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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 · 7 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

Data storage is measured in binary multiples.

Define the term bit.

[2 marks]
Mark scheme

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

  • A binary digit (1)
  • Has the value 0 or 1 (1)
  • The smallest unit of data storage (1)
  • A single 0 or 1 (1)

Allow binary digit/binary place value for binary digit

Allow a single binary digit for a single 0 or 1

Do not accept a group of 8 bits/half a nibble of 4 bits, which describe a byte or a nibble

Do not accept a binary number/binary on its own

Example answers and how they were marked
Full marks2/2

A binary digit that can have the value 0 or 1.

Why: A binary digit (1) and has the value 0 or 1 (1).

Some marks1/2

It is the smallest unit of data storage.

Why: Smallest unit of data storage (1), but does not state that it is a binary digit or that it has the value 0 or 1.

No marks0/2

A group of 8 bits.

Why: Describes a byte, not a bit; no creditworthy point and named in the Do not accept line.

3Calculate3 marksAdding Binary Numbers Swap Remove

Binary patterns can represent numbers.

(a)

Complete the table to show the 8-bit result of the binary addition.

[2 marks]
1286432168421
11011011
11011100
Mark scheme

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

  • 1011 (1)
  • 0111 (1)

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

Ignore a carry shown beyond the eighth bit.

Example answers and how they were marked
Full marks2/2

1011 0111

Why: 219 + 220 = 439, which needs 9 bits; the 8 bits that are kept are both correct (2).

Some marks1/2

0000 0111

Why: Forgets to carry the 1s. The second nibble is correct (1); the first is not.

No marks0/2

1101 1111

Why: Never carries. Neither nibble is correct.

(b)

State the name of the error that occurs when this addition is carried out.

[1 mark]
Mark scheme

Award one mark for:

  • overflow (error) (1)
Example answers and how they were marked
Full marks1/1

Overflow

Why: The result needs more bits than are available (1).

No marks0/1

Underflow

Why: A near miss: the result is too big for 8 bits, not too small.

No marks0/1

A syntax error

Why: A syntax error is a mistake in a program's code, not in arithmetic.

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