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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2021
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Usability considerations in mobile cell phone design

System fundamentals

2 marks

Compact screen size constraints: Small physical displays make touch targets and small fonts difficult to activate accurately for users with large fingers or visual impairments.
Bright ambient glare: Using a mobile device in outdoor sunlight can severely reduce display contrast, impairing screen readability.

Mentor tip

Explain touch accuracy on compact screens or visibility in bright daylight.

Question 2

2 marks

Primary storage vs secondary storage distinctions

Computer organisation

2 marks

Volatility: Primary storage (RAM) is volatile and loses all stored data when power is disconnected, whereas secondary storage (SSD, HDD) is non-volatile and permanently retains data without power.
Direct CPU access and speed: Primary memory is directly addressable by the CPU over the system bus with nanosecond access latencies, whereas secondary storage cannot be directly addressed by the CPU and operates orders of magnitude slower.

Mentor tip

Contrast volatility and direct CPU addressability/access speed.

Question 3

2 marks

Underlying rationale for binary digital computing

Computer organisation

2 marks

Electronic circuit physics: Digital computer circuits are constructed from microscopic transistors operating reliably as bi-stable electronic switches (representing on/high voltage and off/low voltage).
Noise tolerance and reliability: Distinguishing between two discrete voltage levels (1 and 0) provides maximum tolerance to electrical noise compared to distinguishing between continuous analog states.

Mentor tip

Explain bi-stable transistor states (on/off) and electrical noise immunity.

Question 4

3 marks

Network transmission advantages of data compression

Networks

3 marks

Reduced bit volume: Compression algorithms eliminate data redundancy and encode files into significantly fewer bits.
Faster transmission duration: Smaller byte payloads transmit across the network in less time, reducing latency for end users.
Bandwidth conservation: Lower aggregate traffic minimizes network congestion and reduces data transfer costs over metered connections.

Mentor tip

Connect file size reduction to shorter transfer times and conserved network capacity.

Question 5

7 marks

Subprograms as abstraction, concurrent development, and update notifications

Computational thinking, problem-solving and programming
System fundamentals

(a)

2 marks

Hiding implementation complexity: A subprogram encapsulates complex multi-step algorithms behind a simple, well-defined identifier and parameter interface.
Callers only need to understand what arguments to pass and what result is returned, without needing to know internal logic details.

Mentor tip

Explain that subprograms conceal complex inner mechanics behind clean procedural interfaces.

(b)

3 marks

Advantage: Accelerates time-to-market by allowing specialized developer teams to build distinct modules (e.g. database, UI, network layer) simultaneously in parallel.
Disadvantage / Risk: Requires strict interface contract adherence; if inter-module API standards are ambiguous or change mid-development, complex integration defects and costly rework occur during system assembly.

Mentor tip

Balance accelerated development velocity against the risk of integration mismatches.

(c)

2 marks

In-application pop-up notification / splash modal: The application queries a central version repository on launch and prompts the user with release notes and an instant update button.

Mentor tip

Mention in-app update banners or automated email notification to registered accounts.

Question 6

5 marks

Selection sort mechanics and Bubble sort contrast

Computational thinking, problem-solving and programming

(a)

3 marks

Iterative search for minimum: In each pass, the algorithm scans the unsorted partition of the array to find the smallest element.
Single swap into position: It swaps the identified minimum element with the first element of the unsorted partition.
Partition expansion: The boundary between the sorted and unsorted partitions advances by one, repeating until the entire array is sorted.

Mentor tip

Describe finding the minimum in the unsorted portion and swapping it into place.

(b)

2 marks

Swap frequency: Selection sort performs at most one swap per pass (O(N) total swaps), whereas bubble sort can perform numerous swaps per pass as it compares adjacent elements.

Mentor tip

Contrast one swap per pass in selection sort against frequent adjacent swaps in bubble sort.

Question 7

1 mark

Boolean logic expression evaluation

Computer organisation

1 mark

Given X = 5, Y = 3.
Expression: (X <= 5) XOR (Y > X)
- (X <= 5): 5 <= 5 is TRUE.
- (Y > X): 3 > 5 is FALSE.
- TRUE XOR FALSE evaluates to TRUE.

(5 \leq 5) \oplus (3 > 5) = TRUE \oplus FALSE = TRUE

Mentor tip

Evaluate both inequalities before applying the XOR logic operator.

Question 8

3 marks

Trace table execution for while loop accumulation

Computational thinking, problem-solving and programming

3 marks

Trace algorithm:
Initial state: K = 0, I = 0.
- Pass 1: 6 > 0 is TRUE. K = 0 + 0 = 0. I = 0 + 2 = 2.
- Pass 2: 6 > 2 is TRUE. K = 0 + 2 = 2. I = 2 + 2 = 4.
- Pass 3: 6 > 4 is TRUE. K = 2 + 4 = 6. I = 4 + 2 = 6.
- Pass 4: 6 > 6 is FALSE. Loop terminates.
Final output(K) prints 6.

Pass 1: K = 0, I = 2 Pass 2: K = 2, I = 4 Pass 3: K = 6, I = 6 Output: 6

Mentor tip

Notice the loop condition 6 > I stops when I reaches 6.

Questions 9–11 · 45 marks

Question 9

15 marks

Requirement gathering methodologies, prototyping, and rigorous testing

System fundamentals

(a)(i)

2 marks

Scalability and quantitative reach: Surveys can be distributed digitally to hundreds of dispersed stakeholders simultaneously, gathering broad quantitative feedback at minimal cost.

Mentor tip

Highlight wide geographic reach and low distribution cost.

(a)(ii)

2 marks

Inflexible questions and lack of depth: Fixed survey questions prevent dynamic follow-up probing, often yielding superficial responses and low completion rates.

Mentor tip

Point out the inability to ask clarifying follow-up questions.

(b)

1 mark

Direct stakeholder interviews (or direct workplace observation).

Mentor tip

State interviews or workplace observation.

(c)

4 marks

1) Early requirement clarification: Provides users with an interactive, tangible model to test workflows, exposing misunderstandings and missing specifications early.
2) Higher stakeholder engagement and confidence: Users actively participate in design iterations, fostering buy-in and reducing post-launch user resistance.

Mentor tip

Detail two distinct benefits like early requirement validation and reduced design risk.

(d)

3 marks

Iterative refinement: Initial prototypes frequently reveal unexpected user needs or technical constraints that were not anticipated in the first analysis pass.
Subsequent analysis and design cycles allow developers to adapt specifications based on real user interaction data, converging on a robust final design.

Mentor tip

Explain that real user interaction exposes hidden requirements that necessitate iterative revision.

(e)

3 marks

Preventing catastrophic failure and data corruption: Rigorous testing under varied operational loads detects edge-case defects and vulnerabilities before deployment.
Unchecked errors in production can cause business disruption, severe financial losses, and irreversible data corruption.

Mentor tip

Connect rigorous pre-deployment testing to operational reliability and risk mitigation.

Question 10

15 marks

Enterprise data governance, network security, and disaster recovery

System fundamentals
Networks

(a)

4 marks

Role-Based Access Control (RBAC): Assigning permissions strictly according to job roles, ensuring staff can only view or edit files essential to their daily tasks.
Multi-Factor Authentication (MFA) and biometric verification: Requiring two or more independent credentials before granting access to sensitive databases.

Mentor tip

Describe two concrete access control measures such as RBAC and multi-factor authentication.

(b)

4 marks

Next-generation firewall and intrusion prevention systems (IPS): Inspecting inbound and outbound packet headers and payloads to filter malicious traffic.
End-to-end data encryption across network segments: Encrypting internal data transmissions (via TLS/IPsec) to prevent packet eavesdropping.

Mentor tip

Cover firewall/IPS implementation and cryptographic traffic encryption.

(c)

4 marks

Automated periodic backups: Restoring intact records from scheduled differential or immutable off-site backups created before corruption occurred.
Database transaction logging and rollback: Using write-ahead transaction logs to roll back incomplete or damaged states to the last known consistent checkpoint.

Mentor tip

Explain restoring from off-site backups and replaying/rolling back transaction logs.

(d)

3 marks

Secure remote connectivity: Employees traveling or working from home can access confidential corporate servers across public internet connections through an encrypted tunnel, safeguarding corporate secrets.

Mentor tip

Highlight encrypted remote access to internal resources over public networks.

Question 11

15 marks

Bus fare calculation subprograms and passenger ticketing algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Construct subprogram costperkm(AGE):
Base rate = 0.20 EUR.
- If AGE < 5: free -> return 0.0
- Else if AGE <= 15: 50% discount -> return 0.20 * 0.5 = 0.10
- Else if AGE >= 65: 30% discount -> return 0.20 * 0.7 = 0.14
- Else: standard adult -> return 0.20

subprogram costperkm (AGE) if AGE < 5 then return 0.0 else if AGE \leq 15 then return 0.10 else if AGE \geq 65 then return 0.14 else return 0.20 end if end subprogram

Mentor tip

Verify calculations: 50% off 0.20 is 0.10; 30% off 0.20 is 0.14.

(b)

1 mark

Kiko is index 3, Endsley is index 4. DISTANCES[4] = 1.3 km.

DISTANCES [4] = 1.3 km

Mentor tip

Look at DISTANCES[4] connecting station 3 to station 4.

(c)

1 mark

Oppox is index 0, Brinkley is index 2. Sum segment distances: DISTANCES[1] + DISTANCES[2] = 1.2 + 1.0 = 2.2 km.

1.2 + 1.0 = 2.2 km

Mentor tip

Sum DISTANCES[1] and DISTANCES[2] along the consecutive segments.

(d)

3 marks

Initialize a distance accumulator to 0.0.
Loop an index K from P1 + 1 up to P2.
Add DISTANCES[K] to the accumulator on each iteration.
Return the final accumulated total distance.

Distance = K = P_{1} + 1 \sum P_{2} DISTANCES [K]

Mentor tip

Sum values from index P1 + 1 to P2 inclusive.

(e)

6 marks

Construct main ticketing algorithm:
1) Input STAT1, STAT2, AGE.
2) Find indices IDX1 and IDX2 in NAMES array (size 10):
- If either is not found, output 'One or more stations not found.' and terminate.
3) Order indices: P1 = min(IDX1, IDX2), P2 = max(IDX1, IDX2).
4) DIST = calcdistance(P1, P2).
5) RATE = costperkm(AGE).
6) FARE = DIST * RATE.
7) Output FARE formatted with currency.

input STAT1, STAT2, AGE IDX1 = - 1, IDX2 = - 1 loop I from 0 to 9 if NAMES [I] = STAT1 then IDX1 = I if NAMES [I] = STAT2 then IDX2 = I end loop if IDX1 = - 1 or IDX2 = - 1 then output "One or more stations not found." else if IDX1 < IDX2 then P 1 = IDX1, P 2 = IDX2 else P 1 = IDX2, P 2 = IDX1 end if KM = calcdistance (P 1, P 2) RATE = costperkm (AGE) TOTAL = KM \times RATE output "EUR ", TOTAL end if

Mentor tip

Ensure index ordering P1 < P2 so calcdistance receives valid bounds.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals30 marks · 43%
Q1, Q5, Q9, Q10
Computational thinking, problem-solving and programming25 marks · 36%
Q5, Q6, Q8, Q11
Networks10 marks · 14%
Q4, Q10
Computer organisation5 marks · 7%
Q2, Q3, Q7

Mentor insights

Traps to watch for

01
Loop termination boundary on strict inequality · Q8
The loop condition is 6 > I. When I reaches 6, 6 > 6 evaluates to false and the loop terminates. Do not execute a 4th iteration.
02
Index ordering requirement for calcdistance · Q11
calcdistance(P1, P2) strictly requires P1 < P2. If the user enters destination before departure (e.g. Allapay to Kiko), the indices must be swapped before calling calcdistance.
03
XOR evaluation on opposite truth values · Q7
XOR returns true when exactly one operand is true. Here 5 <= 5 is true and 3 > 5 is false, so true XOR false is true.

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 2021
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Usability considerations in mobile cell phone design

System fundamentals

2 marks

Compact screen size constraints: Small physical displays make touch targets and small fonts difficult to activate accurately for users with large fingers or visual impairments.
Bright ambient glare: Using a mobile device in outdoor sunlight can severely reduce display contrast, impairing screen readability.

Mentor tip

Explain touch accuracy on compact screens or visibility in bright daylight.

Question 2

2 marks

Primary storage vs secondary storage distinctions

Computer organisation

2 marks

Volatility: Primary storage (RAM) is volatile and loses all stored data when power is disconnected, whereas secondary storage (SSD, HDD) is non-volatile and permanently retains data without power.
Direct CPU access and speed: Primary memory is directly addressable by the CPU over the system bus with nanosecond access latencies, whereas secondary storage cannot be directly addressed by the CPU and operates orders of magnitude slower.

Mentor tip

Contrast volatility and direct CPU addressability/access speed.

Question 3

2 marks

Underlying rationale for binary digital computing

Computer organisation

2 marks

Electronic circuit physics: Digital computer circuits are constructed from microscopic transistors operating reliably as bi-stable electronic switches (representing on/high voltage and off/low voltage).
Noise tolerance and reliability: Distinguishing between two discrete voltage levels (1 and 0) provides maximum tolerance to electrical noise compared to distinguishing between continuous analog states.

Mentor tip

Explain bi-stable transistor states (on/off) and electrical noise immunity.

Question 4

3 marks

Network transmission advantages of data compression

Networks

3 marks

Reduced bit volume: Compression algorithms eliminate data redundancy and encode files into significantly fewer bits.
Faster transmission duration: Smaller byte payloads transmit across the network in less time, reducing latency for end users.
Bandwidth conservation: Lower aggregate traffic minimizes network congestion and reduces data transfer costs over metered connections.

Mentor tip

Connect file size reduction to shorter transfer times and conserved network capacity.

Question 5

7 marks

Subprograms as abstraction, concurrent development, and update notifications

Computational thinking, problem-solving and programming
System fundamentals

(a)

2 marks

Hiding implementation complexity: A subprogram encapsulates complex multi-step algorithms behind a simple, well-defined identifier and parameter interface.
Callers only need to understand what arguments to pass and what result is returned, without needing to know internal logic details.

Mentor tip

Explain that subprograms conceal complex inner mechanics behind clean procedural interfaces.

(b)

3 marks

Advantage: Accelerates time-to-market by allowing specialized developer teams to build distinct modules (e.g. database, UI, network layer) simultaneously in parallel.
Disadvantage / Risk: Requires strict interface contract adherence; if inter-module API standards are ambiguous or change mid-development, complex integration defects and costly rework occur during system assembly.

Mentor tip

Balance accelerated development velocity against the risk of integration mismatches.

(c)

2 marks

In-application pop-up notification / splash modal: The application queries a central version repository on launch and prompts the user with release notes and an instant update button.

Mentor tip

Mention in-app update banners or automated email notification to registered accounts.

Question 6

5 marks

Selection sort mechanics and Bubble sort contrast

Computational thinking, problem-solving and programming

(a)

3 marks

Iterative search for minimum: In each pass, the algorithm scans the unsorted partition of the array to find the smallest element.
Single swap into position: It swaps the identified minimum element with the first element of the unsorted partition.
Partition expansion: The boundary between the sorted and unsorted partitions advances by one, repeating until the entire array is sorted.

Mentor tip

Describe finding the minimum in the unsorted portion and swapping it into place.

(b)

2 marks

Swap frequency: Selection sort performs at most one swap per pass (O(N) total swaps), whereas bubble sort can perform numerous swaps per pass as it compares adjacent elements.

Mentor tip

Contrast one swap per pass in selection sort against frequent adjacent swaps in bubble sort.

Question 7

1 mark

Boolean logic expression evaluation

Computer organisation

1 mark

Given X = 5, Y = 3.
Expression: (X <= 5) XOR (Y > X)
- (X <= 5): 5 <= 5 is TRUE.
- (Y > X): 3 > 5 is FALSE.
- TRUE XOR FALSE evaluates to TRUE.

(5 \leq 5) \oplus (3 > 5) = TRUE \oplus FALSE = TRUE

Mentor tip

Evaluate both inequalities before applying the XOR logic operator.

Question 8

3 marks

Trace table execution for while loop accumulation

Computational thinking, problem-solving and programming

3 marks

Trace algorithm:
Initial state: K = 0, I = 0.
- Pass 1: 6 > 0 is TRUE. K = 0 + 0 = 0. I = 0 + 2 = 2.
- Pass 2: 6 > 2 is TRUE. K = 0 + 2 = 2. I = 2 + 2 = 4.
- Pass 3: 6 > 4 is TRUE. K = 2 + 4 = 6. I = 4 + 2 = 6.
- Pass 4: 6 > 6 is FALSE. Loop terminates.
Final output(K) prints 6.

Pass 1: K = 0, I = 2 Pass 2: K = 2, I = 4 Pass 3: K = 6, I = 6 Output: 6

Mentor tip

Notice the loop condition 6 > I stops when I reaches 6.

Questions 9–11 · 45 marks

Question 9

15 marks

Requirement gathering methodologies, prototyping, and rigorous testing

System fundamentals

(a)(i)

2 marks

Scalability and quantitative reach: Surveys can be distributed digitally to hundreds of dispersed stakeholders simultaneously, gathering broad quantitative feedback at minimal cost.

Mentor tip

Highlight wide geographic reach and low distribution cost.

(a)(ii)

2 marks

Inflexible questions and lack of depth: Fixed survey questions prevent dynamic follow-up probing, often yielding superficial responses and low completion rates.

Mentor tip

Point out the inability to ask clarifying follow-up questions.

(b)

1 mark

Direct stakeholder interviews (or direct workplace observation).

Mentor tip

State interviews or workplace observation.

(c)

4 marks

1) Early requirement clarification: Provides users with an interactive, tangible model to test workflows, exposing misunderstandings and missing specifications early.
2) Higher stakeholder engagement and confidence: Users actively participate in design iterations, fostering buy-in and reducing post-launch user resistance.

Mentor tip

Detail two distinct benefits like early requirement validation and reduced design risk.

(d)

3 marks

Iterative refinement: Initial prototypes frequently reveal unexpected user needs or technical constraints that were not anticipated in the first analysis pass.
Subsequent analysis and design cycles allow developers to adapt specifications based on real user interaction data, converging on a robust final design.

Mentor tip

Explain that real user interaction exposes hidden requirements that necessitate iterative revision.

(e)

3 marks

Preventing catastrophic failure and data corruption: Rigorous testing under varied operational loads detects edge-case defects and vulnerabilities before deployment.
Unchecked errors in production can cause business disruption, severe financial losses, and irreversible data corruption.

Mentor tip

Connect rigorous pre-deployment testing to operational reliability and risk mitigation.

Question 10

15 marks

Enterprise data governance, network security, and disaster recovery

System fundamentals
Networks

(a)

4 marks

Role-Based Access Control (RBAC): Assigning permissions strictly according to job roles, ensuring staff can only view or edit files essential to their daily tasks.
Multi-Factor Authentication (MFA) and biometric verification: Requiring two or more independent credentials before granting access to sensitive databases.

Mentor tip

Describe two concrete access control measures such as RBAC and multi-factor authentication.

(b)

4 marks

Next-generation firewall and intrusion prevention systems (IPS): Inspecting inbound and outbound packet headers and payloads to filter malicious traffic.
End-to-end data encryption across network segments: Encrypting internal data transmissions (via TLS/IPsec) to prevent packet eavesdropping.

Mentor tip

Cover firewall/IPS implementation and cryptographic traffic encryption.

(c)

4 marks

Automated periodic backups: Restoring intact records from scheduled differential or immutable off-site backups created before corruption occurred.
Database transaction logging and rollback: Using write-ahead transaction logs to roll back incomplete or damaged states to the last known consistent checkpoint.

Mentor tip

Explain restoring from off-site backups and replaying/rolling back transaction logs.

(d)

3 marks

Mentor tip

Highlight encrypted remote access to internal resources over public networks.

Question 11

15 marks

Bus fare calculation subprograms and passenger ticketing algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Construct subprogram costperkm(AGE):
Base rate = 0.20 EUR.
- If AGE < 5: free -> return 0.0
- Else if AGE <= 15: 50% discount -> return 0.20 * 0.5 = 0.10
- Else if AGE >= 65: 30% discount -> return 0.20 * 0.7 = 0.14
- Else: standard adult -> return 0.20

subprogram costperkm (AGE) if AGE < 5 then return 0.0 else if AGE \leq 15 then return 0.10 else if AGE \geq 65 then return 0.14 else return 0.20 end if end subprogram

Mentor tip

Verify calculations: 50% off 0.20 is 0.10; 30% off 0.20 is 0.14.

(b)

1 mark

Kiko is index 3, Endsley is index 4. DISTANCES[4] = 1.3 km.

DISTANCES [4] = 1.3 km

Mentor tip

Look at DISTANCES[4] connecting station 3 to station 4.

(c)

1 mark

Oppox is index 0, Brinkley is index 2. Sum segment distances: DISTANCES[1] + DISTANCES[2] = 1.2 + 1.0 = 2.2 km.

1.2 + 1.0 = 2.2 km

Mentor tip

Sum DISTANCES[1] and DISTANCES[2] along the consecutive segments.

(d)

3 marks

Initialize a distance accumulator to 0.0.
Loop an index K from P1 + 1 up to P2.
Add DISTANCES[K] to the accumulator on each iteration.
Return the final accumulated total distance.

Distance = K = P_{1} + 1 \sum P_{2} DISTANCES [K]

Mentor tip

Sum values from index P1 + 1 to P2 inclusive.

(e)

6 marks

Construct main ticketing algorithm:
1) Input STAT1, STAT2, AGE.
2) Find indices IDX1 and IDX2 in NAMES array (size 10):
- If either is not found, output 'One or more stations not found.' and terminate.
3) Order indices: P1 = min(IDX1, IDX2), P2 = max(IDX1, IDX2).
4) DIST = calcdistance(P1, P2).
5) RATE = costperkm(AGE).
6) FARE = DIST * RATE.
7) Output FARE formatted with currency.

input STAT1, STAT2, AGE IDX1 = - 1, IDX2 = - 1 loop I from 0 to 9 if NAMES [I] = STAT1 then IDX1 = I if NAMES [I] = STAT2 then IDX2 = I end loop if IDX1 = - 1 or IDX2 = - 1 then output "One or more stations not found." else if IDX1 < IDX2 then P 1 = IDX1, P 2 = IDX2 else P 1 = IDX2, P 2 = IDX1 end if KM = calcdistance (P 1, P 2) RATE = costperkm (AGE) TOTAL = KM \times RATE output "EUR ", TOTAL end if

Mentor tip

Ensure index ordering P1 < P2 so calcdistance receives valid bounds.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals30 marks · 43%
Q1, Q5, Q9, Q10
Computational thinking, problem-solving and programming25 marks · 36%
Q5, Q6, Q8, Q11
Networks10 marks · 14%
Q4, Q10
Computer organisation5 marks · 7%
Q2, Q3, Q7

Mentor insights

Traps to watch for

01
Loop termination boundary on strict inequality · Q8
The loop condition is 6 > I. When I reaches 6, 6 > 6 evaluates to false and the loop terminates. Do not execute a 4th iteration.
02
Index ordering requirement for calcdistance · Q11
calcdistance(P1, P2) strictly requires P1 < P2. If the user enters destination before departure (e.g. Allapay to Kiko), the indices must be swapped before calling calcdistance.
03
XOR evaluation on opposite truth values · Q7
XOR returns true when exactly one operand is true. Here 5 <= 5 is true and 3 > 5 is false, so true XOR false is true.

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