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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: May 2019
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2019 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–12 · 25 marks

Question 1

1 mark

Define the term peripheral

Computer organisation

(a)

1 mark

Define peripheral in computing hardware terms:
A hardware device connected externally (or attached peripherally) to a computer system to provide auxiliary input, output, or storage functionality (e.g., keyboard, printer, external drive).

Mentor tip

State clearly that it is an external/attached hardware component used for input, output, or auxiliary storage.

Question 2

2 marks

Features of a graphical user interface

System fundamentals

(a)

2 marks

Identify two characteristic features of a GUI:
1) Visual interaction elements such as windows, icons, menus, and pointers (WIMP components).
2) Visual feedback elements including dialog boxes, buttons, toolbars, and drag-and-drop support.

Mentor tip

Citing any two components of WIMP (windows, icons, menus, pointing devices) earns both marks directly.

Question 3

2 marks

Purpose of the memory address register (MAR)

Computer organisation

(a)

2 marks

Describe the role and operation of the MAR:
1) The MAR holds the physical memory address (in RAM) currently being read from or written to.
2) It communicates this address across the address bus to specify the target memory location for data retrieval into the MDR or storage from the MDR.

Mentor tip

Specify that it stores memory addresses, NOT data words.

Question 4

2 marks

Single processor limitations in 3D graphics rendering

Computer organisation

(a)

2 marks

Explain processing demands and architecture limitations:
1) 3D graphics rendering involves millions of intensive geometric calculations (floating-point matrix transforms, vertex shaders, polygon rasterization) and lighting calculations per frame.
2) A single processor executes instructions sequentially and must multitask with the operating system, creating a processing bottleneck and resulting in low frame rates or latency without specialized parallel hardware (GPUs/coprocessors).

Mentor tip

Emphasize high volume of simultaneous floating-point calculations vs sequential single-core processing.

Question 5

4 marks

Advantages and disadvantages of virtual memory

Computer organisation

(a)

2 marks

Outline one advantage of virtual memory:
Allows programs larger than the physical capacity of RAM to run, or enables greater multitasking concurrency by allocating virtual address spaces and paging inactive pages to secondary storage.

Mentor tip

Mention running larger software or accommodating more concurrent programs.

(b)

2 marks

Outline one disadvantage of virtual memory:
Secondary storage (HDD/SSD) is substantially slower to access than physical RAM, causing noticeable latency; excessive page swapping can lead to thrashing where the system spends more time swapping pages than executing instructions.

Mentor tip

Identify speed disparity between secondary storage and RAM, or define thrashing.

Question 6

1 mark

Hexadecimal conversion

Computer organisation

(a)

1 mark

Split the 8-bit binary number into two 4-bit nibbles:
Upper nibble: 1101 = 8 + 4 + 0 + 1 = 13 in decimal = D in hex.
Lower nibble: 1111 = 8 + 4 + 2 + 1 = 15 in decimal = F in hex.
Concatenate nibbles: DF (or 0xDF).

110 1_{2} = 1 3_{10} = D_{16} 111 1_{2} = 1 5_{10} = F_{16} 1101111 1_{2} = DF_{16}

Mentor tip

Always group into 4-bit nibbles from right to left.

Question 7

3 marks

Logic circuit truth table

Computer organisation

(a)

3 marks

Inspect the logic circuit connections:
Intermediate gate: S is generated by B AND C: S = B AND C.
Final output gate: Z is generated by A OR S: Z = A OR (B AND C).
Compute row by row for all 8 combinations of A, B, C:
Row 1: A=0, B=0, C=0 -> S=0 -> Z=0
Row 2: A=0, B=0, C=1 -> S=0 -> Z=0
Row 3: A=0, B=1, C=0 -> S=0 -> Z=0
Row 4: A=0, B=1, C=1 -> S=1 -> Z=1
Row 5: A=1, B=0, C=0 -> S=0 -> Z=1
Row 6: A=1, B=0, C=1 -> S=0 -> Z=1
Row 7: A=1, B=1, C=0 -> S=0 -> Z=1
Row 8: A=1, B=1, C=1 -> S=1 -> Z=1

S = B \land C Z = A \lor S = A \lor (B \land C) A 00001111 B 00110011 C 01010101 S 00010001 Z 00011111

Mentor tip

Notice that whenever A = 1, the OR gate guarantees Z = 1 regardless of B and C.

Question 8

2 marks

Stakeholder information gathering

System fundamentals

(a)

2 marks

Select one standard requirements elicitation method:
Interviews / Questionnaires / Observations / Document analysis.
Outline how it gathers information: Conducting structured interviews with stakeholders allows direct discussion to explore detailed user expectations, clarify ambiguous requirements, and gather qualitatively rich operational insights.

Mentor tip

Name the technique and explicitly describe how it captures requirement data.

Question 9

2 marks

Characteristics of a personal area network (PAN)

Networks

(a)

2 marks

Identify two defining properties of a PAN:
1) Very short operational range (typically within 10 metres, centered around a single individual).
2) Interconnects personal wearable and mobile devices (smartphones, headphones, smartwatches) typically using short-range wireless protocols like Bluetooth or Zigbee.

Mentor tip

Mention range (< 10m) and typical protocols/devices (Bluetooth, wearables).

Question 10

2 marks

Purpose of network protocols

Networks

(a)

2 marks

Explain why communication requires common rules:
1) Protocols establish an agreed set of formal rules and data formatting standards.
2) This ensures heterogeneous devices with different hardware, operating systems, and architectures can successfully exchange data, handle error detection, and synchronize communication without misinterpreting transmitted signals.

Mentor tip

Focus on standardization enabling interoperability and error-checked data transfer.

Question 11

2 marks

Binary tree postorder traversal

Abstract data structures

(a)

2 marks

Recall postorder traversal order: Left subtree -> Right subtree -> Root node.
Examine tree nodes:
Root is 79.
Left child of 79 is 72, with left child 68 and right child 76 (which has left child 75 and right child 70... traversal of left subtree gives: 68, 70, 75, 76, 72).
Right child of 79 is 83.
Traversing left subtree, then right subtree, then root yields:
68, 70, 75, 76, 72, 83, 79.

Postorder: Left \to Right \to Root Sequence: 68, 70, 75, 76, 72, 83, 79

Mentor tip

Root (79) must strictly appear as the final element in postorder output.

Question 12

2 marks

Binary tree suitability for an address book

Abstract data structures

(a)

2 marks

Outline search and sorting efficiencies:
1) Fast searching: In a balanced binary search tree, searching for a contact takes logarithmic time O(log N) as half the remaining entries are eliminated at each comparison step.
2) Natural sorted order: An inorder traversal produces all contact names in alphabetical order without requiring separate sorting passes, alongside efficient dynamic insertion and deletion of contacts.

Mentor tip

Highlight O(log n) search efficiency or in-order alphabetical traversal.

Questions 13–17 · 75 marks

Question 13

15 marks

Data security and system testing

System fundamentals

(a)

4 marks

Describe two distinct causes of data loss:
1) Human error: Accidental deletion, file overwriting, or formatting of drives by staff.
2) Malicious cyber activity / malware: Unauthorized infiltration, ransomware encrypting files, or deliberate tampering/deletion by intruders.
Alternative valid points include hardware failure (head crash, disk corruption) or environmental disasters (fire/flood/power outages).

Mentor tip

Give two clearly distinct causes and explain the mechanism of loss for each.

(b)

4 marks

Outline two backup strategies to limit data loss:
1) Offsite / cloud storage backups: Regularly transferring copies of data to a geographically separate facility or cloud provider, ensuring data can be restored if the primary site experiences physical damage or local system failure.
2) Incremental backup schedule: Backing up only blocks/files modified since the last backup, saving storage space and network bandwidth while enabling recovery up to the latest checkpoint.

Mentor tip

Pair each strategy with how it mitigates data loss.

(c)

3 marks

Explain the role and value of beta testing:
1) Deployment to external real-world users outside the organization allows testing under authentic, diverse operating environments and hardware setups.
2) Real users interact with the interface in unexpected ways, exposing usability flaws, edge-case bugs, and design oversights that internal developers did not anticipate.
3) Provides actionable user feedback so developers can refine, debug, and optimize the user interface before final general release.

Mentor tip

Highlight testing in real operational contexts by target users.

(d)

2 marks

Outline consequences of omitting end-users from design and testing:
The resulting system may fail to align with real user workflows, leading to poor usability, missing vital features, a steep learning curve, user dissatisfaction, or high rejection rates requiring costly redesigns.

Mentor tip

Focus on usability misalignment, rejected adoption, and costly rework.

(e)

2 marks

Identify two accessibility features for the user interface:
1) Text-to-speech / screen reader compatibility and adjustable font sizing / high-contrast color modes (for visually impaired users).
2) Voice control / speech recognition and keyboard-only navigation options (for users with motor impairments).

Mentor tip

Name specific features such as high contrast modes, screen readers, or voice input.

Question 14

15 marks

Hardware architecture and networking

Computer organisation
Networks

(a)

2 marks

Identify two core characteristics of RAM:
1) Volatility: Contents are lost when electrical power is turned off.
2) Fast read/write access: Allows direct random access to any memory cell at high speeds compared to secondary storage.

Mentor tip

State volatile and high-speed direct access.

(b)

1 mark

State the purpose of persistent secondary storage:
To provide non-volatile, long-term storage for operating system files, application programs, and user data so they are retained when the system is powered down.

Mentor tip

Mention non-volatile retention of data and programs.

(c)

2 marks

Identify two types of general application software suitable for training:
1) Word processor (e.g., text document editor).
2) Web browser (or spreadsheet / presentation software / email client).

Mentor tip

Name software categories (e.g., word processor, spreadsheet, web browser).

(d)

4 marks

Evaluate BYOD (bring your own device) policy in a training environment:
(i) Advantage to company (2 marks): Reduces hardware capital and maintenance expenditure (company does not need to purchase or upgrade 15 laptops), and trainees work comfortably on familiar machines and personal configurations.
(ii) Disadvantage to company (2 marks): Introduces network security risks (unmanaged personal devices may harbor malware or unauthorized software) and compatibility issues where training materials or software fail on varied OS versions.

Mentor tip

Split answer clearly between cost/convenience benefits and security/compatibility risks.

(e)

2 marks

Describe one wireless network security method:
WPA2/WPA3 Encryption: Transmitted data packets are cryptographically scrambled using a pre-shared key or enterprise authentication server; intercepted radio transmissions cannot be decrypted or read without the valid security key.

Mentor tip

Name the security method (encryption, MAC filtering, firewall) and explain its protective mechanism.

(f)

4 marks

Explain why wireless data transmission speed fluctuates in a training room:
1) Physical distance and physical obstructions: Wireless signal strength attenuates over distance and is degraded by walls, furniture, and metallic partitions, reducing modulation rates.
2) Radio frequency interference: Interference from adjacent Wi-Fi channels, Bluetooth peripherals, microwave equipment, or electrical conduits degrades signal-to-noise ratio.
3) Shared bandwidth and network congestion: As multiple devices transmit simultaneously, available radio spectrum is shared, collision rates rise, and CSMA/CA back-off delays lower throughput.

Mentor tip

Explain at least two distinct causes with technical mechanisms (e.g., interference, shared channel bandwidth, physical obstacles).

Question 15

15 marks

Algorithms and sorting

Computational thinking, problem-solving and programming
Abstract data structures

(a)

5 marks

Trace the outer and inner loops of the selection sort algorithm on array [20, 6, 38, 50, 40]:
Initial state: VALUES = [20, 6, 38, 50, 40], LIMIT = 4.
Pass 0 (COUNTER1 = 0):
- MINIMUM initialized to 0 (val 20).
- COUNTER2=1: VALUES[1]=6 < 20 -> MINIMUM=1.
- COUNTER2=2: VALUES[2]=38 > 6.
- COUNTER2=3: VALUES[3]=50 > 6.
- COUNTER2=4: VALUES[4]=40 > 6.
- MINIMUM(1) != COUNTER1(0) -> Swap VALUES[0] and VALUES[1]: TEMPORARY=6, VALUES becomes [6, 20, 38, 50, 40].
Pass 1 (COUNTER1 = 1):
- MINIMUM initialized to 1 (val 20).
- COUNTER2=2,3,4: all elements > 20 -> MINIMUM remains 1.
- No swap. VALUES remains [6, 20, 38, 50, 40].
Pass 2 (COUNTER1 = 2):
- MINIMUM initialized to 2 (val 38).
- COUNTER2=3 (val 50), COUNTER2=4 (val 40): all > 38 -> MINIMUM remains 2.
- No swap. VALUES remains [6, 20, 38, 50, 40].
Pass 3 (COUNTER1 = 3):
- MINIMUM initialized to 3 (val 50).
- COUNTER2=4: VALUES[4]=40 < 50 -> MINIMUM=4.
- MINIMUM(4) != COUNTER1(3) -> Swap VALUES[3] and VALUES[4]: TEMPORARY=40, VALUES becomes [6, 20, 38, 40, 50].
Final array state: [6, 20, 38, 40, 50].

Mentor tip

Carefully track MINIMUM index changes and record TEMPORARY whenever swaps occur.

(b)

4 marks

(i) Construct pseudocode for the flowchart algorithm (3 marks):

LIMIT = 4

FLAG = TRUE

loop while FLAG == TRUE

FLAG = FALSE

loop COUNTER from 0 to LIMIT - 1

if VALUES[COUNTER] > VALUES[COUNTER + 1] then

TEMPORARY = VALUES[COUNTER]

VALUES[COUNTER] = VALUES[COUNTER + 1]

VALUES[COUNTER + 1] = TEMPORARY

FLAG = TRUE

end if

end loop

(ii) State the type of sort (1 mark): Bubble sort.

Mentor tip

Use a while loop checking FLAG combined with an inner loop traversing 0 to LIMIT - 1.

(c)

2 marks

Construct an algorithm fragment to output all elements in array VALUES[]:
loop I from 0 to LIMIT
output VALUES[I]
end loop

Mentor tip

Verify loop bounds match the 5 elements (0 to 4).

(d)

3 marks

Contrast dynamic and static data structures:
1) Memory sizing: Static data structures (such as fixed arrays) have a predetermined, immutable size allocated at compile time. Dynamic data structures (such as linked lists, binary trees) can grow and shrink dynamically during program execution.
2) Memory allocation: Static structures use contiguous memory blocks and can waste space if partially filled or overflow if full; dynamic structures allocate memory from the heap as needed using pointers, though each node incurs pointer memory overhead.

Mentor tip

Focus on size mutability at runtime and pointer memory management.

(e)

1 mark

Identify one alternative data structure for storing class marks:
Linked list (or dynamic array / collection / stack / binary search tree).

Mentor tip

Linked list or Collection is an ideal answer.

Question 16

13 marks

Operating systems and control systems

Resource management
Control systems

(a)

2 marks

Identify two functions of an operating system:
1) Memory management (allocation, paging, virtual memory).
2) Processor scheduling / process management (or peripheral management, file system management, security/access control).

Mentor tip

Name core OS duties: processor management, memory management, peripheral management.

(b)

3 marks

Explain one benefit of a dedicated operating system for a mobile phone:
1) Resource optimization: A dedicated OS includes only necessary software drivers and modules tailored specifically to the phone's hardware architecture.
2) It strips out unused legacy code, resulting in a significantly smaller memory and storage footprint, faster boot and execution speed, and improved energy/battery efficiency compared to a bloated generic OS.

Mentor tip

Link customized hardware drivers to performance gains, lower storage footprint, and power efficiency.

(c)

5 marks

Explain the closed-loop control mechanism for constant temperature:
1) User sets the target reference temperature via the interface.
2) Temperature sensors continuously measure ambient room temperature.
3) Analog sensor readings are converted to digital signals via an Analogue-to-Digital Converter (ADC) and transmitted to the microprocessor.
4) The microprocessor compares the sampled temperature against the user's setpoint.
5) If the temperature deviates beyond a tolerance threshold, the microprocessor sends control signals (via a DAC if needed) to actuators (compressor, fan, heat pump) to adjust cooling/heating output until equilibrium is restored.

Mentor tip

Structure the answer chronologically: Sensing -> ADC -> Microprocessor comparison -> Actuator adjustment.

(d)

3 marks

Contrast distributed vs centralized air conditioning control:
1) Granularity & Efficiency: A distributed system uses local microprocessors and temperature zones, allowing individual rooms to be cooled independently based on local occupancy, reducing energy waste compared to a centralized system that cools the entire house uniformly.
2) Reliability & Redundancy: If a local controller fails in a distributed system, only that single zone is affected; in a centralized system, failure of the central processor disables cooling for the whole house.

Mentor tip

Contrast on two distinct axes: zoned temperature control/energy efficiency, and single point of failure.

Question 17

17 marks

Two-dimensional arrays and linked lists

Computational thinking, problem-solving and programming
Abstract data structures

(a)

5 marks

Identify array layout: CUSTOMERS has 512 rows, columns: 0=LASTNAME, 1=FIRSTNAME, 2=ADDRESS1, 3=ADDRESS2, 4=ADDRESS3, 5=CITY, 6=POSTCODE.
Construct pseudocode to count and display customers in Cardiff:
COUNT = 0
loop ROW from 0 to 511
if CUSTOMERS[ROW][5] == "Cardiff" then
COUNT = COUNT + 1
end if
end loop
output "Number of customers in Cardiff: ", COUNT

Mentor tip

Initialize COUNT = 0 before the loop and output the result after the loop completes.

(b)

5 marks

Construct pseudocode to print mailing labels for customers named Jones in Cardiff:
loop ROW from 0 to 511
if CUSTOMERS[ROW][0] == "Jones" and CUSTOMERS[ROW][5] == "Cardiff" then
output CUSTOMERS[ROW][1] + " " + CUSTOMERS[ROW][0]
output CUSTOMERS[ROW][2]
output CUSTOMERS[ROW][3]
output CUSTOMERS[ROW][4]
output CUSTOMERS[ROW][5]
output CUSTOMERS[ROW][6]
end if
end loop

Mentor tip

Check both LASTNAME (col 0) and CITY (col 5) in the conditional statement.

(c)

5 marks

List steps to insert 'Jones' into alphabetical singly linked list [Bale -> Cousens -> Davies -> Pugh -> Williams]:
1) Create a new node containing the data value 'Jones' and a null pointer.
2) Traverse the list from head pointer (Bale), comparing 'Jones' against each node's data.
3) Locate the insertion point between 'Davies' and 'Pugh' (since Davies < Jones < Pugh).
4) Set the next pointer of the new 'Jones' node to point to 'Pugh'.
5) Update the next pointer of 'Davies' to point to the new 'Jones' node.

Mentor tip

Always specify pointing the new node to the next node BEFORE updating the previous node's pointer.

(d)

2 marks

Identify an application where a circular linked list is preferred:
Round-robin CPU process scheduling (or a media playlist / multiplayer turn rotation): Once the last process or song is served, the list seamlessly cycles back to the first node without requiring null checks or pointer resets to the head.

Mentor tip

Name round-robin task scheduling or continuous playlists.

Our analysis

Where the marks are

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

Computational thinking, problem-solving and programming21 marks · 21%
Q15, Q17
System fundamentals19 marks · 19%
Q2, Q8, Q13
Computer organisation18 marks · 18%
Q1, Q3, Q4, Q5, Q6, Q7, Q14
Abstract data structures15 marks · 15%
Q11, Q12, Q15, Q17
Networks14 marks · 14%
Q9, Q10, Q14
Control systems8 marks · 8%
Q16
Resource management5 marks · 5%
Q16

Mentor insights

Traps to watch for

01
Order of logic operations for compound output Z · Q7
Intermediate signal S is generated by B AND C. Final output Z is A OR S. Whenever A is 1, output Z is 1 regardless of B and C.
02
Selection sort vs bubble sort inner loop execution · Q15
Selection sort scans the unsorted subarray to find the minimum index and swaps at most once per outer iteration. Bubble sort swaps adjacent elements immediately if out of order.
03
Singly linked list pointer update order on insertion · Q17
When inserting Jones between Davies and Pugh, you must set Jones.next = Pugh BEFORE setting Davies.next = Jones. Overwriting Davies.next first causes the remainder of the linked list to be lost in memory.

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: May 2019
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 17

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2019 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–12 · 25 marks

Question 1

1 mark

Define the term peripheral

Computer organisation

(a)

1 mark

Define peripheral in computing hardware terms:
A hardware device connected externally (or attached peripherally) to a computer system to provide auxiliary input, output, or storage functionality (e.g., keyboard, printer, external drive).

Mentor tip

State clearly that it is an external/attached hardware component used for input, output, or auxiliary storage.

Question 2

2 marks

Features of a graphical user interface

System fundamentals

(a)

2 marks

Identify two characteristic features of a GUI:
1) Visual interaction elements such as windows, icons, menus, and pointers (WIMP components).
2) Visual feedback elements including dialog boxes, buttons, toolbars, and drag-and-drop support.

Mentor tip

Citing any two components of WIMP (windows, icons, menus, pointing devices) earns both marks directly.

Question 3

2 marks

Purpose of the memory address register (MAR)

Computer organisation

(a)

2 marks

Describe the role and operation of the MAR:
1) The MAR holds the physical memory address (in RAM) currently being read from or written to.
2) It communicates this address across the address bus to specify the target memory location for data retrieval into the MDR or storage from the MDR.

Mentor tip

Specify that it stores memory addresses, NOT data words.

Question 4

2 marks

Single processor limitations in 3D graphics rendering

Computer organisation

(a)

2 marks

Explain processing demands and architecture limitations:
1) 3D graphics rendering involves millions of intensive geometric calculations (floating-point matrix transforms, vertex shaders, polygon rasterization) and lighting calculations per frame.
2) A single processor executes instructions sequentially and must multitask with the operating system, creating a processing bottleneck and resulting in low frame rates or latency without specialized parallel hardware (GPUs/coprocessors).

Mentor tip

Emphasize high volume of simultaneous floating-point calculations vs sequential single-core processing.

Question 5

4 marks

Advantages and disadvantages of virtual memory

Computer organisation

(a)

2 marks

Outline one advantage of virtual memory:
Allows programs larger than the physical capacity of RAM to run, or enables greater multitasking concurrency by allocating virtual address spaces and paging inactive pages to secondary storage.

Mentor tip

Mention running larger software or accommodating more concurrent programs.

(b)

2 marks

Outline one disadvantage of virtual memory:
Secondary storage (HDD/SSD) is substantially slower to access than physical RAM, causing noticeable latency; excessive page swapping can lead to thrashing where the system spends more time swapping pages than executing instructions.

Mentor tip

Identify speed disparity between secondary storage and RAM, or define thrashing.

Question 6

1 mark

Hexadecimal conversion

Computer organisation

(a)

1 mark

Split the 8-bit binary number into two 4-bit nibbles:
Upper nibble: 1101 = 8 + 4 + 0 + 1 = 13 in decimal = D in hex.
Lower nibble: 1111 = 8 + 4 + 2 + 1 = 15 in decimal = F in hex.
Concatenate nibbles: DF (or 0xDF).

110 1_{2} = 1 3_{10} = D_{16} 111 1_{2} = 1 5_{10} = F_{16} 1101111 1_{2} = DF_{16}

Mentor tip

Always group into 4-bit nibbles from right to left.

Question 7

3 marks

Logic circuit truth table

Computer organisation

(a)

3 marks

Inspect the logic circuit connections:
Intermediate gate: S is generated by B AND C: S = B AND C.
Final output gate: Z is generated by A OR S: Z = A OR (B AND C).
Compute row by row for all 8 combinations of A, B, C:
Row 1: A=0, B=0, C=0 -> S=0 -> Z=0
Row 2: A=0, B=0, C=1 -> S=0 -> Z=0
Row 3: A=0, B=1, C=0 -> S=0 -> Z=0
Row 4: A=0, B=1, C=1 -> S=1 -> Z=1
Row 5: A=1, B=0, C=0 -> S=0 -> Z=1
Row 6: A=1, B=0, C=1 -> S=0 -> Z=1
Row 7: A=1, B=1, C=0 -> S=0 -> Z=1
Row 8: A=1, B=1, C=1 -> S=1 -> Z=1

S = B \land C Z = A \lor S = A \lor (B \land C) A 00001111 B 00110011 C 01010101 S 00010001 Z 00011111

Mentor tip

Notice that whenever A = 1, the OR gate guarantees Z = 1 regardless of B and C.

Question 8

2 marks

Stakeholder information gathering

System fundamentals

(a)

2 marks

Select one standard requirements elicitation method:
Interviews / Questionnaires / Observations / Document analysis.
Outline how it gathers information: Conducting structured interviews with stakeholders allows direct discussion to explore detailed user expectations, clarify ambiguous requirements, and gather qualitatively rich operational insights.

Mentor tip

Name the technique and explicitly describe how it captures requirement data.

Question 9

2 marks

Characteristics of a personal area network (PAN)

Networks

(a)

2 marks

Identify two defining properties of a PAN:
1) Very short operational range (typically within 10 metres, centered around a single individual).
2) Interconnects personal wearable and mobile devices (smartphones, headphones, smartwatches) typically using short-range wireless protocols like Bluetooth or Zigbee.

Mentor tip

Mention range (< 10m) and typical protocols/devices (Bluetooth, wearables).

Question 10

2 marks

Purpose of network protocols

Networks

(a)

2 marks

Explain why communication requires common rules:
1) Protocols establish an agreed set of formal rules and data formatting standards.
2) This ensures heterogeneous devices with different hardware, operating systems, and architectures can successfully exchange data, handle error detection, and synchronize communication without misinterpreting transmitted signals.

Mentor tip

Focus on standardization enabling interoperability and error-checked data transfer.

Question 11

2 marks

Binary tree postorder traversal

Abstract data structures

(a)

2 marks

Recall postorder traversal order: Left subtree -> Right subtree -> Root node.
Examine tree nodes:
Root is 79.
Left child of 79 is 72, with left child 68 and right child 76 (which has left child 75 and right child 70... traversal of left subtree gives: 68, 70, 75, 76, 72).
Right child of 79 is 83.
Traversing left subtree, then right subtree, then root yields:
68, 70, 75, 76, 72, 83, 79.

Postorder: Left \to Right \to Root Sequence: 68, 70, 75, 76, 72, 83, 79

Mentor tip

Root (79) must strictly appear as the final element in postorder output.

Question 12

2 marks

Binary tree suitability for an address book

Abstract data structures

(a)

2 marks

Outline search and sorting efficiencies:
1) Fast searching: In a balanced binary search tree, searching for a contact takes logarithmic time O(log N) as half the remaining entries are eliminated at each comparison step.
2) Natural sorted order: An inorder traversal produces all contact names in alphabetical order without requiring separate sorting passes, alongside efficient dynamic insertion and deletion of contacts.

Mentor tip

Highlight O(log n) search efficiency or in-order alphabetical traversal.

Questions 13–17 · 75 marks

Question 13

15 marks

Data security and system testing

System fundamentals

(a)

4 marks

Describe two distinct causes of data loss:
1) Human error: Accidental deletion, file overwriting, or formatting of drives by staff.
2) Malicious cyber activity / malware: Unauthorized infiltration, ransomware encrypting files, or deliberate tampering/deletion by intruders.
Alternative valid points include hardware failure (head crash, disk corruption) or environmental disasters (fire/flood/power outages).

Mentor tip

Give two clearly distinct causes and explain the mechanism of loss for each.

(b)

4 marks

Outline two backup strategies to limit data loss:
1) Offsite / cloud storage backups: Regularly transferring copies of data to a geographically separate facility or cloud provider, ensuring data can be restored if the primary site experiences physical damage or local system failure.
2) Incremental backup schedule: Backing up only blocks/files modified since the last backup, saving storage space and network bandwidth while enabling recovery up to the latest checkpoint.

Mentor tip

Pair each strategy with how it mitigates data loss.

(c)

3 marks

Explain the role and value of beta testing:
1) Deployment to external real-world users outside the organization allows testing under authentic, diverse operating environments and hardware setups.
2) Real users interact with the interface in unexpected ways, exposing usability flaws, edge-case bugs, and design oversights that internal developers did not anticipate.
3) Provides actionable user feedback so developers can refine, debug, and optimize the user interface before final general release.

Mentor tip

Highlight testing in real operational contexts by target users.

(d)

2 marks

Outline consequences of omitting end-users from design and testing:
The resulting system may fail to align with real user workflows, leading to poor usability, missing vital features, a steep learning curve, user dissatisfaction, or high rejection rates requiring costly redesigns.

Mentor tip

Focus on usability misalignment, rejected adoption, and costly rework.

(e)

2 marks

Identify two accessibility features for the user interface:
1) Text-to-speech / screen reader compatibility and adjustable font sizing / high-contrast color modes (for visually impaired users).
2) Voice control / speech recognition and keyboard-only navigation options (for users with motor impairments).

Mentor tip

Name specific features such as high contrast modes, screen readers, or voice input.

Question 14

15 marks

Hardware architecture and networking

Computer organisation
Networks

(a)

2 marks

Identify two core characteristics of RAM:
1) Volatility: Contents are lost when electrical power is turned off.
2) Fast read/write access: Allows direct random access to any memory cell at high speeds compared to secondary storage.

Mentor tip

State volatile and high-speed direct access.

(b)

1 mark

State the purpose of persistent secondary storage:
To provide non-volatile, long-term storage for operating system files, application programs, and user data so they are retained when the system is powered down.

Mentor tip

Mention non-volatile retention of data and programs.

(c)

2 marks

Identify two types of general application software suitable for training:
1) Word processor (e.g., text document editor).
2) Web browser (or spreadsheet / presentation software / email client).

Mentor tip

Name software categories (e.g., word processor, spreadsheet, web browser).

(d)

4 marks

Evaluate BYOD (bring your own device) policy in a training environment:
(i) Advantage to company (2 marks): Reduces hardware capital and maintenance expenditure (company does not need to purchase or upgrade 15 laptops), and trainees work comfortably on familiar machines and personal configurations.
(ii) Disadvantage to company (2 marks): Introduces network security risks (unmanaged personal devices may harbor malware or unauthorized software) and compatibility issues where training materials or software fail on varied OS versions.

Mentor tip

Split answer clearly between cost/convenience benefits and security/compatibility risks.

(e)

2 marks

Describe one wireless network security method:
WPA2/WPA3 Encryption: Transmitted data packets are cryptographically scrambled using a pre-shared key or enterprise authentication server; intercepted radio transmissions cannot be decrypted or read without the valid security key.

Mentor tip

Name the security method (encryption, MAC filtering, firewall) and explain its protective mechanism.

(f)

4 marks

Explain why wireless data transmission speed fluctuates in a training room:
1) Physical distance and physical obstructions: Wireless signal strength attenuates over distance and is degraded by walls, furniture, and metallic partitions, reducing modulation rates.
2) Radio frequency interference: Interference from adjacent Wi-Fi channels, Bluetooth peripherals, microwave equipment, or electrical conduits degrades signal-to-noise ratio.
3) Shared bandwidth and network congestion: As multiple devices transmit simultaneously, available radio spectrum is shared, collision rates rise, and CSMA/CA back-off delays lower throughput.

Mentor tip

Explain at least two distinct causes with technical mechanisms (e.g., interference, shared channel bandwidth, physical obstacles).

Question 15

15 marks

Algorithms and sorting

Computational thinking, problem-solving and programming
Abstract data structures

(a)

5 marks

Trace the outer and inner loops of the selection sort algorithm on array [20, 6, 38, 50, 40]:
Initial state: VALUES = [20, 6, 38, 50, 40], LIMIT = 4.
Pass 0 (COUNTER1 = 0):
- MINIMUM initialized to 0 (val 20).
- COUNTER2=1: VALUES[1]=6 < 20 -> MINIMUM=1.
- COUNTER2=2: VALUES[2]=38 > 6.
- COUNTER2=3: VALUES[3]=50 > 6.
- COUNTER2=4: VALUES[4]=40 > 6.
- MINIMUM(1) != COUNTER1(0) -> Swap VALUES[0] and VALUES[1]: TEMPORARY=6, VALUES becomes [6, 20, 38, 50, 40].
Pass 1 (COUNTER1 = 1):
- MINIMUM initialized to 1 (val 20).
- COUNTER2=2,3,4: all elements > 20 -> MINIMUM remains 1.
- No swap. VALUES remains [6, 20, 38, 50, 40].
Pass 2 (COUNTER1 = 2):
- MINIMUM initialized to 2 (val 38).
- COUNTER2=3 (val 50), COUNTER2=4 (val 40): all > 38 -> MINIMUM remains 2.
- No swap. VALUES remains [6, 20, 38, 50, 40].
Pass 3 (COUNTER1 = 3):
- MINIMUM initialized to 3 (val 50).
- COUNTER2=4: VALUES[4]=40 < 50 -> MINIMUM=4.
- MINIMUM(4) != COUNTER1(3) -> Swap VALUES[3] and VALUES[4]: TEMPORARY=40, VALUES becomes [6, 20, 38, 40, 50].
Final array state: [6, 20, 38, 40, 50].

Mentor tip

Carefully track MINIMUM index changes and record TEMPORARY whenever swaps occur.

(b)

4 marks

(i) Construct pseudocode for the flowchart algorithm (3 marks):

LIMIT = 4

FLAG = TRUE

loop while FLAG == TRUE

FLAG = FALSE

loop COUNTER from 0 to LIMIT - 1

if VALUES[COUNTER] > VALUES[COUNTER + 1] then

TEMPORARY = VALUES[COUNTER]

VALUES[COUNTER] = VALUES[COUNTER + 1]

VALUES[COUNTER + 1] = TEMPORARY

FLAG = TRUE

end if

end loop

(ii) State the type of sort (1 mark): Bubble sort.

Mentor tip

Use a while loop checking FLAG combined with an inner loop traversing 0 to LIMIT - 1.

(c)

2 marks

Construct an algorithm fragment to output all elements in array VALUES[]:
loop I from 0 to LIMIT
output VALUES[I]
end loop

Mentor tip

Verify loop bounds match the 5 elements (0 to 4).

(d)

3 marks

Contrast dynamic and static data structures:
1) Memory sizing: Static data structures (such as fixed arrays) have a predetermined, immutable size allocated at compile time. Dynamic data structures (such as linked lists, binary trees) can grow and shrink dynamically during program execution.
2) Memory allocation: Static structures use contiguous memory blocks and can waste space if partially filled or overflow if full; dynamic structures allocate memory from the heap as needed using pointers, though each node incurs pointer memory overhead.

Mentor tip

Focus on size mutability at runtime and pointer memory management.

(e)

1 mark

Identify one alternative data structure for storing class marks:
Linked list (or dynamic array / collection / stack / binary search tree).

Mentor tip

Linked list or Collection is an ideal answer.

Question 16

13 marks

Operating systems and control systems

Resource management
Control systems

(a)

2 marks

Identify two functions of an operating system:
1) Memory management (allocation, paging, virtual memory).
2) Processor scheduling / process management (or peripheral management, file system management, security/access control).

Mentor tip

Name core OS duties: processor management, memory management, peripheral management.

(b)

3 marks

Explain one benefit of a dedicated operating system for a mobile phone:
1) Resource optimization: A dedicated OS includes only necessary software drivers and modules tailored specifically to the phone's hardware architecture.
2) It strips out unused legacy code, resulting in a significantly smaller memory and storage footprint, faster boot and execution speed, and improved energy/battery efficiency compared to a bloated generic OS.

Mentor tip

Link customized hardware drivers to performance gains, lower storage footprint, and power efficiency.

(c)

5 marks

Explain the closed-loop control mechanism for constant temperature:
1) User sets the target reference temperature via the interface.
2) Temperature sensors continuously measure ambient room temperature.
3) Analog sensor readings are converted to digital signals via an Analogue-to-Digital Converter (ADC) and transmitted to the microprocessor.
4) The microprocessor compares the sampled temperature against the user's setpoint.
5) If the temperature deviates beyond a tolerance threshold, the microprocessor sends control signals (via a DAC if needed) to actuators (compressor, fan, heat pump) to adjust cooling/heating output until equilibrium is restored.

Mentor tip

Structure the answer chronologically: Sensing -> ADC -> Microprocessor comparison -> Actuator adjustment.

(d)

3 marks

Contrast distributed vs centralized air conditioning control:
1) Granularity & Efficiency: A distributed system uses local microprocessors and temperature zones, allowing individual rooms to be cooled independently based on local occupancy, reducing energy waste compared to a centralized system that cools the entire house uniformly.
2) Reliability & Redundancy: If a local controller fails in a distributed system, only that single zone is affected; in a centralized system, failure of the central processor disables cooling for the whole house.

Mentor tip

Contrast on two distinct axes: zoned temperature control/energy efficiency, and single point of failure.

Question 17

17 marks

Two-dimensional arrays and linked lists

Computational thinking, problem-solving and programming
Abstract data structures

(a)

5 marks

Identify array layout: CUSTOMERS has 512 rows, columns: 0=LASTNAME, 1=FIRSTNAME, 2=ADDRESS1, 3=ADDRESS2, 4=ADDRESS3, 5=CITY, 6=POSTCODE.
Construct pseudocode to count and display customers in Cardiff:
COUNT = 0
loop ROW from 0 to 511
if CUSTOMERS[ROW][5] == "Cardiff" then
COUNT = COUNT + 1
end if
end loop
output "Number of customers in Cardiff: ", COUNT

Mentor tip

Initialize COUNT = 0 before the loop and output the result after the loop completes.

(b)

5 marks

Construct pseudocode to print mailing labels for customers named Jones in Cardiff:
loop ROW from 0 to 511
if CUSTOMERS[ROW][0] == "Jones" and CUSTOMERS[ROW][5] == "Cardiff" then
output CUSTOMERS[ROW][1] + " " + CUSTOMERS[ROW][0]
output CUSTOMERS[ROW][2]
output CUSTOMERS[ROW][3]
output CUSTOMERS[ROW][4]
output CUSTOMERS[ROW][5]
output CUSTOMERS[ROW][6]
end if
end loop

Mentor tip

Check both LASTNAME (col 0) and CITY (col 5) in the conditional statement.

(c)

5 marks

List steps to insert 'Jones' into alphabetical singly linked list [Bale -> Cousens -> Davies -> Pugh -> Williams]:
1) Create a new node containing the data value 'Jones' and a null pointer.
2) Traverse the list from head pointer (Bale), comparing 'Jones' against each node's data.
3) Locate the insertion point between 'Davies' and 'Pugh' (since Davies < Jones < Pugh).
4) Set the next pointer of the new 'Jones' node to point to 'Pugh'.
5) Update the next pointer of 'Davies' to point to the new 'Jones' node.

Mentor tip

Always specify pointing the new node to the next node BEFORE updating the previous node's pointer.

(d)

2 marks

Identify an application where a circular linked list is preferred:
Round-robin CPU process scheduling (or a media playlist / multiplayer turn rotation): Once the last process or song is served, the list seamlessly cycles back to the first node without requiring null checks or pointer resets to the head.

Mentor tip

Name round-robin task scheduling or continuous playlists.

Our analysis

Where the marks are

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

Computational thinking, problem-solving and programming21 marks · 21%
Q15, Q17
System fundamentals19 marks · 19%
Q2, Q8, Q13
Computer organisation18 marks · 18%
Q1, Q3, Q4, Q5, Q6, Q7, Q14
Abstract data structures15 marks · 15%
Q11, Q12, Q15, Q17
Networks14 marks · 14%
Q9, Q10, Q14
Control systems8 marks · 8%
Q16
Resource management5 marks · 5%
Q16

Mentor insights

Traps to watch for

01
Order of logic operations for compound output Z · Q7
Intermediate signal S is generated by B AND C. Final output Z is A OR S. Whenever A is 1, output Z is 1 regardless of B and C.
02
Selection sort vs bubble sort inner loop execution · Q15
Selection sort scans the unsorted subarray to find the minimum index and swaps at most once per outer iteration. Bubble sort swaps adjacent elements immediately if out of order.
03
Singly linked list pointer update order on insertion · Q17
When inserting Jones between Davies and Pugh, you must set Jones.next = Pugh BEFORE setting Davies.next = Jones. Overwriting Davies.next first causes the remainder of the linked list to be lost in memory.

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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