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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 514 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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7 questions · 12 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.

2Convert1 markBinary to Denary Conversion Swap Remove

Computers manipulate binary patterns.

Convert the binary pattern 1010 0100 to denary.

[1 mark]
Mark scheme

Award one mark for:

  • 164 (1)
Example answers and how they were marked
Full marks1/1

164

Why: Correct: 128 + 32 + 4 = 164 (1).

No marks0/1

37

Why: Reads the place values from the wrong end (the left-hand bit as 1), so the value is wrong.

No marks0/1

3

Why: Counts the 1s instead of adding their place values.

3Short recall2 marksImages Swap Remove

Bitmap images are stored as a grid of pixels.

Define the term pixel.

[2 marks]
Mark scheme

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

  • The smallest element/area of a bitmap image (1)
  • The smallest identifiable part of an image (1)
  • Stores/represents a single colour as a binary value (1)
  • Stores/represents a single colour using a number of bits/colour depth (1)

Allow 'picture element' for element/part in the first two bullets, but not on its own

Allow 'bit depth' for colour depth

Do not accept 'picture element'/'pixel' on its own

Do not accept 'a group of pixels' or 'a dot' on its own; allow 'dot' only when it is described as the smallest element/area of the image

Do not accept 'single colour' on its own for the second mark; it must be linked to a binary value/number of bits

Do not accept 'lots of coloured dots' (it contradicts a single colour)

Example answers and how they were marked
Full marks2/2

The smallest identifiable area of a bitmap image, which has one colour stored as a binary value.

Why: Smallest identifiable area (1) and single colour stored as a binary value (1)

Some marks1/2

The smallest part of a bitmap image, made of one colour.

Why: Smallest part (1); no credit for the second mark because the single colour is not linked to a binary value or number of bits

No marks0/2

A picture element made from lots of coloured dots.

Why: 'Picture element' on its own is not accepted and 'lots of coloured dots' contradicts a single colour

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.

5Describe2 marksUser Management Swap Remove

An operating system manages the users of a computer.

Describe how an operating system checks that a user is allowed to log in.

[2 marks]
Mark scheme

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

  • The user enters a username/user ID and password/PIN (1) which is compared with the account details the operating system has stored (1)
  • The operating system keeps a file/list/database of users and their passwords (1) to check the login details the user enters against (1)
  • The operating system compares the entered details with the details stored for that user (1) and if they do not match, access is refused/denied (1)
  • The entered details are checked against the access rights stored for that user (1) so access is only allowed when the details are correct (1)

Do not award for 'it asks for a password' on its own, without saying what the password is checked against.

A stored list/file of users and passwords with no check of the login details earns one mark only.

Do not accept responses about file management, such as finding or opening the user's files.

Example answers and how they were marked
Full marks2/2

The user types in a username and password. The operating system compares these with the details it has stored for that account, and if they match the user is allowed to log in.

Why: Username and password entered (1) which is compared with the account details the operating system has stored (1).

Some marks1/2

The operating system has a list of all the users and their passwords.

Why: The stored list of users and passwords (1), but no check of the login details against it, so the development mark is not earned (0); the mark scheme states a list alone earns one mark only.

No marks0/2

The operating system finds the user's files on the hard drive and opens the last one they saved.

Why: Describes file management, not checking a user's login details; the mark scheme says not to accept responses about file management, so there is no creditworthy point.

6Describe2 marksLow and High Level Languages Swap Remove

A device driver is software that allows the operating system to control a piece of hardware. Device drivers are often written in a low-level programming language.

Describe one reason why a low-level programming language is suitable for writing a device driver.

[2 marks]
Mark scheme

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

  • Instructions can manipulate/control the hardware/memory/registers directly (1) so the driver can control the specific device it is written for (1)
  • The code is written for one specific processor/architecture (1) so it works with the exact hardware it will control (1)
  • The programmer controls exactly how memory/registers are used (1) so the code is efficient and needs little memory, which suits a device that has limited memory (1)
  • Each instruction maps to one machine code instruction (1) so the driver code is small enough to fit in the memory of the device (1)

Extra detail, such as naming the device the driver controls, is not needed to earn the marks.

Do not accept 'it is closer to machine code' on its own: it must be developed to show how this helps the driver, e.g. it can control the hardware directly.

Do not accept 'it is faster/quicker' with no link to the hardware/device, e.g. the device responds quickly.

Do not accept 'it is easier to learn', 'it is easier to debug' or 'it takes less time to write'.

Example answers and how they were marked
Full marks2/2

Low-level instructions can manipulate the hardware and memory directly (1) so the driver can control the specific device it was written for, such as a sound card (1).

Why: Point about direct control of hardware/memory (1) linked to controlling the specific device (1). The named device is extra detail and is not needed.

Some marks1/2

Low-level instructions can manipulate the hardware directly.

Why: Correct point (1) but no linked development saying how this helps the device driver.

No marks0/2

It is closer to machine code so it runs faster than a high-level language.

Why: 'Closer to machine code' is not developed into how this helps the driver, and 'runs faster' is not linked to the hardware/device, so no mark.

7Convert2 marksDenary to Binary Conversion Swap Remove

Binary patterns can represent numbers.

Convert the denary number 199 to 8-bit binary.

[2 marks]
Mark scheme

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

  • 1100 (1)
  • 0111 (1)

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

Example answers and how they were marked
Full marks2/2

1100 0111

Why: 199 = 128 + 64 + 4 + 2 + 1: both nibbles are correct (2).

Some marks1/2

1100 1000

Why: Converts 200 instead (a slip in adding up the place values). The first nibble is correct (1); the second is not.

No marks0/2

1110 0011

Why: The bits are written in reverse, smallest place value first. Neither nibble is correct.

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