Loading…

Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2025
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2025 Global 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 1–10 · 25 marks

Question 1

2 marks

Fundamental features of programming languages

Computational thinking, problem-solving and programming

2 marks

Fixed vocabulary (keywords): A programming language possesses a reserved set of standardized, unambiguous terms that dictate specific operations.
Rigid grammar (syntax): Strict syntactic rules govern statement composition, delimiter usage, and token ordering so the compiler or interpreter parses instructions consistently.

Mentor tip

Distinguish between vocabulary (reserved keywords) and syntax (grammatical structure and delimiter rules).

Question 2

2 marks

Failover redundancy architectures

System fundamentals

2 marks

A failover system is an automated backup mechanism where a secondary server, system component, or network channel seamlessly assumes operations upon the unexpected failure of the primary system.
Its primary objective is ensuring high operational availability and business continuity with near-zero service downtime.

Mentor tip

Define failover as automated operational switchover from a failed primary system to a standby secondary system.

Question 3

2 marks

Purpose and utility of Data Flow Diagrams

System fundamentals

2 marks

A Data Flow Diagram (DFD) provides a graphical model mapping the movement, transformation, storage, and external origins and destinations of data through an information system.
It clarifies process boundaries and data dependencies for analysts, developers, and stakeholders without exposing underlying hardware or programming details.

Mentor tip

Describe a DFD as visualizing how data moves between external entities, processes, and data stores.

Question 4

3 marks

Byte representation capacity and hexadecimal conversions

Computer organisation

(a)

1 mark

An 8-bit integer can represent $2^{8} = 256$ distinct values (e.g. 0 to 255 in unsigned binary).

2^{8} = 256

Mentor tip

Total distinct values for 8 bits is $2^{8} = 256$ .

(b)

1 mark

Convert each hex digit of $7A_{16}$ into its 4-bit binary equivalent:
$7_{16} = 011 1_{2}$
$A_{16} = 101 0_{2}$
Combining nibbles yields $0111101 0_{2}$ .

7A_{16} = 0111101 0_{2}

Mentor tip

7 is 0111 and A is 1010 in binary.

(c)

1 mark

Partition the byte $0001110 0_{2}$ into two 4-bit groups:
$000 1_{2} = 1_{16}$
$110 0_{2} = 1 2_{10} = C_{16}$
Combining yields $1C_{16}$ .

0001110 0_{2} = 1C_{16}

Mentor tip

Split into 0001 (1) and 1100 (C in hex).

Question 5

5 marks

Trace table execution for prefix sum accumulator loop

Computational thinking, problem-solving and programming

5 marks

Initial state: $K = 0, NUM = [7, 1, 5, 9, 6]$ .
Iteration 1 ( $K < 4$ is true): $K = 0 + 1 = 1$ . $NUM [1] = NUM [0] + NUM [1] = 7 + 1 = 8$ .
Iteration 2 ( $K < 4$ is true): $K = 1 + 1 = 2$ . $NUM [2] = NUM [1] + NUM [2] = 8 + 5 = 13$ .
Iteration 3 ( $K < 4$ is true): $K = 2 + 1 = 3$ . $NUM [3] = NUM [2] + NUM [3] = 13 + 9 = 22$ .
Iteration 4 ( $K < 4$ is true): $K = 3 + 1 = 4$ . $NUM [4] = NUM [3] + NUM [4] = 22 + 6 = 28$ .
Loop terminates since $K = 4$ is not strictly less than 4.
Post-loop statement executes: $NUM [0] = NUM [4] + NUM [0] = 28 + 7 = 35$ .

K 01234 - NUM [0] 7 - - - - 35 NUM [1] 18 - - - - NUM [2] 5 - 13 - - - NUM [3] 9 - - 22 - - NUM [4] 6 - - - 28 -

Mentor tip

Each loop iteration adds the previous element to the current element. Remember to execute the post-loop statement updating NUM[0] last.

Question 6

4 marks

Logic diagram construction for compound expressions

Computer organisation

4 marks

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

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

Mentor tip

Place NOT gates on A and B, combine NOT B with C through an AND gate, then join with NOT A using an OR gate.

Question 7

2 marks

System accessibility enhancements

System fundamentals

2 marks

Method 1: Integration of assistive output technologies, such as integrated screen readers and text-to-speech engines paired with high-contrast display modes.
Method 2: Support for alternative physical input peripherals, such as Braille displays, speech recognition input, and single-switch accessibility devices.

Mentor tip

State concrete assistive technologies: text-to-speech/screen readers, voice control, and Braille keyboards.

Question 8

1 mark

Role of the Memory Data Register in instruction execution

Computer organisation

1 mark

The Memory Data Register (MDR) holds the actual data value or machine instruction currently being fetched from or written to primary memory (RAM) via the system data bus.

Mentor tip

Remember that the MAR holds addresses, while the MDR holds the contents/data transferred to or from those addresses.

Question 9

2 marks

Physical characteristics of fibre-optic transmission lines

Networks

2 marks

High transmission bandwidth: Uses pulses of light transmitted along glass or plastic cores, achieving very high data transfer rates with low signal attenuation over long physical distances.
Electromagnetic immunity: Because data propagates optically rather than electrically, fibre cables are immune to electromagnetic interference (EMI) and radio frequency interference (RFI).

Mentor tip

Highlight optical transmission, superior bandwidth over distance, and complete immunity to electrical noise.

Question 10

2 marks

Virtual machine emulation in app development

System fundamentals

2 marks

Cross-platform compatibility testing: Developers can simulate multiple guest operating systems, screen resolutions, and API levels concurrently on a single physical workstation.
Sandboxed isolation: Testing unstable builds or unvetted third-party packages in an isolated guest environment protects the host operating system from crashes or malware contamination.

Mentor tip

Highlight testing diverse target operating systems on one host machine and sandboxed isolation.

Section B · Structured and algorithmic problems

Questions 11–13 · 45 marks

Question 11

15 marks

WLAN infrastructure, benefits, security protocols, and VPN tunneling

Networks

(a)

3 marks

Wireless Access Points (WAPs) to broadcast and receive radio frequency frames.
Wireless Network Interface Cards (NICs) / wireless adaptors in client endpoints.
Network switches or wireless LAN controllers (WLC) to bridge wireless traffic onto the wired backbone.

Mentor tip

List physical WLAN infrastructure elements: wireless access points, wireless NICs, and wireless switches/controllers.

(b)

4 marks

Advantage 1: Workplace mobility and operational flexibility. Employees can move laptops and handheld devices freely between meeting spaces and desks while sustaining continuous network sessions.
Advantage 2: Reduced physical cabling infrastructure and installation costs. Adding new workstations requires zero physical cabling runs through walls and floors, enabling rapid organizational scaling.

Mentor tip

Highlight mobility of workforce and cost/installation reduction compared to structured copper cabling.

(c)

4 marks

Method 1: Implementation of robust encryption protocols such as WPA3 enterprise, utilizing 192-bit cryptographic keys to render intercepted radio frames unreadable.
Method 2: Network access control via MAC address whitelisting, disabling SSID broadcast, and mandating 802.1X certificate-based multi-factor authentication.

Mentor tip

Explain strong encryption (WPA3) and perimeter access authentication (802.1X / MAC filtering).

(d)

4 marks

Functionality: The VPN client establishes a virtual network interface, assigning the remote machine an internal IP address on the enterprise subnet, enabling seamless access to file shares and intranet services as if physically plugged in.
Security: Encapsulates all payload packets within an encrypted IPsec or TLS tunnel across the public internet, ensuring data confidentiality and packet integrity against eavesdropping.

Mentor tip

Explain both dimensions: full functional access (virtual subnet presence) and data security (encrypted tunneling).

Question 12

15 marks

Requirements gathering, prototyping paradigms, and storage architectures

System fundamentals

(a)(i)

2 marks

Captures unbiased, ground-truth workflows in their authentic operating environment.
Uncovers informal workarounds and unspoken operational subtleties that users forget to mention during formal questionnaires.

Mentor tip

Direct observation captures authentic real-world behavior and reveals undocumented procedures.

(a)(ii)

1 mark

Structured interviews (or user survey questionnaires / document examination).

Mentor tip

Name one alternative requirements gathering technique (interviews, surveys, focus groups).

(a)(iii)

2 marks

Enables deep qualitative probing into user motivations and specific pain points.
Allows the analyst to clarify ambiguities in real time and adapt follow-up questions dynamically.

Mentor tip

Highlight interactive clarification and the ability to gather deep qualitative insights.

(b)(i)

2 marks

Early user feedback: Delivers an interactive, tangible visual model early in the development lifecycle, allowing stakeholders to test usability and refine requirements before writing core backend code.

Mentor tip

Focus on early stakeholder validation and early detection of requirement misunderstandings.

(b)(ii)

2 marks

Scope creep and false expectations: Stakeholders may mistake an unoptimized UI mock-up for a finished system, demanding immediate delivery or requesting endless minor visual revisions that delay core architectural work.

Mentor tip

Note risks of scope creep and misleading stakeholders regarding backend completeness.

(c)

6 marks

Data Security: Local storage provides physical sovereignty and absolute internal control over sensitive assets, but relies entirely on in-house backup discipline. Cloud storage offers enterprise automated replication, geo-redundancy, and 24/7 cybersecurity teams, but introduces external third-party compliance and transmission interception risks.
Financial Cost: Local storage requires significant capital expenditure (CapEx) for server hardware, RAID controllers, UPS units, and maintenance personnel. Cloud storage operates on an operational expense model (OpEx), offering predictable recurring fees that scale with consumption.
Access and Reliability: Cloud storage affords anywhere, anytime access across global devices with high service-level availability, yet remains strictly dependent on constant internet connectivity. Local storage allows high-speed internal LAN access independent of external internet outages.

Mentor tip

Address all three required dimensions: security (physical sovereignty vs cloud redundancy), cost (CapEx vs OpEx), and availability (LAN independence vs ubiquitous remote access).

Question 13

15 marks

Collection data structures and statistical iteration algorithms

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

A collection is an abstract container data structure that groups zero or more related items of arbitrary type.
It provides a standardized set of methods (such as resetNext, hasNext, and getNext) to traverse and manipulate elements sequentially without exposing underlying storage mechanisms.

Mentor tip

Define a collection by its ability to hold grouped elements and its iterator primitives (resetNext, hasNext, getNext).

(b)(i)

2 marks

The loop calls `PLTNT.getNext()` twice within each iteration (once to assign $X$ , and a second time inside the output call).
This advances the collection pointer two positions per cycle, skipping every alternate value and attempting to read past the end of the collection if there is an odd number of items, causing a runtime exception.

Mentor tip

Explain that calling getNext() twice advances the iterator cursor twice, outputting alternate elements and risking a runtime crash.

(b)(ii)

2 marks

Either output variable $X$ directly: `X = PLTNT.getNext()`, followed by `output(X)`, or eliminate the assignment line and output `PLTNT.getNext()` directly.

PLTNT.resetNext () loop while PLTNT.hasNext () X = PLTNT.getNext () output (X) end loop

Mentor tip

Call getNext() once per iteration and output the stored variable.

(c)

9 marks

Step 1: Initialize iterator with `PLTNT.resetNext()`. Set `SUM = 0`, `COUNT = 0`, `MIN = 500` (or above 500), and `MAX = 0` (or below 0).
Step 2: Traverse with `while PLTNT.hasNext()`. Retrieve `VAL = PLTNT.getNext()`. Add `VAL` to `SUM` and increment `COUNT` by 1.
Step 3: If `VAL > MAX`, update `MAX = VAL`. If `VAL < MIN`, update `MIN = VAL`.
Step 4: After loop termination, calculate `AVE = SUM / COUNT`. Output the maximum, minimum, and average values.

PLTNT.resetNext () MIN = 500 MAX = 0 SUM = 0 COUNT = 0 loop while PLTNT.hasNext () VAL = PLTNT.getNext () SUM = SUM + VAL COUNT = COUNT + 1 if VAL > MAX then MAX = VAL if VAL < MIN then MIN = VAL end loop output ("Maximum pollutant concentration: ", MAX) output ("Minimum pollutant concentration: ", MIN) output ("Average pollutant concentration: ", SUM / COUNT)

Mentor tip

Reset the collection pointer first, retrieve each element once per loop, update running sum/count/min/max, and calculate average after the loop.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals23 marks · 33%
Q2, Q3, Q7, Q10, Q12
Computational thinking, problem-solving and programming20 marks · 29%
Q1, Q5, Q13
Networks17 marks · 24%
Q9, Q11
Computer organisation8 marks · 11%
Q4, Q6, Q8
Abstract data structures2 marks · 3%
Q13

Mentor insights

Traps to watch for

01
Array index sequencing in prefix sum · Q5
Notice that $K$ increments before modifying $NUM [K]$ . The loop updates index 1 through 4, then finishes by updating $NUM [0] = 28 + 7 = 35$ outside the loop.
02
Operator precedence in Boolean logic · Q6
AND takes precedence over OR. In $\neg A \lor \neg B \land C$ , evaluate $\neg B \land C$ before applying OR with $\neg A$ .
03
Virtual machine purpose vs host machine · Q10
Focus on developer benefits: testing multiple OS environments simultaneously on single hardware and avoiding host system corruption.
04
Double getNext() cursor advance · Q13
Every call to `getNext()` consumes an element from the iterator. Calling it once for an assignment and once in an output call advances the iterator by two items per iteration.

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

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers

Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2025
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2025 Global 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 1–10 · 25 marks

Question 1

2 marks

Fundamental features of programming languages

Computational thinking, problem-solving and programming

2 marks

Fixed vocabulary (keywords): A programming language possesses a reserved set of standardized, unambiguous terms that dictate specific operations.
Rigid grammar (syntax): Strict syntactic rules govern statement composition, delimiter usage, and token ordering so the compiler or interpreter parses instructions consistently.

Mentor tip

Distinguish between vocabulary (reserved keywords) and syntax (grammatical structure and delimiter rules).

Question 2

2 marks

Failover redundancy architectures

System fundamentals

2 marks

A failover system is an automated backup mechanism where a secondary server, system component, or network channel seamlessly assumes operations upon the unexpected failure of the primary system.
Its primary objective is ensuring high operational availability and business continuity with near-zero service downtime.

Mentor tip

Define failover as automated operational switchover from a failed primary system to a standby secondary system.

Question 3

2 marks

Purpose and utility of Data Flow Diagrams

System fundamentals

2 marks

A Data Flow Diagram (DFD) provides a graphical model mapping the movement, transformation, storage, and external origins and destinations of data through an information system.
It clarifies process boundaries and data dependencies for analysts, developers, and stakeholders without exposing underlying hardware or programming details.

Mentor tip

Describe a DFD as visualizing how data moves between external entities, processes, and data stores.

Question 4

3 marks

Byte representation capacity and hexadecimal conversions

Computer organisation

(a)

1 mark

An 8-bit integer can represent $2^{8} = 256$ distinct values (e.g. 0 to 255 in unsigned binary).

2^{8} = 256

Mentor tip

Total distinct values for 8 bits is $2^{8} = 256$ .

(b)

1 mark

Convert each hex digit of $7A_{16}$ into its 4-bit binary equivalent:
$7_{16} = 011 1_{2}$
$A_{16} = 101 0_{2}$
Combining nibbles yields $0111101 0_{2}$ .

7A_{16} = 0111101 0_{2}

Mentor tip

7 is 0111 and A is 1010 in binary.

(c)

1 mark

Partition the byte $0001110 0_{2}$ into two 4-bit groups:
$000 1_{2} = 1_{16}$
$110 0_{2} = 1 2_{10} = C_{16}$
Combining yields $1C_{16}$ .

0001110 0_{2} = 1C_{16}

Mentor tip

Split into 0001 (1) and 1100 (C in hex).

Question 5

5 marks

Trace table execution for prefix sum accumulator loop

Computational thinking, problem-solving and programming

5 marks

Initial state: $K = 0, NUM = [7, 1, 5, 9, 6]$ .
Iteration 1 ( $K < 4$ is true): $K = 0 + 1 = 1$ . $NUM [1] = NUM [0] + NUM [1] = 7 + 1 = 8$ .
Iteration 2 ( $K < 4$ is true): $K = 1 + 1 = 2$ . $NUM [2] = NUM [1] + NUM [2] = 8 + 5 = 13$ .
Iteration 3 ( $K < 4$ is true): $K = 2 + 1 = 3$ . $NUM [3] = NUM [2] + NUM [3] = 13 + 9 = 22$ .
Iteration 4 ( $K < 4$ is true): $K = 3 + 1 = 4$ . $NUM [4] = NUM [3] + NUM [4] = 22 + 6 = 28$ .
Loop terminates since $K = 4$ is not strictly less than 4.
Post-loop statement executes: $NUM [0] = NUM [4] + NUM [0] = 28 + 7 = 35$ .

K 01234 - NUM [0] 7 - - - - 35 NUM [1] 18 - - - - NUM [2] 5 - 13 - - - NUM [3] 9 - - 22 - - NUM [4] 6 - - - 28 -

Mentor tip

Each loop iteration adds the previous element to the current element. Remember to execute the post-loop statement updating NUM[0] last.

Question 6

4 marks

Logic diagram construction for compound expressions

Computer organisation

4 marks

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

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

Mentor tip

Place NOT gates on A and B, combine NOT B with C through an AND gate, then join with NOT A using an OR gate.

Question 7

2 marks

System accessibility enhancements

System fundamentals

2 marks

Method 1: Integration of assistive output technologies, such as integrated screen readers and text-to-speech engines paired with high-contrast display modes.
Method 2: Support for alternative physical input peripherals, such as Braille displays, speech recognition input, and single-switch accessibility devices.

Mentor tip

State concrete assistive technologies: text-to-speech/screen readers, voice control, and Braille keyboards.

Question 8

1 mark

Role of the Memory Data Register in instruction execution

Computer organisation

1 mark

The Memory Data Register (MDR) holds the actual data value or machine instruction currently being fetched from or written to primary memory (RAM) via the system data bus.

Mentor tip

Remember that the MAR holds addresses, while the MDR holds the contents/data transferred to or from those addresses.

Question 9

2 marks

Physical characteristics of fibre-optic transmission lines

Networks

2 marks

High transmission bandwidth: Uses pulses of light transmitted along glass or plastic cores, achieving very high data transfer rates with low signal attenuation over long physical distances.
Electromagnetic immunity: Because data propagates optically rather than electrically, fibre cables are immune to electromagnetic interference (EMI) and radio frequency interference (RFI).

Mentor tip

Highlight optical transmission, superior bandwidth over distance, and complete immunity to electrical noise.

Question 10

2 marks

Virtual machine emulation in app development

System fundamentals

2 marks

Cross-platform compatibility testing: Developers can simulate multiple guest operating systems, screen resolutions, and API levels concurrently on a single physical workstation.
Sandboxed isolation: Testing unstable builds or unvetted third-party packages in an isolated guest environment protects the host operating system from crashes or malware contamination.

Mentor tip

Highlight testing diverse target operating systems on one host machine and sandboxed isolation.

Section B · Structured and algorithmic problems

Questions 11–13 · 45 marks

Question 11

15 marks

WLAN infrastructure, benefits, security protocols, and VPN tunneling

Networks

(a)

3 marks

Wireless Access Points (WAPs) to broadcast and receive radio frequency frames.
Wireless Network Interface Cards (NICs) / wireless adaptors in client endpoints.
Network switches or wireless LAN controllers (WLC) to bridge wireless traffic onto the wired backbone.

Mentor tip

List physical WLAN infrastructure elements: wireless access points, wireless NICs, and wireless switches/controllers.

(b)

4 marks

Advantage 1: Workplace mobility and operational flexibility. Employees can move laptops and handheld devices freely between meeting spaces and desks while sustaining continuous network sessions.
Advantage 2: Reduced physical cabling infrastructure and installation costs. Adding new workstations requires zero physical cabling runs through walls and floors, enabling rapid organizational scaling.

Mentor tip

Highlight mobility of workforce and cost/installation reduction compared to structured copper cabling.

(c)

4 marks

Method 1: Implementation of robust encryption protocols such as WPA3 enterprise, utilizing 192-bit cryptographic keys to render intercepted radio frames unreadable.
Method 2: Network access control via MAC address whitelisting, disabling SSID broadcast, and mandating 802.1X certificate-based multi-factor authentication.

Mentor tip

Explain strong encryption (WPA3) and perimeter access authentication (802.1X / MAC filtering).

(d)

4 marks

Functionality: The VPN client establishes a virtual network interface, assigning the remote machine an internal IP address on the enterprise subnet, enabling seamless access to file shares and intranet services as if physically plugged in.
Security: Encapsulates all payload packets within an encrypted IPsec or TLS tunnel across the public internet, ensuring data confidentiality and packet integrity against eavesdropping.

Mentor tip

Explain both dimensions: full functional access (virtual subnet presence) and data security (encrypted tunneling).

Question 12

15 marks

Requirements gathering, prototyping paradigms, and storage architectures

System fundamentals

(a)(i)

2 marks

Captures unbiased, ground-truth workflows in their authentic operating environment.
Uncovers informal workarounds and unspoken operational subtleties that users forget to mention during formal questionnaires.

Mentor tip

Direct observation captures authentic real-world behavior and reveals undocumented procedures.

(a)(ii)

1 mark

Structured interviews (or user survey questionnaires / document examination).

Mentor tip

Name one alternative requirements gathering technique (interviews, surveys, focus groups).

(a)(iii)

2 marks

Enables deep qualitative probing into user motivations and specific pain points.
Allows the analyst to clarify ambiguities in real time and adapt follow-up questions dynamically.

Mentor tip

Highlight interactive clarification and the ability to gather deep qualitative insights.

(b)(i)

2 marks

Mentor tip

Focus on early stakeholder validation and early detection of requirement misunderstandings.

(b)(ii)

2 marks

Mentor tip

Note risks of scope creep and misleading stakeholders regarding backend completeness.

(c)

6 marks

Data Security: Local storage provides physical sovereignty and absolute internal control over sensitive assets, but relies entirely on in-house backup discipline. Cloud storage offers enterprise automated replication, geo-redundancy, and 24/7 cybersecurity teams, but introduces external third-party compliance and transmission interception risks.
Financial Cost: Local storage requires significant capital expenditure (CapEx) for server hardware, RAID controllers, UPS units, and maintenance personnel. Cloud storage operates on an operational expense model (OpEx), offering predictable recurring fees that scale with consumption.
Access and Reliability: Cloud storage affords anywhere, anytime access across global devices with high service-level availability, yet remains strictly dependent on constant internet connectivity. Local storage allows high-speed internal LAN access independent of external internet outages.

Mentor tip

Address all three required dimensions: security (physical sovereignty vs cloud redundancy), cost (CapEx vs OpEx), and availability (LAN independence vs ubiquitous remote access).

Question 13

15 marks

Collection data structures and statistical iteration algorithms

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

A collection is an abstract container data structure that groups zero or more related items of arbitrary type.
It provides a standardized set of methods (such as resetNext, hasNext, and getNext) to traverse and manipulate elements sequentially without exposing underlying storage mechanisms.

Mentor tip

Define a collection by its ability to hold grouped elements and its iterator primitives (resetNext, hasNext, getNext).

(b)(i)

2 marks

The loop calls `PLTNT.getNext()` twice within each iteration (once to assign $X$ , and a second time inside the output call).
This advances the collection pointer two positions per cycle, skipping every alternate value and attempting to read past the end of the collection if there is an odd number of items, causing a runtime exception.

Mentor tip

Explain that calling getNext() twice advances the iterator cursor twice, outputting alternate elements and risking a runtime crash.

(b)(ii)

2 marks

Either output variable $X$ directly: `X = PLTNT.getNext()`, followed by `output(X)`, or eliminate the assignment line and output `PLTNT.getNext()` directly.

PLTNT.resetNext () loop while PLTNT.hasNext () X = PLTNT.getNext () output (X) end loop

Mentor tip

Call getNext() once per iteration and output the stored variable.

(c)

9 marks

Step 1: Initialize iterator with `PLTNT.resetNext()`. Set `SUM = 0`, `COUNT = 0`, `MIN = 500` (or above 500), and `MAX = 0` (or below 0).
Step 2: Traverse with `while PLTNT.hasNext()`. Retrieve `VAL = PLTNT.getNext()`. Add `VAL` to `SUM` and increment `COUNT` by 1.
Step 3: If `VAL > MAX`, update `MAX = VAL`. If `VAL < MIN`, update `MIN = VAL`.
Step 4: After loop termination, calculate `AVE = SUM / COUNT`. Output the maximum, minimum, and average values.

PLTNT.resetNext () MIN = 500 MAX = 0 SUM = 0 COUNT = 0 loop while PLTNT.hasNext () VAL = PLTNT.getNext () SUM = SUM + VAL COUNT = COUNT + 1 if VAL > MAX then MAX = VAL if VAL < MIN then MIN = VAL end loop output ("Maximum pollutant concentration: ", MAX) output ("Minimum pollutant concentration: ", MIN) output ("Average pollutant concentration: ", SUM / COUNT)

Mentor tip

Reset the collection pointer first, retrieve each element once per loop, update running sum/count/min/max, and calculate average after the loop.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals23 marks · 33%
Q2, Q3, Q7, Q10, Q12
Computational thinking, problem-solving and programming20 marks · 29%
Q1, Q5, Q13
Networks17 marks · 24%
Q9, Q11
Computer organisation8 marks · 11%
Q4, Q6, Q8
Abstract data structures2 marks · 3%
Q13

Mentor insights

Traps to watch for

01
Array index sequencing in prefix sum · Q5
Notice that $K$ increments before modifying $NUM [K]$ . The loop updates index 1 through 4, then finishes by updating $NUM [0] = 28 + 7 = 35$ outside the loop.
02
Operator precedence in Boolean logic · Q6
AND takes precedence over OR. In $\neg A \lor \neg B \land C$ , evaluate $\neg B \land C$ before applying OR with $\neg A$ .
03
Virtual machine purpose vs host machine · Q10
Focus on developer benefits: testing multiple OS environments simultaneously on single hardware and avoiding host system corruption.
04
Double getNext() cursor advance · Q13
Every call to `getNext()` consumes an element from the iterator. Calling it once for an assignment and once in an output call advances the iterator by two items per iteration.

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

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers