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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 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.

2Explain2 marksTopologies Swap Remove

A bus topology uses one main cable shared by all devices.

Explain why a bus topology can be a security risk.

[2 marks]
Mark scheme

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

  • All devices receive every data packet (1) because the main cable is shared by all nodes (1)
  • Data can be read by any device on the network (1) because there is no central switch to send it only to the destination (1)
  • A device can intercept/capture data not meant for it (1) because all transmissions travel along the same cable (1)
  • Any device can view all data on the cable (1) so data is not private/secure (1)

For both marks, the expansion must follow/associate with the statement.

Do not accept 'it is easy to hack' without saying data can be seen/intercepted on the shared cable.

Do not accept 'the central switch can be hacked' because a bus topology has no central switch.

Example answers and how they were marked
Full marks2/2

Any device on the bus can see all the data (1) because every packet travels along the same shared cable (1).

Why: Linked point (1) and reason (1): data is visible to all devices because the cable is shared.

Some marks1/2

All devices can see the data.

Why: Awarded 1 mark for stating that all devices can see the data, but no linked reason about the shared cable is given.

No marks0/2

It is less secure because the central switch can be hacked.

Why: No credit: a bus topology has no central switch; the answer gives a misconception and no valid point about the shared cable.

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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