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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 · 6 marks

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1Short recall2 marksPrimary Memory Swap Remove

A computer is designed using the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • The program instructions and data are stored together in main memory/RAM (1)
  • The same memory is used for both instructions and data (1)
  • The instructions are fetched from main memory and executed by the CPU (1)
  • The instructions are fetched and executed one at a time / in sequence (1)
  • The program can be changed/replaced by loading new instructions into memory, without changing the hardware (1)

Do not accept the program is stored in secondary storage / on the hard disk.

Do not accept the program is stored in the CPU / in registers.

Example answers and how they were marked
Full marks2/2

The program instructions and the data are stored together in main memory, and the CPU fetches the instructions one at a time and executes them.

Why: Creditworthy point that instructions and data are stored in the same main memory (1) and the linked point that the CPU fetches and executes them from memory (1).

Some marks1/2

The program is held in the main memory (RAM) so that the computer can run it.

Why: Earns the point that the program instructions are stored in main memory (1), but does not say that the CPU fetches and executes them, so the second mark is not earned.

No marks0/2

The program is stored on the hard disk and loaded into the CPU when it is run.

Why: Both parts are misconceptions named in the Do not accept lines: the stored program concept keeps the instructions in main memory, not on secondary storage or in the CPU.

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.

3Calculate2 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.

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