Computer Science 9618/32 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 3 Advanced Theory, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Data types can be defined using pseudocode.
The composite record data type, Departure, is used to represent flights from Cambridge Airport and is defined in pseudocode as:
TYPE Departure
DECLARE FlightNumber : STRING
DECLARE Destination : STRING
DECLARE FlightDate : DATE
DECLARE Gate : STRING
DECLARE Airline : STRING
ENDTYPE
A variable, Flight1, is declared in pseudocode as:
DECLARE Flight1 : Departure
(a) Write pseudocode to store the following details to Flight1:
Field Data
FlightNumber SB2789
Destination Dublin
FlightDate 30/07/2025
Gate N03
Airline Cambridge Airways
…… [3]
(b) The data type for Gate is changed to an enumerated data type, GateID.
(i) Write a pseudocode statement to declare GateID to hold the identity codes for the airport gates:
N01, N02, N03, W01, W02, W03, W04
…… [2]
(ii) Write the new pseudocode statement required to replace the declaration of Gate in
Departure.
…… [1]
Mark scheme
1(a) One mark for each mark point 3
MP1 Use of correct Flight1 variable with given field names
MP2 Correct assignments of four string data values
MP3 Correct assignment of date data value
Example answer
Flight1.FlightNumber ← "SB2789"
Flight1.Destination ← "Dublin"
Flight1.FlightDate ← 30/07/2025
Flight1.Gate ← "N03"
Flight1.Airline ← "Cambridge Airways"
1(b)(i) One mark per mark point 2
MP1 TYPE GateID =
MP2 (N01, N02, N03, W01, W02, W03, W04)
Example answer
TYPE GateID = (N01, N02, N03, W01, W02, W03, W04)
1(b)(ii) DECLARE Gate : GateID 1
Numbers are stored in a computer 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.
(a) Calculate the normalised binary floating-point representation of +124.4375 in this system.
Show your working.
Mantissa Exponent
Working ……
[3]
(b) Calculate the denary value of the following normalised binary floating-point number.
Show your working.
Mantissa Exponent
1 0 1 0 0 0 1 0 1 0 1 1 0 1 1 0
Working ……
Denary value ……
[3]
Mark scheme
2(a) Two marks for working 3
• number converted to binary (0)1111100.0111
• use of exponent = 7 // Moving binary point the correct number (7) of
places
One mark for correct answer
Mantissa Exponent
0 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1
2(b) Two marks for working 3
• correct use of exponent seen
• correct conversion method from binary to denary
One mark for correct answer
Working:
1010001.01011 // moving bp 6 places to right
Evaluation of two’s complement −64 +16 + 1 + 0.25 + 0.0625 + 0.03125 //
−64 + 16 + 1 + 1/4 + 1/16 + 1/32 // Converting two’s complement back and
evaluating positive binary number 32 + 8 + 4 + 2 + 0.5 + 0.125 + 0.03125 // 32
+ 8 + 4 + 2 + 1/2 + 1/8 + 1/32
Fractions methods - award both working marks for either
(– 2048 + 512 + 32 + 8 + 2 + 1) / 2048 x 26 = -1493 / 32
OR
– 1 + 1/4 + 1/64 + 1/ 256 + 1/1024 + 1/2048 x 26 = -1493 / 32
Answer:
−46.65625 // −4621/32
(a) Identify two different layers of the TCP/IP protocol suite.
…… [1]
(b) Describe how the TCP/IP protocol suite is applied when a message is sent through the internet from one host to another. Do not describe the function of individual layers of the
TCP/IP protocol suite.
…… [4]
Mark scheme
3(a) Two from 1
• Application Layer
• Transport Layer
• Internet Layer
• Link Layer
3(b) One mark per mark point (Max 4) 4
MP1 The TCP/IP suite can be viewed as layers within a stack
MP2 Each layer can only accept input from the next higher or the next
lower layer/adjacent layer
MP3 The user/sender/computer interfaces with the top layer/application
layer to send/receive a message
MP4 The message when sent passes from the top layer/application layer
to the bottom layer/link layer
MP5 The message when received passes from bottom layer/link layer to
the top layer/application layer
MP6 The link layer interfaces directly with the network and sends/receives
the message.
Circuit switching may be used as a method of data transmission.
State two benefits and two drawbacks of circuit switching.
Benefit 1 ……
Benefit 2 ……
Drawback 1 ……
Drawback 2 ……
[4]
Mark scheme
4 One mark for each benefit (Max 2) 4
MP1 Suitable for long continuous transmission once the connection is
made, it is available until the end of the transmission
MP2 No time lost in resending/rearranging packets as no loss of packets
or out of order packets
MP3 Steady/high rate of transmission because the whole of the bandwidth
is available
MP4 Data loss unlikely as all data follows the same path
One mark for each drawback (Max 2)
MP5 Delays due to dedicated connection required to be set up/established
before transmission can begin
MP6 The dedicated connection cannot be used to transmit any other data
MP7 System resources may be underutilised/inefficient/not very flexible
because the bandwidth can’t be shared/high bandwidth is required //
Bandwidth may be wasted // May send empty frames // the circuit is always
there whether or not used
MP8 There are no alternative routes if there is a failure or fault on the line
MP9 Reduced security due to use of single path
MP10 Scalability is difficult
The management and scheduling of processes are tasks carried out by an operating system.
(a) Identify three process states.
1 ……
2 ……
3 ……
[3]
(b) Describe the function of the shortest job first scheduling routine and give a benefit of this routine.
Function ……
Benefit ……
[4]
Mark scheme
5(a) One mark per point 3
• Running state
• Ready state
• Blocked state
5(b) One mark per mark point (Max 3) 4
MP1 The processes are queued as they arrive
MP2 The process with the shortest time to complete/burst time is selected
first and executed
MP3 The process will continue until complete or put in a waiting state once
execution has begun // it is non-pre-emptive
MP4 The scheduler will continue to choose shorter processes over longer
processes if they continue to be added to the queue can cause starvation for
longer jobs
One mark for a benefit (Max 1)
MP5 Shorter jobs don’t have to wait for longer jobs to complete before
processing // Significantly reduces the average overall waiting time for
processes // Ensures starvation doesn’t occur for process with shorter burst
times
MP6 Higher throughput of processes
(a) Describe the structure of a graph as used in an Artificial Intelligence (AI) system.
…… [2]
(b) Explain how supervised learning and unsupervised learning differ from each other.
…… [4]
Mark scheme
6(a) One mark per point (Max 2) 2
MP1 A graph uses vertices/nodes to identify/represent entities such as
destinations, people, etc
MP2 Edges are used to connect nodes and can represent possible paths
between them // a path is the list of nodes connected by edges between two
given nodes
MP3 Nodes/edges can be labelled/weighted, and this is a weighting that
can be applied and used in the context of the application
MP4 A cycle is a list of nodes that return to the same node.
6(b) One mark per mark point (Max 4) 4
MP1 Supervised learning uses labelled data // Unsupervised learning
makes use of unlabelled data.
MP2 Labelled data means that known outcomes are applied to specific
inputs to help the AI predict outcomes.
MP3 Supervised learning requires initial human input/training //
Unsupervised learning does not require human input/training.
MP4 With unlabelled data in unsupervised learning, outcomes are not
known
MP5 … the AI has to search for hidden patterns/structures/clusters
MP6 … within the data in order to predict outcomes.
The diagram shows a logic circuit.
A Q
B
S
P
C
R Z
D
(a) Complete the truth table for the given logic circuit.
Show your working.
Working space
A B C D P Q R S Z
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
[3]
(b) Write the Boolean logic expression that corresponds to the given logic circuit as the sum-of-products.
Z = ……
…… [1]
(c) Use Boolean algebra including De Morgan’s laws to simplify the following expression.
Show all working.
(A + B) . (A . B + B . C)
Working ……
Simplified expression ……
[4]
Mark scheme
7(a) One mark for working, (all four columns P, Q, R and S) 3
One mark for first eight rows of column Z (Shaded)
One mark for second eight rows of column Z (Unshaded)
Working space
A B C D P Q R S Z
0 0 0 0 1 0 0 1 0
0 0 0 1 1 0 0 1 0
0 0 1 0 0 1 0 1 0
0 0 1 1 0 1 0 1 0
0 1 0 0 1 0 0 1 0
0 1 0 1 1 0 0 1 0
0 1 1 0 0 0 0 0 1
0 1 1 1 0 0 1 0 0
1 0 0 0 1 0 0 1 0
1 0 0 1 1 0 0 1 0
1 0 1 0 0 0 0 0 1
1 0 1 1 0 0 0 0 1
1 1 0 0 1 0 0 1 0
1 1 0 1 1 0 0 1 0
1 1 1 0 0 0 0 0 1
1 1 1 1 0 0 1 0 0
7(b) One mark for correct answer (terms may be in any order) 1
Z = A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D
7(c) One mark for each mark point 4
• Correct application of De Morgan’s laws
• Correct application of Idempotent, Distributive or Absorption laws
• Correct application of Idempotent, Distributive or Absorption laws
• Correct final answer
(A+B).(A.B+B.C)
(A.B).(A+B+B+C) DeMorgan’s laws
(A.B).(A+B+C) Idempotent laws
(A.A+A.B+A.C).(B.A+B.B+B.C) Distributive laws
(A+A.B+A.C).(A.B+B+B.C) Idempotent laws
(A+A.C).(B+B.C) Absorption laws
A.B // A + B Absorption laws / Final Answer
Explain what is meant by lexical analysis during program compilation.
…… [4]
Mark scheme
8 One mark per mark point (Max 4) 4
MP1 It is the first stage of compilation
MP2 White space and comments are removed
MP3 It takes modified source code and breaks it into a series of tokens
MP4 Each token is categorised and assigned types
MP5 Identifiers are stored in a symbol table
MP6 If the lexical analyser finds invalid tokens, it generates an error
MP7 Acceptable data from the lexical analyser passes to the syntax
analyser
Several syntax diagrams are shown.
letter digit
A 0
B 1
C 2
D 3
E 4
F 5
G 6
H 7
J 8
K 9
operator
+
–
*
/ letter
variable letter digit ^
(a) State why 9K is not a valid variable for the given syntax diagrams.
…… [1]
(b) Complete the Backus-Naur Form (BNF) for <operator>.
<operator> ::= ……
…… [1]
(c) An expression is defined as follows:
• A variable is assigned to a variable followed by an operator followed by another variable.
• The operator and final variable stage can be repeated as many times as necessary.
Complete the syntax diagram for an expression.
expression variable =
[3]
(d) A character can be a letter or a digit.
An additional constraint has been applied to the definition of variable. It must comply with the given syntax diagram, but it will only pass validation if it has at least four characters.
State one example of a valid variable.
…… [1]
Mark scheme
9(a) Variables must begin with a letter / not a digit 1
9(b) <operator> ::= + | - | * | / | ^ 1
9(c) One mark per mark point 3
MP1 Two variable boxes added to diagram
MP2 One operator box added to diagram and all boxes in correct order
MP3 Connections, arrows and return loop correctly added and no
additional boxes or connections
expression
variable = variable operator variable
9(d) Answer must begin with a valid letter. It can then be followed by any number 1
of valid digits and/or letters, as long as it is at least four characters in length.
Example answer
AC768
Identify the two main protocols that form Transport Layer Security (TLS) and state the purpose of each.
Protocol 1 ……
Purpose ……
Protocol 2 ……
Purpose ……
[4]
Mark scheme
10 One mark for identifying a protocol and one mark for stating its purpose 4
MP1 Handshake protocol
MP2 To establish a secure and reliable connection between two devices,
systems or networks // Permits the web server and client to authenticate each
other to make use of encryption algorithms
MP3 Record protocol
MP4 Provides a secure and reliable way to send and receive data over a
network // To exchange records between the client and server // Responsible
for securing application data end ensuring its integrity and authenticity during
transmission // Encrypts and authenticates data exchanged between a client
and a server // Deals with the format for data transmission.
(a) A linked list of nodes is used to store an ordered list of strings. Each node consists of the data, a left pointer and a right pointer.
Left pointer Data Right pointer
Pp
The linked list will be organised as a binary tree.
0 is used to represent a null pointer.
Complete the binary tree, including null pointers, to show how the data will be organised after the following strings have been added:
Aa, Mm, Ss, Xx
Root pointer
Pp
Gg Rr
0 Kk
[4]
(b) A binary tree can be used to implement recursion.
Identify one feature of an algorithm that makes it beneficial to use recursion.
Give one example of an application that could use a recursive algorithm.
Feature ……
Example ……
[2]
Mark scheme
11(a) One mark per mark point 4
MP1 Any two nodes added correctly with correct data (Aa, Mm, Ss, Xx)
MP2 Remaining two nodes added with correct data (Aa, Mm, Ss, Xx) and
all nodes with connecting arrows starting from correct pointer boxes
MP3 Correct null pointers (0) added throughout
MP4 … with no entries in other pointer boxes and all nodes correctly
positioned and connected.
Root pointer
Pp
Gg 0 Rr
0 Aa 0 0 Kk 0 Ss
0 Mm 0 0 Xx 0
11(b) One mark for feature and one mark for example (Max 2) 2
• Recursion is beneficial for algorithms when a problem naturally breaks
down into smaller versions of itself.
• … such as calculating a factorial / mathematical series / Fibonacci /
compound interest / tree traversal / evaluation of RPN expressions.
(a) A medical clinic uses objects of the class Patient to assign a priority and a doctor to a patient. Some of the attributes required in the class are listed in the table.
Attribute Data type Description
PatientID STRING Unique identifier of the patient
Name STRING Patient’s full name, surname first
DoctorID STRING ID of doctor administering treatment
Treatment is prioritised with a numeric scale of 1 to 5.
Complete the class diagram for Patient, to include:
• attribute and data type for the date of birth
• attribute and data type for the priority
• methods to assign the patient ID, priority and doctor ID
• methods to return the patient ID, patient date of birth and the priority.
Patient
PatientID : STRING
Name : STRING
…… : ……
…… : ……
DoctorID : STRING
SetName(FullName : STRING)
SetDateOfBirth(DOB : DATE)
GetName()
GetDoctorID()
[5]
(b) (i) Identify the object-oriented programming (OOP) term described as ‘an occurrence of an object’.
…… [1]
(ii) Describe what is meant by the OOP term polymorphism.
…… [2]
Mark scheme
12(a) One mark per mark point 5
MP1 Two correct attributes (DateOfBirth : Date) and priority with a
sensible name and integer data type
MP2 SetPatientID(…) and SetDoctorID(…) seen
MP3 … and appropriate parameters, with string data types
MP4 GetPatientID() and GetDateOfBirth() seen
MP5 Priority setter and getter fully correct with name matching Priority
attribute, with appropriate parameter and correct data type.
Patient
PatientID : STRING
Name : STRING
DateOfBirth : DATE
Priority : INTEGER
DoctorID : STRING
SetPatientID(PatientNumber : STRING)
SetName(FullName : STRING)
SetDateOfBirth(DOB : DATE)
SetPriority(Urgency : INTEGER)
SetDoctorID(DocID : STRING)
GetPatientID()
GetName()
GetDateOfBirth()
GetPriority()
GetDoctorID()
12(b)(i) Instance // instantiation 1
12(b)(ii) One mark for each mark point 2
MP1 Polymorphism is when methods with the same name are redefined /
behave differently …
MP2 … in derived / inherited classes / subclasses
The pseudocode algorithm below uses random file access to copy 50 records from a live file
CurrentResults.dat to a stored file StoredResults.dat one record at a time. It uses the user-defined type StudentResult.
TYPE StudentResult
DECLARE LastName : STRING
DECLARE FirstName : STRING
DECLARE ExamGrade : STRING
ENDTYPE
If any grades are missing in CurrentResults.dat, the text "Missing grade" is added to the
ExamGrade field in StoredResults.dat
Complete this file handling pseudocode algorithm.
DECLARE Grade : StudentResult
DECLARE Position : INTEGER
OPENFILE "StoredResults.dat" FOR RANDOM
SEEK "CurrentResults.dat", Position
GETRECORD "CurrentResults.dat", Grade
IF Grade.ExamGrade = "" THEN
ENDIF
NEXT Position
CLOSEFILE "CurrentResults.dat"
CLOSEFILE "StoredResults.dat"
[5]
Mark scheme
13 One mark for each correctly completed line 5
DECLARE Grade : StudentResult
DECLARE Position : INTEGER
OPENFILE "CurrentResults.dat" FOR RANDOM
OPENFILE "StoredResults.dat" FOR RANDOM
FOR Position ← 1 TO 50
SEEK "CurrentResults.dat", Position
GETRECORD "CurrentResults.dat", Grade
IF Grade.ExamGrade = "" THEN
Grade.ExamGrade ← "Missing grade"
ENDIF
SEEK "StoredResults.dat", Position
PUTRECORD "StoredResults.dat", Grade
NEXT Position
CLOSEFILE "CurrentResults.dat"
CLOSEFILE "StoredResults.dat"
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On this page, free: the mark scheme for AS & A Level Computer Science 9618/32, May/June 2025 (9618_s25_ms_32), and the question paper (9618_s25_qp_32), to read or download, with each question's marking points written out underneath.
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Computer Science 9618/32 for May/June 2025 is out of 75 marks, with 1 hour 30 minutes to answer it.
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Yes. Upload your answers to 9618/32/M/J/25 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.