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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2023
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Compatibility hurdles in corporate information systems integration

System fundamentals

2 marks

Data format discrepancies: Incompatible database schemas, character encodings, or date formatting conventions.
Platform divergence: Different underlying operating systems, network protocols, or legacy software APIs.

Mentor tip

Highlight data format mismatches and differing operating system/software architectures.

Question 2

1 mark

Definition of constant in computer programming

Computational thinking, problem-solving and programming

1 mark

A constant is an immutable named identifier whose assigned value cannot be altered or overwritten during program execution.

Mentor tip

State that a constant's value cannot change during runtime.

Question 3

2 marks

Usability challenges in voice recognition interfaces

System fundamentals

2 marks

Background acoustic noise: Environmental noise from crowds, wind, or traffic disrupts voice signal processing.
Dialect and accent variations: Regional accents, slang, or speech impediments can cause phoneme misinterpretation.

Mentor tip

Mention ambient acoustic noise and regional accents or pronunciation variations.

Question 4

3 marks

Essential components of network data packets

Networks

3 marks

Source and destination IP addresses for end-to-end packet routing across networks.
Packet sequence numbers to enable reassembly into original order.
Error-detection checksum / CRC to verify that data was not corrupted during transit.

Mentor tip

List source/destination IP addresses, sequence numbers, and checksums/payload.

Question 5

3 marks

Truth table construction for compound logic expression

Computer organisation

3 marks

Expression: $Z = A \lor (\neg B \land C)$ .
Precedence order: evaluate NOT first, then AND, then OR.
Rows with $A = 1$ : Output is 1 for all four rows $(1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)$ .
Rows with $A = 0$ :
$(0, 0, 0) ⟹ 0 \lor (1 \land 0) = 0$ .
$(0, 0, 1) ⟹ 0 \lor (1 \land 1) = 1$ .
$(0, 1, 0) ⟹ 0 \lor (0 \land 0) = 0$ .
$(0, 1, 1) ⟹ 0 \lor (0 \land 1) = 0$ .

A 00001111 B 00110011 C 01010101 \neg B 11001100 \neg B \land C 01000100 A \lor (\neg B \land C) 01001111

Mentor tip

Evaluate NOT B first, then AND with C, and finally OR with A.

Question 6

2 marks

Architecture and purpose of virtual machines

Resource management

2 marks

Software emulation: A virtual machine emulates physical hardware in software, enabling guest operating systems to execute inside an isolated container.
Sandboxed security: Isolates untrusted software testing and legacy systems from the underlying host computer.

Mentor tip

Describe software emulation of hardware providing an isolated guest OS environment.

Question 7

3 marks

Primary functions of the Central Processing Unit Control Unit

Computer organisation

3 marks

Instruction fetch: Retrieves instructions from RAM into the instruction register.
Instruction decode: Decodes opcodes to configure internal data pathways.
Control signal dispatch: Sends timing signals and control bus pulses to coordinate the ALU, registers, and memory.

Mentor tip

List instruction fetch, instruction decode, and control/timing signal dispatch.

Question 8

3 marks

Principles and operational advantages of concurrent processing

Resource management

(a)

2 marks

Concurrent processing involves executing multiple processes simultaneously or interleaved across the same time span on single-core or multi-core hardware.

Mentor tip

Concurrent processing means multiple processes make progress over the same time interval.

(b)

1 mark

Increased throughput: Prevents CPU idle time during I/O operations, improving overall application responsiveness.

Mentor tip

State higher throughput or better CPU resource utilization.

Question 9

6 marks

Iterative digit summation algorithm: trace table and purpose deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Initial state: $N = 1216, X = 0$ . Condition $N > 0$ is true.
Iteration 1: $X = 0 + (1216 mod 10) = 0 + 6 = 6$ . $N = 1216 div 10 = 121$ . Condition $N > 0$ is true.
Iteration 2: $X = 6 + (121 mod 10) = 6 + 1 = 7$ . $N = 121 div 10 = 12$ . Condition $N > 0$ is true.
Iteration 3: $X = 7 + (12 mod 10) = 7 + 2 = 9$ . $N = 12 div 10 = 1$ . Condition $N > 0$ is true.
Iteration 4: $X = 9 + (1 mod 10) = 9 + 1 = 10$ . $N = 1 div 10 = 0$ . Condition $N > 0$ is false.
Loop terminates. Output displays: 10.

N 12161211210 X 067910 Output 10

Mentor tip

Track N and X through each iteration until N reaches 0.

(b)

2 marks

The algorithm calculates and outputs the sum of all individual decimal digits of the integer $N$ .

Purpose: Sum of base-10 digits of N = 1 + 2 + 1 + 6 = 10

Mentor tip

State clearly that it calculates the sum of the digits.

Questions 10–12 · 45 marks

Question 10

15 marks

Medical practice data security, mail merge automation, and mobile VPN access

System fundamentals
Networks

(a)

2 marks

Role-based access control (RBAC): Ensure staff have tiered permissions so administrative workers cannot view confidential clinical charts.
Strong authentication: Require complex passwords and multi-factor authentication (MFA) to prevent unauthorized logins.

Mentor tip

Mention role-based access control and multi-factor authentication.

(b)(i)

1 mark

Physical storage failure, malware/ransomware infection, or accidental user deletion.

Mentor tip

Identify a specific cause such as hardware storage failure.

(b)(ii)

2 marks

Regular offsite backups: Automate periodic encrypted database backups to a secure remote cloud repository for disaster recovery.

Mentor tip

Describe automated regular backups stored in an offsite location.

(c)

3 marks

Database integration: Word processor connects to patient database records using query filters for target patients.
Merge field placeholders: The document template contains merge tags (e.g. `<<Name>>`, `<<Address>>`).
Batch output: The mail merge engine replaces merge fields with record data to produce personalized letters.

Mentor tip

Explain linking a database query to template merge fields.

(d)

4 marks

Data encryption: Encrypts patient medical records across public Wi-Fi or cellular connections, ensuring confidentiality.
Direct intranet access: Securely bridges the doctor's remote tablet to the clinic's internal server as if on the local network.

Mentor tip

Highlight encryption over untrusted networks and private clinic server access.

(e)

3 marks

Physical distance and obstructions: Concrete walls, terrain, and distance from cell towers attenuate radio signals.
Network cell congestion: Multiple mobile devices sharing the same base station reduce available bandwidth per user.

Mentor tip

Mention physical obstructions and cell tower congestion.

Question 11

15 marks

Requirements elicitation, software testing lifecycle, image ethics, and email transfer

System fundamentals
Networks

(a)(i)

1 mark

Questionnaires / Online surveys / Direct workplace observation.

Mentor tip

Name questionnaires, surveys, or direct observation.

(a)(ii)

2 marks

Wide reach and standardization: Questionnaires gather structured data from many respondents simultaneously at low cost.

Mentor tip

Surveys gather data from a large user base quickly and economically.

(b)

2 marks

User validation: Prototypes provide early visual mockups of user interfaces, allowing users to validate features before costly development.

Mentor tip

Prototypes allow early feedback to clarify requirements before full coding.

(c)

3 marks

Early bug identification: Finding bugs early during unit testing minimizes repair costs and project delays.
Integration verification: Ensures components interface seamlessly without communication mismatches.
Requirement fulfillment: Validates that the delivered software matches customer specifications.

Mentor tip

Explain that testing at each stage detects errors early when they are cheapest to resolve.

(d)

3 marks

Copyright violation: Using images without licensing or attribution violates intellectual property laws.
Privacy infringement: Publishing photos of individuals without consent violates personal privacy.

Mentor tip

Discuss copyright infringement or unauthorized use of images.

(e)(i)

3 marks

Cloud file sharing: Upload the 60 MB image to a cloud hosting platform (e.g. Google Drive) and email a shared download link.
Archive splitting: Use file compression tools to split the file into smaller sub-10 MB archive parts sent across separate emails.

Mentor tip

Explain uploading to cloud storage with a link or splitting into multi-part archives.

(e)(ii)

1 mark

High RAM/storage consumption / Slow rendering latency on low-end hardware.

Mentor tip

Mention memory usage or rendering latency.

Question 12

15 marks

Gymnastics score analysis: array lookup, average calculation, threshold filter, and binary search

Computational thinking, problem-solving and programming

(a)

1 mark

Index 5 in the parallel arrays corresponds to competitor 'Labar, Tanya'.

NAMES [5] = Labar, Tanya

Mentor tip

Read index 5 directly from the parallel arrays table.

(b)

3 marks

Initialize accumulator `SUM = 0`.
Loop index $I$ from 0 to 9, adding $SCORES [I]$ to `SUM`.
Compute average: $AVERAGE = SUM /10.0$ .
Output the calculated average.

SUM = 0 loop I from 0 to 9 SUM = SUM + SCORES [I] end loop AVERAGE = SUM /10.0 output (AVERAGE)

Mentor tip

Sum all 10 scores and divide by 10.0.

(c)

4 marks

Initialize counter `COUNT = 0`.
Loop index $I$ from 0 to 9.
If $SCORES [I] > AVERAGE$ , increment `COUNT = COUNT + 1`.
Output the total count of qualifying gymnasts.

COUNT = 0 loop I from 0 to 9 if SCORES [I] > AVERAGE then COUNT = COUNT + 1 end if end loop output (COUNT)

Mentor tip

Compare each score against the pre-calculated AVERAGE variable.

(d)

7 marks

Binary search on sorted NAMES array:
Input target name `SEARCHNAME`.
Initialize search boundaries: `LOW = 0`, `HIGH = 9`, `FOUND = false`.
While `LOW <= HIGH` and not `FOUND`:
Compute midpoint: $MID = div (LOW + HIGH, 2)$ .
If $NAMES [MID] == SEARCHNAME$ , set `FOUND = true` and output $SCORES [MID]$ .
Else if $SEARCHNAME < NAMES [MID]$ , narrow to left half: $HIGH = MID - 1$ .
Else narrow to right half: $LOW = MID + 1$ .
If loop terminates without match, output 'this name is not found'.

input SEARCH_NAME LOW = 0 HIGH = 9 FOUND = false loop while LOW \leq HIGH and not FOUND MID = div (LOW + HIGH, 2) if NAMES [MID] == SEARCH_NAME then FOUND = true output (SCORES [MID]) else if SEARCH_NAME < NAMES [MID] then HIGH = MID - 1 else LOW = MID + 1 end if end loop if not FOUND then output ("this name is not found") end if

Mentor tip

Binary search requires sorted data; halve the search space on each comparison.

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 fundamentals24 marks · 34%
Q1, Q3, Q10, Q11
Computational thinking, problem-solving and programming22 marks · 31%
Q2, Q9, Q12
Networks13 marks · 19%
Q4, Q10, Q11
Computer organisation6 marks · 9%
Q5, Q7
Resource management5 marks · 7%
Q6, Q8

Mentor insights

Traps to watch for

01
Logic operator precedence · Q5
NOT has highest precedence, followed by AND, then OR. In A or not B and C, evaluate (not B) and C first before performing the OR with A.
02
Iterative modulo extraction order · Q9
X accumulates N mod 10 at each step while N is divided by 10. The digits are processed from right to left (least significant to most significant).
03
Binary search boundary adjustments · Q12
When narrowing the search space, always adjust by 1: HIGH = MID - 1 or LOW = MID + 1. Forgetting the offset can cause infinite loops when the target is absent.

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

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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2023
Marks: 70
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 12

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2023 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Compatibility hurdles in corporate information systems integration

System fundamentals

2 marks

Data format discrepancies: Incompatible database schemas, character encodings, or date formatting conventions.
Platform divergence: Different underlying operating systems, network protocols, or legacy software APIs.

Mentor tip

Highlight data format mismatches and differing operating system/software architectures.

Question 2

1 mark

Definition of constant in computer programming

Computational thinking, problem-solving and programming

1 mark

A constant is an immutable named identifier whose assigned value cannot be altered or overwritten during program execution.

Mentor tip

State that a constant's value cannot change during runtime.

Question 3

2 marks

Usability challenges in voice recognition interfaces

System fundamentals

2 marks

Background acoustic noise: Environmental noise from crowds, wind, or traffic disrupts voice signal processing.
Dialect and accent variations: Regional accents, slang, or speech impediments can cause phoneme misinterpretation.

Mentor tip

Mention ambient acoustic noise and regional accents or pronunciation variations.

Question 4

3 marks

Essential components of network data packets

Networks

3 marks

Source and destination IP addresses for end-to-end packet routing across networks.
Packet sequence numbers to enable reassembly into original order.
Error-detection checksum / CRC to verify that data was not corrupted during transit.

Mentor tip

List source/destination IP addresses, sequence numbers, and checksums/payload.

Question 5

3 marks

Truth table construction for compound logic expression

Computer organisation

3 marks

Expression: $Z = A \lor (\neg B \land C)$ .
Precedence order: evaluate NOT first, then AND, then OR.
Rows with $A = 1$ : Output is 1 for all four rows $(1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)$ .
Rows with $A = 0$ :
$(0, 0, 0) ⟹ 0 \lor (1 \land 0) = 0$ .
$(0, 0, 1) ⟹ 0 \lor (1 \land 1) = 1$ .
$(0, 1, 0) ⟹ 0 \lor (0 \land 0) = 0$ .
$(0, 1, 1) ⟹ 0 \lor (0 \land 1) = 0$ .

A 00001111 B 00110011 C 01010101 \neg B 11001100 \neg B \land C 01000100 A \lor (\neg B \land C) 01001111

Mentor tip

Evaluate NOT B first, then AND with C, and finally OR with A.

Question 6

2 marks

Architecture and purpose of virtual machines

Resource management

2 marks

Software emulation: A virtual machine emulates physical hardware in software, enabling guest operating systems to execute inside an isolated container.
Sandboxed security: Isolates untrusted software testing and legacy systems from the underlying host computer.

Mentor tip

Describe software emulation of hardware providing an isolated guest OS environment.

Question 7

3 marks

Primary functions of the Central Processing Unit Control Unit

Computer organisation

3 marks

Instruction fetch: Retrieves instructions from RAM into the instruction register.
Instruction decode: Decodes opcodes to configure internal data pathways.
Control signal dispatch: Sends timing signals and control bus pulses to coordinate the ALU, registers, and memory.

Mentor tip

List instruction fetch, instruction decode, and control/timing signal dispatch.

Question 8

3 marks

Principles and operational advantages of concurrent processing

Resource management

(a)

2 marks

Concurrent processing involves executing multiple processes simultaneously or interleaved across the same time span on single-core or multi-core hardware.

Mentor tip

Concurrent processing means multiple processes make progress over the same time interval.

(b)

1 mark

Increased throughput: Prevents CPU idle time during I/O operations, improving overall application responsiveness.

Mentor tip

State higher throughput or better CPU resource utilization.

Question 9

6 marks

Iterative digit summation algorithm: trace table and purpose deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Initial state: $N = 1216, X = 0$ . Condition $N > 0$ is true.
Iteration 1: $X = 0 + (1216 mod 10) = 0 + 6 = 6$ . $N = 1216 div 10 = 121$ . Condition $N > 0$ is true.
Iteration 2: $X = 6 + (121 mod 10) = 6 + 1 = 7$ . $N = 121 div 10 = 12$ . Condition $N > 0$ is true.
Iteration 3: $X = 7 + (12 mod 10) = 7 + 2 = 9$ . $N = 12 div 10 = 1$ . Condition $N > 0$ is true.
Iteration 4: $X = 9 + (1 mod 10) = 9 + 1 = 10$ . $N = 1 div 10 = 0$ . Condition $N > 0$ is false.
Loop terminates. Output displays: 10.

N 12161211210 X 067910 Output 10

Mentor tip

Track N and X through each iteration until N reaches 0.

(b)

2 marks

The algorithm calculates and outputs the sum of all individual decimal digits of the integer $N$ .

Purpose: Sum of base-10 digits of N = 1 + 2 + 1 + 6 = 10

Mentor tip

State clearly that it calculates the sum of the digits.

Questions 10–12 · 45 marks

Question 10

15 marks

Medical practice data security, mail merge automation, and mobile VPN access

System fundamentals
Networks

(a)

2 marks

Role-based access control (RBAC): Ensure staff have tiered permissions so administrative workers cannot view confidential clinical charts.
Strong authentication: Require complex passwords and multi-factor authentication (MFA) to prevent unauthorized logins.

Mentor tip

Mention role-based access control and multi-factor authentication.

(b)(i)

1 mark

Physical storage failure, malware/ransomware infection, or accidental user deletion.

Mentor tip

Identify a specific cause such as hardware storage failure.

(b)(ii)

2 marks

Regular offsite backups: Automate periodic encrypted database backups to a secure remote cloud repository for disaster recovery.

Mentor tip

Describe automated regular backups stored in an offsite location.

(c)

3 marks

Database integration: Word processor connects to patient database records using query filters for target patients.
Merge field placeholders: The document template contains merge tags (e.g. `<<Name>>`, `<<Address>>`).
Batch output: The mail merge engine replaces merge fields with record data to produce personalized letters.

Mentor tip

Explain linking a database query to template merge fields.

(d)

4 marks

Data encryption: Encrypts patient medical records across public Wi-Fi or cellular connections, ensuring confidentiality.
Direct intranet access: Securely bridges the doctor's remote tablet to the clinic's internal server as if on the local network.

Mentor tip

Highlight encryption over untrusted networks and private clinic server access.

(e)

3 marks

Physical distance and obstructions: Concrete walls, terrain, and distance from cell towers attenuate radio signals.
Network cell congestion: Multiple mobile devices sharing the same base station reduce available bandwidth per user.

Mentor tip

Mention physical obstructions and cell tower congestion.

Question 11

15 marks

Requirements elicitation, software testing lifecycle, image ethics, and email transfer

System fundamentals
Networks

(a)(i)

1 mark

Questionnaires / Online surveys / Direct workplace observation.

Mentor tip

Name questionnaires, surveys, or direct observation.

(a)(ii)

2 marks

Wide reach and standardization: Questionnaires gather structured data from many respondents simultaneously at low cost.

Mentor tip

Surveys gather data from a large user base quickly and economically.

(b)

2 marks

User validation: Prototypes provide early visual mockups of user interfaces, allowing users to validate features before costly development.

Mentor tip

Prototypes allow early feedback to clarify requirements before full coding.

(c)

3 marks

Early bug identification: Finding bugs early during unit testing minimizes repair costs and project delays.
Integration verification: Ensures components interface seamlessly without communication mismatches.
Requirement fulfillment: Validates that the delivered software matches customer specifications.

Mentor tip

Explain that testing at each stage detects errors early when they are cheapest to resolve.

(d)

3 marks

Copyright violation: Using images without licensing or attribution violates intellectual property laws.
Privacy infringement: Publishing photos of individuals without consent violates personal privacy.

Mentor tip

Discuss copyright infringement or unauthorized use of images.

(e)(i)

3 marks

Cloud file sharing: Upload the 60 MB image to a cloud hosting platform (e.g. Google Drive) and email a shared download link.
Archive splitting: Use file compression tools to split the file into smaller sub-10 MB archive parts sent across separate emails.

Mentor tip

Explain uploading to cloud storage with a link or splitting into multi-part archives.

(e)(ii)

1 mark

High RAM/storage consumption / Slow rendering latency on low-end hardware.

Mentor tip

Mention memory usage or rendering latency.

Question 12

15 marks

Gymnastics score analysis: array lookup, average calculation, threshold filter, and binary search

Computational thinking, problem-solving and programming

(a)

1 mark

Index 5 in the parallel arrays corresponds to competitor 'Labar, Tanya'.

NAMES [5] = Labar, Tanya

Mentor tip

Read index 5 directly from the parallel arrays table.

(b)

3 marks

Initialize accumulator `SUM = 0`.
Loop index $I$ from 0 to 9, adding $SCORES [I]$ to `SUM`.
Compute average: $AVERAGE = SUM /10.0$ .
Output the calculated average.

SUM = 0 loop I from 0 to 9 SUM = SUM + SCORES [I] end loop AVERAGE = SUM /10.0 output (AVERAGE)

Mentor tip

Sum all 10 scores and divide by 10.0.

(c)

4 marks

Initialize counter `COUNT = 0`.
Loop index $I$ from 0 to 9.
If $SCORES [I] > AVERAGE$ , increment `COUNT = COUNT + 1`.
Output the total count of qualifying gymnasts.

COUNT = 0 loop I from 0 to 9 if SCORES [I] > AVERAGE then COUNT = COUNT + 1 end if end loop output (COUNT)

Mentor tip

Compare each score against the pre-calculated AVERAGE variable.

(d)

7 marks

Binary search on sorted NAMES array:
Input target name `SEARCHNAME`.
Initialize search boundaries: `LOW = 0`, `HIGH = 9`, `FOUND = false`.
While `LOW <= HIGH` and not `FOUND`:
Compute midpoint: $MID = div (LOW + HIGH, 2)$ .
If $NAMES [MID] == SEARCHNAME$ , set `FOUND = true` and output $SCORES [MID]$ .
Else if $SEARCHNAME < NAMES [MID]$ , narrow to left half: $HIGH = MID - 1$ .
Else narrow to right half: $LOW = MID + 1$ .
If loop terminates without match, output 'this name is not found'.

input SEARCH_NAME LOW = 0 HIGH = 9 FOUND = false loop while LOW \leq HIGH and not FOUND MID = div (LOW + HIGH, 2) if NAMES [MID] == SEARCH_NAME then FOUND = true output (SCORES [MID]) else if SEARCH_NAME < NAMES [MID] then HIGH = MID - 1 else LOW = MID + 1 end if end loop if not FOUND then output ("this name is not found") end if

Mentor tip

Binary search requires sorted data; halve the search space on each comparison.

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 fundamentals24 marks · 34%
Q1, Q3, Q10, Q11
Computational thinking, problem-solving and programming22 marks · 31%
Q2, Q9, Q12
Networks13 marks · 19%
Q4, Q10, Q11
Computer organisation6 marks · 9%
Q5, Q7
Resource management5 marks · 7%
Q6, Q8

Mentor insights

Traps to watch for

01
Logic operator precedence · Q5
NOT has highest precedence, followed by AND, then OR. In A or not B and C, evaluate (not B) and C first before performing the OR with A.
02
Iterative modulo extraction order · Q9
X accumulates N mod 10 at each step while N is divided by 10. The digits are processed from right to left (least significant to most significant).
03
Binary search boundary adjustments · Q12
When narrowing the search space, always adjust by 1: HIGH = MID - 1 or LOW = MID + 1. Forgetting the offset can cause infinite loops when the target is absent.

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