Menu

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.

Build a paper

Topics (none ticked: all of them)
Question types (none ticked: all of them)

Practice paper

4 questions · 8 marks

Pick again
Print
1Short 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.

2Describe2 marksMemory Management Swap Remove

A program is stored on secondary storage before it is run.

Describe how the operating system manages memory when a program is started.

[2 marks]
Mark scheme

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

  • The program (and its data) is copied/loaded from secondary storage into main memory/RAM (1) so that the CPU can fetch/execute its instructions (1)
  • An area of main memory is allocated/reserved for the program (1) that is large enough to hold the program and its data (1)
  • The operating system checks that there is enough free main memory (1) before the program can be started/loaded, otherwise it cannot run/an error is reported (1)
  • The operating system keeps a record of the memory addresses used by the program (1) so that its instructions and data can be found when it runs (1)

Allow 'loaded into RAM', 'stored in memory' and 'put into main memory' as equivalent to copied into main memory.

A single statement that the program is loaded/stored in main memory, with no linked development, can be given one mark only.

Do not accept descriptions of the CPU fetching/executing instructions that do not refer to the operating system loading or allocating memory.

Do not accept responses about virtual memory (moving processes to secondary storage) or about which process runs next on the CPU.

Do not accept the program being run from secondary storage/hard disk.

Example answers and how they were marked
Full marks2/2

The operating system copies the program and its data from secondary storage into main memory so that the CPU can fetch and execute its instructions.

Why: Copying the program and data into main memory (1) linked to the CPU fetching/executing instructions (1).

Some marks1/2

The operating system loads the program into main memory.

Why: The program is copied/loaded into main memory (1), but there is no linked development of how or what this leads to, so only one mark is given.

No marks0/2

The CPU fetches the program's instructions from main memory and executes them one at a time.

Why: This describes the fetch-decode-execute cycle, not the operating system loading or allocating memory, which is the Do not accept line; no creditworthy point is made.

3Describe2 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.

4Calculate2 marksBinary Shifts Swap Remove

Computers manipulate binary patterns.

Complete the table with the result of applying the shift to the binary pattern.

[2 marks]
Binary patternShift8-bit binary result
1000 0001Logical shift right by 1
Mark scheme

Award one mark for each correct nibble in the correct order:

  • 0100 (1)
  • 0000 (1)

0 marks if the response is not 8-bit binary.

Example answers and how they were marked
Full marks2/2

0100 0000

Why: Every bit moves 1 place right, with 0s in from the left: both nibbles are correct (2).

Some marks1/2

0100 0001

Why: A slip in the last bit. The first nibble is correct (1); the second is not.

No marks0/2

0000 0010

Why: Shifts the bits the wrong way. Neither nibble is correct.

Questions by topic

More ways to revise Edexcel GCSE Computer Science (1CP2)

Teachers: build a paper, print it with the mark schemes at the end, and sign your school up to set your classes activities on every topic.

Fetch Decode Explain isn't affiliated with or endorsed by Pearson Edexcel.