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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2016
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 2016 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

Usability challenges with mobile smartphones

System fundamentals

(a)

3 marks

Identify issue 1: Small physical screen display and compact on-screen virtual keyboard, leading to frequent typing errors and visual strain.
Identify issue 2: Battery life limitations, necessitating frequent recharging or leading to device shutdown during resource-intensive operations.
Identify issue 3: Intermittent mobile network connectivity and dead zones, causing packet loss, dropped calls, and delayed synchronization when moving between cellular base stations.

Mentor tip

State three distinct usability limitations such as small touch screens, battery depletion, or fluctuating cellular coverage.

Question 2

2 marks

Role and CPU architecture placement of cache memory

Computer organisation

(a)

1 mark

State the purpose of cache memory: providing high-speed staging storage close to the CPU core to minimize wait times when accessing instructions and data from slower main RAM.

Mentor tip

State that cache reduces RAM access latency by holding frequently referenced data.

(b)

1 mark

Describe the physical architecture relationship:
The processor (CPU) is connected directly to Cache Memory, which is in turn connected to Random Access Memory (RAM):
CPU <---> Cache Memory <---> Main Memory (RAM).
Cache acts as a high-speed intermediary buffer between the processor core and system RAM across the memory bus.

Mentor tip

Place cache directly in between the processor and RAM.

Question 3

4 marks

Comparative analysis of wireless networks

Networks

(a)

4 marks

Advantage of wireless networks: Mobility and physical convenience — users can roam freely throughout an office, campus, or home while maintaining continuous network access without being tethered to physical Ethernet wall outlets.
Elaboration on advantage: Work patterns become significantly more flexible and deployment costs are reduced since running physical cabling through walls and ceilings is avoided.
Disadvantage of wireless networks: Vulnerability to radio interference and transmission security risks — radio signals attenuate through solid structures, face channel congestion from competing devices, and can be intercepted from outside building premises.
Elaboration on disadvantage: Strong encryption protocols (WPA3/WPA2-Enterprise) and careful access point placement are mandatory to mitigate unauthorized packet eavesdropping and signal degradation.

Mentor tip

State one advantage (mobility/no cabling) and one disadvantage (security risks/radio interference), providing a clear elaboration for each.

Question 4

4 marks

Truth table construction for NOT (A XOR B) AND C

Computer organisation

(a)

4 marks

Set up the 8 standard binary input permutations for A, B, and C.
Column 4 - Evaluate (A XOR B): outputs 1 when A and B differ, and 0 when they are identical.
Column 5 - Evaluate NOT (A XOR B): inverts column 4, outputting 1 when A == B (XNOR equivalence).
Column 6 - Evaluate NOT (A XOR B) AND C: outputs 1 only when column 5 is 1 AND input C is 1.
Row 1: A=0, B=0, C=0 -> A XOR B = 0 -> NOT = 1 -> AND C = 0
Row 2: A=0, B=0, C=1 -> A XOR B = 0 -> NOT = 1 -> AND C = 1
Row 3: A=0, B=1, C=0 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 4: A=0, B=1, C=1 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 5: A=1, B=0, C=0 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 6: A=1, B=0, C=1 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 7: A=1, B=1, C=0 -> A XOR B = 0 -> NOT = 1 -> AND C = 0
Row 8: A=1, B=1, C=1 -> A XOR B = 0 -> NOT = 1 -> AND C = 1

A 00001111 B 00110011 C 01010101 A \oplus B 00111100 \neg (A \oplus B) 11000011 \neg (A \oplus B) \land C 01000001

Mentor tip

Follow the specified column headings exactly: A, B, C, A xor B, NOT (A xor B), not (A xor B) and C.

Question 5

4 marks

Ethical considerations of using unverified internet scientific data

System fundamentals

(a)

4 marks

Issue 1 - Intellectual Property and Academic Integrity (Plagiarism): Using research data, graphs, or written conclusions published by online researchers without explicit attribution or academic citation constitutes intellectual theft and misrepresents authorship.
Elaboration 1: Scientific ethics mandate transparent sourcing; failing to credit original scientists undermines reproducible research and breaches copyright protections.
Issue 2 - Data Accuracy and Peer-Review Verification: Internet repositories may host unverified datasets, biased methodologies, or fabricated results that have not passed peer review.
Elaboration 2: Relying uncritically on unvalidated data risks propagating false scientific claims, distorting project outcomes, and misleading future readers who build upon the conclusions.

Mentor tip

Identify two distinct ethical concerns (e.g., plagiarism/copyright and unverified/fabricated data) and explain the consequence of each.

Question 6

6 marks

Iterative exponentiation algorithm trace and purpose

Computational thinking, problem-solving and programming

(a)

1 mark

With positive integer exponent N, the loop runs while N > 0, decrementing N by 1 on each iteration.
Multiplication P = P * X is executed exactly once per iteration.
Therefore, multiplication is executed exactly N times.

Total multiplications = N

Mentor tip

Each loop pass executes one multiplication, running N times from N down to 1.

(b)

4 marks

Trace execution for X = 2, N = 4 with initial P = 1:
Before loop: X = 2, N = 4, P = 1.
Row 1: Check N > 0 (4 > 0) -> true. Execute loop: P = 1 * 2 = 2; N = 4 - 1 = 3.
Row 2: Check N > 0 (3 > 0) -> true. Execute loop: P = 2 * 2 = 4; N = 3 - 1 = 2.
Row 3: Check N > 0 (2 > 0) -> true. Execute loop: P = 4 * 2 = 8; N = 2 - 1 = 1.
Row 4: Check N > 0 (1 > 0) -> true. Execute loop: P = 8 * 2 = 16; N = 1 - 1 = 0.
Row 5: Check N > 0 (0 > 0) -> false; loop terminates. Output P = 16.

X 2 N 43210 P 124816 N > 0 true true true true false output 16

Mentor tip

Follow headings X, N, P, N>0, output; trace each multiplication and decrement until N reaches 0.

(c)

1 mark

State the algorithmic purpose: Computes power exponentiation, calculating X raised to the power of N (X^N).

P = X^{N}

Mentor tip

Identify that the algorithm calculates X to the power of N (X^N).

Question 7

2 marks

Core features and security of a Virtual Private Network (VPN)

Networks

(a)

2 marks

Feature 1: Encryption and secure tunneling — encapsulates data packets inside an encrypted tunnel across public networks, ensuring confidentiality against eavesdropping.
Feature 2: Mutual authentication and IP masking — verifies client and server identities prior to tunnel creation and masks the client's actual geographic IP address.

Mentor tip

Highlight encrypted tunneling and user/server authentication to establish a secure private connection over public internet.

Questions 8–10 · 45 marks

Question 8

15 marks

Point of sale and central inventory database system

System fundamentals
Computer organisation
Networks

(a)

5 marks

Construct system flowchart elements for POS and central database interaction:
1) Input symbol: Barcode scanner inputs book barcode at point of sale terminal.
2) Transmission/Process: POS sends barcode request across network to central computer.
3) Storage/Lookup: Central computer queries central disk database to look up price for matching barcode.
4) Return dataflow: Central server transmits retrieved unit price back to POS terminal.
5) Local process: POS computer calculates total transaction price (applying taxes/discounts).
6) Local storage: Transaction record written to local terminal disk storage.
7) Output symbol: POS thermal printer generates physical paper receipt for customer.

Mentor tip

Include barcode input, network transmission, central database lookup, POS calculation, local disk storage, and receipt printout.

(b)

2 marks

Identify one additional peripheral device at POS: Electronic magnetic card reader / chip-and-PIN terminal.
State its purpose: To read encrypted credit/debit card data and process card payments electronically with financial payment clearing networks.

Mentor tip

Name a realistic POS device (card reader, customer display screen, cash drawer) and explain its functional role.

(c)

2 marks

Define network protocols: Standardized communication rules governing data packaging, transmission handshaking, error checking, and packet routing.
Explain their role in customer ordering: Secure protocols such as HTTPS/TLS ensure that sensitive customer information (name, address, payment card) is transmitted reliably and encrypted against man-in-the-middle packet tampering.

Mentor tip

Explain that protocols provide standardized rules for reliable data transfer and secure encryption via HTTPS/TLS.

(d)

6 marks

Sub-part (i) [2 marks] - Identify two sources of risk to personal data:
1) Unauthorized network interception (packet sniffing) while data travels over the public internet during order submission.
2) Database breach or unauthorized administrator access while data is stored at rest on the central server disk.
Sub-part (ii) [2 marks] - State two protective measures the book shop can take:
1) Enforce robust encryption both in transit (TLS/HTTPS) and at rest (AES encryption on database columns holding payment data).
2) Implement strict role-based access control (RBAC), multi-factor authentication for database access, and scheduled offsite backup systems.
Sub-part (iii) [2 marks] - Outline consequences to the customer:
1) Identity theft: Cybercriminals use stolen personal names, addresses, and credit card numbers to open fraudulent accounts or make unauthorized purchases.
2) Financial loss and privacy violation: Exposure of personal contact details leading to targeted phishing scams and direct monetary damages.

Mentor tip

Address risks (transmission interception, storage breach), countermeasures (encryption, access controls), and consequences (fraud, identity theft).

Question 9

15 marks

High-level programming language design and tooling

Computational thinking, problem-solving and programming
System fundamentals
Computer organisation

(a)

2 marks

Reason 1: Ease of learning and developer productivity — predictable syntax allows programmers to master language conventions quickly without encountering confusing irregularities.
Reason 2: Reduction in compiler and semantic errors — consistent grammatical rules simplify compiler parsing and prevent subtle logical bugs during codebase maintenance.

Mentor tip

Highlight developer productivity/learnability and simplified compilation with fewer programmer errors.

(b)

2 marks

Feature 1: Syntax highlighting and visual code formatting — automatically color-coding keywords, strings, variables, and comments to maximize code readability and catch typing mistakes.
Feature 2: Code auto-completion (IntelliSense) and toolbars/menus — offering context-aware popups of available class methods and attributes when typing an object identifier.

Mentor tip

Cite features of an IDE user interface such as syntax color-coding, auto-completion, or graphical toolbars.

(c)

1 mark

State one method of providing user documentation: Context-sensitive built-in help files, API reference documentation, or interactive online support manuals.

Mentor tip

Name online documentation, built-in help files, or developer API guides.

(d)

4 marks

Sub-part (i) [2 marks] - Need for an interpreter or compiler:
High-level programming languages are written in human-readable abstraction that computer CPUs cannot execute directly.
An interpreter or compiler is mandatory to translate high-level statements into low-level machine code (binary opcodes) that the CPU architecture can process.
Sub-part (ii) [2 marks] - Advantage of providing both tools:
An interpreter translates and executes code line-by-line, providing instant feedback and dynamic debugging without lengthy build pauses during development.
A compiler produces an optimized, standalone executable binary that runs at maximum speed and can be distributed to end users without exposing source code.

Mentor tip

Explain translation to machine code in (d)(i); in (d)(ii), contrast rapid debugging via interpreter with fast production executables via compiler.

(e)

4 marks

Construct pseudocode algorithm for sumOdd(N):
sumOdd(N)
if N <= 0 then
output -1
else
TOTAL = 0
loop K from 1 to N
TOTAL = TOTAL + (2 * K - 1)
end loop
output TOTAL
end if
end sumOdd
Trace check for N = 4: K=1 -> 1; K=2 -> 3; K=3 -> 5; K=4 -> 7. Sum = 1 + 3 + 5 + 7 = 16 = 4^2.

k = 1 \sum N (2 k - 1) = N^{2}

Mentor tip

Handle N <= 0 check returning -1, loop from 1 to N generating odd numbers (2*K - 1), accumulate sum, and output.

(f)

2 marks

Outline the need for predefined sub-programs and collections: Code reusability and software reliability.
Elaborate on benefit: Using pre-tested, highly optimized standard library functions and data collections saves substantial development time, prevents developers from re-implementing complex algorithms from scratch, and reduces defect rates in software systems.

Mentor tip

Explain that standard libraries and collection classes provide reusable, tested, reliable components that reduce development time.

Question 10

15 marks

School locker color assignment and painted status audit

Computational thinking, problem-solving and programming

(a)

1 mark

The color cycle repeats in 4-locker periods: 1=red, 2=white, 3=yellow, 4=blue.
Calculate 442 mod 4:
442 = 4 * 110 + 2 -> remainder is 2.
Remainder 2 corresponds to white.
Locker number 442 will be painted white.

442 (mod 4) = 2 ⟹ white

Mentor tip

Calculate the locker number modulo 4: remainder 1 is red, 2 is white, 3 is yellow, 0 is blue.

(b)

5 marks

Construct pseudocode algorithm to determine locker color:
LOCKER = input('Enter locker number:')
REM = LOCKER mod 4
if REM == 1 then
COLOUR = 'red'
else if REM == 2 then
COLOUR = 'white'
else if REM == 3 then
COLOUR = 'yellow'
else
COLOUR = 'blue'
end if
output(COLOUR)

Mentor tip

Input the locker number, compute modulo 4, branch through if-else statements for each color, and output the result.

(c)

5 marks

Sub-part (i) [1 mark] - Identify student responsible for locker 811:
From the provided array snippet, index [2] shows LOCKER_NO[2] = 811.
The corresponding student name in STUDENT_NAME[2] is Ivy Baber (Baber, Ivy).
Sub-part (ii) [4 marks] - Construct pseudocode algorithm to count painted lockers:
COUNT = 0
loop K from 0 to 2399
if PAINTED[K] == true then
COUNT = COUNT + 1
end if
end loop
output(COUNT)

Mentor tip

In (c)(i) match locker 811 to Baber, Ivy; in (c)(ii) initialize a counter, loop from 0 to 2399, increment when PAINTED[K] is true, and output count.

(d)

4 marks

Efficient lookup algorithm: Binary Search on STUDENT_NAME[] array.
1) Why Binary Search is efficient: The array STUDENT_NAME[] is pre-sorted in alphabetical order. Binary search locates a target student in O(log_2 2400) ~ 12 comparisons, whereas a linear search requires up to 2400 comparisons.
2) Procedure:
Set LOW = 0, HIGH = 2399, FOUND = false.
While LOW <= HIGH and NOT FOUND:
MID = (LOW + HIGH) div 2
If STUDENT_NAME[MID] == TARGET_NAME then:
FOUND = true
INDEX = MID
Else if TARGET_NAME < STUDENT_NAME[MID] then:
HIGH = MID - 1
Else:
LOW = MID + 1
3) Cross-array access:
If FOUND is true, retrieve the corresponding locker details using the discovered INDEX:
LOCKER = LOCKER_NO[INDEX]
STATUS = PAINTED[INDEX]
Output LOCKER, STATUS.

O (lo g_{2} 2400) \approx 12 comparisons vs O (2400)

Mentor tip

Explain that binary search takes advantage of the alphabetically sorted STUDENT_NAME[] array (O(log N)), using the found index to look up LOCKER_NO and PAINTED.

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 programming29 marks · 41%
Q6, Q9, Q10
System fundamentals21 marks · 30%
Q1, Q5, Q8, Q9
Computer organisation12 marks · 17%
Q2, Q4, Q8, Q9
Networks8 marks · 11%
Q3, Q7, Q8

Mentor insights

Traps to watch for

01
Modulo remainder mapping for cyclical locker colors · Q10
When locker number mod 4 == 0 (such as locker 4, 8, 2400), the remainder is 0, which corresponds to the 4th color (blue). Do not assume remainders run 1 to 4.
02
Loop condition N > 0 termination in trace table · Q6
When N decrements to 0, N > 0 evaluates to false. Do not execute another multiplication step when N = 0; output the accumulated product (16).
03
Truth table evaluation order for NOT (A XOR B) AND C · Q4
In the heading not (A xor B) and C, the negation binds tightly to (A xor B). Rows where A and B match produce NOT = 1; among those, only rows with C = 1 output 1.

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 2016
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 2016 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

Usability challenges with mobile smartphones

System fundamentals

(a)

3 marks

Identify issue 1: Small physical screen display and compact on-screen virtual keyboard, leading to frequent typing errors and visual strain.
Identify issue 2: Battery life limitations, necessitating frequent recharging or leading to device shutdown during resource-intensive operations.
Identify issue 3: Intermittent mobile network connectivity and dead zones, causing packet loss, dropped calls, and delayed synchronization when moving between cellular base stations.

Mentor tip

State three distinct usability limitations such as small touch screens, battery depletion, or fluctuating cellular coverage.

Question 2

2 marks

Role and CPU architecture placement of cache memory

Computer organisation

(a)

1 mark

State the purpose of cache memory: providing high-speed staging storage close to the CPU core to minimize wait times when accessing instructions and data from slower main RAM.

Mentor tip

State that cache reduces RAM access latency by holding frequently referenced data.

(b)

1 mark

Describe the physical architecture relationship:
The processor (CPU) is connected directly to Cache Memory, which is in turn connected to Random Access Memory (RAM):
CPU <---> Cache Memory <---> Main Memory (RAM).
Cache acts as a high-speed intermediary buffer between the processor core and system RAM across the memory bus.

Mentor tip

Place cache directly in between the processor and RAM.

Question 3

4 marks

Comparative analysis of wireless networks

Networks

(a)

4 marks

Advantage of wireless networks: Mobility and physical convenience — users can roam freely throughout an office, campus, or home while maintaining continuous network access without being tethered to physical Ethernet wall outlets.
Elaboration on advantage: Work patterns become significantly more flexible and deployment costs are reduced since running physical cabling through walls and ceilings is avoided.
Disadvantage of wireless networks: Vulnerability to radio interference and transmission security risks — radio signals attenuate through solid structures, face channel congestion from competing devices, and can be intercepted from outside building premises.
Elaboration on disadvantage: Strong encryption protocols (WPA3/WPA2-Enterprise) and careful access point placement are mandatory to mitigate unauthorized packet eavesdropping and signal degradation.

Mentor tip

State one advantage (mobility/no cabling) and one disadvantage (security risks/radio interference), providing a clear elaboration for each.

Question 4

4 marks

Truth table construction for NOT (A XOR B) AND C

Computer organisation

(a)

4 marks

Set up the 8 standard binary input permutations for A, B, and C.
Column 4 - Evaluate (A XOR B): outputs 1 when A and B differ, and 0 when they are identical.
Column 5 - Evaluate NOT (A XOR B): inverts column 4, outputting 1 when A == B (XNOR equivalence).
Column 6 - Evaluate NOT (A XOR B) AND C: outputs 1 only when column 5 is 1 AND input C is 1.
Row 1: A=0, B=0, C=0 -> A XOR B = 0 -> NOT = 1 -> AND C = 0
Row 2: A=0, B=0, C=1 -> A XOR B = 0 -> NOT = 1 -> AND C = 1
Row 3: A=0, B=1, C=0 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 4: A=0, B=1, C=1 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 5: A=1, B=0, C=0 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 6: A=1, B=0, C=1 -> A XOR B = 1 -> NOT = 0 -> AND C = 0
Row 7: A=1, B=1, C=0 -> A XOR B = 0 -> NOT = 1 -> AND C = 0
Row 8: A=1, B=1, C=1 -> A XOR B = 0 -> NOT = 1 -> AND C = 1

A 00001111 B 00110011 C 01010101 A \oplus B 00111100 \neg (A \oplus B) 11000011 \neg (A \oplus B) \land C 01000001

Mentor tip

Follow the specified column headings exactly: A, B, C, A xor B, NOT (A xor B), not (A xor B) and C.

Question 5

4 marks

Ethical considerations of using unverified internet scientific data

System fundamentals

(a)

4 marks

Issue 1 - Intellectual Property and Academic Integrity (Plagiarism): Using research data, graphs, or written conclusions published by online researchers without explicit attribution or academic citation constitutes intellectual theft and misrepresents authorship.
Elaboration 1: Scientific ethics mandate transparent sourcing; failing to credit original scientists undermines reproducible research and breaches copyright protections.
Issue 2 - Data Accuracy and Peer-Review Verification: Internet repositories may host unverified datasets, biased methodologies, or fabricated results that have not passed peer review.
Elaboration 2: Relying uncritically on unvalidated data risks propagating false scientific claims, distorting project outcomes, and misleading future readers who build upon the conclusions.

Mentor tip

Identify two distinct ethical concerns (e.g., plagiarism/copyright and unverified/fabricated data) and explain the consequence of each.

Question 6

6 marks

Iterative exponentiation algorithm trace and purpose

Computational thinking, problem-solving and programming

(a)

1 mark

With positive integer exponent N, the loop runs while N > 0, decrementing N by 1 on each iteration.
Multiplication P = P * X is executed exactly once per iteration.
Therefore, multiplication is executed exactly N times.

Total multiplications = N

Mentor tip

Each loop pass executes one multiplication, running N times from N down to 1.

(b)

4 marks

Trace execution for X = 2, N = 4 with initial P = 1:
Before loop: X = 2, N = 4, P = 1.
Row 1: Check N > 0 (4 > 0) -> true. Execute loop: P = 1 * 2 = 2; N = 4 - 1 = 3.
Row 2: Check N > 0 (3 > 0) -> true. Execute loop: P = 2 * 2 = 4; N = 3 - 1 = 2.
Row 3: Check N > 0 (2 > 0) -> true. Execute loop: P = 4 * 2 = 8; N = 2 - 1 = 1.
Row 4: Check N > 0 (1 > 0) -> true. Execute loop: P = 8 * 2 = 16; N = 1 - 1 = 0.
Row 5: Check N > 0 (0 > 0) -> false; loop terminates. Output P = 16.

X 2 N 43210 P 124816 N > 0 true true true true false output 16

Mentor tip

Follow headings X, N, P, N>0, output; trace each multiplication and decrement until N reaches 0.

(c)

1 mark

State the algorithmic purpose: Computes power exponentiation, calculating X raised to the power of N (X^N).

P = X^{N}

Mentor tip

Identify that the algorithm calculates X to the power of N (X^N).

Question 7

2 marks

Core features and security of a Virtual Private Network (VPN)

Networks

(a)

2 marks

Feature 1: Encryption and secure tunneling — encapsulates data packets inside an encrypted tunnel across public networks, ensuring confidentiality against eavesdropping.
Feature 2: Mutual authentication and IP masking — verifies client and server identities prior to tunnel creation and masks the client's actual geographic IP address.

Mentor tip

Highlight encrypted tunneling and user/server authentication to establish a secure private connection over public internet.

Questions 8–10 · 45 marks

Question 8

15 marks

Point of sale and central inventory database system

System fundamentals
Computer organisation
Networks

(a)

5 marks

Construct system flowchart elements for POS and central database interaction:
1) Input symbol: Barcode scanner inputs book barcode at point of sale terminal.
2) Transmission/Process: POS sends barcode request across network to central computer.
3) Storage/Lookup: Central computer queries central disk database to look up price for matching barcode.
4) Return dataflow: Central server transmits retrieved unit price back to POS terminal.
5) Local process: POS computer calculates total transaction price (applying taxes/discounts).
6) Local storage: Transaction record written to local terminal disk storage.
7) Output symbol: POS thermal printer generates physical paper receipt for customer.

Mentor tip

Include barcode input, network transmission, central database lookup, POS calculation, local disk storage, and receipt printout.

(b)

2 marks

Identify one additional peripheral device at POS: Electronic magnetic card reader / chip-and-PIN terminal.
State its purpose: To read encrypted credit/debit card data and process card payments electronically with financial payment clearing networks.

Mentor tip

Name a realistic POS device (card reader, customer display screen, cash drawer) and explain its functional role.

(c)

2 marks

Define network protocols: Standardized communication rules governing data packaging, transmission handshaking, error checking, and packet routing.
Explain their role in customer ordering: Secure protocols such as HTTPS/TLS ensure that sensitive customer information (name, address, payment card) is transmitted reliably and encrypted against man-in-the-middle packet tampering.

Mentor tip

Explain that protocols provide standardized rules for reliable data transfer and secure encryption via HTTPS/TLS.

(d)

6 marks

Sub-part (i) [2 marks] - Identify two sources of risk to personal data:
1) Unauthorized network interception (packet sniffing) while data travels over the public internet during order submission.
2) Database breach or unauthorized administrator access while data is stored at rest on the central server disk.
Sub-part (ii) [2 marks] - State two protective measures the book shop can take:
1) Enforce robust encryption both in transit (TLS/HTTPS) and at rest (AES encryption on database columns holding payment data).
2) Implement strict role-based access control (RBAC), multi-factor authentication for database access, and scheduled offsite backup systems.
Sub-part (iii) [2 marks] - Outline consequences to the customer:
1) Identity theft: Cybercriminals use stolen personal names, addresses, and credit card numbers to open fraudulent accounts or make unauthorized purchases.
2) Financial loss and privacy violation: Exposure of personal contact details leading to targeted phishing scams and direct monetary damages.

Mentor tip

Address risks (transmission interception, storage breach), countermeasures (encryption, access controls), and consequences (fraud, identity theft).

Question 9

15 marks

High-level programming language design and tooling

Computational thinking, problem-solving and programming
System fundamentals
Computer organisation

(a)

2 marks

Reason 1: Ease of learning and developer productivity — predictable syntax allows programmers to master language conventions quickly without encountering confusing irregularities.
Reason 2: Reduction in compiler and semantic errors — consistent grammatical rules simplify compiler parsing and prevent subtle logical bugs during codebase maintenance.

Mentor tip

Highlight developer productivity/learnability and simplified compilation with fewer programmer errors.

(b)

2 marks

Feature 1: Syntax highlighting and visual code formatting — automatically color-coding keywords, strings, variables, and comments to maximize code readability and catch typing mistakes.
Feature 2: Code auto-completion (IntelliSense) and toolbars/menus — offering context-aware popups of available class methods and attributes when typing an object identifier.

Mentor tip

Cite features of an IDE user interface such as syntax color-coding, auto-completion, or graphical toolbars.

(c)

1 mark

State one method of providing user documentation: Context-sensitive built-in help files, API reference documentation, or interactive online support manuals.

Mentor tip

Name online documentation, built-in help files, or developer API guides.

(d)

4 marks

Sub-part (i) [2 marks] - Need for an interpreter or compiler:
High-level programming languages are written in human-readable abstraction that computer CPUs cannot execute directly.
An interpreter or compiler is mandatory to translate high-level statements into low-level machine code (binary opcodes) that the CPU architecture can process.
Sub-part (ii) [2 marks] - Advantage of providing both tools:
An interpreter translates and executes code line-by-line, providing instant feedback and dynamic debugging without lengthy build pauses during development.
A compiler produces an optimized, standalone executable binary that runs at maximum speed and can be distributed to end users without exposing source code.

Mentor tip

Explain translation to machine code in (d)(i); in (d)(ii), contrast rapid debugging via interpreter with fast production executables via compiler.

(e)

4 marks

Construct pseudocode algorithm for sumOdd(N):
sumOdd(N)
if N <= 0 then
output -1
else
TOTAL = 0
loop K from 1 to N
TOTAL = TOTAL + (2 * K - 1)
end loop
output TOTAL
end if
end sumOdd
Trace check for N = 4: K=1 -> 1; K=2 -> 3; K=3 -> 5; K=4 -> 7. Sum = 1 + 3 + 5 + 7 = 16 = 4^2.

k = 1 \sum N (2 k - 1) = N^{2}

Mentor tip

Handle N <= 0 check returning -1, loop from 1 to N generating odd numbers (2*K - 1), accumulate sum, and output.

(f)

2 marks

Outline the need for predefined sub-programs and collections: Code reusability and software reliability.
Elaborate on benefit: Using pre-tested, highly optimized standard library functions and data collections saves substantial development time, prevents developers from re-implementing complex algorithms from scratch, and reduces defect rates in software systems.

Mentor tip

Explain that standard libraries and collection classes provide reusable, tested, reliable components that reduce development time.

Question 10

15 marks

School locker color assignment and painted status audit

Computational thinking, problem-solving and programming

(a)

1 mark

The color cycle repeats in 4-locker periods: 1=red, 2=white, 3=yellow, 4=blue.
Calculate 442 mod 4:
442 = 4 * 110 + 2 -> remainder is 2.
Remainder 2 corresponds to white.
Locker number 442 will be painted white.

442 (mod 4) = 2 ⟹ white

Mentor tip

Calculate the locker number modulo 4: remainder 1 is red, 2 is white, 3 is yellow, 0 is blue.

(b)

5 marks

Construct pseudocode algorithm to determine locker color:
LOCKER = input('Enter locker number:')
REM = LOCKER mod 4
if REM == 1 then
COLOUR = 'red'
else if REM == 2 then
COLOUR = 'white'
else if REM == 3 then
COLOUR = 'yellow'
else
COLOUR = 'blue'
end if
output(COLOUR)

Mentor tip

Input the locker number, compute modulo 4, branch through if-else statements for each color, and output the result.

(c)

5 marks

Sub-part (i) [1 mark] - Identify student responsible for locker 811:
From the provided array snippet, index [2] shows LOCKER_NO[2] = 811.
The corresponding student name in STUDENT_NAME[2] is Ivy Baber (Baber, Ivy).
Sub-part (ii) [4 marks] - Construct pseudocode algorithm to count painted lockers:
COUNT = 0
loop K from 0 to 2399
if PAINTED[K] == true then
COUNT = COUNT + 1
end if
end loop
output(COUNT)

Mentor tip

In (c)(i) match locker 811 to Baber, Ivy; in (c)(ii) initialize a counter, loop from 0 to 2399, increment when PAINTED[K] is true, and output count.

(d)

4 marks

Efficient lookup algorithm: Binary Search on STUDENT_NAME[] array.
1) Why Binary Search is efficient: The array STUDENT_NAME[] is pre-sorted in alphabetical order. Binary search locates a target student in O(log_2 2400) ~ 12 comparisons, whereas a linear search requires up to 2400 comparisons.
2) Procedure:
Set LOW = 0, HIGH = 2399, FOUND = false.
While LOW <= HIGH and NOT FOUND:
MID = (LOW + HIGH) div 2
If STUDENT_NAME[MID] == TARGET_NAME then:
FOUND = true
INDEX = MID
Else if TARGET_NAME < STUDENT_NAME[MID] then:
HIGH = MID - 1
Else:
LOW = MID + 1
3) Cross-array access:
If FOUND is true, retrieve the corresponding locker details using the discovered INDEX:
LOCKER = LOCKER_NO[INDEX]
STATUS = PAINTED[INDEX]
Output LOCKER, STATUS.

O (lo g_{2} 2400) \approx 12 comparisons vs O (2400)

Mentor tip

Explain that binary search takes advantage of the alphabetically sorted STUDENT_NAME[] array (O(log N)), using the found index to look up LOCKER_NO and PAINTED.

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 programming29 marks · 41%
Q6, Q9, Q10
System fundamentals21 marks · 30%
Q1, Q5, Q8, Q9
Computer organisation12 marks · 17%
Q2, Q4, Q8, Q9
Networks8 marks · 11%
Q3, Q7, Q8

Mentor insights

Traps to watch for

01
Modulo remainder mapping for cyclical locker colors · Q10
When locker number mod 4 == 0 (such as locker 4, 8, 2400), the remainder is 0, which corresponds to the 4th color (blue). Do not assume remainders run 1 to 4.
02
Loop condition N > 0 termination in trace table · Q6
When N decrements to 0, N > 0 evaluates to false. Do not execute another multiplication step when N = 0; output the accumulated product (16).
03
Truth table evaluation order for NOT (A XOR B) AND C · Q4
In the heading not (A xor B) and C, the negation binds tightly to (A xor B). Rows where A and B match produce NOT = 1; among those, only rows with C = 1 output 1.

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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