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IB Computer Science SL · Paper 1

Computer Science, Standard Level

Timezone

TZ2TZ3

Session: May 2025
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short answer

Questions 1–8 · 25 marks

Question 1

2 marks

Application interface features

System fundamentals

2 marks

Identify two distinct visual or functional components that an application user interface provides to facilitate interaction, such as menus or toolbars, and clear visual controls like buttons and input dialogue boxes.

Mentor tip

Focus on standard graphical user interface (GUI) components (e.g. menus, buttons, dialogue windows) rather than abstract concepts.

Question 2

3 marks

Virtual Private Network mechanisms and use cases

Networks

3 marks

Scenario: Remote staff accessing an enterprise internal network across public or untrusted Wi-Fi connections.
Mechanism: The VPN establishes an encrypted virtual tunnel between the client device and the secure remote gateway.
Impact: Even if intermediate communications on the public network are intercepted, payload encryption prevents unauthorised inspection of credentials and proprietary data.

Mentor tip

Link the usage scenario (remote access over public network), the technical feature (tunnelling and encryption), and the security outcome (confidentiality).

Question 3

4 marks

Requirements engineering via direct observation

System fundamentals

4 marks

Advantage 1: Captures unmediated real-world workflows in their natural working context. Observers witness first-hand how operators navigate actual environmental constraints and routine friction points, uncovering unstated operational requirements that users omit during interviews.
Advantage 2: High empirical reliability and elimination of recall bias. Rather than relying on self-reported estimates or subjective memory, direct observation gathers objective telemetry on how tools and software are truly utilized.

Mentor tip

Contrast unmediated real-time observation against the subjectivity and recall bias of questionnaires or interviews.

Question 4

4 marks

Logic diagram construction for compound expressions

Computer organisation

4 marks

Analyze operator precedence in the Boolean expression $X = A \lor (\neg B \land \neg C)$ :
Gate 1: NOT gate applied to input $B$ , yielding $\neg B$ .
Gate 2: NOT gate applied to input $C$ , yielding $\neg C$ .
Gate 3: 2-input AND gate receiving inverted inputs $\neg B$ and $\neg C$ , producing $\neg B \land \neg C$ .
Gate 4: 2-input OR gate receiving input line $A$ and the output of the AND gate $(\neg B \land \neg C)$ , driving the output line $X$ .

X = A \lor (\neg B \land \neg C)

Mentor tip

Connect NOT gates to inputs B and C, route both outputs into an AND gate, and finally combine with A using an OR gate.

Question 5

6 marks

Trace table execution for modulo accumulator loop

Computational thinking, problem-solving and programming

6 marks

Initial state: $K = 1, S = 0, Z = 0$ .
Iteration 1 ( $K = 1$ ): $A = 1 (mod 3) = 1$ . Since $A = 1$ , $S = S - K = 0 - 1 = - 1$ . Then $Z = Z + S = 0 + (- 1) = - 1$ , and $K = 1 + 1 = 2$ .
Iteration 2 ( $K = 2$ ): $A = 2 (mod 3) = 2$ . Since $A = 2$ , $S = S + K = - 1 + 2 = 1$ . Then $Z = Z + S = - 1 + 1 = 0$ , and $K = 2 + 1 = 3$ .
Iteration 3 ( $K = 3$ ): $A = 3 (mod 3) = 0$ . Since $A \neq = 2$ and $A \neq = 1$ , $S = S + 1 = 1 + 1 = 2$ . Then $Z = Z + S = 0 + 2 = 2$ , and $K = 3 + 1 = 4$ .
Iteration 4 ( $K = 4$ ): $A = 4 (mod 3) = 1$ . Since $A = 1$ , $S = S - K = 2 - 4 = - 2$ . Then $Z = Z + S = 2 + (- 2) = 0$ , and $K = 4 + 1 = 5$ .
Iteration 5 ( $K = 5$ ): $A = 5 (mod 3) = 2$ . Since $A = 2$ , $S = S + K = - 2 + 5 = 3$ . Then $Z = Z + S = 0 + 3 = 3$ , and $K = 5 + 1 = 6$ .
Loop terminates as $K = 6$ is not less than 6. Final output printed: 'The result: 3'.

K 123456 S 0 - 1 12 - 2 3 Z 0 - 1 0203 A - 12012 Output - - - - - "The result: 3"

Mentor tip

Pay close attention to the branching conditions: when A=2 add K, when A=1 subtract K, and when A=0 increment S by 1.

Question 6

2 marks

System implementation methods

System fundamentals

2 marks

Identify two recognized methods of rolling out a new software system across an organisation, such as parallel running (running old and new systems concurrently to verify consistency) and phased conversion (introducing the system module by module).

Mentor tip

Recall the four classic changeover strategies: direct (big bang), parallel, pilot, and phased.

Question 7

2 marks

Domain Name System resolution

Networks

2 marks

A Domain Name System (DNS) server functions as a distributed directory service on the internet, maintaining databases mapping alphanumeric domain names to their respective numerical IP addresses.
When a client browser requests a hostname, the DNS translates that human-readable domain into a routable IP address so network packets can locate the destination host.

Mentor tip

Explain the bidirectional translation between human-readable domain names and machine-readable IP addresses.

Question 8

2 marks

Register capacity and hexadecimal conversion

Computer organisation

(a)

1 mark

Partition the 12-bit register $00011011001 0_{2}$ into 4-bit nibbles:
$000 1_{2} = 1_{16}$
$101 1_{2} = 1 1_{10} = B_{16}$
$001 0_{2} = 2_{16}$
Concatenating the hexadecimal digits yields $1B2_{16}$ .

00011011001 0_{2} = 1B2_{16}

Mentor tip

Group binary digits into nibbles of 4: 0001 is 1, 1011 is B (11 in decimal), 0010 is 2.

(b)

1 mark

A register of $n$ bits can represent $2^{n}$ unique bit combinations.
For a 12-bit register: $2^{12} = 4096$ distinct binary numbers.

2^{12} = 4096

Mentor tip

State $2^{12} = 4096$ as the total distinct numbers representable in a 12-bit register.

Section B · Structured and algorithmic problems

Questions 9–11 · 45 marks

Question 9

15 marks

CAD software, data compression, and business continuity

System fundamentals
Computer organisation

(a)(i)

1 mark

Computer-Aided Design (CAD) refers to specialized software tools used to digitally draft, model, visualize, and test 2D architectural drawings and 3D geometric designs.

Mentor tip

Define CAD as computer-assisted creation, modification, and digital analysis of technical drawings and 3D models.

(a)(ii)

4 marks

Benefit 1: Rapid revision and iterative editing. CAD models can be modified parametrically without redrawing blueprints from scratch, drastically accelerating design turnarounds.
Benefit 2: High dimensional precision and automated structural simulation. Digital models calculate angles, stress loads, and clearances mathematically, eliminating drafting errors and enabling direct export to CAM or 3D rendering engines.

Mentor tip

Contrast digital editing and mathematical precision against manual drafting boards.

(b)

3 marks

Architectural CAD files incorporate extensive high-resolution geometries and layered vector data, leading to large file footprints.
Compression algorithms reduce redundant bit sequences, shrinking total file size.
This conserves local and cloud storage capacity and shortens upload and download transmission times over network channels.

Mentor tip

Explain both the storage reduction and the network throughput/transfer speed advantages.

(c)(i)

3 marks

Strategy: Implement automated periodic backups to redundant offsite or cloud storage, supplemented by a RAID array or standby mirrored hardware.
Recovery: In the event of primary disk failure, the redundant drive array or standby system takes over immediately (failover), while cloud repositories allow rapid data restoration.
Result: Business operations resume with minimal loss of billable hours or design data.

Mentor tip

Cover the backup mechanism, the switchover/restoration procedure, and how downtime is minimized.

(c)(ii)

4 marks

Cause 1: Human error. Unintentional file deletion, overwriting critical revisions, or closing applications without committing changes.
Cause 2: Malicious cyberattacks or ransomware. Malware, cryptolocker infections, or unauthorised access that encrypts or corrupts files, alongside application software crashes during file write cycles.

Mentor tip

Distinguish non-hardware causes: human operational errors versus malicious cyber threats or software write crashes.

Question 10

15 marks

Campus network access control, connectivity, and communication risks

Networks

(a)

6 marks

Role-Based Access Control (RBAC): Segment user groups into defined tiers (Administrators, Faculty, Students, Guests), enforcing distinct permissions through centralised directory authentication.
Access Matrix Implementation: Administrators hold unrestricted read, write, and execute privileges to system services and domain policies. Teachers maintain read/write privileges to gradebooks and curriculum folders but are restricted from peer financial records.
Student and Guest Isolation: Students receive read access strictly to their individual assignments and reports, while guest users are isolated on a firewalled VLAN with rate-limited internet access and zero access to internal subnets.

Mentor tip

Structure access levels systematically: role separation, authentication credentials, and concrete directory permissions for each user tier.

(b)(i)

2 marks

Communication Link: Dedicated fibre-optic broadband connection.
Justification: Fibre-optic lines provide high bandwidth and low latency over long distances with immunity to electromagnetic interference, handling simultaneous concurrent traffic from thousands of school devices.

Mentor tip

Name the physical medium (fibre optic) and justify with bandwidth, latency, and noise immunity.

(b)(ii)

4 marks

Measure 1: Network edge firewalls with Deep Packet Inspection (DPI). Monitors inbound and outbound traffic, blocking unauthorised port scans, intrusions, and blacklisted domains.
Measure 2: Network segmentation and MAC address filtering / 802.1X network access control. Isolates internal server subnets from student-accessible Wi-Fi and validates connecting physical hardware identifiers.

Mentor tip

Select two perimeter security layers (e.g. firewall filtering and network segmentation/access control) and justify each.

(c)

3 marks

Problem 1: Vulnerability to phishing exploits and social engineering, where students or staff inadvertently disclose network credentials or open attachments bearing malware.
Problem 2: Storage exhaustion and network bandwidth saturation caused by heavy attachments and bulk spam emails.
Problem 3: Pastoral and institutional risks, including misuse of institutional email for cyberbullying, academic dishonesty, or policy violations requiring active administrative monitoring.

Mentor tip

Cover diverse threat vectors: cyber risks (phishing), infrastructure load (bandwidth/storage), and behavioral misuse.

Question 11

15 marks

Grade distribution calculation and linear array processing

Computational thinking, problem-solving and programming

(a)(i)

1 mark

Calculate $GRADES [0]$ using the mark of Ximena Alba ( $MARKS [0] = 5$ ):
$GRADES [0] = 1 + (5 div 10) = 1 + 0 = 1$ .

GRADES [0] = 1 + (5 div 10) = 1

Mentor tip

Integer division (div) truncates any fractional component: 5 div 10 = 0.

(a)(ii)

1 mark

Calculate $GRADES [1]$ using Boris Mount's mark ( $MARKS [1] = 88$ ):
$1 + (88 div 10) = 1 + 8 = 9$ .
Since $9 \geq 7$ , the clamping conditional sets $GRADES [1] = 7$ .

GRADES [1] = min (1 + (88 div 10), 7) = 7

Mentor tip

Remember to apply the capping conditional check: if grade >= 7, clamp to 7.

(a)(iii)

1 mark

Calculate $GRADES [2]$ using Hugh Parr's mark ( $MARKS [2] = 45$ ):
$GRADES [2] = 1 + (45 div 10) = 1 + 4 = 5$ .

GRADES [2] = 1 + (45 div 10) = 5

Mentor tip

45 div 10 = 4, so 1 + 4 = 5.

(a)(iv)

1 mark

To achieve Grade 7, we require $1 + (MARKS [K] div 10) \geq 7$ , which means $MARKS [K] div 10 \geq 6$ .
The smallest non-negative integer satisfying $M div 10 \geq 6$ is $M = 60$ .

M_{m i n} = 60

Mentor tip

60 div 10 = 6, yielding 1 + 6 = 7. Marks 0-59 yield grades 1-6.

(b)

4 marks

Initialize an insertion pointer $I = 0$ for the RESIT array.
Iterate index $K$ from 0 to 199 through the MARKS (or GRADES) array.
If $MARKS [K] < 10$ (or equivalently $GRADES [K] = 1$ ), assign $STUDENTS [K]$ to $RESIT [I]$ and increment pointer $I$ by 1.

I = 0 loop K from 0 to 199 if MARKS [K] < 10 then RESIT [I] = STUDENTS [K] I = I + 1 end if end loop

Mentor tip

Keep a separate index pointer I for the destination RESIT array and only advance it when a matching student is appended.

(c)

7 marks

Pass 1: Accumulate the sum of all 200 marks and divide by 200 to establish the class average.
Pass 2: Traverse index $K$ from 0 to 199. Compare each student's mark against the threshold bands $AVE + 20$ and $AVE - 20$ .
Assign 'A' if mark $> AVE + 20$ , 'C' if mark $< AVE - 20$ , and 'B' otherwise, recording each in $LETTERGRADES [K]$ .

SUM = 0 loop K from 0 to 199 SUM = SUM + MARKS [K] end loop AVE = SUM /200 loop K from 0 to 199 if MARKS [K] > AVE + 20 then LETTERGRADES [K] = "A" else if MARKS [K] < AVE - 20 then LETTERGRADES [K] = "C" else LETTERGRADES [K] = "B" end if end loop

Mentor tip

Calculate the average first across all 200 elements before running the second grading loop.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Computational thinking, problem-solving and programming carries the most marks on this paper.

Computational thinking, problem-solving and programming21 marks · 30%
Q5, Q11
System fundamentals20 marks · 29%
Q1, Q3, Q6, Q9
Networks20 marks · 29%
Q2, Q7, Q10
Computer organisation9 marks · 13%
Q4, Q8, Q9

Mentor insights

Traps to watch for

01
Logic gate ordering · Q4
AND takes precedence over OR. In $A \lor \neg B \land \neg C$ , combine $\neg B$ and $\neg C$ in an AND gate before joining with $A$ in the final OR gate.
02
Trace table subtraction condition · Q5
When $A = 1$ , $S = S - K$ . For $K = 4$ , $A = 1$ leads to $S = 2 - 4 = - 2$ . Ensure the negative sign is retained in accumulator $Z$ .
03
Register capacity vs maximum value · Q8
A 12-bit register holds $2^{12} = 4096$ distinct values. The maximum unsigned value is $2^{12} - 1 = 4095$ , but the number of representations is 4096.
04
Conditional grade capping · Q11
Remember that the scoring algorithm explicitly clamps any calculated grade of 7 or higher to 7: $1 + (88 div 10) = 9$ , which becomes 7.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Timezone

TZ2TZ3

Session: May 2025
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short answer

Questions 1–8 · 25 marks

Question 1

2 marks

Application interface features

System fundamentals

2 marks

Mentor tip

Focus on standard graphical user interface (GUI) components (e.g. menus, buttons, dialogue windows) rather than abstract concepts.

Question 2

3 marks

Virtual Private Network mechanisms and use cases

Networks

3 marks

Scenario: Remote staff accessing an enterprise internal network across public or untrusted Wi-Fi connections.
Mechanism: The VPN establishes an encrypted virtual tunnel between the client device and the secure remote gateway.
Impact: Even if intermediate communications on the public network are intercepted, payload encryption prevents unauthorised inspection of credentials and proprietary data.

Mentor tip

Link the usage scenario (remote access over public network), the technical feature (tunnelling and encryption), and the security outcome (confidentiality).

Question 3

4 marks

Requirements engineering via direct observation

System fundamentals

4 marks

Advantage 1: Captures unmediated real-world workflows in their natural working context. Observers witness first-hand how operators navigate actual environmental constraints and routine friction points, uncovering unstated operational requirements that users omit during interviews.
Advantage 2: High empirical reliability and elimination of recall bias. Rather than relying on self-reported estimates or subjective memory, direct observation gathers objective telemetry on how tools and software are truly utilized.

Mentor tip

Contrast unmediated real-time observation against the subjectivity and recall bias of questionnaires or interviews.

Question 4

4 marks

Logic diagram construction for compound expressions

Computer organisation

4 marks

Analyze operator precedence in the Boolean expression $X = A \lor (\neg B \land \neg C)$ :
Gate 1: NOT gate applied to input $B$ , yielding $\neg B$ .
Gate 2: NOT gate applied to input $C$ , yielding $\neg C$ .
Gate 3: 2-input AND gate receiving inverted inputs $\neg B$ and $\neg C$ , producing $\neg B \land \neg C$ .
Gate 4: 2-input OR gate receiving input line $A$ and the output of the AND gate $(\neg B \land \neg C)$ , driving the output line $X$ .

X = A \lor (\neg B \land \neg C)

Mentor tip

Connect NOT gates to inputs B and C, route both outputs into an AND gate, and finally combine with A using an OR gate.

Question 5

6 marks

Trace table execution for modulo accumulator loop

Computational thinking, problem-solving and programming

6 marks

Initial state: $K = 1, S = 0, Z = 0$ .
Iteration 1 ( $K = 1$ ): $A = 1 (mod 3) = 1$ . Since $A = 1$ , $S = S - K = 0 - 1 = - 1$ . Then $Z = Z + S = 0 + (- 1) = - 1$ , and $K = 1 + 1 = 2$ .
Iteration 2 ( $K = 2$ ): $A = 2 (mod 3) = 2$ . Since $A = 2$ , $S = S + K = - 1 + 2 = 1$ . Then $Z = Z + S = - 1 + 1 = 0$ , and $K = 2 + 1 = 3$ .
Iteration 3 ( $K = 3$ ): $A = 3 (mod 3) = 0$ . Since $A \neq = 2$ and $A \neq = 1$ , $S = S + 1 = 1 + 1 = 2$ . Then $Z = Z + S = 0 + 2 = 2$ , and $K = 3 + 1 = 4$ .
Iteration 4 ( $K = 4$ ): $A = 4 (mod 3) = 1$ . Since $A = 1$ , $S = S - K = 2 - 4 = - 2$ . Then $Z = Z + S = 2 + (- 2) = 0$ , and $K = 4 + 1 = 5$ .
Iteration 5 ( $K = 5$ ): $A = 5 (mod 3) = 2$ . Since $A = 2$ , $S = S + K = - 2 + 5 = 3$ . Then $Z = Z + S = 0 + 3 = 3$ , and $K = 5 + 1 = 6$ .
Loop terminates as $K = 6$ is not less than 6. Final output printed: 'The result: 3'.

K 123456 S 0 - 1 12 - 2 3 Z 0 - 1 0203 A - 12012 Output - - - - - "The result: 3"

Mentor tip

Pay close attention to the branching conditions: when A=2 add K, when A=1 subtract K, and when A=0 increment S by 1.

Question 6

2 marks

System implementation methods

System fundamentals

2 marks

Mentor tip

Recall the four classic changeover strategies: direct (big bang), parallel, pilot, and phased.

Question 7

2 marks

Domain Name System resolution

Networks

2 marks

A Domain Name System (DNS) server functions as a distributed directory service on the internet, maintaining databases mapping alphanumeric domain names to their respective numerical IP addresses.
When a client browser requests a hostname, the DNS translates that human-readable domain into a routable IP address so network packets can locate the destination host.

Mentor tip

Explain the bidirectional translation between human-readable domain names and machine-readable IP addresses.

Question 8

2 marks

Register capacity and hexadecimal conversion

Computer organisation

(a)

1 mark

Partition the 12-bit register $00011011001 0_{2}$ into 4-bit nibbles:
$000 1_{2} = 1_{16}$
$101 1_{2} = 1 1_{10} = B_{16}$
$001 0_{2} = 2_{16}$
Concatenating the hexadecimal digits yields $1B2_{16}$ .

00011011001 0_{2} = 1B2_{16}

Mentor tip

Group binary digits into nibbles of 4: 0001 is 1, 1011 is B (11 in decimal), 0010 is 2.

(b)

1 mark

A register of $n$ bits can represent $2^{n}$ unique bit combinations.
For a 12-bit register: $2^{12} = 4096$ distinct binary numbers.

2^{12} = 4096

Mentor tip

State $2^{12} = 4096$ as the total distinct numbers representable in a 12-bit register.

Section B · Structured and algorithmic problems

Questions 9–11 · 45 marks

Question 9

15 marks

CAD software, data compression, and business continuity

System fundamentals
Computer organisation

(a)(i)

1 mark

Computer-Aided Design (CAD) refers to specialized software tools used to digitally draft, model, visualize, and test 2D architectural drawings and 3D geometric designs.

Mentor tip

Define CAD as computer-assisted creation, modification, and digital analysis of technical drawings and 3D models.

(a)(ii)

4 marks

Benefit 1: Rapid revision and iterative editing. CAD models can be modified parametrically without redrawing blueprints from scratch, drastically accelerating design turnarounds.
Benefit 2: High dimensional precision and automated structural simulation. Digital models calculate angles, stress loads, and clearances mathematically, eliminating drafting errors and enabling direct export to CAM or 3D rendering engines.

Mentor tip

Contrast digital editing and mathematical precision against manual drafting boards.

(b)

3 marks

Architectural CAD files incorporate extensive high-resolution geometries and layered vector data, leading to large file footprints.
Compression algorithms reduce redundant bit sequences, shrinking total file size.
This conserves local and cloud storage capacity and shortens upload and download transmission times over network channels.

Mentor tip

Explain both the storage reduction and the network throughput/transfer speed advantages.

(c)(i)

3 marks

Strategy: Implement automated periodic backups to redundant offsite or cloud storage, supplemented by a RAID array or standby mirrored hardware.
Recovery: In the event of primary disk failure, the redundant drive array or standby system takes over immediately (failover), while cloud repositories allow rapid data restoration.
Result: Business operations resume with minimal loss of billable hours or design data.

Mentor tip

Cover the backup mechanism, the switchover/restoration procedure, and how downtime is minimized.

(c)(ii)

4 marks

Cause 1: Human error. Unintentional file deletion, overwriting critical revisions, or closing applications without committing changes.
Cause 2: Malicious cyberattacks or ransomware. Malware, cryptolocker infections, or unauthorised access that encrypts or corrupts files, alongside application software crashes during file write cycles.

Mentor tip

Distinguish non-hardware causes: human operational errors versus malicious cyber threats or software write crashes.

Question 10

15 marks

Campus network access control, connectivity, and communication risks

Networks

(a)

6 marks

Role-Based Access Control (RBAC): Segment user groups into defined tiers (Administrators, Faculty, Students, Guests), enforcing distinct permissions through centralised directory authentication.
Access Matrix Implementation: Administrators hold unrestricted read, write, and execute privileges to system services and domain policies. Teachers maintain read/write privileges to gradebooks and curriculum folders but are restricted from peer financial records.
Student and Guest Isolation: Students receive read access strictly to their individual assignments and reports, while guest users are isolated on a firewalled VLAN with rate-limited internet access and zero access to internal subnets.

Mentor tip

Structure access levels systematically: role separation, authentication credentials, and concrete directory permissions for each user tier.

(b)(i)

2 marks

Communication Link: Dedicated fibre-optic broadband connection.
Justification: Fibre-optic lines provide high bandwidth and low latency over long distances with immunity to electromagnetic interference, handling simultaneous concurrent traffic from thousands of school devices.

Mentor tip

Name the physical medium (fibre optic) and justify with bandwidth, latency, and noise immunity.

(b)(ii)

4 marks

Measure 1: Network edge firewalls with Deep Packet Inspection (DPI). Monitors inbound and outbound traffic, blocking unauthorised port scans, intrusions, and blacklisted domains.
Measure 2: Network segmentation and MAC address filtering / 802.1X network access control. Isolates internal server subnets from student-accessible Wi-Fi and validates connecting physical hardware identifiers.

Mentor tip

Select two perimeter security layers (e.g. firewall filtering and network segmentation/access control) and justify each.

(c)

3 marks

Problem 1: Vulnerability to phishing exploits and social engineering, where students or staff inadvertently disclose network credentials or open attachments bearing malware.
Problem 2: Storage exhaustion and network bandwidth saturation caused by heavy attachments and bulk spam emails.
Problem 3: Pastoral and institutional risks, including misuse of institutional email for cyberbullying, academic dishonesty, or policy violations requiring active administrative monitoring.

Mentor tip

Cover diverse threat vectors: cyber risks (phishing), infrastructure load (bandwidth/storage), and behavioral misuse.

Question 11

15 marks

Grade distribution calculation and linear array processing

Computational thinking, problem-solving and programming

(a)(i)

1 mark

Calculate $GRADES [0]$ using the mark of Ximena Alba ( $MARKS [0] = 5$ ):
$GRADES [0] = 1 + (5 div 10) = 1 + 0 = 1$ .

GRADES [0] = 1 + (5 div 10) = 1

Mentor tip

Integer division (div) truncates any fractional component: 5 div 10 = 0.

(a)(ii)

1 mark

Calculate $GRADES [1]$ using Boris Mount's mark ( $MARKS [1] = 88$ ):
$1 + (88 div 10) = 1 + 8 = 9$ .
Since $9 \geq 7$ , the clamping conditional sets $GRADES [1] = 7$ .

GRADES [1] = min (1 + (88 div 10), 7) = 7

Mentor tip

Remember to apply the capping conditional check: if grade >= 7, clamp to 7.

(a)(iii)

1 mark

Calculate $GRADES [2]$ using Hugh Parr's mark ( $MARKS [2] = 45$ ):
$GRADES [2] = 1 + (45 div 10) = 1 + 4 = 5$ .

GRADES [2] = 1 + (45 div 10) = 5

Mentor tip

45 div 10 = 4, so 1 + 4 = 5.

(a)(iv)

1 mark

To achieve Grade 7, we require $1 + (MARKS [K] div 10) \geq 7$ , which means $MARKS [K] div 10 \geq 6$ .
The smallest non-negative integer satisfying $M div 10 \geq 6$ is $M = 60$ .

M_{m i n} = 60

Mentor tip

60 div 10 = 6, yielding 1 + 6 = 7. Marks 0-59 yield grades 1-6.

(b)

4 marks

Initialize an insertion pointer $I = 0$ for the RESIT array.
Iterate index $K$ from 0 to 199 through the MARKS (or GRADES) array.
If $MARKS [K] < 10$ (or equivalently $GRADES [K] = 1$ ), assign $STUDENTS [K]$ to $RESIT [I]$ and increment pointer $I$ by 1.

I = 0 loop K from 0 to 199 if MARKS [K] < 10 then RESIT [I] = STUDENTS [K] I = I + 1 end if end loop

Mentor tip

Keep a separate index pointer I for the destination RESIT array and only advance it when a matching student is appended.

(c)

7 marks

Pass 1: Accumulate the sum of all 200 marks and divide by 200 to establish the class average.
Pass 2: Traverse index $K$ from 0 to 199. Compare each student's mark against the threshold bands $AVE + 20$ and $AVE - 20$ .
Assign 'A' if mark $> AVE + 20$ , 'C' if mark $< AVE - 20$ , and 'B' otherwise, recording each in $LETTERGRADES [K]$ .

SUM = 0 loop K from 0 to 199 SUM = SUM + MARKS [K] end loop AVE = SUM /200 loop K from 0 to 199 if MARKS [K] > AVE + 20 then LETTERGRADES [K] = "A" else if MARKS [K] < AVE - 20 then LETTERGRADES [K] = "C" else LETTERGRADES [K] = "B" end if end loop

Mentor tip

Calculate the average first across all 200 elements before running the second grading loop.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Computational thinking, problem-solving and programming carries the most marks on this paper.

Computational thinking, problem-solving and programming21 marks · 30%
Q5, Q11
System fundamentals20 marks · 29%
Q1, Q3, Q6, Q9
Networks20 marks · 29%
Q2, Q7, Q10
Computer organisation9 marks · 13%
Q4, Q8, Q9

Mentor insights

Traps to watch for

01
Logic gate ordering · Q4
AND takes precedence over OR. In $A \lor \neg B \land \neg C$ , combine $\neg B$ and $\neg C$ in an AND gate before joining with $A$ in the final OR gate.
02
Trace table subtraction condition · Q5
When $A = 1$ , $S = S - K$ . For $K = 4$ , $A = 1$ leads to $S = 2 - 4 = - 2$ . Ensure the negative sign is retained in accumulator $Z$ .
03
Register capacity vs maximum value · Q8
A 12-bit register holds $2^{12} = 4096$ distinct values. The maximum unsigned value is $2^{12} - 1 = 4095$ , but the number of representations is 4096.
04
Conditional grade capping · Q11
Remember that the scoring algorithm explicitly clamps any calculated grade of 7 or higher to 7: $1 + (88 div 10) = 9$ , which becomes 7.

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