machine code (1).

Some marks1/2

Each assembly instruction matches one machine code instruction.

Why: Correct point about the one-to-one match (1) but no development saying that this is because the processor only executes machine code.

No marks0/2

Because an assembler translates assembly language into machine code.

Why: Names the translator but gives no reason tied to the processor only being able to execute machine code, so no mark; this is in the 'Do not accept' guidance.

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