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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2024
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 15

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2024 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

2 marks

Core computational features of spreadsheet software

System fundamentals

2 marks

Automatic recalculation: When cell inputs change, dependent formulas update dynamically without manual recalculation.
Built-in statistical and mathematical functions: Provides predefined routines (e.g. `SUM`, `AVERAGE`, `LOOKUP`) operating across defined cell ranges.

Total = i = 1 \sum n Cell_{i}

Mentor tip

Focus on computational capabilities rather than cosmetic formatting (like fonts or colours).

Question 2

2 marks

Role and value of beta testing in software release cycles

System fundamentals

2 marks

Diverse operating environments: External users test the pre-release software on varied real-world hardware, operating systems, and network configurations that developers cannot replicate in laboratory environments.
Unanticipated user behaviours: Identifies rare edge cases, usability friction, and intermittent bugs prior to general market release.

Mentor tip

Distinguish beta testing (external target users in real operating conditions) from alpha testing (internal developers and in-house QA).

Question 3

2 marks

Involvement of end users during system changeover

System fundamentals

2 marks

Requirements validation and feedback: End users articulate day-to-day workflow requirements and identify functional gaps between the legacy system and the new solution.
Testing and operational sign-off: End users perform validation testing to ensure day-to-day business processes function smoothly before the legacy system is decommissioned.

Mentor tip

End users provide domain knowledge and practical validation; technical architecture is handled by systems analysts.

Question 4

2 marks

Purpose and architectural role of system flowcharts

System fundamentals

2 marks

High-level process flow: System flowcharts represent the overall sequence of data processing across hardware components, software modules, and physical media.
Hardware and storage boundary mapping: Distinguishes physical inputs (e.g. keyboards, barcode scanners), storage media (hard drives, tapes, cloud stores), and output devices.

Mentor tip

Do not confuse system flowcharts (system-wide data flow across hardware/files) with program flowcharts (step-by-step algorithmic logic).

Question 5

3 marks

Traffic filtering and perimeter defence via firewalls

Networks

3 marks

Packet inspection: Examines incoming and outgoing network packets at network boundaries against predefined security rule sets.
Filtering criteria: Evaluates source/destination IP addresses, port numbers, protocols (e.g. TCP, UDP, ICMP), and connection states.
Access enforcement: Automatically permits legitimate traffic while blocking or dropping unauthorized packets to protect the private intranet from external intrusions.

Rule: (Source IP = trusted) \land (Port = 443) ⟹ ALLOW

Mentor tip

Explain both what a firewall inspects (headers/ports/IPs) and the action it takes (allow or block based on rule tables).

Question 6

2 marks

Modular benefits of subprograms in software engineering

Computational thinking, problem-solving and programming

2 marks

Code reusability: A single subprogram can be invoked from multiple locations in the codebase without duplicating source code.
Isolation and modular testing: Complex programs are decomposed into independent, manageable modules that can be written, tested, and debugged in isolation.

Mentor tip

Frame your answers around software maintainability, abstraction, and code reuse.

Question 7

4 marks

Trace of Euclidean algorithm implemented as recursive method

Computational thinking, problem-solving and programming

4 marks

Initial call: $rec (20, 12)$ . Since $B = 12 \neq = 0$ , evaluate $rec (B, A mod B) = rec (12, 20 mod 12) = rec (12, 8)$ .
Second call: $rec (12, 8)$ . Since $B = 8 \neq = 0$ , evaluate $rec (8, 12 mod 8) = rec (8, 4)$ .
Third call: $rec (8, 4)$ . Since $B = 4 \neq = 0$ , evaluate $rec (4, 8 mod 4) = rec (4, 0)$ .
Base case: $rec (4, 0)$ . Here $B = 0$ , satisfying the base condition; the method returns $A = 4$ .
Return stack: The base return value 4 bubbles up through all nested recursive calls, producing final output 4.

rec (20, 12) \to rec (12, 8) \to rec (8, 4) \to rec (4, 0) = 4

Mentor tip

Notice this is Euclid's algorithm for Greatest Common Divisor (GCD): $g cd (20, 12) = 4$ .

Question 8

2 marks

Robotics and automation in automotive assembly

Control systems

2 marks

Consistent precision and reduced defect rates: Robotic actuators perform welding, spray painting, and component placement with sub-millimetre repeatability 24/7 without fatigue.
Hazardous environment mitigation: Removes human workers from toxic chemical exposure (paint fumes) and extreme physical hazards (heavy lifting, sparks).

Mentor tip

State both the operational benefit (e.g. 24/7 precision/repeatability) and the underlying rationale.

Question 9

2 marks

Binary search tree traversals: postorder and inorder

Abstract data structures

(a)

1 mark

Postorder traversal follows the rule: Left Subtree $\to$ Right Subtree $\to$ Root.
Traversal sequence: Visit left leaf X, then right subtree rooted at C (traverse left child E, then right subtree Y visiting left child H, then Y, then C), and finally visit root A.
Result: X E H Y C A.

Postorder: X, E, H, Y, C, A

Mentor tip

In postorder traversal, the root node is always processed last.

(b)

1 mark

Inorder traversal follows the rule: Left Subtree $\to$ Root $\to$ Right Subtree.
Traversal sequence: Visit left child X $\to$ root A $\to$ right subtree (visit left child E $\to$ node C $\to$ left child H $\to$ node Y).
Result: X A E C H Y.

Inorder: X, A, E, C, H, Y

Mentor tip

Inorder traversal visits the leftmost node first, then moves upward through parents to right subtrees.

Question 10

4 marks

Queue abstract data structure primitive operations

Abstract data structures

4 marks

Enqueue (insert/add): Inserts an incoming element at the rear (tail) of the queue. If the queue is implemented with fixed capacity and is full, this triggers an overflow condition.
Dequeue (remove/serve): Retrieves and removes the element at the front (head) of the queue, shifting the head pointer forward. If the queue is empty, this triggers an underflow condition.

Enqueue \to [Rear] \dots [Front] \to Dequeue

Mentor tip

Queues strictly adhere to First-In, First-Out (FIFO) ordering. Always mention front/rear pointers.

Questions 11–15 · 75 marks

Question 11

15 marks

Mobile computing, operating system roles, and packet protocols

System fundamentals
Resource management
Networks

(a)

1 mark

Full-disk storage encryption: Implementing storage encryption (e.g. BitLocker, FileVault) ensures that if the physical device is stolen, stored files remain unreadable without cryptographic authorization keys.

Mentor tip

A simple login password only protects OS access; full storage encryption protects data even if the drive is removed.

(b)

2 marks

Memory management: The operating system allocates distinct blocks of primary memory (RAM) to the program's code, stack, and heap, tracking freed memory upon process termination.
Process scheduling: The OS scheduler allocates CPU time slices via preemptive multitasking, switching execution context among active threads.

Mentor tip

Identify a specific OS subsystem (memory manager or scheduler) and outline its direct role.

(c)

4 marks

Wired Ethernet advantages: Provides high sustained bandwidth, ultra-low latency, and immunity to radio frequency interference, ideal for high-throughput tasks like large data transfers and gaming.
Wireless Wi-Fi advantages: Provides roaming mobility and physical convenience, allowing users to connect to local area networks and hotspots without physical cabling.
Complementary utility: Supporting both interfaces enables the device to adapt seamlessly across stationary office workstations and mobile environments.

Mentor tip

Structure comparisons by balancing throughput/stability (wired) against convenience/mobility (wireless).

(d)

2 marks

Packet header: Contains routing metadata including source and destination IP addresses, sequence numbering, protocol type, and packet length.
Packet payload: Contains the actual segmented chunk of application data being transported.
Packet trailer: Contains error-detection checksum bits (such as a cyclic redundancy check, CRC) to verify transmission integrity.

Data Packet = [Header ∣ Payload ∣ Trailer]

Mentor tip

Detail the three structural parts of a standard packet: header, payload, and trailer.

(e)

6 marks

Standardization and interoperability: Protocols define universal syntactic and semantic conventions so hardware and software from different manufacturers communicate seamlessly.
Error detection and correction: Transmission protocols (e.g. TCP) verify integrity using checksums and initiate retransmission requests upon packet corruption or packet loss.
Flow control and congestion avoidance: Manages packet transmission rates between fast senders and slower receivers to prevent buffer saturation and network deadlock.

Mentor tip

Provide distinct technical rationales: interoperability, error handling, and data flow synchronization.

Question 12

15 marks

Boolean logic evaluation, security alarm design, and CPU execution cycle

Computer organisation

(a)(i)

1 mark

The Boolean logic operators present in the given expression are `or` and `and`.

Operators: or, and

Mentor tip

Relational operators ( $>, <$ ) compare numerical values; logical operators (`or`, `and`, `not`) combine truth values.

(a)(ii)

1 mark

The numeric constant values explicitly stated in the expression are 6, 3, and 20.

Constants: 6, 3, 20

Mentor tip

Constants are immutable literal values, whereas X and Y are variable identifiers.

(a)(iii)

2 marks

Substitute $X = 6$ and $Y = 6$ into each relational term:
Term 1: $(X > 6) ⟹ (6 > 6)$ is false.
Term 2: $(Y > 3) ⟹ (6 > 3)$ is true.
Term 3: $((X + Y) < 20) ⟹ (12 < 20)$ is true.
Apply operator precedence (`and` binds before `or`):
Evaluate conjunction: $true \land true = true$ .
Evaluate disjunction: $false \lor true = true$ .

(6 > 6) \lor ((6 > 3) \land (12 < 20)) = false \lor (true \land true) = true

Mentor tip

Remember logical precedence: AND has higher precedence than OR.

(b)

4 marks

Siren condition: The siren activates when the door is locked ( $DOOR = 1$ ) AND either movement is detected ( $MOTION = 1$ ) OR the window is open ( $WINDOW = 1$ ).
Logic equation: $SIREN = DOOR \land (MOTION \lor WINDOW)$ .
Evaluate all 8 binary input combinations:
Rows with $DOOR = 0$ : Output is 0 for (0,0,0), (0,0,1), (1,0,0), (1,0,1).
Rows with $DOOR = 1$ :
$(0, 1, 0) ⟹ 1 \land (0 \lor 0) = 0$ .
$(0, 1, 1) ⟹ 1 \land (0 \lor 1) = 1$ .
$(1, 1, 0) ⟹ 1 \land (1 \lor 0) = 1$ .
$(1, 1, 1) ⟹ 1 \land (1 \lor 1) = 1$ .

MOTION 00001111 DOOR 00110011 WINDOW 01010101 SIREN 00010011

Mentor tip

When DOOR = 0, the siren can never sound, zeroing out all four corresponding truth table rows.

(c)(i)

1 mark

The Memory Data Register (MDR), also called the Memory Buffer Register (MBR), temporarily holds the data word fetched from or waiting to be written into primary storage.

Register: Memory Data Register (MDR)

Mentor tip

MAR holds the destination address; MDR holds the data payload itself.

(c)(ii)

1 mark

The Control Unit (CU) contains the instruction decoder circuitry that decodes the opcode bits in the current instruction register.

CPU component: Control Unit (CU)

Mentor tip

The CU coordinates instruction fetching, decoding, and generating timing signals.

(c)(iii)

1 mark

All arithmetic operations (addition, subtraction) and logical comparisons are executed within the Arithmetic Logic Unit (ALU).

CPU component: Arithmetic Logic Unit (ALU)

Mentor tip

Calculations and Boolean evaluations are exclusively handled by the ALU.

(c)(iv)

4 marks

Address Bus (unidirectional): Carries the memory address generated by the MAR from the CPU to the RAM address decoders.
Data Bus (bidirectional): Carries instruction opcodes and operand data between the CPU (MDR) and primary memory.
Control Bus (bidirectional): Transmits control and synchronization signals from the Control Unit (such as Memory Read, Memory Write, clock pulses, and interrupt lines).
Coordinated execution: During fetch, the address bus identifies the memory location, the control bus signals a read operation, and the data bus transports the fetched instruction word into the processor.

Buses = {Address Bus, Data Bus, Control Bus}

Mentor tip

Detail the directionality and specific payload of all three internal system buses.

Question 13

15 marks

Stack interrupts, display physics, geolocation systems, and dedicated operating systems

Abstract data structures
System fundamentals
Networks
Resource management

(a)

2 marks

Context save upon interrupt: When an incoming call triggers a hardware interrupt, the CPU pushes current execution state (Program Counter, general-purpose registers, call stack pointers) onto an interrupt call stack.
Context restoration: Once the phone call concludes, the CPU pops the saved register values off the top of the stack, restoring the Program Counter and resuming app execution seamlessly.

Interrupt: Push CPU context \to Handle Call \to Pop CPU context

Mentor tip

Stacks follow LIFO semantics, making them ideal for handling nested interrupts and subroutine calls.

(b)

2 marks

Sensor-driven physical tracking: Smartphone health apps read internal sensors (e.g. accelerometers, gyroscopes, optical heart rate monitors) to log steps, calories, and cardiac metrics.
User feedback loops: Visualizes progress against targets, providing nudges and reminders that encourage healthier daily activity levels.

Mentor tip

Connect technical hardware sensors to actionable user behavioral feedback.

(c)

2 marks

Higher pixel density: Higher display resolution packs more physical pixels into the same screen area, resulting in a higher pixels-per-inch (PPI) count.
Visual smoothness: Smaller individual pixels reduce visible pixelation and jagged stair-stepping (aliasing), rendering curved vector glyphs and fine image textures with superior visual fidelity.

PPI = \frac{w ^{2} + h ^{2}}{d}

Mentor tip

Explain that pixel size decreases as resolution increases, making individual pixel boundaries indistinguishable to the human eye.

(d)(i)

4 marks

Satellite constellation: The Global Positioning System consists of a network of satellites orbiting Earth, each transmitting precise radio signals with atomic clock timestamps.
Distance calculation: The smartphone receiver compares signal transmission time with arrival time, multiplying the delay by the speed of light to calculate distance to the satellite.
Trilateration: By computing distances from at least four separate satellites, the receiver mathematically solves geometric equations to calculate its exact 3D coordinates (latitude, longitude, and elevation).

d_{i} = c \cdot (t_{received, i} - t_{transmitted, i})

Mentor tip

Mention satellites, time-of-flight distance calculation, and trilateration using at least four satellites.

(d)(ii)

2 marks

Privacy violation and stalker tracking: Sharing precise real-time location can expose a user's residential address, routines, and physical presence to malicious third parties without informed consent.
Targeted profiling: Aggregated telemetry data can be harvested by data brokers for unconsented behavioural profiling.

Mentor tip

Highlight concrete security or personal privacy risks associated with continuous real-time geolocation.

(e)

3 marks

Hardware optimization and energy efficiency: Dedicated mobile operating systems are specifically engineered for battery-powered, ARM-based architectures, aggressively managing CPU power states to maximize battery longevity.
Specialized input and sensor support: Natively integrates touch gesture subsystems, accelerometers, cellular radios, and camera APIs without unnecessary generic desktop driver overhead.

Mentor tip

Contrast low-power mobile optimization against the resource-heavy footprint of generic desktop operating systems.

Question 14

15 marks

Linked list management, sequential traversal, and double linked list architecture

Abstract data structures

(a)

4 marks

Dynamic memory allocation: Linked lists allocate and free nodes on demand from the heap during runtime, avoiding fixed array size limits and preventing wasted pre-allocated memory.
Constant-time node insertion and deletion: Inserting or removing a student name simply requires updating pointer references ( $O (1)$ linking), eliminating the expensive $O (n)$ element-shifting required in static arrays.

Array insert: O (n) shifting vs Linked list insert: O (1) pointer updates

Mentor tip

Identify both dynamic resizing (no waste) and pointer-based insertion/deletion (no shifting).

(b)

5 marks

Pointer initialization: Initialize a traversal pointer `CURRENT` to point to `HEAD`.
Iterative traversal: While `CURRENT != null` and `CURRENT.data <= TARGET`:
Check match: If `CURRENT.data == TARGET`, the student name is located; return success or node reference.
Advance pointer: Otherwise, advance to the next node by setting `CURRENT = CURRENT.next`.
Alphabetical early termination: If `CURRENT == null` or `CURRENT.data > TARGET`, terminate traversal; since the list is sorted alphabetically, the target name is confirmed absent.

CURRENT = HEAD loop while CURRENT \neq = null and CURRENT.data < TARGET CURRENT = CURRENT.next end loop return (CURRENT \neq = null and CURRENT.data == TARGET)

Mentor tip

Because the list is maintained in alphabetical order, stop searching as soon as the node data exceeds the target string.

(c)(i)

4 marks

Bidirectional node structure: Each node in a double linked list contains three fields: a data payload, a `next` pointer referencing the successor node, and a `prev` pointer referencing the predecessor node.
Boundary terminal pointers: The `prev` pointer of the head node points to `null`, and the `next` pointer of the tail node points to `null`.

Node: [prev ∣ data ∣ next] ⟺ [prev ∣ data ∣ next]

Mentor tip

Explicitly mention all three components of a DLL node: data value, forward next pointer, and backward prev pointer.

(c)(ii)

2 marks

Web browser navigation: Maintaining web history to allow instantaneous forward and backward page transitions.
Media player playlists / Undo-redo stacks: Enabling backward/forward navigation across audio tracks or editing command histories.

Mentor tip

Any workflow requiring seamless two-way traversal is a prime application for a double linked list.

Question 15

15 marks

2D matrix frequency tabulation, absence detection, and mode extraction

Computational thinking, problem-solving and programming

(a)

4 marks

Iterate through all rows of the 2D array MAT from index 0 to 6.
Within each row, iterate through all columns from index 0 to 6.
Read integer value $VAL = MAT [R] [C]$ .
Increment the corresponding frequency tally in the 1D count array: $COUNT [VAL] = COUNT [VAL] + 1$ .

loop R from 0 to 6 loop C from 0 to 6 VAL = MAT [R] [C] COUNT [VAL] = COUNT [VAL] + 1 end loop end loop

Mentor tip

Use nested loops over row and column dimensions, indexing directly into COUNT using the matrix values.

(b)

5 marks

Initialize a boolean tracking flag `ALLPRESENT = true`.
Loop through all integer digits $I$ from 0 to 9 in the COUNT array.
If $COUNT [I] == 0$ , output a message indicating digit $I$ is missing, and toggle `ALLPRESENT = false`.
After inspecting all 10 digits, if `ALLPRESENT` remains true, output that every number from 0 to 9 is present.

ALLPRESENT = true loop I from 0 to 9 if COUNT [I] == 0 then output ("Number ", I, " is not present") ALLPRESENT = false end if end loop if ALLPRESENT then output ("All numbers from 0 to 9 are present") end if

Mentor tip

The boolean flag ensures you only print the 'all numbers present' confirmation once at the end.

(c)

6 marks

Pass 1: Find the maximum occurrence frequency by scanning $COUNT [I]$ from $I = 0$ to 9.
Track the highest frequency in variable `MAXFREQ`.
Pass 2: Scan $COUNT [I]$ again from 0 to 9; whenever $COUNT [I] == MAXFREQ$ , output digit $I$ as a mode.
This two-pass design naturally accommodates bimodal or multimodal datasets where multiple numbers share the peak frequency.

MAXFREQ = COUNT [0] loop I from 1 to 9 if COUNT [I] > MAXFREQ then MAXFREQ = COUNT [I] end if end loop output ("Modes with frequency: ", MAXFREQ) loop I from 0 to 9 if COUNT [I] == MAXFREQ then output (I) end if end loop

Mentor tip

Because multiple numbers can share the top count (multimodal), find the maximum frequency first before printing all matching digits.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Abstract data structures carries the most marks on this paper.

Abstract data structures23 marks · 23%
Q9, Q10, Q13, Q14
Computational thinking, problem-solving and programming21 marks · 21%
Q6, Q7, Q15
Networks19 marks · 19%
Q5, Q11, Q13
System fundamentals15 marks · 15%
Q1, Q2, Q3, Q4, Q11, Q13
Computer organisation15 marks · 15%
Q12
Resource management5 marks · 5%
Q11, Q13
Control systems2 marks · 2%
Q8

Mentor insights

Traps to watch for

01
Modulus parameter order in Euclidean recursion · Q7
Ensure rec(B, A mod B) passes B as the first parameter and remainder as the second. Swapping parameters reverses the divisor and dividend.
02
Operator precedence in compound Boolean expressions · Q12
Logical AND has higher precedence than OR. In (X > 6) or (Y > 3) and ((X + Y) < 20), the AND conjunction is evaluated before the OR disjunction.
03
Double linked list bidirectional pointer maintenance · Q14
When inserting or deleting nodes in a double linked list, both forward next and backward prev pointers must be re-linked. Updating only one direction breaks reverse traversal.
04
Multimodal frequency scanning · Q15
Never assume a dataset has only a single mode. Scan the array first to find the peak frequency MAXFREQ, then execute a second pass to output all values whose frequency matches MAXFREQ.

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 2024
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 15

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2024 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

2 marks

Core computational features of spreadsheet software

System fundamentals

2 marks

Automatic recalculation: When cell inputs change, dependent formulas update dynamically without manual recalculation.
Built-in statistical and mathematical functions: Provides predefined routines (e.g. `SUM`, `AVERAGE`, `LOOKUP`) operating across defined cell ranges.

Total = i = 1 \sum n Cell_{i}

Mentor tip

Focus on computational capabilities rather than cosmetic formatting (like fonts or colours).

Question 2

2 marks

Role and value of beta testing in software release cycles

System fundamentals

2 marks

Diverse operating environments: External users test the pre-release software on varied real-world hardware, operating systems, and network configurations that developers cannot replicate in laboratory environments.
Unanticipated user behaviours: Identifies rare edge cases, usability friction, and intermittent bugs prior to general market release.

Mentor tip

Distinguish beta testing (external target users in real operating conditions) from alpha testing (internal developers and in-house QA).

Question 3

2 marks

Involvement of end users during system changeover

System fundamentals

2 marks

Requirements validation and feedback: End users articulate day-to-day workflow requirements and identify functional gaps between the legacy system and the new solution.
Testing and operational sign-off: End users perform validation testing to ensure day-to-day business processes function smoothly before the legacy system is decommissioned.

Mentor tip

End users provide domain knowledge and practical validation; technical architecture is handled by systems analysts.

Question 4

2 marks

Purpose and architectural role of system flowcharts

System fundamentals

2 marks

High-level process flow: System flowcharts represent the overall sequence of data processing across hardware components, software modules, and physical media.
Hardware and storage boundary mapping: Distinguishes physical inputs (e.g. keyboards, barcode scanners), storage media (hard drives, tapes, cloud stores), and output devices.

Mentor tip

Do not confuse system flowcharts (system-wide data flow across hardware/files) with program flowcharts (step-by-step algorithmic logic).

Question 5

3 marks

Traffic filtering and perimeter defence via firewalls

Networks

3 marks

Packet inspection: Examines incoming and outgoing network packets at network boundaries against predefined security rule sets.
Filtering criteria: Evaluates source/destination IP addresses, port numbers, protocols (e.g. TCP, UDP, ICMP), and connection states.
Access enforcement: Automatically permits legitimate traffic while blocking or dropping unauthorized packets to protect the private intranet from external intrusions.

Rule: (Source IP = trusted) \land (Port = 443) ⟹ ALLOW

Mentor tip

Explain both what a firewall inspects (headers/ports/IPs) and the action it takes (allow or block based on rule tables).

Question 6

2 marks

Modular benefits of subprograms in software engineering

Computational thinking, problem-solving and programming

2 marks

Code reusability: A single subprogram can be invoked from multiple locations in the codebase without duplicating source code.
Isolation and modular testing: Complex programs are decomposed into independent, manageable modules that can be written, tested, and debugged in isolation.

Mentor tip

Frame your answers around software maintainability, abstraction, and code reuse.

Question 7

4 marks

Trace of Euclidean algorithm implemented as recursive method

Computational thinking, problem-solving and programming

4 marks

Initial call: $rec (20, 12)$ . Since $B = 12 \neq = 0$ , evaluate $rec (B, A mod B) = rec (12, 20 mod 12) = rec (12, 8)$ .
Second call: $rec (12, 8)$ . Since $B = 8 \neq = 0$ , evaluate $rec (8, 12 mod 8) = rec (8, 4)$ .
Third call: $rec (8, 4)$ . Since $B = 4 \neq = 0$ , evaluate $rec (4, 8 mod 4) = rec (4, 0)$ .
Base case: $rec (4, 0)$ . Here $B = 0$ , satisfying the base condition; the method returns $A = 4$ .
Return stack: The base return value 4 bubbles up through all nested recursive calls, producing final output 4.

rec (20, 12) \to rec (12, 8) \to rec (8, 4) \to rec (4, 0) = 4

Mentor tip

Notice this is Euclid's algorithm for Greatest Common Divisor (GCD): $g cd (20, 12) = 4$ .

Question 8

2 marks

Robotics and automation in automotive assembly

Control systems

2 marks

Consistent precision and reduced defect rates: Robotic actuators perform welding, spray painting, and component placement with sub-millimetre repeatability 24/7 without fatigue.
Hazardous environment mitigation: Removes human workers from toxic chemical exposure (paint fumes) and extreme physical hazards (heavy lifting, sparks).

Mentor tip

State both the operational benefit (e.g. 24/7 precision/repeatability) and the underlying rationale.

Question 9

2 marks

Binary search tree traversals: postorder and inorder

Abstract data structures

(a)

1 mark

Postorder traversal follows the rule: Left Subtree $\to$ Right Subtree $\to$ Root.
Traversal sequence: Visit left leaf X, then right subtree rooted at C (traverse left child E, then right subtree Y visiting left child H, then Y, then C), and finally visit root A.
Result: X E H Y C A.

Postorder: X, E, H, Y, C, A

Mentor tip

In postorder traversal, the root node is always processed last.

(b)

1 mark

Inorder traversal follows the rule: Left Subtree $\to$ Root $\to$ Right Subtree.
Traversal sequence: Visit left child X $\to$ root A $\to$ right subtree (visit left child E $\to$ node C $\to$ left child H $\to$ node Y).
Result: X A E C H Y.

Inorder: X, A, E, C, H, Y

Mentor tip

Inorder traversal visits the leftmost node first, then moves upward through parents to right subtrees.

Question 10

4 marks

Queue abstract data structure primitive operations

Abstract data structures

4 marks

Enqueue (insert/add): Inserts an incoming element at the rear (tail) of the queue. If the queue is implemented with fixed capacity and is full, this triggers an overflow condition.
Dequeue (remove/serve): Retrieves and removes the element at the front (head) of the queue, shifting the head pointer forward. If the queue is empty, this triggers an underflow condition.

Enqueue \to [Rear] \dots [Front] \to Dequeue

Mentor tip

Queues strictly adhere to First-In, First-Out (FIFO) ordering. Always mention front/rear pointers.

Questions 11–15 · 75 marks

Question 11

15 marks

Mobile computing, operating system roles, and packet protocols

System fundamentals
Resource management
Networks

(a)

1 mark

Mentor tip

A simple login password only protects OS access; full storage encryption protects data even if the drive is removed.

(b)

2 marks

Memory management: The operating system allocates distinct blocks of primary memory (RAM) to the program's code, stack, and heap, tracking freed memory upon process termination.
Process scheduling: The OS scheduler allocates CPU time slices via preemptive multitasking, switching execution context among active threads.

Mentor tip

Identify a specific OS subsystem (memory manager or scheduler) and outline its direct role.

(c)

4 marks

Wired Ethernet advantages: Provides high sustained bandwidth, ultra-low latency, and immunity to radio frequency interference, ideal for high-throughput tasks like large data transfers and gaming.
Wireless Wi-Fi advantages: Provides roaming mobility and physical convenience, allowing users to connect to local area networks and hotspots without physical cabling.
Complementary utility: Supporting both interfaces enables the device to adapt seamlessly across stationary office workstations and mobile environments.

Mentor tip

Structure comparisons by balancing throughput/stability (wired) against convenience/mobility (wireless).

(d)

2 marks

Packet header: Contains routing metadata including source and destination IP addresses, sequence numbering, protocol type, and packet length.
Packet payload: Contains the actual segmented chunk of application data being transported.
Packet trailer: Contains error-detection checksum bits (such as a cyclic redundancy check, CRC) to verify transmission integrity.

Data Packet = [Header ∣ Payload ∣ Trailer]

Mentor tip

Detail the three structural parts of a standard packet: header, payload, and trailer.

(e)

6 marks

Standardization and interoperability: Protocols define universal syntactic and semantic conventions so hardware and software from different manufacturers communicate seamlessly.
Error detection and correction: Transmission protocols (e.g. TCP) verify integrity using checksums and initiate retransmission requests upon packet corruption or packet loss.
Flow control and congestion avoidance: Manages packet transmission rates between fast senders and slower receivers to prevent buffer saturation and network deadlock.

Mentor tip

Provide distinct technical rationales: interoperability, error handling, and data flow synchronization.

Question 12

15 marks

Boolean logic evaluation, security alarm design, and CPU execution cycle

Computer organisation

(a)(i)

1 mark

The Boolean logic operators present in the given expression are `or` and `and`.

Operators: or, and

Mentor tip

Relational operators ( $>, <$ ) compare numerical values; logical operators (`or`, `and`, `not`) combine truth values.

(a)(ii)

1 mark

The numeric constant values explicitly stated in the expression are 6, 3, and 20.

Constants: 6, 3, 20

Mentor tip

Constants are immutable literal values, whereas X and Y are variable identifiers.

(a)(iii)

2 marks

Substitute $X = 6$ and $Y = 6$ into each relational term:
Term 1: $(X > 6) ⟹ (6 > 6)$ is false.
Term 2: $(Y > 3) ⟹ (6 > 3)$ is true.
Term 3: $((X + Y) < 20) ⟹ (12 < 20)$ is true.
Apply operator precedence (`and` binds before `or`):
Evaluate conjunction: $true \land true = true$ .
Evaluate disjunction: $false \lor true = true$ .

(6 > 6) \lor ((6 > 3) \land (12 < 20)) = false \lor (true \land true) = true

Mentor tip

Remember logical precedence: AND has higher precedence than OR.

(b)

4 marks

Siren condition: The siren activates when the door is locked ( $DOOR = 1$ ) AND either movement is detected ( $MOTION = 1$ ) OR the window is open ( $WINDOW = 1$ ).
Logic equation: $SIREN = DOOR \land (MOTION \lor WINDOW)$ .
Evaluate all 8 binary input combinations:
Rows with $DOOR = 0$ : Output is 0 for (0,0,0), (0,0,1), (1,0,0), (1,0,1).
Rows with $DOOR = 1$ :
$(0, 1, 0) ⟹ 1 \land (0 \lor 0) = 0$ .
$(0, 1, 1) ⟹ 1 \land (0 \lor 1) = 1$ .
$(1, 1, 0) ⟹ 1 \land (1 \lor 0) = 1$ .
$(1, 1, 1) ⟹ 1 \land (1 \lor 1) = 1$ .

MOTION 00001111 DOOR 00110011 WINDOW 01010101 SIREN 00010011

Mentor tip

When DOOR = 0, the siren can never sound, zeroing out all four corresponding truth table rows.

(c)(i)

1 mark

The Memory Data Register (MDR), also called the Memory Buffer Register (MBR), temporarily holds the data word fetched from or waiting to be written into primary storage.

Register: Memory Data Register (MDR)

Mentor tip

MAR holds the destination address; MDR holds the data payload itself.

(c)(ii)

1 mark

The Control Unit (CU) contains the instruction decoder circuitry that decodes the opcode bits in the current instruction register.

CPU component: Control Unit (CU)

Mentor tip

The CU coordinates instruction fetching, decoding, and generating timing signals.

(c)(iii)

1 mark

All arithmetic operations (addition, subtraction) and logical comparisons are executed within the Arithmetic Logic Unit (ALU).

CPU component: Arithmetic Logic Unit (ALU)

Mentor tip

Calculations and Boolean evaluations are exclusively handled by the ALU.

(c)(iv)

4 marks

Address Bus (unidirectional): Carries the memory address generated by the MAR from the CPU to the RAM address decoders.
Data Bus (bidirectional): Carries instruction opcodes and operand data between the CPU (MDR) and primary memory.
Control Bus (bidirectional): Transmits control and synchronization signals from the Control Unit (such as Memory Read, Memory Write, clock pulses, and interrupt lines).
Coordinated execution: During fetch, the address bus identifies the memory location, the control bus signals a read operation, and the data bus transports the fetched instruction word into the processor.

Buses = {Address Bus, Data Bus, Control Bus}

Mentor tip

Detail the directionality and specific payload of all three internal system buses.

Question 13

15 marks

Stack interrupts, display physics, geolocation systems, and dedicated operating systems

Abstract data structures
System fundamentals
Networks
Resource management

(a)

2 marks

Context save upon interrupt: When an incoming call triggers a hardware interrupt, the CPU pushes current execution state (Program Counter, general-purpose registers, call stack pointers) onto an interrupt call stack.
Context restoration: Once the phone call concludes, the CPU pops the saved register values off the top of the stack, restoring the Program Counter and resuming app execution seamlessly.

Interrupt: Push CPU context \to Handle Call \to Pop CPU context

Mentor tip

Stacks follow LIFO semantics, making them ideal for handling nested interrupts and subroutine calls.

(b)

2 marks

Sensor-driven physical tracking: Smartphone health apps read internal sensors (e.g. accelerometers, gyroscopes, optical heart rate monitors) to log steps, calories, and cardiac metrics.
User feedback loops: Visualizes progress against targets, providing nudges and reminders that encourage healthier daily activity levels.

Mentor tip

Connect technical hardware sensors to actionable user behavioral feedback.

(c)

2 marks

Higher pixel density: Higher display resolution packs more physical pixels into the same screen area, resulting in a higher pixels-per-inch (PPI) count.
Visual smoothness: Smaller individual pixels reduce visible pixelation and jagged stair-stepping (aliasing), rendering curved vector glyphs and fine image textures with superior visual fidelity.

PPI = \frac{w ^{2} + h ^{2}}{d}

Mentor tip

Explain that pixel size decreases as resolution increases, making individual pixel boundaries indistinguishable to the human eye.

(d)(i)

4 marks

Satellite constellation: The Global Positioning System consists of a network of satellites orbiting Earth, each transmitting precise radio signals with atomic clock timestamps.
Distance calculation: The smartphone receiver compares signal transmission time with arrival time, multiplying the delay by the speed of light to calculate distance to the satellite.
Trilateration: By computing distances from at least four separate satellites, the receiver mathematically solves geometric equations to calculate its exact 3D coordinates (latitude, longitude, and elevation).

d_{i} = c \cdot (t_{received, i} - t_{transmitted, i})

Mentor tip

Mention satellites, time-of-flight distance calculation, and trilateration using at least four satellites.

(d)(ii)

2 marks

Privacy violation and stalker tracking: Sharing precise real-time location can expose a user's residential address, routines, and physical presence to malicious third parties without informed consent.
Targeted profiling: Aggregated telemetry data can be harvested by data brokers for unconsented behavioural profiling.

Mentor tip

Highlight concrete security or personal privacy risks associated with continuous real-time geolocation.

(e)

3 marks

Hardware optimization and energy efficiency: Dedicated mobile operating systems are specifically engineered for battery-powered, ARM-based architectures, aggressively managing CPU power states to maximize battery longevity.
Specialized input and sensor support: Natively integrates touch gesture subsystems, accelerometers, cellular radios, and camera APIs without unnecessary generic desktop driver overhead.

Mentor tip

Contrast low-power mobile optimization against the resource-heavy footprint of generic desktop operating systems.

Question 14

15 marks

Linked list management, sequential traversal, and double linked list architecture

Abstract data structures

(a)

4 marks

Dynamic memory allocation: Linked lists allocate and free nodes on demand from the heap during runtime, avoiding fixed array size limits and preventing wasted pre-allocated memory.
Constant-time node insertion and deletion: Inserting or removing a student name simply requires updating pointer references ( $O (1)$ linking), eliminating the expensive $O (n)$ element-shifting required in static arrays.

Array insert: O (n) shifting vs Linked list insert: O (1) pointer updates

Mentor tip

Identify both dynamic resizing (no waste) and pointer-based insertion/deletion (no shifting).

(b)

5 marks

Pointer initialization: Initialize a traversal pointer `CURRENT` to point to `HEAD`.
Iterative traversal: While `CURRENT != null` and `CURRENT.data <= TARGET`:
Check match: If `CURRENT.data == TARGET`, the student name is located; return success or node reference.
Advance pointer: Otherwise, advance to the next node by setting `CURRENT = CURRENT.next`.
Alphabetical early termination: If `CURRENT == null` or `CURRENT.data > TARGET`, terminate traversal; since the list is sorted alphabetically, the target name is confirmed absent.

CURRENT = HEAD loop while CURRENT \neq = null and CURRENT.data < TARGET CURRENT = CURRENT.next end loop return (CURRENT \neq = null and CURRENT.data == TARGET)

Mentor tip

Because the list is maintained in alphabetical order, stop searching as soon as the node data exceeds the target string.

(c)(i)

4 marks

Bidirectional node structure: Each node in a double linked list contains three fields: a data payload, a `next` pointer referencing the successor node, and a `prev` pointer referencing the predecessor node.
Boundary terminal pointers: The `prev` pointer of the head node points to `null`, and the `next` pointer of the tail node points to `null`.

Node: [prev ∣ data ∣ next] ⟺ [prev ∣ data ∣ next]

Mentor tip

Explicitly mention all three components of a DLL node: data value, forward next pointer, and backward prev pointer.

(c)(ii)

2 marks

Web browser navigation: Maintaining web history to allow instantaneous forward and backward page transitions.
Media player playlists / Undo-redo stacks: Enabling backward/forward navigation across audio tracks or editing command histories.

Mentor tip

Any workflow requiring seamless two-way traversal is a prime application for a double linked list.

Question 15

15 marks

2D matrix frequency tabulation, absence detection, and mode extraction

Computational thinking, problem-solving and programming

(a)

4 marks

Iterate through all rows of the 2D array MAT from index 0 to 6.
Within each row, iterate through all columns from index 0 to 6.
Read integer value $VAL = MAT [R] [C]$ .
Increment the corresponding frequency tally in the 1D count array: $COUNT [VAL] = COUNT [VAL] + 1$ .

loop R from 0 to 6 loop C from 0 to 6 VAL = MAT [R] [C] COUNT [VAL] = COUNT [VAL] + 1 end loop end loop

Mentor tip

Use nested loops over row and column dimensions, indexing directly into COUNT using the matrix values.

(b)

5 marks

Initialize a boolean tracking flag `ALLPRESENT = true`.
Loop through all integer digits $I$ from 0 to 9 in the COUNT array.
If $COUNT [I] == 0$ , output a message indicating digit $I$ is missing, and toggle `ALLPRESENT = false`.
After inspecting all 10 digits, if `ALLPRESENT` remains true, output that every number from 0 to 9 is present.

ALLPRESENT = true loop I from 0 to 9 if COUNT [I] == 0 then output ("Number ", I, " is not present") ALLPRESENT = false end if end loop if ALLPRESENT then output ("All numbers from 0 to 9 are present") end if

Mentor tip

The boolean flag ensures you only print the 'all numbers present' confirmation once at the end.

(c)

6 marks

Pass 1: Find the maximum occurrence frequency by scanning $COUNT [I]$ from $I = 0$ to 9.
Track the highest frequency in variable `MAXFREQ`.
Pass 2: Scan $COUNT [I]$ again from 0 to 9; whenever $COUNT [I] == MAXFREQ$ , output digit $I$ as a mode.
This two-pass design naturally accommodates bimodal or multimodal datasets where multiple numbers share the peak frequency.

MAXFREQ = COUNT [0] loop I from 1 to 9 if COUNT [I] > MAXFREQ then MAXFREQ = COUNT [I] end if end loop output ("Modes with frequency: ", MAXFREQ) loop I from 0 to 9 if COUNT [I] == MAXFREQ then output (I) end if end loop

Mentor tip

Because multiple numbers can share the top count (multimodal), find the maximum frequency first before printing all matching digits.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Abstract data structures carries the most marks on this paper.

Abstract data structures23 marks · 23%
Q9, Q10, Q13, Q14
Computational thinking, problem-solving and programming21 marks · 21%
Q6, Q7, Q15
Networks19 marks · 19%
Q5, Q11, Q13
System fundamentals15 marks · 15%
Q1, Q2, Q3, Q4, Q11, Q13
Computer organisation15 marks · 15%
Q12
Resource management5 marks · 5%
Q11, Q13
Control systems2 marks · 2%
Q8

Mentor insights

Traps to watch for

01
Modulus parameter order in Euclidean recursion · Q7
Ensure rec(B, A mod B) passes B as the first parameter and remainder as the second. Swapping parameters reverses the divisor and dividend.
02
Operator precedence in compound Boolean expressions · Q12
Logical AND has higher precedence than OR. In (X > 6) or (Y > 3) and ((X + Y) < 20), the AND conjunction is evaluated before the OR disjunction.
03
Double linked list bidirectional pointer maintenance · Q14
When inserting or deleting nodes in a double linked list, both forward next and backward prev pointers must be re-linked. Updating only one direction breaks reverse traversal.
04
Multimodal frequency scanning · Q15
Never assume a dataset has only a single mode. Scan the array first to find the peak frequency MAXFREQ, then execute a second pass to output all values whose frequency matches MAXFREQ.

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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