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IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2018
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 10

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2018 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–7 · 25 marks

Question 1

2 marks

Definition of a computer network

Networks

(a)

2 marks

Define a computer network:
1) An interconnected system of computers and computing hardware peripherals linked through physical cabling or wireless transmission channels.
2) Established to facilitate data communication, information exchange, and shared hardware and software resource access among users.

Mentor tip

Mention multiple devices connected via communication channels to share data and resources.

Question 2

4 marks

Database management systems and benefits of beta testing

System fundamentals

(a)

2 marks

Outline what is meant by a Database Management System (DBMS):
1) A specialized software application suite designed to define, construct, manipulate, query, and manage relational or non-relational databases.
2) Provides controlled administrative access, data integrity constraints, concurrent multi-user transactional support, and automated backup/recovery mechanisms.

Mentor tip

Define DBMS as system software that creates, maintains, and provides controlled user access to structured data.

(b)

2 marks

Outline one advantage of beta testing prior to public commercial release:
1) Real-world user feedback: Dispatches pre-release software to a sample of actual end users outside the internal development team operating in diverse physical computing environments.
2) Defect discovery: Identifies unpredicted software bugs, edge-case usability bottlenecks, and hardware compatibility issues prior to public deployment, improving overall product quality.

Mentor tip

Explain that real users test in live diverse environments, exposing defects before general release.

Question 3

2 marks

Direct observation advantages and disadvantages in requirements analysis

System fundamentals

(a)

1 mark

Identify an advantage of direct observation:
Allows the systems analyst to observe end-users executing operational workflows in their natural working environment, capturing firsthand, unbiased reality rather than what users claim they do.

Mentor tip

Focus on observing real operational behavior in natural setting without reliance on subjective recollection.

(b)

1 mark

Identify a disadvantage of direct observation:
The Hawthorne effect: Users who are aware of being watched may alter their standard behavior, follow procedures more rigidly, or work unnaturally slowly; additionally, the technique is time-consuming for the analyst.

Mentor tip

Note observer bias or behavioral alteration (Hawthorne effect) when subjects know they are being observed.

Question 4

3 marks

Logic diagram representation of Boolean expression

Computer organisation

(a)

3 marks

Construct logic diagram for Boolean expression: \text{not } A \text{ or } (A \text{ and } B)
1) Connect input line A to a NOT gate (inverter) to produce intermediate signal \neg A.
2) Connect input lines A and B to a 2-input AND gate to produce intermediate signal (A \land B).
3) Connect intermediate outputs \neg A and (A \land B) into a 2-input OR gate to generate final output Z.

Z = \neg A \lor (A \land B) = \neg A \lor B

Mentor tip

Diagram includes 2 inputs (A, B), 3 gates (NOT, AND, OR), and 1 output.

Question 5

6 marks

International data migration compatibility, VPN communication, and social impact

System fundamentals
Networks

(a)

2 marks

Identify two data migration compatibility issues when moving headquarters between Europe and Asia:
1) Character encoding and language differences: Incompatible text standards (e.g., legacy ASCII/ISO vs Unicode/UTF-8 for Asian non-Latin character scripts).
2) Cultural data formatting conventions: Discrepancies in date representations (DD/MM/YYYY vs YYYY/MM/DD), address hierarchy structures, and currency symbol and precision standards.

Mentor tip

Highlight character set / encoding mismatches and differing date, currency, or unit formatting conventions.

(b)

2 marks

Outline how a Virtual Private Network (VPN) enables secure cross-continent communication:
1) Establishes an encrypted point-to-point tunnel over the public internet connecting European branch offices with Asian headquarters.
2) Encrypts all transmitted IP data packets, guaranteeing data confidentiality and integrity while masking internal IP addresses from external interception.

Mentor tip

Explain encrypted tunneling over the public internet providing secure, confidential communication between remote offices.

(c)

2 marks

Outline one social issue associated with relocating corporate headquarters:
Workplace disruption and job redundancy: Employees unable or unwilling to relocate to Asia face job displacement, loss of income, and career disruption. Conversely, relocating staff face cultural adaptation, family relocation stresses, and language barriers.

Mentor tip

Focus on human impact: job loss/redundancy or relocation stress and cultural adjustment.

Question 6

4 marks

Operational mechanism of packet switching across wide area networks

Networks

(a)

4 marks

Describe data transmission via packet switching:
1) Packetization: The source computer breaks the raw message stream into discrete, standardized units called packets, attaching header metadata (source IP, destination IP, packet sequence number, error checksum).
2) Independent routing: Intermediate routers evaluate packet destination addresses dynamically, forwarding individual packets along different optimal pathways across the network based on real-time traffic conditions.
3) Reassembly and error recovery: Upon reaching the destination, packets are reordered into their original sequence using their sequence numbers. If a packet is corrupt or missing according to checksum validation, a retransmission request is triggered.

Mentor tip

Describe breaking data into packets with headers, dynamic routing along independent paths, and destination reassembly with error checking.

Question 7

4 marks

Trace table construction for iterative counter algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Trace the execution of variables A and B across loop iterations:
Initial state: A = 3, B = 7.
Iteration 1: Condition (B >= A) -> 7 >= 3 (True).
- A = 3 + 1 = 4
- output(B - A) -> output(7 - 4) = 3
- B = 7 - 1 = 6
Iteration 2: Condition (B >= A) -> 6 >= 4 (True).
- A = 4 + 1 = 5
- output(B - A) -> output(6 - 5) = 1
- B = 6 - 1 = 5
Iteration 3: Condition (B >= A) -> 5 >= 5 (True).
- A = 5 + 1 = 6
- output(B - A) -> output(5 - 6) = -1
- B = 5 - 1 = 4
Iteration 4: Condition (B >= A) -> 4 >= 6 (False). Loop terminates.
Complete trace table values:
Row 1 (init): A=3, B=7
Row 2 (iter 1): A=4, B=6, output=3
Row 3 (iter 2): A=5, B=5, output=1
Row 4 (iter 3): A=6, B=4, output=-1
Outputs produced in sequence: 3, 1, -1.

7 - 4 = 3, 6 - 5 = 1, 5 - 6 = - 1

Mentor tip

Carefully trace variable updates: A increments BEFORE output(B - A), then B decrements. When B=4 and A=6, the loop terminates.

Questions 8–10 · 45 marks

Question 8

15 marks

Primary memory hierarchy, system bus architecture, and character encodings

Computer organisation

(a)

5 marks

Part (i) Distinguish between RAM and ROM (3 marks):
- Volatility: RAM is volatile (contents are lost upon power interruption), whereas ROM is non-volatile (contents persist permanently without power).
- Read/Write capability: RAM supports both read and write operations at runtime, whereas ROM is strictly read-only in standard operation.
- Contents and purpose: RAM holds running application instructions, operating system kernel modules, and live working data. ROM stores immutable system firmware, such as BIOS/UEFI bootloader routines required to initialize hardware at startup.
Part (ii) Primary memory management role of operating system (2 marks):
- Dynamic address allocation: Allocates contiguous physical memory frames or virtual pages to application processes upon request, and reclaims/de-allocates memory when processes terminate.
- Memory protection and isolation: Enforces boundary checks so processes cannot corrupt adjacent memory segments, utilizing paging to secondary storage when RAM is exhausted.

Mentor tip

Cover volatility, read/write ability, and boot vs runtime roles for RAM/ROM. For OS memory management, detail allocation, deallocation, and virtual memory.

(b)

4 marks

Explain the distinct roles of the address bus and data bus during instruction execution:
1) System bus definition: Parallel electronic communication channels connecting the CPU registers to primary RAM.
2) Address bus: Unidirectional bus driven by the CPU (MAR). Carries the physical memory address specifying the target RAM location to read from or write to.
3) Data bus: Bidirectional bus carrying the actual instruction word or data operand between primary memory and the processor's Memory Data Register (MDR).
4) Execution cycle coordination: During fetch, the CPU places the instruction address on the address bus, signals a memory read, and retrieves the returned opcode across the data bus into the MDR.

Mentor tip

Highlight unidirectional address bus carrying location pointers from MAR vs bidirectional data bus carrying actual content to/from MDR.

(c)

4 marks

Evaluate the binary byte 0101 1110:
Part (i) Convert byte to hexadecimal (1 mark):
Split into two 4-bit nibbles: 0101_2 = 5_{16}, and 1110_2 = E_{16}.
Hexadecimal value = 5E.
Part (ii) State total distinct integers representable in 1 byte (1 mark):
2^8 = 256 distinct integer values (0 to 255 unsigned, or -128 to +127 in two's complement).
Part (iii) Explain why 1 byte cannot encode Chinese characters (2 marks):
A single byte provides at most 2^8 = 256 unique code points, which is sufficient for small alphabetic alphabets (Latin alphabet, punctuation, control codes). Chinese ideographic script contains tens of thousands of unique characters, necessitating multi-byte encodings (such as UTF-8 or UTF-16, requiring at least 2 to 3 bytes per character).

010 1_{2} = 5_{16}, 111 0_{2} = 1 4_{10} = E_{16} ⟹ 5 E_{16} 2^{8} = 256 ≪ Number of Chinese logograms (> 50,000)

Mentor tip

Split into nibbles 0101 (5) and 1110 (E). 1 byte gives 256 combinations, vastly insufficient for thousands of ideographic logograms.

(d)

2 marks

Construct truth table for equivalence condition (Output is True when inputs are equal):
Input A | Input B | Output C (XNOR)
0 | 0 | 1 (True)
0 | 1 | 0 (False)
1 | 0 | 0 (False)
1 | 1 | 1 (True)

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

Mentor tip

This represents the XNOR equivalence logic gate: output is 1 when both inputs match (0,0 or 1,1).

Question 9

15 marks

High-level languages, modular sub-procedures, and collection interval filtering

Computational thinking, problem-solving and programming
Abstract data structures

(a)

2 marks

Outline the necessity for high-level programming languages:
1) Human abstraction and productivity: High-level languages use human-readable syntax and structured abstractions (loops, conditional logic, complex data structures) rather than cryptic binary opcodes, drastically accelerating developer productivity.
2) Hardware portability and machine independence: High-level source code compiles or interprets across diverse CPU architectures without requiring knowledge of machine-level register sets and instruction encoding.

Mentor tip

Focus on programmer productivity, readability, and machine independence.

(b)

4 marks

Explain two distinct benefits of sub-procedures (methods/functions) in program construction:
Benefit 1: Code Reusability and Maintainability: A procedure written and verified once can be invoked from multiple call sites across the program, reducing redundant code and isolating modifications to a single modular location.
Benefit 2: Problem Decomposition and Team Collaboration: Complex software tasks are broken into manageable sub-tasks (top-down functional decomposition). Individual team members can develop, debug, and unit-test separate modules concurrently without interfering with other modules.

Mentor tip

Explain code reuse/easier maintenance and top-down problem decomposition/collaborative development.

(c)

3 marks

Identify three key characteristics of a collection data structure:
1) Abstract container of elements: Groups discrete data items into a single container entity, dynamically managing element storage.
2) Encapsulated access methods: Provides standardized iterator and mutation methods (e.g., resetNext(), hasNext(), getNext(), addItem(), isEmpty()) hiding underlying memory layout.
3) Dynamic sizing and type flexibility: Can grow and shrink dynamically at runtime, holding primitive values or heterogeneous complex object references.

Mentor tip

Mention encapsulated methods (hasNext, getNext), dynamic container capacity, and abstraction over underlying physical memory structures.

(d)

6 marks

Construct pseudocode to iterate collection NUMBERS and count elements within interval [-1, 1]:
COUNTER = 0
NUMBERS.resetNext()
while NUMBERS.hasNext()
ELEMENT = NUMBERS.getNext()
if ELEMENT >= -1.0 and ELEMENT <= 1.0 then
COUNTER = COUNTER + 1
end if
end while
output COUNTER

- 1 \leq x \leq 1 ⟺ ∣ x ∣ \leq 1

Mentor tip

Remember to call NUMBERS.resetNext() before the loop, fetch each value using getNext(), check interval bounds inclusive, and print the counter.

Question 10

15 marks

Body mass index calculation, categorization algorithm, and parallel array searching

Computational thinking, problem-solving and programming

(a)

1 mark

Calculate Boris's BMI using calcBMI(2.00, 104):
H = 2.00, W = 104.
X = H * H = 2.00 * 2.00 = 4.00.
B = W / X = 104 / 4.00 = 26.0.
Value of BorisBMI = 26.0.

BMI = \frac{W}{H ^{2}} = \frac{104}{2.0 0 ^{2}} = \frac{104}{4.0} = 26.0

Mentor tip

Evaluate W / (H * H): 104 / 4 = 26.0.

(b)

4 marks

Construct pseudocode algorithm to output weight category from BMI value B:
procedure outputCategory(B)
if B < 18.5 then
output "underweight"
else if B < 25.0 then
output "normal weight"
else if B < 30.0 then
output "overweight"
else
output "obese"
end if
end procedure

Mentor tip

Use sequential if/else-if branches checking boundary thresholds (< 18.5, < 25.0, < 30.0, else).

(c)

1 mark

Identify person whose height is held in HEIGHT[3]:
Corresponding person in parallel array NAME[3] is 'Paul'.

Mentor tip

Look up index 3 in parallel array NAME, which corresponds to Paul.

(d)

3 marks

Analyze search methodology on HEIGHT array:
Part (i) Reason why binary search cannot be applied (1 mark):
Binary search requires the target array to be sorted in ascending or descending numerical order. The HEIGHT array is unordered (values: 1.56, 2.00, 2.03, 1.75, 1.80, ...).
Part (ii) Describe how to locate and output person's name by height (2 marks):
Execute a linear (sequential) search iterating through HEIGHT from index 0 to 29. Compare each element against the target height. When a match is found at index K, output the corresponding element in parallel array NAME[K] and terminate search.

Mentor tip

State that binary search strictly requires sorted data. Describe linear sequential search through HEIGHT to find the matching index K, then output NAME[K].

(e)

6 marks

Construct algorithm to output names of all people whose BMI exceeds the group average:
TOTAL_BMI = 0
// Pass 1: Accumulate total BMI for all 30 individuals
loop K from 0 to 29
PERSON_BMI = calcBMI(HEIGHT[K], WEIGHT[K])
TOTAL_BMI = TOTAL_BMI + PERSON_BMI
end loop
AVERAGE_BMI = TOTAL_BMI / 30
// Pass 2: Identify and output names of individuals above average
loop K from 0 to 29
if calcBMI(HEIGHT[K], WEIGHT[K]) > AVERAGE_BMI then
output NAME[K]
end if
end loop

Average BMI = \frac{1}{30} K = 0 \sum 29 calcBMI (HEIGHT [K], WEIGHT [K])

Mentor tip

Execute two passes: first loop calculates the sum and divides by 30 to get average BMI, second loop compares each individual's BMI against the average and prints their name.

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 programming31 marks · 44%
Q7, Q9, Q10
Computer organisation18 marks · 26%
Q4, Q8
System fundamentals10 marks · 14%
Q2, Q3, Q5
Networks8 marks · 11%
Q1, Q5, Q6
Abstract data structures3 marks · 4%
Q9

Mentor insights

Traps to watch for

01
Loop variable update sequence in trace table · Q7
In the trace loop, A increments BEFORE the output expression (B - A) evaluates, and B decrements AFTER the output. When B becomes 4 and A becomes 6, 4 >= 6 evaluates to False and the loop terminates.
02
Binary search on unsorted arrays · Q8
Binary search cannot be executed on an unsorted array. If the HEIGHT array is unsorted, a linear sequential search must be used to locate values.
03
Two-pass processing for group average comparisons · Q10
You cannot filter individuals above average BMI in a single pass without knowing the average beforehand. Pass 1 must accumulate the total and calculate the mean; Pass 2 compares each person against that computed mean.

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

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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2018
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 10

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2018 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–7 · 25 marks

Question 1

2 marks

Definition of a computer network

Networks

(a)

2 marks

Define a computer network:
1) An interconnected system of computers and computing hardware peripherals linked through physical cabling or wireless transmission channels.
2) Established to facilitate data communication, information exchange, and shared hardware and software resource access among users.

Mentor tip

Mention multiple devices connected via communication channels to share data and resources.

Question 2

4 marks

Database management systems and benefits of beta testing

System fundamentals

(a)

2 marks

Outline what is meant by a Database Management System (DBMS):
1) A specialized software application suite designed to define, construct, manipulate, query, and manage relational or non-relational databases.
2) Provides controlled administrative access, data integrity constraints, concurrent multi-user transactional support, and automated backup/recovery mechanisms.

Mentor tip

Define DBMS as system software that creates, maintains, and provides controlled user access to structured data.

(b)

2 marks

Outline one advantage of beta testing prior to public commercial release:
1) Real-world user feedback: Dispatches pre-release software to a sample of actual end users outside the internal development team operating in diverse physical computing environments.
2) Defect discovery: Identifies unpredicted software bugs, edge-case usability bottlenecks, and hardware compatibility issues prior to public deployment, improving overall product quality.

Mentor tip

Explain that real users test in live diverse environments, exposing defects before general release.

Question 3

2 marks

Direct observation advantages and disadvantages in requirements analysis

System fundamentals

(a)

1 mark

Identify an advantage of direct observation:
Allows the systems analyst to observe end-users executing operational workflows in their natural working environment, capturing firsthand, unbiased reality rather than what users claim they do.

Mentor tip

Focus on observing real operational behavior in natural setting without reliance on subjective recollection.

(b)

1 mark

Identify a disadvantage of direct observation:
The Hawthorne effect: Users who are aware of being watched may alter their standard behavior, follow procedures more rigidly, or work unnaturally slowly; additionally, the technique is time-consuming for the analyst.

Mentor tip

Note observer bias or behavioral alteration (Hawthorne effect) when subjects know they are being observed.

Question 4

3 marks

Logic diagram representation of Boolean expression

Computer organisation

(a)

3 marks

Construct logic diagram for Boolean expression: \text{not } A \text{ or } (A \text{ and } B)
1) Connect input line A to a NOT gate (inverter) to produce intermediate signal \neg A.
2) Connect input lines A and B to a 2-input AND gate to produce intermediate signal (A \land B).
3) Connect intermediate outputs \neg A and (A \land B) into a 2-input OR gate to generate final output Z.

Z = \neg A \lor (A \land B) = \neg A \lor B

Mentor tip

Diagram includes 2 inputs (A, B), 3 gates (NOT, AND, OR), and 1 output.

Question 5

6 marks

International data migration compatibility, VPN communication, and social impact

System fundamentals
Networks

(a)

2 marks

Identify two data migration compatibility issues when moving headquarters between Europe and Asia:
1) Character encoding and language differences: Incompatible text standards (e.g., legacy ASCII/ISO vs Unicode/UTF-8 for Asian non-Latin character scripts).
2) Cultural data formatting conventions: Discrepancies in date representations (DD/MM/YYYY vs YYYY/MM/DD), address hierarchy structures, and currency symbol and precision standards.

Mentor tip

Highlight character set / encoding mismatches and differing date, currency, or unit formatting conventions.

(b)

2 marks

Outline how a Virtual Private Network (VPN) enables secure cross-continent communication:
1) Establishes an encrypted point-to-point tunnel over the public internet connecting European branch offices with Asian headquarters.
2) Encrypts all transmitted IP data packets, guaranteeing data confidentiality and integrity while masking internal IP addresses from external interception.

Mentor tip

Explain encrypted tunneling over the public internet providing secure, confidential communication between remote offices.

(c)

2 marks

Outline one social issue associated with relocating corporate headquarters:
Workplace disruption and job redundancy: Employees unable or unwilling to relocate to Asia face job displacement, loss of income, and career disruption. Conversely, relocating staff face cultural adaptation, family relocation stresses, and language barriers.

Mentor tip

Focus on human impact: job loss/redundancy or relocation stress and cultural adjustment.

Question 6

4 marks

Operational mechanism of packet switching across wide area networks

Networks

(a)

4 marks

Describe data transmission via packet switching:
1) Packetization: The source computer breaks the raw message stream into discrete, standardized units called packets, attaching header metadata (source IP, destination IP, packet sequence number, error checksum).
2) Independent routing: Intermediate routers evaluate packet destination addresses dynamically, forwarding individual packets along different optimal pathways across the network based on real-time traffic conditions.
3) Reassembly and error recovery: Upon reaching the destination, packets are reordered into their original sequence using their sequence numbers. If a packet is corrupt or missing according to checksum validation, a retransmission request is triggered.

Mentor tip

Describe breaking data into packets with headers, dynamic routing along independent paths, and destination reassembly with error checking.

Question 7

4 marks

Trace table construction for iterative counter algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Trace the execution of variables A and B across loop iterations:
Initial state: A = 3, B = 7.
Iteration 1: Condition (B >= A) -> 7 >= 3 (True).
- A = 3 + 1 = 4
- output(B - A) -> output(7 - 4) = 3
- B = 7 - 1 = 6
Iteration 2: Condition (B >= A) -> 6 >= 4 (True).
- A = 4 + 1 = 5
- output(B - A) -> output(6 - 5) = 1
- B = 6 - 1 = 5
Iteration 3: Condition (B >= A) -> 5 >= 5 (True).
- A = 5 + 1 = 6
- output(B - A) -> output(5 - 6) = -1
- B = 5 - 1 = 4
Iteration 4: Condition (B >= A) -> 4 >= 6 (False). Loop terminates.
Complete trace table values:
Row 1 (init): A=3, B=7
Row 2 (iter 1): A=4, B=6, output=3
Row 3 (iter 2): A=5, B=5, output=1
Row 4 (iter 3): A=6, B=4, output=-1
Outputs produced in sequence: 3, 1, -1.

7 - 4 = 3, 6 - 5 = 1, 5 - 6 = - 1

Mentor tip

Carefully trace variable updates: A increments BEFORE output(B - A), then B decrements. When B=4 and A=6, the loop terminates.

Questions 8–10 · 45 marks

Question 8

15 marks

Primary memory hierarchy, system bus architecture, and character encodings

Computer organisation

(a)

5 marks

Part (i) Distinguish between RAM and ROM (3 marks):
- Volatility: RAM is volatile (contents are lost upon power interruption), whereas ROM is non-volatile (contents persist permanently without power).
- Read/Write capability: RAM supports both read and write operations at runtime, whereas ROM is strictly read-only in standard operation.
- Contents and purpose: RAM holds running application instructions, operating system kernel modules, and live working data. ROM stores immutable system firmware, such as BIOS/UEFI bootloader routines required to initialize hardware at startup.
Part (ii) Primary memory management role of operating system (2 marks):
- Dynamic address allocation: Allocates contiguous physical memory frames or virtual pages to application processes upon request, and reclaims/de-allocates memory when processes terminate.
- Memory protection and isolation: Enforces boundary checks so processes cannot corrupt adjacent memory segments, utilizing paging to secondary storage when RAM is exhausted.

Mentor tip

Cover volatility, read/write ability, and boot vs runtime roles for RAM/ROM. For OS memory management, detail allocation, deallocation, and virtual memory.

(b)

4 marks

Explain the distinct roles of the address bus and data bus during instruction execution:
1) System bus definition: Parallel electronic communication channels connecting the CPU registers to primary RAM.
2) Address bus: Unidirectional bus driven by the CPU (MAR). Carries the physical memory address specifying the target RAM location to read from or write to.
3) Data bus: Bidirectional bus carrying the actual instruction word or data operand between primary memory and the processor's Memory Data Register (MDR).
4) Execution cycle coordination: During fetch, the CPU places the instruction address on the address bus, signals a memory read, and retrieves the returned opcode across the data bus into the MDR.

Mentor tip

Highlight unidirectional address bus carrying location pointers from MAR vs bidirectional data bus carrying actual content to/from MDR.

(c)

4 marks

Evaluate the binary byte 0101 1110:
Part (i) Convert byte to hexadecimal (1 mark):
Split into two 4-bit nibbles: 0101_2 = 5_{16}, and 1110_2 = E_{16}.
Hexadecimal value = 5E.
Part (ii) State total distinct integers representable in 1 byte (1 mark):
2^8 = 256 distinct integer values (0 to 255 unsigned, or -128 to +127 in two's complement).
Part (iii) Explain why 1 byte cannot encode Chinese characters (2 marks):
A single byte provides at most 2^8 = 256 unique code points, which is sufficient for small alphabetic alphabets (Latin alphabet, punctuation, control codes). Chinese ideographic script contains tens of thousands of unique characters, necessitating multi-byte encodings (such as UTF-8 or UTF-16, requiring at least 2 to 3 bytes per character).

010 1_{2} = 5_{16}, 111 0_{2} = 1 4_{10} = E_{16} ⟹ 5 E_{16} 2^{8} = 256 ≪ Number of Chinese logograms (> 50,000)

Mentor tip

Split into nibbles 0101 (5) and 1110 (E). 1 byte gives 256 combinations, vastly insufficient for thousands of ideographic logograms.

(d)

2 marks

Construct truth table for equivalence condition (Output is True when inputs are equal):
Input A | Input B | Output C (XNOR)
0 | 0 | 1 (True)
0 | 1 | 0 (False)
1 | 0 | 0 (False)
1 | 1 | 1 (True)

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

Mentor tip

This represents the XNOR equivalence logic gate: output is 1 when both inputs match (0,0 or 1,1).

Question 9

15 marks

High-level languages, modular sub-procedures, and collection interval filtering

Computational thinking, problem-solving and programming
Abstract data structures

(a)

2 marks

Outline the necessity for high-level programming languages:
1) Human abstraction and productivity: High-level languages use human-readable syntax and structured abstractions (loops, conditional logic, complex data structures) rather than cryptic binary opcodes, drastically accelerating developer productivity.
2) Hardware portability and machine independence: High-level source code compiles or interprets across diverse CPU architectures without requiring knowledge of machine-level register sets and instruction encoding.

Mentor tip

Focus on programmer productivity, readability, and machine independence.

(b)

4 marks

Explain two distinct benefits of sub-procedures (methods/functions) in program construction:
Benefit 1: Code Reusability and Maintainability: A procedure written and verified once can be invoked from multiple call sites across the program, reducing redundant code and isolating modifications to a single modular location.
Benefit 2: Problem Decomposition and Team Collaboration: Complex software tasks are broken into manageable sub-tasks (top-down functional decomposition). Individual team members can develop, debug, and unit-test separate modules concurrently without interfering with other modules.

Mentor tip

Explain code reuse/easier maintenance and top-down problem decomposition/collaborative development.

(c)

3 marks

Identify three key characteristics of a collection data structure:
1) Abstract container of elements: Groups discrete data items into a single container entity, dynamically managing element storage.
2) Encapsulated access methods: Provides standardized iterator and mutation methods (e.g., resetNext(), hasNext(), getNext(), addItem(), isEmpty()) hiding underlying memory layout.
3) Dynamic sizing and type flexibility: Can grow and shrink dynamically at runtime, holding primitive values or heterogeneous complex object references.

Mentor tip

Mention encapsulated methods (hasNext, getNext), dynamic container capacity, and abstraction over underlying physical memory structures.

(d)

6 marks

Construct pseudocode to iterate collection NUMBERS and count elements within interval [-1, 1]:
COUNTER = 0
NUMBERS.resetNext()
while NUMBERS.hasNext()
ELEMENT = NUMBERS.getNext()
if ELEMENT >= -1.0 and ELEMENT <= 1.0 then
COUNTER = COUNTER + 1
end if
end while
output COUNTER

- 1 \leq x \leq 1 ⟺ ∣ x ∣ \leq 1

Mentor tip

Remember to call NUMBERS.resetNext() before the loop, fetch each value using getNext(), check interval bounds inclusive, and print the counter.

Question 10

15 marks

Body mass index calculation, categorization algorithm, and parallel array searching

Computational thinking, problem-solving and programming

(a)

1 mark

Calculate Boris's BMI using calcBMI(2.00, 104):
H = 2.00, W = 104.
X = H * H = 2.00 * 2.00 = 4.00.
B = W / X = 104 / 4.00 = 26.0.
Value of BorisBMI = 26.0.

BMI = \frac{W}{H ^{2}} = \frac{104}{2.0 0 ^{2}} = \frac{104}{4.0} = 26.0

Mentor tip

Evaluate W / (H * H): 104 / 4 = 26.0.

(b)

4 marks

Construct pseudocode algorithm to output weight category from BMI value B:
procedure outputCategory(B)
if B < 18.5 then
output "underweight"
else if B < 25.0 then
output "normal weight"
else if B < 30.0 then
output "overweight"
else
output "obese"
end if
end procedure

Mentor tip

Use sequential if/else-if branches checking boundary thresholds (< 18.5, < 25.0, < 30.0, else).

(c)

1 mark

Identify person whose height is held in HEIGHT[3]:
Corresponding person in parallel array NAME[3] is 'Paul'.

Mentor tip

Look up index 3 in parallel array NAME, which corresponds to Paul.

(d)

3 marks

Analyze search methodology on HEIGHT array:
Part (i) Reason why binary search cannot be applied (1 mark):
Binary search requires the target array to be sorted in ascending or descending numerical order. The HEIGHT array is unordered (values: 1.56, 2.00, 2.03, 1.75, 1.80, ...).
Part (ii) Describe how to locate and output person's name by height (2 marks):
Execute a linear (sequential) search iterating through HEIGHT from index 0 to 29. Compare each element against the target height. When a match is found at index K, output the corresponding element in parallel array NAME[K] and terminate search.

Mentor tip

State that binary search strictly requires sorted data. Describe linear sequential search through HEIGHT to find the matching index K, then output NAME[K].

(e)

6 marks

Construct algorithm to output names of all people whose BMI exceeds the group average:
TOTAL_BMI = 0
// Pass 1: Accumulate total BMI for all 30 individuals
loop K from 0 to 29
PERSON_BMI = calcBMI(HEIGHT[K], WEIGHT[K])
TOTAL_BMI = TOTAL_BMI + PERSON_BMI
end loop
AVERAGE_BMI = TOTAL_BMI / 30
// Pass 2: Identify and output names of individuals above average
loop K from 0 to 29
if calcBMI(HEIGHT[K], WEIGHT[K]) > AVERAGE_BMI then
output NAME[K]
end if
end loop

Average BMI = \frac{1}{30} K = 0 \sum 29 calcBMI (HEIGHT [K], WEIGHT [K])

Mentor tip

Execute two passes: first loop calculates the sum and divides by 30 to get average BMI, second loop compares each individual's BMI against the average and prints their name.

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 programming31 marks · 44%
Q7, Q9, Q10
Computer organisation18 marks · 26%
Q4, Q8
System fundamentals10 marks · 14%
Q2, Q3, Q5
Networks8 marks · 11%
Q1, Q5, Q6
Abstract data structures3 marks · 4%
Q9

Mentor insights

Traps to watch for

01
Loop variable update sequence in trace table · Q7
In the trace loop, A increments BEFORE the output expression (B - A) evaluates, and B decrements AFTER the output. When B becomes 4 and A becomes 6, 4 >= 6 evaluates to False and the loop terminates.
02
Binary search on unsorted arrays · Q8
Binary search cannot be executed on an unsorted array. If the HEIGHT array is unsorted, a linear sequential search must be used to locate values.
03
Two-pass processing for group average comparisons · Q10
You cannot filter individuals above average BMI in a single pass without knowing the average beforehand. Pass 1 must accumulate the total and calculate the mean; Pass 2 compares each person against that computed mean.

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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