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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 · 8 marks

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1Multiple choice1 markCharacter Sets Swap Remove

A text file stores each character using 7-bit ASCII.

Identify the denary ASCII code for the space character.

[1 mark]
  • A32
  • B33
  • C48
  • D65
Mark scheme
  • A 32 (1)
  • The only correct answer is A. The correct answer is 32.
  • B is not correct because 33 is the ASCII code for the exclamation mark (!)
  • C is not correct because 48 is the ASCII code for the character 0
  • D is not correct because 65 is the ASCII code for the uppercase letter A
Example answers and how they were marked
Full marks1/1

A 32

Why: Correct option: the space character has the ASCII code 32 (1).

No marks0/1

C 48

Why: Near miss scoring 0: 48 is the ASCII code for the character 0, not the space character.

No marks0/1

D 65

Why: Misconception scoring 0: 65 is the ASCII code for the uppercase letter A, so a well-known code is given but not the code for a space.

2Short recall2 marksFetch Decode Execute Cycle Swap Remove

Von Neumann architecture uses the 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:

  • Instructions/programs are held/stored in main memory (RAM) (1)
  • The data the program uses is stored in the same memory as the instructions (1)
  • Instructions are fetched (one at a time) from main memory by the CPU (1)
  • Instructions (fetched from main memory) are decoded and executed (1)

Allow 'memory' for main memory/RAM.

Do not award two marks for a single statement that only says the program/instructions are stored in memory.

Do not accept answers that state the program is stored in secondary storage (such as a hard disk).

Example answers and how they were marked
Full marks2/2

The program instructions and the data it uses are both held in main memory, and the CPU fetches each instruction from main memory in turn to decode and execute it.

Why: Instructions/programs are held/stored in main memory (RAM) (1) and the data the program uses is stored in the same memory as the instructions (1). The fetching and executing points are also present but the maximum is two marks.

Some marks1/2

The instructions of a program are stored in main memory.

Why: Instructions/programs are held/stored in main memory (RAM) (1). Only one part of the definition is given, so one mark.

No marks0/2

The program is kept on the hard disk and is loaded when it is needed.

Why: The answer places the program in secondary storage, which the mark scheme says not to accept, so no creditworthy point is given.

3Short 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.

4Calculate3 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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