A Level Computer Science 9618 Paper 32, May/June 2026

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Question paper

9618/32 · May/June 2026

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1

A user-defined data type is required to store the types of animals in a children’s zoo.

a

Write the pseudocode statement to declare a non-composite data type, Animals, that holds the following data: Alpaca, Chicken, Donkey, Duck, Goat, Pony, Sheep

[2]
b(i)

State the type of user-defined non-composite data type created in 1(a).

[1]
b(ii)

Give two features that would apply to any list used in the data type in 1(a). 1 2

[2]
2

Complete the following paragraph using the most appropriate words from the given list. Some words are not used. application controls disconnects four interconnects internet link lists many packets seven transport The TCP/IP protocol suite consists of ....................................... layers which can be thought of as a stack. It is designed to send ....................................... of data across the internet. It is also used as a communications protocol to ensure accurate delivery of data over private networks. The lowest layer of the stack is the ....................................... layer. This is the layer that ....................................... nodes in the network.

[4]
3

Numbers are stored in a computer system using binary floating-point representation with: • 12 bits for the mantissa • 4 bits for the exponent • two’s complement form for both the mantissa and the exponent.

Figure 1 of 1 for question 3
a

Calculate the denary value of the given normalised binary floating-point number. Show your working. mantissa 1 0 1 1 0 1 0 0 0 1 1 0 exponent 0 1 1 0 working denary value

[3]
b

Describe one example of a situation that could cause underflow to occur and the consequences of this underflow.

[2]
4
a

State one example of where sequential file organisation could be used.

[1]
b

Describe how records are organised in sequential files.

[2]
c

Outline the process of sequential file access to find a record in a sequential file.

[3]
5

Explain what is meant by the term virtual machine.

[4]
6
a

Write this infix expression in Reverse Polish Notation (RPN). 16 + ((9 – 7) / (18 + 6))

[2]
b

Write this RPN expression in infix form. 6 15 + 20 3 - - 27 9 * /

[3]
c

The RPN expression c a b - * b c + / d * is to be evaluated, where: a = 8, b = 2, c = 10 and d = 18. Show the changing contents of the stack as the RPN expression is evaluated.

[4]
7

This truth table represents a logic circuit. INPUT | OUTPUT A B C D | Z 0 0 0 0 | 0 0 0 0 1 | 1 0 0 1 0 | 0 0 0 1 1 | 0 0 1 0 0 | 1 0 1 0 1 | 1 0 1 1 0 | 1 0 1 1 1 | 1 1 0 0 0 | 0 1 0 0 1 | 1 1 0 1 0 | 0 1 0 1 1 | 0 1 1 0 0 | 1 1 1 0 1 | 1 1 1 1 0 | 1 1 1 1 1 | 1

Figure 1 of 1 for question 7
a(i)

Complete the Karnaugh map (K-map) for the given truth table.

[2]
a(ii)

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

[2]
a(iii)

Write the Boolean expression from your answer to 7(a)(ii) as a simplified sum-of-products.

[1]
b

Simplify the following expression using De Morgan’s laws and Boolean algebra. Show all the stages in your simplification. X=A.B.C.D‾+A.D‾‾

[3]
8
a

Describe the differences between symmetric key and asymmetric key cryptography.

[3]
b

Outline the process that would allow a business to use encryption keys to send a verified message to a group of people.

[4]
9
Figure 1 of 1 for question 9
a

State the subset of Artificial Intelligence (AI) which makes use of an artificial neural network.

[1]
b(i)

The A* and Dijkstra’s algorithms are used to perform searches on a graph. Outline how the A* algorithm is different from Dijkstra’s algorithm.

[3]
b(ii)

Calculate the best path between the Home node and each of the nodes in the diagram using Dijkstra’s algorithm. Show your working and write your answers in the table provided.

[5]
10
a

Two examples of searching algorithms are a linear search and a binary search. Describe how increasing the number of data items in the file to be searched affects the time efficiency of these search algorithms. Refer to Big O notation in your answer.

[3]
b

A temperature in degrees Celsius (°C) is recorded every minute for one day in a random file, Temperatures.dat. A function is required to find the temperature for a specific time in the day to check if it was freezing. The position of each temperature in the file corresponds to the number of minutes after midnight, for example, the temperature at 10:00am would be stored at position 600 (10 × 60). Write the pseudocode to complete the algorithm to check a temperature stored in Temperatures.dat, by passing its position in the file as a parameter and using file handling commands to find the relevant data item. Once the temperature is found, it is checked to see if it is above or below freezing, and an appropriate message is returned. Some of the blank lines may not be used. FUNCTION TempCheck( ......................................................................................................... DECLARE Temperature : REAL DECLARE Response : STRING OPENFILE .......................................................................................................................... ............................................................................................................................................ ENDFUNCTION

[6]
11
a

Object-Oriented Programming (OOP) is used to design a class called Node. A partial design of the Node class includes the STRING attribute Content and the INTEGER attribute Pointer, together with their getters and setters. The class also includes a constructor, with two parameters, to set the values of these attributes. Write pseudocode statements to declare the Node class, its attributes and the constructor.

[5]
b

Complete the table by filling in the missing OOP terms and descriptions. OOP term | description .......................................... | A method used to change values in private attributes in a class. .......................................... | An occurrence of a class during the execution of a program. aggregation | ................................................................................................. inheritance | .................................................................................................

[4]

Mark scheme

9618/32 · May/June 2026

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1a · 2 marks

  • 1 mark

    TYPE Animals =

  • 1 mark

    (Alpaca, Chicken, Donkey, Duck, Goat, Pony, Sheep)

1b(i) · 1 mark

  • 1 mark

    Enumerated

1b(ii) · 2 marks

  • 1 mark

    The list is complete / contains all possible values // Cannot change values of list after it is defined/declared // The list is of fixed length

  • 1 mark

    The list is ordinal / ordered / in alphabetical order

  • 1 mark

    Elements are unique / not repeated

2 · 4 marks

  • 1 mark

    four

  • 1 mark

    packets

  • 1 mark

    link

  • 1 mark

    interconnects

3a · 3 marks

  • 1 mark

    correct use of exponent seen

  • 1 mark

    correct conversion method from binary to denary

  • 1 mark

    −37.8125 // −37 13/16

3b · 2 marks

  • 1 mark

    When a calculation is performed on a normalised floating-point number // allow a specific example

  • 1 mark

    … so that the result obtained is less than the lowest possible value that can be stored in the system an underflow will occur // the exponent is too small / too negative for the number of bits in the mantissa, so the result becomes zero

4a · 1 mark

  • 1 mark

    Batch processing tasks / Recording log files / Generating bills/payroll / Backup / A valid example where the data is stored in a specified order

4b · 2 marks

  • 1 mark

    Records are stored in a file one after the other / linearly

  • 1 mark

    … in a given order

  • 1 mark

    … usually based on a key field / unique identifier

4c · 3 marks

  • 1 mark

    Each record in the file is searched, one after the other / linearly // Each record in the file is searched from the physical start of the file

  • 1 mark

    … using the record key

  • 1 mark

    … until the record is found.

  • 1 mark

    … until the value of the record key being searched has been exceeded // … until the end of the file is reached.

5 · 4 marks

  • 1 mark

    A VM is a standalone environment running on a physical computer // A VM is not a physical machine

  • 1 mark

    It can emulate the hardware/software of another system

  • 1 mark

    VM software has all the components of a physical computer system

  • 1 mark

    The host machine provides all the physical resources e.g. RAM/ROM

  • 1 mark

    … and is controlled via a host operating system

  • 1 mark

    The hypervisor runs between the host and guest operating systems

  • 1 mark

    … and creates and runs virtual machines / allocates resources from the host to the guest OS.

  • 1 mark

    A guest OS runs within the VM/host OS

  • 1 mark

    Multiple guest OS can run on the same VM // Multiple VMs can run on the same host OS

  • 1 mark

    The guest operating system controls/accesses the virtual hardware/software.

6a · 2 marks

  • 1 mark

    16 9 7 –

  • 1 mark

    18 6 + / +

6b · 3 marks

  • 1 mark

    (6 + 15) // 6 + 15

  • 1 mark

    - (20 – 3)

  • 1 mark

    / (27 * 9) and the pair of brackets around the whole numerator

6c · 4 marks

  • 3 marks

    One mark per ring (Max 3)

  • 1 mark

    One mark for the interim totals (60) and (5).

7a(i) · 2 marks

  • 2 marks

    Two marks if no errors present / One mark if one error present

7a(ii) · 2 marks

  • 1 mark

    First correct loop (column 01 and 11, across all rows)

  • 1 mark

    Second correct loop (row 01, across all columns)

7a(iii) · 1 mark

  • 1 mark

    B + C̄.D // C̄.D + B

7b · 3 marks

  • 2 marks

    (Max 2) for correct working from points shown (De Morgan's laws, Double negation law, Absorption law / Distributive law)

  • 1 mark

    Correct answer: (X =) A + B̄ + C̄ + D̄

8a · 3 marks

  • 1 mark

    Symmetric key encryption uses only one key but asymmetric encryption uses two keys // Asymmetric public and private keys are a mathematically linked pair, whereas symmetric uses the same key.

  • 1 mark

    Asymmetric key encryption uses one key to encrypt and the other key to decrypt, whereas symmetric key encryption uses the same key for both.

  • 1 mark

    Symmetric key encryption suffers from the key distribution problem, which the matched pair of keys for asymmetric has resolved.

  • 1 mark

    Asymmetric keys must be much larger/longer to achieve the same level of security as symmetric key encryption.

  • 1 mark

    Asymmetric key encryption allows verification of the sender, which symmetric key encryption does not.

8b · 4 marks

  • 1 mark

    The organisation generates a matched pair of encryption keys

  • 1 mark

    The organisation makes its public key available to the group of people / via its website

  • 1 mark

    The organisation encrypts its message using its private key

  • 1 mark

    … and sends it to the group of people

  • 1 mark

    Anyone receiving the message can decrypt it using the organisation’s public key

  • 1 mark

    Therefore, verifying the sender // Providing non-repudiation.

9a · 1 mark

  • 1 mark

    Deep learning

9b(i) · 3 marks

  • 1 mark

    The A* algorithm tries to find the shortest path using heuristics / intelligent guesses

  • 1 mark

    The A* algorithm doesn’t necessarily pass through all nodes/paths

  • 1 mark

    Dijkstra’s algorithm finds the shortest path to all nodes // Dijkstra’s algorithm simply checks each vertex looking for the shortest path

  • 1 mark

    Dijkstra’s algorithm always finds the shortest path

  • 1 mark

    The A* algorithm may not find the shortest path if the heuristics are inaccurate.

9b(ii) · 5 marks

  • 1 mark

    Initialisation – setting Home to 0

  • 1 mark

    … and the rest of the nodes to ∞

  • 1 mark

    Evidence to show values at nodes being updated

  • 1 mark

    Evidence to show ‘visited node(s)’

  • 1 mark

    Evidence to show a correct calculation of at least one route

  • 1 mark

    Evidence to show more than one route has been calculated for at least one node

  • 2 marks

    Correct Answers: Two marks for all six correct values (A: 6, B: 13, C: 9, D: 20, E: 17, F: 14) / One mark for four or five correct values

10a · 3 marks

  • 1 mark

    The time it takes to execute a linear search will increase linearly as the number of search items increases linearly

  • 1 mark

    Linear search worst case Big O notation is O(n)

  • 1 mark

    The time it takes to execute a binary search will increase logarithmically as the number of search items increases linearly // The time it takes to execute a binary search will increase linearly as the number of search items increases exponentially

  • 1 mark

    Binary search worst case Big O notation is O(log n) / O(log2 n).

10b · 6 marks

  • 1 mark

    Correct completion of FUNCTION header

  • 1 mark

    Correct OPENFILE statement

  • 1 mark

    Correct use of SEEK

  • 1 mark

    Correct use of GETRECORD

  • 1 mark

    Correct complete selection statement(s) structure IF … THEN …. ELSE … ENDIF OR multiple IF … THEN … ENDIF

  • 1 mark

    Correct application of logic between condition and action to produce / return appropriate messages

  • 1 mark

    Correct use of CLOSEFILE – must be after GETRECORD but do NOT award this mark if RETURN statements(s) come before this line

  • 1 mark

    Correct RETURN statement(s) - allow use of other variables in this statement as the question does not specifically ask for Response to be used.

11a · 5 marks

  • 1 mark

    Correct CLASS Node and ENDCLASS statements

  • 1 mark

    Correct attribute declaration section before the constructor

  • 1 mark

    Correct statement to begin and end the constructor as PUBLIC PROCEDURE NEW … ENDPROCEDURE

  • 1 mark

    Two parameters identified in the constructor with appropriate names and data types.

  • 1 mark

    Values passed using the constructer are assigned correctly – allow use of setters.

11b · 4 marks

  • 1 mark

    setter

  • 1 mark

    instance(s) // object(s)

  • 1 mark

    aggregation

  • 1 mark

    inheritance

Questions

  1. 1
    Question 15 marks
    Mark scheme ›
  2. 2
    Question 24 marks
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  3. 3
    Question 35 marks
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  4. 4
    Question 46 marks
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  5. 5
    Question 54 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 78 marks
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  8. 8
    Question 87 marks
    Mark scheme ›
  9. 9
    Question 99 marks
    Mark scheme ›
  10. 10
    Question 109 marks
    Mark scheme ›
  11. 11
    Question 119 marks
    Mark scheme ›

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