A Level Computer Science 9618 Paper 32, Oct/Nov 2025

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Question paper

9618/32 · Oct/Nov 2025

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1

The composite data type, Car, is defined in pseudocode as: TYPE Car DECLARE RegNumber : STRING DECLARE Make : STRING DECLARE Model : STRING DECLARE BodyStyle : STRING DECLARE Colour : STRING DECLARE IntoStock : DATE DECLARE Price : REAL ENDTYPE

a(i)

Write the pseudocode statement to set up a variable for one record of the composite data type, Car.

[1]
a(ii)

Write the pseudocode statements to assign the following values to the variable set up in part (a)(i): • "Blue" to Colour • 21/10/2025 to IntoStock

[2]
b(i)

The data type for BodyStyle is changed to an enumerated type, Body. Write the pseudocode statement for the type declaration of Body to hold the names of the available choices: Convertible, Hatchback, Saloon, SUV

[2]
b(ii)

Write the new pseudocode statement required to update the declaration of BodyStyle in the definition of Car.

[1]
2

Numbers are stored in a computer system using binary floating-point representation with: • 10 bits for the mantissa • 6 bits for the exponent • two’s complement form for both the mantissa and the exponent.

a

Calculate the denary value of the given normalised binary floating-point number. Show your working. Mantissa: 0 1 1 1 1 0 0 1 0 1 Exponent: 0 0 1 0 1 1 Working Denary value

[3]
b

Calculate the normalised binary floating-point representation of +26.6875 in this system. Show your working. Mantissa: [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] Exponent: [ ][ ][ ][ ][ ][ ] Working

[3]
3
a

HTTP and IMAP are examples of protocols used in the Application Layer of the TCP/IP protocol suite. State the purpose of the HTTP and IMAP protocols. HTTP IMAP

[2]
b

Describe how files are shared using the BitTorrent protocol.

[4]
4
a

Identify one benefit of circuit switching and one benefit of packet switching. Circuit switching Packet switching

[2]
b

Identify two differences between circuit switching and packet switching. 1 2

[2]
5
a

Describe how interrupt handling is used in low-level scheduling.

[2]
b

In process management, a process can be in one of three process states: running, ready or blocked. Complete the table to identify one reason why a process could be in each of the three states.

[3]
6
Figure 1 of 1 for question 6
a

The diagram shows a logic circuit. Complete the truth table for the given logic circuit. Show your working.

[3]
b(i)

Complete the Karnaugh map (K-map) for the Boolean expression: A‾.B‾.C+A‾.B.C+A.B‾.C+A.B.C‾

[2]
b(ii)

Draw loop(s) around appropriate group(s) in the K-map to produce an optimal sum-of-products.

[2]
b(iii)

Write the Boolean expression from your answer to part b(ii) as a simplified sum-of-products. Do not carry out any further simplification.

[2]
7
a

Asymmetric encryption is a type of cryptography. Identify one other type of cryptography.

[1]
b

An organisation holds two asymmetric encryption keys, which they intend to use to receive secure transmissions. Explain how the organisation makes use of the two keys to receive a secure transmission.

[4]
8
a

State the purpose of the optimisation stage in the compilation of a program.

[1]
b

Convert this Reverse Polish Notation (RPN) back to its original infix form: a b - c + c a - * d /

[3]
c

The RPN expression: c a / b d – * b + is to be evaluated, where: a = 3, b = 16, c = 9 and d = 6. Show the changing contents of the stack as the RPN expression is evaluated.

[4]
9

Deep Learning is a form of Machine Learning.

a(i)

State one example where Deep Learning is used.

[1]
a(ii)

Identify how Deep Learning can be made more effective.

[1]
b

Describe the back propagation of errors method in Machine Learning.

[4]
10

An exception is an error that may cause a program to halt unexpectedly.

a

Describe how program termination due to an exception can be avoided.

[2]
b

Identify two possible causes of exceptions. 1 2

[2]
11

The table shows assembly language instructions for a processor that has one register, the Accumulator (ACC). Instruction Label Opcode Operand Explanation LDM #n Load the number n to the ACC LDD <address> Load the contents of the location at the given address to ACC LDI <address> The address to be used is at the given address. Load the contents of this second address to the ACC. ADD <address> Add the contents of the given address to the ACC SUB <address> Subtract the contents of the given address from the ACC STO <address> Store the contents of the ACC at the given address <label>: <data> Gives a symbolic address <label> to the memory location with the contents <data> # denotes a denary number, e.g. #123 <label> can be used in place of <address> The current contents of memory are: Address Contents 150 26 300 86 420 150 Write assembly language code, using only the given instruction set to: • store the contents of location 300 as labelled variable A • store the contents of location 420 as labelled variable B • add the value stored in the address contained in variable B to the value contained in variable A • store the result in variable Answer. Show the initialisation and values of the variables A, B and Answer in the table provided.

[7]
12

A stack has been implemented using pseudocode to store a maximum of 100 string items using the global variables in the following table: Identifier Data type Description Initialisation value Base INTEGER pointer for the bottom of the stack 0 Top INTEGER pointer for the top of the stack -1 StackArray STRING 1D array to implement the stack [0:99] Max INTEGER maximum number of items in the stack 100 The value of Top is incremented each time a data item is added to the stack and decremented each time a data item is removed. If the stack is full, an appropriate error message is output.

a(i)

Complete the pseudocode for the procedure to add a data item onto the stack. PROCEDURE Push(.........................................................................................................) IF Top < Max – 1 THEN Top ← ............................................................................................................... ...................................................................................................... ← NewData ELSE OUTPUT ............................................................................................................... ENDIF ENDPROCEDURE

[4]
a(ii)

Write pseudocode to input a new data item and add it to the stack using Push().

[2]
b

Explain the reasons why a stack is used when a recursive algorithm is executed.

[3]

Mark scheme

9618/32 · Oct/Nov 2025

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1a(i) · 1 mark

  • 1 mark

    DECLARE Car1 : Car

1a(ii) · 2 marks

  • 1 mark

    Car1.Colour ← "Blue"

  • 1 mark

    Car1.IntoStock ← 21/10/2025

1b(i) · 2 marks

  • 1 mark

    MP1 TYPE Body =

  • 1 mark

    MP2 (Convertible, Hatchback, Saloon, SUV)

1b(ii) · 1 mark

  • 1 mark

    DECLARE BodyStyle : Body

2a · 3 marks

  • 1 mark

    correct calculation/application of exponent seen

  • 1 mark

    correct method to find the final answer

  • 1 mark

    Denary value: 1940

2b · 3 marks

  • 1 mark

    MP1 correct method to find the binary number

  • 1 mark

    MP2 correct use of exponent

  • 1 mark

    MP3 correct answer in the space provided

3a · 2 marks

  • 1 mark

    HTTP to send and receive / transfer web pages / hypertext / images / videos

  • 1 mark

    IMAP allows users to access/read their emails from any device without removing the message from the mail server // synchronises emails on any device

3b · 4 marks

  • 1 mark

    MP1 Files are shared over a peer-to-peer network

  • 1 mark

    MP2 A small file, a torrent (descriptor file), is initially created by a peer

  • 1 mark

    MP3 The torrent (descriptor file) contains metadata about the file to be shared

  • 1 mark

    MP4 The whole file must initially be located on at least one peer

  • 1 mark

    MP5 The file is broken into equal sized pieces

  • 1 mark

    MP6 Other peers who wish to download the file first obtain the torrent (descriptor file) and connect to a tracker

  • 1 mark

    MP7 A tracker acts like a server that holds all the data about all the computers connected to it / the swarm

  • 1 mark

    MP8 Each peer downloads parts of the file from other peers until the entire file has been downloaded

  • 1 mark

    MP9 As each peer receives a piece of the file, they become a source for that piece of the file

  • 1 mark

    MP10 A peer who has a complete file and is sharing it is a seed // Once a peer has completely downloaded the file and made the file available to others in the swarm, they become a seed.

4a · 2 marks

  • 1 mark

    One mark for a benefit of circuit switching (Max 1): MP1 Once the connection is made, it is available until the end of the transmission (suitable for long continuous transmission) / MP2 The dedicated path ensures a steady rate of data transmission, once connection has been made (because the whole bandwidth is available) / MP3 The dedicated path ensures data is less likely to be lost. / MP4 No intermediate delays once the circuit is established, enables real time transmission / MP5 No additional time/delay to reorder needed because data arrives in the order it was sent / data sent as a continuous stream

  • 1 mark

    One mark for a benefit of packet switching (Max 1): MP6 No need to tie up a communication line // path available to multiple users // bandwidth can be shared by multiple users / MP7 Possible to overcome failed / faulty / busy lines by re-routing packets / MP8 Individual packets can be resent if lost/damaged / MP9 Users only charged for the duration of connectivity / MP10 High rate of data transmission possible / MP11 Always uses digital networks so data is transmitted directly to the destination / MP12 Good security because all packets can take different routes // A more secure method because all packets can take different routes

4b · 2 marks

  • 1 mark

    MP1 Circuit switching requires a dedicated line to be connected before the data transfer takes place, in packet switching, data transfer commences directly/ straight away

  • 1 mark

    MP2 In circuit switching, each data unit knows the entire path address, but in packet switching, each data unit only knows the final address (routers decide intermediate paths)

  • 1 mark

    MP3 Circuit switching uses the whole bandwidth of the transmission path, but packet switching shares it with other users // circuit switching uses a constant high bandwidth. Packet switching has variable bandwidth

  • 1 mark

    MP4 In circuit switching each data unit follows the same route, but in packet switching, packets can follow any route

  • 1 mark

    MP5 In circuit switching data arrives in the correct order but in packet switching it has to be reordered

  • 1 mark

    MP6 Data remains intact / a continuous stream in circuit switching but is segmented in packet switching

  • 1 mark

    MP7 Lost data in circuit switching requires the whole transmission to be resent, but individual packets can be resent in packet switching

  • 1 mark

    MP8 Circuit switching doesn’t suffer data loss but packets can be lost in packet switching.

5a · 2 marks

  • 1 mark

    MP1 The low-level scheduler manages the handling of interrupts based on priority

  • 1 mark

    MP2 … ensuring that critical events are handled without delay

  • 1 mark

    MP3 It uses an Interrupt Vector Table (IVT) / Interrupt Descriptor/Despatch Table (IDT) / Interrupt Service Routine (ISR) lookup

  • 1 mark

    MP4 … to map the interrupt to the specific handling routine / ISR.

5b · 3 marks

  • 1 mark

    running: CPU time has been allocated and the process is being executed.

  • 1 mark

    ready: The process is waiting (in the ready queue) for a slice of CPU time. It could run. // It has been displaced by a higher priority process and could otherwise still run.

  • 1 mark

    blocked: The process is waiting for an I/O operation / some event to take place / be completed.

6a · 3 marks

  • 1 mark

    One mark for working, (all four columns P, Q, R and S)

  • 1 mark

    One mark for first four rows of column Z

  • 1 mark

    One mark for second four rows of column Z

6b(i) · 2 marks

  • 2 marks

    Correct Karnaugh map: row 0: 0, 1, 1, 0; row 1: 0, 1, 0, 1

6b(ii) · 2 marks

  • 1 mark

    Correct loop for horizontal pair of 1s at (0,01) and (0,11)

  • 1 mark

    Correct loop for vertical pair of 1s at (0,01) and (1,01)

6b(iii) · 2 marks

  • 1 mark

    One correct Boolean term with a + / OR sign

  • 1 mark

    All Boolean terms and operators correct and no other terms present

7a · 1 mark

  • 1 mark

    Symmetric (cryptography / encryption) / Quantum (cryptography)

7b · 4 marks

  • 1 mark

    MP1 The two keys held by the organisation are a private key and a public key

  • 1 mark

    MP2 The organisation makes the public key available to anyone who wishes to send them secure transmissions // The sender obtains the organisation’s public key

  • 1 mark

    MP3 The sender uses the organisation’s public key to encrypt the message / plain text // The sender uses the organisation’s public key to turn the message into cipher text

  • 1 mark

    MP4 The organisation uses its private key to decrypt the message.

8a · 1 mark

  • 1 mark

    To improve the code by making it use minimum resources (CPU, memory/storage, time) // Answer by example: To improve the code by minimising program storage / minimising CPU time / minimising memory use / minimising program execution time (includes peripheral use as well as CPU)

8b · 3 marks

  • 1 mark

    (a – b + c)

  • 1 mark

    * (c - a)

  • 1 mark

    / d

8c · 4 marks

  • 1 mark

    First ring

  • 1 mark

    Second ring

  • 1 mark

    Third ring

  • 1 mark

    Fourth ring

9a(i) · 1 mark

  • 1 mark

    Virtual assistants // AI Agents / Chatbots / Composing music / Self-driving cars / Natural language processing / Fraud detection / Image colouring // photograph enhancement / Visual/Image recognition/facial recognition / Text mining / Computer assisted translation / Large language models (LLMs) / Playing chess / Grading student exams / Weather forecasting

9a(ii) · 1 mark

  • 1 mark

    Use / introduce more hidden layers

9b · 4 marks

  • 1 mark

    MP1 Initial outputs are compared to expected outputs

  • 1 mark

    MP2 … weightings are adjusted to minimise the difference between actual and expected outputs

  • 1 mark

    MP3 Calculus is used to find the error gradient in the obtained outputs

  • 1 mark

    MP4 … the results are fed back into the neural network

  • 1 mark

    MP5 … weightings of each neuron / node are adjusted as a result of the feedback

  • 1 mark

    MP6 … the process repeats until results are more accurate

10a · 2 marks

  • 1 mark

    MP1 Use an exception handling routine (to respond to unwanted / unexpected events when the program is running)

  • 1 mark

    MP2 Use of try … except / catch // Generate some form of error message

10b · 2 marks

  • 1 mark

    MP1 Coding errors

  • 1 mark

    MP2 User errors

  • 1 mark

    MP3 Hardware failure // Losing connection to a device e.g. a printer

11 · 7 marks

  • 1 mark

    MP1 LDD 300 seen

  • 1 mark

    MP2 Correct use of STO seen (at least once)

  • 1 mark

    MP3 Correct use of LDD 420 seen

  • 1 mark

    MP4 Correct use of LDI B

  • 1 mark

    MP5 Correct use of ADD A

  • 1 mark

    MP6 Correct labelling of three addresses A:, B: and Answer:

  • 1 mark

    MP7 Correct contents in A: and B:

  • 1 mark

    MP8 Correct value in Answer:

12a(i) · 4 marks

  • 1 mark

    IF Top < Max – 1 THEN

  • 1 mark

    Top ← Top + 1

  • 1 mark

    StackArray[Top] ← NewData

  • 1 mark

    OUTPUT "Stack full; new data cannot be added"

12a(ii) · 2 marks

  • 1 mark

    MP1 Input with variable, with or without prompt

  • 1 mark

    MP2 Procedure call for Push with parameter used matching input variable

12b · 3 marks

  • 1 mark

    MP1 Stacks store data in Last In First Out (LIFO) / First In Last Out (FILO) order

  • 1 mark

    MP2 Each time a recursive algorithm calls itself data is pushed onto the stack

  • 1 mark

    MP3 When the recursive algorithm reaches its base case / starts to unwind

  • 1 mark

    MP4 … data is popped from the stack in the reverse order to which it was pushed onto it.

Questions

  1. 1
    Question 16 marks
    Mark scheme ›
  2. 2
    Question 26 marks
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  3. 3
    Question 36 marks
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  4. 4
    Question 44 marks
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  5. 5
    Question 55 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 75 marks
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  8. 8
    Question 88 marks
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  9. 9
    Question 96 marks
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  10. 10
    Question 104 marks
    Mark scheme ›
  11. 11
    Question 117 marks
    Mark scheme ›
  12. 12
    Question 129 marks
    Mark scheme ›

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