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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2022
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 2022 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

Essential functional capabilities of a web browser

Networks

2 marks

HTML/CSS rendering engine: Interprets HTML markup, CSS style rules, and JavaScript code to visually format and render web pages.
URL navigation and address bar: Allows users to input Uniform Resource Locators (URLs), request web resources via HTTP/HTTPS, and navigate browsing history (back, forward, bookmarking).

Mentor tip

Name two core features such as web page rendering and URL navigation/history management.

Question 2

3 marks

Truth table evaluation for composite Boolean expression (A xor B) and B

Computer organisation

3 marks

Expression: (A XOR B) AND B
Evaluate row by row for inputs A and B:
1) A=0, B=0 -> (0 XOR 0) = 0 -> 0 AND 0 = 0
2) A=0, B=1 -> (0 XOR 1) = 1 -> 1 AND 1 = 1
3) A=1, B=0 -> (1 XOR 0) = 1 -> 1 AND 0 = 0
4) A=1, B=1 -> (1 XOR 1) = 0 -> 0 AND 1 = 0
Notice the expression evaluates to true strictly when A=0 and B=1 (equivalent to (NOT A) AND B).

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

Mentor tip

Compute the intermediate XOR column before applying the logical AND with B.

Question 3

4 marks

Prototyping software development methodology evaluation

System fundamentals

(a)

2 marks

Early stakeholder feedback: A functional prototype gives clients an interactive representation of the proposed user interface, uncovering ambiguous or missing requirements early in the lifecycle.

Mentor tip

Highlight user feedback and early requirement clarification.

(b)

2 marks

Safety-critical or deeply algorithmic systems: In software where formal mathematical correctness, absolute real-time timing, or faultless reliability is paramount (e.g. avionics fly-by-wire or medical pacemaker firmware), quick surface-level prototyping can encourage superficial interface design while neglecting rigorous formal verification.

Mentor tip

Mention safety-critical systems or scenarios where requirements are already strictly fixed.

Question 4

5 marks

Fundamental computer operations and higher-level programming languages

Computer organisation

(a)

2 marks

1) Arithmetic operations: Performing basic addition, subtraction, and comparison using the Arithmetic Logic Unit (ALU).
2) Data movement: Transferring bit patterns between CPU registers, cache, and primary memory locations.

Mentor tip

Other valid fundamental operations include logical comparisons (AND, OR, NOT) or input/output transfers.

(b)

3 marks

Abstraction and readability: High-level languages allow developers to express algorithms using human-readable syntax, structured control flows, and meaningful identifiers rather than cryptic binary opcodes.
Platform portability: High-level source code can be compiled or interpreted to run across varied hardware architectures without rewriting machine-specific code.
Developer productivity and error reduction: Built-in type checking, modular functions, and abstraction substantially reduce programming mistakes and maintenance costs.

Mentor tip

Cover abstraction/readability, cross-platform portability, and software development productivity.

Question 5

4 marks

Ethical and technical implications of unauthorized online media publication

System fundamentals
Networks

(a)

2 marks

Invasion of personal privacy and lack of consent: Publishing photographs and videos of individuals without their explicit permission violates their personal autonomy and can cause social embarrassment or reputational harm.

Mentor tip

Focus on lack of informed consent and potential personal or professional harm.

(b)

2 marks

Incompatible media codecs or excessive file sizes: The viewer's client browser may lack the necessary video decompression codec, or high video bitrates may fail to load over low-bandwidth or unstable network connections.

Mentor tip

Mention codec incompatibility, network bandwidth constraints, or browser format restrictions.

Question 6

1 mark

Formal definition of a variable in computer programming

Computational thinking, problem-solving and programming

1 mark

A named storage location in primary memory (RAM) designated by an identifier, holding a data value that can be read or modified during program execution.

Mentor tip

Highlight both the named memory location and the capacity for its contents to change during execution.

Question 7

2 marks

24-bit Truecolor RGB pixel representation and colour gamut

Computer organisation

(a)

1 mark

Pure blue has hex representation 0000FF:
- Red byte 00 = 00000000
- Green byte 00 = 00000000
- Blue byte FF = 11111111
Binary value: 000000000000000011111111 (or 00000000 00000000 11111111).

0000 FF_{16} = 00000000000000001111111 1_{2}

Mentor tip

Expand each of the three color channels into 8 binary bits.

(b)

1 mark

With 24 bits allocated per pixel, the total number of distinct addressable colors is 2^24 = 16,777,216.

2^{24} = 16777216 colors

Mentor tip

Calculate 2 to the power of 24 (or 256 * 256 * 256).

Question 8

4 marks

Modulo loop execution and array output trace

Computational thinking, problem-solving and programming

4 marks

Array NAMES: [0]: 'Zixan', [1]: 'Murali', [2]: 'Eli', [3]: 'Kim'.
Algorithm:
K = 3
loop while K >= 0
A = K mod 3
output(NAMES[A])
K = K - 1
end while
Trace table passes:
- Pass 1: K = 3. A = 3 mod 3 = 0. Output: NAMES[0] = 'Zixan'. K becomes 2.
- Pass 2: K = 2. A = 2 mod 3 = 2. Output: NAMES[2] = 'Eli'. K becomes 1.
- Pass 3: K = 1. A = 1 mod 3 = 1. Output: NAMES[1] = 'Murali'. K becomes 0.
- Pass 4: K = 0. A = 0 mod 3 = 0. Output: NAMES[0] = 'Zixan'. K becomes -1.
Loop terminates because -1 >= 0 is FALSE.
Sequence of outputs:
Zixan
Eli
Murali
Zixan

K = 3 K = 2 K = 1 K = 0 ⟹ A = 3 mod 3 = 0 ⟹ NAMES [0] = Zixan ⟹ A = 2 mod 3 = 2 ⟹ NAMES [2] = Eli ⟹ A = 1 mod 3 = 1 ⟹ NAMES [1] = Murali ⟹ A = 0 mod 3 = 0 ⟹ NAMES [0] = Zixan

Mentor tip

Carefully evaluate integer modulo: 3 mod 3 = 0, 2 mod 3 = 2, 1 mod 3 = 1, 0 mod 3 = 0.

Questions 9–11 · 45 marks

Question 9

15 marks

Wired vs Wireless Local Area Networks and VPN security

Networks

(a)

1 mark

Wireless Access Point (WAP) / wireless network interface controller (WNIC).

Mentor tip

A Wireless Access Point is the required hardware bridge connecting wireless devices to the wired backbone.

(b)

4 marks

Wired network transmission reliability:
- Data travels through shielded physical cables (twisted pair or fiber optic) isolated from atmospheric noise, delivering consistent bandwidth, predictable latency, and minimal packet loss.
Wireless network transmission reliability:
- Radio frequency signals attenuate rapidly over distance and are vulnerable to physical obstructions (walls, metal furniture) and electromagnetic interference (microwaves, adjacent Wi-Fi channels), causing variable transmission speeds and packet retransmissions.

Mentor tip

Contrast cable shielding and physical consistency against radio attenuation and environmental interference.

(c)

2 marks

Broadcast medium exposure: Radio waves propagate omnidirectionally through physical walls into public spaces outside the office perimeter.
An eavesdropper within physical range can intercept data packets using a packet sniffer without needing physical access to the building's wiring.

Mentor tip

Emphasize that radio signals cross physical walls and can be captured remotely without physical connection.

(d)

4 marks

Data encryption: All payload data passing through the public network is encrypted using robust cryptographic ciphers (e.g. AES-256), rendering intercepted packets unreadable without the secret decryption key.
Tunneling protocol with IP masking: The original IP packets are encapsulated inside new network packet headers, masking the user's real IP address and origin location from intermediate observers.

Mentor tip

Explain two separate features such as cryptographic payload encryption and packet encapsulation/tunneling.

(e)

4 marks

1) Packetization: The sender segments the original data stream into standardized packets, affixing header metadata (source IP, destination IP, packet sequence number, checksum).
2) Dynamic routing: Each packet travels independently across intermediate routers, which select optimal paths dynamically based on current network congestion.
3) Out-of-order arrival handling: Packets may arrive at the destination via different paths and in scrambled chronological order.
4) Reassembly and verification: The destination host uses packet sequence numbers to reassemble the data in correct sequence and checks error codes, requesting retransmission of any corrupted or dropped packets.

Mentor tip

Detail the full progression: division into packets with headers -> independent dynamic routing -> reassembly by sequence numbers.

Question 10

15 marks

International data migration, change management, and legacy systems

System fundamentals

(a)(i)

1 mark

Data migration is the process of transferring digital data from one computer storage system, format, or database architecture to another.

Mentor tip

State the transfer of data between storage systems, schemas, or environments.

(a)(ii)

4 marks

1) Data corruption and loss: Transmission interruptions, truncated records, or character encoding mismatches during mass migration can compromise data integrity.
2) Schema and format incompatibility: Divergent database structures, missing relational fields, or inconsistent primary keys between legacy and target systems require complex transformation logic, risking record misinterpretation.

Mentor tip

Detail two distinct risks such as data corruption/loss and schema incompatibility.

(b)

4 marks

User training and re-skilling: Delivering comprehensive instruction programs to ensure employees acquire the necessary competence and confidence to navigate the new system effectively.
Stakeholder communication and organizational alignment: Maintaining transparent dialogue regarding project milestones, revised business workflows, and temporary transition schedules to minimize employee disruption and resistance.

Mentor tip

Cover user education and proactive organizational communication.

(c)

3 marks

Scarcity of maintenance expertise: Older legacy programming languages and architectures (e.g. COBOL or proprietary mainframe databases) have dwindling pools of skilled developers, making bug fixes and routine maintenance exceptionally expensive.
Additionally, legacy vendors frequently discontinue security patches, exposing the organization to critical vulnerabilities and severe integration hurdles with modern cloud infrastructure.

Mentor tip

Highlight lack of vendor support, scarce specialized developer skills, and rising maintenance expenses.

(d)

3 marks

Redundant workforce labor: Operating two systems simultaneously requires personnel to duplicate all operational transactions in both platforms, effectively doubling staffing hours.
Infrastructure maintenance overhead: The organization must simultaneously fund operating licenses, dual server hosting environments, electricity, and continuous synchronization maintenance for both systems.

Mentor tip

Explain the dual expenditure of staffing duplication and concurrent infrastructure operational costs.

Question 11

15 marks

Proper divisors and number classification algorithms

Computational thinking, problem-solving and programming

(a)(i)

1 mark

The integer 2 has exactly 1 proper divisor: the number 1.

Mentor tip

The divisors of 2 are 1 and 2; proper divisors exclude 2, leaving only 1.

(a)(ii)

1 mark

4 is not a proper divisor of 10 because dividing 10 by 4 yields 2.5 with a non-zero remainder of 2 (10 mod 4 != 0).

10 mod 4 = 2 \neq = 0

Mentor tip

State that 4 does not divide 10 evenly without leaving a remainder.

(b)

6 marks

Construct algorithm taking integer K > 0 as input:
1) Initialize SUM = 0.
2) Loop D from 1 to K - 1:
- If K mod D = 0, add D to SUM.
3) Evaluate SUM against K:
- If SUM = K, output 'Perfect'.
- Else if SUM > K, output 'Abundant'.
- Else output 'Deficient'.

input K SUM = 0 loop D from 1 to K - 1 if K mod D = 0 then SUM = SUM + D end if end loop if SUM = K then output "Perfect" else if SUM > K then output "Abundant" else output "Deficient" end if

Mentor tip

Accumulate the proper divisor sum up to K - 1 before branching on the three classification categories.

(c)

7 marks

Algorithm to count odd and even abundant numbers in array DATA of size X:
1) Initialize counters: COUNT_ODD = 0, COUNT_EVEN = 0.
2) Iterate loop index I from 0 to X - 1 across DATA.
3) For element VAL = DATA[I]:
- Call subprogram isAbundant(VAL).
- If true:
* If VAL mod 2 != 0, increment COUNT_ODD.
* Else increment COUNT_EVEN.
4) After loop completes, output COUNT_ODD and COUNT_EVEN.

COUNT_ODD = 0 COUNT_EVEN = 0 loop I from 0 to X - 1 VAL = DATA [I] if isAbundant (VAL) = true then if VAL mod 2 \neq = 0 then COUNT_ODD = COUNT_ODD + 1 else COUNT_EVEN = COUNT_EVEN + 1 end if end if end loop output "Abundant odd count: ", COUNT_ODD output "Abundant even count: ", COUNT_EVEN

Mentor tip

First check whether the element is abundant using isAbundant(), then test parity via mod 2.

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 fundamentals21 marks · 30%
Q3, Q5, Q10
Computational thinking, problem-solving and programming20 marks · 29%
Q6, Q8, Q11
Networks19 marks · 27%
Q1, Q5, Q9
Computer organisation10 marks · 14%
Q2, Q4, Q7

Mentor insights

Traps to watch for

01
Logic precedence in (A XOR B) AND B · Q2
Ensure XOR is evaluated first within parentheses. When A=1 and B=1, 1 XOR 1 = 0, so 0 AND 1 = 0. Do not confuse exclusive OR with inclusive OR.
02
Modulo evaluation on 0 · Q8
When K=0, 0 mod 3 evaluates cleanly to 0 without error, indexing NAMES[0] ('Zixan') a second time before K decrements to -1.
03
Proper divisor exclusion of K · Q11
Proper divisors do not include the integer K itself. Summing up to K instead of K - 1 will improperly inflate the sum.

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.

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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: November 2022
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 2022 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

Essential functional capabilities of a web browser

Networks

2 marks

HTML/CSS rendering engine: Interprets HTML markup, CSS style rules, and JavaScript code to visually format and render web pages.
URL navigation and address bar: Allows users to input Uniform Resource Locators (URLs), request web resources via HTTP/HTTPS, and navigate browsing history (back, forward, bookmarking).

Mentor tip

Name two core features such as web page rendering and URL navigation/history management.

Question 2

3 marks

Truth table evaluation for composite Boolean expression (A xor B) and B

Computer organisation

3 marks

Expression: (A XOR B) AND B
Evaluate row by row for inputs A and B:
1) A=0, B=0 -> (0 XOR 0) = 0 -> 0 AND 0 = 0
2) A=0, B=1 -> (0 XOR 1) = 1 -> 1 AND 1 = 1
3) A=1, B=0 -> (1 XOR 0) = 1 -> 1 AND 0 = 0
4) A=1, B=1 -> (1 XOR 1) = 0 -> 0 AND 1 = 0
Notice the expression evaluates to true strictly when A=0 and B=1 (equivalent to (NOT A) AND B).

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

Mentor tip

Compute the intermediate XOR column before applying the logical AND with B.

Question 3

4 marks

Prototyping software development methodology evaluation

System fundamentals

(a)

2 marks

Early stakeholder feedback: A functional prototype gives clients an interactive representation of the proposed user interface, uncovering ambiguous or missing requirements early in the lifecycle.

Mentor tip

Highlight user feedback and early requirement clarification.

(b)

2 marks

Mentor tip

Mention safety-critical systems or scenarios where requirements are already strictly fixed.

Question 4

5 marks

Fundamental computer operations and higher-level programming languages

Computer organisation

(a)

2 marks

1) Arithmetic operations: Performing basic addition, subtraction, and comparison using the Arithmetic Logic Unit (ALU).
2) Data movement: Transferring bit patterns between CPU registers, cache, and primary memory locations.

Mentor tip

Other valid fundamental operations include logical comparisons (AND, OR, NOT) or input/output transfers.

(b)

3 marks

Abstraction and readability: High-level languages allow developers to express algorithms using human-readable syntax, structured control flows, and meaningful identifiers rather than cryptic binary opcodes.
Platform portability: High-level source code can be compiled or interpreted to run across varied hardware architectures without rewriting machine-specific code.
Developer productivity and error reduction: Built-in type checking, modular functions, and abstraction substantially reduce programming mistakes and maintenance costs.

Mentor tip

Cover abstraction/readability, cross-platform portability, and software development productivity.

Question 5

4 marks

Ethical and technical implications of unauthorized online media publication

System fundamentals
Networks

(a)

2 marks

Mentor tip

Focus on lack of informed consent and potential personal or professional harm.

(b)

2 marks

Mentor tip

Mention codec incompatibility, network bandwidth constraints, or browser format restrictions.

Question 6

1 mark

Formal definition of a variable in computer programming

Computational thinking, problem-solving and programming

1 mark

A named storage location in primary memory (RAM) designated by an identifier, holding a data value that can be read or modified during program execution.

Mentor tip

Highlight both the named memory location and the capacity for its contents to change during execution.

Question 7

2 marks

24-bit Truecolor RGB pixel representation and colour gamut

Computer organisation

(a)

1 mark

Pure blue has hex representation 0000FF:
- Red byte 00 = 00000000
- Green byte 00 = 00000000
- Blue byte FF = 11111111
Binary value: 000000000000000011111111 (or 00000000 00000000 11111111).

0000 FF_{16} = 00000000000000001111111 1_{2}

Mentor tip

Expand each of the three color channels into 8 binary bits.

(b)

1 mark

With 24 bits allocated per pixel, the total number of distinct addressable colors is 2^24 = 16,777,216.

2^{24} = 16777216 colors

Mentor tip

Calculate 2 to the power of 24 (or 256 * 256 * 256).

Question 8

4 marks

Modulo loop execution and array output trace

Computational thinking, problem-solving and programming

4 marks

Array NAMES: [0]: 'Zixan', [1]: 'Murali', [2]: 'Eli', [3]: 'Kim'.
Algorithm:
K = 3
loop while K >= 0
A = K mod 3
output(NAMES[A])
K = K - 1
end while
Trace table passes:
- Pass 1: K = 3. A = 3 mod 3 = 0. Output: NAMES[0] = 'Zixan'. K becomes 2.
- Pass 2: K = 2. A = 2 mod 3 = 2. Output: NAMES[2] = 'Eli'. K becomes 1.
- Pass 3: K = 1. A = 1 mod 3 = 1. Output: NAMES[1] = 'Murali'. K becomes 0.
- Pass 4: K = 0. A = 0 mod 3 = 0. Output: NAMES[0] = 'Zixan'. K becomes -1.
Loop terminates because -1 >= 0 is FALSE.
Sequence of outputs:
Zixan
Eli
Murali
Zixan

K = 3 K = 2 K = 1 K = 0 ⟹ A = 3 mod 3 = 0 ⟹ NAMES [0] = Zixan ⟹ A = 2 mod 3 = 2 ⟹ NAMES [2] = Eli ⟹ A = 1 mod 3 = 1 ⟹ NAMES [1] = Murali ⟹ A = 0 mod 3 = 0 ⟹ NAMES [0] = Zixan

Mentor tip

Carefully evaluate integer modulo: 3 mod 3 = 0, 2 mod 3 = 2, 1 mod 3 = 1, 0 mod 3 = 0.

Questions 9–11 · 45 marks

Question 9

15 marks

Wired vs Wireless Local Area Networks and VPN security

Networks

(a)

1 mark

Wireless Access Point (WAP) / wireless network interface controller (WNIC).

Mentor tip

A Wireless Access Point is the required hardware bridge connecting wireless devices to the wired backbone.

(b)

4 marks

Wired network transmission reliability:
- Data travels through shielded physical cables (twisted pair or fiber optic) isolated from atmospheric noise, delivering consistent bandwidth, predictable latency, and minimal packet loss.
Wireless network transmission reliability:
- Radio frequency signals attenuate rapidly over distance and are vulnerable to physical obstructions (walls, metal furniture) and electromagnetic interference (microwaves, adjacent Wi-Fi channels), causing variable transmission speeds and packet retransmissions.

Mentor tip

Contrast cable shielding and physical consistency against radio attenuation and environmental interference.

(c)

2 marks

Broadcast medium exposure: Radio waves propagate omnidirectionally through physical walls into public spaces outside the office perimeter.
An eavesdropper within physical range can intercept data packets using a packet sniffer without needing physical access to the building's wiring.

Mentor tip

Emphasize that radio signals cross physical walls and can be captured remotely without physical connection.

(d)

4 marks

Data encryption: All payload data passing through the public network is encrypted using robust cryptographic ciphers (e.g. AES-256), rendering intercepted packets unreadable without the secret decryption key.
Tunneling protocol with IP masking: The original IP packets are encapsulated inside new network packet headers, masking the user's real IP address and origin location from intermediate observers.

Mentor tip

Explain two separate features such as cryptographic payload encryption and packet encapsulation/tunneling.

(e)

4 marks

1) Packetization: The sender segments the original data stream into standardized packets, affixing header metadata (source IP, destination IP, packet sequence number, checksum).
2) Dynamic routing: Each packet travels independently across intermediate routers, which select optimal paths dynamically based on current network congestion.
3) Out-of-order arrival handling: Packets may arrive at the destination via different paths and in scrambled chronological order.
4) Reassembly and verification: The destination host uses packet sequence numbers to reassemble the data in correct sequence and checks error codes, requesting retransmission of any corrupted or dropped packets.

Mentor tip

Detail the full progression: division into packets with headers -> independent dynamic routing -> reassembly by sequence numbers.

Question 10

15 marks

International data migration, change management, and legacy systems

System fundamentals

(a)(i)

1 mark

Data migration is the process of transferring digital data from one computer storage system, format, or database architecture to another.

Mentor tip

State the transfer of data between storage systems, schemas, or environments.

(a)(ii)

4 marks

1) Data corruption and loss: Transmission interruptions, truncated records, or character encoding mismatches during mass migration can compromise data integrity.
2) Schema and format incompatibility: Divergent database structures, missing relational fields, or inconsistent primary keys between legacy and target systems require complex transformation logic, risking record misinterpretation.

Mentor tip

Detail two distinct risks such as data corruption/loss and schema incompatibility.

(b)

4 marks

User training and re-skilling: Delivering comprehensive instruction programs to ensure employees acquire the necessary competence and confidence to navigate the new system effectively.
Stakeholder communication and organizational alignment: Maintaining transparent dialogue regarding project milestones, revised business workflows, and temporary transition schedules to minimize employee disruption and resistance.

Mentor tip

Cover user education and proactive organizational communication.

(c)

3 marks

Scarcity of maintenance expertise: Older legacy programming languages and architectures (e.g. COBOL or proprietary mainframe databases) have dwindling pools of skilled developers, making bug fixes and routine maintenance exceptionally expensive.
Additionally, legacy vendors frequently discontinue security patches, exposing the organization to critical vulnerabilities and severe integration hurdles with modern cloud infrastructure.

Mentor tip

Highlight lack of vendor support, scarce specialized developer skills, and rising maintenance expenses.

(d)

3 marks

Redundant workforce labor: Operating two systems simultaneously requires personnel to duplicate all operational transactions in both platforms, effectively doubling staffing hours.
Infrastructure maintenance overhead: The organization must simultaneously fund operating licenses, dual server hosting environments, electricity, and continuous synchronization maintenance for both systems.

Mentor tip

Explain the dual expenditure of staffing duplication and concurrent infrastructure operational costs.

Question 11

15 marks

Proper divisors and number classification algorithms

Computational thinking, problem-solving and programming

(a)(i)

1 mark

The integer 2 has exactly 1 proper divisor: the number 1.

Mentor tip

The divisors of 2 are 1 and 2; proper divisors exclude 2, leaving only 1.

(a)(ii)

1 mark

4 is not a proper divisor of 10 because dividing 10 by 4 yields 2.5 with a non-zero remainder of 2 (10 mod 4 != 0).

10 mod 4 = 2 \neq = 0

Mentor tip

State that 4 does not divide 10 evenly without leaving a remainder.

(b)

6 marks

Construct algorithm taking integer K > 0 as input:
1) Initialize SUM = 0.
2) Loop D from 1 to K - 1:
- If K mod D = 0, add D to SUM.
3) Evaluate SUM against K:
- If SUM = K, output 'Perfect'.
- Else if SUM > K, output 'Abundant'.
- Else output 'Deficient'.

input K SUM = 0 loop D from 1 to K - 1 if K mod D = 0 then SUM = SUM + D end if end loop if SUM = K then output "Perfect" else if SUM > K then output "Abundant" else output "Deficient" end if

Mentor tip

Accumulate the proper divisor sum up to K - 1 before branching on the three classification categories.

(c)

7 marks

Algorithm to count odd and even abundant numbers in array DATA of size X:
1) Initialize counters: COUNT_ODD = 0, COUNT_EVEN = 0.
2) Iterate loop index I from 0 to X - 1 across DATA.
3) For element VAL = DATA[I]:
- Call subprogram isAbundant(VAL).
- If true:
* If VAL mod 2 != 0, increment COUNT_ODD.
* Else increment COUNT_EVEN.
4) After loop completes, output COUNT_ODD and COUNT_EVEN.

COUNT_ODD = 0 COUNT_EVEN = 0 loop I from 0 to X - 1 VAL = DATA [I] if isAbundant (VAL) = true then if VAL mod 2 \neq = 0 then COUNT_ODD = COUNT_ODD + 1 else COUNT_EVEN = COUNT_EVEN + 1 end if end if end loop output "Abundant odd count: ", COUNT_ODD output "Abundant even count: ", COUNT_EVEN

Mentor tip

First check whether the element is abundant using isAbundant(), then test parity via mod 2.

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 fundamentals21 marks · 30%
Q3, Q5, Q10
Computational thinking, problem-solving and programming20 marks · 29%
Q6, Q8, Q11
Networks19 marks · 27%
Q1, Q5, Q9
Computer organisation10 marks · 14%
Q2, Q4, Q7

Mentor insights

Traps to watch for

01
Logic precedence in (A XOR B) AND B · Q2
Ensure XOR is evaluated first within parentheses. When A=1 and B=1, 1 XOR 1 = 0, so 0 AND 1 = 0. Do not confuse exclusive OR with inclusive OR.
02
Modulo evaluation on 0 · Q8
When K=0, 0 mod 3 evaluates cleanly to 0 without error, indexing NAMES[0] ('Zixan') a second time before K decrements to -1.
03
Proper divisor exclusion of K · Q11
Proper divisors do not include the integer K itself. Summing up to K instead of K - 1 will improperly inflate the sum.

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