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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2023
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 14

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Compatibility hurdles in corporate information systems integration

System fundamentals

2 marks

Data format and schema differences: Merging organizations often store data in incompatible schemas, date formats, or character encodings, preventing seamless relational joins.
Operating system and software platform divergence: Different hardware architectures, legacy middleware, and operating systems may not support common communication protocols or APIs.

Mentor tip

Focus on data format discrepancies and differing software/hardware architectures.

Question 2

1 mark

Formal definition of the NAND Boolean logic operator

Computer organisation

1 mark

The NAND (NOT AND) operator is a Boolean logic function that evaluates to false (0) if and only if all input operands are true (1); in all other input conditions, it evaluates to true (1).

A nand B = \neg (A \land B)

Mentor tip

State that NAND produces 0 strictly when all inputs are 1.

Question 3

3 marks

Primary functions of the Central Processing Unit Control Unit

Computer organisation

3 marks

Instruction fetching: Directs the retrieval of machine language instructions from primary memory into the Instruction Register using the Program Counter.
Instruction decoding: Decodes opcode bit patterns to determine the required arithmetic, logical, or data transfer operation.
Control and timing signal generation: Emits synchronization clock pulses and bus control signals (e.g. Memory Read/Write, ALU control lines) to direct data flow across CPU registers.

Mentor tip

Cover the three core roles: fetching instructions, decoding opcodes, and issuing control/timing signals.

Question 4

3 marks

Essential components of network data packets

Networks

3 marks

Source and destination IP addresses: Identifies origin and target network hosts for routing across intermediate networks.
Packet sequence number: Enables the destination host to reassemble packets into their original message order and identify missing chunks.
Error-detection checksum: A CRC code stored in the packet header or trailer to verify data integrity upon receipt.

Mentor tip

List source/destination IP addresses, packet sequence numbers, and checksums/payload.

Question 5

3 marks

Binary search tree construction for alphabetical data

Abstract data structures

3 marks

Insert keys in sequential order using alphabetical comparison (left < root <= right):
1. Gladiolus: First inserted element becomes tree root.
2. Lantana: $Lantana > Gladiolus ⟹$ inserted as right child of Gladiolus.
3. Hyacinth: $Hyacinth > Gladiolus$ , but $Hyacinth < Lantana ⟹$ left child of Lantana.
4. Iris: $Iris > Gladiolus$ , $Iris < Lantana$ , $Iris > Hyacinth ⟹$ right child of Hyacinth.
5. Columbine: $Columbine < Gladiolus ⟹$ left child of Gladiolus.
Verification via inorder traversal yields alphabetical ordering: Columbine, Gladiolus, Hyacinth, Iris, Lantana.

Gladiolus ↙ ↘ Columbine Lantana ↙ Hyacinth ↘ Iris

Mentor tip

Follow standard binary search tree insertion order comparing names alphabetically.

Question 6

2 marks

Architecture and purpose of virtual machines

Resource management

2 marks

Software virtualization: A virtual machine is a software-based emulation of a physical computer architecture running an isolated guest operating system atop host hypervisor software.
Sandbox isolation: Provides a secure, self-contained environment for running untrusted applications, testing legacy software, or running multiple OS instances on a single physical server.

Mentor tip

Describe software emulation of physical computer hardware providing an isolated guest environment.

Question 7

2 marks

Usability challenges in voice recognition interfaces

System fundamentals

2 marks

Ambient background noise interference: Acoustic interference from crowded rooms, traffic, or echo degrades phoneme extraction accuracy.
Dialect, accent, and speech variations: Differences in regional pronunciation, pitch, cadence, or speech impediments can lead to high word error rates.

Mentor tip

Mention ambient acoustic noise and regional accents or pronunciation variations.

Question 8

3 marks

Principles and operational advantages of concurrent processing

Resource management

(a)

2 marks

Concurrent processing is the execution of multiple computational tasks or threads over overlapping time periods, either via interleaved time-slicing on a single CPU core or parallel execution across multiple processor cores.

Mentor tip

Concurrent processing means multiple processes make progress over the same time interval.

(b)

1 mark

Improved throughput and reduced latency: Prevents fast computational tasks from being blocked while another task awaits slow I/O operations, maximizing hardware utilization.

Mentor tip

State increased throughput or responsiveness.

Question 9

6 marks

Recursive digit summation trace and algorithmic deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Call 1: $fun (1216)$ . $1216 > 0 ⟹ (1216 mod 10) + fun (1216 div 10) = 6 + fun (121)$ .
Call 2: $fun (121)$ . $121 > 0 ⟹ (121 mod 10) + fun (121 div 10) = 1 + fun (12)$ .
Call 3: $fun (12)$ . $12 > 0 ⟹ (12 mod 10) + fun (12 div 10) = 2 + fun (1)$ .
Call 4: $fun (1)$ . $1 > 0 ⟹ (1 mod 10) + fun (1 div 10) = 1 + fun (0)$ .
Call 5 (Base Case): $fun (0)$ . Condition $0 > 0$ is false $⟹$ returns 0.
Unwinding the call stack: $1 + 0 = 1 ⟹ 2 + 1 = 3 ⟹ 1 + 3 = 4 ⟹ 6 + 4 = 10$ .
Final returned value is 10.

fun (1216) = 6 + 1 + 2 + 1 + 0 = 10

Mentor tip

Show each recursive step extracting the least significant digit via mod 10 and truncating via div 10.

(b)

2 marks

The recursive method calculates and returns the arithmetic sum of all the individual decimal digits that make up the integer $N$ .

Purpose: i \sum d_{i} for base-10 digits of N

Mentor tip

State clearly that it computes the sum of the digits of N.

Questions 10–14 · 75 marks

Question 10

15 marks

Medical practice data security, mail merge automation, and mobile VPN access

System fundamentals
Networks

(a)

2 marks

Role-based access control (RBAC): Configure database access permissions such that receptionists can only view appointment scheduling and billing tables, while sensitive medical histories require authenticated doctor credentials.
Authentication enforcement: Require multi-factor authentication (MFA) and automatic session timeout locks on unattended terminals.

Mentor tip

Explain role-based access control and strong authentication.

(b)(i)

1 mark

Physical storage failure (head crash / SSD controller death), accidental user deletion, or ransomware attack.

Mentor tip

Identify a specific cause such as hardware storage failure or ransomware.

(b)(ii)

2 marks

Automated off-site backup: Regularly generate encrypted backup archives copied to a secure remote cloud repository, enabling point-in-time database restoration after hardware failure.

Mentor tip

Describe automated regular backups stored in an offsite or cloud location.

(c)

3 marks

Data source connection: The word processing mail merge tool connects to the practice database via an SQL query to filter patients eligible for vaccination.
Template placeholder mapping: A standard template letter incorporates dynamic data placeholders (e.g. `<<Title>>`, `<<LastName>>`, `<<Address>>`).
Batch generation: The mail merge engine merges each record into the template, generating customized invitation letters for each eligible patient.

Mentor tip

Explain linking a database query to template merge fields to generate individualized letters.

(d)

4 marks

Cryptographic confidentiality across untrusted networks: Doctor uses public cellular or home Wi-Fi networks; a VPN encrypts all transmitted medical records using end-to-end encryption, ensuring legal compliance (e.g. HIPAA/GDPR).
Secure intranet resource access: The VPN provides an authenticated private IP tunnel, enabling the doctor's remote tablet to query the on-premises database server as if directly cabled inside the clinic.

Mentor tip

Highlight data encryption across public wireless networks and private intranet access.

(e)

3 marks

Signal attenuation and physical obstacles: Distance from cellular towers, terrain, and building materials (thick concrete, metallized glass) attenuate radio signal strength.
Cell tower congestion: Heavy concurrent cellular traffic on a local base station reduces allocated time-frequency resource blocks per user.
Cellular handover and band switching: Switching between 5G, 4G LTE, and 3G bands causes fluctuations in available throughput.

Mentor tip

Discuss physical obstacles/distance from towers and network congestion from other mobile subscribers.

Question 11

15 marks

Requirements elicitation, software testing lifecycle, image ethics, and email transfer

System fundamentals
Networks

(a)(i)

1 mark

Questionnaires / Online surveys / Direct observation / Document review.

Mentor tip

Name questionnaires, surveys, or direct observation.

(a)(ii)

2 marks

Scalability and quantitative analysis: Questionnaires collect standardized responses from large user bases simultaneously with minimal analyst time, producing structured statistical data that can be graphed immediately.

Mentor tip

Surveys reach many users simultaneously and produce quantitative data.

(b)

2 marks

Iterative user validation: Prototypes provide an early interactive demonstration of user interface layout and workflow, allowing end-users to validate design decisions and clarify requirements before expensive development occurs.

Mentor tip

Prototypes allow early stakeholder feedback to detect flaws before full production.

(c)

3 marks

Early defect detection: Unit testing during coding uncovers algorithmic errors when they are simplest to fix.
Interface verification: Integration testing ensures modules exchange parameters correctly without type or schema mismatches.
User validation: System and sign-off testing ensure the final application fulfills business requirements without regressions.

Mentor tip

Explain that testing across each development phase catches bugs early when remediation cost is lowest.

(d)

3 marks

Copyright infringement and intellectual property theft: Using creative images or graphics without proper licensing or creator attribution violates intellectual property laws.
Consent and privacy violations: Distributing photographs featuring identifiable people without model releases violates personal privacy rights.

Mentor tip

Discuss copyright infringement or unauthorized use of images without consent.

(e)(i)

3 marks

Cloud storage hosting: Upload the 60 MB high-resolution image to a cloud hosting platform (e.g. Google Drive, OneDrive) and email an authenticated download URL link.
Multi-part archive splitting: Compress the image using lossless archival tools (e.g. 7-Zip) and split into multi-volume parts under 10 MB each, emailing parts across separate messages.

Mentor tip

Explain uploading to cloud storage with a shared link or splitting into multi-part archives.

(e)(ii)

1 mark

High GPU/RAM memory consumption during image manipulation / Rendering latency over low-speed graphics cards.

Mentor tip

Mention memory consumption or display rendering latency.

Question 12

15 marks

Crossroads traffic control: sensors, ROM firmware, interrupts, and surveillance cameras

Control systems
Computer organisation
Resource management
System fundamentals

(a)(i)

2 marks

Inductive loop sensors (embedded beneath the road surface to detect metallic vehicle bodies via magnetic inductance changes).
Infrared, radar, or ultrasonic motion sensors (mounted overhead to detect vehicle presence via wave reflection).

Mentor tip

Name inductive loops and optical/radar/ultrasonic sensors.

(a)(ii)

2 marks

Autonomous real-time monitoring: Sensors operate continuously 24/7 without requiring manual human oversight, providing instant digital signals to trigger adaptive light cycles.

Mentor tip

Explain autonomous 24/7 real-time detection without human intervention.

(b)

2 marks

Read-Only Memory (ROM) or Flash EEPROM: The control software is permanent firmware that must be retained during power outages and execute immediately upon boot.

Mentor tip

ROM or Flash memory is non-volatile and boots instantly upon power restoration.

(c)(i)

2 marks

A hardware interrupt is an electrical signal sent to the microprocessor that prompts it to pause its current execution flow and prioritize an urgent external event.

Mentor tip

An interrupt suspends the current program cycle to handle a high-priority external signal.

(c)(ii)

3 marks

State preservation: The processor completes its current instruction, pushing the Program Counter and active registers onto the stack.
Interrupt Service Routine (ISR): Looks up the interrupt vector table and jumps to the pedestrian light ISR, changing traffic lights to amber/red and activating the pedestrian walk signal.
State restoration: Pops the saved registers off the stack, restoring the Program Counter to resume normal traffic cycling.

Pedestrian Button Interrupt Push Context \to Execute ISR \to Pop Context

Mentor tip

Detail saving state on the stack, executing the light-change ISR, and restoring state.

(d)(i)

2 marks

Dynamic traffic management: Feeds live visual data to traffic control centers, allowing computerized adaptive light timing to relieve bottlenecks and prioritize emergency vehicles.

Mentor tip

Mention real-time traffic flow optimization or emergency vehicle priority.

(d)(ii)

2 marks

Civil privacy and surveillance concerns: Continuous video capture enables tracking of vehicle number plates and individual driver movements without explicit consent.

Mentor tip

Focus on public privacy concerns regarding continuous automated surveillance.

Question 13

15 marks

Emergency triage data structure: linked list pointer manipulation versus static queues

Abstract data structures

(a)

2 marks

Dynamic node sequence: A linked list is composed of independent nodes where each node contains a data payload and a pointer holding the memory address of the next node.
Head pointer: Access begins at a `HEAD` pointer referencing the first node, with the final node referencing `null`.

Node: [Patient Data ∣ Next Pointer] ⟶ Next Node

Mentor tip

Describe nodes containing data and pointer references to the next node.

(b)

3 marks

Strict FIFO limitation: A standard queue enforces strict arrival-order processing (First-In, First-Out), adding new patients only at the tail.
Clinical triage incompatibility: Patients with critical, life-threatening injuries cannot wait behind earlier patients with minor ailments; priority insertion anywhere in the queue is required.

Mentor tip

Contrast rigid FIFO queue mechanics with clinical triage where high-severity patients must jump to the front.

(c)

3 marks

Node creation: Allocate a new node `NEW_PATIENT` storing the critical patient details.
Link to current head: Set `NEW_PATIENT.next = HEAD`.
Head update: Reassign the `HEAD` pointer to point to `NEW_PATIENT`.
This achieves $O (1)$ constant-time insertion at the beginning of the list.

NEW_PATIENT.next = HEAD; HEAD = NEW_PATIENT

Mentor tip

Describe standard insertion at the head of a linked list in three clear pointer steps.

(d)

4 marks

Pointer traversal: Initialize traversal pointers `CURRENT = HEAD` and `PREVIOUS = null`.
Locate insertion point: Loop while `CURRENT != null` and `CURRENT.severity >= NEW_PATIENT.severity`:
Advance pointers: `PREVIOUS = CURRENT`, `CURRENT = CURRENT.next`.
Splice new node: When the position is reached, set `NEW_PATIENT.next = CURRENT` and set `PREVIOUS.next = NEW_PATIENT`.
This preserves the descending severity order across all triage nodes.

PREVIOUS.next = NEW_PATIENT; NEW_PATIENT.next = CURRENT

Mentor tip

Explain linear traversal to find the priority position, followed by pointer splicing between PREVIOUS and CURRENT.

(e)

3 marks

Memory allocation timing: Static structures (e.g. arrays) allocate fixed blocks of contiguous memory at compile time; dynamic structures (linked lists) allocate nodes on demand from the heap during runtime.
Size mutability: Static arrays cannot expand beyond their pre-set capacity; dynamic structures grow and contract elastically with zero wasted pre-allocated memory.

Mentor tip

Compare fixed compile-time size (static) with elastic runtime heap allocation (dynamic).

Question 14

15 marks

Gymnastics score tabulation: peak scores, 2D matrix scanning, and trimmed mean calculation

Computational thinking, problem-solving and programming

(a)

1 mark

Inspect Tea Fox's scores in row 4: `[7.2, 6.9, 6.9, 7.9, 6.9, 6.9, 6.9, 6.9]`.
The highest score given is 7.9, which appears at index 3, given by Judge 3.

Highest score in row 4: SCORES [4] [3] = 7.9 ⟹ Judge 3

Mentor tip

Locate row 4 in the table and find the column with the highest score.

(b)

4 marks

Initialize counter `COUNT = 0`.
Loop row $R$ from 0 to 19 (across all 20 gymnasts).
Nested loop column $C$ from 0 to 7 (across all 8 judges).
If $SCORES [R] [C] == 10.0$ , increment `COUNT = COUNT + 1`.
Output final tally `COUNT`.

COUNT = 0 loop R from 0 to 19 loop C from 0 to 7 if SCORES [R] [C] == 10.0 then COUNT = COUNT + 1 end if end loop end loop output (COUNT)

Mentor tip

Iterate through the full 2D array and increment the counter whenever the score equals 10.0.

(c)

10 marks

Loop row $R$ from 0 to 19 for each competitor.
Initialize accumulator $SUM = 0$ , $MIN_SCORE = SCORES [R] [0]$ , and $MAX_SCORE = SCORES [R] [0]$ .
Iterate column $C$ from 0 to 7: add each mark to $SUM$ , updating $MIN_SCORE$ and $MAX_SCORE$ .
Subtract the extreme marks: $TRIMMED_SUM = SUM - MIN_SCORE - MAX_SCORE$ .
Compute the average of the remaining 6 scores: $FINAL_SCORE = TRIMMED_SUM /6.0$ .
Output competitor name $NAMES [R]$ and their calculated $FINAL_SCORE$ .

loop R from 0 to 19 MIN_SCORE = SCORES [R] [0] MAX_SCORE = SCORES [R] [0] SUM = 0 loop C from 0 to 7 SCORE = SCORES [R] [C] SUM = SUM + SCORE if SCORE < MIN_SCORE then MIN_SCORE = SCORE end if if SCORE > MAX_SCORE then MAX_SCORE = SCORE end if end loop FINAL_SCORE = (SUM - MIN_SCORE - MAX_SCORE) /6.0 output (NAMES [R], FINAL_SCORE) end loop

Mentor tip

Accumulate the sum and find min and max in a single pass over the 8 scores, then divide the trimmed sum by 6.0.

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 fundamentals26 marks · 26%
Q1, Q7, Q10, Q11, Q12
Computational thinking, problem-solving and programming21 marks · 21%
Q9, Q14
Abstract data structures18 marks · 18%
Q5, Q13
Networks13 marks · 13%
Q4, Q10, Q11
Resource management10 marks · 10%
Q6, Q8, Q12
Computer organisation6 marks · 6%
Q2, Q3, Q12
Control systems6 marks · 6%
Q12

Mentor insights

Traps to watch for

01
Alphabetical tree insertion order · Q5
Compare string values strictly alphabetically when placing children. For example, Lantana > Gladiolus goes to the right, but Hyacinth < Lantana branches left.
02
Recursion base case return value · Q9
When N reaches 0, the function returns 0. Forgetting the base case or returning 1 would add an unintended offset to the digit sum.
03
Linked list priority insertion pointer splicing · Q13
When inserting a node between PREVIOUS and CURRENT, you must first set NEW_NODE.next = CURRENT, then PREVIOUS.next = NEW_NODE. Inverting these assignments breaks the chain.
04
Olympic trimmed mean divisor · Q14
After dropping the highest and lowest scores from the 8 judges, divide the remaining sum by exactly 6.0, not 8.

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 HL · Paper 1

Computer Science, Higher Level

Session: November 2023
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 14

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Compatibility hurdles in corporate information systems integration

System fundamentals

2 marks

Data format and schema differences: Merging organizations often store data in incompatible schemas, date formats, or character encodings, preventing seamless relational joins.
Operating system and software platform divergence: Different hardware architectures, legacy middleware, and operating systems may not support common communication protocols or APIs.

Mentor tip

Focus on data format discrepancies and differing software/hardware architectures.

Question 2

1 mark

Formal definition of the NAND Boolean logic operator

Computer organisation

1 mark

The NAND (NOT AND) operator is a Boolean logic function that evaluates to false (0) if and only if all input operands are true (1); in all other input conditions, it evaluates to true (1).

A nand B = \neg (A \land B)

Mentor tip

State that NAND produces 0 strictly when all inputs are 1.

Question 3

3 marks

Primary functions of the Central Processing Unit Control Unit

Computer organisation

3 marks

Instruction fetching: Directs the retrieval of machine language instructions from primary memory into the Instruction Register using the Program Counter.
Instruction decoding: Decodes opcode bit patterns to determine the required arithmetic, logical, or data transfer operation.
Control and timing signal generation: Emits synchronization clock pulses and bus control signals (e.g. Memory Read/Write, ALU control lines) to direct data flow across CPU registers.

Mentor tip

Cover the three core roles: fetching instructions, decoding opcodes, and issuing control/timing signals.

Question 4

3 marks

Essential components of network data packets

Networks

3 marks

Source and destination IP addresses: Identifies origin and target network hosts for routing across intermediate networks.
Packet sequence number: Enables the destination host to reassemble packets into their original message order and identify missing chunks.
Error-detection checksum: A CRC code stored in the packet header or trailer to verify data integrity upon receipt.

Mentor tip

List source/destination IP addresses, packet sequence numbers, and checksums/payload.

Question 5

3 marks

Binary search tree construction for alphabetical data

Abstract data structures

3 marks

Insert keys in sequential order using alphabetical comparison (left < root <= right):
1. Gladiolus: First inserted element becomes tree root.
2. Lantana: $Lantana > Gladiolus ⟹$ inserted as right child of Gladiolus.
3. Hyacinth: $Hyacinth > Gladiolus$ , but $Hyacinth < Lantana ⟹$ left child of Lantana.
4. Iris: $Iris > Gladiolus$ , $Iris < Lantana$ , $Iris > Hyacinth ⟹$ right child of Hyacinth.
5. Columbine: $Columbine < Gladiolus ⟹$ left child of Gladiolus.
Verification via inorder traversal yields alphabetical ordering: Columbine, Gladiolus, Hyacinth, Iris, Lantana.

Gladiolus ↙ ↘ Columbine Lantana ↙ Hyacinth ↘ Iris

Mentor tip

Follow standard binary search tree insertion order comparing names alphabetically.

Question 6

2 marks

Architecture and purpose of virtual machines

Resource management

2 marks

Software virtualization: A virtual machine is a software-based emulation of a physical computer architecture running an isolated guest operating system atop host hypervisor software.
Sandbox isolation: Provides a secure, self-contained environment for running untrusted applications, testing legacy software, or running multiple OS instances on a single physical server.

Mentor tip

Describe software emulation of physical computer hardware providing an isolated guest environment.

Question 7

2 marks

Usability challenges in voice recognition interfaces

System fundamentals

2 marks

Ambient background noise interference: Acoustic interference from crowded rooms, traffic, or echo degrades phoneme extraction accuracy.
Dialect, accent, and speech variations: Differences in regional pronunciation, pitch, cadence, or speech impediments can lead to high word error rates.

Mentor tip

Mention ambient acoustic noise and regional accents or pronunciation variations.

Question 8

3 marks

Principles and operational advantages of concurrent processing

Resource management

(a)

2 marks

Mentor tip

Concurrent processing means multiple processes make progress over the same time interval.

(b)

1 mark

Improved throughput and reduced latency: Prevents fast computational tasks from being blocked while another task awaits slow I/O operations, maximizing hardware utilization.

Mentor tip

State increased throughput or responsiveness.

Question 9

6 marks

Recursive digit summation trace and algorithmic deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Call 1: $fun (1216)$ . $1216 > 0 ⟹ (1216 mod 10) + fun (1216 div 10) = 6 + fun (121)$ .
Call 2: $fun (121)$ . $121 > 0 ⟹ (121 mod 10) + fun (121 div 10) = 1 + fun (12)$ .
Call 3: $fun (12)$ . $12 > 0 ⟹ (12 mod 10) + fun (12 div 10) = 2 + fun (1)$ .
Call 4: $fun (1)$ . $1 > 0 ⟹ (1 mod 10) + fun (1 div 10) = 1 + fun (0)$ .
Call 5 (Base Case): $fun (0)$ . Condition $0 > 0$ is false $⟹$ returns 0.
Unwinding the call stack: $1 + 0 = 1 ⟹ 2 + 1 = 3 ⟹ 1 + 3 = 4 ⟹ 6 + 4 = 10$ .
Final returned value is 10.

fun (1216) = 6 + 1 + 2 + 1 + 0 = 10

Mentor tip

Show each recursive step extracting the least significant digit via mod 10 and truncating via div 10.

(b)

2 marks

The recursive method calculates and returns the arithmetic sum of all the individual decimal digits that make up the integer $N$ .

Purpose: i \sum d_{i} for base-10 digits of N

Mentor tip

State clearly that it computes the sum of the digits of N.

Questions 10–14 · 75 marks

Question 10

15 marks

Medical practice data security, mail merge automation, and mobile VPN access

System fundamentals
Networks

(a)

2 marks

Role-based access control (RBAC): Configure database access permissions such that receptionists can only view appointment scheduling and billing tables, while sensitive medical histories require authenticated doctor credentials.
Authentication enforcement: Require multi-factor authentication (MFA) and automatic session timeout locks on unattended terminals.

Mentor tip

Explain role-based access control and strong authentication.

(b)(i)

1 mark

Physical storage failure (head crash / SSD controller death), accidental user deletion, or ransomware attack.

Mentor tip

Identify a specific cause such as hardware storage failure or ransomware.

(b)(ii)

2 marks

Automated off-site backup: Regularly generate encrypted backup archives copied to a secure remote cloud repository, enabling point-in-time database restoration after hardware failure.

Mentor tip

Describe automated regular backups stored in an offsite or cloud location.

(c)

3 marks

Data source connection: The word processing mail merge tool connects to the practice database via an SQL query to filter patients eligible for vaccination.
Template placeholder mapping: A standard template letter incorporates dynamic data placeholders (e.g. `<<Title>>`, `<<LastName>>`, `<<Address>>`).
Batch generation: The mail merge engine merges each record into the template, generating customized invitation letters for each eligible patient.

Mentor tip

Explain linking a database query to template merge fields to generate individualized letters.

(d)

4 marks

Cryptographic confidentiality across untrusted networks: Doctor uses public cellular or home Wi-Fi networks; a VPN encrypts all transmitted medical records using end-to-end encryption, ensuring legal compliance (e.g. HIPAA/GDPR).
Secure intranet resource access: The VPN provides an authenticated private IP tunnel, enabling the doctor's remote tablet to query the on-premises database server as if directly cabled inside the clinic.

Mentor tip

Highlight data encryption across public wireless networks and private intranet access.

(e)

3 marks

Signal attenuation and physical obstacles: Distance from cellular towers, terrain, and building materials (thick concrete, metallized glass) attenuate radio signal strength.
Cell tower congestion: Heavy concurrent cellular traffic on a local base station reduces allocated time-frequency resource blocks per user.
Cellular handover and band switching: Switching between 5G, 4G LTE, and 3G bands causes fluctuations in available throughput.

Mentor tip

Discuss physical obstacles/distance from towers and network congestion from other mobile subscribers.

Question 11

15 marks

Requirements elicitation, software testing lifecycle, image ethics, and email transfer

System fundamentals
Networks

(a)(i)

1 mark

Questionnaires / Online surveys / Direct observation / Document review.

Mentor tip

Name questionnaires, surveys, or direct observation.

(a)(ii)

2 marks

Mentor tip

Surveys reach many users simultaneously and produce quantitative data.

(b)

2 marks

Mentor tip

Prototypes allow early stakeholder feedback to detect flaws before full production.

(c)

3 marks

Early defect detection: Unit testing during coding uncovers algorithmic errors when they are simplest to fix.
Interface verification: Integration testing ensures modules exchange parameters correctly without type or schema mismatches.
User validation: System and sign-off testing ensure the final application fulfills business requirements without regressions.

Mentor tip

Explain that testing across each development phase catches bugs early when remediation cost is lowest.

(d)

3 marks

Copyright infringement and intellectual property theft: Using creative images or graphics without proper licensing or creator attribution violates intellectual property laws.
Consent and privacy violations: Distributing photographs featuring identifiable people without model releases violates personal privacy rights.

Mentor tip

Discuss copyright infringement or unauthorized use of images without consent.

(e)(i)

3 marks

Cloud storage hosting: Upload the 60 MB high-resolution image to a cloud hosting platform (e.g. Google Drive, OneDrive) and email an authenticated download URL link.
Multi-part archive splitting: Compress the image using lossless archival tools (e.g. 7-Zip) and split into multi-volume parts under 10 MB each, emailing parts across separate messages.

Mentor tip

Explain uploading to cloud storage with a shared link or splitting into multi-part archives.

(e)(ii)

1 mark

High GPU/RAM memory consumption during image manipulation / Rendering latency over low-speed graphics cards.

Mentor tip

Mention memory consumption or display rendering latency.

Question 12

15 marks

Crossroads traffic control: sensors, ROM firmware, interrupts, and surveillance cameras

Control systems
Computer organisation
Resource management
System fundamentals

(a)(i)

2 marks

Inductive loop sensors (embedded beneath the road surface to detect metallic vehicle bodies via magnetic inductance changes).
Infrared, radar, or ultrasonic motion sensors (mounted overhead to detect vehicle presence via wave reflection).

Mentor tip

Name inductive loops and optical/radar/ultrasonic sensors.

(a)(ii)

2 marks

Autonomous real-time monitoring: Sensors operate continuously 24/7 without requiring manual human oversight, providing instant digital signals to trigger adaptive light cycles.

Mentor tip

Explain autonomous 24/7 real-time detection without human intervention.

(b)

2 marks

Read-Only Memory (ROM) or Flash EEPROM: The control software is permanent firmware that must be retained during power outages and execute immediately upon boot.

Mentor tip

ROM or Flash memory is non-volatile and boots instantly upon power restoration.

(c)(i)

2 marks

A hardware interrupt is an electrical signal sent to the microprocessor that prompts it to pause its current execution flow and prioritize an urgent external event.

Mentor tip

An interrupt suspends the current program cycle to handle a high-priority external signal.

(c)(ii)

3 marks

State preservation: The processor completes its current instruction, pushing the Program Counter and active registers onto the stack.
Interrupt Service Routine (ISR): Looks up the interrupt vector table and jumps to the pedestrian light ISR, changing traffic lights to amber/red and activating the pedestrian walk signal.
State restoration: Pops the saved registers off the stack, restoring the Program Counter to resume normal traffic cycling.

Pedestrian Button Interrupt Push Context \to Execute ISR \to Pop Context

Mentor tip

Detail saving state on the stack, executing the light-change ISR, and restoring state.

(d)(i)

2 marks

Dynamic traffic management: Feeds live visual data to traffic control centers, allowing computerized adaptive light timing to relieve bottlenecks and prioritize emergency vehicles.

Mentor tip

Mention real-time traffic flow optimization or emergency vehicle priority.

(d)(ii)

2 marks

Civil privacy and surveillance concerns: Continuous video capture enables tracking of vehicle number plates and individual driver movements without explicit consent.

Mentor tip

Focus on public privacy concerns regarding continuous automated surveillance.

Question 13

15 marks

Emergency triage data structure: linked list pointer manipulation versus static queues

Abstract data structures

(a)

2 marks

Dynamic node sequence: A linked list is composed of independent nodes where each node contains a data payload and a pointer holding the memory address of the next node.
Head pointer: Access begins at a `HEAD` pointer referencing the first node, with the final node referencing `null`.

Node: [Patient Data ∣ Next Pointer] ⟶ Next Node

Mentor tip

Describe nodes containing data and pointer references to the next node.

(b)

3 marks

Strict FIFO limitation: A standard queue enforces strict arrival-order processing (First-In, First-Out), adding new patients only at the tail.
Clinical triage incompatibility: Patients with critical, life-threatening injuries cannot wait behind earlier patients with minor ailments; priority insertion anywhere in the queue is required.

Mentor tip

Contrast rigid FIFO queue mechanics with clinical triage where high-severity patients must jump to the front.

(c)

3 marks

Node creation: Allocate a new node `NEW_PATIENT` storing the critical patient details.
Link to current head: Set `NEW_PATIENT.next = HEAD`.
Head update: Reassign the `HEAD` pointer to point to `NEW_PATIENT`.
This achieves $O (1)$ constant-time insertion at the beginning of the list.

NEW_PATIENT.next = HEAD; HEAD = NEW_PATIENT

Mentor tip

Describe standard insertion at the head of a linked list in three clear pointer steps.

(d)

4 marks

Pointer traversal: Initialize traversal pointers `CURRENT = HEAD` and `PREVIOUS = null`.
Locate insertion point: Loop while `CURRENT != null` and `CURRENT.severity >= NEW_PATIENT.severity`:
Advance pointers: `PREVIOUS = CURRENT`, `CURRENT = CURRENT.next`.
Splice new node: When the position is reached, set `NEW_PATIENT.next = CURRENT` and set `PREVIOUS.next = NEW_PATIENT`.
This preserves the descending severity order across all triage nodes.

PREVIOUS.next = NEW_PATIENT; NEW_PATIENT.next = CURRENT

Mentor tip

Explain linear traversal to find the priority position, followed by pointer splicing between PREVIOUS and CURRENT.

(e)

3 marks

Memory allocation timing: Static structures (e.g. arrays) allocate fixed blocks of contiguous memory at compile time; dynamic structures (linked lists) allocate nodes on demand from the heap during runtime.
Size mutability: Static arrays cannot expand beyond their pre-set capacity; dynamic structures grow and contract elastically with zero wasted pre-allocated memory.

Mentor tip

Compare fixed compile-time size (static) with elastic runtime heap allocation (dynamic).

Question 14

15 marks

Gymnastics score tabulation: peak scores, 2D matrix scanning, and trimmed mean calculation

Computational thinking, problem-solving and programming

(a)

1 mark

Inspect Tea Fox's scores in row 4: `[7.2, 6.9, 6.9, 7.9, 6.9, 6.9, 6.9, 6.9]`.
The highest score given is 7.9, which appears at index 3, given by Judge 3.

Highest score in row 4: SCORES [4] [3] = 7.9 ⟹ Judge 3

Mentor tip

Locate row 4 in the table and find the column with the highest score.

(b)

4 marks

Initialize counter `COUNT = 0`.
Loop row $R$ from 0 to 19 (across all 20 gymnasts).
Nested loop column $C$ from 0 to 7 (across all 8 judges).
If $SCORES [R] [C] == 10.0$ , increment `COUNT = COUNT + 1`.
Output final tally `COUNT`.

COUNT = 0 loop R from 0 to 19 loop C from 0 to 7 if SCORES [R] [C] == 10.0 then COUNT = COUNT + 1 end if end loop end loop output (COUNT)

Mentor tip

Iterate through the full 2D array and increment the counter whenever the score equals 10.0.

(c)

10 marks

Loop row $R$ from 0 to 19 for each competitor.
Initialize accumulator $SUM = 0$ , $MIN_SCORE = SCORES [R] [0]$ , and $MAX_SCORE = SCORES [R] [0]$ .
Iterate column $C$ from 0 to 7: add each mark to $SUM$ , updating $MIN_SCORE$ and $MAX_SCORE$ .
Subtract the extreme marks: $TRIMMED_SUM = SUM - MIN_SCORE - MAX_SCORE$ .
Compute the average of the remaining 6 scores: $FINAL_SCORE = TRIMMED_SUM /6.0$ .
Output competitor name $NAMES [R]$ and their calculated $FINAL_SCORE$ .

loop R from 0 to 19 MIN_SCORE = SCORES [R] [0] MAX_SCORE = SCORES [R] [0] SUM = 0 loop C from 0 to 7 SCORE = SCORES [R] [C] SUM = SUM + SCORE if SCORE < MIN_SCORE then MIN_SCORE = SCORE end if if SCORE > MAX_SCORE then MAX_SCORE = SCORE end if end loop FINAL_SCORE = (SUM - MIN_SCORE - MAX_SCORE) /6.0 output (NAMES [R], FINAL_SCORE) end loop

Mentor tip

Accumulate the sum and find min and max in a single pass over the 8 scores, then divide the trimmed sum by 6.0.

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 fundamentals26 marks · 26%
Q1, Q7, Q10, Q11, Q12
Computational thinking, problem-solving and programming21 marks · 21%
Q9, Q14
Abstract data structures18 marks · 18%
Q5, Q13
Networks13 marks · 13%
Q4, Q10, Q11
Resource management10 marks · 10%
Q6, Q8, Q12
Computer organisation6 marks · 6%
Q2, Q3, Q12
Control systems6 marks · 6%
Q12

Mentor insights

Traps to watch for

01
Alphabetical tree insertion order · Q5
Compare string values strictly alphabetically when placing children. For example, Lantana > Gladiolus goes to the right, but Hyacinth < Lantana branches left.
02
Recursion base case return value · Q9
When N reaches 0, the function returns 0. Forgetting the base case or returning 1 would add an unintended offset to the digit sum.
03
Linked list priority insertion pointer splicing · Q13
When inserting a node between PREVIOUS and CURRENT, you must first set NEW_NODE.next = CURRENT, then PREVIOUS.next = NEW_NODE. Inverting these assignments breaks the chain.
04
Olympic trimmed mean divisor · Q14
After dropping the highest and lowest scores from the 8 judges, divide the remaining sum by exactly 6.0, not 8.

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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