Loading…

Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: May 2021
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 2021 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

Roles a computer can perform within a networked environment

Networks

2 marks

Server role: Centralizes and provides shared resources, services, or authentication to other network hosts (e.g. web server, database host, DNS server, or file server).
Client role: Initiates requests to network servers to consume services, retrieve web pages, or access remote files on behalf of an end user.

Mentor tip

Identify client and server roles, or peer-to-peer node roles.

Question 2

2 marks

Implementation methodology for deploying a new computer system

System fundamentals

2 marks

Direct changeover: The organization terminates the legacy system at an agreed cutover time and immediately switches all operations to the new system.
While cost-effective and swift with no duplication of data entry, it introduces high operational risk if critical errors emerge post-launch.

Mentor tip

Choose one clear deployment strategy (direct, parallel, phased, or pilot) and describe its operational mechanism.

Question 3

2 marks

Logic circuit deduction from truth table

Computer organisation

2 marks

Examine truth table outputs for inputs A and B:
- A=0, B=0 -> Z=1
- A=0, B=1 -> Z=0
- A=1, B=0 -> Z=0
- A=1, B=1 -> Z=1
The output Z is 1 strictly when inputs A and B have identical values.
This corresponds directly to an XNOR (exclusive NOR) gate, or an XOR gate followed by an inverter (NOT gate).

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

Mentor tip

Recognize that equality (same input bits yielding 1) defines the XNOR logic operation.

Question 4

4 marks

Operating system patch management necessity and acquisition channels

Resource management

(a)

2 marks

Security vulnerability remediation: Patches fix discovered exploits and security holes, preventing malicious infiltration and zero-day attacks.
Bug fixes and system stability: Patches resolve code defects, memory leaks, and driver conflicts, enhancing operating system stability and performance.

Mentor tip

Highlight security vulnerability mitigation and bug/performance fixes.

(b)

2 marks

Automatic operating system update service: Built-in background updater client that polls vendor servers and downloads patches automatically.
Direct manual download from vendor portal: System administrators manually download installer executable packages from the authorized vendor support repository.

Mentor tip

Give two distinct delivery channels: automated OS update daemon and manual vendor download.

Question 5

2 marks

Hexadecimal to denary number conversion

Computer organisation

2 marks

Hexadecimal BF consists of two hex digits: 'B' in the 16^1 place and 'F' in the 16^0 place.
Hex digit B represents denary 11.
Hex digit F represents denary 15.
Compute the positional sum: 11 * 16 + 15 * 1 = 176 + 15 = 191.

BF_{16} = (11 \times 1 6^{1}) + (15 \times 1 6^{0}) = 176 + 15 = 19 1_{10}

Mentor tip

Convert B to 11 and F to 15 before multiplying by positional powers of 16.

Question 6

2 marks

Advantages of fibre optic cabling over wireless connectivity

Networks

2 marks

Immunity to electromagnetic interference: Fibre optics transmit pulses of light through glass strands rather than radio frequencies, making signals completely immune to EMI, radio noise, and atmospheric disturbance.
Significantly greater bandwidth and lower attenuation: Fibre optic cables support vastly higher data transmission speeds over much longer distances without requiring frequent signal repeaters.

Mentor tip

Focus on high bandwidth over distance and complete immunity to electrical interference.

Question 7

2 marks

Distinction between variable and constant data items

Computational thinking, problem-solving and programming

2 marks

Mutability during execution: A variable is a named memory location whose stored value can be reassigned or modified dynamically as the program runs.
Immutability: A constant is a named identifier bound to a fixed value upon declaration that cannot be altered or overwritten during program execution.

Mentor tip

Contrast runtime mutability (variables can change) against immutability (constants remain fixed).

Question 8

3 marks

Trace of divisibility filtering algorithm

Computational thinking, problem-solving and programming

3 marks

Input sequence: 2, 6, 8, 9, 12, 15, 18, 20.
Condition 1: NUMBER div 2 = NUMBER / 2. This checks if NUMBER is divisible by 2 with no fractional remainder (i.e. NUMBER is even).
Condition 2: NUMBER div 3 = NUMBER / 3. This checks if NUMBER is divisible by 3 with no fractional remainder.
Therefore, a number is output if and only if it is divisible by both 2 and 3 (i.e. divisible by 6):
- 2: Divisible by 2, but not 3 -> Not output
- 6: Divisible by 2 and 3 -> OUTPUT 6
- 8: Divisible by 2, but not 3 -> Not output
- 9: Not divisible by 2 -> Not output
- 12: Divisible by 2 and 3 -> OUTPUT 12
- 15: Not divisible by 2 -> Not output
- 18: Divisible by 2 and 3 -> OUTPUT 18
- 20: Divisible by 2, but not 3 -> Not output
Final output list in order: 6, 12, 18.

Condition: (NUMBER mod 2 = 0) \land (NUMBER mod 3 = 0) ⟺ NUMBER mod 6 = 0

Mentor tip

Notice that comparing integer division 'div' to real division '/' tests whether the division produces a whole integer with zero remainder.

Question 9

1 mark

Formal definition of a network data packet

Networks

1 mark

A standardized, formatted unit of data transmitted across a packet-switching network, consisting of control metadata (headers and trailers) and user payload.

Mentor tip

State that a packet contains control metadata (source/destination IP) and payload data.

Question 10

1 mark

User interface design features enhancing system usability

System fundamentals

1 mark

Intuitive visual hierarchy and clear feedback: Clear icons, consistent navigation menus, and immediate visual confirmation of actions (e.g. progress bars, confirmation tooltips) reduce cognitive load.

Mentor tip

Mention tooltips, breadcrumbs, consistent menus, or keyboard shortcuts.

Question 11

4 marks

Truth table evaluation for composite logic circuit

Computer organisation

4 marks

Circuit inputs: A, B, C.
Intermediate gate 1: P = A AND B.
Intermediate gate 2: Q = NOT B.
Intermediate gate 3: R = Q AND C = (NOT B) AND C.
Final gate: Z = P OR R = (A AND B) OR ((NOT B) AND C).
Evaluate all 8 rows:
- A=0, B=0, C=0 -> P=0, Q=1, R=0 -> Z = 0
- A=0, B=0, C=1 -> P=0, Q=1, R=1 -> Z = 1
- A=0, B=1, C=0 -> P=0, Q=0, R=0 -> Z = 0
- A=0, B=1, C=1 -> P=0, Q=0, R=0 -> Z = 0
- A=1, B=0, C=0 -> P=0, Q=1, R=0 -> Z = 0
- A=1, B=0, C=1 -> P=0, Q=1, R=1 -> Z = 1
- A=1, B=1, C=0 -> P=1, Q=0, R=0 -> Z = 1
- A=1, B=1, C=1 -> P=1, Q=0, R=0 -> Z = 1

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

Mentor tip

Trace intermediate nodes P, Q, and R carefully before combining them in the final OR gate.

Questions 12–14 · 45 marks

Question 12

15 marks

Campus wireless network expansion and VPN security

Networks

(a)

4 marks

Wireless Access Points (WAPs): Transceivers positioned across buildings that broadcast Wi-Fi radio frequencies and bridge wireless client traffic onto the school's wired Ethernet backbone.
Wireless Network Interface Controllers (WNICs): Hardware adapters embedded inside student and staff laptops/devices that modulate and demodulate radio frequency transmissions.

Mentor tip

Describe two separate hardware components such as WAPs, wireless routers, or client WNICs.

(b)

2 marks

Mobility and flexible learning: Students can access digital learning materials and collaborate from classrooms, libraries, or outdoor study spaces without physical cable tethers.
Device versatility (BYOD): Enables students to connect diverse personal portable devices (tablets, smartphones, laptops) seamlessly.

Mentor tip

Highlight physical mobility and multi-device support.

(c)

4 marks

Robust wireless encryption protocol (WPA3 / 802.1X enterprise): Encrypts all over-the-air communication using advanced cryptographic ciphers with unique per-user keys, preventing packet eavesdropping.
MAC address filtering or hidden SSID with captive portal authentication: Restricting network access solely to whitelisted device hardware addresses and requiring student portal login credentials before granting DHCP IP leases.

Mentor tip

Cover wireless encryption (WPA2/WPA3) and network admission controls (802.1X, captive portal, MAC filtering).

(d)

2 marks

VPN Gateway / Server software: Dedicated network server running tunneling protocols (e.g. IPsec, OpenVPN, WireGuard) to manage remote cryptographic handshakes.
VPN Client software: Application installed on remote staff devices that encrypts outbound packets and manages secure virtual network adapters.

Mentor tip

Identify the VPN server/gateway and client-side tunneling software.

(e)

3 marks

Secure remote access to intranet resources: Staff working off-campus can securely retrieve sensitive student grading records, internal curriculum repositories, and administration portals over public internet connections as though physically connected to the on-campus LAN.
Encryption protects confidential student data from exposure on untrusted external residential or hotel Wi-Fi connections.

Mentor tip

Emphasize secure off-campus access to internal file shares and administrative databases.

Question 13

15 marks

Employee name collection parsing and parallel array bubble sort

Computational thinking, problem-solving and programming

(a)

4 marks

Algorithm to parse alternating surname and first name from collection NAMES into SURNAMES[] and FIRSTNAMES[]:
1) Initialize index I = 0.
2) NAMES.resetNext() to reset collection iterator.
3) loop while NAMES.hasNext():
- SURNAMES[I] = NAMES.getNext()
- FIRSTNAMES[I] = NAMES.getNext()
- I = I + 1
4) end loop.

I = 0 NAMES.resetNext () loop while NAMES.hasNext () SURNAMES [I] = NAMES.getNext () FIRSTNAMES [I] = NAMES.getNext () I = I + 1 end loop

Mentor tip

Fetch twice per iteration because the collection contains alternating surname and first name values.

(b)

5 marks

Bubble sort algorithm for 600 staff members:
1) Outer loop from 0 to 598 (representing N - 1 passes).
2) Inner loop J from 0 to 598 - pass (or 598).
3) If SURNAMES[J] > SURNAMES[J + 1] then:
- Swap SURNAMES[J] and SURNAMES[J + 1] using TEMP_SUR.
- Concurrently swap FIRSTNAMES[J] and FIRSTNAMES[J + 1] using TEMP_FIRST to preserve parallel alignment.
4) Close conditionals and loops.

loop P from 0 to 598 loop J from 0 to 598 - P if SURNAMES [J] > SURNAMES [J + 1] then TEMP_SUR = SURNAMES [J] SURNAMES [J] = SURNAMES [J + 1] SURNAMES [J + 1] = TEMP_SUR TEMP_FIRST = FIRSTNAMES [J] FIRSTNAMES [J] = FIRSTNAMES [J + 1] FIRSTNAMES [J + 1] = TEMP_FIRST end if end loop end loop

Mentor tip

Always execute matching swaps on both SURNAMES and FIRSTNAMES arrays.

(c)

4 marks

Binary search procedure on sorted SURNAMES array:
1) Initialize LOW = 0, HIGH = 599, and FOUND = FALSE.
2) Repeat while LOW <= HIGH and not FOUND:
- Calculate MID = (LOW + HIGH) div 2.
- If SURNAMES[MID] == TARGET_NAME, set FOUND = TRUE.
- Else if TARGET_NAME < SURNAMES[MID], set HIGH = MID - 1.
- Else set LOW = MID + 1.
3) When found, the record index is MID; if loop finishes without match, report not found.

MID = ⌊(LOW + HIGH) /2 ⌋

Mentor tip

Describe the bisection process: compare target to midpoint, halve search interval accordingly.

(d)

2 marks

Modularity and reusability: Encapsulating array population and sorting into named subprograms allows them to be tested independently and invoked repeatedly without code duplication.

Mentor tip

Emphasize modularity, ease of debugging, and code reusability.

Question 14

15 marks

Decimal to binary conversion algorithm analysis and optimization

Computational thinking, problem-solving and programming

(a)

4 marks

Trace algorithm for input NUMBER = 19:
- Pass 1: DIGIT = 19 mod 2 = 1. NUMBER = 19 div 2 = 9. OUTPUT DIGIT: 1. Is 9 = 1? No.
- Pass 2: DIGIT = 9 mod 2 = 1. NUMBER = 9 div 2 = 4. OUTPUT DIGIT: 1. Is 4 = 1? No.
- Pass 3: DIGIT = 4 mod 2 = 0. NUMBER = 4 div 2 = 2. OUTPUT DIGIT: 0. Is 2 = 1? No.
- Pass 4: DIGIT = 2 mod 2 = 0. NUMBER = 2 div 2 = 1. OUTPUT DIGIT: 0. Is 1 = 1? Yes.
Loop exits on Yes branch. OUTPUT NUMBER: 1.
Sequence of outputs in order: 1, 1, 0, 0, 1.

1 9_{10} = 1001 1_{2} ⟹ outputs LSB to MSB: 1, 1, 0, 0, 1

Mentor tip

Notice the algorithm outputs the binary digits in reverse order (least significant bit first) until reaching 1.

(b)

1 mark

Converts a positive denary integer into its binary equivalent (outputting bits from least significant to most significant).

Mentor tip

State denary to binary conversion.

(c)

6 marks

Construct pseudocode with positive integer input validation:
1) Input validation loop: prompt NUMBER; while NUMBER <= 0 or NUMBER != int(NUMBER): re-prompt.
2) Loop while NUMBER != 1:
- DIGIT = NUMBER mod 2
- NUMBER = NUMBER div 2
- output DIGIT
3) output NUMBER (final remaining 1).

output "Enter a positive integer:" input NUMBER loop while NUMBER \leq 0 output "Invalid. Enter a positive integer:" input NUMBER end loop loop while NUMBER \neq = 1 DIGIT = NUMBER mod 2 NUMBER = NUMBER div 2 output DIGIT end loop output NUMBER

Mentor tip

Ensure the loop structure replicates the decision condition 'Is NUMBER = 1?' faithfully.

(d)

4 marks

1) Early termination / break conditions: Avoiding redundant loop iterations once a solution or match has been confirmed (e.g. adding a flag to terminate a search early).
2) Choosing optimal data structures and memory reuse: Selecting an array with direct O(1) index access rather than repeatedly traversing a linear collection, and minimizing auxiliary variable allocations.

Mentor tip

Discuss minimizing iteration count and selecting appropriate data structures.

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 programming35 marks · 50%
Q7, Q8, Q13, Q14
Networks20 marks · 29%
Q1, Q6, Q9, Q12
Computer organisation8 marks · 11%
Q3, Q5, Q11
Resource management4 marks · 6%
Q4
System fundamentals3 marks · 4%
Q2, Q10

Mentor insights

Traps to watch for

01
Integer division divisibility test · Q8
The expression NUMBER div 2 = NUMBER / 2 only evaluates to true when NUMBER is evenly divisible by 2. For an odd number, integer division truncates, failing the equality check.
02
Truth table row tracing for multiple gates · Q11
P = A AND B is 1 only when both A and B are 1. R = (NOT B) AND C is 1 only when B=0 and C=1. Be careful when combining them with OR.
03
Binary output bit reversal · Q14
Repeated division by 2 outputs the least significant bit (remainder of initial number) first. 19 yields 1, 1, 0, 0, then 1, representing 10011 in binary.

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

Want someone to go through your paper with you?

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

Find your maths mentor More past papers

Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: May 2021
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 2021 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

Roles a computer can perform within a networked environment

Networks

2 marks

Server role: Centralizes and provides shared resources, services, or authentication to other network hosts (e.g. web server, database host, DNS server, or file server).
Client role: Initiates requests to network servers to consume services, retrieve web pages, or access remote files on behalf of an end user.

Mentor tip

Identify client and server roles, or peer-to-peer node roles.

Question 2

2 marks

Implementation methodology for deploying a new computer system

System fundamentals

2 marks

Direct changeover: The organization terminates the legacy system at an agreed cutover time and immediately switches all operations to the new system.
While cost-effective and swift with no duplication of data entry, it introduces high operational risk if critical errors emerge post-launch.

Mentor tip

Choose one clear deployment strategy (direct, parallel, phased, or pilot) and describe its operational mechanism.

Question 3

2 marks

Logic circuit deduction from truth table

Computer organisation

2 marks

Examine truth table outputs for inputs A and B:
- A=0, B=0 -> Z=1
- A=0, B=1 -> Z=0
- A=1, B=0 -> Z=0
- A=1, B=1 -> Z=1
The output Z is 1 strictly when inputs A and B have identical values.
This corresponds directly to an XNOR (exclusive NOR) gate, or an XOR gate followed by an inverter (NOT gate).

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

Mentor tip

Recognize that equality (same input bits yielding 1) defines the XNOR logic operation.

Question 4

4 marks

Operating system patch management necessity and acquisition channels

Resource management

(a)

2 marks

Security vulnerability remediation: Patches fix discovered exploits and security holes, preventing malicious infiltration and zero-day attacks.
Bug fixes and system stability: Patches resolve code defects, memory leaks, and driver conflicts, enhancing operating system stability and performance.

Mentor tip

Highlight security vulnerability mitigation and bug/performance fixes.

(b)

2 marks

Automatic operating system update service: Built-in background updater client that polls vendor servers and downloads patches automatically.
Direct manual download from vendor portal: System administrators manually download installer executable packages from the authorized vendor support repository.

Mentor tip

Give two distinct delivery channels: automated OS update daemon and manual vendor download.

Question 5

2 marks

Hexadecimal to denary number conversion

Computer organisation

2 marks

Hexadecimal BF consists of two hex digits: 'B' in the 16^1 place and 'F' in the 16^0 place.
Hex digit B represents denary 11.
Hex digit F represents denary 15.
Compute the positional sum: 11 * 16 + 15 * 1 = 176 + 15 = 191.

BF_{16} = (11 \times 1 6^{1}) + (15 \times 1 6^{0}) = 176 + 15 = 19 1_{10}

Mentor tip

Convert B to 11 and F to 15 before multiplying by positional powers of 16.

Question 6

2 marks

Advantages of fibre optic cabling over wireless connectivity

Networks

2 marks

Immunity to electromagnetic interference: Fibre optics transmit pulses of light through glass strands rather than radio frequencies, making signals completely immune to EMI, radio noise, and atmospheric disturbance.
Significantly greater bandwidth and lower attenuation: Fibre optic cables support vastly higher data transmission speeds over much longer distances without requiring frequent signal repeaters.

Mentor tip

Focus on high bandwidth over distance and complete immunity to electrical interference.

Question 7

2 marks

Distinction between variable and constant data items

Computational thinking, problem-solving and programming

2 marks

Mutability during execution: A variable is a named memory location whose stored value can be reassigned or modified dynamically as the program runs.
Immutability: A constant is a named identifier bound to a fixed value upon declaration that cannot be altered or overwritten during program execution.

Mentor tip

Contrast runtime mutability (variables can change) against immutability (constants remain fixed).

Question 8

3 marks

Trace of divisibility filtering algorithm

Computational thinking, problem-solving and programming

3 marks

Input sequence: 2, 6, 8, 9, 12, 15, 18, 20.
Condition 1: NUMBER div 2 = NUMBER / 2. This checks if NUMBER is divisible by 2 with no fractional remainder (i.e. NUMBER is even).
Condition 2: NUMBER div 3 = NUMBER / 3. This checks if NUMBER is divisible by 3 with no fractional remainder.
Therefore, a number is output if and only if it is divisible by both 2 and 3 (i.e. divisible by 6):
- 2: Divisible by 2, but not 3 -> Not output
- 6: Divisible by 2 and 3 -> OUTPUT 6
- 8: Divisible by 2, but not 3 -> Not output
- 9: Not divisible by 2 -> Not output
- 12: Divisible by 2 and 3 -> OUTPUT 12
- 15: Not divisible by 2 -> Not output
- 18: Divisible by 2 and 3 -> OUTPUT 18
- 20: Divisible by 2, but not 3 -> Not output
Final output list in order: 6, 12, 18.

Condition: (NUMBER mod 2 = 0) \land (NUMBER mod 3 = 0) ⟺ NUMBER mod 6 = 0

Mentor tip

Notice that comparing integer division 'div' to real division '/' tests whether the division produces a whole integer with zero remainder.

Question 9

1 mark

Formal definition of a network data packet

Networks

1 mark

A standardized, formatted unit of data transmitted across a packet-switching network, consisting of control metadata (headers and trailers) and user payload.

Mentor tip

State that a packet contains control metadata (source/destination IP) and payload data.

Question 10

1 mark

User interface design features enhancing system usability

System fundamentals

1 mark

Mentor tip

Mention tooltips, breadcrumbs, consistent menus, or keyboard shortcuts.

Question 11

4 marks

Truth table evaluation for composite logic circuit

Computer organisation

4 marks

Circuit inputs: A, B, C.
Intermediate gate 1: P = A AND B.
Intermediate gate 2: Q = NOT B.
Intermediate gate 3: R = Q AND C = (NOT B) AND C.
Final gate: Z = P OR R = (A AND B) OR ((NOT B) AND C).
Evaluate all 8 rows:
- A=0, B=0, C=0 -> P=0, Q=1, R=0 -> Z = 0
- A=0, B=0, C=1 -> P=0, Q=1, R=1 -> Z = 1
- A=0, B=1, C=0 -> P=0, Q=0, R=0 -> Z = 0
- A=0, B=1, C=1 -> P=0, Q=0, R=0 -> Z = 0
- A=1, B=0, C=0 -> P=0, Q=1, R=0 -> Z = 0
- A=1, B=0, C=1 -> P=0, Q=1, R=1 -> Z = 1
- A=1, B=1, C=0 -> P=1, Q=0, R=0 -> Z = 1
- A=1, B=1, C=1 -> P=1, Q=0, R=0 -> Z = 1

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

Mentor tip

Trace intermediate nodes P, Q, and R carefully before combining them in the final OR gate.

Questions 12–14 · 45 marks

Question 12

15 marks

Campus wireless network expansion and VPN security

Networks

(a)

4 marks

Wireless Access Points (WAPs): Transceivers positioned across buildings that broadcast Wi-Fi radio frequencies and bridge wireless client traffic onto the school's wired Ethernet backbone.
Wireless Network Interface Controllers (WNICs): Hardware adapters embedded inside student and staff laptops/devices that modulate and demodulate radio frequency transmissions.

Mentor tip

Describe two separate hardware components such as WAPs, wireless routers, or client WNICs.

(b)

2 marks

Mobility and flexible learning: Students can access digital learning materials and collaborate from classrooms, libraries, or outdoor study spaces without physical cable tethers.
Device versatility (BYOD): Enables students to connect diverse personal portable devices (tablets, smartphones, laptops) seamlessly.

Mentor tip

Highlight physical mobility and multi-device support.

(c)

4 marks

Robust wireless encryption protocol (WPA3 / 802.1X enterprise): Encrypts all over-the-air communication using advanced cryptographic ciphers with unique per-user keys, preventing packet eavesdropping.
MAC address filtering or hidden SSID with captive portal authentication: Restricting network access solely to whitelisted device hardware addresses and requiring student portal login credentials before granting DHCP IP leases.

Mentor tip

Cover wireless encryption (WPA2/WPA3) and network admission controls (802.1X, captive portal, MAC filtering).

(d)

2 marks

VPN Gateway / Server software: Dedicated network server running tunneling protocols (e.g. IPsec, OpenVPN, WireGuard) to manage remote cryptographic handshakes.
VPN Client software: Application installed on remote staff devices that encrypts outbound packets and manages secure virtual network adapters.

Mentor tip

Identify the VPN server/gateway and client-side tunneling software.

(e)

3 marks

Secure remote access to intranet resources: Staff working off-campus can securely retrieve sensitive student grading records, internal curriculum repositories, and administration portals over public internet connections as though physically connected to the on-campus LAN.
Encryption protects confidential student data from exposure on untrusted external residential or hotel Wi-Fi connections.

Mentor tip

Emphasize secure off-campus access to internal file shares and administrative databases.

Question 13

15 marks

Employee name collection parsing and parallel array bubble sort

Computational thinking, problem-solving and programming

(a)

4 marks

Algorithm to parse alternating surname and first name from collection NAMES into SURNAMES[] and FIRSTNAMES[]:
1) Initialize index I = 0.
2) NAMES.resetNext() to reset collection iterator.
3) loop while NAMES.hasNext():
- SURNAMES[I] = NAMES.getNext()
- FIRSTNAMES[I] = NAMES.getNext()
- I = I + 1
4) end loop.

I = 0 NAMES.resetNext () loop while NAMES.hasNext () SURNAMES [I] = NAMES.getNext () FIRSTNAMES [I] = NAMES.getNext () I = I + 1 end loop

Mentor tip

Fetch twice per iteration because the collection contains alternating surname and first name values.

(b)

5 marks

Bubble sort algorithm for 600 staff members:
1) Outer loop from 0 to 598 (representing N - 1 passes).
2) Inner loop J from 0 to 598 - pass (or 598).
3) If SURNAMES[J] > SURNAMES[J + 1] then:
- Swap SURNAMES[J] and SURNAMES[J + 1] using TEMP_SUR.
- Concurrently swap FIRSTNAMES[J] and FIRSTNAMES[J + 1] using TEMP_FIRST to preserve parallel alignment.
4) Close conditionals and loops.

loop P from 0 to 598 loop J from 0 to 598 - P if SURNAMES [J] > SURNAMES [J + 1] then TEMP_SUR = SURNAMES [J] SURNAMES [J] = SURNAMES [J + 1] SURNAMES [J + 1] = TEMP_SUR TEMP_FIRST = FIRSTNAMES [J] FIRSTNAMES [J] = FIRSTNAMES [J + 1] FIRSTNAMES [J + 1] = TEMP_FIRST end if end loop end loop

Mentor tip

Always execute matching swaps on both SURNAMES and FIRSTNAMES arrays.

(c)

4 marks

Binary search procedure on sorted SURNAMES array:
1) Initialize LOW = 0, HIGH = 599, and FOUND = FALSE.
2) Repeat while LOW <= HIGH and not FOUND:
- Calculate MID = (LOW + HIGH) div 2.
- If SURNAMES[MID] == TARGET_NAME, set FOUND = TRUE.
- Else if TARGET_NAME < SURNAMES[MID], set HIGH = MID - 1.
- Else set LOW = MID + 1.
3) When found, the record index is MID; if loop finishes without match, report not found.

MID = ⌊(LOW + HIGH) /2 ⌋

Mentor tip

Describe the bisection process: compare target to midpoint, halve search interval accordingly.

(d)

2 marks

Modularity and reusability: Encapsulating array population and sorting into named subprograms allows them to be tested independently and invoked repeatedly without code duplication.

Mentor tip

Emphasize modularity, ease of debugging, and code reusability.

Question 14

15 marks

Decimal to binary conversion algorithm analysis and optimization

Computational thinking, problem-solving and programming

(a)

4 marks

Trace algorithm for input NUMBER = 19:
- Pass 1: DIGIT = 19 mod 2 = 1. NUMBER = 19 div 2 = 9. OUTPUT DIGIT: 1. Is 9 = 1? No.
- Pass 2: DIGIT = 9 mod 2 = 1. NUMBER = 9 div 2 = 4. OUTPUT DIGIT: 1. Is 4 = 1? No.
- Pass 3: DIGIT = 4 mod 2 = 0. NUMBER = 4 div 2 = 2. OUTPUT DIGIT: 0. Is 2 = 1? No.
- Pass 4: DIGIT = 2 mod 2 = 0. NUMBER = 2 div 2 = 1. OUTPUT DIGIT: 0. Is 1 = 1? Yes.
Loop exits on Yes branch. OUTPUT NUMBER: 1.
Sequence of outputs in order: 1, 1, 0, 0, 1.

1 9_{10} = 1001 1_{2} ⟹ outputs LSB to MSB: 1, 1, 0, 0, 1

Mentor tip

Notice the algorithm outputs the binary digits in reverse order (least significant bit first) until reaching 1.

(b)

1 mark

Converts a positive denary integer into its binary equivalent (outputting bits from least significant to most significant).

Mentor tip

State denary to binary conversion.

(c)

6 marks

Construct pseudocode with positive integer input validation:
1) Input validation loop: prompt NUMBER; while NUMBER <= 0 or NUMBER != int(NUMBER): re-prompt.
2) Loop while NUMBER != 1:
- DIGIT = NUMBER mod 2
- NUMBER = NUMBER div 2
- output DIGIT
3) output NUMBER (final remaining 1).

output "Enter a positive integer:" input NUMBER loop while NUMBER \leq 0 output "Invalid. Enter a positive integer:" input NUMBER end loop loop while NUMBER \neq = 1 DIGIT = NUMBER mod 2 NUMBER = NUMBER div 2 output DIGIT end loop output NUMBER

Mentor tip

Ensure the loop structure replicates the decision condition 'Is NUMBER = 1?' faithfully.

(d)

4 marks

1) Early termination / break conditions: Avoiding redundant loop iterations once a solution or match has been confirmed (e.g. adding a flag to terminate a search early).
2) Choosing optimal data structures and memory reuse: Selecting an array with direct O(1) index access rather than repeatedly traversing a linear collection, and minimizing auxiliary variable allocations.

Mentor tip

Discuss minimizing iteration count and selecting appropriate data structures.

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 programming35 marks · 50%
Q7, Q8, Q13, Q14
Networks20 marks · 29%
Q1, Q6, Q9, Q12
Computer organisation8 marks · 11%
Q3, Q5, Q11
Resource management4 marks · 6%
Q4
System fundamentals3 marks · 4%
Q2, Q10

Mentor insights

Traps to watch for

01
Integer division divisibility test · Q8
The expression NUMBER div 2 = NUMBER / 2 only evaluates to true when NUMBER is evenly divisible by 2. For an odd number, integer division truncates, failing the equality check.
02
Truth table row tracing for multiple gates · Q11
P = A AND B is 1 only when both A and B are 1. R = (NOT B) AND C is 1 only when B=0 and C=1. Be careful when combining them with OR.
03
Binary output bit reversal · Q14
Repeated division by 2 outputs the least significant bit (remainder of initial number) first. 19 yields 1, 1, 0, 0, then 1, representing 10011 in binary.

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

Want someone to go through your paper with you?

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

Find your maths mentor More past papers