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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: May 2022
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 14

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–11 · 25 marks

Question 1

2 marks

Compatibility issues arising from international corporate mergers

System fundamentals

2 marks

Character encoding and language barriers: Merging entities across countries frequently store records in divergent character sets (e.g. ASCII vs UTF-8 vs local character encodings) or differing localized date and currency representations, inhibiting seamless database integration.
Disparate software frameworks and legal compliance: Branch offices often employ distinct enterprise software platforms, database engines, or legacy hardware that cannot exchange structured data without complex middleware, compounded by varying regional data retention regulations.

Mentor tip

Focus on data format discrepancies and differing software platforms across international boundaries.

Question 2

2 marks

Necessity of comprehensive user documentation

System fundamentals

2 marks

Operational efficiency: Accurate documentation guides users through daily features, reducing workflow errors and minimizing staff onboarding time.
Reduced technical support overhead: Clear troubleshooting procedures allow users to resolve common operational issues independently, minimizing IT support tickets.

Mentor tip

Explain how accurate documentation benefits both the end user and organizational productivity.

Question 3

2 marks

Structural distinctions between WAN and LAN

Networks

2 marks

Geographic scope: A LAN spans a localized physical area such as a single room, school campus, or office building, whereas a WAN connects multiple dispersed sites across cities, countries, or continents.
Infrastructure ownership: LAN cabling and switches are typically owned and maintained by the private organization, whereas WAN communication relies on leased telecommunications infrastructure and public carrier networks.

Mentor tip

Highlight differences in geographic span and infrastructure ownership.

Question 4

2 marks

Role of data compression in network transmission

Networks

2 marks

Data compression reduces the total number of bits required to represent a payload, minimizing file size before transmission.
This reduction lowers transmission duration, conserves limited network bandwidth, and mitigates congestion across communication channels.

Mentor tip

Connect the reduction in bit volume directly to faster transmission speed and reduced bandwidth consumption.

Question 5

2 marks

Need for program translation from high-level source to machine code

Computer organisation

2 marks

Processors execute instructions strictly in binary machine code consisting of binary opcodes and operands directly interpretable by hardware logic gates.
High-level programming languages use human-readable abstractions, mathematical notation, and English keywords that the CPU cannot parse without a compiler or interpreter translating them into native machine instructions.

Mentor tip

Emphasize that the CPU only understands binary voltage patterns, necessitating translation of human-readable abstractions.

Question 6

4 marks

Truth table evaluation for composite logic circuit

Computer organisation

4 marks

The circuit receives inputs A, B, and C. The intermediate node P is produced by an XOR gate receiving inputs A and B: P = A XOR B.
The intermediate node Q is generated by an inverter (NOT gate) applied to input C: Q = NOT C.
The final output X is generated by a NAND gate combining intermediate signals P and Q: X = NAND(P, Q) = NOT(P AND Q).
Row-by-row evaluation for all 8 input combinations:
1) A=0, B=0, C=0 -> P=0, Q=1 -> X = NOT(0 AND 1) = 1
2) A=0, B=0, C=1 -> P=0, Q=0 -> X = NOT(0 AND 0) = 1
3) A=0, B=1, C=0 -> P=1, Q=1 -> X = NOT(1 AND 1) = 0
4) A=0, B=1, C=1 -> P=1, Q=0 -> X = NOT(1 AND 0) = 1
5) A=1, B=0, C=0 -> P=1, Q=1 -> X = NOT(1 AND 1) = 0
6) A=1, B=0, C=1 -> P=1, Q=0 -> X = NOT(1 AND 0) = 1
7) A=1, B=1, C=0 -> P=0, Q=1 -> X = NOT(0 AND 1) = 1
8) A=1, B=1, C=1 -> P=0, Q=0 -> X = NOT(0 AND 0) = 1

P = A \oplus B Q = \neg C X = \neg (P \land Q)

Mentor tip

Track intermediate columns P and Q systematically before resolving the NAND gate.

Question 7

4 marks

Number base conversions: Denary to Binary and Hexadecimal

Computer organisation

(a)

2 marks

Convert denary 105 to 8-bit unsigned binary by decomposing into powers of 2:
105 >= 64: bit 6 is 1 (remainder 105 - 64 = 41)
41 >= 32: bit 5 is 1 (remainder 41 - 32 = 9)
9 < 16: bit 4 is 0
9 >= 8: bit 3 is 1 (remainder 9 - 8 = 1)
1 < 4: bit 2 is 0; 1 < 2: bit 1 is 0; 1 >= 1: bit 0 is 1
Combining bits: 01101001 (or 1101001).

10 5_{10} = 64 + 32 + 8 + 1 = 0110100 1_{2}

Mentor tip

List powers of 2 (128, 64, 32, 16, 8, 4, 2, 1) and subtract greedily.

(b)

2 marks

Convert denary 200 to hexadecimal by successive division by 16:
200 / 16 = 12 with remainder 8.
In base 16, quotient 12 corresponds to hex digit C.
The remainder 8 corresponds to hex digit 8.
Combining digits yields C8.

200 = 12 \times 1 6^{1} + 8 \times 1 6^{0} = C8_{16}

Mentor tip

Remember that denary 10 to 15 map to hex letters A through F; 12 maps to C.

Question 8

1 mark

Core role of a single-user operating system

Computer organisation

1 mark

Hardware resource allocation and user interface provisioning: Managing local hardware components (CPU, memory, storage, peripherals) while providing an interactive desktop environment dedicated to one user at a time.

Mentor tip

State resource management or hardware abstraction for a single concurrent user.

Question 9

1 mark

Definition of system usability

System fundamentals

1 mark

Usability denotes the degree of effectiveness, efficiency, and user satisfaction with which specific users can achieve specified goals in a system.

Mentor tip

Mention effectiveness, efficiency, or learnability/satisfaction.

Question 10

2 marks

Significance of international standards in networking

Networks

2 marks

Interoperability: Standards (such as IEEE 802.3 Ethernet and TCP/IP) enable hardware and software from competing manufacturers to communicate seamlessly.
Scalability and market competition: Common protocols allow organizations to upgrade infrastructure modularly without proprietary vendor lock-in.

Mentor tip

Emphasize multi-vendor interoperability and universal compatibility.

Question 11

3 marks

Impact of CPU cache memory on overall system performance

Computer organisation

3 marks

Architectural placement: Cache memory is ultra-fast static RAM (SRAM) positioned directly on or adjacent to the CPU die.
Latency reduction: It stores frequently executed instructions and recently accessed data, operating significantly faster than dynamic main RAM.
Throughput enhancement: When the CPU experiences a cache hit, memory fetch latency is drastically minimized, eliminating idle wait states during instruction fetch cycles.

Mentor tip

Explain speed disparity between CPU and RAM and how cache hits eliminate processor idle time.

Questions 12–14 · 45 marks

Question 12

15 marks

Enterprise system upgrade planning and implementation

System fundamentals

(a)(i)

1 mark

End-users / operational staff who interact with the system interface on a daily basis.

Mentor tip

Other valid stakeholders include system administrators, department managers, or clients.

(a)(ii)

2 marks

Excluding key stakeholders leads to incomplete requirement specifications, causing the finished system to lack essential practical features.
This discrepancy results in user resistance, low adoption rates, and costly post-deployment rework.

Mentor tip

Connect stakeholder absence to unmet functional requirements and operational disruption.

(b)

4 marks

Interviews with departmental representatives: Conduct structured one-on-one sessions to gather qualitative insights into workflow bottlenecks, specific data requirements, and usability pain points.
Direct observation of workplace operations: Shadow employees during active shifts to witness actual system usage, identifying unstated workarounds and inefficiencies that users might omit in interviews.

Mentor tip

Pair a direct inquiry method (interviews/surveys) with an empirical observation technique.

(c)

2 marks

Continuous testing detects architectural flaws, logic bugs, and specification deviations early in development.
Resolving bugs during initial module construction prevents errors from compounding into subsequent integration phases, substantially reducing overall remediation expenses.

Mentor tip

Emphasize early defect detection and reduced remediation cost compared to post-release fixes.

(d)

6 marks

Method 1: Direct Changeover (Immediate cutover)
- Benefit: Minimizes operational duplication; immediate migration eliminates running costs of maintaining two parallel systems.
- Drawback: High risk; if unforeseen critical defects emerge in production, operations can halt with no immediate fallback.
Method 2: Parallel Running (Concurrent operation)
- Benefit: Maximum safety; the legacy system remains fully operational alongside the new platform, providing a secure safety net and seamless data verification.
- Drawback: Costly and labor-intensive; staff must enter duplicate transactions into both platforms simultaneously, doubling workload and server expenses.

Mentor tip

Evaluate benefits and drawbacks for two distinct strategies (e.g. direct vs parallel or phased vs pilot).

Question 13

15 marks

Parallel array data manipulation and grade assignment algorithm

Computational thinking, problem-solving and programming

(a)

2 marks

A relational or Boolean condition: An expression that evaluates to either true or false by comparing variables or constants using relational operators.
Branching execution blocks: Instructions enclosed in THEN and ELSE blocks that execute selectively based on the outcome of the Boolean evaluation.

Mentor tip

Identify the Boolean conditional expression and the corresponding conditional execution path.

(b)

5 marks

Loop index I from 0 to 29 across all 30 students.
Examine MARK[I] against grade thresholds in descending order:
- If MARK[I] >= 80, set GRADE[I] = 'Distinction'
- Else if MARK[I] >= 60, set GRADE[I] = 'Merit'
- Else if MARK[I] >= 40, set GRADE[I] = 'Pass'
- Else set GRADE[I] = 'Fail'
Close conditional branches and loop.

loop I from 0 to 29 if M A R K [I] \geq 80 then GR A D E [I] = Distinction else if M A R K [I] \geq 60 then GR A D E [I] = Merit else if M A R K [I] \geq 40 then GR A D E [I] = Pass else GR A D E [I] = Fail end if end loop

Mentor tip

Check boundaries in descending order (80, then 60, then 40) so each student lands in the correct tier.

(c)

2 marks

Positional indexing: The arrays are aligned so that elements stored at identical index positions across NAME[], MARK[], and GRADE[] describe the exact same student.
For any index K, NAME[K] holds the student name, MARK[K] stores their numeric score, and GRADE[K] records their assigned classification.

Mentor tip

State clearly that common subscripts link corresponding data items across parallel arrays.

(d)

3 marks

Construct a loop iterating I from 0 to 29.
Within the loop, evaluate whether GRADE[I] equals 'Merit' OR GRADE[I] equals 'Distinction'.
If true, output NAME[I] alongside GRADE[I].

loop I from 0 to 29 if GR A D E [I] = Merit or GR A D E [I] = Distinction then output N A M E [I], ": ", GR A D E [I] end if end loop

Mentor tip

Use a logical OR condition comparing GRADE[I] against both required string values.

(e)

3 marks

Prompt and read user input into a target string variable, e.g. SEARCH_GRADE.
Iterate through the array checking the condition GRADE[I] = SEARCH_GRADE instead of a hardcoded string.
When a match occurs, output NAME[I] and MARK[I] as required.

Mentor tip

Outline three modifications: receiving user input, updating the comparison condition, and changing the output fields.

Question 14

15 marks

Searching algorithms: Sequential search vs Binary search analysis

Computational thinking, problem-solving and programming

(a)

4 marks

Trace Algorithm 1 (Linear Search) on DATA_ARR for TO_FIND = 39:
Array values: [0]:27, [1]:29, [2]:36, [3]:39, [4]:40, [5]:42, ...
Initialization: LIMIT = 11, LOC = FALSE, ITERATE = 0
- Iteration 0: DATA_ARR[0] = 27 != 39 -> LOC remains FALSE, ITERATE becomes 1
- Iteration 1: DATA_ARR[1] = 29 != 39 -> LOC remains FALSE, ITERATE becomes 2
- Iteration 2: DATA_ARR[2] = 36 != 39 -> LOC remains FALSE, ITERATE becomes 3
- Iteration 3: DATA_ARR[3] = 39 == 39 -> LOC becomes TRUE, ITERATE becomes 4
Loop condition 'not LOC' fails because LOC is TRUE.
Output: '39 is in the array'.

Mentor tip

Record values across all columns for each iteration until the match terminates the loop.

(b)

1 mark

The constant identifier in Algorithm 1 is LIMIT (set to 11).

Mentor tip

Identify LIMIT as the explicitly assigned unchanging boundary value.

(c)

4 marks

Trace Algorithm 2 (Binary Search) on DATA_ARR for TO_FIND = 50:
Array indices 0 to 11: [27, 29, 36, 39, 40, 42, 45, 50, 58, 60, 61, 68]
Initial state: LOW_LIM = 0, UP_LIM = 11, LOC = FALSE
- Pass 1: MID_VAL = (0 + 11) div 2 = 5.
DATA_ARR[5] = 42. Since 50 > 42, LOW_LIM = 5 + 1 = 6.
- Pass 2: LOW_LIM = 6, UP_LIM = 11.
MID_VAL = (6 + 11) div 2 = 17 div 2 = 8.
DATA_ARR[8] = 58. Since 50 < 58, UP_LIM = 8 - 1 = 7.
- Pass 3: LOW_LIM = 6, UP_LIM = 7.
MID_VAL = (6 + 7) div 2 = 13 div 2 = 6.
DATA_ARR[6] = 45. Since 50 > 45, LOW_LIM = 6 + 1 = 7.
- Pass 4: LOW_LIM = 7, UP_LIM = 7.
MID_VAL = (7 + 7) div 2 = 7.
DATA_ARR[7] = 50. Since 50 == DATA_ARR[7], LOC = TRUE.
Loop terminates. Output: '50 is in the array'.

MID_VAL = ⌊(LOW_LIM + UP_LIM) /2 ⌋

Mentor tip

Use integer division (div) to calculate MID_VAL accurately on each pass.

(d)

1 mark

MID_VAL dynamically recalculates on every iteration of the binary search loop as LOW_LIM and UP_LIM shift to bisect the search interval.

Mentor tip

Explain that its value must vary across iterations based on bounding pointers.

(e)

5 marks

Sequential Search:
- Advantage: Operates on unsorted data without preprocessing; straightforward to construct.
- Disadvantage: Inefficient on large datasets; worst-case time complexity is linear O(N).
Binary Search:
- Advantage: Extremely efficient for large collections; logarithmic time complexity O(log N) halves the search space on each comparison.
- Disadvantage: Strictly requires the dataset to be sorted beforehand; maintaining sorted order incurs overhead during insertions and deletions.

O (N) vs O (lo g_{2} N)

Mentor tip

Contrast the requirement of pre-sorted data against execution efficiency across varying dataset scales.

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 programming30 marks · 43%
Q13, Q14
System fundamentals20 marks · 29%
Q1, Q2, Q9, Q12
Computer organisation14 marks · 20%
Q5, Q6, Q7, Q8, Q11
Networks6 marks · 9%
Q3, Q4, Q10

Mentor insights

Traps to watch for

01
NAND truth evaluation with inverted input · Q6
Remember that Q is the inverse of C. When C=0, Q=1. The NAND gate outputs 0 only when P=1 and Q=1.
02
Binary search integer midpoint truncation · Q14
Always use integer floor division: (0 + 11) div 2 = 5, not 5.5 or 6. Rounding up shifts indices and disrupts midpoint evaluations.
03
Parallel array index synchronization · Q13
Every operation modifying or reading student attributes must use the identical loop index I across all three parallel arrays.

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: May 2022
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 14

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–11 · 25 marks

Question 1

2 marks

Compatibility issues arising from international corporate mergers

System fundamentals

2 marks

Character encoding and language barriers: Merging entities across countries frequently store records in divergent character sets (e.g. ASCII vs UTF-8 vs local character encodings) or differing localized date and currency representations, inhibiting seamless database integration.
Disparate software frameworks and legal compliance: Branch offices often employ distinct enterprise software platforms, database engines, or legacy hardware that cannot exchange structured data without complex middleware, compounded by varying regional data retention regulations.

Mentor tip

Focus on data format discrepancies and differing software platforms across international boundaries.

Question 2

2 marks

Necessity of comprehensive user documentation

System fundamentals

2 marks

Operational efficiency: Accurate documentation guides users through daily features, reducing workflow errors and minimizing staff onboarding time.
Reduced technical support overhead: Clear troubleshooting procedures allow users to resolve common operational issues independently, minimizing IT support tickets.

Mentor tip

Explain how accurate documentation benefits both the end user and organizational productivity.

Question 3

2 marks

Structural distinctions between WAN and LAN

Networks

2 marks

Geographic scope: A LAN spans a localized physical area such as a single room, school campus, or office building, whereas a WAN connects multiple dispersed sites across cities, countries, or continents.
Infrastructure ownership: LAN cabling and switches are typically owned and maintained by the private organization, whereas WAN communication relies on leased telecommunications infrastructure and public carrier networks.

Mentor tip

Highlight differences in geographic span and infrastructure ownership.

Question 4

2 marks

Role of data compression in network transmission

Networks

2 marks

Data compression reduces the total number of bits required to represent a payload, minimizing file size before transmission.
This reduction lowers transmission duration, conserves limited network bandwidth, and mitigates congestion across communication channels.

Mentor tip

Connect the reduction in bit volume directly to faster transmission speed and reduced bandwidth consumption.

Question 5

2 marks

Need for program translation from high-level source to machine code

Computer organisation

2 marks

Processors execute instructions strictly in binary machine code consisting of binary opcodes and operands directly interpretable by hardware logic gates.
High-level programming languages use human-readable abstractions, mathematical notation, and English keywords that the CPU cannot parse without a compiler or interpreter translating them into native machine instructions.

Mentor tip

Emphasize that the CPU only understands binary voltage patterns, necessitating translation of human-readable abstractions.

Question 6

4 marks

Truth table evaluation for composite logic circuit

Computer organisation

4 marks

The circuit receives inputs A, B, and C. The intermediate node P is produced by an XOR gate receiving inputs A and B: P = A XOR B.
The intermediate node Q is generated by an inverter (NOT gate) applied to input C: Q = NOT C.
The final output X is generated by a NAND gate combining intermediate signals P and Q: X = NAND(P, Q) = NOT(P AND Q).
Row-by-row evaluation for all 8 input combinations:
1) A=0, B=0, C=0 -> P=0, Q=1 -> X = NOT(0 AND 1) = 1
2) A=0, B=0, C=1 -> P=0, Q=0 -> X = NOT(0 AND 0) = 1
3) A=0, B=1, C=0 -> P=1, Q=1 -> X = NOT(1 AND 1) = 0
4) A=0, B=1, C=1 -> P=1, Q=0 -> X = NOT(1 AND 0) = 1
5) A=1, B=0, C=0 -> P=1, Q=1 -> X = NOT(1 AND 1) = 0
6) A=1, B=0, C=1 -> P=1, Q=0 -> X = NOT(1 AND 0) = 1
7) A=1, B=1, C=0 -> P=0, Q=1 -> X = NOT(0 AND 1) = 1
8) A=1, B=1, C=1 -> P=0, Q=0 -> X = NOT(0 AND 0) = 1

P = A \oplus B Q = \neg C X = \neg (P \land Q)

Mentor tip

Track intermediate columns P and Q systematically before resolving the NAND gate.

Question 7

4 marks

Number base conversions: Denary to Binary and Hexadecimal

Computer organisation

(a)

2 marks

Convert denary 105 to 8-bit unsigned binary by decomposing into powers of 2:
105 >= 64: bit 6 is 1 (remainder 105 - 64 = 41)
41 >= 32: bit 5 is 1 (remainder 41 - 32 = 9)
9 < 16: bit 4 is 0
9 >= 8: bit 3 is 1 (remainder 9 - 8 = 1)
1 < 4: bit 2 is 0; 1 < 2: bit 1 is 0; 1 >= 1: bit 0 is 1
Combining bits: 01101001 (or 1101001).

10 5_{10} = 64 + 32 + 8 + 1 = 0110100 1_{2}

Mentor tip

List powers of 2 (128, 64, 32, 16, 8, 4, 2, 1) and subtract greedily.

(b)

2 marks

Convert denary 200 to hexadecimal by successive division by 16:
200 / 16 = 12 with remainder 8.
In base 16, quotient 12 corresponds to hex digit C.
The remainder 8 corresponds to hex digit 8.
Combining digits yields C8.

200 = 12 \times 1 6^{1} + 8 \times 1 6^{0} = C8_{16}

Mentor tip

Remember that denary 10 to 15 map to hex letters A through F; 12 maps to C.

Question 8

1 mark

Core role of a single-user operating system

Computer organisation

1 mark

Mentor tip

State resource management or hardware abstraction for a single concurrent user.

Question 9

1 mark

Definition of system usability

System fundamentals

1 mark

Usability denotes the degree of effectiveness, efficiency, and user satisfaction with which specific users can achieve specified goals in a system.

Mentor tip

Mention effectiveness, efficiency, or learnability/satisfaction.

Question 10

2 marks

Significance of international standards in networking

Networks

2 marks

Interoperability: Standards (such as IEEE 802.3 Ethernet and TCP/IP) enable hardware and software from competing manufacturers to communicate seamlessly.
Scalability and market competition: Common protocols allow organizations to upgrade infrastructure modularly without proprietary vendor lock-in.

Mentor tip

Emphasize multi-vendor interoperability and universal compatibility.

Question 11

3 marks

Impact of CPU cache memory on overall system performance

Computer organisation

3 marks

Architectural placement: Cache memory is ultra-fast static RAM (SRAM) positioned directly on or adjacent to the CPU die.
Latency reduction: It stores frequently executed instructions and recently accessed data, operating significantly faster than dynamic main RAM.
Throughput enhancement: When the CPU experiences a cache hit, memory fetch latency is drastically minimized, eliminating idle wait states during instruction fetch cycles.

Mentor tip

Explain speed disparity between CPU and RAM and how cache hits eliminate processor idle time.

Questions 12–14 · 45 marks

Question 12

15 marks

Enterprise system upgrade planning and implementation

System fundamentals

(a)(i)

1 mark

End-users / operational staff who interact with the system interface on a daily basis.

Mentor tip

Other valid stakeholders include system administrators, department managers, or clients.

(a)(ii)

2 marks

Excluding key stakeholders leads to incomplete requirement specifications, causing the finished system to lack essential practical features.
This discrepancy results in user resistance, low adoption rates, and costly post-deployment rework.

Mentor tip

Connect stakeholder absence to unmet functional requirements and operational disruption.

(b)

4 marks

Interviews with departmental representatives: Conduct structured one-on-one sessions to gather qualitative insights into workflow bottlenecks, specific data requirements, and usability pain points.
Direct observation of workplace operations: Shadow employees during active shifts to witness actual system usage, identifying unstated workarounds and inefficiencies that users might omit in interviews.

Mentor tip

Pair a direct inquiry method (interviews/surveys) with an empirical observation technique.

(c)

2 marks

Continuous testing detects architectural flaws, logic bugs, and specification deviations early in development.
Resolving bugs during initial module construction prevents errors from compounding into subsequent integration phases, substantially reducing overall remediation expenses.

Mentor tip

Emphasize early defect detection and reduced remediation cost compared to post-release fixes.

(d)

6 marks

Method 1: Direct Changeover (Immediate cutover)
- Benefit: Minimizes operational duplication; immediate migration eliminates running costs of maintaining two parallel systems.
- Drawback: High risk; if unforeseen critical defects emerge in production, operations can halt with no immediate fallback.
Method 2: Parallel Running (Concurrent operation)
- Benefit: Maximum safety; the legacy system remains fully operational alongside the new platform, providing a secure safety net and seamless data verification.
- Drawback: Costly and labor-intensive; staff must enter duplicate transactions into both platforms simultaneously, doubling workload and server expenses.

Mentor tip

Evaluate benefits and drawbacks for two distinct strategies (e.g. direct vs parallel or phased vs pilot).

Question 13

15 marks

Parallel array data manipulation and grade assignment algorithm

Computational thinking, problem-solving and programming

(a)

2 marks

A relational or Boolean condition: An expression that evaluates to either true or false by comparing variables or constants using relational operators.
Branching execution blocks: Instructions enclosed in THEN and ELSE blocks that execute selectively based on the outcome of the Boolean evaluation.

Mentor tip

Identify the Boolean conditional expression and the corresponding conditional execution path.

(b)

5 marks

Loop index I from 0 to 29 across all 30 students.
Examine MARK[I] against grade thresholds in descending order:
- If MARK[I] >= 80, set GRADE[I] = 'Distinction'
- Else if MARK[I] >= 60, set GRADE[I] = 'Merit'
- Else if MARK[I] >= 40, set GRADE[I] = 'Pass'
- Else set GRADE[I] = 'Fail'
Close conditional branches and loop.

loop I from 0 to 29 if M A R K [I] \geq 80 then GR A D E [I] = Distinction else if M A R K [I] \geq 60 then GR A D E [I] = Merit else if M A R K [I] \geq 40 then GR A D E [I] = Pass else GR A D E [I] = Fail end if end loop

Mentor tip

Check boundaries in descending order (80, then 60, then 40) so each student lands in the correct tier.

(c)

2 marks

Positional indexing: The arrays are aligned so that elements stored at identical index positions across NAME[], MARK[], and GRADE[] describe the exact same student.
For any index K, NAME[K] holds the student name, MARK[K] stores their numeric score, and GRADE[K] records their assigned classification.

Mentor tip

State clearly that common subscripts link corresponding data items across parallel arrays.

(d)

3 marks

Construct a loop iterating I from 0 to 29.
Within the loop, evaluate whether GRADE[I] equals 'Merit' OR GRADE[I] equals 'Distinction'.
If true, output NAME[I] alongside GRADE[I].

loop I from 0 to 29 if GR A D E [I] = Merit or GR A D E [I] = Distinction then output N A M E [I], ": ", GR A D E [I] end if end loop

Mentor tip

Use a logical OR condition comparing GRADE[I] against both required string values.

(e)

3 marks

Prompt and read user input into a target string variable, e.g. SEARCH_GRADE.
Iterate through the array checking the condition GRADE[I] = SEARCH_GRADE instead of a hardcoded string.
When a match occurs, output NAME[I] and MARK[I] as required.

Mentor tip

Outline three modifications: receiving user input, updating the comparison condition, and changing the output fields.

Question 14

15 marks

Searching algorithms: Sequential search vs Binary search analysis

Computational thinking, problem-solving and programming

(a)

4 marks

Trace Algorithm 1 (Linear Search) on DATA_ARR for TO_FIND = 39:
Array values: [0]:27, [1]:29, [2]:36, [3]:39, [4]:40, [5]:42, ...
Initialization: LIMIT = 11, LOC = FALSE, ITERATE = 0
- Iteration 0: DATA_ARR[0] = 27 != 39 -> LOC remains FALSE, ITERATE becomes 1
- Iteration 1: DATA_ARR[1] = 29 != 39 -> LOC remains FALSE, ITERATE becomes 2
- Iteration 2: DATA_ARR[2] = 36 != 39 -> LOC remains FALSE, ITERATE becomes 3
- Iteration 3: DATA_ARR[3] = 39 == 39 -> LOC becomes TRUE, ITERATE becomes 4
Loop condition 'not LOC' fails because LOC is TRUE.
Output: '39 is in the array'.

Mentor tip

Record values across all columns for each iteration until the match terminates the loop.

(b)

1 mark

The constant identifier in Algorithm 1 is LIMIT (set to 11).

Mentor tip

Identify LIMIT as the explicitly assigned unchanging boundary value.

(c)

4 marks

Trace Algorithm 2 (Binary Search) on DATA_ARR for TO_FIND = 50:
Array indices 0 to 11: [27, 29, 36, 39, 40, 42, 45, 50, 58, 60, 61, 68]
Initial state: LOW_LIM = 0, UP_LIM = 11, LOC = FALSE
- Pass 1: MID_VAL = (0 + 11) div 2 = 5.
DATA_ARR[5] = 42. Since 50 > 42, LOW_LIM = 5 + 1 = 6.
- Pass 2: LOW_LIM = 6, UP_LIM = 11.
MID_VAL = (6 + 11) div 2 = 17 div 2 = 8.
DATA_ARR[8] = 58. Since 50 < 58, UP_LIM = 8 - 1 = 7.
- Pass 3: LOW_LIM = 6, UP_LIM = 7.
MID_VAL = (6 + 7) div 2 = 13 div 2 = 6.
DATA_ARR[6] = 45. Since 50 > 45, LOW_LIM = 6 + 1 = 7.
- Pass 4: LOW_LIM = 7, UP_LIM = 7.
MID_VAL = (7 + 7) div 2 = 7.
DATA_ARR[7] = 50. Since 50 == DATA_ARR[7], LOC = TRUE.
Loop terminates. Output: '50 is in the array'.

MID_VAL = ⌊(LOW_LIM + UP_LIM) /2 ⌋

Mentor tip

Use integer division (div) to calculate MID_VAL accurately on each pass.

(d)

1 mark

MID_VAL dynamically recalculates on every iteration of the binary search loop as LOW_LIM and UP_LIM shift to bisect the search interval.

Mentor tip

Explain that its value must vary across iterations based on bounding pointers.

(e)

5 marks

Sequential Search:
- Advantage: Operates on unsorted data without preprocessing; straightforward to construct.
- Disadvantage: Inefficient on large datasets; worst-case time complexity is linear O(N).
Binary Search:
- Advantage: Extremely efficient for large collections; logarithmic time complexity O(log N) halves the search space on each comparison.
- Disadvantage: Strictly requires the dataset to be sorted beforehand; maintaining sorted order incurs overhead during insertions and deletions.

O (N) vs O (lo g_{2} N)

Mentor tip

Contrast the requirement of pre-sorted data against execution efficiency across varying dataset scales.

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 programming30 marks · 43%
Q13, Q14
System fundamentals20 marks · 29%
Q1, Q2, Q9, Q12
Computer organisation14 marks · 20%
Q5, Q6, Q7, Q8, Q11
Networks6 marks · 9%
Q3, Q4, Q10

Mentor insights

Traps to watch for

01
NAND truth evaluation with inverted input · Q6
Remember that Q is the inverse of C. When C=0, Q=1. The NAND gate outputs 0 only when P=1 and Q=1.
02
Binary search integer midpoint truncation · Q14
Always use integer floor division: (0 + 11) div 2 = 5, not 5.5 or 6. Rounding up shifts indices and disrupts midpoint evaluations.
03
Parallel array index synchronization · Q13
Every operation modifying or reading student attributes must use the identical loop index I across all three parallel arrays.

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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