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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2024
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 13

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 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 values are modified, all dependent formulas across the sheet automatically recalculate.
Predefined formula libraries: Built-in library of statistical, mathematical, and logical functions (e.g. `SUM`, `AVERAGE`, `COUNTIF`).

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

Mentor tip

Focus on data calculation capabilities rather than visual font styling.

Question 2

2 marks

Role and value of beta testing in software release cycles

System fundamentals

2 marks

Diverse operating environments: Exposes software to thousands of hardware, OS, and networking configurations that in-house developers cannot simulate.
Unanticipated user interactions: Real-world user behaviour uncovers obscure edge cases and user interface flaws prior to final release.

Mentor tip

Beta testing is conducted by real end users in their own environments.

Question 3

2 marks

Involvement of end users during system changeover

System fundamentals

2 marks

User feedback on requirements: End users verify that the replacement system satisfies actual day-to-day workflow tasks.
Validation and user training: Users participate in operational testing and identify gaps before the legacy system is retired.

Mentor tip

End users are key stakeholders for operational validation and workflow feedback.

Question 4

4 marks

Evaluation of phased conversion strategy for system implementation

System fundamentals

4 marks

Advantage - Reduced system failure risk: Implementing the system in discrete stages allows unforeseen issues to be contained within one operational module without halting the entire organization.
Advantage - Structured staff adaptation: Staff adjust to new interfaces and workflows incrementally rather than undergoing overwhelming organizational disruption.
Disadvantage - Temporary bridging overhead: Requires developing temporary data conversion bridges and synchronization routines between legacy modules and upgraded modules.
Disadvantage - Prolonged implementation timeline: Transitioning across multiple consecutive phases extends the changeover schedule compared to a direct switch.

Mentor tip

Provide both positive points (risk containment, gradual learning) and negative trade-offs (data synchronisation overhead, prolonged timeline).

Question 5

2 marks

Characteristics of Media Access Control (MAC) addresses

Networks

2 marks

Permanent physical identifier: A unique 48-bit hardware identifier burned into the Network Interface Card (NIC) at manufacture time.
Data Link layer addressing: Identifies physical nodes within a local area network (LAN), expressed as 12 hexadecimal digits (e.g. `00:1A:2B:3C:4D:5E`).

Format: XX:XX:XX:XX:XX:XX (48 bits / 6 octets)

Mentor tip

Distinguish MAC addresses (fixed physical Layer 2 hardware address) from IP addresses (dynamic logical Layer 3 network address).

Question 6

2 marks

Purpose and architectural role of system flowcharts

System fundamentals

2 marks

High-level process flow: Visualizes the broader movement of data and operations across hardware devices, storage media, and processing units.
System boundary overview: Identifies how physical inputs, external databases, processing programs, and printed/electronic outputs interact.

Mentor tip

System flowcharts map systems and hardware media, while program flowcharts detail line-by-line algorithm logic.

Question 7

3 marks

Traffic filtering and perimeter defence via firewalls

Networks

3 marks

Packet inspection: Intercepts and analyses incoming and outgoing network packets at network boundaries.
Filtering rules: Assesses packet headers against access control criteria (source/destination IP addresses, port numbers, protocols).
Access enforcement: Blocks suspicious packets and drops unauthorized connection attempts while allowing legitimate communication.

Rule: (Port = 80 \lor Port = 443) ⟹ PERMIT

Mentor tip

Explain packet header inspection and filtering decisions based on security policies.

Question 8

2 marks

Benefits of low-power mobile electronic devices

System fundamentals

2 marks

Extended battery runtime: Enables prolonged operational use throughout the day without requiring frequent recharging or bulky power banks.
Thermal comfort and portability: Generates minimal internal heat, permitting silent fanless designs and thinner, lightweight form factors.

Mentor tip

Focus on real user benefits: longer runtime between charges and lighter/cooler hardware.

Question 9

4 marks

Trace table for Euclidean greatest common divisor algorithm

Computational thinking, problem-solving and programming

4 marks

Initialization: $A = 20$ , $B = 12$ . Condition $B > 0$ is true.
Iteration 1: $TEMP = 12$ . $B = 20 mod 12 = 8$ . $A = 12$ . Condition $B > 0$ ( $8 > 0$ ) is true.
Iteration 2: $TEMP = 8$ . $B = 12 mod 8 = 4$ . $A = 8$ . Condition $B > 0$ ( $4 > 0$ ) is true.
Iteration 3: $TEMP = 4$ . $B = 8 mod 4 = 0$ . $A = 4$ . Condition $B > 0$ ( $0 > 0$ ) is false.
Loop terminates. Final output is $A = 4$ .

A 201284 B 12840 TEMP - 1284 Output 4

Mentor tip

Track variables column by column through each iteration until the loop condition $B > 0$ fails.

Question 10

2 marks

Structural characteristics of collection data structures

Abstract data structures

2 marks

Dynamic resizing: Collections grow and contract dynamically during runtime as elements are inserted or removed, without fixed capacity constraints.
Iterator-based access: Elements are accessed sequentially using standardized iterator methods (`getNext()`, `hasNext()`, `resetNext()`) rather than direct numeric indexing.

Mentor tip

Mention dynamic size and sequential iterator traversal without index arithmetic.

Questions 11–13 · 45 marks

Question 11

15 marks

Laptop architecture, network interfaces, and packet protocols

System fundamentals
Resource management
Networks

(a)

1 mark

Full-disk encryption: Enabling disk encryption ensures that all saved files remain cryptographically locked if the physical laptop is stolen.

Mentor tip

Full disk encryption ensures data security even if the internal drive is extracted.

(b)

2 marks

Primary memory allocation: Allocates RAM for code instructions and heap/stack variables during execution.
CPU process scheduling: Distributes processor execution time slices across active tasks via multitasking.

Mentor tip

Name memory management or process scheduling and outline its purpose.

(c)

4 marks

Wired Ethernet utility: Offers maximum data throughput, lower latency, and reliable connection stability unaffected by wireless interference.
Wireless Wi-Fi utility: Provides untethered mobility and convenient network connectivity across home, school, and mobile workspaces.
Hybrid rationale: Equipping both interfaces accommodates demanding data transfers when docked and mobile portability on the move.

Mentor tip

Contrast wired throughput and stability against wireless convenience and mobility.

(d)

2 marks

Packet header: Contains addressing and sequencing metadata (source IP, destination IP, packet sequence number, protocol).
Packet payload and trailer: The payload holds the application data chunk; the trailer contains checksum / CRC bits for error detection.

Data Packet = [Header ∣ Payload ∣ Trailer]

Mentor tip

List the header, payload, and trailer along with their contents.

(e)

6 marks

Universal communication standards: Protocols provide agreed syntax and semantics so disparate hardware and operating systems can exchange data.
Data integrity and error control: Detects corrupted packets via checksums and manages acknowledgments and retransmissions.
Congestion management and flow control: Regulates transmission speed to match receiver processing capacity and prevent buffer overflow.

Mentor tip

Cover standardization across vendors, error handling/retransmission, and flow rate regulation.

Question 12

15 marks

Boolean logic design, sensor truth table, and CPU fetch-decode-execute cycle

Computer organisation

(a)(i)

1 mark

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

Operators: or, and

Mentor tip

Logical operators combine Boolean truth values.

(a)(ii)

1 mark

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

Constants: 6, 3, 20

Mentor tip

Constants are unchanging literal numerical values.

(a)(iii)

2 marks

Substitute $X = 6$ and $Y = 6$ :
$(6 > 6)$ evaluates to false.
$(6 > 3)$ evaluates to true.
$(6 + 6 < 20) = (12 < 20)$ evaluates to true.
Combine using precedence (`and` before `or`):
$true \land true = true$ .
$false \lor true = true$ .

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

Mentor tip

Remember that AND takes precedence over OR.

(b)

4 marks

Siren condition: Activates when the door is locked ( $DOOR = 1$ ) AND either movement is detected ( $MOTION = 1$ ) OR the window is open ( $WINDOW = 1$ ).
Formula: $SIREN = DOOR \land (MOTION \lor WINDOW)$ .
Rows with $DOOR = 0$ : Output is 0 for all four combinations.
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

The siren is only 1 when DOOR is 1 and at least one of MOTION or WINDOW is 1.

(c)(i)

1 mark

The Memory Data Register (MDR) holds data transferred between the CPU and primary memory.

Register: Memory Data Register (MDR)

Mentor tip

MDR holds data contents; MAR holds memory addresses.

(c)(ii)

1 mark

The Control Unit (CU) decodes fetched machine instructions.

Component: Control Unit (CU)

Mentor tip

The CU manages the fetch-decode-execute cycle.

(c)(iii)

1 mark

Arithmetic calculations and logical operations are performed in the Arithmetic Logic Unit (ALU).

Component: Arithmetic Logic Unit (ALU)

Mentor tip

All calculations take place inside the ALU.

(c)(iv)

4 marks

Address Bus: Transmits the memory address from the MAR to RAM (unidirectional).
Data Bus: Transmits data and instruction words between the MDR and primary memory (bidirectional).
Control Bus: Transmits command, timing, and status signals (e.g. read/write, clock pulses) between the Control Unit and memory/peripherals.
Execution coordination: The CPU places addresses on the address bus, asserts control signals on the control bus, and exchanges data over the data bus.

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

Mentor tip

Outline the role and transmission direction of each of the three buses.

Question 13

15 marks

Array statistical processing: mean calculation and median sorting algorithms

Computational thinking, problem-solving and programming

(a)

4 marks

Initialize an accumulator variable `SUM = 0`.
Iterate through the array from index 0 to $N - 1$ , adding each element `DATA[I]` to `SUM`.
Divide the total sum by the count of elements $N$ to obtain the arithmetic mean.
Output the calculated mean value.

SUM = 0 loop I from 0 to N - 1 SUM = SUM + DATA [I] end loop MEAN = SUM / N output (MEAN)

Mentor tip

Accumulate the sum first, then perform the division once outside the loop to avoid redundant divisions.

(b)(i)

1 mark

Bubble Sort (or Selection Sort / Insertion Sort) can sort the array into ascending order.

Algorithm: Bubble Sort

Mentor tip

Name any standard $O (n^{2})$ or $O (n lo g n)$ sorting algorithm from the syllabus.

(b)(ii)

10 marks

Step 1 - Sort the array: Use Bubble Sort with a boolean swapped flag to arrange the $N$ elements into non-decreasing order.
Step 2 - Determine median position: Check if $N$ is odd or even using modular division: $(N mod 2) == 1$ .
Odd length: If $N$ is odd, the middle element is located at index $div (N, 2)$ ; assign $MEDIAN = DATA [div (N, 2)]$ .
Even length: If $N$ is even, find the two middle elements at indices $(N /2) - 1$ and $N /2$ ; compute their arithmetic average: $MEDIAN = (DATA [N /2 - 1] + DATA [N /2]) /2.0$ .
Step 3 - Output result: Print the calculated median.

SWAPPED = true loop while SWAPPED SWAPPED = false loop I from 0 to N - 2 if DATA [I] > DATA [I + 1] then TEMP = DATA [I] DATA [I] = DATA [I + 1] DATA [I + 1] = TEMP SWAPPED = true end if end loop end loop if (N mod 2) == 1 then MID = div (N, 2) MEDIAN = DATA [MID] else MID2 = N /2 MID1 = MID2 - 1 MEDIAN = (DATA [MID1] + DATA [MID2]) /2.0 end if output (MEDIAN)

Mentor tip

Handle both odd and even parity cases: single middle element for odd $N$ , average of two middle elements for even $N$ .

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 programming19 marks · 27%
Q9, Q13
Networks17 marks · 24%
Q5, Q7, Q11
System fundamentals15 marks · 21%
Q1, Q2, Q3, Q4, Q6, Q8, Q11
Computer organisation15 marks · 21%
Q12
Abstract data structures2 marks · 3%
Q10
Resource management2 marks · 3%
Q11

Mentor insights

Traps to watch for

01
Temporary variable caching in Euclidean loop · Q9
Cache B in TEMP before calculating A mod B. Overwriting B prematurely corrupts subsequent variable assignments.
02
Operator precedence in compound Boolean expressions · Q12
Logical AND takes precedence over OR. The condition (Y > 3) and ((X + Y) < 20) evaluates first before being OR-combined with (X > 6).
03
Even-length array median calculation · Q13
When array size N is even, the median is the arithmetic mean of the two middle elements at indices (N/2) - 1 and N/2, requiring floating-point division by 2.0.

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

Want someone to go through your paper with you?

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

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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2024
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 13

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 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 values are modified, all dependent formulas across the sheet automatically recalculate.
Predefined formula libraries: Built-in library of statistical, mathematical, and logical functions (e.g. `SUM`, `AVERAGE`, `COUNTIF`).

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

Mentor tip

Focus on data calculation capabilities rather than visual font styling.

Question 2

2 marks

Role and value of beta testing in software release cycles

System fundamentals

2 marks

Diverse operating environments: Exposes software to thousands of hardware, OS, and networking configurations that in-house developers cannot simulate.
Unanticipated user interactions: Real-world user behaviour uncovers obscure edge cases and user interface flaws prior to final release.

Mentor tip

Beta testing is conducted by real end users in their own environments.

Question 3

2 marks

Involvement of end users during system changeover

System fundamentals

2 marks

User feedback on requirements: End users verify that the replacement system satisfies actual day-to-day workflow tasks.
Validation and user training: Users participate in operational testing and identify gaps before the legacy system is retired.

Mentor tip

End users are key stakeholders for operational validation and workflow feedback.

Question 4

4 marks

Evaluation of phased conversion strategy for system implementation

System fundamentals

4 marks

Advantage - Reduced system failure risk: Implementing the system in discrete stages allows unforeseen issues to be contained within one operational module without halting the entire organization.
Advantage - Structured staff adaptation: Staff adjust to new interfaces and workflows incrementally rather than undergoing overwhelming organizational disruption.
Disadvantage - Temporary bridging overhead: Requires developing temporary data conversion bridges and synchronization routines between legacy modules and upgraded modules.
Disadvantage - Prolonged implementation timeline: Transitioning across multiple consecutive phases extends the changeover schedule compared to a direct switch.

Mentor tip

Provide both positive points (risk containment, gradual learning) and negative trade-offs (data synchronisation overhead, prolonged timeline).

Question 5

2 marks

Characteristics of Media Access Control (MAC) addresses

Networks

2 marks

Permanent physical identifier: A unique 48-bit hardware identifier burned into the Network Interface Card (NIC) at manufacture time.
Data Link layer addressing: Identifies physical nodes within a local area network (LAN), expressed as 12 hexadecimal digits (e.g. `00:1A:2B:3C:4D:5E`).

Format: XX:XX:XX:XX:XX:XX (48 bits / 6 octets)

Mentor tip

Distinguish MAC addresses (fixed physical Layer 2 hardware address) from IP addresses (dynamic logical Layer 3 network address).

Question 6

2 marks

Purpose and architectural role of system flowcharts

System fundamentals

2 marks

High-level process flow: Visualizes the broader movement of data and operations across hardware devices, storage media, and processing units.
System boundary overview: Identifies how physical inputs, external databases, processing programs, and printed/electronic outputs interact.

Mentor tip

System flowcharts map systems and hardware media, while program flowcharts detail line-by-line algorithm logic.

Question 7

3 marks

Traffic filtering and perimeter defence via firewalls

Networks

3 marks

Packet inspection: Intercepts and analyses incoming and outgoing network packets at network boundaries.
Filtering rules: Assesses packet headers against access control criteria (source/destination IP addresses, port numbers, protocols).
Access enforcement: Blocks suspicious packets and drops unauthorized connection attempts while allowing legitimate communication.

Rule: (Port = 80 \lor Port = 443) ⟹ PERMIT

Mentor tip

Explain packet header inspection and filtering decisions based on security policies.

Question 8

2 marks

Benefits of low-power mobile electronic devices

System fundamentals

2 marks

Extended battery runtime: Enables prolonged operational use throughout the day without requiring frequent recharging or bulky power banks.
Thermal comfort and portability: Generates minimal internal heat, permitting silent fanless designs and thinner, lightweight form factors.

Mentor tip

Focus on real user benefits: longer runtime between charges and lighter/cooler hardware.

Question 9

4 marks

Trace table for Euclidean greatest common divisor algorithm

Computational thinking, problem-solving and programming

4 marks

Initialization: $A = 20$ , $B = 12$ . Condition $B > 0$ is true.
Iteration 1: $TEMP = 12$ . $B = 20 mod 12 = 8$ . $A = 12$ . Condition $B > 0$ ( $8 > 0$ ) is true.
Iteration 2: $TEMP = 8$ . $B = 12 mod 8 = 4$ . $A = 8$ . Condition $B > 0$ ( $4 > 0$ ) is true.
Iteration 3: $TEMP = 4$ . $B = 8 mod 4 = 0$ . $A = 4$ . Condition $B > 0$ ( $0 > 0$ ) is false.
Loop terminates. Final output is $A = 4$ .

A 201284 B 12840 TEMP - 1284 Output 4

Mentor tip

Track variables column by column through each iteration until the loop condition $B > 0$ fails.

Question 10

2 marks

Structural characteristics of collection data structures

Abstract data structures

2 marks

Dynamic resizing: Collections grow and contract dynamically during runtime as elements are inserted or removed, without fixed capacity constraints.
Iterator-based access: Elements are accessed sequentially using standardized iterator methods (`getNext()`, `hasNext()`, `resetNext()`) rather than direct numeric indexing.

Mentor tip

Mention dynamic size and sequential iterator traversal without index arithmetic.

Questions 11–13 · 45 marks

Question 11

15 marks

Laptop architecture, network interfaces, and packet protocols

System fundamentals
Resource management
Networks

(a)

1 mark

Full-disk encryption: Enabling disk encryption ensures that all saved files remain cryptographically locked if the physical laptop is stolen.

Mentor tip

Full disk encryption ensures data security even if the internal drive is extracted.

(b)

2 marks

Primary memory allocation: Allocates RAM for code instructions and heap/stack variables during execution.
CPU process scheduling: Distributes processor execution time slices across active tasks via multitasking.

Mentor tip

Name memory management or process scheduling and outline its purpose.

(c)

4 marks

Wired Ethernet utility: Offers maximum data throughput, lower latency, and reliable connection stability unaffected by wireless interference.
Wireless Wi-Fi utility: Provides untethered mobility and convenient network connectivity across home, school, and mobile workspaces.
Hybrid rationale: Equipping both interfaces accommodates demanding data transfers when docked and mobile portability on the move.

Mentor tip

Contrast wired throughput and stability against wireless convenience and mobility.

(d)

2 marks

Packet header: Contains addressing and sequencing metadata (source IP, destination IP, packet sequence number, protocol).
Packet payload and trailer: The payload holds the application data chunk; the trailer contains checksum / CRC bits for error detection.

Data Packet = [Header ∣ Payload ∣ Trailer]

Mentor tip

List the header, payload, and trailer along with their contents.

(e)

6 marks

Universal communication standards: Protocols provide agreed syntax and semantics so disparate hardware and operating systems can exchange data.
Data integrity and error control: Detects corrupted packets via checksums and manages acknowledgments and retransmissions.
Congestion management and flow control: Regulates transmission speed to match receiver processing capacity and prevent buffer overflow.

Mentor tip

Cover standardization across vendors, error handling/retransmission, and flow rate regulation.

Question 12

15 marks

Boolean logic design, sensor truth table, and CPU fetch-decode-execute cycle

Computer organisation

(a)(i)

1 mark

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

Operators: or, and

Mentor tip

Logical operators combine Boolean truth values.

(a)(ii)

1 mark

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

Constants: 6, 3, 20

Mentor tip

Constants are unchanging literal numerical values.

(a)(iii)

2 marks

Substitute $X = 6$ and $Y = 6$ :
$(6 > 6)$ evaluates to false.
$(6 > 3)$ evaluates to true.
$(6 + 6 < 20) = (12 < 20)$ evaluates to true.
Combine using precedence (`and` before `or`):
$true \land true = true$ .
$false \lor true = true$ .

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

Mentor tip

Remember that AND takes precedence over OR.

(b)

4 marks

Siren condition: Activates when the door is locked ( $DOOR = 1$ ) AND either movement is detected ( $MOTION = 1$ ) OR the window is open ( $WINDOW = 1$ ).
Formula: $SIREN = DOOR \land (MOTION \lor WINDOW)$ .
Rows with $DOOR = 0$ : Output is 0 for all four combinations.
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

The siren is only 1 when DOOR is 1 and at least one of MOTION or WINDOW is 1.

(c)(i)

1 mark

The Memory Data Register (MDR) holds data transferred between the CPU and primary memory.

Register: Memory Data Register (MDR)

Mentor tip

MDR holds data contents; MAR holds memory addresses.

(c)(ii)

1 mark

The Control Unit (CU) decodes fetched machine instructions.

Component: Control Unit (CU)

Mentor tip

The CU manages the fetch-decode-execute cycle.

(c)(iii)

1 mark

Arithmetic calculations and logical operations are performed in the Arithmetic Logic Unit (ALU).

Component: Arithmetic Logic Unit (ALU)

Mentor tip

All calculations take place inside the ALU.

(c)(iv)

4 marks

Address Bus: Transmits the memory address from the MAR to RAM (unidirectional).
Data Bus: Transmits data and instruction words between the MDR and primary memory (bidirectional).
Control Bus: Transmits command, timing, and status signals (e.g. read/write, clock pulses) between the Control Unit and memory/peripherals.
Execution coordination: The CPU places addresses on the address bus, asserts control signals on the control bus, and exchanges data over the data bus.

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

Mentor tip

Outline the role and transmission direction of each of the three buses.

Question 13

15 marks

Array statistical processing: mean calculation and median sorting algorithms

Computational thinking, problem-solving and programming

(a)

4 marks

Initialize an accumulator variable `SUM = 0`.
Iterate through the array from index 0 to $N - 1$ , adding each element `DATA[I]` to `SUM`.
Divide the total sum by the count of elements $N$ to obtain the arithmetic mean.
Output the calculated mean value.

SUM = 0 loop I from 0 to N - 1 SUM = SUM + DATA [I] end loop MEAN = SUM / N output (MEAN)

Mentor tip

Accumulate the sum first, then perform the division once outside the loop to avoid redundant divisions.

(b)(i)

1 mark

Bubble Sort (or Selection Sort / Insertion Sort) can sort the array into ascending order.

Algorithm: Bubble Sort

Mentor tip

Name any standard $O (n^{2})$ or $O (n lo g n)$ sorting algorithm from the syllabus.

(b)(ii)

10 marks

Step 1 - Sort the array: Use Bubble Sort with a boolean swapped flag to arrange the $N$ elements into non-decreasing order.
Step 2 - Determine median position: Check if $N$ is odd or even using modular division: $(N mod 2) == 1$ .
Odd length: If $N$ is odd, the middle element is located at index $div (N, 2)$ ; assign $MEDIAN = DATA [div (N, 2)]$ .
Even length: If $N$ is even, find the two middle elements at indices $(N /2) - 1$ and $N /2$ ; compute their arithmetic average: $MEDIAN = (DATA [N /2 - 1] + DATA [N /2]) /2.0$ .
Step 3 - Output result: Print the calculated median.

SWAPPED = true loop while SWAPPED SWAPPED = false loop I from 0 to N - 2 if DATA [I] > DATA [I + 1] then TEMP = DATA [I] DATA [I] = DATA [I + 1] DATA [I + 1] = TEMP SWAPPED = true end if end loop end loop if (N mod 2) == 1 then MID = div (N, 2) MEDIAN = DATA [MID] else MID2 = N /2 MID1 = MID2 - 1 MEDIAN = (DATA [MID1] + DATA [MID2]) /2.0 end if output (MEDIAN)

Mentor tip

Handle both odd and even parity cases: single middle element for odd $N$ , average of two middle elements for even $N$ .

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 programming19 marks · 27%
Q9, Q13
Networks17 marks · 24%
Q5, Q7, Q11
System fundamentals15 marks · 21%
Q1, Q2, Q3, Q4, Q6, Q8, Q11
Computer organisation15 marks · 21%
Q12
Abstract data structures2 marks · 3%
Q10
Resource management2 marks · 3%
Q11

Mentor insights

Traps to watch for

01
Temporary variable caching in Euclidean loop · Q9
Cache B in TEMP before calculating A mod B. Overwriting B prematurely corrupts subsequent variable assignments.
02
Operator precedence in compound Boolean expressions · Q12
Logical AND takes precedence over OR. The condition (Y > 3) and ((X + Y) < 20) evaluates first before being OR-combined with (X > 6).
03
Even-length array median calculation · Q13
When array size N is even, the median is the arithmetic mean of the two middle elements at indices (N/2) - 1 and N/2, requiring floating-point division by 2.0.

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

Want someone to go through your paper with you?

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

Find your maths mentor More past papers