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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2019
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 2019 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

3 marks

Fundamental and compound operations

Computer organisation

(a)

1 mark

Identify one fundamental operation of a computer processor:
Addition / subtraction (ALU arithmetic), logical comparison (AND/OR/NOT), loading data from memory into a register, or storing a register value into memory.

Mentor tip

Name add, compare, load, or store.

(b)

2 marks

Distinguish between fundamental and compound operations:
1) A fundamental operation is an atomic, low-level instruction executed directly by the processor in a single machine cycle (e.g., ADD, LOAD, CMP).
2) A compound operation is a complex operation constructed by combining multiple fundamental operations in sequence (e.g., matrix multiplication, sorting, searching) requiring multiple machine cycles.

Mentor tip

Contrast atomic single-cycle instructions with composite multi-instruction routines.

Question 2

3 marks

Data loss causes and offsite storage prevention

System fundamentals

(a)

1 mark

Identify one cause of data loss:
Human error (accidental deletion/overwriting), hardware component failure (disk head crash), malware infection, or power failure/natural disaster.

Mentor tip

Name accidental deletion or hardware failure.

(b)

2 marks

Describe offsite data storage protection:
1) Backup copies of critical data files are physically transferred to a separate remote geographic facility or securely uploaded to remote cloud servers.
2) In the event of a local catastrophic disaster (such as a fire, flood, hardware theft, or localized ransomware incident), the primary facility's lost data can be restored from the undamaged offsite copy.

Mentor tip

Explain geographical isolation protecting against local disasters.

Question 3

4 marks

Control unit function and cache memory

Computer organisation

(a)

1 mark

State the function of the Control Unit (CU) in the CPU:
Directs and coordinates all hardware operations within the CPU by fetching, decoding, and managing the execution of instructions, and generating timing/control signals across the control bus.

Mentor tip

Mention fetching, decoding instructions, or generating control signals.

(b)

3 marks

Explain the role and architecture of cache memory:
1) Cache is extremely fast, small SRAM located directly on (or very close to) the CPU core.
2) It buffers frequently accessed instructions and recently referenced data copied from slower main memory (RAM).
3) When the processor needs data, it checks cache first (cache hit); accessing cache takes vastly fewer clock cycles than querying RAM across the system bus, substantially minimizing processor idle wait states and boosting overall throughput.

Mentor tip

Mention high-speed SRAM, storing frequently used instructions/data, and reducing RAM access latency.

Question 4

3 marks

Colour representation and hexadecimal

Computer organisation

(a)

1 mark

State why hexadecimal is frequently used in computing:
Hexadecimal provides a much shorter, human-readable representation of binary values (each hex digit represents exactly 4 binary bits / 1 nibble), making memory addresses, color codes, and machine bytes easier to read, write, and debug without transcription errors.

Mentor tip

State that it is a compact, human-friendly representation of binary.

(b)

1 mark

Determine bits used for a non-primary colour (such as yellow, made from red + green + blue channels):
A 24-bit true color representation allocates 8 bits (1 byte) for Red, 8 bits for Green, and 8 bits for Blue: 8 + 8 + 8 = 24 bits.

8 bits (Red) + 8 bits (Green) + 8 bits (Blue) = 24 bits

Mentor tip

State 24 bits (or 3 bytes).

(c)

1 mark

Calculate the maximum number of unique colours in a 24-bit pixel:
Each channel has 256 levels (0 to 255): 256 x 256 x 256 = 2^24 = 16,777,216 distinct colours.

2^{24} = 16777216 unique colours (or 25 6^{3})

Mentor tip

Write 16,777,216 or 2^24.

Question 5

4 marks

Compound logic truth table

Computer organisation

(a)

4 marks

Formulate Boolean expressions for intermediate and output signals:
1) X = A XOR B (1 when A and B differ, 0 when equal).
2) Y = A NOR C = NOT (A OR C) (1 only when both A=0 and C=0).
3) Z = X OR (NOT Y). Note that NOT Y is 1 whenever Y=0.
Evaluate row by row:
Row 1: A=0, B=0, C=0 -> X=0, Y=1 -> NOT Y=0 -> Z = 0 OR 0 = 0
Row 2: A=0, B=0, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 3: A=0, B=1, C=0 -> X=1, Y=1 -> NOT Y=0 -> Z = 1 OR 0 = 1
Row 4: A=0, B=1, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 5: A=1, B=0, C=0 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 6: A=1, B=0, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 7: A=1, B=1, C=0 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 8: A=1, B=1, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1

X = A \oplus B, Y = \overline{A \lor C}, Z = X \lor \overline{Y} A 00001111 B 00110011 C 01010101 X 00111100 Y 10100000 Z 01111111

Mentor tip

Notice that Y=0 for 6 out of 8 rows; whenever Y=0, NOT Y=1, forcing Z=1.

Question 6

3 marks

Abstraction in algorithm design

Computational thinking, problem-solving and programming

(a)

3 marks

Explain abstraction in algorithmic problem solving:
1) Filtering non-essential detail: Abstraction removes irrelevant real-world complexities and physical specifics, isolating only the core logical properties necessary to solve the problem.
2) Managing cognitive complexity: Decomposing large problems into generalized abstract components (procedures, objects, classes) enables programmers to focus on high-level logic without being overwhelmed by low-level implementation details.
3) Reusability and generalization: Abstract algorithms can be applied across different applications and hardware platforms without rewriting fundamental logic.

Mentor tip

Focus on hiding irrelevant details, managing complexity, and creating reusable generalized models.

Question 7

5 marks

Trace table for iterative loop

Computational thinking, problem-solving and programming

(a)

5 marks

Trace the execution step by step:
Initial values: K = 1, N = 1, M = 2.
Iteration 1 (K = 1 < 5):
- output(1, 2)
- K = 1 + 1 = 2
- N = 1 + 2 = 3
- M = 2 * 2 = 4
Iteration 2 (K = 2 < 5):
- output(3, 4)
- K = 2 + 1 = 3
- N = 3 + 2 = 5
- M = 4 * 2 = 8
Iteration 3 (K = 3 < 5):
- output(5, 8)
- K = 3 + 1 = 4
- N = 5 + 2 = 7
- M = 8 * 2 = 16
Iteration 4 (K = 4 < 5):
- output(7, 16)
- K = 4 + 1 = 5
- N = 7 + 2 = 9
- M = 16 * 2 = 32
Loop check: K = 5 is not < 5. Loop terminates.
Final variables: K=5, N=9, M=32. Outputs produced: (1,2), (3,4), (5,8), (7,16).

K 12345 N 13579 M 2481632 OUTPUT 1, 2 3, 4 5, 8 7, 16

Mentor tip

Record values on the exact rows they are updated, and list outputs in sequence.

Questions 8–10 · 45 marks

Question 8

15 marks

System implementation and testing

System fundamentals

(a)

2 marks

Identify two organizational issues related to new system implementation:
1) User resistance to change and requirement for comprehensive employee retraining programs.
2) Temporary disruption to operational workflows, data migration overhead, or business restructuring.

Mentor tip

Mention user retraining and workflow disruption/resistance.

(b)

5 marks

Evaluate phased conversion vs direct changeover:
1) Phased conversion:
- Advantage: Implemented module by module or department by department; allows staff to adapt gradually, and if an issue occurs in a newly deployed module, the rest of the business continues operating normally.
- Disadvantage: Prolonged implementation period; complex temporary data bridging and communication interfaces must be maintained between old and new subsystems.
2) Direct changeover:
- Advantage: Immediate transition at a fixed cut-over date; low operational cost with zero duplication of effort and no bridging interfaces needed.
- Disadvantage: Extremely high risk; if the new system contains catastrophic flaws, operations halt immediately with no fallback system available.
Conclusion: Phased conversion is much safer for critical enterprise systems, whereas direct changeover is only suitable for non-critical systems with complete confidence in testing.

Mentor tip

Provide clear advantages and disadvantages for both strategies and give a concluding comparative judgment.

(c)

4 marks

(i) State one type of testing that involves users (1 mark):

Beta testing (or End-user operational testing / UAT).

(ii) Identify three consequences of inadequate testing (3 marks):

1) Undetected critical software bugs causing frequent system crashes, data corruption, or permanent data loss.

2) Severe financial losses due to operational downtime, customer churn, and emergency development bug-fixing costs.

3) Reputational damage, loss of client trust, and potential legal liabilities from failing service level agreements (SLAs).

Mentor tip

Name Beta or UAT for part (i), then list 3 distinct business/technical impacts for (ii).

(d)

4 marks

Discuss social and ethical issues associated with implementing a new computer system:
1) Social issues: Potential workforce redundancies or role displacements due to automation, leading to employee stress and job insecurity; digital divide challenges for staff struggling to adapt to digital tooling.
2) Ethical issues: Privacy concerns surrounding employee monitoring (e.g., tracking keystrokes, active hours, or location); ethical responsibility to protect sensitive corporate/customer data gathered by new platforms against data breaches.

Mentor tip

Address both social impacts (job loss, retraining stress) and ethical impacts (privacy, monitoring).

Question 9

15 marks

Network hardware, protocols, and security

Networks

(a)

4 marks

Outline two advantages of a school using a computer network:
1) Resource and hardware sharing: Multiple classrooms can share expensive peripheral equipment (centralized high-volume printers, scanners) and central network storage drives, substantially lowering hardware costs.
2) Centralized data management and collaboration: Central servers allow automated administrative backups, centralized software licensing, collaborative document editing, and controlled student access to shared educational resources from any school workstation.

Mentor tip

Detail how sharing hardware or centralized administration directly benefits school operations.

(b)

4 marks

Describe the purpose of network hardware components:
(i) Router (2 marks): An internetworking device that connects disparate networks (e.g., school LAN to the external Internet); it inspects destination IP addresses in incoming packets and forwards them along the optimal path using routing tables.
(ii) Network Interface Card / NIC (2 marks): A hardware controller inside a device that provides the physical and electrical interface to the network medium (Ethernet cable or Wi-Fi radio); it formats data into frames and contains the unique physical MAC address.

Mentor tip

Associate Router with IP routing between networks, and NIC with physical media connection and MAC addressing.

(c)

2 marks

Outline why protocols are necessary in network communications:
1) Protocols establish standardized syntactic, semantic, and synchronization rules for data transmission.
2) This guarantees that disparate computing hardware and operating systems can exchange data reliably, detect transmission errors, and manage flow control without misinterpreting transmitted signals.

Mentor tip

Mention standard rules, interoperability across different vendors, and error detection.

(d)

1 mark

Define data encryption:
The process of encoding plaintext data into ciphertext using a mathematical algorithm and an encryption key, such that unauthorized parties intercepting the transmission cannot read or decipher the data without the corresponding decryption key.

Mentor tip

Mention transforming plaintext into ciphertext using a key.

(e)

4 marks

Evaluate trusted MAC address filtering as a network security method:
1) How it works & Advantages (2 marks): The network access point maintains an authorized whitelist of physical MAC addresses; devices not on the approved whitelist are denied connection, providing a simple first-line perimeter defense against casual unauthorized users.
2) Limitations & Vulnerabilities (2 marks): MAC addresses are transmitted in unencrypted plaintext in every frame header; attackers running network sniffers can easily capture an authorized MAC address and spoof their own device's MAC address to bypass the filter. Additionally, administrative overhead is high when adding new legitimate devices.

Mentor tip

Contrast simple access restriction against MAC address spoofing vulnerabilities.

Question 10

15 marks

Integer arithmetic and array sorting algorithms

Computational thinking, problem-solving and programming

(a)

3 marks

Trace the DIV and MOD algorithm for N = 372:
1) X = N DIV 100 = 372 DIV 100 = 3 (integer division quotient).
2) Y calculation:
- N MOD 100 = 372 MOD 100 = 72.
- 72 DIV 10 = 7.
- Y = X + 10 * 7 = 3 + 70 = 73.
3) Z calculation:
- N MOD 10 = 372 MOD 10 = 2.
- Z = Y + 2 * 100 = 73 + 200 = 273.
Result: X = 3, Y = 73, Z = 273.

X = 372 DIV 100 = 3 Y = 3 + 10 \times (72 DIV 10) = 3 + 70 = 73 Z = 73 + (372 MOD 10) \times 100 = 73 + 200 = 273

Mentor tip

Evaluate sub-expressions inside parentheses first.

(b)

8 marks

Construct algorithm to extract 3-digit numbers with strictly increasing digits (hundreds < tens < units) into array THREE:
COUNT = 0
NUMBERS.resetNext()
loop while NUMBERS.hasNext()
NUM = NUMBERS.getNext()
// Verify NUM is a 3-digit positive integer (100 to 999)
if NUM >= 100 and NUM <= 999 then
H = NUM DIV 100
T = (NUM MOD 100) DIV 10
U = NUM MOD 10
if H < T and T < U then
THREE[COUNT] = NUM
COUNT = COUNT + 1
end if
end if
end loop
if COUNT == 0 then
output "No such numbers"
end if

Mentor tip

Use NUM >= 100 and NUM <= 999 to isolate 3-digit numbers before extracting digits.

(c)

4 marks

Describe how selection sort sorts array THREE in ascending order:
1) An outer loop traverses the array from index 0 to N - 2, defining the boundary of the sorted portion.
2) For each position I, assume the minimum value is at index I (MIN_INDEX = I).
3) An inner loop scans the remaining unsorted subarray from index I + 1 to N - 1, comparing each element with THREE[MIN_INDEX] and updating MIN_INDEX whenever a smaller value is found.
4) After scanning the unsorted segment, if MIN_INDEX != I, swap the element at THREE[I] with the element at THREE[MIN_INDEX], placing the minimum item into its final sorted position.

Mentor tip

Clearly describe tracking the minimum index through an inner loop and performing a single swap per outer iteration.

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 programming23 marks · 33%
Q6, Q7, Q10
System fundamentals18 marks · 26%
Q2, Q8
Networks15 marks · 21%
Q9
Computer organisation14 marks · 20%
Q1, Q3, Q4, Q5

Mentor insights

Traps to watch for

01
Truth table NOR gate and NOT gate precedence · Q5
Remember that Y = A NOR C is 1 only when A = 0 and C = 0. For any other combination, Y = 0 so NOT Y = 1, forcing Z = 1.
02
Loop termination condition in trace table · Q7
The loop condition is K < 5. When K reaches 5, the loop terminates immediately without outputting values for K = 5.
03
Integer division (DIV) vs modulus (MOD) in digit extraction · Q10
DIV produces the integer quotient (truncating remainder), while MOD yields the remainder. For N = 372, N DIV 100 gives 3 (hundreds digit), while N MOD 10 gives 2 (units digit).

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 2019
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 2019 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

3 marks

Fundamental and compound operations

Computer organisation

(a)

1 mark

Identify one fundamental operation of a computer processor:
Addition / subtraction (ALU arithmetic), logical comparison (AND/OR/NOT), loading data from memory into a register, or storing a register value into memory.

Mentor tip

Name add, compare, load, or store.

(b)

2 marks

Distinguish between fundamental and compound operations:
1) A fundamental operation is an atomic, low-level instruction executed directly by the processor in a single machine cycle (e.g., ADD, LOAD, CMP).
2) A compound operation is a complex operation constructed by combining multiple fundamental operations in sequence (e.g., matrix multiplication, sorting, searching) requiring multiple machine cycles.

Mentor tip

Contrast atomic single-cycle instructions with composite multi-instruction routines.

Question 2

3 marks

Data loss causes and offsite storage prevention

System fundamentals

(a)

1 mark

Identify one cause of data loss:
Human error (accidental deletion/overwriting), hardware component failure (disk head crash), malware infection, or power failure/natural disaster.

Mentor tip

Name accidental deletion or hardware failure.

(b)

2 marks

Describe offsite data storage protection:
1) Backup copies of critical data files are physically transferred to a separate remote geographic facility or securely uploaded to remote cloud servers.
2) In the event of a local catastrophic disaster (such as a fire, flood, hardware theft, or localized ransomware incident), the primary facility's lost data can be restored from the undamaged offsite copy.

Mentor tip

Explain geographical isolation protecting against local disasters.

Question 3

4 marks

Control unit function and cache memory

Computer organisation

(a)

1 mark

State the function of the Control Unit (CU) in the CPU:
Directs and coordinates all hardware operations within the CPU by fetching, decoding, and managing the execution of instructions, and generating timing/control signals across the control bus.

Mentor tip

Mention fetching, decoding instructions, or generating control signals.

(b)

3 marks

Explain the role and architecture of cache memory:
1) Cache is extremely fast, small SRAM located directly on (or very close to) the CPU core.
2) It buffers frequently accessed instructions and recently referenced data copied from slower main memory (RAM).
3) When the processor needs data, it checks cache first (cache hit); accessing cache takes vastly fewer clock cycles than querying RAM across the system bus, substantially minimizing processor idle wait states and boosting overall throughput.

Mentor tip

Mention high-speed SRAM, storing frequently used instructions/data, and reducing RAM access latency.

Question 4

3 marks

Colour representation and hexadecimal

Computer organisation

(a)

1 mark

State why hexadecimal is frequently used in computing:
Hexadecimal provides a much shorter, human-readable representation of binary values (each hex digit represents exactly 4 binary bits / 1 nibble), making memory addresses, color codes, and machine bytes easier to read, write, and debug without transcription errors.

Mentor tip

State that it is a compact, human-friendly representation of binary.

(b)

1 mark

Determine bits used for a non-primary colour (such as yellow, made from red + green + blue channels):
A 24-bit true color representation allocates 8 bits (1 byte) for Red, 8 bits for Green, and 8 bits for Blue: 8 + 8 + 8 = 24 bits.

8 bits (Red) + 8 bits (Green) + 8 bits (Blue) = 24 bits

Mentor tip

State 24 bits (or 3 bytes).

(c)

1 mark

Calculate the maximum number of unique colours in a 24-bit pixel:
Each channel has 256 levels (0 to 255): 256 x 256 x 256 = 2^24 = 16,777,216 distinct colours.

2^{24} = 16777216 unique colours (or 25 6^{3})

Mentor tip

Write 16,777,216 or 2^24.

Question 5

4 marks

Compound logic truth table

Computer organisation

(a)

4 marks

Formulate Boolean expressions for intermediate and output signals:
1) X = A XOR B (1 when A and B differ, 0 when equal).
2) Y = A NOR C = NOT (A OR C) (1 only when both A=0 and C=0).
3) Z = X OR (NOT Y). Note that NOT Y is 1 whenever Y=0.
Evaluate row by row:
Row 1: A=0, B=0, C=0 -> X=0, Y=1 -> NOT Y=0 -> Z = 0 OR 0 = 0
Row 2: A=0, B=0, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 3: A=0, B=1, C=0 -> X=1, Y=1 -> NOT Y=0 -> Z = 1 OR 0 = 1
Row 4: A=0, B=1, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 5: A=1, B=0, C=0 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 6: A=1, B=0, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 7: A=1, B=1, C=0 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 8: A=1, B=1, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1

X = A \oplus B, Y = \overline{A \lor C}, Z = X \lor \overline{Y} A 00001111 B 00110011 C 01010101 X 00111100 Y 10100000 Z 01111111

Mentor tip

Notice that Y=0 for 6 out of 8 rows; whenever Y=0, NOT Y=1, forcing Z=1.

Question 6

3 marks

Abstraction in algorithm design

Computational thinking, problem-solving and programming

(a)

3 marks

Explain abstraction in algorithmic problem solving:
1) Filtering non-essential detail: Abstraction removes irrelevant real-world complexities and physical specifics, isolating only the core logical properties necessary to solve the problem.
2) Managing cognitive complexity: Decomposing large problems into generalized abstract components (procedures, objects, classes) enables programmers to focus on high-level logic without being overwhelmed by low-level implementation details.
3) Reusability and generalization: Abstract algorithms can be applied across different applications and hardware platforms without rewriting fundamental logic.

Mentor tip

Focus on hiding irrelevant details, managing complexity, and creating reusable generalized models.

Question 7

5 marks

Trace table for iterative loop

Computational thinking, problem-solving and programming

(a)

5 marks

Trace the execution step by step:
Initial values: K = 1, N = 1, M = 2.
Iteration 1 (K = 1 < 5):
- output(1, 2)
- K = 1 + 1 = 2
- N = 1 + 2 = 3
- M = 2 * 2 = 4
Iteration 2 (K = 2 < 5):
- output(3, 4)
- K = 2 + 1 = 3
- N = 3 + 2 = 5
- M = 4 * 2 = 8
Iteration 3 (K = 3 < 5):
- output(5, 8)
- K = 3 + 1 = 4
- N = 5 + 2 = 7
- M = 8 * 2 = 16
Iteration 4 (K = 4 < 5):
- output(7, 16)
- K = 4 + 1 = 5
- N = 7 + 2 = 9
- M = 16 * 2 = 32
Loop check: K = 5 is not < 5. Loop terminates.
Final variables: K=5, N=9, M=32. Outputs produced: (1,2), (3,4), (5,8), (7,16).

K 12345 N 13579 M 2481632 OUTPUT 1, 2 3, 4 5, 8 7, 16

Mentor tip

Record values on the exact rows they are updated, and list outputs in sequence.

Questions 8–10 · 45 marks

Question 8

15 marks

System implementation and testing

System fundamentals

(a)

2 marks

Identify two organizational issues related to new system implementation:
1) User resistance to change and requirement for comprehensive employee retraining programs.
2) Temporary disruption to operational workflows, data migration overhead, or business restructuring.

Mentor tip

Mention user retraining and workflow disruption/resistance.

(b)

5 marks

Evaluate phased conversion vs direct changeover:
1) Phased conversion:
- Advantage: Implemented module by module or department by department; allows staff to adapt gradually, and if an issue occurs in a newly deployed module, the rest of the business continues operating normally.
- Disadvantage: Prolonged implementation period; complex temporary data bridging and communication interfaces must be maintained between old and new subsystems.
2) Direct changeover:
- Advantage: Immediate transition at a fixed cut-over date; low operational cost with zero duplication of effort and no bridging interfaces needed.
- Disadvantage: Extremely high risk; if the new system contains catastrophic flaws, operations halt immediately with no fallback system available.
Conclusion: Phased conversion is much safer for critical enterprise systems, whereas direct changeover is only suitable for non-critical systems with complete confidence in testing.

Mentor tip

Provide clear advantages and disadvantages for both strategies and give a concluding comparative judgment.

(c)

4 marks

(i) State one type of testing that involves users (1 mark):

Beta testing (or End-user operational testing / UAT).

(ii) Identify three consequences of inadequate testing (3 marks):

1) Undetected critical software bugs causing frequent system crashes, data corruption, or permanent data loss.

2) Severe financial losses due to operational downtime, customer churn, and emergency development bug-fixing costs.

3) Reputational damage, loss of client trust, and potential legal liabilities from failing service level agreements (SLAs).

Mentor tip

Name Beta or UAT for part (i), then list 3 distinct business/technical impacts for (ii).

(d)

4 marks

Discuss social and ethical issues associated with implementing a new computer system:
1) Social issues: Potential workforce redundancies or role displacements due to automation, leading to employee stress and job insecurity; digital divide challenges for staff struggling to adapt to digital tooling.
2) Ethical issues: Privacy concerns surrounding employee monitoring (e.g., tracking keystrokes, active hours, or location); ethical responsibility to protect sensitive corporate/customer data gathered by new platforms against data breaches.

Mentor tip

Address both social impacts (job loss, retraining stress) and ethical impacts (privacy, monitoring).

Question 9

15 marks

Network hardware, protocols, and security

Networks

(a)

4 marks

Outline two advantages of a school using a computer network:
1) Resource and hardware sharing: Multiple classrooms can share expensive peripheral equipment (centralized high-volume printers, scanners) and central network storage drives, substantially lowering hardware costs.
2) Centralized data management and collaboration: Central servers allow automated administrative backups, centralized software licensing, collaborative document editing, and controlled student access to shared educational resources from any school workstation.

Mentor tip

Detail how sharing hardware or centralized administration directly benefits school operations.

(b)

4 marks

Describe the purpose of network hardware components:
(i) Router (2 marks): An internetworking device that connects disparate networks (e.g., school LAN to the external Internet); it inspects destination IP addresses in incoming packets and forwards them along the optimal path using routing tables.
(ii) Network Interface Card / NIC (2 marks): A hardware controller inside a device that provides the physical and electrical interface to the network medium (Ethernet cable or Wi-Fi radio); it formats data into frames and contains the unique physical MAC address.

Mentor tip

Associate Router with IP routing between networks, and NIC with physical media connection and MAC addressing.

(c)

2 marks

Outline why protocols are necessary in network communications:
1) Protocols establish standardized syntactic, semantic, and synchronization rules for data transmission.
2) This guarantees that disparate computing hardware and operating systems can exchange data reliably, detect transmission errors, and manage flow control without misinterpreting transmitted signals.

Mentor tip

Mention standard rules, interoperability across different vendors, and error detection.

(d)

1 mark

Define data encryption:
The process of encoding plaintext data into ciphertext using a mathematical algorithm and an encryption key, such that unauthorized parties intercepting the transmission cannot read or decipher the data without the corresponding decryption key.

Mentor tip

Mention transforming plaintext into ciphertext using a key.

(e)

4 marks

Evaluate trusted MAC address filtering as a network security method:
1) How it works & Advantages (2 marks): The network access point maintains an authorized whitelist of physical MAC addresses; devices not on the approved whitelist are denied connection, providing a simple first-line perimeter defense against casual unauthorized users.
2) Limitations & Vulnerabilities (2 marks): MAC addresses are transmitted in unencrypted plaintext in every frame header; attackers running network sniffers can easily capture an authorized MAC address and spoof their own device's MAC address to bypass the filter. Additionally, administrative overhead is high when adding new legitimate devices.

Mentor tip

Contrast simple access restriction against MAC address spoofing vulnerabilities.

Question 10

15 marks

Integer arithmetic and array sorting algorithms

Computational thinking, problem-solving and programming

(a)

3 marks

Trace the DIV and MOD algorithm for N = 372:
1) X = N DIV 100 = 372 DIV 100 = 3 (integer division quotient).
2) Y calculation:
- N MOD 100 = 372 MOD 100 = 72.
- 72 DIV 10 = 7.
- Y = X + 10 * 7 = 3 + 70 = 73.
3) Z calculation:
- N MOD 10 = 372 MOD 10 = 2.
- Z = Y + 2 * 100 = 73 + 200 = 273.
Result: X = 3, Y = 73, Z = 273.

X = 372 DIV 100 = 3 Y = 3 + 10 \times (72 DIV 10) = 3 + 70 = 73 Z = 73 + (372 MOD 10) \times 100 = 73 + 200 = 273

Mentor tip

Evaluate sub-expressions inside parentheses first.

(b)

8 marks

Construct algorithm to extract 3-digit numbers with strictly increasing digits (hundreds < tens < units) into array THREE:
COUNT = 0
NUMBERS.resetNext()
loop while NUMBERS.hasNext()
NUM = NUMBERS.getNext()
// Verify NUM is a 3-digit positive integer (100 to 999)
if NUM >= 100 and NUM <= 999 then
H = NUM DIV 100
T = (NUM MOD 100) DIV 10
U = NUM MOD 10
if H < T and T < U then
THREE[COUNT] = NUM
COUNT = COUNT + 1
end if
end if
end loop
if COUNT == 0 then
output "No such numbers"
end if

Mentor tip

Use NUM >= 100 and NUM <= 999 to isolate 3-digit numbers before extracting digits.

(c)

4 marks

Describe how selection sort sorts array THREE in ascending order:
1) An outer loop traverses the array from index 0 to N - 2, defining the boundary of the sorted portion.
2) For each position I, assume the minimum value is at index I (MIN_INDEX = I).
3) An inner loop scans the remaining unsorted subarray from index I + 1 to N - 1, comparing each element with THREE[MIN_INDEX] and updating MIN_INDEX whenever a smaller value is found.
4) After scanning the unsorted segment, if MIN_INDEX != I, swap the element at THREE[I] with the element at THREE[MIN_INDEX], placing the minimum item into its final sorted position.

Mentor tip

Clearly describe tracking the minimum index through an inner loop and performing a single swap per outer iteration.

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 programming23 marks · 33%
Q6, Q7, Q10
System fundamentals18 marks · 26%
Q2, Q8
Networks15 marks · 21%
Q9
Computer organisation14 marks · 20%
Q1, Q3, Q4, Q5

Mentor insights

Traps to watch for

01
Truth table NOR gate and NOT gate precedence · Q5
Remember that Y = A NOR C is 1 only when A = 0 and C = 0. For any other combination, Y = 0 so NOT Y = 1, forcing Z = 1.
02
Loop termination condition in trace table · Q7
The loop condition is K < 5. When K reaches 5, the loop terminates immediately without outputting values for K = 5.
03
Integer division (DIV) vs modulus (MOD) in digit extraction · Q10
DIV produces the integer quotient (truncating remainder), while MOD yields the remainder. For N = 372, N DIV 100 gives 3 (hundreds digit), while N MOD 10 gives 2 (units digit).

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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