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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 631 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 · 5 marks

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1Multiple choice1 markAddressing Swap Remove

A packet of data is sent across the internet to a web server. The packet has a header that stores information about the packet.

Identify the information in the packet header that a router uses to forward the packet towards its destination.

[1 mark]
  • AThe destination domain name
  • BThe destination IP address
  • CThe MAC address of the sender
  • DThe name of the file being sent
Mark scheme
  • B The destination IP address (1)
  • The only correct answer is B. The correct answer is the destination IP address.
  • A is not correct because a domain name is converted to an IP address by a DNS server before routing begins
  • C is not correct because the sender's MAC address is only used to deliver the packet on the local network
  • D is not correct because the file name is not used to route packets
Example answers and how they were marked
Full marks1/1

B The destination IP address

Why: Correct option (1): a router reads the destination IP address in the header to forward the packet towards its destination.

No marks0/1

A The destination domain name

Why: A domain name is converted to an IP address by a DNS server before routing begins, so it is not the routing information in the header (mark scheme line A); no mark.

No marks0/1

C The MAC address of the sender

Why: The sender's MAC address is only used to deliver the packet on the local network, so it is not used by the router to forward the packet (mark scheme line C); no mark.

2Short recall1 markTopologies Swap Remove

A bus topology uses terminators at each end of the main cable.

State the purpose of terminators in a bus topology.

[1 mark]
Mark scheme

Award one mark for:

  • Absorb signals (at the ends of the bus cable) / prevent signals reflecting (back along the cable) (1)

Allow 'stop data/signals bouncing back' for prevent signals reflecting

Do not accept 'terminate the cable' / 'close the end of the cable' on its own

Do not accept boost/repeat/strengthen the signal

Example answers and how they were marked
Full marks1/1

To absorb signals so that they do not reflect back along the bus cable.

Why: Absorbing signals and preventing reflection are alternatives within the one creditworthy point (1).

No marks0/1

To terminate the cable at each end.

Why: Near miss: this is rejected by the 'terminate the cable' guidance because it does not say signals are absorbed or prevented from reflecting.

No marks0/1

To make the signal stronger at the end of the cable.

Why: Misconception: terminators absorb signals; they do not boost, repeat or strengthen them.

3Short recall1 markImages Swap Remove

Each pixel in a bitmap image is stored using a number of bits.

Define the term colour depth.

[1 mark]
Mark scheme

Award one mark for any of the following:

  • The number of bits per pixel (1)
  • The number of bits used to represent/store the colour of each pixel (1)

Allow bit depth

Do not accept the number of colours a pixel can have on its own

Do not accept the number of pixels in the image (this is the resolution/image size)

Example answers and how they were marked
Full marks1/1

The number of bits used to store the colour of one pixel.

Why: Gives the number of bits per pixel (1)

No marks0/1

The number of colours that each pixel can use.

Why: Gives the number of colours, not the number of bits; the mark scheme does not accept this on its own

No marks0/1

The number of pixels in the image.

Why: Confuses colour depth with resolution/image size, which the mark scheme does not accept

4Short 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.

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